Analysis of Structural Member Systems

JEROME J. CONNOR, Sc.D., Massachusetts Institute ofTechnology, is Professor of Civil Engineering at Massa-chusetts Institute of Technology. He has been active inteaching and research in structural analysis and mechanicsat the U.S. Army Materials and Mechanics ResearchAgency and for some years at M.I.T. His primary inter-est is in computer based analysis methods, and his currentresearch is concerned with the dynamic analysis of pre-stressed concrete reactor vessels and the development offinite element models for fluid flow problems. Dr. Connoris one of the original developers of ICES-STRUDL, andhas published extensively in the structural field.

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ANALYSIS OF

STRUCTURAL MEMBER

SYSTEMS

JEROME J. CONNORMassachusetts Institute of Technology

THE RONALD PRESS COMPANY • NEW YORK


Copyright © 1976 by

Ttrn RONALD PRESS COMPANY

All Rights Reserved

No part of this book may be reproducedin any form without permission in writing

from the publisher.

Library of Congress Catalog Card Number: 74—22535

PRINTED IN ThE UNITCD STATES OF AMERICA


Preface

With the development over the past decade of computer-based analysismethods, the teaching of structural analysis subjects has been revolutionized.The traditional division between structural analysis and structural mechanicsbecame no longer necessary, and instead of teaching a preponderance of solu-tion details it is now possible to focus on the underlying theory.

What has been done here is to integrate analysis and mechanics in a sys-tematic presentation which includes the mechanics of a member, the matrixformulation of the equations for a system of members, and solution techniques.The three fundamental steps in formulating a problem in solid mechanics—.enforcing equilibrium, relating deformations and displacements, and relatingforces and deformations—form the basis of the development, and the centraltheme is to establish the equations for each step and then discuss how the com-plete set of equations is solved. In this way, a reader obtains a more unifiedview of a problem, sees more clearly where the various simplifying assumptionsare introduced, and is better prepared to extend the theory.

The chapters of Part I contain the relevant topics for an essential back-ground in linear algebra, differential and matrix transformations.Collecting this material in the first part of the book is convenient for the con-tinuity of the mathematics presentation as well as for the continuity in thefollowing development.

Part II treats the analysis of an ideal truss. The governing equations forsmall strain but arbitrary displacement are established and then cast intomatrix form. Next, we deduce the principles of virtual displacements andvirtual forces by manipulating the governing equations, introduce a criterionfor evaluating the stability of an equilibrium position, and interpret the gov-erning equations as stationary requirements for certain variational principles.These concepts are essential for an appreciation of the solution schemes de-scribed in the following two chapters.

Part III is concerned with the behavior of an isolated member. For com-pleteness, first are presented the governing equations for a deformable elasticsolid allowing for arbitrary displacements, the continuous form of the princi-ples of virtual displacements and virtual forces, and the stability criterion.Unrestrained torsion-flexure of a prismatic member is examined in detail andthen an approximate engineering theory is developed. We move on to re-strained torsion-flexure of a prismatic member, discussing various approachesfor including warping restraint and illustrating its influence for thin-walled

iii


PREFACE

open and closed sections. The concluding chapters treat the behavior ofplanar and arbitrary curved members.

How one assembles and solves the governing equations for a member sys-tern is discussed in Part IV. First, the direct stiffness method is outlined;then a general formulation of the governing equations is described. Geo-metrically nonlinear behavior is considered in the last chapter, which dis-cusses member force-displacement relations, including torsional-flexuralcoupling, solution schemes, and linearized stability analysis.

The objective has been a text suitable for the teaching of modern structuralmember system analysis, and what is offered is an outgrowth of lecture notesdeveloped in recent years at the Massachusetts Institute of Technology. Tothe many students who have provided the occasion of that development, I amdeeply appreciative. Particular thanks go to Mrs. Jane Malinofsky for herpatience in typing the manuscript, and to Professor Charles Miller for hisencouragement.

JEROME J. CONNOR

Cambridge, Mass.January, 1976


Contents

I—MATHEMATICAL PRELIMINARiES

1 Introduction to Matrix Algebra

1—i Definition of a Matrix 3

1—2 Equality, Addition, and Subtraction of Matrices 5

1—3 Matrix Multiplication 5

1—4 Transpose of a Matrix 8

1—5 Special Square Matrices 10

1—6 Operations on Partitioned Matrices 12

1—7 Definition and Properties of a Determinant 16

1—8 Cofactor Expansion Formula 19

1—9 Cramer's Rule 21

1—10 Adjoint and Inverse Matrices 221—11 Elementary Operations on a Matrix 24

1—12 Rank of a Matrix 27

1—13 Solvability of Linear Algebraic Equations 30

2 Characteristic-Value Problems and Quadratic Forms 46

2—1 Introduction 462—2 Second-Order Characteristic-Value Problem 482—3 Similarity and Orthogonal Transformations 52

2—4 The nth-Order Symmetrical Characteristic-ValueProblem 55

2—5 Quadratic Forms 57

3 Relative Extrema for a Function 66

3—1 Relative Extrema for a Function of One Variable 66

3—2 Relative Extrema for a Function of n IndependentVariables 71

3—3 Lagrange Multipliers 75

4 Differential Geometry of a Member Element 81

4—1 Parametric Representation of a Space Curve 81

4—2 Arc Length 82

V


CONTENTS

4—3 Unit Tangent Vector 85

4—4 Principal Normal and Binormal Vectors 86

4—5 Curvature, Torsion, and the Frenet Equations 88

4—6 Summary of the Geometrical Relations for a SpaceCurve 91

4—7 Local Reference Frame for a Member Element 92

4—8 Curvilinear Coordinates for a Member Element 94

5 Matrix Transformations for a Member Element 100

5—1 Rotation Transformation 100

5—2 Three-Dimensional Force Transformations 103

5—3 Three-Dimensional Displacement Transformations 109

Il—ANALYSIS OF AN IDEAL TRUSS

6 Governing Equations for an Ideal Truss

6—1 General 115

6—2 Elongation—Joint Displacement Relation for a Bar 116

6—3 General Elongation—Joint Displacement Relation 120

6—4 Force-Elongation Relation for a Bar 125

6—5 General Bar Force—Joint Displacement Relation 130

6—6 Joint Force-Equilibrium Equations 130

6—7 Introduction of Displacement Restraints;Governing Equations 132

6—8 Arbitrary Restraint Direction 134

6—9 Initial Instability 137

7 Variational Principles for an Ideal Truss 152

7—1 General 1527—2 Principle of Virtual Displacements 153

7—3 Principle of Virtual Forces 1597—4 Strain Energy; Principle of Stationary Potential

Energy 1627—5 Complementary Energy; Principle of Stationary

Complementary Energy 165

7—6 Stability Criteria 169

8 Displacement Method—Ideal Truss 178

8—1 General 178

8—2 Operation on the Partitioned Equations 178

8—3 The Direct Stiffness Method 180


CONTENTS

8—4 Incremental Formulation; Classical StabilityCriterion 191

8—5 Linearized Stability Analysis 200

9 Force Method—Ideal Truss

9—1 General 210

9—2 Governing Equations—Algebraic Approach 211

9—3 Governing Equations—Variational Approach 2169—4 Comparison of the Force and Mesh Methods 217

Ill—ANALYSIS OF A MEMBER ELEMENT

10 Governing Equations for a Deformable Solid 229

10—1 General 22910—2 Summation Convention; Cartesian Tensors 230

10—3 Analysis of Deformation; Cartesian Strains 23210—4 Analysis of Stress 24010—5 Elastic Stress-Strain Relations 24810—6 Principle of Virtual Displacements; Principle of

Stationary Potential Energy; Classical StabilityCriteria 253

10—7 Principle of Virtual Forces; Principle ofStationary Complementary Energy 257

11 St. Venant Theory of Torsion-Flexure ofPrismatic Members 271

11—1 Introduction and Notation 271

11—2 The Pure-Torsion Problem 273

11—3 Approximate Solution of the Torsion Problem forThin-Walled Open Cross Sections 281

11—4 Approximate Solution of the Torsion Problem forThin-Walled Closed Cross Sections 286

11—5 Torsion-Flexure with Unrestrained Warping 293

11—6 Exact Flexural Shear Stress Distribution for aRectangular Cross Section 303

11—7 Engineering Theory of Flexural Shear StressDistribution in Thin-Walled Cross Sections 306

12 Engineering Theory of Prismatic Members 330

12—1 Introduction 33012—2 Force-Equilibrium Equations 331


CONTENTS

12—3 Force-Displacement Relations; Principle ofVirtual Forces 333

12—4 Summary of the Governing Equations 339

12—5 Displacement Method of Solution—Prismatic Member 340

12—6 Force Method of Solution 349

13 Restrained Torsion-Flexure of a Prismatic Member 371

13—1 Introduction 371

13—2 Displacement Expansions; Equilibrium Equations 372

13—3 Force-Displacement Relations—Displacement Model 37513—4 Solution for Restrained Torsion—Displacement Model 37913—5 Force-Displacement Relations—Mixed Formulation 38313—6 Solution for Restrained Torsion—Mixed Formulation 389

13—7 Application to Thin-Walled Open Cross. Sections -39513—8 Application to Thin-Walled Closed Cross Sections 405

13—9 Governing Equations—Geometrically NonlinearRestrained Torsion 414

14 Planar Deformation of a Planar Member 425

14—1 Introduction; Geometrical Relations 425

14—2 Force-Equilibrium Equations 42714—3 Force-Displacement Relations; Principle of

Virtual Forces 42914—4 Force-Displacement Relations—Displacement

Expansion Approach; Principle of VirtualDisplacements 435

14—5 Cartesian Formulation 44514—6 Displacement Method of Solution—Circular Member 44914—7 Force Method of Solution 45814—8 Numerical Integration Procedures 473

15 Engineering Theory of an Arbitrary Member 485

15—1 Introduction; Geometrical Relations 48515—2 Force-Equilibrium Equations 48815—3 Force-Displacement Relations—Negligible Warping

Restraint; Principle of Virtual Forces 49015—4 Displacement Method—Circular Planar Member .49315—5 Force Method—Examples 49915—6 Restrained Warping Formulation 50715—7 Member Force-Displacement Relations—Complete

End Restraint 511

15—8 Generation of Member Matrices 517


CONTENTS

15—9 Member Matrices—Prismatic Member 52015—10 Member Matrices—Thin Planar Circular Member 52415—11 Flexibility Matrix—Circular Helix 531

15—12 Member Force-Displacement Relations—PartialEnd Restraint 535

tV—ANALYSIS OF A MEMBER SYSTEM

16 Direct Stiffness Method—Linear System 545

16—1 Introduction 54516—2 Member Force-Displacement Relations 54616—3 System Equilibrium Equations 54716—4 Introduction of Joint Displacement Restraints 548

17 General Formulation—Linear System 554

17—1 Introduction 55417—2 Member Equations 55517—3 System Force-Displacement Relations 55717—4 System Equilibrium Equations 55917—5 Introduction of Joint Displacement Restraints;

Governing Equations 56017—6 Network Formulation 56217—7 Displacement Method 56517—8 Force Method 56717—9 Variational Principles 57017—10 Introduction of Member Deformation Constraints 573

18 Analysis of Geometrically Nonlinear Systems 585

18—1 Introduction 58518—2 Member Equations—Planar Deformation 58518—3 Member Equations—Arbitrary Deformation 591

18—4 Solution Techniques; Stability Analysis 597

Index 605



Part IMATHEMATICALPRELIMINARIES



1

Introduction toMatrix Algebra

1—1. DEFINITION OF A MATRIX

An ordered set of quantities may be a one-dimensional array, such as

a two-dimensional array, such as

a11, a12, . . . ,a1,

a21, a22, . . . ,

ami,

a two-dimensional array, the first subscript defines the row location of anelement and the second subscript its column location.

A two-dimensional array having ,n rows and n columns is called a matrixof order m by n if certain arithmetic operations (addition, subtraction, multi-plication) associated with it are defined. The array is usually enclosed in squarebrackets and written as*

a11 a12 - - a1,,

a21 a22 - a2,,= = a

a,,,1 am2 a,,,,,

Note that the first term in the order pertains to the number of rows and thesecond term to the nuiñber of columns. For convenience, we refer to the orderof a matrix as simply m x n rather than of order m by n.

* In print, a matrix is represented by a boldfaced letter.

3


4 INTRODUCTION TO MATRIX ALGEBRA CHAP. 1

A matrix having only one row is called a row matrix. Similarly, a matrixhaving only one column is called a column matrix or column vector.* Bracesinstead ofbrackets are commonly used to denote a column matrix and thecolumn subscript is eliminated. Also, the elements are arranged horizontallyinstead of vertically, to save space. The various column-matrix notations are:

C11 C1

C21 C2{c1, c2,. . . , {c1} = c

If the number of rows and the number of columns are equal, the matrix is saidto be square. (Special types of square matrices are discussed in a later section.)Finally, if all the elements are zero, the matrix is called a null matrix, and isrepresented by 0 (boldface, as in the previous case).

Example 1—1

3 x 4 Matrix

4 2—1 2

3 —7 1 —8

2 4 —3 1

1 x 3 Row Matrix

[3 4 2]

3 x 1 Column Matrix

f3]or 4Jor{3,4,2}

2 2 Square Matrix

5

[2 7

2 x 2 Null Matrix

[0 0

[o o

* This is the mathematical definition of a vector. In mechanics, a vector is defined as a quantityhaving both magnitude and direction. We will denote a mechanics vector quantity, such as forceor moment, by means of an italic letter topped by an arrow, e.g., F. A knowledge of Vector algebrais assumed in this text. For a review, see Ref. 2 (at end of chapter, preceding Problems).


SEC. 1—3. MATRIX MULTIPLICATION

1—2. EQUALITY, ADDITION, AND SUBTRACTION OF MATRICES

Two matrices, a and b, are equal if they are of the same order and if cor-responding elements are equal:

a = b when

If a is of order m x n, the matrix equation

a=bcorresponds to mn equations:

= 1, 2,. . . , m=

= 1, 2,.. . ,

Addition and subtraction operations are defined only for matrices of the sameorder. The sum of two m x n matrices, a and b, is defined to be the m x nmatrix +

+ = +Similarly,

— = — bLJ]

For example, if[1 2 ii [0 —1 —1

—db=[3

ithen

[1 1 0

1 —1

and [1 3 2

—1 —1

It is obvious from the example that addition is commutative and associative:

a+b=b+a (1—6)

a+(b+c)=(a+b)+c (1—7)

1—3. MATRIX MULTIPLICATION

The product of a scalar k and a matrix a is defined to be the matrixin which each element of a is multiplied by k. For example, if

k=5 and

then[—10 +35ka=[

10 5



Scalar multiplication is commutative. That is,

ka = ak = {ka11]

To establish the definition of a matrix multiplied by a column matrix, weconsider a system of m linear algebraic equations in n unknowns, x1, .x2

+ a12x2 + + C1

a21x1 + a22x2 + + = C2

l2miXi + am2x2 + +

This set can be written as

alkxk C1 i = 1, 2, . . . ,rn

where k is a dummy index. Using column matrix notation, (1—9) takes the form

i =

Now, we write (1—9) as a matrix product:

i= 1,2,..,,rn= {c1}

1,2(1—11)

Since (1—10) and (1—Il) must be equivalent, it follows that the definitionequation for a matrix multiplied by a column matrix is

ax = ulkxk} j = 1, 2,. . . , m

This product is defined only when the column order of a is equal to the roworder of x. The result is a column matrix, the row order of which is equal tothat of a. In general, if a is of order r x s, and x of order s x 1, the productax is of orderr x 1.

Example 1—2

1 11 2a= 8

-4jx={3}

1(1)(2) + (—1)(3)

4

+ (3)(3) 9


SEC. 1—3. MATRIX MULTIPLICATION

We consider next the product of two matrices. This product is associatedwith a linear transformation of variables. Suppose that the n original variablesx1, x2,. . . ,x,, in (1—9) are expressed as a linear combination of s new variablesY1,Y2, . . . ,ys:

Xk = k = 1, 2,. . . , n (1—13)1=

Substituting for Xk in (1—10),

= i 1, 2,. . . , m

Interchanging the order of summation, and letting

i = 1,2 in(1—14)

k=i j —

the transformed equations take the form

= i 1,2,.. .,

Noting (1—12), we can write (1—15) as

py = C

where p is in x .s and y is S x 1. Now, we also express which definesthe transformation of variables, in matrix form,

x = by

where b is n x s. Substituting for x in (1—11),

aby=cand requiring (1—16) and (1—18) to be equivalent, results in the followingdefinition equation for the product, ab:

=ab = [bkJ] = [pt,] k 1,2,. . . , n

This product is defined only when the column order of a is equal to the roworder of b. In general, if a is of order r x n, and b of order n x q, the productab is of order r x q. The element at the ith row and jth column of the productis obtained by multiplying corresponding elements in the ith row of the firstmatrix and the jth column of the second matrix.



Example 1—3

(1)(1) + (0)(O) (IXI) + (O)(1) (1)(O) + 1) (1)(— 1) + (01(3)

ab = (—l)(1) + (1)(O) (—1)(l) + (1)(l) (—1)(O) + (—1)(—1) + (1)(3)

(O)(1) + (2)(O) (O)(1) + (2)(l) (0)(0) + (2)(—1) (0)(—1) + (2)(3)

[+1 +1 0 —l

ab=J_1 0 —1 +4

[ 0 +2 —2 +6

If the product ab is defined, a and b are said to be confbrmable in the orderstated. One should note that a and b will be conformable in either order onlywhen a is in x n and b is n x in. In the previous example, a and b are con-formable but b and a are not since the product ha is not defined.

When the relevant products are defined, multiplication of matrices is as-sociative,

a(bc) = (ab)c (1—20)

and distributive,a(b + c) = ab + ac(b + c)a = ha + Ca

but, in general, not commutative,

ab ba (1—22)

Therefore, in multiplying b by a, one should distinguish preinultiplication, ab,from postrnultiplication ha. For example, if a and b are square matrices of order2, the products are

[a11 a121[bij b121 — [aitbji + a12b21 a11b12 + a12b22

[a21 a22j[b21 b22j [a21b11 + a22b21 a21b12 + a22b22

[b11 b121[aji aizl — [bjjaj1 + b12a21 b11a12 + b12a22

[b21 b22j[a21 a22] [b21a11 + b22a21 b21a12 + b22a22

When ab = ha, the matrices are said to commute or to be permutable.

1—4. TRANSPOSE OF A MATRIX

The transpose of a = is defined as the matrix obtained from a byinterchanging rows and columns. We shall indicate the transpose of a by


SEC. 1—4 TRANSPOSE OF A MATRIX 9

aT = {a79]:a11 a12 a1,

021 a22 a2,a

= = (1—23)

amj am2 a,,,

a21

= [a79] = 012 022 am2

a,,,

The element, a79, at the ith row and jth column of aT, where now i varies from 1to n and j from 1 to m, is given by

a79 = (1—24)

where is the element at the jth row and ith column of a. For example,

[3 2T r3 7 5

1 a =[21 4

Since the transpose of a column matrix is a row matrix, an alternate notationfor a row matrix is

[a1, a2 a,] = (1—25)

We consider next the transpose matrix associated with the product of twomatrices. Let

p==ab (a)

where a is m x n and b is n x s. The product, p, is m x s and the element,Pu,

m= Ilukbkf — 1 (b).1 —

The transpose of p will be of order s x m and the typical element is

p79 = (c)

where now I = 1, 2 s and j = 1, 2,. . . ,m. Using (1—24) and (b), we canwrite (c) as

p79 = = =1, 2,. S

(d)k1 k1 j —

It follows from (d) that= (ab)T = bTaT

Equation (1—26) states that the transpose of a product is the product of the



transposed matrices in reversed order. This rule is also applicable to multipleproducts. For example, the transpose of abc is

(abc)T = cT(ab)T cTbTaT (1—27)

Example 1—4

ab = 13 (ab)T = [4 13 6]

6

Alternatively,

aT

== [2 —1]

(ab)T = bTaT = [2 —1] = [4 13 6]

1—5. SPECIAL SQUARE MATRICES

If the numbers of rows and of columns are equal, the matrix is said to be squareand of order n, where n is the number of rows. The elements (i = 1, 2,. .. , n)

lie on the principal diagonal. If all the elements except the principal-diagonalelements are zero, the matrix is called a diagonal matrix. We will use d fordiagonal matrices. If the elements of a diagonal matrix are all unity, the diagonalmatrix is referred to as a unit matrix. A unit matrix is usually indicated bywhere n is the order of the matrix.

Example 1—5

Square Matrix, Order 2[1 7

[3 2

Diagonal Matrix, Order 3[2 0 0

5 0

[o 0 3

Unit Matrix, Order 2

12[ 0

LO I


SEC. 1—5. SPECIAL SQUARE MATRICES

We introduce the Kronecker delta notation:

oij=0(1—28)

+1 i—j

With this notation, the unit matrix can be written as

= i,j = 1, 2 n (1—29)

Also, the diagonal matrix, d, takes the form

d = (1—30)

where d1, d2,. . . , are the principal elements. If the principal diagonal elementsare all equal to k, the matrix reduces to

= = (1—31)

and is called a scalar matrix.Let a be of order rn x n. One can easily show that multiplication of a by a

conformable unit matrix does not change a:

a(1—32)

Ima = a

A unit matrix is commutative with any square matrix of the same order.Similarly, two diagonal niatrices of order n are commutative and the productis a diagonal matrix of order a. Premultiplication of a by a conformablediagonal matrix d multiplies the ith row of a by and postmultiplicationmultiplies the jth column by

Example 1—6

[2 01[3 01[2 01 [6 0

[o —i][o 5j[O 5j[O _ij[o —5[2 01[3 'l_[ 6 2

—ij [2 7] — [—2 —7

[3 11[2 01 [6 —'

[2 7j[0 _1j[4 —7

A square matrix a for which = is called symmetrical and has theproperty that a = If = (i j) and the principal diagonal elementsall equal zero, the matrix is said to be skew-symmetrical. In this case, aT = — a.

Any square matrix can be reduced to the sum of a symmetrical matrix and askew-symmetrical matrix:

a=b+c= +

(1-33)= —



The product of two symmetrical matrices is symmetrical only when the matricesare commutative.* Finally, one can easily show that products of the type

(aTa) (aaT) (aTba)

where a is an arbitrary matrix and b a symmetrical matrix, result in symmetricalmatrices.

A square matrix having zero elements to the left (right) of the principaldiagonal is called an upper (lower) triangular matrix. Examples are:

Upper Triangular Matrix

352071004

Lower Triangular Matrix

300570214

Triangular matrices are encountered in many of the computational proceduresdeveloped for linear systems. Some important properties of triangular matricesare:

1. The transpose of an upper triangular matrix is a lower triangular matrixand vice versa.

2. The product of two triangular matrices of like structure is a triangularmatrix of the same structure.

[a11 0 1[b11 0 1 [aijbij 0I I =

[a21 b22j [a21b11 + a22b21 a22b22

1-6. OPERATIONS ON PARTITIONED MATRICES

Operations on a matrix of high order can be simplified by considering thematrix to be divided into smaller matrices, called .subina.trices or cells. Thepartitioning is usually indicated by dashed lines. A matrix can be partitionedin a number of ways. For example,

a11 012 0131 a11 a12 013 a11 a12 a13

a a21 a22 023 = a1=

031 a32 a33J 031 032 a33 a31 a32 a33

Note that the partition lines are always straight and extend across the entirematrix. To reduce the amount of writing, the submatrices are represented by

* See Prob. 1—7.


SEC. 1—6. OPERATtONS ON PARTITIONED MATRICES

a single symbol. We will use upper case letters to denote the submatriceswhenever possible and omit the partition lines.

Example 1-1

We represent

[au a12 a13

a=Ia,i a22 a23

as

a[A11 A121 or a = [A11 A12

[A21 A22J [A21 A22

where

Ia11 a121 Ia13A11 = I A12 = I[a21 La23

A21 = [a31 a32] A22 = [a33]

If two matrices of the same order are identically partitioned, the rules ofmatrix addition are applicable to the submatrices. Let

[A11 A121 [B11 8121I

(134)[A23 A22J [823 B22j

where BLJ and A13 are of the same order. The sum is

[A11 + 8fl A12 + B121a + b = (1-35)

LA2I + B21 A22 + B22j

The rules of matrix multiplication are applicable to partitioned matricesprovided that the partitioned matrices are conformable for multiplication. Ingeneral, two partitioned matrices are conformable for multiplication if thepartitioning of the rows of the second matrix is identical to the partitioning ofthe columns of the first matrix. This restriction allows us to treat the varioussubmatrices as single elements provided that we preserve the order of mul-tiplication. Let a and b be two partitioned matrices:

a [A131t = 1, 2,.. , M

b = [B1d I = 1,2 M(1—36)

k= 1,2,...,SWe can write the product as

C = ab = [CIk]M

C1 ,,...,

— ik i . i —1,

when the row partitions of b are consistent with the column partitions of a.



As an illustration, we consider the product

au a12 a13 h1

ab = 1221 a22 a23 h2

1233 a32 033 b3

Suppose we partition a with a vertical partition between the second and thirdcolumns,

1211 1212 a13

a = a21 a22 a23 = [A11A12]

a31 a32 a33

For the rules of matrix multiplication to be applicable to the submatrices of a,we must partition b with a horizontal partition between the second and thirdrows. Taking

the product has the form

= [A,1A12] = A11B11 + A12B21

The conformability of two partitioned matrices does not depend on thehorizontal partitioning of the first matrix or the vertical partitioning of thesecond matrix. To show this, we consider the product

a12 a13 b11 b12

ab £121 1222 a23 1322

1231 a32 1233 b31 b32

Suppose we partition a with a horizontal partition between the second andthird rows:

a11 1212 C1j3 rA11a 1221 a22 1223 =

1231 a32 a33

Since the column order of A11 and A21 is equal to the row order of b, nopartitioning of b is required. The product is

[A111 [A11bab = LA2ijb = [A21b

As an alternative, we partition b with a vertical partition.

b12

b = b21 b22 = [811B12]

b31

In this case, since the row order of B11 and B12 is the same as the column


SEC. 1—6. OPERATIONS ON PARTITIONED MATRICES

order of a, no partitioning of a is necessary and the product has the form

ab = a[B11B12] = [aBj1 aBi2]

To transpose a partitioned matrix, one first interchanges the off-diagonalsubmatrices and then transposes each submatrix. If

A11 A12 A1,,

a = A21 A22

Arnt Am2

thenAT1 AT1AT AT . . . AT

AT AT . . . AT

A particular type of matrix encountered frequently is the quasi-diagonalmatrix. This is a partitioned matrix whose diagonal submatrices are square ofvarious orders, and whose off-diagonal submatrices are null matrices. Anexample is

a11 0 0

a= 0 a22

0 a32 a33

which can be written in partitioned form as

a= [Ai

A2]where

A1 = [a11] A2 = [a22 a23]a32 a33

and 0 denotes a null matrix. The product of two quasi-diagonal matrices oflike structure (corresponding diagonal submatrices are of the same order) isa quasi-diagonal matrix of the same structure.

A1 0 ... 0 B1 0 ... 0 A1B1 0 ... 0

0

A 0

A and are of the same order.We use the term quasi to distinguish between partitioned and unpartitioned

matrices having the same form. For example, we call

(1—40)

a lower quasi-triangular matrix.


16 INTRODUCTION TO MATRIX ALGEBRA Cl-lAP. 1

1—7. DEFINITION AND PROPERTIES OF A DETERMINANT

The concept of a determinant was originally developed in connection withthe solution of square systems of linear algebraic equations. To illustrate howthis concept evolved, we consider the simple case of two equations:

a11x1 + a12x2 =a21X1 + a22x2 = C2

Solving (a) for x3 and x2, we obtain

(a11a22 — a12a21)x1 c2a12

(a11a22 — a12a21)x2 = —c1a21 + c2a11

The scalar quantity, a1 1a22 — a21 a2 defined as the determinant of the second-order square array (i,j 1, 2). The determinant of an array (or matrix) isusually indicated by enclosing the array (or matrix) with vertical lines:

a11 a12= al = a31a22 — a12a21

a21 a22

We use the terms array and matrix interchangeably, since they are synony-mous. Also, we refer to the determinant of an eth-order array as an nth-orderdeterminant. It shou'd be noted that determinants are associated only withsquare arrays, that is, with square matrices.

The determinant of a third-order array is defined as

a11 a12 a13 +a11a22a33

a21 a22 a23 = —a12a21a33 + a12a23a31 (1—42)

a31 a32 a33 +a13a21a32 — a13a22a31

This number is the coefficient of x1, x2, and x3, obtained when the third-ordersystem ax c is solved successively for x1, x2. and x3. Comparing (l—41) and(1—42), we see that both expansions involve products which have the followingproperties:

1. Each product contains only one clement from any row or column andno element occurs twice in the same product. The products differ onlyin the column subscripts.

2. The sign of a product depends on the order of the column subscripts,e.g., +a11a22a33 and —a11a23a32,

These properties are associated with the arrangement of the column subscripts

and can be conveniently described using the concept of a permutation, which

is discussed below.A set of distinct integers is considered to be in natural order if each integer

is followed only by larger integers. A rearrangement of the natural order iscalled a permutation of the set. For example, (1, 3, 5) is in natural order and


SEC. 1—7. DEFINITION AND PROPERTIES OF A DETERMINANT

(1,5,3) is a permutation of(1, 3,5). If an integer is followed by a smaller integer,the pair is said to form an inversion. The number of inversions for a set is definedas the sum of the inversions for each integer. As an illustration, we considerthe set (3, 1, 4, 2). Working from left to right, the integer inversions are:

Integer Inversions Total3 (3, 1)(3, 2) 2

1 None 04 (4,2) 1

2 None 0

3

This set has three inversions. A permutation is classified as even (odd) if thetotal number of inversions for the set is an even (odd) integer. According tothis convention, (1, 2, 3) and (3, 1, 2) are even permutations and (1, 3, 2) is anodd permutation. Instead of cbunting the inversions, we can determine thenumber of integer interchanges required to rearrange the set in its natural ordersince an even (odd) number of interchanges corresponds to an even (odd)number of inversions. For example, (3,2, 1) has three inversions and requiresone interchange. Working with interchanges rather than inversions is practicalonly when the set is small.

Referring back to (1—41) and (1—42), we see that each product is a permutationof the set of column subscripts and the sign is negative the permutationis odd. The number of products is equal to the number of possible permutationsof the column subscripts that can be formed. One can easily show that thereare possible permutations for a set of n distinct integers.

We let . . . , ce,,) be a permutation of the set (1, 2,. . . , n) and define• as

• + I when . . , is an even permutation(1—43)

— 1 when . .. , a,,) is an odd permutation

Using (1—43), the definition equation for an ,ith-order determinant can bewritten as

a11 a12 a1,,

a21 a22 a2,,= (1—44)

1

where the summation is taken over all possible permutations of (1, 2, . . , n).

Factorial n = = n(n — 1)(n — 2) . • (2)(1).



Example 1—8

The permutations for n = 3 are

a1=1 a33 e123=+1cxi—1 x23 a32 e132=—1

= 2 1 = 3z1=2 a3=1 e231=+1

a32 e312=+1a3=1 e321—-—1

Using (1—44), we obtain

a11 a12 a13 a11a22a33 — a11a23a32

a21 a22 a23 = —a12a21a33 + a12a23a31

a32 a33 +a13a21a32 — a13a22a31

This result coincides with (1—42).

The following properties of determinants can be established* from (1—44):

1. If all elements of any row (or column) are zero, the determinant is zero.2. The value of the determinant is unchanged if the rows and columns are

interchanged; that is, aT! = a!.3. If two successive rows (or two successive columns) are interchanged, the

sign of the determinant is changed.4. If all elements of one row (or one column) are multiplied by a number k,

the determinant is multiplied by k.5. If corresponding elements of two rows (or two columns) are equal or in

a constant ratio, then the determinant is zero.6. If each element in one row (or one column) is expressed as the sum of

two terms, then the determinant is equal to the sum of two determinants,in each of which one of the two terms is deleted in each element of thatrow (or column).

7. If to the elements of any row (column) are added k times the cor-responding elements of any other row (column), the determinant isunchanged.

We demonstrate these properties for the case of a second-order matrix. Let

a= [a31

[a21 a22

The determinant isa! = a11a22 — a12a21

Properties 1 and 2 are obvious. It follows from property 2 that laTl a!. We

* See Probs. 1—17, 1—18, 1—19.


SEC. 1—8. COFACTOR EXPANSION FORMULA

illustrate the third by interchanging the rows of a:

a' = [a21 a22

a12

a'! = a21a12 — a11a22 = —Ia!

Property 4 is also obvious from (b). To demonstrate the fifth, we take

a21 = ka11 a22 = ka12Then

a! = a11(kaj2) a12(ka1j) = 0Next, let

a11 + c11 a12 = b12 + c12

According to property 6,al hi + ci

where

ibib11 b12

ci=

a21 a22 a21 a22

This result can be obtained by substituting for O.ii and a12 in (b). Finally, toillustrate property 7, we take

b12 = a12 + ka22b21 = a21

b22 = a7,Then,

ibi = (a11 + ka21)a22 — (a12 + ka22)a21 = a!

1-8. COFACTOR EXPANSION FORMULA

If the row and column containing an element, in the square matrix, a,are deleted, the determinant of the remaining square array is called the minorof and is denoted by The cofactor of denoted by is related tothe minor of by

= (— (1—45)

As an illustration, we take 328a= 1 7 4

531The values of and associated with a23 and a22 are

M23. = = —1 A23 = (— 1)5M23 = + 1

M22 = = —37 A22 = (—1)4M22 = —37



Cofactors occur naturally when (.1 —44) is expanded9 in terms of the elementsof a row or column. This leads to the following expansion formula, calledLaplace's expansion by cofactors or simply Laplace's expansion:

=a1kAIk akJAkJ (1 —46)

Equation (1—46) states that the determinant is equal to the sum of the productsof the elements of any single row or column by their cofactors.

Since the determinant is zero if two rows or columns are identical, if followsthat

= 0k1

(147)0 s

k I

The above identities are used to establish Cramer's rule in the following section.

Example 1—9

(1) We apply (1—46) to a third-order array and expand with respect to the first row:

a11 a12 a13

(121 a23 a23

a31 a32 a33

2a22 023 023 a22

= + + 0j3(— 1)a33 a31 (133 035 a32

a11(a22a33 — a23a32) + a52(—a21a33 + a23a31) + a53(a21a32 — 022035)

To illustrate (1 —47), we take the cofactors for the first row and the elements of the secondrow:

= a21(a22a33 — a23a32) + a22(—a21a33 + a23a31) + a23(a21a32 — a22a31) 0

(2) Suppose the array is triangular in form, for example, lower triangular. Expandingwith respect to the first row, we have

0 0(122 0

a21 a22 0 = a11 = (a51)(a22a33) = a11a22a33032 033

031 a32 033

Generalizing this result, we find that the determinant of a triangular matrix is equal tothe product of the diagonal elements. This result is quite useful.

* See Probs. 1—20, 1—21.f See Ref. 4, sect. 3 15, for a discussion of the general Laplace expansion method. The expansion

in terms of cofactors for a iow Or a COlUmn is a special case of the general method.


SEC. 1—9. CRAMER'S RULE

The evaluation of a determinant, using the definition equation (1—44) or thecofactor expansion formula (1—46) is quite tedious, particularly when the arrayis large. A number of alternate and more efficient numerical procedures forevaluating determinants have been developed. These procedures are describedin References 9—13.

Suppose a square matrix, say c, is expressed as the product of two squarematrices,

c='aband we want cJ. It can be shown* that the determinant of the product of twosquare matrices is equal to the product of the determinants:

ci = a! hi (1—48)

Whether we use (1—48) or first multiply a and b and then determine lab! dependson the form and order of a and b. If they are diagonal or triangular, (1—48)is quite efficient. t

Example 1—10

[1 31 r2 3

5] 4

a! = hi = Ic! = —20

Alternatively,

c [ and cj = —20[11 29J

[1 31 r2 0a=[0 5]

b__[1

a! = 5 bi = 8 Ic! = +40

Determining c first, we obtain

rs 121

= [5 20]and ci = +40

1—9. CRAMER'S RULE

We consider next a set of n equations in n unknowns:

= j = 1, 2, . . . , ii (a)

* See Ref. 4, section 3—16.t See Prob. 1 —25 for an important theoretical application of Eq. 1—48.



Multiplying both sides of (a) by Air, where r is an arbitrary integer from 1 to n,and summing with respect to j, we obtain (after interchanging the order ofsummation)

Xk =k1 j=1

Now, the inner sum vanishes when r k and equals al when r = k. Thisfollows from (1—47). Then, (b) reduces to

lalxr =

The expansion on the right side of (c) differs from the expansion

al=

ajrAj.

only in that the rth column of a is replaced by c. Equation (c) leads to Cramer'srule, which can be stated as follows:

A set of n linear algebraic equations in n unknowns, ax = c, has aunique solution when 0. The expression for Xr (r = 1, 2 n) isthe ratio of two determinants; the denominator is al and the numeratoris the determinant of the matrix obtained from a by replacing the rthcolumn by c.

If jaf = 0, a is said to be singular. Whether a solution exists in this ease willdepend on c. All we can conclude from Cramer's rule is that the solution, ifit exists, will not be unique. Singular matrices and the question of solvabilityare discussed in Sec. 1 —13.

1—10. ADJOINT AND INVERSE MATRICES

We have shown in the previous section that the solution to a system of nequations in n unknowns,

i,j 1, 2,..., n

can be expressed as

1 1, 2,. . ., ii

(note that we have taken r = I in Eq. c of Sec. 1—9). Using matrix notation,(b) takes the form

[Au]T{cj}

Equation (e) leads naturally to the definition of adjoint and inverse matrices.


SEC. 1—10. ADJOINT AND INVERSE MATRICES 23

We define the adjoint and inverse matrices for the square matrix a of order n as

adjoint a = Adj a = (1—49)

inverse a = a1 Adj a (1—50)

Note that the inverse matrix is defined only for a nonsingular square matrix.

Example 1—11

We determine the adjoint and inverse matrices for

123a= 2 3 1

412The matrix of cofactors is

5 0 —10

—1 —10 +7—7 +5 —1

Also, al = —25. Then5 —i —7

Adja 0 —10 +5—10 +7 —1

—1/5 + 1/25 +7/25= —-- Adj a = 0 + 2/5 — 1/5

a+2/5 —7/25 + 1/25

Using the inverse-matrix notation, we can write the solution of (a) as

x =

Substituting for x in (a) and c in (d), we see that a1 has the property that

a1a = aa' =

Equation (1—51) is frequently taken as the definition of the inverse matrixinstead of (1—50). Applying (1—48) to (i—Si), we obtain

It follows that (1—Si) is valid only when 0. Multiplication by the inversematrix is analogous to division in ordinary algebra.

If a is symmetrical,, then a 1 is also symmetrical. To show this, we take thetranspose of (1—5 1), and use the fact that a =. aT:

(a_la)T =



Premultiplication by a' 1 results in

— a"'and therefore a1 is also symmetrical. One can also show* that, for anynonsingular square matrix, the inverse and transpose operations can be inter-changed:

bT,_t = (1—52)

We consider next the inverse matrix associated with the product of two squarematrices. Let

c = ab

where a and b are both of order n x n and nonsingular. Premultiplicationby and then b1 results in

a'c = b

(b'a'')c =It follows from the definition of the inverse matrix that

(ab)1 = (1—53)

In general, the inverse of a multiple matrix product is equal to the product ofthe inverse matrices in reverse order. For example,

=

The determination of the inverse matrix using the definition equation (1 —50)is too laborious when the order is large. A number of inversion proceduresbased on (1—51) have been developed. These methods are described in Ref. 9—13.

1—11. ELEMENTARY OPERATIONS ON A MATRIX

The elementary operations on a matrix are:'

1. The interchange of two rows or of two columns.2. The multiplication of the elements of a row or a column by a number

other than zero.3. The addition, to the elements of a row or column, of k times the cor-

responding element of another row or column.

These operations can be effected by premultiplying (for row operation) orpostmultiplying (for column operation) the matrix by an appropriate matrix,called an elementary operation matrix.

We consider a matrix a of order x n. Suppose that we want to interchangerowsj and k. Then, we premultiply a by an rn x in matrix obtained by modifyingthe mth-order unit matrix, I,,,, in the following way:

1. Interchange and 5k•2. Interchange and

* See Prob. 1—28.


SEC. 1—11. ELEMENTARY OPERATIONS ON A MATRIX 25

For example, if a is 3 x 4, premultiplication by

001010100

interchanges rows 1 and 3 and postmultiplication by

1000000100100100

interchanges columns 2 and 4. This simple example shows that to interchangerows, we first interchange the rows of the conformable unit matrix and pre-multiply. Similarly, to interchange columns, we interchange columns of theconformable unit matrix and postmultiply.

The elementary operation matrices for operations (2) and (3) are also obtainedby operating on the corresponding conformable unit matrix. The matrix whichmultiplies row j by is an mth order diagonal matrix having d1 = 1 for i jand = Similarly, postmultiplication by an nth order diagonal matrixhaving = 1 for i j and = will multiply thejth column by Supposethat we want to add times row jto row k. Then, we insert in the kth rowand jth column of and premultiply. To add z times column jto column k,we put in the jth row and kth column of and postmu-ltiply.

We let e denote an elementary operation matrix. Then, ea represents theresult of applying a set of elementary operations to the rows of a. Similarly,ac represents the result of applying a set of elementary operations to thc columnsof a. In general, we obtain e by applying the same operations to the conformableunit matrix. Since we start with a unit matrix and since the elementary opera-tions, at most, change the value of the determinant by a nonzero scalar factor,*it follows that e will always be nonsingular.

Example 1—12

We illustrate these operations on a third matrix:

1 1/2 1/5

a= 3 7 2

—2 1 5

We first:

1. Add (—3) times the first row to the second row.2. Add (2) times the first row to the third row.

* See properties of determinants (page 18).



These operations are carried out by premultiplying by

100—3 1 0

201and the result is

1 1/2 1/5

0 11/2 7/5

0 2 27/5

Continuing, we multiply the second row by 2/11:

1 0 0 1 1/2 1/5 1 1/2 1/5

0 2,'Il 0 0 11/2 7/5 = 0 1 14/55

0 0 1 0 2 27/5 0 2 27/5

Next, we add (—2) times the second row to the third row:

1 0 0 1 1/2 1/5

1 1 1 14/55

1 0 0 269/55

Finally, we multiply the third row by 55/269. The complete set of operations is

100 1 0 0110 0 100 11/21/5010 0 1 012/110 —310 3720 0 55/269 0 —2 1 0 0 1 2 0 1 —2 1 5

1 1/2 1/5

= 0 1 14/55 =b00 1

This example illustrates the reduction of a square matrix to a matrix usingelementary operations on rows, and is the basis for the Gauss elimination solution scheme(Refs. 9, 11, 13). We write the result as

ea = b

where e is the product of the four operation matrices listed above:

0 0

e —6/11 2/11 0

+ 1870/2959 —220/2959 55/269

We obtain e by applying successive operations, starting with a unit matrix. This is moreconvenient than listing and then multiplying the operation matrices for the various steps.The form of e after each step is listed below:

Initial Step 1 Step 2

100 100 1 0 0

0 1 0 —3 1 0 —6/11 2/11 0

001 201 2 0 1


SEC. 1—12. RANK OF A MATRIX 27

Step 3 Step 4

1 0 0 [1 0 0

—6/11 2/11 0}

—6/11 2/11 0

+34/11 —4/11 0 L+187o/2959 —220/2959 55/269

Two matrices are said to be equivalent if one can be derived from the otherby any finite number of elementary operations. Referring to Example 1 —12, thematrices

1 1/2 1/5 1 1/2 1/5

3 7 2 and 0 1 14/55

—21 5 00 1

are equivalent. Tn general, a and b are equivalent if

b = paq (1—54)

where p and q are nonsinqular. This follows from the fact that the elementaryoperation matrices are nonsingular.

1-12. RANK OF A MATRIX

The rank, r, of a matrix is defined as the order of the largest square array,formed by deleting certain rows and columns, which has a nonvanishing deter-minant. The concept of rank is quite important since, as we shall see in thenext section, the solvability of a set of linear algebraic equations is dependenton the rank of certain matrices associated with the set.

Let a be of order in x n. Suppose the rank of a is r. Then a has r rowswhich are linearly independent, that is, which contain a nonvanishing deter-minant of order r, and the remaining rn — r rows are linear combinations ofthese r rows. Also, it has n — r columns which are linear combinations of rlinearly independent columns.

To establish this result, we suppose the determinant associated with the first rrows and columns does not vanish. If a is of rank r, one can always rearrangethe rows and columns such that this condition is satisfied. We consider the(r + 1)th-order determinant associated with the first r rows and columns, rowp, and column q where r < p in, r < q n.

a11 a12 aIr 01q

a21 a22 azq

(1—55)

an ar2 arr 0rq

apq

We multiply the elements in rowj by (j 1, 2,. . . ,r) and subtract the resultfrom the last row. This operation will not change the magnitude of Ar+t (seeSec. 1—7). In particular, we determine the constants such that the first r elements



in the last row vanish:

a11 021 -

012 022 a,2 = 0p2(1—56)

a1, a2r apr

Equation (1—56) has a unique solution since the coefficient matrix is non-singular. Then (1 —55) reduces to

a11 012 a1,

a21 022

Ar+i (1—57)a,2 0rr

0 0 0where

= apq — Orq] (1—58)

Applying Laplace's expansion formula to (1—57), we see that A,÷1 vanisheswhen

a is of rank r, A,÷1 vanishes for all combinations of p and q. Itfollows that

apq = [aiq. 02p ., : m(1—59)

Apr

Combining (1—56) and (1—59), we have

a11 a21 0r1 4p1 0p1

012 022r r + 2 m (1—60)

Equation (1—60) states that the last m — r rows of a are linear combinationsof the first r rows. One can also show* that the last n — r columns of a arelinear combinations of the first r columns.

Example 1—13

Consider the 3 x 4 matrix 1234a=21 32

5 7 12 14

? See Prob. 1—39.


SEC. 1—12. RANK OF A MATRIX 29

We see that a is at least of rank 2 since the determinant associated with the first two rowsand columns is finite, Then, the first two rows are linearly independent. We consider thedeterminant of the third-order array consisting of columns 1, 2, and q:

1 2 ajq

2 1 a2q

5 7 a3q

Solving the system,+ 223 = 5

22 = 7

we obtain

If a is of rank 2, A3 must vanish. This requires

a3q = 2 101q + 22a2q = 3ajq + a3q

q = 3,4

Since a33 and (134 satisfy this requirement, we conclude that a is of rank 2. The rows arerelated by

(third row) = + 3 (first row) + (second row)

One can show* that the elementary operations do not change the rank ofa matrix. This fact can be used to dctcrmine the rank of a matrix. Suppose bdefined by (1—61) is obtained by applying elementary operations to a. We knowthat band a have the same rank. It follows that a is of rank p. A matrix havingthe form of b is called an echelon matrix. When a is large, it is more efficientto reduce it to an echelon matrix rather than try to find the largest nonvanishingdeterminant:

(pxpt

a11 I b12 ...(121

0 II, b2p B12

(1—61)

0 0

Example 1—14

[i 2 3 4

a=)2 1 3 2

[5 7 12 12

First, we eliminate and a31, using the first row:

1 2 3 4

0 —3 —3 —60 —3 —3 —8

* See Prob. 1—40.



Next, we eliminate aW, using the second row:

—1 2 3 4

0 —3 —3 —60 0 0 —2

At this point, we see that r = 3. To obtain b, we multiply the second row by — 1/3, the thirdrow by — 1/2, and interchange the third and fourth columns:

1243b= 0 1 2 1

0010

Suppose a is expressed as the product of two rectangular matrices:(rnxn) (nxs)

a = b c (1—62)

One can show* that the rank of a cannot be greater than the minimum valueof r associated with b and C:

r(a) ruin [r(b), r(c)] (1—63)

As an illustration, consider the product

[—1/2 +1/2 01a

— [—1/2 +1/2 1]0 I

Since each matrix is of rank 2, the rank of a will be Evaluating the product,we obtain

[0 0

1

It follows that a is of rank 1.

1—13. SOLVABILITY OF LINEAR ALGEBRAiC EQUATIONS

We consider first a system of two equations in three unknowns:

[:: :: (1-64)

Suppose a is of rank 2 anda11

0 (1—65)a21 a22

* See Prob. 1—44.


SEC. 1—13. SOLVABILITY OF LINEAR ALGEBRAIC EQUATIONS

If a is of rank 2, we can always renumber the rows and columns such that (1—65)is satisfied. We partition a and x,

a[a11 a12 a131 [A1 A2] (1—66)[a21 a22 a23j

X1çx1

1x2

and write (1—64) as A1X1 + A2X2 = c. Next, we transfer the term involvingX2 to the right-hand side:

A1X1 = c — A2X2 (1—67)

Since jA1j 0, it follows from Cramer's rule that (1—67) has a unique solutionfor X1. Finally, we can write the solution as

= Aj'(c — A2X2) (1—68)

Since X2 is arbitrary, the system does not have a unique solution for a given c.The order of X2 is generally called the defect of the system. The defect for thissystem is 1.

If a is of rank 1, the second row is a scalar multiple, say A, of the first row.Multiplying the second equation in (1—64) by 1/A, we have

a11x1 + a12x2 + a13x3 = C1(1—69)

a11x1 + a12x2 + a13x3 = c2/A

Tf c2 2cr, the equations are inconsistent and no solution exists. Then, whena is of rank 1, (1—64) has a solution only if the rows of c are related in the samemanner as the rows of a. If this condition is satisfied, the two equations in (1—69)are identical and one can be disregarded. Assuming that 0, the solution is

x1 = (1/a11)(c1 — a12x2 — a13x3) (1—70)

The defect of this system is 2.The procedure followed for the simple case of 2 equations in 3 unknowns is

also applicable to the general case of in equations in n unknowns:

a11 a12 a1,, x1 C1

(221

a is of rank in, there exists an mth order array which has a nonvanishingdeterminant. We rearrange the columns such that the first in columns are



linearly independent. Partitioning a and x,

a11 a12 ai,m+1az,,, a2m÷1 az,,

= [ A1 A2 ](mxm)

am2 amni am,m÷1 a,,,, (1—72)

{x1 X2 Xm Xm+i •.. x4 = { X1 X2 }(m 1) ((n—rn)x 1)

we write (1—71) asA3X1 c — A2X2 (1—73)

Since IA1I 0, (1—73) can be solved for X1 in terms of c and X2. The defectof the set is n — m, that is, the solution involves n — m arbitrary constantsrepresented by X2.

Suppose a is of rank r where r < m. Then, a has r rows which contain anrth-order array having a nonvanishing determinant. The remaining m — r rowsare linear combinations of these r rows. For (1—71) to be consistent, that is,have a solution, the relations between the rows of c must be the same as thosefor a. The defect for this case is n — r.

Example 1—15

As an illustration, consider the third-order system

a11x1 + a12x2 + a13x3 = C1

a21x1 + a22x2 + a73x3 = (a)

a31x1 + a32x3 + = C3

Suppose that r = 2 and the rows of a are related by

(third row) = (first row) + (second row) (b)

For (a) to be consistent, the elements of c must satisfy the requirement,

A1C1 + 22C2 (c)

To show this, we multiply the first equation in (a) by the second by —A2, and add tothese equations the third equation. Using (b), we obtain

0 = C3 — — I12C2 (d)

Unless the right-hand side vanishes, the equations are contradictory or inconsistent and nosolution exists. When e 0, (c) is identically satisfied and we see that (a) has a nontrivialsolution (x 0) only when r < 3. The general case is handled in the same manner.*

See Prob. 1—45.


REFERENCES 33

In general, (1 —71) can be solved when r < ,n if the relations between the rowsof a and c are identical. We define the augmented matrix, ci, for (1—71) as

a11 a12 C1

= a2,,[a cJ (1—74)

afl,2 a,flfl Cm

When the rows of a and c are related in the same way, the rank of tz is equal tothe rank of a. It follows that (1—71) has a solution only if the rank of the aug-mented matrix is equal to the rank of the coefficient matrix:

= r(a) (1—75)

Note that (1—75) is always satisfied when r(a) = m for arbitrary c.We can determine r(cz) and i(a) simultaneously using elementary operations

on provided that we do not interchange the elements in the last column. Thereduction can be represented as

(1-76)

where is of rank r(a). If has a nonvanishing element, > r(a) andno solution exists.

When r(a), (1—71) contains r independent equations involving n un-knowns. The remaining m — r equations are linear combinations of theser equations and can be disregarded. Thus, the problem reduces to first findingr@) and then solving a set of r independent equations in n unknowns. Thecomplete problem can be efficiently handled by using the Gauss elimmationprocedure (Refs. 9, 11, 13).

REFERENCES

1. R. A., W. J. DUNCAN and A. R. cOLLAR: Elementary Matrices, cambridgeUniversity Press, London, 1963.

2. THOMAS, G. B., JR.: Calculus and Analytical Geometry. Addison-Wesley PublishingCo., Reading, Mass., 1953.

3. BODBWIG, E.: Matrix calculus, Interscience Publishers, New York, 1956.4. HOUN, F. E.: Elementary Matrix Algebra, Macmillan Co., New York, 1958.5. HADLEY, G.: Linear Algebra, Addison-Wesley Publishing Co., Reading, Mass., 1961.6. HOUSEHOLDER, A. S.: The Theory of Matrices in Numerical Analysis, Blaisdell,

Waltham, Mass., 1964.7, NOBLE, B.: Applied Linear Algebra, Prentice-Hall, New York, 1969.8. HIL DEBRAND, F. B.: Methods of Applied Mathematics, Prentice-Hall, New York,

1952.9. Faddeeva, V. N.: Computational Methods of Linear Algebra, Dover Publications,

New York, 1959.



10. RALSTON, A. and H. S. WILF: Mathematical Methods for Digital Computers, Vol. 1,

Wiley, New York, 1960.11. RALSTON, A. and H. S. WILF: Mathematical Methods for Digital Computers, Vol. 2,

Wiley, New York, 1967.12. BEREZIN, I. S. and N. P. ZHIDKOV: Computing Methods, Vols. I and 2, Addison-Wesley

Publishing Co., Reading, Mass., 1965.13. FORSYTHE, G. E., and C. B. MALER: Computer Solution of Linear Algebraic Systems,

Prentice-Hall, New York, 1967.14. VARGA, R. S.: Matrix Iterative Analysis, Prentice-Hall, New York, 1962.15. CONTE, S. D.: Elementary Numerical Analysis, McGraw-Hill, New York, 1965.

PROBLEMS

1—1. Carry out the indicated operations:

(a)1 4 0 2 .3 5

321 +713510 056

(b)[2 1

1 6j[ 3 —1 2

(c)

[1 21 [—i ii [2 33[3 4j+2[0 3j4[l

(d) [i —2152[—3 4J 55

(e)[—i l][4 1

[ 2 —3j [2 3

[4 11 [—i I

[2 3j [ 2 —3

1—2. Expand the following products:

[a1, a2, . . . , {b1, b2, . . . , b,j

(b){ai, a2, . . . , [b1, b2

(c)[c1 01 [a11 a12

[o c2j [a21 a22


PROBLEMS 35

(d)[Cii a121 0

[021 022j [0 C2

1—3. Show that the product of

Sl= a1+ a2 + 03 =

S2 = + b2 + b3=

bk

can be written as3 3

S1S2I 1

Generalize this result for the sum of n elements.1—4. Suppose the elements of a and b are functions of y. Let

cia — db — [dblk]dy [dv j dy — [dy j

Using (1—19), show that ifc = ab

thendc db da

= a — + — bdy dy dy

1—5. Consider the triple product, abc. When is this product defined? Let

p = abc

Determine an expression for What is the order of p? Determine for thecase where c aT.

1—6. Evaluate the following products:

(a)

F 41 [1 21 [5 1

[—2 ij [2 sj [4 1

(b)

(a + b)(a + b) where a is a square matrix.

1—7. Show that the product of two symmetrical matrices is symmetricalonly when they are commutative.

1—8. Show that the following products are symmetrical:

(a)aTa

(b)aTba where b is symmetrical

(c)bTaTcab where c is symmetrical


36 INTRODUCTION TO MATRtX ALGEBRA CHAP. 1

1-9. Evaluate the following matrix product, using the indicated sub-matrices:

1

3 1 3

4 511—10. Let c = ab. Show that the horizontal partitions of c correspond to

those of a and the vertical partitions of c correspond to those of b. Hint: SeeEq. (1—37).

1—11. A matrix is said to be symmetrically partitioned if the locations ofthe row and column partitions coincide. For example,

a11 012 0131

a21 a22 a23

a31 a32 a33

is symmetrically partitioned and

a11 a12 a13

a31 a32 033

is unsymmetrically partitioned. Suppose we partition a square matrix withN — 1 symmetrical partitions.

a i,j = 1, 2,. . ., N

(a) Deduce that the diagonal submatrices are square and Ars, havethe same order.

(b) If a = aT, deduce thatArs =

1—12. Consider the product of two square nth order matrices.

c = ab

(a) If a and b are symmetrically partitioned, show that CJk, AJk, are ofthe same order. Illustrate for the case of one partition, e.g.,

[A11 A12a

— [A2, A22

(b) Suppose we symmetrically partition c. What restrictions are placedon the partitions of a and b? Does it follow that we must also partitiona and b symmetrically? Hint: See Prob. 1—10.

1—13. Consider the triple product,

C =

a symmetric rth-order square matrix and a is of order r x ii. Supposewe symmetrically partition c. The order of the partitioned matrices are indicatedin parentheses.

(pxp) (pxq)

(nxn) — [C11 C12

— [c21 c22(q><p) (qxg)


PROBLEMS 37

(a) Show that the following partitioning of a is consistent with that of c.(r x n) (r X p1 (r x q)

a ={A1 A2]

(b) Express (j, k = 1,2) in terms of A1, A2, and b.1—14. Let d = [Di] be a quasi-diagonal matrix. Show that

da = [DJAJk]bd = [BJkDk]

when the matrices are conformably partitioned.

1—iS. Determine the number of inversions and interchanges for thefollowing sets.

(a) (4,3,1,2)(b) (3, 4, 2, 1)

1—16. How many permutations does (1,2, 3, 4, 5) have?

1—17. Consider the terms

1a1122a1h3

The first subscripts in (a) are in natural order. We obtain (b) by rearranging (a)such that the second subscripts are in natural order. For example, rearranging

e231 a12 a23 a31 = (2, 3, 1)we obtain

e312a31a12a23 fi = (3, 1,2)

Show that if is an even permutation, (fl1, is also an evenpermutation. Using this result, show that

=p

and, in general,al =

1—18. Consider the terms

,

Suppose that= = =

Then, (b) takes the form

Show that(c) = —(a)

Generalize this result and establish that the sign of a determinant is reversedwhen two rows are interchanged.

1—19. Consider the third-order determinant

a! =


3 3

—

=

Following this approach, establish Laplace's cofactor expansion formula foran nth-order determinant.

1—22. Use Laplace's expansion formula to show that

(pxp) (pxn)

- = albFa(nXn)

1—23. Consider the quasi-diagonal matrix,(pxp)

d_[D1 0

[o D2(qxp) (qxq)

By expressing d as0

[o Iqj[0 D2


Suppose the second row is a multiple of the first row:

=

Show that = 0. (Hint: = — Generalize this result and establishproperties 5 and 7 of Sec. 1—7.

1—20. Suppose all the integers of a set are in natural order except for oneinteger, say n, which is located at position p. We can put the set in naturalorder by successively interchanging adjacent integers. For example,

3 l2—* 132—* 123231 —*21 3—* 123

Show that In — p1 adjacent interchanges (called transpositions) are required.It follows that the sign of the resulting set is changed by

1) In —

1—21. We can write the expansion for the third-order determinant as\

( ) . . —i=1 \j k / t,J, —

Using the result of the previous problem,

=and (a) reduces to

1 0

9

o o ...i

0

0 0

a

b11 b12

b22

bni k12

b2P

a11 012 (un

a22 (12,7

0n1 0n2


PROBLEMS

show that dl = D11 D21. Verify this result for

1100d—

2 3 0 0

00210053

Generalize ford [A,

1—24. Let(pxp) (pxq)

[Gii 0g

— [G2, G22(qxq)

Show that= jG1 G221

Generalize for a quasi-triangular matrix whose diagonal submatrices are square,of various orders.

1—25. Suppose we express a as the product of a lower triangular matrix, g,and an upper triangular matrix, b.

a11 a12 g11 0 0 b11 b12

a21 a22 az,, = g21 g22 0?

b22

'1n2 1n2 Yflfl 0 0 bflfl

We introduce symmetrical partitions after row (and column) p and write theproduct as

(pxp) (pxq) (pxp) (pxq) (pxp) (pxq)[A1, A121 [G1, 0 1[Bi, B12

LA21 A22J — [G2, G22J[0 B22(qxp) (qxq) (qxp) (qxq) (qxp) (qxq)

Note that the diagonal submatrices of g and b are triangular in form.(a) Show that

= A.11

G11B12 = A12

G21B11 = A21

G21B12 + G22B22 = A22

(b) Show thatA111 = 1G111 BijI

andal = 1G111 G22j 1B111 1B221

(c) Suppose we require that

By taking p = 1, 2,. . . , n — 1, deduce that this requirement leads to the



following n conditions on the elements of a:

a11 a12

2

aj1 aj2

The determinant of the array contained in the firstj rows and columns is calledthe jth-order discriminant.

1—26. Does the following set of equations have a unique solution?

1 2 3 x1 2

1 3 5 X2 = 3

3 7 ii x3 5

1—27. Determine the adjoint matrix for

123a= 1 3 5

3 7 11

Does a exist?1—28. Show that b1' T = bT,

Find the inverse of(a)

[1 3

[3 2

(b)[2 4

[i S

(c)[1 31[2 4

[3 2J[l s(d)

[2 O1[2[o 3j[l s

Let

a12[A11 A12

a= = [A2, A22

31 32 33and

— B32

[8,, 82,

where the order of BJk is the same as AJ1. Starting with the condition

aa' 13


PROBLEMS 41

determine the four matrix equations relating BJk and Aft (j, k 1, 2). Use thisresult to find the inverse of 124

212121

1—3!. Find the inverse ofA

L0 lq

Note that A is (p x q).1-32. Find the inverse of

0

D2

1—33. Use the results of Probs. 1—31 and 1—32 to find the inverse of

b— [B11 B12

[o B22

where B11, B22 are square and nonsingular. Hint: write has

b[B11 01[1 0

[o ij[o 1 j[o B221—34. Consider the 3 x 4 matrix

1121a= 1 3 2 2

1211Determine the elementary row operation matrix which results in a21 = a31 =

= Oand a11 022 033 = +1.1—35. Let

(pxp) (p'<q)[A11 A12

a [A2, A22(qxq)

where 0 and 0. Show that the following elementary operationson the partitioned rows of a reduce a to a triangular matrix. Determine

°1 [ 01 01 [A11 Aizi — AW

L0 cj [A21 lqj [0 Iqj [A21 A22J — [o Iq

A AIA \1—

1—36. Suppose we want to rearrange the columns of a in the following way:

1 2 3

a= 2 1 3 col2—+coll

3 4 5col3—*col2



(a) Show that postmultiplication byIl(which is called a permutation matrix)results in the desired column rearrangement:

o 0 11

H = 1 0 01

o i oj

Note that we just rearrange the corresponding columns of 13.(b) Show that pre;nultiplication by rearranges the rows of a in the

same way.(c) Show that 11TH 13.(d) Generalize for the case where a is n x n.(e) Show that

= a!

1—37. Let a be of order 2 x n, where n 2.. Show that a is of rank 1 whenthe second row is a multiple of the first row. Also, show that when r = 1, thesecond, third nth columns are multiples of the first column.

1—38. Determine the rank of

(a)

1 3 7

5 2 4

3 —4 —10

(b)1 2 3 —1

2 4 6 —2

—1 —2 —3 1

1—39. Let a be of order in x ii and rank r. Show that a has n — r columnswhich are linear combinations of r linear independent columns. Verify for

1234a= 2 1 3 2

5 7 12 14

1—40. Using properties 3, 4, and 7 of determinants (see Sec. 1—7), deducethat the elementary operations do not change the rank of a matrix. For con-venience, consider the first r rows and columns to be linearly independent.

1—41. Find the rank of a by reducing it to an echelon matrix.

11222132

a—4 2 1

7797


PROBLEMS 43

1—42. Show that c is at most of rank 1.

a1

a2C

:

When will r(c) 0?1—43. Consider the product,

a11 a12 r b11b12

b is of rank 1 and b11 0. Then, we can write

cblk)

(j)2kj

Show that the second, third,. . , nth columns of c are multiples of thefirst column and therefore r(c) 1. When will r(c) = 0?

(b) Suppose r(a) = 1 and a11 0. In this case, we can write

= j = 2,3,...,m

Show that the second, third mth rows of c are multiples of the firstrow and therefore r(c) 1. When will r(c) = 0?

1 —44. Consider the product

a11 a12 ' a1., b11 b12 -

a21 a22 b21 h22 ..a,,1 a,,,2 h2 ..

Let= - - -

B1 {b11b21 - -

Using (b), we can write (a) as

A1

A2

Am

Suppose r(a) = r4, r(b) rb. For convenience, we assume the first r0 rowsof a and the first r1 columns of b are linearly independent. Then,

A3 = j = ra + 1, ra + 2, -. . , mp= 1



rb

Bk = XkqBq k — + 1, Tb + 2, . . . ,

q 1

(a) Show that rows ra + 1, + 2,. . . , in of c are linear combinations ofthe first rows.

(b) Show that columns Tb + 1, + 2,.. . , n of c are linear combinationsof the first columns.

(c) From (a) and (b), what can you conclude about an upper bound onr(c)?

(d) To determine the actual rank of e, we must find the rank of

A1 A1B1 A1B2 A1Br1,

A2 B B •.. A2B1 A2B2 •. A2Brb[1 2 rb]

AraB2

Suppose ra Tb. What canyou conclude about A1 is orthogonaltoB1,B2 Br?

(e) Utilize these results to find the rank of

—1/2 1/2 0 1 0 1

—1/2 1/2 1 1 0 1

—1 1 1 1 1 2

(f) Suppose ra = rb = s. Show that r(e) = s. Verify for

U

1—45. Consider the m x n system

a12 X1 C1

021 022 0 C2

Gm i 2 0R111 X,,

Let= . . . j 1, 2,. . . , fl2

Using (b), we write (a) as

A3x = j = 1, 2 in (c)

Now, suppose a is of rank r and the first r rows are linearly independent. Then,

A1=

k = r + 1, r + 2 m

(a) Show that the system is consistent only if

Ck k = i• + 1, r + 2,.. . , inp=1

Note that this requirement is independent of whether in < n or in > n.


PROBLEMS 45

(b) If m < n and r = m, the equations are consistent for an arbibrary c.Is this also true when rn > n and r = n? Illustrate for

[i —1 11 fc1

[i 2 4]X2

and1 1 C1

—1 2 c2

1 4 c3

1—46. Consider the following system of equations:

x1 + X2 + 2x3 + 2x4 4

2x1+x2+3x3+2x4=63x1-l-4x2+2x3+x4=9

7x1 +7x2+9.x3+7x4=23(a) Determine whether the above system is consistent using elementary

operations on the augmented matrix.(b) Find the solution in terms of x4.


2

Characteristic-ValueProblems andQuadratic Forms

2—i. INTRODUCTION

Consider the second-order homogeneous system,

(ajj 2)x1 + at2xz 0

a25x1 + (a22 — A)x2 = 0

where A is a scalar. Using matrix notation, we can write (2—i) as

ax Ax (2—2)

or(a — 212)x 0 (2—3)

The values of 2. for which nontrivial solutions of (2—i) exist are called thecharacteristic values of a. Also, the problem of finding the characteristic valuesand corresponding nontrivial solutions of (2—i) is referred to as a second-ordercharacteristic-value problem,*

The problem occurs naturally in the free-vibrationanalysis of a linear system. We illustrate for the system shown in Fig. 2—1.

The equations of motion for the case of no applied forces (the free-vibrationcase) are

d2y2+ k2(y2 — 0

d2ym1 + k1y1 — k2(y2 — y1) = 0

* Also called "eigenvalue" problem in some texts. The term "cigenvalue" is a hybrid of theGerman term Elgenwerte and English "value."

46


SEC. 2—1. NTRODUCT$ON 47

Assuming a solution of the form— A iwt Ay2zize

and substituting in (a) lead to the following set of algebraic equations relatingthe frequency, w, and the amplitudes, A1, A2:

(k1 + k2)A1 — k2A2 = in1w2A1

—k2A1 + k2A2 = m2w2A2

We can transform (c) to a form similar to that of(2—1) by defining new amplitudemeasures,*

2—co2

A2 =and the final equations are

k1+k2_ k2 —A — —== 2 =rn1

k2A1 + A2 = 2A2

'fl2

The characteristic values and corresponding nontrivial solutions of (e) arerelated to the natural frequencies and normal mode amplitudes by (d). Notethat the coefficient matrix in (e) is symmetrical. This fact is quite significant,as we shall see in the following sections.

]

IFig. 2—1. A system with two degrees of freedom.

Although the application to dynamics is quite important, our primary reasonfor considering the characteristic-value problem is that results obtained for thecharacteristic value problem provide the basis for the treatment of quadratic

* See Prob. 2—i.


48 PROBLEMS CHAP. 2

forms which are encountered in the determination of the relative extrema of afunction (Chapter 3), the construction of variational principles (Chapter 7), andstability criteria (Chapters 7, 18). This discussion is restricted to the case wherea is real. Reference 9 contains a definitive treatment of the underlying theoryand computational procedures.

2—2. SECOND-ORDER CHARACTERISTIC-VALUE PROBLEM

We know from Cramer's rule that nontrivial solutions of

— 2)x1 + a12x2 — 0(2—4)

a21x1 + (a22 — A)x2 = 0

are possible only if the determinant of the coefficient matrix vanishes, that is,when

a11 — A a120 (2-5)

a21 a22—).

Expanding (2—5) results in the following equation (usually called the charac-teristic equation) for 1:

22 — (a11 + a22)). + (a11a22 — a21a12) = 0 (2—6)

We let= a11 + a22

(2—7)= a11a22 — a12a21 = H

and the characteristic equation reduces to

22 — + P2 = 0 (2—8)

The roots of (2—8) are the characteristic values of a. Denoting the roots by22, the solution is

21,2 = (P1 ± (2—9)

When a is symmetrical, a12 = a21, and

— = (a11 a22)2 + 4(a12)2

Since this quantity is never negative, it follows that the characteristic values for asymmetrical second-order matrix are always real.

Example 2—i

[2 2a={2

= 2 ÷ 5 = 7

P2 = (2)(5) — (2)(2) = 6

The characteristic equation for this matrix is

22 — 72 + 6 = 0


SEC. 2—2. SECOND-ORDER CHARACTERISTIC-VALUE PROBLEM 49

Solving (a),

[1 —2

—!

131=0A.1 2 — ± j where i =

By definition, nontrivial solutions of (2—4) exist only when 2 = or 22.In what follows, we suppose the characteristic values are real. We considerfirst the case where A A.1. Equation (2—4) becomes

(a11 — A.1)x1 + a12x2 = 0

a21x1 + (a22 — = 0

The second equation is related to the first by

second eq. times the first eq.

This follows from the fact that the coefficient matrix is singular.

(Oii — %1)(a22 — — a12a21 = 0

Since only one equation isindcpendcnt and there are two unknowns, the solu-tiOn is not unique. We define as the solution for A = Assuming*that a12 0, the solution of the first equation is

xi') = C1

1) =012

where c1 is an arbitrary constant. Continuing, we let

and take c1 such that = 1. This operation is called normalization,and the resulting column matrix, denoted by Q1, is referred to as the charac-teristic vector for

= (2—10)a12

= + —_2112

L JBy definition,

Q1Q1 = 1- (2—11)

if a12 0, we work with the second equation.


50 CHARACTERISTIC-VALUE PROBLEMS CHAP. 2

Since Qi is a solution of (2—4) for A = we see that

aQ1 = A1Q1 (2—12)

Following the same procedure for A = A2, we obtain

Q2 = c2 (2—13))

where

= +L j

Also,

(2—14)aQ2 =

It remains to discuss the case where A1 = A2. If a is symmetrical, the char-acteristic values will be equal only when a11 a22 and a12 = c121 = 0. Equa-tion (2—4) takes the form

(a11 — A)xi + (0)xz 0

(O)x1 + (a11 A)x2 = 0

These equations are linearly independent, and the two independent solutions

{c1, 0}

{0, c2}

The corresponding characteristic vectors are

Q1 = {+1,0}(2—15)

Q2 = {0, +1}

If a is not symmetrical, there is only one independent nontrivial solution whenthe characteristic values are equal.

It is of interest to examine the product, QfQ2. From (2—10) and (2—13),we have

— / (a11 — A1)(a11 — A2)Q1Q2 — —C1C2 1 +

- 2a12

Now, when a is symmetrical, the.right-hand term vanishes since

a11 — A2 —(a22 — A1) = —a11 —

and we see that QrQ2 = 0. This result is also valid when the roots are equal.In general, QIQ2 0 when a is unsymmetrical. Two nth order column vectorsU, V having the property that

UTV=VTU=0 (2—16)

are said to be orthogonal. Using this terminology, Q1 and Q2 are orthogonalfor the symmetrical case.


SEC. 2—2. SECOND-ORDER CHARACTERISTIC-VALUE PROBLEM

Example 2—2

[2 2a=[2A1=+6 A2=+1

The equations for A = +6 are

—4x1 + 2x2 = 0

2x1 — = 0

We see that the second equation is times the first equation. Solving the first equation,we obtain

= c1 = = 2c1Then,

2}

and the normalized solution is

Q1

Repeating forA = A2 = +1, we find

= c2{1,and

Q2

One can easily verify that= j = 1,2

and— ——

Ii 8a=[t3

The characteristic values and corresponding normalized solutions for this matrix are

1

Q2 = {4, — l}

We see that 0. Actually,

QTQ2 =

[1 —2

—l

A1—_+i A2=—i

We have included this example to illustrate the case where the characteristic values are


52 PROBLEMS CHAP. 2

complex. The equations corresponding to 2 are

(1 — i)x1 — 2x2 0

x5 (1 + i)x2 = 0

Note that the second equation is (1 — i) times the first equation. The general solution is

x(1t = c1

Repeating for 2 = 12, we find

= c21-3-i}

When the roots are complex, 12 is the complex conjugate of Now, we take c2 = c1.Then, xt2> is the complex conjugate of We determine c1 such that

= I

Finally, the characteristic values and characteristic vectors are

21,2 =

Qt,2 =

In general, the characteristic values are complex conjugate quantities when the elementsof a are real. Also, the corresponding characteristic vectors are complex conjugates.

2—3. SIMILARITY AND ORTHOGONAL TRANSFORMATIONS

The characteristic vectors for the system satisfy the followingrelations:

aQ1 =(a)

aQ2 = 22Q2

We can write (a) as

Q2] = EQ1 2j (b)

Now, we letq=[Qj Q2]

=(2—17)

Column j ofq contains the normalized solution for We call q the normalizedmodal matrix* for a. With this notation, (b) takes the form

aq = (2—18)

* This terminology has developed from dynamics, where the characteristic vectors define thenormal modes of vibration for a discrete system.


SEC. 2—3. AND ORTHOGONAL TRANSFORMATIONS 53

We have shown that the characteristic vectors are always linearly indepen-dent when a is symmetrical. They are also independent when a is unsym-metrical, provided that 11.2. Then, 0 except for the case where a isunsymmetrical and the characteristic values are equal. If 0, q1 existsand we can express (2—18) as

q'aq = (2—19)

The matrix operation, p is arbitrary, is called a similarity trans-formation. Equation (2—19) states that the similarity transformation, q1( )q,reduces a to a diagonal matrix whose elements are the characteristic valuesof a.

If a is symmetrical, the normalized characteristic vectors are orthogonal,that is,

— —— —

Also, by definition,— —

Using these properties, we see that

[Qfl [1 0q q = [QTJ [Qi Q2] [o 1

and it follows thatqT (2—20)

A square matrix, say p, having the property that = p' is called anorthogonal matrix and the transformation, pT( )p, is called an orthogonaltransformation. Note that an orthogonal transformation is also a similaritytransformation. Then, the modal matrix for a symmetrical matrix is orthog-onal and we can write

qTaq = (2—21)

Example 2—3

[2 2

5

+6

Q2{2,-1}

—[0 0 +1]

q=[Qi


54 PROBLEMS CHAP. 2

We verify that qT = q' and qTaq =

1 [1 21 [1 21 1 [5 01 [1 0

—lj[2 sj[o i1 [2 21[1 21 1 [6 2

aq— [2 —ij — [12 —1

— I [1 21[6 01—ij[o

T i[i 21[6 21 [6 01q aq

= 5L2 _-1][12 —ii = [o ij =

(2)

[1 8

3

= +5 Qi = +1}

01 —

Lo —ij

Since a is not symmetrical, qTq Actually,

-1 ( 1— — [,15/6q

[— j — [\/17/6

One can easily verify that[5 01

q

[1 —2

01

Lo —iJ[1 1

q involves complex elements. Since the characteristic vectors are complexconjugates, they are linearly independent and q -' exists. We find q — 1, using the defi-nition equation for the inverse (Equation (1—50)):


SEC. 2—4 THE nth-ORDER SYMMETRICAL CHARACTERISTICS 55

One can easily verify that

-1 [+i 01q

0

2—4. THE nth-ORDER SYMMETRICAL CHARACTERISTIC-VALUEPROBLEM

The nth order symmetrical characteristic-value problem involves deter-mining the characteristic values and corresponding nontrivial solutions for

a11x1 + a12x2 + + = Ax1

a12x1 + + + — AX2(2—22)

+ + +

We can write (2—22) asax = AX

(2—23)(a — 0

In what follows, we suppose a is real.For (2—23) to have a nontrivial solution, the coefficient matrix must be

singular.a — AI,4 0 (2—24)

The expansion of the determinant is

+ + = 0

where= a11 + a22 ± +

(2—25)-

and is the sum of all the jth order minors that can be formed on the diag-onal.* Letting 22,. . . , denote the roots, and expressing the characteristicequation in factored form, we see that

= 2522 + 2523 + ... + (2-26)

We summarize below the theoretical results for the real symmetrical case.The proofs are too detailed to be included here (see References 1 and 9):

1. The characteristic values 22,. . . , are all real.2. The normalized characteristic vectors Q1, Q2,. . . , Q,,, are orthogonal:

QTQJ = i,j = 1,2

* Minors having a diagonal pivot (e.g.. delete the kth row and column). They are generally calledprincipal minors.


56 CHARACTERiSTIC-VALUE PROBLEMS CHAP. 2

3. a is diagonalized by the orthogonal transformation involving the nor-malized modal matrix.

qTaq

where

=

Example 2—4

5—2 0

a= —2 3 —.1

0 —1 1

Since a is symmetrical, its characteristic values are all real. We first determine /3k, $2, f33,using (2—25):

5 + 3 + 1 +9$2 +11 + 5 + 2 +18/33 = 5(2) — (—2)(—2) = +6

The characteristic equation is

182—6=0

and the approximate roots are+0.42

22 +2.30+6.28

To determine the characteristic solutions, we expand ax = 2x,

(5 — 2)x1 = 2x2

—x3 = —(3 — 2)x2

(l—,t)x3= x2

Solving the first and third equations for x1 and x3 in terms of x2, the general solution is

j=l,2,3

Finally, the modal matrix (to 2-place accuracy) is

+0.22 +0.51 —0.84

q = [Q1Q2Q3] = +0.50 +0.68 +0.54

+0.85 —0.52 —0.10

(2) 120a= 2 1 0

0 .0 3


SEC. 2—5. QUADRATIC FORMS 57

The expansion of Ia — 213J = 0 is

and the roots are22—3 23=—i

Writing out ax = 2x, we have

(1—2)x1 +2x2 =02x1 +(1 — 1)x2 = 0 (a)

(3 — 2)x3 = 0

When 2 = 3, (a) reduces to—2x3 +2x2 = 0

—2x2 = 0 (b)

(0)x3=0

We see from (b) that (a — 213) is of rank 1 when 2 = 3. The general solution of (b) is

xI=c1 x2=cl x3=cz

By specializing the constants, we can obtain two linearly independent solutions for therepeated root. Finally, the characteristic vectors for 22 = 3 are

Q2 = (0,01)

When 2 = 23 = —1, (a) reduces to

2x3 + 2x2 = 0

2x1 + 2x2 0

4x3 = 0

The general solution and characteristic vector for are

and = 0

- o}

This example illustrates the case of a symmetrical matrix having two equal characteristicvalues. The characteristic vectors corresponding to the repeated roots are linearly inde-pendent. This follows from the fact that a — 213 is of rank I for the repeated roots.

2-5. QUADRATIC FORMS

The homogeneous second-degree function

F + 2a12x1x2 +

a quadratic form in x2. Using matrix notation, we can expressF as

[a11 ajal TF=[xix2]t[a12



In general, the function

F = = (2—27)j=1

where afk = for] k, is said to be a quadratic form in xj, x2, .. . , x,,.If F = x xa we a positive definite

for is zero for some x 0, we say that F is positivesemidefinite. We define negative definite and negative semidefinite quadraticforms in a similar manner. A quadratic form is negative definite if F 0 forall x and F = 0 only when x = 0. The question as to whether a quadratic formis positive definite is quite important. For example, we will show that anequilibrium position for a discrete system is stable when a certain quadraticform is positive definite.

Consider the quadratic form

F =b1

b2 0 x2 (2—28)LXIX2

13 1

When F involves only squares of the variables, it is said to be in canonical form.According to the definition introduced above, F is positive definite when

b1 >0 b2 >0It is positive semidefinite when

b1 0

and at least one of the elements is zero.Now, to establish whether is positive definite, we first reduce a to a

diagonal matrix by applying the transformation, q '()q, where q is the orthog-onal normalized modal matrix for a. We write

xTax = (xTq)(q_taq)(qlx)=

Then, lettingy = qTx x qy (2—29)

(a) reduces to a canonical form in y:

F = xTax = (2—30)

It follows that F is positive definite with respect to y when all the characteristicvalues of a are positive. But y is uniquely related to x and y = 0 only whenx = 0. Therefore, F is also positive definite with respect to x. The problem ofestablishing whether xTax is positive definite consists in determining whetherall the characteristic values of a are positive.


SEC. 2—5. QUADRATIC FORMS

We consider first the second-order symmetric matrix

[a11 cz12

Laiz a22

Using (2—26), the characteristic values are related by

+ =aii -F a22,t122 = 132 = a11a22 — = aJ

We see from (a) that the conditions

132>0

are equivalent to,t2>0

Suppose we specify thatau > 0

at —. a11a22 — > 0

Since a1 > 0, it follows from the second requirement in (d) that a22 > 0.Therefore, (d) is equivalent to (b). We let

= aj1j = a11

A2 = a11a12= at (2—31)

a12a12

Then, a is positive definite when

132>0(2—3 2)

A1>O A2>0The quantities and are called the invariants and discriminants of a.

The above criteria also apply for the case. That is, one can showthat a is positive definite when all its invariants are greater than zero.

131 > 0 > 0 •.. /3,, > 0 (2—33)

where is the sum of all the jth-order principal minors. Equivalent conditionscan be expressed in terms of the discriminants. Let represent the deter-minant of the array consisting of the first j rows and columns.

a11 a12 aU

A = £112 a22(2—34)

a2J

The conditions,A1 > 0 A2 > 0 ... A,, > 0 (2—35)

are sufficient for a to be positive definite.*

* See Ref. 1 for a detailed proof. Also see Prob. 2—15.



Example 2—5

111122123

The discriminants are

A1 = +1= 2 — 1 = +1= 1(6—4)— 1(3—2) + 1(2—2) = +1

Since all the discriminants are positive, this matrix is positive definite. The correspondinginvariants are

= 1 + 2 + 3 = +6$2(2—1)+(3—1)+(64) +5/13 = A3 = +1

1 1 1

1 —2 2

1 2 3

Since A2 is negative = —3), this matrix is nor positive definite.

Suppose b is obtained from a by an orthogonal transformation:

b = pTap p1ap (2—36)

If a is symmetrical, b is also symmetrical:

bT = pTaTp pTap (2—37)

Now, b and a have the same characteristic values.* This follows from

— = — a — (2—38)

Then, if a is positive definite, b is also positive definite. In general, the positivedefinite character of a matrix is preserved under an orthogonal transformation,

REFERENCES

1. HILDSBRAND, F. B.: Methods of Applied Mathematics, Prentice-Hall, New York,1952.

2. BODEWIG, E.: Matrix Calculus, Interscience Publishers, New York, 1956.3. SMiRNOV, V. I.: Linear Algebra, Addison-Wesley Publishing Co., Reading, Mass.,

1964.4. TURNBULL, H. W., and A. C. AITKEN: An Introduction to the Theory of Canonical

Matrices, Dover Publications, New York.5. HADLEY, G.: Linear Algebra, Addison-Wesley Publishing Co., Reading, Mass., 1961.

* See Prob. 2—5.


PROBLEMS

6. CRANDALL, S. H.: Engineering Analysis, McGraw-Hill, New York, 1956.7. NOBLE, B.: Applied Linear Algebra, Prentice-Hall, New York, 1969.8. FRAZER, R. A., W. J. DUNCAN and A. R. COLLAR: Elementary Matrices, Cambridge

University Press, London, 1963.9. WILKINSON, 3. H.: The Algebraic Eigenvalue Problem, Oxford University Press,

London, 1965.10. FADDEVA, V. N.: Computational Methods of Linear Algebra, Dover Publications,

New York, 1953.11. RALSTON, A., and H. S. WILF: Mathematical Methods for Digital Computers, Vol. 2,

Wiley, New York, 1967.12. FORSYTHE, G. E., and C. B. MALER: Computer Solution of Linear Algebraic Systems,

Prentice-Hall, New York, 1967.13. PETERS, G., and J. H. WILKINSON: "Eigenvalues of AX = with Band Synimetric

A and B," Comput. J., 12, 398—404, 1969.

