Contents - ZU · PDF filecontents 2- buckling 1 ... 13-maxwell's law of reciprocal deflections; 93 betti' s law 14-influence line for deflection 95 15-problems 97

Contents

2- Buckling

107 1-INTRODUCTION 108 2-SLENDERNESS RATIO 109 3-END CONDITIONS 110 4-EULER'S FORMULA 113 5-COLUMN WITH ONE END FIXED AND THE OTHER

FREE 114 6-CRITICAL STRESSES 115 7-LIMITATION OF EULER'S FORMMULA 117 8-FACTOR OF SAFETY 117 9-EMPIRICAL FORMULAE 120 10-MAXIMUM FIBER STRESS 131 11-COLUMN WITH INITIAL CURVATURE 135 12-LATERALLY LOADED COLUMN 139 13-PROBLEMS

3- statically indeterminate structures

147 1-INTRODUCTION 149 2-FORCE METHODS ( flexibility Approach) 150 3-DISPLACEMENT METHODS (stiffness approach )

Page 1- Deflections 1 1-INTRODUCTION

2-DEFINITIONS

33-THE DIFFERENTIAL EQUATION OF THE ELASTIC

LINE 7 4-THE DOUBLE INTEGRATION METHOD

21 5-MOMENT- AREA METHOD 28 6-ELASTIC – LOAD METHOD 39 7-THE CONJUGATE BEAM METHOD 51 8-THEORY OF REAL WORK 61 9-METHOD OF VIRTUAL WORK 66

10-EVALUATION OF INTEGRAL dxMM 10 84 11-DEFLECTION OF TRUSSES 89 12-CASTIGLIANO' S SECOND THEOREM 93 13-MAXWELL'S LAW OF RECIPROCAL DEFLECTIONS;

BETTI' S LAW 95 14-INFLUENCE LINE FOR DEFLECTION 97 15-PROBLEMS

151 4-DEGREE OF STATIC INDETERMINACY 151 4.1-Degree of Indeterminacy For Beams (Line Structures)

153 4.2-Degree of Indeterminacy for Plane Frames

155 4.3-Degree of Indeterminacy of Plane Trussed

4- METHOD OF CONSISTENT

DEFORMATIONS 159 INTRODUCTION

160 ONCE STATICALLY INDETERMINATE STRUCTURES

171 TWICE STATICALLY INDETERMINATE STRUCTURES

176 3-TIME STATICALLY INDETERMINATE

182 CASE OF n- STATICALLY IN DETERMINATE

STRUCTURES 186 SETTLEMENT OF SUPPORTS

187 CHANGES OF TEMPERATURE

194 BEAM ON ELASTIC SUPPORTS

209 FORCED DEFORMATIONS

227 TRUSSED BEAM AND TRUSSED FRAME

235 DEFLECTION OF STATICALLY INDETERMINATE

STRUCTURES (REDUCTION THEORY) 248 ANALYSIS OF STATICALLY INDETERMINATE TRUSSES

252 DEFLECTION OF INDETERMINATE TRUSSES

258 GENERAL REMARKS CONCERNING SELECTION OF

REDUNDANTS

260 CHOICE OF MAIN SYSTEM IN CASE OF SYMMETRY

AND ANTISYMMETRY

263

ANALYSIS OF STATICALLY INDETERMINATE

STRUCTURES USING CASTIGLIANO SECOND

THEOREM; THEOREM OF LEAST WORK 268 TORSION OF FIXED BEAM

5-Influence lines of statically indeterminate beams

and frames 278 1-INTRODUCTION

279 2-INFLUENCE LINES FOR DETERMINATE BEAMS BY

MULLER-BRESLAU PRINCIPLE

286

3-MULLER-BRESLAU’S PRINCIPLE FOR

INDETERMINATE STRUCTURES

308 4-INFLUENCE LINE OF TRUSSED BEAM

322 5-INFLUENCE LINE FOR FRAME

324 6-ANALYSIS OF STATICALLY INDETERMINATE

FRAMES 341 7-MAXIMUM EFFECT USING INFLUENCE LINE

343 8-PROBLEMS

6- INFLUENCE LINES OF STATICALLY

INDETERMINATE TRUSSES 346 1-INTRODUCTION

346 2- THE DISPLACEMENT DIAGRAM

347 3- WILLIOT DIAGRAM

357 4- WILLIOT-MOHR DIAGRAM

362 5- INFLUENCE LINES FOR STATICALLY

INDETERMINATE TRUSSES 370 6- PROBLEMS

7- MOMENT DISTRABUTION METHOD

374 1-Introduction

376 2-Sign Convention.

376 3-Stiffness

380 4-Carry over factor

381 5-Distribution factor

383 6-Fixed End Moment

387 7-Non-Prismatic Members

391 8-A continuous Beams with simply supported ends

395 9-Beams with end overhanging

10-Beams with a settlement supports.

383 11-Frame without sideways symmetry & anti-symmetry

416 12-Structures subjected to sideways.

464 13-Temperature Effect

474 14-Frame with n- degrees of Freedom.

494 15-Influence lines by moment distribution

496 16-Case of continuous beam

497 17-Procedures

500 18-Examples

509 19-Problems

APPROXIMATE ANALYSIS OF STATICALLY

INDETERMINATE STRUCTURES516 1-INTRODUCTION

517 2-Trusses

522 3-Building Frames Subjected to Vertical Loads

527 4-Portal Frames

533 5-Portal Method

539 6-Cantilever Method

548 7-Problem

551 8-References

1 DEFLECTIONS

1._INTRODUCTION

The computation of elastic deformation for structures, either the linear

deformations of points or the rotational deformations of lines (slopes)

from their original position, is of great importance, not only in the design,

construction of structures but also in the analysis and solution of statically

indeterminate structures. In structural design the dimensions of beams

and girders are sometimes governed by the allowable deflections. It must

be noted that, if the deflections of beams or frames is excessive, some

problems will be produced as cracking of plaster, drainage problems,

damage of walls. Most important, the stress analysis for statically

indeterminate structures is based largely upon an evaluation of their

elastic deformations under load. By a statically indeterminate structure

we mean a structure in which the number of unknown elements involved

is greater than the number of static equilibrium equations available for

solution of the equations. The elastic deformations liable to accure in the

structure resulting in the external loads, temperature variation and

differential settelements between supports cause various deformations,

but such elastic deformations vanish when the loads disappearent as

shown in Fig. 1.

The following assumptions are made for the computation of

deformations:

1. Bernoulli's law is valid, plan sections before deformations remain

plane after the deformations.

2. The materials obey Hook's low.

E =

2 Chapter (1) - Deflections

3. The depth/span ratio is very small, i.e. the external loads act on the

non-deformed member.

Numerous methods of computing elastic deformations have been

developed. Among them the following are considered;

1. The double integration method.

2. The conjugate beam method (Moment area, elastic load methods)

3. The method of virtual work

4. Castigliano's theorem

The first two methods are used for beams and frames, whose members

subject to bending strain, while the method of virtual work is used for all

types of strain; bending, axial and shear.

Figure (1)

Chapter (1) - Deflections 3

2. DEFINITIONS

Deflection

Is the displacements of various points from their original positions.

Stiffness

Is the resistance of a structural element to deflection.

The deflection of a member (beam) depend on;

1- The value of loads acting on the beam.

2- The stiffness of the member, which depend on the type of material and its

modulus of elasticity (E).

3- The dimensions of elements (span, cross section)

3. THE DIFFERENTIAL EQUATION OF THE ELASTIC

LINE

3.1 Curvature of elastic line (from mathematic)

The mathematical definition for curvature is the rate at which a curve is

changing direction. To derive the expression for curvature, we shall

consider a curve such as the one shown in Fig. 2


The average rate of change of direction between points P1 and P2 is S

.

The limiting value of this ratio as S approaches zero is called curvature

(K) and the radius of curvature (R) is the reciprocal of the curvature, we

have,

K = R

1

=

S

SLim

= S d

d

but

tan = x d

y d

x d

d tan =

2

2

x d

y d

or

(1 + tan 2 )

x d

d =

2

2

x d

y d

Then:

2

2

x d

y d = (1 + (

x d

y d)

2)

x d

d

where

x d

d =

2

2

2

dx

dy 1

xd

yd

also s d

x d =

dx

ds

1


s d

x d =

2

1

2

22 dy

1

dx

dx

=

2

12

1

1

dx

dy

hence,

R

1 =

S d

d =

x d

d

s d

x d

R

1 =

2

32

2

2

dx

dy 1

xd

yd

(1)

For a loaded beam with its longitudinal axis taken as the X-axis, we

may set dy/dx in formula (1) equal to zero if the deflection of beam is

small. Then we obtain;

R

1 =

S d

d

2

2

x d

y d (2)

R

h

M

h

ds

d

M

1

1h-d

h-d

1

1

Figure (3)


3.2. Curvature from elastic bending:

In general, except for very deep beams with a short span, the

deflcetion due to the shearing force is considered, In order to develop a

formula for the curvature due to elastic bending, let us consider a small

element of a beam shown in Fig. 3. Owing to the action of bending

moment M, the two originally parallel sections 1-1 and 1\ - 1

\ will change

directions. This angle change is denoted by d. If the length of the

element is dS and the maximum bending stress, which occur at the

extreme fibers, is called , the total elongation at the top or bottom fiber

is h. d which equals to . ds/E, E being the modulus of elasticity. Thus,

E = /

h. d = E

ds .

or s

d

d =

h

E

=

R

1

Replacing with I

h.M, I being the moment of inertia of the cross

sectional area of the beam about the axis of bending, gives.

R

1 =

s

d

d =

EI

M (3)

which expresses the relationship between the curvature and bending

moment. Now equating eqs. (2) and (3), we obtain the approximate

curvature for a loaded beam as;

2

2

x d

yd =

EI

M (4)

Note that eqn. (4) involves four major assumptions;

1. Small deflection of beam

2. Elastic material

3. Only bending moment considered significant


4. Plane section remaining plane after bending

The curvature, established in the coordinate axes of Fig. 2, clearly has

the same sign as M, but the sign may be reversed if the direction of the y

axis is reversed. In that case, we have

2

2

x d

y d = -

EI

M (4

)

This Eqn.4

is the differential equation of elastic line. In our next

computation we consider Eqn. 4.

4. THE DOUBLE INTEGRATION METHOD:

This method is so named because the successive integrations of the

second order differential equation;

2

2

x d

y d =

EI

M

resulting the equation of elastic curve, the values of deflection, can be

obtained as follows

1- Consider The Case of Simple Beam;

From fig.4;

w t /m

C

y

y

x

Figure (4)


2

2

x d

yd = -

EI

M

M = 2

Wx - x .

2

WL 2

2

2

xd

yd = -

2

Wx - x .

2

WL 2

EI

1

= dx

y d

=

1

32

C 6

Wx -

4

x.WL

EI

1

EI = 1

32

C 6

4

x.

WxWL (1)

EI y = 21

43

C X . C 24

Wx

12

X .WL (2)

To get C1 & C2, from boundary conditions :

At x= 0, y = 0 C2 = 0

And At x= L, y = 0 C1 = WL3/24

At x= L /2, = 0

Hence

= EI

1

24

WL

4

x.WL -

6

Wx

323

= EI

W

24 L Lx 6 - x 4 323

y = EI

1 x.

24

WL

12

x.WL -

24

Wx

334


ymax

y = EI 24

W Lx 2 - x 334 L

y maximum at = 0 or at x = 2

L

ymax = EI 843

WL 5 4

Example (1):

Get ymax. for the given beam shown in the Fig.5

E = 210 t/cm2

I = 12

50 25 3 = 260416.6 cm

4

Solution:

ymax = 260416.6 210 843 100

(500) 1 5 4

= 0.1488 cm.

= EI 24

WL

EI 2

1 L

3

2

8

32

WL

elastic

curve

25

50

a

L = 5 m

W = 1 t/m\

Figure (5)

b

a


= 260416.6 210 24 100

(500) 1 3

= 0.000952 rad.

to draw the elastic line, calculate the values of y at different point

Example (2):

Find ymax for the given beam show in Fig.6. Also get the equation of

Figure (6)

Solution

Reactions

RA = 6

WL

RB = 3

WL

MX = x. 6

L . W-

3

x .

2

x. 2

L

W

= x. 6

L W-

L 6

. Wx3


2

2

xd

yd = -

EI

M

= EI

1 x.

6

.WL -

L 6

Wx

3

2

2

xd

yd =

EI

W

6

..x L -

L 6

x

3

= dx

dy =

EI

1 C

12

x.WL -

L . 24

Wx 1

24

y = EI

1 C x . C

36

x.WL -

L 201

Wx 21

35

Boundary Conditions

at x = 0, y = 0 C2 = 0

at x = L, y = 0

L . 36

- 120

1

44

CWLWL

= 0

C1 = 120

WL -

36

WL

33

= 360

WL 7

3

i.e.

y = EI

1 x .

360

WL 7

36

x.WL -

L 201

Wx

335

ymax. at dx

dy = 0

i.e. WL . 360

7

12

x.L. W -

L 24

Wx 3

24

= 0


x = 0.519 L

Hence

ymax = EI

WL .00650 4

If L = 5 m

W = 2 t/m\

E = 210 t/Cm2

I = 260416 Cm4

ymax = 260416 210 100

500 2 00665.0 4

= 0.152 cm.

Example (3)

Find the slope and deflection

equations for the given

cantilever beam shown in Fig. 7

Solution

M = - P (L – x)

2

2

xd

yd= -

EI

M

= + EI

P (L – x)

Figure (7)

= dx

dy =

EI

P C

2

x - X 1

2

L

y = EI

P C .x C

6

x -

2 21

32

Lx

x


at x = 0 , y = 0, = 0 C1 = 0 , C2 = 0

i.e.:

= EI

P

2

x - x

2

L

y = EI

P

6

xX -

2

32

Lx

max . & ymax.. at x = L. i.e.

ymax .= EI 3

PL3

, max. = EI

PL

2

2

If p= 2t , L = 4m , EI = 3000 t.m2

Hence

max = 3000

2

2

4 - 4 4

2

= 0.00534 radian

ymax.. = 3000

2

6

4 -

2

4 4

32

= 0.014 m = 1.4 cm.

Example (4)

Find the expression of

slope and deflection.

Locate the position of

max. deflection for the

given beam shown in

Fig.8, EI = constant

Figure (8)

Solution

Part ac :… x a

M = x. L

b. .P


2

2

xd

yd = - x.

LEI

b. .P

dx

dy = - 1

2

C 2

x .

LEI

b. .P

yac = - 21

3

C x . C 6

x .

LEI

b. .P

Part cb x a

M = P - x . b. .

L

P (x – a) =

L

a .P(L – x)

2

2

xd

yd = -

EIL

a .P(L – x)

dx

dy = - 3

2

C 2

x -Lx

L I E

a. .P

ycb = - 43

32

C x . C 6

x -.

2

Lx

EIL

a. .P

Boundary Conditions

Part a-c:

at x = 0, yac = 0,

at x = a, yac = ycb


Part c-b:

at x = L, yb = 0

at x = a, ac = cb

Hence:

C2 = 0, C4 = 0

and

C1 = C3 = 22 b - L . EIL 6

b. .P

Part a-c:

y = 222 b - L x . EIL 6

bx P

y = 222 b - L . 6

bx .x

EIL

P

= 222 x3 b - L . EIL 6

b .P

Part c-b:

=

2222 )(

33x-b L .

6

bax

b

L

EIL

P

y =

3223

x- b - L a -x b

L .

6

bx

EIL

P

In case of a = b = 2

L

ymax =EI 84

PL3


Note

In case of beam with variable moment of inertia; the moment equation

were different for each part have the same I. each part should be

considered separately.

Example (5)

For the shown beam in Fig. 9

under the given loads it is required

to sketch the elastic line and

calculate the maximum deflection

EI = 10000 t.m2

Solution

To sketch the elastic line it is

sufficient to determine the values

of deflections at different points B

and D and the angle of rotation at

points A, C and D.

Similarly by the same method one

can get the following:

yB = 1.0246 cm

yD = 0.078 cm

A = 0.0021 radian (0.12)

C = 0.003 radian (0.172)

Figure (9)

To get ymax.

For part AB

ymax. at = 0 or dx

dy = 0

x = 6.94 m

ymax. = 1.0632 cm.


Example (6):

For the given frame, it is required to sketch the elastic line due to given

load. (Fig. 10)

Figure (10)

Solution:

After drawing the M.D. the frame can be divided into 4 parts for each the

equation of bending moment can be easily written.

a) Equation of the bending moment,

part A-B M1 = 2 x1

B-C M2 = 1.33 x2 + 12

C-D M3 = 2.67 x3

D-E M4 = 0


Differential equations of the elastic line. Part A-B;

EI 21

12

dx

yd = -2 x1,

EI 1

1

dx

dy = 1

21 C x- (1)

EI y1 = - 211

31 C x. C

3

x (2)

Part B – C (2EI)

(2EI) 22

22

dx

yd = -1.33 x2 – 12

EI dx

yd

22

22

= - .67 x2 – 6

EI dx

dy

2

2 = - .33 3222 C x6 - x (3)

EI y2 = - .11 42322

32 C xC 3x - x (4)

Part C – D (2 EI)

2EI dx

yd

23

32

= - 2.67x3

EI dx

yd

23

32

= - 1.33 x3

EI dx

dy

3

3 = - 0.67 5

23 C x (5)

EI y3 = - .228 63533 C xC x (6)


Part D – E (EI)

EI dx

yd

24

42

= 0

EI dx

dy

4

4 = 7C (7)

EIy4 = 847 C X C (8)

To obtain the 8 constants C1 to C8 we consider the boundary conditions;

(1) at x1 = 0 y1 = 0

(2) at x1 = 6 x2 = 0, dx

dy

1

1 = dx

dy

2

2

(3) at x2 = 0 y2 = 0

(4) at x2 = 9 x3 = 9, 2dx

2dy =

dx

dy

3

3

(5) at x2 = 9 x3 = 9, y2 = y3

(6) at x3 = 0 y3 = 0

(7) at x3 = 0 x4 = 0, dx

dy

3

3 = dx

dy

4

4

(8) at x4 = 0 x1 = 6, y1 = - y4

By solving these equation one can sketch the following elastic line.

(y4)E

1

2

y1=y3

3

(y4)D

3

1 1`

y1 max.


5. MOMENT- AREA METHOD

This method is used for any case of loading which cause bending

moment. It is more convenient when the deformation is caused by

concentrated rather than distributed loads. This method is based on a

consideration of the geometry of the elastic curve of the beam and the rate

of change of slope and the bending moment at a point on the elastic curve

0A

Ay

By

0B

AByB

A

AxBx

B A

mAEI

M

Figure (11)

Referring to above Fig. 11, consider a portion A B of elastic curve of a

beam that was initially straight and continuous in position A0B0 in the

unloaded condition. Draw the tangents to the elastic curve at points A and

B. The tangent at A and B intersect the vertical at origin 0 at A- and B

-.

The angle AB is the change in slope between the tangents at points A

and B. Consider a differential equation of the elastic curve:


i.e. 2

2

dx

yd =

EI

M

Multiply by dx

i.e.

2

2

dx

yd dx =

EI

Mdx

Integrating the two sides, hence

dx d

2

2

B

A dx

y = dx

B

AEl

M

BA ds

dy-

ds

dy

=

A

B

EI

Monemt Bending of Area

AB = ABA

= EI

Am

Where Am is the area of bending moment diagram between points A and

B. Let a b the bending- moment diagram for portion AB after it is

modified by dividing every ordinate by EI of the beam at that point. Such

diagram is called the M/EI diagram. It is evident that the integral

dx d

2

2

B

A dx

ycan be interpreted as the area under the M/EI diagram between

A and B. Hence we may state:

5.1 First moment – area theorem

"The angle or change in slope in radians of the tangents of the elastic

curve between two points A and B is equal to the area under the M/ EI

diagram between these two points."

In fact the deformation and slopes are actually small. From equation (1)


The left hand side represent the vertical ( yab) through the origin between

the tangents to elastic curve at points A and B and this distance is equal to

(Fig.12)

yab = cm x

EI

A.

The right hand side term in equation (2) is the static moment about axis

thought origin 0, of the area under the EI

Mdiagram between the points A

and B. therefore we may state:

5.2. Second Moment – area Theorem

"The deflection of any point B on the elastic curve from the tangent to

this curve at other A is equal to the static moment about an axis through

origin 0 of the under the EI

M diagram between points A and B" (Fig.13)

If the origin at point a

yab = the deflection of point

A . W. r. t. the tangent at B

yab = EI

Am xc

as shown in Fig.13.a.

By

ABy

mA

EI

M

Cx

Figure (13.a)


2

2

dx

yd =

EI

M

ya – yb = EI

Mx d x

ya = x.

or

dxxdx

ydB

A

.2

2

= dxxEI

MB

A

.

i.e.

BAy

dx

dyx )( = c

m xEI

A.

(2)

0A

By

0B

ABy

Ax

Bx

AAy

AAx

mA

EI

M

cx

Figure (12)


Similarly if the origin at B

yba = c xEI

Am

as shown in Fig.13.b.

Ay

BAy

mA

Cx

EI

M

Figure (13.b)

Note that this deflection is measured in a direction normal to the original

position of beam. The two theorems can be used directly to find the

slopes and deflections of beams simply by drawing the moment diagram

for the loads causing deformation and then computing the area and static

moments of all or part of the corresponding EI

Mdiagram. The procedure

is illustrated in the following examples,


Example (7)

Calculate the max. deflection of the given simple beam Fig.14 and sketch

the elastic curve

Solution

ymax. at mid span from elastic curve.

ymax = 2

yba - yca

yba = 3

2 16 8 4

1

El

= 0.1137 m

x

y

yba

yca

maxy

Ao

maxM =16 t.m

83 x4

2EI = 3000 t.m

CBA

2 t /m

a) B.M.D.

b) Elastic Curve

Figure (14)

yca = 3

2 16 4 1.5

1

El = 0.0213 m

ymax = 2

1155.0- .0216 = 0.036 m


Example (8)

Find Yc for the given beam shown in Fig.15 and find ymax

EI = 3000 t.m2

Solution

8

bay =

3

cac yy

3yba = 8yc + 8yca

yc = 8

2yba - yca

15 t.m

452 2

75

C d

x

A B

A B

8 t

B

yba

CA

ca

c

y

y

Ao

a) B.M.D.

b) Elastic Curve

Figure (14)

8

3yba =

3000

1

2

75

3

10+

2

45 6)

8

2

= 3000

5.97100 = 3.25 cm


yca = 3000

100 1

2

45 = 0.75 cm

yc = 3.25 – 0.75

= 2.5 cm

To get ymax

d = 0; at x from B

ydB

dy

A Bd

oBy

x

yab

c. Elastic Curve (yab)

d. Elastic Curve (ydb)

Figure (15)

yd = yAB . 8

x- ydB

ydB = 2

3 2

EI

xx (Fig.15.d)

B -d = 2

..3

EI

xx =

5.1 2

EI

x

B = 5.1 2

EI

x =

8

ABy

x = 19.37 = 4.04 from B,

yd = 2.8 cm. =ymax

Example (9)

Calculate yB and B for the

given cantilever beam

shown in Fig.16

(EI = 2500 t.m2)

Solution

a - b = EI

1

48

3

1

b = EI3

32

= 25003

32

ob

by

8 tm 3EI75

b

1 t /m

a

Figure (16)


= 0.0043 radian

yb = 100

EI

3

3

32 = 1.28 cm

Example (10)

Find yB &B for the given cantilaver

in Fig .17

EI = 3000 t. m2

Solution

A-B = EI2

00.9= 0.0015 radian = B

yb = 200 EI2

9

= 0.3cm

b

yb

1t

a

92EI3

bo

Figure (17)

It will become apparent from these examples that these computations can

be facilitated by introducing some new ideas. The analogy based on these

ideas, discussed in the next method.

6. ELASTIC – LOAD METHOD

The ideas involved in the elastic – Load method can be developed by

considering the beam AB is loaded by elastic load which is equal to M/EI

diagram, the elastic load produce elastic reactions, elastic shear and

elastic bending moment at any section. Let beam AB which was

originally straight line and has been bent as shown in Fig.18.

Applying the second moment – area theorem givens;

yba = dxEI

MB

A

x)- (L


x = x)- (L EI

Am

then

A = L

yba = RA

A = L

x- L

EI

Am = RB

a) B.M.D.

b) Elastic Curve

c) Elastic Load

AmAR BR

BA

oc

c

oByba

BA

oA cay

cy

MEI

MA B

M

BA

Figure (18)

If we imagine that the M/EI diagram represents a distributed vertical load

applied to a simple beam AB as shown in Fig.18. The computation for the

RA RB


vertical reaction at A of the imaginary beam would yield a value exactly

equal to the value of A computed above, then:

A = RA

Where RA is the elastic reaction at support A. in the same time equal to

elastic shear at A with the same sign. Similarly, B = RB = elastic shear at

B. This analogy can be carried still further if we consider the form of

computation for c and yca . Note that c given the slope of the tangent

with reference to the direction of the chord AB of the elastic curve and

that yca gives the deflection of point c from the same chord AB. Consider

first c , then

c = A - (A - c)

= RA – (Area of EI

M digram between A and C

= Eastic shear at C from left

Further, "The slope of the tangent to the elastic curve at any point is

equal to the corresponding ordinate of the elastic shear diagram for

the imaginary beam AB loaded with the M/EI diagram."

Similarly; from Fig.18.d

yc = A . a - yca

= RA . a - yca

Where yca is the first

moment of M/EI diagram

between points A and C

about point C. hence

Figure (18.d)

yc is the elastic bending moment at point C form left side or;

"The deflection at any point along an end supported beam AB is

equal to the value of the elastic bending moment at that point'.

Sign Convention

In order to take full advantage of the elastic, load method, it is desirable

to follow the same sign convention and principles as those used in

drawing regular load, shear, and bending moment diagrams. Since


downward elastic loads are considered as positive in such computations,

positive M/EI ordinate downward loads. Plotting shear and bending

moment diagrams for the imaginary beam according to the usual beam

convention, positive bending moment indicate deflection below the chord

AB. Likewise, positive shear indicate that the slopes with clockwise or;

"Positive elastic shear mean positive angle of slope and positive

elastic bending moment mean positive deflections".

Example (11)

Calculate the deflection and

angle of rotation at point c, for

the given beam shown in Fig.

19. Sketch the elastic curve

(EI = 6000 t.m2)

Solution:

by using principle of supper

position i.e B.M.D. as shown

in fig. due to uniform load +

due to concentrated load.

EI9

+6

2EI6x3

36EI

+

x23 EI

3x18

18

4t

B

4 t /m

CA

B.M.Ds.

Figure (19)

c = Elastic shear at c from left side

= 0 9

- 36

- 9

36

EIEIEIEI

yc = Moment of elastic load at c

=

3

3

1 - 3

9 3

8

3 - 3

36

EIEI

= EI

2 9

9

15

36

EI


= cm 100 5.85

EI

= 1.425 cm

Example (12):

Find yc & c for the

given beam shown

in fig.20 sketch the

elastic curve

(EI = 6000 t.m2)

Solution:

W1 =

EI

16

EI 2

8 4

ra1 = EI 2

16

W2 =

EI 2

3

2

3

= EI 4

9

W3 =

EI

8 16

3

2

Figure (20)

ra2 = EI 3

8 16

W4 = EI 2

3 15

W5 = EI 8

9 2 3

3

2

= EI 2

9

c =

EI 4

9

EI 3

16 -

EI 3

8 16

-

EI 2

9 -

EI 2

45

= EI 12

151 = 0.0020 radians

1y

w 45w

x 233

w =16x8

2w =162w

2ar

=ra1163 16x2

3

316x8

w =2 t /m

4t

Bw =2 t /m

C

A

16 t.m


yc = EI

1

2

3

2

9 - 1

2

45 - 3

3

8 16 1

4

9 3

3

16

= (- 16 + 2.25 + 128 – 22.5 – 6.75) EI

100 cm.

= 1.416 cm

FRAME DEFLECTION:

The moment–area, and elastic load methods can be used advantageously

in the computation of frame deflections. The frame deflections computed

in this manner, however, do not include the effect of axial changes in

length of the members. It is usually permissible to neglect the effect of

axial deformation in most frame–deflection problems.

Example (13)

Compute the deflections of the shown frame in Fig.21. Hence sketch the

elastic curve

Solution

The deflection at points A and C are equal to zero then;

Angles of slopes

W1 = 2

7.2

EI 2

6.21 = 38.88/EI

W2 = 2

7.2 .

EI 2

4.14 = 25.92/EI

RA = (7.2) EI

2.4 25.92 - 3.6 38.88

= 10.8 / EI

Rc = 2.16 / EI

A = RA

= 10000

80.10 = 1.08 10

-3 rad

= 0.0619


c = 10000

16.2 = 2.16 10

-4 rad

= 0.012

EIM

deformed frame

Diagram

B.M.D.

RC

A

AR

C

CA

0.24

y =2.3cmb

4= 21.6X10

cocoA

+

2EI21.6

2EI21.6

2EI7.2

-

+

2EI14.4

+

14.4

14.4

EI = 10000 t.m2

B

I

2I

12t12t

AD

C

E

c

E

Figure (21)


Deflections:

yB = (10.8 (3.6) – (19.44 – 6.48) 1.2) EI

1

= 10000

328.23

= 0.0023 m = 0.23 cm

To get ymax at = 0, assume the position at distance x from A then

= EI) 2( 2

x

7.2

4.14

8.10

EI

2

x

EI 2 3.6

21.6 -

2

= 10.80 + .5 x2 – 1.5 x

2

i.e;

x = 3.28 m ,

ymax = 3.28 8.10

EI

3

3.28

EI

3.28 -

2

= 0.24 cm

2

1w

10.8

y =2x

19.44

6.48

BA

10.8

BA C

RCAR

w

Since the joint at c is rigid, the tangents to the elastic curves of all

members meeting at that joint rotate through the same angle. Since there

is no bending moment in the column, the elastic curve is a straight line

inclined at the same angle as the tangent of the elastic curve of the beam.

E = c. 3.6 = .078 cm


Example (14):

Compute the slopes and

deflection of the given

compound beam shown in

Fig.22; compute the

maximum deflection.

EI = 3000 t.m2

Solution

After studying the sketch of

the elastic curve, it is

apparent that yb can be

computed by applying the

second moment – area

theorem to the part AB. This

deflection establishes the

position of chord BC. The

deflections and slopes of

beam BC can be calculated

by the elastic load method

applied to an imaginary

beam of span Bc.

C

Figure (22)

EIyb = 1.2 2

1.8 36

1.5

2

.9 18

= 1.7 cm

B = EI

5.40

Rotation of chord BC

BC = 540

7.1 = 13.510

-3 rad.

The point of maximum y occurs where the tangent to elastic curve is

horizontal i.e., where the tangent slopes down to the right w.r.t. the chord

BC or


EI m = 4.5

77.24 = 7.92

7.92 = 110.4 - 2

x

6.3

72

x2 = 10.248

x = 3.20 m

EI my = 110.4 (3.20)

- 3

(3.2) 10 3

= 244.05

EI ym = 244 + 7.92 3.2

ym = 3000

39.269

= 8.98 cm

64.8

mEI0

my

EIt.m72

EI

40.5EI

EI

A

18

B

54

Figure (22)

Example (15)

Compute the deflections at point B for the given frame shown in Fig.23

Solution

Joint c is rigid hence;

c = 3

2

EI 2

7.2 108

= EI

259 = 25.92 10

-3 rad.

EIB = c (3.6) (3.6) 3

2

2

3.6 108

= (259.2 3.6 + 466.56)


B = 13.99 cm

Note: Compare the elastic curve with that given in example 13.

B

C

2EI = 10000 t.m

108tm.

108tm.

30t

co

co

co

A

A

a) B.M.D.

b) Elastic Curve

Figure (23)


7. THE CONJUGATE BEAM METHOD

Form the previous examples; it is apparent that deflection computations

for any beam can be handled effectively by a proper combination of the

use of the moment area theorems and of the elastic load method. The

procedure involved in this combination, however, may be identified as

nothing more than a slight extension and variation of the elastic load

method. This extension is called the conjugate – beam method. In order to

develop these new ideas, consider the beam shown in figure 24.

Figure( 24)


Fig.24.a shows the real beam and its loading. Also shown by the dashed

line is the deflection curve, which has certain notable characteristics

determined by the type of supports and by the hinge at B:

1. At support A both deflection and slope of the elastic curve are zero.

2. At support C the deflection is Zero but the elastic curve is free to

assume any slope required; and

3. At hinge B, there can be a deflection, and in addition, a sudden

change in slope can occur between the left and right sides of the

hinge.

The objective is to select a conjugate beam i.e., a corresponding beam,

which has the same length as the real beam but is supported and detailed

in such a manner that when the conjugate beam is loaded by the

EI

Mdiagram of the real beam as an (elastic) load, "the elastic shear in the

conjugate beam at any location is equal to the slope of the real beam at

the corresponding location and the elastic bending moment in the

conjugate beam is equal to the corresponding deflection of the real

beam". Note that these slopes and deflections of the real beam are

measured with respect to its original position; i.e. they are the true slopes

and deflections. It is always possible to select the proper supports for the

conjugate beam to achieve the desired objective by simply noting the

known characteristics of the elastic curve of the real beam at its supports

or at any special construction features, such as the hinge at B. To

illustrate the selection of the supports of the conjugate beam, consider the

beam in the following Fig.25


Figure (25) Real Loaded Beam

At A, there is neither slope nor deflection of the real beam; therefore,

must be neither shear nor moment at this point of the conjugate beam.

That is, point A of the conjugate beam be free and unsupported. At C,

there is slope but no deflection of the real beam; therefore, there must be

shear but no moment in the conjugate beam. That is, point C of the

conjugate beam must be provided with the vertical reaction of a roller

support. At B, there is deflection and a discontinuous slope in the real

beam; there for, a vertical reaction must be provided to create the sudden

change in shear in the conjugate beam, and it must be capable of resisting

bending moment. That is, point B of the conjugate beam is loaded and

purported as shown in fig.26.

Figure (26) Conjugate Loaded Beam

These and similar considerations lead to the rules shown in Fig.27, 28,

and 29.

Illustration of selection of the supports and details of typical conjugate

beams are shown in Fig.28, and 29.

Not that statically determinate real beams always have corresponding

conjugate beams which are also determinate. Statically indeterminate real


beams appear to have unstable conjugate beams. However, such

conjugate beams turn out to be in equilibrium since they are stabilized be

the elastic loading corresponding to the M/EI diagram for the

corresponding real beam.

Figure (27) Selection of Supports and Details of Conjugate Beam


a b

dcba

b c

b c d

a

ab

fd eca cfe

d

d

d

c

c

c

c

b

b

b

b

a

a

a

a

a

a

bcb

a

b ca

cb

ba

a

b c

cba

ba

ba

stabilized by elastic loadUnstable equilibrium but

Ind

eter

min

ate

Str

uct

ure

sD

eter

min

ate

Str

uct

ure

s

Figure (28)

a b

dcba

b c

b c d

a

ab

fd eca cfe

d

d

d

c

c

c

c

b

b

b

b

a

a

a

a

a

a

bcb

a

b ca

cb

ba

a

b c

cba

ba

ba

stabilized by elastic loadUnstable equilibrium but

Inde

term

inat

e S

truc

ture

sD

eter

min

ate

Str

uctu

res

Figure (29)


Example 17

Calculate the deflection and the slope angle at points A, C, B, D, for the

given beam shown in Fig.30, draw elastic curve.

El = 4000 t.m2

Solution

1. Draw B. M . D. For actual beam

2. Construct the conjugate beam

3. Calculate Q & M at points A, B, C, & D as

BeamCONJ

M . D

EI

2EI

14EI

CA BD

D

EI6

4EI

EI10

10EI

aQ

36EI

12EI

EI1212

EI

EI36

9 t.m

4 t.m

c

2t

a 2 t /mb

EI3

18

Figure (30)

point A, To get A at conj. Beam

MB = O


El

12 2 -

El

36 3 + QA. 6 = O

A = El

14

MA = 0.0

Hence

QA = A

= El

14 = 0.0035 radians

MA = zero

Point B

B = QB

MA = 0

QB = (El

12 4 +

El

36 3 )

6

1

QBleft = El

10 = -0.0025 radians

MB = YA = zero

Point C, From left side

QC = EI

14+

EI2

32-

3

2

EI

93

= EI

1

c = - 0.0025 radians (anticlockwise)

yc = Mc


= EI

1( 14 3 + 3 1 – 18 3

8

3

= EI

75.24 m

= 0.618 cm (downward)

Point D, from part B D

D = Q

= El

6 = 0.0015 radians (antilock wise)

yD = MD = (EI

10 2 +

EI

4

3

2 2)

= EI

67.14

= - 0.36 cm ( upward)

bLeft

cY

=0.0035a Ac

b

D

DY

Figure (30) Elastic Curve


Example 18

Find A , C , B ,yd , draw

elastic curve and locate the

deformations on the elastic

curve. For the given beam

fig.31 El = 8000 t.m2

Solution

By using conjugate beam

Method

A = 8000

23 rad.

