Large deflection theory for arches

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Authors Callan, Michael Dolan, 1940-

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LARGE DEFLECTION .THEORY FOR ARCHES

by.Michael D Callan

A Thesis Submitted to the Faculty of theDEPARTMENT OF CIVIL ENGINEERING

In Partial Fulfillment of the Requirements For the Degree ofMASTER OF SCIENCE

In the Graduate CollegeTHE UNIVERSITY OF ARIZONA

1 9 6 3

STATEMENT BY AUTHOR

This thesis has been submitted in partial fulfillment of requirements for an advanced degree at the University of Arizona and is deposited in The University Library to be made available to borrowers under the rules of the Library,

Brief quotations from this thesis are allowable without special permission, provided that accurate acknowledgment of the source is made, Requests for permission for extended quotation from or reproduction of this manuscript in whole or in part may be granted by the head of the major department or;the Dean of the Graduate College when in their judgment the proposed use of the material is in the interests of scholarship. In all other instances, however, permission must be obtained from the author,

SIGNED;

APPROVAL BY THESIS DIRECTOR

This thesis has been approved on the date shown below;

R.M. kZchard : "scs" DateAs-sodlate Professor of Civil Engineering

11

ACKNOWLEDGEMENT

The author wishes to express his gratitude to his thesis director Dr. R .Mc Richard for his advice and suggestions which aided so materially in the completion of this thesis.

Special thanks are also due to the staff of the Numerical Analysis Laboratories of the University of Arizona, .

ill

TABLE OP CONTENTS

CHAPTER PAGETo iHTRODUOTION 12.. FORMULATION OF THE LARGE DEFLECTION PROBLEM:V-\ FOR ARCHES .......... 3 :

General Description and Assumptions. . . . 3Strain-Displacement Relationships. . . . . 5Deformed Radius of Curvature in Terms of Undeformed Quantities. . . . . . . . . . . 7Strain at Any Point . . . . . . . . . . . 8Stress-Strain Relationships . , 8Equilibrium. . . . . . . . . . . ........ 9Summary of Equations . .......... 12

3Q: -FORMULATION OF THE LARGE DEFLECTION PROBLEMFOR CIRCULAR CANTILEVERED ARCHES . . . . . . . 13

Introduction . . . . . . . . . . . . . . . . 13Transformation from the General Equations. 14Boundary Conditions. . . . . . . . . . . . 16Loading Not Dependent on Rotation of Cross- section (Case 1). . . . . . . . . . . . . . 16Loading Dependant on Rotation of Cross- section (Case 2) . . . . . . . . . . . . . 18

4. DIGITAL COMPUTER SOLUTION OF THE PROBLEM . . . 20The Method of Runge-Kutta.......... 20Application to Case 1. . . ........... 21Mathematical Check for Case 1 . . . . . . . 22Application to Case 2. ............... 23Relation Between Vertical End Load andStress Resultants. . . . ............ . 24Vertical and Horizonal Deflections . . . . 2 4

5. COMPARISON TO KNOWN RESULTS . . . . . . . . . . 2?The Method of Seames and Conway. . . . . . 27Comparison 1 . . . . . . . . . . . . . . . 28Comparison #2. . . . . . . . . . . . . . . 28 .Comparison #3. . . . . . . . . . . . . . . 29Advantages of the Runge-Kutta Approach . . 36

6. APPLICATIONS. . . . . . . . . . . . . . . . . . 37Application to Rings and Simply Supported Arches . . . . . . . . . . . . . . . . . . 37The Correction Equation. . . . . . . . . . 37

iv

OE AFTER

6 cont Examples: Simply Supported Arch Examples: Ring . . . . . ,

6 0 O O O O O O

7.. FINITE DEFLECTIONS OF FRAMES AND TRUSSES Introduction Method of Analysis Matrix Formulation Linear Theory Nonlinear Approach .Limitations Method of Solution .Compatibility Equations in Coordinate NotationExamples and Discussion of Limitations

0 6 9 0 0 0 0

O O O O & O O O O O

0 6 0 0 0 0 0

0 0 0 0 0 9 0 0 0 0

O 0 0 O 0

in Coordin;9 0 0 9 0 0 0 0 6 0 0 0 9

:,8v CONCLUSIONS . . . . . . . .APPENDICES

. Appendix A - Flow diagram and computer program for solution of problem of cantilever arch where loading remains normal to the arch after deformation. o . . . . . . . . . . . . .

Appendix B - Flow diagram and computer program for solution of problem of cantilever arch where loading remains in its original direction after deformation . .

Appendix 0 - Flow diagram and computer program for solution of finite deflection problems a o o e e e o e o o o a

Appendix D - Notation: Chapters 2 through 6. . Appendix B - Notation: Chapter 7

REFERENCES . . . . . . . . . . . . . . . . . . . .

PAGE

4044505052525555575960 6068

70

75

81

889092

v

CHAPTER 1 INTRODUCTION

The classical theory of structures is a linear theory which presupposes that the deflections of the structural systems which are being analyzed are small,Woday, this theory is still of the utmost importance a# applied to the bulk of engineering problems On the Other hand, modern structural design is now concerned. .. With the analysis of systems which are capable of undergoing large deflections without complete failure For 'example, this phenomena occurs in thin aircraft structures where large distortions can occur without exibiting plasti behavior. Also in the design of underground structures to resist nuclear blasts, it is known that flexible structures, as opposed to rigid structures, are more Capable of resisting an underground shock due to deformation which transfers loads to the soil around the structure through soil arching phenomena.

At the present, the theory governing the behavior of systems undergoing large deflections is still in a fundamental stage. On the other hand, all interest in this problem cannot be relegated to this present century.

Euler and many others dealt with the large deflection problem under the name of the Elastica* or the elastic curve(1) Euler's solution to the axially loaded cantilever beam was the first and one of the few closed form solutions which have been developed up to this present date On the other hand, it was not until II that considerable practical interest was developed concerning this problem.

The reason for the lack of solutions in this field is due mainly to the fact that the. governing equations are nonlinear and, in general, extremely difficult to solve. For the most part, numerical methods must be used to obtain solutions. However, the development of high speed computation facilities has considerably broadened the scope of these numerical techniques. This thesis will utilize these techniques in the formulation of a computer approach to a specific problem which is the larg deflections of cantilevered arches. In addition, an iterative technique for calculating finite deflections, of frames and arches will be investigated.

1 Euler, Xis, "Methodus Inveniendi Llneas OurvasMaxi mi Minimive Proprietate Gaudentes,11 Lausanne and Geneva, 1744.

CHAPTER 2FORMULATION OF THE LARGE DEFLECTION PROBLEM FOR ARCHES

general Description and AssumptionsIn developing a small deflection theory8 the

^assumption that the undeformed geometry can be used to ilhlte the governing equations is Introduced This procedure results in a system of linear equations whichg #cr the most part9 are easily resolved This is not the case whn the system undergoes large deflections In writing the governing equations the deformed geometry must.be taken into account, resulting in a system of non-*-.-#lhear equations which in general are extremly difficult. to solve. In this section the governing equations are developed with three inital assumptions s 1) Plane sections.;., remain plane after deformation; 2) The materials in question are of a linear elastic nature; and 3) The #0,1 seen Effect is negligible.

the deformed and undeformed states. It defines the following quantities: z denotes the distance from the

length; w and v refer to normal and tangential displacements respectively; is the position angle of the arch,

Figure 1 represents a general arch element in

neutral surface of the arch; ds is the differential arc

y the angle of rotation3 '

4

Ptv+dvds w+dw

ds

dO

FIGURE 1

Pt

GENERAL ARCH ELEMENT (Neutral Surface)

* s

radius of curvature of the arch. The subscript o refers a quantity to the neutral surface. The asterisk superscript (*) refers to deformed quantities. All primed quantities denote differentiation with respect to the arc length, s.Straln-Dlsplacement Relationships

To derive the strain-displacement relations, refer to Figure 1 and the definition of engineering strain, 0 = . At the neutral surface, it isfound convenient to sum the components of the displacements and arc lengths in the directions of v+dv andw+dw. These components result in the vector sum of the

* ? 2arc length ds0 ; or, dsQ = x + y , where x is in the direction of v+dv, and y is in the direction of w+dw.From Figure 1, the following can be written:

= dsQ + dv2 + w2d2 + 2ds0dv - 2wdds0 y2 * dw2 + 2vdwd + v2d2

Using the above expressions and the definition of engineering strain, and noting that ds0 = y

6

Grouping terms In the preceding equation,(c + 1)2 = (1-w/|O0+dv/ds0 )2 + (dw/dso+v//0o ) ? (la)

This equation relates the strain at the neutral surface to the displacement quantities.

Referring again to Figure 1, and observing the right triangle formed by x, y, and ds*, it can be written that

tan(^> - dG) = y/x.In the limit as dO approaches zero, and since dsQ = /^dQ, it follows that

= w + (2a)1 + v'-v/fo

The form of equations (la) and (2a) makes them very inconvenient to use directly. It was found that the following substitutions separate linear and nonlinear terms, making the equations much easier to work with.Let

V = \ - v/f>0 + dv/ds0; and ( = dw/ds0 + v

dsds

(Undeformed) FIGURE 2

dOd

(Deformed)

GENERAL ARCH ELEMENT (Calculation of Deformed Radius of Curvature)

Deformed Radius of Curvature in Terms of Undeformed quantities

In the derivations to follow, it will be neces- sary to have the deformed radius of curvatureyO0 in terms of undeformed quantities, This can be done in the following manner. Referring to Figure 2, the following relationship can be written:

d* = + d@ + (d^/ds0 )ds0 = d + (d^/ds0 )ds0 .Since ds* =(1+fo)dSQ, and ^ojd* = ds*,the following expression for the deformed radius of curvature results:

1^>o = (1^0 + y / d s 0 )/( H 6 0 ). ---------- (A)

Strain at Any PointAt this point it will be possible to develop

an expression for the strain at any arbitrary distance from the neutral surface. The subscript z will refer to a quantity located at a distance z from the neutral surface. From the definition of engineering strain,

z =(ds*/dsz ) - 1.Referring to Figure 2,

Substituting equation (A) results in the following equation:

/ o - zThis equation can be simplified to read

= *o (B)' 1 - < o >

Stress-Straln RelationshipsUsing equation (3), it is now possible to derive

relationships between the stress resultants and the quan titles 4=o and Here, M is defined as the bending moment in the arch, while N denotes the axial force.

In terms of the normal stresses 0~z, the stress resultants may be written as follows:

M = J t (TyZdA ;

and Nwhere dA is a differential element of the area A of the cross-section.

It was previously assumed that the arch materialXobeys linear elastic laws and that the Poisson Effect can be neglected. In this case, Hooke1s Law may be written, where E is the Modulus of Elasticity.