PROBLEMS

2—1. Consider the systemAy =

where A and B arc symmetrical nth-order matrices and is a scalar. SupposeB can be expressed as (see Prob. 1—25)

B = brb

where b is nonsingular. Reduce (a) to the form

ax =

where x by. Determine the expression for a in terms of A and b.2—2. Let x1, x2 be two nth-order column matrices or column vectors and

let c1, c2 be arbitrary scalars. If

c1x_I + c2x2 0

only when c1 = = 0, x1 and x2 are said to be linearly independent. It followsthat x1 and x2 are linearly dependent when one is a scalar multiple of the other.Using (2—10) and (2—13), show that and Q2 arc linearly independent when

2—3. Determine the characteristic values and the modal matrix for

[3 2(a)

[2 7

[2 0 3

(b) to s 0

[3 0 2

2—4. Following the procedure outlined in Prob. 2—I, determine the charac-teristic values and modal matrix for

+ 12Y2

l2Y1 + =



.2—5. Suppose that b is derived from a by a similarity transformation.

b = p'apThen,

lb — t1,4 = a —

and it follows that b and a have the same characteristic equation.(a) Deduce that

2(b) — '(a)k

a(b) — p(a) k = 1, 2,.. ., nYk Pk

Demonstrate for[1 —21 [1 1P[2

The fact that . . , /3,, are invariant under a similarity trans-formation is quite useful.

(b) Show thatQ U') _. —1 (a)

k P k

2—6. When a is symmetrical, we can write

qTaq =

Express in terms of q and Use this result to find the inverse of

[3 2a=[2

2—7. Positive integral powers of a square matrix, say a, arc defined as

a2 = aa

a3 = aa2

ar =

If al # 0, exists, and it follows from the definition that

= a''(a) Show that ar is symmetrical when a is symmetrical.(b) Let 2L be a characteristic value of a. Show that is a characteristic

value of ar and is the corresponding characteristic vector.

Hint: Start with = 2,Q, and premultiply by a.2—8. A linear combination of nonnegative integral powers of a is called a

polynomial function of a and written as P(a). For example, the third orderpolynomial has the form

P(a) = c0a3 + c1a2 + c2a + c3L

Note that P(a) is symmetrical when a is symmetrical.


PROBLEMS

Let F(1) 0 be the characteristic equation for a. When the characteristicvalues of a are distinct, one can show that (see Ref: 1)

F(a) = 0

where 0 is an nth-order null matrix. That is, a satisfies its own characteristicequation. This result is known as the Cayley-Hamilton Theorem.

(a) Verify this theorem for[2 1

2

Note: F(a) = a2 f31a + /3212.

(b) Show that

= '(a2 — /31a + /3213) forn = 3

(c) Establish a general expression for a —' using (2—25).2—9. Determine whether the following quadratic forms are positive definite.

(a) F = + 4x1x2 +(b) F = 34 + + — 4x1x2 + 6x1x3 — 8x2x3

2—10. Show that a necessary but not sufficient condition for a to be positivedefinite is

a11>O,a22>O(Hint: = Oforj 1,] = 1,2,.. .,n)

2—11. If = 0, ax = 0 has a nontrivial solution, say x1. What is. thevalue of xfax1? Note that 2 0 is a characteristic value of a when a is singular.

2—12. Let C be a square matrix. Show that CTC is positive definite whenIC! 0 and positive sernidefinite when CI = 0.(hint: Start with F = xT(CTC)x and let y = Cx. By definition, F can equalzero only when x 0 in order for the form to be positive definite.)

2—13. Consider the product CTaC, where a is positive definite and C issquare. Show that C1aC is positive definite when CI 0 and positive semi-definite when CI = 0. Generalize this result for the multiple product,

. . . . .

2—14. Let a be an mth-order positive defInite matrix and let C be of orderm x n. Consider the product,

b = CTaC

Show that b is positive definite only when the rank of C is equal to n. Whatcan we say about b when r(C) < n?

2—15. Consider the quadratic form

a11 a12 x1

a12 a22 x2: :

a27, x,,



We partition a symmetrically,(pxp) (pXl)

NzTvTl [A11 A121 f XtAT A MV("2(qxp) (qxq)

where q = n — p. The expansion of F = XTaX has the form

F = XfA11X1 -i- 2XTA12A2 +

Now, we take X2 0 and denote the result by

XTAIIX1

For > 0 for arbitrary X1, A11 must be positive definite. Since 1A111 is equalto the product of the characteristic values of A11, it follows that Ajj must bepositive.

(a) By taking p = 1, 2,.. . , n, deduce that

p=1,2,...,nare necessary conditions for a to be positive definite. Note that itremains to show that they are also sufficient conditions.

(b) Discuss the case where = 0.2-46. Refer to Prob. t—25. Consider a to be symmetrical.(a) Deduce that one can always express a as the product of nonsingular

lower and upper triangular matrices when a is positive definite.(b) Suppose we take

Show that a is positive definite when

j=1,2,...,nand positive semi-definite when

j=1,2,...,nand at least one of the diagonal elements of g is zero.Suppose we take g = hT. Then,

=and

A — A ..h2i.2 ,..i.2ti.11 — Upp

Show that the diagonal elements of b will always be real when a ispositive definite.

2—17. If a quasi-diagonal matrix, say a, is symmetrically partitioned, thesubmatrix A11 is also a quasi-diagonal matrix. Establish that

a = i,j = 1, 2,. . ., N

is positive definite only when A, (i = 1, 2,. . . , N) are positive definite.


PROBLEMS 65

Hint: Use the result of Prob, 1—23. Verify for

1100230000210052

2—18. Suppose we express a as the product of two quasi-triangular matrices,for example,

(pxp) (pxq)(nxn) [G11 1) 1 [B11 B12

a= [G2, G22j[O B22

(qxp)

where p + q = n. We take

B11 1P 8221qShow that the diagonal submatrices of g are nonsingular for arbitrary p whena is positive definite.


3

Relative Extremafor a Function

3—1. RELATIVE EXTREMA FOR A FUNCTION OF ONE VARIABLE

Letf(x) be a function of x which is defined for the interval x1 x x2. Iff(x) — f(a) 0 for all values of x in the total interval x1 x x2, exceptx a, we say the function has an absolute minimum at x a. If f(x) — f(a)> 0for all values of x except x = a in the subinterval, x containing x = a,

we say that f(a) is a relative minimum, that is, it is a minimum with respect toall other values of f(x) for the particular subinterval. Absolute and relativemaxima are defined in a similar manner. The relative maximum and minimumvalues of a function are called relative extrema. One should note thatf(x) mayhave a number of relative extreme values in the total interval x1 x x2.

As an illustration, consider the function shown in Fig. 3—1. The relativeextrema are [(a), f(h), f(c), f(d). Using the notation introduced above, we saythat f(b) is a relative minimum for the interval x fib. The absolutemaximum and minimum values of f occur at x = a and x = d, respectively.

Fig. 3—i. Stationary points at points A, 8, C, and 0.

66

x

f(x)

x1 a b c X2


SEC. 3—1. RELATIVE EXTREMA FOR A FUNCTION OF ONE VARIABLE 67

In general, values of x at which the slope changes sign correspond to relativeextrema. To find the relative extrema for a continuous function, we first deter-mine the points at which the first derivative vanishes. These points are calledstationary points. We then test each stationary point to see if the slope changessign. If the second derivative is positive (negative) the stationary point is arelative minimum (maximum). If the second derivative also vanishes, we mustconsider higher derivatives at the stationary point in order to determine whetherthe slope actually changes sign. In this case, the third derivative must alsovanish for the stationary point to be a relative extremum.

Example 3—1

Setting the first derivative equal to zero,

x2 + 4x + I = 0dx

and solving for x, we obtain

x1,2 = —2 ±The second derivative is

d2f= 2x + 4 = 2(x + 2)

Thcn,x = x1 = —2 + = x2 = —2—corresponds to a relative maximum.

J(x) = (x — a)3 + cThe first two derivatives are

= 3(x —

a)

Since both derivatives vanish at x = a, we must consider the third derivative:

d3fdx3

6

The stationary point, x = a, is neither a relative minimum nor a relative maximum sincethe third derivative is finite. We could have also established this result by considering theexpression for the slope. We see from (a) that the slope is positive on both sides of x = a.

The general shape of this function is shown in Fig. E3—l.


66 RELATIVE EXTREMA FOR A FUNCTION CHAP. 3

Fig. E3—1

f(x)

I xa

The sufficient condition for a stationary value to be a relative extremum(relative minimum (maximum) when d2f/dx2 > 0 (< 0)) follows from a con-sideration of the geometry of the f(x) vs. x curve in the vicinity of the stationarypoint. We can also establish the criteria for a relative extremum from the Taylorseries expansion of f(x). Since this approach can he readily extended to func-tions of more than one independent variable we will describe it in detail.

Suppose we know the value of f(x) at x = a and we want f(a + Ax) whereAx is some increment in x. If the first n + 1 derivatives off(x) are continuousin the interval, a x a + Ax, we can express f(a + Ax) as

f(a + Ax) f(a)

=Ax + (Ax)2 + (Ax)" + (3.1)

where denotes the jth derivative of f(x) evaluated at x a, and theremainder is given by

1(3-2)

where is an unknown number between a and a + Ax. Equation (3—1) is calledthe Taylor series expansion* of f(x) about x = a. If f(x) is an eth-degreepolynomial, the (n + 1)th derivative vanishes for all x and the expansion willyield the exact value off(a + Ax) when n terms are retained. In all other cases,there will be some error, represented by due to truncating the series at nterms. Since depends on we can only establish bounds on The fol-lowing example illustrates this point.

See Ref. 1, Article 16—8.


SEC. 3—1. RELATIVE EXTREMA FOR A FUNCTION OF ONE VARIABLE 69

Example 3—2

We expand sin x in a Taylor series about x = 0 taking n = 2. Using (3—1) and (3—2),and noting that a = 0, we obtain

sin Ax = Ax + R2

The bounds on R2j are

cos Ax < R21

If we use (a) to find sin (0.2), the upper bound on the truncation error is 0.0013.

If Ax is small with respect to unity, the first term on the right-hand side of(3—1) is the dominant term in the expansion. Also, the second term is moresignificant than the third, fourth nth terms. We refer to df/dx Ax asthe first-order increment in f(x) due to the increment, Ax. Similarly, we call4d2f/dx2(Ax)2 the second-order increment, and so on. Now, f(a) is a relativeminimum when f(a + Ax)— f(a) > 0 for all points in the neighborhood ofx a, that is, for all finite values of Ax in some interval, — Ax e, where

and e are arbitrary small positive numbers. Considering Ax to be small, thefirst-order increment dominates and we can write

f(a + Ax) — f(a) = Ax + (second- and higher-order terms) (3—3)

For f(a + Ax) — f(a) to be positive for both positive and negative values ofAx, the first order increment must vanish, that is, df(a)/dx must vanish. Notethat this is a necessary but not sufficient condition for a relative minimum, ifthe first-order increment vanishes, the second-order increment will dominate:

f(a + Ax) — f(a) = (Ax)2 + (third- and higher-order terms) (3—4)

It follows from (3—4) that the second-order increment must be positive forf(a + Ax) — > 0 to be satisfied. This requires d2f(a)/dx2 > 0. Finally,the necessary and sufficient conditions for a relative minimum at x = a are

df(a) — d2f(a)3 5

dx0

dx2 >

lithe first two derivatives vanish at x = a, the third-order increment is nowthe dominant term in the expansion.

f(a + Ax) + f(a) = (Ax)3 + (fourth- and higher-order terms) (3—6)

Since the third-order increment depends on the sign of Ax, it must vanish for


RELATIVE EXTREMA FOR A FUNCTION CHAP. 3

f(a) to be a relative extremum. The sufficient conditions for this case are asfollows:

Relative Minimum

d3f d4fdX4>

(3—7)

Relative Maximum

d3f d4f

The notation used in the Taylor series expansion off(x) becomes somewhatcumbersome for more than one variable. In what follows, we introduce newnotation which can be readily extended to the case of 11 variables. First, wedefine to be the total increment in f(x) due to the increment, Z\x.

41 = f(x + Ax) — f(x) (3—8)

This increment depends on Ax as well as x. Next, we define the differentialoperator, d,

(3—9)

The result of operating onf(x) with d is called the first and is denotedby df:

df=-1Ax = df(x,Ax) (3—10)

The first differential off(x) is a function of two independent variables, namely,x and Ax. Iff(x) = x, then df/dx = 1 and

c/f = dx = Ax (3—11)

One can use dx and Ax interchangeably; however, we will use Ax rather than dx.Higher differentials of f(x) are defined by iteration. For example, the seconddifferential is given by

d2f = d(df)=

(3—12)

Since Ax is independent of x,

(Ax) = 0

and d2f reduces to

d2f = (Ax)2 = d2f(x, Ax) (3—13)

In forming the higher differentials, we take d(Ax) = 0.


SEC. 3—2. FUNCTION OF n INDEPENDENT VARIABLES

Using differential notation, the Taylor series expansion (3—1) about x can bewritten as

(3-14)

The first differential represents the first-order increment in f(x) due to theincrement, Ax. Similarly, the second differential is a measure of the second-order increment, and so on. Then,f(x) is a stationary value when df = 0 for allpermissible values of Ax. Also, the stationary point is a relative minimum(maximum) when d2f> 0 (<0) for all permissible values of Ax. The abovecriteria reduce to (3—5) when the differentials are expressed in terms of thederivatives.

Rules for forming the differential of the sum or product of functions are listedbelow for reference. Problems 3—4 through 3—7 illustrate their application.

f= u(x) + v(x)

df=du+dv3 15

d2f = d(df) d2u + d2v —

f= u(x)v(x)

df = u dv + v du2 2 2

(3—16df = ud v + 2dudv + vd u

f = fly) where y = y(x)

dfdy dy

3-2. RELATIVE EXTREMA FOR A FUNCTION OF n INDEPENDENTVARIABLES

Let f(x1, be a continuous function of n independent variables(x1, x2 We define Af as the total increment in f due to increments inthe independent variables (Ax1, Ax2

Af = f(x1 + Ax1, x2 + Ax2,. .. , + — f(x1, x2,.. . , (3—18)

If Af> 0 (<0) for all points in the neighborhood of(x1, x2,. . . , we saythat f(x1, x2 is a relative minimum (maximum). We establish criteriafor a relative extremum by expanding f in an n-dimensional Taylor series. Theprocedure is identical to that followed in the one-dimensional case. Actually,we just have to extend the differential notation from one to n dimensions.



We define the n-dimensional differential operator as

d Ax1 + Ax2 + + = (3-19)ox2

where the increments Ax2 are independent of (x1, x2,. . ,

The result obtained when d is applied to f is called the first

df=

(3—20)

Higher differentials are defined by iteration. For example, the second differen-tial has the form

d2f=d(df)= (3—21)

Since are considered to be independent, (3—21) reduces to

d2f Axk (3—22)k=i.

Now, we let

r 1f(2) j, k = 1, 2, . . . , n (3—23)

LOXJOXkJ

Ax =

and the expressions for the first two differentials simplify to

df = AXTf(l)(3—24)

d2f = AXTf(2) Ax

The Taylor series expansion forf about (x1, x2, . . . , when expressed interms of differentials, has the form

Af= df + + + + (3-25)

We say that f(x1, x2 is stationary when df 0 for arbitrary Ax. Thisrequirement is satisfied only when

= 0 (3—26)

Equation (3—26) represents n scalar equations, namely,

= 0 j = 1, 2,. . . , n (3—27)

The scalar equations corresponding to the stationary requirement are usually


SEC. 3—2. FUNCTION OF n INDEPENDENT VARIABLES 73

called the Euler equations for f, Note that the number of equations is equalto the number of independent variables.

A stationary point corresponds to a relative minimum (maximum) of fwhen d2f is positive (negative) definite. It is called a neutral point when d2fis either positive or negative semidefinite and a saddle point when d2f is indif-ferent, i.e., the eigenvalues are both positive and negative. This terminologywas originally introduced for the two dimensional case where it has geometri-cal significance.

To summarize, the solutions of the Euler equations correspond to points atwhich f is stationary. The classification of a stationary point is determined bythe character (definite, semidefinite, indifferent) of evaluated at the point.We are interested in the extremum problem since it is closely related to thestability problem. The extremum problem is also related to certain other prob-lems of interest, e.g., the characteristic-value problem. In the following exam-ples, we illustrate various special forms of f which are encountered in membersystem analysis.

Example 3—3

f = y,)

= x2

df=1 1 ek k 1 j 1 Xk

Now,

>-—&Xkk1 OXk

It follows that

df=j=l

Repeating leads to

d2f=

Consider the double sum,

f >k=t

The first differential (see Prob. 3—9) has the form

df = (duJwJkvk + dwfkvk + dv1)

Introducing matrix notation,

u = w = [w31] v = {v1}=

and lettingdu



and so forth, we can write df as

df = d(urwv)

= duTwv + nTdWv + uTwdv

One operates on matrix products as if they were scalars, but the order must be preserved.As an illustration, consider

f = — x7c

where a, c are constant and a is symmetrical. Noting that da = dc = 0 and dx Ax,the first two differentials are

df AxT(ax — c)

d2f = AxTa Ax

Comparing (g) and (3—24), we see that

fO) ax — c

= a

The Euler equations are obtained by setting equal to 0:

ax c

The solution of (i) corresponds to a stationary value of (f). If a is positive definite, thestationary point is a relative minimum. One can visualize the problem of solving thesystem ax = c, where a is symmetrical from the point of view of finding the stationaryvalue of a polynomial having the form f = — XTC.

Suppose f u/v. Using the fact that

3j7u\ ldu\,vJ V UX1 :X3

1(6u uôv— vax1

we can write

df = = —f dv)

We apply (b) to

1. = xxwhere a is symmetrical, and obtain (see Prob. 3—5)

2 AXT= —f---- (ax — 2x)

(d)

(122 = Ax A AxT Ax — 2 dA AxTx)xx

Setting dA = 0 leads to the Euler equations for (c),

ax—Ax=0 (e)

which we recognize as the symmetrical problem.


SEC. 3—3. LAGRANGE MULTIPLIERS 75

The quotient xTax/xTx, where x is arbitrary and a is symmetrical, is called Rayleigh'squotient. We have shown that the characteristic values of a are stationary values ofRayleigh's quotient. This property can he used to improve an initial estimate for acharacteristic value. For a more detailed discussion, see Ref. 6 and Prob. 3—11.

3-3. LAGRANGE MULTIPLIERS

Up to this point, we have considered only the case where the function isexpressed in terms of independent variables. In what follows, we discuss howone can modify the procedure to handle the case where some of the variablesare not independent. This modification is conveniently effected using Lagrangemultipliers.

Suppose f is expressed in terms of n variables, say x1, x2 some ofwhich are not independent. The general stationary requirement is

df= > (3—28)j=1

for all arbitrary differentials of the independent variables. We use insteadof to emphasize that some of the variables are dependent. In order toestablish the Euler equations, we must express df in terms of the differentialsof the independent variables.

Now, we suppose there are r relations between the variables, of the form

g5(x1, x2 = 0 It = 1, 2, . . , r (3—29)

One can consider these relations as constraint conditions on the variables.Actually, there are only n — r independent variables. We obtain r relationsbetween the n differentials by operating on (3—29). Since = 0, it followsthat 0. Then,

"a k=1,2,...,r (3—30)j=t

Using (3—30), we can express r differentials in terms of the remaining n — rdifferentials. Finally, we reduce (3—28) to a sum involving the n — r indepen-dent differentials. Equating the coefficients to zero leads to a system of n — requations which, together with the r constraint equations, are sufficient todetermine the stationary points.

Example 3—4

We illustrate the procedure for n = 2 and r = 1:

f =g(x1, x2) = 0

The first variation is

ox1 Ox2



Operating on g(x1, x2) we have

ax1

Now, we suppose ag/ax2 0. Solving (b) for dx2 (we replace dx1 byx1 is the independent variable.)

/ag\dx2= —t—i-----itSx1

8x2/and substituting in (a), we obtain

df = [PL —ax1 8x1 ox2 ox2

Finally, the equations defining the stationary points arc

ax1 \ox1fOx2/ox2g(x1, x2) = 0

To determine whether a stationary point actually corresponds to a relative extremum,we must investigate the behavior of the second differential. The general form of d2f fora function of two variables (which are not necessarily independcnt) is

d2f =

2 2 2 Of= dx1 + —i—

k1

a quadratic form in the independent differential, using (c), and notingd2x1 = 0,

d2f + + +ax1ax.2 ox2 \0x1 ax2j Ox2

where

Og Jagu=

The character of the stationary point is determined from the sign of the bracketed term.

An automatic procedure for handling constraint conditions involves the useof Lagrange multipliers. We first describe this procedure for the case of twovariables and then generalize it for n variables and r restraints. The problemconsists in determining the stationary values of f(x1, x2) subject to the con-straint condition, g(x1, x2) = 0. We introduce the function H, defined by

H(x1, x2, A.) = f(x1, x2) + Ag(x1, x2) (3—31)

where A. is an unknown parameter, referred to as a Lagrange multiplier. We


SEC. 3—3. LAGRANGE MULTIPLIERS 77

consider x1, x2 and ,% to be independent variables, and require H to be sta-tionary. The Euler equations for H are

OH —+ A

Og—

Ox, Ox1 —

(3-32)Ox2 Ox2 Ox2

OHg(x1, x2) = 0

We suppose Og/0x2 0. Then, solving the second equation in (3—32) for A,and substituting in the first equation, we obtain

A = (3-33)0x2/ Ox2

and

— = 0 (3-34)Ox1

g(x1, x2) = 0

Equations (3—34) and (e) of the previous example are identical. We see thatthe Euler equations for II are the stationary conditions for f including theeffect of constraints.

Example 3—5

f= + + 2x1 + 7x2

g = — x2 = 0

We form H f + 2g,

H = + + 2x1 + 7x2 + A(x1 — x2)

The stationary requirement for H treating x1, x2, and 2 as independent variables is

6x1 + 2 + 2 = 0

4x2 + 7 — 2 = 0

x1 — x2 = 0

Solving this system for x1, x2 and A we obtain

A 4x2 + 7= = —9/10

This procedure can be readily generalized to the case of n variables andr constraints. The problem consists of determining the stationary values Ofj(x1, x2, . . , subject to the constraints gk(xl, x2,. . . , = 0, wherek = 1, 2, . , r. There will be r Lagrange multipliers for this case, and H has



the form

H = f +k1

2k9k H(x1, x2 . . , (3—35)

The Euler equations for H are

+ 0 i = 1,2,. ..,n (3—36)k=1

9k = 0 k = 1, 2,. . ., r (3—37)

We first solve r equations in (3—36) for the r Lagrange multipliers, and thendetermine the n coordinates of the stationary points from the remaining n — requations in (3—36) and the r constraint equations (3—37). The use of Lag-range multipliers to introduce constraint conditions usually reduces the amountof algebra.

REFERENCES

1. THOMAS, G. B., JR., C'alculus and Analytical Geometry, Addison-Wesley PublishingCo., Reading, Mass., 1953.

2. COURANT; R., Differential and Integral Calculus, Vol. 1, Blackie, London, 1937.3. COURANT, R., Differential and Integral Calculus, Vol. 2, Interscience Publishers,

New York, 1936.4. HANCOCK, H., Theory of Maxima and Minima, Dover Publications, New York, 1960.5. APOSTOL, T. M,, Mathematical Analysis, Addison-Wesley Publishing Co., Reading,

Mass., 1957.6. CRANDALL, S. H., Engineering Analysis, McGraw-Hill, New York, 1956.7. HILDEBRAND, F. B., Methods of Applied Mathematics, Prentice-Hall, New York,

1952.

PROBLEMS

3—1. Determine the relative extrema for(a) f(x) = 2x2 + 4x + 5(b) f(x) = —2x2 + 8x + 10(c) f(x) = ax2 + 2bx + c(d) f(x)=x3+2x2+x+10(e) f(x)=1x3+2x2+4x+15(f) f(x) = (x — + (x — a)2

(g) f(x) = 4ax3 + 4bx2 + cx + d3—2. Expand cos x in a Taylor series about x = 0, taking n = 3. Determine

the upper and lower bounds on R3.3—3. Expand(1 + x)112inaTaylorseriesaboutx = Otakingn = 2. Deter-

mine upper and lower bounds on R2.3—4. Find df and d2f for(a) f=x2+2x+5(b) f=3x3+2x2+5x+6(c) f=x2sinx(d) f = cosywhcrey = x3


PROBLEMS 79

3—5. Let f = u(x)/v(x). Show that

df = (du — f dv)

d2f = — fd2v) —

3—6. Let u1, u2, u3 be functions of x and f = f(u1, u2, u3). Determine df.3—7. Suppose f = u(x)w(y) where y = y(x). Determine expressions for df

and d2f. Apply to(a) u=x3—x(b) w=cosy(c) y=x2

3—8. Find the first two differentials for the following functions:(a) f = + +(b) f' = + 6x1x2 + + 5x1 — 4x2

3—9. Consider f = uv, where

u = u(y1, .v2) v = v(y1, Y2)and

Yi y1(x1, x2) = y2(x1, x2)Show that

df = d(uv) u dv + v du d2f = ud2v + 2 du dv + vd2u

Note that the rule for forming the differential of a product is independent ofwhether the terms are functions of the independent variables (x1, x2) or ofdependent variables.

3—10. Classify the stationary points for the following functions:(a) f 3xl + — 9x1 + 12x2 — 10(b)

f + 6x1x2f 6x1x2 + 2x1 +f = 6x1x2 + 34 — 3x1

3—11. Consider Rayleigh's quotient,

xTax

x is arbitrary. Since a is symmetrical, its characteristic vectors are linearlyindependent and we can express x as

x =

where Q3 (j = 1, 2,. . . , n) are the normalized characteristic vectors for a.(a) Show that

= j=j-



(b) Suppose x differs only slightly from Qk. Then, ICjI << for j k.Specialize (a) for this case. Hint: Factor out 2k and

(c) Use (b) to obtain an improved estimate for A.

[3a=[ix {1, —3}

The exact result is2=1 x={1,—2}

3—12. Using Lagrange multipliers, determine the stationary values for thefollowing constrained functions:

(a)g + x2 0

(b)

g1 = x1 + x2 + X3 — 1 = 0g2=x1—x2+2x3+2=0

3—13. Consider the problem of finding the stationary values of f = =xrarx subject to the constraint condition, = 1. Using (3—36) we write

H =f+ Ag = — 2(XTX —1)

(a) Show that the equations defining the stationary points off are

ax=Ax xTx=1

(b) Relate this problem to the characteristic value problem for a symmetri-cal matrix.

3—14. Supposef = and g = I — xTax = 0 where aT a. Show thatthe Euler equations for H have the form

xTax=1

We see that the Lagrange multipliers are the reciprocals of the characteristicvalues of a. How are the multipliers related to the stationary values of f?


4

Differential Geometryof a Member Element

The geometry of a member element is defined once the curve correspondingto the reference axis and the properties of the normal cross section (such asarea, moments of inertia, etc.) are specified. In this chapter, we first discuss the.differential geometry of a space curve in considerable detail and then extendthe results to a member element. Our primary objective is to introduce theconcept of a local reference frame for a member.

4—1. PARAMETRIC REPRESENTATION OF A SPACE CURVE

A curve is defined as the locus of points whose position vector* is a functionof a single parameter. We take an orthogonal cartesian reference frame havingdirections X1, and X3 (see Fig. 4—1). Let F he the position Vector to a point

x1

X2(y)

x2

Fig. 4—1. Cartesian reference frame with position vector ?(y).

* The vector directed from the origin of a fixed reference frame to a point is called the positionvector. A knowledge of vectors is assumed. For a review, see Ref. 1.

81

x3

i3 X3(y)


82 DIFFERENTiAL GEOMETRY OF A MEMBER ELEMENT CHAP. 4

on the curve having coordinates (j = 1, 2, 3) and let y be the parameter. Wecan represent the curve by

=3

Since F = an alternate representation isi—i

= (j = 1,2, 3) (4—2)

Both forms are called the parametric representation of a space curve.

Example 4—i

(1) Consider a circle in the X1-X2 plane (Fig. E4—1A). We take y as the polar angle andlet a = The coordinates are

x1 = a cos y

x2 = a sin yand F = + asiny'12

(2) Consider the curve (Fig. E4—1B) defined by

= a cos y

x2 = bsiny (4—3)

X3 = CY

where a, b, care constants. The projection on the X, -X2 plane is an ellipse having semiaxesa and b. The position vector for this curve has the form

F = a cos + b sin Y12 + CYI3

4—2. ARC LENGTH

Figure 4—2 shows two neighboring points, P and Q, corresponding to y andy + The cartesian coordinates are and + (j 1, 2, 3) and thelength of the chord from P to Q is given by

As Exy —* 0, the chord length approaches the arc length, In the limit,

ds2=

Noting that

dx1 = dy

we can express ds as

ds+ + +

dy (4-4)


Q(+Ay)

P(y)

Fig. 4—2. Differentia' segment of a curve.

83

Fig. E4—1A

Fig. E4—1B

I LISI


84 DIFFERENTIAL GEOMETRY OF A MEMBER ELEMENT CHAP. 4

Finally, integrating (4—4) leads to

dx 2 dx 2 dx 21/2s(y)

= + +dy (4—5)

We have defined ds such that s increases with increasing y. It is customary tocall the sense of increasing s the positive sense of the curve.

To simplify the expressions, we let

dx 2 1/2=

+(4-6)

Then, the previous equations reduce to

ds = dy

(4—7)

One can visualize as a scale factor which converts dy into ds. Note thatx > 0. Also, if we take y = s, then +1.

Example 4—2

Consider the curve defined by (4—3). Using (4—6), the scale factor is

[a2 sin2 y + b2 cos2 y + c2]'12

We suppose that b a. One can always orient the axes such that this condition is satisfied.Then, we express as

= (b2 + c2)"2 [1 — k2 sin2 y]"2

whereb2 — 2

k2 =b2 + c2

The arc length is given by

s dy = (b2 + [1 — k2 sin2 yJ112 dv

The integral for s is called an elliptic integral of the second kind and denoted by E(k, y).

Then,s (b2 + c2)'12 E(k, y)

Tables for E(k, y) as a function of k and y are contained in Ref. 3. When b = a, the curveis called a circular helix and the relations reduce to

= (a2 + const.

S = ny


* See Ref. 1, p. 401.

— — dP dy — 1 dF

— ds — dy ds — dy

—dy

/df dP\"2\dy dy

Fig. 4—3. Unit tangent vector at P(y).

SEC. 4—3. UNIT TANGENT VECTOR 85

4—3. UNIT TANGENT VECTOR

We consider again the neighboring points, P(y) and Q(y + shown inFigure 4—3. The corresponding position vectors are P(y + ky), and

As L\y -+ 0, approaches the tangent to the curve at P. Then, the unit tangentvector at P is given by*

- . PQ d1t = Jim --=—- (4—8)

•

ds

Using the chain rule, we can express I as

(4—9)

Since > 0, 1 always points in the positive direction of the curve, that is, in thedirection of increasing s (or y). It follows that dP/dy is also a tangent vector and

Equation (4—10) reducescoordinates.

(4—10)

to (4—6) when is expressed in cartesian

Q(y+6.y)+s

+

r(y)



Example 4—3

We determine the tangent vector for the curve defined by (4—3). The position vector is

F = a cos + b sin Y12 + cyi3

Differentiating P with respect to y,

dP= —a sin + h cos Y12 +

dy

and using (4—9) and (4—10), we obtain

a = +[a2 sin2 y + b2 cos2 y + c2]"2

=1[—asinyT1 + bcosyi2 + c13]

When a b, a [a2 + =' coast, and the angle between the t?ngent and the X3direction is constant. A space curve having the property that the angle between the tangentand a fixed direction (X3 direction for this example) is constant is called a helix.*

4-4. PRINCIPAL NORMAL AND BINORMAL VECTORS

Differentiating = 1 with respect to y, we have

-= 0

dy

It follows from (a) that di/dy is orthogonal to f. The unit vector pointing in thedirection of di/dy is called the principal normal vector and is usually denoted by ii.

ldtH =

dywhere (4..-lt)

d (1 dF

The binormal vector, h, is defined by

b='?xh (4-12)

We see that b is also a unit vector and the three vectors. ñ, b comprise a right-handed mutually orthogonal system of unit vectors at a point on the curve.Note that the vectors are uniquely defined once y) is specified. The frameassociated with b_and ii is called the moving trihedron and the planes deter-mined by (1, ñ), (ii, b) and are referred to as the osculating normal, andrectifying planes (see Fig. 4—4).

* See Ref. 4, Chap. 1.


SEC. 4—4. PRiNCIPAL NORMAL AND BINORMAL VECTORS 87

Normal plane

Rectifying

plane

Fig. 4—4. Definition of local planes.

Example 4—4

We determine fi and b for the circular helix. We have already found that

a — [a2 +and

= sin VT1 + a cos + c13]

Differentiating t with respect to y, we obtain

di a— — [cos ytj + Sm

Then,i dt

fl — C05 — Smdt dydy

The principal normal vector is parallel to the plane and points in the inward radialdirection. It follows that the rectifying plane is orthogonal to the X1-X2 plane. We candetermine b using the expansion for the vector product.

—asiny acosy Ca

This reduces to- C. C ab sin — — cos + £3

a a a

The unit vectors are shown in Fig. E4—4.


88 GEOMETRY OF A MEMBER ELEMENT CHAP. 4

Fig, E4—4

The derivative of the tangent vector with respect to arc length is called thecurvature vector, K.

Using (4—11), we can write

dt d2FK

c/s i/s2

Kic/i

c/s2

— — Kuds

(4—13)

(4—14)

Note that K points in the same direction as Ft since we have taken K 0. Thecurvature has the dimension L1 and is a measure of the variation of the tangentvector with arc length.

We let R be the reciprocal of the curvature:

R = K1 iS)

In the case of a plane curve, R is the radius of the circle passing through threeconsecutive points* on the curve, and K = JdO/dsj where 6 is the angle betweenI and To show this, we express I in terms of 0 and then differentiate withrespect to s. From Fig. 4—5, we have

cos + Sifl 012

* See Ref. 4, p. 14, for a discussion of the terminokgy 'three consecutive points."

4—5. CURVATURE, TORSION, AND THE FRENET EQUATIONS


— . dOK [—sin + cos 617] a—

dO 1

Kds R

dO/ds[— sin + cos 612]

In the case of a space curve, the tangents at two consecutive points, say P and Q,are in the osculating plane at F, that is, the plane determined by and ñ at. P.We can interpret R as the radius of the osculating circle at P. It should be notedthat the osculating plane will generally vary along the curve.

Fig. 4—S. Radius of curvature for a plane curve.

The binormal vector is normal to both and ñ and therefore is normal to theosculating plane. A measure of the variation of the osculating plane is givenby db/ds. Since his a unit vector, db/ds is orthogonal to h. To determine whetherdb/ds involves we differentiate the orthogonality condition I b 0, withrespect to s.

- db - dlds ds

But dl/ds Kñ and b ii = 0. Then, db/ds is also orthogonal to I and involvesonly ñ. We express db/ds as

= —tn (4—16)

where r is called the torsion and has the dimension, L

SEC. 4—5. CURVATURE, TORSION, AND THE FRENET EQUATIONS

Then

and

x2

i2

\R

R

+

t

it

db


90 DiFFERENTIAL GEOMETRY OF A MEMBER ELEMENT CHAP. 4

It remains to develop an expression for a. Now, h is defined by

xnDifferentiating with respect to s, we have

db di dñ= x n + t

This reduces to -

db diii—=t

x ii = 0. Finally, using (4—16), the torsion is given by

dñ l-dfl—— — b — (4—17)ds dy

Note that a can be positive or negative whereas K is always positive, accordingto our definition. The torsion is zero for a plane curve since the osculating planecoincides with the plane of the curve and b is constant.

Example 4—5

The unit vectors for a circular helix are

= [—a sin vij + a cos Yti + cT3]

—cos — sin

b = yl' — ccosyi3 + at3]

wherea = (a2 + c2)112

Then,a aK=-—

adv a a +cand

1— dñ c c2

2—consta dy a

We have developed expressions for the rate of change of the tangent andbinormal vectors. To complete the discussion, we consider the rate of changeof the principal normal vector with respect to arc length. Since fi is a unitvector, dñ/ds is orthogonal to ñ. From (4—17),

- dñb — a

ds


SEC. 4—6. GEOMETRICAL RELATIONS FOR A SPACE CURVE 91

To determine the component of dñ/ds in the I direction, we differentiate theorthogonality relation, I n = 0.

(b)ds ds

it follows from (a) and (b) that

dñ -— = —I(t + tb (4—18)us

The differentiation formulas for 1, ii, and b are called the Frenet equations.

4—6. SUMMARY OF THE GEOMETRICAL RELATIONS FORA SPACE CURVE

We summarize the geometrical relations for a space curve:

Orthogonal Unit Vectors

di ldit = = — = tangent vector

thu exdy

1 dia = —i-- i— principal normal vector

(4—19)

= I x ñ binormal vectordi ds—dy dy

Di:fferenriation Formulas Equations)

dl lull— -— = Knds adydb 1db— = —— = —rnds ctdydñ 1 dñ -. (4—20)

—Kt +tbds ady

1 diK = — curvature

a dy1—dñ= — b — = torsiona dy

We use the orthogonal unit vectors (I, ñ, b) to define the local reference framefor a member element. This is discussed in the following sections. The Frenet



equations are utilized to establish the governing differential equations for amember element.

4—7. LOCAL REFERENCE FRAME FOR A MEMBER ELEMENT

The reference frame associated with ñ, and b at a point, say P, on a curveis uniquely defined once the curve is specified, that is, it is a property of thecurve. We refer to this frame as the natural frame at P. The componentsof the unit vectors b) are actually the direction cosines for the naturalframe with respect to the basic cartesian frame which is defined by the ortho-gonal unit vectors (1k, 12, 13). We write the relations between the unit vectors as

ft £12 133 11

n = t22 12 (4—21)

£32 e33

One can express* the direction cosines in terms of derivatives of the cartesiancoordinates (x1, x2, x3) by expanding (4—19). Since (1, b) are mutually or-thogonal unit vectors (as well as 13) the direction cosines are related by

1jm6m = j, k = 1, 2, 3 (4—22)

Equation (4—22) leads to the important result

[ljk]T = (4—23)

and we see that is an orthogonal matrix.fThe results presented above arc applicable to an arbitrary continuous curve.

Now, we consider the curve to be the reference axis for a member clement andtake the positive tangent direction and two orthogonal directions in the normalplane as the directions for the local member frame. We denote the directionsof the local frame by (Y1, Y2, 1'3) and the corresponding unit vectors by (t1, 2,We will always take the positive tangent direction as the Y1 direction = 1)and we work only with right handed systems x t3). This notationis shown in Fig. 4-6.

When the centroid of the normal cross-section coincides with the origin ofthe local frame (point P in Fig. 4—6) at every point, the reference axis is calledthe centroidal axis for the member. It is convenient, in this case, to take Y2,Y3 as the principal inertia directions for the cross section.

In general, we can specify the orientation of the local frame with respect tothe natural frame in terms of the angle between the principal normal directionand the I'2 direction. The unit vectors defining the local and natural frames

* See Prob. 4-5.t See Prob. 4—6.


SEC. 4—7 LOCAL REFERENCE FRAME FOR A MEMBER ELEMENT 93

are related by -tl —t2 = COS 4)11 + sin 4)b (4—24)

çbn + cos 4th

Combining (4—21) and (4—24) and denoting the product of the two directioncosine matrices by the relation between the unit vectors for the local and basicframes takes the concise form

t = (4—25)

where

[£21cos4)+€31sin4)

[21sin4)+€31cos4)

Note that the elements of fi are the direction cosines for the local frame withrespect to the basic frame.

fJjk = Xk) (4—26)

Since both frames are orthogonal, J1'. We will utilize (4—25) in the next

chapter to establish the transformation law for the components of a vector.

€12

€22cos4)+ €32sin4)

—€22si+C32cos4)

€j3

€23cosçb+ €33sinqS

—€23 sin 41+ £33cos41

x3

Normat

Y1

Fig. 4—6. Definition of local reference frame for the normal cross section.



Example 4—6

We determine for the circular helix. The natural frame is related to the basic frame.by

Using (4-.25)

a. a cI ——slay —cosy —

a a a= — cos y — sin y 0 12 = {Ik}

C. c ab —slay —--cosy —

a a a

a

a

C.+cosysm4 + —sinycos4

= r + Y2t2 + y3t3

COS + Sifl (

t3 = = + cos4b

We consider 4 to be a function of y1.

a

a

aI.

a

a

acos y

a

—sinycosçb ——cosysin4

sin y sin — — cos y cos

4—8. CURVILINEAR COORDINATES FOR A MEMBER ELEMENT

We take as curvilinear coordinates (yi, Yz' y3) for a point, say Q, the parameterof the reference axis and the coordinates (Y2, of Q with respect to the

orthogonal directions (Y2, 1'3) in the normal cross section (see Fig. 4—7). LetR(y1, Y2' Y3) be the position vector for Q(Yl, y3) and F(y1) the

position vector for the reference axis. They are related by

where

(4—27)

Y2

y3 — — —

Y2

Fig. 4—7. Curvilinear coordinates for the cross section.


SEC. 4—8. CURVILft4EAR COORDiNATES FOR A MEMBER ELEMENT 95

The curve through point Q corresponding to increasing Yj with Y2 and y3held constant is called the parametric curve (or line) for yj. In general, thereare three parametric curves through a point. We define as the unit tangentvector for the parametric curve through Q. By definition,

13RUi =

(4—28)

Operating on (4—27), the partial derivatives of R are

0R dt2— + Y2 + Y3dy1 dy1 dy1

aR= t2

aR= t3

aIR

=The differential arc length along the curve is related to by

aIR

= = (4—29)

This notation is illustrated in Fig. 4—8. One can consider the vectors(or to define a local reference frame at Q.

x3

y2t2 +y3t3

x2

Fig. 4—8. Vectors defining the curvilinear directions.



We see thatt2 g2 1

(4—30)ü3=t3 g3=1It remains to determine ü1 and g1.

Now,

= =dy1

Also, -d12 (dñ dçb . 1db+ —bj + —dy1 \dy1 dy1 j '\dv1 dy1

dt3 . (dii dq5\ 1db= + b—)+ —11——

\dy1 dy1j \dy1 dy1

We use the Frenet equations to expand the derivatives of ñ and h. Then,

cit2

dy1 dyjI d4)'\.

and finally,/ d4)\ -

= cc(1 — Ky'2)!1 + + _)(Y2t3 y3t2)(4—31)

Y2 COS 4) J73 sin 4)

We see from Fig. 4—9 that y'2 is the coordinate of the point with respect to theprincipal normal direction.

y3

\y3

Fig. 4—9. of y.

Since 13R/ay1 (and therefore ii1) involve and the reference frame definedby iii, u2, will not be orthogonal. However, we can reduce it to an orthogonal


REFERENCES 97

system by taking

= (4—32)dy

which requires

= 150cer dy (4—33)

When (4—32) is satisfied,aR

= — Ky'2)t1

and= (4—34)

= cx(l —

In this case, the local frame at Q coincides with the frame at the centroid. Oneshould note that this simplification is practical only when ccc can be readilyintegrated.

Example 4—7

The parameters a and t are constant for a circular helix:

a = (a2 + c2)112

C

Then,C

at = —a

and integrating (4—33), we obtain

— Yo) tS

For this curve, varies linearly with y (or arc length). The parameter g1 follows from (4—34).

ds1hi = = a(1 — Ky2)

/ ax(l —

'\ cc-

REFERENCES

1. THOMAS, G. B., JR.: Analytical Geometry and Calculus, Addison-Wesley PublishingCo., Inc., Reading, Mass., 1953.

2. HAY, U. 13.: Vector and Tensor Analysis, Dover Publications, New York, 1953.3. JA}INKE, E., and F. EMDE: Tables of Functions, Dover Publications, New York, 1943.4. STRUm, D. J.: Differential Geometry, Addision-Wesley Publishing Co., Reading,

Mass., 1950.


DIFFERENTIAL GEOMETRY OF A MEMBER ELEMENT CHAP. 4

PROBLEMS

4—1. Determine Il, b, ; K, x for the following curves:(a) x1 = 3 cos y x2 = 3 sin y x3 = 5y(b) x1 3 cos y x2 = 6 sin y x3 = 5y(c) = + + p313(d) x1 = cos y

x2 = sin yx3 = cywhere a, /3, c are real constants.

4—2. If 0, the curve lies in the plane. Then,r Oandb ±i3.The sign of b will depend on the relative orientation of ñ with respect to 1.Suppose the equation defining the curve is expressed in the form

x2=J(x1) x3=0Equation (a) corresponds to taking x1 as the parameter for the curve.

(a) Determine the expressions for 7, ñ, b, and K corresponding to thisrepresentation. Note that y and + f(x1)12 + Let

_Lf' .f"etc(b) Apply the results of (a) to

4a2= — x1)

where a and b are constants. This is the equation for a parabola sym-metrical about x1 = b/2.

(c) Let 9 be the angle between and

cos0 = I •11.

Deduce that = sec 0. Express t, h, !, and K in terms of 0.(d) Specialize for the case where 02 is negligible with respect to unity.

This approximation leads to

sin 0 tan 0 0

cos 0 1

A curve is said to be shallow when 02 << 1.

4—3. Let K = l/R and t Show that (see (4—20))

di

dh— = —— 12dy.

dñ


PROBLEMS 99

4—4. The equations for an ellipse can be written as

x1 = a cos y x2 = b sin yor

2 2x1 x2

Determine 1, n for both parametric representations. Take x1 as the parameterfor (b). Does y have any geometrical significance?

4—5. Show that

dx,.

72 IA A1k C

dy dv2C2k [3 (d€

=[,.=,. \ dyj J \ dy dy dyj j

€311 42 1€21

€32 €13 0

€33j °

4—6. Let

42 43]Then,

€1

[4k]€3

Using (4—22), show thatIi' iT i—i

—

4—7. Determine D for Prob. 4—la.4—8. Determine JI for Prob. 4—lb.4-9. Specialize for the case where the reference axis is in the X1 — X2

plane. Note that b = €3313 where 431 = 1. When the reference axis is a planecurve and = 0, we call the member a "planar" member.

4—10. We express the differentiation formulas for as

dt— = atds

(a) Show that a is, in general, skewsymmetric for an orthogonal systemof unit vectors, ie., c5Jk. Determine a.

(b) Suppose the reference axis is a plane curve but çb 0. The memberis not planar in this case. Determine a.


5

Matrix Transformationsfor a Member Element

5—1. ROTATION TRANSFORMATION

Suppose we know the scalar components of a vector with respect to a ref-erence frame and we want to determine the components of the vector corre-sponding to a second reference frame. We can visualize the determination ofthe second set of components from the point of view of applying a transforma-tion to the column matrix of initial components. We refer to this transforma-tion as a rotation transformation. Also, we call the matrix which defines thetransformation a rotation matrix.

Let (j = 1, 2, 3 and n = 1, 2) be the directions and corresponding unitvectors for reference frame n. (See Fig. 5—1.) We will generally use a super-script to indicate the reference frame for directions, unit vectors, and scalar

Fig. 5—1. Directions for reference frames "1" and "2."

100

a


SEC. 5—1. ROTATION TRANSFORMATION

components in this text. We consider a vector, a. The scalar components ofa with respect to frame n are a

a is independent of the reference frame. Then,

a = = (a2)Ti2

To proceed further, we must relate the two reference frames. We write, therelations between the unit vectors as

i2 = Ili'where is the scalar component of with respect to The transformationmatrix, is nonsingular when the unit vectors are linearly independent. Sub-stituting for and equating the coefficients of i' leads to

a' =a2

Finally, we letR'2 (53)R21 =

With this notation, the relations between the component matrices take theform

a2 =a1 = R21a2

=

The order of the superscripts on R corresponds to the direction of the trans-formation. For example, R'2 is the rotation transformation matrix corre-sponding to a change from frame 1 to frame 2. We see that the transformationmatrix for the scalar components of a vector is the inverse transpose of thetransformation matrix governing the unit vectors for the reference frames.

Example 5—i

We consider the two-dimensional case shown in Fig. E5-1. The relations between theunit vectors are

= cos + sin

= —sin + cos

We write (a) according to (5—2).

cos6 sinO


102 MATRIX TRANSFORMATIONS FOR A MEMBER ELEMENT CHAP. 5

Then,

R21—sin4

Lsinfl cos4

R'2 = (fV) 1 =L—sjn 6 cos U

0 cos + o sin

andJafl — I [ cos sin

— ?J L — sin 0 cos oJ

When both frames are orthogonal, q5 = 0 and ir

Fig. E5—1

The result obtained in the preceding example can be readily extended to thecase of two orthogonal reference frames. When both frames areorthogonal, the change in reference frames can be visualized as a rigid bodyrotation of one frame into the other, f3jk is the direction cosine for withrespect to and the rotation transformation matrix is an orthogonal matrix:

R'2 — 1— LI'jki (55)'X2 A'Pjk — COSt j, k

In Sec. 4—7, we defined the orientation of the local frame (1k, 13, at apoint on the reference axis of a member element with respect to the naturalframe (1, ñ, b) at the point. This frame, in turn, was defined with respect to afixed cartesian frame 12, 13). In order to distinguish between the threeframes, we use superscripts p and p' for the local and natural frames at p anda superscript 1 for the basic cartesian frame:

= t2,

= {Z, ii, b} (5—6)

=

12


SEC. 5—2. THREE-DIMENSIONAL FORCE TRANSFORMATIONS 103

With this notation, the relations between the unit vectors and the variousrotation matrices are:

t" = R141=

1. From(4—21),

— (5—7)

2. From (4—24),1 0 0

= 0 cos4; sin 4;

0 —sin4; cos4;3.

defined by (4—25).

5—2. THREE-DIMENSIONAL FORCE TRANSFORMATIONS

The equilibrium analysis of a member element involves the determinationof the internal force and moment vectors at a cross section due to externalforces and moments acting on the member. We shall refer to both forces andmoments as "forces." Also, we speak of the force and moment at a point, sayP, as the "force system" at P. The relationship between the external forcesystem at P and the statically equivalent internal force system at Q hasa simpleform when vector notation is used. Consider a force F and moment M actingat P shown in Fig. 5—2. The statically equivalent force and moment at Q are

Feqrnv.-- (5-8)

Mequiv = M + X F

One can visualize (5—8) as a force transformation in which the force systemat P is transformed into the force system at Q. This transformation will be

Fcquiv.

Q

5—2. Equivalent force system.

linear if is constant, that is, if the geometry of the element does not changeappreciably when the external loads are applied. We will write (5—8) in matrixform and treat force transformations as matrix transformations.


— v' ri 1 ._ '1PL.. P

= (jl)TM1


We develop first the matrix transformation associated with the moment ofa force about a point. Let be a force vector acting at point P and MQ themoment vector at point Q corresponding to We will always indicate thepoint of application of a force or moment vector with a subscript. The relationbetween MQ and is —

= QP x (5—9)

We work with an orthogonal reference frame (frame 1) shown in Fig. 5—3and write the component expansions as

(5—10)

The scalar components of QP are 4, — Expanding the vector crossproduct leads to

= (5—11)

0 —(1 1\ (1 (1. 1

—

—(42 XQ2) — 0

Note that is a matrix. One can interpret it as a force-translation transformation matrix. The force at P is transformed by into

,1

P3F1 /'I —

2k

i—-iM1

43

I' /'i/ xP1/

Xp3

Fig. 5—3. Notation for orthogonal reference frame.



a moment at Q. Note that the order of the subscripts for the translation trans-formation matrix, corresponds to the order of the translation (from Pto Q). Also, and must be referred to the same frame, that is, the super-scripts must be equal.

Up to this point, we have considered only one orthogonal reference frame.In general, there will be a local orthogonal reference frame associated witheach point on the axis of the member, and these frames will coincide only whenthe member is prismatic. To handle the general case we must introduce rota-tion transformations which transform the components of F and M from thelocal frames to the basic frame (frame 1) and vice versa. We use a superscriptp to indicate the local frame at point P and the rotation matrix correspondingto a transformation from the local frame at P to frame 1 is denoted by R'1.With this notation,

=(5—12)=

and the general expression for takes the form

= (5—13)

We consider next the total force transformation. The statically equivalentforce and moment at Q associated with a force and moment at P are given by

— (a)MQ = + QP x

When all the vectors are referred to a common frame, say frame 1, the matrixtransformation is

— I(b)

'Q) L"PQ L3J

We let

=(5—14)

The 6 x 1 matrix is called the force system at Q referred to frame 1. Usingthis notation, (b) simplifies to

d1 —

When the force systems are referred to local frames, we mast first transformthem to a common frame and then apply (5--15). Utilizing the generalmatrix,

I 0 1

=_ (5—16)

and applying.

= (a)



we obtain= (5—17)

Equation (5—17) states that when the matrix transformation is appliedto we obtain its statical equivalent at Q. Actually, we could leave off thesubscripts and superscripts on when we write (5—17). However, appearsalone, we must include them. Note that the force transformation generallyinvolves both translation and rotation. The order of the subscripts correspondsto the direction of the translation, e.g., from P to Q. Similarly the order of thesuperscripts defines the direction of the rotation or change in reference frames,e.g., from frame p to frame q.

In general, the geometry of a member element is defined with respect to a basicreference frame which we take as frame 1. To evaluate we must determineRn', and from the geometrical relations for the member. We havealready discussed how one determines R1 in Secs. 4—7 and 5—f.

When the member is planar* and the geometry is fairly simple (such as astraight or circular member), we can take frame I parallel to one of the localframes. This eliminates one rotation transformation. For example, suppose wetake frame I parallel to frame p. Then, R1 = 1 and reduces to

— [R = (S—18)

Similarly, if 1 and q are parallel,

- - (5-19)— I

When both p and q arc parallel to 1, reduces to

= 91i'PQ (5—20)

By transforming from P to Q and back to F, we obtain

=

and it follows that(5—21)

If the transformation from P to is carried out in the order P S1, S1

S2,..., —÷ Q, where S2, . . , S,, are intermediate points, the transforma-tion matrix, is equal to the product of the intermediate transformationmatrices.

— (5—22)

* If the reference axis isa plane curve and the local frame coincides with the natural frame = 0)we say the member is planar.



where s1, s2, .. . , SN are arbitrary reference frames. It is convenient to take acommon reference frame for the intermediate transformations.

Example 5—2

We consider the plane circular member shown in Fig. ES—2. We take frame 1 parallel toframe p. Then,

4 — = {a sin 0, —a(1 — cos 0), O}

0 0 —a(1—cos6)= = 0 0 — a Sin 6

a(1 — cos 0) a sin 0 0

cos6 —sinO 0= = sin 0 cos 0 0

0 0

9— — I— —

0 0 +a(1 —cos6)= 0 0 —asinO

a(1 — cos 0) asin0 0

As an illustration of the case where the geometry is defined with respect to a basicCartesian frame, we consider the problem of finding for a circular helix. The general

The transformation matrix has the form

where

P

Fig. E5—2

p2

t7

Example 5—3



expansion for has the form

[RPS

The parametric representation for a circular is given in Sec. 4—i:

= a cos y

= a sin y= cy

where (j = 1,2,3) are the cartesian coordinates with respect to the basic frame (frame 1).Let yp and y corresponding to points P and Q. The coordinate matricesfor P and Q are

4 = {a cos Vp, a Sifl Cyp)

= {a cos yQ, a sin YQ, cy0}

Then,

0 — C(yp —

= C(yp y0) 0

—a(sin — sin yQ) a(cos Yp C05 Y0) 0

To simplify the algebra, we suppose the local frame coincides with the natural frame atevery point along the reference axis, that is, we take = 0. Using the results of Sec. 4—7,the rotation matrices reduce to

a(sin y,, — sin YQ)

—a(cos ,, — cos

a

a

C a

C.

a C

cc cc

0

where cc2 a2 + c2.Evaluating the product, we obtain

(a\2 /c\2I—I

a——sin ij

cc

ac—i (1 —

I cc

cos

I

c

cc

/a\2 /c'\2\ccJ


SEC. 5—3. THREE-DIMENSIONAL DISPLACEMENT TRANSFORMATIONS

where = yp y0. Also,

=

a2c ac—3-(2 — I — sinq)a a

(IC .

a

I

—I a a

ac2 a a2 ci2c— — c2) —

a a a a a a

Note that we can specialize the above general results for the case of a plane circular member(Example 5-2) by taking c = 0 and = 0.

5—3. THREE-DIMENSIONAL DISPLACEMENT TRANSFORMATIONS

Let P and Q be two points on a rigid body. Suppose that the body experiencesa translation and a rotation. We define Up and Ui,, as the translation and rota-tion* vectors for point P. The corresponding vectors for point Q are given by

UQ = Up + (Up X(5—23)

= Wp

Equation (5—23) is valid only when is negligible with respect to unity.Since PQ — QP and x PQ — PQ x an alternate form for is

UQ = Up + QP X (5—24)

We define

=(5—25)

as the displacement matrix for P referred to frame 1. The displaccment at Qresulting from the rigid body displacement at P is given by

— [13 XPQ1i —L"l 3J

consider next the case where the local frames at P and Q do not coincide.The general relation between the displacements has the form

=0 13

- (5—27)

— d14[o j

One can showt that alternate forms of (5—27) are= T (yqp)T

The units are radians. -

t See Prob. 5--7.

a— cosn)(a5 — c2)]



We see that the displacement transformation matrix is the inverse transpose ofthe corresponding force transformation matrix. This result is quite useful.

REFERENCES

1. HALL, A. S., and R. W. WOODS-LEAD: Frame Analysis, 2d ed., Wiley, New York, 1967.2. MORLCE, P. B.: Linear Structural Analysis, Ronald Press. New York, 1959.3. PESTEL, E., and LECKIE, F.: Matrix Methods in Elastomechanics, McGraw-Hill, New

York, 1963.4. Livasr.Ey, R. K.: Matrix Methods of Structural Analysis, Pergamon Press, London,

1964.5. MARTLS.t, H. C.: Introduction to Matrix Methods of Structural Analysis, McGraw-Hill,

New York, 1966.

PROBLEMS

5—1. Consider the two-dimensional cartesian reference frames shown. Ifa1 = {50, —100}, find a2.

Prob. 5—i

5—2. The orientation of two orthogonal frames is specified by the directioncosine table listed below.

1/2

1/21/21/2

\/2/2 o

(a) Determine R12. Verify that (R12)T = (R12y1


PROBLEMS

(b) If a1 = {10, 5, 10}, find a2.(c) If a2 {5, 10, 10), find a'.

5—3. Consider two points, P and Q, having coordinates (6, 3,2) and (—5, 1,4)with respect to frame 1. The direction cosine tables for the local reference framesare listed below.

—1 —,

S

1/2 1/2

1/2 1/2

0

1/2 1/21/2 1/2

0 — 12/2(a) Determine 91'),Q and(b) Determine(c) Suppose {l00, —50, 100, 20, —40, +60). Calculate

5--4. Consider the planar member consisting of a circular segment and astraight segment shown in the sketch below. Point P is at the center of the circle.

Prob,. 5—4

Q

(a) Determine by transforming directly from P to Q. Also find(b) Determine by transforming from P to S and then from S to Q.(c) Find corresponding to = {0, 0, 1, 0, 0, 0}.

5—5. Consider the circular helix,

= 2 cos + 2 sin + 13. -

(a) Suppose 4(y) = 0. Determine Take = ir/2, YQ =(b) Suppose = —y. Determine

t'i

Ic



5—6. Refer to Problem 5—3. Determine corresponding to = {1/2,— 1/4, 1/3, — 1/10, 1/10, 0}. Verify that

Q Q_ P P

5—7. Verify that (5—27) and (5—28) are equivalent forms. Note that

rI ii r niT•1 — I _T___ I — c ortIA I I

L" 3 1 3J

5—8. Consider the plane member shown. The reference axis is defined byx2 = f(xj).

Prob. 5—8

x1

(a) Determine Note that the local frame at P coincides with thebasic frame whereas the local frame at Q coincides with the naturalframe at Q.

(b) Specialize part (a) for the case where

4a2— (x1b — Xj)

and the x1 coordinate of point Q is equal to h/4. Use the results ofProb. 4—2.


XC

I)C C

l)

z



6

Governing Equationsfor an Ideal Truss

6—i. GENERAL

A system of bars* connected at their ends by frictionless hinges to joints andsubjected only to forces applied at the joint centers is called an ideal truss.tThe bars arc assumed to be weightless and so assembled that the line con-necting the joint centers at the ends of each bar coincides with the centroidalaxis. Since the bars are weightless and the hinges are frictionless, it follows thateach bar is in a state of direct stress. There is only one force unknown asso-ciated with each bar, namely, the magnitude of the axial force; the directionof the force coincides with the line connecting the joint If the bars liein one plane, the system is called a plane or two-dimensional truss. Thereare two displacement components associated with each joint of a plane truss.Similarly, a general system is called a space or three-dimensional truss, andthere are three displacement components associated with each joint.

We suppose there are in bars (members) and j joints. We define i as

= 2 for a plane truss= 3 for a space truss

Using this notation, there are if displacement quantities associated with the]joints. In general, some of the joint-displacement components are prescribed.Let r be the number of prescribed displacement components (displacementrestraints) and nd the total number of unknown joint displacements. It followsthat

= if — r (6—2)

Corresponding to each joint displacement restraint is an unknown joint force

A prismatic member is conventionally referred to as a bar in truss analysis.± See Ref. 1.

115


116 GOVERNING EQUATIONS FOR AN iDEAL TRUSS CHAP. 6

(reaction). We let flf be the total number of force unknowns. Then,

n1—m+r (6—3)

Finally, the total number of unknowns, n, for an ideal truss is

fl—flj+flj—(j+tfl (6—4)

The equilibrium equations for the bars have been used to establish the factthat the force in each bar has the direction of the line connecting the jointcenters at the ends of the bar. There remains the equilibrium equations for thejoints. Since each joint is subjected to a concurrent force system, there are iJscalar force-equilibrium equations relating the bar forces, external joint loads,and direction cosines for the lines connecting the joint centers in the deformedstate. In order to solve the problem, that is, to determine the bar forces, reac-tions, and joint displacements, m additional independent equations are required.These additional equations are referred to as the bar force—joint displacementrelations and are obtained by combining the bar force—bar elongation relationand bar elongation—joint displacement relation for each of the in bars.

In this chapter, we first derive the elongation—joint displacement relation fora single bar and then express the complete set of in relations as a single matrixequation. Thisprocedure is repeated for the bar force-elongation relations andthe joint force-equilibrium equations. We then describe a procedure for in-troducing the joint-displacement restraints and summarize the governing equa-tions. Finally, we briefly discuss the solvability of the governing equations forthe linear case. In this case, the question of initial instability is directly relatedto the solvability.

In Chapter 7, we develop variational principles for an ideal truss. Thetwo general procedures for solving the governing equations are described inChapters 8 and 9. We refer to these procedures as the displacement and forcemethods. They are also called the stWhess and flexibility methods in sometexts.

The basic concepts employed in formulating and solving the governingequations for an ideal truss are applicable, with slight extension, to a membersystem having moment resisting connections. Some authors start with thegeneral system and then specialize the equations for the case of an ideal truss.We prefer to proceed from the truss to the general system since the basicformulation techniques for the ideal truss can be more readily described. Toadequately describe the formulation for a general system requires introducinga considerable amount of notation which tends to overpower the reader.

6—2. ELONGATION—JOINT DISPLACEMENT RELATION FOR A BAR

We number the joints consecutively from 1 through j. It is convenient torefer the coordinates of a joint, the joint-displacement components, and theexternal joint load components to a common right-handed cartesian referenceframe. Let (j = 1, 2, 3) be the axes and corresponding orthogonal unit


SEC. 6—2. ELONGATION—JOINT DISPLACEMENT RELATION

vectors for the basic frame. The initial coordinates, displacement components,and components of the resultant external force for joint k are denoted by

(j = 1, 2, 3) and the corresponding vectors are written as

rk =j= 1

Uk = Ukl

Pk

(6—5)

The coordinates and position vector for joint k in the deformed state are

1k + 11k

= + Uk

Figure 6—1 illustrates the notation associated with the joints.

Fig. 6—1. Notation for joints.

x2

(6—6)

We number the bars from 1 through m and consider bar n to be connectedto joints k and s. The centroidal axis of bar n coincides with the line con-necting joints k and s. From Fig. 6—2 the initial length of bar n, denoted by

is equal to the magnitude of the vector = —

= (6—7)

14313

Deformed positionof joint k

x1

/// flkl

11k2

/


116 GOVERNING EQUATIONS FOR AN IDEAL TRUSS CHAP. 6

Since the basic frame is orthogonal, (6—7) reduces to

= — xk)T(xs — xk)=

— xkf) (6—8)

Before the orientation of the bar can be specified, a positive sense or direc-tion must be selected. We take the positive sense for bar n to be from joint k

13

///

Fig. 6—2. Undeformed position of Bar n.

12

to joint s and define as the direction cosine for the positive sense of bar nin the undeformed state with respect to the direction:

1 1= (Ar = — XkJ)

It is convenient to list the direction cosines in a row matrix,

= L— Xk)

(6—9)

(6—10)

Note that 1, due to the orthogonality of the reference frame. Finally,we let be the unit vector associated with the positive direction of bar n inthe undeformed state. By definition,

- 1= Ar =

12

The deformed position of bar n is shown in Fig. 6—3. The length and direc-tion cosines for bar n are equal to the magnitude and direction cosines for thevector, = — Pk. Let L12 + e12 be the deformed length, the unit vector

s

x.,3

I // XkI

Xk2

Xg3


SEC. 6—2. ELONGATION—JOINT DISPLACEMENT RELATION

x3

Fig. 6—3. Deformed position of Bar n.

x2

associated with the positive direction in the deformed state, and the corre-sponding direction cosine matrix. These quantities are defined by

+ eh)

— 1 —.

——

We consider first (6—12). Substituting for

= A1 + — Uk)

and noting (6—7), (6—11), we obtain, after dividing both sides by

e 2 2. 1(1 + = 1 + — Uk) + — uk)T(US ilk)

The expression for the direction cosine, expands to

1 [ 1

=[cx,,j + E UkJ)

I +-;:-

We list the fl's in a row matrix,

(6—12)

(6—13)

(6—14)

(6—15)

(6—16)

Ap Joints

x1t?&2



[cm + uk)T] (6—17)

By definition, is the change in length of bar n. Then, e,,/L7, is the extensionalstrain which is considerably less than unity for most engineering materials. Forexample, the strain is only for steel at a stress level of 3 x ksi. Therelations simplify if we introduce the assumption of small strain,

<< I

Expanding the left-hand side of (6—15), and noting (6—18), we obtain

e,, — Uk) + (ii,, — uk)T(u$ — Uk) (6—19)

The direction cosines for the deformed orientation reduce to

+ — Uk) (6—20)

To simplify the expression for further, we need to interpret the quadraticterms. Using (6—20), we can write (6—19) as

/ 1= + — (u, Uk)

This form shows that the second-order terms arc related to the change inorientation of the bar. If the initial geometry is such that the bar cannot ex-perience a significant change in orientation, then we can neglect the nonlinearterms. We use the term linear geometry for this case. The linearized relationsare

— Uk)(6—21)

We discuss this reduction in greater detail in Chapter 8. Since we are concernedin this chapter with the formulation of the governing equations, we will retainthe nonlinear rotation terms. However, we will assume small strain, i.e., wework with (6—19), (6—20).

6—3. GENERAL ELONGATION—JOINT DISPLACEMENT RELATION

We have derived expressions for the.direction cosines and elongation of a barin terms of the initial coordinates and displacement components of the jointsat the ends of the bar. By considering the truss as a system or network, thegeometric relations for all the bars can be expressed as a single matrix equation.The relations for bar n, which is connected to joints s and k (positive direction


SEC. 6—3. GENERAL ELONGATION—JOINT DISPLACEMENT RELATION 121

from k to s) are summarized below for convenience:

= (x. — — Xk)

T= (x., — xk)

= — Uk)

= + (ii, — uk)T

= + Uk)

Up to this point, we have considered joints s and k as coinciding with the.positive and negative ends of member n. Now we introduce new notation whichis more convenient for generalization of the geometric relations. Let n÷, n_denote the joint numbers for the joints at the positive and negative ends ofmember n. The geometric relations take the form (we replace s by and kbynJn(a)):

= — x,,.)

=—

= — (6—22)

= ci,, + (u,,, 11)T

= ci,, + — u,,)

To proceed further, we must relate the bars and joints of the system, that is,we must specify the connectivity of the truss. The connectivity can be definedby a table having m rows and three columns. In the first column, we list the barnumbers in ascending order, and in the other two columns the correspondingnumbers, and n, of the joints at the positive and negative ends of the mem-bers. This table is referred to as the branch-node incidence table in networktheory.* For structural systems, a branch corresponds to a member and a nodeto a joint, and we shall refer to this table as the member-joint incidence tableor simply as the connectivity table. It should be noted that the connectivitydepends only on the numbering of the bars and joints, that is, it is independentof the initial geometry and distortion of the system.

Example 6—1

As an illustration, consider the two-dimensional truss shown. The positive directions ofthe bars are indicated by arrowheads and the bar numbers are encircled. The conneêtivity

See Ref. 8.


122 GOVERNING EQUATIONS FOR AN DEAL TRUSS CHAP. 6

table (we list it horizontally to save space) for this numbering scheme takes the followingform:

Bar,n 1 2 3 4 5 6 7 8 9 10 11

+Joint(n÷) 1 2 4 5 1 2 3 1 2 4 5

—Joint(n..) 2 3 5 6 4 5 6 5 6 2 3

0

3

6

2

(x1 — x)T(x * x5)

= — x5)T

4

Fig. E6—1

0 0

With the connectivity table, the evaluation of the initial length and direction cosines canbe easily automated. The initial data consists of the j coordinate matrices, x1, x2To compute and a,,, we first determine n÷ and n_ from the connectivity table and thenuse the first two equations of(6—22). For example, for bar 8, 8÷ 1, 8.. 5, and

xs• — x1 — x5

We define e and qj as the system elongation and joint-displacement matrices,

e = {e1, C2,...,6—23

= {u1,u2

and express the m elongation-displacement relations as a single matrix equationeu = (6—24)

where d is of order m x ij. The elements in the uth row of d involve onlythe elements of Then, partitioning d into submatrices, of order 1 x 1,where k = 1, 2 in and £ = 1, 2 j, it follows that the only nonvan-ishing submatrices for row n are the two subinatrices whose column numbercorresponds to the joint number at the positive or negative end of member it,


SEC. 6—3. GENERAL ELONGATION—JOINT DISPLACEMENT RELATION

namely, n÷ and n.:= +777

= (6—25)

= 0 when t' n÷ orn

Example 6—2

The .& matrix can be readily established by using the connectivity table. For row n,one puts +y,, at column at column n_, and null matrices at the other locations.The general form of the d matrix for the truss treated in Example 6—1 is listed below.We have also listed the elongations and joint displacement matrices to emphasize thesignificance of the rows and partitioned columns of .&.

Uj U2 U3 U4 U5 U6

e1 Ii 0 0 0 0

e2 0 12 0 0 0

e3 0 0 0 13 13 0

e4 0 0 0 0 14 14

e5 15 0 0 0 0

e6 0 is 0 0 —Ys 0

0 0 0 0

e5 is 0 0 0 0

e9 0 19 0 0 0 —19

e10 0 1io 0 ho 0 0

e11 0 0 0 0

The d matrix depends on both the geometry and the topology. It is ofinterest to express d in a form where these two effects are segregated. The formof (6—25) suggests that we list the y's in a quasi-diagonal matrix,

71

= 72 (6—26)


124

and define C as

GOVERNING EQUATIONS FOR AN IDEAL TRUSS CHAP. 6

Example 6—3

The connectivity matrix for Example 6—1 is listed below. The unit matrices are of order2 since the system is two-dimensional.

Joint Numbers

1 2 3 4 5

1 12

2 +12

12

+12

12

+12

+12 12

+12 12

+12

12 +12

12 +12

One can consider row n of C to define the two joints associated with bar n. It followsthat column k of C defines the bars associated with joint k. This association is usually

See Prob. 6—6. See also Ref. 8.

C = [Ck(ik 1,2 in

e= 1,2,...,j* = + Cr,, - = —Ij

= 0 n÷ or n...Then,

d = yC

The network terminology* for C is augmented branch-node incidence matrix.We shall refer to it simply as the connectivity matrix.

(6—27)

(6—28)

6

BarNumbers


SEC. 6—4. FORCE-ELONGATION RELATION FOR A BAR 125

referred to as incidence. We say.that a joint is positive incident on a bar when it is atthe positive end of the bar. Similarly, a bar is positive incident on a joint when its positiveend is at the joint. For example, we see that joints 1 and 4 are incident on bar 5 andbars 3, 4, 6, 8, and 11 are incident on joint 5. We will use this property of the connectivitymatrix later to generalize the joint force-equilibrium equations.

6-4. FORCE-ELONGATION RELATION FOR A BAR

By definition, each bar of an ideal truss is prismatic and subjected only toaxial load applied at the centroid of the end cross sections. It follows that theonly nonvanishing stress component is the axial stress, a, and also, a is con-stant throughout the bar. We will consider each bar to be homogeneous butwe will not require that all the bars be of the same material. The strain, s, willbe constant when the bar is homogeneous and the force-elongation relationwill be similar in form to the uniaxial stress-strain curve for the material.

A typical a-c curve is shown in Fig. 6—4. The initial portion of the curve isessentially straight for engineering materials such as steel and aluminum. Amaterial is said to be elastic when the stress-strain curve is unique, that is,when the curves corresponding to increasing and decreasing a coincide (OABand BAO in Fig. 6—4). If the behavior for decreasing a is different, the materialis said to be inelastic. For ductile materials, the unloading curve (BC) is essen-tially parallel to the initial curve.*

0

Fig. 6—4. Stress-strain curves for elastic and inelastic behavior.

We introduce the following notation:

A = cross sectional areaF axial force, positive when tension

= initial elongation, i.e., elongation not associatedwith stress

* A detailed discussion of the behavior of engineering materials is given in Chap. 5 of Ref. 2.

Elasticbehavior

B

C



Since the stress and strain are constant throughout the bar,

F =e = Lg

e0 = Le0

(6—29)

We convert the a-c relation for the material to the force-elongation relationfor the bar by applying (6—29).

Fig. 6—5. Linear elastic behavior.

We consider first the case where the stress-strain relation is linear, as shownin Fig. 6—5. A material having this property is called Hookean. The initial andtransformed relations are

a = E(r —

AEF

=(e — e0) k(e — a0)

La F + e0 = fF + e0

(6—30)

We call k, f the stiffness and flexibility factors for the bar. Physically, k is theforce required per unit elongation and f, which is the inverse of k, is the elon-gation due to a unit force.

We consider next the case where the stress-strain relation is approximatedby a series of straight line segments. The material is said to be piecewise linear.Figure 6—6 shows this idealization for two segments. A superscript (j) isused to identify the modulus and limiting stress for segment j. The force-elongation relation will still be linear, but now we have to determine what

C0


SEC. 6-4. FORCE-ELONGATION RELATION FOR A BAR 127

Fig. 6—6. Piecewiselinear approximation.

segment the deformation corresponds to and also whether the strain is in-creasing (loading) or decreasing (unloading). For unloading, the curve is as-sumed to be parallel to the initial segment.* The relations for the variouspossibilities are listed below.

1. Loading or Unloading—initial Segment

F = F>1>

F = k">(e —

2. Loading—Second Segment

F>l) < F ,4a12> F>2>

F k'2>(e (6—32)

= + (f°> —

3. Unloading—Second Segment

F k>1>(e — (6—33)

One can readily generalize these relations for the nth segment.f

* We are neglecting the Bauschinger etlect. See Ref. 2, Sec. 5.9. or Ref. 3, Art. 74.t See Prob. 6—8.

(6—3 1)



Example 6—4

We consider a bilinear approximation, shown in Fig. E6—4.

= = 83.3 kips/in. f"> = 1/k"> = 12 x in./kipL

AE>2>k>2> = —.— = 41.7 kips/in. f"> = 24 x in./kip

L

F"> = = 3okips42> = + ([1)

— 120 — 0.36 in.

40

Fig. E6—4

3041.7

(in./in.)

TakingL=lQft=l2Oin ,4=lin.2

we obtain

Segment 1

Segment 2

F (83.3)(e — 120

F (41.7)(e —

Suppose a force of 35 kips is applied and the bar is unloaded. The equivalent initialstrain is (see Equation 6—33 and Fig. 6—6):

= —

= + (f>2) — = + 0.06 in.

The procedure described above utilizes the segment stiffness, which can beinterpreted as an average tangent stiffness for the segment. We have to modifythe stiffness and equivalent initial elongation only when the limit of the seg-


F — — — A= kU)(e — eo,eq)

eo,eq =A = — —

=

SEC. 6-4. FORCE-ELONGATION RELATION FOR A BAR 129

ment is reached. An alternate procedure is based on using the initial linearstiffness for all the segments. In what follows, we outline the initial st(ffnessapproach.

.4

/1'II'

-eo,eq.

Fig. 6—7. Notation for the initial stiffness approach.

Consider Fig. 6—7. We write the force-elongation relation segment 2 as

where e0, eq is interpreted as the equivalent linear initial strain and is given by

(6—34)

(6—35)

The equivalent initial strain, eoeq, depends on e, the actual strain. Since e inturn depends on F, one has to iterate on eoeq regardless of whether the seg-ment limit has been exceeded. This disadvantage is offset somewhat by the useof for all the segments.

The notation introduced for the piecewise linear case is required in order todistinguish between the various segments and the two methods. Rather thancontinue with this detailed notation, which is too cumbersome, we will dropall the additional superscripts and write the force-deformation relations for barn in the simple linear form

(6—36)= +



where k, f, and e0 are defined by (6—31) through (6—35) for the physicallynonlinear case.

6—5. GENERAL BAR FORCE—JOINT DISPLACEMENT RELATION

The force-deformation and deformation-displacement relations for bar n aregiven by (6—22) and (6—36). Combining these two relations leads to an expres-sion for the bar force in terms of the displacement matrices for the joints atthe ends of the bar. The two forms are:

F,, = — e0, = F0,,, + — k,,y,,u,, (6—37)

F0,,, = —k,,e0,,,and

— u,_) = e,, e0,, + f,,F,, (6—38)

We can express the force-displacement relations for the "m" bars as a singlematrix equation by defining

k1(6 39)

k= k2

km

and noting (6—24). The generalized forms of (6—37) and (6—38) are:

F = k(e — e0) = F0 + (6—40)and

d°1I = e0 + fF (6—41)

6—6. JOINT FORCE-EQUILIBRIUM EQUATIONS

Let F,, be the axial force vector for bar n (see Fig. 6—8). The force vectorhas the direction of the unit vector, i,,, which defines the orientation of the barin the deformed state. Now, = fi,,i. Then,

F,, = (6—42)

When F,, is positive, the sense of F,, is the same as the positive sense for thebar. Continuing, we define F,,,, as the forces exerted by bar n onthe joints at the positive and negative ends of the bar. From Fig. 6—8,

= — F,, = — F,,fi,,i(6—43

F,,,, +F,, =


SEC. 6—6. JOINT FORCE-EQUILIBRIUM EQUATIONS

Joint n_

Fig. 6—8. Notation for barforce.

We consider next joint k. The external joint load vector is Pk, wherePk = For equilibrium, the resultant force vector must equal zero. Then,

Pk - —j+=k

The first summation involves the bars which are positive incident on joint k(positive end at joint k) and the second the bars which are negative incident.Using (6--43), the matrix equilibrium equation for joint k takes the form:

Pk = — (6—44)j+kLet be the general external joint load matrix:

= P2, ,(if x 1) (6—45)

We write the complete set of joint force-equilibrium equations as:

= (6—46)

Note that the rows of pertain to the joints and the columns to the bars.We partition into submatrices of order i x 1.

= (if x m)(6—47)

= 1, 2,.. . ,j and k = 1, 2. . . , m

Since a bar is incident only on two joints, there will be only two elements inany column of From (6—44), we see that, for column n,

== (6—48)

= 0 when e orn_

The matrix can be readily developed using the connectivity table. It willhave the same form as dT with y, replaced by n,,. When the geometry is linear,

= = and



Example 6—5

The matrix for the truss of Example 6—1 has the following general form:

Bar Numbers

1 +llç

RT T+ 2nT nT ftT

—plo

-' DTP2

OF+p7 OT

A oT±1J3

nT

C-I

oTJ3

oT+p4 P6øT oT+pU

I,•r

6—7. INTRODUCTION OF DISPLACEMENT RESTRAINTS;GOVERNING EQUATIONS

We have developed the following equations relating F, e, and qj,

e = d°l1 = e0 + IF=

where the elements of and are the external joint-displacement and externaljoint-load matrices arranged in ascending order. Also, in our derivation, wehave considered the components to be referred to a basic reference frame. Now,

* See Sec. 6—3, Eq. 6—27.

1 2 3 4 5 6 7 8 9 10 11

IzaC

We could have also utilized the connectivity* matrix C to develop ft waspointed out in Example 6—3 that the elements of the kth column of C definethe incidence of the bars on joint k. Using this property, we can write thegeneralized form of (6—44) as

where0 -. 0

(rn x Lm) (6—49)

o oFinally, we have

= (6—50)


SEC. 6—7. INTRODUCTtON OF DISPLACEMENT RESTRAINTS 133

when joint displacement restraints are imposed, there will be a reduction inthe number of joint displacement unknowns and a corresponding increase inthe number of force unknowns. This will require a rearrangement of d,

andLet r be the number of displacement restraints and 11d the number of displace-

ment unknowns. There will be n4 prescribed joint loads and r unknown jointloads (usually called reactions) corresponding to the na unknown joint displace-nients and the r known joint displacements. We let U1, U2 be the columnmatrices of unknown and prescribed joint displacement components and P1,P2 the corresponding prescribed and unknown joint load matrices. The re-arranged system joint displacement and joint load matrices are written as U, P:

(fld x 1)

(r x 1)6—51

—>< 1)

- x 1)+ V = 13

We point out that the components contained in U (and P) may be referred tolocal reference frames at the various joints rather than to the basic frames.This is necessary when the restraint direction at a joint does not coincide withone of the directions of the basic frame. Finally, we let A and B be the trans-formation matrices associated with U and P. Then, (a) takes the form:

e = = AU e0 + fFP BF

We partition A, B consistent with the partitioning of U, P:

A2](fflxnd) (mxr)

B [Bil (nd x in) (6—52)

LB2i(r x m)

and write (b) in expanded form:

e A1U1 + A2tJ2 = e0 + fF (6—53)

= B1F (6—54)

P2 B2F (6—55)

Equation (6—53) represents equations relating the in unknown bar forces,the nd unknown displacements, and the r prescribed displacements. Equation(6—54) represents equations involving the in unknown bar forces and the

prescribed joint loads. Lastly, Equation (6—55) represents r equations. forthe r reactions in terms of the m bar forces. When the geometry is nonlinear,A and B involve the joint displacements. If the geometry is linear, A J3T, and

= AT j = 1, 2 (6—56)



We have introduced the displacement restraints into the formulation byreplacing d, with A, B. It remains to discuss how one determines A, Bfrom d, In the following section, we treat the case of an arbitrary restraintdirection. We also describe how one can represent the introduction of displace-ment restraints as a matrix transformation.

6—8. ARBITRARY RESTRAINT DIRECTION

When all the restraint directions are parallel to the direction of the globalreference frame, we obtain U from by simply rearranging the rows ofsuch that the elements in the first rows are the unknown displacements andthe last r rows contain the prescribed displacements. To obtain A, we performthe same operations on the columns of d, Finally, since P corresponds to U,we obtain B by operating on the rows of or alternately, by operating on thecolumns and then transposing the resulting matrix.

When the restraint at a joint does not coincide with one of the directions ofthe basic frame, it is necessary first to transform the joint displacement andexternal load components from the basic frame to a local frame associated withthe restraint at the joint. Suppose there is a displacement restraint at joint k.Let (j = 1, 2, 3) be the orthogonal directions for the local reference frameassociated with the displacement restraint at joint k. Also, let and bethe corresponding displacement and external joint load components. Finally,let R0k be the rotation transformation matrix for the local frame at joint k withrespect to the basic frame (frame o). The components are related by:

k_ okUk— Uk

k.. ak —Pk— Pk

whereROk = [cos (6—58)

We have omitted the frame superscript (o) for quantities referred to the basicframe (ut, to simplify the notation.

We define CU', as the system joint-displacement and -force matricesreferred to the local joint reference frames,

= ...= . . .

(6-59)

and as the system joint-rotation matrix,

R°1

R02=

. (6—60)

R0j

Then,== (a)


SEC. 6—8. ARBITRARY RESTRAINT DIRECTION 135

Operating on the initial equations with (a),

b(IP =leads to

= (6—61)

The transformation to is the same as for the case where the restraintdirections are parallel to the directions of the basic frame, that is, it will involveonly a rearrangement of the rows of Similarly, we obtain A by rearrangingthe columns of .cifi. The steps are

A2]

-+ -* BLB2

Example 6—6

To obtain the submatrices in column kof we postmultiply the submatrices in columnk of ri by R°" T We can perform the same operation on and then transpose theresulting matrix or, alternately, we can premultiply the submatrices in row k of by R°".As an illustration, see the matrix for Example 6—5 on page 136. The matrix can bedetermined by transposing and replacing il,, by y,.

One can visualize the introduction of displacement restraints as a matrixtransformation. We represent the operations

U and P (6—62)

asU=D°lI

and call D the displacement-restraint transformation matrix.When the restraint directions are parallel to the directions of the basic frame,

D is a permutation matrix which rearranges the rows of We obtain D byapplying the same row rearrangement to a unit matrix of order ij. Postmulti-plication by Dr effects the same rearrangements on the columns. Also,*Dr D1.

For the general case of arbitrary restraint directions, we first determineand then U. Now,

= (a)

The step, —* U, involves only a permutation of the rows of

U = (6—63).

where H is the permutation matrix corresponding to the displacement restraints.

* See Prob. I —36 for a discussion of permutation matrices.


for

Exa

mpl

e 6—

6

F1F2

F3F4

F5F6

F7F8

F9F1

0F1

1

Pt pt

R°'

pb

——

—R

04P

1

RO

SDT

I

C) 0 m z z m 0 C —

1 0 z C,) m 0 :13 > 2 C m -1 C (I)

Cl)


SEC. 6—9. INITIAL INSTABILITY 137

Combining (a) and (6—63), we have

U =and it follows that

I) = HPII°' (6—64)

Since both H and are orthogonal matrices, D is also an orthogonal matrix.Using (6—62),

A dDT

and then substituting for d, P.s, and D in terms of the geometrical, connectivity,local rotation matrices lead to

B6 65A = ( — )

Equation (6—65) is of interest since the various terms are isolated. However,one would not generate A, B with it.