= 2.875 10-3

rad.

clockwise

B =8000

19

= -2.37510-3

rad.

anticlockwise

D = EI

15

= -1.875 10 -3

rad.

anticlockwise

c = EI

391823

4

12

18

9

84

9

9 t.m

4t 2t

D

4 t.m

bc

6 t.m

a 2 t /m

O aa

0.53

c

23

a

18 9

c

D

O

bd

O b Y

a

23

3

18 36

b

19 4 d

18

Figure (31)


= EI

1 =-0.125 10

-3

yc = EI

10016125.118323 = 0.53 cm Downward

yd = -( EI

3

44219

) =-0.408 cm Upward

Example (19)

Find yc , c , A , B for the shown beam in Fig.32

EI = 3000 t. m2

L = 4m P = 4t

Solution

A = EI

5.2

= 0.83 10 -3

B = .83 10 -3

c = zero

( from symmetry )

pL/8pL/8pL/4

B.M.D

EIB.M.D

modefied

2.5/EI 2.5/EI

1/EI1/EI 1/EI 5/EI 1/EI

ba

2/EI 2/EI2/EI

I

2I

I

P

ba

Figure (32)

yc = 3000

1( 2.5 2 -1 1.33 – 1 0.5 - 0.5 - 0.5 0.33)

= 3000

3 = 0.001 m

= 0.10 cm


Example (20)

Calculate the deflections and the angles of rotations at the given points

for the shown beam in fig. 33 . ( EI = 8000 t.m2 )

AO

Oe

Oe eO edcA b

10t.m

4t.m

4t.m

2t

2t

4t

4t.m

4t.m

10t.m

2II

5t4t

c d ebA

__EI8 20

10-2.67

7.3310-5.334.67

8.33

8

2.33

EI __

EI __

EI __

EI __

EI __

EI __

EI __2

Figure (33)

Solution

Elastic reactions:

rc = 10 – 5.33 = 4.67 / EI


re = 10 – 2.67 = 7.33 / EI

rA = EI

33.2

rB = EI

33.8

Point A , B

QA = rA = A = 0.25 10-3

radians( clockwise)

QBleft = EI

833.2

= -5.67 / EI = - 0.75 10-3

rad.

QBright = + EI

67.2

yA = MA = zero

yB = MB = (- 4.67 2 + 2 3

4)

EI2

1

= -0.10 cm (Upward)

Point C

c left = cright

= EI

67.4 = 0.58 10

-3 radians

yc = Mc = zero

Point d

Qd = EI

33.78

d = 0.085 10 -3

rad.


yd = ( 7.33 4 – 8 3

4)

EI

1

= 0.235 cm (downward)

Point e :

e =Qe = - EI

33.7 = 0.915 10

-3

ye = Me = zero

8. THEORY OF REAL WORK

If a variable force F moves along its direction dL, the real work done is F

dL. The total work done by F during a period of movement may be

expressed by

W = 2

1

L

L

dLF

Where L1 and L2 are the initial and final values of position. Consider a

load gradually applied to a structure. Its point of application deflects and

reaches a value as the load increase from 0 to N. As long as the

principle of superposition holds, a linear relationship exist, between the

load and the deflection a represented by the line Oa in the following

figure (Fig.34)


C

da

b

FLoad

deflectionO

Fig. 34

Figure(34)

The total work performed by the applied load during this period is given

by:

W =

0

FdL

= 2

1N.

Which is equal to the area of the triangle Oab in Fig.34. If further

deflection d, caused by an agent other than N, occurs to the structure in

the action line of N, then the additional amount of work done by the

already existing load P will be;

dW = N.d

Which equals the rectangular area abcd shown in Fig.34. Similarly the

work done by a couple M to turn an angular displacement d is M.d .

The total work done by M is:-


W = 2

1

Md

Also, the work performed by a gradually applied couple M accompanied

by a rotation increasing from O to is given by :=

W = 2

1 M.

Now consider a beam subjected to gradually applied force. As long as the

linear relationship between the load and the deflection maintains, all the

external work will be converted into internal work or elastic strain

energy. Let dW be the strain energy restored in an infinitesimal element

of the beam as shown in the following figure (Fig.35)

Figure(35)

We have dW = Md2

1

If only the bending moment M produced by the forces on the element is

considered significant. Using;

dx

d =

EI

M ,

EI

M

dx

yd2


Or d = dx EI

M

We have

dW = EI

M

2

xd . 2

For the loaded beam, its longitudinal axis taken as the x-axis, we let dL =

dx. The total strain energy restored in the beam of span L is, therefore,

given by:

W = L

EI

M

0

2

2

xd .

For a truss subjected to gradually applied loads, the internal work

performed by a member with constant cross sectional area A, length L,

and internal axial force N is:

dW = EA 2

L d . N2

, ( E = LL

AN

/

,

)

The total internal work or elastic strain energy for the entire truss is:

W = EA 2

L . N

2

In some special cases deformations of structures can be found by equation

of conservation of energy:

External work (WE) = Internal work (WI)

Or WE = WI

Example (19)

Find the deflection at free end of the loaded cantilever beam shown in fig.

36.


Solution

WE = 2

1 P. b

WI = L

EI

M

0

2

2

xd .

Figure(36)

= L

EI0

2

dx 2

Px -

= EI 6

P

32 L

Setting

WE = WI

b = EI 3

3pL

Note that the method illustrated is quite limited application since it is only

applicable to deflection at a point of concentrated load. Furthermore, if

more than one load is applied simultaneously to a structure, then more

than one unknown deformation will apear in one equation, and a solution

becomes impossible. Thus, we do not consider this as a general method.

8.1. Deformation and Work due to Normal Force N

From hook's law

E =

=

/

dx

d

AN

d = EA

dxN .

i.e. dW = 2

. dN

dW = EA 2

.2 dxN


8.2. Deformation and work due to M

d = change in the slope

of elastic curve

d = EI

M xd .

dW = 2

d . M

dW = EI

M

2

xd . 2

8.3. Deformation and work due to shearing force Q

= G

,

G = ) (1 2

E

where:

G = shear modulus, or shear

rigidity

dy

dy

Q

= poisson's ratio (3

1for steel and

6

1for concrete)

Using

= rA

Q , =

dx

dy

dy = .dx = rGA

Q. dx

Where Ar reduced area of cross section. It depends on the shape of cross

section:-

Ar = 0.9 A for rectangular section (steel)

Ar = 0.83 A for rectangular section (conc.)

i.e. dW = 2

dy . Q or dW =

G . A 2

dx .

r

2Q


8.4. Deformation and work due to torsion

dW = Mt. d/2

= G

where

= PI

R .Mt

= . G

= G p, I

R .Mt

but = dx

dR

R.d = .dx

i.e. d = p

t

GI

M dx . ,

i.e. dW = 2

d . tM, d =

6 .

dx . t

pI

M

dW = p

t

I

M

G . 2

dx . 2

Where

Ip = polar moment of inertia

Ip = Ix + Iy

For rectangular cross sections;

d = t

t

IG

M

.

dx .

Where

It = torsional moment of inertia of the cross section.

For rectangular cross sections:


It = 3

1a b

3(1–0.63

a

b

+ 0.052 5

5

a

b )

Where, a > b

For I – beams (steel) or channels;

It = 3

ab

3

For Hollow cross sections

It =

t

U

4F

2m

Where

Fm = closed area (dashed) between

the center line of the perimeter

of the area

U = length tangential to the

perimeter of the cross section

t = thickness of cross walls

For Hollow rectangular sections

It =

21

22

t

b

t

a 2

ba 4

For hollow sections with more than one

cell, neglect the interior webs.


Hence for an element subjected to N, M, Q; and Mt

WI = EA 2

2dxN +

EI

M

2

dx2

+ r

2

GA 2

dxQ +

p

t

I

M

G . 2

dx 2

This is the general equation of real work where

EI = Flexural (bending) rigidity

EA = Normal rigidity

GAr = Shear rigidity

8.5 Deformation due to temperature

a- uniform temperature change:

The strain due to temperature change t is

t = . t

Where:

= Coefficient of temperature deformation

= 1.2 10-5

for steel

= 1.0 10-5

for concrete

L = . t. L

t = temperature change

L = length of member

L = free elongation

If a member is not free to deform, then the stresses would arise in the

member with value. From Hook's low

= E. t = E. . t

For example if two hinged concrete beam with span L and

t = 40,

= 510

40 210000

= 84 kg/ cm2 (big)


b- Non- uniform temperature change

The uniform temperature change 2

t t 21 causes a uniform strain

t = . 2

t t 21

And the non-uniform change of t causes a rotation angle:

dt = h

dL t. .

Note

In case of the length is dL

Work due to Uniform rise of Temperature

W = L N. L

0

= L

0

dL . N. t

= dL .N t L

O

= t (area of normal force diagram)

Work due to Nonuniform rise of Temperature

W = L N. L

O

= d M. L

O

= t (Area of N. diag.)

+ h

t (Area of M. diag.)


9. METHOD OF VIRTUAL WORK

Bernoulli's principle of virtual work for rigid bodies is the most general

and direct method for computing the deflections of all types of structures.

This method is based on an application of an alternate form of the

principle of virtual displacements, which was originally formulated by

John Bernoulli in 1717. This alternate form of these ideas can be

developed from the following considerations.

Figure(37) Translation of Rigid Body

Consider a rigid body shown in fig 37 which is in static equilibrium under

a system of forces Q. In this sense, a rigid body is intended to mean an

undeformable body in which there can be no relative movement of any of

its particles. Suppose first that, as shown in the above fig., this rigid body

is translated without rotation a small amount by some other cause which

is separate from, and independent of, the Q-force system. Upon selecting

an origin O and two coordinate reference axes x and y, this translation

may be defined by o, the actual translation of the origin O, or by the two

components ox and oy in the x and y directions, since the body is rigid,

every point on the body will translate through exactly the same distance

at point O. All the Q–forces can be resolved into x and y components,


designated as Qnx and Qny for any particular force Qn. Since these Q

forces are in static equilibrium, the following equations are satisfied by

the components of these forces:

Qnx = 0

Qny = 0

(Qnx. yn – Qny. xn) = 0

Consider now work WQ done by only these Q forces. All Q forces may be

assumed to maintain the same position and direction relative to the body

and to each other and hence to remain in equilibrium during the

translation. Then we can write,

WQ = (Qnx. ox – Qny. oy) = 0

= o Qnx + oy Qny = 0

The total work done by the Q forces in such a case is equal to zero.

Similarly, the total work done by the p forces during a small of the rigid

body above point 0 also equal zero. Hence the following Bernoulli's

principle, it may be stated as:

"If a rigid body is in equilibrium under a system of loads, and remain so

when it is subjected to any small displacement, the virtual work done by

the P- force system is equal to zero".

Bernoulli's principle of virtual work for rigid bodies can now be used to

develop the basis for the method of virtual work for computing the real

deflections of structures. This method is applicable to any type of

structures-beam, truss or frame, planar or space frameworks. For

simplicity, however, consider any planar structure such as shown in the

figure 38.


Figure(38) Planar Structure in Equilibrium under Q-force System

For the shown structure, it may by stated that;

"If a deformable structure is in equilibrium under a virtual P-load system

and remains so while it is subjected to a small virtual deformation, then,

the external virtual work (We) done by the P–load system as it moves

through the virtual displacement is equal to the internal virtual work (Wi)

done by the internal straining actions produced by Q–load system as they

move through the virtual deformation"

We = Wi

The term virtual deformation means that the action producing the

deformation is independent of the Q–load system or that it is caused by

some additional action. Such action may be another load system, hence,

referred to a P- load system, temperature, error in lengths of members, or

other causes or whether the material follows Hook's law or not. Also, the

virtual work refers to the work by the P-load system during the virtual

deformation.

It should be noted that the external virtual work (We) is the work

produced by the P-load system only as it moves through the virtual

displacement, and that the internal virtual work (Wi) is the work done by

the internal forces produced by the Q- load system as they move through

the virtual deformations. The equation We = Wi is the basis of the method

of virtual work for deflection computation. With some assumptions, it

may be used to calculate any deflection component at any point of a given

structure. This is done by choosing a P-load system consisting of a single

unit load and placed on the structure at the point where the deflection is

required and in its direction before the structure is subjected to the actual


Q–load system. The Q-load system will then be considered as the source

of the virtual displacement, which in effect is the actual displacement

required.

9. APPLICATION

9.1. In cases of Beams and Frames

Figure (39)

The frame shown in Fig.39.a loaded by a Q-load system, the deflection at

point C is required. The procedure of solutions as follows;

a) The Q – load system is removed and a P–load system, consist of unit

load at point C, is placed.

b) The internal force diagram produced from 1t at C are drawn and

denoted by N1 , Q1 , and M1 diagrams

c) The Q–load system is added to the frame already loaded by the unit

load. The Q–load system produces another internal forces are denoted

by No, Qo, and Mo. The deformation in element of frame, dx Length

are as follow;

d = EA

.0 dLN

dy = L d . GA r

0Q

d = EI

.0 dLM

dtor = G .I

.

p

dLM t


The actual load (Q–load system) causes a deformed shape with different

deflection at any point along the beam span. The deflection at C denoted

by c.

d) Consider that the unit load at C is the original load and that the Q–

load system to be the source producing the virtual displacement, then

We = 1 c

And the internal virtual work in the length dx of beam is

dWi = M1 d + Q1 . dy + N1 . d.

And the total internal work is

Wi = L

O

M1 d + L

O

Q1 . dy + L

O

N1 . d.

Hence form We = Wi and substitute for the value of d , dy , and d;

1c = L

O

EI

01MM d x +

L

O

EA

01NN d x +

L

O

r

0 1

GA

QQ d x

For the most member subjected to bending

moment, however, the deflection due to shear

force is small, then:-

1c = EI

01 MM d x

If the angle of rotation is required we can

applied by unit moment at desired point

(Fig.40).

1c = EI

01 MM d x

Figure (40)


Where M1 is the bending moment diagram due to 1 t.m at point C.

If the horizontal

displacement is required at

point C, we can apply unit

load at C as shown in

Fig.40.c.

1.C = dxMM 01

Figure (40.c) hl. Displacement

If the relative vertical displacement is required, the unit loads are as

shown in Fig.41.a and b.

a. Relative angle of slope at C b. Relative displ. Of c and D CD

Figure (41)

10.EVALUATION OF INTEGRAL dxMM 10

This integral which appears in the computing of deflection, is actually an

integral of two bending moment diagrams M1 and Mo. At least either the

M1 or Mo – diagrams are a straight line while the other may be a single

straight line, broken line or curve. It can be readily evaluated by use of

the following simple formulae;

10.1. Mo–diagram is a curve and M1–diagram is linear


L

O

M1 Mo dx = A.C

where:

A = Area of Mo.D

C = ordinate of linear M1 .D opposite to centroid of Mo.D

10.2. Mo. D and M1. D are Both linear

a) L

O

M1 . Mo dx = (Area of M1 or Mo) ordinate of Mo or M1

respectively opposite to centroid. L

O

M1 Mo dx = 3

L (ac + bd +

2

1 ad +

2

1 cb)


b) L

O

21M d x =

3

L (a

2 + b

2 + ab)

c) L

O

21M d x =

3

L a

2

d) L

O

21M d x = L a

2

M . D1

1M . D

1M . D

L

a a

L

a

C a

b

1- Mo and M1 Diagrams are both second degree Parabolas

L

O

M1 Mo dx = 5

4 A1 .b

= 5

4 A2 .a

= 15

8 Lab

where

A1 = area of Mo. D

A2 = area of M1 . D

2- Composite M1-Diagram

L

O

21M d x =

L

O

(M2+M3)2dx

= L

O

22M d x +

L

O

23M d x +

L

O

2 23 M M d x

Every integral evaluated by the same above methods.


Example (21)

Calculate the vertical deflection at

point d and the angle of rotation at

point A for the given beam. Shown

in Fig.42 (EI = 8000 t.m2)

Solution: Draw MO . D

deflection at d, draw M1 .D hence;

1 yd = EI

dx . . 1MM o

yd = 2

2 8 2 (-

8000

1

2) 3

2

2

2 4 8

= 0.33 cm

angle of rotation at A

put 1 t.m at A; draw M1-diagram

Figure(42)

1 A = dx . EI

. 1MM o

=

5.0

2

8 8

3

1

2

8 4 -

8000

1

= 1.33 103 radians (clockwise)

Example (22)

Determine the vertical deflection at point n, the horizontal displacement

at B and the angle of rotation at C. for the given frame shown in Fig. 43

(consider effects of M,N and Q)

EI = 25200 tm2 , EA = 63000 t, GAr = 49200 t.


Figure(43)

Solution:

1- Draw Mo, No and Qo

Figure (43.a)

a) Vertical deflection at n

Figure (43.b)


1n = 30 2

2 1

3

2

2

2 30

25200

2

2 49200

.5 2 15

63000

0.5 5 15

= .0099 m = 0.99 cm

b) Horizontal displacement at B: (put 1t at B)

Figure (43.c)

1B = EI

01

MM .d x +

EA 0

1

NN . d x +

r

10

GA

QQ d x

= 0 0.625 63000

5 15 -

2

5

2

4 8 30

25200

1

= 0.017 m =1.7 cm

c) The angle of rotation at C (put 1 t.m at c)

Figure (43.c)

t


1 c = EI

. 10 MM .d x +

A.E

. 10 NN .d x +

G A

.

r

10 QQ .dx

1 c = 0 8

1

63000

5 15 - 0.5

2

4 8 30

25200

1

= 0.0035 radian clockwise

Note.

a) due to load

1. = M1 d + N1 . d. + Q1 . dy

= M1. EI

.0 dLM +

EA

dL 01NN

+ p

0 1

GI

QQ

b) due to temperature

1. = N1 . d. + M1 d

= . t. N1.d L +h

t . M1. d L

Example (23)

Find the horizontal displacement of support B (in example 22) due to

uniform rise in temperature 20 C (Fig.44) = 1 10-5

WE = wi

Solution

Mo = zero

No = zero

Qo = zero

d = . tL

d = zero

dy = zero

1 . B = N1.d. + M1d

= N1 . d.

= N1 . d.

= . t. N1 .L.

Figure (44.a)


= 1 10-5

20 (1 8 + 0.625 5)

= 0.22 cm

Example (24)

For the frame shown in Fig.44, calculate the horizontal displacement at B

due to vertical downward displacement at A equal to 1.0 cm as shown in

Fig.44.c.

Figure (44.b)

Solution

a) From the external work WE equal to internal work WI equal zero, then

WE = WI = 0

i.e. WE = 0

1.B + 0.625 1.0 = 0

B = -0.625 cm (to left)

b) Due to vertical and horizontal movements = 0.5 cm to outward. at A.

From WE = 0

1.B + 0.6250.5 +10.5 = 0

B = - 0.8125 cm (to left)


Figure (44.c)

Example (25)

Find the change of the angle at

C due to a rise in temp. of 20 C

at interior fibers and 40 C at

the exterior fibers. For the

shown frame in Fig. 45

(Section 30 100)

= 1 10-5

/C

Solution:

d = 2

21

= . 2

)t (t 12 . dL

= 30 10-5 dL

Figure(45)

d = .h

)t (t 12 . dL

1t

1t

0.625 t


d = 00.1

01 1 5- (40 -20) . dL

( -ve sign because it is in the same

sense of – ve B.M)

change of the angle at C = c

1xc = M1 d + N1 . d

= -20 10-5

M1 dL + 30 10-5 N1 . dL

= 2 – 20 10-5

5-10 30 10.44 .991 10.44 2

1 7.0

2

7 7.0

= - 4.8 10-3

rad.

i.e. increase in angle at C = .0048 rad.

Example (26)

Find B , c , d , B, yc, yd , xd for the shown frame in Fig.46

EI= 10000 t.m2

Figure(46)

Solution:

1- Draw Mo. D


Figure (46.a)

2- Draw M1-diagram for each case.

d bc

Figure (46)


a) angle of rotations

1 = EI

dx.M. M o1

1b = (12 8) EI

1 = 9.6 10

-3 radians

1 c =

2EI 2

1 3 12

EI

90 =

EI

89

= -310 8.9 rad.

d = c =8.9 10-3

rad.

b) displacement

1 = dx . EI

M. M o1

1 b = 100

8 0.5

EI

8 12 = 0.384 cm (xb = xc)

1 yc =

2EI

5 4 12 6

EI

4/2 8 - 8 12

= 5.4 cm. (yc = yd)

1 d = 6 2EI

4 5 4 - 5

3 EI

8 2 - 8 12

= 1.4 cm,

Example (27): Find the horizontal displacement at point b for the given frame shown in

Fig.47.

El1 = 4000 t.m2

El2 = 8000 t.m2

Figure (47)

Solution

B1 = dx . EI

M. M o1


= 4000

1 4

3

2

2

5 18( 2

8000

4- ). 6 4

3

2 4 (18

= - (.06 + 0.044)

= - 0.104 m

= - 10.4 cm (to right)

Example (28)

For the given hinged arch

shown in Fig.48 calculate

the vertical deflection at C

due to horizontal`

displacement at B = 0.5

cm (to Left)

Figure (48)


Solution:

Apply 1 ton at c

We = Wi = 0

1. yc – 1.5 0.5 = 0

yc = 1.5 (.5)

= 0.75 cm

Figure (48)

Example (29):

For the given frame shown in Fig.49

with variable moment of inertia,

compute the displacement of point B

relative to point A. EI = 10000 m2 . t

Solution:

1) Drow Mo.D due to given loads.

2) To find BA , apply a unit loads

at B and A in direction of AB, and

draw M1. D.

BA = EI

. 1 dxMM o

Figure (49)

A

12


= 1.8) 3

6 9

3 3

6 3.33 6

3

6(

EI

1

= 0.599 cm

Positive sign, which mean that B moves away from A in direction of AB.

Example (30):

For the shown arched–frame shown

in Fig.50, has a parabolic arch,

Determine the horizontal

displacement at B.

EI = 100000 t.m2

Solution:

1) Draw Mo.D due to given load.

2) Drow M1. D due to unit load

at B

Figure (52)

B = dxMM o

EI

. 1

= - 1.4 cm (to right)


Example (31):

For the shown cantilever, Fig.51, it is

required to calculate yb.

E = 200 t/cm2,

Solution:

EIo.B = M1 Mo dxIo I

= M1 Mo )( I

dLIo

Assume

Io = 12

1 3.0 3

Figure (51)

Io = .025 m4

(Io = I at point A)

EIo = 50000 t.m2

I

oI =

3

d

od

The following Table gives the results

Part D

(m) I

oI

Mo

(t.m.)

M1

t.m. Mo . M1

I

oI

1

2

3

4

1.0

0.85

0.70

0.55

0.40

1

1.628

2.915

6.011

15.62

-50.0

-37.5

-25.0

-12.50

0

-5.0

-3.75

-2.50

-1.25

0

250.00

194.59

127.53

51.65

0

623.785

B = 0EI

1.25 785.623 100 = 1.56 cm


Example (32):

For the shown cantilever in Fig.52 with variable

moment of inertia, calculate the vertical defection

and rotation of point C.

Solution:

The structure is divided into number of divisions

as shown

depth

cm

Point

50

80

120

C1

B

A

Figure (52)

E = 200 t/ cm2

B = 30 cm

Choosing

Io = Ic

= 12

50 03 3

= 6250

EIo = 6250 t. m2

1.yc = EI) / (L EI

M . M

o

o1

= L I

I M M

EI

1 oo1

o

1.c = L I

I M M

EI

1 oo2

o

Figure (52)

Calculations are shown in the follow Table

Sec L

m

d

cm

3

1

oo

d

d

I

I

M0.

t.m. M1 yc M2 c

C

1

2

B

0.5

1.0

1.0

0.5

50

60

70

80

1

0.578

0.364

0.244

0

-3

-8

-15

0

-1

-2

-3

0

1.73

5.83

5.50

-1

-1

-1

-1

0

1.736

2.916

1.83


B

3

4

5

A

0.75

1.5

1.5

1.5

0.75

80

90

100

110

120

0.244

0.172

0.125

0.094

0.072

-15

-18

-21

-24

-27

-3

-3

-3

-3

-3

8.20

13.9

11.81

10.14

4.39

-1

-1

-1

-1

-1

2.146

4.63

3.937

3.38

1.46

61.53 22.64

yc = 100 6250

53.61 =0.985 cm

c = 6250

64.22 = 0.003622 rad. = 0.003622

180

= 12\ 27

\\

Example (33):

For the given beam, Fig.53, fixed at

A and spring support at B. If the

vertical deflection at B = 2 cm

draw the S.F and B.M. Ds EI =

2000 t.m2

(Hint: this beam is satically

indeterminate)

Solution:

1) Draw Mo. D due to given

load. By assume the reaction at B =

RB

(If we know RB, one can Draw

B.M.D)

2) Draw M1. D due to unit load

at B

1B = EI

1 . 1 dMM o

100

2 1 = 4.5 6 36

3

1(

1

EI

) 4 6 2

y 6 - 4 6

2

12 B

100 = 468 – 72yB

yB = 4.94 t.

Figure (53)

A

M = - 48 + 6 4.94 = - 18.33 t.m


DEFLECTION OF TRUSSES

Figure (54.a)

For the given truss shown in Fig.54:

1- Given Q load system then calculate No for each member

2- Apply P–load system unit load and calculate N1 for each member

The truss members have generally normal forces only (N), from the

virtual work equation we have generally:

1. = N1 . d + N1 . .t dx

= case of load + temp. effect

= all

members EA

L NN o1 +

affected

members

N1 . . t .L

3- To calculate displacement of any point in a truss, one apply the

virtual unit load in the direction of the required displacement at this point.

4- Calculation of Relative Displacement


Figure (54.b)

The calculation relative displacement between two points c and d in a

truss, shown in Fig.54.b; apply the virtual unit loads in the two points c, d

in the direction c-d, calculate for this case of loading normal forces in

each member N1, and from the virtual work equation, calculate the

relative displacement.

1.cd = EA

L 01NN

5- Calculation of Member Rotation

Figure (54.c)

For a member rotation, apply unit moment at the member as shown in

Fig.54.c, to produce unit moment, the two virtual loads should be equal to

h

1 in case of member BE or (

cdL

1) in case of member c d.

Example (34)

For the given truss shown in Fig.55, compute the vertical deflection due

to given loads and due to rise in temperature t = 30 of the upper chord

members at point 7. (A

L = 20), E = 2000 t / cm

2.


Figure (55)

Deflection at point 7

a) due to given load

1- It is easy for the given load 20t at point 7 to determine the forces in

all members (No)

2- For 1t at point 7 the forces in members (N1) is 20

1 from (No) then.

3- 1. 7 = L . EA

1NNo (

EA

L = 0.01 cm/t)

= 0.01 No. N1

The calculation is tabulated in the following table;

Chord Member No

tons

N1

tons EA

L

cm/t EA

L . 1NNo

Cm

Lower

chord

A-2

2-4

4-6

0

10

20

0

0.5

1

.01

.01

.01

0

.05

.20

upper

chord

1-3

3-5

5-7

-10

-20

-30

-.5

-1.0

-1.5

.01

.01

.01

.02

.20

.45

diagonals 1-2

3-4

5-6

14

14

14

.7

.7

.7

.01

.01

.01

.098

.098

.098

Verticals 1-A

3-2

5-4

-10

-10

-10

-10

-.5

-.5

-.5

-.5

.01

.01

.01

.01

.05

.05

.05

.05


2

1 (7-6)

2

1 = 1.40 cm

7 = 2 1.4 = 2.8 cm down ward

b) due to temperature change

17 = N1 . .t.L

= N1 . L (10-5

) (30)

The calculation of 7 due to temperature change of upper chord is given

in the following table.

Member L

(cm) . t .L

(cm)

N1

(t) N1. . t .L

1-3

3-5

5-7

200

200

200

0.06

0.06

0.06

-0.5

-1.0

-1.5

-0.03

-0.06

-0.09

2

1 = -0.18

1 7 = 2 (-0.18) = - 0.36 cm (upwards)


Example (35)

For the given truss shown in

Fig.56, calculate the vertical

deflection at joint 2 and the

horizontal displacement, at

the roller support.

A

L = 10 cm,

E = 2000 t/cm2.

Figure (56)

Solution

a) Vertical 2

The calculation may be tabulated in the following table. In this case (N1 =

5

No )

b) Horizontal displacement at A

Calculate 1N due to horizontal unit load at A

Member No N1 1N No. N1 No.

1N

A-B

B-1

A-1

A-2

1-2

-5

5

-5

-5

5

-1

1

-1

-1

1

-1.16

0

0

0

0

5

5

5

5

5

5.8

0

0

0

0

25 5.8

1. 2 = EA

L )N . N( 1o

= 2000

10 25

= 0.125 cm

1. A = EA

L )N . N( 1o

= 2000

10 .85

= 0.29 cm


9. CASTIGLIANO' S SECOND THEOREM

In 1897, castigliano published the results of an elaborate research on

statically indeterminate structures in which he used two theorems which

bear his name.

Castigliano's second theorem

"In any structure the material of which is elastic and follows Hook's law

and in which the temperature is constant and the supports unyielding, the

first partial derivative of the strain energy with respect to any particular

force is equal to the displacement of the point of application of that force

in the direction of its line of action".

In this statement, the words force and

displacement should be interpreted also to

mean couple and angle of rotation,

respectively. Consider the beam shown in

fig.57, loaded gradually by forces P1, P2, ...

, Pn.

Figure (57)

Then the external work done by these forces is some function of these

forces. According to the principle of the conservation of energy, we know

that in any elastic structure at rest and in equilibrium under a system of

loads, the internal work or strain energy stored in the structure is equal to

the external work done by these loads during their gradual application.

Wi = We = F (P1 , P2 , ... , Pn) (1)

Suppose now that the force Pn is increased by a small amount (dPn), the

internal work will be increased by, and the new amount will be.

iW = Wi +

nP

Wi

. d Pn (2)

The magnitude of the total internal work, however, does not depend upon

the order in which the forces are applied; it depends on the final value of

these forces. Further, if the material follows Hook's law, the deformation

and deflection caused by P1 , P2 , ... , Pn. and hence the work done by

them are the same whether these forces are applied to a structure already

acted upon by other forces or not, as long as, the total stresses within the

elastic limit. If therefore, the infinitesimal force dPn is applied first and

the forces P1, P2, ..., Pn. are applied produces an infinitesimal


displacement (dn), so that the corresponding external work done during

the application of (dPn) is a small quantity of the second order and can be

neglected. If the forces P1, P2, ... , Pn. are now applied, the external work

done just by them will not be modified owing to the presence of dPn and

hence will be equal to the value given in eqn.(1). However, during the

application of these forces, the point of application of dPn is displaced an

amount n. let the total amount of external work done by the entire system

during this loading sequence by We . Then,

eW = We + dPn. n (3)

But, according to the principle of conservation of energy, We must be

equal toiW ,

We + dPn. n = Wi + n

i

P

W

. dPn (4)

However,

We = Wi

n = n

i

P

W

(5)

This equation is the mathematical statement of castigliano's second

theorem.

Castigliano's first theorem

" In any structure the material of which is linearly or nonlinearly elastic

and in which the temperature is constant and the supports are unyeilding,

the first partial derivative of the strain energy with respect to any

particular deflection component is equal to the force applied at the point

and in the direction corresponding to that deflection component"

n

Wi = pn (6)

Castigliano's theorems are used principally in the analysis of statically

indeterminate structures, although it is sometimes used to solve deflection

problems.


Example (36)

Compute the deflection of point b at

cantilever beam shown in Fig.58

Solution

Wi = L

O

2

EI2

dx M

Figure (58)

p

w i

= b =

L

OEI

M.

p

M

dx

M = -Px

p

M

= -x

Therefore

b = L

O

P x (-X ) EI

dx =

L

O

3

EI3

Px

hence

b = EI3

PL3

Example (37)

Compute the slope at A for the given structure (Fig .59).

Figure (59)

Solution

Suppose M1 was applied at a, considering this as part of load system


Wi = EI2

M2

dx

a

i

M

w

= a = EI

M

aM

M

. dx

From a to b

M = Ma + ( 7 - 20

Ma ) x

aM

M

= 1-

20

x

From d to c

M = ( 13 + 20

aM)x – 10 (x-4)

aM

M

=

20

x

From c to b

M = ( 13 + 20

aM)x- 10(x - 4)

aM

M

=

20

X

Then, (and put Ma = 0 since is imaginary load)

Ela

i

M

w

= EI a =

10

O

7 x( 1- 20

x) d x +

4

O

1320

xdx

+ 10

4

( 3x+ 40 ) 20

xdx

= (21

7 3x

60

x7 3

)0

10 +(

60

x13 3

)0

4 + (

60

x3 3

+x2)

0

10

a =EI

378 radians


11.MAXWELL'S LAW OF RECIPROCAL

DEFLECTIONS; BETTI' S LAW:

Maxwell's law is special case of the more general Betti's law. Both laws

are applicable to any type of structures, whether beam , truss, or frame, to

simplify this discussion, however, these ideas will be developed by

considering the simple truss shown in Fig.60. Suppose that the truss is

subjected to separate and independent systems of forces, the system of

forces Pm and Pn. The Pm system develops the normal forces Nm in the

various members of the truss, while the Pn system develops Nn. Let us

imagine two situations. First, suppose that the Pm system is at rest on the

truss and that we then further deform the truss by applying the Pn system.

As a second situation, suppose that

just the reverse is true, i.e. that the

Pn system is acting on the truss and

that then we further deform the truss

by applying the Pm system. In both

situations, we may apply the law of

virtual work and thereby come to a

very useful conclusion known as

Bettis law.

Pm mn = Nm . L

where

Pm1 m2P

m3P

n2P

n1P

n3P

Fig. 59

Figure (60)

mn is the deflection of point of application of one of the forces Pm (in

direction and sense of this force) caused by application of Pn force

system.

L = EA

L . Nn

then:

Pm mn = Nm Nn EA

L

In the second situation, however the Pn forces will now be in the role of

the virtual Q forces, the deformation caused by Pm


Pn .nm = Nn . L

Where nm is the deflection of point of application of one of the force Pn

caused by application of Pm forces.

L = Nm . L/EA

then:

Pn .mn = Nn Nm EA

L

Hence

Pm mn = Pn .mn

Which when stated in words is called Bett's law.

BETTI'S LAW

"In any structure, the material of which is elastic and obey hook's law

and in which the supports are unyielding and the temperature constant,

the external virtual work done by a system of forces Pm during the

deformation caused by a system of forces Pn is equal to the external

virtual work done by the Pn system during the deformation caused by the

Pm system"

11. MAXWELL'S LAW

This suggests that Maxwell's

law of reciprocal deflection can

be derived directly from Betti's

law consider a beam as shown

in Fig.61, hence

P .12 = P .21

12 = 21

2a = a2

figure (61)


Maxwell's law of reciprocal deflections:

"In any structure as shown in fig. 62 the material of which is elastic and

follows Hooke's law in which the supports are unyielding and the

temperature constant, the deflection of point 1 in the direction ab due to

P at ponint 2 acting in a direction cd is numerically equal to the

deflection of point 2 in the direction cd due to a load P at point 1 acting

in direction ab"

12 = 21

Figure (62)

12. INFLUENCE LINE FOR DEFLECTION:

Suppose that we wish to draw the influence line for the vertical deflection

at point a on the given beam Fig.63. The ordinates of such an influence

line can be computed and plotted by placing a unit vertical load

successively at various points along the beam and in each case computing

the resulting vertical deflection of point A. In this manner, when the unit

load is placed at any point m, it produce a deflection am at point a, or

when placed at some other point n, it produces a deflection an at point a.

Note however, the advantage of applying Maxwell's law to this problem.


Figure (62)

If we simply placed unit load at a the deflection ma and na at points m

and n will be equal to am and an, respectively. In other words,

"The elastic curve of the beam when unit load is placed at point (a) is the

influence line for the vertical deflection at point a".

To obtain the influence line for the deflection of a certain point, simply,

place a unit load at that point and compute the resulting elastic curve.


PROBLEMS

(1) By using the double integration method, Moment – area Method, and

elastic load Method Determine:

a- deflection at points C, d

b- Slope at pint a, b, d

c- Maximum deflection for cases 3 and 5

d- Draw elastic line

E = 2100 t/cm2

I = 70000 cm4

(2) Determine by using conjugate beam method for the following

beams;

a- Slopes at points a, b

b- Deflections at points c, d, e, f

c- Relative slopes at intermediate hinges

d- Sketch the elastic line for the given beams.