EquilibriumIn this developement of a large deflection theory,

the only remaining step is the derivation of the equilibrium equations. In Figure 3, the general arch element for the deformed configuration is shown with the stress resultants and applied loads acting in their positive senses. V is defined as the shear force; and p* and p* are the normal and tangential loads respectively, with dimensions of force per unit arc length. It should be noted that the loading in the deformed state is in general not equal to the undeformed loads pn and p^. The relation ships governing these quantities can be seen by referring to Figure 1, where it is found that

Thus,

and

Making use of equation (B),

pn003/ ~ Ptsln/ (C)

10and p* = pnslny + p^cosy. (D)These relationships hold true with the limitation that the change in arc length after deformation is quite small, in other words, if 6 0 // 1. This assumption will be true in the majority of cases, but this limitation should be kept in mind when applying equations (G) and (D) to a specific problem. Further discussion on this topic can be found in a master's thesis by N.E. Kesti(2).

*

dON+dNN

FIGURE 3EQUILIBRIUM OF DEFORMED ARCH ELEMENT

2. Kesti, N.E., "The Developement of a General Buckling Theory for Rings and Arches with Applications to Cir cular Arches, Master's Thesis, Univ. of Arizona, 1962.

11The equilibrium equations are developed by re

ferring to Figure 3. By summing forces in the normal direction, the following equation is obtained:

p*ds*cosd@^2 + P*ds*slndG/2 + (V>dV)cosd6* - V+ (N+dN)sindO* = 0.

*As dsQ tends to zero, the above equation is reduced to:P*|0* + (dV/dSQ ) Oq + N = 0. (5a)

Also, by summing forces in the tangential directionand taking moments about the left-hand side of the element,

*the following equations can be written as ds0 approaches zero:

pt/o + (aN/ds*)jO* - V = 0; (6a)*

and dM/ds0 - V = 0. (7a)These equations, being in terms of the deformed geometry, can be expressed in terms of the undeformed quantities by making use of the following expressions:

and ds* = (1+0 )ds0 .Thus, equations 5a, 6a, and 7a can be written:

Pn + (dV/ds0 )/(U^0 ) + N(/' + 1/^) = 0 ; ----(5)

P* + (dN/dso )/(U

Summary of EquationsThe governing equations may be summarized as

follows:1 - + dv/dsQ = ( 1 + 0 )cos^;------------ (1 )dw/dsQ + v/o0 = (1 + 0 )sln^;------ ---------- (2 )K = [ [ zdA - Ev1 / / z2dA ; (3)

V j k i - W p 0 ) / J J p , t Tz^50')N = E f f dA - Ey'i f zdA (4)

J j a" r - (z/^o) * J J k i - (z!P a )p* + (dV/ds0 )/(H^0 ) + N(y' + 1 / ^ ) = 0; (5)

1 +

CHAPTER 3FORMULATION OF THE LARGE DEFLECTION PROBLEM

FOR CIRCULAR CANTILEVERED .ARCHES

IntroductionThe governing equations which were, derived In

the previous section are in general quite difficult to solve. In certain specific cases namely the cantilever beam or simply supported beam loaded with concentrated forces or couples, solutions of the nonlinear equations have been found in the form of elliptic integrals (3),(4), (5).

The method presented herein will cover the cases mentioned above, but will not be limited to these types of problems only. This procedure is capable of analyzing the general case of the cantilevered arch and is not restricted to circular arches and cases in which only con- cfffbrat'ed forces and moments are applied. Also, it

3. Barten, H.J., "On the Deflection of a CantileverBeam, " Quarterly of Applied Mathematics, vol. 2, 1944, vol. 3, 1945.4. Bisshopp, K.E. and Drucker, D.C., "Large Deflectionsof Cantilever Beams," Quarterly of Applied Mathematics, vol. 3, 1945.5. Conway, H.D., "Large Deflections of Simply Supported Beams, Philosophical Magazine, vol. 38, 1947.

13

14should be noted that by making use of the property of symmetry, this method can be extended to specific cases of full rings and simply supported arches.

For the purpose of illustration, the problems dealt with in this thesis will be limited to structures of circular shape. Also, since most structures undergoing large deflections must be quite thin in order that -stress remain within the elastic range, it will be as- . sumed that the ratio of thickness (t) to the radius of the arch (R) is very small,Transformation from the General Equations

The fact that the system in question is a cir- n u l W arch makes it convenient to use polar coordinates, . Hence all dotted quantities will represent differentiation with respect to the position angle (9). See Figure

*Referring to Figure 4, 0 denotes the position

angle of the arch measured from the free end; R represents the radius of curvature of the undeformed arch, VQ, MQ, and N0 are the stress resultants at the free end; q* is the resultant of the loadings p^ and p on the deformed -system; and ^ is the total angle of the arch.

Putting the governing equations in terms of polar coordinates, and noting that since t/R / 1/50, where tdenotes the thickness of the arch, it is permissible to to neglect the term z/R in equations (4) and (5) since it

is a quantity much less than unity. Thusf f zdA = [ f zdA;JJk 1 - ( z / R ) ' 'A

A study of these integrals results in some simplifications. It should be noted that the first integral is the first moment of the area of the cross-section. In the case of a thin arch, if z is measured from the neutral axis, this integral will be identically zero. The second integral is the moment of inertia of the cioss-section; and the third integral represents the area of the cross-section.

CIRCULAR CANTILEVER ARCH UNDER ARBITRARY LOADING

Undef ormi

FIGURE A

16These facts make It possible to write equations

(3) and (4) as follows:M = -(El/R)(dy/d6) = -(EI^)/R; (3b)

and N = E0A. ---------- (4b)Changing the equilibrium equations to polar

coordinates, these can be written as follows:p*(1 + 0 ) + V R - NM/EI + N/R = 0; (5b)p*(W

17case might be referred to as the statically determinate system. In other words, the equations of equilibrium can be solved independently of the displacements or rotations. This case occurs when the loading is not a function of the rotationy . In a physical sense, this system exists if the loading on the undeformed arch remains normal and tangential to the arch after deformation.

With these restrictions, it is now possible torestate the boundary value problem. Eliminatingfrom equations (5b), (6b), and (7b) with the use ofequation (4b) enables the following system of equationsto be written:

dV/de = KMN/BI - (qR/EA + 1)N - qR; -------(5c)dN/dO = -RMV/SI + V; (6c)dM/dQ = VR + RVB/EA; (7c)

Where pn = pn = q = constant;*and pt = Pt = 0.

Note that the tangential load is eliminated only for the purpose of Illustration of solution.

The boundary conditions are:At 0 = 0, V=V0 ;

N=No ;and M=Mq .

The method of solution of these equations which was used in this paper was the fourth-order Runge-Kutta

18method.

After solving these equations, and therefore knowing the values of M, N, and V along the arch, equation (3b) can be solved for the rotation of the cross-section by numerical Integration using the condition that when 0 = the rotation will be Identically zero.

Alien these steps are completed, the displacements can then be calculated. Referring to equations (1b) and (2b), and eliminating 6 Q with equation (4b), the following expressions can be written:

dv/dO = w - R + R (N/EA + 1 )cosy; -----------(1c)dw/dQ = R (N/EA + 1 )siny - v; (2c)

Since the rotation and axial forces are known at this point, it is possible to calculate the tangential and normal deflections by numerical integration with the conditions that at the fixed end.(Q = ) , w =v = 0.In summary, it is seen that the solution to this case consists in solving three separate but dependant problems.

The method just presented is quite limited in that it deals only with cases in which the load remains in the same direction with respect to the arch after deformation.Loading Dependant on Rotation of Cross-sectlon(Case 2 )

With a slight change in the procedure, the system of equations can be solved for a larger variety of problems, one of which is the case when the loads

19remain in their original directions after deformation,This makes it possible to treat problems which Include the case of dead (or gravity) loading, which is common in engineering application.

It should be noted that if equations (3b), (5b),(6b), and (7b) are solved simultaneously, it will be /possible to approach problems in which the loading is a function .of the rotation .of. the cross-section,. This procedure presents a problem in that to apply an Initial .value method such as the Runge-Kutta method, the boundary . conditions at.the free end must be known. However, the rotation at the free end may be assumed at the start and then adjusted so that the condition that^ = 0 is satisfied at the fixed end, When this section of the problem is solved, the same method as outlined previously may be utilized for the calculation of the displacements,

In the next chapter, a computer solution of both methods will be outlined. The former case will be a direct form of solution, while the latter is an iterative process,

CHAPTER 4DIGITAL COMPUTER SOLUTION OF THE PROBLEM

' The Method o f Rurxge-Kutta

Application of the method of Runge-Kutta, even in the simplest of casesP is extremely tedious unless Calculations can be carried out in an automatic manner vSueh as that afforded by electronic digital computer . .facilities. The computer used for these problems dealt "- Vsfi'th in this thesis was the IBM 7074. .

The Runge-Kutta method is an algorithm formulated to..approximate the Taylor series solution. The approx- ..'Imations to the functions being integrated are obtained :'through several evaluations of the expressions for the . A^lrst derivatives. Although the best known and most ;widely used Runge-Kutta process is the fourth-order pro-. r.cesSg wherein four evaluations of the first derivatives "are used to obtain agreement with the Taylor series sol-

Autlon through the terms of order .A, lower order Runge- Kutta schemes are well known. The principle of the method is presented in detail in numerous publications on numerical integration techniques (6 )$,(?) and will

6. Levy and Baggott, Numerical Solution of Differential Equations. Dover Publications IncphN,Y. 9 '1950.

20

' 21not be presented here because of its availability in literature.

It should be noted, however, that the application; of the Runge-Kutta scheme requires that the problem be of an Initial value nature, and that the system of equations to be solved be a group of first order differential equa-. lions. In both cases described previously, these con ditions hold true. In the first case, the stress result: ants at the free end correspond to initial conditions , :and the second case provides an additional initial con ditlon with the assumed end rotation Application to Oase 1

The Runge-Kutta method was applied to both : in / .the previous, section in separate programs. In the case of loading remaining normal after deformation, the; /program was direct. In other words, no iterative process.; - Was necessary. . . The flow diagram and corresponding com iputer program are given in Appendix A..

The program was written in the following manner. SirSt, the quantities describing the arch, boundary eon c -ditions, and increments of integration were read into the/ computer. Next, the stress resultants were calculated by ::|tungeICutta and were then used to compute the rotation of . r ih e crossection. The next step was to use the calculated

7. Ince, E,.L. .Ordinary Differential.Equations0 DoverPublications, N.Y., 1926.

22

values to solve for displacements. It should be noted that only the first two operations must be changed to provide for the case in which the loading is a function of the rotation. In both cases, the last step, that is, calculation of displacements, is identical.Mathematical Check for Case 1

As no literature was found to be available for comparison to this case, one must rely on the validity of the second case for which comparisons were found.

Using small deflection theory as a check on mathematics, an arch of the following dimensions and properties was utilized:

R = 50 ft;ip = 1.2 radians ;EI = 180 x 106 lb-ft2 ;A = 0.5 ft2;

Mo = No 3 vo = q = 60.0 lb/ft. of arc length;Increments of Integration = 40.

The free end deflections for both methods were as follows:FREE END DEFLECTIONS

Runge-Kutta Method Small Defl. Theory0.4322 ft. 0.4068 ft.