6—9. INITIAL INSTABILITY

The force equilibrium equations relating the prescribed external joint forcesand the (internal) bar forces has been expressed as (see Equation 6—54):

P1=B1F

where P1 is < I) and F is (m x 1). When the geometry is nonlinear, B1depends on the joint displacements as well as on the initial geometry andrestraint directions. In this section, we are concerned with the behavior underan infinitesimal loading. Since the nonlinear terms depend on the load intensity,they will be negligible in comparison to the linear terms for this case, wetake B1 as constant. Then, (a) represents linear equations in in unknowns.If these equations are inconsistent for an arbitrary infinitesimal loading, we saythe system is initially unstable.

When the geometry is linear, B1 is independent of the loading and the initialstability criterion is also applicable for a finite loading. This is not true for anonlinear system. We treat stability under a finite loading in Chapter 7.

Consider a set of j linear algebraic equations in k unknowns.

ax=c (b)

In general, (b) can be solved only if a and [a c] have the same rank,* It followsthat the equations are consistent for an arbitrary right-hand side only whenthe rank of a is equal to], the total number of equations. Applying this condition

* See Sec. 1—13; see also Prob. 1—45.



to (a), we see that the truss is initially unstable when the rank of B1 is less

than na.For the truss to be initially stable under an arbitrary loading, B1 must be

of rank This requires m That is, the number of bars must be at leastequal to the number of unknown displacement components. Since the rankmay still be less than this condition is necessary but not sufficient for initialstability. In order to determine whether a truss is initially stable, one mustactually find the rank of B1. The following examples illustrate various casesof initial instability.

Example 6—7

The force-equilibrium equations for the accompanying sketch are:

x2

x1

Pu —1

P12 +1—P21 +1

B1

P22 +1

P31 +1

Row 3 is (— 1) times row 1. The equations are consistent only if P21 = — PuSince m < we know the system is unstable for an arbitrary loading withoutactually finding r(B1).

Fig. E6—7

in = 4na = 5

F

F1 F4F2 F3

)


SEC. 6—9.

Example 6—8

INITIAL INSTAB1LITY 139

We first develop the matrix for the truss shown in Fig. E6—8A and then specializeit for various restraint conditions.

There are three relations between the rows

(1) row®+row®+row®= —row®(2) row® + row® + row = —row®(3) (sin 8)(row ® + row ©) — cos U (row ®) = cos U (row ®)

Fig. E6—8A

'1

0M

J

I4 3

F

F

F3F 2 F4 F5 F6

® Pit —1 —cosU

© Piz +1 sinO

® Psi +1—a--—-—

cosO.

P22 ±1 sinO

® Psi +1 cosO

® P32 —1 —sinO

© P41 —1 —cosO

—1 —sinO



The first two relations correspond to the scalar force equilibrium conditions for the externaljoint loads:

Pkl = P11 + P21 + P31 + P41 = 0

Pk2 = P12 + P22 + P32 + P42 = 0

The third relation corresponds to the scalar moment equilibrium condition:

k1

Mk is the moment of the external force vector acting at joint k with respect to point0, the origin of the basic frame. We obtain relation (3) by taking Oat joint 4. Equation (b)reduces to

—d(p11 + P21) + b(p22 + P32) = 0

Using

ci = L sin 8

b = L cos 0

we can write (c) as

cos 0p32 sin O(pii + P21) — cos

which is relation 3.We see that rows 2 and 5 arc independent. Thc remaining set (rows 1, 3, 4, 6, 7, 8)

contains only three independent rows. Now, we obtain B1 from by first taking a linearcombination of the rows (when the restraints are not parallel to the basic frame) and thendeleting the rows corresponding to the joint forces associated with the prescribed jointdisplacements. Since has three linear dependent rows, it follows that we must introduceat least three restraints. Initial instability will occur if—

1. An insufficient number of restraints are introduced (n4 > 5).2. A sufficient number of restraints are introduced (/24 = 5) but the rows of B1 are

not linearly independent. We say the restraints are not independent in this case.These cases are illustrated below.

Case 1

Fig. E6—8B

x2

m6

x1

1 2


SEC. 6—9. INITIAL INSTABILITY

We obtain B1 by deleting rows 6 and 8 (corresponding to P32 and P42). The system isstable only when the applied joint loads satisfy the condition

Case 2

x2

xl

Pu + P21 + P31 = P41

We delete rows 4, 6, and 8. The number of restraints is sufficient (fld = 5) but the restraintsare not independent since r(B1) < 5. Actually, r(81) = 4. To make the system stable, atleast onc horizontal restraint must be introduced.

In Example 6—8, we showed that there are three relations between the rowsfor a two-dimensional truss. These relations correspond to the force- and

moment-equilibrium conditions for the complete truss.To establish the relations for the three-dimensional case, we start with the

equilibrium equations,3 (jxl)

o

where is the moment of with respect to an arbitrary moment center, 0.For convenience, we take 0 at the origin of the basic reference frame. Par-titioning

(6—66)

where is of order (i x m) and using the matrix notation introduced in

Fig. E6—8C

rn = 6—5

(2i—3)x 1

0



Sec. 5—2 for the moment,* the equilibrium equations take the form

PA,. = 0 (6—67)

= 0 (6—68)

Equation (6—67) represents i relations between the rows of PA,

row q + row (q + I) + ... + row [i(j — 2) + q] = row [i(j 1) + q]q=l,2,...i

(6—69)

and (6—68) corresponds to (2i — 3) relations.We have shown that there are at least 3(i — 1) relations between the tows

of PA. Now, we obtain B by combining and rearranging the rows of PA. Itfollows that B will also have at least 3(i — 1) relations between its rows. Finally,we obtain by deleting the rows corresponding to the restraints. For thesystem to be initially stable, we must introduce at least 3(i — 1) restraints:

r no. of restraints 3(i — 1) (6—70)

Note that this requirement is independent of the number of bars. Also, it is anecessary but not sufficient condition for initial stability.

The number of restraints must also satisfy the necessary condition in.

This requiresr = ((j — — m) (6—71)

Both (6—70) and (6—71) must be satisfied. Either condition may control r,depending on the arrangement of the bars.

REFERENCES

1. NORRIS, C. H., and J. B. WILBUR: Elementary Structural Analysis, McGraw-Hill,New York, 1960.

2. CRANDALL, S. H,, and N. C. DAHL: An Introduction to the Mechanics of Solids,McGraw-Hill, New York, 1959.

3, TIMOSNENKO, S.: Strength of Materials, Part 2, Van Nostrand, New York. 1941.4. TIMoSISENKO, S., and D. H. YOUNG: Theory of Structures, McGraw-Hill, New York,

1945.5. MCMINN, S. I.: Matrices jbr Structural Analysis, Wiley, New York, 1962.6. MARTIN, H. C.: Introduction to Matrix Methods of Structural Analysis, McGraw-

Hill, New York, 1966.7. LIVIISLEY, R. K.: Matrix Methods of Structural Analysis, Pergamon Press, London,

1964.8. FENVES, S. J., and F. H. BRANIN: "Network-Topological Formulation of Structural

Analysis," J. Struct. Div., ASCE, Vol. 89, No. ST4, pp. 483—514, 1963.

* See Eq. 5—11.


PROBLEMS 143

PROBLEMS

6—1. Determine in,J, r, and for the following plane trusses:

Prob. 6—1

6—2. Suppose bar n is connected to joints s and k where

Xk = {l, 1,0] (ft) = {5, —5. —2] (ft)

(a) Take the positive direction of bar n from k to s. Determineand

(b) SupposeUk {1/10, 1/20, 1/l0} (inches)

= — 1/10, — 1/30} (inches)

Find 1k and Note that the units of x and u must be consistent.Determine and ji,,, using the exact expressions (Equations 6—15,6—17), the expressions specialized for the case of small strain (Equa-tions 6—19, and 6—20), and the expressions for the linear geometriccase (Equation 6—21). Compare the results for the three cases.

6—3. Discuss when the linear geometric relations are valid and develop theappropriate nonlinear elongation-displacement relations for the trusses shown.Assume no support movements.6—4. Consider the truss shown:

(a) Establish the connectivity table.(b) List the initial direction cosines. Do we have to include nonlinear

geometric terms for this truss?(c) Locate the nonzero submatrices in .sd, using the connectivity table.

Determine the complete form of d.

(a)

(b)


x2

X1 3

x2

I'

(d) Determine C.(e) Verify that d = cxC.

6—5. Determine d for the three-dimensional truss shown.6—6. Consider the d-c network shown. The Junctions are generally called

nodes, and the line connecting two nodes is called a branch. The encirclednumbers refer to the branches and the arrowheads indicate the positive sense(of the current) for each branch.


Prob. 6—3

2 .3

(a)

2

Ib)

Prob. 6—4


PROBLEMS 145

Let (j = 1, 2,. . . , 5) denote the potential at node j. Also, let and n.denote the nodes at the positive and negative ends of branch n. The potential

drop for branch n, indicated by is given by

We define v and e as

v = {v1, v2,.. . , v5} = general node potential matrix

(0, 1,0)

Prob. 6—5

Prob. 6—6

e = {e1, e2, . . . , = general branch potential difference matrix

and write the system of branch potential difference—node potential relations as

e = .cjv

Determine d, using the branch-node connectivity table. Discuss how the trussproblem differs from the electrical network problem with respect to the, formof ad. How many independent columns does ad have? In network theory, adis called the augmented branch node incidence matrix.

x3

xI

4

(1,0,0) (1,1,0)

03

0'= v,,,



Take L 20 ft, A = 2 in2, and the a-s curve shown.Develop the piecewise linear force-elongation relations.Suppose a force of + 60 kips is applied and then removed. I)eterminethe force-elongation relation for the inelastic case.Suppose the bar experiences a temperature increase of 1000 F. Deter-mine the initial elongation. Consider the material to be aluminum.

6—8. Generalize Equation 6—32 for segmentj. Start with

e +

Prob. 6—7

and express eb0 in terms of quantities associated with segment (J — 1).6—9. Generalize Equation 6—35 for segmcntj.6—10. Suppose the stress-strain relation for initial loading is approximated,

as in the sketch, by

Prob. 6—10

6—7.(a)(b)

(c)

20 ksi

6 X ksi

a = E(s — be3)

GA

Ee

da Et


PROBLEMS 147

(a) Determine expressions for ES and E', the secant and tangent moduli.(b) Determine expressions for k5 and kt.(c) Suppose the material behaves inelastically for decreasing 4 Consider

the unloading curve to be parallel to the initial tangent. Determine theforce-elongation relation for AB.

6—li. Repeat Prob. 6—10, using the stress-strain relation

= (u +

where E, c, and n are constants.6—12. For the accompanying sketch:

Prob. 6—12

p

(a) Locate the nonzero submatrices in(b) Assemble for the linear geometric case.

6—13. Repeat Prob. 6—12 for the three-dimensional truss shown.

Prob. 6—13

LX2

I 'I.

6—14. Consider the electrical network of Prob. 6—6.(a) Let be the current in branch n. The positive sense of is from, node

n_ to node n÷. Now, the total current flowing into a node must equalthe total current flowing out of the node. This requirement leads toone equation for each node involving the branch currents incident on



the node. Let

= {i1, j2, . . . , = general branch-current matrix

Show that the complete system of node equations can be written as(Sxl)

0

where d is given in Prob. 6—6.(b) How many independent equations does (a) represent?

(Hint: d has only four independent columns).When the resistance is linear, the current and potential drop for abranch are related by

=

e0 is the branch emf and R,, is the branch resistance. Analternate form is

i,, = — e0 ,,)

Note the similarity between (b) and the linear elastic member force-elongation relation. Show that the complete system of branch cur-

potential relations can be written as

e = = e0 + Ri= R1(e — e0) = R1dv — R1e0

Equations (a) and (c) are the governing unpartitioned equations for alinear-resistance d-c network. The partitioned equations are developedin Prob. 6—23. It should be noted that the network problem is one-dimensional, that is, it does not involve geometry. The d matrixdepends only on the topology (connectivity) of the system. Actually,d corresponds to the C matrix used in Sec. 6—3 with i = 1.

6—15. Refer to Prob. 6—12, Suppose u11, u42, is52 are prescribed. IdentifyB1 and B2.

6—16. Refer to Prob. 6—12.(a) Develop the general form of(b) Suppose is21, u42, are prescribed. The orientation of the local

frame at joint 5 is shown in the sketch. Determine B1 and B2.

Prob. 6—16x2


PROBLEMS

6—17. Refer to Prob. 6—13(a) Develop the general form of(b) Determine B1 and B2 corresponding to the following prescribed dis-

placements:U11, U12, U3j, U-33, U23, U13

The local frame at joint 2 is defined by the following direction cosinetable.

x1 x2 x3

0

1/2 1/2

1/2 1/2

6—48. Consider the two-dimensional truss shown. The bars are of equallength and 0 is the center of the circumscribed circle. The restraint directionis degrees counterclockwise from the tangent at each joint. Investigate theinitial stability of this system. Repeat for the case of four bars.

11 = 13

Prob. 6—18

6—19. Suppose na = in.equations for P1 0 are

Then, B1 is of order tn x m. The equilibrium

(,nxrn) (mxl) rnXlB1 F=0

If (a) has a nontrivial solution, the rank of B1 is less than m and the system isinitially unstable (see Prob. 1—45). Rather than operate on B1, to determiner(B1), we can proceed as follows:

(1) We take the force in some bar, say bar k, equal to C:

Fk = C

r (restraint direction)

t (tangent)

I.


150 GOVERNING EQUATIONS FOR AN DEAL TRUSS CHAP. 6

(2) Using the joint force-equilibrium equations, we express the remainingbar forces in terms of C.

(3) The last equilibrium equation leads to an expression for Fk in termsof C. If this reduces to an identity, r(B1) < since a nontrivial solution forF exists. This procedure is called the zero load test.

(a) Apply this procedure to Prob. 6—18. Take F1 = C and determineF2, F3, and then F1 using the equilibrium condition (summation offorces normal to r must equal zero) for joints 1, 2, 3.

(b) When n4 = m and the geometry is linear, the truss is said to be staticallydeterminate. In this case, we can determine F, using only the equationsof static equilibrium, since the system, P1 = B1F, is square. Do initialelongations and support settlements introduce forces in the bars ofa statically determinate truss?

6—20. Modify the zero load test for the case where na < in. Note that thegeneral solution of B1F 0 involves m — r(B1) arbitrary constants.

6—21. Investigate the initial stability of the two-dimensional truss shown.Use the zero load test.

Prob. 6—21

6—22. Investigate the initial stability of the systemdirections are indicated by the slashed lines.

shown. The restraint

5

4

Prob. 6—22

-j

-t

I3

c


PROBLEMS

6—23. We generalize the results of Probs. 6—6. and 6—14 for a networkhaving b branches and n nodes. Let

e branch potential duff, matrix = {e1, e2, -. . ,

= branch current matrix = {i1, j2

v = node potential matrix = {v1, v2,. . . ,

The general relations are (1) node equations (n equations)(nxb) (bxl) (nxl)&T =

and (2) branch equations (b equations)

e = dv = e0 + Ri

Now, dT has only n — 1 independent rows. One can easily show that therows of are related by

row ii=

row k

It follows that (a) represents only n — 1 independent equations, and oneequation must be disregarded. Suppose we delete the last equation. Thiscorresponds to deleting the last column of d (last row of dT). We partition d,

(bxn) bx(n—1) bxld2]

and let d1 = A. The reduced system of node equations has the form

ATj = 0

Note that AT corresponds to B1 for the truss problem.Equation (e) represents (n — 1) equations. Since v is of order n, one of the

node potentials must be specified. That is, we can only determine the potentialdifference for the nodes with respect to an arbitrary node. We have deletedthe last column of d which corresponds to node n. Therefore, we take asthe reference potential.

(a) Let= {v1 — v2 — . . . , —

Show thatdv = AV

Summarize the governing equations for the network.(b) The operation

corresponds to introducing displacement restraints in the truss pro-blem. Compare the necessary number of restraints required for thenetwork and truss problems;


7

Variational Principlesfor an Ideal Truss

7-1. GENERAL

The formulation of the governing equations for an ideal truss describedin Chapter 6 involved three steps:

1. The elbngation of a bar was related to the translations of the joints atthe end of the bar.

2. Next, the bar force was expressed in terms of the elongation and then interms of the joint translations.

3. Finally, the equilibrium conditions for the joints were enforced, re-sulting in equations relating the external joint loads and internal barforces.

The system equations were obtained by generalizing the member force-displacement and joint force equilibrium equations and required defining onlytwo additional transformation matrices Later, in Chapter 10, we shallfollow essentially the same approach to establish the governing equations foran elastic solid.

In this chapter, we develop two variational principles and illustrate theirapplication to an ideal truss. The principle of virtual displacements is treatedfirst. This principle is just an alternate statement of force equilibrium. Next,we discuss the principle of virtual forces and show that it is basically a geo-metrical compatibility relation. Both principles are then identified as the sta-tionary requirements for certain functions. For this step, we utilize the materialpresented in Chapter 3, which treats relative extremas of a function. Finally,we discuss the question of stability of an elastic system and develop the stabilitycriterion for an ideal truss.

Why bother with variational principles when the derivation of the governingequations for an ideal truss is straightforward? Our objective in discussingthem at this time is primarily to expose the ieader to this point of view. Also,we can illustrate these principles quite easily with the truss. Later, we shall

152


SEC. 7—2. PRINCIPLE OF VIRTUAL DISPLACEMENTS 153

use these principles, particularly the principle of virtual forces, to constructapproximate formulations for a member.

7-2. PRINCIPLE OF VIRTUAL DiSPLACEMENTS

The principle of virtual displacements is basically an alternate statement offorce equilibrium. We will establish its form by treating first a single particleand then extending the result to a system of particles interconnected withinternal restraints. The principle utilizes the concept of incremental work and,for completeness, we review briefly the definition of work before starting withthe derivation.

Let v be the displacement of the point of application of a force F in thedirection of F. The work done by F (see Fig. 7—1) is defined as

w w0 + JFdv = W(v)

where v0 is an arbitrary reference displacement. Since W is a function of v,the increment in W due to an increment Ar can be expressed in terms of thedifferentials of W when F is a continuous function of yr f

= dW + 4d2W +dWdW —Ar = F Audv (7—2)

d2W = d(dW)

We refer to dW as the first-order work. Similarly, we call d2W the second-order work. If dF/dv is discontinuous, as in inelastic behavior, we must usethe value of dF/dv corresponding to the sense of Av. This is illustrated in

F

w—rv0

Fig. 7—1. Work integral for the one-dimensional force-displacement relation.

t Differential notation is introduced in Sec. 3—I.


154 VARIATIONAL PRINCIPLES FOR AN IDEAL TRUSS CHAP 7

Fig. 7—2. We use dF/dv = +k1 for > 0, and dF/du = —k2 for < 0.

Note that W is not a single-valued function of v when there is a reversal in the

F-v curve.

vs V

Fig. 7—2. Work integral for direction-dependent force.

We consider first a single mass particle subjected to a system of forces (seeFig. 7—3), Let R be the resultant force vector. By definition, the particle is inequilibrium when R = 0. We visualize the particle experiencing a displacementincrement Au from the initial position. The first-order work is

dW=R'Afi (7—3)

If the initial position is an equilibrium position, dW 0 for arbitrary Au sinceR = 0. Therefore, an alternate statement of the equilibrium requirement is:

The first-order work is zero for an arbitrary displacement 7 4of a particle from an equilibrium position. —

The incremental displacement is called a virtual displacement; this state-ment is the definition of the principle of virtual displacements.

R +

Fig. 7—3. Virtual displacement of a single mass particle.

One can readily generalize (7—4) for the case of S particles. Let be thefirst-order work associated with the forces acting on particle q and the

We consider the forces to be continuous functions of Au.

A



corresponding virtual-displacement vector. If particle q is in equilibrium,= 0 for arbitrary It follows that the scalar force-equilibrium equations

for the system are equivalent to the general requirement,

dW=

= 0 for arbitrary

Equation (7—5) is the definition of the principle of virtual work for a systemof particles.

In general, some of the forces acting on the particles will be due to internalrestraints. We define dW5 as the first-order work done by the external forcesand dW1 as the work done by the internal restraint forces acting on the particles.Substituting for dW, (7—5) becomes

dW5 + dW1 = 0 for arbitraryq=l,2 S

Now. let be the work done by the internal restraint forces acting onthe restraints. We use the subscript D for this term since it involves the

F1 /(Deformedl F1

I .

El

(Initial)

(Deformed)

F1 F1 IIFig. 7—4. Work done on the mass particles and internal restraints.

deformation of the restraints. The restraint force acting on a particle is equal inmagnitude, but opposite in sense, to the reaction of the particle on the restraint.Since the points of application coincide, it follows that

As an illustration, consider the simple system shown in Fig. 7—4. For thiscase, we have

dW,, = —F1 Au1 + F1 Au2dW1 = F1 Au1 — F1 Au2


156 VARIATIONAL PRINCIPLES FOR AN IDEAL TRUSS CHAP. 7

Using (b), we can write (a) as:

dWE = dWD for arbitrary(7—6)q=1,2,...,S

Also, the general principle of virtual displacements can be expressed as follows:

The first-order work done by the external forces is equal to thefirst-order work done by the internal forces acting on the restraintsfor any arbitrary virtual displacement of a system of particles from anequilibrium position.

We emphasize again that (7—6) is just an alternate statement of the forceequilibrium conditions for the system. Some authors refer to (7—6) as thework equation.

To apply the principle of virtual displacements to an ideal truss, we con-sider the joints to be mass points and the bars to be internal restraints. Wehave defined and as the column matrices of external joint loads andcorresponding joint displacements, Then,

=

where &W contains the virtual joint displacements. The first-order work doneby the restraint forces acting on bar n due to the virtual displacements is

= F,, deH

Generalizing (b), we havedW0 = FraC

Finally, the work equation for an ideal truss has the form

Ml = FT de for arbitrary MIt (7—7)

The scalar force-equilibrium equations are obtained by substituting for de interms of Mt.

It is convenient to first establish the expression for the differential elongationof an individual bar and then assemble de. Operating on

— (— — — u,

U

and noting the definition of (see (6—22)), we obtain

de, = + — u,)T] (Au — Au,)

= — Au,,) (7—8)

1- = J,,, F, de, = Wd(e,). We must use the rules for forming the differentials of a compoundfunction since e, depends on the joint displacements. Using (3—17), we can write

dWddW4 = de, = F, de,

de

d2 = d(F, de,,) = (de,)2 + F, d2e,de,



The assembled form follows from (6—25). We just have to replace y,, with il,,:

de MI (7—9)

Substituting for de in (7—7),?,TMI =

and requiring (a) to be satisfied for arbitrary results in the joint forceforce equilibrium equations.

For the geometrically linear case, e where d is constant and de =dM1 follows directly from e. We have treated the geometrically nonlinearcase here to show that the principle of virtual displacements leads to force-equilibrium equations which are consistent with the geometrical assumptionsassociated with the deformation-displacement relations.

Example 7—1

We consider a rigid member subjected to a prescribed force, P. and reactions R,, R2,as in the diagram. There is no internal work since the body is rigid. Introducing the virtual

Fig. E7—1

b

displacements shown above, and evaluating the first-order work,

dW = = R, Au1 + R2 Au2 — P

Now, is not independent:( d\ 7d

= Au,—

+ Au2

Then,

dW = —

—

+ Au2 {R2 — = 0

Requiring (c) to be satisfied for arbitrary Au,, Au2 leads to

R2 = P

which are the force and moment equilibrium equations, in that order.



Example 7—2

We consider the outside bars to be rigid (see sketch). To obtain the force equilibriumequation relating P and the internal bar forccs F1, F2, we introduce a virtual displacement,Au1, of the point of application of F:

dWE = PAu1

dW0 F1 de1 + F2 de2

The first-order increments in the elongations are

de1 = Au1 cos El de2 = —Au2 cos El —Au1 cos U

where U defines the initial position. Then, equating dW5 and dW0,

dWE = (1WD for arbitrary Au1

P = (F1 — Fjcos 0

The force in bar 3 does not appear explicitly in the equilibrium equation, (c). It is possible

Fig. E7—2

to include F3 even though bar 3 is rigid by treating it as a Lagrange multiplier.f Weconsider Au3 as independent in the work equation:

P Au1 — (F1 cos 0)Au1 + (F, cos 0)Au2 = 0 (d)

Now,Au1 — Au2 = 0 (e)

Multiplying the constraint relation by —2, adding the result to (d), and collecting terms, weobtain

Au1(P — F1 cos 0 — 2) + Au2(F2 cos 0 + 2) = 0 (f)

Finally, we require (f) to be satisfied for arbitrary Au1 and Au2. The equilibrium equations

See See, 3—3.

Bars 3,4, 5, 6 are rigid


SEC. 7—3. PRINCIPLE OF VIRTUAL FORCES 159

areP=F1cosO+AF2cos6+2=O

and we recognize 2 as the force in bar 3.

7—3. PRINCJPLE OF VIRTUAL FORCES

The principle of virtual forces is basically an alternate statement of geo-metrical compatibility. We develop it here by operating on the elongation—joint displacement relations. Later, in Chapter 10, we generalize the principle.for a three-dimensional solid and describe an alternate derivation.

We restrict this discussion to geometric linearity. The governing equations

—

e = =

Now, we visualize a set of bar forces AF, and joint loads, which satisfythe force-equilibrium equations:

=

A force system which satisfies the equations of static equilibrium is said to bestatically permissible. Equation (b) relates the actual elongations and joint dis-placements. If we multiply the equation for Ck by sum over the bars, andnote (c), we obtain the result

AFTe =

=

which is the definition of the principle of virtual forces:

The actual elongations and joint displacements satisfy the conditionAFTC — = 0 (7-10)

for any statically permissible system of bar forces and joint loads.

The principle of virtual forces is independent of material behavior but isrestricted to the geometrically linear case. The statically permissible system(AF, is called a virtual-force system.

To illustrate the application of this principle, we express cW and inpartitioned form,

where U2 contains the prescribed support movements. Using (a), (7—10) takesthe form:

AFTe — U2 = APr U1


==

APf U1 (l)ukj

Ukf = eTFJ;,k1_ U2TP2IPkJ1


If the elongations are known, we can determine the unknown displacementsby specializing AP1. To determine a particular displacement component, say

we generate a force system consisting of a unit value of PkJ and a set of barforces and reactions which equilibrate Pkj = 1.

(7—11)

The internal bar forces and reactions are obtain from an equilibrium analysisof a statically determinate structure. Since only one element of is finite,

and (b) reduces to(7—12)

The principle of virtual forces is also used to establish geometric compati-bility relations required in the force method which is discussed in Chapters9 and 17. We outline the approach here for completeness. One works withself-equilibrating virtual-force systems, i.e., statically permissible force systemswhich involve only bar forces and reactions.

By definition, a self-equilibrating force system F*, f)* satisfies

B1F* = P7 = 0(7—13)

P7 = B2F*For this case, (b) reduces to

—. = 0 (7—14)

Equation (7—14) represents a restriction on the elongations and is called ageometric compatibility equation.

Example 7—3

The truss shown (Fig. E7—3A) has support movements and is subjccted to aloading which results in elongations (e1, e7) in the diagonal bars. We are coniidering theoutside bars to be rigid.

Fig. E7—3A

u,P

Bars 3,4, 5, 6 are rigid


SEC. 7—3. PRINCIPLE OF VIRTUAL FORCES

To determine the translation, u, we select a statically determinate force system consistingof a unit force in the direction of u and a set of bar forces and reactions required to equilibratethe force. One possible choice is shown in Fig. E7.-3B. Evaluating (7—12) leads to

u = + — tan —cos 6

This truss is statically indeterminate to the first degree. A convenient choice of forceredundant is one of the diagonal bar forces, say F2. The equation which determines F2 is

Fig. E7—3B

derived from the gcometric compatibility relation, which, in turn, is obtained by takinga self-equilibrating force system consisting ofF2 = + I and a set of bar forces and reactionsrequired for equilibrium. The forces are shown in Fig. E7--3C.

Evaluating (7—14), we obtaine1 + e2 = 0

Fig. E7—3C

To show that (a) represents a geometrical compatibility requirement, we note that theelongation-displacement relations for the diagonal bars are

ucos6 e2 = —ucosO

Specifying e1 determines u and also e2. We could have arrived at Equation (a) startingfrom Equation (b) rather than (7—14). Flowever, (7—14) is more convenient since it doesnot involve any algebraic manipulation. We discuss this topic in depth later in Chapter 9.

0

— ens 6

0



7—4. STRAIN ENERGY; PRINCIPLE OF STATIONARY POTENTIALENERGY

In this section, we specialize the principle of virtual displacements for elasticbehavior and establish from it a variational principle for the joint displacements.

We start with the general form developed in Sec. 7—2,

FT de = gpT for arbitrary

If we consider all the elements of to be arbitrary, i.e., unrestrained,

de = A'W

and (a) leads to the complete set of force-equilibrium equations in unpartitionedform,

=

We can obtain the equation for P1 by rearranging (c) or by starting with thepartitioned form of

and noting that U2 is prescribed. The reduced form is

FT de — iW1 = 0 for arbitrary AU1 (7—15)where now

In what follows, we will work with (7—15).Our objective is to interpret (7—15) as the stationary requirement for a

function of U1. We consider F to be a function of e, where e = e(U1). The formof F = F(e) depends on the material behavior. We could express F in termsof U1 but it is more convenient to consider F as a compound function of e. Theessential step involves defining a function, V1- = VT(e), according to

FT dc = (7—16)

With this definition, and letting

= VT — = (7—17)

we can write (7—15) as

= 0 for arbitrary (7—18)

We call the total strain energy function and the total potential energy.One should note that VT exists only when F is a continuous single-valued functionof e. This requirement is satisfied when the material is elastic.

Equation (7—18) states that the joint force-equilibrium equations (P1 = B1F)expressed in terms of the unknown displacements are the Euler equations for the

t See Secs. 6-4, 6—5.


SEC. 7—4. STRAIN ENERGY 163

total potential energy. It follows that the actual displacements, i.e., the dis-placements which satisfy the equilibrium equations, correspond to a stationaryvalue of 11,'

It remains to discuss how one generates the strain-energy function. Bydefinition,

anddVT

=where is the strain energy for bar). Since we are considering to be a com-pound function of e1, Equation (b) is equivalent to

=(7—19)

That is, the strain energy function for a bar has the property that its derivativewith respect to the elongation is the bar force expressed in terms of the elonga-tion. Finally, we can express as

= (7—20)

where e0 is the initial elongation, i.e., the elongation not associated with theforce. Actually, the lower limit can be taken arbitrarily. This choice correspondsto taking as the area between the F-c curve and thee axis, as shown in Fig.7—5.

Fig. 7—5. Graphical representation of strain energy and complementary energy.

We consider the linearly elastic case. Using (6—30),

F1 — e0,

= — e0, )2Then

(7—21)

The total strain energy is obtained by summing over the bars. We can expressVTas

VT = 4(e — e0)Tk(e — eo)j=1

(7—22)


164 VARIATIONAL PRNCIPLES FOR AN IDEAL TRUSS CHAP. 7

Finally, we substitute for e in terms of U1, U2, using

e = A1U1 + A2U2 (7—23)

When the geometry is linear, A1, A2 are constant and is a quadratic function.If the geometry is nonlinear, is a fourth degree function of the displacements.

Up to this point, we have shown that the displacements defining an equilib-rium position correspond to a stationary value of the potential energy function.To determine the character (relative maximum, relative minimum, indifferent,neutral) of the stationary point, we must examine the behavior of the seconddifferential, in the neighborhood of the stationary point.

Operating on and noting that AP1 = 0 leads to

d2

(7—24)

= +

The next step involves expressing d2VT as a quadratic form in AU1. We restrictthis discussion to linear behavior (both physical and geometrical). The generalnonlinear case is discussed in Sec. 17.6 When the geometry is linear, we canoperate directly on (7—23) to generate the differentials of e,

de A1AU1

d2e = 0

since A1 is constant. When the material is linear,

dF=kdewhere k is a diagonal matrix containing the stiffness factors (AE/L) for the bars.Then, d2VT reduces to

d2VT = dFT dc = deTk de 7 25AUT(ATkA1)AU1 -

If de 0 for all nontrivial AU1, d2VT is positive definite and the stationarypoint is a relative minimum. This criterion is satisfied when the system isinitially stable, since de = 0 for AU1 0 would require that

A1 AU1 = 0 (m equations in unknowns)

have a nontrivial solution. But a nontrivial solution of (a) is possible only whenr(A1) < ne,. However, A1 = for the geometrically linear case and r(B1) =when the system is initially stable. Therefore, it follows that the displacementsdefining the equilibrium position for a stable linear system correspond to anabsolute minimum value of the potential energy.

Example 7—4

We establish the total potential energy function for the truss considered in Example 7—2.For convenience, we assume no initial elongation or support movement. The strain


SEC. 7—5. COMPLEMENTARY ENERGY 165

energy isVT = +

Substituting for the elongations in terms of the displacement,

e1 =u1cosO e2= —u2cosO= —u1cosOresults in

= + cos2 0and finally

= + cos2 0 — P1u1

The first differential of is

= {{(k1 + k2)cos2 Ojuj — P1}Au1

Requiring to be stationary leads to the Euler equation,

P1 = [(k1 + k2)cos2 0]u1

which is just the force-equilibrium equation

P1 = (F1 — F2)cos 0

with the bar forces expressed in terms of the displacement using

F1 = k1e1 = ku1 cos 0 F2 = k2e2 = —k2u1 cos 0

The second differential of is

= [(k1 + k,)cos2 0](Au1)2

and we see that the solution,P1

Ul= k1 +

corresponds to an absolute minimum value of H,, when 0 0. Thc truss is initially unstablewhen 0 0.

7—5. COMPLEMENTARY ENERGY; PRINCIPLE OFSTATIONARY COMPLEMENTARY ENERGY

The principle of virtual forces can be transformed to a variational principlefor the force redundants. We describe in this section how one effects the trans-formation and utilize the principle later in Chapter 9. This discussion isrestricted to linear geometry.

We start with Equations (7—13) and (7—14), which we list below for con-venience:

eTAF — = 0

where AF, represent a self-equilibrating forcesystem, i.e., they satisfy thethe following constraint relations:

B1 iW = 0B2 AF



Our objective is to establish a function of F, whose Euler equations are (a) and(b). We cannot work only with (a) since F is not arbitrary but is constrainedby the force-equilibrium equations,

P1 = B1F (fld equations in m variables)

We interpret as the first differential of a function

= eT = dV7 (7—26)

and call the complementary energy function for bar j. By definition,

= (7—27)

That is, the complementary energy function for a bar has the property that itsderivative with respect to the bar force is the elongation expressed in terms ofthe force. We express as

dF1 (7—28)

This definition corresponds to taking Vj' as the area bounded by the F-e curveand the F axis as shown in Fig. 7—5. Also, the strain and complementaryenergy functions are related by

+ = (7—29)

When the material is linear elastic,

= e0, +11 r 1.1r2j — j T 2Jj' 3 —

= + 4FTfF

Next, we define II. as:1-Ic =

7—31)= —

We call the total complementary energy function. With these definitions,Equations (a), (b), and (c) can be interpreted as

0

subject to the constraint condition

d(P1 — 81F) = 0

We can combine (e) and (f) into a single equation by introducing Lagrangemultipliers. Following the procedure described in Sec. 3—3, we add to (7—31)the joint force equilibrium equations and write the result as:

+ (P1 — (7—32)

where. ,

contains the Lagrange multipliers. The Euler equa-


SEC. 7—5. COMPLEMENTARY ENERGY 167

tions for treating F and as independent variables are

dfl, = 0 for AF, arbitrary

e(F) + B102 (7—33)

B1F = P1

We recognize the first equation in (7—33) as the member force-displacementrelation, and it follows that = U1.

An alternate approach involves first solving the force-equilibrium equation,(d). There arena equations in m variables. Since B1 is of rank n4 when the systemis initially stable, we can solve for na bar forces in terms of P1 and the remaining(rn — bar forces. One can also work with a combination of bar forces andreactions as force unknowns. We let

q = m — number of redundant forces (734)X = {X1, X2 Xq} = matrix of force redundants

and write the solution of the force-equilibrium equations as

F = F0 + (735)P2 P2,0 +

The force system corresponding to is self-equilibrating, i.e.,

= 0 for arbitrary X (7—36)

We substitute for F in (7—3 1) and transform to Then,

eT AF — UI AP2= (eTFx — UIP2,

and the Euler equations are

eTFx — UIP2 x 0 (737)

Note that (7—37) is just a reduced form of (7—33). Also, we could have obtainedthis result by substituting directly in (a).

Up to this point, we have shown that the force redundants which satisfythe geometric compatibility equations correspond to a stationary value of thetotal complementary energy. To investigate the character of the stationarypoint, we evaluate the second differential. Operating on (g),

= deTFx AX

d2 is positive definite with regard to AX, the stationary point is a relativeminimum. This requirement is satisfied for the linear elastic case. To showthis, we note that

de = fAF = WXAX=



Since f contains only positive elements, is positive definite with regard toAX provided that there does not exist a nontrivial solution of

AX = = 0

For (j) to have a nontrivial solution, there must be at least one relation betweenthe columns of But this would correspond to taking force redundants whichare not independent, and the solution scheme would degenerate. Therefore,we can state that the actual force redundants correspond to an absolute mini-mum value of for the linear elastic case.

Example 7—5

We consider the truss treated in Example 7—3. It is statically indeterminate to the firstdegree with respect to the bars (statically detcrminate with respect to the reactions) and wetake

X = F2

The force influence matrices defined by (7—35) follow from the force results listed on thesketches:

F0 = 0; 0; 0; 0; —tan 0; 0}

= {+1; +1; —cosU; —sin 0; —sin 0; —cos0}

P20 = P{—l; —tanG; +(an0}= 0

Assuming a bar is rigid is equivalent to setting f = 0 for the bar. Then, the comple-mentary energy is due only to the diagonal bars:

= yr +

= ± e02F2 + + f2Fi')

We convert to a function of X by substituting

F1 = ± Xcos 0

F2 = +XFinally, has the form

= e0,1 + + tan 0 — tan 0)P +

+ (eoi + e0,2 + f1 + 4(f1 + f2)X2

Differentiating (e) leads to

dIlC = {[eoi + + (fl

d211, = (f1 + (g)


SEC. 7—6. STABILITY CRITERIA

The Euler equation follows from (f):

e01 + C02 +f1 (f1 + f2)X = 0

Comparing (h) with (a) of Example 7—3, we see that the Euler equation for is thegeometric compatibility equation expressed in terms of the force redundant.

7—6. STABILITY CRITERIA

Section 6—9 dealt with initial stability, i.e., stability of a system under in-finitesimal load. We showed there that initial stability is related to rigid bodymotion. A system is said to be initially unstable when the displacement restraintsare insufficient to prevent rigid body motion. In this section, we develop criteriafor stability of a system under finite loading. If a linear system is initiallystable, it is also stable under a finite loading. However, a nonlinear (eitherphysical or geometrical) system can become unstable under a finite load.

We consider first a single mass particle subjected to a system of forces whichare in equilibrium. Let Il be the displacement vector defining the equilibriumposition. We introduce a differential displacement All, and let AW he thework done by the forces during the displacement All. if A W > 0, the particleenergy is increased and motion would ensue. It follows that the equilibriumposition (ll) is stable only when AW < 0 for arbitrary All.

We consider next a system of particles interconnected by internal restraints.Let AWE be the incremental work done by the external forces and AW1 theincremental work done by the internal restraint forces acting on the particles.The total work, AW, is given by

= AWE + AW1

The system is stable when A W < (I for all arbitrary permissible displacementincrements, that is, for arbitrary increments of the variable displacements. Now,we let AW11 be the work done by the internal restraint forces acting on therestraints. Since —AW1, we can express the stability requirement as

(7-38)

One can interpret AW0 as the work required to deform the system to the alternateposition and as the actual work done on the system.

When the behavior is continuous, we can express and AWE as Taylorseries expansions in terms of the displacement increments (see (7—2)):

+ += dW0 + +

We have shown that the first-order work is zero at an equilibrium position:

dWD — dWE 0



If we retain only the first two terms in (b), the general stability condition re-duces to

d2 WD — d2 W5 > 0 for all arbitrary permissible displacement increments

(7—39)

Equation (7—39) is called the "classical stability criterion." Retaining onlythe first two differentials corresponds to considering only infinitesimal displace-ment increments. If (7—39) is satisfied, the equilibrium position is stable withrespect to an infinitesimal disturbance. In order to determine whether it isstable with respect to a finite disturbance, one must use (7—38).

Ifd2WD = d2WE (7—40)

for a particular set of displacement increments, the equilibrium position is saidto be neutral, and there exists an alternate position infinitesimallydisplaced from the first position. One can interpret (7—40) as the necessarycondition for a bifurcation of equilibrium positions.

To show this, suppose U and U represent the displacement componentsfor the two possible equilibrium positions of a system where

Also, let R and represent the resultant forces corresponding to U and 0.We can express as

= R + dR + +

Now, the second-order work for the initial equilibrium position is given by

d2W d2W5 — d2WD = txUT dR

If d2 W = 0 for some finite it follows that

dR = R0AU = 0

The condition= 0

is equivalent to (7—40). Finally, if we consider to he infinitesimal,

R=R+dRand (7—40) implies R = 0.

To apply the classical stability criterion to an ideal truss, we note that thefirst-order work terms have the form

dW5 = P1 AU1

dWD =

where U2, P1 are prescribed. Operating on (a) yields

d2W5 0 (7-41)d2WD >jFjd2ej +


SEC. 7—6. STABILITY CRITERIA

and the stability criterion reduces tostable d2WD > 0 for arbitrary nontrivial AU1neutral d2 WD = 0 for a particular nontrivial AU1 (7—42)

unstable d2 WD < 0 for a particular nontrivial AU1

where d2WD is a quadratic form in AU1. We postpone discussing how onetransforms (7—41) to a quadratic form in AU1 until the next chapter.

When the material is elastic, we can identify (7—39) as the requirement thatFir, be a relative minimum. By definition,

drIp = dVT — dWE

For elastic behavior,dVT = dWD

and it follows thatd2WD — d2WE = (7—43)

Finally, we can state:An equilibrium position for an elastic system is stable (neutral, un-

stable) if it corresponds to a relative minimum (neutral, indifferent)stationary point of the total potential energy.

Example 7—6

The system shown in Fig. E7—6A consists of a rigid bar restrained by a linear elasticspring which can translate freely in the .x2 direction. Points A and A' denote the initial anddeformed positions. We will first employ the principle of virtual displacements to establishthe equilibrium relations and then investigate the stability of the system.

Fig. E7—6A

j2

x1

The first-order work terms are

(a)dWD = F de

= P2 du2


172 VARIATIONAL PRINCIPLES FOR AN iDEAL TRUSS CHAP. 7

where F, e are the spring force and extension. Since the bar is rigid, the system has only onedegree of freedom, i.e., only one displacement measure is required to define the configura-tion. It is convenient to take 0 as the displacement measure. The deformation-displacementrelations follow from the sketch:

e = u1 = L(sin 0 — sin

0)

0 — sin

andde = (cos 0)L

du, = (sin 0)L

Using (a) and (d), the principle of virtual displacements takes the form

dW0 — dWa = {F cos 0 — P2 sin 0) (L AO) = 0 for arbitrary AG

Finally, (e) leads to the equilibrium relation,

F cos 0 = P2 sin 0

which is just the moment equilibrium condition with respect to point 0. We transform(1) to an equation for ()by substituting for F using (c). The result is

sin 0—

tan 0 = sin 00

Since the system is elastic,— dW5

and (e) is equivalent to= 0 for arbitrary AU

The potential energy function for this system has the form

= — P2u2

= 4kL2(sin 0 — sin 0)

and (g) can be interpreted as

=

0 of 00 are plotted in Fig. E7—6B. The resultfor 0 consists of two curves, defined by

0 = 0 for arbitrary P2/kLcos 0 = P2/kl for (P2/kL) 1

To investigate the stability of an equilibrium position, we have to evaluate the second-order work at the position. After some algebraic manipulation, we obtain

= d2W0 — = k(L AU)2[cossU —P2/kL]

cos 0

Let 0* represent a solution of(g). Applying (m) to 0* results in the following classification:


stable

neutral

unstable

One can show that (n) is equivalent to

REFERENCES

P2COS 0* >

cos3

3P2

cos 0'

dP20

dP2o

dO

dP2

stable

neutral

unstable

A transition from stable to unstable equilibrium occurs at point A, the peak of thedeflection curve. The solution for 0 is different in that its stable segment is the linearsolution and the neutral equilibrium point (P2 kL) corresponds to a branch point,Both the linear and nonlinear branches are unstable.

Fig. E7—6B

0

REFERENCES

1 WANG, C. T.: Applied Elasticity, McGraw-Hill, New York, 1953.2. LANGHAAR, H. L,: Energy Methods in Applied Mechanics, Wiley, New York, 1962.3. REISSNeR, E.: "On a Variational Theorem in Elasticity," J. Math. Phys., Vol. 29,

pages 90-95, 1950.4. ARGYRIS, i. H., and S. KIsLseY: Energy Theorems and Structural Analysis, Butter-

worths, London, 1960.5. CI.IARLTON, T. M.: Energy Principles in Applied Statics, Blackie, London, 1959.6. HOFF, N. J.: The Anal vsis of Structures, Wiley, & New York, 1956.7. K.: Variational Methods in Elasticity and Plasticity, Pergamon Press, 1968.



PROBLEMS

7—1. Consider the two-dimensional symmetrical truss shown. Assume= 03 = 0.(a) Determine the first two differentials of e1 and ez by operating on the

expanded expression (equation 6—19) for e.(b) When a b, we can neglect the nonlinear term involving u12 in the

expressions for e and Specialize (a) for this case.(c) When a b, we can neglect the nonlinear term involving u11 in the

expressions for e and Specialize (a) for this case.

T21

Prob. 7—1

7—2. Refer to the figure of Prob. 7—1. Assume = u3 = 0 and a> b.Using the principle of virtual displacements, determine the scalar force-equili-brium equations for joint 1.

7—3. Suppose a force F is expressed in terms of e,

F = C1e + 4C2e3

where a is related to the independent variable u by

a u + 1u2

(a) Determine the first two differentials of the work function, W = W(u),defined by

W = F de

(b) Suppose (a) applies for increasing e and

F = C1(e —

e decreasing from e*. Determine d2 W at a = e*.

7—4. Refer to Prob. 6—23. The n — 1 independent node equations relatingthe branch currents are represented by

ATI U

Now, the branch potential differences, e, are related to the n — 1 independentnode potentials, V, by

e = AV

x2

3


PROBLEMS 175

Deduce that the requirement,

1T de = 0 for arbitrary

is equivalent to (a). Compare this principle with the principle of virtual dis-placements for an ideal truss.

7—5. Consider the two-dimensional truss shown. Assume u2 = = 0.(a) Using (7—14), obtain a relation between the elongations and ü32. Take

the virtual-force system as LxF2 and the necessary bar forces and re-actions required to equilibrate AF2.

(b) Using (7—12), express u11, u12 in terms of e1, e3. Note that bar 2 isnot needed. One should always work with a statically determinatesystem when applying (7—12).

Prob. 7—5

- 7—6. Refer to Prob. 6—23. One can develop a variational principle similarto the principle of virtual forces by operating on the branch potential differ-ence—node potential relations. Show that

AiTe=0

for any permissible set of current increments. Note that the currents must satisfythe node equations

ATi 0

Deduce Kirchhoff's law (the sum of the voltage drops around a closed loopmust equal zero) by suitably specializing Lsi in (a). Illustrate for the circuitshown in Prob. 6—6, using branches 1, 2, 4, and 6.

7—7. By definition, the first differential of the strain-energy function due toan increment in U1 has the form

= dV. = F,, de,,n1

We work with expressed as a compound function of e = e(U) since it ismore convenient than expressing V directly in terms of U1. One can also

x2

2


176 VARIATIONAL PRINCIPLES FOR AN IDEAL TRUSS CHAP 7

write as

(a) Using (b), show that the system of if joint force-equilibrium equationsexpressed in terms of the joint displacements can be written as:

ÔU(k k =

Equation c is called Castigliano's principle, part I.(b) Show that an alternate form of (c) is

P(k=

Note that (d) is just the expansion of (c). Rework P rob. 7—2, using (d).7—8. Determine V(e), dv, and d2V for the case where the stress-strain

relation has the form (see Prob. 6—10)

a = — be3)

7—9. Determine V*(F), dV*, and d2V* for the case where the stress-strainrelation has the form

= (a + ca3)

7—10. Show that (7—12) can be written as

UkJ =0Pkj

where = is defined by (7—31). This result specialized for U2 = 0 iscalled Castigliano's principle, part H. Apply it to Prob. 7—5, part b. Assumelinear elastic material and f1 = = = f.

7—11. The current and potential drop for a linear resistance are related by

ef e0,j +

Inverting (a), we can express as a function of e1.

= —

(a) Suppose we define a function, which has the property that

=

Determine corresponding to (b).b

(b) Let W where h = total number of branches. Considering

the branch potential drops to be functions of the node potentials,deduce that the actual node potentials V correspond to a stationaryvalue of W. Use the results of Prob. 7—4. The Euler equations for


PROBLEMS 177

W = W(V) are the node current equilibrium equations expressed interms of the node potentials.

(c) Suppose we define a function which has the property that

= (d)

Determine corresponding to (a).b

(d) Let W* = W7. Show that the Euler equations for

H = iTe — = 1T(AV) — = H(i,V) (e)

are the governing equations for a d-c network.(e) Show that the actual currents correspond to a stationary value of

One can either introduce the constraint condition, An = 0, in (e) oruse the result of Prob. 7—6.

7—12. Investigate the stability of the system shown below. Take k, = aL2k5

PProb. 7—12

Lineartranslationalrestraint

Rigid rod

ICr (Linear rotational restraint)

and consider a to range from 0 to 6.


8

Displacement MethodIdeal Truss

8—1. GENERAL

The basic equations defining the behavior of an ideal truss consist of force-equilibrium equations and force-displacement relations. One can reduce thesystem to a set of equations involving only the unknown joint displacementsby substituting the force-displacement relations into the force-equilibriumequations. This particular method of solution is called the displacement or

method. Alternatively, one can, by eliminating the displacements,reduce the governing equations to a set of equations involving certain barforces. The latter procedure is referred to as the ,fin'ce or flexibility method.We emphasize that these two methods are just alternate procedures for solvingthe same basic equations. The displacement method is easier to automatethan the force method and has a wider range of application, However, it isa computer-based method, i.e., it is not suited for hand computation. Incontrast, the force method is more suited to hand computation than to machinecomputation.

In what follows, we first develop the equations for the displacement methodby operating on the governing equations expressed in partitioned form. Wethen describe a procedure for assembling the necessary system matrices usingonly the connectivity table. This procedure follows naturally if one first operateson the unpartitioned equations and then introduces the displacement restraints.The remaining portion of the chapter is devoted to the treatment of nonlinearbehavior. We outline an incremental analysis procedure, apply the classicalstability criterion, and finally, discuss linearized stability analysis.

8—2. OPERATION ON THE PARTITIONED EQUATIONS

The governing partitioned equations for an ideal truss are developed inSec. 6—7. For convenience, we summarize these equations below.

178


SEC. 8—2. OPERATION ON THE PARTITIONED EQUATIONS 179

P1 = B1F eqs.)

P2 = B2F (r eqs.)

F = F, + kA1U1 (in eqs.)

F, = k(—e0 + A2tJ2)

The unknowns are the in bar forces (F), the r reactions (P2), and the na jointdisplacements (U1). One can consider F, to represent the initial bar forces,that is, the bar forces due to the initial elongations and support movementswith U1 0. The term kA1U1 represents the bar forces due to U1. When thematerial is linear elastic, k and e0 are constant. Also, = BT when the geometryis linear.

We obtain a set of equations relating the flj displacement unknowns, U1,by substituting for F in (a). The resulting matrix equation has the form

(B1kA1)U1 = — B1F1

We solve (8—i) for U1, determine F from (e), and P2 from (b). The coefficientmatrix for U1 is called the system stiffness matrix and written as

K11 = B1kA1 (8—2)

One can interpret as representing the initial joint forces due to the initialelongations and support movements with U1 = 0. Then — B1F1 representsthe net unbalanced joint forces.

When the geometry is linear, K1 1 reduces to

K11 = B1kBT = AfkA1

If the material is linear, k is constant and positive definite for real materials.Then, the stiffness matrix for the linear case is posiLive definite when the systemis initially stable, that is, when r(B1) Conversely, if it is not positivedefinite, the system is initially unstable.

If the material is nonlinear, k and e0 depend on e. We have employed apiecewise linear representation for the force-elongation curve which results inlinear relations. However, one has to iterate when the limiting elongationfor a segment is exceeded.

The geometrically nonlinear case is more difficult since both A and B dependon U1. One can iterate on (8—1), but this requires solving a nonsymmetricalsystem of equations. It is more efficient to transform (8—1) to a symmetricalsystem by transferring some nonlinear terms to the right-hand side. Nonlinearanalysis procedures are treated in Sec. 8—4.

Even when the behavior is completely linear, the procedure outlined abovefor generating the system matrices is not efficient for a large structure, since

f See Prob. 2—14.


180 DISPLACEMENT METHOD: IDEAL TRUSS CHAP. 8

it requires the multiplication of large sparse matrices. For example, one obtainsthe system stiffness matrix by evaluating the triple matrix product,

=AfkA1

One can take account of symmetry and the fact that k is diagonal, but A1 isgenerally quite sparse. Therefore, what is needed is a method of generating Kwhich does not involve multiplication of large sparse matrices, A methodwhich has proven to be extremely efficient is described in the next section.

8—3. THE DIRECT STIFFNESS METHOD

We start with (6—37), the force-displacement relation for bar ii:

= F0, +=

where n4, n. denote the joints at the positive and negative ends of barn. Onecan consider F0, as the bar force due to the initial elongation with the endsfixed (un, = 0). Now, we let be the external joint force matricesrequired to equilibrate the action of Noting (6—43), we see that

=(8—4)

p,I_ =

Substituting for (8—4) expands to

= + —

pn_ =

One can interpret (b) as end action—joint displacement relations since theelements of ± are the components of the bar force with respect to thebasic frame.

Continuing, we let(8—5)

Note that is of order i x i where I = 2 or 3 for a two or three-dimensionaltruss, respectively. When the geometry is linear, = y,, and is sym-metrical. With this notation, (b) takes a more compact form,

= + —(8—6)

= — +

We refer to as the bar stiffness matrix. Equation (8—6) defines the jointforces required for bar n. The total joint forces required are obtained bysumming over the bars.


+k, in row column— k,, in row column—ku in row n_, column

n

The connectivity table and general form of if and for the numbering shown inFig. E8—l are presented below:

Fig. E8—l

SEC. 8—3. THE DIRECT STIFFNESS METHOD

We have defined

— p2, . . , (U X 1)

= {u1, u2 (ii x 1)

as the general external joint force and joint displacement matrices. Now, wewrite the complete system of if joint force-equilibrium equations, expressed interms of the displacements, as

= + (8—7)

We refer to if, which is of order if x as the unrestrained system stiffnessmatrix. The elements of are the required joint forces due to the initialelongations and represents the required joint forces due to the jointdisplacements.

We assemble if and in partitioned form, working with successive mem-bers. The contributions for member n follow directly from (8—6).

(Partitioned Form is j x 1)

in row8 8

in row n. —

if (Partitioned Form is j x j)

Example 8—1

0

4 3



Bar 1 2 3 4 5

+joint 1 2 2 4 2

—joint 4 1 3 3 4

U1 U2 U3 U4

P2

k1+k2 —k2 —k1

—k2 k2 + k3 + k5 —k3

--k3 k3 + k4 —k4

—k1 —k5 —k4 k1 + k4 + k0

Example 8—2

L' UT ftTPai 0.1PI — 0,2P2L' pT L' nT r øT

P0.2 0,212 r 0,3P3 0,SPS,10 = UT t' UT

Po, 3 — 0. 3P3 — 0, 4P4uT L'

Po, 4 — 0, 1I'I 1 0, 41'4 0. 5P5

The external force matrix, involves and the displacement matrices for those jointsconnected to joint j by bars. Now, corresponds to row j and ii,, to column j of ir. Bysuitably numbering the joints, one can restrict the finite elements of X' to a zone aboutthe diagonal. This is quite desirable from a computational point of view.

Sect. 1

0

Fig. E8—2

fs

ft® -.(-I2:

/-..- —--

(71

©3\%._ —--L6 ;

',. /

Consider the structure shown. We group the vertical joints into sections. The equi-librium equations for section k involve only the joints in section k and the adjacent sections.For example, the equations for section 3 (which correspond to P6) will involve onlythe displacement matrices for sections 2, 3, 4. This suggests that we number the joints bysection. The unpartitioned stiffness matrix corresponding to the above numbering scheme


SEC. 8—3. THE DIRECT STIFFNESS METHOD

is listed below. Note that has the form of a quasi-tridiagonal band matrix when it ispartitioned according to sections rather than individual joints. The submatrices for thistruss are of order 4 x 4,

The introduction of displacement restraints involves first transforming thepartitioned elements and to local frames associated with the restraints,permuting the actual rows, and finally partitioning the actual rows. The stepsare indicated below.

-+ U-+

We write the system of joint force-equilibrium equations referred to the localjoint frames as

= + (8—10)

The transformation la'vs for the submatrices of and follow from (6—57).

== T

€,n= 1,2,...,j(8—11)

U' U2 U3 U4 U6 U7 U8

k, +k2 —k, I—k2

P2 —k,

—k2

k,+k3 —k3+k4

—k3 k2+k3

+k6

—k4

—k5 —k6

I

—k7

—k4 —k,

—k6

k4+k51

k6+k9+k,0

—k8

—k9 —k10

p6 —k7 —k8 —k9

I

k8+k7 —k1, —k,2

+k9+k,2

—k,, k10+k,, —k,,3

JJ7 —k,0

p8 —k,, —k,3 k,2+k,,



The step, —+ P, involves only a rearrangement of the rows of Weobtain the corresponding stiffness matrix, K, by performing the same operationson both the rows and columns of The rearranged system of equations iswritten as

P = KU + P0 (8-12)

Finally, we express (8—12) in partitioned form:

= K11U1 + K1202 + P0,18 13

P2 = + K2202 + P0,2

The first equation in (8—13) is identical to (8—1).

Example 8—3

It is of interest to express the partitioned elements of K in terms of the geometrical,connectivity, and displacement transformation We start with the general Un-partitioned equations(6—28), (6—40), and (6—44), (6—50):

F F0 + kda/1 = F0 + kyC'W

Then, substituting for F in (a) and equating the result to (8—7) leads to

==

The matrix, DTkY, is a quasi-diagonal matrix of order im. The diagonal submatrices arc oforder i, and the submatrix at location n has the form, We have defined this productas k,,. Then, if we let

k5

= [ki

we can express as

Carrying out (8—9) for n = 1, 2 m is the same as evaluating the triple matrix product.Obviously, (8—9) is more efficient than (f).

The introduction of displacement restraints can be represented as

P =11

= D1dP(g)

=and

= DTU = DfU1 + (h)

I. r,T2p272

= CTk5c


SEC. 8—3. THE DIRECT STIFFNESS METHOD 185

Substituting (g) and (h) in (8—7) and equating the result to (8—13), we obtain

K,, = =t — 1 2

P0. = DsCTDke0 —

In order to obtain (8—13), we must rearrange the rows and columns ofand then partition. This operation is quite time-consuming. Also, it leads torectangular submatrices. In what follows, we describe a procedure for intro-ducing displacement restraints which avoids these difficulties.

We start with the complete system of equations referred to the basic frame,

(8—14)

We assemble and using (8—8) and (8—9). Then, we add to theexternal force matrices for those joints which are unrestrained. It remains tomodify the rows and columns corresponding to joints which are either fully orpartially restrained.

Case A: Fit!! Restraint

Suppose uq = Uq. Then is unknown. We replace the equation for Pq by

= Uq

This involves the following operations on the submatrices of X and

1. On X. Set off diagonal matrix elements in row q and column q equalto 0 and the diagonal matrix element equal to I,.

I = 0

(8—15)

= Ii

2. On Add terms in due to

t CX(qUq

(8—16)

j

ease B: Partial Restraint—Local Frame

We suppose the rth element in is prescribed.

= = prescribed= unknown



We have to delete the equation corresponding to and replace it with

4 =

Step I —Assemblage of Basic Matrices

We assemble Eq, Gq, ui', according to the following:

1. Eq and Gq. We start with

G=O,and we set

G,r +1

2. us'. We start with an ith-order column vector having zero elements andwe set the element in the rth row equal to Note that this matrix involvesonly the prescribed displacements (local frame) in their natural locations.

3. We start with an ith-order column vector having zero elements andwe insert the values of the prescribed joint forces (local frame) in their naturallocations. Note that the elements corresponding to the reactions are zero.

When the joint is fully restrained,

E=O, G=11

Suppose joint 5 is partially restrained, The data consist of:

(a) The rotation matrix, R°5, defining the direction of the local frameat 5 with respect to the basic frame.

(b) The direction (or directions) of the displacement restraint and thevalue (or values) of the prescribed displacement.

direction r,

(c) The values of the prescribed joint forces:

j=1,...,iAs an illustration, suppose r = 2. Then, in (b), we read in

In (c), we read in—5 —5Psi P53

The four basic matrices are (for r = 2)

1 0 0 [0 0 0

E5= 0 0 0 Gs=IO 1 0

0 0 1 [0 0 0


SEC. 8—3. THE DIRECT STIFFNESS METHOD 187

1°

In —5

Step 2—Operation on Jr and

1. Premultiply row q of it" and by

=e — 1 2

— ' ' .

" N, q '-'q 1% " N, q

2. Postmultiply column q of it" by — T11* and add to PPN.

=

3. Postmultiply column q of it" by (Eq

irtq = 1, 2, . . .

4. Add Gq to irqq= it"qq + Gq

5. Add and to

= P'N,q + U +

The operation on row q and column q are summarized below.

On Jr

= .YV'eq(E9R0")T— 1 2

X'qq = (Eq R°").Y(qq(Eq + —' I

Oiz

2PN, — R0q,

q 1, 2,. . . ,j (8—18)

I?'N,q = + + 0

When ir is symmetrical (this will be the case when the system is geometricallylinear), we can work only with the submatrices on and above the diagonal. Thecontracted operations for, the symmetrical case are threefold:

= —

it"eq(EqR°")T (8—19)

€= 1,2,...,q.— 1



q q — Ttlq* + +(8—20)it'qq = + Gq

— *—

= (8—21)

The operations outlined above are carried out for each restrained joint.Note that the modifications for joint q involve only row q and column q. Wedenote the mOdified system of equations by

= (8—22)

Equation (8—22) represents if equations. The coefficient matrix will benonsingular when K1 1 is nonsingular. To show this, we start with the firstequation in (8—13) and an additional set of r dummy equations:

[K11 Olfuil —f—P0,1 —

—-fJ - +- N

Equation (a) represents 1/equations. This system is transformed to (8—22) whenwe permute U, to d?tJ, They are related by (sec (6—63))

U [1°?, J

= rVp

where H is a permutation matrix. It follows that

= HT[K11

and, since H is an orthogonal matrix,

= IK11I (8—23)

It is more convenient to work with (8—22) rather than (a) since the solutionof (8—22) yields the joint displacement matrices listed in their natural order,that is, according to increasing joint number. Once ciii' is known, we convertthe joint displacement matrices to the basic frame, using

uq =

The bar forces are determined from

F,, = F0,,, + —

Next, we calculate F,, and assemble in partitioned form by summing the


SEC. 8—3. THE DiRECT STIFFNESS METHOD 189

contribution for each member. For member n, we put (see (8—4))

+ FOIIPf

—

Once is known, we convert the force matrix for each partially restrained jointto the local joint reference frame, using

=

The final result is required to equilibrate the bar forces. This operationprovides a static check on the solution in addition to furnishing the reactions.

When the problem is geometrically nonlinear, y,, and depend on thejoint displacements. In this case, it is generally more efficient to apply anincremental formulation rather than iterate on (8—22).

Example 8—4

We illustrate these operations for the truss shown in Fig. E8—4.

/Fig. E8—4

/50

Ii

Bar(n) 1 2 3 4 5 6 7 8 9 10 11

+joint(n+) 1 3 1 3 4 3 5 3 5 6 5

—joint(n...) 2 1 4 2 2 4 3 6 4 4 6

2. Assemblage of

We consider the geometry to be linear. Then, = and = Applying(8—9) results in listed below.

in row n+in row n_

1. Member-Joint Connectivity Table


190 DISPLACEMENT METHOD: DEAL TRUSS CHAP. 8

Note that i( is symmetrical and quasi-tridiagonal. with submatrices of order 4 x 4.

3. Introduction of Joint Displacement Restraints

The original equations are= =

where contains the external joint forces. We start with i?PN — If joint q is un-restrained, we put in row q If joint q is fully restrained, we modify andaccording to (8—15) and (8—16). Finally, if joint q is partially restrained, we use (8—19)through (8—21). Since is symmetrical, we have to list only the submatrices on and abovethe diagonal. It is convenient to work with successive joint numbers. For this system,joint 2 is fully restrained and joints 4, 6 are partially restrained. The basic matrices forjoints 4,6 and the initial and final forms of.Yt', are listed below. Note that this proceduredoes not destroy the banding of the stiffness matrix.

Joint 4 (u42 is prescribed)

R°4 2

ri 01 [0 0E4=[oj

G4=[0

= {O,ii42j =

Joint 6 (t42 is prescribed)

1 [ 1 iiii

ri 01E6=[0oJ

——

N 1 2 3 4

k1+k2+k3 —k1 —k2 —k3

3

51 6

—k2 ---k4 k2+k4+k6+k7+k8

—k6 —k7

—2 —k1 k1+k4+k5 —k4 —k5

0

4 —k3 —k5 —k6 k3+k5±k6+k9+k10

—k9 —k10

5

—6

0

—k7

—k8

—k9

.

—k10

+k7+k9+k11

—k11

—k11

+k8+k10+k11

-— k3

=


SEC. 8—4. INCREMENTAL FORMULATION

(ui) (u,) (U3) (U4)

o ir,4E4

Initial matrices (ir and =

Final matrices (ir* and

8—4. INCREMENTAL FORMULATION; CLASSICAL STABILITYCRITERION

Equations (8—13), (8—22) are valid for both linear and nonlinear behavior.However, it is more efficient with respect to computational effort to employan incremental formulation when the system is nonlinear. With an incrementalformulation, one applies the load in increments and determines the corre-sponding displacements. The total displacement is obtained bysumming the displacement increments. An incremental loading procedure canalso be used with (8—13) but, in this case, one is working with total displacementrather than with incremental displacement. In this section, we develop a set ofequations relating the external load and the resulting incrementaldisplacements. These equations are also nonlinear, but if one works with smallload increments, the equations can be linearized. Our approach will besimilar to that followed previously. We first establish incremental memberforce-displacement relations and then apply the direct stiffness method to

("1) (Us) (U4)

ir22

)r33 ir34.X44 ir4,5

Sym

(Us)

1,0 0U

.K13 .K34E4 J35

+ G4

E4ir46(E6R°6)T — —

+ + --

ir56(E6R°6)T —

, (E6R°6)iq66(E6R°6)T

+ G6

E6R°6(—


192 DISPLACEMENT METHOD: IDEAL TRUSS CHAP. S

generate the incremental system equations. We complete the section with adiscussion of the classical stability criterion.

We start with (8—4), which defines the external joint forces required toequilibrate the action of the force for bar n,

Pn* = p,, =

Equations (a) are satisfied at an equilibrium position. We suppose an in-cremental external load AP is applied and define AU as the resulting incrementaldisplacement for the new equilibrium position. Since F and depend on U,their values will change. Letting AF, AD be the total increments in F, D due toAU, and requiring (a) to be satisfied at both positions, leads to the followingincremental force-equilibrium equations:

= Afif + +(8—24)

Ap,,.. =

To proceed further, we need to evaluate the increments in e and D. The exactrelations are given by (6—22):

= — u,,_) + — —

— ci,, = —

To allow for the possibility of retaining only certain nonlinear terms, we write(a) as

fi,, — = (u,,÷ — u)Tg,= — u,,..) + — u,,) (8—25)

= Yh(ufl. — u,,_)

If all the nonlinear terms are retained,

g,, =

To neglect a particular displacement component, we delete the correspondingelement in For geometrically linear behavior, = 0. Operating on (8—25),we obtain

dv,, = — Au,,)Tg,, (8—26)

andAe,, = +

= — Au,,) (827)d2e,, —

It remains to evaluate AF,,. We allow for a piecewise linear material andemploy the relationst developed in Sec. 6—4. For convenience, we drop all the

t See (6—31), (6—32), and (6—33).


SEC. 8—4. INCREMENTAL FORMULATION 193

notation pertaining to a segment and write the "generalized" incrementalexpression in the simple form

= k(Ae — (8—28)

where k, are constant for a segment. They have to be changed if the limitof the segment is exceeded or the bar is unloading. Since is unknown, onehas to iterate, taking the values of k, Ae0 corresponding to the initial equilibriumposition as the first estimate. This is equivalent to using the tangent stiffness.The initial elongation, Ae0, is included to allow for an incremental temperaturechange. Substituting for (8—28) takes the form

At' _AC' I if 72— Q + + Ct

(8—29)

Finally, we substitute for Lw,, in (8—24) and group the terms as follows:

LXPn+ = — &i,..) + + (8—30)

where= Fag,, += (8—31)

= -i- + +

We interpret k7 as the tangent stiffness matrix. The vector, L\p9, contains linear,quadratic, and cubic terms in We have included the subscript g to indicatethat it is a nonlinear geometric term.

We write the total set of incremental joint equilibrium equations as

= + M'0 + (8-32)

where is assembled using (8—9) and MPO + with (8—8). Note thatis symmetrical. Finally, we introduce the displacement restraints by ap-

plying (8—19)—(8—21). The modified equations are

= — — (8-33)

It is convenient to include the prescribed incremental support displacementterms in so that involves only the incremental temperature and

the variable displacement increments. The contracted equations are

K1,11 AU1 = — AP0,1 — AP9,1 — 1(1,12 AU2 (8-34)

where K1, is symmetrical.We cannot solve (8—33) directly for since contains quadratic and

cubic terms in MI. There are a number of techniques for solving nonlinearalgebraic equations. t We describe here the method of successive substitutions,

t See Ref. 12.



which is the easiest to implement, but its convergence rate is slower in com-parison to most of the other methods.

First, we note that and are independent of A'1/1. They de-pend only on the initial equilibrium position and the incremental loading. Wecombine and and write (8—33) as

X7K MI1 = — (8-35)

Now, we let represent the nth estimate for LXa/IJ and determine the(n + 1)th estimate by solving

)p* L\cW(n+ = — &?P (8—36)

The iteration involves only evaluation of and back-substitution onceis transformed to a triangular matrix. The factor method is particularly con-venient since X7 is symmetrical. With this method,

STS (8—37)

where S is an upper triangular matrix. We replace (8—36) with

S = (8—38)STQ = A9* —

In linearized incremental analysis, we delete in (8—35) and take thesolution of

. (8-39)

as the "actual" displacement increment. One can interpret this scheme as onecycle successive substitution.

The solution degenerates when the tangent stiffness matrix becomes singular.To investigate the behavior in the neighborhood of this point, we apply theclassical stability criterion developed in Sec. 7—6. The appropriate form for atruss is given by (7—41):

± den)> 0 for arbitrary AU1 with AU2 = 0 (a)

We have already evaluated the above terms. Using (8—26), (8—27), and (8—29)with Ae0 = 0,

+ — — Au,,) (b)

and (a) can be written as

d2WD ALT ICE, AU1 > 0 for arbitrary AU1 (8—40)

It follows that must be positive definite for a stable equilibrium position.

¶ Iterative techniques are discussed in greater detail in Secs. 18—7, 18—8, 18—9.


SEC. 8—4. iNCREMENTAL FORMULATION

But K1, and are related by

HT[Kt.11(8—41)

where H is a nonsingular permutation matrix, which rearranges the elementsof according to

= H (8-42)

Then, and K1 have the same definiteness Finally, we canclassify the stability of an equilibrium position in terms of the determinant ofthe tangent stiffness matrix:

D = = 1K1, iiiD>OD = 0 (8—43)

We illustrate the application of both the total (8—13) and incremental (8—34) formulationsto the truss shown in Fig. E8—5A. To simplify the analysis, we suppose the material islinearly elastic, k1 = k2 = k, and there are no initial elongations or support movement.

x2

The initial direction cosines for the bars are

—b] —b]

The deformed geometric measures are defined by (8—25). They reduce to

+== cc, +

fl = 1,2

Fig. E8—5A

for this example. Since b cc d, we can neglect the nonlinear terms due to u11, i.e., we cantake

t See Prob. 2—13 for a proof.See Sec. 2—5.

1 [0 0

1

stableneutralunstable D < 0

Example 8—5

b <<d

_____

d d_____



Using (c),

1I

Pi —b + u12]

12 —b +

112 —b +

Continuing, the bar force-displacement relations are

= = ky,,u1 n = 1, 2

Finally, the force-equilibrium equation for joint I follows by applying (8—6) to both bars.

= (k1 + k2)u1 = k(11r71 ± 11212)U1

Equations (e) and (1) expand to

Jusi

0 — u12)(b

and

F1 = — (h —

F2 = — (b —

The diagonal form of the coefficient matrix is due to the fact that we neglected u11 in theexpressions for y and This approximation uncouples the equations. Note that (g) isthe first equation in (8—13) with U2 and P0 set to 0.

Solving the first equationt in (g). we obtain

I (J\2Pti

The corresponding bar forces are

=

Pi

This result is actually the solution for the linear geometric case.The expression for and the corresponding bar forces follow from the second equation

[2k 1P12 — u12)(b — 1u12)j 012

t Equation (g) is (8—1) with F1 = 0.


SEC. 5—4. INCREMENTAL FORMULATION 197

k 1 P12F1 = F2 = — — (b — 2u12)u12 =

L 2b—u12L

We can write (k) asL2

U12 =—— 12k (b — u12)(b

and solve (m) by iteration. Alternatively, one can specify u12 and evaluate from (k).

The latter approach works only when there is one variable. The solution is plotted inFig. E8-5B.

P12 Fig. E8—5B

9

B

We describe next the generation of the incremental equations which follow from (8—26)—(8—32). Applying (8—26), (8—27) to (b)—(d) results in

= d[i,, = = [o

de, = a,1 Au11 + +

d2e, = Au1 = (Au1 )2

n=l,2We arc assuming no initial elongation. Then,

k = k(de,, +

The tangent stiffness matrix and incremental geometric load term are defined by (8—31).Using (n), we obtain

+ u12)

= Sym + u12)2 +

(Au12)2

A

0(I +

b


198 DISPLACEMENT METHOD: DEAL TRUSS CHAP. 8

Finally, we assemble the incremental equilibrium equations for joint 1 using (8—30).

Ap1 = + k, 2)Au1 + Ap1, i + Al)9. 2 (r)

o ?4(_b + u12)2 + + F2)

0

— [3(—b + u12) + Au12]}

Note that (s) is (8—34). Also, the incremental equations are uncoupled.We restrict the analysis to only 112 loading. Setting F1 = F2 in (s) results in

+ + u17)2)] Au12 = Ap12 [Au12 + 3(—b + (t)

where F is determined from (e). The coefficient of Au12 is the tangent stiffness with respectto u12.

= + + u12)2) (u)du12 L L

Applying the classical stability criterion (8—43) to (t), we see that

> 0 stable

0 neutral (v)du12

< 0 unstable

Points A, B are stability transition points and the segment A-B is unstable.If k = 0, the truss is neutral with respect to Au11. Now there is a discontinuity in k

at F = — Feb, the pin-ended Euler load, when the material is linearly elastic:

k=0 (w)

,t2EI3

To determine whether the members buckle before point A is reached, we compare F4With — Feb. Using (u),

AE( \2 AEb2F4 = —

) = (x)

Then, for system instability rather than member instability to occur, b must satisfy

b = (y)

where p is the radius of gyration of the section.


INCREMENTAL FORMULATION

Lastly, we outline how one applies the method of successive substitution to (t). Forconvenience, we drop the subscripts and write (t) as

Ltu = —

In the first step, we take Ap9 = 0.

= — (aa)

Generalizing (bb),

=

= —

(bb)

The convergence is illustrated in Fig. E8—5C. Case (b) shows how the scheme diverges

Ib)

Fig. ES—5C

in the vicinity of a neutral point = 0). Convergence generally degenerates as 0and one has to resort to an alternate method.

SEC. 8—4. 199

The second estimate is determined from

— (a)

Lip


200 DISPLACEMENT METHOD: IDEAL TRUSS CHAR 8

8—5. LINEARIZED STABILITY ANALYSIS

In the previous section, we illustrated the behavior of a geometrically non-linear system. The analysis involves first solving the nonlinear equilibriumequations for the displacements and then applying the classical stability crite-rion to determine the stability of a particular equilibrium position. Once thenonlinear equilibrium equations are solved, the stability can be readily deter-mined. Now, if a geometrically nonlinear system is loaded in such a way thatit behaves as if it were geometrically linear, we can neglect the displacementterms in that is, we can take = in the expression for This ap-proximation is quite convenient since we have only to solve the linear problemin order to apply the stability criterion. We refer to this procedure as linearizedstability analysis.

According to (8—40), an equilibrium position is stable ineutral, unstable)when the tangent stiffness matrix is positive definite (positive semi-definite, in-different). We generate transform to and test for positive defi-niteness. We have shown that and K,, u have the same definiteness prop-erty, i.e., K1, is positive definite if,K7 is positive definite. Working with i(7rather than K,, avoids having to permute the rows and columns.

In linearized stability analysis, we approximate k, with

k,, = + F,,g,, (8—44)

The first term is the linear stiffness matrix. We interpret the second term as ageometric stiffness. The bar forces are determined from a linear analysis ofthe truss. If the loading is defined in terms of a single load parameter, we

can write (8—44) asF,, =

(8—45)k,,,, = k1,,, + ).k9,

The tangent stiffness matrix is generated by applying the Direct Stiffness Methodto each term in (8—45). We express the actual and modified matrices as

K,, = K1, ,i + Kg, ii (8—46)

and= + (8—47)

where K1 is the system stiffness matrix for linear behavior. It is symmetricaland positive definite when the system is initially stable. The geometrical stiff-ness, K9, is also symmetrical but it may not be positive definite.

Equation (8—46) shows that the tangent stiffness matrix varies linearly withthe load parameter. If the system is initially stable, K,, is positive definitefor 2 = 0. As 2 is increased, a transition from stable to neutral equilibriummay occur at some load level, say 2cr To determine we note that neutralequilibrium (see (8—43)) corresponds to K,, = 0 which, in turn, can beinterpreted as the existence of a 'non-trivial solution of

K,, = 0 (a)


SEC. 8—5. LINEARIZED STABILITY ANALYSIS 201

Substituting for K1, transforms (a) to a characteristic value problem,

AU1 = —).K5,11 AU1 (8—48)

and is the smallest eigenvaluet of (8—48).Since 1K1, itt = we can work with

Mt' = 0

1,1

Ault' = Mlii (8—49)

instead of (8—48). Both equations lead to the same value of However,(8—49) has r additional characteristic values equal to — I since we have addedr dummy equations. To show this, we substitute for using (8—41) andnote (8—42).

11T [ICe, 0 1JAU51— AHT [Kg, ii 01 5Au1

[o U — [o i,j l,,AU2

Premultiplying by H = (a) becomes

K1. ii AU1 = '2Kg. ii AU1 (na eqs.) (b)

AU2 = —2 AU2 (r eqs.) (c)

The solution of (c) is

1 0 0

AU2C1 +C2

o o i

This solution must be disregardcd since AU2 is actually a null matrix.

Example 8—6

Consider the system shown. We suppose the bars are identical, the material is linearlyelastic, and there is no support movement,

The geometry change is negligible under a vertical load and we can use the linearizedstability criterion. Working with the undeformcd geometry, we have

F1 = F2 = —

= = —

F Matrix iteration (Ref. 1) is a convenient computational scheme for determining 2,,. We applyjtto

(—K9,,1)AU1 = AU1

which satisfies the restrictions on the method,


202 DISPLACEMENT METHOD: IDEAL TRUSS CHAP. &

Fig. £8—B

12

We let k1 = k2 = k. The system stiffness matrices follow from (8—44) and (8—45).

= + k€ 2 = +

— 21

0(b)

-and

= k8,1 + k9,2 = — (g5 + g2) (c)

It remains to determine g1 and g2 which are defined by

= + ufg,, (d)

We neglect u12 in the general expression for This is reasonable when d K< b. Theapproximate expressions for fi, and g, arc

—b] —b]L L (e)

i[l 0

0

0

0

(d)2

= 11e,ii + AKçj,ji = 2k

0(b)2

0 0

(g)

A

d <<b


SEC. 8—5. LINEARIZED STABILITY ANALYSIS 203

Neutral equilibrium (K1, is semidefinite) occurs at

2kb\\L) L \,LJ

Note that (g) has only one eigenvalue instead of two. This is a consequence of our usingapproximate expressions for Equations (e) instead of the exact expressions. At 2 =the system is neutral with respect to Au1 i.e., the buckling mode is antisymmetric.

Neutral equilibrium also occurs when the bars either buckle or yield. The value of 2for Euler buckling of the bars is

2b 2b k2EIl 2AEb/irp'\2eb = — = IL LLL2J L \L

Comparing (h) and (i), we see that Euler buckling of the bars controls when

d> irpThe exact expression for g, is

I,.If we work with (k),

Kg,jj =

(d)2

K1, = K,, + = 2k(b)2

—2

0

In this case, there are two characteristic values and therefore two critical values of 2.

(d\21 = 2kb

2cr. 2 2kbC)

Acri(Ii)

The second root corresponds to neutral equilibrium with respect to Au12. For this example,d b, and the first root defines the critical load.

It is of interest to compare 2cr 2 with the buckling load found in Example 8—5. Therewe considered d b and followed the nonlinear behavior up to the point at which theslope of the P12 — u12 curve vanished (neutral with respect to Au12):

1(1 = 0 = (0)du12 mex

The linearized result is significantly higher than the true buckling load. In general, thelinear buckling load is an upper bound. How close it is to the actual value will depend onthe geometry and loading. When d << b, it is quite close, while it considerably overestimatesthe true load for d b.



REFERENCES

1. C. FL, and J. B. WILCUR Elementary Structural Analysis, McGraw-Hill,New York, 1960.

2. HALL, A. S.. and R. W. W00DHEAD: Frame Analysis. Wiley, New York, 1967,3. ARGYRIS, J. H. and S. Kelsey: Energy Theorems and StruciuralAnalysis, Butterworths,

London, 1960.4. LIVESLEY, R. K.: Matrix Methods of Structural Analysis, Pergamon Press, 1964.5. DC VEUBEKE, B. F., Matrix Methods of Structural Analysis, Pergamon Press, 1964.6. MARTIN, H. C. Introduction to Matris Methods of StructuralAnalysis, McGraw-Hill,

New York, 1965.7. ARGYSIS, J. H.: Recent Advances in Matrix Methods of Structural Analysis, Pergamon

Press, 1964.8. RUBINSTEIN, M. F.: Matric Computer Analvsis of Structures, Prentice-Hall, 1966.9, PRZEMIENIECKI, J. S.: Theory of Matrix Structural Aizah'sis, McGraw-Hill, 1968.

10. THOMPSON, J. M. T. and A. C. WALKER: "The Nonlinear Perturbation Analysis ofDiscrete Structural Systems," Jut. J. Solids Structures, Vol. 4, 1968, pp. 757—768.

11. RUBINSTEIN, M. F.: Structural ,S'vsse,'ns—Statics. Dynamics, and Stability, Prentice-Hall, 1970.

12. RALSTON, A.: A First Course in Numerical Analysis, McGraw-Hill, 1965.

PROBLEMS

8—1. Consider U2 and P1 to he prescribed and the behavior to be physicallylinear. —

(a) Express = Vr — PTU1 in terms of U1, U2. Use

VT=

— e0,1)2 = eo)Tk(e e0)

(b) Show that (8-4) are the Euler equations for Note that

dVT = FTde de = BfAT.11

(c) Express as a quadratic form in AU1. Hint: Obtain d2e byoperating on (7—8).

8—2. For the structure sketched:(a) Determine K11.(b) Determine and F due to a temperature increase of 100°F for all

the bars. Assume no support movements at joints 2, 3, 4.8—3. For the structure sketched:

Determine the displacements, bar forces, and reactions.8—4. Refer to Example 8—2. Suppose we number the joints as shown.

Develop the general form of Example 8—2.

8—5. For the structure sketched, determine\EaJ

8—6. For the structure sketched:Develop the general form of Indicate how you would obtain K11.


PROBLEMS 205

Prob. 8—2

E = 3 X ksi

Bar areas = 3

Coefficient ofthermal expansion =6 X 106/°F

Prob. 8—3

20' 10'

I

I

0

E=3X ksi

Initial elongation of = in.

Horizontal displacement of joint 2 = to the left

6 kips

x2

Prob. 8—4

©

807 0 605



E constant for all bars

Bar Area

I 3a2 4a3 3a4 4a5 2.5a

Prob. 8-6

8—7. Determine the load-deflection relation for the system shown. Considerthe material to be linearly elastic and the bars to bc identical, Assume no initialelongation or support movement.

8—8. Investigate the elastic stability of the system shown. Assume thematerial is linearly elastic and no support movements. Use the linearizedstability criterion and work with the exact expression for Rework theproblem, considering d b and using the corresponding approximate expres-sion for

8—9. Determine the lowest critical load for the truss shown. Assume thematerial is linearly elastic and all bars have the same stiffness.