(1)

(2)

(3)

(3) a- A horizontal paneled floor beam composed of three simply

supported main girders and across girder as shown. Draw

B.M.Ds for all beams due to concentrated load P at the center of

the floor,

EI = constant for all beams.

(3) b- Determine the slope at A and deflection at A , B for the shown

cantilever


(3) c- For the given beam, if the deflection at the spring support = 0.5

cm, EI = 2000 m2t, Draw S.F.D. & B.M.D. and find the stiffness

of spring k

(4) By using virtual work method, Determine;

a- Horizontal displaement at support A.

b- Vertical deflection at c, D.

EA = 45000 t

EI = 30000 m2.t


EI = 18000 t.m

2 EI = 15000 t.m

2

(5) For the given frame, determine

a- Relative rotation,

horizontal displacement,

vertical displacement at

the intermediate hinge (at

c)

b- Relative displacement db.

c- Rotation at a

El = 10000 m2t.

(6) For the shown frame, calculate

a- Relative vertical deflection between b, b-

b- Relative rotation

between b, b-

c- Relative horizontal

displacement bet. c,


c-

El = 20000 m2t.

(7) For the following trusses it is required to determine

a- Vertical deflection at c, d, e

b- Horizontal displacement at h

c- Relative displacement gf in truss (a)

(EA = 30000 t, EI = 25000 m2t.)

(8) a- For the given frame, it is required:-

1- Horizontal displacement at B

2- Rotation at A

3- Relative displacement between c, A due to

a- Uniform rise of temperature 20

b- Non-uniform rise of temperature from – 20 to + 20 for the

inner and outer sides respectively

= 1 10-5

, h = 1.0 m , EI = 12000 t.m2


B

C

A

(9) Draw B.M.D., S.F. D. for beam C D if the deflection at k = 0.8 cm

EI = 3000 m2t.


(10) For the shown structure, determine the vertical deflection at G and

the horizontal movement at roller B

El = 2 104 m

2t.

5t 5t

5t 5t 5t

10t

B

A

(11) Using the method of virtual work, compute the vertical component

of the deflection of joint d due to the given load, for the shown

trusses also horizontal component at roller support

E = 2100 t/ cm2 , A = 50 cm

2

a)

b

10t

d

e

c

a


b)

c) e

cb

d

12t

d)

fa

5t

c

b e

d

e)

A

20t

5t

e

c d

b


f)

a b2 t /m

2EI = 3000 t.m

g)

ba

b

b

3t

3t

3t

3th

(12) Draw the influence lines for the displacement at the roller support

and the relative rotation at the intermediate hinge

EI = 20000 t.m2

A B

D

B

A

C P

(13) Draw the influence lines for the relative displacement between c and d cd. For

the given trussed beams braced by a system of link members as shown in the

following figures.

EI = 30000 t.m2


a) ba

b)

2 BUCKLING

1. INTRODUCTION

A vertical member whose cross section dimensions are small as compared

to its height and subjected to compressive force is known as a column.

Horizontal or inclined members subjected to compressive force are

known as struts

Along column when subjected to direct load, deflects in lateral direction

which is known as buckling. The effect of lateral deflection is quite

considerable in long columns. In contrast to long columns the effect of

lateral deflections is negligible in short columns. In very long columns

the effect of direct stresses is small as compared with bending stresses.

The main causes of bending in the columns are lack of straightness in

member, i.e. initial curvature in the member, eccentricity of the load and

non –homogeneity in the material of construction of the column. Every

column will have at least small degree of eccentricity.

From mechanics, it is known that a body may be in three types of

equilibrium; stable, neutral, and unstable. A stable equilibrium is one in

which body returns to its original position on being displaced from its

position of equilibrium e.g. a ball resting on a concave surface as shown

in Fig.1.a. A neutral equilibrium is one in which a body does not return

to original position on being displaced but its motion stops e. g a ball on a

horizontal plane as shown in Fig.1.b. In unstable equilibrium the body

continues to move further away from its position of equilibrium on being

displaced e.g. a ball on convex surface as shown Fig.1.c.

A long column subjected to small loads is in a state of stable equilibrium.

If it is displaced slightly by lateral forces, it returns to its original position

on the removal of the force. When the axial load (P) on column reaches

certain critical value ( crP ), the column will be in a state of neutral

equilibrium. When it is displaced slightly from its straight position, it

108 Chapter (2) - Buckling

remains in deflected position, Fig.1.b. If the force (P) exceeds the critical

load ( crP ), the column becomes in unstable equilibrium (Fig.1.c). The

column either collapses or undergoes large deflection (Fig.1.c). The

critical load of column is defined as the load at which column is in neutral

equilibrium.

a) Stable b) Neutral c) Unstable

Figure (1)

2. SLENDERNESS RATIO

In the long column as shown in Fig.2, the effect of bending is to be

considered while designing. The resistance of any member to bending is

governed by its flexural rigidity (EI), where;

2.iAI

Where i is the radius of gyration and A is the

cross sectional area. Every structural member

will have two principal moments of inertia

one is maximum and the other is minimum.

Resistance to bending is determined by least

moment of inertia minI ( Iy ).

2minmin .iAI

Where minI is the least radius of gyration

A

Ii min2min

Figure (2)

Chapter (2) - Buckling 109

The ratio

mini

L=

gyrationofradiusleast

memberoflength=

is known as the slenderness ration of the member. Whether a column is

short or long is according to the numerical value of the of the slenderness

ratio.

3. END CONDITIONS

The various end conditions encountered in columns are; fixed, hinged,

roller or free as shown in Fig.3. Thus considering end conditions, we

have the following categories of columns:-

1. One end free and other fixed

2. Both ends are hinged

3. Both ends are fixed

4. One end is fixed and the other end is hinged

5. One end roller and the other end fixed

6. Both ends are elastically

Figure (3)


4. EULER'S FORMULA

Euler found out the failure load for various end conditions,

considering stability of columns on assumptions that column is initially

straight, homogeneous of uniform cross section throughout, axially

loaded, the material is linearly elastic and lateral deflections of the

column remain small in relation to its length.

Figure (4)

Consider a strut (or beam) as shown in Fig.4 carrying an axial

compressive load P. Hence the buckling will occur in the plane of the

least rigidity, the critical load (Pcr) can be calculated by using the

differential equation of elastic curve as follows:-

min2

2

EI

M

dx

yd

The bending moment (M) at any distance x from the left hand end is ;

M = yp.

Hence;

yEI min = yp.

yEI

py .

min

=0

The solution of this differential equation is:-

CxCCxCy cossin 21

Where:


minEI

PC

C1 and C2 are integration constants and can be determined from the

boundary conditions as follows:-

(1) At x = 0 → y= 0

(2) At x = L → y=0

From first condition C2 = 0 and CxCy sin1 , from second condition:

CLC sin0 1

If C1=0, this case of no buckling , then the strut remains straight . If

sin(CL)=0, hence

CL = n (n=0, 1, 2, …… etc.)

If n=0 → then; no buckling

If n=1 → then ; c=L

i.e. P = 2 .2

min

L

EI …..(1)

And y = 1C .sinL

x …..(2)

The deflection form is a half sine curve of undefined magnitude.

The value of P given by Equation (1) is the smallest value for which this

non–straight equilibrium form can exist. If P is smaller than this value,

then the strut can be in equilibrium in the straight form only (C1=0). The

value P given by Equation (1) is therefore a very special value in that it

marks the boundary between the straight equilibrium state and the non–

straight given in Equation (2) and it is termed the critical load (Pcr) or

buckling load. this case was first solved by the Swiss mathematician

Euler, and is therefore frequently termed the Euler load.

2

2 .

L

EIPcr

…..(3)

Pcr is the lowest critical load for hinged ended strut (Euler strut) by taking

n=1. If n is taken equal to 2 , 3, 4 etc . , series of critical load is obtained


EIL2

24, EI

L2

29, EI

L2

216, . . . .etc.), each corresponding to a more

complex equilibrium form as shown in Fig.5.

(a) 2

2

L

EIPcr

(b)

2

24

L

EIPcr

(c)

2

29

L

EIPcr

Figure (5)

The behavior of the strut can be compared to that of the sphere shown in

Fig.1. In Fig.5.a, a sphere is shown resting on a concave surface; it is in a

state of stable equilibrium. If displaced from its initial "at rest" position in

the center of the surface, it will return to that position. This is similar to

the column carrying an axial load less than the Euler load (Pcr). Fig.1.c is

in state of unstable equilibrium, and is similar to that of the column

loaded beyond the Euler load. Fig.1.b shows a state intermediate between

the stable and unstable. Fig.6 shows a graph of axial load P against

central lateral deflection.

crP 2min

2

L

EIPcr

Figure (6)


5. COLUMN WITH ONE END FIXED AND THE OTHER

FREE

As shown in Fig.7, bending Moment at

distance x

M = )y - ( P -

yEI = M

yEI = )( yP

yEI = )( yP

yPyEI . = P

Putting

2c

EI

P

Figure (7)

22 cycy

The solution of this differential equation is

)sin()cos( 21 cxCcxCy

Taking x from the fixed end,

At x = 0 → y = 0

i.e. 0 = C1 + 0 +

C1 = -

y = )cos(..)sin(.. 21 cxcCcxcC

At x = 0 → y = 0

0.2 cC → 02 C

Hence )cos(cxy

))cos(1( cxy

At x = L → y =

))cos(1( cL

0)cos( cL


cL = .,.........2

3,

2

Taking least value of cL

cL = 2

crP = min2

2

4EI

L

crP = min2

2

)(EI

Lb

i .e. LLb 2

Similarly as shown in Fig.8, in general Euler critical load crP equal to

crP = 2

min2

)( bL

EI ….(4)

Where bL is the buckling length according to end conditions Fig.3 and

Fig.7 and minI is the minimum moment of inertia

2

min2

04.2L

EIPcr

2

min2

4L

EIPcr

2

min2

L

EIPcr

Figure (8)

6. CRITICAL STRESSES

The proceeding analysis have given values of critical load ( crP ) for

various types of column, but the designer usually works with stresses. In

the fundamental case of the hinged column, the critical stress is given by,

A

Pcrcr ….(5)


Where A is the cross section area of column in case of short column, the

failure occur when the stress reaches the ultimate value of column

material )( u and.

A

Puu (6)

In case of slender column if P < crP , the column in state of equilibrium ,

and if P = crP ,the column in state of neutral equilibrium and the buckling

has occured if P > crP , the column become unstable hence;

IL

iEI

b

cr2

22 .

2

min

2

)(i

L

E

b

cr

2

2

Ecr ….(7)

The critical stress in a column is therefore dependent only on the young's

modulus of column material, and the slenderness ratio i

L of the column .

7. LIMITATION OF EULER'S FORMMULA

In the Euler formula it was assumed that the member absolutely straight

and the load is axial. The critical load was derived from the differential

equation of the elastic curve which is based on Hook's law. Hook's law is

valid as long as stresses do not exceed proportional limit. The critical

stress cr was found by;

2

min

2

)(i

L

E

bcr

p (8)

Where p is the proportional limit of the material of the column. For

Euler's formula to be valid; i.e;

p

b E

i

L

min


According to Egyptian Code of Practice for steel, 1989, and to avoid

inelastic buckling, then assume; P=0.8y, where y is the yield stress.

For steel, the modulus of elasticity E=2100 t/cm 2 , hence;

104i

Lb for steel 37 (y=2400 kg/cm2, P=1920 kg/cm

2)

96i


2)

85i


2)

Euler's formula is valid for steel 37 if pcr and 104/ iLb . Fig.9

shows the relation between cr and min/ iLb and the curve is asymptotic

to both axes. As the ratio iLb / increase the column will buckle at a lower

value of cr . In the other hand the value of cr tend to infinity as the ratio

iLb / become very small, and the column actually become a short one

and the failure is governed by the ultimate strength of column material

u which is represented in Fig.9 by horizontal line c-e. Euler's formula is

valid only for part a-b ( 104/ iLb ). If 104/ iLb , (for steel 37),

empirical formulae, which represent the experimental results used to

calculate the buckling stress, according to part b-c, this part may be

parabolic or straight line. Hence from the above discussion, the buckling

stress is therefore represented by the curve a-b-c as shown in Fig.9.

i

Lb

cr

p

pb

pc

u

Figure (9)


8. FACTOR OF SAFETY

In the design of columns a factor of safety (n) is used, and the permissible

or working buckling stress ( )pb is represented by curve e-f-g as shown

in Fig.9; where:

n

crpb

(9)

where n is the factor of safety.

9. EMPIRICAL FORMULAE

a) For steel 37

The ultimate strength u =3.7 t/cm 2 ; and the yield stress

is y =2.4 t/cm 2 ; and the modulus of elasticity E = 2100 t/cm 2 . The

permissible or working stresses in tension and compression are;

2/2.1 cmtpt

2/1.1 cmtpc (no buckling )

The permissible buckling stress pb value for steel 37 as follows;

For 104i

Lb 2

2

/

)(

6000cmt

i

Lbpb

For 104i

Lb

Empirical formula is: 22 /)(00005.0 cmti

Lbpcpb

i.e. 22 /)(00005.01.1 cmti

Lbpb

a) For steel 44 (PC=1.30 t/cm2)

For 96i

Lb

Use Euler's formula 2

2

/

)(

6000cmt

i

Lbpb


For 96i

Lb

The parabolic formula is: 22 /)(00007.030.1 cmti

Lbpb

a) For steel 52 (PC=1.40 t/cm2)

For 85i

Lb

use Euler's formula 2

2

/

)(

7500cmt

i

Lbpb

For 85i

Lb

The parabolic formula is: 22 /)(000065.04.1 cmti

Lbpb

Example (1)

For the shown steel column (steel 37) in Fig.10 determine the

permissible buckling load pbP for two end conditions;

a) Cantilever

b ) Hinged-fixed

E=2100 t/cm 2 , A=77.8 cm 2 , 412510 cmI x

4782 cmI y , cmix 7.12 , cmiy 67.2

c) Calculate the factor of safety for two cases Solution

a) case of cantilever

bL = 2502 = 500 cm

mini

Lb = 67.2

500 = 26.187 >104

use Euler formula

pb = 2)26.187(

6000 = 0.171 2/ cmt

The permissible buckling load is:

pbpbp

= 0.171 8.77

= 13.31 ton

Figure (10.a)


b) in case of hinged-fixed column

Lb = 0.70 250 = 175 cm

mini

lb =

67.2

175 =65.54 <104

Use parabolic formula:

2

min

)(00005.010.1i

lbpb

Figure (10.b)

= 1.10 2)54.65(00005.0

= 1.10 2/8852.02148.0 cmt

The permissible buckling load is:

pbpbP

=0.8852 8.77

=68.86 ton

c)Factor of safety (n) for (case a)

n = pb

cr

or n=pb

cr

P

P

= 171.0

)//( 2min

2 iLE b

= 171.0

)26.187/()2100( 22

= 3.45

Factor of safety for (case b)

n = 8852.0

)54.65/()2100( 22

= 5.445

pb

pc

i

Lb


10. MAXIMUM FIBER STRESS

The value of permissible buckling stress represents the average stress in

column cross section area where:

pb = av = A

Ppb

On the other hand, the value of maximum compressive stress not exceed

the value of permissible compressive stress of material ( )pc as shown

in Fig.11. Hence ;

max = kav . pc

i.e.

pc = kpb .

Where k is buckling coefficient and

depend on slenderness ratio. Hence

pb

pc

i

Lb

pb

Figure (11)

K =pb

pc

(10)

In case of column subjected to compression force P which is less

than pbP , hence the maximum stress in case of buckling is equal to :

kA

p.max (11)

In case of eccentrically loaded columns subjected to axial load and

bending moment, the maximum stress is equal to;

max = maxmax xI

My

I

Mk

A

P

y

y

x

x pb (12)


Example (2)

Fig.12 shows steel column (steel44) subject to eccentric load b and

horizontal load 1.0t. the cross section is B.F.I no 40 with;

A=209 cm 2

60640xI cm4 , 0.17xi cm

11710yI cm4, 49.7yi cm

Determine the maximum value of P if 2/3.1 cmtpc

Figure (12)

Solution :

Buckling length = cm8004002

2

22min

min

52908106

60006000

10480106497

800

497

t/cm.).()(lb/i

σ

..

/iL

.iI

pb

yb

y

Also M )4(1004001100 PPx t.cm

The maximum value of P depend on maximum stress then :

maxmax .. yI

M

A

P

xpb

pc

3.1


=60640

20100)4(

5259.0

3.1

209

pp 3.1

i.e. P(0.0118+0.033)-0.13 3.1

0448.0

43.1P 92.31 t

P 92.31 t

hence the maximum allowable value of P 92.31max t

Example (3)

For the given steel column (steel 52)

Fig.13 find the allowable buckling load if:

a) Both ends are hinged with height 10m

b) One end is hinged and the other end

fixed.

c) Both ends are hinged but the column is

laterally supported in plane x-x at middle

height as shown in Fig.13, calculate the

factor of safety for each case.

2)//(7500 ilbpb for 85i

lb

2)2

(000065.04.1 bwb

l for 85

i

lb

Column section B.F.I No 300

A=154cm2, 425760cmI x , 49010cmI y

E=2100 t/cm 2

Figure (13)

Solution

a) Both ends are hinged: (lb=10m)

cr =2

min2

bL

EI

=2

2

)1000(

)9010)(2100( =186.74 t

cr =A

Pcr = 154

74.186 =1.21 t/cm

2


2

2

min

/437.0)89.130(

7500

10489.13064.7

1000

64.7154

9010

93.12154

25760

cmt

i

l

cmA

Ii

cmA

Ii

pb

b

y

y

xx

Safe or permissible Load ( AP pbpb )

154437.0 =67.268 t

factor of safety = b

cr

=

b

cr

= 437.0

21.1 = 76.2

b) hinged- fixed column

lb = 0.7L

= 700 cm

cr = 2

2

)700(

90102100)14.3( =381.10 t

cr =A

Pcr =154

1.381

= 2.47 t/cm2

mini

lb = 64.7

700 =

= 8562.91

pb =2)62.91(

7500

= 0.893 t/cm2

safe load = b

b = 154893.0 =117.50 t


factor of safety =pb

cr

pb

cr

= 89.2516.131

1.381

b) hinged – hinged column with laterally

support at middle height

lb about x-x =1000cm

lb about y-y = 10005.0

= 500cm

t.)(

)(πp

t.)(

))((πP

..i

l

..i

l

cm.A

Ii

.A

Ii

cr

cr

y

yyb

x

xxb

y

y

xx

97746500

90102100

95331000

254602100

1044465647

500

10433779312

1000

647

9312

2

2

2

2

Hence

2/46.3,9.533 cmtA

t crcrcr

22 /011.1)33.77(000065.040.1 cmtpb

Permissible safe load tpb 694.155154011.1

factor of safety =pb

cr

=

694.155

97.746 =3.42


Example (4) For the shown steel column (steel 37),

calculate;

1. Max compressive stress in the column

due to given load

2. The max value of P which can be

safety carried by the column

section properties ; (B.F.I 28)

A =144 cm2

Ix = 20720 cm4

Iy =7320 cm4 ix = 12cm

Iy = 7.14 cm

Solution

buckling length (lb) =600 cm

least radius of gyration iy = 7.14 cm

greatest slenderness ratio is:

= 14.7

600 = 84.03 < 104

tmL

M

K

cmt

b

c

pb

.5.74

65

4

473.1747.0

10.1

/747.0)03.84(00005.010.1 22

Max compressive stress

maxmax yI

Mk

A

P

x

x 1.10

Figure (14)

1420720

)100(5.7473.1

144

0.30max =0.3068 + 0.5067

= 0.8135 t /cm2 < 1.10 (ok)

To get Pmax , 5067.0)473.1(144

1.1 max P

Hence ; Pmax = 58.655 t


Example (5)

For the shown column (Fig.15) and

link members, determine the

maximum stress in the column and

marked truss member due to

buckling

2

2

/20.1

/10.1

cmt

cmt

t

c

B.F.I No 32 42 32250,171 cmIcm x

49910 cmI y

Figure (15)

Ch .No 10 25.13 cm 4206 cmI x 43.29 cmI y

cme 55.1

cmb 5

Solution

Column ABC

bl =2L =1600 cm

xI = )1723612

236(232250 2

3

=73914 cm4

12

3629910

2yI = 17686 cm

4


)236(2171 = 315 cm

2

315

17686min I =7.49 cm

mini

lb = 49.7

1600 =213.5 >104

pb =25.213

6000 =0.1316 t/cm

2

N = - 20 t Mx=40 m.t for part AB

max = yI

M

A

P

x

x

pb

pc..

(comp)

= 1873914

2000

1316.0

10.1.

315

10 = - (0.2653 + 0.4870 )

= - 0.7523 t/cm2 <1.10 o.k

Member B-d, ch. No. 10

A =213.5 =27 cm2

bl =L = 10022 =282.84 cm

xI =2(206) =412 cm4

yI =2(13.51.552 +29.3) =123.47 cm4

mini =27

47.123 =2.138 cm

mini

lb =138.2

84.282 =132.26 >104

pb =226.132

6000 =0.343 t/cm

2

max =pb

pc

A

N

=

343.

1.1

27

210 =-1.679 t/cm

2 >1.10 Unsafe

The cross section dimensions must be increased

The other member c-d is tension and the buckling does not occur, hence

t =27

10 =+0.37 t/cm

2 <1.2 Safe

The other members d-e and e-b are zero force and stresses


Example (6)

Figure (16) shows a

frame has the given cross

section for link member AC;

find:-

1) the max value of P

2) the max value of stress for

the

given section

3) the value of safe load P

4) factor of safety.

Figure (16)

pc =1.1 t/cm2, E =2000 t/cm

2

xI =4670 cm4, yI =139000 cm

4

A =254 cm2 ,

xi =4.287 cm

yi =23.39 cm

Solution

lb = 600 cm

1- Value of P/2 = 0.306 254 =77.80 t

i.e. P =155.60 t

2- Max. stress max = pb

pc

A

P

.

2/

i.e. max =306.

1.1.

254

8.77 =1.10 t/cm

2 Just Safe

3- factor of safety =pb

cr

=

306.0

)/( 22 ilE b =

306.0

06.1 =3.45

b pb

b

p

cm t

cm i

lb

2

/ 306 . 0 ) 95 . 139 (

6000

104 95 . 139 287 . 4

600

2 2

min


Example (7)

For the shown column in Fig.17 determine Pcr. The cross section is an

angle 130x130x12cm

xI = yI =472 cm4

A =30.0 cm2, e =3.64 cm

E =2100 t/cm2, pc =1.1 t/cm

2

Calculate the factor of safety and max stress

Figure (17)

Solution

Get Ixy = 312 cm4

Lb = L = 600 cm

minI = vI =22)

2(

2xy

yxyxI

IIII

minI =472–312 =160 cm4

30

160min i =2.3 cm , 104

3.2

600

min

i

lB

crP =2

min2

)( bl

EI

= 2

2

)600(

)160)(2100( =9.20 t


mini

lb = 30.2

600 =259.8 >104

pb =2

min

)(

6000

i

lb

=0.0889 t/cm2

Safe load (Ppb)

tpbpb 67.2)30(0889.0

Factor of safety (n)

N = pb

cr

P

P =3.45

Maximum Stress

45.330

67.2max =0.30 t/cm

2 < 1.10 (o.k)

Example (8)

For the shown column (steel 44) in Fig.8, a rise

in temp. accrued by a value t find the

maximum value of t which make a column still

stable. Cross section is SIBNo.30:-

A =64.1 cm2

yI =451 cm4

E =2000 t/cm2

=1.2 10-5

Solution

mini =A

I y = 2.55 cm

Lb = 0.51000 = 500 cm ,

104

3.2

600

mini

lB

Compressive stress occur due to rise of temp t

is:

Figure (18)


E =

E

tt

L

LtE

L

LE

0252.0102.12100

...

.

5

Axial force = .

= 69.1 (0.0252t)

= 1.741 t

The column is stable if the axial force crP = 2

min2

)( bl

EI

Pcr = 2

2

500

4512000 =35.57 t

Hence 35.57 1.741 t, i.e. t ≤ 20.40C

11. COLUMN WITH INITIAL CURVATURE

Let ABC be the shape of the column before

loading with central deflection e as shown in

Fig.19 and let AC B be the shape after loading

with total central deflection y0. At a point

distance x from top hinge, let y1 be the

deflection before loading and y be the total

deflection after loading. As the curvature is very

small, the curve may be assumed to be sin curve

such that

1y =L

xe

sin

It will be assumed that bending is uniplanar,

M = P . y

Figure.19


2

12 )(

dx

yydEI

=-M = -P.y

2

2

dx

ydEI =-P.y +

2

12

dx

ydEI

L

xey

sin1

dx

dy1 =L

x

L

e cos

2

12

dx

yd =

L

x

Le

sin

2

2

2

2

dx

ydEI =-P.y =

L

x

LeEI

sin

2

2

yEI

P

dx

yd

2

2

=L

x

Le

sin

2

2

Solution of this differential equation is

2

2

2

2

2

21

sin

sincos

cL

L

x

Lecxccxcy

= 1/

/sinsincos

22221

Lc

Lxecxccxc

From boundary conditions :-

(1) at x = o y = o

i.e c1 = o

then y = c2 sin cx- 1/

/sin222

Lc

Lxe

1

cos

cos

2

222

Lc

L

x

Le

cxccdx

dy


(2) At x = 2

L, 0

dx

dy

i.e. 0 = 2

cos.12

CLCC

Then C2 = 0, Put EI

PC 2

y =

2

22

1

sin

Lc

L

xe

=

2

2

1

sin

EI

PL

L

xe

but Pcr = 2

2

L

EI

i.e. y =

crP

PL

xe

1

sin

(3) At x = L/2, y = y0

i.e. y0 = PP

ePP

cr

cr

..

Maximum B.M. = P .y0

=PP

ePP

cr

cr

.. =

PP

ePP

cr

cr

..

Maximum compressive stress = Z

M

A

P

= ))(

..1(

2iPP

yeP

A

P

cr

ccr

Where Z =cy

I =

cy

Ai2


Example (9) The shown column (steel 37) in

Fig.20;has initial curvature e =10 cm.

find the safe value of P if the cross

section Is B.F.I No 38

E=2100 t/cm2

A=194 cm2

xI =50950 cm4

yI =10810 cm4

xi =16.2 cm

yi =7.46 cm

Find also the factor of safety

Solution

mini

lb =46.7

800

=107.23 >104

Figure (20)

cr =2

min2

bL

EI =

2

2

)800(

)10810)(2100( =349.72 t

y0 = ePP

P

cr

cr .

y0 = )10(72.349

72.349

P

i.e. M =P.y0

max = ))(

..1(

miniPP

yeP

A

P

cr

ccr

< 1.10 t/cm

2

i.e. 1.1 = )46.7)72.349(

)15)(10(72.3491(

194 P

P

1.1 ≥ P

PP

72.349

25.36

194

If we assume P =10 t ( trial and error )


i.e. 1.1 = 0.0500+ 1.06 (o.k) i.e. P ≤ 10 t

The safe value offload: P ton10

12. LATERALLY LOADED COLUMN (beam – column ) Column with a concentrated load at mid height is considered. Let F

be the concentrated load applied

Figure.21

M = yPxF

.2

.

2

12

dx

ydEI =-M

= yPxF

.2

.

yEI

P

dx

yd

2

2

=2

.x

EI

F

Where:

C2 =

EI

P

ycdx

yd 2

2

2

= xEI

F.

2

)( 22 cD = xEI

F.

2


The solution of this equation from mathematics is:-

y = ..sincos 21 IPcxCcxC

Where P.I . is the particular integral .

P.I. = )2

(1

22x

EI

F

cD

= ).2

(2

xcEI

F

Then:-

y = xcEI

FcxCcxC

221.2

sincos

At x = 0 ; y = 0 i.e. C1=0

dx

dy =

22.2

cos.cEI

FcxCc

At x =2

L ;

dx

dy = 0

i.e. C2 =

2cos.2 3 cL

cEI

F

then; y =P

Fxcx

cLcEI

F

2sin

2cos.2

1

3

Where

EI

P = c

2

At x =2

L ; y = y0

i.e. y0 =P

FLcL

cP

F

42tan

2

Max. B.M is:-

= 04

PyFL


=2

tan2

cL

c

F

Max. stress max

=Z

M

A

P

=2..2

)2

tan(..

1(icP

cLyF

A

P c

Example (10)

Determine the maximum uniformly distributed lateral load which can be

carried by a given cross section as shown in Fig.22

y =140 kg / cm2 , E = 100000 kg / cm

2 (Wood)

Figure(22)

Solution

M =2

.

2

...

2xwxLwyP

yEI =2

.

2

...

2xwxLwyP

From mathematics;

y = )2

.(2

sincos2

221

cxLx

P

wcxCcxC

where C2 = P/EI


Form boundary conditions

1C =2.cP

w

2C =2

tan. 2

cL

cP

w

0y =P

wLcL

cP

w

8)1

2(sec

.

2

2

Max. B.M.:-

= P . 0y +8

2wL

= )12

(sec2

cL

c

w

2

cL =

2.3

3 rad. =31

o

2

seccL

=1.165

Max. B. M = 22.5 w m.t

Max. stress =I

My

A

P max

= w31608.0

80.05.22

08.016.0

200

w < 0.525 t/m


PROBLEMS

1. A steel column (steel 37) 6 m height, has an B.F.I –section with

properties:

A = 110 cm 42 11700, cmIx

I y = 4200 cm 24 /2100, cmtE

Determined the permissible value of the axial Load P for this column,

in the following cases

a- both ends hinged

b- one end hinged and the other fixed

c- both ends hinged but the column is laterally supported in

the X – plane at middle height. Find the factor of safety

for each case.

2. The steel column (steel 44) AB is hinged at A&B and carries an axial

load P and a lateral load H = 2 t as shown the cross–section

B.F.I.No.16 with properties

A = 58.6 cm2, 42634cmI x ,

4958cmI y

Determine the permissible value of load P

2/2100,30.1 cmtEpc

Find the factor of safety


3. In the shown structure; if the column AC is restricted in direction x –

x at c, Calculate:

a) the max value of uniform load w if the column properties :

B.F.I.No.36, A = 190 cm 2

Ix = 45000 cm4, 10800yI cm

4

E = 2000 t/cm 22 /4.1, cmtc (steel 52)

4. Find the max. stresses in the column given

A = 77.8 cm2, 12510Ix cm

4 , 555Iy cm

4

E =2100 t/cm2

5. For the shown column truss element, find the max. fiber stresses for

column and link members

)44(/2100./30.1 22 steelcmtEcmtb

Column section is B.F.I No 30 and 2PL 2 cm34 as shown and link

member section is 2 channels No. 10. Find the factors of safety.


6. For the given indeterminate frame. ABCD determine the factor of

safety against buckling if;

96)(00007.030.1

96)/(

6000

2

2

i

lfor

i

l

i

lfor

ilb

bbb

bb

7. For the given column section which consist of 4 angles

100 )37(10100 steelmm latticed as shown. If the height of

column is 5.0 ms, and is fixed at base and hinged at top find the safe

load; critical load and the factor of safety.


8. A Latticed column shown in Fig. has 8.0 high, with hinged ends,

find:

a) The safe load if the eccentricity in direction x, e = 5cm

b) The factor of safety

c) Maximum stress. ( 22 /1.1,/2100 cmtcmtE b )

d) Show how to increase the safe load to its double value?

9. A Shown column (steel 44) hinged at A and B and the load as given

in fig, calculate;

a) Max stress in the column.

b) Safe load.

2

44

111

,4150,11690

cmA

cmIcm y

10. For the shown fixed frame a rise of temperature occur t. find the

max. value of t which make the frame still stable.

A =53.4 , I 45740 cmx ,I4288 cmy


Coefficient of thermal expansion of steel is 5102.1 2/2100 cmtE

11. For the given fixed column (steel 44) with initial curvature e= 12cm

and the cross section is S.I.B No 40

A =118 cm 42 29210, cmI x

E =2100 t/cm 42 1160 cmI y

a) Find the safe value of P.

b) Maximum fiber stress.

c) Factor of safety.

12. Determine the value of P which causes instability of the structure

shown in fig.

E = 2100 t/cm 42 270300, cmI x

24 324,15350 cmAcmI y

22 /2.1/1.1 cmtcmt tc (Steel 37)


13. Determine the buckling load for member AB for the given frame

shown in fig for one channel No. 30

A=58.8 cm 42 8030, cmI x

cmI y 495 cme 70.2,4

E=2000 t/cm ,2 2/1.1 cmtc

14. For the shown truss find the value of safe load P considering

buckling effect. Section; of member is 2 angle mm14100100

with the following Properties, for one angle:

)34(/2.1

,/1.1,/2000

98.2

2.26

235

2

22

2

4

steelcmtF

cmtRPccmtE

cme

cmA

cm

pt

y

Find the factor of safety and maximum stress for all members.


15. For the following statically indeterminate structures; and for the

marked member; find the critical buckling load, safe buckling load;

and value of factor of safety;

E=2100t/cm 224 /2.1/1.1,37, cmtcmtsteel ptc

(a)

(b)

(c)

(d)


(e)

(f) All members are2 angles 120x120x11 mm

(g) All members are2 angles 90x90x9 mm

(h) All truss members are 2 angles 80x80x8 mm

STATICALLY

INDETERMINATE

STUCTURES 3

1. INTRODUCTION

Statically indeterminate structure means that the system cannot be

determined by means of the given equations of equilibrium and the

conditional equations. In other words, when the equilibrium conditions

are not sufficient to determine the reactions and the internal forces of a

structure, one say that this structure is statically indeterminate. A

structure can be externally, internally or externally and internally

statically indeterminate. As shown in figs.1, 2 and 3, if the equilibrium

conditions are not sufficient to determine the reactions, but when the

reactions are obtained we can determine the internal forces, then the

structure is externally statically indeterminate, if the equilibrium

conditions are sufficient to determine the reactions, but we can not

determine the internal forces, then the structure is internally statically

indeterminate. The methods of solving statically indeterminate structures

are categorized into two groups, the first includes force methods and the

second includes the displacement methods.

For plane structures, we have three equilibrium conditions as follows;

M =0, yF =0, M =0

For space structures, the conditions of equilibrium are:

xF =0, yF =0, zF =0

And

xM =0, yM =0, tM =0

The following figures (Figs.1, 2 and 3.) show statically indeterminate

structures:

148 Chapter (3) STATICALLY

INDETERMINATE

STUCTURE

Frames

Beams

Trusses

Figure (1) externally statically indeterminate

Beams

Frames

Trusses


INDETERMINATE

STUCTURE

Figure (2) Internally statically indeterminate

Frames

Trusses

Arched Frame

Figure (3) Externally and internally statically in terminate

2. FORCE METHODS ( flexibility Approach)

Which assume that the reactions and internal forces are the

unknowns. They are determined from deflection equations. From these

methods; consistent deformations, three moments equation, and column

analogy method. For example (Fig.4) the shown once statically

indeterminate beam, the deflection at point 1 is equal to zero, if the

support at 1 is removed, the deflection at 1 due to given load equal to 10.

If 1t is applied at point 1; the deflection is 11. Then, the compatibility

equation is

0. 11110 X


INDETERMINATE

STUCTURE

Hence calculate the reaction X1 and the other reaction can be

calculated. Term 11 is called flexibility of the maim system at point 1.

10

11

Figure (4)

3. DISPLACEMENT METHODS (stiffness Approach )

In these methods, the unknowns are the deformations that are

determined from equilibrium conditions. From these methods; moment

distribution, slope deflection methods, and stiffness matrix methods. For

example, fig.5 shows statically indeterminate beam. In the stiffness

approach, deformation (rotation & translation) at joints A and B are

obtained and used to obtain the internal forces. .To determinate A at

support A, assume fixed support at A, hence;

aM = 0AM + 1AAM =0

From the above equation, calculate the value of A and then

calculate the internal forced at any section.


INDETERMINATE

STUCTURE

A

Figure (5)

4. DEGREE OF STATIC INDETERMINACY

The first step to solve any statically indeterminacy structures is

determination of the degree of indeterminacy. From the degree of

indeterminacy one knows how many redundant forces and moments

could be considered in solution by using the force methods.