A second comparison was carried out with an arch of thefollowing dimensions and properties:

R = 60 ft;

2j= 1.2 radians;

EI = 180 x 106 lb-ft3 ;A = 0.3 ft2; q = 0.0 lb/ft.;M = N_ = 0.0 o oVQ = 2000.0 lb.;Increments of integration = 20

The free end deflections were as follows:FREE END DEFLECTIONS

Runge-Kutta Method Small Defl. Theory1.879 ft. 1.885 ft.

Although there was good correspondence with small deflection theory, the burden of the proof for the validity of this method will have to rest with the comparisons applied to the next case.Application to Case 2

The second case considered was that in which the loads remain in their original directions after deformation. It will be shown later in this thesis that the use of symmetry will permit extending 'this method to solve certain ring problems and also the case in which a simply- supported arch is acted upon by a dead load and a vertical concentrated force.

In this case, equations (3b), (4b), (5b), (6b), and (7b) were solved simultaneously. To make this a complete boundary value problem, it was necessary to assume

a rotation at the free end. Then the equations were Integrated and the rotation at the fixed end calculated. Since it was required that no rotation occur at the fixed end, the assumed free end rotation was corrected by adding or subtracting some percentage of the residual fixed end rotation to or from this value. This procedure was followed until the fixed end rotation was small enough so that zero rotation was assumed to exist at that point. In this study, a value of 10^ radians at the fixed end was chosen as a limit of rotation.

The flow diagram and corresponding program for this problem is given in Appendix B .Relation Between Vertical End Load and Stress Resultants

It should be noted that if the arch is acted upon by concentrated loads at the free end, these loads will be a function of the rotation at this end. Hence the vertical force at the free end was broken into the following shear and normal forces: (See Figure 5)

v0 = -Pcosty-yo );N0 = Psln(-i|

25

-V

Undeformed Shape

FIGURE 5CANTILEVERED ARCH ACTED UPON

BY VERTICAL END LOAD

FIGURE 6RESOLUTION OF TANGENTIAL AND NORMAL DISPLACEMENTS

INTO HORIZONAL AND VERTICAL COMPONENTS

26This transformation Is shown In Figure 6. The relationship between these displacements is as follows:

Y = wcosty-G) - vsinty-Q); and X = wsinty-O) + vcos fy-Q),where X is the horizonal displacement, positive to the right, and Y is the vertical displacement, positive downward.

Another interesting problem which can be approached with this iterative technique is the case when the free end rotation is a known quantity ( i.e. a boundary condition ); and one of stress resultants at the free end is an unknown. In this case, the unknown stress resultant must be adjusted each time to enable the fixed end rotation to approach zero. This is the case in a symmetrically loaded ring, where one quadrant may be treated as a cantilevered arch, provided that the equilibrium of this segment is satisfied in the transformation. This case will be treated in the next section.

CHAPTER 5 COMPARISON TO KNOWN RESULTS

The Method of Seantes and ConwayThe comparison which will now he presented Is

.based on a work by A.E. Seames and HD ConWay (8). v/Jn this article, a numerical method was presented which Was found to correspond well with certain exact solutions. .(9),(10). The method of Conway and Seames consists in assuming that the deformed arch can be broken into a series of circular elements which are then analyzed sue- . cessively for equilibrium, beginning with the free end.

Three of the examples given by Conway were used to.compare with the Runge-Kutta method of solution. The first example consists of a circular cantilever beam with a concentrated vertical end load; the second deals with a circular cantilever acted upon by a load distributed uniformly along the arc length; the third case considers a ring acted upon by diametrically opposing

8., Seames, A.E. and Conway, H.D., "A Numerical Procedure for Calculating the Large Deflections of Straight and Curved Bars, Journal of App. Mech., vol. 24, 1957%9. Conway, H.D., "The Nonlinear Bending of Thin Circular Rods," Journal of App. Mech., vol. 78, 1956.10. Sonntag, R., "Die Spiralringfeder," Ingenieur- Archiv, vol. 14, 1943.

27

28loads.Comparison /? 1

In the first case, the parameters of the system are as follows:

R = 10 In.;P/EI = 0.05 In-2;'4' = 0.427 radians;Assumed end rotation = -0.373 rad. (Results of Conway); Increments of integration = 10;A = 0.00625 In2 .

It took seven iterations to cause the rotation at the fixed end to converge to a value less than 10"^ radians. ' The adjustment formula used to correct the rotation was as follows:

New free end rotation = Old free end rotation -Residual fixed end rot.

The correction factor, in this case ( - Residual fixed end rotation), must be changed when a different type of loading is applied. This Is illustrated in the next example. Comparison /2

In the second comparison, the dimensions and properties are as follows:

R = 10 in; q/EI = 0.05 In-3;\|/= 0.509 radians;A = 0.00794 In"2;Assumed end rotation = -0.691 rad. (Results of Conway);

29Increments of integration = 10.

The computer program was then set up with an absolute limit of 75 iterations. It was found that using theprevious adjustment formula, convergence to a residual fixed

=4.end rotation less than 10" radians was quite slow, and in fact did not reach that value through seventy-five iterations . The adjustment equation was then changed to the following:

New free end rotation = Old free end rotation -(Residual fixed end rot)/2

Making this change caused the rotation at the fixed end .to converge to its limiting value In 3 iterations. ;, : The results of comparisons 1 and 2 are shownin Figures 7, 8, and 9. Figure 7, Case 1 shows the de- ..flection curves for the case of the concentrated endload. Figure 7 Case 2 represents the deflection curves-for the case of a uniform load per unit arc length. In .loth eases, a close agreement was found between the twomethods,

In Figures 8 and 9, the ratio M/BI and the deformed radius of curvature R* are plotted against the . position angle 0 for both cases using the methods of Conway and Seames and the Runge-Kutta integration. A close correspondence between both methods is also observed here. Comparison #3

The last comparison makes use of the property of

30FIGURE 7

COMPARISON OF METHOD OF CONWAY AND SEAMES TO RUNGE-

KUTTA INTEGRATION

ELASTIC CURVES

Undeformed Shape

Deflected Shape

10 InSCALE (Concentrated Load at Free End)

Undeformed Shape

In CASE 2(Uniform Load/Unit Arc Length)

Deflected ShapeA Conway-SeamesO Runge-Kutta

0) Xi o d'6.0Pd

iW*H>gO2sM9Pd

.0

.0

w

sI

e-t55

sHA55Wro

>494MAMC5MPd

9XpoAP4

0 1

.20

R vs

o-A- Runge-Kut'.aGonway-SeamesFIGURE 3

Bending Momen'- and Raddue of Curvature

vs .Position Angle

(COMPARISON- Cantilever- ed Arch with Vertical End

M/EIvsO

Load)

(Radians)POSITION ANGLE 0

vj

RADIUS

OP CURVATURE

- R

- (i

nche

s) 8.0

6.0

4.0

2.0

SCI

EH%

sgAgPQ

EHHPH0)H

.4

.3

.2

.1

\ ix7

r vs e

< /

r oA

Runge-KuttaConway-SeamesFIGURE 9

/ \u c n u x ju g , n u iu e u u emu.Radius of Curvature

vs.Position Angle

/k (COMPARISON- Uniform- X. ly Loaded Cantilever-

M/El v s o ----- v VA

id Arch)

\ 3

POSITION ANGLE - 0 - (Radians)

V4ro

sym m etry i n t h e cas e o f a r i n g a c te d upon by d i a m e t r i c a l l y

o p p o s in g f o r c e s . I s o l a t i n g o n e - q u a r t e r o f th e r i n g , as

shown i n F ig u r e 1 0 , i l l u s t r a t e s t h a t th e r i n g can be b ro k e n

i n t o f o u r i d e n t i c a l c a n t i l e v e r p r o b le m s . To s a t i s f y e q u i l

i b r i u m , th e c a n t i l e v e r must have a v e r t i c a l end lo a d o f

o n e - h a l f t h e m a g n itu d e o f th e d i a m e t r i c a l l y o p p o s in g l o a d s .

At t h e f r e e end , by sym m etry , r o t a t i o n must be r e s t r a i n e d

(yQ te 0 ) . The o n l y unknown w i l l be th e b e n d in g moment a t t h e f r e e e n d . H e n ce , t h i s i s a t r i a l and e r r o r p r o c e s s ,

w i t h th e b e n d in g moment a t t h e f r e e end as th e q u a n t i t y

w h ic h m ust be a d ju s t e d a f t e r t h e i n i t i a l g u e s s .

2P

A x is o f Symmet r y

2P iA t f i x e d >

e n d , i w=v=p*=0

A x is o f Symmet r y/ o =

FIGURE 10

REDUCTION OF RING PROBLEM TO 0 ANTILEVERED ARCHES

34The exam ple f o r c o m p a r is o n i s a r i n g o f th e f o l

lo w in g d im e n s io n s and p a r a m e t e r s :

R = 10 i n . ;

E I = 3 2 .7 8 7 l b - i n 2 ;

P = 0 .3 2 7 8 7 l b . ;A = .01 i n 2 ;

Aj/= 90Assumed f r e e end b e n d in g moment = 1 .0 i n - l b (C o n w ay );

In c r e m e n ts o f i n t e g r a t i o n = 1 0 .

The d e f l e c t i o n c u rv e s o b t a in e d by b o th m ethods a r e

shown i n F ig u r e 1 1 . E x c e l l e n t a g re e m e n t i s a p p a r e n t .

The a d ju s t m e n t e q u a t io n was as f o l l o w s :

Mnew = Mo0 l d + ( R e s id u a l f i x e d end r o t a t i o n ) / l O .

T h is f o r m u la was a r r i v e d a t a f t e r u s in g t h e same e x p r e s s io n ,

o n ly w i t h a minus s i g n i n f r o n t o f th e c o r r e c t i o n t e r m . I n

t h i s c a s e , due to th e n e g a t i v e c o r r e c t i o n te r m , th e i t e r

a t i o n p r o c e d u r e was fo u n d t o d i v e r g e . As i t i s q u i t e d i f

f i c u l t to p r e d i c t t h e b e h a v io r o f th e r o t a t i o n a t th e f i x e d

end w i t h r e s p e c t t o a change i n t h e f r e e end r o t a t i o n o r

s t r e s s r e s u l t a n t s , i t i s u s u a l l y n e c e s s a r y to assume a c o r

r e c t i o n f o r m u la and s tu d y th e r e s u l t s o f t h i s f o r m u la f o r

each i t e r a t i o n . These r e s u l t s w i l l show w h e th e r t h e assumed

f o r m u la causes d iv e r g e n c e , o r p e rh a p s s lo w c o n v e r g e n c e . As

a r e s u l t , th e a d ju s t m e n t e q u a t io n may be changed t o compen

s a t e . F u r t h e r d i s c u s s io n on t h i s t o p i c i s t a k e n up i n th e

f o l l o w i n g c h a p t e r .

35

FIGURE 11 i 2P = 0 .6 5 5 7 4 l b .