20' Prob. 8—5

IY3


w p Prob.8—7

20'

d <<b

— _AEk1 —Ic2 —-•y-

x1

K2

H

Prob. 8—S

Prob. 8—9

0

H2¼

T



8—10. The governing equations for geometrically nonlinear behavior of alinearly elastic discrete system such as a truss are nonlinear algebraic equationscontaining up to third-degree displacement terms. We have expressed them as

= + = ±

where d, contain linear displacement terms. This form is dictated by ourchoice of matrix notation. In order to expand (a), we must shift from matrixto indicial notation.

For convenience, we employ the summation convention. If a subscript isrepeated in a term, it is understood the term is summed over the range of therepeated subscript. An example is

(j = 1, 2 n)

We write the ith equilibrium equation for tile system as (this representationis suggested in Ref. 8-40):

+ KIJkU,, + = 2P1

where i, j, k, range over the total number of unknowns, U1 is the total valueof the jth displacement unknown, 2 is a load parameter, defines the loaddistribution, and the K's are constants which can be interpreted as second-,third, and fourth-order tensors. The second-order tensor, is the linearstiffness matrix.

(a) We generate the system tensors by superimposing the contribution ofeach bar. The first step involves converting the matrix expressions

p,, = p,, - = — p,, (d)

where

F,, k,,e,, + F0,,,e,, = 'y,,(u,,÷ —

+ — u)T

= + —

over to indicial form. We drop the n subscript, define p and u as

(f)l.Pn_j (U,,J

and write (d) in the form

= (k,1 + + + Po,


Show that

where c is defined by

PROBLEMS 209

k11 =

=

[1 {::} = cu (i)

Discuss how you would locate the appropriate addresses for the barstiffness tensors in the system tensors. What symmetry properties dothe k's exhibit? Do these properties also apply for the system tensors?

(b) Develop the incremental equations relating Au, AA and compare with(8—30).Specialize the incremental equations for linearized stability analysis.For the structure sketched:

(a) Determine the nonlinear incremental equilibrium equations at the equi-librium position corresponding to Pi = 0, P2 P2, Cr, the linearizedcritical load.

(b) Take Ap1 = 0 and solve for Ap2 as a function of Au1. Comment onhow the system behaves when a small horizontal load, Pi ±ep2, isapplied in addition to

= L + (Ii)

(c)

8—11.

Prob. 8—11

Linearly elastic material.No support movement orinitial elongation.

Ic1 = = k


9

Force MethodIdeal Truss

9—i. GENERAL

The basic equations for the linear geometric case have the form

P1 = B1F

e = BfU1 + e0 + fF

P2 = B2F

where the elements of B1 and B2 are constants. Equation (a) represents linearequations relating the na prescribed joint forces and the in unknown bar forces.For the system to be initially stable, r(B1) = that is, the rows of B1 mustbe linearly independent. This requires in In what follows, we consideronly stable systems. If in = the system is said to be statically determinatesince one can find the bar forces and reactions using only the equations ofstatics. The defect of (a) is equal to in — nd = q, and is called the degree ofindeterminacy. One can solve (a) for na bar forces in terms of the applied forcesand q bar forces. We refer to the system defined by the na bars as the primarystructure and the q unknown forces as force redundants. In order to determineF, q additional equations relating the bar forces are required. These equationsare called compatibility conditions and are obtained by operating on (b) whichrepresents m relations between the na unknown displacements and the barforces.

The general procedure outlined above is called the Jbrce or flexibilitymethod. This procedure is applicable only when the geometry is linear. Inwhat follows, we first develop the governing equations for the force methodby operating on (a)—(c). We then show how one can establish the compatibilityequations using the principle of virtual forces and discuss the extremal characterof the force redundants. Finally, we compare the force method for a truss withthe mesh method for an electrical network.

210


SEC. 9—2. GOVERNING EQUATIONS—ALGEBRAIC APPROACH 211

9-2. GOVERNING EQUATIONS—ALGEBRAIC APPROACH

We consider the first columns of B1 to be linearly independent (if thesystem is initially stable, one can always renumber the bars such that thiscondition is satisfied) and partition B1, B2 and F as follows:

B1 =[B11 B12]x nil ii "a) I

(na x q)

B2 [B21 B22](rxm) (rxn4) (rxq)

(na x 1)

=F2

(qx 1)

The bars corresponding to F1 comprise the primary structure and F2 containsthe q redundant bar forces. Using (9—1), the force-equilibrium equations ((a)and (c)) take the form

B11F1 = P1 — B12F2 (ad eqs.) (9—2)

P2 = B21F1 + B22F2 (reqs.) (9—3)

Since Bit! 0, we can solve (9—2) for F1, considering P1, —B12 as right-hand sides. The complete set of q + 1 solutions is written as

F1 = F1,0 + Fl,F,F2 (9—4)

where F1.0 and F1, F2 satisfy1111F10 = P1

B1IFI,F2 = —B12

Note that the kth column of F1, F2 contains the bar forces in the primarystructure due to a unit value of the kth element in F2. Also, F1, containsthe bar forces in the primary structure due to the applied joint loads, P1, withF2 = 0. The reactions follow from (9—3):

P2 = P2,0 +P2,o = B21F1,0 (9—6)

= B21FI,F2 + B22

We consider next (b). Partitioning e, e0, and f,

(fl4X 1)(nixl)

c e1e = ——— = (9—7)

C2 j(q< 1)

ii

-L 0 f2

(q x q)


212 FORCE METHOD: IDEAL TRUSS CHAP. 9

and using (9—1), the force-displacement relations expand to

B11U1 + B21U2 = = e1, o + f1F1 (ne, eqs.) (9-8)Bf2U1 + Bf2U2 e2 = e2, + 12F2 (q eqs.) (9—9)

Once e1 is known, (9—8) can be solved for U1.We obtain the equation for F2 by eliminating U1 in (9—9). First (see (9—5))

we note that

BT 12 \T_ T T12 — 1,F21 — — IF2 11

Then, premultiplying (9—8) by Ff, adding the result to (9—9), and using(a), (9—6) leads to

P2, F2U2 = ez + Ff (9—10)

= e2,0 + f2F2 + Ff,FZ(CLO + f1F1) (9—11)

The first form, (9—10), shows that the equations are actually restrictions onthe elongations. One can interpret (9—10) as a compatibility condition, i.e.,it must be satisfied in order for the bars tofit in the deJbrmed structure definedby U1. The second form, (9—11), follows when we express the elongations interms of the bar forces. Finally, we substitute for F1 and write the result as

122F2 d2 (9—12)

wheref C t'T122 T 1, F2t1' t,F2 1O...43

.1 — r'T ( v nT f'rU2 — —e2,0 — + +

The coefficient matrix, f22, is called the flexibility matrix for F2. One can showthat f22 is positive definite when the bar flexibility factors in f2 are all positive.t

If the material is physically nonlinear, f. and e0,, depend on F1. Iterationis minimized by applying the loading in increments and approximating theforce-elongation relation with a piecewise linear representation. The incre-mental equations are similar in form to the total equations4 We just haveto replace the force, displacement, and elongation terms with theit incrementalvalues and interpret las a segmental (tangent) flexibility.

At this point, we summarize the steps involved in the force method.

I. Determination of F1, 0, P2. 0, F1. and P2, F2

We select a stable primary structure F1 and determine the bar forces andreactions due to P1 and a unit value of each force redundant. This step involvesq + 1 force analyses on the primary structure. Note that we obtain the primarystructure by deleting q = ni — bars. The selection of a primary structureand solution of the force equilibrium equations can be completely

t See Prob. 9—i.X See Prob. 9—4.

§ We reduce to an echelon matrix. See (1—61).


SEC. 9—2. GOVERNfNG APPROACH 213

2. Deterininatio,z of F2, F,, aizdP2

We assemble d2, and solve f,,F, = d, for F,. Then, we determine F,and P2 by combining the q + basic solutions.

F1 = F10 +P2 = P2,0 + P2,F2F2

3. Determination of U,

Once F1 is known, we can evaluate e,,

e1 e1,0 + f1F1and then solve (9—8).

= e,for U,.

If only a limited number of displacement components are desired, one candetermine these components without actually solving (9—8). To show this,we write U, as

U —' —hT1 — t e1 — 21 111 2

We see from (9—15) that the kth column of Br,' contains the bar forces in theprimary structure due to a unit value of the kth element in P1. Also, it followsfrom (9—6) that the kth column of B2 contains the reactions due to a unitvalue of the kth clement in P,. Now, we obtain the kth element in U1 (whichcorresponds to the kth element in P,) by multiplying the kth column of Br1' by

the kth column of B21Br,' by Of. and adding the two scalars. Then, letting

F,pJk = F, due to an unit value of PJJ, with F2 = 0(9—14)

P2.pj,, = P2 due to an unit value of with F2 0

we can write the expression for as

nT fl0,ik I — —

Note that one works with the statically determinant primary structure tomine the displacements.

Example 9—1 —

Step 1: Determination of F,, o, P2, F,, F2, and P2 F2

For the truss shown in Fig. E9—IA,

= 2 in 3 q = 1

We take F3 as the redundant b,tr force:

F1={F,,F2} F2={F3}

The primary structure consists of bars I and 2. Note that all force analyses are performedon the primary structure. The forces and reactions corresponding to P, and F3 = + 1 canbe readily obtained using the method ofjoints. The results are shown in Fig. E9—1B.


214 FORCE METHOD: DEAL TRUSS CHAP. 9

10 kips

20 kips

2

I(1) A1 = 1.Oin,2 A2 0.51n.2 A3 =0.5 in.2(2) Material is linearly elastic. E = 3 )< i04 ksi for all bars.

(3) e0,1 —1/16 in. eo,2 = 0.

(4) u32 + 1/10th. u41 =—1/I5in.

10 kips

We could have obtained the above results for F1 by solving

B11F1 = P1 B12F'2

which, for this system, has the form

[—.6 +61 (F1) 110) [01[-.8 +.8flFj = + [1j{F3}

Step 2: Determination of f22, d2, F1, and F2

Since only u32 and u41 are finite, we can contract U2 and P2,

== {u3,.u41}

and writePT fl /D'

2 2

Fig. E9—1A

3

3.33

2.50

Fig. E9—1B

1/2

—20.83

3/8

20 kips


SEC. 9—2. GOVERNING EQUATIONS—ALGEBRAIC APPROACH 215

The force matrices follow from step 1:

F1,0 {—2083, —4.17} (kips)

F1, F2 { — (kips)

= (kips)Also, we are given that

e1, o {eo, e0, 2} = { T'6, 0} (inches)

e2,0 = {eo,3} = 0

(inches)

It remains to assemble f2 and evaluate f22 and d2.The flexibility factors are (in./kip)

12(25) 12(25) 12(20)

= 3 x 12 = 13 =

Then,

= =x

f2 [f3] = 0.8(2 x

Evaluating the various products in (9-43), (9—12) reduces to

1.3SF3 = —7.31 (a)Solving (a), we obtain

F2 = {F3} = —5.27kips

1— 17.53kipsF1 = F1,0 +

= 0.S7kips

Equation (a) actually represents a restriction on the elongations. The original form of (a)follows from (9—10).

a —— 8e1 — —.

Equation (b) reduces to (a) when we substitute for the elongations in terms of the bar forces.

Step 3: Determination of the Displacements

Suppose only u11 is desired. Using (9—15),

= Ff, p,1e1 — )T (c)

Now,fr' (1 1

2 — UO' 15

= e1,0 + f1F1 {-.-.24, —.018}

We apply a unit load at joint 1 in the X1 direction and determine the bar forces in theprimary structure and the reactions (P32, P41) corresponding to the nonvanishing prescribeddisplacements

1



Substituting in (c), we obtain

a11 = +185 — .033 = +15 in

If both displacement components are desired, we apply (9—15) twice. This is equivalentto solving (9—8).

9—3. GOVERNING EQUATIONS—VARIATIONAL APPROACH

We obtained the elongation compatibility equations (9—10) by operating onthe elongation-displacement equations. Alternatively, one can use the principleof virtual forces developed in Sec. 7—3. It is shown there (see Equation (7—14))that the true elongations satisfy the condition,

AFTe — 0

for any statically permissible system of virflial bar forces and reactions whichsatisfy the constraint condition,

B1 AF = AP1 = 0

Equation (b) states that the virtual bar forces cannot lead to increments in theprescribed jOint loads, i.e., they must be self-equilibrating.

Now, using (9—4), (9—5), we can write

F= {Fio}

+where

B1

and

B1 = 0

Then

=AF2

satisfies (b) for arbitrary AF2. The reactions due to AF2 are obtained from(9—6):

A?2 = B2 AF = P2 F2

Substituting for AF and A?2, (a) expands to

F2e1 + C2 — F2U2) 0

Equation (h) must be satisfied for arbitrary AF2. Finally, it follows that

FT,F,el + C2 P2 F2U2 = 0 (i)

Equation (i) is identical to (9—10). Note that the elongation compatibility


SEC. 9—4. COMPARtSON OF THE FORCE AND MESH METHODS 217

equations are independent of the material behavior. If the material is physicallylinear, (i) leads to a set of q linear equations in F2 when we substitute for theelongations in terms of the bar forces.

We determine the displacements by applying the general form of the principleof virtual forces (see (7—10))

AFTe — = APfU1

where the virtual forces satisfy the force-equilibrium equations,

=

Since only F1 is required to equilibrate P1, we can take

=AF2 = 0

and (j) leads toU1 = —

Note thatI — 1)11

One can interpret the compatibility equations expressed in terms of F2 as theEuler cquations for the total complementary energy function,

— Pf02 =

This approach is discussed in sec. 7—5. We take X = F2 in (7—35). Then,

== P2,F,

and (7—37) coincides with (i). We have written the expanded form of (i) as

f22F2 = d2

Since (i) are the Euler equations for

= d2)

and it follows that H _I Tc — 2 221 2 — 2U2

for the linearly elastic case. One can show that the stationary point correspondsto a relative minimum value of H. when the tangent flexibility factors for theredundant bars are all positive. t

9-4. COMPARISON OF THE FORCE AND MESH METHODS

It is of interest to compare the force method for a truss with the procedurefollowed to find the currents in an electrical network. The latter involves the

t See Prob. 9—8.



application of Kirchhoff's laws and is called the mesh method. Various phasesof the electrical network formulation are discussed in Probs. 6—6, 6—14, and thegoverning equations for a linear resistance d-c network are developed in Probs.6—14, 6—23. We list the notation and governing equations for convenience:

b = number of branchesn number of nodesN=n—1M—b—N=b—n+1

= potential at node j with respect to the reference potential,k+, k = nodes at positive and negative ends of branch k

= current in branch k, positive when directed from node k_ to node

ek = potential drop for branch k = Vk. — Vk+

e0, k = emf for branch k= resistance for branch k

The governing equations expressed in matrix notation are (see Prob. 6—23):

An = 0 (N eqs.) (9—16)

e = AV = e0 + Ri (b eqs.) (9—17)

wherei

e = {ej, e2, . . , (9—18)

v={V1,V2

R1

R= R2

Rb

and A is obtained by deleting the last column of the branch-node connec-tivity matrix d. Note that d has only two entries1 in any row. For row k(Ic =

jlkk = +1dkk+ = —1 1 k+ ork. (9—19)

dkj =0 1= 1,2,...,NActually, d is just the matrix equivalent of the branch-node connectivity table.

Example 9—2

A network can be represented by a line drawing consisting of curves interconnected atvarious points. The curves and intersection points are conventionally called branches andnodes respectively. Each branch is terminated at two different nodes and no two brancheshave a point in common which is not a node. Also, two nodes are connected by at leastone path. A collection of nodes and branches satisfying the above restrictions is calleda linear connected graph, If each branch is assigned a direction, the graph is said to be


SEC. 9—4. COMPARISON OF THE FORCE AND MESH METHODS 219

2

N

A

3

Fig. E9—2

Now, A has N linearly independent columns. Therefore, it is possible tosolve (9—16) for N branch currents in terms of b — N = M branch currents.We suppose the branches are numbered such that the first N rows of A containa nonvanishing determinant of order N and partition A, i after row N.

(N x N)(bxN)

— [ A1[A2

(MXN)

(Nx 1)

= JJi__

(Mx))

oriented. The connectivity relations for a network are topological properties of the cor-responding oriented graph.

Consider the oriented graph shown. We list the branch numbers vertically and the nodenumbers horizontally. We assemble d working with successive branches. Finally, weobtain A by deleting the last column (cot. 4) of d.

Branch I

Node2 3

1 —1 +1

2 +1

3 +1 —.1

4

—1

—1



Introducing (9—20) in (9—16) leads to

ATi1 = (9—21)

Since tAil 0, we can solve for i1 in terms of i2. We write the solution of the

node equations asi = Ci2

[Cii. (9—22)

= I I'2

Note that C1 is of order N by M and is related to A1, A2 by

C1 = (9—23)

It remains to determine a set of M equations for i2.One can express (9—17) in partitioned form and then eliminate V, or alterna-

tively, one can use the variational principle developed in Prob. 7—6. Usingthe first approach, we write (9—17) as

= A1V = e1 + R1i1 (N eqs)(9—24)

e2 A2V = e20 + R212 (M eqs)

Once i1 is known, we can find V from

A1V e1 = e1 + R1i1 (9—25)

Eliminating V from the second equation in (9—24) and using (9—23), we obtain

e2 + CTe1 = 0 (9—26)

Equation (9—26) represents M equations relating the branch potential differ-ences (voltages). Finally, substituting for in terms of leads to

(R2 + CTR1C1)i2 —e2,0 — Cfe10 (9—27)

The coefficient matrix for i2 is positive definite when the branch resistancesare positive. This will be the case for a real system.

The essential step in the solution involves solving (9—21), that is, finding C1.Note that C1 corresponds to for the truss problem. Also, the branchescomprising A1 (and i1) correspond to the primary structure. Although theequations for the truss and electrical network are similar in form, it should benoted that the network problem is one dimensional whereas the truss probleminvolves the geometry as well as the cpnnectivity of the system. One can as-semble C1 using only the topological properties of the oriented graph whichrepresents the network. To find the corresponding matrices (F1, 0 andfor a truss, one must solve a system of linear equations. In what follows, wedescribe a procedure for assembling C1 directly from the oriented graph.

A closed path containing only one repeated node that begins and ends atthat node is called a mesh. One can represent a mesh by listing sequentiallythe branches traversed. A tree is defined as a connected graph having no


SEC. 9—4. COMPARISON OF THE FORCE AND MESH METHODS 221

meshes. Let hT be the number of branches in a tree connecting 11 nodes. Onecan easily show that

bT=n—l=N (9—28)

We reduce a graph to a tree by removing a sufficient number of branches suchthat no meshes remain. The branches removed are generally called chords.The required number of chords is equal to b — = b — N = M. Now, weassociate the branches comprising a tree with the rows of A1. Selecting a treeis equivalent to selecting N linearly independent rows in A. The M chordscorrespond to the redundant branches, that is, the rows of A2. Note that onecan always number the branches such that the first N branches define a tree.

Chord j and the unique path (in the tree) connecting the terminals of chordj define a mesh, say mesh j. We take the positive direction of mesh j (clockwiseor counterclockwise) such that the mesh direction coincides with the positivedirection for chord j. Now, the current is constwit in a niesh. Suppose branchr is contained in mesh j. Then, the current in branch r due to a unit value of

is equal to + 1 (—1) if the positive directions of branch rand meshj coincide(are opposite in sense).

We have expressed the solution of the node equations as

I) (NXM) (Mxl)i1=C1 12

Now, we take the elements of i2 as the chord (mesh) currents. Then i1 representsthe required branch currents in the tree. We assemble C1 working with thecolumns. The column corresponding to involves only those branches of thetree which are contained in mesh j. We enter (+1, — 1,0) in row k of thiscolumn if branch k is (positively, negatively, not) included in mesh j.

Example 9—3

For the graph in example 9—2, N = n — 1 = 3 and b = 6. Then M = b — N = 3

and we must remove 3 branches to obtain a tree. We take branthes 4 , 5 , and 6 as thechords. The resulting tree is shown in Fig. E9—3. We indicate the chords by dashed lines.

Fig. E9—3

1 3

2



For this selection of a tree,

ij = {i1, i2, 13} i2 {i4, i5, i6}

The meshes associated with the chords follow directly from the sketch:

mesh4mesh5mesh6

To assemble C1 we list the branches of the tree vertically and the chord numbers hori-zontally. We work with successive columns, that is, successive chords. The resulting matrixis listed below. Note that C1 is just the matrix equivalent of(a).

4 5 6

1Branches

ofthe

1

—2

—l• 0 +1

+1 +1 0tree

13 0 —1 —1

The matrices, A1 and A2, follow from Example 9—2:

—! +1 0

A1= 0 +1 0

0 +1 —l

—1 0 0

A2= 0 0 —1

+1 0 —1

One can readily verify thatC1 =

The matrix, C = {C1, Im}, is called thc branch-mesh incidence matrix. Using(9—23), we see that A and C have the property

(N x Ill)ATC = 0 (9—29)

Also, we can express the compatibility equations, (9—26), as

(SIx 1)CTe = 0 (9—30)

The rows of CT define the incidence of the meshes on the branches. Equation(9—30) states that the sum of the potential drops around each mesh must bezero and is just Kirchhoff's voltage law expressed in matrix form. The matrix


PROBLEMS 223

formulation of the network problem leads to the same system of equationsthat one would obtain by applying Kirchhoff's current and voltage laws to thevarious nodes and meshes. This, of course, also applies to the truss problem.The two approaches differ only with respect to the assemblage of the governingequations. In the conventional approach, one assembles the equations in-dividually. This involves repeated application of the basic laws. When theequations are expressed in matrix form, the steps reduce to a sequence of matrixmultiplications.

REFERENCES

1. C. H., and J. B. WILBUR: Elementa,'v Structural Analysis, McGraw-Hill,New York, 1960.

2. HALL, A. S., and R. W. W000HEAD: Frame Analysis, Wiley. New York, 1967.3. MORICE, P. B.: Linear Structural Analysis, Ronald Press, New York, 1969.4. RUBINSTEIN, M. F.: Matrix computer Analyris of Structures, Prentice-Hall, 1966.5. PRZEMIENIECKI, J. S.: Theory of Ma/rLr Structural Analysis, McGraw-Hill, 1968.6. RUBINSTEIN, M. F.: Structural Systems—Statics, Dynamics, and Stability, Prentice-

FlaIl, 1970.7. Di MAGGIO, F. D., and W. R. SNLLERS: "Network Analysis of Structures," Eng.

Mech. Div.. A.S,C.E., Vol. 91, No. EM 3, June 1965.8. FF.NVE.S, S. .1., and F. H. BRANIN, JR.: "Network-Topological Formulation of Struc-

tural Analysis," J. Structures Div., A.S.C.E. Vol. 89, No. ST4, August 1963.

PROBLEMS

9—1. Show that the coefficient matrix f22 is positive definite for arbitraryrank of F1 P2 when is positive definite. Use the approach suggested inProblems 2—12 through 2—14.

9—2. Solve the following system using the procedure outlined in Sec. 9—2.TakeX1 {xi,x2}

2 1 0 0 0 x1 3

2 2 0 0 0 x2+

3 4 l.Y25 00 1 0 X3 1

12 0002 x4 2

9—3. Consider a system of in equations in n unknowns, ax = c, wherein> n. Suppose r(a) = a and the first a rows of a are linearly independent.Let q = m — n.

(a) Show that the consistency requirement. for the system leads to qrelations between the elements of c.

(b) Interpret (9—10) from this point of view.


224 FORCE METHOD: DEAL TRUSS CHAP. 9

9—4. Develop an incremental "force" formulation starting with

— B1 AF

= B2 AF

Ae Bf AU1 + AU2 = + ft AF

where f', represent the flexibility factor and incremental initial elongationfor the segment corresponding to the initial value of F. One has to modifybothft and if the limit of the segment is exceeded (see sec. 6—4 for a detailedtreatment).

Consider the case where the loading distribution is constant, i.e., where onlythe magnitude is increased. Let P1 1,/i where 2 is the load parameter and

defines the loading distribution. Discuss how you would organize thecomputational scheme. Also discuss how you would account for either yieldingor buckling of a bar. Distinguish between a redundant bar and a bar in theprimary structure.

9—5. Solve Prob. 8—3 with the force method: Take F3 as the force re-dundant.

9—6. Assemble theshown.

equations for F2 = (F8, F9, F10, F, for the truss

(1) Material is linear elastic and the flexibility factors are equal.(2) Only u42 is finite. Take = {u42 }(3) Only initial elongation for bar 4.

9—7. For the truss shown:

Prob, 9—6

x2

xl

(a) Using (9—10), determine the elongation-compatibility relations. Taktbars ©, as the redundant bars.

(b) Express u52 in terms of the elongations and support movements.9—8. By definition (see (7—26) and (7—31))

= AFTe —

15' 15'


Then

PROBLEMS 225

— = AFT de

Prob. 9—i

x2

x1

Express d2fl, as a quadratic form in AF2. Consider the material to be nonlinearelastic and establish criteria for the stationary point to be a relative minimum.

9—9. Consider the oriented linear graph shown.

0

(a) Determine A.(b) Determine C.(c) Verify that ATC = 0.

Prob. 9—9

2



Part IIIANALYSIS OF AMEMBER ELEMENT



10

Governing Equations fora Delormable Solid

10—1. GENERAL

The formulation of the governing equations for the behavior of a deformablesolid involves the following three steps:

1. Study of deformation. We analyze the change in shape of a differentialvolume element due to displacement of the body. The quantities re-quired to specify the deformation (change in shape) are conventionallycalled strains. This step leads to a set of equations relating the strainsand derivatives of the displacement components at a point. Note thatthe analysis of strain is purely a geometrical problem.

2. Study of forces. We visualize the body to consist of a set of differentialvolume elements. The forces due to the interactions of adjacent volumeelements are called internal forces. Also, the internal force per unitarea acting on a differential area, say dAd, is defined as the stress vector,

this step, we analyze the state of stress at a point, that is, weinvestigate how the stress vector varies with orientation of the areaelement. We also apply the conditions of static equilibrium to thevolume elements. This leads to a set of differential equations (calledstress equilibrium equations) which must be satisfied at each point inthe interior of the body and a set of algebraic equations (called stressboundary conditions) which must be satisfied at each point on thesurface of the body. Note that the study of forces is purely an equilibriumproblem.

3. Relate forces and displacements. In this step, we first relate the stressand strain components at a point. The form of these equations dependson the material behavior (linear elastic, nonlinear elastic, inelastic, etc.).Substitution of the strain-displacement relations in the stress-strainrelations leads to a set of equations relating the stressand derivatives of the displacement components. We refer to thissystem as the stress-displacement relations.

229


230 GOVERNING EQUATIONS FOR A DEFORMABLE SOLID CHAP. 10

The governing equations for a deformable solid consist of the stress equilib-rium equations, stress-displacement relations, and the stress and displacementboundary conditions.

In this chapter, we develop the governing equations for a linearly elastic solidfollowing the steps outlined above. We also extend the variational principlesdeveloped in Chapter 7 for an ideal truss to a three-dimensional solid.

In Chapter 11, we present St. Venarit's theory of torsion-fiexure of prismaticmembers and apply the theory to some simple cross sections. St. Venant'stheory provides us with considerable insight as to the nature of the behaviorand also as to how we can simplify the corresponding mathematical problemby introducing certain assumptions. The conventional engineering theory ofprismatic members is developed in Chapter 12 and a more refined theoryfor thin walled prismatic members which includes the effect of warping of thecross section is discussed in Chapter 13. In Chapter 14, we develop the engi-neering theory for an arbitrary planar member. Finally, in Chapter 15, wepresent the engineering theory for an arbitrary space member.

1O—2. SUMMATION CONVENTION; CARTESIAN TENSORS

Let a and b.represent nth-order column matrices:

a = {Oj, 0210—1

b = {h1,b2

Their scalar (inner) product is defined as

aTb = bTa = a1b1 + a2h2 + ... +=

To avoid having to write the summation sign, we introduce the conventionthat when an index is repeated in a term, it is understood the term is summedover the range of the index. According to this convention

(i = 1,2 ii) (10—2)

and we write the scalar product as

aTb = a1b1 (10—3)

The summation convention allows us to represent operations on multi-dimensional arrays in compact form. It is particularly convenient for formu-lation, i.e., establishing the governing equations. We illustrate its applicationbelow.

Example 10—i

1. Consider the product of a rectangular matrix, a, and a column vector, x.

c=ax aisrnx n


SEC. 10—2. SUMMATION CONVENTION; CARTESIAN TENSORS 231

The typical term is

c• = > (b)

2. Let a, b be square matrices, x a column vector, and f, g scalars defined by

frxTaxg = xrbx (c)

The matrix form of the product, fg, is

fg =(xTax)(xTbx)

One could expand (d) but it is more convenient to utilize (b) and write (c) as

f =g = bk(xkxe

fg = alJbk,xlxfxkx(

= DIJk(XIXJXkX(

3. We return to part 1 The inner product of c is a scalar, H,

II = xT(aTa)x

Using (b),

H = c-c1 = 0f50

The outer product is a second-order array, il,

d = ccT axxraTand can be expressed as

= = alkaf,xkx,.

= AIJk(XkXe

According to the summation convention,

= d11 + d22 + = trace of d

Then, we can write (h) asH = d11 = AIIk(XkX(

4. Let represent square second-order arrays. The inner product is defined as thesum of the products of corresponding elements:

Inner product ç,) =

= + + + + g21e21 + (m)

In order to represent this product as a matrix product, we must convert cki, ejj over to

one-dimensional arrays.

Let represent a one-dimensional set of elements associatedwith an orthogonal reference frame having directions If the



corresponding set for a second reference frame is related to the first set by—— k

(10—4)= cos

1, 2, 3

we say that the elements of b comprise a first-order cartesian tensor. Noting(5—5), we can write (10—4) as

= (10—5)

and it follows that the set of orthogonal components of a vector are a first-ordercartesian tensor. We know that the magnitude of a vector is invariant. Then,the sum of the squares of the elements of a first-order tensor is invariant.

(10—6)

A second-order cartesian tensor is defined as a set of doubly subscriptedelements which transform according to

= (10—7)j. k. ,n.

1.2. 3An alternate form is

= (10—8)

The transformation (10—8) is orthogonal and the trace, sum of the principalsecond-order minors, and the determinant are invariant.t

=fl(2)

where=

=b12 b22 b23 b11 b13

L 7+

7 7+

021 022 032 033 1733

In the cases we encounter, b will be symmetrical.

10-3. ANALYSIS OF DEFORMATION; CARTESIAN STRAINS

Let P denote an arbitrary point in the undeformed state of a body and theposition vector for P with respect to 0, the origin of an orthogonal cartesianreference frame. The corresponding point and position vector in the deformedstate are taken as F'; and the movement from P to P' is represented by thedisplacement vector, fl. By definition,

(10—10)

This notation is shown in Fig. 10—1.

f See Prob. 2—5.


SEC. 10—3. ANALYSIS OF DEFORMATION; CARTESIAN STRAINS 233

Excluding rigid body motion, the displacement from the initial undeformedposition will be small for a solid, and it is reasonable to take the initial Cartesiancoordinates (xi) as the independent variables. This is known as the Lagrange

Fig. 10—1. Geometric notation.

2

approach. Also, to simplify the derivation, we work with cartesian componentsfor ü. Then,

ii

=

We consider a differential line element at P represented by the vector dii.(See Fig. 10—1). The initial length and direction cosines are ds and We areusing the subscript notation for partial differentiation.

= / = (10—12)

The corresponding line in the deformed state is Since we arefollowing the Lagrange approach, p = arid we can write

= = (10—13)

The extensional strain, r, is defined as the relative change in length with respect

Undeformeddp

F' (Deformed)

i3

112



to the initial length.t= (1 + (10— 14)

Finally, we write (a) as

and the strain is defined as

(1

e(1 + 4c) = ap.kejk

= —

+ = e0 (no sum)

= -1)

=

= (1 —

— ds2 = 2eJkdxJdxk

(10—15)

(10—16)

— '/12

p.-

dp'1

Using the dot product, (10—14) becomes

One can readily establish that (eJk) is a second-order symmetrical Cartesiantensor4

Taking the line element to be initially parallel to the direction and lettingrepresent the extensional strain, we see that

To interpret the off-diagonal terms, we consider 2 initially orthogonalline elements represented by (see Fig. 10—2) and having direction cosines

d r',xa

I

x2

Fig. 10—2. Notation for shearing strain.

t This is the definition of Lagrangian strain. In the Eulerian approach, the cartesian coordinatesfor the deformed state are taken as the independent variables,

See Prob. 10—4. It is known as Green's strain tensor. The elements, are also called thecomponents of finite strain. They relate the difference between the square of the initial and deformedlengths of the line element, i.e., an alternate definition of Cjk



We define as the angle between the lines in the deformedstate. The expression for which is called the shearing strain, follows bytaking the dot product of the deformed vectors.

(it , .COS — — J = Y12 =

)Substituting for

k)dsf(sum on k only)

= (1 +

and noting that the lines are initially orthogonal,

=(a) takes the form

(1 + + = (10—17)

Specializing (10—17) for lines parallel to X,, shows that is related to theshearing strain.

(I + + = 2e13 = (10—18)

Equations (10—15) and (10—17) are actually transformation laws for exten-sional and shearing strain. The state of strain is completely defined once thestrain tensor is specified for a particular set of directions. To generalize theseexpressions, we consider two orthogonal frames defined by the unit vectors

and (see Fig. 10—3), take the initial frame parallel to the global frame= ti), and let = 15 tk. With this notation:

+ = )(10—19)

(1 + + =

The strain measures (e, y) are small with respect to unity for engineeringmaterials such as metals and concrete. For example, e for steel.Therefore, it is quite reasonable (aside from the fact that it simplifies theexpressions) to assume r, y 1 in the strain expressions. The relations for"small" strain are:

(10—20)

It remains to expand eJk. Now,

= + ii + u,,Ji,,

Differentiating with respect to

S OP —

== + Un, j)l,n


236 GOVERNING EQUATIONS FOR A DEFORMABLE SOLID Cl-lAP. 10

and substituting into the definition of (Equation (10—15)) leads to

eJk = k + Uk. + 4Um, ,u,,,, k (sum on m only) (10—21)

In order to simplify (10—21), we must establish the geometrical significance ofthe various terms.

x3

t3

/ X2

Fig. 10—3. Unit vectors defining transformation of orthogonal directions.

With this objective, we consider a line element initially parallel to the X1axis. Figure 10—4 shows the initial and deformed positions, and the angles012, which define the rotation of the line toward the X2, K3 directions.The geometrical relations of interest to us are

033sin 0j3 =

1 +1421

sine12(1 + 81)cos 0j3

(1 + )2 + + uj 1Also, by definition,

+ = e11 = U1, 1 + i + i4, i +

We solve (a), (b) for u2, and 03,

03,1 (1 +1 013 —

t2

t,3



and then solve (c) for u1,

1 = (1 + {1 — A}112 — 1

(10—23)A = sin2 013 + cos2 013 sin2 012

Applying the binomial expansion,

(1 — x)"2 = I — + + (10—24)

to (1 — we can write (10—23) as

— + + — + + (10—25)

In what follows, we assume small strain and express the derivatives and exten-sional strain (see Equation (d)) as

u3. 1 = 0(013) 1 = 0(012,n2

U1 1 — t/12, "13a11 = u1, + + (f)

The various approximate theories are obtained by specializing (f).

Fig. 10—4. Initial and deformed positions of a line element.

In the linear geometric case, the rotations are neglected with respect to strain.Formally, one sets 012 = 613 = 0 in (f) and the result is a linear relation betweenstrain and displacement, -

a11 (g)

Note that, according to this approximation, the deformed orientation coincides

'U3

dx1

dx1

1123 dx1

X1,u1


238 GOVERNING EQUATIONS FOR A DEFORMA8LE SOLID CHAP. 10

with the initial orientation. The general relations for the linear geometric case(small strain and infinitesimal rotation) are

= en = (no sum)(10—26)

= = + ui,,

The next level of approximation is to consider 62 to be of the same orderas strain.

02 = 0(s) << 1

sin 0 0 (10—27)

cos 6 1

We can neglect with respect to 1 in (f), but we must retain 1and 1

since they are of 0(62).

ci u11 + + (h)

The complete set of strain-displacement relations for small strain and small-finite rotation are listed below for reference.

= = + + (no sum)(10—28)

-= + + Uk iUk. j

I k

We utilize these expressions to develop a geometrically nonlinear formulationfor a member in Chapter 18.

Lastly, if no restrictions are imposed on the magnitude of the rotations,one must use (10—21). The relations for finite rotation and small strain are

= = + + + (no sum)10—29

= = + + u11(l + + Uk.iUk,jk

Note that the truss formulation presented in Chapter 6 allows for arbitrarymagnitude of the rotations.

We have shown that linear strain-displacement relations are based on thefollowing restrictions:

1. The strains are negligible with respect to unity, and2. Products of the rotations are negligible with respect to the strains.

The first condition will always be satisfied for engineering materials such asmetals, concrete, etc. Whether the second restriction is satisfied depends onthe configuration of the body and the applied loading. If the body is massivein all three directions, the rotations are negligible with respect to the strainsfor an arbitrary loading. We have to include the nonlinear rotation terms inthe strain displacement relations only if the body is thin (e.g., a thin plate orslender member) and the applied loading results in a significant change in thegeometry. As an illustration, consider the simply supported member shown



in Fig. 10—5. We can neglect the change in geometry.if only a transverse loadingis applied (case 1). However, if both axial and transverse loads are applied(case 2), the change in geometry is no longer negligible and we must include thenonlinear rotation terms in the strain-displacement relations.

Case 2 (Q,P) Case1 (Q)

Fig. 10—5. Example of linear and geometrically nonlinear behavior.

To treat a geometrically nonlinear problem, we must work with the deformedgeometry rather than the initial geometry. This can be defined by trackingthe movement of a triad of line elements initially parallel to the global directions.We let be the initial set and the deformed set (see Fig. 10—6). By definition,

= (no sum)

= (no sum)= (1 +

The unit vector pointing in the direction of dTh is denoted by Using (a),we can write

vi= p.

for small strain

Finally, we express in terms of the unit vectors for the initial frame.

/3jklk

I3jk — +(10—30)

+ for small strain

We will utilize (10—30) in the next section to establish the stress equilibriumequations for the geometrically nonlinear case.

Equations (10—30) reduce to(10—31)

for the geometrically linear case and to

+ 13j1Jk + (no sum)[0—32

for the case of small strain and small-finite rotations.



10—4. ANALYSIS OF STRESS

The effects of the surroundings on a body such as contact pressure, gravita-tional attraction, etc.,result in internal forces. In this section, we establish theequilibrium conditions for the internal forces in a body. This step is generallycalled the analysis of stress.

Consider a body subjected to some effect which results in internal forces.We pass a cutting plane through the deformed body and separate the twosegments as shown in Fig. 10—7. We let in denote the outward normal directionfor the internal face of body and refer to this face as the + in face. In general,the subscript, m, is used for quantities associated with the + m face. Now, weconsider a differential area element AAm, and let A Fm be the resultant internalforce vector acting on this element. The stress vector, is defined as

hmAA, -. I)

(10—33)

Note that has the units of force/area. Also, it depends on the orientationof the area element, i.e., on the direction of the outward normal. We do notallow for the possibility of the existence of a moment acting on a differentialarea element. One can include this effect by defining a vectortin addition to a stress vector.

f See Ref. 6, p. 68.

f

/=dxji1

ii

Fig. 10—6. Initial and deformed geometries.


SEC. 10—4. ANALYSIS OF STRESS 241

We consider next the corresponding area element in the —rn face. FromNewton's law,

=and it follows that

= —ö,,, (1O—34)

The stress vector has the same magnitude and line of action but it's sense isreversed.

Body 2

Note: Deformed state

Fig. 10—7. Notation for internal force.

In order to analyze the state of stress at a point, say Q, we need an expressionfor the stress vector associated with an arbitrary plane through Q. With thisobjective, we consider the tetrahedron shown in Fig; 10—8. The orientation ofthe arbitrary plane is defined by q, the outward normal direction. The outwardnormals for the other three faces are parallel to the reference axes (X1; j = 1,2, 3). To simplify the notation, we use a subscript j for quantities associatedwith the X face, that is, the face whose outward normal points in the +direction. For example, we write

== = (10—35)

=etc.

The force vectors acting at the centroids of the faces are shown in Fig. 10—8.The term M0 represents the change in due to translation from Q to thecentroid.

For equilibrium, the resultant force and moment vectors must vanish. In thelimit (as P Q), the force system is concurrent and therefore we have to

Body I


- - - —

crq + &Tq = (a3 +

Fig. 10—8. Differential tetrahedral element.

Equation (10—37) is the transformation law for the Stress vector. The com-ponent of 5q in a particular direction is equal to the scalar product of 6q anda unit vector pointing in thedesired direction. Now, we express the stressvectors in terms of their components with respect to the coordinate axes(j = 1, 2, 3).

j = 1,2,3- (10—38)

= aqklk


consider only the force equilibrium condition. From Fig. 10—8, we have

Now, AA1 is the projection of AAq on the X2-X3 plane. Noting that the projec-tion of LxAq on a plane is equal to AAq times the scalar product of and theunit normal vector for the plane, and letting aqj be the direction cosine for theq direction with respect to the direction, we can write

AA.= = cos(q, X1) = (10—36)

Finally, in the limit, Equation (a) reduces to

= (1037)

Once the stress vectors for three orthogonal planes at Q are known, we can deter-mine the stress vector for an arbitrary plane through Q with (10—37).

x2

+

i2

I+

x3

+ 2)LXA2



Note that the first subscript on a stress component always refers to the flice,and the second to the direction. For example. a12 acts on the X1 face andpoints in the X2 direction. The positive sense of the components for a negativeface is reversed since The normal (a13) and in-plane (afk) com-ponents are generally called normal and shearing stresses. This notation isillustrated in Fig. 10—9.

FIg. 10—9. Notation for stress components.

Substituting for the stress vectors in (10—37) results in

= cxqjajk (10—39)

The component of 5q with respect to an arbitrary direction, m,is determined from

Letting

and noting (10—38), (a) expands to

aqn,

We generalize (c) for two orthogonal frames specified by the unit vectors(see Fig. 10—3) where

t. = -ii

t3 = cJ3k:k

Defining as the component acting on the 1 face in the direction andidentifying with i,,,, (c) takes the form

= (10—41)

This result shows that the set is a second-order cartesian tensor.

x2ta22

4

T

(10—40)


244 GOVERNING EQUATIONS FOR A DEFORMABLE SOUD CHAP. 10

It remains to establish the equilibrium equations for a differential volumeelement. The equilibrium equations relate to the deformed state, i.e., we mustconsider a differential element on the deformed body. Since we have definedthe stress components with respect to the global Cartesian directions, it is naturalto work with a rectangular parallelepiped having sides parallel to the globaldirections. This is shown in Fig. 10—10. Point 0 is at the centroid of the element.

+dfl3

+ (— di73

Fig. 10—10. Differential volume element in Eulerian representation.

The stress vectors are considered to be functions of the deformedt coordinatesWe obtain the forces acting on the faces by expanding the stress vectors

about 0 and retaining only the first two terms4 Letting b denote the externalforce per unit volume and enforcing the equilibrium conditions leads to

(10-42)

and= 0 x = 0 (10—43)

The scalar force equilibrium equations are obtained by expanding the vectorequations using (10—38).

Force equilibrium + = 0 k = 1, 2, 3 (10—44)

Moment equilibrium =k = 1 2 3

(10—45)

Moment equilibrium requires the shearing stress components to be symmetrical.Then, the stress tensor is symmetrical and there are only six independent stressmeasures for the three-dimensional case and three for the two-dimensional case.

t We arc following the Eulerian approach here. Later we will shift back to the Lagrange approachSecond- and higher-order terms will vanish in the limit, i.e.. when the element is shrunk to a

point.

i3

dr13

ant



Equations (10—44) must be satisfied at each point in the interior of the body.Also, at the boundary, the stress components must equilibrate the appliedsurface forces.

We define as the outward normal vector at a point on the deformed surfaceand write

=The external force per unit deformed surface area is denoted by

= pnjlj

(10—46)

(10—47)

Applying (10—37) leads to the stress-boundary force-equilibrium relations:

Ph =

Pnj i3nk0kj f = 1, 2, 3(10—48)

When p,0 is prescribed, i.e., (10—48) represent boundary conditionson the stress components. If is prescribed, is a reaction.

Our derivation of strain-displacement relations employed the Lagrangeapproach, i.e., we considered the displacements (and strains) to be functionsof the initial coordinates (xi). The analysis of stress described above is basedon the Eulërian approach, where the deformed coordinates are taken as theindependent variables. This poses a problem since the strain and stress measuresare referred to different volume elements. Figure 10—Il shows the initial and

f See Prob. 10—12.

/ Deformed

(I + €2)dx2

Initia'

dX2[

dx1

(I

(I

(1 +e1)dxj

Lagrange

4U12

Eu'erdrj1

x2

Fig. 10—11. Comparison of Eulerian and Lagrangian representations for a volumeelement.



deformed area elements corresponding to the two viewpoints. To be consistentwith the Lagrange strains, we must work with a nonorthogonal parallelepipedwhose sides are parallel to the deformed line elements in the analysis of stress.Conversely, to be consistent with the Eulerian stresses, we have to refer thestrain measures to nonorthogonal directions in the initial state.

In the linear geometric case, we assume small strain and neglect the changein orientation due to rotation. The two approaches coalesce and we just haveto replace with and with where is the direction cosine for theinitial direction of the exterior normal. The linear equilibrium equations are:

+ hk 0CXj

(10—49)

For the geometrically nonlinear case, we work with stress measures referredto the deformed directions (see Fig. 10—6) defined by the unit vectors, Wedefine as the stress vector per unit initial area acting on the face whichinitially is normal to the direction, b* as the force per unit initial volume, and

as the force per unit initial surface area. Figure .10—12 shows this notationfor the two-dimensional case. The stress and force vectors are considered to befunctions of the initial coordinates (xe).

The equilibrium equations at an interior point are

+ b* = 0 (10—50)(X1

(1 + x 0

We express the body force and stress vectors as

—k (10-5 1)

= +

The set, is called the Kirchhoff stress tensor. Substituting for using(10—30), results in the following scalar equations, which correspond to (10—44)and (10—45): -

+ b7 = 0 1,2,3 (10-52)

1, 2, 3(10—53)

The boundary equilibrium equations are obtained by expanding

= (10—54)

and have the form*_ k

pnj — —



These equations apply for arbitrary strain and finite rotation. For smallstrain, we neglect the change in dimensions and shape of the volume element.This assumption is introduced by taking

(10—56)

Since the deformed unit vectors are orthogonal (toe 1), the Kirchhoff stressesnow comprise a second-order cartesian tensor and they transform according

x2

dx2

dx2

—oj dx2

(1 e2)dx2

p2

p,,ds

Fig. 10—12. Definition of stress components in Lagrangian representation.

to (10—41). The equations simplify somewhat if we assume small-finiteFor infinitesimal (linear geometry), and

the equations reduce to (10—49), (10—50).In what follows, we will work with the Kirchhoff stress components to keep

the treatment general. However, we will assume small strain.

t See Prob. 10—16.

1,, = 4 /pn = ij

dx1

c4 dx1

/dx (1 + €1)dx1

x1



10—5. ELASTIC STRESS-STRAIN RELATIONS

A body is said to be elastic if it returns to its initial dimensions and shapewhen the applied forces are removed. The work done during the deformationprocess is independent of the order in which the body is deformed. We treatfirst an arbitrary elastic material and then specialize the results for a linearlyelastic material.

Our starting point is the first law of thermodynamics:

5W = OVT + OQ

where OW = first-order work done by the forces acting on the body0 VT = first-order change in the total strain energy (also called internal

energy)= first-order change in the total heat content.

When the deformation process is isothermal or adiabatic, OQ = 0, and (a) re-duces to

OW

Now, we apply (b) to a differential volume clement in the deformed state(see, e.g., Fig. 1Q—12). We define V as the strain energy per unit initial volume.In general, V is a function of the deformation measures.

V = = Y12' . . .) (10—57)

The material is said to be hyperelastic (Green-type) when V is a continuousfunction. This requires

= (10—58)oek, cejj ciek(

By definition,= OV(dx1 dx2 dx3)

cW= (10—59)cetj

where is the first-order changet in due to an incremental displacement,M. Also, one can show that the first order work done by the force vectorsacting on the element is

OW = . + + . + h Ltfl)dx1 dx2 dx3(10—60)

dx2 dx3

Equating OVT and OW leads to the general form of the stress-strain relation fora Green-type material,

(3•

=(10—61)

tSeeProb. 10—11.See Prob. 10—18. The forces are in equilibrium. is. they satisfy (10—50).


SEC. 10—5. ELASTIC STRESS-STRAIN RELATIONS 249

This definition applies for arbitrary strain. Once V is specified, we can obtainexpressions for the stresses in terms of the strains by differentiating V. Since Vis continuous, the stress-strain relations must satisfy (10—58), which requires

(10—62)43e1j

In what follows, we restrict the discussion to small strain and a linearly elas-tic material, i.e., where the stress-strain relations are linear. We also shift fromindicial notation to matrix notation, which is more convenient for this phase.

We list the stress and strain components in column matrices and drop thesuperscript k on the Kirchhoff stress components:

== {e11, e22, e33, 2e12, 2e23, 2e31} (10—63)

=

With this notation,(10—64)

The matrix transformation laws are

=(10—65)

Since ÔV is invariant under a transformation of reference frames, the trans-formation matrices are related by

1 (10—66)

The total strain, e, is expressed as

a a° ± (10—67)

where a° contains the initial strains not associated with stress, e.g., strain dueto a temperature increment, and A is called the material compliance matrix.We write the inverted relations as

= D(a — a°) (10—68)

where D = A' is the material rigidity matrix. Equation (10—62) requires D(and A) to be symmetrical. The elements of A are determined from materialtests, and D is generated by inverting A. Substituting for in (10—64), weobtain the form of the strain energy density for the linear case,

V = — a°)TD(a — a°) (10—69)

Since V > 0 for arbitrary (E — a°), D and A are positive definite matrices.There are 21 material constants for a linearly elastic Green-type material.

The number of independent constants is reduced if the material structure

t See Prob. 10—6, 10—13.



exhibits In what follows, we describe the transition from an aniso-tropic material to an isotropic material.

A material whose structure has three orthogonal axes of symmetry is calledorthotropic. The structure of an orthotropic material appears identical after a1800 rotation about a symmetry axis. To determine the number of indepen-dent constants for this case, we suppose X1, X2, X3 are axes of symmetry andconsider a 180° rotation about X2. We use a prime superscript to indicate therotated axes. From Fig. 10—13,

= —x1= -x3= x2

The stress and deformation quantities are related by (we replace 1 by — I and3 by —3 in the shear terms)

=a12 = —a12 a23 —a23

= 1,2.3a13 = a13

Y12 = Y12 = Y23 Y13 Y13

Now, the stress-strain relations must be identical in form. We expand e =a' Acv', and substitute for using (b). Equating the expressions for

Fig. 10—13. Rotation of axes for symmetry with respect to the X2-X3 plane.

t A material whose structure has no symmetry is said to be anisotropic.


SEC. 10—5. ELASTIC STRESS-STRAIN RELATIONS 251

and leads to the following relations between the elements of A,

+ = 314(T12 — ti15a33

a24a12 + a25a23 = —a24a12 — a25a23

a34a12 + a35a23 = —a34u12 —

For (c) to be satisfied, the coefficients must vanish identically. This requires

£434 = a15 = 0

a24 = £435 = 0

a34a350We consider next the expansions for The symmetry conditions requirea46 = a56 = 0. By rotating 1800 about X1, we find

a16 = = a36 = a45 0

A rotation about the X3 axis will not result in any additional conditions.Finally, when the strains are referred to the structural symmetry axes, thestress-strain relations for an orthotropic material reduce to

a11 a12 0j3a12 a22 a23 0 a1,

—- (10-70)

-—£444 0 0 a12

0 0 0 a23

Y31 0 0 a66 (733

We see that A is quasi-diagonal and involves 9 independent constants. Thereis no interaction between extension and shear. Also, the shearing effect isuncoupled, i.e., cr12 leads only to

An alternate form of the orthotropic stress-strain relations is

a1 = AT +—

— !(733

1 V32—-----a33

E2(10—71)

1= /13 AT + a33 — — —i—-

1 1 1

Y12 = Y23 = Y31 =

where E4 are extensional moduli, are shear moduli, Vjk are coupling coeffi-cients, and AT is the temperature increment. The coupling terms are related by

(10-72)E2 E1 E3 E1 E3 E2



It is relatively straightforward to invert these relations:t One should note that(10—71) apply only when X, coincide with the material symmetry directions4

If the stress-strain relations are invariant for arbitrary directions in a plane,the material is said to be transversely orthotropic or isotropic with respect tothe plane. We consider the case where the X1 direction is the preferred direc-tion, i.e., where the material is isotropic with respect to the X2-X3 plane. Bydefinition, A is invariant when we transform from X1-X2-X3 toThis

c,'—'12'-'31'--'v32 v23 v 1 2(1 + v)

F --

and the relations reduce to

= AT + — (a22 +

= PAT + — — (10—74)

= pAT + — va27)

1 1 2(l+v)Yi2 Y31 y23

There are five independent constants (F, v, E1, v1, G1).Lastly, the material is called isotropic when the stress-strain relations are

Invariant for arbitrary directions, For this case, A = A' for arbi-trary The relations are obtained by specializing (10—74):

= p AT + (at, — + akk))

2(1 + v)(10—75)

F

Note that now there are only two independent constants (F, v). The couplingcoefficient, v,is called Poisson's ratio.

The inverted form of (10—75) is written as

a = a0 + + + +a0 = (10—76)

= + 2G)pAT

t See Prob. 10—19 for the inverted form of (10—7 1).IO--21.

§ See Prob. 10—22.



where G are called Lamé constants and are related to E, v by

EG = shear modulus =

2(1 + v)(10—77)

AyE

— (1 + v)(1 — 2v)

Since D must be positive definite, v is restricted to — 1 < v < 1/2. The limitingcase where v = + 1/2 is discussed in Problem 10—24.

10—6. PRINCIPLE OF VIRTUAL DISPLACEMENTS; PRINCIPLE OF.STATIONARY POTENTIAL ENERGY; CLASSICAL STABILITYCRITERIA

Chapter 7 dealt with variational principles for an ideal truss. For com-pleteness, we derive here the 3-dimensional form of the principle of virtualdisplacements, principle of stationary potential energy, and the classical stabilitycriterion. The principle of virtual forces and stationary complementary energyare treated in the next section.

The principle of virtual displacements states that the Iirst-order work doneby the external forces is equal to the first oidcr work done by the internalforces acting on the restraints for an arbitrary virtual displacement ofthe body from an equilibrium position. f In the continuous case, the externalloading consists of body (b) and surface loads and the internal forces arerepresented by the stress vectors.

We follow the Lagrange approach, i.e., we work with Lagrange finitestrain components (eJ,j, Kirchhoff stresses and external force measures perunit initial volume or area (b*, p*). This is consistent with our derivation ofthe equilibrium equations. Let Au denote the virtual displacement. The first-order external work is

= dx1 dx2 dx3 + JJj3* Au10—78

= dx1 dx2 dx3 ± dfI

where fI is the initial surface area. The total internal deformation work isobtained by summing the first-order work done by the stress vectors acting ona differential volume element. *

= dx2dx3(10—79)

= dx1 dx2 dx3

Equating (a) and (b), we obtain the 3-dimensional form of the principle of

See Sec. 7—2.See Fig. 10—12.

§ See (10—60).



virtual displacements,

5WD =

dx1 dx2 dx3 = fJJh* dx1 dx2 dx3 + (10—80)

dx1 dx2 dx3 = dx1 dx2 dx3 +

Requiring (10—80) to be satisfied for arbitrary (continuous) is equivalent toenforcing the equilibrium equations.

To show this, we work with the vector form and utilize the following inte-gration by parts formula: t

= J — dx2dx3 (10-81)

where is the direction cosine for the initial outward normal (n) with respectto the direction. Operating on the left-hand term and equating coefficients

in the volume and surface integrals leads directly to (10—50) and (10—54).The principle of virtual displacements applies for arbitrary loading (static

or dynamic) and material behavior. When the behavior is elastic and the loadingis independent of time, it can be interpreted as a variational principle for thedisplacements. The essential steps required for the truss formulation are de-scribed in Sec. 7—4. Their extension to a continuous body is straightforward.

When the behavior is elastic,

=

Letting VT denote the total strain energy, the left-hand side of (10—80) reduces to

dx1 dx2 dx3 fJJ öVdx1 dx2 dx3 =

We consider the surface area to consist of 2 zones as shown in Fig. 10—14.

+

where displacements are prescribed on

U1 on c�d (10—82)

and surface force intensities arc prescribed on

pni Pni on

The displacement variation, L\u1, is admissible if it is continuous and satisfies

= 0 on (10—83)

We also consider the surface and body forces to be independent of the displace-ments. With these definitions, the principle of virtual displacements is trans-




formed to= 0 for arbitrary admissible (1084)

fl,, = VT cIx1 dx2 dx3 —

where is the total potential energy functional. According to (10—84), thedisplacements defining an equilibrium position correspond to a stationaryvalue of the total potential energy functional. Note that this result applies forarbitrary strain and finite rotations. The only restrictions are elastic behaviorand static loading.

PH

Example 10—2

Direct methods of variational calculus such as Rayleigh-Ritz, Galerkin, weighted resid-uals, and others are applied to fl,, to determine approximate solutions for the displacements.In the Rayleigh-Ritz method, one expresses the displacements in terms of unknown param-eters, q, and prescribed functions, x2, x3),

U1 = +

where

>= 0 forj = 1. 2

The displacement boundary conditions on f�d are called "essential" boundary conditions.Substituting for transforms to a function of the q's. When the material is linearlyelastic, V is a quadratic function of the strains. Then, V will involve up to fourth-degreeterms for the geometrically nonlinear case. If the behavior is completely linear, reduces

H, = Const. + qTQ +

q = . . . . . . (3N x 1)K is symmetrical

Fig. 10—14. Classification of boundary zones.



Finally, requiring to be stationary for arbitrary c5q leads (for linear behavior) to

Kq = Q

The strains are evaluated by operating on (a) and the stresses are determined from the

stress-strain relations.Polynomials and trigonometric functions are generally used to construct the spatial

distribution functions. The mathematical basis for direct methods is treated in numeroustexts (see Refs. 9, 10).

The "classical" stability criterion for a stable equilibrium position ist

— o2WE > 0 for arbitrary Ad

where = is the second-order work done by the external forcesduring the incremental displacement, Ad, and WD = ó(ö WD) is the second-order work done by the internal forces acting on the restraints during theincremental deformation resulting from Ad. The form of the work terms fora continuous body are obtained by operating on (10—78) and

= Sfl Ad dx1 dx2 dx3 + j( Ad

= Au1 dx1 dx2 dx3 + J$ Au,(10—85)= dx2 dx3

= ie11 + dx2 dx3

If = Ô2WE for a particular Ad, the equilibrium position is neutral. Theposition is unstable if ö2 WD < o2 Note that öb, are itull vectors whenthe forces are prescribed.

For elastic behavior, the incremental deformation work is equal to theincrement in strain energy = and (10—84) can be written as

= > 0 for arbitrary Ad (10—86)

It follows that a stable equilibrium position corresponds to a relative minimumvalue of the total potential energy. Bifurcation (neutral equilibrium) occurswhen = 0 for some Ad, say Ada. If the loading is prescribed,and ö2VT = 0 at bifurcation.

The governing equations for bifurcation can be obtained by expandingö2WD = This involves transforming the integrand of by applying(10—81). Since bifurcation corresponds to the existence of an alternate equili-brium position, it is more convenient to form the incremental equations directly.The equations for the case of linearly elastic material and prescribed externalforces are listed below.

f See Sec. 7—6 for a derivation of the classical stability criterion.See Probs. 10—11, 10—18.



L Equilibrium Equation in the Interior

+ + = 0 = 1,2,3

2. Stress-Boundary Force Equations on

+ + Au1 0 J = 1, 2, 3 (10—87)

3. Stress-Strain Relations

= D

4. Strain-Displacement Relations

= 3 + AUJ, + AIIm, j + Urn, j Am,

Au1 = 0

10—7. PRINCIPLE OF VIRTUAL FORCES; PRINCIPLE OF STATIONARYCOMPLEMENTARY ENERGY

Let u1 be the actual displacements in a body due to some loading and thegeometrically linear strain measures corresponding to u1. The strain anddisplacement measures are related by

= +

u1=fl

Once the strains are known, we can find the displacements by solving (a) andenforcing (b). The principle of virtual forces is basically a procedure for deter-mining displacements without having to operate on (a). It applies only forlinear geometry. We developed its form for an ideal truss in Sec. 7—3. We willfollow the same approach here to establish the three-dimensional form.

The essential step involves selecting a statically permissible force system,i.e., a force system which satisfies the linear equilibrium equations. For thecontinuous case, the force system consists of stresses, surface forces,

on and reactions, on Static permissibility requires

Ac31,3 = 0

= on (10—88)

= on

If we multiply e13 by Ac13, integrate over the volume using (10—81), and note thestatic relations, we obtain the following identity,t

Acr13 dx1 dx2 dx3 = u1 + $ Th (10—89)0,,




which is referred to as the principle of virtual forces (or stresses). This result isapplicable for arbitrary material behavior. 1-lowever, the geometry must belinear.

Suppose the translation at a point Q on in the direction defined by isdesired (see Fig. 10—15). Let d0 be the displacement. We apply a unit forceat Q in the tq direction and generate a statically permissible stress field.

(1) 1q at point Q Acr and

The integral on reduces to (l)dQ, and it follows that

= dx1 dx2 dx1 — (10—90)

A second application is in the force method, where one reduces the governingequations (stress equilibrium and stress displacement) to a set of equations

Fig. 10—15. Notation for determination of the translation at point Q.

involving only force unknowns. We start by expressing the stress field in termsof a prescribed distribution and a "corrective" field

+ cit, (10—91)

where is a particular solution of the equilibrium equations which satisfiesthe boundary conditions on

+ 0 (10-92)= Thu on

and satisfies= 0

= 0 on (10—93)

on

Stress fields satisfying (10—93) are called seljequilibrating stress fields. For theideal truss, a-° corresponds to the forces in the primary structure due to theprescribed loading and ? represents the contribution of the force redundants.



The governing equations for the force redundants were obtained by enforcinggeometric compatibility, i.e., the bar elongations are constrained by the require-ment that the deformed bar lengths fit in the assembled structure.

Geometric compatibility for a continuum requires the strains to lead tocontinuous displacements. One can establish the strain compatibility equationsby operating on the strain- displacement relations. This approach is describedin Prob. 10—10. One can also obtain these equations with the principle ofvirtual forces by taking a self-equilibrating force system. Letting Aox, Apc denotethe virtual stress system, (10--89) reduces to

dx1 dx2 dx3 = (10—94)

The compatibility equations are determined by expressing in terms of stressfunctions and integrating the left-hand term by parts. We illustrate its applica-tion to the plane stress problem.

Example 10—3

If the stress components associated with the normal direction to a plane are zero, thestress state is called planar. We consider the case where = = 033 = 0. Theequilibrium equations and stress-boundary force relations reduce to

+ + b1 = 0

012 5 + 2 + b2 = 0

+ a,,2o21

= + •z,,2a22

The stress field, oi,, must satisfy (a) with h1 = h2 = 0 and also p,,1 = = 0 onWe can satisfy the equilibrium equations by expressing in terms of a function, as

follows:t= 033 =

=

The boundary forces corresponding to are

=Pa OS

where s is the arc length on the boundary (sense is from X1 —* X2).Substituting for crC, pC in terms of i/i, (10-94) expands to

if+ a2 11 — y52 12)dx1

— f 0CS Ps ,i

There is no loss in generality by taking = 0 on S. Then, integrating (e) by parts,

22 + a2, — 712, dx1 dx2 0




and requiring (f) to be satisfied for arbitrary results in the strain compatibility equation,

+ — 0

which is actually a continuity requirement

U1 122 + 211 — (u1 212 + 112 112) = 0

We express (g) in terms of by substituting for the strains in terms of the stresses.t

The principle of virtual forces is also employed to generate approximatesolutions for the stresses. It is convenient to shift over to matrix notation forthis discussion, and we write (10—94) as

dx1 dx2 dx3 if ApC

We express the stress matrix in terms of prescribed stress states and unknownparameters, a1,

= += + + (12(l)2 + ' + 04,,

where satisfies (10—92) and (1 1, 2,...,r) are self-equilibrating stressstates, i.e., They satisfy the homogenous equilibrium equations and boundaryconditions on The corresponding surface forces arc

p = p° + 0101 + 0209 + +p° = p

= 0 (i = 1, 2,

Taking virtual-force systems corresponding to (i = 1, 2, ., r) results in requations for the parameters.

dx1 dx2 dx3 = jjT9.1 1, 2,...,r (10—97)

In order to proceed, we need to introduce the material properties. When thematerial is linearly elastic,

+ Ai'= + +

and the equations expand to

= d1 i,j = 1, 2,....rf,j = dx1 dx2 dx3 (10—98)

d, = — r$J1T(a° + A6°)dx1 dx2 dx3

One should note that (10—97) are weighted compatibility conditions. Thetrue stresses must satisfy both equilibrium and compatibility throughout the




domain. We call the corrective stress field since it is required to correct thecompatibility error due to

For completeness, we describe here how one establishes a variational principlefor Our starting point is (10—94) restricted to elastic behavior. We define

= according to

c5V* = = (10—99)

and call V* the complementary energy density. The form of V* for a linearlyelastic material is

= + (10—100)

By definition, V* complements V, i.e.,

V + = (10—101)Then, letting

= cjx1 dx2 dx3 (10—102)

we can write (10—94) as

(511. 0 for arbitrary* (10—103)

TiC — $$ =

This form is called the principle of stationary complementary energy and showsthat the true stresses correspond to a stationary value of

Since is linear in the second variation of reduces to(52fl = = dx1 dx2 dx3 (10—104)

We shift over to matrix notation and express öe as

= A, (10— 105)

where represents the tangent compliance matrix. Now, must be positivedefinite in order for the material to be stable.t Then, (5211. > 0 for arbitrary

and we see that the solution actually corresponds to a relative minimumvalue of

The approximate method described earlier can be applied to 11g. Substitutingfor given by (10—95) converts to a function of the stress parameters(a1. a2, . . . , ar). When the material is linearly elastic,

H, = — ard + const (10—106)

The equations for the stress parameters follow by requiring H. to be stationaryfor arbitrary

(511, = — ci) = 0

(10—107)fa=d

The classical stability criterion specialized for elastic material and linear geometry requiresSCTD, & > 0 for arbitrary Sc which, in turn, requires D, to be positive definite. Since

A, D1', it follows that A must be positive definite for a stable material.



Operating onc52fl = LtaTfLui (10—108)

and noting that ö211. > 0, we conclude that f is positive definite.

REFERENCES

I. CRANDALL, S. J., and N. C. DAHL: An Introduction to the Mechanics of Solids,McGraw-Hill, New York, 1959.

2. BISPLINGHOFF, R. L., MAR., J. W., and T. H. H. PlAN: of Deformable Solids.Addison-Wesley, Reading, Mass., 1965.

3. WANG, C. T.: Applied Elasticity, McGraw-I-jill, New York, 1953.4. TIMOSHENKO, S. J., and J. N. GooDiag: Theory of Elasticity, 3d ed., McGraw-Hill,

New York, 1970.5. SOKOLNIKOFF, I. S.: Mathematical Theory of Elasticity, 2d ed. • McGraw-Hill. New

York, 1956.6. FUNG, Y. C.: Foundations of Solid Mechanics, Prentice-Hall, 1965.7. LEKIINITSKU, S. G. Theory ofElasticity of an Anisotropic Elastic Body, Holden-Day,

San Francisco, 1963.8. WAsmzu, K. Variational Methods in Elasticity and Plasticity, Pergamon Press, 1968.9. HLDEBRAND, F. B.: Methods of Applied Mathematics, Prentice-Hall, 1965.

10. CRANDALL, S. J.: Engineering Analysis, McGraw-Hill. New York, 1956.

PROBLEMS

10—1. Write out the expanded form of the following products. Considerthe repeated indices to range from 1 to 2.

(a)(b) + u1, where =

± Urn, + Urn. k) —10—2. Let f be a continuous function of x1, x2, x3. Establish the trans-

formation laws for and (3Xk.

10—3. Establish the transformation law for Jbk where are cartesiantensors.

10—4. Prove thateJk = — ôJk)

is a second-order cartesian tensor. Hint: Expand(3/3 (If)

P. P.

10—5. Equations (10—19) are the strain transformation laws. Since isa symmetrical second-order cartesian tensor, there exists a particular set ofdirections, say Xi', for which is a diagonal array. What are the strain com-ponents for the frame? Consider a rectangular parallelepiped having sidesdXy in the undeformed state. What is its deformed shape and relative changein volume, with respect to its initial volume? Specialize the expression for

for small strain. Then determine for the initial (Xi) directions and smallstrain. Finally, show that r.., is invariant.


10—6.

(a)

PROBLEMS 263

Specialize (10—19) for small strain and write out the expressions forin terms of ei, 62, . . P13•

(b) Let = P12, P23, y31}. We can express the strain trans-formation (small strain) as

=

Develop the form of using the results of part a.(c) Evaluate TE in terms of cos 0, sin 0 for the rotation shown below.

Comment on the transformation law for the out-of-plane shear strainsP32.

Prob. 10—6

10—7. Tn the Eulerian approach, the cartesian coordinates for thedeformed state are taken to be the independent variables, i.e.,

Ui = Xj(f/k)

Almansi's strain tensor is defined as

— (ds)2 = 2Efk thik

Determine the expression for EJk in terms of the displacements. Compare theresult with (10—21).

10—8. Consider the case of two-dimensional deformation in the X1-X2plane (83 = P13 = P23 = 0). Let 6b, be the extensions in the a, b, c direc-tions defined below and let 6N = {8a, 6b, We can write

= BE

C =

(a) Determine the general form of B.(b) Determine for = 0, 9b = 90c.

(c) Determine B1 for Oa = 0, 6h = 60°, = 120°.(d) Extend (a) to the three-dimensional case. Consider six directions

having direction cosines GJ2, with respect to X1, X2, X3. Canwe select the six directions arbitrarily?

x2



Prob. 10—8xz

10—9. For small strain, the volumetric strain is

= + + C3 = eti + &22 +

Rather than work with one can express it as the sum of two tensors,

= +

where is called the spherical strain tensor,

=

and is the deviator strain tensor.(a) Write out the expanded form for and(b) Determine the first invariant of and compare with the invariant

of ejj.10—10. This question concerns strain compatibility equations.(a) Show that

+ = +CX,, cXj, — 8X,,, (?X,,

where1 (CII,,

eflk = ek, = — +2 \CXk

and k, in, n range from I to 3. This expression leads to six indepen-dent conditions, called geometric compatibility relations, on the strainmeasures.

(b) Show that for two-dimensional deformation in the X1-X2 plane= 813 = = 0; this called plane strain) there is only one com-

patibility equation, and it has the following form:

22 + 83 11 = Y12. 12

Is the following strain state permissible?

= +82 = kx2

Y12 = 2kx1x2

k = constant

-a


PROBLEMS 265

10—11. Equation (10—21) defines the strain measures due to displacements,To analyze geometrically nonlinear behavior, one can employ an incre-

mental formulation. Let represent the displacement increment and Ae1kthe incremental strain. We write

= +

where contains linear terms (Aug) and öeJk involves quadratic terms. The5-symbol denotes the first-order change in a functional and is called the varia-tional operator (see Ref. 8). We refer to 5e as the first variation of e. Determinethe expressions for

10—12. Let i,, be the unit vector defining the initial orientation of thedifferential line element d1,, at a point.

= dsi,, 1, =

The unit vector defining the orientation in the deformed state is

= (1 + =

Determine the general expression for Then specialize it for small strain.10—13. The several parts of this question concerns stress transformation.(a) Starting with (i0--41). write out the expressions for in terms of

all, a22, . . ,

(b) Let = a22, a33, ai2, = stress matrix. We express thestress transformation as a matrix product.

a' = T,a

Develop the form of T,, using the results of part a.(c) Evaluate 1',, in terms of cos (9, sin (9 for the axes shown.

(d) Plane stress refers to the case where a13 = a23with reduced stress and strain matrices,

Er {a11, a22,

=

and write the transformations in the same form as the three-dimensionalcase:

a' =a' =

x2

x,t

Prob. 1O—13

xl

= a33 = 0. We work



Evaluate T,. from part c above and T, from Prob. 10—6. Verify that

13

10—14. This question develops a procedure for generating self-equilibratingstress fields.

(a) Expand the linear equilibrium equations, (10—49) and (10—50).(b) Specialize the equilibrium equations for plane stress (a13 = a23 =

a33 = 0).(c) Suppose we express the two-dimensional stress components in terms

of a function = as follows:

a11 = t1'.22 — b1 dx1

a22 = tI'. ii — fx7 b2 dx2

a12 = —1//,12

The notation for body and surface forces is defined in the followingsketch.

Prob. 10—14

x2

Verify that this definition satisfies the equilibrium equations in theinterior. Show that the expressions for and P2 ifl terms of derivativeswith respect to x1, x2, and s are

Pi — b1 dx1

P2 = T t/"1 — b2 dx2

10—15. The mean stress, a,,,, is defined as

am = + a22 + a33)

Rather than work with we can express it as the sum of two tensors,

— L— aU

x1


PROBLEMS 267

where is called the spherical stress tensor,

= óijOrn

and is the deviator stress tensor.(a) Write out the expanded forms for and(b) Determine the first invariant of

10-46. Establish the stress-equilibrium equations for small-finite rotationand small strain.

10—17. Starting with (10—52), (10—55) specialized for small strain, establishthe incremental equilibrium equations in terms of Au, Ab*, andGroup according to linear and quadratic terms. Specialize these equationsfor the case where the initial position is geometrically linear, i.e., where canapproximate with in the incremental equations.

10—18. Prove (10—60). Hint:

= + P.k/ =

10—19. Verify that the inverted form of(l0—71) is

D(e —

where

D11 = E1/C3 D12 D13 C4D11

D22 = E2/C1 + (C2/C1)D12D23 = v32E2/C1 + (C2/C1)D13D31 = E3 + v31D13 + v32D23

and

C1 = 1 —

C2 = v21 + v31v32(E2/E3)

E2C1

C4 = v31 + "32

Specialize for plane strain = = = 0)10—20. Consider 2 sets of orthogonal directions defined by the unit vectorsand The stress-strain relations for the two frames are

= +(a°)' + A'&

Express A' in terms of A and Also determine D'.1O--21. Consider the three-dimensional stress-strain relations defined by

(10—71).(a) Specialize for plane stress = = = 0).


268 GOVERMNG EQUATIONS FOR A DEFORMABLE SOUD CHAP. 10

(b) Leta22, cri2}

C = 62, Y12}

C =

Verify that D has the following form:

V2t 0

1

G(1 —

n =E2

Assuming X1-X2 in the sketch are material symmetry directions,determine D' for the X'1-X'2 frame. Use the results of Prob. 10—13,10—20. What relations between the properties are required in orderfor D' to be identical to D?

x2

xI

Prob. 10—21

10—22. Verify (10—73). Start by requiring equal properties for the X2 andX3 directions. Then introduce a rotation about the X1 axis and consider theexpression for Isotropy in the X2-X3 plane requires

I.

Y23 = 7023

10—23. Verify that the directions of principal stress and strain coincide foran isotropic material. Is this also true for an orthotropic material?

10—24. Equations (10—76) can be written as

a11 = a°&1 + + 2Ge11

where is the volumetric strain. Using the notation introduced in Probs. 10—9and 10—15——

(a) Show that= Ka,, + a0

I


PROBLEMS 269

where K is the bulk modulus = (E/3(1 — 2v)). Discuss the case where

(b) Show that=

(c) Verify that the strain-energy density can be written as

V —

= — += +

Determine and for the isotropic case.(d) When v = = We must work with 7 stress measures ('u' Urn)

and the mean stress has to be determined from an equilibrium con-sideration. Summarize the governing equations for the incompressiblecase.

.10—25. Prove (l0--81) for the two-dimensional case. Is this formula re-stricted to a specific direction of integration on the boundary? Does it applyfor a multi-connected region, such as shown in the figure below?

Prob. 10—25

10—26. Verify Equation (10—89).10—27. Refer to Example 10—3. Express (g) in terms of Consider the

material to be orthotropic.10—28. Verify that the stationary requirement

= 0 for arbitrarywhere

= — — dx2— — —

= Kirchhoff stress= Lagrange strain = + + 1u,,,,

= complementary energy density (initial volume)= prescribed force measures (initial dimensions)

leads to the complete set of, governing equations for an elastic solid, i.e.,1. stress equilibrium equations2. stress-displacement relations3. stress boundary conditions on4. displacement boundary conditions on5. expressions for the reaction surface forces on



This variational statement is called Reissner's principle (see Ref. 8).(a) Transform HR to by requiring the stresses to satisfy the stress

displacement relations. Hint: Note (10—101).(b) Transform 11R to — by restricting the geometry to be linear =

and (ui, + and requiring the stresses to satisfy the stressequilibrium equations and stress boundary conditions on Hint:Integrate by parts, using (10—8 1).

10—29. Interpret (10—90) as

dQ ==

where PQ is a force applied at Q in the direction of the displacement measure, dQ.


11

St. Venant Theory ofof

Prismatic Members11—1. INTRODUCTION AND NOTATION

A body whose cross-sectional dimensions are small in comparison with itsaxial dimension is called a member. If the centroidal axis is straight and theshape and orientation of the normal cross section are the memberis said to be prismatic. We define the member geometry with respect to aglobal reference frame (X1, X2, X3), as shown in Fig. 11—1. The X1 axis istaken to coincide with the centroidal axis and X2, X3 are taken as the principalinertia directions. We employ the following notation for the cross-sectionalproperties:

A = if dx2 dx3 = dA

12 — Sj(x3)2 dA

13 = fl(x2)2 dA

Since X2, X3 pass through the centroid and are principal inertia directions,the centroidal coordinates and product of inertia vanish:

'23 jJx2x3 dA = 0

One can work with an arbitrary orientation of the reference axes, but this willcomplicate the derivation.

St. Venant's theory of torsion-flexure is restricted to linear behavior. It is anexact linear formulation for a prismatic member subjected to a prescribed

t The case where the cross-sectiona' shape is constant but the orientation varies along thecentroidal axis is treated in Chapter 15.

271


272 TORSION-FLEXURE OF PRISMATIC MEMBERS CHAP. it

distribution of surface forces applied on the end cross sections. Later, inChapter 13, we modify the St. Venant theory to account for displacementrestraint at the ends and for geometric nonlinearity.

Fig. 11—i. Notation for prismatic member.

The distribution of surface forces on a cross section is specified in terms ofits statically equivalent force system at the centroid. Figure 11—1 shows theStress components on a positive face. We define M.. as the force andmoment vectors acting at the centroid which are statically equivalent to thedistribution of stresses over the section. The components of F.., M÷ are calledstress resultants and stress couples, respectively, and their definition equationsare

F1 = ffcrij c/A F2 c/A F3 = JJcc13 c/A

M1 = JJ(x2cr13 — x3c12)dA

M2 = JJx3crj1 dA

M3 = dA

The internal force and moment vectors acting on the negative face are denotedbyF_,M_. Since

F_ = —F÷ M_ = (11—4)

it follows that the positive sense of the stress resultants and couples for thenegative face is opposite to that shown in Fig. 11 —1.

We discuss next the pure-torsion case, i.e., where the end forces are staticallyequivalent to only M1. We then extend the formulation to account for flexure

x2

F3

(11—3)


SEC. 11—2. THE PURE-TORSION PROBLEM 273

and treat torsional-flexural coupling. Finally, we describe an approximateprocedure for determining the flexural shear stress distribution in thin-walledsections.

11—2. THE PURE-TORSION PROBLEM

Consider the prismatic member shown in Fig. 11—2. There are no boundaryforces acting on the cylindrical surface. The boundary forces acting on the endcross sections arc statically equivalent to just a twisting moment M1. Also,there is no restraint with respect to axial (out-of-plane) displacement at the ends.The analysis of this member presents the pure-torsion problem. In whatfollows, we establish the governing equations for pure torsion, using theapproach originally suggested by St. Venant.

Rather than attempt to solve the three-dimensional problem directly, weimpose the following conditions on the behavior and then determine whatproblem these conditions correspond to.

1. Each cross section is rigid with respect to deformation in its plane,i.e., = = 723 = 0.

2. Each cross section experiences a rotation w5 about the X3 axist andan out-of-plane displacement u1.

These conditions lead to the following expansions for the in-place displace-ments:

112 = —C01X3

03 +W3.3.2

The corresponding linear strains are

0

Fig. 11—2. Prismatic member in pure torsion.

(11—5)

13 = a3 = Y23 0

Li —Ui 1(11—6)

712 U12 + 012 —

713 = U5, + U3, = 05,3 + X20)1, 1

t Problem 11—i treats the general case where the cross section rotates about an arbitrary point.


274 TORSION-FLEXURE OF PRISMATIC MEMBERS CHAP. 11

Now, the strains must be independent of x1 since each cross section is

subjected to the same moment. This requires

= const = k1

= u1(x2, x3)

We consider the left end to be fixed with respect to rotation and expressco1,u1aS

= k1x1(11—8)

U.1 =

where = x3) defines the out-of-plane displacement (warping) of a crosssection. The strains and stresses corresponding to this postulated displacementbehavior are

Cl = = C3 = 0

712 = x3)

713 + x2)and

a11 = a22 = a33 O'23 = 0

a12 = Gy12 = 2 — x3) = a12(x2, x3) (11—10)

U13 = Go'13 + x2) a1 3(x2, x3)

We are assuming that the material is and there are no initial strains.One step remains, namely, to satisfy the stress-equilibrium equations and

stress boundary conditions on the cylindrical surface. The complete system oflinear stress-equilibrium equations, (10—49), reduces to

U21,2 + = 0 (11—11)

Substituting for the shearing stresses and noting that Gk1 is constant lead tothe differential equation

(11-12)

which must be satisfied at all points in the cross section.The exterior normal n for the cylindrical surface is perpendicular to the X1

direction. Then 0, and the stress boundary conditions, (10—49), reduce to

Pfli = + = 0 (11—13)

Using (11—10), the boundary condition for is

2 — x3) + + x2) = 0

(11—14)

— (on S)

t Problem 11 —3 treats the orthotropic case.



The pure-torsion problem involves solving V2q5, = 0 subject to (11—14). Onceçb, is known, we determine the distribution of transverse shearing stresses from(11—10). Note that depends oniy on the shape of the cross section.

The shearing stress distribution must lead to no shearing stress resultants:

F2 = dA = 0

F3 = J$a13 dA 0This requires

J J OX2 JJ ('X3

To proceed further, we need certain integration formulas. We start with

if dA = (IS

which is just a special case of (10—81). Applying (11—15) to dA leads toGreen's theorem,

JJV2VJdA +0X2

11—16

.1 ôn

If is a harmonic function (i.e., = 0), Green's theorem requires

dS = 0

Now, /, is a harmonic function. For the formulation to he consistent, (11—14)must satisfy (c). Usiiig (11—15), (c) transforms to

#(XH2x3— = 0

Since is specified on the boundary, we cannot apply (11—15) directlyto (b). In this case, we use the fact that = 0 and write

(j=2,3)cxi ax2 ax2j ox3 \ ax31

Integrating (e),

(j=2,3)

and then substituting for the normal derivative, verifies (h).The constant k1 is determined from the remaining boundary condition,

= J$(x2c13 — x3c12)dA (11—17)



We substitute for the shearing stresses and write the result as

Gk1J

where J is a cross-sectional property,

= if ++ — X3 dA

At this point, we summarize the results for the pure-torsion problem.

1. Displacements

=02

U3 = W1X2

= k1x1

k1 = (if

2. Stresses

M3

J \(;X2(11—20)

A'110j3 = —H + X2

J

3. Governing Equations

mA:on S: —

It is possible to obtain the exact solution for for simple cross sections.The procedure outlined above is basically a displacement method. One

can also use a force approach for this problem. We start by expressing theshearing stresses in terms of a stress function so that the stress-equilibriumequation (Equation 11—li) is identically satisfied. An appropriate definition is

012X3

(11—21)

013

The shearing stresses for the 2, v directions, shown in Fig. 11—3, follow directlyfrom the definition equation

01Acv

(11—22)

0lv = —CA.


SEC. 11-2. THE PURE-TORSION PROBLEM 277

Taking S 900 counterclockwise from the exterior normal direction, and notingthat the stress boundary condition is = 0, lead to the boundary conditionfort/i,

= const on S (11—23)

We establish the differential equation for t/i by requiring the warping functionbe continuous. First, we equate the expressions for a in terms of t/i and

M1a12 = — x3)

a13 = = + x2)

Now, for continuity,=

Operating on (a), we obtain

=

It is convenient to express t/.' as

(11—24)

The governing equations in terms of aret

M1 dt/a12 =

a

=(11--25)

13 jand

= —2 (mA)tJi = (on boundary S1) (11—26)

Substituting (11—25) in the definition equation forM1 leads to the followingexpression for J:

cr7—

JJ \. CX3J

Applying (10—8 1) to (a) and that

—dS = A1 = area enclosed by the interior boundary curve, S1 (b)

= C1 = const

t Equations (11—26) can be interpreted as the governing equations for an initially stretchedmembrane subjected to normal pressure. This interpretation is called the "membraneSee Ref. 3.

The S direction is always taken such that n — S has the same sense as X2 — X3. Then, the+ S direction for an interior boundary is opposite to the + S direction for an exterior boundarysince the direction for n is reversed. This is the reason for the negative sign on the boundary integral.



we can writeJ = dA + (11—27)

where = 0 on the exterior boundary.To determine the constants C, for the multiply connected case, we use the

fact that is continuous. This requires

dS 0Js ('IS

for an arbitrary closed curve in the cross section,

x2

Fig. 11—4. Graphical representation of sector area.

(11—28)

x

x

0

Fig. 11—3. Definition of n-s and A.-vdirections.

x3



Consider the closed curve shown in Fig. 11 —4. The shearing strain is

given byYis = ct52y12 + 0t53y13

Using (11—9), we can write (a) as

Yis = 2 + 3 — xacls2 +

= k1 + (11-29)

where p is the projection of the radius vector on the outward normal.t Themagnitude of p is equal to the perpendicular distance from the origin to thetangent. Integrating between points P, Q, we obtain

= — + 2APQ) (11—30)

wherer50

APQ =J

p dS = sector area enclosed by the arc PQ and theradius vectors to P and Q.