4.1. Degree of Indeterminacy For Beams (Line Structures) (Degree Of

Redundancy)

If we assume

n = degree of redundancy

r = number of reactions

c = number of equations of condition

( c = 1 for a hinge, c = 1 for roller )

h = degree of indeterminacy

if r < c + 3 → the beam is unstable

if r = c + 3 → the beam is statically determinacy

if r > c + 3 → the beam is statically indeterminate

i.e. n = r - c - 3


INDETERMINATE

STUCTURE

Figure (6)

For the above beam shown in fig. 6

r = 6

c = 1

i.e. n = 6 – 1 -3

= 2 → Twice statically indeterminate

Another example, for beam shown in fig. 7

Figure (7)

R = 6

C = 2

n = 6 – 2 - 3

= 1 → Once statically indeterminate

For beam shown in fig. 8

n = n1 + n2 = 3

Figure (8)

n1 = 4 - 2

= 2


INDETERMINATE

STUCTURE

n2 = 2 - 1

= 1

4.2. Degree of Indeterminacy for Plane Frames

If we assume the frame shown in fig. 9


b = number of members

(3 unknowns / member )

j = number of rigid joints

(3 eqn. / joint )

c = number of equations of conditions

n = degree of indeterminacy

if 3b + r < 3j + c → the frame is unstable

if 3b + r = 3j + c → the frame is statically determinate

if 3b + r > 3j + c → the frame is statically

indeterminate

hence;

n = 3b + r – 3j – c

For example the above frame shown in fig. 9

b = 6

r = 6

j = 6

c = 0

i.e.

3 6 + 6 > 3 6 + 0

Hence

n = 6

→ degree of indeterminacy

or number of redundants.

Figure (9)


INDETERMINATE

STUCTURE

Example (1) For the given frames shown in figs. 10, 11, 12 and 13, determine the

degree of indeterminacy

B = 10

r = 9

j = 9

C = 0

n = 310+9

–39–0

= 12

(12t h degree)

Figure (10)

B = 10

r = 9

j = 9

C = 4

n = 8

(Eighth degree)

Figure (11)

b = 10

r = 9

j = 9

c = 1

the overhanging portion a b

should not be counted in the

number of members

n = 310+9

-39–1

= 11

(11th degree)

Figure (12)


INDETERMINATE

STUCTURE

b = 10

r = 9

j = 9

c = 3

c is the number of member

meeting at the internal hinge

minus one

c = 4 – 1

= 3

n = 30 +9

- 39–3

= 9

(ninth degree)

Figure (13)

Note:

For stable structure at least

three components of external

reactions which must not be

parallel or intersecting at a point.

The shown frame in fig.14 is

unstable because

M , at A ≠ 0.

Figure (14)

4.3 Degree of Indeterminacy of Plane Trussed

If we assume the truss shown in fig.15

b = number of members (1 unknown / member )


j = number of joints ( 2eqn. / joint )

n = number of redundant

n = b + r – 2 j


INDETERMINATE

STUCTURE

Figure (15)

Hence:

1) b + r < 2j → the truss Is unstable

2) b + r = 2j → the truss is stable and determinate

3) b + r > 2j → the truss is statically indeterminate

i.e. n = b + r – 2j

Example (2)

For the above truss (Fig.15)

B = 19

R = 3

J = 10

n = 19 + 3 – 2 10

= 2 (Twice degree of indent.)

Example (3)

For the given truss in fig.16

B = 7

R = 3

J = 5

n = 7 +3–52

= 0

(determinate and stable)

Figure (16)


INDETERMINATE

STUCTURE

See the following examples shown in table1

truss

b

r

J

n

Type

7

3

5

0

unstable *

( we must avoid three hinges on one line)

7

3

5

0

unstable **

6

4

5

0

unstable ***

8

4

5

2

Indeterminate to

2nd

degree

* Internal geometric instability due to three hinges a, b, c, on a link,

possible displacement as shown by dotted link.

4 METHOD OF CONSISTENT DEFORMATIONS

1. INTRODUCTION

Statically indeterminate structure may be analyzed by direct use

of the theory of elastic deformations. Any statically indeterminate

structure can be made statically determinate and stable by removing the

extra restraints called "Redundant Forces" or simply redundant, that is,

the force elements which are more than the minimum necessary for the

static equilibrium of the structure. The statically determinate and stable

structure that remains after removal of the extra restraints is called the

"Primary Structure". The original structure is then equivalent to the

primary structure subjected to the combined action of the original loads

plus unknown redundancy. The conditional equations for geometric

consistence of the original structure at redundant points are called the

"Compatibility equations" are then obtained from the primary structure by

superposition of the deformations caused by the original loads and

redundancies. We can have as many compatibility equations as the

number of unknown redundant so that the redundant can be determined

by solving these simultaneous equations. This method known as

"Consistent Deformations" is generally applicable to the analysis of any

types of structures, whether it is being analyzed for the effect of loads,

support settlement, temperature change, or any other case. However, there

is only one restriction on the use of this method: the principle of super

positions must hold.

Chapter (3) - Consistent Deformations 159

2. ONCE STATICALLY INDETERMINATE

STRUCTURES

Figure (1) shows once statically

indeterminate structures. The unknown

reaction components at supports are

four, while the conditions of static

equilibrium are only three. Hence one

more equation, depending on the elastic,

behavior of the structure should be

added in order to determine the

reaction components. The application of

consistent deformation method to solve

such beam or frame, may be carried out

as follows:-

a. Beam with fixed and Roller

Supports

b. Two Hinged Frame

Figure (1)

1.

Choose a main system Which is

determinate, by removal of the

redundant reaction component. For

the given beam in fig. 2 the main

system is simple beam by assuming a

hinged support at A, thus eliminating

the reaction component X1. Draw for

the main system Mo. D as shown in

fig.2.b.

Original Beam

PA B10

X1 = 0

a.main system

2.

Apply a virtual load

X1 =1 t.m at A and draw M1 .D

PA B

+

b. Mo Diagram

3.

The rotation 10 (or 10)at support A

of main system, can be calculated

from the equation:

X1 = 1 t.m

11

c. Elastic Curve

160 Chapter (3) - Consistent Deformations

10 = 10 = B

AEI

dlMM 10 .

Similarly, the rotation 11 (or 11)

due to unit load 1 t.m at A is equal

to:

11 = 11 = B

AEI

dlM .2

1

-1 t.m

d. M1 Diagram

-

+

e. B.M.D

.

Figure (2)

4. Since the rotation at fixed support A must be equal zero in the original

indeterminate beam hence (A =0) then; the compatibility equation as

follow:

10 + X1 . 11 = 0

5. The total bending moment or shearing force at any support or section of

the indeterminate beam is equal to:

M = M0 +X1 .M1

Q = Q0 +X1 .Q1

The bending moment diagram may also be obtained by the

superposition of Mo and X1.M1 diagrams.

Example (1)

For the given indeterminate beam

shown in Fig.3, Draw B.M, and

S.F.Ds.

Solution

6 t

3 3

A B

BA6 t

3 3X1 = 0

A B

9 t.m

a. Mo D.

1.

choose the main system as

shown and draw M0.D.

(Fig.3.a)


2.

3.

4.

Draw M1.D as shown (Fig 3.b )

Write the compatibility

equation at A :-

10 + X1 . 11 = 0

Calculate 10 and 11

10 = EI

dlMM 10.

= EI2

5.069

= EI4

54

11 = EI2

5.061

= EI4

6

i.e.

X1 = 11

10

S

S

= 6

54 = 9 t.m

Hence

M = M0 + X1 M1

MA = 0+ -91 = -9 t.m

Mc = 9-9 0.5 = + 4.5

Hence B.M, S.F.Ds, reactions,

free body, and main steel RFT.

As shown

BA6 t

3 3X1 = 0

1 t.m

-

X1 = 1 t.m

b. M1 D.

+

-

9 t.m

4 t.m

9 t.m

c. B.M.D.

6 t9 t.m

3 t3 t

1.5 t 1.5 t

4.5 t 1.5 t

d. Reactions

6 t9 t.m

4.5 t 1.5 t

e. Free Body Diagram

g. Main steel Reinforcement.

4.5 t

1.5 t

4.5 t

1.5 t

+

-

f. S.F.D.

Figure (3).


Example (2)

For the shown two hinged

frame in fig.4 Draw B. M. D

Solution:

a) Main System

b) M0 Diagram

1.

2.

3.

4.

Choose the main system

as shown in Fig 4.a

Draw Mo.D (Fig4-b)

Draw M1. D (Fig4.c)

The compatibly equation

at B is

10 + X1. 11 = 0

EI.10 =EI

dlMM .. 10

= 43283

2

+ 42

816

+ 43

2

2

416

= -341.33

EI.11 =

EI

dlM1

= 484

+2

4

3

2

2

44

= 170.67

c) M1 Diagram

X1 = t

S

S2

11

10

M = M0 t X1M1

Mc = -16-42 = -24

m.t

Md = 0 -42 = -8

m.t

d) B.M.D.


Example (3)

Draw B.M.D and free body

diagram for the given frame in

Fig. (5)

EI =10000 2.mt

Solution

4

A

I

16

B

I

C I D

10 t

1.

2.

3.

4.

Choose main system as

shown in Fig.5.a

Draw M0. D

Draw M1. D

The compatibility eqn. is

10 + X1 11 = 0

10 t

X1 = 0

10

a) Main System

40 t.m

+

b) M0 Diagram

-4 t.m

X1 = 1 t

-

4 t.m

4 t.m

-

111 t

4 t.m

c) M1 Diagram

17.16 t.m17.16 t.m

-

17.16 t.m

-

17.16 t.m- -+

22.84 t.m

8.57 t

5 t

8.57 t

5 t

d) B.M.D.

10 t

5 t5 t

9 t.m 9 t.m

5 t 5 t

9 t.m 9 t.m8.57 t

8.57 t

8.57 t

8.57 t

5 t5 t

8.57 t8.57 t

e) Free Body Diagram

Figure (5)

10 =EI

MdlM 0 =

4

2

8402

EI


= EI

1280 = 128 m

11 =

484)4(

3

2

2

442

EI =

EI3

)38464(2

= EI3

896 =0.0299 m

X1 = 896

31280 = 4.29 t

5. Draw M.D

M = M0 +X1 M1

Mc = Mc = 29.44 = -17.16 t.m

Example (4)

Draw B.M,S.F.D s for the given

continues beam in fig.6

EI = 8000 t.m 2

Solution

The structure is once statically

indeterminate

a) Main System

1.

2.

3.

4.

Choose main system or primary

structure as shown the figure

(remove support B)

Draw M0. D

Draw M1. D

Compatibility equation

10 +X1. 11 = 0

10 = EI

dlMM 10

b) M0.D

c) M1.D

Figure (6)


=

8752.1636

3

22

EI

= EI3

1404 =

EI

540

= 0.0675 m d) B.M.D.

11 =

32

32632

EI

= EI

36 = m0045.0

X1 = 36

468 = -15 t

M = M0 + X1. M1

MB = +39 - 45

= -9 m . t

Sketch the steel reinforcement as

shown in fig. 6.g

e) Reactions

f) S.F.D

g) Steel reinforcement

Figure (6)

Example (5)

Determine the reactions and

draw S.F and B.M.Ds for the

following structure shown in

Fig.7

(EI constant )

Solution ( using the deflection at 1)

Figure (7)


One of the reactions may be considered as being extra. In this case let us

first choose the vertical reaction at 1 as the redundant assumed to be

acting downward. By the principle of superposition we may consider the

beam as being subjected to the sum of the effects of the original uniform

loading and the unknown redundant X1, as shown in Fig.7.b and 7.c we

should multiply 11 by X1 because 11 due to unit load only. The vertical

deflection at point 1 due to uniform load w t/m for primary structure is

given by

10 = EI

Lw

384

)2(5 4

And that the vertical deflection 11 due to 1 t at point 1 is given by

11 = EI

L

48

)2.(1 3

Note that 10and 11 can be determined by the method described in

deflection chapter.

Applying compatibility equation:-

10 +X1. 11 = 0

We obtain

EI

Lw

384

)2(5 4

+ X1 EI

L

48

)2( 3

= 0

From which

X1 = -8

10 w L

The minus sign indicates an upward reaction. With reaction at 1

determinate, we find that the beam reduces to a statically determinate one.

We can obtain reaction component at 0 and 2 from the equilibrium

equations:

Y = 0


and from symmetry

y0 = y2

= (w(2L) - 2

1)

8

10WL

= wL8

3

The shearing force and bending

moment diagrams are as shown in

Fig.7.d and Fig.7.e

Figure (7)

Example(6)

Solve the given beam in Fig.8 using

different main systems.

Solution (1) (using conjugate beam)

10 + X1. 11 = 0

10 = 333.21

E

= 1

64

E

11 = EI

8)4

3

2(

= EI3

64

1.3

6464X

EIEI = 0


X1 = YB = 3 t

i.e.

YA = 342 = 5 t

MA = 24243

= - 4 m. t

Figure (8)

solution (2) (using elastic load)

11110 .X = 0

10 = EI2

44

3

2

= EI3

16

11 = EI2

41

3

2

= EI3

4

13

4

3

16X

EIEI

= 0

i.e

X1 = MA = 4 m.t.

Figure (8) cont.

Solution(3) (adding intermediate hinge )

From the previous solutions we come to recognize that we are free to

select redundant in analyzing a statically indeterminate structure, the only

restriction being that the redundant should be so selected that the structure

remains stable. Let us cut the beam at mid-span section C and introduce a

pair of redundant couples called Mc, together with the original loading are

then applied to the main system as shown in the following Fig.8.c.


The redundant Mc is solved by the condition of compatibility that the

rotation of the left side relative to the right side at section C must be zero.

Using the method of virtual load.

10 = co

= LL d

EI

MM 100

.

11 = cl

= LL d

EI

M2

10

The compatibility equation is

10 +Mc . 11 = 0,

knowing

Mc = + 16

2WL

Then Mfinal can be drawn by

knowing Mc +ve at mid span,

and:

M = M0 +X1 .M1

Figure (8) cont.

MA = - 4

2WL +

16

2WL(2)

= 8

2WL (ok)


3. TWICE STATICALLY INDETERMINATE

STRUCTURES Consider the continuous beam shown in fig. 12 the first step is to remove

supports 1and 2 and put the two redundants X1 and X2. The original

structure is now considered as a simple beam (the primary structure )

subjected to the combined action of number of external loads and two

redundants X1 and X2 as shown in Fig.9.b. The resulting structure in

Fig.9.b can be regarded as the superposition of those shown in Fig.9.c and

Fig.9.d. multiplied by X1 and fig.9.e multiplied by X2 consequently. Any

deformation of of the structure can be obtained by the superposition of

these effects. Referring to Fig. 9.a, for unyielding supports, we find that

compatibility requires

1 = 0

2 = 0

Where

1 = deflection at redundant point (1) in the direction of reaction X1

due to external loads of the original structure

2 = deflection at point (2) in the direction of X2 due to external

loads of the original structure

By the principle of superposition the compatibility equations at supports 1

and 2 are as follow and referring to Figs. 9.c, 9.d and 9.e

10 + 11. X1 + 12 . X2 = 0

20 + 21. X1 + 22 . X2 = 0

Where

10 = deflection at redundant point (1) in the direction of reaction X1

due to external loads of the main system.

20 = deflection at point (2) in the direction of x2 due to external loads

of the main system.

11 = deflection at point (1) due to a unit load at point 1. (see Fig.9.d.)

12 = deflection at point (1) due to a unit load at point 2.( see Fig. 9.e.)

And so on.


(a)

(b)Main System

(c)Deflections 10

and 20

(d) )Deflections

11 and 21

(e) Deflections 12

and 22

f) B.M.D.

Figure (9)

Example (7) For the shown continues beam in fig.10, Draw S.F, B.M.Ds. due to given

loads.

Solution Choose the main system as shown in fig.10.a., then the compatibility

equations are:

10 + X1. 11+ X2. 22 = 0

20 + X1. 21 +X2. 22 = 0

a) Main system

Figure (10)

The Flexibility coefficients are :-

10 = EI

dlMM 10

= 23

2

2

448(

2

EI


)4842

42

= EI

1280

20 = EI

dlMM 02

20 =

4882

7525

)83253

2(

2

448

1

EI

= EI

288

11 = EI

dlM 2

1

11 =

4

3

2

2

842

EI

48 t.m 48 t.m

+

b. M0 Diagram

+

X1 = 1t

4 t.m

11

c. M1 Diagram

X2 = 1t.m

1 t.m

22

-

d. M2 Diagram

e. Reactions

Figure (10)

11 = EI3

256

21 = 12 = EI

dlMM 21 =

5

2

1641

EI =

EI

16

22 = EI

dlM

2

2 = 13

2

2

161

EI =

EI3

16

then

1280+ 2163

2561 X = 0 --------(1)

and

864 + 48X1 +16X2 = 0 ---------(2)

Solve equations (1) and (2)

X1 = -11.14 t

X2 = 20.57 m .t

To get reaction Yc 0 aM

0 = 814.1157.20

+ )121294)16( cY

Yc = 5.14 t

f. S.F.D.

Figure (10)


From; y = 0

Y C +Y B +YA = 122

5.14 +11.14+ YA = 24

YA = 7.72 t

Mb = 412814.5

= -6.88 m .t

g. B.M.D.

Figure (10)

Hence S. F and B. M. Ds as shown in fig.10.f and g

Example(8) For the shown frame fig.11 draw N.F, S.F, and B.M.Ds due to given

loads.

a) Main System

Figure (11)

Solution (Twice statically Indeterminate)

Choose the main system as shown in Fig 11.a, hence the compatibility

equations are:

10 + X1 11 + X2 12 = 0

20 + X1 21 + X2 22 = 0

The value of flexibility coefficients are as Follows:

11 = EI

dlMM 11

=

5455

3

2

2

552

EI =

EI3

850


21 = 12 = EI

dlMM 21

=

1

2

55585

2

EI

= EI

5.32 =

EI3

5.97

22 = EI

dlM 2

2

=

3

2

2

81151

1

EI =

EI3

23

10 = EI

dlMM 10

=

5

3

2840

1

EI=

EI

3200

20 = EI

dlMM 20

=EI3

320

hence

-3200+ 850 X1 + 23 X2 = 0 ----(1)

-320 + 97.5 X1 + 23 X2 = 0 ----(2)

solving equation (1) and (2)

-1843.48+ 436.68 X1 =0

i.e. X1 = 4.22 t Figure (11)

And X2 = -3.98 m. t

MF = Mo + X1. M1 - X2. M2

MF = Mo + 4.22 M1 - 3.98 M2

MB = + 3.98 m.t

Mc = 22.45 = 21.10 m.t

Md = 98.3522.4 = -17.12m.t 4

Reactions

YF = Yo + X1. Y1 - X2. Y2


i.e.

YA = YA0 + X1. YA1 + X2 YA2

= 8

198.3020 = 19.5 t

e) Reactions f) B.M.D.

g) N.F.D. h) S.F.D.

Figure (11)

4. 3-TIME STATICALLY INDETERMINATE The following structures shown in Fig.12 are 3-time statically

indeterminate structures:

X3 = 0

X1 = 0

X2 = 0

Main System Original Structure

(Fixed Bexm)

Case (1)

X3 = 0

X1 = 0

X2 = 0

Main System Original Structure

Case (2)

Figure (12)


X2

X3

X1

X3

X1

X2

Main System (2) Main System (1) Original Structure

Case (3)

Figure (12)

In this case the redundant are; X1, X2, and X3 and the compatibility

equations are:

10 + X1 11 + X2 12 + X3 13 = 0

20 + X1 21 + X2 22 + X3 23 = 0

30 + X1 31 + X2 32 + X3 33 = 0

Example (9) For the given frame shown in fig.13 find the reactions and draw B.M.D.

Solution To solve this frame we start by removing support A and introducing in its

place three redundant reaction components X1, X2, and X3 as shown in

Fig.13.a, hence:-

10 m

10 m

A

C D

B

1.2 t / m

1.2 t / m

X2

X3

X1

a) Main System

Figure (13)

10 +X1 11 + X2 12 + X3 13 = 0

20 +X1 21 + X2 22 + X3 23 = 0

30 +X1 31 + X2 32 + X3 33 = 0


60 t.m

60 t.m

60 t.m

-

-

Mo.D

X1 = 1t

10 t.m

10 t.m

10 t.m

10 t.m-

- -

M1.D

X2 = 1t

10 t.m

10 t.m

10 t.m

+

+

M2 .D

X3 = 1t.m

+

+ +

1t.m

1t.m 1t.m

1t.m

1t.m

M3 .D

Figure (13) cont.

Values of flexibility coefficients:-

11 = EI

dLM .21 =

EI

1667

12 = dLEI

MM 21 = EI

1000

13 = dLEI

MM 31 = EI

200

21 = 12

22 = dLEi

M 22 =

EI

1333

23 = 32

= dLEI

MM 23 = EI

150

31 = 13 = EI

200

33 = dLEI

M 23 .

= EI

30


10 = dLEI

MM 01 = 5000

20 = dLEI

MM 02 = -7500

30 = dLEI

MM 03 = -800

hence

0

0

0

30150200

15013331000

20010001667

800

7500

5000

3

2

1

X

X

X

by Solving the above equations; we obtain

3

2

1

X

X

X

=

m33.3

6

1

tons

Final B.M.D

M = M0 + X1 M1 + X2 M2 +X 3. M3

6 t

3.33 t.m

1 t

1.2 t / m

6.67 6.67

6.67 6.67

3.33 3.33

8.33

+- -

b) Reactions c) B.M.D.

A

C D

B

d) Elastic Curve

Figure (13)


The final results are shown in Fig. 13.b, c and d at which the moment

diagram for the whole frame is given. A sketch of elastic curve of the

frame due to bending distortion is also shown by the dashed line in

Fig.13.d. Note that in this case there is one point of inflection in each

column and two point of inflection in the beam CD.

By referring to the previous example, we see that by using the method of

consistent deformations in analyzing a rigid frame, we encounter hard

calculations of the flexibility coefficient. The work, if done by hand, will

become complex if the problem involves as many redundants as a rigid

frame usually does. As a matter of fact, the method of consistent

deformations is often used by hand calculation, since a solution can be

much move easily obtained by any other method. However, with the

development of high-speed electronic computers, this method has

regained considerable strength in the scope of structural analysis.

Another solution

By using the symmetry; the frame becomes twice statically indeterminate

(fig.14); because X2= X3; hence:

X1X3 = X2

X2

1.2 t / m

10 10

Main System

Figure (14)

10 + X1 11 + X2 12 = 0

20 +X1 21 + X2 22 = 0

Calculate the displacement coefficients as follows:

15 t.m

1t

-

-10 t.m

10 t.m

-

10 t.m

10 t.m

X1 = 1t

X2 = 1t.m

1t.m

1t.m

1t.m

1t.m

X2 = 1t.m

Mo .D. M1 .D. M2 .D.

Figure (14) cont.


10 = EI

dlMM 10

= 103

21015 =

EI

1000

20 = EI

100

11 = EI

dlM

2

1

=

)10

3

2

2

1010(2101010

1

EI

= EI3

5000

12 = EI

dlMM 21 = 21

=

)1

2

1010(211010

1

EI

= EI

200

22 = EI

dlM 2

2

= 1101101101

EI

= EI

30

Hence;

-1000 + 22003

50001 X = 0 ------ (1)

and

- 230200100 1 X = 0 -------(2)

Solve(1) ,(2)

- 213

255 XX = 0

- 213

20

3

10XX = 0

13

5X =

3

5


X1 = 1 m.t, X2 = - 3

10

Mf =M0 + X1M1 + X2 M2

MA = 0 + 0- 13

10 = +3.33 m.t OK

Mc = 3

1010 = -6.67 m.t OK

3.333.33

-+

8.33

6.67

6.67

-

6.67

6.67

B.M.D.

Figure (14) cont.

The same result in Fig.13.c.

5.CASE OF n- STATICALLY IN DETERMINATE

STRUCTURES

Both the deflections resulting from the original external loads and the

flexibility coefficients for the primary structure can be obtained by the

method of virtual work. The remaining redundant unknowns are then

solved by simultaneous equations. This process can be generalized. Thus,

for a structure with n redundants, we have:


10 + 11X1 + 12 X2 +………….. +1n X n = 0

20 + 21X1 + 22 X2 +………….. +2n X n = 0

. .

. .

. .

. .

. .

n0 + n1 X1 + n2 X2 +………... + nn X n = 0

..(5)

Equation (5) in matrix from is:

0

.

.

.

.

0

0

.

.

.

.

.

.......

...........

..........

...........

..........

.......

.......

.

.

.

.

2

1

21

22221

11211

0

20

10

nnnnn

n

n

n X

X

X

……(6)

or simply

110 nnxnn

X = 0

Example (10)

For the shown fixed beam in

fig.15 draw B.M & S.F.Ds due

to the shown applied loads

(w=2 t/m)

Solution

From symmetry

X1 = X3

X2 = 0

hnce the beam is once st. Ind.

10 + 11.X1 = 0

Figure (15)


10 = EI

dlMM 10

= 183

2 2

Lwl

= EI

wl

12

3

11 = EI

dlM .21

= EI

L 11 =

EI

L

X1 = 11

10

S

S =

12

2WL

= 12

)6(2 2

= 6 t.m

YA =2

wl =

2

62 = 6t

Figure (15)

Example (11)

For the given fixed beam

shown in fig. 16; find the fixed

end moments MA and MB. Hence

draw B.M and S. F. Ds (use the

method of conjugate beam)

Figure (16)

Solution (using conjugate beam)

The fixed beam shown is

statically indeterminate to the

second degree since the

horizontal force does not exist.

End Moments MA and MB are

selected as redundants. By

applying the conjugate beam

method to determine MA and MB

based on the slope and

deflection at A and B must be

zero.


Y = 0

EI

LM

EI

LM

EI

abp BA

222

. = 0

MA +MB = L

bap ..

BM = 0

2MA+MB = 2

2....

L

bap

L

bap

hence

MA = 2

2..

L

bap, MB =

2

2 ..

L

bap

YA = L

bPMM BA . ,

YB = P-YA

Example (12)

For the given fixed beam

shown in fig.17 draw S.F

and B.M.Ds.

A

b

B

a

L

M

A B

BA

+

-M.a/L

M.b/L

2EILM.a

2

2M.b

2EIL

a)M0.D

2EI

M

M

b

a

M .La

2EIbM .L

-

+

b) Ma and Mb

Figure (17)

Solution The deflected curve will

be somewhat like that

shown by the dotted line,

which gives the sense of

the end moments as

indicated. Now we choose

MA and MB as redundants.

The elastic load based on

the moment diagrams

divided by EI plotted for

external moment M and

redundants MA and MB as

shown Must be in

equilibrium from Y = 0;

then:-


EIL

aM

EI

M

EIL

bM

EI

LM LBA

2

.

22

.

2

. 2.

2

= 0

MA – MB = 2

22 )(

L

baM …….(1)

From BM = 0; then:-

)3

.(2

.

3.

2

.

3

2.

2

.

3

2.

2

. 22 ab

EIL

aML

EI

LMb

EIL

bML

EI

LM BA =0 ….(2)

2MA – MB = 2

22 22(

L

bbaaM

Solving (1) and (2)

MA = )2(.2

baL

bM

MB = )2(.2

abL

aM

Ma

bM- -

+ +

c) B.M.D.

-Ya bY

d) S.F.D.

Figure (17)

Note that MA and MB are the same sense as externally applied M, if a>3

L

and b>3

L.

6. SETTLEMENT OF SUPPORTS

In a more general from, we may include the prescribed displacements

(other than zero ) occurring at the releases of the original structures. Then

these values ,......., 21 (Fig.18) must be substituted for the zeros on the

right-hand side of equation (6)

21P0Q

3w

1 2

Figure (18)

t/m-


Thus

nnnnnn

n

n

n X

X

X

.

.

.

.

.

.

......

.........

.........

.........

......

......

.

.

.

2

1

2

1

20

22221

11211

0

20

10

In which the column matrix n0 on the left hand side represents the

displacements at redundant points of the released structure due to the

original loads; the square matrix nn represents the structure flexibility,

each column of which gives various displacements at redundant points

due to a certain unit redundant forces; and the column matrix on the

right hand side contains the actual displacements redundant points

(settlements) of the original structure.

Settlements of supports (or temperature changes) in case of indeterminate

structures will develop additional internal force, and reactions. But in

case of determinate structures the settlement of supports (or temperature

changes), no additional internal force and no reactions will develop.

7. CHANGES OF TEMPERATURE:

The change of temperature has no effect on internal forces and internal

deformations at centerline of statically determinate structures. On the

other hand, in statically indeterminate structures, the changes of

temperature will result in the development of internal force and reactions.

a. Case of Uniform Rise of Temperature:

For the given once statically indeterminate beams and frame in Fig.19,

the compatibility equation is

10 +X1 11 = 0


t co

ot c

a. Frame

t co

ot c

b. Beam

Figure (19)

And the main system as shown in Fig.19.c and d.

t co

ot c

10

c. Main system

11

t co

ot c

X1 = 1t

d. get N1.D

Figure (19)Cont.

We = Wi

i.e.

10 = dlN .1 = )..(1 dltN

= dlNt .. 1

10 = t. ( Area of normal force diagram )

Mf = X1 M1 (M0 = 0 )

Nf = X1 N1 (N0 = 0 )

b. Non–Uniform Rise of Temperature

In case of non-uniform rise of temperature shown in Fig.20, also the

compatibility equation is

10 +X1.11 = 0


t co

ot c1

2

=

2

1t c

ot c

10

a. Main system

X1 = 1t11

b. get M1.D

C. L.h/2

h/2

.t .dL1

2.t .dL

d

d

d - ve M

c. Elongation & Rotations

Figure (20)

And from

We = Wi

hence

1. S10 = dMsdN .11

= )).(

()2

.( 211

211

h

ttMdl

ttN

= dlMh

ttdlN

tt.

21

211

21

Negative sign for d because the elongation is opposite to the M. Get X1

and hence

Mf = X1. M1

Nf = X1. N1

Knowing that:-

S11 = EA

dLN

EI

dLM 2

121


Example (13)

For the given continuous beam

shown in Fig.21; draw S.F and

B.M.Ds due to

1) given load

2) Settlement at support B of

2cm

3) t temp. = 20

b =25 cm

I = 7.146 310 m4

E = 2.1 26 /10 mt

= 1 510

h1 = 70 cm

h2 = 80 cm

Solution

1. Due to Uniform Load

The compatibility equation is:

10 +X1.11 = 0

10 = dlEI

MM 10 .

10 = -46.59 / EI

11 = dlEI

M.

12

Figure (21)

= 3.80 / EI

10 +X1 11 = 0

i.e.

X1 = 11

10

= 12.26 t. m

B.M.D. as shown in Fig.21.c.


2. Due to 2 cm Settlement at B

10 +X1 11 = 600

2

800

2

10 = 0

11 = EI

dLM 2

1 = 3.8/ EI

X1 = 8.31200

7 EI

= 23.0 t.m

as shown in Fig.21.f

3. Anther Solution for B. M, S. F. Ds due to 2cm Settlement at B:

10 +11 .X1 = 100

2

10 = 0

11 = EI

dlM 2

=

2.12

8

2

643.3

3

2

1

43.3

EI

= 44.45/ EI

X1 = 45.44

10146.7)2100000(02 3

= 6.75 t

Mf = X1 M1

Qf = X1 .Q1

Figure (21)

4. Due to t = +20o

= 1 510

10 +X1 11 = 0

10 = dMdN 11


= dlMh

tdlN

tt.

211

21

= 0 + h

t (area of M1.D)

10 = 1 510 20 ( )8.0

8

7.0

6(

2

43.3 )

= - 6.37 310 m

i.e.

X1 = 11

10

11 = EI

dlM 2

1

= EI

45.44

Figure (21)

X1 = 45.44

10146.7)210000(1037.6 33

= 2.15 t

Mf = X1 M1

Qf = X1 .Q1

hence

Mb = 2.153.43 = 7.37 m. t

Reaction at A

= 6

37.7 = 1.23 t

Reaction at B

=8

36.7

6

36.7 = 2.15 t

Reaction at C

= 8

36.7 = 0.92 t

Figure (21)


Example (14)

Compute the reactions and draw the Bending Moment diagram for the

beam shown in fig.22 due to the following support settlement:-

- point A . 5 cm downward

- point B 1.0 cm downward

- point C 1.5 cm downward

- point D 0

EI = 5000 t .m

Solution

A B C D

I 3 I 2 I

866

a) Main

System

b) M1.D.

X1 X1 X2X2

1 t.m

+

c) Settlement

Condition

d M2D.

0.5 1.0 1.5

1 t.m

+

e. B.M.D.

f. S.F.D.

0.86 t.m

6.89 t.m

+

-

+

--0.14

1.29

0.86 0.86

Figure (22)


The compatibility equations are:

12211110 . XX = 1 …….(1)

22221120 .. XX = 2 …….(2)

From either the method of virtual work and the geometry of settlement

condition the values of 1 and

2 are:

1) 1 = 0

2) 2800

5.1

600

5. = 0.0027 rad.

And;

11 = dlEI

M 12

= )3

1

3

21

3

2(

2

61

EIEI

=

EI3

8

12 = 21 = dlEI

MM 21

= 13

1

)3(2

61

EI =

EI3

1

22 = dlEI

M 22 =

3

2

22

81

3

2

)3(2

61

EIEI =

EI

2

Hence;

cB XX3

1

3

40 = 0

cB XX 23

10 = EI (0.0027)

XB = -0.86 t. m

Xc = 6.89 t. m

The find B.M.D are shown in Fig.22.e.

8- BEAM ON ELASTIC SUPPORTS:

For the given structure shown in fig.23, support at b is an elastic support.

The spring stiffness is K (force / unit displacement). To determine

reaction and deflection at b (b), assume the spring load is X1. The

downward deflection 10 at b plus that caused by upward reaction X1

should be equal to the spring contraction (b). Hence;


10 +11. X1 +K

1 .X1 = 0

The last term in the equation is the

spring contraction

X1 =

kS

111

10

10 = EI

WL

8

4 =

EI

dLMM 10

11 = EI

L

3

3

= EI

dLM 2

1

X1 =

3

31

1

8

3

KL

EIWL

1 = .1

K

X

1

Figure (23)

For non-yielding support (k = ) , the above equation gives;

X1 = WL8

3 o.k. (Case of roller support)

Example (15)

For the shown beam in Fig.24,

draw S.F and B.M.Ds in case

the support with stiffness K=

10t/m. compare the results of

the two examples

2 t / m

B

6

A

6

C

Figure (24)

Solution K = 1000 t/m

The compatibility equation is;

10 +X1 (11+ )1

K = 0

10 = EI3

1404 = 0.0675 m

36 t.m

+

a) M0.D

Figure (24)


11 = EI

36 = 0.0045 m

3 t.m

1t

+

b) M1.D

Hence

X1 =

1000

10045.0

0675.

= -12.27 t

Mf = M0 + X1 M1

Mb = 327.1236

= -0.82 t.m

Qf = Q0 +X1.Q1

QA = 6

82.

2

62

= 5.86

QB = 6

282.

2

622

= 12.28 t

0.82 t.m

8.59 t.m

++

8.59 t.m

c) B.M.D.

+ +- -

5.86 6.14

6.14 5.86

d) S.F.D.

e) Main Steel reinforcement

Figure (24)

The value of –ve moment at spring support B is very small if we compare

with –ve moment at B for example 5 (Fig.6). If a beam is provided with n

redundant elastic supports with stiffness K1, K2, …. K n; then the general

compatibility equations in matrix form are:-

0

.

.

.

.

0

0

.

.

.

.

.

)1

(

...

...

...

...

..)1

(

..)1

(

.

.

.

.

2

1

21

2

2

2221

112

1

11

0

20

10

n

n

nnnn

n

n

n X

X

X

K

K

K


Example (16) For the given frame shown in fig.25 Draw B.M.D. due to

1) Given loads

2) Horizontal displacement at C = 1 cm to right

EI = 12000 t.m2

Kspring = 100 t/m

Solution

1. Due to given loads

4 4

5 t 5 t

4 4

6

A

B CD

Spring

X2= 0

X1= 0

a) Main System

+

20 t.m 20 t.m

b) M0.D.

X1= 1

1t

6 t.m

6 t.m -

-

b) M1.D

X2= 1

4 t.m

-

b) M2D

Figure (25)

10 + X1 11 + X2 12 = 0

20 +X1 21 + X2 (22 + )1

KB = 0

10EIS = dlMM 01 = 3)2

168(20

= -720

20EIS = dlMM 02 = )2042

422

3

2

20

420(2

= -586.7


12EIS = dlMM 21 = )66

5

2

84(2

= 96

11EIS = dlM 21 = 6

3

2

2

666

3

2

2

166

= 264

K

EIEIS 22 =

100

11200024

3

2

2

84 2

= 205.33

264 X1 + 96 X2 = 720 …..(1)

96 X1 + 205.33 X2 = 586.70 …..(2)

96 + 34.91 X2 = 261.8 …..(3)

170.42 X2 = 324.9

X2 = 1.9 t

X1 = 2.03 t

M = M0 + X1 M1 +X2 M2

- ++

12.18

12.18

6.14

e) B.M.D.