(COMPARISON - R in g w i t h TD i a m e t r i c a l l y O pposing F o r c e s )

ELASTIC CURVES

10 I n . D e f le c te d Shape

U ndeform ed Shape

SCALE

O A

2P = 0.65574 lb. T O R u n g e -K u t taA Seames-Conway

sb of the Runge-Kutta Approach In generals, the Bunge-Kutta approach seems to

have numerous advantages over the procedure used by Seames and Conway. In the latter method, it is necessary to assume a deflected shape as well as end rotations Also, the case of double curvature or the case where the curvature changes rapidly would make this approach extremely tedious to apply * If a small enough increment of integration is used, the Runge-Kutta approach will not be affected by a large change in curvature Also, the systematic nature of the method of Runge-Kutta enables this procedure to be easily programmed for solution on the digital computer.

CHAPTER 6 APPLICATIONS

Application to Rings and Simply Supported ArchesCertain problems by the use of symmetry, may

be reduced to the problem of the cantilevered arch.This simplification, as illustrated previously, arises in the case of the simply supported arch and certain ring problems. By the use of symmetry, it is possible to use half of a symmetrically loaded simply supported arch as the mathematical model for analysis. Since in the case of symmetrical loading the rotation at the center point of the arch is zero, this problem is, in effect, one of a cantilevered arch if the loadings correspond to that of the complete structure, and if the center point is used as a reference point for displacements.This is shown in Figure 12. Case 1 represents a simply supported arch acted upon by a uniform load per unit arc length. Case 2 is the same arch acted upon by a concentrated load at mid-span.The Correction Equation

Using the cantilevered model and the Iterative computer program outlined in the previous sections, it is possible to solve both of these cases. In doing this,

37

38

T h is i s i d e n t i c a l to th e f o l l o w i n g :

2P

T h is i s i d e n t i c a l to t h e f o l l o w i n g :

CASE 2

FIGURE 12

REDUCTION OF SIMPLY SUPPORTED ARCH TO CANTILEVERED ARCH(S y m m e tr ic L o a d in g )

the biggest apparent difficulty was the-selection of the proper assumed rotation and adjustment formula.It should be noted that a solution of equations (3b)s (5b), (6b), and (7b) is not necessarily unique. To prove that this is true would be extremely difficult due to the nonlinearity of the equations involved,

vHowever, this property was observed in attempting to . analyze the cases which will soon be illustrated. It .was found that if the assumed rotation or correction (.term was too large, a deflection curve could result phieh did not correspond to the case which should occur.;:(for this type of loading. That is, the curve found dif-' fered from any elastic curve found in the case which physically occurs by increasing the magnitude of the

vloading from the initial state. For these cases, it was found that an adjustment formula which used only small percentages of the residual fixed end rotation was best suit ed to begin the iteration. Also, an initial guess of zero for the free end rotation was found to work well in most cases. Since no direct procedure for finding the correction equation was apparent at this time, it was necessary to arrive at an adjustment relationship through a trial and error process.

After assuming an equation, the program was put into the computer with a reduced limit on the number of iterations,, say ten. At both ends, for each iteration, the rotations and stress resultants were printed in the output

40

data. If the rotation at the fixed end diverges continuously s it is possible that the sign in front of the correction term must be changed. If the rotation at the fixed end seems to converge and then diverges, it is possible that the incremental change or correction term is too large. When the fixed end rotation seems to be converging to zero, the convergence' can usually be speeded Up by increasing the magnitude of the correction term. This should be small enough to keep the solution from converging into a deflection pattern which is physically Improb- ; able/bxsMbles: Simply Supported Arch

The following two examples illustrate the large deflection patterns of a simply supported arch. In both examples, an arch of the following dimensions and parameters was selected:

R = 100 ft.;^ - 1.0 radians;El = 405 x 106 Ib-in2;A = 18 in^;E = 30 x 106 lb/in2sincrements of integration = 10.

In Case 1 (See Figure 12.), the arch was analyzed for the following magnitudes of .uniform loading:

1) q = 0.25 lb/in.?2) q = 1.0 lb/in.;3) q = 2.5 lb/in.

FIGURE 13DEFLECTION CURVES FOR SIMPLY SUPPORTED CIRCULAR ARCH

ACTED ON BY UNIFORM DEAD LOAD (q )

42The e l a s t i c c u rv e f o r each m a g n itu d e o f lo a d i n g I s i l

l u s t r a t e d I n F ig u r e 13 . The number o f i t e r a t i o n s and

c o r r e s p o n d in g c o r r e c t i o n fo r m u la e f o r each l o a d i n g w ere

as f o l l o w s :

q ( l b / i n . ) C o r r e c t i o n F o rm u la No. o f I t e r a t i o n s

2*5 /knew = A old - /n/20 52U 0 /onew = /oold - /n/10 56-25 /Onew = /oold - /n/'0 64

^ r e f e r s to th e r e s i d u a l r o t a t i o n a t th e f i x e d en d .

I n a l l c a s e s , th e c o r r e c t i o n te rm s c o u ld have been

in c r e a s e d to in s u r e a more r a p i d c o n v e rg e n c e .

Case 2 d e a ls w i t h th e s i t u a t i o n w here th e a r c h

was a c t e d upon by a c o n c e n t r a te d lo a d a t m id s p a n . The

m a g n itu d e s o f th e s e f o r c e s w ere as f o l l o w s :

1 ) 2P = 1000 l b . ;

2 ) 2P = 400 l b . ;

3 ) 2P = 100 l b .

The r e d u c t i o n o f t h i s case to a c a n t i l e v e r a r c h p ro b lem i s

i l l u s t r a t e d i n F ig u r e 12 , Case 2 . The e l a s t i c c u rv e s

f o r th e s e m a g n itu d e s o f lo a d i n g a r e shown i n F ig u r e 13 .

The c o n v e rg e n c e d a t a c o r r e s p o n d in g to t h i s exam ple a r e

shown b e lo w .

2P ( l b . ) C o r r e c t i o n F o rm u la No. o f I t e r a t i o n s

1000 / n e w = / o 0 i d ~ M l / 5 . --------

4 0 0 / o new = / o 0 i d - / n / 5 . 44

100 A n e w V o old " / n/5. 32

2P

Undeform edShape 2P = 100 l b

2P = 400 l b

SCALE

1" = 300

E = 1200 i n .

2P = 1 k .

FIGURE 14

DEFLECTION CURVES FOR SIMPLY SUPPORTED CIRCULAR ARCH ACTED UPON BY CONCENTRATED LOAD AT MID-SPAN

vj

Examples: RingRings acted upon by symmetric loads are easily

analyzed. The two cases presented are as follows: 1) A ring with diametrically opposing concentrated loads;2) A ring acted upon by opposing uniform loads each acting on one-half of the ring. The ring. In both cases, has the following dimensions and properties:

R = 10 In.;EI = 32.787 lb-In2 ;B = 30 x 106 lb/ln2 ;A = 0.005259 In2.

The reduction of the full ring to a cantilever problem is shown in Figure 14. As was illustrated in . a previous example, the moment at the free end must be adjusted for convergence of the fixed end rotation to zero.

Both cases were subjected to various magnitudes of respective loadings. In Figures 15 and 16, the variation of the deflections d on the vertical diameter of the ring are plotted against the loading. The load anddisplacement parameters were found with the use of theBuckingham Pi Theorem which yielded the following functional relationship:

d/R = d/R( q/ER, R^/l, A^I ) for the uniform load q; and d/R = d/R( P/ER2, R4/l, A2/l )for the case of concentrated loads.

This can be reduced to: q

M0 (Assumed)=- (Hirq )/2

UNIFORM LOAD

This can be reduced to:

/o = (M (Assumed) ^

JN0 = -P/2CONCENTRATED LOAD

FIGURE 14REDUCTION OF RING PROBLEMS TO CANTILEVERED ARCHES

46

In both cases9 the rate of deflection with load Increases initially. This is due to a flattening of the ring which reduces the stiffness of the system. At a deflection d of approximately three-quarters of the radius 0 the ring reaches its most flexible configuration.From then on an increase in loading will tend to increase the stiffness due to the more significant load transferring effect of the axial forces.

It is Interesting to note that when the deflection tatio d/a of the ring approaches unity the ring should exibit the stiffness characteristics of a fixed end beam. This correspondence is illustrated in the following discussion.

The horizonal expansions of the rings were found %:;to be 8.98 in. for the uniformly loaded case* and 7.98 in. for the case of diametrically opposing concentrated loads. /Ifsing these values, the effective lengths of a fixed end beam were assumed to be 2.89R for the uniform load, and 2,8011 for the concentrated load, where it = 10 in.Observing the slopes of the load-deflection diagram, the following load-deflection ratios were calculated:

q/d = 5.04EI/R4 (Uniform Load)p/d = 9.15EI/R'. (Concentrated Load)

The load-deflection relationships for fixed end beams are as follows:

q/d = 384EI/14 (Uniform Load)

47P/d = 192EI/13 (Concentrated Load)

1 is the length of the fixed end beam.Using the effective lengths previously calculated,

q/d = 5.44EI/R4; and P/d = 8,?6ei/r 3.

(Uniform Load) (Concentrated Load)

These results should not be expected to correspond exactly since the actual length of the ring segment (fvR) may increase the effective length; and, conversely, the curvature of the deflected segment may cause the effective length to decrease through the effect of axial load transmission. Still, a close * correspondence is evident.

(q/ER) x

10

(Dim

ensi

onle

ss) 5

4.8120 x 10

FIGURE 15LOAD-DEFLECTION RELATIONSHIP (For Ring with Opposing Uniform Vertical Load/Unit Arc Length)

0 86.2 1 .0d/R (Dimensionless)

(P/ER

) x

10 (D

lmen

sion

less

) R = 0.9153 x 10

2.0

FIGURE 160.5 LOAD-DEFLECTION RELATIONSHIP

(For Ring with Diametrically Opposing Forces)

0.60.0 0.2 0.8d/R (Dimensionless)

CHAPTER 7FINITE DEFLECTIONS OF FRAMES AND TRUSSES

IntroductionThe preceding chapters of this thesis dealt with

the large deflections of a rather limited range of structural systems. On. the other hand, the deflections imposed on the system did not limit the applicability of this method. Using matrix theory, a method can be developed which is more general in relation to the type of system which may be analyzed, but is limited with respect to the magnitude of the displacements encountered. Recent works pertaining to this method usually come under the heading of "Finite.Deflection Theory," The basic idea of the theory is to impose conditions of equilibrium and compatibility on an assumed deformed shape. Since the correct deformed configuration is unknown, an iterative process is necessary. Hence, due to the number of mathematical operation^ necessary,'-this method of analysis 'has been limited to the present decade, during which time the use of matrices in conjunction with digital computers has become widely available.

There are various approaches to the problem. A method of applying loads incrementally to arrive at finite deflections by a series of lineal analyses has been

- ,

suggested by Wilson (11). Richard (12),(13) and Gold- /berg (13) treated the equations developed to describe the force distribution and deflection pattern for nonlinear structural systems as an initial value "problem. The effect of finite deflections on the elastic stability of 'frames was presented by Horne (14). Other methods compare the vector of actual loads to a load matrix calculated on the basis of an assumed configuration. The correct configuration is then approached by iteration*- Articles utilizing this approach have been presented by Saafen and Britton (15),(16).