Finally, taking P =5dS = 2k1A5 (11—31)

where A5 denotes the area enclosed by the curve. Since = we canwrite

2Gk1A5 = (11-32)

Note that the +S direction for (11—32) is from .X2 toward X3. Also, this resultis independent of the location of the origin.

Instead of using (11—9), we could have started with the fact that the crosssection rotates about the centroid. The displacement in the + S direction fol-lows from Fig. 11—4:1

u,5 = x is) = w1p k1x1p (11—33)

Substituting for inYss = Us + s (11—34)

and noting that Ut = lead to (11—29).Using (11—22), we can write

M5==

(11—35)

t This interpretation of p is valid only when S is directed from X2 to X3, i.e., counterclockwisefor this case.

See Prob. 11—14 for an alternate derivation.§ This development applies for arbitrary choice of the +S direction. The sign of p is positive

if a rotation about X1 produces a translation in the +S direction. Equation (11—29) is used todetermine the warping distribution once the shearing stress distribution is known. See Prob. 11—4.



Then, substituting for in (11—32), we obtain

= (11—36).3 s

where n is the outward normal, A5 is the area enclosed by S, and the + S senseis from X2 to X3. This result is valid for an arbitrary closed curve in the crosssection. We employ (11—36) to determine the values of 17 at the interior bound-aries of a multiply connected cross section.

It is of interest to determine the energy functions associated with pure tor-sion. When the material is linearly elastic and there are no initial strains, thestrain and complementary energy densities are equal, i.e., V =We let

V V dA strain energy per unit length (11—37)

The strain energy density is given by

V = +

Substituting for Y12' '/13,

2 2V = X3) + ( + x2)j

and integrating (b) over the cross section, we obtain

V = (11—38)

Since = V, and M1 = GJk1, it follows that

= + (11—39)

WI xldx1

Fig. 11—5. Differential element for determination of the rotational work.

Instead of integrating the strain-energy density, we could have determinedthe work done by the moments acting on a differential element. Consider theelement shown in Fig. 11—5. The boundary forces acting on a face are staticallyequivalent to just a torsional moment. Also, the cross sections are rigid in


SEC. 11—S. THIN-WALLED OPEN CROSS SECTIONS 281

their plane and rotate about X1. The relative rotation of the faces is

/ dw1 '\((01 + —dx1 — = dx1dx1 ,i

and the first-order workdone by the external forces due to an increment in wjreduces to

Now,5WE = M1 ,Xk1 dx1

= = 5jJ dx2dx3 = óVdx1

for an elastic body. Then, expanding ö V.

and it follows that

= M1 = GJk1dk1

V =

11—3. APPROXIMATE SOLUTION OF THE TORSION PROBLEM FORTHIN-WALLED OPEN CROSS SECTIONS

We consider first the rectangular cross section shown in Fig. 11—6. The exactsolution for this problem is contained in numerous texts (e.g., see Art. 5—3 ofRef. 1) and thcrefore we will only summarize the results obtained.

d,

H-FigS 11—6. Notation for rectangular section.

x3

Idl 2

2



When t d, the maximum shearing stress occurs at x2 = ± t/2, x3 = 0

(points 5, 6). The exact expressions are

dt3J = K1—(11-41)

= K2t

where192 (t'\ I

K1 = 1 —

=

tanh

8 1 1

K2 = 1— (2n+1)2 cosh A,,

2n+1 Id

Values of K1, K2 for d/t ranging from 1 to 10 are tabulated below:

d/t K1 K2

1 0.422 0.6752 .687 .9303 .789 .9854 .843 .9975 .873 .999

10 0.936 1.000

If t d, we say the cross section is thin. The approximate solution for athin rectangle is

J 4dt3

(113 2—-—x2 = 2Gk1x2

(11-42)x2x3(t)2

(We take d/t = in the exact solution.) The shearing stress varies linearlyacross the thickness and

M1 3M1

A view of the warped cross section is shown in Fig. 11—7.Since the stress function approach is quite convenient for the analysis of

thin-walled cross sections, we illustrate its application to a thin rectangular


SEC. 11—3. THIN-WALLED OPEN CROSS SECTIONS 283

cross section. Later, we shall extend the results obtained for this case to anarbitrary thin walled open cross section. The governing equations for a simplyconnected cross section are summarized below for convenience (see (11—26),(11—27)):

= —2 (in A)

0 (on the boundary)

=

J = (1A

where the S direction is 900 counterclockwise from the is direction.t Since t issmall and a12, the shearing Stress component in the thickness direction, must

Fig. 11—7. Warping function for a rectangular cross section.

vanish on the boundary faces, it is reasonable to assume = 0 at all pointsin the cross section. This corresponds to taking independent of x3. Theequations reduce to

d2= —2

Solving (b), we obtain -

J=

dx2 =

M1= ——---—-- = 2——x2J

t This applies for X3 counterclockwise from X2. The general requirement is the n — S sensemust coincide with the X2-X3 sense.

_:k,1,



The expression for developed above must be corrected near the ends

(x3 ± d/2) since it does not satisfy the boundary condition,

ti

This will lead to a12 0 near the ends, but will have a negligible effect on J

and Actually, the moment due to the approximate linear expansion for

is equal to only one half the applied moment:

I't/2 P4 /1d x2a13 dx, = dt3) =

J

The corrective stress system (a12) carries M1/2. This is reasonable since, even

though is small in comparison to amax, its moment arm is large.

We consider next the arbitrary thin-walled open cross section shown inFig. 11 —8. The S curve defines the centerline (bisects the thickness) and the ndirection is normal to S. We assume = 0 and take = —n2 + t2/4. Thiscorresponds to using the solution for the thin rectangle and is reasonable whenS is a smooth curve. The resulting expressions for I and are

J = 4 t3 dS

(11—43)

M1a15, ma, = = Gkitrnax

-f-s

t(s)

Fig. 11—8. Notation for thin-walled open cross section.


SEC. 11—3. THIN-WALLED OPEN CROSS SECTIONS 285

The results for a single thin rectangle are also applied to a cross sectionconsisting of thin rectangular elements. Let t1 denote the length and thicknessof element i. We take J as

J = (11—44)

Asan illustration, consider the symmetrical section shown in Fig. 11—9. Apply-ing (11—44), we obtain

3 ..i 31'ff + w4v

The maximum shearing stress in the center zone of an element is taken as

M1= —7t1 = Gk1t1 (11—45)

In general, there is a stress concentration at a reentrant corner (e.g., point A inFig. 11—9) which depends on the ratio of fillet radius to thickness. For the case

bi

+(Iw

IFig. 11—9. Symmetrical wide-flange section.

of an angle having equal flange thicknesses, the formulat

= + (1146)\ 4rf/

where is the fillet radius and 0rn is given by (14—45), gives good results forrf/t < 0.3. The stress increase can be significant for small values of rf/t. Forexample, for Tf = 0,lt. Numerical procedures such as finitedifferences or the finite element must be resorted to in order to obtainexact solutions for irregular sections. -

t See Ref. 2 and Appendix of Ref. 9.See Ref. 4.



11—4. APPROXIMATE SOLUTION OF THE TORSION PROBLEM FORTHIN-WALLED CLOSED CROSS SECTIONS

The stress function method is generally used to analyze thin-walled closedcross sections. For convenience, the governing equations are summarized below(see (11—26), (11—27), (11—36)):

—2 (in A)

(on the exterior boundary)— ci

J — dA +

—

£ dS = —2A5j on

(on the interior boundary, S,)= area enclosed by

S

and +S sense from X2 toward X3,)

We consider first the single cell shown in Fig. 11—10. The curve definesthe centerline. Since there is an interior boundary, we have to add a term

Fig. 11—10. Single closed cell.

involving C1 to the approximate expression for used for the open section.We take as

+

2n\tz (11—47)

where represents the contribution of the interior boundary. This expression

S.

n

E

Sect. E-E


SEC. 11—4. THIN-WALLED CLOSED CROSS SECTIONS 287

satisfies the one-dimensional compatibility equation and boundary conditions,

2

atn— +t/2= C1 at n = — t/2

(a)

and is a reasonable approximation when S is a smooth curve.Differentiating

b(fl

and substituting (b) in the expressions for the shearing stress components lead to0

M1 / C1\+ 7) (11—48)

cr?5 +The tangential shearing stress varies linearly over the thickness and its averagevalue is We let q be the shear stress resultant per unit length along S,positive when pointing in the + S direction,

1/2

q= J

cr15 (11—49)—1/2

and call q the shear flow. Substituting for a we find

q (11—50)

The additional shearing stress due to the interior boundary (i.e., closed cell)corresponds to a constant shear flow around the cell. One can readily verifytthat the distribution, q = const, is statically equivalent to only a torsionalmoment, given by

= (11—51)

The torsional constant is determined from

J dA + 2C1A1 M1/Gk1 (a)

Substituting for using (11—47), we obtain

j = Jo +(11—52)

4 t3 dS =

Equation (a) was established by substituting for the shearing stresses in termsin the definition equation forM1 and then transforming the integrand. We

could have arrived at (11—52) by first expressing the total torsional moment as

M1 = + (11—53)

See Prob. 11-5.


288 TORSION-FLEXURE OF PRfSMATIC MEMBERS CHAP. 11

where MI is the open section contribution and is due to the closure. Next,we write

M1 = GkIJ it'll = Gk1J° = Gk1J' (11—54)Then,

J = J0 + JC (11—55)

and it follows thatJo Jc

(11-56)

Finally, using (11—5 1), we can express JC as

JC = Mu/(M1/J) = (11—57)

This result shows that we should work with a modified shear flow,

C q/(M1/J) (11—58)

rather than with the actual shear flow. Note that C C1 for the single cell.It remains to determine C1 by enforcing continuity of the warping function

on the centerline curve. Applying (11—32) to we have

= (11—59)

Substituting forM1C1

q/t=

leads

(1160)

One should note that C1 is a property of the cross section. Once C1 is known,we can evaluate .J from (11—52) and the shearing stress from

M1 (+ -i—,)

(11—61)

Example 11—1

Consider the rectangular section shown. The thickness is constant and a, b are centerlinedimensions. The various properties are

CdS 2(a+h)

Cl =

t See Prob. 11—6.


SEC. 11-4. THIN-WALLED CLOSED CROSS SECTIONS 289

We express J as

= +

For this section,J° 1 (r'Y /

We consider a > b. Then,Jo (t'Y

= 01—J' \\h

The section is said to be thin-walled when c< b. In this case, it is reasonable to neglectJo vs.

Fig. Eli—i

H- b +s, q I

r+tbThe strcss follows from (11—61),

M1C1/ t2'\ (= —-———ii ± —-I = I ± —s—J t\, C11

where, for this section,

t2 / h'\t (t

If the section is thin-walled, we can neglect the contribution of i.e., we can take

M1= q/t =

We consider next the section shown in Fig. 11—li. Rather than work withit is more convenient to work with the shear flows for the segments. We

number the closed cells consecutively and take the + S sense to coincide withthe X2-X3 sense. The +S sense for the open segments is arbitrary. We defineq3 as the shear flow for çellj and write

(11—62)

Note that is the value of on the interior boundary of cell j and the shearflow is constant along a segment. The total shear flow distribution is obtained



Fig. 11—11. Cross section consisting of closed cellsand open segments; A, and A,, are centerline areas.

by superimposing the individual cell flows. Then, the shear flow in the segmentcommon to cells i andj is the difference between qj and q1. The sign dependson the sense of S.

q = — q2 = — C2) for S1

q=q2—q1 forS2(11—63)

The shearing stress is assumed to vary linearly ovcr the thickness. For con-venience, we drop the subscripts on and write the limiting values as

wherecr = ±a° + Cr"

M1 /Cnetcr=—1-t

It remains to determine C1, C2, and J.We have shown (see (11—55)) that

and

We determine J° from

J = Jo + Jc

=

segments

(11—64)

(a)

(h)

(11—65)

x3 q2. S2



Substituting for

MI = 2qjA1 + 2q2A2

+ C2A2)

in (b) leads to- + A2C2) (11—66)

The constants are obtained by enforcing continuity of on the centerlineof each cell. This can also be interpreted as requiring each cell to have the sametwist deformation, k14

I = 1,2 (11-67)

Substituting for q in terms of C and letting

C dS C dS Cads= a22 = = = — I

JS, t .J52 Jc t

where a12 involves the segment common to cells 1, 2, the continuity equationstake the following form:

+ a12C2 =(11—69)

a12C1 + a22C2 = 2A2

We solve this system of equations for C2, then determine f with (11—66),and finally evaluate the stresses with (11—64).

We can represent the governing equations in compact form by introducingmatrix notation. The form of the equations suggests that we define

c = A a[a11 a121 (11—70)

a22j

With this notation,JC = 2ATC

(11—7.!)aC 2A

Substituting for A in the expression forJC CTaC

and noting that JC is positive, we conclude that a must be positive definite,The complementary energy per unit length along the centroidal axis is defined

by (11—39),

11 2

We apply (11—51) to each cell.See (11—32).



Since ais varies linearly over the thickness, the open and closed stress dis-tributions are uncoupled, i.e., we can write

= + toqwhere

=2G

(11-72)aq)

—i

It is reasonable to neglect the open contribution when the section is thin-walled.

Examp'e 11—2

The open-section torsional constant for the section shown is

= ± 2(b + d + + htfl (a)

Applying (11 —68) to this section, we obtain

= hd

A2=hb

a11 = 1(h + 2d) +11 t2

6012 =

a22 = + 2b) +tl t2

and the following equations for C1. C2 and i.

+ + 2—

C2 = 2 dt1

—

c1 + + ± 2 C2 = 2bt1

J = Jo + Jo

Finally, the shear stress intensities in the various segments are

M1 (C1

= (k— +

M1 /C1 — C2+t2

J t2 /

M1 (C2

= 7 + t1

M1= t3


2bt3Cl = C2 = —s

1 + 2-13

and the section functions as a single cell with respect to shear flow.

Fig. 11—12. Prismatic member in shear loading.

on the cross section at x1 = L is statically equivalent to a single force P212,acting at the centroid. Also, the end cross sections are not restrained againstwarping, i.e., displacement. In what follows, we describe St.Venant's torsion-flexure formulation for this problem. Later, in Chapter 13,we shall modify the theory to include restraint against warping.

SEC. 11—5. TORSION-FLEXURE WITH UNRESTRAINED WARPING 293

____

Fig. E11—2t3 tl

03

1Ii I

I032 A2 h

X3

M1

When d = b.

a'

11—5. TORSION-FLEXIJRE WITH UNRESTRAINED WARPING

Consider the prismatic member shown in Fig. 11—12. There are no boundaryforces acting on the cylindrical surface. The distribution of boundary forces

x2

xI

x2

_____________

if—_____

I +S

P2


294 TORSION-FLEXURE OF PRISMATIC MEMBERS CHAP. il

We start by postulating expansions for the stresses. The stress resultants andcouples required for equilibrium at x1 are

=A4,r0F2 = P2

M3 = P2(L — x1)

Introducing (a) in the definition equations for the stress resultants and couplesleads to the following conditions on the stresses:

dA = JJx3a13 c/A = 0

dA = P2(L — x1)

Jj712 c/A = P2

$fcrj3 c/A = 0

S$(x2a13 — x3cr12)dA —, 0

The expansion,M3 P2

= = —13 13

satisfies the first three conditions (i.e., F1, M3) identically since

JJx2 c/A = jJx2x3 dA = 0

dA = 13

The last three conditions (i.c., F2, F3, M1) require a12, a13 to be independentof x1. This suggests that we consider the following postulated stress behavior:

P2cru = ——--—x2 = — x1)x2

13 13

a1 2 a1 2(x2, x3) (11 —73)

a13 = a13(x2, X3)

a22 a33 0

Introducing (11—73) in the stress-equilibrium equations and stress boundaryconditions for the cylindrical surface leads to

P2a21,2 + a31,3 + 0 (mA)

13 (11—74)

2a21 + = 0 (on S)

At this point, we can either introduce a stress function or express (11—74) interms of a warping function. We will describe the latter approach first.

The displacements can be found by integrating the stress-displacement rela-tions. We suppose the material is linearly elastic, isotropic with respect to theX2-X3 plane, and orthotropic with respect to the axial direction. This is aconvenient way of keeping track of the coupling between axial and in-plane


SEC. 11—5. TORSION-FLEXIJRE WITH UNRESTRAINED WARPING 295

deformation. Substituting for the stresses in (10—74), we obtain

I P2= u1 = (L — xj)x2E1 E113

V1 V1P2= u2,2 = = —(L —. x1)x2

El3v1 v1P263 = u33 = = —7—(L — xj)x2

LI3

I'Y12 = u1, 2 + u2, = = function ofx2, x3

1

Y13 U1 3 + U3, 1 = = function ofx2, x3

Y23 = U2,3 + U3,2 0

Integrating the first three equations leads to

1u1 = (Lx1 — + f1(x2, x3)

E1 13

v1P2

=(L — + .f2(xj, x3) (b)

v1P2u3 = ——- (L — x1)x2x3 + x2)

El3

The functions f1, f2, f3 are determined by substituting (b) in the last threeequations. We omit the details and just list the resulting expressions, whichinvolve seven constants:

= C1 + C5x2 + C6x3 + x3)

f2 = C2 — C5x1 + C4x3 — k1x1x3

v1P2+

2E(c)

— Xi. —--7-(L —,iJ 3

= C3 — C5x1 — C4x2 + k1x1x2

The constants C1, C2, ..., C6 are associated with rigid body motion and k3is associated with the twist deformation.t

We consider the following displacement boundary conditions:

1. The origin is fixed:

u1=u2=u3==0 at(0,0,0)

2. A line element oh the centroidal axis at the origin is fixed:

= u3,1 = 0 at(0,0,O)

fSee Eq. (11—5).


296 TORSION-FLEXURE OF PRISMATIC MEMBERS CHAP. Ii

3. A line element on the X2 axis at the origin is fixed with respect to rotationin the X2-X3 plane:

u2,3 = 0 at (0, 0, 0)

These conditions correspond to the "fixed-end" case and are sufficient toeliminate the rigid body terms. The final displacement expressions are

u1 (Lxj — + 4)(x2, x3)

= — — +—

— (11—75)

Vj 2 -

— x1)x2x3 + k1x1x2El3

One step remains, namely, to satisfy the equilibrium equation and boundarycondition. The transverse shearing stresses are given by

1 v1P241,2 — k1x3 + —. x2)

2L13(11—76)

1 v1P2(7j3 4), + k1x2 — x2x3

LI3

Substituting for the stresses in (11—74), we obtain the following differentialequation and boundary condition for 4):

P2f'2v1 l\(mA)

1 77v1P2 C - -

+ 12) +The form of the above equations suggests that we express 4) as

4) = kjq5t+ — + + (11-78)

where is the warping function for pure torsion and 4)2. and 4)2d are harmonicfunctions which define the warping due to flexure. Substituting for 4) leadsto the following boundary conditions for 4)2. and

2

2

2 )+ (11—79)

One can show, by using (11—15), that

= 0cn

dS = 0



and therefore the formulation is consistent. Terms involving vj/E are due toin-plane deformation, i.e., deformation in the plane of the cross section, andsetting v1/E = 0 corresponds to assuming the cross section is rigid. Then,defines the flexural warping for a rigid cross section and represents thecorrection due to in-plane deformation.

The shearing stress is obtained by substituting for in (11—76). We writethe result as

01j = + 01j,r + (j = 2,3) (11—SO)

where crU, is the pure-torsion distribution and r, d are flexural distri-butions corresponding to and 42d:

=

2 22

P2x2x3)

The pure torsion distribution is statically equivalent to only a torsional mo-ment, = G1k1J. One can show thatt

dA P2 dA — 0(11—82)

J$a12,a dA = 0 dA 0

Note that the shear stress due to in-plane deformation does not contributeto P2.

The total torsional moment consists of a pure torsion term and two flexuralterms,

M1 = G1k1J +

S2r 2 + X24)2r 3)dA

S2d + X242d,3 X34)2a,2)dA

Since and depend only on the shape of the cross section, it follows thatand S2d are properties of the cross section. For convenience, we let

1X3 (1184)13\ £. J

and (11—83) reduces to= — (11—85)

Now, — is the statically equivalent torsional moment at the centroid due

tSeeProb. Il—lO.



to the fiexural shear stress distribution. Then, defines the location of theresultant of the flexural shear stress distribution with respect to the centroid.

The twist deformation is determined from

k1 + (11—86)

where M1 is the applied torsional moment with respect to the centroid. If P2is applied at the centroid, M1 = 0, and

k1 =

The cross section will twist unless = 0. Suppose P2 has an eccentricity e3.In this case (see Fig. 11—13), M1 = —e3P2, and

P2k1 — e3)

For Ilexure alone to occur, e3 must equal

Whether twist occurs depends on the relative eccentricity, e3 — Now, tofind x3, one must determine S2. and S2d. This involves solving two second-order partial differential equations. Exact solutions can be obtained for simplecross sections. In the section following, we present the exact solution for arectangular cross section. If the section is irregular, one must resort to suchnumerical procedures as finite differences to solve the equations. In Sec. 11—7,we describe an approximate procedure for determining the flexural shear stressdistribution in thin walled cross sections.

Suppose the cross section is symmetrical with respect to the X2 axis. Then,is an even function of x3 and is an odd function of x3. The form of the

boundary conditions (11—79) requires and to be even functions of x3

x3

Fig. 11—13. Notation for eccentric load.



for this case. Finally, it Ibilows thatt S2,. = 0 and S2d = 0. Generalizing thisresult, we can state:

The resultant of the shear stress distribution due to fiexure in thedirection passes through the centroid when is an axis of symmetryfor the cross section.

x3

We consider next the case where the member is subjected to P2, P3 andat the right end (see Fig. 11—14). The governing equations for the P3 loadingcan be obtained by transforming the equations for the P2 case according to

> X3 .-

U2 —* U3 U3—> —U2

13 (3 13 (3—---+-— —--*------ox2 Ox3 Ox3 Ox2

U12 U13 —a12

13 '* 12

Two additional flexural warping functions must be determined. The expres-sions defining the flexural shear stress distributions due to P3 are

P3cr12 r -r 413r. 2

12

3r ((/33r, 3

'2 (11—87).v1G1 P3

d = 2 x2x31!L. 12

Vj i= —i--- + + (,b3d, 3]

D

t is even in x>, is odd in x3, and S2r, Ssd involve only integrals of odd functions of

Shear center

Fig. 11—14. Coordinates of the shear center.



where q53r, are harmonic functionst satisfying the following boundaryconditions:

=2

(11—88)(X2 +

=2

Note that the distribution due to leads to no shearing stress resultants.Finally, the total normal stress is given by

M2 M3 (P3 P2 '\(11—89)

12 13 13 J

Superimposing the shearing stresses and evaluating the torsional moment,we obtain

M1 = G1k1J — + (11—90)

where defines the location of the resultant of the flexural shear stress distri-bution due to P3. One can interpret X2, x3 as the coordinates of a point,called the shear center. The required twist follows from (11—90):

k1=

(M1 + —

Since (see Fig. 11—14)

M1 + P2x3 — P3x2(11—91)

= the applied moment with respect to the shear center = MT

we can write (a) ask1 = (11—92)

To determine the twist deformation (and the resulting torsional stresses), onemust work with the torsional moment with respect to the shear center, not thecentroid. For no twist, the applied force must pass through the shear center.In general, the shear center lies on an axis of symmetry. lithe cross sectionis completely symmetrical, the shear center coincides with the centroid.

It is of interest to determine the complementary energy associated withtorsion-flexure. The only finite stress components are 012, aild ThenV* reduces to

= + + dA (a)

The contribution from follows directly by substituting (11 —89) and usingthe definition equations for 13.

t The total flexural warping function for P3 is

P3 ( I '\ v,P3( 1

— + + —4


SEC. 11—5. TORSION-FLEXURE WITH UNRESTRAINED WARPING

1 M2 M2=+

(11-93)

Now, the total shearing stress is the sum of three terms:

1. a pure torsional distribution due to MT2. the flexural distribution due to F23. the fiexural distribution due to F3

Each of the flexural distributions can be further subdivided into—

dr. the distribution corresponding to a rigid cross section (definedby

2. dId, the distribution associated with in-plane deformation of the crosssection (defined by 4)jd)

We combine the flexural distributions and express the total stress as

C33 d12,t + C12,, + C12.d

= 013.1 + C13r + 013,d

where the various terms are defined by (11—81) and (11—87). For example,

F2 • F3r = 2 2

13 13

The complementary energy due to pure torsion follows from (11—38) and(11—92):

+

a as

F2 F3_C12r —

2

3 3

r + r) and integrating over the cross section, we obtaint

F2 2FF F2

JJ+ + +

= + dA (1196)

Jj 2 + 3413r, 3)dA= JJ

CIA

1 See Prob. 11—il.



The coupling term, I/A23, vanishes when the cross section has an axis ofsymmetry.

We consider next the coupling between or., and

JJ(a12,to12,r +

=2 ± 2)

+ + x2) + dA

=+ — + + (11—97)

MT ""F2 F3

— Jj+ dA = 0

13 12

The remaining terms involve a,a, the shearing stress distribution due to in-plane deformation of the cross section, We will not attempt to expand theseterms since we are interested primarily in the rigid cross section case.

Summarizing, the complementary energy for flexure-torsion with unre-strained warping is given by

1 M2 M2 M2 I F2 FF F2(11-98)

+ terms involving v1/E

where M1 + — We introduce the assumption of negligiblein-plane deformation by setting v1/E = 0. Similarly, we introduce the assump-tion of negligible warping due to flexure (&3r by setting 1/A1 =1/A2 1/A23 = 0.

In Sec. 11 —7, we develop an approximate procedure, called the engineeringtheory, for determining the flexural shear stress distribution, which is basedupon integrating the stress-equilibrium equation directly. This approach issimilar to the torsional stress analysis procedure described in the previoussection. Since the shear stress distribution is statically indeterminate when thecross section is closed, the force redundants have to he determined by requiringthe warping function to be continuous. For pure torsion, continuity requires(see (11—32))

= 2G1k1A5

where the integration is carried out in the X2-X3 sense around S. and is thearea enclosed by S. To establish the continuity conditions for flexure, weoperate on (11 —81) and (11 There are four requirements:

j 2,32v1G1P2

(aisd)F2 dS X3 dA(11—99)

2v1G1P3 rrdS = ii X2 dA

El2 ,jj4,


SEC. 11—6. EXACT FLEXURAL SHEAR STRESS DISTRIBUTION 303

In the engineering theory of flex ural shear stress distribution, the cross sectionis considered to be rigid, i.e., the distribution due to in-plane deformation isneglected. The consistent continuity condition on the flexural shearing stress

4sajsdSO (11—100)

One can take the + S direction as either clockwise or counterclockwise. Bydefinition, the positive sense for coincides with the + S direction.

11—6. EXACT FLEXURAL SHEAR STRESS DiSTRIBUTION FOR ARECTANGULAR CROSS SECTION

We consider the problem of determining the exact shear stress distributiondue to F2 for the rectangular cross section shown in Fig. 11—15. For con-venience, we first list the governing equations:

x2

rd3

— dt3'2

A = dt

d

rFig. 11—15. Notation for rectangular cross section.

1. Warping functions

=+

1 2=an

+=

2 )+



2. Shearing stresses

012 = - + + + xi)]

F2 v1G1E2= T 3) + x2x3)

13

Determination of

The boundary conditions for are

l(d'\2 d

= 0 atx3 =

We can take the solution as.i. 112<P2r = X2

The corresponding stresses and warping function are

13 12 132r = 4)2r X2 —

012,r =(11—101)

0j3,r 0

One can readily show thatF2

Finally, we evaluate 1/A2 using (11—96):

= (11—102)

Determinatio,, of

The boundary conditions for /2d are

1(d2 '\ d2 = + at x2 =

atx3 =

Now, the form of (a) suggests that we express q52d as

= — — f(x2, x3) (b)


SEC. 11—6. EXACT FLEXURAL SHEAR STRESS D!STRJBUTON 305

where f is an harmonic function. The shearing stresses and boundary condi-tions expressed in terms of f are

v1G1 P2 2= —f 2)

VLGI F2

=

f3=0

It remains to solve V2f = 0 subject to (d).Since the cross section is symmetricaL f must be an even function of x3 and

an odd function of x2. We express f as

f = B0x2 + B,, cos sinh(2nxx2)

This expansion satisfies V2f = 0 and the boundary condition at x3 = ± t/2.The remaining boundary condition requires

(2nnB0 + B,, co

nicd\sh

—)c

2nicx32os——- = x3 < x3 < (f)

Expanding in a Fourier cosine series and equating coefficients leads to

t2B0

B,, = (±..Ymr1 mtd

cosh ——

The final expressions for the shearing stresses are

2nxx2\ 1cosh1/t\2 I

—

t nndcosh—t /J

[ 2nirx2 1sinh —-—t . 2n7tx3

sin ——- (11—103)n2 mid t

[cosh— j

This system is statically equivalent to zero.

v1G1 P2=.a

cJl3,d =v1G1 F2

E 13

t



To investigate the error involved in assuming the cross section is rigid, wenote that the maximum value of • occurs at x2 0:

F2 22, ci

Specializing d for x2 0,

v1G1 F2 d2 2nnx3(512, d)x20 = C,, COS

13 '1 t

where1

1

2,, cosh 2,,

Now, C,, decreases rapidly with n. Retaining only the first term in (b) leads to thefollowing error estimate,

4 / 1

I5lz,di 1

2,. I cosh

Results for a representative range of d/t and isotropic material are listed below.They show that it is reasonable to neglect the corrective stress system for arectangular cross section. The error decreases as the section becomes thinner,i.e., as d/t becomes large with respect to unity:

d/t 1512,4/512.r

2 0.0241 0.092

0.122

11—7. ENGINEEffiNG THEORY OF FLEXURAL SHEAR STRESSDISTRiBUTION IN THiN-WALLED CROSS SECTIONS

The "exact" solution of the flexure problem involves solving four second-order partial differential equations. If one assumes the cross section is rigidwith respect to in-plane deformation, only two equations have to be solved.Even in this case, solutions can be found for only simple cross sections. Whenthe cross section is irregular, one must resort to a numerical procedure such asfinite differences or, alternatively, introduce simplifying assumptions as to thestress distribution. In what follows, we describe the latter approach for a thin-walled cross section. The resulting theory is generally called the engineeringtheory of shear stress. We apply the engineering theory to typical cross sections


SEC. 11—7. ENGINEERING THEORY OF FLEXURAL SHEAR STRESS 307

and also illustrate the determination of the shear center and the energyefficients, 1/Ai (j 2, 3).

Figure 11—16 shows a segment defined by cutting planes at x1 and x1 + dx1.Since the cross section is thin-walled, it is reasonable to assume that the normalstress, Oh, is constant through the thickness and to neglect Also, we work

xlFig. 11—16. thin-wafled segment.

with the shear flow, q, rather than with Integrating the axial force-equi-

librium equation,

— + —(a11t) 0C2S

with respect to S, we obtain the following expression for q,

('S

JSA

(11—104)

Equation (11—104) is the starting point for the engineering theory of shear

stress distribution. Once the variation of over the cross section is known,

we can evaluate q. Now, we have shown that the normal stress varies linearly

over the cross section when the member is subjected to a constant shear (F2, F3

qA

/



constant) and the end sections can warp freely. Noting that the member isprismatic, the derivative of for this case is

— X3 dM2 x2dM3

— 12 dx1 13 dx1

3 '2X3 + X212 13

and (11—104) expands to

i'S ,' i'S12!q=qA—-—I x2tdS—-7-H x3tdS

13 j54 JSA

The integrals represent the moment of the segmental area with respect to X2, X3and are generally denoted by Q2, Q3:

rsQ2 =j x3tdS = Q2(S,SA)

(11—105)

=x2t c/S = Q3(S, SA)

With this notation, (b) simplifies to

F2 F3q = — — —-Q2 (11—106)

13

Equation (11—106) defines the shear flow distribution for the case of negligiblerestraint against warping, i.e., for a linear variation in normal stress. Notethat q is positive when pointing in the + S direction.

We consider first the open section shown in Fig. 11—17. The end faces areunstressed, i.e.,

= = 0

Taking the origin for S at A, (11—106) reduces to

F2 F3q = f-Q2 (11-107)

Q3 =J5x2tdS Q2 =$x3tdS

We determine Q2, Q3 and then combine according to (11—107).The shearing stress distribution corresponding to F2,

q =

satisfiesffai2 c/A = F2

dA = 0

identically. To show this, we expand

= +



x3

—

B I Shear center

X2

Centroid

Fig. 11—17. Flexural shear flow—open segment.

and evaluate the shear stress resultants:

Jjui2dA =

JdS

Equation (b) requires

rsaJ0

=

= 0Now,

Integrating (e) by parts and noting that X2, X3 are principal centroidal axes,twe obtain

=— J" dS = 13

=— J

x2x3t dS = 0

The shear stress distribution predicted by (a) is statically equivalent to aforce F212. To determine the location of its line of we evaluate themoment with respect to a convenient moment center. By applying the sameargument, one can show that the shear flow corresponding to F3 is statically

t See Eq. (11—2).11—12.


310 TORSJON-FLEXURE OF PRISMATIC MEMBERS CHAP. 11

equivalent to a force, The intersection of the lines of action of the tworesultants is the shear center for the cross section (see Fig. 11—17).

Example 11—3

Consider the thin rectangular section shown. We take + S in the + X2 direction. Then,+ q points in the +X2 direction and q/t = The various terms are

= jtx2 dx2 =

—-d(2 2 4F2 tF2 (d2q=

q F2/'d22

—

This result coincides with the solution for obtained in Sec. 11—6. Actually, the engi-neering theory is exact for a rigid cross section, i.e., for r1/E = 0.

Fig. E11—3x2

I+

Example 11—4

We determine the distribution of q corresponding to F2 for the symmetrical sectionof Fig. Eli —4A. Only two segments, AB and BC, have to be considered since 1Q31 is

symmetrical.

Segment AB

=

q =

According to our definition, + q points in the + S direction (from A to B). Since q isnegative for this segment, it actually acts in the negative S direction (from B to A).


SEC. 11-7. ENGINEERING THEORY OF FLEXURAL SHEAR STRESS 311

Pg. E11—4A

IC d

tw

tf

I,

____________________________

ItH

Segment BC

We measure S from B to C. Then,

= + —

q = [hh1t1 + — xi)]

Note that the actual sense of q is from C to B. The distribution and sense of q are shownin Fig. E1I—413.

It is of interest to evaluate A2. Specializing (11-96) for a thin-walled section,

dS flif dA

5=

and substituting for q yields1 1 2dS=

jWe let

= area of the web =A1 = total flange area = 2b1t1

A2 = kA,.

The resulting expression for kislÀ,.

I 3Af\ 6A1

2A F id, l/bf\2



x3

x2

— X2)

Fig. E11—48

This factor is quite close to unity. For example, taking as typical, for a wide-flange section,

we find

= 2ç1.. 3,(If =

.4f =k = 0.95

The shearing stress corresponding to F3 varies parabolically in the flanges and is zero inthe web. Each flange carries half the shear and

Examp'e 11—5

1 61 1

A3 — 5 A1 — 5 hr1

Cross-Sectional Properties

This section (Fig. Eli —5A) is symmetrical with respect to X2. The shift in the centroidfrom the center of the web due to the difference in flange areas is

b2t2 —(a)

b,t1 + b2t2 +

We neglect the contribution of the web in '2 since it involves

12 + ('2)2 = + (b)

.4

4.


Distribution of q Corresponding to

The shear flow corresponding to F3 is obtained by applying

p3q = Q2

and is shown in Fig. El 1—SB. The shear stress vanishes in the web and varies parabolicallyin each flange.

Integrating the shear flow over each flange, we obtain

= F3

Then, the distribution is statically equivalent to F373 acting at a distance e from the leftflange, where

R2d (12)2e = =

R 12

Since X2 is an axis of symmetry, the shear center is located at the intersection of R and X2,

SEC. 11—7. ENGINEERING THEORY OF FLEXURAL SHEAR STRESS

Determination of

Taking S as shown in the sketch, we have

= t[(b)2

t2 [,"b2\22—X3

= 0 since X2 is an axis of symmetry.

313

(c)

Fig. E11—5Ax3

II Centroid



The coordinates of the shear center with respect to the centroid are

Torsional Shear Stress

,c3 0

(I= e —(1 +

= d —

L12 2

The flexural shear stress distribution is statically equivalent to a torsional moment equalto with respect to the centroid. We have defined M1 as the required torsional momentwith respect to the centroid. Then, the moment which must be balanced by torsion is

— F3; the required torsional moment with respect to the shear center. Usingthe approximate theory developed in Sec. 1 1 —3, the maximum torsional shear stress in asegment is

where

M= ti

.1 = + +

We consider next the closed cross section shown in Fig. 11 —18. We take theorigin for S at some arbitrary point and apply (11-106) to the segment Se-S:

F2 F3q = q1 — -1—Q3 —

13 12

= Q3 =

(11— 108)

where q1 is the shear flow at P. The shear flow distribution is statically indeter-minate since q1 in unknown. We have previously shown that q q1 = con-

Fig. E11--58

R = F31'3t2frbf\2 2q12 X3

R1



stant is statically equivalent to only a torsional moment equal to Thesecond and third terms are statically equivalent to F212 and F313.

The constant q1 is determined by applying the continuity requirement to thecenterline curve. Since the engineering theory corresponds to assuming thecross section is rigid with respect to in-plane deformation, we use (11—100).

The flexural shear stress distribution must satisfy

0 (11—109)

for an arbitrary closed Substituting for q.

C dS F2 C dS f (IS

2

and considering separately the distributions corresponding to F2 and F3, weobtain

q = +

= T (B2 — Q3) (B3 — Q2)

QdS QdS (11-110)

B— B27

dS IEach distribution satisfies (11—109) identically. Also, the distribution isstatically equivalent to a force Fyi,, located Y4 units from the centroid. Notethat q = B3 leads only to a torsional moment equal to

f One can interpret (11-109) as requiring the flexural shear stress distribution to lead to no twistdeformation, See Prob. 11 —14 for the more general expression, which allows for a variable shearmodulus.

P

Fig. 11—18. Notation for closed cell.


TORSION-FLEXURE OF PRISMATIC MEMBERS CHAP. 11

The general expression for 1/Ai follows from (11— 96):

Substituting for

== j = 2,3 (11—111)

2dS dSQkT

42j

which applies for an arbitrary single cell.

Example 11—6

We illustrate the determination of for the square section of Fig. Eli —6A. It is

convenient to take P at the midpoint since the centerline is symmetrical.

= a2

(a3\ a2 3

= + = 4a3t

= (at)(a/2)—

5a1 10

fdS a= 3.5-it t

Fig. E11—6A

1 =— 2BJQk + (j k;j, k 2,3)

'kand noting that

we obtain .1

(11—112)

H

2z

H




Determination of

We start at P and work counterclockwise around the centerline. The resulting distribu-tion and actual sense of q due to arc shown in Fig. Eli —6B. Note that + Q2 correspondsto a negative i.e., clockwise, q.

Evaluation of B3

By definition,

1 " dSB3=

rds.it I

Fig. E11—6B

Using the above results, and noting that the area of a parabola is equal to (2/3) (base) x(height), we obtain

I dS a3+

B3 =

of Flexural Shear Flov for F3

The shear flow is given by

F3 F3(( 4Q2q — — Q2) —

12 a 3at(+ sense clockwise). The two distributions are plotted in Fig. Ell—6C.

To locate the line of action of the resultant, we sum moments about the midpoint (0 inthe sketch:

I /a\ 4F3 /a\ 19(M)0 =

+ =aF

The resultant acts e units to the right of 0, where

19e = a



j F3/21

t

I

Finally, the coordinates of the shear center with respect to the ccntroid (which is A units tothe right of 0) are

16x2 = e — LI = +

X3 = 0

Torsional Shear

The shear flow for pure torsion is due to Mr, the torsional moment with respect to theshear center. For this section,

Mr 5L2F3 + M1 —

We apply the theory developed in Sec. II —4. One just has to replace M1 with MT in

_

Fig. Eli —6C

q — a 1.21

i F36 a

I F330

F3f 4 Q2


SEC. 11—7. ENGtNEERING THEORY OF FLEXURAL SHEAR STRESS

Equation (11—61):

C1— ± ICi = 4at

Determination of 1/A3

Applying (11—112), we find

J = +

1 12

dSB

dS'\ — 1.276

- I - I) - 3atNote that 3at is the total web area.

We consider next the analysis of a two-cell section and include open segmentsfor generality. There is one redundant shear flow for each cell. We select aconvenient point in each cell and take the shear how at the point as the redun-dant for the cell. This is illustrated in Fig. 11—19: qj represents the shear flowredundant for cell j and the + S sense coincides with the X2-X3 sense to beconsistent with the pure-torsion analysis. The + S sense for the open segmentsis inward from the free edge. For convenience, we drop the CL (centerline)subscript on S and A.

Fig. 11—19. Notation for section.

+5, q

x3

x2



The total shear flow is the sum of q0, the open cross-section distribution(q1 = q2 = 0), and

q q0 + (11—113)

We determine q0 by applying (11—107) to the various segments. The redun-dant shear-flow distribution is the same as for pure torsion (see Fig. 11—11).Finally, we obtain a system of equations relating q1, q2 to F2, F3 by applyingthe continuity requirement to each centerline, t

q—--= 0 j = 1,2 (11—114)

where q is positive if it points in the + S direction. Using the aJk notationdefined by (11—68), the equations take the following form:

a11q1 + a12q2 =a12q1 + a22q2 =

11—115

t

The shear flows q2, for pure torsion are related by (we multiply (11—71)by MT/J and note (11—62))

+ a12q2,1 = 2A1

(11--1l6)MT

a12q1,1 + a22q2,1 2A2

Thus, the complete shear stress analysis involves solving aq = b for three differ-ent right-hand sides. The equations developed above can he readily generalized.

Example 11—7

We determine the flexural shear stress distribution corresponding to F3 for the sectionshown in Fig. E11—7A. We locate and P2 at the midpoints to take advantage ofsymmetry.


A1 = 202

A2 a2

=(a3t\ [ a21 7

'2 + 2[(3at)_4_j

Ga 4a a= — a22 = 012 =

See Prob. 1!-- 14 for the more general expression, which allows for a variable shear modulus.



Fig. Eli—lA

x2

+S,q

2a a

Distribution of

This system (Fig. E1l—7B) is statically equivalent to a moment

2a2(2q1 +

Distribution of q0 Due to F3

We apply—--i02

'2

to the various segments starting at points P1. P2. The resulting distribution is shown inFig. Eli —7C.

Deter,niiuztion of q1. q2

=

C dS 2F3q0—=+.322 t

The equations for q1 and q2 are1 F3

6q1 — q2 = 7a2 F3

—q1 + 4q2 = ——70Solving (a), we find

2F3q1 = —

161 a

11 F3q2 = +

The total distribution is obtained by adding qR and q0 algebraically.

_______a _______

I



Fig. El 1 —7B

q2

Location of the Shear Center

Taking moments about the midpoint of the left web, and letting e be the distance to theline action of the resultant, we obtain

M( + 2a2(2q1 + q2) + (2a) + (3a) F3) = eF3

(2 32\e = + a = 1.61a

The shear center is located on the K2 axis and

= c — (2a — = +O.055a

REFERENCES

1. WANG, C. T.: Applied Elasticity, McGraw-Hill, New York. 1953.2, TIMOSHENKO, S. J., and J. N. Oooo!ETt: Theory of Elasticity, 3d ed., McGraw-Hill,

New York, 1970.

q1a

2q1a4

q2a

q1

Fig. Eli —7C

I P314 a


PROBLEMS 323

3. Dmt HARTOG, J. P.: Advanced Strength of Materials, McGraw-Hill, New York,1952.

4. HERRMANN, L. R.: "Elastic Torsional Analysis of Irregular Shapes," I. Eng. Mech.Div., A.S.C.E., December 1965.

5. SOKOLNIKOFF, LS.: Mathematical Theory of Elasticity. McGraw-Hill, New York,1956.

6. TIMOSHFNKO, S. J.: Strength of Materials, Part 2, Van Nostrand, New York, 1941.7. CEiRNICA, J. N.: Strength of Materials, I-Jolt, Rinchart, New York, 1966.8. DABROWSKI, R.: GekrUmmte diinnwan.dige Trëger ("Curved Thin-walled Girders"),

Springer-Verlag, Berlin, 1968.9. KOLLBRUNNER, C. F., and K. Basler: Torsion in Structures, Springer-Verlag, Berlin,

1969.10. VLASOV, V. Z.: Thin- Walled Elastic Beams, Israel Program for Scientific Transla-

tions, Jerusalem, 1961.11. ODEN, J. T.: Mechanics of Elastic Structures, McGraw-Hill, New York, 1967.

PROBLEMS

il—i. The pure-torsion formulation presented in Sec. 11—2 considers thecross section to rotate about the centroid, i.e., it takes

U2 = W1X3 wj == +(01x2 a1 =

Suppose we consider the cross-section to rotate about an arbitrary pointThe general form of (a) is

a2 = —wj(x3 — wj = k1x1 + c1

U3 = +w1(x2 — a1 = k147

(a) Starting with Equation (h), derive the expressions for a13 and thegoverning equations for

(b) What form do the equations take if we write

= + C2 +

Do the torsional shearing stress distribution and torsional constant Jdepend on the center of twist?

11—2. Show that .1 can be expressed as

= + 2) — 3)2]dA

= I,, —)2

+ 3)2]dA

Hint:dA = 0

Compare this result with the solution for a circular cross section and commenton the relative efficiency of circular vs. noncircular cross section for torsion.

11—3. Derive the governing differential equation and boundary conditionfor for the case where the material is orthotropic and the material symmetryaxes coincide with the X1, X2, X3 directions.



11—4. The variation in the warping function along an arbitrary curve Sis obtained by integrating (11—29),

= k1 + (a)

where is the perpendicular distance from the center of twist to the tangent.One selects a positive sense for S. The sign of p is positive when a rotationabout the center of twist results in translation in the +.S direction. We express

as1 lM1

(b)

and (a) reduces to-+ a15 (c)

Determine the variation of along the centerline for the two thin-walled opensections shown.

x3tf

4 Center of twist

b/2 b/2

(a)

11—5. Verify that the distribution. q = const, satisfies

F2 = ja12 = dS = 0

F3 = dA = dS = 0

= x =

for the closed cross section sketched.11—6. Refer to Prob. 11—4. To apply Equation (c) to the centerline of a

closed cell, we note that (see (11—50))

— J Jq Ca15 = = k—— = — (a)

Prob. 11—4

Centerof twist/—

(bi


Then,

PROBLEMS 325

x2

Prob. 11—5

Integrating (b) leads to the distribution of Apply (h) to the section shown.Take = 0 at point P. Discuss the case where a = b.

Prob. 11-6

11—7. Determine the torsional shear stress distribution and torsionalconstant J for the section shown. Specialize for t a.

11—8. Determine the equations for (j = 1, 2, 3) and J for the sectionshown. Generalize for a section consisting of "n." cells.

I 1—9. Determine the distribution of torsional shear stress, the torsionalconstant J, and the distribution of the warping function for the section shown.Take = 0 on the symmetry axis and use the results presented in Prob. 11—6.

11—10. Verify Equation (11—82). Utilize (11—15).11—11. The flexural warping function satisfy

mA

on S

= + £3

C+-Ct

T p.


TORS!ON-FLEXURE OF PRISMA1]C MEMBERS

t t tI I:f

0Si

t 0-

0

Utilizing the following integration formula,

if (fr, 212,2 + 3f2, 3)dx2 dx3 dS — JJII V2f2 dx2 dx3

where 11,12 are arbitrary functions, verify Equation (11—96).11—42. Refer to Fig. 11—17. Starting with (11—107), derive the expressions

for the coordinates of the shear center in terms of the cross-sectional parameters.

0 —

+S1+S2

326

T

t

CHAP. 11

Prob. 11—7

IH

Prob. 11—8

t—.

t t ta a— a

tProb.11—9

II


Td/2

+

I

I-

2/ 2/ 2/

_a

+

(c)

I:L

1 = Ca

(el

IIH-a -j

(dl

PROBLEMS 327

11 —13. Determine the flexural shear flow distributions due to F2, F3 andlocate the shear center for the five thin-walled sections shown.

Prob. 11—13

R

x3

(b)

(a)



11—14. We established the expression for the twist deformation (Equation(11—31) by requiring the torsional warping function to be continuous. Onecan also obtain this result by applying the principle of virtual forces to thesegment shown as part of the accompanying figure.

Prob. 11—14

-44M'11 wi

Arbitraryclosedcurve

44—

LsM1 w1 + dx1

GtE)

2G(2E)

Fa-

(a)

(b)

2G(2E)

x3

X21G(E)

Ic)


PROBLEMS 329

The general principle states that

(J$eT dA)dx1 = (if AbTu dA)dx1 + if ApT x+dx (a)

for a statically permissible force system. Now, we select a force system actingon the end faces which is statically equivalent to only a torsional moment M1.If we consider the cross section to be rigid, the right-hand side of (a) reducesto AM1o,1 1 dx1,and we can write

Next, we select an arbitrary closed curve, S (part b of figure), and consider theregion defined by S and the differential thickness dn. We specialize the virtual-stress system such that AG = 0 outside this domain and only is finiteinside the domain. Finally, using (11 -51), we can write

dn(&riz)=

and Equation (b) reduces to ifk1

The derivations presented in the text arc based on a constant shear modulusG throughout the section, so we replace (d) with

Gk1 =

If G is a variable, say G = fG* (where f = f(x2, x3)), we have to work with

G*k1 =

Also, we define the torsional constant J according to

G*k1J

Consider a thin-walled section comprising discrete elements having differentmaterial properties. Develop the expressions for the torsional and fiexuralshear flow distributions accounting for variable G and E. Determine thenormal stress distribution from the stress-strain relation. Assume a linear varia-tion in extensional strain and evaluate the coefficients of the strain expansionfrom the definition equations for F1, Al2, and M3. Apply your formulationto the section shown in part c of the figure.


Engineering Theory ofPrismatic Members

12—1. INTRODUCTION

St. Venant's theory of fiexure-torsion is restricted to the case where—-

1. There are no surface forces applied to the cylindrical surface.2. The end cross sections can warp freely.

The warping function consists of a term due to flexure (ç&j) and a term due topure torsion Since is independent of x1, the linear expansion

F1 M2 M3a11 =—;—+——X3 ——---Xz

'2 13

is the exact solutiont for The total shearing stress is given by

= + crj (12—2)

where is the pure-torsion distribution (due to and represents theflexural distribution (due to We generally determine by applying theengineering theory of shear stress distribution, which assumes that the crosssection is rigid with respect to in-plane deformation. Using (12—1) leads to thefollowing expression for the flexural shear flow (see (11—106)):

= — — (12—3)

The warping function will depend on x1 if forces are applied to the cylindricalsurface or the ends are restrained with respect to warping. A term due tovariable warping must be added to the linear expansion for This leadsto an additional term in the expression for the flexural shear flow. Since (12—1)

t A linear variation of normal stress is exact for a homogeneous beam. Composite beams (e.g.. asandwich beam) are treated by assuming a linear variation in extensional strain and obtainingthe distributions of from the stress-strain relation. See Probs. 1—14 and 12—1.

330


FORCE-EQUILIBRIUM EQUATIONS

satisfies the definition equations for F1, M2, M3 identically, the normal stresscorrection is self-equilibrating; i.e., it is statically equivalent to zero. Also, theshear flow correction is statically equivalent to only a torsional moment since(12—3) satisfies the definition equations for F2, F3 identically.

In the engineering theory of members, we neglect the effect of variable warpingon the normal and shearing stress; i.e., we use the stress distribution predictedby the St. Venant theory, which is based on cons tant warping and no warpingrestraint at the ends. In what follows, we develop the governing equations forthe engineering theory and illustrate the two general solution procedures. Thisformulation is restricted to the linear geometric case. In the next chapter, wepresent a more refined theory which accounts for warping restraint, and in-vestigate the error involved in the engineering theory.

12—2. FORCE.EQUILIBRIUM EQUATIONS

In the engineering theory, we take the stress resultants and couples referredto the centroid as force quantities, and determine the stresses using (12—1),(12—3), and the pure-torsional distribution due to MT. To establish the force-equilibrium equations, we consider the differential element shown in Fig.12--i. The statically equivalent external force and moment vectors per unit

F—dxi/2 .-I. clxl/2___H

Fig. 12—1. Differential element for equilibrium analysis.

length along X1 are denoted by b, Summing forces and moments about 0leads to the following vector equilibrium equations (note that F = —

+ = odx1

dM÷ — — -+ m + x F+) = 0

dx1

SEC. 12—2. 331

dx1'— dF+ dx1

+ dx1

4L dx1+


332 ENGINEERING THEORY OF MEMBERS CHAP. 12

We obtain the scalar equilibrium equations by introducing the componentexpansions and equating the coefficients of the unit vectors to zero. The re-sulting system uncouples into four sets of equations that arc associated withstretching, flexure in the X1-X2 plane, flexure in the X1-X3 plane, and twist.

StretchingdF1

+ b1 = 0dx1

Flexure in X1-X2 Plane

dF2+ b2 0

dx3

(12—4)+ m3 + F2 = 0

dx1

Flexure in X1 -X3 Plane

--s- + b3 = 0dx1

dM2+ ni2 — F3 = 0

dx1

TwistdM1—— + m1 = 0ax

This uncoupling is characteristic only of prismatic members the equilibriumequations for an arbitrary curved member are generally coupled, as we shallshow in Chapter 15.

The fiexure equilibrium equations can be reduced by solving for the shearforce in terms of the bending moment, and then substituting in the remainingequations. We list the results below for future reference.

Flexure in X1 -A'2 Plane

dM3F2 — ui3

dx1

d2it.13 din32 + — b2 = 0

dx1 dx1

Flexure in X1 -A'3 Plane (1 25)

dM2+ in2

dx1

d2M2 din22 + -— + b3 = 0

dx1 dx1


SEC. 12—3. FORCE-DISPLACEMENT RELATIONS 333

Note that the shearing force is known once the bending moment variation isdetermined.

The statically equivalent external force and moment components acting onthe end cross sections are called end forces. We generally use a bar superscriptto indicate an end action in this text. Also, we use A, B to denote the negativeand positive end points (see Fig. 12—2) and take the positive sense of an end

x2

MA2

2

x3 —L

Fig. 12—2. Notation and positive direction for end forces.

force to coincide with the corrcsponding coordinate axis. The end forces arerelated to the stress resultants and couples by

=

(J 1 2 3) (12—6)FAJ =

MAJ =

A minus sign is required at A, since it is a negative face.

12—3. FORCE-DISPLACEMENT RELATIONS; PRINCIPLE OFVIRTUAL FORCES

We started by selecting the stress resultants and stress couples as forceparameters. Applying the equilibrium conditions to a differential element re-sults in a set of six differential equations relating the six force parameters. Tocomplete the formulation, we must select a set of displacement parameters andrelate the force and displacement parameters. These equations are generallycalled force-displacement relations. Since we have six equilibrium equations,we must introduce six displacement parameters in order for the formulationto be consistent.

Now, the force parameters are actually the statically equivalent forces andmoments acting at the centroid. This suggests that we take as displacement


334 ENGINEERING THEORY OF PRISMATIC MEMBERS CHAP. 12

parameters the equivalent rigid body translations and rotations of the crosssection at the centroid. We define i? and as

= = equivalent rigid body translation vector at the centroid(12—7)

= equivalent rigid body rotation vector

By equivalent displacements, we mean

(force intensity) (displacement) dA = fl + M (.12—8)

Note that (12—7) corresponds to a linear distribution of displacements over thecross section, whereas the actual distribution is nonlinear, owing to shear de-formation. In this approach, we are allowing for an average shear deforma-tion determined such that the energy is invariant.

We establish the force-displacement relations by applying the principle ofvirtual forces to the differential element shown in Fig. 12—3. The virtual-force

+

dx1 Idx1

dudXl

dx1 22

Fig. 12—3. Statically permissible force system.

system is statically permissible; that is, it satisfies the one-dimensional equili-brium equations

ödx1

+ x = Odx1

Specializing the principle of virtual forces for the one-dimensional elasticcase, we can write

dV* dx1 = AP1

where represents a displacement quantity, and is the external force quan-tity corresponding to The term dV* is the first-order change in the one-dimensional complementary energy density due to increments in the stressresultants and couples.


SEC. 12—3. FORCE-DISPLACEMENT RELATfONS 335

Evaluating the right-hand side of (b), we have

=+ A]\+ + dx1

Using the second equation in (a), (c) takes the form

+ x +

Finally, evaluating the products, we obtain

= [AF1u1,1 + /XF2(u2, — co3) + AF3(u3, 1 + (02)

+ + zXM2a2 + j]dx1 (12—9)

Continuing, we expand dV*:

dV* = +\CFJ

3

= +1

The quantities and k1 are one-dimensional measures. Equating(12—9) and (12—10) leads to the following relation between the deformationmeasures and the displacements:

av*Cl = = u1, = WL

e2 —— = 112. j —. (03 k2 = (02, 1 (12—Il)aF2 cM2

(123 23

We see that—

1. e1 is the average extensional strain.2. e2, e3 are average transverse shear deformations.3. k1 is a twist deformation.4. k2, k3 are average bending deformation measures (relative rotations of

the cross section about X2, X3).

Once the form of V" is specified, we can evaluate the partial derivatives. Inwhat follows, we suppose that the material is linearly elastic. We allow for thepossibility of an initial extensional strain, but no initial shear strain. Thegeneral expression for is

=+ + +

where denotes the initial extensional strain. Now, for unrestrainedtorsion-Ilexure is given by (11—98). Since we are using the engineering theory


336 ENGINEERING THEORY OF MEMBERS CHAP. 12

of shear stress distribution, it is inconsistent to retain terms involving in-planedeformation, i.e., v1/E. Adding terms due to F1/A, and neglectingthe coupling between F2, F3 leads to

= Fie? + + + LF22AE 2GA2 2GA3 (12—12)

+ + + + +

whereMT M1 + F2x3 —

e?

= $5 x2e1 cIA

We take (12—12) as the definition of the one-dimensional linearly elastic com-plementary energy density for the engineering theory. One can interprete?, as "weighted" or equivalent initial strain measures.

Differentiating (12—12) with respect to the stress resultants and couples, andsubstituting in (12—11), we obtain the following force-displacement relations:

F1 MT

F2 M2+ U21 (03 k2 =k2 (12—13)

F3X2 = U3 1 + (02 k3 = k3 + = (03,

To interpret the coupling between the shear and twist deformations, we note(see Fig. 12—4) that

U2 X3W1

U3

defines the centroidal displacements due to a rigid body rotation about theshear center. Comparing (a) with (12—13), we see that the cross section twistsabout the shear center, not the centroid. This result is a consequence of neglect-ing the in-plane deformation terms in i.e., of using (12—12).

Instead of working with centroidal quantities (M1, u2, u3), we could havestarted with M1. and the translations of the shear center. This presupposesthat the cross section rotates about the shear center. We replace u2, u3 (seeFig. 12—4) by

U2 + (01X3(12—14)

U3 = t01X2


SEC. 12—3. FORCE-DtSPLACEMENT 337

where um, U53 denote the translations of the shear center. The terms involvingF2, F3, Al1 in (12—9) transform to

1 + 1 — w3) + 1 + w2)

Then, taking as an independent force parameter, we obtain

= a)1,1

F2US2, 1

F3053 1 + (02

(a)

(12— 15)

Since the section twists about the shear center, it is more convenient to workwith and the translations of the shear center. Once 052, 053, and w1 are

x2

Fig. 12—4. Translations of the centroid and the shear center.

known, we can determine 02, 03 from (12—14). We list the uncoupled sets offorce-displacement relations below for future reference.

Stretching

Flexure in X1 -X2 Plane

F1e? + 01:1

F 2 = 1 — (03GA2

+ =El3



Flexure in X1-X3 Plane (12—16)

F3U53 +

k2 + = I

Twist About the Shear Center

MT= Wi,1

The development presented above is restricted to an elastic material. Now,the principle of virtual forces applies for an arbitrary material. Instead of firstspecializing it for the elastic case, we could have started with its general form(see (10—94)),

[ss dA] = AP1 (12—17)

where represents the actual strain matrix, and denotes a system of staticallypermissible stresses due to the external force system, AP1. We express theintegral as

.

ifA j=1

and determine using as defined by the engineering theory. For example,taking

AF1 AM2 AM3+ —— —

A 12 13

leads to

e1 = if dA

k2 ifk3 if x2c1 dA

Once the extensional strain distribution is known, we can evaluate (b).Using (12—18), the one-dimensional principle of virtual forces takes the form

+ AM1)]dxj = AP1 (12—19)

The virtual-force system must satisfy the one-dimensional equilibrium equations(12—4). One should note that (12—19) is applicable for an arbitrary material.When the material is elastic, the bracketed term is equal to dV*, and we canwrite it as

dV's dx1 = >d1 AP1 (12—20)


SEC. 12—4. SUMMARY OF THE GOVERNING EQUATIONS 339

The expanded form for the linearly elastic case is

J[(eq + AF1 + AF2 + AF3 +

(12-21)+ + El2) AM2 + + El) AM3] dx1 =

We use (12—21) in the force method discussed in Sec. 12—6.

12—4. SUMMARY OF THE GOVERNING EQUATIONS

At this point, we summarize the governing equations for the linear engineeringtheory of prismatic members. We list the equations according to the differentmodes of deformation (stretching. flexure, etc.). The boundary conditions reduceto either a force or the corresponding displacement is prescribed at each end.

Stretching (F1, u1)

F1 I + b1 = 0

F1 (12—22)

F1 or u1 prescribed at x1 = 0, L

Flexure in X1-X2 Plane (F2, M3, U2, 0)3)

F2, + b2 = 0

M31 + m3 + F2 = 0

F2= U2 i —

M2

(12—23)

+ (03,1

u2 or F2 prescribed at x1 = 0, LM3 or 0)3 prescribed at .x1 = 0, L

Flexure in the X1-X3 Plane (F3, M2, U3,

F3,1 + b3 0

M2, + — F3 = 0

F3= u3 1 + 0)3

M(12—24)

u3 or F3 prescribed at x1 = 0, L

0)3 or M2 prescribed at x1 = 0, L



Twist About the Shear Center (MT, (01, u2, u3)

MT. 1 + 01T 0

MT=

or a1 prescribed at x1 = 0, L (12—25)

m5. = m1 + b2x'3 — b3g2

U2 X3W1

U3 =

12-5. DISPLACEMENT METHOD OF SOLUTION—PRISMATIC MEMBER

The displacement method involves integrating the governing differentialequations and leads to expressions for the force and displacement parametersas functions of x1. When the applied external loads are independent of thedisplacements, we can integrate the force-equilibrium equations directly andthen find the displacements from the force-displacement relations. If the appliedload depends on the displacements (e.g., a beam on an elastic foundation),we must first express the equilibrium equations in terms of the displacementparameters. This.problem is more difficult, since it requires solving a differentialequation rather than just successive integration. The following examples illus-trate the application of the displacement method to a prismatic member.

Example 12—1

We consider the case where b2 = coast (Fig. This loading will produce flexurein the X1-X2 plane and also twist about the shear center if the shear center does not lie onthe X2 axis. We solve the two uncoupled problems, superimpose the results, and thenapply the boundary conditions.

Flexure in X1 -A'2 Plane

We start with the force-equilibrium equations,

F21M3,1 = —F2

Integrating (a), and noting that b2 = coast, we have

F2 = — b2x1

For convenience, we use subscripts A, B for quantities associated with x1 = 0, L:

= FAJ etc.

With this notation, (c) simplifies to

F2 = FA2 — b2x1

Substituting for F2 in (b), and integrating, we obtain

= MAI — XIFA2 +


SEC. 12—5. DISPLACEMENT METHOD OF SOLUTION

We consider next the force-displacement relations,

Integrating (g) and then (h), we obtain

M3£03, 3

F 2Uz, i —. +

GA2

£03 = WA3 + — (x1MA3 —

4 4 \\2E13/ GA 12Ff)

The general flexual solution (for b2 = const) is given by (e), (f), and (I).

Twist About the Shear Center

The applied torsional moment with respect to thc shear center is

mr =

Substituting for mr in the governing equations,

and integrating, we obtain

=MT

(01,1 =

Fig. E12—1

MT = MAT — b2y3x1

= +

The additional centroidal displacements due to twist are

U2 = X3W1

U3 =

x2

b3 13Shear center

H LCentroid



Cantilever Case

We suppose that the left end is fixed, and the right end is free. The boundary conditionsare

UA2 = = 0FBZ = M53 = MBT

Specializing the general solution for these boundary conditions requires

= b2LAn 11. i2

A3 2

MAT = b2;L

and the final expressions reduce to

F2 = b2(L — x1)

M3 = — Lx1 +

= — x1)

U2 = + b2Lx1 -+

+ ()U3 =

b2 4L 4

a)1

It is of interest to compare the deflections due to bending and shear deformation.Evaluating u2 at x1 = L, we have

I h2LU52 LI

=

lh2L2UB2 =

E 13

GL2A2

As an illustration, we consider a rectangular cross section and isotropic material withv = 0.3 (d = depth):

13 613d2A2 5 A 10

=

By definition, d/L is small with respect to unity for a member element and, therefore, it is


SEC. 12—5. DISPLACEMENT METHOD OF SOLUTION 343

reasonable to neglect transverse shear deformation with respect to bending deformationfor the isotropic case.t Formally, one sets 1/A2 = 0.

Fixed-End Case

We consider next the case where both ends are fixed. The boundary conditions are

0.42 = W.43 = (ISA! = 0= = = 0

Specializing (h), (i), and (k) for this case, we obtain

b2LFAZ=----

2

b2L2M.43 =

12

MAT = b2T3L

The final expressions are

IL2 Lx1M3 = b2

+

MT = b2T3 —

2) +b2

— 2L4 + xi) (u)u2 = + — x1

U3 = —X20)1

b2IL2 L(03 = El

Xj + xi)

b2y3= xi)

Example 12—2

We consider a member (Fig. E12—2) restrained at the left end, and subjected only toforces applied at the right end. We allow for the possibility of support movement at A.The expressions for the translations and rotations at B in tcrms of the end actions at Band support movement at A are called member force-displacement relations. We canobtain these relations for a prismatic member by direct integration of the force-displacement

t For shear deformation to be significant with respect to bending deformation, G/E must be ofthe same order as l/A,L2 where A, is the shear area. This is not possible for the isotropic case.However, it may be satisfied for a sandwich beam having a soft core. See Prob. 12—1.



relations. In the next section, we illustrate an alternative approach, which utilizes theprinciple of virtual forces4'

x2

The boundary conditions at x1 = L are

Fig. E12—2

==

(a)

Integrating the force-equilibrium equations and applying (a) lead to the following expres-sions for the stress resultants and couples:

= (j= 1,2,3)MT

(b)M2 = M52 — (L — x1)F53

M3 MB3 + (L

Using (b), the force-displacement relations take the form


C03,j = + (L — xi)F5]

U3,1 = + +

= El2_

—

U3 j = — 0)2 +—

M5.,-

Col,i =

FA3

MA3/

/x3

M1, WB 1

UB 3

3



Integrating (c) and setting x1 = L, we obtain

L_UB1 UAI +

L_ L2(.053 03.43 + M83 +

L2 LX3_ /L L3\..UA2 + LWA3 + + Mjjr + +

L2_(052 + M82 —

L2 — IL L3\_U83 = UA3 — LWA2 — — + +

L= co,0 + MET

Finally, we replace MET by

MET = + X31B2

and write the equations in matrix form:

LUS'

L L3+

+

L L3

GJ

L

L

L2 L -

+ UA2 + U.43 LOA2, (.0,42, (.0A3} (f)

The coefficient matrix is called the member "flexibility" matrix and is generally denotedby fB.

We obtain expressions for the end forces in terms of the end displacements by invertingf. The final relations are listed below for future reference:

U82

(082

(053



AE

= (um — UA1)

FB2 ——

+ WA3)—

(coB! WA!)

(u53 — UA3) + (co82 + C°A2) +—

(co81 — WAI)

rGJ 12EMB1

=+ + — WA1)

(082 — UA2) + + (0A3)— L3

+ — UA3) + + COA2)

MB2 = — UA3) + — WA!)

+ (4 + a2) + (2 — a2) j_ WA2

M83 — (053 UA2) + — (1)4j)

-1 (4 + + (2 a3)_L_coA3

where12E12 12E13

a2=

a3 =

1+03

We introduce the assumption of negligible transverse shear deformation by setting03 = a3 = 0

The end forces at A and B are related by

(j= 1,2,3)MA! = —!V151

MA2 + LF53

MA3 —M53 — LF52

We list only the expressions for M43:

MA2 = — 0A3) + — WA!)

+ (4 + a2) WA2 + (2 — 02)

MA3 —T-(uB2 — UA2) +L2

— WA!)

+ (4 + a3) C0A3 + (2 — a3)



Example 12—3

We consider next the case where the applied loads depend on the displacements. Tosimplify the discussion, we suppose the shear center is on the X2 axis and the member isloaded only in the plane. The member will experience only flexure in the X1-X2plane under these conditions.

The governing equations are given by (12—23):

F2 = —M3,1 — in3 (b)

M3

(03

An alternate form of(a) isM3 + 1723, b2 0

Once M3 is known, we can, using (b), find F3.Now, we solve (d) for 03 and substitute in (c):

F2

F,1(03,1 p2,11 +

Then,

M3 E13(u3 + b2 —

and

F2 = —m3 — El2 (02 + b2, —

Finally, we substitute forM3 in (e) and obtain a fourth-order differential equation involving02 and the load terms:

d4u2 d2 ( b2 '\ I (din3 \+ — + — h2) = 0

The problem reduces to solving (i) and satisfying the boundary conditions:

F2 or 02 prescribed

or (03 prescribed)

Neglecting transverse shear deformation simplifies the equations somewhat. The re-sulting equations are (we set 1/GA2 = 0)

(03 j

= E13(u2, 11 —

F3 = —in3 — E13(u2 —

d4u3 d2 1 (din3 '\—

+— b2) = 0



As an illustration, consider the case of linear restraint against translation of the centroid,e.g., a beam on a linearly elastic foundation. The distributed loading consists of twoterms, one due to the applied external loading and the other due to the restraint force.We write

= q — ku3

where q denotes the external distributed load and k is the stiffness factor for the restraint.We suppose rn3 = = 0, k is constant, and transverse shear deformation is negligible.Specializing (k) for this case, we have

(03 = U2

M3 = E13u2,

F2 = —E13u2

d4u2 k q+

=F3 or u2 prescribed 1

0 LM3 or (03 prescrihedJ

The general solution of (n) is

+ sin A.x1 + C2 cos Ax1) + sin Ax1 + C4 cos Ax1)

/ k

where u2 represents the particular solution due to q. Enforcement of the boundaryconditions at x 0, L leads to the equations relating the four integration constants.

The function e_2x decays with increasing x, whereas increases with increasing x.For Ax > 3, 0. If the member length L is greater than 2(3/A) = 2Lb (we interpretLb as the width of the boundary layer), we can approximate the solution by the following:

0 x1 < Lb: = + sin Ax1 + C2 cos Ax1)

LB<xl<L—Lb:L — Lb < x1 L: u2 = + sin Ax1 + C4 cos Ax1)

The constants (C1, C3) are determined from the boundary conditions at x1 = 0 and(C3, C4) from the conditions at x1 = L. Note that C3 and C4 must be of order sinceu2isfiniteatx1 = L.

Application .1

The boundary conditions at x1 = 0 (Fig. E12—3A) are

u2 = 0

M3 = E13u2,11 = 0

Since q is constant, the particular solution follows directly from (11),

= q/kThe complete solution is

U2 = (1 cos Ax1)


SEC. 12—6. FORCE METHOD OF SOLUTION 349

Fig. E12—3Aq = const

//////////// ///////////////////////////////////////// X1

x2

Application 2

The boundary conditions at x1 = 0 (Fig. E12—3B) are

U2,1 = 0

F2 = —E13u2,111 = —P12and the solution is

PAU2 = Ax1 + sin Ax1)

The four basic functions encountered are

= Ax + sin Ax)

= sin Ax = —

(12—26)

=

e cos Ax = —

Their values over the range from Ax = 0 to Ax = 5 are presented in Table 12—1.

Fig. E12—3B

x1////////////////////j///

x2

12—6. FORCE METHOD OF SOLUTION

In the force method, we apply the principle of virtual forces to determine thedisplacement at a point and also to establish the equations relating the forceredundants for a statically indeterminate member. We start with the one-dimensional form of the principle of virtual forces developed in Sec. 12—3 (seeEquation 12—19):

+ = d1 AP1



Table 12—i

Numerical Values of the Functions

,tX iJi1 AX

0.0 1.000 0.000 1.000 1.000 00.2 0.965 0.163 0.640 0.802 0.20.4 0.878 0.261 0.356 0.617 0.4

0.6 0.763 0.310 0.143 0.453 0.60.8 0.635 0,322 —0.009 0.313 0.8

1.0 0.508 0.310 —0.111 0.199 1.0

1.2 0.390 0.281 —0.172 0.109 1.2

1.4 0.285 0.243 —0.201 0.042 1.4

1.6 0.196 0.202 —0.208 --0.006 1.6

1.8 0.123 0.161 —0.199 —0.038 1.8

2.0 0.067 0.123 —0.179 —0.056 2.02.2 0.024 0.090 —0.155 —0.065 2.22.4 —0.006 0.061 —0.128 —0.067 2.42.6 —0.025 0.038 —0.102 —0.064 2.62.8 —0.037 0.020 —0.078 —0.057 2.8

3.0 —0.042 0.007 —0.056 —0.049 3.03.2 —0,043 —0.002 —0.038 —0.041 3.23.4 —0.041 —0.009 —0.024 —0.032 3.43.6 —0.037 —0.012 —0,012 —0.024 3.6

3.8

'

—0.031 —0.014 —0.004 —0.018 3.8

4.0 —0.026 —0.014 0.002 —-0.012 4.04.2 —0.020 —0.013 0.006 —0.007 4.24.4 —0.016 —0.012 0.008 —0.004 4.44.6 —0.011 —0.010 0.009 —0.001 4.64.8 —0.008 —0.008 0.009 0.001 4.8

5.0 —0.005 —0.007 0.008 0.002 5.0

where e3, are the actual one-dimensional deformation measures;d, represents a displacement quantity;AP, is an external virtual force applied in the direction of

The relations between the deformation measures and the internal forces dependon the material properties and the assumed stress expansions. The appropriaterelations for the linear elastic engineering theory are given by (12—13). Ifa displacement is prescribed, the corresponding force is actually a reaction.We use AR,. to denote a prescribed displacement and the correspondingreaction increment, and write (a) as

+ AM3)]dx,, — d,. AR,. d, AP, (12—27)

where d, represents an unknown displacement quantity.To determine the displacement at some point, say Q, in the direction defined

by the unit vector we apply a virtual force APQIq, and generate the necessaryinternal forces and reactions required for equilibrium using the one-dimensionalforce-equilibrium equations. We express the required virtual-force system as

= AP0

= MJ,QAPQ (12—28)

ARk==Rk,QAPQ



Introducing (12—28) in (12—27) and canceling leads to

dQ = — Q + + k3M1, Q)]dxj (12—29)

This expression is applicable for an arbitrary material, but is restricted to thelinear geometric case. Since the only requirement on the virtual force systemis that it be statically permissible, one can always work with a statically deter-minate virtual force system. The expanded form of (12—29) for the linearlyelastic case follows from (12—21):

dQ = +$

[(e? + Fj,Q

(F2\ i'F3\ M1.+ +

M3\ 1+ + -JM3,Q 1dx1J..J3/ J

if dA

=7±JJx2s?dA

Finally, we can express (12—29) for the elastic case in terms of V*:

L kdQ — (12—31)

JxLt'AQ ORQ

This form follows from (12—20) and applies for an arbitrary elastic material.

Example 12—4

We consider the channel member shown in Fig. E12--4A. We suppose that the materialis linearly elastic and that there is no support movement. We will determine the vertical

where

+ +El21

(12—30)

Fig. E12—4A

x2

I)Centroid

Q

Shearcenter



displacement of the web at point Q due to—1. the concentrated force P2. a temperature increase AT, given by

AT = a1x1 + a2x1x2 + a3x1x3

Force System Due to P

Applying the equilibrium conditions to the segment shown in Fig. E12—4B leads to

.F2 == +Pe

M3 = —P(L — x1)

F1 = F3 = M2 0

Fig. E12—4B

. •1

M3 ( P Shear center axis

I I FeF2,

System

We take dQ positive when downward, i.e., in the — X, direction. To be consistent,we must apply a unit downward force at Q. The required internal forces follow fromFig. E12—4C:

F2,0 = —1

Mr,0 C

/L \ (b)2 M3,Q = — x1)

F5,0 = F3,0 M2,Q 0

(p. =o(j= 1,2,3)

0

Fig. E12—4C

I Shear center axis/F2,Q

S



hzitial Deformations

The initial extensional strain due to the temperature increase is

= a = + a2x1x, + a3x1x3)

The equivalent one-dimensional initial deformations are

Deter,ninatio,, of (IQ

e? = .1JJ

d,4

= ±.Jf

dA = aa3x1

=dA = —aa2x,

Centroid (andshear center)

(d)

(e)

Fig. E12—5

(f)

Substituting for the forces and initial deformations in (12—30), we obtain

('Lii I P Pc2 r P 1 IL \)dQJL c2L 5 1)) cxa,L2

=

Example 12—5

When the material is nonlinear, we must use (12—29) rather than (12—30). To illus-trate the nonlinear case, we determine the vertical displacement due to P at the right end

x2

P

"I

xi

of the member shown in Fig. E12—5. We suppose that transverse shear deformation isnegligible, and take the relation between k3 and M3 as

k3 = a1M3 + (a)



Noting that only F2, and M3,0 are finite, and letting e2 = 0, the general expression forreduces to

L

d0 k3M30dx1

Now,M3 = —P(L — x1)

M3,Q = —(L — x1)

Then,k3 = —Pa1(L — x1) — P3a3(L — x1)3

Substituting for k3 in (b), we obtain

dQ = Pa1 + P3a3

We describe next the application of the principle of virtual forces in theanalysis of a statically indeterminate member. We suppose that the memberis statically indeterminate to the rth degree. The first step involves selectingr force quantities, Z1, Z2,. . . , Z.. These quantities may be either internal forcesor reactions, and are generally called force redundwns.

Using the force-equilibrium equations, we express the internal forces andreactions in terms of the prescribed external forces and the force redundants.

= +

= M3,0 Mf,kZk (12—32)

= + R1, kZk

The member corresponding to = Z2 = ''' = Z,. = 0 is conventionallycalled the primary structure. Note that all the force analyses are carried out onthe primary structure. The set (F3, 0, M3, R1, represents the internal forcesand reactions for the primary structure due to the prescribed external forces.Also, (F1, k, M3, k, R1, k) represents the forces and reactions for the primarystructure due to a unit value of Zk. One must select the force resultants suchthat the resulting primary structure is stable,

Once the force redundants are known, we can find the total forces from(12—32). It remains to establish a system of r equations relating the forceredundants. With this objective, we consider the virtual-force system consistingof AZ,, and the corresponding internal forces and reactions,

= F3,,,AZ,,

AM3 = MJ,,,AZ,,

AR, =



This system is statically permissible. Substituting (a) in (12—27), and notingthat = 0, we obtain

k + kJMi.k)] dx1 = k (12—33)

Taking k = 1, 2, ..., r results in a set of r relating the actual de-formations. One can interpret these equations as compatibility conditions,since they represent restrictions on the deformations.

To proceed further, we must express the deformations in terms ofIn what follows, we suppose that the material is linearly elastic. The com-patibility conditions for the linearly elastic case are given by

J[(eq + + + +

+ + + + dx1(12—34)

A more compact form, which is valid for an arbitrary elastic material, is

C (7R(k = i,2,.. .,r) (1235)

Ic Ic

The final step involves substituting for M1 using (12—32). We write theresulting equations as

.fIcJZJ = is,, (k = 1, 2. . . . , r) (12—36)j= 1

where

rr1 1 1

= fjk= J

+ +

+ + dx1

=>— j [(eq + F1.o)F + +

+ + + + +

The various terms in (12—36) have geometrical significance. Using (12—30),we see that is the displacement of the primary structure in the direction ofZ1 due to a unit value of ZIc. Since fik = fkJ, it is also equal to the displacementin the direction of Zk due to a unit value of Z3. Generalizing this result, we canwrite

= (12—37)



where i, j are arbitrary points, and corresponds to i.e., i has the samedirection and sense. Equation (12—37) is called Maxwell's law of reciprocaldeflections, and follows directly from (12—30). The term Ak is the actual dis-placement of the point of application of Zk. minus the displacement of theprimary structure in the direction of Zk due to support movement, initial strain,and the prescribed external forces. If we take Zk as an internal force quantity(stress resultant or stress couple), Ak represents a relative displacement (trans-lation or rotation) of adjacent cross sections.

One can interpret (12—36) as a superposition of the displacements dueto the various effects. They are generally called superposition equations inelementary texts.t If the material is physically nonlinear, (12—36) are notapplicable, and one must start with (12—33). The approach is basically thesame as for the linear case. However, the final equations will be nonlinear.The following examples illustrate some of the details involved in applying theforce method to statically indeterminate prismatic members.

Example 12—6

This loading (Fig. E12—6A) will produce flexure in the plane and twist aboutthe shear center; i.e., only F2, M3 and are finite. The member is indeterminate to thefirst degree. We will take the reaction at B as the force redundant.

Fig. E12—6A

x2 x2

X3

Shearcenter

Primary Structure

One can select the positive sense of the reactions arbitrarily. (See Fig. El 2—6B.) Wework with the twisting moment with respect to the shear center. The reactions are relatedto the internal forces by

= Z1

R2 —

R3 =R4 = +[MT]x,=o

t See, for example, Art. 13—2 in Ref. 3.


SEC. 12—6. FORCE METHOD OF SOLUTION

R3,d3

R2, d2

x2

357

E12—68

Force System Due to Prescribed External Forces R1

x2

Mr,oF2,0

q

—__

-.——p-*' bI qe

F2,0 = — x1) R1,0 = 0

MTo = qe(L — x1) R2,0 = qL

q qL2M3,0 = — x1)2 R3,0 =

F1,0 = F3,0 = M2,0 = 0 R4,0 = qeL

Force System Due to Z1 = +

lB

Fig. E12—6C

Shear center axis

Fig. E12—60

ZI = 0

(b)

M3,1

F2,1

Bt

Shear center axis

e



F2,1=+1MT,j = —e R2,1 = —1

M3,1=+(L—x1)F1, = F3, 1 = M2, 1 = 0 R4•1 = —e

Equation for Z1

We suppose that the member is linearly elastic. Specializing (12—36) for this problem,

f 11Z1 = A1

= 1)2 + 1)2 + )2]dX (d)

=

and then substituting for the forces and evaluating the resulting integrals, we obtain

L Le2 L3fit

(e)

—(L — x1)dx1

The value of Z1 for no initial strain or support movement is

z1 =8

Final Forces

The total forces are obtained by superimposing the forces due to the prescribed externalsystem and the redundants:

F2=F20+Z1F2,1 = —q(L—x1)+Z1MT qe(L — x1) — eZ1

M3 = — (L —

(g)

= qL — Z1

L2R3 = LZ1

= e(qL — Z1)



Example 12—7

This loading (Fig. E12—7A) will produce only flexure in the X1-X2 plane. We supposethe material is physically nonlinear and take the expression for k3 as

k3 = + a1M3 + (a)

To simplify the analysis, we neglect transverse shear deformation.

Fig. E12—7A

q

f////

V.

F

Primary Structure

= Z1 R7 = R3 = (b)

Fig. E12—78

x1

R2, d2

R1,2j

Force System Due to Prescribed External Forces (see Example 12—6)

F2,0 = —q(L — x1)

M3,0 = —

qL2R1,0 = 0 R2,0 = qL R3,0 =

z1 =0



Force Due to Z1 = + 1 (see Example 12—6)

F2, + 1 M3, = L —

R1,3=+1 R21=—1R3,1 = —L

Compatibility Equation

Since the material is nonlinear, we must use (12—33). Neglecting the transverse sheardeformation term (e2), the compatibility condition reduces to

Jdx1 =

We substitute for k3 using (a):

J (ajM3 + dx1 =— JL

Now,1W'3 = M3, + Z1M3,

= — x1)2 + Z1(L — xt)

Introducing (g) in (f), we obtain the following cubic equation for Z1:

z?(asLs) +

++

=

+ +— — — x1)dx1

For the physically linear case,

030and (h) reduces to

= + — —— fLko(L

— xi)dxi]

Example 12—8

The member shown (Fig. E12—8A) is fixed at both ends. We consider the case where thematerial is linearly elastic, and there are no support movements or initial strains. We takethe end actions at B referred to the shear Center as the force redundants.

=Z2 =Z3 = MTB

The forces acting on the primary structure are shown in Fig. E12—8B.