Figure (25)

Moment at spring support is +ve moment because the stiffness of spring

support is small value. If the value of stiffness is bigger then 100 t/m the

positive moment decrease and in case of K = , the –ve moment at

support B will be the maximum values of –ve moment

2. Horizontal Displacement 1 cm at C

0 + X1 11 + X2 12 = -1 (-ve sign because displacement

in the opposite side of X1)

0 + X1 21 + X2 (22 + )1

K = 0

Get X1 and X2 and then draw B.M,, S.F, N.F..Ds as shown in Fig.26


Hence (EI = 12000 t.m2)

264 X1 + 96 X2 = 100

1 (12000)

96 X1 + 205.33 X2 = 0

Hence

X1 + 0.36 X2 = -0.45 …..(1)

X1 + 2.14 X2 = 0 …..(2)

Subtraction (1) and (2)

X2 = 78.1

45.0 = 0.25 t

X1 = 96

9.51 = -0.54 t

Hence:-

M = X1 M1 +X2 M2

Q = X1 Q1 +Q2 M2

N = X1 N1 +N2 M2

MD = +3.24 m.t

MB = +0.62 m.t

X2= 0

X1= 1t

3.24

3.24

0.62+

+

f) B..M.D

0.54

0.32

0.08

0.54

0.40

g) Reactions

+

+

0.32

0.32

0.54 0.54

h) N.F.D

+

+

-

0.54

0.32 0.32

0.08

i) S.F.D.

Figure (25) Cont


Example (17)

Draw B.M.D. for the given structure shown in Fig.26

EI = 10000 t.m2, kB = 5 t/cm

Figure (26)

Solution:

The given structure is once statically indeterminate

a. Primary Structure

b. Mo Diagram

c. M1 Diagram

d. Figure (26)


d. B.M.D.

Figure (26)

)1

( 11110

BKX = 0

10EI = 5.125.235.275.383

6.35.1

3

2

2

325.23

+

5.275.386243

3)5.225.235.175.38(5.0

28163

24

3

2

2

8504623)475.385.262(5.0

= -1826.985

11EI = )5.13

25.2(

2

35.1)

2

45.2(35.24344

3

2

2

48

5.13

2

2

35.1)67.5.1(

2

6.31)

2

5.25.1(6.35.1

= 139.87

Hence

)500

1000087.139(98.1826 1 X = 0

X1 = 11.43 t

Mfinal =M0+X1M1

Qf =Q0 +X1 Q1

Nf =N0 +X1N1

Rf =R0 + X1R1

B.M.D. is shown in Fig.26.d.


Example (18):

Draw B.M.D. for the following structures shown in Fig.27

EI = 15000 t.m2, kA = 300 t/cm, kB = 350 t/cm

Figure (27)

Solution:

The structure is twice statically indeterminate; hence:-

10 +X1 (11 + 122)1

XK A

= 0

20 +X1 21+X2 (22 + BK

1) = 0

a. Primary Structure

125

304

26

26

50

24

+

-

-

- -

b. Mo.D.

Figure (27)


c. M1.D.

X2= 1t

+

+

+

10

10

10

10

15

d. M2.D.

Figure (27)

EI.10 = dlMM 01.

= 26610)203

24(

2

1010

)30510125(5.10305125(3

5

42.32455.7525.63

2

= 3344.63

EI 20 = dlMM 02

= 3010125153

51062681050

3

1

+ 5.12525.63

25.0)153010125(

= -21713.28

EI 11 = )53

25(

2

555.7551061010

3

2

2

1010


= 1225.21

EI 22 = )53

210(

2

555.125101061010

3

2

2

1010

= 1725.21

EI 21 = -1225

Hence

3344.63+X1(1225 + 21225)300

15000X = 0 ……..(1)

)350

1500021.1725()21.1225(28.21713 21 XX = 0 ……..(2)

From (1) and (2)

X1 = 5.51 t

X2 = 8.46 t

Hence

M final = M0 +X1.M1 +X2.M2

B.M.D. as shown in Fig.27.e

e. B.M.D.

Figure (27)

Example (19)

Draw B.M.D. due to loads and 10t inter fiber for the given frame

shown in Fig.28

I = 0.048 m4, E = 200 t/cm

2, kC = 400 t/m,

kD = 300 t/m, h=100 cm


4 t / m

o10 c

o0 c

A

B

C D

D

8 8 10

Figure (28)

Solution:

This structure is twice statically indeterminate

1. B. M. D. due to given loads

EI10 = dlMM 01

=

53.585.137(

3

853.5

3

2

2

885.137

+ )53.57.14707.385.137(2

1)07.37.147

2

07.315.6(832

3

207.3

3

2

2

107.147

= -9226.370

17.23

14.77

137..85147.70

32

+

a) M0.D.

X1= 1t

5.53

3.07

-

b M1.D.

Figure (28)


X2= 1t

6.15

3.07

c) M2.D.

d) B.M.D.

Figure (28)

EI20 = dlMM 02

= -

70.1473

2

2

1015.607.3

3

2

2

885.137

15.67.14707.385.137(3

8

))07.37.14715.685.137(2

1

2

15.607.3832

3

2

= -10229.045

EI11 = dlM 21 = 2653.5

3

1 2

= 265.03

EI22 = dlM 22 = 2615.6

3

1 2

= 327.79


12 = 15.63

2

2

1007.307.3

3

2

2

853.5

)07.315.653.5(

2

115.607.307.353.5

3

8 2

= 261.73

But

10 +X1(11+ 122) XK

EI

C

= 0

20 +X1 21 + X2 (22 + )dK

EI = 0

Hence

21 73.261)400

048.200000003.265(37.9226 XX

= 0 ……(1)

)300

046.10279.327()73.261(04.10229

6

21

XX = 0 ….…(2)

Solving equations (1) and (2), hence:-

X1 = 12.75 t

X2 = 10.63 t

then

M final = M0 +X1 .M1 +X2 .M2 , and B.M.D. as shown in Fig.28.d

2) Springs deflections due to loads

c = 100400

75.12 =3.1875 cm

d = 100300

63.10 = 3.546 cm

2. B. M.D. due to t = +100 (inside the frame)

Figure (28)


(10)t = dlMh

ttdlNt .

)(... 1

1210

+ dlM .1

= 10-5

2

2653.5

00.1

)100( (neglect effect of N1)

= -0.72 10-2

(20)t = )2615.62

1(

1

1010 5

(neglect effect of N2)

= -0.8 10-2

-ve sign because the rotation of the cross section is in the opposite sense

of M1or M2.

Hence:-

EI (10)t+X1( 12211 ) XK

EI

c

= 0

EI (20)t+ )()( 222211

dK

EIXX = 0

-691.2+X1 (265.03+240) +X2207.97 = 0 ….. (3)

-768.0+X1 207.97+X2 (327.79+320) = 0 ….. (4)

From (3) and (4)

X1 = 0.99 = 1.0

X2 = 0.90

Then

Mfinal = X1M1+ X2 M2 , hence B.M.D. as shown in Fig.28.f.

-

8.298.60

f. B.M.D.

Figure (28)


In case of t = +10 outside the frame hence the B.M.D. is as follows:

+

8.29

8.60

g) B.M.D.

Figure (28)

9. FORCED DEFORMATIONS: Forced deformation are those not caused by external loads; but caused for

example by sliding or dettelemets of supports, and temperature

changes,….. etc. hence; We = Wi = 0

The procedures are:

1. Choose the main system.

2. Apply the forced deformation to the main system

3. Apply unit redundant, one at a time to main system and find M1, N1,

Q1 for each redundant

4. Calculate the values of displacements or rotation at removed

redundant in the main system Due to applied case of forced

deformation and also all unit redundant. Hence;

We = Wi

5. Calculate 10 from We = 0 or from geometry of deformed shape as

follows:-

10 = d1

M 1dN (for temps change)

Where:

d = . t dl (uniform rise of temp).


and d =

2

2

1

ttdl (non-uniform rise of temps).

d = 0 (uniform rise of temps).

And d =

h

2 t

1t

dl (non-uniform rise of temp).

(t1 inside the frame, and t2 out side the frame)

6. Apply the compatibility equations and get the values of redundants

for once statically indeterminate structures.

10+ X1 11 = 0

Where:

11 = EA

LM

.N

EI

dL ̀

2

12

1

For frames, Arches, and beams with link

11 = EA

LM

.N 2

12

1 for trusses

7. Calculate the internal forces in the statically indeterminate structures

due to force deformation then;

Mf = X1. M1 (M0 = 0)

Qf = X1. Q1 (Q0 = 0)

And Nf = N1.N1 (No = 0)

Internal forces in determinate main system due to forced deformation

are equal to zero. Also, the displacement or rotation at any point (due

to forced deformation is:

n = n0 + X1 n1

8. In case of twice statically indeterminate structures;

10 + X1 11 + X2 12 = 0

20 + X1 21 + X2 22 = 0

And also;

20 = dM .2N d .2

22 = EA

LNM

.22

EI

dL 2

2


12 = 21 = EI

dL

2

1MM

Mf = M1. X1 + M2 . X2

Qf = Q1. X1 + Q2. X2

Nf = N1. X1 + N2. X2

Example (20)

For the shown frame fig.29, draw B.M.D due

1) 3 cm down ward settlements at A

2) Uniform rise of temperature of 40o c.

3) Non uniform rise as given.

EI = 12000 t.m2, = 1 x 10

-5 , EA = 8000 t.

AB

C D

t1= 10°C

t1= 30°C

Figure (29)

Solution

1) 3 cm down ward settlement for main system

S10

X1=0a) main system

Figure (29)


1m.t

X1=1m.t

1m.t

1/121/12

b) M . D1

1m.t

c) N . D11m.t

Figure (29)

We = Wi = 0

or

We = 0

i.e.

1.10 + 12

1 (0.03) = 0.

10 = 12

0.03-= -0.0025 m

11 = EI

dL 2

1M +

EA

dL 2

1N

= (1 x 4 + 2

1x12 x

3

2)

EI

1

= EI

8= 0.00067 m

then

X1 =

11

10-

=

0.00067

0.0025 = 3.73 t

Mf = X1. M1


3.73m.t

3.73m.t

d) B.M.D

Figure (29)

2) Uniform rise of temperature

At main system

1 10 = .d 1

N

= . t. .dl 1

N

= 1 x 10-5

x 40

12

4 -

12

4= 0

hence X1 = 0 no internal forces in this case because the roller support at A

will move due to uniform rise of temp hence B.M.D. is zero if the two

supports at A and B not in the same level the internal forces for this case

has values.

3) Non uniform nise in temps:

1. 10 = .ds 1

.do 1

MM

=

1

30 - 10

EA

dL.

12

30 10

EI

dL.

1

NM

= 10-5

0 x 20

EI

8 20

= EI

31060.1 x

X1 =

11

10

=

8

31060.1 x = 0.0002 t

Mf = X1. M1


0.0002m.t

0.0002m.te) B.M.D

Figure (29)

Example (21)

For the shown frame find B.M.D. due to 4 cm downward settlement at, B,

2 cm slide displacement to right at B, 1cm at B, downward settlement,

and 0.01 rad clockwise at A

EI = 16000 t.m2

C

A

B

0.01 rad

1cm2cm

4 cm

Figure (30)

Solution

We = 0

10 (1)- 4

3

100

1 + 0.01 x 10 +

4

3 x

100

4 = 0

10 = 400

3- 0.10 - 0.03

= - 0.13 + 0.007s

= - 0.1225 m

11 = EI

dL 2

1M


= EI

1 4 x

3

2 x

2

4 x 4 6 x 4 x 6

2

6 x 3

2 x

2

8 x 6 (6)

3

2 x

2

6 x 6

= 0.268 m

X1 =

11

10

=

0.268

1225.0 = 0.46 t

Mf = X1. M1

0.01

X1=0

1cm

S10

1t

3/4t

3/4t

X1=1t

6

1

Figure (30)

The problem can be solved for each deformation and by the principal of

superposition we can get the final result. As shown.

4.6m.t

2.76m.t

2.76m.t

2.76m.t

Figure (30)


Example (22)

For the shown frame Fig.31, draw B.M.D. due to given forced

deformation EI= 50000 t.m2

1cm

C

B

D

2cm

2cm

Figure (31)

Solution

The frame is 3 times statically indeterminate; hence;

10 + X1. 12 + X2. 12 + X3 13 = 0

20 + X1 21 + X2. 22 + X3 23 = 0

30 + X1 31 + X2 32 + X3 33 = 0

We = Wi = 0

i.e.

1. 10 – 1 x 0.01 – 18 x 0.001 = 0

X3=0

S30S10

X1=0

X2=0

S20

a) Main system

b) M1.D.

Figure (31)


10m.t

10m.t

X2=1

M . D2

c) M2.D.

8m.t

8m.t

X3=1

M . D3

d) M3.D.

Figure (31)

Then:-

10 = 0.028

and;

1 x 20 – 1 x .02 – 10 x 0.001 = 0

20 = 0.03

And,

1. 30 – 1 x .02 + 8 x 0.001 = 0

30 = 0.012

Also

11 = EI

dL 2

1M

=

18 x 5 x 18 18 x

3

2 x

2

18 x 8

EI

I = 0.05 m

13 = (-8)5 18EI

I = -0.07128 m

21 = 12 =

10 x

3

2 8

2

10 x 8 18x5x10

EI

I = 0.0326 m

22 =

10 x 5 x 10 10

3

2 x

2

10 x 10

EI

I x = 0.0166 m

32 = 23 = (-8) 5 x 10EI

I = -.8 x 10

-3 m

33 =

8 x

2

8 x 8 8 x 5 x 8

EI

I

3

2 = 9.8 x 10

-3 m


Then the compatibility equations are

0.028 + 0.05 X1 + .0326 X1 – 0.0144 X3 = 0

0.03 + 0.0326 X1 + .0166 X2 – 0.008 X3 = 0

0.012 – 0.0144X1 - .008 X2 + 0.0098 X3 = 0

Solve the 3 equations hence:

X1 = 5.27 t

X2 = -14.186 t

X3 = -5.6 t

Hence

Mf = M0 + X1M1 + X2M2 + X3M3

= 0 + 5.27 M1 – 1486M2 – 5.6M3

e) B.M.D.

Figure (31)

Example (23)

For the shown frame in Fig.32, draw B.M.D. due to given case of forced

displacements at supports

EI = 20000 t.m2

Figure (32)


Solution

Part cgf is determinate then no internal forces due to forced displacement.

Hence the part A de B is twice statically indeterminate, then;

10 + X1 11 + X2 12 = 0

20 + X1 21 + X222 = 0

Choose main system as shown, hence:-

X1

S20

X2

S10

2I

main system

M . D1X1=1

6m.t

1t

6m.t 6m.t

6m.t2I

M . D2

1/101/10

X1=1

1m.t

1m.t

Figure (32)

We = Wi = 0


i.e.

1. 10 + 1 0.03 = 0

10 = - 0.03 m

1 x 20 +1 (0.001) - 0.02 1 = 0

20 = +0.001 m

And

11 =

EI

324 6 x

2

x56 6 x

3

2 x

2

6 x 6 2

EI

I

22 = EI

7.67

3

2

2

1 x 6 x 1

EI

I

2

10 x 1

x

21 =12 EI

33-

Substitute in compatibility eqns., hence:-

324X1 – 33X2 = .03 EI

-33X1 – 7.67X2 =-.001 EI

Hence

X1 = 2.82t

X2 = 9.55t

Mf = 2.82 M1 + 9.55 M2

B .M . D

7.37m.t

16.92m.t

9.55m.t

16.92m.t

Example (24)

For the shown frame in

Fig.33 draw B.M.D. due to

downward settlement at B

of 3cm

EI = 12000 t.m2

Solution

6

4

12

A

C D

B

3cm

Figure (33)


Choose main system as

shown in fig.29.a. The

compatibility equation is:-

10+X1. 11 = 0

To get 10 at support A due

to downward settlement of

3 cm at B apply the force

system hence;

We = wi = 0

i.e

1.10 36

1 = 0

10 = -.05cm

Hence

-0.05+X111 = 0

11 =EI

dlM 2

1

3cmX1 = 0

10

a) Main system

11X1 = 1

3cm

1t

1/6

1/6

b) Force System

6 t.m

6 t.m4 t.m

4 t.m-

--

c) M1.D.

Figure (33)

11 =

2

3

2

2

12241246

3

2

2

664

3

2

2

441

EI

= EI

33.381

= 3.2 cm

i.e

-0.50+3.2X1 =0

X1 = 0.156 t

Mf = X1.M1

Hence B.M.D as shown in fig

33.d

-

-

0.936

-

0.625

0.9360.625

d) B.M.D.

Figure (33)


Example (25)

For the shown frame in Fig.34, draw B.M.D. due to:

1) A anticlockwise rotation at B of 0.003rad.

2) A horizontal slip of support at A of 2cm to right.

EI = 18000t.m2

Figure (34)

Solution

The frame is once statically indeterminate

1) B.M.D. due rotation

a- Choose main system as shown in fig.34.a then calculate the

Reactions and draw M1.D as shown in fig.30.b & c.

b- Compatibility equation:-

10 +X1 11 = 0.003

0+X111 =0.003

X1 = 11

003.0

11 = EI

dlM 2

1

=

22.

2

333.1

EI+( 2)22.

2

633

33.633.

+32

67.026667.0

Hence:-

11 = rad410164.1

X1 = 25.76t.m

Mf = 25.76M1


A

C

B

a. Main system

A

B

C

1/9

1/18

1/18

1/9X1 = 1

11

b. Force system

0.33

0.330.33

0.33

1.00

--

+

+

c. M1 Diagram

8.5

-

-

8.5

8.5+

+

25.76

8.5

d. B.M.D.

Figure (34)

2) B.M.D. due to horizontal slip

Apply the force system at main system, Fig.34.f, hence:-

We = WI = 0

We =1.10 + )100

2(

9

1 = 0

10 = rad900

2

The compatibility equation is:

10 +X1. 11 = 0

X1 = 11

10

11 = rad410164.1


Then:-

X1 = 410)164.1(900

2

= 19m.t

Final B.M.D = 19.0 M1

A

C

B

2 cm

2 cm

10

e) Main System

11

2 cm

1/18

1/18

X1 = 1

1/9

1/9

f) Force System

-

-

6.30

6.30

6.30

+

+

19.0

6.30

g) B.M.D.

Figure (34)

Example (26)

Draw B.M, S.F.Ds for the shown beam in Fig.35 due to

1) applied loads

2) a rotation at A of 003.0 rad

3) rise in temperature in member BC as shown t = -30oc

EI =10000t.m2

c/101 5 ,h = 0.8 m

0 = 0.003 rad

A CB

4 t/m

Figure (35)


Solution

The beam is twice statically indeterminate hence choose the main system

as shown in Fig.35.a. The compatibility equations are:

10+X1. 11+X212 = 0.0003 .…….(1)

20+X1. 21+X2. 22 = 0 ……..(2)

+ve 0.003 because it is in the direction of X1. The flexibility coefficients

are:-

Figure (35)

10 =

C

B

dlMh

t

EI

dlMM 110 .

= 05.3

2618

1

EI = -0.0072 rad.

20 =

C

B

dlMh

tt

EI

dlMM .)( 2

2120

= 2

66)

8

30(103

3

2618

1 5

EI

= 00675.0216.0

= 0.02835 m

22 = EI

dlM 2

1

=

4

2

662

EI = 0.0144 m

12 = 21

= EI

dlMM 21

= 33.2

661

EI = -0.0006 m


11 = EI

167

2

61

= 0.0020 rad.

Then

-.0072 +.002X1 -.0006X2 = 0.0003 …….. (1)

0.02835-.0006X1+.0144X2 = 0 …….. (2)

i. e

-3.6 +X1 - 0.3X2 = 0.15

47.25 - X1 + 24 X2 = 0

23.7X2 = -43.5

X2 = -1.83 t

X1 = 4.30 t.m

The final B.M, S.F.Ds as shown in fig.31.e and g

Mf =M0 + X1 M1+ X2 M2

Figure (35)


10. TRUSSED BEAM AND TRUSSED FRAME (Composite Structures)

Figure (36)

In case of trussed beam or trussed frame the redundant (redundants) is the

force in truss (link) member, as shown in Fig.32. In the shown figure, the

beam and frame are once statically indeterminate. In case of composite

structure (frame with link members) the redundant force is the force in

links

Example (27)

For the shown trussed beam in Fig.36, draw S.F, B.M.Ds

EI = 10000t.m2, EA link = 4000t

A B

31

2 4

4t 4t

Figure (37)

Solution


The beam is once statically indeterminate hence, the compatibility

equation is:

X1=0

X1 X1

4t 4t

a) main system

16m.t 16m.t

b) M . D0

10 +X1. 11 = 0

EA

dlNN

EI

dlMM 10

0110

001 EI

dlMM

)4416

3

2

2

416(

2

EI

mEI

0426.067.426

9.2 9.2

d) B.M.D

EI

dlN

EI

dlM 2

1

2

111

EAEI

24)41.1(441

2444)4

3

2(

2

44 2

EAEI

3267.170

= 0.0171+.008 = 0.0251 m

X1 = tX 7.10251.

0426.1

= -1.7 t

Hence

M = -1.7M1+M0

Figure (36) Cont.

Example (28)


For the shown trussed frame Fig.34 draw B.M.D.

EI =30000 t.m2, EAlink =10000 t

Figure (36)

Solution

The frame is twice statically indeterminate. Choose the main system as

shown in Fig.36.a; hence:-

10 = EA

dlNN

EI

dlMM 2002

= )72

123(

1

EI

= -0.0042 m

20 = EA

dlNN

EI

dlMM 1001

= 0)71260(1

)67.72

660(

2

EIEI

= -0.248 m

22 = EA

dlN

EI

dlM 2

222

= )23

2

2

24262466

3

2

2

668

3

2

2

88(

1

EI

= ))241.16(5.061(1

EA

= 0.038 + 0.0014

= 0.0394 m


12 = EI

dlMM 21

= )72

123(

1

EI

= 0.00126 m

Figure (36)


11 = EI

dlM 2

1

= )33

2

2

36(

2

EI

= 0.0012 m

The compatibility equations are:-

10+X1. 11+X2 12 = 0 ….(1)

20+X1. 21+X2. 22 = 0 ….(2)

Substituting the values of

-0.0042 + 0.0012 X1 +0.00126 X2 = 0 ….(1)

-0.248 + 0.00126 X1 +0.0394 X2 = 0 ….(2)

Subdivide equation (1) by 0.00125 and equation (2) by 0.0394

-3.33 + 0.95 X1 + X2 = 0 ….(1)

-6.29 + 0.032 X1 + X2 = 0 ….(2)

-2.96 + 0.923 X1 = 0

Hence

X1 = -3.22 t

X2 = +6.40 t

Hence

Mf =M0 + X1 M1+ X2 M2

Nf =N0 + X1 N1+ X2 N2

Qf =Q0 + X1 Q1+ X2 Q2

Mc =-8 6.4 = -51.2 t.m

MD = -38.40 t.m

Mf = 5.50 t.m

The B.M.D. is shown in Fig.36.f


Figure (36)

Example (29)

For the frame with link shown in Fig.35, draw B.M.D. due to given loads.

Calculate the relative displacement between points A and d

EI =20000 t.m2, EAlink =8000 t

Figure (37)

Solution


In case of frame with link member or trussed frame or trussed beam,

generally the redundants are the forces in link members as shown in

Fig.37.

Figure (37)

sin = 10

8 = 0.80

cos = 10

6 = 0.60

The compatibility equation is

10 +X1 11 = 0

Figure (37)


Figure (37)

10 = EA

dlNN

EI

dlMM 1001

= )08.13

28192(

1

EI = -0.09216 m

11 = EA

dlN

EI

dlM 2

121

= )8.43

2

2

88.48.4

3

2

2

68.4(

1

EI

EA

110

= EAEI

1052.107 = -0.00663 m

Hence

X1 = 11

10

=

00663.0

09216.0 = 13.91 t

Mfinal =M0 + X1 M1

Mc = 0 +13.90 (-4.8) = -66.768 t.m

Md = 192 +0 = 192 t.m

The B.M.D. as shown in Fig.37.e

Figure (37)


11. DEFLECTION OF STATICALLY INDETERMINATE

STRUCTURES (REDUCTION THEORY)

The deflection of statically indeterminate structures may be calculated

exactly as determinate structures using virtual work method, but in this

case the calculations are very difficult because the indeterminate structure

must be solved twice, one to obtain final moments due to given loads

(Mf.D) and the second to obtain (M1.D) due to unit load at the section,

hence

y = EI

dlMM f 1

The reduction theory method reduce the solution of statically

indeterminate structure to once due to given loads (Mf.D) but the unit

load is applied at the section on the suitable main system or primary

structure chos en to obtain (M1.D), Hence

y = EI

dlmM f 1

Example (30)

Calculate the vertical deflection at n due to given loads for the shown

fixed beam in Fig.38.

EI =10000 t.m2

10t

A Bn

Figure (38)

Solution

1- Solution by traditional virtual work method (Get Mf Diagram)

10 +X1 11 = 0

X1 =11

10

10 =EI

1

2

820

=

EI

80


11 =EI

81

X1 = X2 = 10 m.t

X1

10t

X2=X1

20t.m

a) M . D0

1m.t1m.t

0

b) M . D1

10mt10mt

10mt

c) B. M.D Figure (38)

To get M1 due to 1 t

Figure (38)


10 +X1 11 = 0

10 = EI

dlMM 01

= )2

82(

1

EI =

EI

8

11 = EI

8

X1 = 1 m.t

Hence M1.D as shown in the Fig.36.f

1mt1mt

1mt

f) M1.D

Figure (38. f) M1.D.

yn = EI

dlMM f1

= )3

2

2

210(2

2

EI =

EI3

80

= 100003

10080

= 0.267 cm

2- Solution using reduction theory (Fig.38.h)

Choose any suitable main system (1)

y = EI

dlmM f 1

= ))10203

2(

2

42(

2

EI


= EI3

80 = 0.267 cm

Figure (38)

For main system (2):-

yn = EI

dlmM f 1

= )3

4

2

4202410(

1

EI

= )3

16080(

1

EI

= EI3

80 = 0.267 cm

(the same result of main system 1)

Figure (38)


Example (31)

Compute the reactions and

draw the bending moment

diagram for the frame shown

in Fig.39. due to given loads.

Find the deflection at point n

and rotation at A.

EI = 12000 t.m2

Solution

Draw M1 and M0 diagrams.

This frame is statically

indeterminate to the first

degree. Select X1 as the

redundant, then

Figure (39)

10 +X1 11 = 0

10 = EI

dlMM 01 = ))4()3(

3

2

2

848(

1

EI

= EI2

6848

= EI

1152

11 = EI

dlM 2

1

12t

A Bn

c

A B

c X1=0

10

12t

a) Primary system

Figure (39)

= )63

2

2

666

3

2

2

166(

1

EI =

EI

72192 =

EI

264

Figure (39)

Hence

EI

1152 +X1

EI

264 = 0

X1 =264

1152 = 4.36 t

Mfinal =M0 + Xc M1

48 t.m

Xc = 1t

X1=1t

11

b) M . D 0


Mb = 0 + 4.36 (-6) = -26.16 t.m

Figure (39)

Reactions

XA = XB = 6

16.26 = 4.36 t.m

YA =16

1636.4812 = 4.36 t.m

YB = 7.64 t.m

Deflection at point n:-

Apply 1 ton on the main system at point n as shown in the figure


Figure (39)

yn = EI

dlmM f 1

= )2

16.26

2

164(

1)48

3

2

2

84(

2

EIEI

= 12000

28.2091024 = 0.0678 m

Rotation at A:-

Apply 1 t.m at A and draw B.M.

Figure (39)


A = EI

dlmM f 1

= )3

1

2

1616.26

2

1

2

1648(

1

EI

= 12000

24.122 = 0.01018 rad. Clockwise

Example (32)

Compute the force in the tie rod of the shown composite structure shown

in Fig.40. Draw B.M.D., N.F.D. Find deflection at point G

Figure (40)

E1I1 =12000 t.m2

E1A1 =10000 t

E2A2 =5000 t

Solution

The effect of normal force in link member and beam are taken into

consideration to obtain 10 and 11. The compatibility equation is:-

10 +X1 11 = 0

10

Figure (40)


Figure (40)

10 = EA

dlNN

EI

dlMM 1001

= 4.2365.04.2548.436(3

64.2

3

2

2

654[

01

]363

2

2

128.4)8.4545.0

EI

= 11

6.1900

IE

= -1.584 m

11 = EA

lN

EI

dlM 2

121

= ]1110

[]6.0126.0

[]8.43

2

2

128.4[

2

221111 AEAEIE

= 5000

0.10

10000

32.4

12000

32.184

= 0.156 m t

X1 = 10.15 t

The force in the tie rod equal to 10.15 t


Figure (40)

Deflection at B:-

Apply 1 t at B on the main system. Draw m1.D and n1.D is zero as shown

in Fug.38.g

Figure (40)

1 B = EA

dlnN

EI

dlmM ff 11

= 01 EI

dlmM f

= 72.12)62

36(

2

4672.12[

2

1

EI


]33

2

2

664.294

2

636.42

= EI

84.17732.50872.17164.152

= 12000

8.361 = 0.030 m

= 3 cm

The elastic curve is shown

Figure (40.k) Elastic Curve

Example (33) For the shown trussed structure shown in Fig.41, draw N.F, S.F, and

B.M.Ds. Find deflection at d

Figure (41.a)

Solution The structure is once statically indeterminate. The primary structure is as

shown in the figures.

10+X1. 11 = 0


10 = EA

dlNN

EI

dlMM 1001

= 2

70.477.10100(

2

7.410.98400[

01

)]7.44

3

3

77.10300

EI = -0.28407 m

Figure (41)

11 = EA

dlN

EI

dlM 2

121

= 2

7.410.987.47.4

3

2

2

77.107.4[

4.4

3

2(8

2

4.4

EI

1)]7.4

EA

1)81.177.106.087.181( 222

= 43 1026.4107.4

= 0.00513 m

X1 = 0051.0

4.2844.0

= 55.69 t (tension)

Figure (41)

Figure (41)


Deflection at d:-

1d = EA

dlnN

EI

dlmM ff 11

Apply 1 t at d on main system then draw m1, n1 Ds

Figure (41)

tan = 20

8 = 0.40

sin = 54.21

8 = 0.371

cos = 54.21

20 = 0.9285

1d = 3

5.172/54.2127.38

3

5.72/54.211000[

1

x

EI

]202

8)78.10627.138(20827.138152/54.21100

+

]8134.872

2/54.21)1.3868.30(

2

2/54.2141.75928.0[

1

EA

= 80000

15.1106

100000

26.25117 = 0.2500 + 0.0138 = 0.2638 m

Compare the value of deflection from bending moment and normal force?


12. ANALYSIS OF STATICALLY INDETERMINATE

TRUSSES

The indeterminancy of a truss may be due to redundant supports or

redundant members or both. If it results from redundant supports, the

procedure for attack is is the same as that described for a continuous

beam. If the redundant element is a member, the element is considered to

be cut at a section and replaced by two equal and opposite axial redundant

forces representing the internal force for that member. The condition

equation is such that the relative axial displacement between the two sides

at the cut section caused by the combined effect of the original loading

and the redundants should be zero.

Example (34)

Compute the reactions and member

forces for the given truss shown in

Fig.42.

E = 2100 t/cm2

L/A = 120 m-1

Solution The truss is indeterminate to the first

degree. Select yb as redundant. Then

Figure (42)

b0 +Xb bb = 0

or

10 +X1 11 = 0

a) Primary Structure (N0)

b) N1

Figure (42)


10 = b0 =EA

dlNN 10 = )(10

EA

LNN

11 = bb =EA

dlN 2

1 = )(21

EA

LN

The following table gives the values of 10 and 11 and internal forces N

Member L/A N0 N1 N0 N1 N12 X1 N1 N

ab 120 -6 -0.375 2.25 0.14 +0.82 -5.18

bc 120 -6 -0.375 2.25 0.14 +0.82 -5.18

ad 120 +10 0.625 6.25 0.39 -1.36 8.64

dc 120 -10 0.625 -6.25 0.39 +1.36 11.36

Bd 120 0 -1.0 0 1.0 +2.18 2.18

4.50 2.06

The values of reactions

X1 = Xb =06.2

5.4 = -2.18 t

N = N0 + Xb N1

or

N = N0 + X1 N1

Example (35)

Compute the reactions of the given truss shown in Fig.43

L/EA=const.

Figure (43)


Solution

The truss is indeterminate to the first degree. Select Xb as redundant.

10 +X1 11 = 0

Figure (43)

10 = )(10EA

LNN

11 = )(21

EA

LN

The following table shows the values of 10, 11. Owing to symmetry of

the load, structure and redundant, only one-half of the members need be

included in the following table:-

Member N0 N1 N0 N1 N12 X1 N1 N

a-1 -1.5 +0.75 -1.125 0.56 1.0 -0.5

a-2 0 -1.25 0 1.56 -1.67 -1.67

1-2 +2.5 -1.25 -3.125 1.56 -1.67 +0.83

1-3 -2.0 +1.0 -2.0 1.0 1.34 -0.66

2

1 (2-3)

-3.0 0 0 0 0 -3.0

2

1

-6.25 4.68

X1 =68.4

25.6 = 1.34 t

XA = XB = X1 (From 0XF )

YA = YB = 1.50 t (From symmetry)


Example (36)

Compute the normal forces (N)

in the members of the given

truss shown in Fig.42.

Solution

The truss is once indeterminate.

The analysis and results are

tabulated as follow

Figure (44)

b) N1 a) N0

Figure (44)

Member L

(cm) Area

N0

tons

N1

Tons EA

LNNo 1

EA

LN 2

1

X1 N1 N

2-A 400 - -10 0 0 0 0 -10

A-B 400 2A -10 -0.71 EA

1420 EA

100 1.31 -8.69

B-4 400 - -10 0 0 0 0 -10

1-3 400 2A +10 -0.71 -EA

1420 EA

100 1.31 -11.31

2-1 566 - +14.14 0 0 0 0 14.14

A-1 400 2A -10 -0.71 EA

1420 EA

100 1.31 -8.69

B-3 400 2A -10 -0.71 EA

1420 EA

100 1.31 -8.69

4-3 566 - +14.14 0 0 0 0 14.14

1-B 566 A 0 1 0

EA

566 -1.85 -1.85

A-3 566 A 0 1 0

EA

566 -1.85 -1.85

EA

2840 EA

1532

10 +X1 11 = 0

X1 =1532

2840 = -1.85 t

N = N0 +X1 N1 as shown in the given tables.


13. DEFLECTION OF INDETERMINATE TRUSSES

Figure(45)

The vertical displacement at point d for the truss shown in Fig.43 is

obtained from the following equation:-

yd = LEA

NN.

Where:

N are the normal forces in the members of the indeterminate truss due to

the given external loads and N are the normal forces in the members of

the indeterminate truss when subjected to1 ton at point d. This means that


the indeterminate truss is to be solved two times; due to external loads

and the virtual load. Such procedure may be simplified as follows:-

N = 110 .NXN

and

N = 110 .NXN

Knowing that

dy = LEA

NN.

= EA

LNXNNXN ).)(.( 110110

= )( 1111

011

101

00 LEA

NNXXL

EA

NNXL

EA

NNXL

EA

NN

The term LEA

NN 10 is equal to the vertical deflection at support B due to

external loads which should equal zero.

The term LEA

NN 11 represents the vertical deflection at B due to vertical

load at B which also equal zero.

Then

dy = )( 1100 NXNL

EA

N

= LEA

NN0

Where 0N are the normal forces in the members of the choosen primary

structure due to 1 ton at d.

Example (37)

Compute the horizontal displacement at point 6 for the given truss shown

in Fig.46.

A =50 cm2, E = 2000 t/cm

2

Solution

The procedure and results are given in the following table


Figure(46) Mem. L

(cm)

N0 N1

EA

LNNo 1

EA

LN 2

1

N* 0N

EA

LNN 0.