A finite deflection analysis was presented by

11. Wilson, E.L., "Analysis of Frames with Nonlinear Behavior," ASQE Journal of Engineering Mechanics Division, vol. 86, Dec. i9 6 0.12. Richard,R.M., "A Study of Structural Systems Having Nonlinear Elements," Doctoral Dissertation, Purdue Univ.,1961 o

13. Richard, R.M. and Goldberg,J.B., "Analysis of Nonlinear Structures," ASOE Journal of the Structural Div.,vol. 8 9, Aug. 1963.14. Horne, M.R., The Effect of Finite Deformations in Elastic Stability of Plane Frame," Proceeding of the Royal Society, Series A, London, Feb. 1962.15 Saafen, S.A., "Structural Plane Frameworks," Doctoral,Dissertation, Manchester Univ., Manchester, England16. Saafen, S.A. and Britton, D., "Elastic FiniteDeflection Analysis of Rigid Frameworks by Digital Computers, " International Symposium for the Use of Electronic Computers in Civil Engineering, Oct. 1962, Lisbon, Portugal

52

Saafen (17) at the recent ASOB Conference on Electronic Computation, This consisted of a matrix analysis of frames and trusses exibiting nonlinear geometric properties, and included stability effects as well as the ef-..,feet Of bowing of the structural members. In this article, no mention was made concerning the limitations which must be imposed upon this method.

It was the purpose of this chapter to define the term finite deflections by investigation of the limit- ations upon which this theory is based. Another important aspect of the following presentation is that a basic computer program for linear theory will be changed only slightly to enable finite deflections to be treated.Method of Analysis '

In this analysis, the displacement method for indeterminate structures is utilized. That is, the dis- . placements will be treated as unknowns,Matrix Formulation

The following discussion summarizes the system of matrix equations used in the linear analysis of a structural system. ' "

1) The compatibility equations which relate the generalized displacements x to the element boundary dis-

17. Saafen, S.A., "Nonlinear Behavior of Structural Plane Frames, Journal of the Structural Division, ASOB, vol. 89, August 1963.

53placements y . The relationships between these quantities is illustrated in Figure 17, which represents a general element ij.

Letoc be the angle of orientation, yj denote the element elongation, y2 be the rotation of the element at 1 , y^ be the rotation of the element at j, xj denote the generalized horizonal displacement at 1 , x2 denote the generalized vertical displacement at 1 , x^ denote the generalized rotational displacement at 1 , and x^, x^, and xg equal the corresponding generalized displacements at j . From these definitions, the compatibility relationships for an element can be written as follows:

> v cosoc sin* 0 cos (oc+tt) slnfa+Tr) 0 fxilsina/l -cosa/l 1 sln(o(+7r) ' -COS (

54In more compact form,

y = ax = bTx, ---------- (1 )where = the transpose of b.

2) The equilibrium relationships are:P = bF, (2)

where P is the column vector of external loads associated with the generalized displacements. F denotes the element boundary forces associated with the boundary displacements.

5) The stress-strain ( force-deflection of moment- rotatlon ) relationships are:

F = ky, (3 )where k is a square matrix relating boundary forces toboundary displacements. To comply with the chosen boundary displacements, equation (3 ) may be written in the following form:

5 ' AE/1 0 0 = 0 4EI/1 2EI/1

,*2j 0 2EI/1 4EI/1where S is the axial force in the member 1j, Mj is the internal bending moment at 1 , Mg is the internal bending moment at j, E is the Modulus of Elasticity, A is the area of the crossection of the member, and I is the moment of inertia of the crossection of the member.

Substituting (1) into (3) yieldsTF = kb x.

Then substituting (4) into (2) yields(4)

55P = bkb^x.------------------------- -------- (5)

L i n e a r T h e o ry

To compute deflections using this method, the bkb - matrix is constructed for each element. Then the individual bkb^ matrices are arranged in a manner in which the coefflcents of these matrices correspond to the displacements x and the external loads throughout the entire structure. This matrix will be referred to as the master stiffness matrix K.

The boundary conditions are then imposed by eliminating the columns and rows in K which correspond to the external restraints. The matrix resulting from this operation will be referred to as K.Thus,

P = K x . ( 6 )

Inverting K, the generalized displacements can be solved for with the equation

x = K ~ 1P,- 1where K is the inverse of the matrix K.

N o n l i n e a r A pproach

The procedure thus outlined is perfectly valid provided that the deflections remain small enough so as not to change the geometry appreciably. In order to change this method to make it applicable to finite deflections, it is necessary to make certain changes in the matrix formulation. Before doing this however, certain assumptions

A :v.56:will be made;

1) The material In question Is of a linear elas- itlc nature; 2) the structural system remains in a stabler ..configuration throughout deformation; 3) the members obey tfiei simple theory of bending; 4) stresses, in the material,., remain within the elastic range; and 5) the bowing, effect; W-'the members in flexure is negligible. Besides these rassumptions, the method presented was limited to small rigid body rotations of the members. This point will be illustrated later.

The governing equation for a system in its equilibrium position under a set of applied forces is written as follows:

P = K 6 x,where P is the column matrix of loads on the structure, and x is the generalized displacement matrix corresponding to the external loads. The purpose of the K e or master stiffness matrix is to relate the loads to these displacements, As in small deflection theory, the K 9 patrix is constuoted using, the bT matrix, or compatibility matrix, and the stiffness matrix, k. For structural elements obeying the simple theory of bending, the k matrix remains a constant of the system. Hence it is the compatibility relationships which must be revised to permit a system to be analyzed for finite deflections.

57Limitations

Referring to Figure 18, the vertical and horlzonal projections of the element, Z and Y respectively, are as follows:

Z = l0slntx0 + x2 - and Y = -l0cosa0 + X4 - x1,where 10 Is the undeformed length; Oq is the position angle in the undeformed state; and x^, x^, x^, and x^ are the generalized displacements as defined previously.

Tx5 "

Z

FIGURE 18STRUCTURAL ELEMENT IN UNDEFORMED AND DEFORMED STATES

From these relationships, the elongation yj of a structural element is as follows:

y, = 10 (coscx1 coso

58where ex1 is the position angle in the deformed configuration.

If, as mentioned previously, the rigid body rotations of the individual elements are small, the above equation can be written as a linear combination of the generalized displacements.Thus,

y j = (xj-x^ )cos ex1 + (xg-x^jslno/. ---------(a1)In a similar manner, assuming small element rotations, the boundary rotational displacements can be written as follows:

y2 = (x1-x^)slnot1/l 1 + (x^-x2 )coso

59titles in the compatibility matrix refer to the deformed geometryMethod of Solution

Since 048 and I 8 are unknown quantities, the method of solution is of an iterative nature In essence, it'consists in satisfying the matrix relationship

P = K exThe method used in establishing this equality is

hs follows:1) Assume a deflected shape. This may be done

by .the use of small deflection theory,2) Establish a master stiffness.matrix correspond

ing to this deformed configuration,3) Using the assumed displacements and the master

stiffness matrix, check for equality by computing a load 'vector P#

4) Compare the load vector P* to the actual loads P and compute a residual, ( Pr = P. - P* ) .

5) Apply the residual load vector to the deformed structure and compute a new configuration.

6) Using this new configuration, compute P*, etc,7) Follow this procedure until the residual Pr

approaches zero or some limiting value,' depending on the accuracy required.

Due to the use of matrices, this method is well suited for solution with the aid of the digital computer.

60The computer program and flow diagram are shown In Appendix C. The main part of the program consists of a small deflection analysis. Only a few changes were necessary to convert this to a method capable of handling finite deflections.Compatibility Equations In Coordinate Notation

TIt should be noted that the b matrix Is computed using coordinates rather than trigonometric functions. Here, Z and Y refer to the vertical and horlzonal projections of the members as shown previously In Figure 18. Hence the compatibility relationships can be written as follows:

,y1) -Y/l1 Z/l' 0 Y/l' -z/i' 0y2 2/(1 ' ) 2 Y/(l ' ) 2 1 -2/(1 ' ) 2 -Y/(l' ) 2 0,y3, Z/(1 ' ) 2 Y/(l ' ) 2 0 -2/(1 ' ) 2 -Y/(l ' ) 2 1

f e )x5x4

K

The purpose of the coordinate formulation Is found In the shortening of computation time. Calculation of a trigonometric function in a digital computer Is carried out by a series expansion, which can appreciably affect the running time of the program.Examples and Discussion of limitations

The following examples illustrate the technique for computing finite deflections.

Example 1 consists of a triangular pin-jointed truss as shown In Figure 19. Using the method outlined

61above, the following deflections were computed:Xj = 0.0 in. Xg = 0.0 in. x^ = 15.226 in.

= 1.8?0 in. Xy = 2.739 in. Xq = 0.0 in.15#

log = 100 inEAp = 200#= 100 in. = 200#

Iq - 100 in 60 EA^ = 300,V 60

FIGURE 19EXAMPLE 1

In the deformed state, the lengths of the members were as follows:

1{ = 106.930 in; lg = 92.665 in;1^ = 102.738 in.

Prom these lengths, the axial forces in the members were computed, and from these forces, the horizonal and vertical projections of the joint forces were calculated. The lack of equilibrium in the structure under these loads is illustrated in Figure 20.

The error evident in the above example was due to the fact that the lengths of the members were calculated

62directly from the deformed configuration, while the forces determined from the computer program were found using the approximate compatibility relationship, and requiring the

*calculated load vector P to equal the actual load vector P.

8.455# 5.477#FIGURE 20

FORCES DUE TO ELEMENT ELONGATION

P, = 15#,12.371#| ----- - I 12.371#

'___ -2F = 0.000#.523# 5.477#'ZFH = 15.000#

12.371#

Z-F = 15.000# n (D -CPg = 0.000#9.523# 5.477# 5.477# 5.477#

FIGURE 21EQUILIBRIUM CHECK BY GEOMETRY

OF DEFORMED STRUCTURE (Calculated by Computer Program)

63These forces determined by the computer program are shown in Figure 21, It was observed that equilibrium was indeed present, but referring back to Figure 20, it was evident that compatibility was not. For an approximation to this already approximate method, the compatibility condition may be satisfied if the members are so strained as to bring the forces in the members (Fig. 20) into agreement with the forces calculated by the computer program (Fig. 21). This procedure is valid if the deformed configuration of the structure, and thus the calculated force pattern (Fig. 21), is not changed appreciably by these strains. In the above example, due to the. forces shown in Figure 21, the axial load in member #1 is 15.612 lb. Taking into account the length of the member strictly on the basis of the deformed configuration (Fig. 20), the axial load in member #1 is 13.860 lb. To achieve compatibility, it was necessary to strain member #1 additionally to raise the axial load to that determined by the computer program. Since EA/lo=2.0 lb/in., an elongation of member #1 of less than 1 in. raises its axial load to that calculated by the equilibrium through the geometrical configuration. Similarily, in member #2, the change in length to achieve compatibility is less than i in. Hence, it can be concluded that giving the elements these adjusted lengths does hot appreciably affect the calculated force pattern. Hence compatibility is attained; and the calculated forces and deflections are still quite accurate. The correspondence between the axial load due to geometrical pos-

64ition and that due to element strain should always he Investigated. As a rule, element rotations of greater magnitude than 10 degrees may cause considerable error in the analysis due to the fact that the term

10 [cos (ex' - txe) - 1] cannot be assumed to be negligible with respect to the displacements.