Initial Force System

F2,0 = P = P(a — x1)

MT,0



x3

Fig. E12—8A

tP

Fig. E12—8B

Fig. E12—8C

M30

MTO (F2,0

Shear center axis

x2

Shear

• a b

L

z3

PA

Px3



Z = +1

Fig. E12—8D

Al31

_________________

'( ti

F2,1 Shear center axis

L—x1'-I

F2,1=+1 M31=L—x1(c)

—0

Z2 = +1

Fig. E12—8E

M3,2

M72 (

M3,2 = +1 F22 = MT,2 0

z3 = +1

Fig. E12—8F

Al33

(I ////

I. —x1 Shear center axis

= + 1 F3, M3, = 0 (e)



Goinpatibility Equations

The compatibility equations for this problem have the form

JkJZJ = (k = 1,2,3)

fkj = i: + +

= f + +

Substituting for the various forces and evaluating the resulting integrals lead to the fol-lowing equations:

/ L I) \ 7 V [ a I (a3 a2bl——+-—-—1Z1 +l—1Z2 =\GA2 3E13) [GA2 El3 \ 3 2

(L2\ /L\ Pa2Z1

+Z2 = (g)

'\GJJ GJ

Finally, solving (g), we obtain

6E13

2b1 PICA= — P 1 ±

L2GAZ

Pox3Z3 Z3__

Application

Suppose the member is subjected to the distributed loading shown in Fig. E12—8G.We can determine the force redundants by substituting for P, a, and b in (h),

P = q dx1

a =b = L — x1

and integrating the resulting expressions. The general solution is

CL (2

2 6E13 1)

=+ — x1) + —

z2

X3Z3 — j x1q dx1



where12E13

C = 1 +L2GA2

As an illustration, we consider the case where q iswe obtain

qLz1 = — —

2

z2 =12

q const in (j),

Fig. E12—80

REFERENCES

1. TIMOSHaNKO, S. J. : Advanced Strength of Materials, Van Nostrand, New York, 1941.2. HETENY!, M.: Beams on Elastic Foundation, University of Michigan Press, Ann Arbor,

1946.3. Noiuus, C. H., and J. B. Elementary Structural Analysis, McGraw-Hill,

New York, 1960.4. ASPLUND, S. 0.: Structural Mechanics: Classical and Matrix Methods, Prentice-Hall,

1966.5. DEN HARTOG, J. P.: Advanced Strength of Materials, McGraw-Hill, New York, 1952.6. ODEN, J. T.: Mechanics of Elastic Structures, McGraw-Hill, New York, 1967.7. Geac, J. M. and WEAVER, W.: Analysis of Framed Structures, Van Nostrand, 1965.8. MARTIN, H. C.: Introduction to Matrix Methods of StructuralAnalysis, McGraw-Hill,

New York, 1966.

PROBLEMS

12-1. The accompanying sketch shows a sandwich beam consisting of acore and symmetrical face plates. The distribution of normal stress over thedepth is determined by assuming a linear variation for the extensional strain:

x2

q(xj)

H L H


PROBLEMS 365

= —x2k3

We relate k3 to M3 by substituting for in the definition equation for M3:

M3 = x2a11 dA

M3 + Ef13,f)k3

To simplify the notation, we drop the subscript and write (b) as

iVI

where (EI)equiv is the equivalent homogeneous flexural rigidity.

Prob. 12—1

x2

012 A*

fM3

I

The shearing stress distribution is determined by applying the engineeringtheory developed in Sec. 11—7. Integrating the axial force-equilibrium equationover the area A* and assuming is constant over the width, we obtain

JJ(aii,i + a21,2 + a31, 3)dA = o

(d)ba12 = cru, dA

Then, substituting for cru,

= —(Ek3)x2M

(—Ex2) (e)( I)equiv

and noting that F2 —M3, (d) becomes

a12 = J'J x2E dA



(a) Apply Equations (e) and (f) to the given section.(b) The flange thickness is small with respect to the core depth for a typical

beam. Also, the core material is relatively soft, i.e., and G. aresmall with respect to Ef. Specialize part a for 0 and tf/h 1. Alsodetermine the equivalent shear rigidity which is defined as

1(17*)

= J)dA

2

(c) The member force-deformation relations are

FY2

=

kM3

= (EI)equjv

Refer to Example 12—1. Specialize Equation (q) for this section anddiscuss when transverse shear deformation has to be considered.

12—2. Using the displacement method, determine the complete solutionfor the problem presented in the accompanying sketch. Comment on theinfluence of transverse shear deformation.

Prob. 12—2

12—3. For the problem sketched, determine the complete solution by thedisplacement method.

12—4. Determine the solution for the cases sketched. Express the solutionin terms of the functions defined by (12—26).

q = const

F b

x1


PROBLEMS 367

(0)

Prob. 12—3

q const

Prob. 12—4

/////////////////////////////7///////////////////

Jr

Ib)

(c)

12—5. The formulation for the beam on an elastic foundation is based ona continuous distribution of stiffness; i.e., we wrote

b2 = —ku2 (a)

Note that k has units of force/(Iength)2,We can apply it to the system of discrete restraints diagrammed in part a

of the accompanying sketch, provided that restraint spacing c is small in

x2

Shear

HeH

R



comparison to characteristic length (boundary layer) Lb, which we have taken as

3 3

2 (k/4E1)'14

A reasonable upper limit on c is

c <

Letting k4 denote the discrete stiffness, we determine the equivalent distributedstiffness k from

k = kd/c

Evaluate Lb with (b), and then check c with (c).

Prob. 12—5

1 . (

J> J...—

J J

r r r r

+ C +(a)

IIIC+C

L,E,11

a/2

/7/7 /7/7 7

(b)

Consider the beam of part b, supported by cross members which are fixedat their ends. Following the approach outlined above, determine the distribu-tion of force applied to the cross members due to the concentrated load, P.


PROBLEMS 369

Evaluate this distribution for

a=241t L=64ft c—lit12—6. Refer to Example 12—3. The governing equation for a prismatic

beam on a linearly elastic foundation with transverse shear deformation in-cluded is obtained by setting b2 = q — ku2 in (i). For convenience, we dropthe subscripts:

d4u k d2u k 1 / d2 7 q— + u =

— +—

We let

&112and (a) takes the form

d4 ,12—

4 2 u—qdx dx

Note that is dimensionless and A has units of 1/length. The homogeneoussolution is

u cos bx + C2 sin bx) + cos bx + C4 sin bx)

wherea = 2(1 +b = 2(1 —

To specialize (d) for negligible transverse shear deformation, we set = 0.

(a) Determine the expression for the boundary layer length (e3 0).(b) Determine the solution for the loading shown. Assume L large with

respect to Lb. The boundary conditions at x 0 are0) = 0

PF2

Investigate the variation of Mmax and Umax with Consider to varyfrom 0 to 1.

Prob. 12—6

P

/////////)///////////// ////////////)//////// X

a

L


370 ENGINEERING THEORY OF MEMBERS CHAP, 12

12—7. Refer to the sketch for Prob. 12—3. Determine the reaction R andcentroidal displacements at x1 L/2 due to a concentrated force Pi2 appliedto the web at x1 L/2. Employ the force method.

12—8. Refer to Example 12—7. Assuming Equation (h) is solved for Z1,discuss how you would determine the translation u2 at x1 = L/2.

12—9. Consider the four-span beam shown. Assume linearly elastic be-havior, the shear center coincides with the centroid, and planar loading.

(a) Compare the following choices for the force redundants with respectto computational effort:

1. reactions at the interior supports2. bending moments at the interior supports

(b) Discuss how you would employ Maxwell's law of reciprocal detlectionsto generate influence lines for the redundants due to a concentratedforce moving from left to right.

Prob. 12—9

12—10. Consider a linearly elastic member fixed at both ends and subjectedto a temperature increase

Determine the end actions and displacements (translations and rotations) atmid-span.

12—11. Consider a linearly elastic member fixed at the left end (A) andsubjected to forces acting at the right end (B) and support movement at A.Determine the expressions for the displacements at B in terms of the supportmovement at A and end forces at B with the force method. Compare thisapproach with that followed in Example 12—2.


13

Restrainedof.

a Prismatic Member

13—1. INTRODUCTION

The engineering theory of prismatic members developed in Chapter 12 isbased on the assumption that the effect of variable warping of the cross sectionon the normal and shearing stresses is negligible, i.e., the stress distributionspredicted by the St. Venant theory, which is valid only for constant warpingand no warping restraint at the ends, are used. We also assume the crosssection is rigid with respect to in-plane deformation. This leads to the resultthat the cross section twists about the shear center, a fixed point in the crosssection. Torsion and flexure are uncoupled when one works with the torsionalmoment about the shear center rather than the centroid. The complete set ofgoverning equations for the engineering theory arc summarized in Sec. 12—4.

Variable warping or warping restraint at the ends of the member leads toadditional normal and shearing stresses. Since the St. Venant normal stressdistribution satisfies the definition equations for F'1, M2, M3 identically, theadditional normal stress, must be statically equivalent to zero, i.e., it mustsatisfy

dA = dA = dA = 0 (13—1)

The St. Venant flexural shear flow distribution is obtained by applying theengineering theory developed in Sec. 11—7. This distribution is statically equiva-lent to F2, F3 acting at the shear center. It follows that the additional shearstresses, and due to warping restraint must be statically equivalentto only a torsional moment:

Sfri2 dA 0(13—2)

dA = 0

To account for warping restraint, one must modify the torsion relations. Wewill still assume the cross section is rigid with respect to in-plane deformation.

371


372 RESTRAINED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13

En what follows, we develop the governing equations for restrained torsion.We start by introducing displacement expansions and apply the principle ofvirtual displacements to establish the force parameters and force-equilibriumequations for the geometrically linear case. We discuss next two proceduresfor establishing the force-displacement relations. The first method is a pure-displacement approach, i.e., it takes the stresses as determined from the strain(displacement) expansions. The second method is similar to what we employedfor the engineering theory. We introduce expansions for the stresses in termsof the force parameters and apply the principle of virtual forces. This cor-responds to a mixed formulation, since we are actually working with expansionsfor both displacements and stresses. Solutions of the governing equations forthe linear mixed formulation are obtained and applied to thin-walled open andclosed cross sections. Finally, we derive the governing equations for geomet-trically nonlinear restrained torsion.

13—2. DISPLACEMENT EXPANSIONS; EQUILIBRIUM EQUATIONS

The principle of virtual displacementst states that

JJJaT ös d(vol.) = SSSbT d(vol.) + JJJJT d(surface area)

is identically satisfied for arbitrary displacement, L\u, when the stresses (r) arein equilibrium with the applied body (b) and surface (p) forces. We obtain asystem of one-dimensional force-equilibrium equations by introducing expan-sions for the displacements over the cross section in terms of one-dimensionaldisplacement parameters. This leads to force quantities consistent with the dis-placement parameters chosen.

We use the same notation as in Chapters 11, 12. The X1 axis coincides withthe centroid; X2, X3 are principal inertia axes; and x3 are the coordinatesof the shear center. We assume the cross section is rigid with respect to in-planedeformation, work with the translations of the shear center, and take the dis-placement expansions (see Fig. 13—1) as

U1 = U1 + W2X3 W3X2 +

U2 — w1(x3 — (13—3)

U3 = + w1(x2 — x2)

where 4 is a prescribed function of x2, x3, and—

1. u1, ;2, u53 are the rigid body translations of the cross section.2. W2, £03 are the rigid body rotations of the cross section about the

shear center and the X2, X3 axes.3. f is a parameter definining the warping of the cross section. The

variation over the cross section is defined by

Note that all seven parameters are functions only of x1. For pure torsion

t See Sec. 10—6.


SEC. 13—2. DISPLACEMENT EXPANSIONS; EQUILIBRIUM EQUATIONS 373

(i.e., the St. Venant theory developed in Chapter 11), one sets f = co1, i = constand = For unrestrained variable torsion (i.e., the engineering theorydeveloped in Chapter 12), one sets f = 0. Since there are seven displacementparameters, application of the principle of virtual displacements will result inseven equilibrium equations.

x3

Shear center

— —:: — —e U52

I

I

I x2

Centroid

Fig. 13—1. Notation for displacement measures.

The strain expansioust corresponding to (13—3) are

6j u1, 1 + (02. (03, 1x2 + f

Ytz = us2, 1 — C03 — COj, 1(x3 — x3) + 2 (13—4)= us3, 1 + (02 + cot. 1(x2 — + 3

Using (13—4), the left-hand side of (a) expands to

öe d(voL) Au1, + F2(Au,2, 1 — Aw3)

+ F3(Au33 + Aw2) + M2 Aw2, + M3 Ac)3,1

+ MT Aw5,1 + A1 + MR Af]dxi

where the two additional force parameters are defined by

= dA(13—5)

MR = +

Note that Mç1, has units of (force) (length)2 and MR has units of moment. Thequantity Mci, is called the biinornent.

t This derivation is restricted to linear geometry. The nonlinear strain expansions are detivedin Sec. 13—9.



To reduce the right-hand side of (a), we refer the transverse loading to theshear center, The additional load terms are

m4, = 4picb dS = distributed bimornent(136)

= $5pjcb dA = external bimoment at an end section (x1 = 0, L)

Then

SJJbT Au d(vol.) + SSPT Au d(surface area)

= Au1 + b2 1xu32 + b3 Au,3 ± mr + rn2 Aw2

+ m3 Aw3 + m# Af]dx1 + JF1 Au1 + F2 Au,2+ F3Au,3 + MTLXWI + M2Aw2 + M3 Aw3 +

The definitions of mj, mj', F,,, are the same as for the engineeringtheory.

Finally, we equate (b), (c) and require the relation to be satisfied for arbitraryvariations of the displacement parameters. This step involves first integrating(b) by parts to eliminate the derivatives and then equating the coefficients ofthe displacement parameters. The resulting equilibrium equations and bound-ary conditions are as follows:

Equilibrium Equations

F1 + b1 0

F2, + b2 0

F3, 1 + b3 = 0

MTI+m-I-=OM2, 1 — F3 + rn2 = 0

M3,1 + F2 + m3 = 0

M4, — MR + 0

Boundary Conditions at x1 0 (13—7)

u1 = u1 or F1 = —F1

Us2 or F2 =US3 or F3 —F3

w1=w1 or Mr=—MTor M2 =

w3=co3 or

f=J or

Boundary conditions at x1 = L

These are the same as for x1 = 0 with the minus sign replaced with a plus sign.For example:

f=J or


SEC. 13—3. DISPLACEMENT MODEL 375

We recognize the first six equations as the governing equations for theengineering theory. The additional equation,

O<xj<Lf=f or

is due to warping restraint. Also, we see that one specifies either f or thebimoment at the ends of the member. The condition f = 7' applies when theend cross section is restrained with respect to warping. If the end cross sectionis free to warp, the boundary condition is = ± M4, (+ for x1 = L).

To interpret the equation relating MR and the bimoment, we consider thedefinition for MR,

MR = +

Integrating (e) by parts leads to

= SS4(ail,2 + a13, 3)dA

Utilizing the axial stress equilibrium equation,

a12, 2 + c713, 3 + a11, 1 0we can write

MR = + JJç'au,j dA

We see that (h) corresponds to the axial equilibrium equations weighted withrespect to

$$(a12,2 + a13,3 + + 4(PI — — x,,3a13)çdS = 0(i)

i + — MR = 0

In most cases, there is no surface loading on S. i.e., Pi = 0 on the cylindricalboundary. We will discuss the determination of stresses in a later section. Wesimply point out here that MR involves only the additional shear stresses dueto warping restraint since the St. Venant shearing correspond to

1 3—3. FORCE-DISPLACEMENT RELATIONS—DISPLACEMENT MODEL

To establish the relation between force parameters and the displacementparameters, we consider (13-4) to define the actual (as well as virtual) straindistribution and apply the stress-strain relations. We also consider the materialto be isotropic and suppose there is no initial strain. The stress expansions are

au 1 + i +a12 = Gy12 = — 0)3 cot, 1(x3 X3) + f4, 2] (13—8)

a13 = Gy13 = G[u,3, 1 + + cot, 1(x2 — + f4, 3]

t M5 = = 0 for St. Venant (pure) torsion. We neglect and for unrestrained variab'etorsion.



where denotes the effective modulus. Although our displacement expansionscorrespond to plane strain (&2 = = 0), the in-plane stresses vanish on theboundary. Therefore, it seems more reasonable to use the extensional stress-strain relations for plane stress. In what follows, we will take = Young'smodulus, F.

Consider the expression for The term involving is due to warping ofthe cross section. This additional stress must satisfy (13—1), which, in turn,requires 4 to satisfy the following orthogonality conditions :t

dA dA = dA = 0 (13—9)

Assuming (13—9) is satisfied, and noting that X2, X3 are principal centroidalaxes, the expressions for F1, M2, M3, and the reduce to:

EAu1,1

M2 E12w2,113—10

M3 E13a,3 1

jwhere

cIA

We have included the subscript r on E to keep track of the normal stress dueto warping restraint. Inverting (13—10) and then substituting in the expressionfor lead to

F1 M2 M3'Yii + — + (13—Il)

13

The expressions for F2, F3, and MR expand to

F2 1 — (03 + i) + fS2

F3 = A(u33, i + + fS3

1 (13—12)11c01,1

— + —

= + + +

whereSi =

polar moment of inertia = 12 + 13— x3q5,2)dA

= + +

t F1 = M2 = = 0 for c11 due to warping restraint.


SEC. 13—3. DISPLACEMENT MODEL 377

Also, the expressions for the shearing stresses canbe written as

a12

— + G 1 + f 3

—

The essential step is the selection of which, to this point,must satisfy onlythe ortliogonality conditions (13—9). To gain some insight as to a suitable formfor let us reexamine the St. Venant theory of unrestrained torsion. Wesuppose the section twists about an arbitrary point instead .of aboutthe centroid as in Sec. 11—2. The displacement expansions are

u2 = —w1(x3 — U3 coj(x2

=

where i = M1/GJ = const. Operating on (a) leads to

= 0

a12 = 2 (x3 — x3)]

7 3 +

The equation and boundary condition for follow from the axial equilibriumequation and boundary condition,

mA

— — on S

We can express as

= C — + x'2x3 +

where C is also an arbitrary constant. The boundary condition and expressionsfor tile stresses become

—

a12 = — x3)

M1 -

cr13 3 + x2)

Since depends only on the cross section, it follows that the stress distribution



and torsional constant are independent of the center of twist. Also, one canshowt that

SJcbt,2dA = = 0

3 — 2)dA $S[(4)t. 2) + 3)2]dA

Suppose we take 4' = The constants (C, are evaluated by requiring4' to satisfy (13—9), and we obtain

C=

=

= dA

Now, one can shows that the equations for are identical to the equationsfor the coordinates of the shear center when the cross section is considered tobe rigid with respect to in-plane deformation. That is, the warping functionfor unrestrained torsion about the shear center is orthogonal with respect to1, x2, x3.

Summarizing, we have shown that

4) C — T3x2 + x2x3 + (13—14)

is a permissible warping function. The cross-sectional properties and force-displacement relations corresponding to this choice for 4' are listed below:

Properties

S2 = =14, 14,

'13—15

2) + 3)2]dA

Shear Stresses

F2a12 = + G(—x3co1,1 +

(13—16)

= + G(x2w1,1 + f4)t,

t See Sec. 11—2 and Prob. 11—2.See Prob. 13—1.


SEC. 13—4. RESTRAINED TORSION—DISPLACEMENT MODEL 379

Force-Displacement Relations

MT = G11w1, i — + —

MR = w1, — + .T2F3

us2, j W3 — w1, (13—17)

F3j + W2 + x2(f Wi,

We introduce the assumption of negligible restraint against warping bysetting Er 0. Then, M4, 0, and thc seventh equilibrium equation reducesto MR = 0. Specializing (13—17) for this case, we obtain

—,f — to1, = (x3F2 — x21'3)

(13—18)

MT = GJW1,1

andF2(1 F3( x2x3

\J1(13—19)F2( F3!'!

us3,1 = to2 + + -- +G\ 14,

The shearing stress distributions due to F2, F3 do not satisfy the stress boundarycondition

+ 0 on S

However, one can show that they satisfy

+ = 0

for arbitrary F2, F3. Equations (13—19) are similar in form to the resultsobtained in Chapter 12, which were based on shear stress expansions satisfying(a) identically on the boundary.

Finally, we point out that torsion and flexure are uncoupled only whenwarping restraint is neglected (F. = 0). Equations (13—17) show that restrainedtorsion results in translation of the shear center. We will return to this point inthe next section.

13—4. SOLUTION FOR RESTRAINED TORSION—DISPLACEMENTMODEL

To obtain an indication of the effect of warping restraint, we apply thetheory developed in the previous section to a cantilever member having arectangular cross section. (See Fig. 13—2). The left end (x1 = 0) is fixed with



x3

T

/

2b

A

Sect. A-A

Fig. 13—2. Restrained torsion-cantilever with rectangular cross section.

respect to both rotation and warping while the right end (x1 = L) is free towarp. The boundary conditions are

x1=Ox1=L M1=M (a)

For convenience, we list the governing equations for restrained torsion:

Equilibrium Equations (See (13—7))

M1,1+m1=O (b)

(c)

Force-Displacement Relations (See (13—10) and (13—12). Note thatF2 = F3 = 0)

=M1 = G11w11 + (d)

= +

Boundary Conditions (for this example)

At xj= 0,(e)

Atx1= M

1,1=0We start with (b). Integrating (b) and enforcing the boundary condition at

Xj = L leads toM1 = M (13—20)


SEC. 13—4. RESTRAINED TORSION—DISPLACEMENT MODEL 381

Next, we combine (c) and (d):

G11co11 + = M (f)

i + = t (g)Solving (f) for w1,

G111, (h)

and then substituting in (g) lead to

(i)—

where 2 is defined asG [i,, 13 21-

Note that has units of(1/length)2. The solution of(i) and (h) which satisfiesthe boundary conditions (e) is (we drop the subscript on x for convenience)

f = {l — cosh + tanh sinh

= x±

[sinh + (1 — cosh tL]} (13—22)

=+I,fr '1

The rate of decay of the exponential terms depends on For )L > 2.5,we can take tanh )L 1, and the solution reduces to

—

As a point of interest, the St. Venant solution is

dw1 M(J)

We see that is a measure of the length, Lh, of the interval in which warpingrestraint is significant. We refer to Lb as the characteristic length or boundarylayer. By definition,

0 (13—24)

In what follows, we shall take

(13—25)

The results obtained show that is the key parameter. Now, depends onthe ratio G/ET and on terms derived from the assumed warping function. If



we take 4) the warping functiowt for unrestrained torsion defined by(13—14), the various coefficients are related by

J — (13—26)

— J

At this point, we restrict the discussion to a rectangular section (see Fig. 13—2)and 4) = We evaluate the various integrals defined by (13—15) and writethe results as

J =

= K1a3b (13—27)

=

where the K's are dimensionless functions of b/a. With these definitions, theexpression for takes the form

— /G'\112 12= --- K2-

K -The coefficients are tabulated in Table 13—1. We see that is essentiallyconstant. Assuming E 2.6G and K1 3.2, we find 2/b and Lb 2b.

The influence of warping restraint is confined to a region of the order of thedepth. Although this result was derived for a rectangular cross section, we willshow later that it is typical of solid and also thin-walled closed cross sections.

Table 13—1

b/<4

1 2.25 .0311 .156 3.362 3.66 .165 .450 3.163 4.21 .283 .683 3.23

10 4.99 .425 .964 3.32

We consider next the problem of locating the center of twist. We utilize thesolution corresponding to 4) and large

-,, (13—29)

M1—e

f C = = 0 for a rectangular section and reduces to


SEC. 13—5. MIXEIJ FORMULATION 383

The translations of the shear center follow from (13—17):

= x3(f —(13—30)

Us31 = —x2(f —

By definition, the translations are zero at the center of twist. Setting i12 =0 in (13—3) and letting denote the coordinates of the center of twistlead to

X2 = gx2

x )We see that the center of twist approaches the shear center as x increases. Themaximum difference occurs at x = 0 and the minimum at x = L.

1

= = —

1 —2L!1

For unrestrained warping, E, = 0, = oc, and g = 1.

13—5. FORCE-DISPLACEMENT RELATIONS—MIXED FORMULATION

We first review briefly the basic variational principles for the three-dimen-sional formulation. The principle of virtual displacements requires

5e d(vol.) = JJjbT Au d(vol.) + Au d(surface area)

to be satisfied for arbitrary Au and leads to the stress-equilibrium equationsand stress-boundary force relations. Note that ôg is a function of Au and isobtained using the strain-displacement relations, The stress-strain relationscan be represented as

since, by definition of the complementary energy density,av*

= =

By combining (a) and (b), we obtain a variational principle which leads to bothsets of equations. The stationary requirement,

— liT11 — V*)d(vol.) — d(surface area)] = 0 (13—33)

considering i, u as independent quantities, g E(u), and b prescribed, iscalled Reissner's principle.t

See Ref. 11 and Prob. 10—28. Reissner's principle applies for arbitrary geometry and elasticmaterial. This discussion is restricted to linear geometry. The nonlinear case is treated in Sec. 13—9.


384 RESTRAINED TORSION-FLEXURE OF PRTSMATIC MEMBER CHAP. 13

The essential point to recognize is that Reissner's principle allows one towork with and u as independent quantities. In a displacement formulation(Sec. 13—3), we take as a function of u, using the stress-displacement relations

= = and — V* reduces to V, the strain-energy density. In amixed formulation we start by introducing expansions for the displacements.

The Euler equations for the displacement parameters are obtained by ex-panding (a). This step leads to the definition of force parameters and force-equilibrium equations. We then generate expansions for the stresses in termsof the force-parameters from an equilibrium consideration. The relationsbetween the force and displacement parameters are obtained from the secondstationary requirement:

Tx1[f$(CT — öV*)dA]dxj = 0 (13—34)

The first step was carried out in Sec. 13—2 and the expanded form of 5$ dAis given by (b) of Sec. 13—2. Letting represent the complementary energyper unit length along X1, and using (13—4), the stationary requirement on thestresses (Equation 13—34) expands to

+ 1 — W3) + öF3(u,3, i + co2) + i13—35)

+ + + + — = o

In order to proceed further, we must express in terms of the force parameters(F1, F2, ..., MR). Equating the coefficients of each force variation to zeroresults in the force-displacement relations.

Instead of applying (13—34), one can also obtain (13—35) by applying theprinciple of virtual forces to a differential element. We followed this approachin Chapter 12 and, since it is of interest, we outline the additional steps requiredfor restrained torsion. One starts with (see Fig. 13—3)

5V* dx1 = op + [$$uT ISp (a)

The boundary forces are the stress components acting on the end faces. Takingu according to (13—3) and considering only MR. we have

$5oiiTudA = ±SJISrsTudA

= ±(ISMTWI + ISM4f)

where the plus sign applies for a positive face. The virtual-force system mustbe statically permissible, i.e., it must satisfy the one-dimensional equilibriumequations. This requires

const

=

Then,

= dxi{f,i + w1,1(d)

= dx1{f 0M4 + .[ISMR + 1 ISMT}


SEC. 13—5. MIXED FORMULATION 385

The first procedure (based on (13—34)) is more convenient since it avoidsintroducing the equilibrium equations. However, one has to have the strain-displacement relations. In certain cases, e.g., a curved member, it is relatively

6Mg, +

o,1+w11dx1

f + f,1 dxj

Fig. 13—3. Virtual force system.

easy to establish the force-equilibrium equations by applying the equilibriumconditions to a differential element. We obtain the force-displacement relationsby applying the second procedure (principle of virtual forces) without havingto introduce strain expansions.t

In what follows, we consider the material to be homogeneous, linearly elasticand isotropic. To simplify the treatment, we also suppose there is no initialstrain. The complementary energy density is

=if

dA + if + (13-36)

It remains to introduce expansions for the stress components in terms of theforce parameters such that the definition equations for the force parametersare identically satisfied.

Considering first the normal stress, we can

F1 M2 M3(a)

where satisfies the orthogonality conditions: §

dA = JJx2çb c/A = c/A = 0 (b)

Note that we have imposed a restriction on q5. The complementary energydue to c11 expands to

—, 1 I

t The approach based on the principle of virtual forces is not applicable for the geometricallynonlinear case.

See 1). Problem treats the case of a nonhomogeneous material.§ F1 = = M3 = 0 for due to warping restraint.

—aM,5

WI

f



Finally, substituting for in (13—35), we obtain

F1

W3, I f, —L.13

These expansions coincide with the corresponding relations obtained with thedisplacement model (see (13—10)).

The shearing stress distribution must satisfy thc definition equations forF2, F3, MT and MR identically. We can obtain suitable expansions by addinga term due to warping restraint to the results for unrestrained torsion andfiexure. We write

('If = + + (13—37)

where is the flcxural distribution due to F2, F3; is the unrestrainedtorsion distribution; and is the distribution due to restrained torsion.

Since we are assuming no in-plane deformation, the fiexural distribution fora thin-walled section can he obtained by applying the engineering theorydeveloped in Sec. 11—7. For a solid section, we utilize the results of Sec. 11—5,taking v = 0.

The shear stress distribution for unrestrained torsion is treated in Sees. 11—2through 11—4. Since the restrained-torsion distribution is statically equivalentto a torsional moment, we have to distinguish between the unrestrained andrestrained torsional moments:

MT = += (13—38)

=

It remains to determine We follow the same approach as in the engi-neering theory of flexural shear stress, i.e., we utilize the axial equilibriumequations and stress boundary condition:

('12,2 + mA+ 0 on S

Differentiating the expression for au and noting the equilibrium equations,we obtain

F2 F3 MR('11,1 = + +

13 12

Since satisfies (a) for arbitrary F2, F3 and a" corresponds to = 0, itfollows that r is due to MR:

y r MR('12.2 + ('13.3 = (mA)

4 —-

+ 0 (on 5)


SEC. 13—5. MIXED FORMULAflON 387

The orthogonality conditions on and boundary condition on a' ensure thatt

= dA = 0 dA = 0 (13 —40)

We solve (13—39) and then evaluate from

= J$[—(x3 — x3)a12 + (X2 -.- (c)

Noting (13—40), we see that Finally, we write (c1 as

= (13—41)

where is a cross-sectional property which depends on With this definition,

MT = + (13—42)

When the cross section is thin-walled, we neglect and (a) reduces to

= d= 0 at a free edge

We take and to be constant over the thickness t and work with the shearflow qr Equation (d) becomes

= 1çô1(13—43)

qV = 0 at a free edge

The orthogonality conditions on and boundary condition on ensure that

= (IS — 0(13—44)

= 0

Finally, we determine by evaluating and equating to (13—41).We consider next the complementary energy density. We write the expanded

form of the shear contribution as

V*,hear

JJ+ + + + + (134)

We have evaluated and in Sec. 11—5. For convenience, theseresults are summarized below (See Equation 11—98)

I /z'2I. 112 13

Vf = + +23 3

(a)LI

2GJ11* —V uf

f See Prob. 13—2.



The coupling term, 1/A23, vanishes when the section has an axis of symmetry.Also = 0 is a consequence of our assuming the cross section is rigid withrespect to in-plane deformation.

We evaluate using (13—39) ((13—43) for the thin-walled case), and writethe results as

=if

+ (13—46)

where Cr is a dimensionless factor which depends on q5.The coupling between unrestrained and restrained torsion is expressed as

=if + (13-47)

It is obvious that = 0 for a thin-walled open section since is an oddfunction of n whereas a' is constant over the, thickness. We will show laterthat it is possible to make vanish for a closed section by specializing thehomogeneous solution of (13—43). Therefore, in what follows, we will takeCur 0

Finally, we write the coupling bctwcen flex ural and restrained torsion as

= +

1(13—48)

= + X2rF3Mg)

where Xjr have units of length. if X2 is an axis of symmetry, = 0 sinceis symmetrical and a' is antisymmetrical with respect to the X2 axis.

We substitute for in (13—35). replace Mr with + Ccj,MR, and equatethe coefficients of oF2, OF3, and <5MR. The resulting force-displacementrelations are

I "F2 F3 x3,U521 — 0)3 + —-- + —

Us3,1 + 0)2=

+GA23 A3 J(13—49)

Wi.j=

C, 1

I + I = MR + (x3,F2 + x2,F3)

The corresponding relations for the displacement model are given by (13—12).Up to this point, we have required to satisfy the orthogonality relations

and also determined a' such that there is no energy coupling between au and(C,1, = 0). If, in addition, we take

If" — — ISC5) — — X3X2 -t-- X2X3 -r —


SEC. 13—6. RESTRAINED TORSION—MIXED FORMULATION 389

Cçt, = +1(13—50MrT=+MR

Note that is the warping function for unrestrained torsion about the shearcenter. We discuss the determination of 4> in Secs. 13—7 and 13—8.

One neglects shear deformations due to flexure by setting

(13-51)

Similarly, we neglect shear deformation due to restrained torsion by setting

Cr = X2r = X3r 0 (13—52)

This assumption leads to the center of twist coinciding with the shear center and

1 = (13—53)

One now has to determine from the equilibrium relation,

+ in4,

If is known, it is more convenient to work with

tsr — AS AK:'— —

In what follows, we outline the solution procedure for restrained torsion andlist results for various loadings. We then discuss the application to open andclosed cross section.

13—6. SOLUTION FOR RESTRAINED TORSION—MIXED FORMULATION

We suppose only torsional loading is applied. The force-displacement rela-tions are obtained by setting F2, F3, co2, w3 equal to zero and C,1, +1 in(13—49). For convenience, we summarize the governing equations below.


MT. 1 + 112T = 0

M — AKT

Force-Displacement Relations (4> —

E,.l4,j 1

= GJw1,1GJ

+f)

f See Prob. 13—3. We include the minus sign so that C1 will be positive.



Boundary Conditions

MT oror

prescribed at each end

Translations of the Shear renter

u52, 1 ==

We start by integrating (a):

MT C1 — dx1 = C1 + (13—54)

Substituting (c) in (b) and (13—54) leads to the governing equations for w1 and f:

(1 + Cr)O)i,t +

GJ(w1,1 + f) = 0

After some manipulation, (f) becomes

Eric,, C1x1 I C

f —=

(C1 +

where ,12 is defined ast

cc =1 + Cr (1355)

22 CErI#

Equation (g) corresponds to (h), (i) of Sec. 13—4.The general solution for f and has the following form:

f C3 cosh 2x + C4 sinh 2x — +

Wi = + C2 + (13—56)

— sinh 2x + C4 cosh 2x) —

is the particular solution due to We have dropped the subscripton x1 for convenience.

t The corresponding paramater for the formulation is 2 (see (13—21)).



The significance of A has been discussed in Sec. 13—4. We should expect,on the basis of the results obtained there, that AL will be large with respectto unity for a closed section. We will return to the evaluation of A in the nextsection. In the examples below, we list for future reference the solution forvarious loading and boundary conditions.

Example 13—1

Cantilever—Concentrated Moment

Fig. E13—1

X1

The boundary conditions (Fig. E13—l) are

x=O cv1=f=Ox=L

f. = 0

Starting with (13—54), we set = 0 and C1 = M. The remaining constants are deter-mined from

w1—f——0 atx=0atx=L

and the final solution ist

M[ coshA(L—x)GJ cosh AL

WI [x — {sinh AL — sinh A(L — x)}

r —c 1 (13—57)M sinh ).(L — x)

LA cosh ,,L.

[cosh

M —

t The corresponding solution based on the displacement model is given by (13—22), (13—26).The expressions for! differ by a minus sign. This is due to our choice of We tookin the displacement model and = in the mixed model.



Note that C, 1 when the complementary energy term due to the restrained torsionshear stress (o') is neglected.

The translations of the shear center are obtained by integrating

u,2,, = =

and requiring u,2, 0s3 to vanish at x = 0. We write the result as

U,2 = X3.U = X2,.U

I M (13-58)

u=j M1dx=x—co1

Let denote the coordinates and translations of the center of twist. By definition,

= u,2 — — x3) = 0

=u,3 + —X2)=0

Substituting for and w1, we obtain t

X2 — — — gx3,

g —1 + c (13—59)

x — ——f—-— [sinh ).L — sinh — x1]A. cosh ,.L

The limiting of g occur at x = 0, L.

_______

1

(13—60)

1

+

Note that Xi,. = 0 if X,, (j k) is an axis of symmetry for the cross section. Also, x2. =x3,. = 0 if we neglect shear deformation due to the restrained shear stress and, in thiscase, the center of twist coincides with the shear center throughout the length.

Example 13—2

We consider next the case where warping is restrained at both ends; the left end (x = 0)

is fixed and the right end rotates a specified amount w under the action of a torsionalmoment. The boundary conditions are

x=0 co1=f=0x=L ce1=w [=0

f See (13—31), (13—32) for the displacement model solution.There is no twist or translation at x = 0. We determine g(O) by applying L'I-IOspital's rule

to (13—59).



To simplify the analysis, we suppose there is no distributed load. Starting with thegeneral solution,

MT = C1

f = C3 cosh Ax + C4 sinh —

= + C2 — {C3 sinh Ax + C4 cosh Ax}

and enforcing the boundary conditions leads to the following relations:

11—cC3=—GJ

= cosh AL s = sinh AL

= 4

C1L C, [2(1 — c)11

+= (0

C1( /1—c'\ 1Ax

+sinh Ax —

(1361)

(01 = + (1. — cosh Ax) — sinh Ax]}

= ci {i — [cosh Ax + sirih Ax]}

= —

Mç1, = ErI#A {sinh Ax + (L__f) cosh Ax}

We write the relation between the end rotation, (0, and the end moment M, as

M

=L,5

where L,ff denotes the effective length:

r 2C (c— 1'\1= L

(13-62)

= L(1 — cC3)

The following table shows the variation of with AL. For AL > 4, C2 2/AL. Notethat = 1 if transverse shear deformation due to restrained torsion is neglected.

AL C3

0.5 0.981 .9242 .76

3 .604 .48



Example 13—3

Uniform Distributed Moment-Symmetrical Supports

The general solution for m for convenience) is:

MT = C1 — ?flX

C3 C4. C1 flIXf= —coshAx +

—+

C1 m /x2 c (a)

=x + C2

—+

—(C3 sirth Ax + C4 cosh Ax)

We consider the boundary conditions to be identical at both ends and measure x from themidpoint (Fig. E13—3). Symmetry requires

MT=O} atx=O (b)

and (a) reduces toMT =

[= sinh Ax + x (13-63)

= C2 — + — cosh xx

We treat first the case where the end section is fixed with respect to both rotation andwarping. Requiring (13—63) to satisfy

f = = 0 at x = L/2 (a)

results in

I=

{x — sinh Ax}

mL2 fi[ /x"t21CO1 U — j +

(cosh ,,x — c)

MT = —mx

( x C ) (13—64)= ,nL1— +

AL1

AL . ALc = s =

The solution represents an upper bound. A lower bound is obtained by allowing thesection to wrap, i.e., by taking


SEC. 13—7. APPLICATION TO THIN-WALLED OPEN CROSS SECTIONS 395

Fig. E13—3

—x1

H

and the result is

f = {x — sinhxc

mL2 (1 [ C,— j + — c)

= —mx

( C (13—65)—x+fsinhAx

IC, I cosh Ax= —

ALc = cosh

13—7. APPLICATION TO THIN-WALLED OPEN CROSS SECTIONS

In what follows, we apply the mixed formulation theory to a wide flangesection and also to a channel section. We first determine the cross-sectionalproperties corresponding to = — and then obtain general expressions forthe stresses in terms of dimensionless geometric parameters. Before discussingthe individual sections, we briefly outline the procedure for an arbitrary section.

Consider the arbitrary segment shown in Fig. 13—4. We select a positivesense for S and an arbitrary origin (point P). The unrestrained torsion warpingfunction is obtained by applying (11—29) to the centerline curve and requiringthe section to rotate about the shear center.t

is positive when translation in the + S direction rotates the positionvector about the + X1 direction. The unrestrained torsional shear flow is zero

By definition, Ic1 = We work with q" rather than to facilitate treatment of closedand mixed sections where one generates q" in terms of


396 RESTRAfNED TORSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13

for an open section. Then, taking = — and integrating leads to

pscclSsP

(13—67)

Note that one can select the sense of S arbitrarily. Also, varies linearly withS when the segment is straight. The constant is evaluated by enforcing theorthogonality condition —* F1 = 0),

dS = 0

If the section has an axis of symmetry, = 0, if we take P on the symmetryaxis. The remaining orthogonality conditions (a'j1 ?v.t2 = M3 0),

$4x2tdS = 0

are identically satisfied by definition of the shear center. t

x3

x2

Fig. 13—4. Notation for determination of the warping function.

When the section has branches, we apply (13—67) to each branch. One hasonly to require continuity of 4) at the junction point. As an illustration, considerthe section shown in Fig. 13—5. The distribution of 4) for the three branches isgiven by

A — B 4) 4)p+$gpscdSB—CB—D b4)B+JopscdS

We are taking the origin at B for branches B — C and B — D.

tSeeProb. 13—1.

Shear center

IPsc



The shear flow due to is obtained by integrating (13—43) and noting(13—50). Forconvenience, we let

qr = (13—68)

With this notation, the resulting expression simplifies toS

+ J q5tdS = + (13—69)

We start at a free edge and work inward. A +q points in the +S direction(see Fig. 13—5). Then, a + corresponds to _qr, i.e. qr acting in the —Sdirection, If the section has an axis of symmetry, is an odd function withrespect to the axis and is an even function.

Fig. 13—5. Example of a section with branches.

Once and are known, we can evaluate 1w,, and with (13—10), (13—46):

= JJqYdA = 542tdS

Cr = if+

In order to evaluate XZr, X3r, we need the fiexural shear stress distributions.We let q(J) be the distribution due to and write

qU) (13—71)

145

x3

S.q

p

S,q

j=2j=3

'(=3k= 2



The coupling terms are defined by (13—-48), which reduces to

Ij

q q—

X3r + X21.

for a thin-walled section with 4) = — Substituting for qr and qf results in

J I 3dS

Y —

(13-72)

X3r = t

IfX2 is an axis of symmetry, is an even function of x3, qt2t is an odd function,and X3r = 0. By analogy, x2, = 0 if X3 is an axis of symmetry.

The definition equations for Cr, Ia,, x2r, and apply for an arbitrary thin-walled section. When the section is closed, we have only to modify the equa-tions for 4) and We will discuss this further in the next section.

Example 13—4

Symmetrical I Section

The I section shown (Fig. El 3—4A) has two axes of symmetry; it follows that the shearcenter coincides with the centroid and the warping function is odd with respect to K2, X3.

Fig. E13—4Ax3

1,

Applying (13—67), we obtainq5=O forweb

q5 = S for flange

Note that the sense of S is reversed for the bottom flange.



The shear flow vanishes at S = ± b/2. Applying (13—69) and starting from pt. A, we find

= S-b/2=

(25)2]

The distributions of and q' are shown in Fig. E13—4B, where the arrows indicate thesense of q' for + Ms-.

Fig. E13—48

b2ht

Plot Plot of qr

We express the cross-sectional properties in terms of Ii, t, and a shape factor

= b/h

ht33 = +

th5=

=(t)2

8(1 +=

The dimensionless parameters occurring in the solution of the differential equationsfor the mixed formulations are and AL (see (13—55)). Using (c) and assuming a valueof 1/3 for Poisson's ratio, we write

[3(1 +=

I

AL =


400 RESTRAINED TORSION-FLEXURE OF PR!SMATtC MEMBER CHAP. 13

The coefficients are tabulated below:

2.4 3

0.75 2.66 4.22

0.50 3.2 6.93

Since (t/h)2 << 1 and 0(1), we see that 1. The warping parameter, ),L, dependson t//i as well as L/h. This is the essential difference between open and closed cross sections.For the solid section, we found that AL = 0(L/h) and, since L/h is generally large in com-parison to unity, the influence of restrained warping is Iocalized.f The value of AL for anopen section is O(L//,) 00/1,) and the effect of warping restraint is no longer confined to aregion on the order of the depth at the end but extends further into the interior.

We consider next the determination of the stresses due to restrained warping. Thegeneral expressions are

M4,

r qdTts

= 7Using the distribution for çb and qr shown above, the maximum values of normal and shearstress are

6=

=

The shearing stress due to unrestrained torsion is obtained from

3

To gain some insight as to the relative magnitude of the various strcsses, we considera member fully restrained at one end and subjected to a torsional moment M at the otherend. This problem is solved in Example 13—1. The maximum values of the moments are

tanh

AL atx = 0= C5M J

We substitute for the moments in (f), (g) and write the results in terms the maximum

t We defined the boundary layer length, (sec (13—24). (13—25)) as

Lh 40

L ;.L


SEC. 13—7. APPLICATION TO THIN-WALLED OPEN CROSS SECTIONS

shear stress for unrestrained torsion:

,,, = tanh

(f))Mt

—

The variation of these coefficients with b/h is shown below:

b

1?

1 2 1.5

0.75 2.11 1.670.50 2.31 2

Since and are of 0(1), it follows that

= 0(d)

The additional shearing stress (at) is small in comparison to the unrestrained value.Therefore, it is reasonable to neglect the terms in the complementary energy density dueto ic., to take C, 0 and = 1 for an open section. We will show in the next sectionthat this assumption is not valid for a closed section.

Example 13—5

Channel Section

We consider next the channel section shown in Fig. El3—SA. Since X2 is an axis ofsymmetry, = x3, = 0. The expressions for the location of the centroid, shear center,

Fig. E13—5A

S

Shearcenter

x2

H°



and 12 are

= bI +

e=b =beI +

th3

b

h

The dimensionless coefficient? is essentially constant, as the following table shows:

b?

h

1.00 0.429

0.75 0.409

0.50 0.375

We determine by applying (13—67) to the three segments. Taking S as indicated above,and noting that is odd with respect to X2, we obtain:

Segment 1—2

6Psc =

hh( S= - -

Segment 2—3

bh( 2S\

The distribution is plotted in Fig. E13—5B. Since? < 1/2, the maximum value of q5 occursat point I (and 4).

We generate next the distribution of starting at point 1 (since q = 0 at that point)and using (b):

Segment 1—2

S bin 152

Segment 2—3

/ '\ s2= + +

The distribution of is plotted in Fig. El3--5C.



—e

—(1— e)

—

Distribution of

D1

ID2

0

Fig. E13—5B

Fig. E13—SC

Distribution of qr/1242t

The expressions lbr J, and AL are written in the same form as for the previousexample:

I = (1 ±=

I —h5-+

c(t'\2 ± + + + (t\2

= f =Cs =

AL = (t)=

D2®

)+Mi

D20



The following table shows the variation and with b/h for G/E. = 3/8. i.e., Poisson'sratio equal to 1/3. Note that the comments made for the wide-flange section also applyto the channel section.

bc =—

h

1 2.33 2.550.75 2.65 3.39

0.50 3.4 5.24

In order to evaluate X2r, we need the flexural shear stress distribution due to F3. Applying(11—106) leads to

Segment 1—2

Segment 2—3

4(3)

4(3) =— — S)

The distribution is plotted in Fig. E13—SD; the arrows indicate the sense of q for a +F3.

+

Distribution of lb/it/2

Fig. E13—5D

I

I

t

t÷F3

—1



Substituting for and the cross-sectional constants in (13—72) leads to

(t\2=

(1 + + +=

The coefficient is of order unity, as the following table shows:

1 0.9260.5 1.03

In Example 13—1, we determined expressions for the coordinates of the center of twistin terms of .'c,, and It is of interest to evaluate these expressions for this cross section.The coordinates at x 0(sec (13—59), (13—60)) are

= 0

= X2

——i-—1 +

Substituting for and evaluating

we obtainX2 = —

=

1 0.476 0.8360.5 0.625 0.485

13—8. APPLICATION TO THIN-WALLED CLOSED CROSS SECTIONS

We treat first a single closed cell and then generalize the procedure for multi-cell sections. Consider the section shown in Fig. 13—6. The +S direction isfrom X2 toward X3 (corresponding to a rotation about the +X1 direction).Using the results developed in Sec. 11—4, the shear flow for unrestrained tor-sion is

2Aq =-1-C



where A is the area enclosed by the centerline curve. The shearing stress varieslinearly over the thickness,

C'\= +—) = +

but the open-section term has a zero resultant.

x3

Fig. 13—6. Notation for single closed cell.

Substituting for qU in (13—66), taking 4 — and integrating from pointP lead to

CS

= + dS — C (13—73)

We determine by enforcing

= 0

The two additional orthogonality conditions

4x2gbtdS = 0

are identically satisfied by definition of the shear center. tThe shear flow due to is defined by (13—69),

q=—-7--q14

+ Q4

t Noting that x2t = dQs/ds, we can write

#x24.t =We merely have to identify this term as the moment of the flexural shear stress about the shearcenter. See Prob. 11-12.

q

S

x2



where is indeterminate. Our formulation is based on no energy couplingbetween qU and i.e., we require (see (13 --47))

= (13—74)

Noting that is constant for a single cell, and using (e), we obtain

f dS

=——

(13—75)

The flexural shear flow distributions for F2, F3 are generated with (11—110).We merely point out here that there is no energy coupling between qU and

quqf 0 (f)

One can interpret (13—74) and (f) as requiring qr to lead to no twist deforma-tion, i.e., w1 0. We have expressed the fiexural shear flows as (see (13—71)):

ft —(J) — FJ_0) = 2 kqjij=q j=3 k=2

Finally, the definition equations for the cross-sectional properties have thesame form as for the open-section:

Eq. 13—70 Cr

Eq. 13—72

X2 is an axis of symmetry. Then, is an odd function of x3. If wetake the origin for S (point p) on the X2 axis, = 0. Also, is an even func-tion of x3 and = 0. In what follows, we illustrate the application of theprocedure to a rectangular cross section.

Example 13—6

Rectangular Section—C'onsta,,t Thickness

Applying (13—73) and taking q5 = 0 at point shown in (Fig. E13—6A) leads to

ci + b

fa — b\

The distribution is plotted in Fig. E13—6B. Note that = 0 when a = b, i.e., a squaresection of constant thickness does not warp.


— b'\ S2= at

( —+ bJ 2

(Q4 +(a — b\ /

=a+b1\

and evaluate with (13—75):

for segment 1—2

for segment 2—3

Fig. E13—6B

dS

a—b-.11


Fig. E13-.6A

TCentroid

- x2

2a—

Distribution of

We determine Q4, by integrating (a),

(a—b+ b



The distribution of follows from (b), (c),

2 j( 2a\)

2a2a(S 1(S\2\ j/

2 \a+band is plotted in Fig. E13—6C. Note that corresponds to q' acting in the clockwise(— S) direction for + Ms-. Also, D is negative for b > a.

qr(+

We introduce a shape factor (,

depth b

width a

Fig. E13—6C

x3

Tb

2ab

q'/D

D -2



and express the various coefficients in terms of a, t, and The resulting relations are

J = 16a3t (neglecting the contribution oft)

4a5t [2(1 —

= LATh4 + + +

5(1

9(1

I (G'\ / \) 1/2 L L

=

x2, = x3,. 0

The variation of C,, and with b/a is shown in the table below;

b .' G 3= -- C, L;f for

a \ E 8

cc 0 0.982 10.43 0.0877 1.27

3 4.41 0.185 1.39

We found

(g)

= 0 T)

for an open section. Our results for the single cell indicate that

IL= 0

C,>> 1

C, 1

for a closed section. We obtained a similar result for using the displacement-modelformulation for a solid section. Since is due to the restrained shearing stress (q'), wesee that shear deformation due to q' cannot be neglected for a closed cross section.

We discuss next the determination of the normal and shearing stresses due to warping.The general expressions are

q'°isqt tie



The maximum normal stress occurs at point 2 while the maximum shear stress can occurat either points I or 3.

We consider the same problem as was treated in Example 13—4, i.e., a member fullyrestrained at one end and subjected to a torsional moment M at the other end. We ex-press the stresses in terms of ag,, the maximum shear stress for unrestrained torsion,

M( C

= J+

which reduces toMC M

= 7 =

since we are considering the section to be thin-walled. The maximum stresses are

2 = i tanh 2.L

0$ ,nax,I =[3C,1112

— S.C

The variation of and 2 with height/width is shown below. We are taking Poisson'sratio equal to 1/3.

= h/a c, (point 2) (point 1) (point 3)

1 0 0 02 1.04 —0.35 +0.443 —1.51 —0.46 +0.65

For large tanh I and we see that both the normal and shear stress are of theorder of the unrestrained-torsion stress. In the open section case, we found the restrained-torsion shear stress to be of the order of (thickness/depth) times the unrestrained shearstress.

To illustrate the procedure for a multicell section. we consider the sectionshown in Fig. 13—7. The unrestrained-torsion analysis for this section is treatedin Sec. 11—4 (see Fig. 11—11). For convenience, we summarize the essentialresults here.

We nttmber the cells consecutively and take the +S sense from X2 to X3for the closed segments and inward for the open segments. The total shearflow is obtained by superimposing the individual ccli flows

q' = 0 for an exterior (open) segmentqU = constant for an interior segment

We let(U

— WIT

—



e

Fig. 13—7. Notation for mixed cross section.

The constants C1, C2 are determined by requiring each cell to have the sametwist deformation, w1, Enforcing (11—67),t

for each cell leads to

= =

2A

where a, A are defined as

f dS—

,Jsj t

dS= a21

=A = {A1, A2}

a= Psc — 7

+

"2

q,S

q1 ,S1

The warping function is generated by applying (13—6):

4' =(13—76)

We start at point P1 in cell 1 and integrate around the centerline, enforcingcontinuity of 4, at the junction points b, c, and d. For example, at b, we require

t See also (11—32).



which leads to a relation between and 4),,,:

= 4)e + j Psc dS = +Jb

—

dS

Repeating for points C and d results in the distribution of 4) expressed interms of One can easily verify that 4) is continuous, i.e., & determinedfrom segment ca is equal to determined from segment cdcL. Finally, we

evaluate by enforcing

JJ4)dA=J4)tdS=O

where the integral extends over the total centerline. Note that 0 if P1is taken on an axis of symmetry.

The shear flow for restrained torsion is obtained with (13—69):

a=

as

The steps are the same as for the flexural shear determination discussed inSec. 11—7. We take the shear flow at points P1. P2 as the redundants,

= J = 1, 2 (13—77)

and express the shear flow as+ ii. (13—78)

where Zj0 is the open section distribution and is due to The dis-tribution, has the same form as We just have to replace C withC'S. We generate by integrating (i) around the centerline, and enforcingequilibrium at the junction points. For example, at point b (see Fig. 13—7),

= +

Note that = 0 at points P1, P2. e andfThe redundant shear flows are evaluated by requiring no energy coupling

between qU and qr which is equivalent to requiring qr to lead to no twist de-formation, j. Noting (c), we can write

= 0 j = 1,2 (13—79)

Finally, substituting for we obtain

aCr = B1' — dS

(13—80)

f See footnote on page 385.



Once 4) and zir are known, the cross-sectional properties (1 , Ci., X2r, xar)can be evaluated. Also we can readily generalize the above approach for ann-cell section.

1 3—9. GOVERNING EQUATIONS—GEOMETRICALLY NONLINEARRESTRAINED TORSION

In this section, we establish the governing equations for geometrically non-linear restrained torsion by applying Reissner's principle. This approach is amixed formulation, i.e., one introduces expansions for both stresses and dis-placements. The linear case was treated in Sec. 13—5. To extend the formulationinto the geometrically nonlinear realm is straightforward. One has only tointroduce the appropriate nonlinear strain-displacement relations.

Our starting point is the stationary requirement t

— V*)d(vol.) — d(surface area)] 0

where a, are independent variables, C e(u), = V*(o.), and b areprescribed.

We take the displacement expansions according to (13—3) and use the strain-displacement relations for small strain and small finite rotations4

U1 = U1 + C02X3 W3X2 + /4)

112 — cn1(x3 x3)

U3 = + w1(x2 —(13—81

= + +Yiz = + U2,1 + U3,3U3,2

Y13 l.1j3 + U31 + 1)2

The in-plane strain measures (62, 63, Y23) are of 0(w2), which is negligibleaccording to the assumption of sinai! finite rotations. Actually we assumeO'22 = 1723 0, i.e., plane Stress. Substituting for the displacements andnoting the definition equations for the force parameters, the first term in (a)expands to

d(vol.) = 1 1 1)]+ F2[u,2. i W3 + —

+ F3[u53 S + — t ++ M2{w2, 1 — co1, 1(u52 i + x3w1,+ M3[w3, i — w1, i — T2w1. 1)]

+ 1 M0f1 + MRf+ i + MQW1W1, 1}dxj

f See Eqs. 13—33 and corresponding footnote. We are working with Kirchhoff Stress andLagrangian strain here.

See Sec. 10—3. Eq. 10—28. Tile displacement expansions assume small-finite rotation, i.e.,sin w and cos w 1. To be consistent, we must use (10—28).


SEC. 13—9. GOVERNING EQUATIONS 415

where the two additional force parameters are

= ÷MQ = $J(x2c12 + x3a13)dA

The terms involving the external forces have the same form as for the linearcase, but we list them again here for convenience (see (13—6)):

JJcbTu d(vol.) + jJpTu d(surface area)

+ + + mrwi + in2w2 + m3oj3 + rn4f)dxi(13—83)

+ F1u1 + + F3u33 + MTO1 + M2(02 + M3co3 +

where the end forces (the barred quantities) are defined as previously, forexample,

= (5Jp1 etc.

It remains to introduce expansions for the stresses in terms of the independentforce parameters and to expand V*. In the linear case, there are 8 forcemeasures. F1 M3, and M4, MR. Two additional force measures (Me, MQ)are present for the nonlinear case but they can be related to the previous forcemeasures. We proceed as follows. We use the stress expansions employed forthe linear case with = They are summarized below for convenience(see Sec. 13--5):

F1 M2 M3a11 ± —1—-x3 — -T—X2 +

A 13

=± + &ij

—

+MT +Il _.-

where 4,, f, q, h2 and h3 are functions of x2, x3. Introducing (a) in the definitionequations for and MQ leads to

= f11F1 + fl2M2 + fl3M3 +

= if + =

$2 = if +(13-84)

$3 = if +

/34,

4,



andtMQ — + + +

=$J(x2h2k +xlh3k)dA (k = 2, 3)

(13—85)

= +

Certain coefficients vanish if the cross section has an axis of symmetry4 Onecan readily verify that

fi1F1(13—86)

MQ 0

when the section is doubly symmetric. For generality, we will retain all theterms here.

The complementary energy density function has the same form as for thelinear case:

— 1 1= —----' + + —-—— + ——'

2Ek.,A '2 13)

+ ++ +

((Mw +

+ + X2rF3)

We have shown that it is quite reasonable to neglect transverse shear deforma-tion due to warping (C. X2r X3r = 0) for a thin-walled open section.

Substituting Equations (13—82)—(13—87) in Reissner's functional and re-quiring it to be stationary with respect to the seven displacement and eightforce measures leads to the following governing equations:


F1,1 + b1 = 0

j + + — w1F3 — w1 1M2} + b2 0

— + F3 + w1F2 — wi,1M3} + b3 = 0

(1 + + (1 + 1

1 +

1 — 1 + 2J32w1,

+ + + = 0

M2, 1 — F3 + m2 = 0

M3,1 + F2 + m3 = 0

— +

t See Prob. 13—il.See Prob. 13—12.



where

Relations

= 1 + 1 + 1 + Wj, i — x2uS3, + i)

1FF2 X3r 1 —+ + = — w3 + wi[u53, 1 — Wj, 1/33]

1FF2 F3 1+ MrTJ = + (02 + 1 +

G= +

M(13—88)

= (02,1 + (0l,j(—US2,j + /32(01,1)

= + + /33(01,1)

= 1. + j

[CrM?+ X3rF2 + = I' + +

Boundarp Conditions (+ for x1 = L, — for x1 = 0)

u1 prescribed or F1 =prescribed or + T3w1,1) + F2 — wjF3 (01,1M2 = ±F2prescribed or — x2w1, + F3 + w1F2 — 1M3 ±F3

wi prescribed or 1 1 + 7J1w1,

+ (0j(172F2 + + (1 + + (I ++ M2(—u52,j + + 1 + + = ±MT

(02 prescribed or M2 = ±M2(03 prescribed or M3 = ± M3

f prescribed or = ±

These equations simplify considerably when the cross section is symmetricaland transverse shear deformation is neglected.1' We discuss the general solutionof (13—88) in Chapter 18. The following example treats one of thecases, a member subjected to an axial force and torsional moment.




Example 13—7

We consider a prismatic member (see Fig. E13—7A) having a doubly symmetric crosssection, fully restrained at one end and loaded by an axial force P and torsional momentM. We are interested here in evaluating the influence of axial force on the torsionalbehavior. The linear solution (with no axial force) was derived in Example 13—i.

x2 Fig. E13—7A

Equilibrium Equations (symmetrical cross section and no distributed load)

d

=F1. 0

(M1 + i) = 0dx1


Boundary conditions

= GJw11

= ErI,jiF1 = i +

.xi=Oxj = L F1 P = 0 M1 + J3tF1w1,1 = Al

Integrating the last two equations in (a) and noting the boundary conditions, lead to

F1 = const = P

M1 + /31F1w1, = corlst = M

The first equilibrium equation takes the form

21,11

FL

P M



whereP11

7;:ij GJA

2GJ i±PI + + F)

This expression reduces to Equation (g) of Sec. 13—6 when P = 0. Once f is known,we can determine the rotation by integrating (d), which expands to

+ F+

M — fwhen we substitute for M1 using (b).

The general solution is,

Mf = C1 cosli ,ux + C2 smh

[GJ (i ++

= C3 + Mx {i + — {C1 sinh px + C2 cosh

(We drop the subscript on x1 for convenience.) Finally, specializing (g) for these particularboundary conditions result in

f = { —1 + cosh — tanh sinh

wi = —

—

{sinh jtx + (1 — cosh

These equations reduce to (13-57) when P = 0.

A tensile force (P > 0) increases the torsional stiffness whereas a compressive force(P < 0) decreases the stiffness. Equation (h) shows that the limiting value of P is 1. Welet F, represent the critical axial force and the corresponding axial stress

11

(;J

In order for to be less than the yield stress, (J/11) must be small with respect to unity.As an illustration, consider the section shown in Fig. E13—7B. The various coefficients

(see Example 13—4) are

J = +



Fig. E13—78x3

X2

and

r,, (t'\2(G

REFERENCES

1. VON T., and N. B. CHRISTENSEN: "Methods of Analysis of Torsion withVariable Twist," J. Aero. Sci., pp. 110—124, April 1944.

2. TIMOSHENKO, S. J:: "Theory of Bending, Torsion and Buckling of Thin-WalledMembers of Open Cross Section," J. Franklin Inst., pp.559—609, 1945.

3. VON KARMAN, T.. and W. C. CHIEN: "Torsion with Variable Twist," J. Aero. Sci.,Vol. 13, No. 10, pp. 503—510, October 1946.

4. BENSCOTER, S. U.: "Secondary Stresses in Thin-Walled Beams with Closed CrossSections," NACA—TN 2529, Washington, D. C., 1951.

5. BENSCOTER, S. U.: "A Theory of Torsion Bending for Multiceil Beams," J. Appi.Mech., Vol 21, No. 1, 1954.

6. VLASOV, V. Z.: Thin Walled Elastic Brains, israel Program for Scientific Translations,Office of Technical Services, IJ.S. Dept. of Commerce, Washington. D.C. 1961.

7. HEILIG, R.: "Der Schuberverformungseinfiuss auf die Wölbkrafttorsion Von Stilbenmit offenern Profil," Der Stahlbau, April 1961.

8. HEILIG, R,: "l3eitrag zur Theorie der Kastentrhger beliehigerDer Stahlbau, December 1961.

9. J. T.: Mechanics of Elastic Structures, McGraw-Hill. New York, 1967.10. KOLLORUNNER, C. F., and K. BASLER: Torsion in Structures, Springer-Verlag. Berlin,

1969.Ii. K.: Variational Methods in and Plasticity, Pergarnon Press.

1968.

12, MAISEL, B. I.: "Review of Literature Related to the Analysis and Design of Thin-Walled Beams," Technical Report 440, Cement and Concrete Association, London,July 1970.


PROBLEMS 421

13. DABROWSKi, R.: "Gekrüinmte dUnnwandige Trager," Springer-Verlag, Berlin, 1968.14. GALAMBOS, T. V.: Structural Members and Fiames, Prentice-Hall, 1968.15. BLEICH, F.: Buckling Strength of Metal Structures, McGraw-Hill, New York, 1952.16. BURGERMEISTER, G., and H. STEin': Srabilitar Theorie, Part 1. Akademie Verlag,

Berlin, 1957.17. CHILVER, A. H.: Thin- Walled Structures, Chatto and Windus, London, 1967.18. REISSER, E.: "Note on Torsion with Variable Twist." .J. AppI. Mech., Vol. 23, No. 2,

pp. 315—316, June 1956,

PROBLEMS

13—1. The shear stress distribution due to is given by (see (11—95))

F2 F2= 2 (733 ' 3

13 13

where are fiexural warping functions which satisfy

= — x2 (in A)

(onS)

This result applies when the cross section is assumed to be rigid with respectto in-plane deformation. The coordinate of the shear center is defined by

= X3 if 3

X3

where is the St. Venant torsional warping function. Hint: See Prob. 11—11and Equation (11—97).

13—2. Verify (13—40) and (13—44).13—3. This problem reviews the subject of the chapter in two aspects.(a) No coupling between the unrestrained and restrained torsional dis-

tribution requires+ 0

The unrestrained torsional shear stress distribution for twist aboutthe shear center (see Sec. 13—3, Equation (b)) is given by

,f UIVIT —

O'12=

— X3 + X3]

—= + x2 — x2]

The restrained torsional shear stress distribution is determined from(13—39). Verify that = MR when ç& = and (a) is enforced.



(b) When the cross section is thin-walled, (a) and (b) take the form

•fquqr_ = o

where is the perpendicular distance from the shear center to thetangent at the centerline, Equation (d) follows from (11—29) andProb. 11—4. We determine qf from (13—43). Finally, the force param-eters for the thin-walled case are defined as

= dSMR = Jqrc& dS

Verify that = MR when = Consider the following cases:

1. Open section2. Closed section3. Mixed section

13—4. Specialize (13—57) for .L > I and compare vs. Mu. Also evaluateat x L and compare with the unrestrained value.

13—5. Refer to Examples 12—2 and 13—2. Discuss how you would modifythe member force-displacement relations developed in Example 12—2 to accountfor restrained torsion. Consider 1, X3r = 0, and——

(a) warping restrained at both ends(b) warping restrained only at x L

13—6. Refer to Example 13—2. Determine the translations of the shearcenter. Consider the cross section fixed at x 0. Discuss how the solutionhas to be modified when the cross section at x = L is restrained againsttranslation.

13—7. Starting with the force-deformation relations based on the mixedformulation (13—49), derive the member force-displacement relations (seeExample 12—2). Consider no warping at the end sections and take = + 1.

Specialize for—(a) symmetrical cross section(b) no shear deformation due to restrained torsion and flexure—arbitrary

cross section,13—S. Consider a thin-walled section comprising discrete elements of

material properties (F, G). Discuss how the displacement and mixedformulations haveto be modified to account for variable material properties.Note: The unrestrained torsion and flexural stress distributions are treated inProb. 11—14 and 12—1.

13—9. Determine the distribution of qr, and expressions for Cr,for the cross sections shown in parts a and b and part e—d of the accom-

panying sketch (four different sets of data).13—10. Determine and qr for the section shown.13—11. Using the fiexural shear distributions listed in Prob. 13—1, show

that -

'12 =


F— 0.75k(b)

I II

(d)

/z 2k + 'i—H

(c)

Prob. 13-10

t

0 0t ç1s2

I I

Ha

423

Prob. 13—9

PROBLEMS

I

Ii

T

See part c.


424 RESTRAINED TOIRSION-FLEXURE OF PRISMATIC MEMBER CHAP. 13

Hint: One can write

22(x2 V q52r +13 •JJ

Also show that113 —

13—12. Specialize Equations (13—84) and (13—85) for the case where thecross section is symmetrical with respect to the X2 axis. Utilize

x3)H0(x2, x3)dA = 0

where He is an even function and H,, an odd function of x3. Evaluate the co-efficients for the channel section of Example 13—5. Finally, specialize theequations for a doubly symmetric section.

13—13. Specialize (13—88) for a doubly symmetrical cross Section. Thenspecialize further for negligible transverse shear deformation due to flexure andwarping. The symmetry reductions are

X2 = = 0 X2r X3r = 0i/A23=O

!72 = 'li 0

13—14. Consider the two following problems involving doubly symmetriccross section.

(a) Establish "linearized" incremental equations by operating on (13—88)and retaining only linear terms in the displacement increments.Specialize for a doubly symmetric cross section (see Prob. 13—12).

(b) Consider the case where the cross section is doubly symmetric and theinitial state is pure compression (F1 —P). Determine the criticalload with respect to torsional buckling for the following boundaryconditions:

1. co1 = f 0 at x = 0, L (restrained warping)

2. == = 0 at x 0, L (unrestrained warping)

Neutral equilibrium (buckling) is defined as the existence of a nontrivialsolution of the linearized incremental equations for the same externalload. One sets

F1 = —P

U2 U3 = W1 = (02 = (03 = f 0

and determines the value of P for which a nontrivial solution whichsatisfies the boundary conditions is possible. Employ the notationintroduced in Example 13—7.

13—15. Determine the form of V, the strain energy density function (strainenergy per unit length along the centroidal axis), expressed in terms of displace-ments. Assume no initial strain but allow for geometric nonlinearity. Notethat V = V* when there is no initial strain.


14

Planar Deformation of aPlanar Member

14—i. INTRODUCTION: GEOMETRICAL RELATIONS

A member is said to be planar if—

1. The centroidal axis is a plane curve.2. The plane containing the centroidal axis also contains one of the

principal inertia axes for the cross section.3. The shear center axis coincides with or is parallel to the axis.

However, the present discussion will be limited to the case where theshear center axis lies in the plane containing the centroidal axis.

We consider the centroidal axis to he defined with respect to a global referenceframe having directions X1 and K2. '[his is shown in Fig. 14—i. The orthogonalunit vectors defining the orientation of the local frame (Y1, Y2) at a point aredenoted by where points in the positive tangent direction and x 12 =13. Item 2 requires Y2 to be a principal inertia axis for the cross section.

i2

x2

ii

Fig. 14—i. notation for planecurve.

425

Yl

x1

r2

S

n

A

tl

n

B


426 PLANAR DEFORMATION OF A PLANAR MEMBER CHAP. 14

By definition, tdx1 dx2= = + (14-1)

Since we are taking t2 according to 11 x t2 = 13, it follows that

- dx2 dx5t2 = + (142)

The differentiation formulas for the unit vectors are

dt1 1

(14-3)

where1 dt1 — d2x1 dx2 d2x2 dx1

According to this definition, R is negative when d11/dS points in the negativet2 direction, e.g., for segment AB in Fig. 14—1. One could take t2 = ii, the unitnormal vector defined by

- 1 d11

(14-4)ciS

rather than according to x 12 = 13 but this choice is inconvenient whenthere is a reversal in curvature. Also, this definition degenerates at an inflectionpoint, i.e., when dt/dS = O. If the sense of the curvature is constant, one canalways orient the X1-X2 frame so that coincides with ñ, to avoid workingwith a negative R.

To complete the geometrical treatment, we consider the general parametricrepresentation for the curve defining the centroidal axis,

x1 = x1(y) (j45x2 = x2(y)

where y is a parameter. The differential arc length is related to dy by

d. - 2 2 1/2

dS + + (p)] dv = dy (14—6)

According to this definition, the +S sense coincides with the direction of

t We summarize here for convenience the essential geometric relations for a plane curve whichare developed in Chapter 4.


SEC. 14—2. FORCE-EQUILIBRIUM EQUATIONS 427

increasing y. Using (14—6), the expressions for and 1/R in terms of y are

- I 7dx1 dx2t1 = — ( 1j + —

dy

- if dx2.. dx1t2 = — ( ——-—-- +

dy dy (14_1 if._ (It1

—- = -( t2R dy

— 1 ( d2x1 dx2 d2x2 dx1—

k\ dv2 dy+ dy2 dy

A planar member subjected to in-plane forces plane for our notation)will experience oniy in-plane deformation. In what follows, we develop thegoverning equations for planar deformation of an arbitrary planar member.This formulation is restricted to the linear geometric case. The two basicsolution procedures, namely, the displacement and force methods, are describedand applied to a circular member.

We also present a simplified formulation (Marguerre's equations) which isvalid for a shallow member. Finally, we include a discussion of numericalintegration techniques, since one must resort to numerical integration whenthe cross section is not constant.

14—2. FORCE-EQUILIBRIUM EQUATIONS

The notation associated with a positive normal cross section, i.e., a crosssection whose outward normal points in the + S direction, is shown in Fig. 14—2.We use the same notation as for the prismatic case, except that now the vector

Fig. 14—2. Force and moment components acting on a positive cross section.

- I'3

Centroidal axis

dA

012



components are with respect to the local frame (Y1, Y2, Y3) rather than thebasic frame (X1, X2, X3). The cross-sectional properties are defined by

A = if dy2 dy2 = if dii(14—8)

13 = JJ(y2)2 dii '2 = .iJ(Y3)2

Since Y2, Y3 pass through the centroid and are principal directions, it followsthat

dA = flY3 dA = SSY2Y3 dii = 0 (14—9)

When the member is planar (X1-X2 plane) and is subjected to a planarloading,

F3 M1 M2 0 (14—10)

in this case, we work with reduced expressions for F÷ and M÷ (see Fig. 14—3)and drop the subscript on M3:

= + F212'(14-11)

M+ = M3t3 = Mt3

Note that 13 is constant for a planar member.

x2

x1

Fig. 14—3. Force and moment components in planar behavior,

To establish the force-equilibrium equations, we consider the differentialvolume element shown in Fig. 14—4. We define b and as the statically equiva-lent external force and moment vectors per unit arc length acting at the centroid.For equilibrium, the resultant force and moment vectors must vanish. Theseconditions lead to the following vector differential equilibrium equations:

+ = o— dS

(14—12)dM÷ ,_. — -

+ in + r1 x F+ = 0

)

= t1 x t2



We expand b and in terms of the unit vectors for the local frame:

b = + b212(14—13)

= mt3

Introducing the component expansions in (14—12), and using the differentiationformulas for the unit vectors (14—3), lead to the following scalar differentialequilibrium equations:

dF1 F2— + b1 = 0

(14-14)

dM+ + m 0

that the force-equilibrium equations are coupled due to the curvature.The moment equilibrium equation has the same form as for the prismatic case.

dS


The positive sense of the end forces is shown in Fig. 14—5. We work withcomponents referred to the local frame at each end. The end forces are relatedto the stress resultants and stress couples by

== Mj52

(14-15)=

MA= —MISA j=1,2

14—3. FORCE-DISPLACEMENT RELATIONS; PRINCIPLE OF VIRTUALFORCES

We establish the force-displacement relations by applying the principal ofvirtual forces to a differential element. The procedure is the same as for the

r(S)



prismatic case described in Sec. 12—3, except that now we work with displace-ment components referred to the local frame at each point. We define ü and as

= = rigid-body translation vector at thecentroid. (14—16)

= = equivalent rigid-body rotation vector

For planar deformation, only u1, u2 and 0J3 are finite, and the terms involvingu3, co1, and w2 can be deleted:

u1t1 + U2T2

— C03t3 Wt3

The positive sense of the displacement components is shown in Fig. 14—6.

x2

Fig. 14—5. Convention for end forces.

t "Equivalence" refers to work. See (12—8).

x1

(14—17)

F41

FIg. 14—6. Definition of displacement measures.



We define as the complementary energy per unit arc length. For planardeformation, = (F1, F2, M). One determines by taking expansionsfor the stresses in terms of F1, F2, M, substituting in the complementary energydensity, and integrating with respect to the cross-sectional coordinates Y2, y3.We will discuss the determination of later.

Specializing the three-dimensional principle of virtual forces for the one-dimensional elastic case, and writing

= AF1 + AF2 + AMcF1 0F2 cM

(14—18)= e1 AF1 + e2 AF2 + k AM

lead to the one-dimensional form

Ss(ei AF1 + e2 AF2 + k AM)dS AP1 (14—19)

where is a displacement measure and is the force measure correspondingto d1. The virtual-force system (AF1, AF2, AM, AP1) must be statically permis-sible, i.e., it must satisfy the one-dimensional equilibrium equations.

(

Fig. 14—7. Virtual force system

We apply (14—19) to the differential element shown in Fig. 14—7. The virtualforce system must satisfy the force-equilibrium equations (14—17),

AF÷ = 0dS (a)

Evaluating AP1,

= +AM÷ +(b)

= {AF1—

+ AF2 + — + dS



and then substituting in (14-49) results in the following relations between theforce and displacement parameters:

du1 U2cj

du2 u1(14-20)

keV* dw

dS

We interpret e1 as an average extension, e, as an average transverse sheardeformation, and k as a bending deformation. Actually, k is the relative rotationof adjacent cross sections. In what follows, we discuss the determination of

Consider the differential volume element shown in Fig. 14—8. The vectordefining the arc QQ1 is

ar2 di2 dt-\QQ1 = dy = + + dv

Noting that—

dy

(112—7k-ti

= odv

for a planar member, (a) can be written as

dS2 == —

= —

By definition, is the complementary energy per unit length along thecentroidal axis. Substituting for dS2 in the general definition, we obtain

dS dS2 dv2 dvY2,Y3

(14—2 1)

if.-

In general, V* = V" (ô11, We select suitable expansions for thestress components in terms of F1, F2, M, expand V*, and integrate over thecross section. The only restriction on the stress expansions is that they satisfythe definition equations for the stress resultants and couples identically:

dA = $5c12 dA F2 SSa13 dA = 0

JJy3aii dA = 0 —ify2ci1 dA = MJ$(y2a13 — y3a12)dA = 0


SEC. 14—3. PRINCIPLE OF VURTUAL FORCES 433

The most convenient choice for iH is the linear expansion,t

M— (14—22)

where I 13. A logical choice for (when the cross section is thin-walled)is the distribution predicted by the engineering theory of flexural shear stressdistribution described in Sec. 11—7:

a11 = 1q(F2) q = F2t/i (14—23)

where t denotes the local thickness, and q is the flexural shear flow due to F2.Both expansions satisfy (a).

Fig. 14—8. Differential volume element.

In what follows, we consider the material to be linearly elastic. The comple-mentary energy density is given by

11*..... 0 2 2— a12

where c? is the initial extensional strain. Substituting (a) in (14—21) and takingthe stresses according to (14—22), (14—23) results in the following expression

f This applies for a homogeneous beam. Composite beams are more conveniently treated withthe approach described in the next section.

x2

r +1)212 +Y33 r2

r1ty +dy)

Y2

it



for V*:

= e?Fi + k°M + + + +2GA2*

(14-24)

where

=if (i - dA

55

I — —\\ R}

If the section is symmetrical with respect to the 1'3 axis, 1* 1 and = A2.

The deformation-force relations correspoiiding to this choice for are

F1 M dr,1 u2—ei

F2 du2 U1e,

= =+ - w (14-25)

F1 M dw

Note that the axial force and moment are coupled, due to the curvature.Inverting (14—25) leads to expressions for the forces in terms of the deformations:

LAF1 = — e1)

— R(lk )

M—EI*

/ k°— R(1

Ô)(el — + (14-26)F2

We observe thatI

—

where p is the radius of gyration and d is the depth of the cross section, Forexample,

1 d2

AR2 = i2R2

for a rectangular cross section. Then, is of the order of (d/R2) and can beneglected when (dIR)2 1.

A curved member is said to be thin when O(d/R) 1 and thick when O(d/R)2

1. We set ö = 0 for a thick member. The thinness assumption is introduced



by neglecting y2/R with respect to unity in the expression for the differentialarc length, i.e., by taking

dS14 27-

Assuming a curved member to be thin is equivalent to using the expressionfor V* developed for a prismatic member. The approximate form of (14—25)for a thin member is

F1 dii1 Li2

(14—28)

i—k°

To complete the treatment of the linear elastic case, we list the expandedforms of the principle of virtual forces for thick and thin members. Note thatthese expressions are based on a linear variation in normal stress over the crosssection.

Thick Member

Cit0 F1 M'\ F2+

+AF1 + /XF2

(14—29)

+ (ko + + AM}dS —

Thin

J ++

/ M'\ 1 (14—30)+ (\kO + h-i) dS = d1 AP1

14—4. FORCE-DISPLACEMENT RELATIONS—DISPLACEMENTEXPANSION APPROACH; PRINCIPLE OF VIRTUALDISPLACEMENTS

In the variational procedure for establishing one-dimensional force-displace-ment relations, it is not necessary to analyze the deformation, i.e., to determinethe strains at a point. One has only to introduce suitable expansions for thestress components in terms of the one-dimensional force parameters. Ndw, we•can also establish force-displacement relations by starting with expansions forthe displacement components in terms of one-dimensional displacement pa-rameters and determining the corresponding strain distribution. We express the


436 PLANAR OEFORMA11ON OF A PLANAR MEMBER CHAP. 14

stresses in terms of the displacement parameters using the stress-strain relations,and then substitute the stress expansions in the definition equations for F1, F2,and M. The effect of transverse shear deformation is usually neglected in thisapproach. To determine the strain distribution, we must first analyze thedeformation at a point. This step is described in detail below.

Figure 14—9 shows the initial position of two orthogonal line elements, QQ1and QQ2, at a point (y, Y2' y3). The vectors defining these elements are

QQ1

QQ2 = dy2 dy2 t2

Ia2 =—

a

(14—31)

We use a prime superscript to denote qua Iltities associated with the deformedposition of the member, which is shown in Fig. 14—10; for example:

?'= = position vector to point P(y) in the deformed position (point P').tangent vector to the deformed centroidal axis.

= position vector to Q(y, Y2, y3) in the deformed position (point Q').

x1

Fig. 14—9. Initial geometry for orthogonal curvihnear line elements.

x2

Q2Q1

P(y)

Pj(y +dy)

axis



From Fig. 14—10, and noting (14—3 1):

— /P'P'1 = = + =

dy (14—32)0)) C))J

/or2 — c,u2dy2

&Y2 \.

The analysis of strain consists of determining the extensions and change inangle between the line elements. We denote the extensional strains by(j = 1, 2) and the shearing strain by Y12 The general expressions are

'—123—(1-3)

Sin Y12

Now, we restrict this discussion to small strain, Substituting for the deformedvectors and neglecting strains with respect to unity, (14—33) expands to

Is au2tj, + —.

c'y 2(a2)" ô,V

1 ('U2

2

-' "Y12 tj t2 -±

a2 cc2

The nonlinear terms arc associated with the rotation of the tangent vector.Neglecting these terms corresponds to neglecting the difference between thedeformed and undeformed geometry, i.e.. to assuming linear geometry.

The next step involves introducing an expansion for in terms of y2. Weexpress ü2 as a linear function of

ü wv211 (14—35)

where co = w(y) and

U U1t1 + U2t2 = 1kv) (14—36)

is the displacement vector for a point on the centroidal axis. Equation (14—35)implies that a normal cross section remains a plane after deformation. One caninterpret co as the rotation of the cross section in the direction from towardt2. This notation is illustrated in Fig. 14—1 1.

In what follows, we consider only linear geometry. Substituting for ü2, takingy = S, and evaluating the derivatives lead to the following strain expansions:


PLANAR DEFORMATION OF A PLANAR MEMBER

e2 = + 0) (14—37)

CHAP. 14

du, u2=—

= I61IY20

Q2

438

= 1

— y2k)

doi

The vanishing of c2 is due to our choice for ü2. One could include an addi-tional linear term, This would give = $ and, additional terms in the

x2

x1

Fig. 14—10. Deformed geometry for orthogonal curvilinear line elements.

u2t2

(u1 —Wy2)tl

Centroidal axis

UI tl

Fig. 14—11. Displacement expansion.



expressions for and Y12• Note that the assumption that a normal crosssection remains plane does not lead to a linear variation in extensional strainover the depth when the member is curved.

We introduce the assumption of negligible transverse deformation by settinge2 = 0. The resulting expressions for (0 and k in terms of u1 and u2 are

e2 = 0

du2+

u1

dS R (14—38)

dIui— dS — dS2

When transverse shear deformation is neglected, one must determine F2 usingthe moment-equilibrium equation.

The next step involves expressing F1, F,, and M in terms of the one-dimen-sional deformation parameters e1e2 and k. In what follows, we consider thematerial to be linearly elastic and take the stress-strain relations for c12as:

= E(c1 = Gy12

Substituting for r1, Y12, using (14—37),

F= ———--—-(e1 — y,k) — Fe1

1 —y2/R(14—39)

and then evaluating F1, F2, and M, we obtain

F2 = Ge2 ifd

(14-40)

= —Fe1 + Ekjj +

The various integrals can be expressed in terms of only one integral by usingthe identity

11-F

y2/R

1 — y2/R — 1

and noting that Y is a axis:

$5Y2 dA = 0

f The relation for is exact only when = (733 11 We generally neglect for amember.



One can easily show that

ri c/A I'

dA = L (14—41)

JJ 1 - y2/R

For completeness, we list the inverted form of (14—40),

M= + +

F2

F1 lvik = k° + +

where

= +

= A(14—42

e? = if (i—

c/A

k° = if -dA

The expressions for e1 are identical with the result (see (14—25)) obtained withthe variational approach. However, the result for k differs in the coefficientfor M. This difference (1' or F') is due to the nonlinear expansion used for

Example 14—1

We determine I' for the rectangular cross section shown in Fig. F.14—1.

I' = 11 = h1 y2/R J—a;2 1 — y2/R

=—R2bd+R3bln

To obtain a more tractable form, we expand the log terms, using

(1+x'\ I



This series converges for xI < 1. Then

d d3 I 3(d\2 3(d\4In = + ii + + +

and3 d 2

3 dI' =

{+

+ + ..

Fig. E14—1

I H

Y3

The relations listed above involve exactintegrals. Now, when the memberis thick, we neglect (y2/R)2 with respect to unity. This assumption is introducedby taking

1 —y2/R= 1 + + + ... +

in the expansions for and I':

Co

Y2—.--..e2

- yJR JJ +dA

=i{i


442 PLANAR DEFORMATiON OF A PLANAR MEMBER CHAP. 14

To be consistent, we must also neglect 1'/AR2 with respect to unity in theexpression for A'2 and I". When the member is thin, we neglect y2/R withrespect to unity.

1—y2/R

— y2k — (14—44)

at2

It is of interest to establish the one-dimensional form of the principle of virtualdisplacements corresponding to the linear displacement expansion used in thisdevelopment. The general three-dimensional form for an orthogonal coordinatesystem is (see Sec. 10—6):

SJJ(aii + + a12 öy12 )d(vol.) =

where represents an external force quantity and d1 is the displacement quantitycorresponding to We consider only and Viz to be finitc, and express thedifferential volume in terms of the cross-sectional coordinates Y2' Y3 and arclength along the ccntroidal axes (see Fig. 14—9):

d(vol.) = dS2 dy2 dv3 (i—

dS d7 dy3

Then (a) reduces to

(a11 + a12 (i—

dA] dS = (14—45)

We take (14—45) as the form of the principle of virtual displacements for planardeformation.