A-1 400 80 -1.33 -42560 707.6 5.71 2.67 6098.3

1-3 400 80 -1.33 -42560 707.6 5.71 2.67 6098.3

3-5 400 0 0 0 0 0 0 0

B-2 400 -180 2.67 -192240 2851.6 -30.87 -4 49392

2-4 400 -20 0 0 0 -20 -1.33 10640

4-6 400 -20 0 0 0 -20 -1.33 10640

A-B 300 0 0 0 0 0 0 0

1-2 300 -30 0 0 0 -30 0 0

3-4 300 0 -1 0 300 -55.85 0 0

5-6 300 -15 0 0 0 -15 0 0

A-2 500 125 -1.67 -104375 1394.45 31.73 1.67 52.99

2-3 500 -75 1.67 -62625 1394.45 18.28 -1.67 -30.53

3-6 500 25 0 0 0 25 1.67 41.75

4-C 600 0 1 0 600 0 0 0

EA

444360

EA

7.7955

EA

81.82932


* N = No + X1 N1

X1 = 7.7955

444360 = 55.85 t

6 = LEA

NN 0. =

502000

80.82932

=0.83 cm

Example (38)

For the given truss shown in Fig.47, find the forces in members due to '

a) Given loads

b) Rise in temperature t = 20o in the upper members

EA

L =constant =0.01 cm/t

Figure (47)

Solution

This truss is once indeterminate internally. The chosen main system and

No and N1 are as shown in the given table

a) Due to given loads

10 + X1 11 = 0

X1 = 11

10

Figure (47)

From the table


01NN = 14.14

21N = 4.0

11 = EA

LN

21 = 4 (0.01) = 0.04 cm

0.0002 LN1 = -0.057

Hence:-

a) Due to given loads

X1 = 4

14.14 = -3.54 t = -2.5 2

Normal force N =No + X1.N1

= No - 3.54.N1

b) Due to t = 20o for upper members

10 = dLtN ...1

= LtN ...1

= 110-520 LN .1

= 0.0002 LN .1

The following table shows the results

Due to loads Due to t

Member L N0 N1 N0 N1 N12 0.0002L 0.0002N1.L

A-2

400

0 0 0 0 - -

2-4 10 0.71 7.1 0.5 - -

4-B 0 0 0 0 - -

1-3 -10 0 0 0 0.08 0

3-5 -10 -0.71 -7.1 0.5 0.08 -0.057

5-6 -10 0 0 0 0.08 0

A-1 -10 0 0 0 - -

2-3 -10 -0.71 -7.1 0.5 - -

4-5 -10 0.71 -7.1 0.5 - -

B-6 -10 0 0 0 - -

1-2 14.14 0 0 0 - -

2-5 0 1 0 1 - -

3-4 0 1 0 1 - -

4-6 14.14 0 0 0 - -

-0.057


i.e.10 = -0.057

X1 =11

10

=

04.0

057.0 = 1.41t

Hence:-

N = X1 N1

= 1.41 N1

Example (39)

For the shown truss in Fig.48, determine the forces in members due

settlement at B = 2 cm

L/EA =0.012 cm/t for all members.

Solution

Figure (48)

10 = 0.0

and

10 + X1.11 = -2.0 cm

The given truss is once statically indeterminate and since the truss is

symmetric, calculations are obtained for half the truss. The calculations

and results of 11 are shown in the following table:-


Member N1 N12 N= X1.N1 (tons)

1-3 0.5 0.25 -5.95

3-5 1.0 1.0 -11.9

5-7 1.5 2.25 -17.85

a-2 0 0 0

2-4 -0.5 0.25 5.95

4-6 -1.0 1.0 11.90

a-1 0.5 0.25 -5.95

3-2 0.5 0.25 -5.95

5-4 0.5 0.25 -5.95

0.5(7-b) 0 0 0

1-2 -0.71 0.5 8.45

3-4 -0.71 0.5 8.45

5-6 -0.71 0.5 8.45

2

1

7.0

11 =7 2 0.012 =0.168 cm

= -2.0 cm

10 + X1.11 = -2

X1 = 168.0

2 = -11.9 t

N = X1.N (as shown in the table)

14. GENERAL REMARKS CONCERNING SELECTION

OF REDUNDANTS

From the previous examples, we recognize that there is considerable

latitude in selecting redundants. The only restriction being that they shall

be selected so that a stable primary structure remains. By proper selection

of the redundants, however we can minimize the numerical computations.

This objective can be achieved by adhering to the following polices.

1. Take advantage of any symmetry of the structure

2. Select the primary structure so that the effect of any of the various

loading conditions is localized as much as possible

Considerations of several alternative selections of the redundants for the

continuous truss shown in Fig.49 will illustrate the validity of these


statements. This structure is indeterminate to the second degree. Any

selection of redundants will involve two equations of the following form:-

10+X1. 11+X2 12 = 0

20+X1. 21+X2. 22 = 0

Primary Structure (1)



Figure (49)

Knowing that:-

i21 = 12

10 = EA

LNN 01.

11 = EA

LN 2

1

20 = EA

LNN 02 .

21 = EA

LNN 12 .

Before these terms can be evaluated, N0, N1, N2 member forces must be

computed. If the structure is symmetrical, and if symmetric redundants

are selected, the X2 forces can be obtained from the X1 forces by

symmetry. Further 22 will be equal to 11 in such a cases, leaving only


four deflection terms to be evaluated. The evaluation of these terms will

involve less computations if the redundants are selected to as to restrict

the effect of the various loading conditions to as few members as

possible. The latter will be true whether or not the structure is

symmetrical.

All the three alternative selections of the redundants show in the previous

figure take advantage of symmetry. The various loading conditions effect

the portions of the structure indicated in each case. Comparison of these

primary structures shows clearly that selection 3 is the best since it is

most effective in localizing the effects of various loading conditions. In

selection 3,if such a member is cut, only its axial force may be considered

as a redundant.

15. CHOICE OF MAIN SYSTEM IN CASE OF

SYMMETRY AND ANTISYMMETRY

1) Case of beam symmetry

The fixed beam shown in Fig.48 is three times statically indeterminate. If

one takes the symmetry into consideration, the beam becomes once

statically indeterminate; hence at center line shearing force and normal

force are zero; then:-

10+X1. 11 = 0

10 = EI

dlMM 01

11 = EI

dlM 2

1

Main System (1)

Main System (1)

Figure (50)


2) Case of beam anti-symmetry

The shown fixed beam is anti-symmetry, hence bending moment at C.L.

and normal force are zero:-

10+X1. 11 = 0

10 = EI

dlMM 01

11 = EI

dlM 2

1

Main System

Figure (51)

3) Case of Frame Symmetry

Structure symmetry mean that the structure must has complete symmetry

in shape, loads, dimensions, supports and momet of ineria. The main

system in case of frame shown in Fig.49 may chosen as shown in main

system 1 or main system 2. The frame in this case is a twice statically

indeterminate not three times statically indeterminate,because the shear at

center line is equal to zero.

From symmetry, the frame has two

redundants only; X1 and X2

(Shearingforce at center line is

zero)

Figure (52)


4) Case of Frame Anti-Symmetry

In case of anti-symmetry,at centerline

moment and normal force are equal to

zero. As shown in Fig.50, the frame is

once statically indeterminat (Bending

moment at center line is zero)

Figure (53)

5) Case of Closed Frame Symmetry

From symmetry, the frame is twice statically indeterminate because the

shear at centrline (cut section) is zero (Fig.51). Hence:-

10+X1. 11+X2 12 = 0

20+X1. 21+X2. 22 = 0

Figure (54)


6) Case of Closed Frame Anti-Symmetry

The frame is anti-symmetry and

once staticallyindeterminate.Hence

10+X1. 11 = 0

and

Mf = M0 +X1. M1

Figure (55)


STRUCTURES USING CASTIGLIANO SECOND

THEOREM; THEOREM OF LEAST WORK

The previous approach to the analysis of indeterminate structures

involved writing superposition equations of the deflections of the points

of application of the redundants. Instead of doing this, however,

expressions for these deflections can be set up using Castigiliano's second

theorem.

The latter approach is actually very similar to the former. It is a somewhat

more automatic procedure, however, and is therefore preferred by some

engineers. Since Castigiliano's theorem should really limited to the

computation of the deflection produced simply by loads on the structure,

this method lacks the generality of the superposition- equation approach.

Consider the following indeterminate beam shown in Fig.53. The

deflection at point b can be evaluated by using Casigiliano's second

theorem. In this case only bending deformations are involved.

u = EI

dlM

2

2

……(1)


But

bx

u

= b

But point b in the actual structure

does not deflect, b on the primary

structure must equal zero. As a

result,

bx

u

=

EI

dl

x

MM

b

= 0 ……(2)

Figure (56)

However, M, being equal to the total bending moment in the primary

structure due to all causes, may be expressed as being the superposition

of the contribution of the applied load only and the contribution of the

redundant Xb, thus

M = M0 + X1 M1 ……(3)

bx

M

= Mb

Equation (2) becomes

EI

dlMX

EI

dlMM bbb

20 = 0 ……(4)

But

b0 = EI

dlMM b0

……(5)

bb = EI

dlM b

2

From Eqn.5 it is immediately apparent that Eqn.4 is actually a statement

that:-

b0 + Xb bb = 0 ……(6)

Thus it is possible to use Castigiliano's theorem somewhat more

automatically and effectively in certain problems.


If in the analysis of indeterminate structures the deflection of the point of

application of a redundant is zero, then applying Castigiliano's theorem as

in Eqn.1 reduces to the statement that the first partial derivative of the

strain energy with respect to that redundant is equal to zero. This is

equivalent to stating that the value of the redundant must be such as to

minimize the strain energy. This special case of Castigiliano's second

theorem is often called the theorem of least work and may be stated as

follows:-

"In a statically indeterminate structure, if there are no support movement

and no change of temperature, the redundants must be such as to make

the strain energy a minimum".

Castigiliano's theorem is applicable only when the deflection of the

structure is caused by loads. It is possible to solve the cases of settlement

and temperature but the procedure is often not as straight forward as that

of the superposition- equation approach.

Example (40)

Solve the following frame shown in Fig.57 using Castigiliano's theorem.

Figure (57)

Solution

This frame is indeterminate to the third degree. Cut the girder at mid-span

and select the moment, axial force, and shear as the three redundants X1,

X2, X3.

u = EA

dlN

EI

dlM

22

22


3X

u

= 0

2X

u

= 0

1X

u

= 0

Figure (57)

Hence:-

A

L

X

NN

I

dL

X

MM

33

= 0 ……(1)

A

L

X

NN

I

dL

X

MM

22

= 0 ……(2)

A

L

X

NN

I

dL

X

MM

11

= 0 ……(3)

From F to C (0 < X < 2)

M = X3 – x .X2

3X

M

= 1,

2X

M

= -x,

1X

M

= 0

N = X1

3X

N

= 0,

2X

N

= 0,

1X

N

= 0

From C to B (2 < X < 6)

M = X3 – x .X2 – 10(x-2)

3X

M

= 1,

2X

M

= -x,

1X

M

= 1

N = X1

3X

N

= 0,

2X

N

= 0,

1X

N

= 1


From B to A (0 < y < 6)

M = X3 – y .X1 – 6 .X2 -40

3X

M

= 1,

2X

M

= -6,

1X

M

= -y

N = -10 – X2

3X

N

= 0,

2X

N

= -1,

1X

N

= 0

From F to d (0 < x < 6)

M = X3 + x .X2

3X

M

= 1,

2X

M

= x,

1X

M

= 0

N = – X1

3X

N

= 0,

2X

N

= 0,

1X

N

= -1

From d to E (0 < y < 6)

M = X3 + 6 .X2 - y .X1

3X

M

= 1,

2X

M

= 6,

1X

M

= -y

N = X2

3X

N

= 0,

2X

N

= 1,

1X

N

= -1

Setting up Eqn.1

6

2 1

23

2

0 1

232

)1)(2010.(2

)1)(.(E

dxxXxX

E

dxXxX


6

0 1

23

6

0 1

2132

)1)(.()1)(406.(I

dxXxX

E

dyXXyX

6

0 1

123 )1)(.6(I

dyXyXX = 0

We obtain 6

2

2

2

2

3

2

0

2

2

3 )202

102

.(2

1)

2.(

2

1

X

xX

xxXX

xxX

6

2

2

2

3

6

0

21

2

3 )2

.(2

1)40.6

2.(

X

xxXyyXX

yyX

6

0

2

123 )2

6.(

yXyXyX = 0

Then

X3 - 2 X1 = 0 ……..(1)

Setting up Eqn.2 and 3and solving 1, 2 and 3, we can get X1, X2, X3. The

results show that the effect of axial forces is very small and maybe

neglected.

17. TORSION OF FIXED BEAM

The fixed beam shown in Fig.55

with torsion force. The statically

indeterminate beam fixed at two

ends (6 times indeterminate) but in

case of torsion moment (t.m/m) the

solution independent on bending

moment only while (10 = 0). Hence

for torsion moments for fixed beam

is once statically indeterminate, the

primary structure (Mt0) and Mt1 due

to X1 = 1t.m/m are shown in Fig.58.

Then

10+X1. 11 = 0

X1 =11

10

Figure (58)


10 = L

t

ttGI

dxMM

010 .

Mt0 = L

t dxxm0

)(

Mt1 = 1

Hence:-

10 = L x

t

t

dxdxxmGI 0 0

)).((1

= t

c

GI

XLA )(

11 = L

t

tGI

dxm

0

21 =

tGI

L

Hence

X1 = L

XLA )(

It must be noted that the X1 is equal to the reaction at this point of a

simple beam loaded with the original tm diagram as a load, the torsion

moment at end A is then.

MtA = A - X1 = A - L

XLA )( =

L

Axc

The final torsion moment diagram is then similar to the S.F.D. of a simple

beam.

Example (41)

Draw T.M.D. for the shown

fixed beam in Fig.56

Solution

Figure (59)

MtA = 6

5.12)75.4)(5.24(

= 7.42 t.m

MtB =0.58 t.m


The S.F.D. for simple beam which is the T.M.D.

Figure (59)


PROBLEMS

1. Draw the S.F. and B.M.Ds for the following structures due to

a) Given loads

b) Settlement and temperature changes as goven

c) Calculate the deflection at n.

a) B = 2 cm

EI =10000 t.m2 4 t/m

5t

nA

Bc

b) Kb = 200 t/m

EI =12000 t.m2 3t 3t

A B c

springsupport

c) t = 20o

h = 120 cm

E =210 t/cm2

B = 3 cm

A = 0.003 rad

4 t/mA B

n

d) B = 1 cm

EI =8000 t.m2

A = 0.002 rad

e) t = 30o

h = 0.8 m

EI =8000 t.m2

f) I =32000 cm

4

E =2000 t/cm2

K = 6 t/cm

t = 20o

h = 0.5 m


g) A = 0.004 rad

C = 0.002 rad

EI = 9000 t.m2 B

A C

3t/m

I

8t

2I In

h) EI =10000 t.m

2

K = 3 t/cm

i) A = 0.001 rad

EI = 6000 t.m2

B = 3 cm

12t.m

n

0 5t

A B

3I I

A

j) B = 2 cm

A = 0.01 rad

EI =14000 t.m2

0A

A B

A

2. For the following frames

a) Find the reactions and draw B.M, S.F. and N.F.Ds.

b) Compute the vertical deflection at point n and horizontal displacement

at m and B

a) EIframe =10000 t.m2

EAtie =10000 t


b) EIframe=20000 t.m2

EAtie =30000 t

3t

tie

A B

3t

3t

3t

c) EI =10000 t.m

2

3t/m

A B

I

m

4I

4t/m

d) EI =16000 t.m

2

EA =8000 t

GAr = 4000 t

Uniform rise in temp

t = 20o

E = 200 t/cm2

h = 100 cm

e) Find the

deflection at

springs D & E

EI =12000 t.m2

KD = 400 t/m

KE = 350 t/m


3) Find the reaction and Draw B.M.d., S.F.D., N.F.D. for the following

frames: due to:

1. Given loads

2. Vertical downward settlement at B of 2 cm.

3. Rise in temp. t = 200 in CD

4. Rotation of 0.003 rad at fixed support A

a) EI =

30000t.m2

b) EI = 20000 t.m2

EA = 10000 t

c)


d)

4. Find the reactions and Draw B.M.D., S.F.D., N.F.D. for the following

frames.

a) EI = 30000 t.m2

EA = 10000 t (For link members)

2t/m2t/m

b) EI =16000 t.m2


c) EI = 40000 t.m2

EA = 10000 t (For links)

2t/m

I

2I 2I

d)

2t/m

10t

AB C

e) EI = 10000 t.m2

EAlink = 4000 t


f) EI = 20000 t.m2

5. Draw B.M.D., S.F.D., N.F.D. for the following frames due to:

a. Given load

b. Rotation at A of 0.003 rad.

a) EI = 8000 t.m2

A

B

m

10t

6t

b) EI = 40000 t.m2

EA = 10000 t

c) For girder EI = 3000 t.m2

For links A = 20 cm2, E = 2100 t/cm

2


d) Rise in temp.

At external

fibers t1 = 200

and at internal

fiber t2 = 100

4 t/m

4 t/m

3I I2I

E=200t/cm2

3I3I

6. Draw B.M.D., S.F.D., N.F.D. for the following frames:

a) EI = 20000 t.m2

EAlink = 10000 t

10t

5t

A B

10t

5t

EI=20000 t.m

EA=10000 tLinse

b) Vertical settlement at A of 2 cm, rotation at B of 0.001 rad. clockwise,

horizontal displacement at c of 3 cm to left

EI = 16000 t.m2

c) Due to rise of

temperature t= 20o of

the outer side. and

rotation at B of 0.001

rad. Clockwise


7- Analyze each of the following trusses due to

1. Given loads

2. Rise of temperature t = 40o at top chord.

3. Settlement at B = 2 cm .

E = 20 x 107 t.m

2,

A

L = 10

-1

a)

A

1

2 4 B 6 8 10 12 14 16

19171513119753

20t

C

b) 1 3 5 7 9

BA2 4 6

15t

c) 5t 5t10t 10t

B

1 3 7

CA2 4 6

5

45°

d) 4t 8t

AB

C

4t8t 8t


8. Analyze each of the following trusses due to given loads and rise of

temperature t = 30o at top chord and due to settlement as shown.

A

L is assumed constant, E = 21 x 10

7 t.m

2,

A

L = 10

-1

a)

A

1

2 4 B 6 8 10 12 14

19171513119753

30t

CD

b)

A B

1 973 5

10t

2 4 6

c) 10t 10t

31 5 9

2 4

B

6

C

45°

d)


e)

f)

A

1

2 4

B

6

973 7

10t 10t

g)

h)

5 INFLUENCE LINES OF

STATICALLY INDETERMINATE

BEAMS AND FRAMES

1. INTRODUCTION

An influence Line for a certain function such as reaction, moment, shear,

normal forces, or deflection; is line or curve the ordinate of which, at any

point gives the value of the function when a unit load is placed at that

point.

For example Fig.1 shows the influence lines for the reaction at supports A

& B shear at n in a continuous beam. The ordinate Y in that influence line

at point n is equal to the value of the reactions at A & B when a unit load

is placed at n.

Figure (1)

To construct the influence line for any function one method is to place the

unit load at several positions along the structure, and compute the value Y

of the function for each position of the unit load, and line joining these

ordinates Y is the required influence line.

Chapter (5) - I.L.of Indeterminate Beams and Frames 279

Because this method is time consumed time, then the most general

method for constructing influence lines is presented by Muller Breslau’s

principle method. This method is based on Maxwell’s reciprocal theorem

and Betti’s low of deflection.

It must be noted that, the influence lines for reactions and internal forces

of statically determinate structures are straight lines which are calculated

from the equilibrium conditions or by virtual work method. For statically

indeterminate structures the influence lines are generally curves. Which

are determined from the equilibrium conditions and deformation

conditions.

The influence lines can be obtained also experimentally by using a model

of beam made of any suitable material as steel. The influence lines be

identical in shape as deflected structures.

2- INFLUENCE LINES FOR DETERMINATE BEAMS BY

MULLER-BRESLAU PRINCIPLE

By the principle of Muller-Breslau very simple method for constructing

the influence lines for beams. It can be stated as follows;

1- To obtain an influence Line for the reaction of any statically

determinate beam, remove the support and make a positive

unit displacement of its point of application. The deflected

beam is the influence line for the reaction. Fig.2.a.

2- To obtain an influence line for the shear at a section of any

statically determinate beam, cut the section and induce a unit

relative transverse sliding displacement between the portion to the

left of the section and the portion to the right of the section keeping

all other constrains (both external and internal) intact. The

deflected beam is the influence line for the shear at the section,

Fig.2.b.

3- To obtain the influence line for the moment at a section of any

statically determinate beam, cut the section and induce a unit

rotation between the portion to the left of the section and the

portion to the right of the section keeping all other constrains (both

external and internal) intact. The deflected beam is the influence

line for the moment at the section, Fig.2.c.

280 Chapter (5) - I.L.of Indeterminate Beams and Frames

The proof of Muller-Breslau principle can be establish using virtual work

method the procedure of proof is generally applicable to more

complicated beams. Fig.2 shows a simple beam subjected to a unit

moving load. To find the reaction at A by the method of virtual work, we

remove the constrain at A, substitute YA for it, and let A travel a small

virtual displacement A along YA. We then have a deflected beam A- B,

as shown in fig.2, 2.a, where (Y) indicates the transverse displacement at

the point of unit load. Applying the virtual-work equation, we obtain

Figure (2)

YA. A - 1 (y) =0

i.e. YA = A

y

If we put A = 1

Then YA = y

Since y is, on the one hand, the ordinate of the deflected beam at the point

where the unit load stands and is, on the other mean at the point where the

unit load stands and is, on the other hand, the value of function YA due to

the unit moving load (i.e., the influence ordinate at the point), we


conclude that the deflected bean A- B of Fig.2.a is the influence line for

YA if A is set to be unity.

To determine the shearing force at any beam cross section S, we cut the

beam at S and let the two parts AS and SB have a relative virtual

transverse displacement S at S without causing relative rotation between

the two parts. This is equivalent to rotating AS1 and BS2 the same small

angle about A and B respectively. Applying the virtual work equation, we

obtain:-

QS (S)-1 (y) = 0

i.e. QS = S

y

If we put S = 1

Then Q S = y

This proves that the deflected beam AS1, S2B of Fig.2.b is the influence

line for QS. It should be pointed out that the virtual displacement is

supposed to be vanishing small and that, when we say S =1, we do not

mean that S = 1 meter. Or cm., but one unit of very small distance for

which the expressions.

SS1 = L

a

SS2 =L

b

Shown in Fig.2.b are justified.

To determine the moment at any beam cross section S by the method of

virtual work, we cut the beam at S and induce a relative virtual rotation

between the two parts AS and SB at S without producing relative

transverse sliding between the two. Thus by virtual work;

(MS) () – (1) (y ) = 0

i.e. MS =

y

If we put = 1 rad.

MS = y


This prove give that; the deflected beam ASB at Fig.2.c is the influence

line for MS. Note that when we say S = 1, we do not mean that S =1

radian. One unit of S may be small as 100

1radian, for which it is

justified to write

AA- = a. S = a. units

BB- = b. S =b. units, as indicated in Fig.2.c

Example (1)

Figure (3) shows a simple beam with an overhang. Construct the I.L. for

YA , YB, (QA)L, (QA)L, (QA)r, QS, MS and MA by the virtual work method.

Figure (3)

Solution:

To construct the I.L. for YA, we remove the support at A and let it move a

unit distance upward. The deflected beam C- A

- B, shown in Fig.3.a is the

influence line of YA. The I.L. for YB is obtained in a similar manner, as

shown in Fig.3.a.

To construct the influence line for (QA)r, we cut the section just to the

right of A and let the two parts of the beams have a unit relative

transverse displacement without producing relative rotation. To do this,

end A of Ac cannot move because of the presence of the support at A,

however, it is allowed to rotate about the hinge at A. we therefore, first

raise the end A of AB a unit distance to A-

and then turn AC to AC-

parallel to A-B The deflected beam so arrived at, as shown in Fig.3.b by

C--A A

--B, is the I.L. (QA)r.

To draw the I.L. for (QA)L, we cut the section just to the left of A and let

the left part have a unit relative displacement with respect to the right part

at


Figure (3)


The cut point without causing relative rotation between the two. Since the

part to the right of the cut section is a simple beam; it remains stable and

rigid. Only the left part can go down a unit distance. Thus, the influence

line for (QA)L is C- A

- A C Fig.3.c.

To construct the I.L. QS, we cut the beam at S and let the two parts SA

and SB have a relative transverse displacement equal to unity the two are

kept parallel to each other so that no relative rotation is introduced. Then

the deflected beam C- S1 S2 B of Fig.3.e is the I.L. QS the slope of the

deflected beam is 6

1, with which the influence ordinates can easily be

calculated.

The influence line for MS, as shown in Fig.3.f by the deflected beam C- S

-

B. it results from cutting the beam at S and letting SC and SB have a unit

rotation at S without allowing relative translation between the two. We

find the rotation of part SB about support B is 3

1 and the rotation of part

CS about A is 3

2.

Similarly, the I.L. MA, as shown in Fig.3.g by C-

AB by cutting the

section just to left of A and putting the left part have a unit relative

rotation with respect to the right part of the beam at A.


Example (2)

Draw influence lines for YA, Q S , MS , Qn, for a compound beam shown in

Fig.4.

Figure (4)

Figure (4)


Solution

To construct the I.L. for YA, we remove support A and move end A up a

unit distance. The deflected beam A- C B shown in Fig.4.a is the I.L. Note

that the part CB is a cantilever and will remain un-removed.

To construct the I.L. for QS, we cut the beam through S and let the left

part of beam have a relative transverse displacement equal to unity w.r.t.

the right portion of beam at S without causing relative rotation between

the two. The deflected beam AS1, S2 CB shown in Fig.4.b is the I.L.QS.

To construct the I.L. MS, we cut the beam through S and let the left part

of beam rotate a unit angle with respect to w.r.t. the right portion at S. the

deflected beam AS CB of Fig.4.c is the I.L. for MS.

The I.L. Qn is shown in Fig. 4.d by AC- n

- n B, which results from cutting

the beam through n and moving the left part of beam down a unit distance

w.r.t. right part of beam at n, while keeping the deflected protein c- ,n

-

parallel to B n.

The I.L. Mn is shown Fig.4.e by A C- n B, which results from cutting the

beam through n and rotating the left protein of beam a unit angle w.r.t. the

right part of beam at n. point n is kept fixed in the original position.

3. MULLER-BRESLAU’S PRINCIPLE FOR INDETERMINATE

STRUCTURES:

Muller-Breslau principle`s states that “for a beam, the influence line for

reaction, shear, or moment is the same as deformed shape of the beam

when that beam is loaded by the reaction, shear, or moment respectively,

but after removed; the capacity of the beam to resist the applied force, so

the beam can deformed when the force is applied. For statically

indeterminate beams, the influence lines for reactions, shear, and moment

are generally curves.

3.1. I.L. For Reactions:

This method is considered an easy for construct the influence lines. If its

required to draw the influence line for YA (as shown in Fig.5.e), then the

support at A must be removed and YA is applied with a unit value. The

elastic line shown in Fig.6.a with ordinate AA at as shown in Fig.6.a

divided by AA in order to that the deflection at A becomes equal to

unity


Figure (5)

and the figure become the influence line of YA. If applying the

Maxwell’s reciprocal theorem (as previous example) then; (we=0).

YA. AA + 1. 1n = 0 …..(1)

Then

YA= AA

n1

…..(2)

If AA = 1.0, YA = - n1

The above equation proves that the ordinate of the elastic line of Fig.6.b

at point n means that this elastic line is the influence line for YA.

3.2. Sign convention of reaction

In general an ordinate of the influence line is positive when elastic line is

deflected upwards and vice versa. A positive reaction means a reaction in

the chosen assumed direction.


3.3. Influence Lines For Shear at m

By Muller-Breslau’s principle the beam is assumed cut at m as shown in

Fig.6 and by positive unit shear force at m, with equal slope on each side

of m, the deflected curve represents to I.L. of Qm. Using the force

method and Maxwell’s theorem of reciprocal displacement, however, it

can be shown that: (we=0).

Qm. mm + 1. n1 = 0

i.e. Qm = n1 mm

1 , if m 1, then : Qm = n1

Figure (6)

3.3. Influence Lines For Bending Moment

Muller-Breslau’s principle can be used to construct the influence line for

bending moment. For example it is required to draw the influence line

for bending moment at support B for the given continuous beam shown in

Fig.8. two moments MB are applied at left and right of point B. after

introducing of hinge at B as shown in Fig.7.a.

The continuous beam is analyzed to find the value of MB which produces

a unit rotation at B, B=1 rad. The corresponding elastic line is the

influence line for MB.

Mb. bb + 1.Yn = 0

Mb = - bb

1

. Yn

Mb = -Yn , i.e. the elastic curve is I.L.MB


Figure (7)

Example (3)

For the given beam shown in Fig.8. It is required to draw the influence

lines of YB, YA, MB, Mn and Qn .

Figure (8)

Solution

1- I.L YB:-

Remove the support at B and apply by a load YB = 1t , draw the elastic

line for this case. The ordinates of elastic line shown in Fig.8.e are

determined by using the conjugate beam method. The ordinates of elastic

line can be obtained as follow; (using conjugate beam method), (EI=1).

1 = (-6.4 1 + 2

16.0

3

1) = -6.3 (upward)

2 = (- 6.4 2 + 3

2

2

22.1

) = -12 (upward)

3 = (- 6.4 3 + 3

3

2

38.1

) = -16.5 (upward)

b = (- 6.5 4 + 3

4

2

44.2

) = -19.2 (upward)

4 = (- 5.6 5 + 3

5

2

52

) = -19.67 (upward)


5 = (-5.6 4 + 3

4

2

46.1

) = -18.13 (upward)

6 = (- 5.6 3 + 3

3

2

32.1

) = -15 (upward)

7 = (-5.6 2 + 3

2

2

28.0

) = -10.67 (upward)

8 = (-5.6 1 + 3

1

2

14.0

) = -5.53 (upward)

9 = ( 5.6 1) = 5.6 (downward)

d = (5.6 2) = 11.2 (downward)

By dividing the ordinate of elastic line by ( b = 19.2) to obtain the

ordinate of influence line for (YB) as shown in Fig.9.e.

2- I.L. YA :-

YA = YAo + x1. YA1

The determination of the influence line of (YA) can be obtain by knowing

I.L. of YB . It is known that from consistent deformation method:

YA = YAo + (YB) YA1 [or YA = Yo + X1 Y1]

Where:

(YAo) is I.L of reaction of (A) on the main system and (YA1) is value of

reaction at (A) due to YB = 1t at (B) and using the superposition of two

diagrams hence;

I.L. YA = I.L. YAo – 0.6 I.L.YB

3- I.L. MB :-

I.L. MB = I.L. Mb0 + MB1 I.L. YB ( YB = X1)

= I.L. Mb0 – 2.4 I.L.X1

as shown in Fig.9.h and Fig.8.i.

4- I.L. Mn :-

I.L. Mn = I.L Mno + Mn1 I.L.YB

= I.L. Mno – 1.6 I.L. YB

as shown in Fig.8.k and Fig.8.l.

5- I.L. Qn :-

I.L Qn = I.L. Qno + Qn1 I.L. YB

= I.L. Qno + 0.4 I.L . YB


as shown in Fig.8.m and Fig.8.n.

Figure 1

Figure (8)


Figure (8)


Example (4)

For the shown beam in Fig.9 required :-

a) I.L. Ya

b) I.L. Mb

c) I.L. Yd.

Figure (9)

Solution (1)

1- I.L. YA :- (Fig.9.a) 1. 111 X , )( 1 AYX

11 = (EI

M21 )dl = 1

1152

3

29)12(111

2

X

EIEI

1152

1X1

1)1152(1152

1a (upward)

7.03

3

2

3331201152

1152

11

(upward)

406.03

6

2

6661201152

1152

12

(upward)

.168 03

9

2

9991201152

1152

13

(upward)

08.03

9

2

99929

1152

14

(downward)

093.03

6

2

66629

1152

15

(downward)

058.03

3

2

33324

1152

16

(downward)

06.03241152

1

c (upward)


I.L. YA as shown in Fig.9.a.

Figure (9)

2- I.L. Mb:- (Fig.9.b)

I.L. Mb = I.L. Mbo + Mb1 I.L. YA (X1 = YA)

= I.L. Mbo + 12 I.L. YA

Figure (9)


3- I.L. YD as determinate structure

Figure (9)

Solution (2)

Another Main system, Fig.10.a

1- I.L. Mb:-

EI

82

3

12112

11

6.03

3

2

35.032125.0

1

(upward)

125.13

6

2

65.062125.0

2

(upward)

984.03

9

2

975.092125.0

3

(upward)

75.0)32(125.0 e

(downward)

I.L. Mb as shown in Fig.10.b.


2- I.L. YA:- I.L. YA = I.L. YAo + YA1 I.L. x 1 (X1 = MB)

= I.L. YAo + (-1/12) I.L x 1 , as shown in Fig.10.e.

2- I.L. YD as determinate beam, Fig.10.d

Figure (10)

Figure (10)


Example (5):

For the shown continuous beam in Fig.11, Construct the I.L. for Mn and

Qn

Figure (11)

Solution:

1- Choose the main system as shown in Fig.11.a. with the

reaction at C and moment at A as redundant. Draw M1,

M2.Ds

2- Draw the elastic curve due to unit displacement at C

EI

5128

3

168

3

8 22

11 = EI

dLM 2

1

EI

33.21

2

18

3

1612

EI

33.51

3

16 2

22

The compatibility equations at c are:-

1212111

XX

0222121

XX

EI X X 21

33.21512

033533.2121 X .X

Solve to get EI

X67.426

1 ,

EIX

67.1062

Assume EI = 426.67

X 11 , X 4

2

MXMXM21F 21

Additional moment = 426.67


0562.03

4

2

42

3

42

2

41243

67.426

11

150.03

8

2

84

3

82

2

82482

67.426

12

1687.03

12

2

126

3

122

2

1236121

67.426

13

425.067.4263

4

2

44464

67.426

14

00.167.426

67.426

C

The I.L.Yc is shown in Fig.11.b.

Figure (11)


Figure (11)

3- Draw the elastic curve due to unit rotation at C

212111 XX = 0

222121 XX = 1.0

EI X X 21

33.21512

033533.2121 X .X

Solve to get EI

X67.106

1 ,

EIX

44.42

Assume EI = 106.67

X 11 , X 025.24

2

MXMXM21F 21

Additional shear at (A) = 106.67

4.243

2

2

462418

3

4

2

42467.106

67.106

11

4.283

2

2

8124812

3

8

2

84867.106

67.106

12

2.182

121861264

2

1261267.106

67.106

13


30.03

4

2

44467.106

67.106

14

I.L.MA is shown in Fig.11.c.

Figure (11)

I.L.Mn= I.L.Mno + 4 I.L.YC – 0.50 I.L.MA

I.L.Mn as shown in Fig.11.d.

4. I.L. Qn:-

I.L.Qn= I.L.Qno + 0.50 I.L.YC – 16

1 I.L.MA

I.L.Qn as shown in Fig.11.e.


Figure (11)


Figure (11)

Example (6)

For the shown Continuous beam in Fig.12, it is required to draw the

influence lines for Ya , Yb , Ma & Mc

Figure (12)

Solution First we choose a main system by introducing hinges at supports B, C, D.

From M1 , M2 & M3 Diagrams Find δ11 ,δ22 ,δ33 ,δ12=δ21 ,δ23=δ32 &

δ13=δ31, hence;

δ11= δ22 = δ33 =1*1*10/3 *2 = 6.6666667

δ12=δ21=δ23=δ32=1*1*10/6 = 1.6666667

δ13=δ31=0.0

For I.L. X1 (Mb)

Δ1=1,Δ2=Δ3=0

6.6667 X1 + 1.6667 X2 = 1

1.6667 X1 + 6.6667 X2 +1.6667 X3 = 0

1.6667 X2 + 6.6667 X3 = 0


solving we get

X1 =0.1607 , X2=-0.043 , X3 = 0.0107

Figure (12)

M1. D.

M2. D.

M3. D.

Mb. D.