Example 2 is a rectangular frame as shown in Figure 22. In this case, the properties of the structural system were as follows:

(In-) El (k-ln2) AE(k)300 1,692,000 150,000150 1,692,000 150,000150 1,692,000 150,000300 1,692,000 150,000

Member1234

180 180 270P8WF17

1 2 ' - 6

8WF17

90

FIGURE 22EXAMPLE 2

10

65When P, = 40 kips and P = 40 kips, the dlsplace- 1 5

ments of the system were as follows:= 0 .1324 rad

Xy = 41.585 In.= 0 . 0 In.

Xg = 0 . 0 In.Xj = 0 . 0 In. Xg = 12 .972 in.

x^j = 5 .8 5 0 in. x = .0368 rad. x = 0 .0 in.

Xg = -0.0430 rad.x4 = 41.935 in.77 in.Due to these deflections, the equilibrium of the

x._ = 5.977 in.5 x 10 = 41.225 in.x 14 = 0 .0 in.x = 0 .0 rad 15

system can be represented by Figure 23. ? ' ' *k .786*

17.473!1839k1

.841k1738*22,520k*-

nrl) V1839k"

.786k

2114k

0520k

39.214k 39.269k2114k"

3332k"

17 47 3k

22.5

39.214k

4328k"22

841k FIGURE 23 39.269kEXAMPLE 2: EQUILIBRIUM DUE TO DISPLACEMENTS

Taking the new deformed shape of the structure into account, the following checks can be obtained by summation of forces at each joint. The joints were numbered with respect to the horlzonal generalized displacement at that point.

66Joint 4

Fh = 0 : 40.000 - 22 .52 0 - 17.473 = 0.007 k.Fy = 0 : 0.841 - 0 .7 8 6 = 0.053 k.

Joint 7F = 0 : 2 2 .5 2 0 - 22.524 = -0.004 k.HFy a 0 : 40.000 - 0 .7 8 6 - 39.214 = 0 .0 0 0 k.

Joint 10F = 0: 22.524 - 22.516 = 0.008 k.n '

Fy = 0 : 39.214 - 39.269 = 0.055 k.Hence, In this position, with this loading,

the structural system was found to be in equilibrium.Checking for compatibility, the deformed lengths

of the members are as follows:lj = 296 .998 In. 1 ' = 149.810 In.1 ' = 149.814 In. = 297.025 In.

These correspond to axial loads S of the following magnitudes :

5 1 = - 1501.00 k. S = - 190.00 k.52 = -186.00 k. S4 = -1487.50 k.

Here, as in the previous example, a poor correlation results. On the other hand, note that the product AE is 150,000 kips. Hence, for members #1 and #4,

(10/AE ) 1 4 = 0.002 in/kip.For members #2 and #3,

(I0/A E)2>3 = 0.001 in A l p .From these stiffness constants, it is seen that only a

small elongation will cause a significant change in the axial load Hence, making the axial forces caused by elongation and by equilibrium correspond will only require a small change in length which will do little to .alter the calculated geometry and equilibrium.

CHAPTER 8 CONCLUSIONS

The methods presented herein illustrate the potential problem solving capabilities of the digital computer . The operations which were utilized in this thesis would be much too lengthy and tedious to attempt manually. In both cases9 the computer storage necessary was well (within the limits of the IBM 7074 system.

It should be noted that the cantilevered arch analysis is not limited to circular arches alone. This 'analysis is applicable to any system for which the quan- (titles E Aj, R, or the loading parameters can be represented as a function of the arc length s. It is then onlynecessary to rewrite the governing equations which are again solved by the Runge-Kutta integration. Hence,(this method may be of considerable value in predicting.the behavior of rings which are acted upon by various soil . loadings which are of a symmetric nature.

To insure an optimum computer analysis, it is pf interest to investigate the relationship governing the convergence of the iterative - method for various states of loading.

The finite deflection analysis which was presentedappears to be somewhat limited, since large deflections

68

tisually entail considerable rotation of the members. On the other hand, this method is well suited for problems %:zi whlch the rotation of the members Is not significant..This usually occurs in frames in which the sidesway is As not a significant factor. It should be noted again tliat only a slight change in the computer program for small deflection theory is necessary to convert the prob- lini: to a finite deflection analysis.

By applying certain assumptions to the general governing equations for arches, it may be possible to develop a quite useful theory. From results generated in this work. It was found that in many cases, the axial strain.was quite small in comparison to unity. Therefore, one possible avenue of approach may be to neglect this strain when small in comparison to unity. Also, and improvement in small deflection theory may be possible by the assumption of small rotations of the cross-section. Investigation of these and other simplifications may result in theories well suited for a wide Variety of large deflection problems.

APPENDIX A

H o FLOW DIAGRAM (loads remain normal and tangential after

A) Definition of variables CM s bending momentON s axial force T ss position angle V shearing forceG =5 angle of rotation of crossection W = normal deflection VM = tangential deflection D = increment of integration F = BI = flexural rigidity A = area of crossection R = undeformed radiusN = 1 + the number of increments of integration Q = uniform load per unit arc length

B) Main flow diagram*

(sTART^

(over)

f The parenthesis behind a Variable refers to subscripts denoting points between increments of integration starting with (1) at the free end, and going to (N)'; at the fixed end.

70

71

LOOP J = 1 to N

LOOP L = 1 to N

LOOP L a 1 to N

(Flow diagram, cont,)

(over)

T ( 5 0 ) , V ( 5 0 ) , G ( 5 o ) , W ( 5 0 )

GENERATE G(L) from numerical integration of equation (3b) using the boundary cond. that G(N)=0, where

LET

GENERATE VM(L) and W(L) using the Runge-Kutta method applied to equationsusing the B.C. that W(N)=VM(N)=0, where

and (2c)

GENERATE CN(J),CM(J), V(J),T(J) using the Runge-Kutta method applied to the equilibrium equations and boundary(initial) conditions, where

(Flow diagram, cont.)72

PRINT T(I),CM(I),CN(I), V(I),G(I),VM(I),W(I), where I = 1 to N.

n n n

n73

2 . OOMPUTER PROGRAM (FORTRAN C om puter L an g u ag e )( lo a d s re m a in n o rm a l and t a n g e n t i a l a f t e r d e fo r m a t io n )

LARGE DEFLECTIONS OF CANTI LEVERED ARCH CM-MOMENT CN^AXIAL FORCE, V = SHEAR, T=ANGl E ,

' GaUNIFORM LOAD , F= EI=FLEXURAL RIGIDITY0 = R0T ATI ON OF X-S EC T ION ,Vv = NORMAL OEFL., VM = T ANGLNTI AL DEFL.

DIMENSION CM ( 50 ) ,CN ( 50 ) , T ( 50) , V( bC ) ,G ( SC ) , V. ( 50) , VM( 50 )READ 11,D,F,E,A,R,N

11' FORMAT ( 5F12.L ,12)READ 10,C M (1) ,C N (1),V t 1 ) , 1 (1) , VM (N ) G , G (N ) , W (N )

10 FORMAT(3F10.0)NK=N-1 S = k/FU=Q*R/E*A+1.Z=R/E*A Y=0*R

C COMPUTE STRESS RESULTANTSDO 30 1=1,NMAK1 = ( S4CM ( I ) *CN( 1 )-U-*CN( I)-Y)*U AL1 = (-S*CM fI )*V(I )+V( L) )*D AM1 = (V ( I)*k + Z*V(I )* C N ( I ) )*D DV=V(I1+AK1/2.DN=CN(1I+AL1/2.DM=CM(1)+AMl/2.AK2=(S*DM*DN-J*UN-Y)*D AL2=(-S*DM*DV+DV)*D A M 2 =(DV*RHZ*DV*uN;*D EV = V ( I )+AK2/2.EN=CN(1)+AL2/2.EM = C M (I)+AM2/2.AK2=(S*EM*EN-U*EN-Y)*0AL3=(-S*EM*EV+EV)*DAM3=(EV*R+Z*LV*EN)*DFV = V( I)+AK 3FN=CN(I)+AL3FM=CM(1)+AM3AK4=(S*FM*FN-U*FN-Y)*0AL4= (~S->FM*FV + Fv ) *DAViA = ( F V * R + Z * F V * F N ) * DJ=i + 1C M J)=CN( I ) + {ALl + 2.*AL2 + 2.*AL3+AL4)/6.CM ( J ) = CM ( I ) + ( AM 1 -r 2 AM2 + 2 * AM 3 +AM4 ) / 6 .V (J )= V (I)+(AKl+2.*AK2+2.*AK3+AK4)/6.T ( J ) =T { 1 ) -rD

30 CONTINUE C COMPUTE ROTATION OF CROSSECTION

DO 31 1=1,NM K =N~I + 1 L = N-1BK1=CM(K)*S*D

G ( U =G(K)+UK131 COM I 1NUE

COMPUTE DEFLECTIONS DO 32 I=1 *NM K = N-I-rl L = N- IX = R*(CN(Kj/L*A+1.)*COSFi U(KJ )P=R*(CN(K)/E*A+1.)*SIMF(G (K ) )

. Citi=-(W(K)-R + X)*D CL1=-( P-VM (,x ) ) *D DW=W(K)+CLl/2.DVM = VM (K ) -CKl/2 .CK2=-(DW-R+X)*D CL2 = ~ (P-DVM)*o EW=W(K)+CL2/2.EVM=VM(K)+CK2/2.CK3=-(EW-R+X)*D CL3=-(P-EVM)FW=W(K)+CL3 FVM = VM ( ) +CK3 CK4=-(Fw-%+X)*D CL4=-(P-FVM)*UV M (L )=VM(K)+(CK1+2.*CK2+2.*CK3+CK4)/6.W ( L )=W(K)4(CLl+2.*CL2+2.*CL3+CL4)/6.

32 CONTINUE PRINT 20

2GOFORMAT ( 2X 4HANGl 5X , 6iiMO,v.E,NT4X , lOhAXl AL LOAD3X 3riSHtAP.3X 1 ShROTATION3X,9H1ANG DErL3X9UN0RM DEFL///)PRINT21,(Til) ,GM( I),CN( I)V ( 1 )G ( I )*VM( I ) ,W( i ) I=1,N)

21 FORMAT (F6.2,3F11.2,3E12.4/)END

APPENDIX B

1. FLOW DIAGRAM (Large Deflections of 0antilevered Arches with Loads Remaining in Their Original Directions Throughout Deformation)

A) Definition of variablesSAME AS IN APPENDIX A WITH THE FOLLOWING ADDITION YD = vertical deflection XD = horizonal deflection SV = vertical end load (positive downward)EE = allowable maximum of G(N)

B ) Main flow diagram

START

(over)

LOOP |Max. No. of Iterations = 75

GENERATE NORMAL AND TANGENTIAL END LOADS

READ D,F,E,A,R,TMAX,N, CM(1),T(1),VM(N),G(1), G(N),Q,EE,W(N),SV.