The strains corresponding to a linear expansion for displacements and lineargeometry are defined by (14—37), which are listed below for convenience:

ci —

Y12— y2/R

e2

du1 U2

du2 u1= + — U)

dok

dS

Substituting for e1, Y12 and using the definition equations for F1, F2, and M,



we obtainJs[Fi + F2 + M ök]dS = Ad1 (14—46)

This result depends only on the strain expansions, i.e., (c). One can apply itfor the geometrically nonlinear case, provided that (cS) are taken as definingthe strain distribution over the cross section.

We use the principle of virtual displacements to establish consistent force-equilibrium equations. One starts with one-dimensional deformation-displace-ment relations, substitutes in (14—46), and integrates the left-hand side by parts.Equating coefficients of the displacement parameters leads to a set of forceequilibrium equations and boundary conditions that are consistent with the geo-metrical assumptions introduced in establishing the deformation-displacementrelations. The following example illustrates this application.

Example 14—2

The assumption of negligible transverse shear deformation is introduced by setting e2equal to zero. This leads to an expression for the rotation. w, in terms of the translationcomponents,

(In2= +

and the relations for negligible transverse shear deformation reduce to

hEFt ôe1 + M ök]dS

d (du2 u11< =

= +

Substituting for Aw and the strain variations,

Au1 d= —i- Au2

d 1

Aui—

AU2

d2 I d(5k = Au2 +

and integrating by parts, the left- and right-hand sides of (b) expand to

j [F1 + M (5k]dS54

/ dM d= F1 + Au1 — Au2 + M An2

Rj dS uS

/ M\ dM d— I F1 + J

Au1 — —-- Au2 + M —\ Rj dS dS

I r dF1 1 dM1 [ F1

+ j1Aui

[— —+ Au2

[— +dS



and

Ad,=

(b1 + + (b2 dS

+ (PB! + + (p22 + ma) Au32 + A

+ (p41 +MA)

A + (r42 014) AUA2 + M4 A

The consistent equilibrium equations and boundary conditions for negligible transverseshear deformation follow by equating corresponding coefficients of the displacementvariations in (e) and (f):

S4<S<S3dF1 1dM 01

+ + + b1 + 0

F1 d2M drn+ - — + — = 0

s—snu1 prescribed or F1 —

u2 prescribed or p42 — m

do2prescribed or M = —MA

S =UI prescribed or F1 = F21

U2 prescribed or = —F32 — in

du2prescribed or M

One can obtain (g) by solving the last equation in (14—14) for F2 and substituting in thefirst two equations.

Suppose we neglect u1/R in the expression for w:

do2CD

d2u2k

This assumptiont is generally referred to as Mush tori's approximation. The equilibriumequations for the tangential direction reduce to

dF1

t See Ref. 5.


SEC. 14—5. CARTESIAN FORMULATION 445

The other equilibrium equation and the boundary conditions are not changed. Using(h) instead of (a) eliminates the shear term, F2/R, in the tangential force-equilibriumequation.

14—5. CARTESIAN FORMULATION

We consider the case where the equation defining the centroidal axis has theform x2 = f(x1). The geometrical relations for this parametric representationare obtained by taking y x1 in (14—7). They are summarized belowt forconvenience and the notation is shown in Fig. 14—12:

dS =

dx1

r 7df\21112 1

[ ylx1jj. cos0

i[ f'df'\ 1=I + I—)

j

- I [ / df \ - - (14—47)t2 + '2

= t1 X t2 = 13d2f

Ici:

In the previous formulation, we worked with displacement components andexternal force components referred to the local frame. An alternate approach,originally suggested by involves working with components re-ferred to the basic frame rather than the local frame. The resulting expressionsdiffer, and it is therefore of interest to describe this approach in detail. Westart with the determination of the force-equilibrium equations.

Consider the differential clement shown in Fig. 14—13. The vector, equilibriumequations are

dF+ - -- ax1

(14—48)

+ x + = 0dx1 dx1

See Prob. 14—1.See Ref. 6.


x2

PLANAR DEFORMATION OF A PLANAR MEMBER

'I

Y2

YI

dx1

x1

Fig. 14—12. Notation for Cartesian formulation.


446 CHAP. 14

X2

p-

dx1 2

N2 12

1

F1t1

lj


SEC. 14—5. CARTESIAN FORMULATION 447

where fl, h are the external applied force and moment vectors per unit projectedlength, i.e., per unit x1. They are related to b and (see Fig. 14—4) by

dx1 = b dS = (cth)dx1 (1449)hdx1 = iñdS = (cthi)dx1

Substituting for the force and moment vectors,

= F1t1 + F2t2 = N171 + N212it—hi3

P = + (14—50)

N1 = F1 cos 0 — F2 sin 0N2 = F1 sin 0 + F2 cos 0

the equilibrium equations expand to

dN1 d—- = ——(F1 cosP — sin 0) =dx1 dx1

= (F1 sin0 + F2 cos 0) P2 (14-51)dx1 dx1

—1(dM '\——-(—-- + hJ=' F2 —N1 sm0 + N2cosO

\dx1 ,i

We restrict this treatment to an elastic material and establish the force-displacement relations, using the principle of virtual forces,

dx1 = [e1 AF1 + e2 AF2 + k dx1 = d1 (a)

where V* V* (F1, F2 M) is the complementary energy per unit arc length.Consider the differential element shown in Fig. 14—14. The virtual-force sys-tem is statically permissible, i.e., it satisfies the force-equilibrium equationsidentically:

= odx1

+ J1 x =dx1

Expanding d1

x

and then substituting for the displacement and rotation vectors,

O = v111 + 1)212(14—52)

(0 (013 = (0t3


= + +dx1 dx1 dx1;

1)2

2

V1


Marguerre's equations are obtained by assuming the member is shallowand, in addition, neglecting the contribution of F2 in the expression for N1.


we obtain

Finally, substituting for N1, N2, in terms of F1, F2 and equating coefficients ofthe force increments result in

(d)

oV 2 do1 do2cos 0-— + sinOcosO—-

. dx1

a v* do1 , do2U +cosO——we2 = —sin 0 COS

-- dx1(14—53)

dwk = = ——— cos 0

dx1

The member is said to be shallow when 02 << 1. One introduces this assump-tion by setting

4fcos U 1 Sm 0 tan 0 =

in (14—50), which relate the cartesian and local forces.

(14—54)



Marguerre starts withN, F1

N2 F2 +

and the resulting equations are

dF,—— + Pi = 0dx,

dF2 d 7 df\+ (F1

J+ P2 = 0

dx, dx, dx,jdM

F2 —— — indx1 (14—55)

dv' df dv2e, = + ——d.x, dx, dx,dv2

e2 = = —- — co0F2 dx,

kdw

OM dx,

One step remains, namely, to establish the boundary conditions. The generalconditions are

v1 or N,v2 or N2 prescribed at each end (14—56)

M or w

We obtain the appropriate boundary conditions for the various cases consideredabove by substituting for N,, N2 and ox For example, the boundary conditionsfor the Marguerre formulation are

or F2 + F, prescribed at each end (14-57)

w or M

14—6. DISPLACEMENT METHOD OF SOLUTION—CIRCULAR MEMBER

The displacement method involves solving the system of governing dif-ferential equations which, for the planar case, consist of three force-equilibriumequations and three force-displacement equations. If the applied loads areindependent of the displacements, we can first solve the force equilibriumequations and then integrate the force-displacement relations. This methodis quite straightforward for the prismatic case since stretching and flexure areuncoupled. However, it is usually quite difficult to apply when the member is



curved (except when it is circular) or the cross section varies. In what follows,we illustrate the application of the displacement method to a circular memberhaving a constant cross section, starting with—

1. the exact equations (based on stress expansions) for a thick member2. Marguerre's equations for a thin member

The results obtained for this simple geometry provide us with some insight as tothe relative importance of transverse shear deformation and stretching deforma-tion versus bending deformation.

When the centroidal axis is a circular segment, R = const, and the equationssimplify somewhat. It is convenient to take the polar angle 8 as the independentvariable in this case. We list the governing equations below for convenienceand summarize the notation in Fig. 14—15:

dF1 dM / m

— RF1 = R2b2 — (14—58)

1dM

F1 M 1 f'du1= ej + + = —

Eq. (14—25) e2=

+ ui) — 0) (14—59)

F1 M Idwk = k° + + =

Solution of the Force-Equilibrium Equations

We consider the external forces to be independent of the displacements.Integrating the first equilibrium equation, we have

RF1 = —M — R2 j (b1 + +

where C1 is an integration constant. Substituting for F1 in the second equationresults in a second-order differential equation for M:

+ M C1 + R2[b2 — +

The general solution of (b) is

M = C1 + C2cosO + C3sinO + (14—60)



where denotes the particular solution due to the external distributed loadingand C2, C3 are constants. Once M is known, we find F1 using (a) and F2 fromthe moment equilibrium equation. The resulting expressions are

F1 = cosO + C3 sinO + Mr)— R j(b1 ± ;)dU(14—61)

F2 = sine + c3 cos 9 + — in

dS RdO

F

To determine u1

F1

Fig. 14—15. Notation for circular member.

Integration of the Force-Displacement Relations

We start with (14—59) written in a slightly rearrangcd form:

du1— u2 Re? + + RF1)

du2 RF2+ u1 + Rco

= Rk° + [M + (RF1)]

and u2, transform the first two equations to

= u2 + Re? + (M + RF1) (14—62)



andd2u2

+ = 1/1

=+ Re? — + RF1)

I

± R2k° — Re?+ a1R2 (14—63)

a1 = —

We have previously shownt that is of the order of (d/R)2. It is reasonableto neglect with respect to 1 but we will retain it in order to keep track of theinfluence of extensional deformation. We solve (14—63) for u2, determine u1from (14—62), and w from the second equation in (a),

F2 1 (du, '\W (14-64)

This leads to three additional integration constants. The six constants aredetermined by enforcing the three boundary conditions at each end. Variousloading conditions are treated in the following examples.

Example 14—3

Consider a member (Fig. El4—3) fixed at the negative end (A) and subjected onlyto at the right end (B). The boundary conditions for this case are

F1=Fa1; F2=M=O atO=011u1 =u2=w=O atO=O

Specializing the force solution for no external distributed loading and enforcing theboundary conditions at B, we obtain

F1 = F81 cos(08 — 0)

F2 = FRI Sifl(08 8)

M RF51(l — cos(08 — 0))

To simplify the analysis, we suppose there is no initial deformation. Using (b). takesthe form —

F81 R''I' [a1 — 02 COS(08 0)1

whereEI* (d\'

=

f See Sec. 14—3, Eq. (14—26).


+ {o + [o — 0) + sin(66 — 0)]}

Next, we determine w using (14—64),

C6U) = + {O + sm(OB — O)}

Finally, the constants are found by enforcing the displacement boundary conditionsat 0 = 0:

F R3C4 =

(05C5 = — sin

C6 =

To determine the relative importance of stretching and shear deformation versus bendingdeformation, we evaluate the displacements at 0 and write the resulting expressions

Fig. E14—3


Note that is associated with transverse shear deformation. Substituting for tJ' in (14—63)and integrating, we obtain

c4 cos 0 + c5 sin o + [ai + 0 — 0)]

The solution for u1 follows from (14—62):

= sin 0 — C5 cos 0 + C6

A

FBI

Constant cross section



in the following form:

WE = — sin + b1 Oe)

UBI = —2 sin 63 + 4sin 03 COS ± b2 + b3 &)

— COS 0B 4Sin —

sin(I)

— Slfl 63

b— —4 03 + 2 sin 03 — 4 sin 03 COS 013

2 63 + sin 03 Co

b4(03 — sin_03 cos 03)

T

b — 4sin2 63 — I + COS 63-

The coefficients (b1,..., b4) are of order unity or less when is not small with respectto unity, i.e., when the segment is not shallow. Also, öe and ó, are of order (d/R)2. It followsthat the displacements due to stretching and shear deformation are of order (dIR)2 timesthe displacement due to bending deformation for a nonshallow member.

To investigate the shallow case, we replace the trigometric terms in (i) by their Taylorseries expansions,

/ 02sin 0 = — + —

02cos 6 = I — + —

sin0cos6 0(1 — +—

and neglect with respect to unity. The resulting expressions are

P8152 1

1 1*U3j

P31S3 fOfl [1 EI*032

= +Now,

1*

El" (d'12

=



For example,1 (d\2

AS2

EI* E (d\2 (d'\2

= lOG= 0,26

for a rectangular section and v = 0.3. Since (cl/S)2 << I for a member, we can neglect thetransverse shcar terms in UBL, UB2 and the stretching term in co8. However, we must retainthe stretching term in u51 since it is of the same order as the bending term. The appropriateexpression for is

PatS3

In sum, we have shown that the percentage of error due to neglecting stretching andtransverse shear deformation is of the order of (d/R)2 for a nonshallow circular member.If the member is shallow < I 5°), we catnwt neglect stretching deformation. Actually,the stretching term dominates when the member is quite shallow. The error due to ne-glecting transverse shear deformation for the shallow case is still only of the order of(d/R)2.

Example 14—4

The internal force distributions due to acting on the cantilever member shownin Fig. E14—4 are given by

F1 sin(011 — 0)

F2 = F52 cos(05 0)

M R sin(02 0)

We suppose the member is not shallow and neglect stretching and shear deformation.The force-displacement relations reduce to (we set A = = in (14—59))

du1— = Re?

dO

du2+ u1 = Rw

dw RM= Rk° +

dO EJ*

Eliminating u1 from the first two equations, we obtain

12u R2+ u2 = R2k° — Re? + M

du1= u2 + Re?

I (du2

We determine u2, then u1, and finally oi. Note that (c) corresponds to (14—62), (14—63)and (14—64) with A = A2 = cc. The final expressions (for no initial deformation or support



movement) are

02

={(O COS(OB — 0) — sin 0 cos

= cos 0 0 Sjfl(OB — 0) + COS(OB 0)}

F52R0) = { COS(OB — 0) — cos

Example 14—5

We analyze the shallow parabolic member shown in Fig. E14—5 using Marguerre'sequations. We consider the member to be thin and neglect transverse shear deformation.Taking f = and = rn = 0, the governing equations (see (14—55) and (14—57))reduce to

dx1

d2Mal'1 — P2 = 0

dx1

dMF2

F1 dv1= e? + = -— + ax1

AL dx1 dx1

dv2Cl)

dx1

M d2v2k = k + =

v1, v2, w prescribed at x1 0

N2 = — + ax1F1 = 0 at x1 = I.dx1

M =0

Integrating (a) and using the boundary conditions at x1 = L, we obtain

M = — xi)2 — — x?)

F2 = p2(L — x1) — ax1N51

We suppose e? = k° = 0 to simplify the discussion. Integrating the moment-curvaturerelation,

= M = x1)2 CINBI(L2 4)


tP2 = COflSt

SEC. 14—6. DISPLACEMENT METHOD OF SOLUTION

B

457

Fig. E14—4

Fig. E14—5

l.A const

Al—,.- F1

a =L2

(h/L)2<(1

BNB! j


458 PLANAR DEFORMATfON OF A PLANAR MEMBER CHAP. 14

and noting that v2 dv2/dxj = 0 at x1 = 0 lead to the solution for v2,

Ely2 = — + — —

The axial displacement is determined by integrating the extensional strain displacementrelation,

dv1 F1 dv2= —

= - - + + ÷ -We express the last term in (g) as

NB1X1+

1

AE 6 ) 1) [VT)Now,

a 2h,/L2

Thena2L4(A'\ — 2(h'\2

6

and we see that this term dominatcs when h is larger with respect to the cross-sectionaldepth.

14—7. FORCE METHOD OF SOLUTION

Our starting point is the principle of virtual forces restricted to planardeformation,

Js(ei AF1 + e2 AF2 + k AM)dS — AR1 = d1 (14—65)

where the virtual-force system is statically permissible, represents a supportmovement, and AR1 is the corresponding reaction increment. The relationsbetween the deformation measures (e1, e2, k) and the internal forces (F1, F2, M)depend on the material properties and on whether one employs stress or dis-placement expansions. This discussion is limited to a linearly elastic materialbut one should note that (14--65) is valid for arbitrary material. For con-venience, we list the force-deformation relations below. The notation forinternal force quantities is shown in Fig. 14—3.

Arbitrary Linearly Elastic Menther

F1 M— e1 + +

F2e2 = (14—66)

F1 Mk = + + -=AER. El



k°, A2, and] are defined by(14—24) for the stress-expansion approachand (14—42) for the displacement-expansion approach.

Thin Linearly Elastic Member

F1e1 = e? +

(14-67)

where A2, a?, k° are the same as for a prismatic member.When the member is nor shallow, it is reasonable to neglect stretching and

transverse shear deformation. As shown in ExampLe 14—3, this approximationintroduces a percentage error of O(cl/R)2. Formally, one sets A = A2 =If the member is shallow, we can still neglect transverse shear deformationbut we must include stretching deformation.

The basic steps involved in applying the force method to a curved memberarc the same as for the prismatic case. However, the algebra is usually morecomplicated, due to the geometry. We will discuss first the determination ofthe displacement at a point.

To determine the displacement at Q in the direction defined by tQ, we applyan external virtual force, APQ7Q, generate a statically determinate system ofinternal forces and reactions corresponding to

= FJ,QAPQ (j = 1,2)

AM = M AP0 (14-68)ARk = Rk. Q APQ

and substitute in (14—65):

dQ = SS(eIFI,Q + e2F2Q + kM,Q)dS — (14—69)

This expression is valid for an arbitrary material. We set e2 = 0 if transverseshear deformation is negligible and a1 = a? if stretching deformation is

negligible.

Example 14—6

We consider the thin linearly elastic circular segment shown in Fig. E14—6A. Wesuppose the member is not shallow and neglect stretching and franslerse shear deformation.The reactions are the end forces at A for this example, and (14—69) expands to

= + (k0 + + OA1FAIQ + UA2FA2.Q + WAMA,Q (a)

In what follows, we illustrate tile application of (a)



Fig. E14—6A

U2L

Expressions Displacements at B

To determine u81, we take = AF51. The internal virtual-force system correspondsto F81 = ±1. It is convenient to work with = 0 as tlie independent variablerather than 0.

The force-influence coefficients (F10, F2,Q, M.Q) follow directly from Fig. E14—6B:

F10 = = COSq

F20 = (b)

= R(1 cos

Substituting (b) in (a) results in the following general expression for um

= RJo3

{e? cos + R (ko + (1 — cos

cos + sin + — cos

Once the loading is specified, we can evaluate the integral. Terms involving the supportdisplacements define the rigid body displacement at B.

Taking = SF82, leads to expression for u82 and We list them belowfor future reference:

"u82 = R

j(d)

Jo (. Elj

COB = W.4 + R f3(ko +

Solutionfor a Concentrated Loading at an Arbitrary Interior Point

We consider an arbitrary force vector, and moment, Mc, applied at point C as shownin Fig. E14—6C.


SEC. 14—7.

ill

A.

FORCE METHOD OF SOLUTION 461

Fig. E14—6B

Fig. El 4—SC

Ffi2 ,t482

RI — eQS



= +

=

The expressions for the displacements at B due to an external loading are obtained byspecializing (c) and (d) for no initial deformation or support movement and noting that

M=OM = RPc1[1 — iic)] + sin(1i — ilc) + ivIc

The solution for constant I is

Pc1R3( . . .

=Sifl 0c — sin + Sill 1k + 1k + Sill 0c cos

PC2R( Or . I+ — cos + sin — sin sin

MCR2+ (Oc + Sin Sin

PciR3I 1 1= OR + cos — Oc sin — sm Oc sin

PC2R3 /1 1

+ —--h-— cos 1k — sin 9c cos

+ —-—- (ens 1k — cos

. R2Pc2Sin Oc) + cos Oc) +

If we take point C to coincide with B, = 0 and = OR. The resulting equationsrelate the displacement at B due to forces applied at B in the directions of the local frameat B and can be interpreted as member force-deformation relations. It is convenient toexpress these relations in matrix form:

=— 2 sin R2[l — cos OR — sin OR] F81

+ 4sin 08 cos 08] OR]

_______________

= Symmetrical — sin °B cos OR] R[l — cos 98] F82

We call the member flexibility matrix.

We describe next the application of the principle of virtual forces in theanalysis of a statically indeterminate planar member. Let the member be in-determinate to the rth degree and let Z1, . . , Z, represent the force redundants.Using the equilibrium equations, we express the internal forces and reactions in



terms of the applied loads and the force redundants:

F1 = F10 +k-= 1

F2 = F2,0 +(14—70)

M = M0 +k= 1

R1 = R10 +k1

R1.kZk

Substituting the virtual force system corresponding to (which is staticallypermissible and in (14—65) and letting j range from 1 to rlead to the compatibility equations relating the actual deformations:

+ + kM,1)dS = 0(a)

1= l,...,rWhen the material is linearly elastic, the compatibility equations take the

form

k1 (j = ...,r) (14—71)

where

=fkj = f + + Fl.kM,f)

+ +

=—

+ +

+ + Fl,OM.J) + ±

Wesetl = I,A2 = ,42,and 1/AR Oforathinmember.Note that fik is the displacement of the primary structure in the direction of

Z1 due to a unit value of Zk. Also, is the actual displacement of the point ofapplication of minUs the displacement of the primary structure in the direc-tion of Z, due to support movement, initial deformation, and the prescribedexternal forces.

Example 14—7

Consider the symmetrical closed ring shown in Fig. E14—7. From symmetry,

at6r=0 (a)

F2 = 0 j



We take the moment at 0 = 0 as the force redundant. To simplify the algebra, we supposethe member is thin and neglect stretching and shear deformation. The compatibility equa-tions reduces to

f11Z1 =

fii =

& =

Note that is the relative rotation due to a unit value of Z1 and is the relativerotation (X) due to the applied load. Equation (h) states that the net relative rotationmust vanish.

Now,

M = R(t — cos 0)

fm/F

1'

E14—7

We consider I to be constant. Then, (b) reduces to

1 dS — —J(1 — cos 8)dO /PR\

JM21 cIS

PR1 2\= TI\1

——)

M F1


FORCE METHOD OF SOLUTION

Because of symmetry, we need to integrate over only a quarter of the ring. Finally, thetotal moment is

The axial and shear force variations are given by

F1

F2 =

When the equation defining the centroidal axis is expressed in the formx2 = f(x1), it is more convenient to work with force and displacementtities referred to the basic frame rather than to the local frame, i.e., to use thecartesian formulation developed in Sec. (14—5). The cartesian notation is sum-marized in Fig. 14—16.

Ag. 14—16. Notation for Cartesian formulation.

The geometrical quantities and relations between the internal force com-ponents are

tan 6 =dx1

dS =cos 6

F1 = N1 cos 6 + N2 sin UF2 = —N1 sin (9 + N2 cos 0

SEC. 14—7. 465

xl

N2,

F2

——--'-

i2

it

(14—72)


PLANAR DEFORMATION OF A PLANAR MEMBER CHAP. 14

We first find N1, N2 and then determine F1, F2. To obtain the equations forthe Cartesian case, we just have to replace dS by dx1/cos U in the general ex-pressions ((14—69) and (14—71)). In what follows, we suppose the member isthin and linearly elastic.

When the member is not shallow, we can neglect the stretching and transverseshear deformation terms. The equations for this case reduce to:

Displacement at Point Q

dQ

= L + (ke + (14-73)

C'oinpatibility Equations

\dx1= j (14—74)

= — 5[e?Fi,i + (ko +

We can the terms involving F1 in terms of N1 and N2 since

Plo = cos 0 N10j + sin 0Then,

j'+ ,f'N2, ( (14—75)

One must generally resort to numerical integration in order to evaluate theintegrals, due to the presence of the term 1/cos 9.

When the member is shallow, 02 1, and we can approximate (14—72) with

cos 0 1

sm 0 tan 0 =cls dx1 (14—76)

F1 N1 +f'N2F2 —f'N1 + N2

We cannot neglect the stretching deformation term in this case. However, it isreasonable to take F1 N1. We also introduced this assumption in the devel-opment of Marguerre's equations. The equations for the shallow case withnegligible transverse shear deformation and F1 N1 have the forms listedbelow:


dQ

= J[(ei° + N1, + (ko + M, dx1 — R1, (14—77)




Example 14—8

Consider the two-hinged arch shown in Fig. E14—8A. We work with reaction com-ponents referred to the basic frame and take the horizontal reaction at B as the forceredundant.

x2,

Primary

Fig. E14—8A

We must carry out two force analyses on the primary structure (Fig. E14—SB), one forthe external forces (condition Z1 = 0) and the other for Z1 1. The results are displayedin Figs. E14—8C and D, respectively.

R242

>j1k4 =

fjk= Jx,

+ dx1 (14—78)

=— f + + (k0 +

P

V2

Fig. E14—8B

= Z1

zI


N2

KM

(+)

(+)

We suppose the member is not shallow. The compatibility equations for Z1 followfrom (14—74):

dxI tJ11—I El cosO

(ko= + f'N2,1) + +

Jo L


Fig. E14—8C

0

Fig. E14—8D

H

'VI,'

ii

N2,,

M,1




Using the results listed above, the various terms in (a) expand to

f= .ft

— h

0E1 L cost?

= ± +

s: + f'N2 + k° dx1

JL[(+ + k0(f_JLIJLI(h)(h)

+ !L (+P(xi - a))dxiLIcosO L

Once the integrals are evaluated, we can determine Z1 from

(c)

Finally, the total forces are obtained by superposition of the two loadings:

M = M•0 + Z1M1(d)

R. = + Z1R, = 1, 2, 3

R4 =

To evaluate the vertical displacement at point Q, we apply a unit vertical load at Q onthe prinwry structure and determine the required internal forces and reactions plotted inFig. E14—8E.

Fig.

F—

XQ 1/L

Q

(i_IL)FQ=+l



Applying (14—73), we obtain

VQ2 = VA2 + (13B2 — VA2)

+ —

('XQ I' M\ dx1 CL / dx1— I x1Ik°+—1—---—---—x1 I (k°+—'l—-—

Jo \ El)cosO El) cosO

XQ1 1L( M'\ dx1+— f x1(k°+-—J-——L jo \ EljcosO

A numerical procedure for evaluating these integrals is described in the next section.

Example 14—9

The symmetrical nonshallow two-hinged parabolic arch shown in Fig. E14—9A is sub-jected to a uniform load per unit horizontal length, that is, per unit x1. The equation forthe centroidal axis is

4h(

where h is the elevation at mid-span = L/2). We take the horizontal reaction at theright end as the force redundant and consider only bending deformation. Figures E14—9Band C carry through an analysis parallel to that of the preceding example,

Deternzination of Z1 and Total Forces

The equation for Z1 follows from (14—74):

1L

Jo EJcosO pL2

ElcosO

Note that this result is valid for an arbitrary variation of El. Finally, the total forces are

N1 — N10 + Z1N1 =

N2 = N20 + Z1N21 =—

M=M,0+Z1M,1 =0Since M = 0, the deformed shape of the arch coincides with the initial shape when axialdeformation is neglected. It follows that (c) also apply for the fixed nonshollolv case.

When the arch is shallow, the effect of axial deformation cannot be neglected. Theexpression for Z1 follows from (14—78):

— +



Pnmary structure

R2,d2

Force System Due to p

N2

xI

p = Coflst

Fig. E14—9A

Bxi

Fig. E14—9B

L

N5,00 pL( Xj\ pLM0 )



If E is constant, (d) reduces to

pL2

JLdx1

— pL2 1

— 8h+

1L1— 8h I +

jL1

-

The parameter ö is a measure of the influence of axial deformation. As an illustration, weconsider A and I to be constant and evaluate ö for this geometry. The result is

I 15(p'\2— 8 Ah2 — 8

where p is the radius of gyration for the cross section.

A'2 Fig.

N11 = +1 N2,1 = 0 M,1 = +fForce System Due to Z1 = + I

One should note that (e) applies only for the shallow case, in, for (f')2 K< 1. Now,

4/1 / 2x1

For the assumption of shallowness to be valid, 16(h/L)2 must he small with respect tounity. The total forces for the shallow case are

pL2 1N1 = =

p1? ( ô \ PL(M = f =

—


SEC. 14—8. NUMERICAL INTEGRATION PROCEDURES 473

It is of interest to determine the rotation at B. The "Q" loading consists of a unit momentapplied at B to the primary structure (see Fig. E14—9D). Applying (14—77) (note that

Fig. E14—9D

PQ +l

the stretching terms vanish since Nj,Q = 0), we obtain

1L M x1 P / \ / ,

jWhen El is constant, (i) reduces to

pL3( (5

2

Since 0, the results for the fixed end shallow case will differ slightly from (h).

14—8. NUMERICAL INTEGRATION PROCEDURES

One of the steps in the force method involves evaluating certain integralswhich depend on the member geometry and the cross-sectional properties.Closed-form solutions can be obtained for only simple geometries, and oneusually must resort to a numerical integration procedure. In what follows, wedescribe two proceduresi which can be conveniently automated and illustratetheir application in deflection computations.

We consider the problem of evaluating

J (14—79)

t See Ref. 8 for a more detailed treatment of numerical integration schemes.



where f(x) is a reasonably smooth function in the interval XA x Wedivide the total interval into n equal segments, of length h:

h = — XA(14—80)

If f(x) is discontinuous, we work with subintervals and use a different spacingfor each subinterval. For convenience, we let x0, x1, . . , x,, represent thecoordinates of the equally spaced points on the x axis, and f0, thecorresponding values of the function. This notation is shown in Figure 14—17.

12 1;,

__________________

a 11 B

XO X1 X2 X,,

Fig. 14—17. Coordinate discretization for numerical integration.

The simplest approach consists in approximating the actual curve by a setof straight lines connecting etc., as shown in Fig. 14—18. With thisapproximation,

('xk hEXJk1,k = f(x)dx + (14—81)

= dx = +

If only the total integral is desired, we use,

Jf(x)dx

=+ L)

+(14—82)

which is called the trapezoidal rule.A more accurate formula is obtained by approximating the curve connecting

three consecutive points with a second-degree polynomial, as shown in Fig.14—19. This leads to

hAJk,k+2

= J

fdx + 4fk+1 + (14—83)

Jk÷2= Jk +To apply (14—83), we must take an even number of segments, that is, n mustbe an even integer. If the values of J at odd points are also desired, they can



be determined usingh

= {Sfk + — Jk+2] (14—84)Jo 12

Finally, one can express as

in = + + 4(11 + .13 + +(14—85)

+ 2(f2 +f4 +Equation (14—85) is called Simpson's rule.

f

N

fk—1 fk fk+1

h Sx

Xk_1 Xk

Fig. 14—18. Linear approximation.

I

fk fk+1 fk÷2

S S

Xk Xkf.1

Fig. 14—19. Parabolic approximation.



Example 14—10

Consider the problem of determining the vertical displacement at Q for the straightmember of Fig. E14—lO. We suppose shear deformation is negligible. The deflection due

Fig. E14—10

XQ

________________M

M

PQ +1

XQ(1t)

to bending deformation (we consider the material to be linear elastic) is given by

dQJ

MQ dx

where M is the actual moment and M0 is due to the "Q" loading. Substituting for M,(a) expands to

/ 1L M CXQ M '\ M M=

— J+

J —



To evaluate (b), we divide the total length into ii equal segments of length h, numberthe points from 0 to n, and let

MI —dxJ0 ElCx MI x—dx

Jo El

With this notation, (b) takes the form

= Xk — + "k Xkjk

If, in determining we also evaluate the integrals the interior points, then we canreadily determine the displacement distribution using cd).

Example 14—11

Consider the simply supported nonshallow arch shown. We suppose there is somedistribution of It'! and we want to determine the vertical deflection at Q. Considering

A

1WdQ = M, ds

J1W, dx

L

Fig. E14—11

illEl

only bending deformation, dQ is given by



Now, the distribution of M is the same as for the straight member, Then, the procedurefollowed in Example 14—10 is also applicable here. We just have to replace M/EI withM/EI cos 0 in Equation (c) of Example 14—10.

REFERENCES

1. TIMOSHENKO, S. 1.: Advanced Strength of Materials, Van Nostrand, New York, 1941.2. Boan, S. F., and J. J. GENNARO: Advanced Structural Analysis, Van Nostrand,

New York, 1959.3. REISSNER, E.: "Variational Considerations for Elastic Beams and Shells," J. Lag.

Mech. Div., A.S.C.E, Vol. 88, No. EM 1, February 1962.4. MARTIN, H. C.: introduction to Matrix lvi etliods of Structural Analysis, McGraw—Hill,

New York, 1966.5. MUSRTARI, K, M., and K. Z. (IALIMOV: "Nonlinear Theory of Thin Elastic Shells,"

Israel Program for Scientific Translations. Jerusalem, 19626. MARGUERRE, K.: "Zur Theoric der gekriimmten Platte grosser Formanderung,"

Proc. 5th mt. Congress App!. Mccli. pp. 93—101. 1938.7. Onai'i, J. 1.: Mechanics of Elastic Structures. McGraw-Hill, New York, 1967.8. HILDEBRAND, F. J.: introduction to Numerical Analysis, McGraw-Hill, New York,

1956.

PROBLEMS

14—1. Specialize (14—7) for the case where Yi = x1. Let x2 = f(x1) andlet 9 be the angle from X1 to Y1 as shown below. Evaluate the various termsfor a parabola

f = +

Finally, specialize the relations for a shallow curve, i.e., where

Prob.14—132

14—2. Evaluate 1* and (see Equation 14-24) for the section defined bythe sketch.

14—3. Verify (14—34).14—4. Verify (14—41) and (14—42).14—5. Discuss the difference between the deformation-force relations based

on stress and displacement expansions (Equations (14—25) and (14—42)).Illustrate for the rectangular section treated in Example 14—i. Which set ofrelations would you employ?


h=O.75dt=d120

14—7. Consider a circular sandwich member comprised of three layers,as shown. The core layer is soft (E 0). and the face thickness is small incomparison to the depth d). Establish force-deformation relations basedon strain expansions (see (14—37)).

Prob. 14—i

14—8. Starting with (14—34) and (14—35), derive a set of nonlinear straindisplacement relations for a thin member. Assume small finite rotation, and

PROBLEMS 479

Prob. 14—22t

I Tb=O.75d

d

I14—6. Evaluate I' and 1" for the symmetrical section shown.

Prob. 14—6

I III



linearize the expressions with respect to Yz' i.e., take

= e1 —

Y12

Determine the corresponding force-equilibrium equations with the principle ofvirtual displacements.

14—9. Refer to Fig. 14—10 and Equation (14—31). If we neglect transverseshear deformation, is orthogonal to t'1 and we can write

(1 + =—

ldF'tj

-= fl1t1 + fl2t2

= —/32t1 + f31t2

dt'1 1 + e1 .-, di'2 I + e1= R'

= + = (1 +

(a) Verify that can be expressed as

1 c (i+e1 ICl

- y,k}

Also determine e1 and R' for small strain. Express ü in terms of theinitial tangent vectors,

ü = U1t1 + U2t2

and take y S (i.e., 1).

(b) Derive the force-equilibrium equations, starting with the vector equa-tions (see (14—12) and Fig. 14—4),

+ = 0dS

dM÷ —+ m + x F÷ = 0

and expanding the force vectors in terms of components referred tothe deformed frame:

= F11'I + h = b11'1 += Mi3

Assume small strain.(c) Derive the force-equilibrium equations with the principle of virtual

displacements. Take the strain distribution according to Equation (b).


PROBLEMS 481

(d) Derive the nonlinear deformation-displacement and equilibrium equa-tions for the cartesian formulation. Refer the translations and loadingto the basic frame, i.e., take

= + V212

P Pi'i + P2t2

Specialize the equations for the case of a shallow member.14—10. The accompanying sketch applies to both phases of this problem.

h2 = constProb. 14—10

(a) Determine the complete solution for the circular member shown.Utilize symmetry at point A = co = F2 = 0) and work with (14—58),(14—59). Discuss the effect of neglecting extensional and shear de-formation, i.e., setting (1/A) (1/42) = 0.

(b) Repeat (a), using Mushtari's equations for a thin member with notransverse shear deformation, which are developed in Example 14—2.Show that Mushtari's approximation (u1 << du2/dO) is valid when thesegment is shallow.

14—11. The sketch presents the information relevant to the problem:

Prob. 14—11

2

________

L L 'j L2

x2

P2 = cOnSt



(a) Apply the Cartesian formulation to the symmetrical parabolic archshown. Consider the member to be thin and neglect transverse sheardeformation.

(b) Specialize (a) for negligible extensional deformation (set 1/A = 0).(c) Specialize (a) for the shallow case and investigate the validity of

Marquerre's approximation.14-42. Refer to Example 14—6. Determine UB2 due to a uniform distributed

loading, b2 constant.14—13. Determine the displacement measures at B (see sketch). Consider

only bending deformation. Note: It may be more convenient to integrate thegoverning equations rather than apply (14—69).

14—14. Solve two problems with the information sketched:

Prob. 14—13

Prob. 14—14

(a) Determine the fixed end forces and radial displacement at point Bwith the force method. Consider only bending deformation and utilizesymmetry at B.

(b) Generalize for an arbitrarily located radial force.

A


PROBLEMS 483

14—15. Refer to Example 14—7.(a) Determine the radial displacement at B defined in Fig. E14—7.(b) Determine the force solution for the loading shown.

14—16. The sketch defines a thin parabolic two-hinged arch.

Prob. 14—15

Prob. 14—16

(a) Determine the horizontal reaction at B due to the concentrated load.Consider the arch to be nonshallow.

(h) Utilize the results of(a) to obtain the solution for a distributed loadingp2(x) per unit x1.

(c) Determine the reactions resulting from a uniform temperature increase,T.

(d) Suppose the horizontal support at B is replaced by a prismatic memberextending from A to B. Assume the connections are frictionless hinges.Repeat parts a and c.

P

P

x2



14—17. Consider the arbitrary two-hinged arch shown. Discuss how you

Prob. 14—17

would generate the influence line for the horizontal reaction. Utilize the resultscontained in Examples 14—10 and 14—11.


15

Engineering Theory ofan Arbitrary Member

15—i. INTRODUCTION; GEOMETRICAL RELATIONS

In the first part of this chapter, we establish the governing equations for amember whose centroidal axis is an arbitrary space curve. The formulation isrestricted to linear geometry and negligible warping and is referred to as the

theory. Examples illustrating the application of the displacementand force methods are presented. Next, we outline a restrained warping for-mulation and apply it to a planar circular member. Lastly, we cast the forcemethod for the engineering theory in matrix form and develop the memberforce-displacement relations which are required for the analysis of a systemof member elements.

The geometrical relations for a member are derived in Chapter 4. Forconvenience, we summarize the differentiation formulas here. Figure 15—1shows the natural and local frames. They are related by

—

cos çbñ + sin çt'b

= —sin4iii +

where = Differentiating (a) and using the Frenet equations (4—20),we obtain

dt= at

0

0 t (15-1)

01t3

Note that a is skew-symmetric.

485


486 ENGINEERING THEORY OF AN ARBITRARY MEMBER CHAP. 15

Fig. 15-i. Natural and local reference frames for a member element.

The curvilinear of a point, say Q, are taken as S and Y2, y3.Letting K be the position vector to Q (see Fig. 15—2),

and differentiating, we find

R = + y2t2(S) + (15—2)

13R -= (t — y2a12 — y3a23t2 + y2a23t3

3K —— = t2

3K

3y3

d(vol.) (1 Y2a12 — y3a13)dS dy2 dy3

= (i E dS dy2 dy3

where is the coordinate of Q in the normal (11) directionthe radius of curvature. Also,

3R .. (1Y3 +

3R /1 dcby2a23 = Y2 +

n

I

b

Local reference directions

The differential volume at Q is

(15—3)

(15—4)

and = 1/K is

(15—5)

f See Sec. 4—8.


dq5

dS

ay2

SEC. INTRODUCTION; GEOMETRICAL RELATIONS 487

and the local vectors at Q are orthogonal when a23 0, which requires

a23 =0

d/ (15—6)

It is reasonable to neglect y/R terms with respect to unity when the memberis thin, i.e., when the cross-sectional dimensions are small in comparison to

Fig. 15—2. Curvilinear directions.

and We express d4,/dS as

where L is the total arc length and is the total increment in The non-orthogonality due to can be neglected when the member is only slightlytwisted, i.e., when

<< 1 (15—8)

where b is a typical cross-sectional dimension. In what follows, we will assumethe member is thin, (15—8) is satisfied, and defines the orientation of theprincipal inertia directions.

—

(15—7)



Example 15—i

The curvature and torsion for a circular helix are derived in Example 4—5:

1

R, R (H'\2

1

where R is the radius of the base circle and H is the rise in one full revolution. The helixis thin when b/R c< 1, where b is a typical cross-sectional dimension.

Example 15—2

By definition, a member is planar if r = 0 and the normal direction (Il) is an axis ofsymmetry for the cross section. We take the centroidal axis to be in the X1-X2 planeand define the sense oft2 according tot2 x 13 = Theangle is constant and equal toeither 0° (12 or 180° (12 = —ii). Only a12 is finite for a planar member:

a13=a23=0

Example 15—3

Consider the case where the centroidal axis is straight and varies linearly with S.

The member is said to he naturally twisted. Only 023 is finite for this case:

= a13 = 0

(Ic!)

T = const IC

If bk < 1, we can assume aR/aS is orthogonal to F2,

15—2. FORCE-EQUILIBRIUM EQUATIONS

To establish the force-equilibrium equations, we consider the differentialelement shown in Fig. 15—3. We use the same notation as for the planar case.The vector equilibrium equations follow from the requirement that the resultantforce and moment vectors must vanish:

dS

_. — -+ rn + t1 x F÷ = 0

We express the force and moment vectors in terms of components referredto the local frame,

= = FTt

= (15—10)

m't


SEC. 15—2. FORCE-EQUILIBRIUM EQUATIONS 489

bc/S

inc/S+ c/S 2


where F = F2, F3} etc. The vector derivatives are

dF÷ dFT'T= -------t + at

tIS dS(a)

— +dS dS

Also,x = FJ3 — = {O, _F3,F2}Tt (b)

Substituting in (15—9), and noting that = —-a, lead to the following equi-librium equations:

aF + b = 0

dM

+F0J

— a12F2 —a33F3 + b1 = 0

+ a12F1 — a23F3 + b2 = 0. (15—11)

dF3+ a13F1 + a23F2 + b3 = 0

dM1a12M2 — a13M3 + m1 = 0

+ a12M1 — a23M3 + m2 — F3 0

When the member is planar, a13 = a23 = 0 and the equations uncouplenaturally into two systems, one associated with in-plane loading (b1, b2, rn3,

+ c/S 2

+ dS2



F1, F2, M3) and the other with out-of-plane loading (b3, in1, m2, F3, M1, M2).The in-plane equations coincide with (14—14) when we set a12 = 1/R and theout-of-plane equations take the form

dF3

dM1 1.

+ in1 = 0 (15—12)

dM2 1

+ + m2 — F3 = 0

15—3. RELATIONS—NEGLIGIBLE WARPINGRESTRAINT; PRINCIPLE OF VIRTUAL FORCES

We consider the material to be elastic and define V" as the complementaryenergy per unit arc length. Since we are neglecting warping restraint, is afunction only of F and M. We let

Iav*. -

— 1, 2, 3 (15 13)

k = {kJ

and write the one-dimensional principle of virtual forces as

dS = Js(eT AF + kT AM)dS (15—14)

It

dS

(0



SEC. 15—3. NEGLIGIBLE WARPING RESTRAINT 491

Now, we apply the principle of virtual forces to the element shown inFig. 15—4. We define ü and ii5 as

= = uTt = equivalent rigid body translationvector at the centroid (15—15)

= = equivalent rigid body rotation vector

The virtual system satisfies the equilibrium equations (15—5) identically andtherefore is statically permissible. Evaluating APi,

AFT an + + AMT - aw) dS

and substituting in (15—14) lead to the following force-displacement relations:

10du

e — an +

(15—16)dO)

k=

aY* du1e1 = = — —

du2C2 = + a12u1 — —

Cj2

a v* du3e3 = = + + 023u2 +

(113

aV* dw1k1

= =— 0120)2 — (1130)3

DV" dw2k2

= =+ 0120)1 — 023C03

(* d0)3k3 = = + + 0230)2

Once is specified, the left-hand terms can be expanded. The form. ofdepends on the material properties, the particular stress expansions selected,and the member geometry. In what follows, we consider the material to belinearly elastic and approximation with the complementary energy function


492 ENGINEERiNG THEORY OF AN ARBITRARY MEMBER CHAP. 15

for the prismatic case, which is developed in Sec. 1 2—3:

= + + + +(15-17)

+ + + +

where

MT = M1 + — F3y2 = torsional moment with respect tothe shear center

Y2, coordinates of the shear center with respect tothe centroid

= if dA

=—j-'- if dA

Note that (lST-i7) is based on a linear expansion for the normal stress,

F1 M-, M3= 7 +

13

and using the shear stress distribution predicted by the engineering theory,

= +

where is the unrestrained torsional distribution due to MT and is theflexural distribution due to F2, In addition to these approximations, weare also neglecting the effect of curvature, i.e., we are considering the memberto be thin. The approximate force -displacement relations for a linearly elasticthin curve member are

F1 du1+ = — —

F2 du2

=+ = + a12u1 — C21U3 — -

F3 MT du3e3 = + a13u1 + a21tc2 + (03

15 18MT dü1

k1 = = a12co2 — a13w3

M2 dw2—a23co3

M3 dw3k3 = + + a13co1 + 1223(02

UI3 U,)


SEC. 15—4. CIRCULAR PLANAR MEMBER 493

When the member is planar, the shear center is on the Y2 axist and there isno coupling between in-plane (u1, U2, and out-of-plane (u3, (02) dis-placements. That is, an out-of-plane loading will produce only out-of-planedisplacements. The approximate force-displacement relations for out-of-planedeformation for a thin planar member are

F3 MT — du3e3 = = +

MT dw1 I

(15-19)

M2 1k2 = k2 +

where = M1 — y2F3. Note that flexure and twist are coupled, due to thecurvature, even when the shear center coincides with the centroid.

15—4. DiSPLACEMENT METHOD—CIRCULAR PLANAR MEMBER

Since the displacement method involves integrating the governing differentialequations, its application is restricted to simple geometries. In what follows,we apply the displacement method to a circular planar member subjected toout-of-plane loading. We suppose the cross section is constant and the shearcenter coincides with the centroid. It is convenient to take the polar angle eas the independent variable. The governing equations are summarized belowand the notation is defined in Fig. 15—5.

Equilibrium Equations (sec (15—12))

+ Rh3 0

— M2 + R,n1 = 0

dM2 +Rrn2—RF3=O

Force-Displacement Relations (see (15—19))

F3 ldu3e3

= =+

M1 I /dw1k1

o M2 l/dw2k2 = k2 + = +

t The shear center axis lies in the plane containing the centroidal axis, which, by definition, isa plane of symmetry for the cross Section.


494 ENGiNEERING THEORY OF AN ARBiTRARY MEMBER CHAP. 15

Boundary Conditions

F3 or u3

M1 or prescribed at each end (pts. A, B)

M2 or

The solution of the equilibrium equations is quite straightforward. Weintegrate the first equation directly:

F3=C1—RJ0b3d0 (15—20)

The remaining two equations can be transformed to

(d)

dM1= + Rin1 (e)

We solve (d) for M1 and determine M2 from (e). The resulting expressions are

M1 = C2cosO + C3 sinG +

M2 = —C2sinO + C3cosO + + Rm1

where M1 ,, is the particular solution of (d).

x2P3

the X1 —X2 plane.

are axes of symmetry.

i313

B

F3

Fig. 15—5. Notation for circular member.



The solution of the force-displacement relations is also straightforward.First, we transform (b) to

d2co1 R2 R+ w1 = Rk2 — in1 + (1 +

RM5(f)

=- Rw2

where is a dimensionless parameter,

=(15—22)

which is an indicator for torsional deformation. Solving the first equation forand then determining W2 and u3 from the second and third equations lead

to= C4cosO + C5 sin fJ + wi,,

d RM1= —C4 sin U + C5cosU +

C6 — + j +dO

where is the particular solution for wi.The complete solution involves six integration constants which are deter-

mined by enforcing the boundary conditions. The following examples illustratethe application of the above equations.

Example 15—4

The member shown is Sxed at A and subjected to a uniform distributed loading. Taking

Fig. E15—4

h3 = coast in (15—20), we obtain

F3 = C1 — Rb30 (a)

b3 = const



The equation for M3 reduces to

+ M1 = RF3 = RC1 — R250

Then,= RF3 = — R2b30

and the solution for M3 and M2 follows from (15—21),

= C2 cos 0 + C3 sin 0 + RF3M2 —C2.sinO + C3cosO — R2b3

The boundary conditions at B require

F3=M1=M2=O at 0=08

= R5308

C2 = —R2h3 sin

C3 R2b3 cos

Replacing — 0 by the final solution is

F3 = Rb30

= R2b3[q — sin ?q]

M2 .—R2h3[1 — cos

Example 15-5

The force system due to the end action, can be determined by applying thelibrium conditions directly to the segment shown in Fig. E15—SA. This leads to

F3 == FB3R(l — cos F83R[l •— — 0)] (a)

M2 = — sin = — sin(Oa 9)

We suppose there is no initial deformation. Using (a), the equation for w3 becomes

+ w1 = (1 + — 0) (b)

The particular solution of (b) is

—(1 + [0 cos(OB 9)]

Using the above results and specializing (15—23) for this support condition lead to thefollowing expressions for the displacements:

WAI COS 0 + sin 0

+ C05 + sin 0 — 9 cos(08 —



ISlflO + cosO + + cosO]

. . . 1

—

Sm 0—

Osln(Oa — (d)

133 UA3 + RthAI(l — cos £1) — Rö5A2 Sifl 0

RFB3 IF I+

1LCOS C1j sin P SW

0 cos(05 — 0) + c, + + sln(OR —

Terms involving and define the rigid body displacements due to supportmovement. Also, terms involving c1 are due to twist deformation. The rotations and

F3

translation at B are listed below:

COS + sin 0a

1R2P83 f+ COS ± c1] sin

sin + cos

1—c,2+ —h--— 033 -. 1] — —--—- sin

13)33 + Roi41(l — cos Rw42 Sm 0)3

1—c3+ C05 sin — 2c1 sin

3 El2 Th+ ++

Fig. E15—5A

A

M2

I?



To investigate the relative importance of the various deformation terms, we considerthe rectangular cross section shown in Fig. El 5—5)3. The properties aret

1 61 _6 I

A3 5A — 5d2d3

J = (for d3)

Then,(d3\2

El2 — E [1 (di'\21GA3R2 — G [io k,i2) J

The values of 4k and for d3/d2 = 1, 2, 3 and v = 0.3 are tabulated below:

= E12/GJd3/d2 4k (Ibr v 0.3)

1 1,69 1.54

2 2.75 3.8

3 3.16 7.4

Fig. E15—5B

T

1(3

t The torsional constant for a rectangular cross Section is developed in Sec. 11—3.


SEC. 15—5. FORCE METHOD—EXAMPLES 499

Since (d2/R)2 << 1, we see that it is reasonable to neglect transverse shear deformation. Ingeneral, we cannot neglect twist deformation when the member is not shallow. For theshallow case, we can neglect in the expressions for UB3.

Example 15—6

Consider a closed circular ring (Fig. El 5—6) subjected to a uniformly distributed twistingmoment. From symmetry, F3 0 and M1, M2 are constant. Then, using (15—16), we find

The displacements follow from (15—18)

M1 = 0

1142 Rm1

U3 = (13; — 0

RM2 R2m1WI

= =

b3 = m2 0

Fig. E15—6

15—5. FORCE METHOD—EXAMPLES

In this section, we illustrate the application of the principle of virtual forcesto curved members. The steps involved are the same as for the prismatic orplanar case and therefore we will not reiterate them here. We restrict thisdiscussion to the case where the material is linearly elastic, the member is thinand slightly twisted, and warping is neglected. The general form of theexpression for the displacement at an arbitrary point and the compatibilityequations corresponding to these restrictions (see (15—14), (15—17)) follow.

x2

const


ENGINEERING THEORY OF AN ARBITRARY MEMBER CHAP. 15


= —i + j [(ei° + Fj,Q + (a-)

+ + (15—24)

+ (k20 + + +

Compatibility Equations

Z1, Z2 Zr = force redundants

= F30 +

= M3,0 + kA

R1 = R1,0 + kZk

.fkJZJ (k = 1, 2 r) (15—25)3= 1

where

+ JMT.JMT.k + + ETM31M3,kldS

= — + + +

+ ++

+

= M1 + y3F2 — v2F3

The reduced form for out-of-plane deformation is obtained by setting F1 =F2 + = 0.

Example 15—7

Consider the nonprisinatic member shown below. The centroidal axis is straight but theorientations of the principal inertia axes vary. We take X1 to coincide with the ceniroidalaxis and X2, X3 to coincide with the principal inertia directions at the left end (point A).The principal inertia directions are defined by the unit vectors t2, t3.

= cos + sin çbt3

= —sin + cos

at x1=O


SEC. 15—5. FORCE METHOD—EXAMPLES

Now, we consider the problem of determining the translations of the centroid at B dueto the loadingshown in Fig. E15—7A. It is convenient to work with translationcomponents(v52, v53) referred to the basic frame, i.e., the X2, X3 directions. We suppose that the shear

Fig. E15—7A

x1

center coincides with the centroid and shear deformation is negligible. Spe-cializing (15—24), and noting that M1 = 0 for a transverse load applied at the centroid,the displacement expression reduces to

Force Systems

dQ =11L(i

+ (b)

The moment vectors acting on a positive cross section due to P2, P3 applied at B (Fig.E15—7B) are —

P2(L —

= — P3(L — x1)12

To find M2, M3, we must the components of It,! with respect to the local frame.These follow from Fig. E15—7C:

For P3.M2 = P2(L —

M3 == P2(L — x1)cos q5

x2

axts



Fig. E15—7B

L-x,

'I1'2(L—x1)i3

—P3(L

Fig. E15—7C

ForM2 = — 1)cos 4)

( )M3 = +P3(L — x1)sin4)C

Determination of Due to P2

The virtual-force system for corresponds to P2 = + 1. Introducing (d) in (b),we obtain

(f)

Determination of CR3 Due to P2

The virtual-force system for corresponds to P3 = + 1. Using (e) leads to

VB3

/BP3

M2r2

——P2(L—x1)z1

M3t3


SEC. 15—5.

Example 15—8

FORCE METHOD—EXAMPLES 503

We rework Example 15—6 with the force method. Using symmetry, we see that

M1 = 0

=

Suppose the rotation w1 in the direction of in1 is desired. The virtual loading for thisdisplacement is rn1 = + 1. Starting with

and substituting for M2, we obtain

TO R2COl =

Example 15—9

Consider the closed ring shown. Only M1 and M2 arc finite for this loading. Also, thebehavior is symmetrical with respect to X1 and we have to analyze only one half the ring.

Fig. E15—9

x2

42

4 4

T

12,J are constant

We take the torsional moment at 0 = 0 as the force redundant. The moment distributions

M1

T



Specializing (l5—25) for this problem,

f11Z1 = A1

A1 = —2R[M1 0M1 0M2,

dO+El2

,,/21

f11 = 2RJ +

dO

and then substituting for M1, M2,

A1

= ' 12 E12) sin cos 0

= - [sin2 0

and it follows that Z1 = 0. We could have arrived at this result by noting that the behavioris also symmetrical with respect to X2. This requires M2 to be an even function of 0.

The virtual-force system for WAI is T = + 1. Using (15—24) and (a) leads to

Example 15-10

r/T sin 0\sin 0 fT cos O'\ cos 012wAi = 2RJ

+

RTa[l ICOAl

We analyze the planar circular mcmber shown in Fig. El 5— bA. The loading is out-of-plane, and only F3, M1, and M2 are finite. To simplify the algebra, we consider the shearcenter to coincide with the centroid and neglect transverse shear deformation. It is con-venient to take the reaction at B as the force redundant.

B

Fig. E15—1OA

A

P13 (Displacement restraint

C iriX1 direction at B)


SEC. 15—5. FORCE METHOD—EXAMPLES 505

Primary Structure

The primary structure is defined in Fig. El 5—lOB:

B1 = R2 —MA2

d1 — UA3 d2 =R4 = Z1 = F53 d4 =

The force solutions for the loadings shown in Fig. El 5— [OC are:

For F:F'3,0 = +P

M1, = PR[l COS(ii — llc)]M2,0 = —PR sin(lJ — 'Ic)

'Ic

F3,1 = +1= R(1 — cos ii)

M21 —Rsini1

= —MAI

=

M1 ,

E15—1OB

(b)

(c)

Force Analyses

xl

ForZ1 = +1:



Fig. E15—1OC

=+i

B

Compatibility Equation (15—25)

IL,j =fit

!ii =

R + + M2

Substituting for the internal force and reactions, we obtain the following expressionsfor and

R3 [(1 + (1 —— Sin

—

Sill COS

— UA3 + RrOA2 sin — — cos

+ R2 sin(05 — O)dG

—

{o. [i + cos(05 — Or)] — sin — sin Oc

+

sin

El2=GJ

Note that we could have determined A1 and using the results of Example 15—5.

C


SEC. 15—6. RESTRAINED WARPING FORMULATION 507

15-6. RESTRAINED WARPING FORMULATION

In what follows, we consider the member to be thin and slightly twisted.Referring to Fig. 15—2, these restrictions lead to

dR -tl(15—26)

d(vol.) dS dr2 dy3

Therefore, in analyzing the strain at Q (S, y3), we can treat the differentialline elements as if they were orthogonal. Thc approach followed for the pris-matic case is also applicable here. One has only to work with stress and strainmeasures referred to the local frame 12, 13) rather than the global frame.

Our formulation is based on Reissner's principle (13—33):

bTIi V*)d(vol.)— d(surface area)] = 0

r, ü = independent quantities= e(iI)

p, b = prescribed forces= = complementary energy density

We introduce expansions for fl, in terms of one-dimensional displacementand force measures (functions of S) and integrate over the cross section. Theforce-equilibrium equations follow from the stationary requirement with respectto displacement measures.

We start with the strain measures, = Y12. One can show thatt

S1

au -. (3Uti + t2 (15—27)

-tj + (3

where is the displacement vector for Q (S, Yi' Y2). We use the same displace-ment expansion as for the prismatic case:

+ u212 + 113(3

U1 U1 + W2y3 — WIY2 ± f4iIA2 lAsZ — — y3) (15—28)

= + co1(y2 — Y2)

Y3)Expanding

= + + cr13y13)dy2dy3

t See Prob. 15—5.



leads to

dy2 = F1e1 + F2e2 + F3e3 + MTkI+ M3k3 + MRf + Mj,.

e1, e2,. . . , k3 (defined by (15—16)) (i5—29)

= SSai14) dy2 dy3

MR = JJ[ci2(4),2 + a12çb) + + a134))]dy2dy3

The equilibrium equations consist of(15—11) and the equation due to warpingrestraint,

MR (15—30)

which can be interpreted as the stress equilibrium equation for the directionweighted with respect to 4).

Now, we use the stress expansion developed for the prismatic case. Thederivation is discussed in Sec. 13—5, so we only list the essential results here.The normal stress is expressed as

F1 M2 M3= + Y3 — + -r '1) (15—31)

'2 '3

where 4) = — the St. Venant warping function referred to the shear center.We write the transverse shear stress distribution as

=+ ±

(15—32)

are functions of The corresponding complementary energy function

= $SV* dy2 dy3 =+

+ M32) +

+

+ 2F2F3

++ + (15—33)

+ +

Also, (15—32) satisfies (see(13—50))

JS(a124),2 + cr134),3)dy2dy3

Finally, noting (b), we express MR as

MR = (1 + + b2F2 + b3F3 (15—34)

where the b's involve the curvature (a32, a1 3). If the cross section is symmetrical,

A23 Y3r = Y2r br = 0


t See Prob. 15—6.

dM1=

dM2

dO=

R dOM1 + M'j

SEC. 15—6. RESTRAINED WARPING FORMULATION 509

aej

=

av*k2

=

av*k3

=

and b2, b3 are due to self-equilibrating stress distributions.tIt is reasonable,in this case, to take b2 0 and compute the shear coefficients (A2, A3)based on the primary fiexural shear stress distributions.

Expanding the stationary requirement with respect to force measures yieldsthe force-displacement relations,

e2 + b2f = e3 + b3f = (15—35)aF2

aV* i3V*

where e1, e2, . , k3 are defined by (15—16). The corresponding unrestrainedwarping relations arc (15—18).

Example 15—11

To investigate the influence of warping restraint, we consider a planar circular memberhaving a doubly svrnnietricai cross section (Fig. his—il), clampedat one end and subjected

Fig. E15—11

NM

to a torsional moment at the other end. We neglect transverse shear deformation dueto restrained torsion. The governing equations for this loading (See Sec. 15—4) follow.





El2M2 = E12k2 = +

El4, fR (10

(b)1 (dco1f=

= GJk1

Boundary ('onditions

0—U (01C02J01v11 =M

M2 = 0

One can write the equilibrium solution directly from the sketch:

M1 M — 0) M2 = Al sin(05 — 0)

We substitute for the moments in the force-displacement relations.

M1 = GJk1— = — 0)

1

k1 —

I (dco2 '\ lvi

=+ Wi) — 0)

and solve for k1, and then Wi. The resulting expressions are

2 G.J .,. El22 =———

GJ

= ME1

— 0) — cos — siiih tanhGJ +

= {o cos(05 — 0) sin 0 cos

+ 0 cos(05 — 0) + — sin 0 cos

—{sinh — tanh cosh + cos ü tanli

Warping restraint is neglected by setting Er = 0 and =


SEC. 15—7. COMPLETE END RESTRAINT

The rotation at B is

(RM\ = — COS 0B) +

l\ El2)

K = 1

+ Sin 0B[

1 —

—cos2 tanh 704

If we set

On On

and let (g) reduces to (13—57), the prismatic solution. The influence of warpingrestraint depends on 2 and Values of K vs. 2 for = ir/4, it/2 are tabulated below:

1

for

= +

for = it/4 = ir/2

1 0.179 0.51)0

5 0.786 0.96

10 0.907 0.99

We showed in Chapter 13 that

2 0 (open section)

(j)2 = 0 (closed section)

where t is the wall thickness and h is a depth measure. Since 2 = RA and R/li 1 fora thin curved member, the influence of warping restraint is not as significant as for theprismatic case.

15—7. MEMBER FORCE-DISPLACEMENT RELATIONS—COMPLETEEND RESTRAINT -

In the analysis of a member system, one needs the relations between the forcesand displacements at the ends of the member. For a truss, these equations



reduce to a single relation between the bar force and the elongation. Matrixnotation is particularly convenient for this derivation so we start by expressingthe principle of virtual forces and the complementary energy density in termsof generalized force and deformation matrices.

Referring back to Sec. 15—3, we define

a v*

(1536)

aM1

and write the principle of virtual forces as

Ss dS J dS dT (15—37)

Note that we are working with M1, not MT. We use the complementary energyfunction for a thin slightly twisted member with negligible warping restraint(i.e., (15—17)). With the above notation,

Eq. (15—17) +

where

g = (15-38)gfm

1 Y2Y3gf

A2G+ (7 GJ

ISym +

o ojoJ'3 1

El2

0 0 Sym

The force-deformation relation implied by (15—38) Is

+ (15—39)

We will use these general expressions for planar and out-of-plane deformationas well as for the arbitrary case. One has only to delete the rows and columns


El

for planar loading applied to a planar member.Finally, we substitute fort in (15 —37) and distinguish between prescribed and

unknown displacements. The principle of virtual forces expands to

J5(t° + dS — dT dT (15—41)

where contains prescribed displacements and R are the corresponding re-actions; d contains unknown displacements and are forces correspondingto d. The virtual-force system must satisfy the force-equilibriumequations, (15—11). It is more convenient to generate and R with the equili-brium equations for a finite segment rather than attempt to solve (15—il).

Consider the arbitrary member shown in Fig. 15—6. Each end is completelyrestrained against displacement. The positive sense of S is from A toward B.

SEC. 15—7. COMPLETE END RESTRMNT 513

of g corresponding to the zero force measures. For example,

{F1F2M3}= 0

(15—40)

13

B

Basic member frame

Fig. 15—6. Arbitrary curved member.



We suppose the geometry of the member is defined with respect to a basicframe which we refer to as frame ii, and take the end forces at B as the forceredundants. Then, the primary structure consists of the member cantileveredfrom A.

Throughout the remaining portion of the chapter, we will employ the notationfor force and displacement transformations that is developed in Chapter 5. Asuperscript n is used to denote a quantity referred to (lie basic frame. Whenno frame superscript is used, it is understood the quantity is referred to thelocal frame. For example, represents the internal force matrix at point Qreferred to the local frame at Q. Note that acts on the positive face. Theforce matrix for the negative face is — The end forces at A, B are denotedby and are related to the internal force matrices by

= =15 42

— = —

Also, the displacement matrix at point Q is written as

0//Q = {u1, U2, U3 (Oj, W2, c03}Q{u}

(15—43)

For this system, and are prescribed.We determine for the primary structure, i.e. the member cantilevered

from A, due to displacement of A, temperature, loads applied along the member,and the end forces at B and then equate it to the actual The virtual-forcesystem is

Lw =AR = =

= =

Also,d =d

Introducing (a), (b) in (15—41), we obtain

++

Il

+ +

Next, we express as

+ (15—44)

where is the internal force matrix at Q due to the prescribed externalloading applied to the member cantilevered from A. Finally, substituting for


SEC. 15—7. COMPLETE END RESTRAINT 515

leads to

= +[SB

. T(18 + 0)dS(15—45)

+[JB

The first term is due to rigid body motion of the member about A whereasthe second and third terms are due to deformation of the member. We define

as the member deformation matrix:

= — to rigid body (1546)notion aI,out .4

By definition, is equal to the sum of the second and third terms in (15—45).We also define

JSB+ 0)dS = initial deformation matrix

S-4 (15—47)

fll member flexibility matrix

and (15—45) reduces to— = + (15—48)

Equation is the force-displacement relation for an arbitrary memberwith complete end restraint. It is analogous to the force-elongation relationfor the ideal truss element that we developed in Chapter 6.

The member flexibility matrix, f", is a natural property of the member sinceit depends only on the geometry and material properties. For simple memberssuch as a prismatic member or a planar circular member with constant crosssection, one can obtain the explicit form off. When the geometry is complex,one must generally resort to numerical integration such as described in Sec.14—8 in order to determine f and This problem is discussed in the nextsection. Finally, we point out that the general definitions off, are also validfor in-plane or out-of-plane deformation of a planar member. One simply hasto use the appropriate forms for the various matrices.

Up to this point, we have considered only a simple member. Now supposethe actual member consists of a set of members rigidly connected to each otherand the flexibility matrix for each member is known. We can obtain the totalflexibility matrix by compounding the flexibility matrices for the individualelements. To illustrate the procedure, we consider two members, AA1 andA1B, shown in Fig. 15—7.

The matrix, f", contains the displacements at B due to the end forces at Bwith A fixed:

Now, suppose point A1 is fixed. Then, the displacement at B due to thetion of member A1B is

A1B



where is the flexibility matrix for member A1B referred to frame n. Theadditional displacement at B due to movement of A1 is

—• B,' displaccment at 4 — ' BA, 'Ai

It remains to determine

B

Fig. 15—7. Segmented member.

The force system at A1 due to the end forces at B is given by

— CF-nA, B

and the resulting deformation of member AA1 is— ffl — ffl (TI?

A, mernberAA, — A, AA,' It

Finally, we have

= + = ffl (15—49)

The end forces at B are found by inverting (15—48):= (fn)_ 1 member stiffness matrix

— 1 •,J/'fl— 1550= — + —

The first term is due to external load applied along the member and representsthe initial (or fixed-end) forces at B. For convenience, let

(15—51)

The second and third terms are the end forces at B due to end displacement atB, A. Once is known, we can evaluate the interior force matrix at a pointusing (15—44),

= 0+ (a)

Thus, the analysis of a completely restrained member reduces to a set of matrixmultiplications once the member stiffness and initial deformation matrices areestablished.


SEC. 15—8. GENERATION OF MEMBER MATRICES 517

When analyzing a system of members by the displacement method, expres-sions for the end forces in terms of the end displacements are required. Inaddition to (15—50), we need an expression for Now,

= — 0

Substituting for leads to

i + —

— — — ( )A,0 BA B.i

where represents the initial end forces. In order to express the equationsin a more compact form, we let

in iiKBBin inwn.T

—

k — n — j n —AB L BA

BA BA BA

With this notation, the force-displacement relations simplify to

= + +(15—54)= + +

Note that only and are required in order to evaluate and

15-8. GENERATION OF MEMBER MATRICES

The member flexibility matrix is defined by

= g014)dSNoting that A

=and letting

= (15—55)

we can writeffl (15—56)

If numerical integration is used, the values of the integral at intermediate pointsalong the centroidal axis as well as the total integral can he determined in thesame operation. This is desirable since, as we shall show later, the intermediatevalues can be utilized to evaluate the initial deformation matrix.

We consider next the initial deformation matrix:

= + 0)dS

We transform 4',g, and from the local frame to the basic frame, using (15—55)



and==

The contributions of temperature and external load are

T,O)dS (15—57)

0)dS (15—58)

Suppose there is an external force system applied at an intermediate point,say C. Let denote the force and moment matrices and the totalforce matrix:

=(15—59)

Normally, the external force quantities are referred to the basic frame for themember, i.e., frame n. The initial force matrix at Q due to this loading isgiven by

0 SA Sc (15—60)07fl _II c <C. <CQ.0 — U

Writing— or" or"

CD

and introducing the above relations in (15—59) result in

=(

The bracketed term is an intermediate value of the integral defining Finally,we let

= (15—61)

With this notation, (f) simplifies to

(15—62)Also,

(15—63)

The determination of the member flexibility matrix reduces to evaluatingJ defined by (15—61). One can work with unpartitioned matrices, i.e., g,but it is more convenient to express the integrand in partitioned form. Thepartitioning is consistent with the partitioning of into F, M. Since theformulation is applicable for arbitrary deformation, it is desirable to maintainthis generality when expanding in partitioned form. Therefore, we definea as the row order of F and /3 as the row order of M.

(15—64)

Continuing, we partition and g symmetrically, consistent with (15—64),


SEC. 15—8. GENERATION OF MEMBER MATRICES 519

and simplify the notation somewhat:

0=

(15—65)

=(fixfj)

() ()— g11 g12

gQ— Eg12 i g22

The translation and rotation transformation matrices are developed in Secs.5—1, 5—2 and the form of g for a thin curved member is given by (15—38).

The local ilexihility matrix is defined by (15—55). Using the above notation,the expressions for the submatrices are

=(15—66)

g22 —

Note that g12 = 0 and g22 are diagonal matrices when the shear centercoincides with the centroid. If, in addition, axial and shear deformation areneglected, g11 = 0.

We let

= = (15—67)

The submatrices follow from (15—65):

+ + + XHQgZ2XBQ

'1112 = -F- (15—68)

'1122 =

Next, we partition J consistent with'1':

j — IdS —

—'I'

— Jp22j (15—69)x

JP = F" is

Finally, we partition f":-

ffl ffl

r = (15-70)



The initial deformation matrix due to an arbitrary loading at point C can bedetermined with (15—62). Its partitioned form is

=

r nI1V01 — cli — C,12 BC C.12 Ii C

rT —lYe) L"c. 12 —- 2V'BC 22J ' C

(flxfl)

where denote the initial translation and rotation matrices.The member stiffness matrix, is obtained by inverting We write

= (r)' = I I(15—72)

[ku k22j(11 )< /1)

One can easily show that (we drop the frame superscript on for convenience)

k,' — ( \111 — k ii — 12 22 12)

(15—73)1 n —If çT i

22k $ —

Once is known, the stiffness matriccs and can be generated.Expanding (15—53) leads to the following partitioned forms:

knB — — I

rin I nvn,T ml rnIin I Ii

I

llJtB,! T 12 I iiI (15—74

KBA I = I 1n.!TL 12

— A ] [k';1 I AAA

—

15—9. MEMBER MATRICES—PRISMATIC MEMBER

In Chapter 12, we developed the governing equations for a prismatic memberand presented a number of examples which illustrate the displacement andforce methods of solution. Actually, we obtained the complete set of force-displacement relations and also the initial end forces for concentrated anduniform loading. Now, in this section, we generate the member flexibilitymatrix using the matrix formulation. We also list for future reference thevarious member stiffness matrices.

The notation is summarized in Fig. 15—8. For convenience, we drop theframe reference superscript n, since the basic frame coincides with the localframe, i.e., = I. The positive sense of a displacement, external force, orend forces coincides with the positive sense of the corresponding coordinateaxis.

Starting with (15—66), we have = since R$ are identity matrices.Once XBQ is assembled we can determine the submatriees of from (15—68).


Now,

o o 0

X5Q= 0 0 —(L—.xQ1)

o (L—xQt) 0

Then, using g defined by (15—38), we obtain

0

_________ ______

f12= GJLT2

GJ

L/GJ=

Sym

SEC. 15—9. MEMBER MATRICES—PRISMATIC MEMBER 521

i2 MB!, WB i

UB!

I. /

X1 is centroiclal axis.X2, X3 are principal inertia directions.

Fig. 15—8. Summary of notation for a prismatic member.

L/AE 0

/1 1)L + +

Sym

0

0

L\-2x3

I)+ +

(a)

(15—75)

0

.

0-— +

L2

2E12

10 0

V2E13

0

LIEu


0 0

L3

(change sign of(2, 3) and (3, 2) in k12)


The submatrices of k are generated with (1 5—73), (1 5— 74) and are listed belowfor reference. Transverse shear deformation is neglected by setting a2 = 03 0:

12E12a2 =

'22 1 + a2

12E11a3

=

= —I'---- 1+03

GJ 12Eb1 = + +

AEL

k11=

Sym

0 0

0

0

6E1

L2

0

k12 =

k22

(15—76)

L2 L2

(4 +

0

Sym

0

0

Fl*(4 +

L3

—

L0

0

10L2

L2b1

B=

L2

1! 0

—

L2

0


SEC. 15—9. MEMBER MEMBER 523

b—6E!!Ye2

V L2

FI*=

(4 + a3)_L_

(change sign of(1, 2), (1, 3) in k22)

Finally, the fixed end forces due to a concentrated transverse force and auniform transverse loading are summarized below.

concentrated Force

12E13a3 =

MB3 —

FB2 —(15—77)

MBI = +MA1 =

— F112

MA3 = — L + FB2 +

Force

12E12a2

— GA1L2

—÷

FB3 +(15—78)

+ FB3)

MAI —MB1

— _PC3 — FR3

MA2 L + — MB2)



Concentrated Torque

= —(15—79

MA1 = Tc1(1 —

Uniformly Distributed Load, b2

- - b2LF82 = FA2

— h2L2 (15—80)M83 = —MA3 =

M81 MAI = 0

Uniformly Distributed Load, b3

— b3L.F83 = 1q43 =

— — b3L2 (15—81)M82 =

= ——u—

= = 0

Uniformly Distributed Torque, rn

M81 MAI = (15—82)

15—10. MEMBER MATRICES—THIN PLANAR CIRCULAR MEMBER

In this section, we generate the flexibility and initial deformation matricesfor a thin planar circular member, of constant cross section, using matrixoperations. We include extensional and transverse shear deformation for thesake of generality. Some of the relations have already been obtained as illustra-tive examples of the force and displacement methods. In particular, the readershould review Example 14—6, which treats planar deformation, and Examples15—4, 15—5, 15—10 for out-of-plane deformation.

The notation is summarized in Fig. 15—9. By definition, Y2 and Y3 areprincipal inertia axes and p3 = 0, i.e., the shear center lies in the plane con-taining the centroidal axis. It is convenient to take the basic frame (frame n)to be parallel to the local frame at B. The three-dimensional forms ofand are

cos — sin 01 rRbq I 0= = sin cos 0

=

17] ° 1

= = (15—83)


SEC. 15—ID. THIN PLANAR CIRCULAR MEMBER 525

0I

0

= jOR sin

R sin ,fl

for planar deformation andR0=1

— [—R(1 — cosSQ

for out-of-plane deformation. Since the complete flexibility matrix is desired,it is just as convenient to work with subinatrices of order 3 as to considerseparately the planar and out-of-plane cases.

x;

Centroidal axis

bB2

F;3,U;,

Mw

Fig. 15—9. Summary of notation for a planar circular member.

We consider the member to be thin and use the local flexibility matrixdefined by (15—38). Expanding (15—66), (15—68) leads to the member flexibilitymatrix.


ENGINEERiNG THEORY OF

13ae

El2Ct =

a1 = ae + a7

a2 = ae —

c1 = += —

( Y2\

—

Y2 \—\2

+ c41

— 2c3 05 — C2 Sifl COS

AN ARBITRARY MEMBER CHAP. 15

El3as =

El7Cs =

(15—84)

- sin ± + az)i}

526

— T —t212t) —

Ib22

Symmetrica'

I cosO5-— sin2 05(1 + a2)}

2

R3+r!j)

a2)sin cos

0 0

0 0R2

— sin

0 01,12

—cos05)

+

+ c2 sin cos f15}

0


SEC. 15-10. THIN PLANAR CIRCULAR MEMBER 527

We consider next the determination of the initial deformation matrix due toan arbitrary concentrltted load at an interior point, C. Now, the flexibilitymatrix for the segment AC referred to the local frame at C, which we denoteby is known. We just have to change to and superscript b to c in

When the external load is referred to the local frame at C, the displace-ment at C is given by

('IC

The displacement at B due to rigid body motion about C is— TrI & T011hc.

"B — BC BC 3

Finally, we can write

= =

U

1, — b .. j T c v-I,. T Tçc. T cV0 — UB — c AC,11 /1 .4C.i21 C

+ + (15—85)1' r T ( Tcc \

— Ii 4C,121 C C

The uncoupled expressions follow.

Planar Loading

/ 1+a1= = + cos + 'Ic —

M3

I' 1+02+ Slfl 1 + Cos

R3 I 1+a1+ -'-'--— — cog 0c ± -—-— sin ic

— o4

+ + Sifl 5jfl1:13

(15—86)I I R ç

2 = — —s-—-- Sill 'IcL

1+02. . 1

+ 2

+ j —

+ {cos 'Ic — cos OB}Tc3

RO00,3 WB3 — sin + (1 COS +



Out-of-Plane Loading

R3 (V0. = = COS 'ic + c4) — C2 Sifl 0c cos 0B

+ c3(— sin — sin °B + sinR21

+ —- cos 'ic + C2 6c COS 0J3 + C3 Sifl)

R21+ — CZS1flGCS1flOB — c3(1

£12 ( jR2

1 = COS 'ic + c3(sin 0B — sin 'ic)

+ c2 sin cos

RI+ Cos 'ic — (15—87)

+ 5jfl + C2 Sifl °c o5}Tc2

2 = = Sjfl 'ic + c3(cos — COS 'ic)£12

)— C2 SIfl Oc Sin

+i-i—

)[CIOC sin 'ic + c2 sin sin

+E12

cos 'ic + c2 sin 0c cos

When the loading is symmetrical, one can utilize symmetry to determine thefixed end forces. The most convenient choice of unknowns is the internal forcesat the midpoint, i.e., 6 = OB/2; F1 and M3 are unknown for the planar case andonly M2 is unknown for the out-of-plane case. Explicit expressions for thefixed end forces due to various loading conditions arc listed below.

Planar Loading

Fig. 15—10 defines the notation for the planar case.We consider two loadings: a concentrated radial force P applied at C, and

a uniform distributed radial load b2 applied per unit arc length over the entiresegment. The basic frame is chosen to utilize symmetry. We determine theaxial force and moment at C from the symmetry conditions u1 = w3 = 0.

CASE 1—CONCENTRATED RADIAL FORCE P


(1 + a2\sin cos

RP ci — COS (=—-i-

+—

— 7771 — 77' 131 — Al — Cl

—— 1'

'A2

— PR I 1 — co.s (sin= MA SIfl + 1,/I COS

CASE 2—UNIFORM DISTRIBUTED RADIAL LOAD b2

= 0

—Rb2(1

sin

SEC. 15—10. THIN PLANAR CIRCULAR MEMBER 529

P

MA

R

Fig. 15—10. Notation for planar loading.

1;.flB!

(15—88)

SiflO! 1+Q2 2— cos cc) + —--—— SIfl CL

CL 2

(1 + a1\ sin2 ccCL I I — +

cc

(1 + ai) sin2 cc (1 + a2'\2 —

+ sin cc cos cc

cc



Mc=R2b2aecb(_1

= — a4)= = —Rb2 Sill C(

= —MA = R2b2aecb — cos

Out-of-Plane Loading

(15—89)

Figure 15—11 defines the notation for the out-of-plane case.We consider four loadings: a concentrated force P, and a couple T—both

applied at C; a uniform distributed force h3; and a uniform distributed couplein1. Tile bending moment at C is obtained using the symmetry conditionCD2 = 0.

F3

Fig. 15—li. Notation for out-of-plane loading.

CASE 1—CONCENTRATED FORCE P

P=

= 0PR + c3(l — cosa)

2

Mt

1

R


SEC. 15—11. FLEXIBILITY MATRIX—CIRCULAR HELIX 531

PRMB1 = MA1 --i-- (1 — cos cx)

- - PR.= — = — ----- sin cx

FB = =

CASE 2—CONCENTRATED TORQUE T

= 0

=

T c2sin2cx=

2 cxc1 + c2 sin cx cos cx (15—91)

=

FB=FA=0 -

CASE 3—UNIFORM DISTRIBUTED LOAD h3

0

2 c1(sincx — cx)+ c2 sin cx(i — coscx)+ — cxcoscx)— R b3

+ C2 COS(15—92)

= = R2b3(sin cx — cx cos cx)

= = — R2b3(cx sin cx — 1 + cos cx)

=FA=—PRcx

CASE 4—UNIFORM DISTRIBUTED COUPLE m1

= 0

cj(a — sin cx) + c2 sin cx(cos cx —1)= m1R——--——-—--

cxc1 + C2 SJfl cx cos cx (15—93)

—m1R sin cx

MA2 = — rn1R(1 — cos cx)

15—11. FLEXIBILITY MATRIX—CIRCULAR HELIX

In this section, we develop the flexibility matrix for a member whose centroidalaxis is a circular helix. The notation is shown in Fig. 15—12. The principalinertia direction, Y2, is considered to coincide with the normal direction, i.e.,the inward radial direction, at each point. We also suppose theproperties are constant. For convenience, we summarize the geometrical



relations:x3 = R cos 0x2 R sin 0x3=C8dS a dO

a = [R2 + C2)112 = constant

= + Rcos 012 + Cl3)

fl t2 = -—COS — sin 012

b = t3 = — Ccos 072 + R13)

R. IC—---sinO —cosO

a a

Rc, = R,3 = =C. C R—sin 0 ——-cos 0

a a

0 — C(OB — 0) R(sin — sin 0)

= C(OB — 0) 0 — R(cos — cos 0)

— R(sin — sin 0) R(cos 0

The steps involve only algebraic operations and integration. We first deter-mine using (15—66), then from (15—68), and finally with (15—70). Inwhat follows, we assume the shear center coincides with the centroid and neglectextensional and transverse shear deformation. With these restrictions,

= g12 = ()=

1

GJ

g2=

El3

and the expressions for reduce to

='I'12 'V22

The flexibility matrix for a constant cross section is given below.

t See Examples 4—6 and 5—3.


Xi", X', X31 —directions of basic frame

Y2, Y3—principal inertia directions

Fig. 15—12. Notation for circular helix.

Notation—-Din,ensionless Para,neters

R2E12 C2l2= +

R2 C2 (E12\ 13C12 = + ij T

RC[12 El2a3

Ra3 R2 [12 El2

1+a1 1—a1a5 = a6 =

a4+a6a7=2

a8=2

a6+3a4 a6—3a4a9

2a10

2

SEC. 15—li. 533FLEXIBILITY MATRIX—CIRCULAR HELIX

j

Centroidal axis

Yi



Elements of f71

[fit Sym 1

f22 I

f33j

C2a( a= + + 2a4 sin 08 + a10 sin OB}

R2a2cz 1 (1+

El3+ Sin208) — 2sinO8 +

f a602 a8 0B + a4(OB 0B — COS 05 + COS OB)}B

0B(08 sin 1 + cos —— El3RCr1

f31 = + (a1 — a4)(i — cos0,1) a4 sin2 05 + a9OssinOn}

C2x

=+ — a10 sin 08 Cos — 2a4OB cos

R2a2cc 1 /12+El + COS

=sin + sin 08) + a1005 COS

El2

+ a8 0B + a4 sin COS

R2af33 = {(ai + a5)05 — 2a1 sin a6 sin cos (15—94)

Elements of

[114 1161

f261Lf34 f35 f36j

f14 = sin 08 + a705 + a8 sin COS

02 + a8 sin28

08}

(f16 = — — <08 sin 1 + cosoB}

El2 ( El3

f24 — + a8 sin2 0B a4(1 — cos

f= <p8(08 Sffi 0B C05


SEC. 15—12. PARTIAL END RESTRAINT 535

Ca3cJ .

f26=

= sin OB a1(1 — COS

= — 0B cos

f36 =—

Elements of

{a508 + a6 sin COS 013j Sym

2 a= sin {a508 — a6 sin 08 cos OB}

aa3 . aa2— cos08)

El2 112

15—12. MEMBER FORCE-DISPLACEMENT RELATIONS—PARTIAL ENDRESTRAINT

In Sec. 15—7, we considered an arbitrary member which is completely re-strained at both ends. This led to the definition of the member flexibility matrixand a set of equations relating the end forces and the end displacements. Now,when the member is only partially restrained, there is a reduction in the numberof member force unknowns. For example, if there is no restraint against rotationat B, M8 = 0, and there are only a unknowns (where a is the order of F8), therotation at B has no effect on the end forces. To handle the case of partialrestraint, we first determine the compatibility equations corresponding to thereduced set of force unknowns. Inverting these equations and using the equi-librium relations for the end forces results in force-displacement relations whichare consistent with the displacement releases.

Let Z denote the force redundants. Normally, one would work with theprimary structure corresponding to Z = 0. However, suppose we first expressthe force at a point, say Q, in terms of the end forces at B, using, as a primarystructure, the member cantilevered from A:

(aB

Next, using the primary system corresponding to Z = 0, we express in termsof the applied external load and the force redundants:

= EZ + G (15-95)



The elements of G are the end forces at B (for Z 0) due to the applied externalloads. Note that G = 0 if Z contains only end forces at B.