From the chosen main system each span is considered a simple beam

having one or two concentrated moments at its to ends then the conjugate

beam is solved to get the deflections as follows:-

Beam A-B

0.000

0.628

1.0040.879

0.000

0 2.5 5 7.5 10

Figure (12)

Beam B-C

station X Mrord mlord Y Theta

start 0.0000 0.0000 -0.1607 0.0000 -0.4643

1 2.5000 0.0107 -0.1205 0.7115 -0.1261

2 5.0000 0.0214 -0.0804 0.7366 0.0848

3 7.5000 0.0321 -0.0402 0.3934 0.1685

end 10.0000 0.0429 0.0000 0.0000 0.1250

0.711 0.737 0.393

0.0000 2.5000 5.0000 7.5000 10.0000

Beam C-D

station X mrord mlord Y Theta

start 0.0000 0.0000 0.0429 0.0000 0.1250

1 2.5000 -0.0027 0.0321 -0.1925 0.0346

2 5.0000 -0.0054 0.0214 -0.2009 -0.0223

3 7.5000 -0.0080 0.0107 -0.1088 -0.0458

end 10.0000 -0.0107 0.0000 0.0000 -0.0357

station X Mrord mlord Y Theta

start 0.0000 0.0000 0.0000 0.0000 -0.2679

1 2.5000 -0.0402 0.0000 0.6278 -0.2176

2 5.0000 -0.0804 0.0000 1.0045 -0.0670

3 7.5000 -0.1205 0.0000 0.8789 0.1842

end 10.0000 -0.1607 0.0000 0.0000 0.5357


-0.193 -0.201-0.109

0.0000 2.5000 5.0000 7.5000 10.0000

Beam D-E

station X mrord mlord Y Theta

start 0.0000 0.0000 -0.0107 0.0000 -0.0357

1 2.5000 0.0000 -0.0080 0.0586 -0.0123

2 5.0000 0.0000 -0.0054 0.0669 0.0045

3 7.5000 0.0000 -0.0027 0.0418 0.0145

end 10.0000 0.0000 0.0000 0.0000 0.0179

0.059 0.067 0.042

0.0000 2.5000 5.0000 7.5000 10.0000

Figure (12)

For I.L. X2 (Mc) :-

1=0, 2=1, 3=0

6.6667 X1 + 1.6667 X2 = 0

1.6667 X1 + 6.6667 X2 +1.6667 X3 = 1

1.6667 X2 + 6.6667 X3 = 0

solving we get

X1 =X3 =-0.0429 , X2 =.1714


Applying the same procedure we can draw Mc as shown

Figure (12)

For I.L. X1 (Md)

1=0, 2=0, 3=1

6.6667 X1 + 1.6667 X2 = 0

1.6667 X1 + 6.6667 X2 +1.6667 X3 = 0

Figure (12)

b) MC. D.

d)

c)


1.6667 X2 + 6.6667 X3 = 1

solving we get X3=0.1607 , X2=-0.043 , X1= 0.0107

It is obviously clear that Md is the same as Mb but mirrored about point c

After constructing the three redundant Influence lines we can now

proceed to find any required function as follows

For I.L. Ya

Figure (12)

Ya1=-0.1 & Ya2=Ya3= 0.0

So

For I.L. Yb, Yb1=0.2 , Yb2= -0.1 & Yb3= 0.0

Figure (12)

e)

f)


4. INFLUENCE LINE OF TRUSSED BEAM

It is required to draw the influence line for the normal force X in member

44`, the displacement condition to be satisfied in this case is nn =1

where nn is the relative displacement at point n. The elastic line due to

the force X after cutting the member 44 at center line is the influence line

for X. The elastic line may be straight portions in case of panel floor and

indirect loading.

It is required to draw the influence line for internal forces (Mm and Qm) at

section m in girder AB as shown in Fig.13.

Figure (13)

From virtual work method generally, M = Mo + X1. M1 then;

I.L. Mm = I.L. MO + L.L. X1. M1

Similary;

Qm = Qo + X1. Q1

Hence;

I.L. Qm = I.L. Qo + I.L. X1. Q1

Also

Nm = X1


i.e. L.L. Nm = I.L. X1

Example (7)

For the given Trussed Beam shown in Fig.14, calculate

a) I.L. Nn

b) I.L. Qm

c) I.L. Mm

Figure (14)

Where

E= 2100 t/cm2, Alink= 30 Cm

2, I girder= 200000cm

4

Abeam = 60 cm2.

Solution

EI = 2100 200000 104 = 42000 t.m

2.

EAlink = 2100 30 = 63000 t.

EAbeam = 2100 60 = 126000t.

LEA

N

EA

NdL

EI

M

LinksBeam

21

21

21

11

011587.05.26)6464(3

5

3

)5(42 222

221

dL

EI

M

422

1 10984.1126000

25)1(

dLEA

N

Beam

To get LEA

N

Links

21 for links, as given in the following table :-


Member L/EA N1 21N

EA

LN21

A-3`, B- 3`

410016.163000

4.6 1.282 1.6435 1.67 10

-4

3`- 4` 5105978.863000

385.5 1.077 1.16 10 10

-5

4` – 5` 510863000

5 1 1 8 10

-5

1` – 3` 51035.663000

4 - 0.4 0.16 1.016 10

-5

2` - 4` 510524.963000

6 - 0.4 0.16 1-524 10

-5

6.6 10-4

421 106.6 L

EA

N

link

012445.0106.610 1.984 0.011587 4 -4

11

1111 X

X1 = 80.351

235.03

5.2

2

5.225.250

42000

351.802

(downward)

446.03

5

2

54550

42000

351.803

(downward)

6118.03

5.2

2

5.2125.15.24167.4

2

545.750

42000

351.804

(downward)

717.03

5

2

525.25467.6

2

541050

42000

351.805

(downward)

753.025.165.2167.42

52554167.9

2

545.1250

42000

351.806

(downward)

24.05.25042000

351.801 (upward)


48.055042000

351.800 (upward)

a) I.L Nm = I.L. X1

b) I.L Qm = I.L. Qmo + Q1 I.L. x 1

= I.L. Qmo + 0.4 I.L. x 1

c) I.L. Mm = I.L. Mmo + Mm1 I.L. x 1

= I.L. Mmo + 5 I.L x 1

Influence Lines are shown in Figs.12.a, 12.b & 12.c

Figure (14)


Figure (14)

Example (8)

Figure (15)

For the given trussed Beam (Fig.15), determine:

1- I.L. F1.3 (I.L. Nn)

2- I.L. M4

3- I.L. Q4 right.

(EI beam = 40000 t.m2, EA link = 3000t, EA girder = 20000t)


Solution

1- I.L. X1= I.L. Nn (Fig.15.a)

a) Cut member (1-3) and apply by X1 = 1t, then draw M1. D and get N1

for truss members as shown in Fig.15.a to d.

Hence:-

LdLdLEI

M

LinkEA

N

EA

N

21

Beam

21

21

11

Where: 11 is the relative displacement at cut point get the summation of

EA

LN 2

1 and determine the value of EA

dLa

EI

dLM 2

121 N nd , then get 11

3221 1025.23)3(

3

633

40000

2

dL

EI

M

422

1 1092000

18(1)

dL

EI

N

The values of Llinks

21

EA

N are given in the following table :-

Member L/EA F1 2

1F EA

LF 21

1-3

2 10-3

1 1 2 10-3

1-2 1 10 –3

- 0.5 0.25 0.25 10-3

3-4 1 10-3

- 0.5 0.25 0.25 10-3

1-4 2.236 10-3

1.12 1.2544 2.8 10-3

3-b 2.236 10-3

1.12 1.2544 2.8 10-3

8.103 10-3

linksEA

LN 21 = 8.103 10

-3

01125.010103.81091025.2 343

11

apply Muller- Breslau’s condition at point n, hence:-


1111 X

X1 = 88.88

Calculate the deflection by using elastic load, Fig.15.e.

115.03

3

2

35.1 318

40000

88.881

(up word)

2.03

6

2

63 618

40000

88.882

(up word)

23.05.13352

63 918

40000

88.883

(up word)

Then, I.L.Nn = I.L.X1 , as shown in Fig.13.f.

I.L. M4 = I.L. M4o + (M4)1 I.L.X1

2- I.L. M4 = I.L. (M4)0 – 3 I.L.X1

3- I.L. Q4 night = I.L Q4o Q1 + I.L.X1

= I.L. Q4o + I.L.X1

I.L. of M4 and I.L. of Q4 are shown in Fig.15.h and k

Figure (15)


Figure (15)


Example(9)

For the given arched truss structure which shown in Fig.16. If the load is

transmitted to the horizontal membrs as shown in Fig.16.a.

Figure (16)

Required:

1- I.L. Mn , I.L. Nn , I.L.Qn.

2- I.L. Mm , I.L. Qm , I.L. Nm.

Where:

For arch: EI= 50000 t.m2

EA= 100000 t

For tie: EA= 60000 t

Solution

The given structure is statically indeterminate to the first degree. A

possible main system may be obtained by cutting the tie (ab) at any

section along its length, and the redundant (X1) is then the force in the tie

Fig.16.a.

Cas 1 = 0.8412

Sin 1 = 0.540757

Cos 2 = 0.94174

Sin 2 = 0.33633 , the length of members as shown in Fig.16.a.

Draw the normal force and bending moment diagrams for arch and also

normal force for ties.

Get the value of 11 by using consistence deformation

Where:

LEA

N dL

EA

N dL

EI

M

Link

2

1

Beam

2

1

2

1

11


0.0201999.035 (1) 60000

1

5.3)1(43.7(0.941) 8.32 )841.0(100000

2

5.3)7(5.47)5.4()7(3

43.7

3

32.8)5.4(

5000

2

2

222

2222

11

by using Muller-Breslau condition

11 x 1 =1.0

02.0

11

11

1

X

X1 = 50

The corresponding deflected shape of the arched truss which is also the

influence line for the relative displacement at the cut section it’s value

given by using virtual work method at different joint in the truss.

a) Put (1t) at point (1) and calculate the deflection for the main system

Fig.16.d and e.

dLEA

NnL

EA

Nndl

EI

MM

linkbeam

ff

111

1

M = X1 M1 = 50 M1

zeroEA

NN

link

1

N = X1 N1 = 50 N1 , hence:-

556.00.81 8.32 0.108 0.841 0.4326 8.32100000

50

3

32.84.15.4

1.4) 7 4.5 (2.8 0.5 4.5 1.4 2.8 7 3

7.43 7 7

2

4.2 2.8

7) 5.6 4.2 4.2 ( 0.5 4.2 7 4.5 (5.6 3

7.43

3

8.32 4.5 5.6

50000

50- 1


b) Put (1t) at point (2) and calculate the deflection.

LEA

NL

EA

NndL

EI

Mm

linkbeam

2222

n

8867.0 7.433 0.941 0.1345 7.433 0.941 0.2018

0.216 8.32 0.841 0.841 8.32 3244.0

100000

50

3

32.88.25.4

2.8) 7 4.5 (5.6 0.5 2.8 4.5 5.6 7 3

7.43 7 7

2

8.4 5.6

4.5) 4.8 7 4.2 ( 0.5 8.4 7 4.2 (4.5 3

7.43

3

8.32 4.2 4.5

50000

50- 2

hence the I.L. X1 as shown in Fig.16.f.

Also;

I.L. Mn = I.L. Mno + Mn1 I.L X1

= I.L Mno – 7 I.L. X 1 (Fig.16.g)

I.L. Nn = I.L. Nno + Nn1 I.L. X 1

= I.L. Nno – 1 I.L. X1 (Fig.16.h)

I.L. Nn = - I.L. X1

I.L. Qn = I.L. Qno + Qn1 I.L. X1

= I.L Qno + (Zero) I.L. X1 (Fig.16.i)

I.L. Qn = I.L Qno

I.L. Mm = I.L. Mmo + Mm1 I.L. X1

= I.L. Mmo - 5.75 I.L. X1 (Fig.16.j)

I.L. Nm = I.L. Nmo + Nm1 I.L. X1

= I.L. Nmo - 0.94179 I.L. X1 (Fig.16.k)

I.L. Qm = I.L. Qmo + Qm1 I.L. X1

= I.L. Qmo – 0.33634 I.L X1 (Fig.16.L)


Figure (16)


Figure (16)


Figure (16)


5. INFLUENCE LINE FOR FRAME

It is required to draw the influence lines for YA, Yc , YB , and Mc for the

given frame as shown in Fig.17.

I.L. YA

In this case the movable support at A is removed and YA is applied with

a value just enough to make the vertical displacement A equal to unit,

A =1, the resulting elastic line is the influence line for YA, Fig.17.a.

I.L. YC

Also to find the influence line for vertical component of reaction at C,

YC, as shown in Fig.17.b. , a unit vertical displacement is imposed at c,

without rotation or horizontal displacement. The elastic line of frame is

the influence line for Yc.

Figure (17)


I.L.XB

To construct the influence line for XB the horizontal component of

reaction at B, XB, of the given frame. In this case, the hinge at B is

replaced by a roller support as shown in Fig.17.c and XB is applied with a

value just enough to make a horizontal displacement SB = 1. the elastic

line for this case is the influence line for XB as shown in Fig.17.c.

I.L.MC

To draw the influence line for the moment reaction at c; Mc; the fixed

support at c is replaced by a hinge as shown in Fig.17.d and Mc is applied

at c with a value such that the angle of rotation Sc (or c) equals to unity.

The resulting influence line is the influence line for Mc is shown in

Fig.17.d.

Figure (17)

From the previous cases solved it is clear that the procedures for construct

any influence line of any function are according to the following.

1- The structure is modified in such a way that the function can do work.

2- The function is doing work through a unit displacement.

3- The resulting elastic line is the required influence line.



FRAMES (Using Muller- Breslau’s principle)

The procedures are;

1. Choose the main system and find the redundant reaction components;

X1 and X2 for the given frame as shown in Fig.16.

2. To find the influence line for any function, the influence lines of the

redundant reaction components should be obtain at first. And the

influence line of other functions can then be obtained by

superposition, or static relations.

3. Find the values of , , , and , using the virtual work

method and hence find the magnitudes of redundant X1 and X2 .....

etc; which satisfy the two displacement conditions imposed by

Muller- Breslau principles.

4. To draw influence line of X1, the displacement conditions at A& B

are;

= 1 (at A)

= 0 (at B) , hence


X1 11 + X2 12 = 1

get X1

X1 21 + X2 22 = 0

The above equations give the values of X1 and X2 which are used to

find the B.M.D. as follows :-

M = X1.M1 + X2.M2

Hence, the elastic line produced by this B.M.D, as elastic load; is the

influence line of X1.

5. To draw influence line of X2 the displacement conditions are;

= 0 (at A)

= 1.0 (at B)

Hence:

X1. 11 + X2. = 0

get X2

X1. + X2. = 1

and

M= X1.M1 + X2. M2

The influence line is the elastic line produce by the second M.

Incase of frame with three time indeterminate, in this case the

analysis includes three sets of there simultaneous equations as

follows:

1

33332 2311

233222211

133122111

Xfor

0 = . X .X + . X

0 = . X . X + . X

1 = . X . X + . X

and


2

333322311

233222211

133122111

Xfor

0 = . X . X + . X

1 = . X . X + . X

0 = . X . X + . X

and

3

33332 2311

233222211

13 3122111

Xfor

1 = . X .X + . X

0 = . X . X + . X

0 = .X . X + . X

The B.M.D for the three sets are obtained from the following

equation;

M = X1 . M1 + X2 . M2 + X3 . M3

The influence line can be obtained using M.D.

6. I. L. for any other function (R) such as, reaction, moment, shear, and

normal force can then be obtained by use of conditions of

superposition or by equilibrium as follows;

I.L.R = I.L.Ro + R1 (I.L.X1) + R2 (I.L.X2) + Rn (I.L.Xn)

Where:

I.L.Ro is the influence lime of function R in main system.

and

R1, R2, ..... Rn are the values of R due to:

X1 = 1

X2 = 1

.

.

.

.

Xn = 1 respectively


7. In case of once statically in determinate frames;

X1 11 = 1 and M = X1 . M1

Example(10)

For the given frame shown in Fig.18, determine : -

I.L Xa 2- I.L Mn and Qn

Figure (18)


Solution

1) Choose the determinate main system as shown in Fig.18.a, then find

the M1.D. as shown in Fig.18.b.

by apply X1 = 1t at A.

2) Calculate 11 form M1.D. then;

EI

106.67 4

3

12 4 2

3

4 4 4

EI

1 11

3) Muller Breslau condition at A

1X

106.67

EI

1 X

11

4) The ordinates of elastic line of part CDE are obtain using elastic

equilibrium by knowing the elastic load, M/EI as shown in Fig.18.c

Draw the elastic curve by knowing the slope and deflection using the

elastic load Method, conjugate beam method Fig.18.d as follows :-

0.125- 1 3

2 4 2 8-

106.67

EI 5

(upward)

0.2- 2 3

4 4 4 8-

106.67

1

4

2 (upward)

0.225- 3 3

4 6 6 8-

106.67

1 3

(upward)

0.15- 2 8 106.67

1 e (upward)

I.L Xa = I.L X1 as shown in Fig.18.e, also

I.L Mn = I.L Mno + Mn1 I.L X1

= I.L Mno – 4 I.L X1

I.L Mno shown in Fig.18.f

I.L Mn shown in Fig.18.h, and also


I.L Qn = I.L Qno + Qn1 I.L X1

= I.L Qno + zero I.L X1

= I.L Qno

I.L Qn = I.L Qno as shown in Fig.17.g

Example (11)

For the shown frame in Fig.18, draw I.L. of Ms

EI= 30000 t.m2

Figure (19)

Solution

The frame is once statically indeterminate structure, hence:

EI

.dLM

21

11

=

44 6 4

3

2

2

4 4

EI

2

= EI

67.234

1.0 .X 111 , hence; I.L.X1 as shown in Fig.19.d.

I.L.Ms:-

Ms = Mso + X1.M1s

I.L.Ms = I.L.Mso + I.L.X1 (-4)


I.L.Mso:

1 ton between c-s

Mso = 8Ybo

I.L.Mso = 8 I.L.. Ybo

1 ton between s-d

Mso = 4. Yao

I.L.Mso = 4.I.L.Yao

I.L.MS as shown in Fig.19.e.

Figure (19)


Figure (19)

Example (12)

For the shown frame in Fig.20, Construct the I.L. for YA , MA , Mn , Qn

and Nn .

Figure (20)

Solution:

1- Choose the main system as shown in Fig.20.a. with the vertical

reaction at C and rotation at A as redundant;

EI

3888 218

3

18 211 = EI

dLM 2

1


EI

270 118

3

18118

2

1812

= EI

dLMM 21

EI

24 11

3

18118122 = EI

dLM 2

2

From 11 (to get elastic curve due to unit displaceent) hence:-

1212111 XX

0222121 XX

EI X X 21

2703888

02427021

X X

Solve to get EI

X50.850

1 ,

EIX

60.752

Let EI = 850.50

11 X , 25.112 X

MM 2211 XXM F

Additional moment at A = 850.50

884.06

5.4688.1

3

5.4812.2

2

5.44375.85.850

5.850

1 222

1

607.06

9375.3

3

963.5

2

962.55.850

5.850

1 222

2

2765.06

5.1306.5

3

5.1344.8

2

5.138125.25.850

5.850

1 222

3

1407.0 6

5.1306.550.1325.20

5.850

1 2

4

1607.06

9375.3925.20

5.850

1 2

5

10.06

5.4688.15.425.20

5.850

1 2

6


Figure (20)


Figure (20)

3- 1A

0270388821

X X

EI X X 21

24270

Solve to get EI

X60.75

1 ,

EIX

25.52

Let EI = 75.60

11 X , 40.142 X

MM 2211 XXM F

77.2 3

5.46.3

2

5.48.10

6

5.49.05.46.75

60.75

1 222

1

893.2 3

92.7

2

92.7

6

98.196.75

60.75

1 222

2

567.13

5.138.10

2

5.136.3

6

5.137.25.136.75

60.75

1 222

3

84.06

5.137.25.138.10

60.75

1 2

4

964.06

98.198.10

60.75

1 2

5

603.06

5.49.05.48.10

60.75

1 2

6


I.L.MA as shown in Fig.20.c.

Figure (20)

Example (13)

For the shown frame in Fig.21, it is required to draw the influence

lines for Ya , Ma , Nn & Qn

Figure (21)


Solution

First we choose a main system

Figure (21)

From M1 & M2 Diagrams Find δ11 ,δ22 & δ12=δ21

δ11= 118+1120/3=14.6667

δ22= 8820+2888/3=1621.3333

δ12=δ21=-(818/2+8120/2)=-112

to draw I.L. for X1 (Ma) Put 1=1 & 2=0 then;

δ11X1+δ12X2 =1 ….(1)

δ21X1+δ22X2 =0 ….(2)

14.667X1-112X2=1

-112X1+1621.33X2=0

solving the above equations we get

X1=0.1443 , X2=0.009968

From the Above values we draw the final moment M1 final on the frame

then using Conjugate beam method we find the elastic curve (I.L) for X1

(Ma) as shown in Fig.21.5.

Figure (21)


Figure (21)

Station El. Load ord. B M (Y)

E - -0.65823

1 0.06456 0.00000

2 0.05013 0.20962

3 0.03570 0.21873

4 0.02127 0.08506

5 0.00684 -0.13367

6 -0.00759 -0.37975

7 -0.02203 -0.59544

8 -0.03646 -0.72304

9 -0.05089 -0.70481

10 -0.06532 -0.48304

11 -0.07975 0.00000

F - 1.26582

Similarly to find I.L. for X2 (Xb) put Δ1=0 & Δ2=1


Figure (21)

Solving

we get X1=0.00997 , X2=0.0013

station El. Load ord. B M (Y)

E - 0.15190

1 -0.00047 0.00000

2 -0.00147 -0.07434

3 -0.00247 -0.14278

4 -0.00347 -0.20136

5 -0.00446 -0.24608

6 -0.00546 -0.27294

7 -0.00646 -0.27797

8 -0.00745 -0.25718

9 -0.00845 -0.20658

10 -0.00945 -0.12218

11 -0.01044 0.00000

F - 0.28481

After getting the two redundant Influence lines we can easily draw any

Function by Just substituting in the following relation:-

I.L. F = I.L F0 + F1 I.L.X1+ F2 I.L.X2

So to draw I.L Ya we should construct I.L. Ya0 and find Ya1, Ya2 when

X1=1 & X2 =1 respectively as follows:-

Ya1=-.05, Ya2= 0.0, I.L YA as shown in Fig.21.d.


Figure (21)

station I.L. Ya

e 1.2

1 1

2 0.89

3 0.79

4 0.7

5 0.61

6 0.52

7 0.43

8 0.34

9 0.24

10 0.12

11 0.00

f -0.26


I.L. For Nn :-

Nn1=0.0, Nn2= -1.0 & I.L.Nn0 = 0.0

So I.L.Nn =-1 I.L.X2 = - I.L. Xb, as shown in Fig.21.c.

I.L. For Qn :-

Qn1=-0.05 Qn2= 0.0, I.L. as shown in Fig.21.e.

Figure (21)

station I.L. Qn

e 0.2

1 0

2 0.11

nleft 0.21

nright 0.79

4 0.7

5 0.61

6 0.52

7 0.43

8 0.34

9 0.24

10 0.12

11 0.00

F -0.26


7. MAXIMUM EFFECT USING INFLUENCE LINE:

For the given continuous beam, Fig.22.a and from the sketch of shape of

the influence line for reaction at B (for example) as shown in Fig.22.b;

the maximum reaction due to live loads is as follows :-

Figure (22)

Max. +ve YB will occur at B when spans AB, BC, and DE are loaded as

shown in Fig.22.b. similarly the maximum values of bending moment at

support C, shear at S, moment at S will occur due to the loading shown in

Fig. C, D and S respectively. For multistory building frame Fig.23, the

influence line can be also use very useful in determining the loading

pattern in the case of live.


Figure (23)

Absolute Bending Moment Diagram:

For the design load purpose, the absolute BMD may be required. Fig.24

shows the absolute BMD for the given beam.

Figure (24)


PROBLEMS

Obtain the influence lines for reactions, and indicated internal forces for

each of, the given statically indeterminate structures; hence calculate the

value of Mn due to moving live load 2 t/m`.



For the given structures, draw the I.L. of Ya, Mn, Qn, and Nn.

6 INFLUENCE LINES

OF STATICALLY INDETERMINATE

TRUSSES

1. GRAPHICAL DETERMINATION OF TRUSS

DISPLACEMENTS

1.1. INTRUDUCTION

Truss displacements were obtained by the method of virtual work. The

use of virtual work method is limited, as it gives only the deformations at

certain points. The use of graphical method gives graphically the

deformations of all the truss joints in addition to constructing easily the

influence line of statically indeterminate trusses.

1.2. THE DISPLACEMENT DIAGRAM

Assume the triangle truss shown in Fig.(1.a) the displacement 1 and

2 of

both joints a and b respectively are assumed to be known. The change in

length of any member is calculated as follows;

eTemperatur

Force Axial

LtAE

LN

It is required now to obtain the displacement cc’ of joint c

The procedure could be summarized as follows;

Assume that the triangle abc is opened only at c and member

ac moves parallel to itself to a new position by a’c1 and its

elongation c1c3 is equal to ac

. Similarly, member bc moves to b’c2

parallel to itself with elongation (or contraction) bc

.

To close the triangle, c3 moves on an arc with center at a’ and

c4 moves on an arc with center at b’. The intersection point is c’ and

cc’ is the required displacement. Because the displacements are

very small if compared with the length of the members, the arcs can

be replaced by straight lines normal to the members.

Chapter (6) - I.L.of Indeterminate Trusses 347

The displacement can be drawn as a separate diagram

Fig.(1.b), from the origin point o, the displacement aa’ and bb’ are

drawn as vectors oa’ and ob’. From a’ draw ac

in the direction of

ac, ac

may be elongation or contraction, and a line normal to it is

drawn. Similarly, for bc

. The intersection point c’ of the two

normal determines the displacement cc’ of the joint cc’ of the joint

c which is represented by vector oc’= 3

.

1.3. WILLIOT DIAGRAM

In the case of a truss consists of triangles, the position of one panel point

and a direction of one member remain unchanged. The displacement of

the other panel points can be found by repeating the previous

displacement diagrams several times and the final diagram is called

Williot diagram. In order to draw the Williot diagram, we must know two

points in the Williot as a’ and b’ .This is achieved if we know the

displacements 1

and 2

or we know the relative displacement 21

between the two joints a and b, see Fig.(1).

Figure

(1.a)

Figure (1.b)

348 Chapter (6) - I.L.of Indeterminate Trusses

1.3.1. Cases Of Williot Diagram

a. Trusses with two joints fixed

Cantilever truss, 021 , Fig.(2.a & b)

To construct the Williot diagram for this case apply the following

steps

1- First of all, find out the internal forces N0 in all members due

to the external loads.

2- Calculate the member deformations for all members due

to the internal forces N0. The type of deformations (may be

extension of contraction) should be considered. The usual

practice is to use +ve sign for extensions and –ve sign for

contraction as shown in the given truss.

eTemperatur

Force Axial

0 LtAE

LN

.

Figure (2.a)

Figure (2.b)


3- Select a joint (or joints) in the truss, which is fixed. Now start

the construction of Williot diagram as given.

4- The vertical (V6) as well as the horizontal (H6) displacements

of point 6 (for example) may now be obtained from the Williot

diagram as shown

Example (1)

A given crane truss has two hinged supports at joints A & D. The

values of both internal forces and deformations due to the applied

loads are indicated around each member. Find the horizontal as well

as vertical displacement of joint C using the displacement diagram

method. Given EA=80000 ton for all members.

Solution 1. Find the internal force in each member due to the

given loads.

2. Calculate the member deformations due to the internal

member forces.

Member L (m) EA (ton) F (ton) (cm) Remarks

AB 8.49 80000 34.17 0.36 Tension

BC 7.71 80000 25.71 0.25 Tension

CD 11.28 80000 -37.59 - 0.53 Compression

DA 6.00 80000 0.00 0.00 Zero Member

Figure (3.a) Figure (3.b)


DB 6.00 80000 -15.37 - 0.12 Compression

3. Choose a convenient scale to enlarge the displacement

values.

4. Select a member of fixed direction (member AD) and

fixed joint on either end of the selected member (joint A).

5. Point a’ (corresponding to joint A) is located first on a

sheet paper.

6. Point d’ is drawn relative to a’ horizontally (away to

a’ if the member AD elongates or towards a’ if AD shortens

and parallel to AD), in this case, point d’ coincides to a’

because AD is zero member.

7. Select a triangle such that two joints are previously

located while the third one is required. This is achieved in

ADB in which both joints A & D are known but joint B is

required.

8. From a’ draw a’b1 of length=0.36 cm parallel to AB

and in the direction of AB because FAB is Tension.

9. From d’ draw d’b2 of length=0.12 cm parallel to DB

and in the direction of BD because FBD is Compression.

10. The intersection of the normals to a’b1 and d’b2 at b1

and b2 respectively gives joint B which represented by b’.

11. Select a triangle BCD in which both joints B & D

are known but joint C is required.

12. From b’ draw b’c1 of length=0.25 cm parallel to BC

and in the direction of BC because FBC is Tension.

13. From d’ draw d’c2 of length=0.53 cm parallel to DC

and in the direction of CD because FCD is Compression.

14. The intersection of the normals to b’c1 and d’c2 at c1

and c2 respectively gives joint C which represented by c’.


15. Measure horizontally the distance between a’ and c’

and divide by the scale to find Hc=1.59 cm.

16. Measure vertically the distance between a’ and c’ and

divide by the scale to find Vc=2.02 cm.

b. Williot diagram for three hinged trusses

0BA , Fig.(4)

To draw the Williot diagram, start from A’ or B’ as fixed points,

then find C’. From A’ and C’ find 5’. From A’ and 5’ find 1’. From

1’ and 5’ find 6’. From 5’ and 6’ find 3’ …………etc. The above

procedure is applicable only when AC and BC are straight lines.

c. Williot diagram for symmetrical trusses

Figure (3.c)

Figure (4)


To construct the Williot diagram for symmetrical trusses of

symmetrical loading, consider the given truss in Fig.(5.a). For this

case, assume point 4 (or point 3) as fixed points and direction 3-4 is

fixed, then Williot diagram is as shown in Fig.(5.b), point 4’

coincides to the origin 0. In reality, point A at hinged support

remains in place not point 4.

Hence, the vectors A’1’, A’4’, A’3’, …………etc. Each

displacement may be resolved into a vertical component V and a

horizontal component H. For example, the vertical displacement

component of point 4 is equal to V4 while the horizontal

displacement component of point 4 is equal to H4 as shown in

Fig.(5.b).

For other cases of truss symmetry as shown in the following

Fig.(6.a), assume point A as a fixed point and direction 4-6 is the

fixed direction. Then complete the Williot diagram as shown in

Fig.(6.b)

Figure(5.a

)

Figure (5.b)


Example (2)

For the given truss shown in Fig.(7) find the vertical and horizontal

displacement of point C. Given EA=80000 ton for all members.

Solution

1. Find the internal force in each member due to the

given loads.

2. Calculate the member deformations due to the internal

member forces.


AB 8.00 80000 -13.33 - 0.13 Compression

BC 5.00 80000 16.67 0.10 Tension

Figure(6.a)

Figure (6.b)

Figure(7.a

)

Figure (7.b)


CA 5.00 80000 16.67 0.10 Tension

3. Choose a convenient scale to enlarge the displacement

values.

4. Select a member of fixed direction (member AB) and

fixed joint on either end of the selected member (joint A).

5. Point a’ (corresponding to joint A) is located first on a

sheet paper.

6. From a’ draw a’b’ of length=0.13 cm parallel to AB

and in the direction of BA because FBA is Compression.

7. Select a triangle ABC in which both joints A & B

are known but joint C is required.

8. From a’ draw a’c1 of length=0.10 cm parallel to AC

and in the direction of AC because FAC is Tension.

9. From b’ draw b’c2 of length=0.10 cm parallel to BC

and in the direction of BC because FBC is Tension.

10. The intersection of the normals to a’c1 and b’c2 at c1

and c2 respectively gives joint C which represented by c’.

11. Measure horizontally the distance between a’ and c’

and divide by the scale to find Hc=0.065 cm.

12. Measure vertically the distance between a’ and c’ and

divide by the scale to find Vc=0.25 cm


d. Williot diagram for trusses having only one fixed joints

In the case of unsymmetrical loading, we cannot draw Williot diagram

until we assume either the direction of joint G or joint B with respect to

joint A is fixed. Strictly speaking, either of the two assumptions are

wrong, since the joint B and G are found to have vertical as well as

horizontal displacement, under the action of the external loading, Fig.8.

In order to start the Williot diagram, we have to make either of two

assumptions stated above. The usual practice in such case is that the

direction of G with respect to joint A is first assumed to be fixed and then

the Williot diagram for the whole structure is drawn. After completing the

diagram, we have to apply the necessary correction for the wrong

assumption we made first. This correction is applied by drawing another

Figure (7.c)

Figure (8)


diagram called “Mohr diagram”. The combined Williot and Mohr

diagrams is known as Williot-Mohr diagram.

e. Mohr diagram, Fig.9.

For example, consider the unsymmetrical loaded truss shown in the above

figure with joint A as a fixed point and assume member A2 has a fixed

direction, hence draw Williot diagram. From Williot diagram, the

displacement at roller support B does not satisfy the boundary conditions

at B. This support can only move horizontally to satisfy this condition. So

we must rotate the deformed shape of truss with respect to point A. Each

joint will rotate with a radius from A to the joint by such an amount that

point B’ will reach the horizontal line AB as shown in Fig.(9.d) while

Fig.(9.c) shows the deformed shape of truss according to Williot diagram

(before correction).

The actual displacement of point B’ and other points are obtained from

Fig.(9.b) named A’ 2” B” 3” 1” in which A’B” is taken equal to the

vertical component A’B”. The actual displacement of each joint is then

the distance between the dashed letter (for example B’ in Williot diagram)

and the same double dashed one (for example B” in the rotated deformed

shape of truss).

Figure (9.a)

Figure (9.c)

Figure (9.d)

Figure (9.b)


1.4. WILLIOT-MOHR DIAGRAM

In the case of having a fixed point but no fixed direction, Williot-Mohr

diagram have to be constructed. Consider the unsymmetrical loaded truss

as shown in Fig.(10.a). Assume joint 2 as a fixed point and member 2-4 as

a fixed direction (member 2-4 has minimum change in slope at bottom

chord). Hence draw both Williot and Mohr diagrams to satisfy the

boundary conditions at A and B. The support at B moves horizontally,

then the horizontal displacement at B is equal to Hb at Williot-Mohr

diagram. Similarly, point A’ must be taken as an origin instead of point

2’. The final displacement of any joint, say joint 3, will be equal to the

vector connecting the two points on Williot and Mohr diagrams (3’3” for

joint 3) and similarly the other joints.

Example (3)

Figure (10.a)

Figure (10.b)


A given Warren truss shown in Fig.(11) is subjected to external loads and

the extensions as well as contractions in different members are given as

shown. Draw the Williot-Mohr diagram to find the vertical deflection at

joint C. Given EA=80000 ton for all members.

Solution

1. Find the internal force in each member due to the given loads.

2. Calculate the member deformations due to the internal member

forces.


AB 6.00 80000 23.33 0.17 Tension

BC 6.00 80000 50.00 0.38 Tension

CD 6.00 80000 16.67 0.13 Tension

EF 6.00 80000 -46.67 - 0.35 Compression

FG 6.00 80000 -33.33 - 0.25 Compression

AE 4.24 80000 -33.00 - 0.17 Compression

EB 4.24 80000 33.00 0.17 Tension

BF 4.24 80000 -4.71 - 0.02 Compression

FC 4.24 80000 -23.57 - 0.12 Compression

CG 4.24 80000 23.57 0.12 Tension

Figure (11.a)


GD 4.24 80000 -23.57 - 0.12 Compression

3. Choose a convenient scale to enlarge the displacement values.

4. Since there is no member having a fixed direction, assume that

member AB has a fixed direction and then start with fixed joint on

either end of the selected member (joint A).

5. Point a’ (corresponding to joint A) is located first on a sheet paper.

6. From a’ draw a’b’ of length=0.17 cm parallel to AB and in the

direction of AB because FAB is Tension.

7. Select a triangle ABE in which both joints A & B are known but

joint E is required.

8. From a’ draw a’e1 of length=0.17 cm parallel to AE and in the

direction of EA because FEA is Compression.

9. From b’ draw b’e2 of length=0.17 cm parallel to BE and in the

direction of BE because FBE is Tension.

10. The intersection of the normals to a’e1 and b’e2 at e1 and e2

respectively gives joint E which represented by e’.

11. Select the next triangle BEF in which both joints B & E are

known but joint F is required.

12. From b’ draw b’f1 of length=0.02 cm parallel to BF and in the

direction of FB because FFB is Compression.

13. From e’ draw e’f2 of length=0.35 cm parallel to EF and in the

direction of FE because FFE is Compression too.

14. The intersection of the normals to b’f1 and e’f2 at f1 and f2

respectively gives joint F which represented by f’.

15. Select the next triangle BFC in which both joints B & F are

known but joint C is required.


16. From b’ draw b’c1 of length=0.38 cm parallel to BC and in the

direction of BC because FBC is Tension.

17. From f’ draw f’c2 of length=0.12 cm parallel to FC and in the

direction of CF because FCF is Compression.

18. The intersection of the normals to b’c1 and f’c2 at c1 and c2

respectively gives joint C which represented by c’.

19. Select the next triangle CFG in which both joints C & F are

known but joint G is required.

20. From c’ draw c’g1 of length=0.12 cm parallel to CG and in the

direction of CG because FCG is Tension.

21. From f’ draw f’g2 of length=0.25 cm parallel to FG and in the

direction of GF because FGF is Compression.

22. The intersection of the normals to c’g1 and f’g2 at g1 and g2

respectively gives joint G which represented by g’.