J. v-Li v n \ ) . VIM \ /T(50),V(50),G(50),W(50)VM(50),YD(50),XD(50).

75

76

LOOP

Max. No. of Iterations = 75

(Flow diagram, cont.)

Equal to or less than

END

greater I thanADJUST 0(1)

COMPUTE Horizonal and vertical deflections.

COMPUTE Normal and tangential deflections .

GENERATE Stress resultants and rotations of crossectlon using the Runge-Kutta method.

r\ n. n r\

n n

n n n n n n n

n n n n

n n

n n n n

n77

2 . COMPUTER PROGRAM (FORTRAN C om puter L an g u ag e )( lo a d s re m a in I n t h e i r o r i g i n a l d i r e c t io n s a f t e r d e fo rm

a t i o n )

. l a r g e DEFLECTIONS OF Ca n t il e v e r e d a r c h l o a d s r e m a i n in t h e i r o r i g i n a l d i r e c t i o n sTHIS IS AN ITERa-. i lVE Pk u Cl SSTHIS PROGRAM WILL REo'JiRE OR TO 75 l OMPLETE LOOPS

I AuLES* * > **** * wCM= BENDING MOMENT CN= AXIAL FORCET- ANGLE MEASURED FROM FREE ENDV= SHEARING FORCE6= ROTATION OF CROSSECTIONW= NORMAL DISPLACLMENTVM= TANGENTIAL DISPLACEMENTYD= VERTICAL uISPLACLMENTXD= HOR1 ZONAL DISPLACEMENTD= INCREMENT OF m NGLEF = EI= FLEXURAL RIGID! TYE= YOUNG'S MODULUSA= AREA OF CROSSECTIONR= RADIUS OF ARCHTMAX= TOTAL ANGLE OF ARCHN= 1+ T MAX/DC= NORMAL LOAD PER UNIT LENGTH OF ARCH EE= ERROR CRITERIm ( .0001 RADIANS )SV= VERTICAL CONl LNTRATED END LOADDIMENSION C M

CiN( 1 ) =SV*5I TnF ( ]>iAX-G( 1 ) )COMPUTL 5TRLSS RlSul r^NTS AND ROTAi 10,\ Or

40 DO 50 1=1,NM G1=G(I}CM1=CM(1;n = T ( 1 )V1 = V ( 1 }CN1=CN(1)Cl=(l.-rC,jl/(E-A) )Y=0*RC2 = COSF(TMAX-1 i )C3 = S 1JMF ( TMAX-T1 )C=Y*Ci*(C2*C5-Gj*G4)AVI=(CNi*CMl*S-CNl-b)*D AN1 = 1-V1^CM1*S + V1-C )-D AM 1=(V1*R*C1)*D AG1=-CM1*3*D Tl=Tl+D/2.CNl=CNl+ANl/2.CMl*CMl+AMl/2.VI * Vl+AV1/2 #Cl* (1 .-f-CM/ l l *A) )C2 = C0SF(TWAX-Ti )C3*SINF(TMAX-T1 )C4-C0S F (G 1 )C5=S1NF(G1)B*Y*C1*IC2*C4+C3*C5)C=Y*C1*(C2*C5-C3*C4)AV2= i CINi*CMl*S-CNl-b ) *b AN2= (-Vl*CMl*S-D Gl=Gl+AG2/2.-A01/2.CNl=CNl+AN2/2.~A,Nl/2.CMl=GKl+AM2/2.-ANl/2.vl*Vl+AV2/2.-AVl/2.Cl=(1+CN1/(E*A)}C2 = CCS F ( TV, AX- 11 )C3*SINF(TMAX-T1 )C4=C0SF(G1)C5 = SiNF(G1 )B=Y*C1*(C2*C6+C3*C5)C=Y*C1*(C2*C5-C3*C4)AV3=(CN1*CMI*S-CN1-U)*D AN3=(-V1*0M1*S+V1-0)*D AM3=(V1*R*C1)*D AG3=-CM1*S*L T1 * T1+D/2 Gl=Gl-+AG3-AG2/2.CNl=CNl+AN3-AN2/2.CIYI = CMl+AM3-AM2/2 VI-VI+AV3-AV 2/2.C1=(1.+0N1/(L*A))C2=CCSF(TMAX-T1)C3=SINF(TMAX-T1)C4=C0SF(G1)C 5 = SIN F ( G1 )

CROSSEC11 ON

G=Y*C1*(C2*C4+C3*Cs)C=Y*C1^(C2*C5-C3*C4) 79AV4- ( ) D FW=W(K)+CL3 FVM=VM(K)+CK3 CK4=-(FW-R+X)*D CL4=-(P-FVM)*uVM ( L ) =VM,(K i + ( CK1 + Z.*CK24Z.*CK3 + CK4 )/6.W(L)=W(K)+tCL1+2.*CL2+2.*CL3+CL4)/

32 CONTINUE

POINT 2b2 5 Q F O R M A T { / / / 7 X * 5 H * N u L E i 5 X . 6 H M G ; 4 E N T l l X . i i H A X I A L FORC C12 X. 80

1 5HSHEAR/)P R I N T 2 6 , ( I ( I ) , vri( I ) , C N ( i ) . V ( I ) , I = 1 , N J

26 FORMAT (4 E 20.8 J PRINT 2 7

27 FORMAT (/TXtbhMNCLL 14X , dMNOT AT i ON 1 2X 9nT ANG UL'FL 11X 9HN0RM DCFL/ ) PRINT 2 8 ( T ( I ) Cj ( 1 J * V M ( I ) * V ( I ) * i=l,N)

28 F O R M A T (4 E 20.8 )DO 993 1=1,NY D ( 1 ) = W ( I ) * C O S F t I M A X - 1 ( 1 ) ) - V M ( I ) * S 1 N F t 1M A X - T ( I ) )

998 XD(l)=w(l)*SlNF(TMm X-T( 1))r V M ( I)>COSF(TMAX-I(I ) )PRINT 997

9 9 7 . F O R M A T ( 6 X , 9 H v L R T D t F L 1 0 X 1OnH ORIZ D l F L / )PRINT 999, ( Y O m ,XD(I) ,I = 1,N)

999 FORMAT(2E2U.d )GO 10 14E N D

APPENDIX C

FLOW DIAGRAM (Finite deflections)A) Definition of variables

SEE APPENDIX E

B ) Main flow diagram

READ in loads, points of restraint, and El,AE,1,and c* for each element.

3

CALCULATE Z and Y from and 1E

GENERATE b kb matrix for each element and arrange in a master array K.Use Z and Y in compat. eq.

1 ------IMPOSE BOUNDARY CONDITIONS

*P = Kx(over)

(Plow diagram, cont.)

COMPARE

PRINT RESULTS

COMPUTE new values of Z and Y .

COMPUTE a new displacement vector ( x ) with K and P.

U U

U V

U U

U U

U U

U U

U V

V U

V V

u.

832 0 QOMPPyBR PROGRAM ( FOBT1AH' Oomputer Language ) (finite deflections of frames and trusses)

LARGE D E F L E C T IONS U S I N G A D I S C R E T E S Y S T E MY THIS S H O U L D T A K E ABOUT 20 C O M P L E T E LOOPS

SM= M A S T E R S T I F F N E S S MATRIX' S* INDIVIDUAL S T I F F N E S S M A T R I X FOR EACH M E M B E R^YB= C O M P A T I B I L I T Y M A TRIX FOR EACH M E M B E R

I N C R E M E N T A L LOAD VE C T O R A V M N I D E N T I F I E S P O I N T S OF Z E R O D I S P L A C E M E N T

' Vy-IE = ID E N T I F I E S ENDS OF M E M B E Rtv-.UE ID E N T I F I E S ENDS OF ME M B E R; AE= A R E A TIMES M O D U L U S OF E L A S T I C I T Y v E l * F L E X U R A L R I G I D I T YY EL== LENG T H OF I N D I V I D U A L M E M B E R

8M= B E N D I N G M O MENT (INCREMENTAL) Y= H O R I Z O N A L P R O J E C T I O N OF M E M B E R 2= V E R T I C A L P R O J E C T I O N OF M E M B E R AN* A N G L E D E S C R I B I N G THE P O S I T I O N OF MEMB E R S MOUT= N U M B E R OF R E S T R A I N T S MALL* N U M B E R OF M E M B E R S MAX* 3 T I M E S THE NUMB E R OF N O D E POINTSDI MENS I ON SM( 40 o AO ) S C 3 $>3 ) e>8 < 3 *6 ) P 140 ) PT (40 ) oMN (10) o.Xl (40 ) o

1 XX 6 o 40) oXG(40 ) s SB ( ) .8 SB (6 9 6 ) 9 IE(20)JE( 20 ) AE (10) o-EH 10) 2 E L ( 1 0 ) A N ( 1 0 ) X P ( 4 0 ) 9B M { 6 4 G ) B M T ( S 940 i Y ( 1 0 I (10) X 2 (40)3 X4(40) 9X 5 (40) #PG(40 ) sXT(SolO ) XSUM( 40) vELO( 10 ) e R H (10) s>RVi 10 )D I M E N S I O N V I 1 0 )

716 D O 723 1= 1 e40 P T ( I ) = 0 o X G ( I ) = 0 d XP( I) * Odxsumu) o0P G (I ) * Oo

723 P < I ) = OoDO 724 1= I 9IO M N < I ) = Oo 'AE (I ) =0o EI(I) = Oo

E L t n = Oo AN ( I ) = 0 o Y d ) = Oo2 (1 ) = OoE L O (I ) = Oo RH(I) = Oo R V ( U = Oo

724 V(I) = OqDO 725 1=1920 IC(I) = 00

725 J E U ) = OoDO 726 I = 140 DO 726 J = 1 96

726 X U g I } =0oDO 727 1*1,10 84D O 727 J 1 ,6

727 XT'tJfrI) = Oo .READ 197, M A L L M A X e M O U T E E A M I N

197 FORMAT (3I2o2F10o0)R E A D 1 9 8 < M N t n e I = loMOUT)

.198 F O RMAT (812)M I N * M A X - M O U TREAD 196* .(Ptlle I *

1 9 6 j F O RMAT (8 F I O 0O)ORE A D 1 9 4 * (IE I N U N )* U E ( M U M ) E L (M U M ) E I (N U M )*A M (N U M ), A E ( M O M )* 1 M U M * 1 , M A L L )

194 FORMAT ( 1 2 , 2 X I 2 , 4 X , 4 F 1 0 0 )C CO M P U T E H O R i Z O M A L AMD V E R T I C A L P R O J E C T I O N S OF M E M B E R S

00 940 M U M =1 , MALL . E L O (NUM $ = EL (MUM)

Z (N U M )* E L ( N U M t S I N F (3 0 14 1 5 9 3 - A N ( N U M >)940 Y ( NUM>*ELfNVMl*C0SF(3fll4l593 -AN(NUM>)