Now, the principle of virtual forces requires

+ = + TO/fl

for any self-equilibrating virtual-force system. Taking the system due to AZresults in the compatibility equations for Z. It is convenient to work first withthe virtual force system due to Equation (b) reduces to

T(,fçfl + T(O//,l

=

where ffl are the initial deformation and flexibility matrices for thefull endrestraint case. Substituting for using (15—95), and requiring the resultingexpression to he satisfied for arbitrary AZ, we obtain

(ETIftE)Z + ET( + = = (15—96)

It should be noted that are the displacements of the supports at B, A.We suppose Z is of order q x 1, i.e., there are q force redundants. Also, we

let t be the row order of (and 0?t).

( F= (15—97)

([3 x 1)

With this notation,E is i x qGisi x 1

and (15—96) represents q equations. For convenience, we let

(q x q)(15—98

= + rG (1 1))

and the member force-deformation relations take the form

trZ = — z)

ETJc) n Tcjjgn ( — )— ' A — ' O,Z

We refer to 1. as the reduced flexibility matrix since, in general, q < i. Actually,is the flexibility matrix for Z and it is positive definite since E must be of rank

q, i.e., the force systems corresponding to the redundants must be linearlyindependent. Note that one can determine directly by working with theprimary system corresponding to Z = 0. This is the normal approach. Theapproach that we have followed is convenient when the member flexibilitymatrix is known.


SEC. 15—12. PARTIAL END RESTRAINT 537

At this point, we summarize the force-displacement relations for partial endrestraint:

Z = member force matrix= EZ + G=

0 —

f,. = reduced flexibility matrix (q x q) ETrE— 'I/''

0,z — p

f.Z ,) (15—100)

= —

Note that, for complete end restraint,'7 —. 14'

B15—101

G = 0 =

We will use (15—100) in Chapter 17 when we develop the formulation for amember system.

Continuing, we letkr = (15—102)

The force redundants are obtained by inverting (15—99):

Z = — — (15-103)

Substituting for Z, the end forces at B are given by

= — — + G (e)

We defined as the effective member stiffness matrix:

= EkYET15 104

LET ( — )

In general, k is singular when q < i, since E is only of rank q. Equation (e)takes the form

= + —

= + G (15—105)

= + (I, —

The end forces at A are determined from (a):

= — +— — ar" (

A,0 'BA iLl

Finally, we write the relations in the generalized form

I I n' LB,i RB BA 'A

= + +


538 ENGINEERING THEORY OF AN MEMBER CHAP. 15

where

=

(15—107)

k" — — ar" n — TAA BA BA BA e BA

Comparing (15—107) with (15—53), the corresponding expressions for the com-plete restraint case, we see that one has only to replace by in the partitionedforms for and The equation for is different, however, dueto the presence of the G term.

Example 15—12

Suppose there is no restraint against rotation at B. Then, 0. We take Zand generate E, G with (15—95).

— Fl +G-

For this case, G = 0. The reduced flexibility and stiffness matrices follow from (15—98),(15—102),

=

and the effective stiffness matrix follows from (15—104):

rc". — 1 I

o

Finally, the force-displacement relations are (see (15—99)):

rt, — ,i it vn, T itiiNote that premultiplication of by Er eliminates 9", the relative rotation at B. There isno compatibility requirement for the end rotations in this case; i.e., the support rotationat B, which we have defined as does not introduce any member deformation.

REFERENCES

I. REISSNER, F.: "Variational Considerations for Elastic Beams and Shells." J. Eng.Mech. Div. A.S.C.E., Vol. 88, No. EMI, February 1962.

2. HALL, A. S. and R. W. W000HEAD: Frame Analysis, Wiley, New York, 1967.3. GEaR, J. M. and W. WEAVER: Analysis of Framed Structures, Van Nostrand, New

York, 1965.4. RUEINSTRJN, M. F.: Matrix Computer Analysis of Structures, Prentice-Hall, 1966.5. LIVESLEY, R. K.: Matrix Methods of Structural Analysis, Pcrgamon Press, London,

1964.6. DAEROWSKJ, R.: Gekriimmte dünnwandige Trüger (Curved thin-walled beams),

Springer-Verlag, Berlin, 1968.


PROBLEMS 539

7. V. Z.: Thin Walled Elastic Beams, Israel Program for Scientific Transla-tions, OfiIce of Technical Services, U.S. Dept. of Commerce, Washington, D.C., 1961.

8. BAZANT, Z. P.: "Nonuniform Torsion of Thin-walled Bars of Variable Section,"International Association for Bridge and Structural Engineering Publications, Zurich,Vol. 25, 1965, pp. 17—39.

PROBLEMS

15—1. Refer to Example 15—5. Determine c, for a typical wide-flange sectionand a square single cell. Comment on the relative importance of torsionaldeformation vs. bending deformation (i.e., terms involving in Equation (e)).Distinguish between deep and shallow members.

15—2. Refer to Example 15—7. Consider a rectangular cross section andI/p j3\

varying linearly with x1, as shown in the sketch. Evaluate VB2 /( and/ \3E12J

vB3/ for a range of 4 and a/b.3E12

15—3. Determine the reaction at B and translation (in the direction of F)at C for the member sketched. Neglect transverse shear deformation.

I— Vertical restraint at B

Prob. 15—3

x2Prob. 15—2

b

x3

y3

'2

13

/P

Shear center



15—4. Repeat Prob. 15—3, considering complete fixity at B. Utilize sym-metry with respect to point C.

15—5. Derive (15—27). Start with the definitions for the strain measures(see Fig. 1 5—2),

DR(1 + r,) =

D15 Df)

DS Dy2S[fl Y12 =

E3f)

DS Dy2

neglect second-order terms, and note (15—26).15—6. Summarize the governing equations for restrained torsion. Evaluate

b2 and b3 (see (15—34)) for a symmetrical wide-flange section and a symmetricalrectangular closed cell. Comment on whether one can neglect these terms.— 15—7. Refer to Example 15—11. Specialize the solution (Equations f) for

= 1L 1. Verify that (g) reduces to the prismatic solution, (13—57),when —* 0.

15—8. Consider a member comprising of three segments. Assuming theflexibility matrices for the segments are known, determine an expression forthe member flexibility matrix in terms of the segmental flexibility matrices.Generalize for n segments.

15—9. Discuss how you would apply the numerical integration schemesdescribed in Sec. 14—8 to evaluate defined by (15—69).

15—10. Verify (15—73) and (15—74).15—11. Determine the fixed end forces for the member shown, using (15—77)

and (15—79).

15—12. Solve Prob. 15—3 using (15—84) and (15—87).15—13. Verify (15—90) and (15—91). Apply them to Prob. 15—4.

X2 Prob. 15—11


PROBLEMS 541

15—14. Starting with (15—87), develop expressions for the initial deforma-tions due to an aribitrary distributed loading, h3 = b3(O). Specialize for b3 =constant and verify(15—92).

15—15. Using the geometric relations and flexibility matrix for a circularhelix (constant cross section; Y2 coincides with the normal direction) developedin Sec. 15—11:

(a) 1)evelop a matrix equation for the displacements at B due to a loadingreferred to the global frame and applied at flint: See (15—85).

(b) Evaluate for the loading and geometry shown.

Y2,fl

Y3, b

x1

Prob. 15—15

15—16. Determine the reduced member flexibility matrix for no restraintagainst rotation at an interior point P.

15—17. For the planar member shown, determine E and G corresponding to

Z = MB M4}

Then specialize for rotation releases at A, B and determine kg.

IIP

— —— ir/2

c =R/2G =E/2



Part IVANALYSiS OF AMEMBER SYSTEM



16

Direct Stiffness MethodLinear System

1 6—i. INTRODUCTION

We consider a system comprised of in members which are connected at jjoints. We suppose the geometry of the assembled system is defined withrespect to a global framet and use a superscript o to indicate quantities referredto the global frame. The external force and displacement matrices for joint kare denoted by

(p0) (ax!)

(16—1)

(jOt) (xxi)

(fbi)

where c is the number of translation (force) components, /1 is the number ofrotation (moment) components, and i + fi. Note that = 2, /3 = 1 for aplanar system subjected to in-plane loading and 1, /3 = 2 for a planarsystem subjected to out-of-plane loading. For an arbitrary system, /3 = 3.

In what follows, we assume the material is linearly elastic and the geometryis linear, i.e., we neglect the change in geometry due to deformation. Thegoverning equations consist of joint force-equilibrium equations and memberforce-displacement relations. We have already developed the member force-displacement relations in Chapter 15, so that it remains only to establish thejoint force-equilibrium equations. In this chapter, we apply the direct stiffnessmethod, which consists in assembling the system stiffness and initial forcematrices by superimposing the. contribution of each member. the nextchapter, we present the general formulation for a linear member system andobtain the equations corresponding to the force and displacement solution by

t By global frame, we mean a fixed Cartesian frame.

545


546 DIRECT STIFFNESS METHOD—LINEAR SYSTEM CHAP. 16

matrix operations. Finally, in Chapter 18, we extend the direct stiffness methodto include geometrical nonlinearity.

16—2. MEMBER FORCE-DISPLACEMENT RELATIONS

In developing the relations between the end forces and end displacementsfor a member, we considered the member geometry and loading to be referredto a basic member frame (frame n) and used A, B to denote the negative andpositive ends of the member. The general relations were written (see (15—107))as

= + ++ +

Note that (a) also applies when there is only partial end restraint or internalreleases.

Now, we define n_ as the joints at the positive and negative ends ofmember n. Replacing B by n,

n the member frame take the form

= + +• + +

where

\T— '

We transform the force and displacement quantities from the member frameto the global frame for the system by applying

== (16—2)

to (b). This step is necessary since we are working with joint forces and displace-ments referred to the global frame. The final expressions arc:

= + +(16—3)

= + +

where the global member stiffness and initial force matrices are generated with

k — TI n rzou16—4=

Once the displacements are known, we evaluate using (16—3) and thentransform to the member frame.

Since the initial end force and stiffness matrices are generated in partitionedform, it is natural to express (16—4) in partitioned form. Using the notation


SEC. 16—3. SYSTEM EQUILIBRIUM EQUATIONS 547

introduced in Section 1 5—8, we write

(o

plion

=

ip= (16—5)

t.MoJ (flxl)

()O,22(Sxo)

Expanding (16—4) leads toon

o — on,T n on12 — 00. 12

k on, Ti n on — o,T21 = fi 21 ( )( ), 12 (16—6)

1 0 — ii on11 "(')(),22 P

— Don,L t 0,1

=

Note that is a natural property of the member whereas depends onthe orientation of the member frame with respect to the global frame. Theoperations defined by (16--6) can be considered as the element matrix generationphase.

The member force-displacement relations satisfy the equilibrium conditionsfor the member and compatibility between the restrained end displacementsand the corresponding joint displacements. Actually, the equilibrium condi-tions were used to determine Compatibility is satisfied by setting =

and = When there is only partial restraint at an end, there willbe displacement discontinuities. For example, if there is a rotation release atthe positive end, will not be equal to the end rotation matrix. We havetreated'r partial end restraint by defining an effective member stiffness matrixk0. In the derivation of k0, we consider °1IA to be the displacements of thesupports (i.e., the joints) and enforce continuity of only the restrained enddisplacements.

16—3. SYSTEM EQIJILIBRIIJM EQUATIONS

The equilibrium equations for joint k arc obtained by summing the endforces for the members incident on k:

t See Sec. 16—12.



In general, depends on and the displacements of those joints which areconnected to joint k. We define as the total (or system) external jointforce and joint displacement matrices:

= (if x 1)16—7

(if <

and write the complete set of if joint force-equilibrium equations as

= 91'o + (16-8)

where contains the joint forces required to equilibrate the initial end forcesWe have dropped the reference frame superscript for convenience.

The most efficient way to assemble and is to work with submatricesof order 1, the natural partition size, and superimpose the contributions ofeach member which follow directly from (16—3). This operation requires nomatrix multiplications. The terms due to member n are listed below.

In (Partitioned Form Is] x 1):

in row n÷(16—9)

in row

111 X (Partitioned Form Isj x

in row column n+in row column n_

oT 1

in row n_, columninrown...,columnn_

Since is symmetrical, only the upper or lower half has to be stored.

16—4. INTRODUCTION OF JOINT DISPLACEMENT RESTRAINTS

In this section, we extend the procedure described in Sec. 8—3 for introducingjoint translation restraints in the formulation for an ideal truss to an arbitrarymember system. Actually, only the notation for the joint force and jointdisplacement matrices has to be changed.

The governing equations are:

= — =

1 1

= + =(16—11)

+

The stiffness and initial force matrices are assembled using (16—9) and(16—10). It remains to introduce the prescribed external forces and displace-ment restraints. If joint q is unrestrained, is prescribed, and we just add


SEC. 16—4. INTRODUCTION OF JOINT DISPLACEMENT RESTRAINTS 549

to — q is completely restrained, is unknown. We replacethe matrix equation for with the matrix identity,

Finally, ifjoint q is partially restrained, some of the elements in 19q are unknown.In this case, we replace the scalar equations for the unknown reactions byscalar identities.

We suppose joint q is partially restrained and, for generality, consider thetranslation and rotation restraint directions to he arbitrarily orientated withrespect to the basic frame. We define X'1, .. . , as the orthogonal directionsfor the translational restraint frame and as the orthogonal directionsfor the rotational restraint frame. Quantities referred to the restraint framesare indicated with primes and a single superscript is used for the total matrix:

=L°"i (flxj)

(16—12)

Now,P, =

q (16—13T" —/3 q

We define 9/°" as the total rotation transformation matrix:

0 1= I I

(16—14)L0

With this notation, the transformation laws take the form

9/oqçpr16 15( — )

The modification requires two operations. First, we transform in(16—11) to This is accomplished by premultiplying row q of PPO

with and postmultiplying column q of with 9/°" In the second step,we replace the equations corresponding to the unknown elements in withidentities. This operation can also be represented in matrix form.

Suppose the rth element in is prescribed. We assemble four matrices,Eq, and as follows:

1. EqandGq

We start withE=I, G=O,

and set+1


550 DIRECT STIFFNESS METHOID—.-LtNEAR SYSTEM CHAP. 16

2.

We start with an ith-order column matrix having zero elements and set theelement in row r equal to the prescribed displacement.

We start with an ith-order column matrix having zero elements and enterthe values of the prescribed forces and moments referred to the restraint frames.Note that element r is zero.

Premultiplying transformed row q of 3r, t?P0 with Eq reduces the rth equationto 0 = 0. Then, adding Gq to Eqirqq and + to — q introducesthe identity for the rth element and includes the prescribed external forcesin We also operate on the qth column of to preserve syrnnwtry andinclude the terms due to prescribed displacements in The complete set ofoperations for joint q are listed below:

1. €=1,2,...,q—lT)a/e;

T)F

2. 9N, q q + +

= T)Eq + Gq (l616)

3. €=q+i,q+2,...,j== —

The operations defined by (16—16) are carried out for each joint, workingwith successive joint members. We represent the modified equations as

= (16—17)

The superscript J is placed on % to indicate that the joint displacement matricesare referred to the local joint restraint frames, which may not coincide with theglobal frame. Again we point out that the primary advantage of this modifica-tion procedure is that no row or column rearrangement is required. Solving(16—17) yields the joint displacements (local restraint frame) listed in theirnatural order, i.e., according to increasing joint number. The modified stiffnessmatrix, will be positive definite when the system is stable.

Once QgJ is known, we transform the displacements from the restraint framesto the global frame, using (16—15), and evaluate the member end forces from(16—3). Next, we the total external force matrix, The contribution


SEC. 16—4. INTRODUCTION OF JOINT DISPLACEMENT RESTRAINTS 551

of member n is

(16—IS)in row fl...

Finally, we transform the external joint forces from the global frame to thelocal restraint frames. This step determines the reactions and also provides astatics check on the solution.

Example 16—1

Suppose joint q is completely restrained. Then, "li and = 0.. The formsfor E, G are

EqOi Gqljand (16—16) reduces to

1. €=L2,...,q—l— ,t,.

— q

2.

3. jOi

=

Example 16—2

Suppose joint q is completely restrained against translation. Then, the translationmatrix and external moment matrix are prescribed. The appropriate matrices forthis case are

r0 01 r1= Gg = }

=

Example 16—3

We consider the case where joint q is restrained with respect to translation in one direc-tion and there is no restraint against rotation. This corresponds to a "roller" support.We take to coincide with the restraint direction and X'2, as mutually orthogonaldirections comprising a right-handed system. The translation, is prescribed. Theprescribed forces are P52, and



We first assemble From (16—14),

rRoa'= I

[ 13

where

= r,s 1,2,3=

The forms of E, G, and are

0

==

We specialize the results for a planar system subjected to planar loading. In order foronly planar deformation to occur, the translation restraint direction must lie in the planeof the system, which we take as the plane. It is convenient to select the orientationof X'2 such that X3 coincides with X",. 'Ihe specialized forms are

== [S,.,] r,s = 1,2

0 1

(d)

Eq = Gq = ---H0

Finally, we consider the case of a planar system subjected to an out-of-plane loading.The translational restraint direction must be parallel to the direction in order for onlyout-of-plane deformation to occur. For this case, and arc prescribed. Thespecialized forms are

Eq=

Cq=

qe;= =

Note that (e) is obtained by settinge = 1, = 2 in (a) of Example 16—2.


REFEREt'JCES

REFERENCES

1. R. K.: Matrix Methods of Structural Analysis, Pergamon Press, London,1964.

2. MARTIN, H. C.: Introduction to Matrix Methods of StructuralAnalysis, McGraw-Hill,New York, 1965.

3. RUBINSTEIN, M. F.: Matrix computer Analysis of Structures, Prentice-Hall, NewYork, 1966.

4. Gere, I. M. and W. Weaver: Analysis of Framed Structures, Van Nostrand, NewYork, 1965.


17

General FormulationLinear System

17-1. INTRODUCTION

We consider a system comprising m linear elastic members interconnectedat j joints. We suppose there are i degrees of freedom per joint (i.e., the jointdisplacement and force matrices are of order i x 1) and the geometry and jointquantities are referred to a global frame. Also, we neglect geometry changedue to deformation. In the previous chapter, we applied the direct stillnessmethod, which is actually a displacement method, to this system. Now, in thischapter, we first develop the governing matrix equations and then deduce theequations corresponding to the force and displacement solution procedures.We also establish variational principles for the force and displacement methods.Finally, we discuss how one can introduce member deformation constraints inthe displacement method. Since the basic steps involved in the member systemformulation are the same as for the ideal truss formulation, we recommend thatthe reader review Chapters 6 through 9 before starting this chapter.

Let r be the number of prescribed joint displacements. Then, the total numberof joint displacement unknowns, is

na = if — r (17—1)

The total number of force unknowns, flf, is equal to r (the reactions corre-sponding to the prescribed displacements) plus qT, the total number of memberforce unknowns:

nf=?+qT=r+(qj +q2+ '.. +qm) (17—2)

where represents the number of force unknowns for member n. By definition,is equal to the number of force quantities that have to be specified in order

to be able to determine the total internal force matrix at an arbitrary point.If the member is fully restrained at each end, q1, = i. For partial restraint, q,, isequal to i minus the number of independent force releases. Note that when themember is pinned at both ends, = 1 since there are only five independentmoment releases.

554


SEC. 17—2. MEMBER EQUATIONS 555

There are qT equations relating the member forces and the joint displacements.Also, there are ij equilibrium equations relating the external joint forces andthe member forces. The formulation is consistent, i.e., the number of equationsis equal to the number of unknowns. If flf if, the system is said to be staticallydeterminate since the force unknowns can be determined using only the equi-librium equations. The difference, flf — ii, is generally called the degree ofstatic indeterminacy, and represents the order of the final system of equationsfor the force method. For the displacement method, the final system ofequationsarc of order In what follows, we first establish the member force—jointdisplacement relations by generalizing the results of Sec. 15—12. Then, weassemble the joint force-equilibrium equations. Finally, we introduce the jointdisplacement restraints.

17—2. MEMBER EQUATIONS

The reduced member equations were developed in Sec. 15—12. For conve-nience, we summarize the notation and equations below (see (15—100)):

Z = member force matrix (q, x 1)(ixi)

—

f" member flexibility matrix (i x i)f, reduced member flexibility matrix (q,, x = ETfnE

= member deformation matrix (i x i) = —

= initial member deformation matrix (i x i) = + f"G= — )

These equations include the effect of partial end restraint, internal force releases,and reductions due to symmetry or antisymmetry. We can also use (a) forcomplete end restraint by setting F = and G = 0.

Now, we introduce new notation which is more convenient. First, we notethat G contains the end forces at B due to the external member loads actingon the primary structure defined by Z = 0. Also, — — are theend forces at A. Then we write

= G(17—3)

—

Next, we note that the equation relating Z and K", is a compatibilityrequirement. The term fZ + ETeC?OZ is the relative deformation in the positivesense of Z due to the member loads and the member redundants, Z, whereas

is the relative deformation in the negative sense of Z due to support(joint) movement. The net relative deformation must be zero for continuity.


556 GENERAL FORMULATION—LINEAR SYSTEM CHAP, 17

Then, we define

= reduced member deformation matrix (q,, x 1)= =

—17—4

reduced initial member deformation matrix x 1)= ETI/'O

z= + f"G)

With this notation, the member equations take the form—

B R,o

(175)= + frZ —

We generalize the relations for member n by setting

B—n÷ A=n_E=E,, Z=Z. (17-6)

= "r, n = fr.

Since the joint quantities are referred to the global frame, we must transformthe end forces and displacements from the member frame (frame n) to theglobal frame (frame o), using

=

The final equations follow.

Member Forces—End Forces= TE)Z +

17 7= +

Member Forces—Joint Displacements (q,,

,t n + 17 8—

( — )

The force translation transformation matrix, 2C, is a second-order tensor,i.e., it transforms according tot

q

and itT

T

t See Sec. 10—2.


SEC. 17—3. SYSTEM FORCE-DISPLACEMENT RELATtONS 557

Using (17—JO), we can express and 'r. as.

— TF)Z= — .

We prefer to work with since it is a natural property of the member whereasdepends on the selection of the global frame.

17—3. SYSTEM FORCE-DISPLACEMENT RELATIONS

Equation (17—8) represents the force-deformation relations for member n.By defining general flexibility and deformation matrices, we can express thecomplete set of member force-deformation relations as a single matrixequation. We let

Z total member force matrix x 1)= {Z1, Z2, . . . , Z,,,}

= total reduced member deformation matrix x 1)

= 1, 2 m}

= total reduced initial member deformation matrix (q,. x 1)— '17—12

I ro, , ro, mJ

f total reduced member flexibility matrix x

fr, 2

Note that f is quasi-diagonal, symmetrical, and positive definite. With thisnotation, the force-deformation relations are given by

-V = -V,, + fZ (17—13)

It remains to generalize the deformation-displacement relations.We define as the total joint displacement matrix referred to the global

frame.(ij x 1)

= °/4, . . ., (17—14)

and express 'V as

= (17—15)

The partitioned form is

d12 ... -

1 (17-16)


558 GENERAL FORMULATION—LINEAR SYSTEM CHAP. 17

Row n of d corresponds to member n. The submatrices in row n are of orderx i. Now, we see from (17—8) that there are only two non-zero elements in

row n and they are at columns n_. The assembly of d is defined by

d17 17=

sins = 0S fl_

ii = 1, 2, . . . , in

It is of interest to express si in factored form. First, we define the followingmatrices: (inz x 1)

411 = _, ,..., 411,,,_ } (im x 1)

(mi x un)

(im x q1) ([7—18)

(zinxirn)

(un x im)

Using this notation,the expression for "V takes the form

= -= ET9II(Q71+ — T411) (a)

Next, we relate 411÷, 411_ to using member-joint connectivity matrices forthe positive and negative (C_) ends:

411÷ C÷%17 19-

E= E

Em


SEC. 17—4. SYSTEM EQU)L)BRIUM EQUATIONS 559

Note that rows a of C ÷, C— correspond to member a. There is only one nonzeroelement in a row. For row n, we enter in column n÷ of and columnn of C_. Finally, combining (a) and (17—19), we have

17 20= —

and it follows thatd. =

= —

For an ideal truss, (17—21),reduces to (see Equation 6—28)

= —

where contains the direction cosines for the bars.

17—4. SYSTEM EQUILIBRIUM EQUATIONS

We have used the member force-equilibrium equations in developing themember force-displacement relations, so it remains only to satisfy equilibriumof the joints. There arc i equations for each joint, and a total of if equations.The expressions for the end forces in terms of the member forces are givenby (17—7). Assembling the joint force.equilihrium equations involves onlysumming at each joint the end forces incident on the joint.

We define as the total external joint force matrix referred to the globalframe:

= (17—22)1)

and as the initial (Z = 0) joint force matrix:

= 1' 2 (17—23)(ijx j)

The elements of are the joint forces due to external forces acting on themembers with Z 0. We express the complete set of equations as

+

, p1,1 Z1

P1 = +Z2

(17-24)

IZm

We assemble and working with successive members. The contributionof member a follows from (17—7):

in row


GENERAL FORMULATION—LINEAR SYSTEM CHAP. 17

Column n

TE— Tdyn

n_n — (17—26)= — TE

p171sn = 0

S fl+,s=

Comparing (17—26) with (17—17), we see that

= (17—27)

We let— —

1' 1,, 2, (17—28)=

Then, we can express as

= + C!1..O (17—29)

17—5. INTRODUCTION OF JOINT DISPLACEMENT RESTRAINTS;GOVERNING EQUATIONS

The governing equations for the unrestrained system are

(17-30)

Now, we suppose r joint displacements are prescribed. We rearrange sothat the prescribed displacements arc last. We also rearrange and

cu1)-+ U =

(U2) (rxl)

—* P = (17—31)

Pf(r x

where U2, P1, and P1 are prescribed. We use B, A to represent the rearrangedforms d:

(qr x r)

IA2]

B = [liii = [An (17—32)

[ _J [ A2 j fr'qr)


SEC. 17—5. JOINT DISPLACEMENT RESTRAINTS 561

Finally, we write the equations for the restrai;ied system as

Pi = + B1Z P1 + AfZ (na eqs.) (17—33)

P2 = 2 + B2Z P1, 2 + (r eqs.) (17—34)

= fZ + 1/,, A1U1 + A2U217 35

-The unknowns are the member forces (Z), the displacements (U1), andthe r reactions (P2).

If the restraints are parallel to the directions of the global frame, the trans-formation of d to A (or to B) involve only a permutation of the columns ofd (rows The same permutation is applied to the rows

Suppose joint q is partially restrained and the restraint directions do notcoincide with the global frame directions. We first transform the force and dis-placement matrices for joint q from the global frame to the restraint frame, using

A O/Iq(17—36)=

This step involves postmultiplying column q of d by T and premultiplyingrow q of by We write the transformed equations as

= J)1 + 4L(1737)

+ fZ =

where the superscript J indicates that joint forces and displacements are referredto local restraint frames. The final equations are obtained by permuting thecolumns of d2 (rows of the rows of and then partitioning.

The transformation of a71 to U can be expressed as a matrix product,

U = DQ/ = (17—38)

where contains the rotation matrices for the joint restraint frames,

= ... (17—39)

and H is the row permutation matrix. One can generate H by starting withI and permuting the rows according to the new listing of the joint displace-ments, i.e., with the prescribed displacements last. Now, D is an orthogonalmatrix,

(17—40)Then,

= DTUP =



and it follows thatP, —A = (17—41)

B AT =

The partitioned forms are obtained by partitioning D:

[D 1 (n x (j)D = ., (17—42)

LD2J (r

Finally, we can writeA1 = = BTA2 = = Bf

(17-43)P1 1 =

2 =

To determine the requirement for initial stability, wc consider (17—33),

B1Z = P2 -.

which represents nj equations in unknowns. For the equations to he con-sistent for an arbitrary loading, the rank of 'B1 must equal n,,. Therefore, thestability requirement for the system is

r(B1) = ;(A1) = (17—44)

Since B1 is of order x a necessary hut not sufficient condition for stability

(17—45)

Equation (17—44) is the stability requirement for a geometrically linear system.It is also the initial stability requirement for a geometrically nonlinear system.In the next chapter, we develop the stability criteria for a geometrically non-linear system subjected to a finite loading.

17—6. NETWORK FORMULATION

In the formulation presented in the previous articles, we worked with theactual joint displacements and external joint forces referred to the global frame.The governing equations are given by (17—30), which we list below forconvenience:

= +

where— Tc)

= +

One assembles d, 9PI, using(17—17), (17—25), which are actually the expansionsof(b). By introducing new joint variables, we can express d in terms of only one


SEC. 17—6. NETWORK FORMULATION 563

connectivity matrix, C.4 — C. The rule (17—17) for assembling d stillapplies except that now

Let Y denote some arbitrary point. Suppose we express the actual force anddisplacement matrices for joint k in terms of their equivalents at point Y.We define

= statically equivalent force at Y due to theactual force matrix at joint k. (1746)

= displacement at Y due to rigid body motion aboutjoint k.

The actual and equivalent quantities are related by— r)r°

Y, k kY k (17—47)k

where

0

kY

(17-48)= — 0 — 4i) plane

I" 0 0Xy2; — Xy1

tplanar

We could operate on (b), but it is more convenient to start with (17—11):

'Kr,, = —

Now, by definition,=

Substituting for using (17—47), and noting that

we obtain'Kr,,, = (17—48)

The remaining steps are the same as followed previously. We let

'Wy = {0/t°y, 1, 'Wy 2 (17—49)

and write'K = (17—50)

The generation of follows from (17—48). For now n,

= = (17—51)

= 0 (q,, x 1)S

s = 1, 2 j


564 GENERAL FORMULATION—UNEAR SYSTEM CHAP. 17

To express in factored form, we let

1+Y

= (irn x irn) (17—52)

+ Y

Then,C_)

(17—53)=

We transform the joint forces, using (17—47), and write the resultingequationsas

= + (17-54)= + fZ =

To relate corresponding terms in (a) and (17—54'). we generalize (17—47):

(17—55)

= (if x i/)

• jY

It follows that ,, r— (17—56)

g,YI =

The expression for reduces to (17--53) when (d) and (c) are introduced.The formulation developed above can be interpreted as a network formula-

tion since the connectivity term appears seperately in the factored form of d.A simplified version which does not allow for member force releases has beenpresented by Fenves and Branin (see Ref. 1). The only operational advantageof not working with the actual joint quantities is in the generation øf d,m÷ andd,,,. This advantage is trivial compared to the additional operations requiredto generate ., to introduce the displacement restraints, and finally, totransform to once the solution is obtained. Another serious disadvantageis that the equations tend to become ill-condii.ioned.

Fenves and Branin's primary objective was to show that the governingequations for a member system can be cast in a form such that geometrical andtopological effects are separated, i.e., a network formulation. DiMaggio andSpillars (Ref. 2) have also presented a network formulation for a rigid jointedmember system. Actually their formulation is a special case of our first formula-tion. It is not, strictly speaking, a true network formulation since connectivityis not completely separated from geometry (see (17—21)). The only way that onecan separate connectivity from geometry is to redefine the joint variables. Note


SEC. 17—7. DISPLACEMENT METHOD 565

that the ideal truss is an exception. Connectivity and geometry are naturallyuncoupled for this system.

Whether one interprets the governing equations for a member system from anetwork viewpoint is of academic interest only. In the displacement method,the equations reduce to the equations for the direct stiffness method. Theonly possible advantage of the network interpretation is in the force method.There one can use certain concepts of the mesh methodt to select a primarystructure, provided that there are no member force releases or partial jointrestraints. However, the selection of a primary structure for a rigid-jointedframe having fixed supports is quite simple, and even this advantage is debatable.

17—7. DISPLACEMENT METHOD

The governing equations are given by (17—33), (17—34), and (17—35). Oncethe member forces are known, we can find the reactions from (17—34). Now,we start by solving (17—35) for Z in ternis of the displacements,

Z = Z1 + kA1UE + kA2U2 17-57)where

= initial member force matrix (q1 x 1)= (17—58)

k = = reduced member stiffness matrix x

Note that k is quasi-diagonal, symmetrical, and positive definite. The matrix,Z1, contains the initial member forces due to external loads acting on themembers and initial deformation resulting from fabrication errors or tempera-ture changes.

We substitute for Z in (17—33) and write the result as

= P0 + + K12U2 (17—59)where

= (nd x lid)K12 = ATkA2 x r)

(17—60)= + (lid X 1)

The elements of P0 are the joint forces due to the initial end forces. SinceA1 is of rank (when the system is stable) and k is positive definite, it

f See Sec. 9—5.See Prob. 2—18.



that K1.1 is positive definite. Conversely, jfK1 is not positive definite, the systemis unstable. The joint displacements are determined by solving (17—59) and themember forces are obtained by back substitution in (17—57).

Operating on the restrained equations, as we have done above, is not efficientsince the various coefficient matrices must be generated by matrix multiplica-tion. By first manipulating the unrestrained equations and then introducingthe displacement restraints, one can avoid any matrix multiplication. Thisprocedure corresponds to the direct stiffness method.

Operating on (17—30), we obtain

Z = Z1 + kd°ll (17-61)and

= + c/TZ +(17—62)

= +

Equation (17—62) is identical to (16—8). The of reduces to(16—9), (16—10) when we introduce the factored forms of d, Z1.

First, we review the definitions of the member stiffness matrices,The effective member stiffness matrix (see (16—104)) has been

defined as

k to the global frame and applying (16- 107) reads to

k T ne.n e.H

=and

L0 .._1Ofl+n+ — e,fl

Now, substituting for d using (17—21), the expression for takes the form

= —

where= (17—63)

Finally, we expand (d):

= + T)C+ + —

One can easily show that (17—64) reduces to (16—10) when the properties ofC... are taken into account.

The initial end actions for member n= + TE)( — kr, n'17ro. n)

T,,, TEn(_kr. n)

t See Eqs. (17—7), (17—8), and (17—il).


SEC. 17—8. FORCE METHOD 567

Using the factored forms for d, and 4, the expression for takes theform

= ++ —

(17—65)

The general form defined according to (16—9) is

= + (17—66)

Substituting (e) in (17—65) results in (17—66).In Sec. 16—4, we presented a procedure for introducing joint displacement

restraints and represented the modified equations as

= (ii eqs.)

Now, (f) consists of (17—59) plus r relations for the prescribed displacements.We obtain (f) by starting with

=[0 IrJ L

U2

and permuting the rows and columns. This operation can be represented interms of the permutation matrix, 11, defined by

u=[IdllJ= HTP

Then,

=

It follows that is positive definite when K11 is positive definite, i.e., whenthe system is stable.

17—8. FORCE METHOD

We start with the governing equations for the restrained system:

B1Z = P1 P1, (fld eqs.) (a)

BfU1 + = 'V0 + fZ (qr eqs.) (b)

= 2 + 82Z (r eqs.) (c)

Equation (a) represents equations in unknowns where Also,B1 is of rank The system is statically determinate when n4. We let

be the degree of static indeterminacy, i.e., the number of member forceredundants:

= — (17—67)



Since B1 is of rank we can solve (a) for member forces in terms of the netprescribed joint forces (P1 — P1 i) and r member forces. The compatibilityequations for the member force redundants are obtained by eliminating U1from (b). This is possible since (b) represents qT equations whereas U1 is onlyof order x 1. In the next section, we specialize the principle of virtual forcesfor a member system and utilize it to establish the compatibility equations.

We suppose the first 0d columns of B1 are linearly independent. If the systemis initially stable, the member force matrix Z can always be rearranged so thatthis condition is satisfied. We partition Z after row

1z,)Z (17—68)

1) (ZR) 1)

The elements of ZR are the force redundants for the system. We refer to thesystem obtained by setting ZR = 0 as the primary system. Continuing, wepartition B1 and B2 consistent with (17—68):

(fld 0 qj-) x fld) ("s 'Jo,)

B1 = [ BIR17 69

B2 —

(i (r 0 "d> (r °qo,)

The equilibrium equations take the form

= P1 — BIRZR (17—70)

P1,2 + + B2RZR (17—71)

We write the solution of (17—70) as

ZP + KZR (17—72)

The force influence matrices can be expressed as

("a 1)

Zr,,, = (B1pY'(P1 —

ZPR =

but it is not necessary to determine Actually, the solution procedurecan be completely automated.( The complete solution for is

xqo,)

z (17-74)

Note that the member forces due to are self-equilibrating, i.e., they satisfyB1Z = 0. Finally, we substitute for in the expression for P2 and write theresult as

P2 = P2,0 + P2 RZR (17—75)

t See Sec. 9—2.


SEC. 17—8. FORCE METHOD 569

where(rXI)

= P1,2 + J37PZP,0'17—76)

B2R + B2pZp,ft(r X

It remains to determine Zft.Equation (b) represents qr equations in unknowns, U1. Since = +

qft, there are excess equations. We partition (b) consistent with the partioningof Z,

rDT 1 rDT 1 I.e' 1 1 11 + 2 —

+L"2RJ V R) o,R) L PR *ftftJ ft

and obtain the following two sets of equations relating to U1 and ZR:

BfPU1 + 'V'p + + (fld eqs.) (17—77)

BfftU, + = = 1'O.J( + + fRRZR eqs.) (d)

The joint displacements can he determined from (17—77) once Zft is known.Eliminating U1 from (d) leads to

RU2 I/R + ZP 17 78+ + fftftZft ± + + fpftZR)

—

Equation (17—78) represents the compatibility equations for the force redun-dants. Finally, we substitute for using (17—72) and write the resultingequations as

fZRZR = A (17—79)

where>< qg)

— f 7T 4• 7 7T c 17T— 'RR 1-q', R'PP'—P. ft

—

A = P2, ftU2 — + o) — p +

These equations are similar in form to the corresponding equations for theideal truss developed in Sec. 9—2.

The flexibility matrix, fzR, can be expressed as

[Zp [ZP ft= lil

J 'RRJ— [zp,R1T f[ZP.R1

(17—81)

j jNow, f is positive definite for a deformable system. Then, it follows that fzRis also positive definite. In a later article, we consider the case where certainmember deformations may be prescribed.

Once the preliminary force analyses have been carried out, the remainingsteps are straightforward. We generate A, solve for Zft. and then determine



Zr,, P2 by back substitution. If the displacements are also desired, they can bedetermined by solving (17—77).

The final number of equations for the force method is usually smaller thanfor the displacement method VS. lid). However, the force method requiresconsiderably more operations to generate the equations. The force methodcan be completely automated, but not as conveniently as the direct stiffnessmethod. Also, automating the preliminary force analyses requires solving anadditional set of nd equations. Another disadvantage of the force method isthat the compatibility equations tend to he ill-conditioned unless one is carefulin selecting force redundants.

17—9. VARIATIONAL PRINCIPLES

In Chapter 7, we developed variational principles for the displacement andforce formulations for an ideal truss, Now, in this section, we develop thecorresponding variational principles for a member system. The extension isquite straightforward since the governing equations are almost identical inform.

We start with the force-equilibrium equations,t

P = P, + ATZThe partitioned form is

= P,,1 + AfZ= P1,2 +

To interpret (a) as a stationary rcquircment, we consider the deformation-displacement relation,

"K = AU = A1U1+ A2U2

The first differential of "K due to an increment in U is

d"K = A AU = A1 AU1 + A2 AU, (17-82)

Then, the requirement that

PTAU = + ZTd.K (17-83)

be satisfied for arbitrary AU is equivalent to (a). If we consider U, to be pre-scribed, (17—83) results in only (b). We refer to (17—83) as the principle of virtualdisplacements for a member system.

In the displacement method, we substitute for Z in the joint force-equilibriumequations, using

Z = — "1/',,) = (AU —

The form of (17—83) suggests that we define a scalar quantity, V = V(U),having the property

dV = = dV(U) (17—84)

t We work with the governing equations for the restrained system. See (17—33), (17—34), (17—35).


SEC. 17—9. VARIATIONAL PRINCIPLES 571

One can interpret V as the strain energy function for the members. For thelinear case, V ôan be expressed as

V = —17 85

= — —

Continuing, we define the potential energy function, as

V + — PTU (17-86)

The Euler equations for H,, are the unpartitioned joint force-equilibriumequations expressed in terms of U. Finally, we introduce the joint displacementconstraint condition, U2 =02, by writing (17—86) as

H,, = V + 1U1 + — — — U2) (17-87)

where U1, U2, and P2 are variables. The Euler equations for (17—87) are thepartitioned equilibrium equations (Equations (h), (c)) expressed in terms of thedisplacements with U2 set equal to 02, i.e., they are the governing equationsfor the displacement formulation presented in Sec. 17—7.

If only the equations for P1 are desired, we set U2 = 02 in (17—87),

= V + 1U1 — (17—88)where

V = + A202 — + A202 — (17—89)

The Euler equation for (17—88) is (17—59), and the second differential has theform

i2rr — /tITTIATI A \AIT'4 — 1)(17—90)

= AUrK11 AU1

Since K11 is positive definite, we can state that the displacements defining theequilibrium position correspond to a minimum value of defined by (17—88)or (17—87).

We consider next the force-method formulation. We let AP, AZ be a staticallypermissible virtual-force system. By definition,

= ATAZ = BAZ (17-91)

Premultiplying both sides of (d) with AZT and introducing (17—91) leads tothe principle of virtual forces,

APTU = (17-92)

Note that (17—92) is valid only for a statically permissible virtual-force system,i.e., one which satisfies (17—91).

The compatibility equations follow directly from the principle of virtualforces by requiring the virtual-force system to be self-equilibrating. If AZsatisfies -

AP1 = B1 AZ = 0 (17-93)then (17--92) reduces to

= (17-94)


572 GENERAL FORMULATION—LiNEAR SYSTEM CHAP. 17

This result is valid for an arbitrary self-equilibrating virtual-force system. Theformulation presented in the previous section corresponds to taking

[ZP,RlAAZ

= ZR(17-95)

AP2 =

We define the member complementary energy function, V's' = V*(Z), suchthat

dV* = (17—96)

For the linear case,

and= 4ZTfZ + ZT.K0 (17—97)

We also define the total complementary energy function, as

= — (17—93)

The deformation compatibility equations, (17—94), can he interpreted as thestationary requirement for 11. subject to the following constraints on Z, P2:

= —

= + B2Z

The constraint conditions are the joint force-equilibrium equations. Operatingon (g), and noting that P, 2 are prescribed, lead to the constraintconditions on the force variations

B1 AZ = 0

AP2 = B2 AZ

Note that (h) require the virtual-force system to be statically permissible andself-equilibrating.

In the previous section, we expressed Z, P2 as

Z= +

P20 + P2RZR

This representation satisfies (g) and (h) identically for arbitrary AZR, Sub-stituting for Z, P2 in (17-98) and expanding V* using (17-97), we obtain

= ++

ZR](17-99)

— ZR 2. RU2 + const

The Euler equations for (17—99) are (17—79), and the second differential has theform

= AZR (17—100)


SEC. 17—10. MEMBER DEFORMATION CONSTRAINTS 573

Since is positive definite, it follows that the true forces, i.e., the forces thatsatisfy compatibility as well as equilibrium, correspond to a minimum valueof

Instead of developing separate principles for the displacements and forceredundants, we could have started with a general variational principle whoseEuler equations are the complete set of governing equations. One can easilyshow that the stationary requirement for

rIR = ZT(8TU1 + — — Pfu1IT IT ( 7—10

T 1 —

considering Z, U1, U2, and P2 as variables, lead to the partitioned jointforce-equilibrium equations and the member force-joint displacement relations.This principle is a specialized form of Rcissner's principle.

We obtain (17—87) from (17—101) by introducing the force-displacementrelations as a constraint condition on Z,

Z k(BfU1 + —

= —

and noting that, by definition,

ZT(BTIJ1 ± — V" = y

Introducing the joint force-equilibrium equations as constraint conditionsreduces 11a to —11. as defined by (17—98).

17—10. INTRODUCTION OF MEMBER DEFORMATION CONSTRAINTS

Suppose a member is assumed to he either completely or partially restrainedwith respect to deformation due to force. The rigidity assumption is introducedby setting the corresponding deformation parameters equal to zero in the localflexibility matrix, g. For example, if axial extension is to be neglected, we set1/AE = 0. For complete rigidity, we set g Now, in what follows, wediscuss the case where neglecting member deformation parameters causes themmher flexibility matrix t. to be singular, This happens, for example, whenaxial extension is neglected for a straight member. The rank is decreasedt byI and the axial force-deformation relation degenerates to

-= — = v,, + (a)

where is the initial axial deformation due to temperature and fabricationerror. Note that now the axial force has to be determined from the equilibriumequations. For rigidity, = 0, and the force-displacement relations(see (17—5)) degenerate to

(b)

See (16—75).



One can interpret (a), (b) as either member deformation constraints or as con-straint conditions on the joint displacements. In general, the decrease in rankof the system flexibility matrix f is equal to the number of constraint conditions.

We consider first the force method. The governing equations are given by

fZrZR = A (qR eqs.)where

="Zr L

Suppose these are c deformation constraints. Then, f is of rank — C. Inorder to solve (c), must be nonsingular, i.e., it must be of rank Thisrequires

qT — C (17—402)

That is, there must be at least unconstrained member deformations. Thiscondition is necessary but not sufficient as we will illustrate below. Aside frominsuring that the flexibility matrix is of rank there is no difliculty involved inintroducing member deformation constraints in the force formulation.

Example 17—1

Consider the ideal truss shown. For this system.

qT 4

q4 = 2

We take the forces in bars 3, 4 as the redundants:

cF1) IF3

(F2J ff4Then,

ri 0Zr5[01

and

1 1 0

010

=

We can specify that, at most, two bars are rigid. No difficulty is encountered if only onebar is rigid. However, we cannot specify that bars 1, 3 or 2, 4 are rigid.

We consider next the displacement formulation. Since f is singular, k doesnot exist and, therefore, we cannot invert the complete set of force-displacement



Fig. E17—1

0

0

relations, i.e., (17—57) are not applicable. In what follows, we first develop theappropriate equations by manipulating the original set of governing equations.We then show how the equations can be deduced from the variational principlefor displacements.

The governing equations are

P1 = P1 1 + AfZ eqs.)

+ fZ = A1U1 + A2U2 (qT eqs.)

Now, we suppose there are c deformation constraints and the elements ofare listed such that the last c elements are the prescribed deformations. Wepartition '/7' and Z as follows:

1

= (cx1)z

=(17—103)

where contains the constrained member deformations and Z. the corre-sponding member forces. We use subscripts c, u to indicate quantities asso-ciated with the constrained and unconstrained deformations. Continuing, wepartition

AA1

(q-r 'ia)(cx I'd)

—

(cxr) (17—104)

•) (qr—e)xi

(qT xl) 05 (cx I) -

f = fT= 0

(cxc)



The deformation constraints are introduced by setting = 0. Note that, inorder for f to be singular, there must be no coupling between and i.e.,f must have the form shown above. Using this notation, the governing equationstake the form

P1 + + AfCZC (17—105)

= + = + A2aU2 (17106)

= = A1,U1 + A2,02 (17—107)

Equation (17—107) represents c constraint conditions on the unknown jointdisplacements, U1. The rank of A1, is equal to the number of independentconstraint equations. One can easily demonstrate that c independent constraintconditions are required in order to be able to analyze the system for an arbitraryloading.

Example 17—2

Suppose we specify that bars 1, 3 are rigid for the system considered in Example 17—I.The constraint equations arc (we take = {e1, e1})

— = — U21

e3 = = —u11 +

For (a) to be consistent, we must have

er,., + e30 = + U41

Even if(b) is satisfied, we cannot find the forces in bars 1, 3 due to Pi 1.

In what follows, we assume A1, is of rank c. We solve (17—106) for 4 andsubstitute in (17—105). This is permissible since is nonsingular. The resultingequations are

4 = (17—108)= + —

and

+ AfCZC = P1 — — — (17—109)

A1,U1 = — A2,02 (17—110)

Now, the coefficient matrix, is nonsingular only when the structureobtained by deleting the restraint forces (4) is stable. By suitably redefining4, we can transform (17—109) such that the coefficient matrix is always non-singular. Suppose we write

Z. = 4 + —

= 4 + + A2,02 — (17—ill)

where 4 represents the new force variable and is an arbitrary symmetrical



positive definite matrix of order c. Substituting for in (17—109), we obtain

P1 = P1,1 +

T [k,, 1 ([A1,, A2,,1 (a)+ -

By definingrk 1 rf 11

= = (17-112)L kcj L

and noting (17—104), we can write (a) as

Pi + ATk'(A2tY2 — + (Afk'A1)U1 +

Using the notation introduction in Sec. 17—7 (see (17—60)), we let

Kr, ATk'A,,

= (17—113)

= ±

Finally, the governing equations take the form

Z= =

— (17-114)

K11U1 + = — — K1202 = H1 (17—115)

= — = H2 (17—116)

Since A1 must he of rank for stability and we have required to be positivedefinite, it follows that K1 is positive definite. Also, the solution for U1 mustsatisfy (17—116) and we see from (17—111) that is equal to the actualconstraint force matrix, for arbitrary k'.

The expressions for Z and with Z. deleted, have the same form as theunconstrained expressions (17—57) and (17—59). Now, it is not necessary torearrange Z such that the constraint forces arc last. One can work with thenatural member force listing,

Z = {Z1, . . . , Z,,,}

and take arbitrary values for the member deformation parameters that are to benegelected. We obtain K11 and Fl1 by first generating using thedirect stiffness method and then deleting the rows and columns correspondingto the prescribed displacements. The constrained deformations, 1/v, can belisted arbitrarily. It is only necessary to specify the locations of the constraintforces (elements of in. the natural member force listing. Once thements and constraint forces are known, we can determine the force matrix formember n by first evaluating (see (17—8) and (17—11))

Z =k'r, r, ro. ,,j

— (O/g° — "—fl / '. n + n n -—



where k.,, is the modified stiffness matrix, and then adding the constraint forcesin the appropriate locations. In what follows, we describe two procedures forsolving (17—115) and (17—116).

In the first method, we solve (17—115) for U1,

U1 = — (17—118)

and then substitute in (17—116),

= — 112 (17—119)

The coefficient matrix for is positive definite since K1 is positive definiteand is of rank c. Note that, with this procedure, we must invert an ndthorder matrix and also solve a set of c equations. For the unconstrained case,we have to solve only equations.

Example 17—3

We suppose bar 2 (Fig. E17—3) is rigid. The constraint equation is

e2 = U, =

To simplify the example, we consider only the effect of joint forces. Using the notationintroduced above, the various matrices for this example are

U1 {u1,u2}

P1 Pz}

= e1 4 = F1 k,, =e2 4 = F2 =

11,1=2 c=1(P1 U2, are null matrices.)

Fig. E17—3

Bar is rigid

We start by assembling A1,

=e2

1

0



and then partition according to (17-104):

Ii1]

Note that we cannot invert (17—109), since Af,,k,,A1,, is singular.Now, we assume an arbitrary value for the stiffness of bar 2,

—

a is an arbitrary positive constant, and assemble K1 i:

=0

L kJ LU a

k1•[I' 1

K11 = Afk'A1 —

The governing equations (17—114), (17—115), and (17-116) reduce to

= + [u-.] (h)

K11U1 + = P1 (i)

= 0 (j)

The solution follows from (17—118), (17—119). We just have to take

H1 = H2 = 0 = (k)The inverse of K11 is

1 [1+2a —iiI (1)(1k1[-l +Ij

Then

.L[1 +11

=

and (17—119) reduces to

=ak1 ak1

F'2—p2—p1 (n)

Substituting for F'2 in (17—118), we obtain

2p'U1 =

Id1

U2 = 0

Finally, we substitute for U1, u2 in (h):

F1 =F2 = F'2 = P2 — P1


GENERAL FORMULATION—LINEAR SYSTEM CHAP. 17

Instead of first solving (17—115) for U1 in terms of one can start with(17—116),

= — = H2

which represents c relations between the displacements. Since A is of rankc, we can express c displacements in terms of rid — c displacements, i.e., thereare only nd — c independent displacements.

We suppose the first c columns of are linearly independent. Sinceis of rank c, we can always permute the columns such that this requirement issatisfied. We let

— c (17—120)

and partition U1:

(cx 1)U1 (nxl)

(cxnd) (Cxc) (17—121)

The elements of U are the independent displacements. By definition, is

nonsingular. Then, solving (a) for the constrained displacements, we have

= — 2U (17—122)

Finally, we express U1 asU1 = BU + H3 (17-123)

wherea)

=(cx

I f I (axx)L fl -J

(cxl)H3 =

( 0 j (ax!)

Note that B is of rank n and

0 (17125)

The generation of B, H3 from H2 can be completely automated using thesame procedure as employed in the force method to select the primary structure.

We consider next the joint force-equilibrium equations, (17—115),

K11U1 + = H1 (fld eqs.)

Substituting for U1 leads to

(K11B)U + = -- K11H3 114

We eliminate from (b) by premultiplying by BT and noting (17—125). Theresulting system of n equations for U is

(BTK11B)U = BTH4 (n eqs.) (17—126)

Since B is of rank n, the coefficient matrix is positive definite. One can interpret



BTK1 1B as the reduced system stiffness matrix. We solve for U and thenevaluate U1 from (17—123). It remains to determine the restraint forces,

We consider again Eq. (a). Assuming U.1 is known, we can write

= — K11U1 = 115 (n0 eqs.) (17—127)

The matrix, 115, is the difference between the external applied force, P1, and thejoint force due to member force with the constraint forces deleted, i.e.,

115 = — P1, 1 — ATZ (c)

where (see (17—114))Z —

(d)= k'(A1U1 + A2U2 —

We determine Z using the member force-displacement relations and assembleP1 + AfZ by the direct stiffness method. Now has c independent rows.In determining B, we permuted the columns such that the first c columnsare linearly independent. We apply the same permutation to (17—127) andpartition after row C:

[ATe. i

(17—128)

H52

Considering the first c equations, we have

1 (17—129)

Since is nonsingular, we can solve (17—129) for We obtain the finalmember forces by adding the elements ofZ defined by (17—114) and (d).

In this approach, we have to invert a matrix of order c and solve a systemof no — c equations. Although the final number of equations is less than in thefirst approach, there is more preliminary computation (generation of B) andthe procedure cannot be automated as easily.

Example 17—4

For this example (Fig. E17—4),

c—4 n=1 (a)

The constraint conditions are

e1 U12 n42 e10

= e2 = 021 — e,0 =(b)

e3 U22 — 032 e30

e4 031 144j e4,



Note that (b) corresponds to (17—107). The form corresponding to (17—116) is

+1

1e10 +U12(

Je,0e30 + u32f

..e4,o + U41JU31

1- IA1, U1 FL2

Columns 2,4, 5, and either I or 3 comprise a linearly independent set. Then, we can takeeither u1 or u21 as U. It is convenient to take U = u1

1We permute the columns according

to

U1 = {u12,u21,u22u31 u11}

= (U, U}

We determine U,by applying (17—122). This step is simple for this example since I.

Finally, we assemble U1 defined by (e) and then permute the rows to obtain the initial

U11 +1 0

0 e1,,, -I— 1142

U21 = +1 {u11} + e2,0

U22 0 e30 + 1132

u31 0 e4,0 + U4j

1' IB 113

Fig. E17—4

+1—1 +1

+1

1 —*5

2—. 1

3-.2

The rearranged form of U1 is

2

x2

ears 1,2,3,4 are rigid

listing of U1. The final result is


I=

F4H5,5

SEC. 17—10. MEMBER DEFORMATION CONSTRMNTS 583

The constraint forces are determined from (17—127), which for this example has the form

[

+1+1

+1

I I IH,

We permute the rows of (g) according to (d) and consider only the first four equations.The resulting equations correspond to (17—129).

It is of interest to derive the equations for the constrained case by suitablyspecializing the variational principle for displacements. We start with theunconstrained form of developed in Sec. 17—9,

= V + !5L1U1 —

whereV =

"K = A1U1 + A202

Now, the displacements are constrained by

= + =

Then, V reduces to

V = )Tk('K —

+

We obtain the appropriate form of by substituting for V using (d) andintroducing the constraint condition, "Kr — = 0:

= V + — + — (17-430)

The elements of 4 are Lagrange multipliers. One can easily show that thestationary requirement for (17—130) considering U1 and 4 as independentvariables leads to (17—109) and (17—110).

Since = v", we can add the term

'(1"' —

to (d). TakingV = — — (17—131)

in (17—130) leads to (17—115) and (17—116).In the second approach, we substitute

U1=BU+H3 (f)



in (a) and (17—131):= V + — + H3)

V —— (17—132)

A1BU + A1H3 + A2U2

The variation of considering U as the independent variable is

= AUT[BT(P11

P1) ++ BTATkr(ASH3 + A2U2 — (g)

= — BTH4]

Requiring to be stationary for arbitrary AU results in (.17—126). Note thatwe could have used the reduced form for V, i.e., equation (d). Also, we stillhave to determine the constraint forces.

REFERENCES

1. FENVES, S. J., and F. H. BRANIN, JR., "Network-Topological Formulation of Struc-tural Analysis," J. Structural Div., A.S.C.E., Vol. 89, No, ST4, August, 1963, pp.483—514.

2. DIMAGOT0, F. L., and W. R. SPILLARS. "Network Analysis of Structures," .1. Eng.Mech, Div., 4.S.C.E., Vol 91, No. EM3, June, 1965, pp. 169—188.

3. ARGYRIS, J. H;, "The Matrix Analysis of Structures with Cut-Outs and Modifica-tions," Proc. Ninth International congress App!. Mech., Vol. ô, 1957, pp. 131—142.


18

Analysis ofGeometricallyNonlinear Systems

18-1. INTRODUCTION

In this chapter, we extend the displacement formulation to include geometricnonlinearity. The derivation is restricted to small rotation, i.e., where squaresof rotations are negligible with respect to unity. We also consider the materialto be linearly elastic and the member to be prismatic.

The first phase involves developing appropriate member force-displacementrelations by integrating the governing equations derived in Sec. 13—9. We treatfirst planar deformation, since the equations for this case are easily integratedand it reveals the essential nonlinear effects. The three-dimensional problemis more formidable and one has to introduce numerous approximations in orderto generate an explicit solution. We will briefly sketch out the solution strategyand then present a linearized solution applicable for doubly symmetric cross-sections.

The direct stiffness method is employed to assemble the system equations.This phase is essentially the same as for the linear case. However, the governingequations are now nonlinear.

Next, we described two iterative procedures for solving a set of nonlinearalgebraic equations, successive substitution and Newton-Raphson iteration.These methods are applied to the system equations and the appropriate re-

relations are developed. Finally, we utilize the classical stabilitycriterion to investigate the stability of an equilibrium position.

18-2. MEMBER EQUATIONS—PLANAR DEFORMATION -

Figure shows the initial and deformed positions of the member. Thecentroidal axis initially coincides with the X1 direction and X2 is an axis ofsymmetry for the cross section. We work with displacements (u1, u2, co3),

585


586 ANALYSIS OF GEOMETRICALLY NONUNEAR SYSTEMS CHAP. 18

distributed external force (b2), and end forces (F1, F2. M3) referred to theinitial (X1-X2-X3) member frame. The rotation of the chord is denoted byp3 and is related to the end displacements by

— U,42

L

The governing equations follow from (13—88). For convenience, we drop thesubscript on x1, and M3, w3, 13. Also, we consider h1 = rn3 = 0.

Fig. 18—1. Notation for p'anar bending.

= 0

(F1u2, + F2) + b2 0

F2 =

Force-Displacement Relatio,is

x1 ,

(a)

F1= ULx +

F2= U2. — CO

2

M= (0,

(b)

Deformed position

b2 dxi 1182

Centroidai exis



SEC. 18—2. MEMBER EQUATIONS—PLANAR DEFORMATION 587

Boundary Conditions

Forx = 0:

= UAI or 1F110 = —FAI

U2 = or iF2 +w = WA3 or Mb =

Forx =U1 = or FIlL +FB1

or IF2 + = F112 (d)

= or Ml,.

Integrating (a) leads to

F1 = P

F2 +Pu2 = — C3P — C2Px + Jx(Jx b2 dx)dx

where C2, C3 are integration constants. We include the factor P so that thedimensions are consistent. The axial displacement is determined from thefirst equation in (a),

PL 1('1.

211111 — UAI

= — j(u2 dx (18—2)

Combining the remaining two equations in (a), we obtain

/ P\ riM = El + u2 +

Finally, the governing equation for u2 follows from the third equation in (e),

+ + C3)+ —

where (18—3)2__El

The solutions for u2 and M arc

= C4 cos px + C5 sin + C2x + C3 + U2b

co (i + C4 sin j.tx + C5 cos px)(18—4)

+ C2 + b2 dx + (i + x -

where U2b denotes the particular solution due to b2. If b2 is constant,

b(EI 1/U2b — (18—5)


588 ANALYSIS OF GEOMETRICALLY NONLINEAR SYSTEMS CHAP. 18

Enforcing the boundary conditions on u2, w at x = 0, L leads to four linearequations relating (C2 C5). When the coefficient matrix is singular, themember is said to have buckled. In what follows, we exclude member buck-ling. We also neglect transverse shear deformation since its effect is small for ahomogeneous cross section.

We consider the case where the end displacements are prescribed. The netdisplacements are

u = (u — 1. (18—6)CD' = (a5 U2bX)X_OL

Evaluating (18—4) with A2 = oc, we obtain

C2 = — jzC5

C3 = — C4

— C— —

— sin 1iL ,u sin (18—7)

1 1 . 1— COS/LLC5 = — — 1tL — —

D = 2(1 —

Note that D 4 0 as This defines the upper limit on P, i.e., the memberbuckling load:

PJrnax (18_8)

The end forces can be obtained with (c—e). We omit the algebraic detailssince they are obvious and list the final form below.

MA3 = + + — — uA2)]

MB3 = + +—

UA2)]

+ + — UA2)] — (u52 — uA2)(18—9)

2 1 PFB2 = — + C0A3 — — Unz)] +

P1. ('Lj2 PL

— =— j

(u2, dx = — erL

whereD = 2(1 — cos — pL sin pL

— 1iL cos

Dç62 = — sin iL)

= + cos 1zL)


SEC. 18—2. MEMBER EQUATIONS—PLANAR DEFORMATION 589

The functions were introduced by Livesley (Ref. 7), and are plotted in Fig.18 —2. They degenerate rapidly as —+ 27t. The initial end forces depend onthe transverse loading, b2. If b2 is constant,

bLA2 — 52 —

bL21

1- (18—10)

=

— —

In order to evaluate the stiffness coefficients, P has to be known. If one end,say B, is unrestrained with respect to axial displacement, there is no difficultysince is now prescribed. The relative displacement is determined from

PLUB1 = UA1 + — Le.

('L

=—— j

dx = er( jiL, UA2, U52, WA, w5)

2 (UB2 —+ — w43) +

L jDg54 = sin

(18—11)

(U52 —C5 = — WAS) + — WA3

= sin jiL cos jiL) + 2(1 — cos

= (1 — cos jiL)—

(1 — eos jiL'\ 4 if sin jiL cos jiL

= + + -We call Cr the relative end shortening due to rotation. However, when bothaxial displacements are prescribed, we have to resort to iteration in order toevaluate P since e, is a nonlinear function of P. The simplest iterative scheme is

p(i+ 1) = (u51 — UAI) + (18—12)

and convergence is rapid when jiL is not close to 2it. -

Expressions for the incremental end forces due to increments in the enddisplacements are needed in the procedure and also forstability analysis. If jiL is not close to 2ic, we can assume the stability functions



pL

Fig. 18—2. Plot of the 0 functions.

are constant and equal to their values at the initial position, when operatingon (18—9). The resulting expressions are

(IMA3 + Aco,43 + c&2 — AU42)]

dMB3 = dMç3 + + — (Au,32 —

dP42 + + AWA3 — —

(18—13)P

— — ——h-——- dP

— ,JL'i .4141 112 112 42 1 42

dFB1 = dP dF41 = —dP

dP = — Au41) + AEder

I

+02

—2

—4

—6

—8

—10

Oi

A


SEC. 18—3. MEMBER EQUATIONS—ARBITRARY DEFORMATION 591

where the incremental initial end forces are due to loading, Lxb2. We canobtain an estimate for by assuming Au2 is constant.

Au12.x dxAuA2)

(18—14)

The coefficients in (18—13) arc tangent stiffnesses. They are not exact sincewe have assumed and Au2, constant. To obtain the exact coefficients,we have to add

El3 F , — 1LA2 , 1+— j 4)3] dCuL)

(18-15)

d(,uL) = — dP2b13

to dM4 and similar terms to dM5, . . , , The derivatives of the stabilityfunctions are listed below for reference:

— 2(1iL)2 sin

D

=(18—16)

- 41 + + - pL)

= —

We also have to use the exact expression for

ae tie.der = + ——Au52 + + -—-- + —--—A(1zL)

cIUA2 (WA

in the equation for dP. An improvement on (18—14) is obtained by operatingon (18—11), and assuming ,tL is constant.

18—3. MEMBER EQUATIONS—ARBITRARY DEFORMATION

The positive sense of the end forces for the th case is shownin Fig. 18—3. Note that the force and displacement measures are referred tothe fixed member frame. The governing equations for small rotations werederived in Sec. 13—9. They are nonlinear, and one must resort to an approximatemethod such as the Galerkin scheme,t in which the displacement measures areexpressed in terms of prescribed functions (of x) and parameters. The problemis transformed into a set of nonlinear algebraic equations relating theeters. Some applications of this technique are presented in Ref. 5.

t This method is outlined in Sec. 1O—6.



Note: The centroidal axis coincideswith X1, X2 and X3 are prin-cipal inertia directions.

Fig. 18—3. Notation for three-dimensional behavior.

If we consider b1 = 0, the axial force F1 is constant along the member and thenonlinear terms involve and coupling terms such as co1M2;Neglecting these terms results in linearized equations, called the Kap pusequations. Their form is:


F1 P

+ x3w1,1) + F2]+ b2 0dx1

[P(u,3 i + F3] + b3 = 0

d1 + 1 + rn-1- + — + 711w1 = 0

dx1

1142 1 — F3 + = 0

M, + F2 + in3 = 0

M4,1 + = 0

— 11

=+ +

x2 M52, WB2

!152

x1//I

//

P2

(18—18)




= u1,i + 1

+ u53,1 + + w1, i —

I "F2 F3 X3rU52,1 — CO3 ..— +GA2 A23 J

1/F2 F3 x2U53, + Ui2 = + +GA23 A3 J

M2 M3(Dz,ij7-

f + C01,t + X3rF7 ± X2rF3)

Boundary Goiidif ions (± for x = L, — for x = 0)

P =P(u521 + + F2 =P(u,3

•t — i) + F3 = ±F3+ M!1 + — + 711w1 =

M2 = ±M2 M3 = ±M3 =

To interpret the linearization, we consider (13—81). If one neglects thenonlinear terms in the shearing strains,

Y12 u12 + 112.1 -

Y13 u13 + U3,1

takes the extensional strain as

a1 u1 + zfu2, i + U3, + + 1

and assumes+ = 0

+ = 0

+ = 0

one obtains (13—81). Equations are exact when the section is doubly sym-metric, Assumptions (a) and (b) are reasonable if 1 is small w.r. to u2 andu3, However, they introduce considerable error when co1 is the dominantterm. This has been demonstrated by Black (Ref. 5).

When the cross section is doubly symmetric,

= = X2r = X3r = = 0A23

(18—19)1 2

—= r



(r is the radius of gyration with respect to the centroid) and the problemuncouples to—

1. Flexure in plane2, Flexure in X1-X3 plane3. Restrained torsion

We have already determined the solution for fiexure in the X1-X2 plane.If we introduce a subscript for /L and

P PCu2)2 =

El2 (18—20)

= cbjCu3L)and then replace

(02

U3 —* —U2 —4

(18—21)F2 F3 M'2

F3 —F2 -

in (18—9), (18—13), we obtain the member relations for flexure in the X1-X3frame. For example,

MA3 + + — U.42)]

(18—22)

= +E12

+ + U43)]

andF42 =

F43 = + [_WB2 W42 — U43)] — — U.43)

The expressions for the axial end forces expands to

= P J'Al =AE

P = — U41) + AE(Cri + er2 + er3)

r2 Ci-' 1 CL (18—23)

= jdx1 er2

Jdx1

1

er3 = f(u3 dx1

2L

where is obtained from er2 by applying (18—21).We generate the restrained torsion solution following the procedure described

in Example 13—7. If the joints are moment resisting (i.e., rigid), it is reasonableto assume no warping, which requires f = 0 at x = 0, L. The correspondingsolution is summarized below:



r2P GJ 1+PP=— UGJ Erlcb 1 + + '>

- GJMB1 <1)A1)

MA1 —MB1

________1+P

(18—24)

1 +2

sinh 1uL [iiL(1 + Cr(1 + P))

(— +

GJ(1 + P)1 [ 1—coshuL

— + —-——--—(1 —/1(1 + Cr(1 + P))[ sinh,uL

We neglect shear deformation due to restrained torsion by setting C,. 0.If warping restraint is neglected,

En4, 0 c%)(18—25)

_

I + PAt this point, we summarize the member force-displacement relations for

a doubly symmetric cross section. For convenience, we introduce matrixnotation:

= {F1P2F3M1M2M3}8

{u1u2u3w1w2w3}Betc. (18—26)

+ + kBJl%A += ÷ + —

where contains nonlinear terms due to chord rotation and end shortening

{AE(eni + er2 + e,.3); uA2); — uA3); 0; 0; contains

the initial end forces due to member loads; and

AEL

El3 El3

El2 El2'P33

GJ

Sym

El3


El3

. El2 El2

GJ

Operating on leads to the incremental equations. i.e., the three-dimensional form of (18—13). Assuming the stability functions are constant andtaking

—dP

— UA2)(AU82 + — iiA3)(AuB3 — AuA3)(18—27)

we obtain

+ r2(w81 — — AWAI)}

+ (knE + + (kBA —

= + (knA — Mi11 + (kAA + kr)L\%A(18 —28)


AE

L

k44 =

El3


2 LFB1.f.J3 + P2P3

2r p1p3 0 0

+TAT

—r2p1p2 0 0

Symmetrical (r2p1)2 0 0

0 0

0

p3 = — uA2) p2 = — UA3) Pi = — WA1)

1fF is close to the member buckling load, one mustinclude additional termsdue to the variation in the stability functions and use the exact expression for der

Kappas's equations have also been solved explicitly for a monosymmetricsection with warping and shear deformation neglected. Since the equations arelinear, one can write down the general solution for an arbitrary cross section.It will involve twelve integration constants which are evaluated by enforcing thedisplacement boundary conditions. The algebra is untractable unless oneintroduces symmetry restrictions.

18—4. SOLUTION TECHNIQUES; STABILITY ANALYSIS

In this section, we present the mathematical background for two solutiontechniques, successive substitution and Newton-Raphson iteration, and thenapply them to the governing equations for a nonlinear member system.

Consider the problem of solving the nonlinear equation

= 0 (18—29)

Let represent one of the roots. By definition,

= 0 (18—30)

In the method of successive substitution,t (18—29) is rewritten in an equivalentform,

x = g(x) (18—31)

and successive estimates of the solution arc determined, using

—q(k) (18—32)

where represents the kth estimate.The exact solution satisfies

x=gt See Ref. 9.

SEC. 18-4. SOLUTION TECHNIQUES; STABILITY ANALYSIS 597

where k, is the incremental stiffness matrix due to rotation,

0 P3 Pz rp1 0

AEkr L



Then,= —

Expanding in a Taylor series about

= g(k) + g(T — + — +

and retaining only the first two terms lead to the convergence measure

— — (18—33)

where is between and T.In the Newton-Raphson method,t is expanded in a Taylor series about

= + Ax + + = 0

where Ax is the exact correction toAx = —

An estimate for Ax is obtained by neglecting second- and higher-order terms:

=(18—34)

+

The convergence measure for this method can be obtained by combining (a)and (18—34), and has the form

— — (18—35)

Note that the Newton-Raphson method has second-order convergence whereassuccessive substitution has only first-order convergence.

We consider next a set of n nonlinear equations:

'I' = = 0(18—36)

= x2,...,

An exact solution is denoted by Also, =In successive substitution, is rearranged to

ax = c — g (18—37)

where a, c are constant, g = g(x), and the recurrence relation is taken as

= c — (18—38)The exact solution satisfies

a5 = c —

Then,1)) =

— g(k))

t See Ref. 9.


SEC. 18—4. SOLUTION TECHNIQUES; STABILITY ANALYSIS 599

Expanding g in a Taylor series about= g(k) + — +

9i,i 91,2 91n

== 92,1 92,2 92,n

L0xrJ'

and retaining only the first two terms results in the convergence measure

(x — = a (18—39)

where lies between xk and For convergence, the norm of 'g. must beless than unity.

The generalized Newton-Raphson method consists in first expandingabout

= + + = 0

where

= = [T'—j —

= (18—40)

=

Neglecting the second differential leads to the recurrence relation

= 18 41= + ( — )

The corresponding convergence measure is

— = (18—42)

Let us now apply these solution techniques to the structural problem. Thegoverning equations are the nodal equations referred to theqlobcil system frame,

1?e — = 0 (1843)

where contains the external nodal forces and — is the summation ofthe member end forces incident on node i. One first has to rotate the memberend forces, (18—26), from the member frame to the global frame using

=

k° =

In our formulation, the member frame is fixed, i.e. is constant. We introducethe displacement restraints and write the final equations as -

e m 1844Pm P1 + + KU



Note that K and depend on the axial forces while Pr depends on both theaxial force and the member rigid body chord rotation, if the axial forces aresmall in comparison to the member buckling loads, we can replace K withK1, the linear stiffness matrix.

Applying successive substitution, we write

KU = —

and iterate on U, holding K constant during the iteration:

= P1 — (18—45)

We employ (18—45) together with an incremental loading scheme since K isactually a variable. The steps are outlined here:

1. Apply the first load increment, and solve for U(I), using K K1.

2. Update K using the axial forces corresponding to Pe(l)• Then applyPe(2) anditerate on

K (ni_p _p(n—1)(1) e(1) e(2) r

3. Continue for successive load increments.

A convenient convergence criterion is the relative change in the Euclideannorm, N, of the nodal displacements.

N ='\ . (18—46)

— 1 e (a specified value)Jabs va]uc

This scheme is particularly efficient when the member axial forces are smallwith respect to the Euler loads since, in this case, we can take K = K1 duringthe entire solution phase.

In the Newton-Raphson procedure, we operate on Vi according to (18—41):

= — 4,(fl)

Now, Pe is prescribed so that

= due to+ dPr + K LW + (18—47)

= —

where denotes the tangent stiffness matrix. The iteration cycle is

AU(n) = —

18 48= + ( — )

We iterate on (18—48) for successive load increments. This scheme is moreexpensive since has to be updated for each cycle. However, its convergencerate is more rapid than direct substitution. If we assume the stability functions


SEC. 18—4. SOLUTION TECHNIQUES; STABILITY ANALYSIS 601

are constant in forming due to AU, the tangent stiffness matrix reduces to

dI( 0 dP, 0(18—49)

K + Kr

where K. is generated with (18—28). We include the incremental member loadsin at the start of the iteration cycle. Rather than update at each cycle,one can hold fixed for a limited number of cycles. This is called mod (fledNewton-Raphson. The convergence rate is lower than for regular Newton-Raphson but higher than successive substitution.

We consider next the question of stability. According to the classical stabilitycriterion,t an equilibrium position is classified as:

stable — > 0neutral d2W,,, — d2We 0 (18—50)

unstable d2 W,, d2 < 0

where d2 is the second-order work done by the external forces during adisplacement increment AU, and is the second-order work done by themember end forces acting on the members. With our notation,

d2w, Pe)TAU

= (dAU

(18—51)

= AUTK, AU

and the criteria transform to

/ / \T < 0 stable(AU)TK, AU

—Pe) AU = 0 neutral (18—52)

> 0 unstable

The most frequent case is Pe prescribed, and for a constant loading, the tangentstiffness matrix must be posil.ive definite,

To detect instability, we keep track of the sign of the determinant of thetangent stiffness matrix during the iteration. The sign is obtained at no cost(i.e., no additional computation) if Gauss elimination or the factor methodare used to solve the correction equation, (18—48). When the determinantchanges sign, we have passed through a stability transition. Another indicationof the existence of a bifurcation point (K1 singular) is the degeneration of theconvergence rate for Newton-Raphson. The correction tends to diverge andoscillate in sign and one has to employ a higher iterative scheme.

Finally, we consider the special case where the loading does not producesignificant chord rotation. A typical example is shown in Fig. 18—4. Both the

t See Sees. 7—6 and 10—6.



frame and loading are symmetrical and the displacement is due only to short-ening of the columns. To investigate the stability of this structure, we deletetthe rotation terms in K, and write

K

K is due to a unit value of the load parameter The member axialforces are determined from a linear analysis. Then, the bifurcation problemreduces to determining the value of 2 for which a nontrivial solution of

(K + 2K;)AU 0 (18—54)

exists. This is a nonlinear eigenvalue problem, since K = K(2).

______________________

12X

I I

Fig. 18—4. Example of structure and loading for which linearized stability analysisis applicable.

In linearized stability analysis, K is assumed to be K1 and one solves

K, AU = —2K AU (18-55)

Both K, and K; are symmetrical. Also, K1 is positive definite, Usually, onlythe lowest critical load is of interest, and this can be obtained by applyinginverse iterations to

(—K;)Au1 (18—56)

REFERENCES

1. TIMOSHENKO, S. P., and J. M. GERE: Theory of Elastic Stability, 2d ed., McGraw-Hill,New York, 1961.

2. KOLLBRUNNSR, C. F., and M. MEIsTER: Knicken, Biegedriilknicken, Kippen. 2d ed.,Springer-Verlag, Berlin, 1961.

3. BLEICH, F.: Buckling Strength of Metal Structures, McGraw-Hill, New York, 1952.

t Set Pi P2 = = 0 in (18—28).See Refs. 11 and 12 of Chapter 2.


REFERENCES 603

4: BLYRGERMEISTEa, G.. and F!. STEUP: Stabilidhsrheorie, Part 1, Akademie-Verlag.Berlin, 1957,

5. CFJtLVER, A. H., ed.: Thin- Walled Structures, Chatto & Windus, London, 1967.6. VLASOV, V. Z.: Thin Walled Elastic Beams, Israel for Scientific Transla-

tions, Office of Technical Services: U.S. Dept. of Commerce, Washington, D.C., 1961.7. LIVESLItY, R. K.: Matrix Methods of Structural Anal vsis, Pergamon Press, London,

1964.

8. AROYRIS, J. H.: Recent Advances in Matrix Methods of Structural Analysis, PergamonPress, London, 1964.

9. HILDEBRAND, F. B.: Introduction to Numerical Analysis, McGraw-Hill, New York,1956.

10. GALAMBOS, T. V.: Structural Members and Frames, Prentice Hall, 1968.11. BRUSU, D. and B. ALMROTH: Buckling of Bars, Plates, and Shells, McGraw-Hill,

New York, 1975.



index

Associative multiplication, 8Augmented branch-node incidence ma-

trix, 124, 222Augmented matrix, 33Axial deformation, influence on bending

of planar member, 472

Bar stiffness matrix, 180Bifurcation; Neutral equilibriumBimoment, MqS, 373Branch-node incidence table, 121, 145

Cç1, Cr, C,,r—coefficients appearing incomplementary energy expressionfor restrained torsion, 387, 388, 416

Canonical form, 58Cartesian formulation, principle of vir-

tual forces for a planar member, 465Castigliano's principles, 176Cayley—Hamilton Theorem, 63Center of twist, 383, 389Characteristic values of a matrix, 46Chord rotation. p. 586Circular helix, definition equation, 84, 86Circular segment

out-of-plane loading, 504restrained warping solution, 509

Classical stability criterioncontinuum, 256member system, 603truss, 170

Closed ring, out-of-plane loading, 503Cofactor, 19Column matrix, 4Column vector, 4Complementary energy

continuum, 261member system, 572planar curved member, 434restrained torsion, 385; 387, 388unrestrained torsion-flexure, 301

Conformable matrices, 8, 35Connectivity matrix, member system, 563Connectivity table for a truss, 121, 143.Consistency, of a set of linear algebraic

equations, 31, 44

Constraint conditions treated. with La-grange multipliers, 76, 80

Curved memberdefinition of thin and thick, 434thin, 487slightly twisted, 487

Defect, of a system of linear algebraicequations, 31

Deformationfor out-of-plane loading of a circular

member, transverse shear, twist, andbending, 498

for planar member, stretching andtransverse shear vs. bending, 454

Deformation constraintsforce method, 573displacement method, 576variational approach, 583

Deformed geometry, vector orientation,239

Degree of statical indeterminacymember, 555, 567truss, 210

Determinant, 16, 37, 39Diagonal matrix, 10Differential notation for a function, 70, 72,

79Direction cosine matrix for a bar, 119Discriminant, 40, 59Distributive multiplication, 8

Echelon matrix, 29Effective shear area, cross-sectional prop-

erties, 302Elastic behavior, 125, 248End shortening due to geometrically non-

linear behavior, 589Engineering theory of a member, basic

assumptions, 330, 485Equivalence, of matrices, 27Equivalent rigid body displacements, 334,

414, 430Euler equations for a function, 73Eulerian strain, 234

605


606 INDEX

First law of thermodynamics, 248Fixed end forces

prismatic member, 523thin planar circular member, 528

Flexibility matrixarbitrary curved member, 515circular helix, 534planar member, 462prismatic member, 345, 521thin planar circular member, 526

Flexural warping functions, 296, 300/nFrenet equations, 91

Gauss's integration by parts formula. 254Geometric compatibility equation

arbitrary member, 499continuum, 259, 264member system, 569planar member, 463, 466prismatic member, 355truss, 160, 212, 216, 223unrestrained torsion, 279, 315

Geometric stiffness matrix for a bar, 200Geometrically nonlinear restrained torsion

solution, 595Green's strain tensor, 234

Hookean material, 126, 249Hyperelastic material, 248

Incremental system stiffness matrixmember system, 601truss, 193

Inelastic behavior, 125Initial stability

member system, 562truss, 137

Invariants of a matrix, 59, 62Isotropic material, 252

Kappus equations, 592Kronecker delta notation, 11

Lagrange multipliers, 76, 80, 583Lagrangian strain, 234Lamé constants, 253Laplace expansion for a determinant, 20,

38Linear connected graph, 218Linear geometry, 120, 143, 237Linearized stability analysis, 602Local member reference frame, 92

Marguerre equations, 449, 456Material compliance matrix, 249Material rigidity matrix, 249Matrix iteration, computational method,

201

Maxwell's law of reciprocal defiections.356

Member, definition, 271Member buckling, 588Member force displacement relations. 537,

546, 556Member on an elastic foundation, 384, 369Mesh, network, 220Minor, of a square array, 19Modal matrix, 52Modified Neuton-Raphson iteration, 601Moment, MR,Mushtari's equations, 444

Natural member reference frame, 92Negative definite, 58Negligible transverse shear deformation,

planar member, 443, 454, 498Network, topological, 220Neutral equilibrium, 170, 256, 601Newton-Raphson iteration, member sys-

tem, 598Normalization of a vector, 49Null matrix, 4

One-dimensional deformation measures,335, 338, 432

arbitrary member, 491Orthogonal matrices and trnasformations,

50, 53Orthotropic material, 250, 251

Permutation matrix, 42, 135Permutation of a set of integers, 16, 37Piecewise linear material, 126, 146Plane curve, 98, 425Poisson's ratio, 252Positive definite matrix, 58, 63Positive semi-definite matrix, 58Postmultiplication, matrix, SPotential energy function, member system,

571Premuftiplication, matrix, 8Primary structure

member system, 568planar member, 463prismatic member, 354truss, 211

Principle minors, 55Principle of virtual displacements

member system, 570planar member, 442

Principle of virtual forcesarbitrary member, 490, 492, 512member systens, 571planar member, 435, 458prismatic member, 338, 351

Quadratic forms, 57


INDEX 607

Quasi-diagonal matrix, 15, 38Quasi-triangular matrix, 39

Radius of gyration, 434Rank of a matrix, 27, 42, 43Rayleigh's quotient, 75, 79Reissner's principle

continuum, 270member, 383, 414member system, 573

Relative minimum or maximum value of afunction, relative extrema, 66

Restrained torsion solution, prismaticmember

linear geometry, 391nonlinear geometry, 595

Restrained torsion stress distribution andcross-sectional parameters

channel section, 401multicell section, 411symmetrical I section, 398thin rectangular cell, 407

Rigid body displacement transformation,109

Rotation transformation matrix, 101, 232Row matrix, 4

Self-equilibrating force systems, 160, 211,258

member systems, 568Shallow member, assumptions, 448Shear center, 297, 300, 309, 378, 389Shear flow, 287Shear flow distribution for unrestrained

torsiOn, 308Similarity transformation, 53, 62Simpson's rule, 475Singular matrix, 22Skew symmetrical matrix, 11Small strain, 120, 235Small-finite rotation approximation, 238Square matrix, 4Stability of an equilibrium position, 171,

195Stability functions (4), prismatic member,

589Statically equivalent force system, 103, 106Statically permissible force system, 159,

216, 257Stationary values of a function, 67, 79Stiffness matrix

arbitrary curved member, 516, 520

modification for partial end restraint,535

prismatic member, geometrically non-linear behavior, 588, 595

prismatic member, linear geometry, 522Strain and complementary energy for pure

torsion, 280Strain energy density. 248Stress and strain component trnasforma-

tions. 249Stress components

Eulerian, 242Kirchhoff, 246

Stress function, torsion, 276Stress resultants and stress couples, 272Stress vector, 240Stress vector transformation, 242Submatrices (matrix partitioning), 12, 36Successive substitution, iterative method

member system, 597truss, 193

Summary of system equations, force equi-ibrium and force displacement, 561

Symmetrical matrix, Il, 35System stiffness matrix

member system, 548, 550, 565truss, 179, 180, 188, 206

Tangent stiffness matrixfor a bar, 193prismatic member, 590, 596

Tensor invariants, 232Torsion solution, rectangle, 281Torsional constant, J, 276, 278, 323Torsional warping function, 274, 377Transverse orthotropic material, 252Transverse shear deformation

planar member, 454, 498prismatic member, 355

irapezoidal rule, 474Tree, network, 220Triangular matrix, 12Two-hinged arch solutions, 467, 470

Unit matrix, 10

Variable warping parameter, f, for re-strained torsion, 372

Vector, definition (mechanics), 4/n

Work done by a force, definition, 153, 156


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Analysis of Structural Member Systems