LTR = 0 KTR 1DO 711 1 = 1 , MIN

711 P T ( I) = P ( 1 )C SET M A S T E R S T I F F N E S S M A T R I X EQUAL TO ZERO

20 1 DO 199 1=1,40 DO. 199 J 1,40

199 SM ( I * J e O e 'LTR = LTR 1

. 3 0 0 DO 200 N U M = 1 o M A L LDO 250 1=1,3 DO 250 J = 1,3

250 S

DO 260 1=1*6 DO 260 0 = 1 * 3

- SBC J t n = 0 o 00 260 K = 1 *3

. 260 SBC J I ) =SB < J 1 > 4-'S< J K ) # 8 1K I )DO 400 1=1*3

: "BM.t lNUM)eODO 400 J = 1 *6

' - . B M (I o N U M ) 6 M ( I * M U M )* S B (I J )* X T (J e M U M )400 C O N T I N U E

DO 401 1=1*3401 BMTCZoMUM) = BMCloNUM)262 DO 270 1=1*6

DO 270 J = l * 6r-BSBCJo|)=0

DO 270 K = 1 *3 270 B S B C J * ! > 8 S B C J * I ) + B ( K # J ) # S B t K I )

C- . GEN E R A T E M A S T E R S T I F F N E S S M A T R I X D O 280 1=1*3 IM w IEtMUMI+1-1 DO 280 J*lo3 0 M = 1 E ( N U M ) + J - 1

280 SMC J M o.IM) a SMi J M * I M I FBSB (J* I )DO 281 1 = 1*3 I M = 1 E C M U M H I - 1D O 281 J=4 * &

281 = JM*lM>>eS8< J * I )DO 282 1=4,6 I M = J E ( N U M ) + l - 4DO 282 J = l*3-

JMs> I.E t N U M H J ~ l282 SMC J M i M ) = S M C J M s l M ^ B S B C J , I )

0 0 283 1=4,6

DO 283 J = 4 * 6 D M = J E < M U M > 4 J 4

263 S M f J M * IM> =SM< 3M IM>-t-BSB (J I )200 C O N T I N U E

C EllM'iMAfE R O M S A N D C O L U M N S C O R R E S P O N D I N G TO B O U N D A R Y284 DO 110 K = 1 M 0 U T

M 1 = M N ( K ) - K + 1 DO 111 1 = 1 MAX

. IF.(Ml i ) 1 2 2 , 1 1 2 , 1 2 1 112 DO 113 J2 *MAX

I P * $41 - 123 SMI $ *0 > S M {I P ,J >111 C O N T I N U E 210 C O N T I N U E

. D O 1 M K=2 *MOUT

DO 115 J=2 MAX '$F(Ml*JI 2 1 6 , 1 1 6 * 1 1 5

116 DO 117 1 = 1 ,MAX ' , JP=J+1 -

117 SMi 1,0) =SMri ,-ip i.. ITS C O N T I N U E

85

C O N D I T I O N

lib CONTINUE0 TO ! 708,710 ) s KTR 86710 DO 706 1=1,MIN

PG( U = 0 q BO 70.6 J = 1 ?MIN

706 PG(I)= PG(II + SMt!J)*XSUM(JI V PSUM GoPRES Oc RES..* OoDO 70? I * IfMIN PTM) = P ( I } - P6 (I >

- PRES * PRES + A6SF(PJ)00 59 IkJ oNTNP

59 SM(Jo 11*SM(Jo I)/DD 'DO 60 L1*MIN IP(L-J) 51,60,61

61 CB^SMCLgJ)DO 62 K^JgMIMP

62 SM (L oK ) *5M (L * K ) -SM (J , K) *CB 60. CONTINUE58 CONTINUE

C PUT SUPPORT DISPLACEMENTS BACK INTO DISPLACEMENT VECTOR DO 705 1=1,MIN

70S KSUMUT = SM( l oMIfM) + XSUM( I)MM2=M0UT ' DO 120 1=1,MAX J=MAX-l-i-l

" IF(mi-J) 121,1220123121 L=J-MM2 . "

XPtJT=SMCLMIN-o-l>XOI J) = SM(L,MIN+1T .+ XO( J)GO TO 120122 XP(J)=0o. . . .X6TJ) * Oo'MM2=MM2-1TFtMM2li 120,124,124

124GO TO.120

123 XP(J)=SMCJM1>3^1)' ,XG(J) = SM(JgMi:N-5-l) + XG(J)220 CONTINUE

C CHANGE DISPLACEMENT VECTOR TO 2 SUBSCRIPT HATRIXC68Y MALL)DO 700 m m I MALL .

263 IEO = IE(NUM} '. I1 = 1E(NUMHI.IE221NUM>42 'JE0=JE(WM)JE1=JE(NUM)>1 . -

4 E 2 = J E C ? W M H 2X t 1 s W U M ) = X P (I E O ) 87X ( 2 s M U M ) = X P f I E 1 )X (:3 o N U M ) XP ( 1 2 )XNUM>sXP( J 0 ) x< 5 NUM)*XP( J E D .X(6 i X T i 6 h5UM) = XG < JE2 ). XI (N U M ) XP ( 1EO )

X2 < M U M ) a X P I 1 E 1 ?X 4 ( N U M ) = X P (J E O )X5(NUM)=XP( J E D

700 C O N T I N U E 285 KTR = 2 IBS DO 187 N U M = 1 , M A L L

Z I N U M ) *2 < N U M > -6-X 2 (M U M > -X 5 f M U M )Y ( N U M ) Y ( N U M D X 4 ( N U M ) - X 1 ( N U M ) . E L ( M V M } = S 0 R T P t ( Z ( N U M ) ) * - & 2 + ( Y ( N U M D * * 2 ;

,187 CGMT IMUE- PRINT 9 5 1 s (Z (N U M ) Y (M U M D E U M U M )e M U M = 1 o M A U t )

9 5 i FORMAT ( 3 E 2 0 o 8 IFtLTR ~ 1 5 1 2 0 1 * 7 0 9 9 7 0 9

'709 PR:NT 22 FORMAT t-Z2X8MXDI SPL14X$8HY-D1SPL13X *9MR0TATI0M14X *-Zt NUM.) )/EL INUM)"

712 RM.CNUM) = (Vf NUM)*Z tNUM) * BMT < 1 aNUM 1 ( N U M D / E L (N U M )PRINT 714

714 FORMAT (1 0 X a 3 3 H I N T E R N A L F O R C E S IN XY COO R D I N A T E S )P RI N T 715

715 FORMAT 16X 9 8HVERTI C A L12X aRHHOR I Z O N A L ?PRINT 7139 ( RVtHUM)RH(NUM) N U M * 1 M A L L )

713 FORMAT ( 2E20e8 )GO TO 716END .

APPENDIX D

The following symbols have been selected foruse, in Chapters 2 through 6 of this thesis;

R = radius of circular archs radius of curvature of arch

w = normal deflectionv = tangential deflectionX - horizonal deflectionT - vertical deflectionM = bending momentN = axial forceV = sheafing force = position angle of archy = rotation of cross=sectlonP = vertical concentrated end load

~ total angle of circular cantilever archq - uniform load per unit arc length

' E = Young's Modulus or Modulus of ElasticityI = moment of inertia of arch cross-sectionA = area of arch cross-sectiont = arch thicknessz = distance from neutral surfacex and y = rectangular projections of deformed arc

length in v+dv and w+dw directions respectively

88

arc length strain of the neutral surface,

APPENDIX E

The fallowing symbols have been selected for use In Chapter 7 of this thesis:

y = element boundary displacement x = generalized displacementd = angle of orientation of structural element(linear

analysis% = original angle of orientationto8 = angle of orientation In deflected stateb = compatibility matrixP = column matrix of external loads# 'P = load matrix calculated using deflections

and master stiffness matrix corresponding to the deflected configuration

Pr = difference between P* and the actual loads PF = column matrix of internal loads corresponding

to element boundary displacements: k - stiffness matrix for individual structural

elementS - axial force in elementM = bending momentK = master stiffness matrix without boundary con

ditions ImposedK = master stiffness matrix after imposition of

boundary conditions (linear analysis)K 6 = master stiffness matrix after Imposition of

boundary conditions (nonlinear analysis)Z = vertical projection of deflected element

90

Y = horizonal projection of deflected element 1 = length of element (linear analysis)1@= original length of memberI 8 = length of member in deflected stateE = Youngs Modulus or Modulus of ElasticityI = moment of inertia of cross-section of elementA s area of cross-section of element.

REFERENCES1 Barten, H .J ., "On the Deflection of a Cantilever

Beam," Quarterly of Applied Mathematics, vol. 2,1944.

2. Bisshopp, K.E.and Drucker, B.C., "Large Deflections of Cantilever Beams," Quarterly of Applied Mathematics, vol. 3, 1945.

3= Conway, H.D., "Large Deflections of Simply Supported Beams," Philosophical Magazine, vol. 38, 1947.-

4. Conway, H.D., "The Nonlinear Bending of Thin Circular Rods," Journal of Applied Mechanics, vol. 78,1956.

5. Euler, L ., "Methodus Inveniendi Lineas Curvas Maxim!. Mlni'mive Proprietate Gaudentes," Lausanne and Geneva, 1744.

6. Horne, M.R., "The Effect of Finite Deformations in Elastic Stability of Plane Frame," Proceeding of the Royal Society, Series A, London, Feb. 1962.

7. Ince, E.L., Ordinary Differential Equations, Dover Publications Inc., N.Y., 1926.

8. Johnson, D.E# and Smith, R.E., Fortran Autotester,John Wiley and Sons Inc., N.Y. and London.

9. Kesti, N.E., "The Developement of a General Buckling Theory for Rings and Arches with Applications to Circular Arches," Masters Thesis, Univ. of Arizona, 1962.

10. Levy and Baggott, Numerical Solution of Differential Equations, Dover Publications Inc., N.Y., 1950.

11. McCraken, D.D., A Guide to Fortran Programming, John Wiley and Sons Inc., N.Y. and London, 196^.

12. Pestel, E.G. and Leckie, F.A., Matrix Methods in Elastomechanics, McGraw-Hill Publications, N.Y., San Francisco, Toronto, and London, 1963.

13. Richard, R.M., "A Study of Structural Systems Having Nonlinear Elements," Doctoral Diss., Purdue Univ., 1961.

9 2

14. Richard, R.M. and Goldberg, J.E., "Analysis of Nonlinear Structures," ASCB Journal of the Structural Division, vol. 89, Aug. 1963.

15. Saafen, S.A., "Structural Plane Frameworks," Doctoral Dissertation, Manchester Univ., Manchester, England.

16. Saafen, S.A. and Britton, D.M., "Elastic Finite Deflection Analysis of Rigid Frameworks by Digital Computers," International Symposium for the Use of Electronic Computers in Civil Engineering, Lisbon, Portugal, Oct. 1962.

17. Saafen, S.A., "Nonlinear Behavior of Structural Plane Frames," Journal of the Structural Division, ASCE, August 1963.

18. Seames, A.E. and Conway, H.D., "A Numerical Procedure for Calculating the Large Deflections of Straight and Curved Bars," Journal of Applied Mechanics, vol. 24,1957.

19. Sonntag, R., "Die Spiralringfeder," Ingenieur-Archiv, vol. 14, 1943.

20. Wilson, E.L., "Analysis of Frames with Nonlinear Behavior," ASCE Journal of Engineering Mechanics Div,, vol. 86, Dec. i9 6 0.