23. Select the last triangle CDG in which both joints C & G are

known but joint D is required.

24. From c’ draw c’d1 of length=0.13 cm parallel to CD and in the

direction of CD because FCD is Tension.

Figure (11.b)


25. From g’ draw g’d2 of length=0.12 cm parallel to GD and in the

direction of DG because FDG is Compression.

26. The intersection of the normals to c’d1 and g’d2 at d1 and d2

respectively gives joint D which represented by d’.

27. The vertical distance in the Williot diagram between joints d’ and

a’ must be zero (to satisfy the roller supporting conditions), so, joint d’

is displaced towards a’ vertically (draw D” above a’ such that d’D” is

Figure (1.c)


parallel to the direction of roller displacement) by the value d”a’. All

joints of truss must be displaced a distance proportional to its spacing to

joint A and the length d”a’ , this is achieved by redrawing the whole

truss vertically such that the lower chord AD of the original truss

coincides completely to the length d”a’ in the Williot diagram to obtain

the Williot-Mohr diagram.

28. The actual displacement of each joint is then the distance between

the dashed small letter (for example c’) and the same double dashed

capital one (for example C”).

29. Measure horizontally the distance between c’ and C’’ and divide by

the scale to find Hc=0.55 cm.

30. Measure vertically the distance between c’ and C’’ and divide by

the scale to find Vc=1.43 cm

2. INFLUENCE LINES FOR STATICALLY

INDETERMINATE TRUSSES

The principle of Muller Breslau is applied for the influence lines of

indeterminate trusses. For the given truss shown in Fig.(12.a), the

influence lines for X1 is found as follows;

1. Apply X1 = 1 ton at the main system by cutting the member 4-7,

then get the values of N1 for each member.

2.

3. Compute the values of due to N1 from the equation

Force Axial

1

AE

LN

for each member.

4. Draw the Williot or Williot-Mohr diagram and draw the elastic line

of the loaded chord which is the required I.L.X1 .


5. The influence line for a force in any member can be obtained from

the following equation 110

X.L.INN.L.I.N.L.I , where I.L.N0 is

the influence line of the force at main system and N1 is the force in the

member due to X1 = 1 ton.

6. In cases of trusses with n degree of indeterminacy and

indeterminate beams or frames, the same procedure is followed.

Example (4)

For the shown continuous truss shown in Fig.(13.a) determine:

I. L. X1

I. L. N2-5

Solution

Figure (12)

Figure (13.a)


1. Remove the vertical support at B and apply a force of 1 ton

vertically at B, and then calculate the internal force in each member due

to this external load.


forces.

3. Draw the Williot diagram using these internal deformations as

stated in the previous examples.


A—1 5.00 100 0.50 2.50 Tension

2—3 5.00 100 0.00 0.00 Zero Member

4—5 5.00 100 0.00 0.00 Zero Member

B—6 5.00 100 0.00 0.00 Zero Member

1—3 5.00 100 0.50 2.50 Tension

3—5 5.00 100 0.50 2.50 Tension

5—6 5.00 100 1.50 7.50 Tension

A—2 5.00 100 0.00 0.00 Zero Member

2—4 5.00 100 -1.00 - 5.00 Compression

4—B 5.00 100 -1.00 - 5.00 Compression

1—2 7.07 100 -0.71 - 5.02 Compression

2—5 7.07 100 0.71 5.02 Tension

5—B 7.07 100 -0.71 - 5.02 Compression

Figure (13.b)


Figure (13.c)

4. From the Williot diagram, the elastic curve of the loaded chords

(lower chords) can be drawn and then divided by the value of vertical

deflection at B to get the I. L. X1

5. Draw the I. L. N(2-5)0

i. When 1 ton is right to joint 4,

N(2-5)0 = - YA-Left

I.L.N(2-5)0= - I.L.YA-Left ii. When 1 ton is left to joint 2,

N(2-5)0 = + YA-Right

I.L.N(2-5)0 = + I.L.YA-Right

6. Draw the I. L. N(2-5)

I.L.N(2-5)= I.L.N(2-5)0 + N(2-5)1 I.L.X1

I.L.N(2-5)= I.L.N(2-5)0 + 0.71 I.L.X1


Example (5)

For the given truss shown in Fig.(14) determine:

I. L. X1

I. L. N2-4

Given: EA=100 ton for all members.

: The unit load moves along the lower chords of truss.

Figure (14.a)

Figure (13.d)


Figure (14.a)

Solution

1. Cut the member 4-7 and apply a tensile unit force in it and then

calculate the internal force in each member due to this external load.


forces.


4—6 4.00 100 -0.71 - 2.84 Compression

5—7 4.00 100 -0.71 - 2.84 Compression

4—5 4.00 100 -0.71 - 2.84 Compression

6—7 4.00 100 -0.71 - 2.84 Compression

4—7 5.66 100 1.00 5.66 Tension

5—6 5.66 100 1.00 5.66 Tension

Figure (14.b)


3. Draw the Williot diagram using these internal deformations as

stated in the previous examples.

4. From the Williot diagram, the elastic curve of the loaded chords

(lower chords) can be drawn and then divided by the value of 4-7 =

19.39 to get the I. L. X1

5. Draw the I. L. N(2-4)0

i. When 1 ton is right to joint 2,

N(2-4)0= + YA

I.L.N(2-4)0= + I.L.YA

ii. When 1 ton is left to joint 2,

N(2-4)0 = + 4 YB

I.L.N(2-4)0 = + 4 I.L.YB

6. Draw the I. L. N(2-4)

I.L.N(2-4) = I.L.N(2-4)0 + N(2-4)1 I.L.X1

I.L.N(2-4) = I.L.N(2-4)0 + 0.00 I.L.X1

Figure (14.c)


Figure (14.d)


PROBLEMS

For the given trusses, find graphically the vertical and horizontal

displacements of joints C, D, E and horizontal displacement of roller

support at B. Given: A = 10 cm2 , E = 2000 ton/ cm

2.

(Ex.1.a) (Ex.1.b)

(Ex.1.c)

(Ex.1.d)


2. For the following trusses, draw the I. L. for the marked members.

(Ex.1.e)

(Ex.1.f)

(Ex.2.a)


(Ex.2.b)

(Ex.2.c)

(Ex.2.d)


(Ex.2.e)

(Ex.2.f)

(Ex.2.g)

7 THE MOMENT

DISTRIBUTION METHOD

Introduction

This method was originated by Prof. Hardy Cross in 1930 in a

paper entitled “Analysis of Continouous Frames by Distributing Fixed-

End Moments”. It is the method normally used to analyze all types of

statically indeterminate beams and rigid frames (Fig. 1) in which the

members are primarily subjected to bending. All the methods discussed

previously involve the solution of simultaneous equations, which

constitutes a major part of the computational work. The method of

moment distribution usually does not involve as many simultaneous

equation and is often much shorter than any of the other methods. It has

the further advantage of consisting of a series of cycles, each converging

on the precise final result; therefore the series can be terminated

whenever one reaches the degree of precision required by the problem at

hand. We know that the moment acting on the end of a member is the

sum of four separate effects.

(1) The moment due to the external loads if the member is considered

as a fixed-end beam (F.E.M.).

(2) The moment due to the rotation of the near end while the far end is

fixed.

(3) The moment due to the rotation of the far end.

(4) Them moment due to the relative translation between the two ends

of the member.

375 Chapter (7) - THE

MOMENT DISTRIBUTION

METHOD

In this method, all the members of a structure are first assumed to be

fixed in position and direction and fixed and moments due to external

loads are obtained. Now all the hinged joints are released, by applying an

equal and opposite moment and their effects are evaluated on the opposite


MOMENT DISTRIBUTION

METHOD

joints. The unbalanced moment at a joint, is distributed in the two spans

in the ratio of their distribution factors. This process is considered simple

and understand and has scientific and systematic approach and used in the

design offices. The method of moment distribution can be applied to

structures composed of prismatic members (Uniform I) or no prismatic

members with or without joint translation.

Sign Convention:

Though different types of sign conventions are adopted by different

authors in their books, yet the following sign conventions, which are

widely used and internationally recognized, will be used.

“All the clockwise and moments and member rotations are taken as

positive and vice verse.”

Note:

The F.E.M. represents the action of the joints on the member.

Stiffness of Member:

For a member of uniform section (Constant EI), the stiffiness

(rotational stiffness) is defined as the end moment required to produce a

unit rotation at one end of the member while the other end is fixed.

Consider member AB in Fig. 2 with a constant section. End B is fixed

and end A is allowed to rotate (continuous support). The end moment

required at end A to rotate θA = 1 while θB = 0 is given as follows:


MOMENT DISTRIBUTION

METHOD

By using the method of conjugate beam; and from the condition of

geometry Fig. 3, θB = 0 then :

θB1 = θB2 ---------------- (1)

θA = θA1 - θA2 ---------- (1) (Note θB1 = 2

1θA1)

Therefore:

θB1 = EI6

L.M

EI2

L.M

3

1W3

1 ABAB1

θB2 = EI3

L.M

EI2

L.M

3

2W

3

2 BABA2

Substitute in eqn. 1

EI3

L.M

EI6

L.M BAAB

MBA = 2

1MAB ------------------- (3)


MOMENT DISTRIBUTION

METHOD

Also θA = θA1 – θA2 (Note θA2 = 2

1θB1)

θA = EI6

L.M

EI3

L.M BAAB

θA = EI4

L.MAB

Or

The moment produce θA = 1 is given by

MAB = L

EI4

This moment is defined as “Absolute stiffness” and will be denoted

by K, thus

K = 4E L

I

= 4E S ……………. (4)

Where L

I= s being the “Stiffness Factor” or “Relative Stiffness”

Member with one hinged end

θA = EI3

L.MAB

MAB = L

EI4 θA


MOMENT DISTRIBUTION

METHOD

θAB = L

EI3. θA

K = L

EI3 = 4

3K

And S = 4

3 L

I

The modified stiffness K at end A when end B is hinged support is

three quarters the absolute stiffness K

Member of symmetry:

From symmetry

θA = - θB

M = θL

EI2

K = K2

1

L

EI2

And S = 2

1 * L

I


MOMENT DISTRIBUTION

METHOD

Member of antisymitry:

θA = θB

MAB = M = AθL

EI6

K = K3

2

L

EI6

S = 2

3 L

I

Carry over Factor (C.O.F) and carry over moment (C.O.M)

we have already discussed that the moments are applied on all the

and joints of structure, whose effects are evaluated on the other joints.

The ration of moment produced at a joint to the moment applied at the

other.

Joints (Fig. 4), without displacing it, are called “carry over factor”. The

value of carry over factor for the above beam fig. 4 is equal to 2

1. Then,

the carry over moment MBA is one-half of moments MAB.


MOMENT DISTRIBUTION

METHOD

MBA = CAB. MAB

= 2

1 MAB

= Aθ.L

EI2

Where CAB is the carry-over factor from A to B. for a member of

uniform EI. If we consider end A as the far end (fixed), and B as the near

end (allowed to rotate), we can, in like manner, prove that.

CBA = CAB = 2

1 (6)

Distribution Factors (D.F.)

Referring to the following figure (Fig. 5), which shows a frame composed

of four members, each with one end fixed and the other end rigidly

connected at joint O whose translation is prevented. If a clockwise

moment M is applied to the joint, it will cause the joint to rotate

clockwise through an angular deformation θ, as shown in Fig. 5. Since 0

is a rigid joint, each tangent.


MOMENT DISTRIBUTION

METHOD

(8)

(10)

To the elastic curve of the connected end rotates the same angle θ. The

applied moment M is resisted by four members meeting at the joint. The

resisting moments MoA, MoB, MoC, and MoD, will be induced at the ends of

the four members to balance the effect of the external moment M, as

shown in Fig. 5,c

Equilibrium of the joint 0 requires that Σ Mo = o

and also

MoA = θ.AL

EIoA4

o

= KoA. θ

MoB = θ.BL

EIoB4

o

= KoB. θ

MoC = θ.CL

EIoC4

o

= KoC. θ

MoD = θ.DL

EIoD4

o

= KoD. θ

The above equation 7 shows that, when the external moment is

applied to a joint, the resisting moments developed at the near ends of the

members metting at the joint, while the other ends are all fixed, are in

direct proportion to the rotational stiffness “K”.

Substituting in eqn. 8 in eqn. 7, we obtain

(KoA + KoB + KoC + KoD). θ = M

Thus, θ = K

M

(9)

We see that

MoA = K

KoA

. M = DoA. M

MoB = K

KoB

. M = DoB. M

MoA + MoB + MoC + MoD = M


MOMENT DISTRIBUTION

METHOD

MoC = K

KoC

. M = DoC. M

DoD = K

KoD

. M = DoD. M

In which the ratio K

KoD

or DoA is defined as the “distribution

factor”. Thus; a moment resisted by a joint will be distributed among the

connecting members is proportion to their distribution factors, only the

relative K or S values for connected members are needed. Thus, in most

cases we are concerned with the “relative stiffness” rather than the

absolute stiffness.

Fixed End Moment (F.E.M.)

the application of the mothed of moment distribution requires

knowledge of the moments developed at the ends of loaded beams with

both ends. These moments are called “Fixed-End Moments”, often

denoted by the symbol F.E.M. in the examples.

The determination of fixed-end moment proveiously (3 moment

eqn.). for a straight prismatic member the fixed-end moments due to

common types of loading are given in the following Table (1).


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD

Application to various types of continouous Beams

In the previous articles, we have studied the principles of the

moment distribution mothod. First of all, all the fixed-end moments are


MOMENT DISTRIBUTION

METHOD

found for all the spans. The unbalanced moment at a support is

distributed among the two spans in the ratio of their stiffness factors or

relative stiffness (L

I or

4

3 L

Ior

2

1

L

I or

2

3 L

I) and their effects are

evaluated on the opposite joints. This process is continued, till we reach

the required degree of accuracy.

Non-Prismatic Members

The members have not uniform cross-Section are known, non prismatic

members.

The fixed end moments, stiffness, cary over factor, and distribution

factors for these members are not the same values as prismatic members.

To obtain these values, the method of column analogy or method of

consistant deformation, or conjugate beam method must be used.

Example (1): For the given beam, draw B.M. and S.F. Ds


MOMENT DISTRIBUTION

METHOD

Solution:

At first, let us consider the continuous beam ABC be split into two fixed

beams AB and BC as shown in Figure (6).

1- Fixed end Moments.

Span AB: F.E.MAB = - 12

WL2= -

12

62 2 = - 6 t.m

F.E.MBA = + 12

WL2 = - 6 t.m

Span BC: F.E.MBC = 8

L.P= -

8

612 = - 9.0 t.m

F.E.MCB = + 9.0 t.m

2- Stiffness (or relative stiffness):

SAB = 6

1

L

I

SBC = 6

1

L

I

3- Distribution Factors (D) for member BA and BC will be 2

1and

2

1or

6/16/1

6/1

= 0.5

Now draw the beam and fill up the distribution factors and F.E.M.S

as shown in the following figure.


MOMENT DISTRIBUTION

METHOD

We obtain the F.E.MBA is +6.0 t.m and that in span BC F.E.MBC is 9.0.

thus there is an unbalanced moment at joint B equal to –9 + 6 = -3.0 t.m.

Now distribute this unbalanced moment (equal to –3.0) into the span BA

and BC in the ratio of their distribution factors i.e. + 1.5 and 1.5. Now

carry over the effects of these distributed moments (or Balanced moment)

at A and D equal to 2

1× 1.5 = +.75 then distribute the unbalanced moment

at B. (in this case, there is no carry over moment from A or C at B. So the

distribution of moment at B is zero). Now find out the final moment at A,

B, and C in the spans AB and BC by algebraically adding the respective

values. Calculate the bending moment in the spans AB and BC by


MOMENT DISTRIBUTION

METHOD

considering them as simply supported beams. Also calculate the reactions

at supports A, B, and C by considering the free body diagram for each

span. Now we can draw both shearing force and bending moment

diagrams as shown in the previous figures.


MOMENT DISTRIBUTION

METHOD

A continuous Beams with simply supported ends:

Sometime, a continuous beam is simply supported over one or both

of its ends. We know that the fixing moment on a simply supported end is

zero. Therefore, in such a case, the simply supported ends are released by

applying equal and opposite moments, and their effects are carried over

on the opposite joints as follow.


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD

Or by using the following table:

-

Joint B C D

Member BA BC CB CD DC DE

Relatives 6

1

4

3

10

1

10

1

12

1

12

1

6

1

4

3

D.F. 0.56 .44 .55 .45 0.4 .6

F.E.M +6.75 -6.75 +16.67 0 0 -13.5

D.M. +5.55 +4.37 -9.17 -7.5 +5.4 +8.1

.CO.M. 0 -4.585 +2.185 +2.70 -3.75 0

D.M. +2.56 +2.025 -2.685 -2.20 +1.5 +2.25

C.O.M. 0 -1.34 1.01 +.75 -1.1 0

D.M. +.75 +.59 -.97 -.79 .66 .44

C.O.M. 0 -.48 +.30 +.33 -.40 0

D.M. +.27 +.21 -.35 -.28 +.16 +.24

C.O.M. 0 -.175 +.105 +.08 -.14 0

D.M. .10 .075 -.1 -.085 +.08 +.06

Final

Moment

+15.54 -15.98 +6.99 -6.99 +2.41 -2.41

Average 15.72

Beams with end span overhanging

Sometimes, a beam is overhanging at its one or both the end

supports. In such a case, the bending moment at the supports near the

overhanging end will be due to the load over the cantilever portion and

will remain constant (Determinate), while the moments on the other far

support is one half of the overhanging moment plus or minus the fixed

end moment due to the load acting on the span. It is thus obvious, that the


MOMENT DISTRIBUTION

METHOD

distribution factors over the support having one span overhanging will be

1 and 0.

Moreover this support is considered as a simply supported for the

purpose of calculating the distribution factors in the span, adjoining the

overhanging span.

Example (6): A beam A B C D 9 M long is simply supported at A,

B and C such that the span AB is 3 m, span BC is 4.5

m and the overhanging CD is 1.5 m. It carries a

uniformly distributed load of 1.5 t/m in span AB and a

point load of 1 ton at the free end D. The moment of

inertia of the beam in span AB is I and that in the span

BC is 2I. Draw B.M., S.F.Ds.


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD

Example (8): for the given loaded beam ABCDE fixed at A

and roller at B, C, and D, DE being a

cantilevr. AB = 7 m, BC – 5, CD = 4 m and

DE = 1.5 m. the values of I, are 3I, 2I, I and I

respectively Draw B.M.D.

Solution:


MOMENT DISTRIBUTION

METHOD

3-10 Beams with settlement and rotation of supports:

For some cases one of the supports of a continuous beam

sinks down, with respect to others, as a result of loading. As a

result of differential settlement between the supports, additional

moments are caused on the supports, in addition to the moments

due to loading. Knowing that the moments due to settlement ∆ at

support B for the following figure, at A and B are: M = 2L

EI6 ∆,

while in case of left hand support A.

Sinks ∆ with respect to B the fixed end moment on both supports

is, M = + 2L

EI6. ∆

If support A has a rotation θ A, then F.E.M is

MAB = - θALAB

EI4

Example (9): A continuous beam is fixed at A and is carried

over roller at B and C as shown. Draw B.M.,

S.F.Ds due to given loads and settlement at B,3

cm.


MOMENT DISTRIBUTION

METHOD

E = 2 × 105 kg/cm

2, I = 2 × 10

6 cm

4

Solution:


MOMENT DISTRIBUTION

METHOD

Example (10): A continuous beam ABC shown in the

following Figure. Draw B.M. and S.F. Ds due to given loads and

settlement at support B by 4 cm below A and C. E = 2 × 107

t/m2, I = 3 × 10

-4 m

4

Solution:


MOMENT DISTRIBUTION

METHOD

1-Due to ∆ t = 20o (t2 > t1)

F.E.M.

MBA = - 1.5 h

.t.EI = -

8.0

10120800025.1 5 =-6 t.m

MBC = + h

.t.EI = -

8.0

10208000 5 =+2t.m

MCB = - h

.t.EI = =+2t.m

MCD = + h

.t.EI = +

8.0

102080002 5 =+4t.m


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD

2- Due to loads


MOMENT DISTRIBUTION

METHOD

Example (12): for the given continuous beam ABCDE. Draw

B.M., S.F.Ds due to ∆ t = - 15o at part BC,

EI = 10000 t.m2

H = 1.0 m, α = 1 × 10-5

(t2 < t1).


MOMENT DISTRIBUTION

METHOD

Example (13): for the given continuous beam D.B.M. due to

given loads.


MOMENT DISTRIBUTION

METHOD

Frames without side sway:

A frame without side sway (i.e. without any translation)

way also be solved by the method of moment distribution in the

same procedure as that for a continuous beam.

Example (14):

For the shown frame ABC Draw, N.f., S.F. and B.M.Ds

Prepare the following table


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD

Example (15): Draw B.M.D. for the shown frame

Solution:

1- Stiffnesses:

SAC = 8

1 = 0.125

SCD = 24

2 =

12

1 = 0.083

SDB = SAC = 0.125

2- Distribution Factors:

Joint – C-

D.FCA = 6.0083.125.0

125.0

D.FCD = 4.0083.125.0

083.0

Also at Joint D

D.FDC = 0.4

D.FDB = 0.6

3- Fixed end moments:

MCD = t.m48012

2410

12

WL22

MDC = +480 t.m

MAC = MDB = 0


MOMENT DISTRIBUTION

METHOD

The Previous calculation can be put in the following table:

Joint A C D B

Member AC CA CD DC DB BD

S 8

1

12

1

12

1

8

1

D.F. 0.6 .4 .4 .6

F.E.M. 0 0 -480 +480 0 0

D.M. 0 +288 +192 -192 -288 0

C.O.M +144 0 -96 +96 0 -144

D.M. 0 +57.6 +38.4 -38.4 -57.6 0

C.O.M +28.8 0 -19.2 +19.2 0 -28.8

D.M. 0 +11.52 +7.68 -7.68 -11.52 0

C.O.M +5.76 0 -5.76

F.M. 178.56 357.12 -357.12 +357.12 -357.12 -178.56


MOMENT DISTRIBUTION

METHOD

Symmetrical Portal Frames:

A symmetrical portal frame is that, in which both the

columns are of a same length, moment of inertia, having similar

and conditions, and modulus of elasticity as wall as subjected to

symmetrical loading. The joints of a portal frames will not be

subjected to any translation or side sway. A simple portal frame

consists of a beam, resting over two columns. The joints of the

beam and column frames are either symmetrical (without side

sway) or unsymmetrical (with side sway).

Example (16): For example 15 taking the symmetry in

consideration. Draw B.M.D.

Solution:

Due to symmetry, we may use the half-frame of the

following figure and take into consideration the modified

stiffness s due symmetry as follows.


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD

Notes on symmetry and anti – symmetry of structures :

1- From the previous example 16 it is possible to take the

advantage of any symmetry of a structure if it is subjected

to load systems which are themseelves symmetrical or anti

– symmetrical about the same geometric centre line.on

such a symmetrical structure, symmetical liad systems

produce equal and opposite slopes about the centre line,

whilst anti – symmetrical loads cause deflections of equal

magnitude and different sign at corresponding points about

the centre as shown in the following figures;Example

(16):- Draw B.M.D for the given problem.


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD

Symmetrical loads:

Since these cause deformation, which are mirrored about

the structural centre – line, the slope at that centree line must be

zero. If this occure at a nodal point; that is, if the structure has an

even number of spans; we may regard the structure as being effec

– tively fixed at that point, and complete the calculation for one –

half of the structure only as shown in the shown figures of

beams or frames. If the number of spans is odd, however, as


MOMENT DISTRIBUTION

METHOD

given the figure, which show three span with symmetrical loads,

although the slope is zero at centre line, some vertical

displacement will occur, then this point cannot be consifer a

fived and the solution can be completed on the half of beam, take

account the modified stiffness S = 2

1S.

2- Anti – symmetrical loads:

If a symmetrical structure is subjected to loads which are

anti – symmetrical about the centre line as shown in the given

figure, solution of problem will vary depending on the numbert

of spans. Whether this be even or add as shown, the structure will

behave as if hinged support on the centre line ( case of beam) in

which the stiffness of span is (L

I

4

3). whereas in case of the

centre line at mid span of the member. In which the centre span

has a stiffness factor of ( L

I

2

3).


MOMENT DISTRIBUTION

METHOD

Example (17): Draw B.M.D. for the give

structure.


MOMENT DISTRIBUTION

METHOD

Example (18): find the B.M.D. for the given symmetrical

closed frame.

Solution:


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD

Example (19): draw B.M.D. for the symmetrical frame shown

in the given figure under the given anti – symmetrical loading.

Solution:

Now and from anti – symmetrical loading member EB is

split up to two members, each having half the actual moment of

inertia as shown. Then considering one – half of the frome and it


MOMENT DISTRIBUTION

METHOD

must noted the equivalent structure is symmetri – acl loading. We

get.


MOMENT DISTRIBUTION

METHOD

Example (20):


MOMENT DISTRIBUTION

METHOD

To draw B.M.D ising joints equilibrium;


MOMENT DISTRIBUTION

METHOD

Example (21):


MOMENT DISTRIBUTION

METHOD

Example (22):


MOMENT DISTRIBUTION

METHOD

Example (23):


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD

Example (24):

For the shown horizontal closed frome of a part of water tank.

Draw B.M.D. for tank.

Solution:

The frame is symmetrical, and symmetrically loaded about the

horizontal axis X-X. Then, it is sufficuent to consifer one half of

the frame only as follows:


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD

Example (25):

Draw B.M.D. for the double span frame shown in the following

Figure calculate reaction at F.


MOMENT DISTRIBUTION

METHOD

From Joint Stability:

To get horizontal reaction at F (XF):

From the free body diagram of columns and by applying


MOMENT DISTRIBUTION

METHOD

Example (26): Draw the B.M.D. for the double story frame

shown in the following figure. Calculate the

forces in link members.


MOMENT DISTRIBUTION

METHOD

1-Stiffness factors:

Joint C:

SAC = 4

2 = .5 ; SCD = .25 ; SCE

4

2 = .5

D.FAC = 0.4 ; D.FCD = 0.2 DCE = 0.4

Joint E:

SEC = 4

2 = .5 ; SEF =

16

2 = 0.125

D.FEC = 0.8 , D.FEF = 0.20

Joint D

SDB = 4

1= .25 ; SDC =

16

4 = 0.25 , SDF = 4

1= .25

D.FDB = 0.333 , D.FDC = .333 , D.FDF = 0.333

Joint F

SDB = 4

1= .25 ; SFE =

16

2 = 0.125

D.FFD = .67 D.FFE = .33

2- Fixed end Moments:

MAC = - 12

42 2 = - 2.67 t.m

MCA = + 2.67 t.m

MCD = - 12

164 2 = - 85.33 t.m

MDC = = + 85.33 t.m


MOMENT DISTRIBUTION

METHOD

MCE = - 12

44 2 = - 5.33 t.m

MEC = = 5.33 t.m

MEF = - 8

1612 = - 24 t.m

MFE = = 24 t.m

MFD = MDB = Zero

Forces in Upper Link Member

H1 = 2

44

h

Story)(UpperM


MOMENT DISTRIBUTION

METHOD

= 4

86.3398.203.367.25 + 8 = 9.735 t

Forces in Upper Link Member

H2 = 4

05.1541.3037.780.43 +

2

42

+ 2

44

4

6.253.3686.3321

= 14.18 to


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD

Example (27): Draw B.M.D. the given two – span, two-story

frame.

Solution:

Hj = 6

1 (- 1.06 + .92 – 1.94 + .42) = - 1.66 (to right)


MOMENT DISTRIBUTION

METHOD

HF = 6

1 (3.8 + 3.94 + 1.97 – 4.68 – 2.29 – 5.73) +

2

65.1 - (- 1.66) = 5.66 (to left).

Structures subjected to Side Sway:

In more general case, however, when the joints are not

laterally supported , the equillubrium position at any joint cannot

be reaches by allowing rotation only at the joints, besides rotation

untill the sum of the forces as well as the moments at each joint

becomes zero. This translation is usually called “SIDE SWAY”.

Generally in case of side sway the final benfing moment diagram

not only due to the exterrnal loads but also due to translation of

joints. The solution of the problems is ussually divided into two

parts:

1- A calculation od end moments produced by the given

lodss, assuming that there aree no lateral movements.

2- A calculation od end moments produced by lateral move –

ments. The bending moments produced from this transla –

tion will be called “SIDE SWAY CORRECTION”.

The moment diatribution can be carried out in the

following steps:

1- Restrain all joints against translation, and carry out the

moment distribution as before.


MOMENT DISTRIBUTION

METHOD

2- Calculate horizontal as well as vertical reactions at the

column bases. The algebric sum of the horizontal reaction

will give the magnitude and direction of the sway force F,

which had held the frame against side sway as shown in

the following figures.

3- Remove the force F, which had held the frame, and let the

joints be allowed to sawy as shown in Figure (b). This will

cause a set of fixed end moments. Calculate these fixedenf

moments and distribute them. (This is done fixed by

assuming some suitable arbitrary sway moments and

distributed them). Now find out the horizontal reactions at

the column bases. The algebraic sum of the horizontal

reactions will give the assumed sway force. Find out the

moments and horizontal as well as vertical reactions due

to the actual sway force F, proportionately to the assumed

sway force).


MOMENT DISTRIBUTION

METHOD

4- The final mment at each joint maynow be obtained by

adding algebraically the momentss obtained in steps 1 and

3.

5- The horizontal as well as vertical reactions at the column

bases may also be obtained by adding algeebraically the

reactions obtained in steps 1 and 3 M = MO + fO

f1. M1

6- Knowing that the F.E.Ms. in case of translation

(Settlement) are as follow.

a- Case of fixed beam

b- Case of hinged at far end


MOMENT DISTRIBUTION

METHOD

- MA = Δ2L

EI3Δ.

2L

EI3Δ.

2L

EI6

Fixed – end moments in Frames due to side sway:

a) Members with fixed ends.

2L2.

2EI6

CDM

2L1.

1EI6

BAM

Δ

Then MBA = MCD 2

12

1

2 .)(L

I

L

L


MOMENT DISTRIBUTION

METHOD

b) Members with hinged ends:

2

2

1

2

3

2.

6

1.

EI

LM

EI

LM CDBA

then MBA = MCD 2

2

12

1

2 ).()(L

I

L

L

= 1

2

2

1 .2

3

L

L

S

SM CD

where S1 1

1

L

I and S2

2

2

4

3

L

I

Example (28): Draw the B.M.D. for the shown frame due to the

given loads. Also draw the deformed shape.


MOMENT DISTRIBUTION

METHOD

Solution:

The frame is unsymmetrical,

theregore it will be

analysed first as a frome

without side sway and then

applying a sway correction .

Fixed end moments:

MBC = - 34.24

5.25.142

2

t.m

MCB = + 40.14

5.25.142

2

t.m

MBA = MCD = Zero


MOMENT DISTRIBUTION

METHOD

Distribution factors:

SBA : SBC

4

1:

4

1

4

3

0.43 : 0.57


MOMENT DISTRIBUTION

METHOD

From MO.D XA = 0.32 , XD = 0.24

Then the horizontal force FO = 0.32 - .24 = .08

We have worked out the moments and reactions, under the

assumption, that a horizontal force FO = 0.08 t is acting at C.

Actually there is no such force acting at c. we know that if

a force of 0.08 t is applied at C, it will neutralise the effect

of force fO = 0.08 little con – sideration will dhow that if a

force of 0.08 at C.A Little con – sideration will show that id a

force of 0.08 t is applied at C, it will cause some sway moments

at the joints B and C of the frame.

Let MBA : MCD

.3

:.3

2

2

2

1

1

L

EI

L

EI


MOMENT DISTRIBUTION

METHOD

4

1:

4

1

1 : 1

Assume some arbitrary moments at B and C i .e. (MBA and

MCD) in the ratio 1:1 as already calculated and distribute the same

at other joints also. Let us assume the sway moments as – 2 t.m.

Then complete the following table:

(Use short solution by using antisymitry)

Horizontal reaction at A = 4

33.1

= .33

Horizontal reaction at B = .33

from the free body siagram as dhown we find that the

unbalanced horizontal force at C.

F1 = .33 + .33 = .66

But the actual unbalanced horizontal force is 0.08 t

therefore the actual sway moments in the members due to a sway

force of 0.08 t may be found out by proportion.

Then

M = MO + 1

1

MF

FO


MOMENT DISTRIBUTION

METHOD

MB = - 1.27 + )33.1(66.

08.

= - 1.11 t.m

MC = - .96 + )33.1(66.

08.

= - 1.12 t.m

Reaction at A and D:

XA = 0.32 + 66.

08.(- .33) = .28 t

XD = -.24 + )33.(066.

08. = 0.28


MOMENT DISTRIBUTION

METHOD

Solution:

Assumption, that a horizontal force FO =0.08 t is acting at C.

Actually there is no such force acting at c. we know that if

a force of 0.08 t is applied at C, it will neutralize the effect

of force fO = 0.08 little consideration will dhow that if a

force of 0.08 at C.A Little consideration will show that id a force

of 0.08 t is applied at C, it will cause some sway moments at the

joints B and C of the frame.

Let MBA : MCD

.3

:.3

2

2

2

1

1

L

EI

L

EI

4

1:

4

1


MOMENT DISTRIBUTION

METHOD

1 : 1

Assume some arbitrary moments at B and C i .e. (MBA and

MCD) in the ratio 1:1 as already calculated and distribute the same

at other joints also. Let us assume the sway moments as – 2 t.m.

Then complete the following table:

(Use short solution by using antisymitry)

= .33

Horizontal reaction at B = .33

from the free body diagram as shown we find that the

unbalanced horizontal force at C.

F1 = .33 + .33 =.66


MOMENT DISTRIBUTION

METHOD

But the actual unbalanced horizontal force is 0.08 t

Therefore the actual sway moments in the members due to a

sway force of 0.08 t may be found out by proportion.

Then:

MB = - 1.27 + )33.1(66.

08.

= - 1.11 t.m

M = MO + 1

1

MF

FO


MOMENT DISTRIBUTION

METHOD

MC = - .96 + )33.1(66.

08.

= - 1.12 t.m

Reaction at A and D:

XA = 0.32 + 66.

08.(- .33) = .28 t

XD = -.24 + )33.(066.

08. = 0.28


MOMENT DISTRIBUTION

METHOD

Example 27:

A portal frame shown in the following figure is subjected to a

loading as shown. Draw B.M.D.

Solution:

Since the portal frame is unsymmetrical, therefore, it will

be analyzed first by assuming it without sway and then appiy ing

a sway correction.


MOMENT DISTRIBUTION

METHOD

Horizontal reaction at A and D

XAO = 357.16

75.25.5


MOMENT DISTRIBUTION

METHOD

XDO = 79.06

58.116.3

FO = 1.357 – 7.9 = 0.585

We have worked out the moments and reactions under the

assumption that a horizontal force fo = 0.585 t is acting at c

to prevent the side sway: Actually there is no such force acting at

C. We know that if a force of 0.585 is applied at C, it will

neutralize the effect of do at C. If a force 0.585 is applied at,

it will cause some sway moments at joints B and C of the portal

frame.

MBA : MCD

1 : 1

Now assume some arbitrary moments at B, C, A and D in ratio of

1:1 assume M = -10 t.m


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD

X1A = 6

55.748.8

= 2.67

X1D = 2.67

F1 = 2.67 + 2.67

= 5.34 ton

Final Moment M = Mo + 1

1

MF

Fo

MA = 2.75 + 34.5

585.0 (-8.48) = 1.82 t.m

MB = 5.5 + 34.5

585. (+7.55) = - 4.67 t.m

MC = - 3.16 + 34.5

585.0 (-7.55) = - 3.98 t.m

MD = + 1.58 + 34.5

585. (+8.48) = 2.51 t.m


MOMENT DISTRIBUTION

METHOD

Example 28:

For the given frame.Draw the B.M.D.


MOMENT DISTRIBUTION

METHOD

Moments assuming no joint transition:

XAO = 6

64.829.17 = 4.32

XDO = 6

59.718.15 = 3.80

FO = 4.32 – 3.8 = 0.53

2- Moments due to horizontal Movement:

Assume arbitrary Moment = -10 t.m and using the results

of previous example then;

F1 = 5.34


MOMENT DISTRIBUTION

METHOD

Hence:

M = Mo + 1F

Fo (M1)

XA = 4.1 ton

XD = 4.1 ton

Example (29):


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD

From Mo we get Fo = 1.67 t

From M1 we get F1 = 3.686

Then M = Mo + 1F

FoM1


MOMENT DISTRIBUTION

METHOD

Example (30):


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD

Example 31:

Draw B.M., N.F., and S.F.D. for the given frame due to given

loads.

Solution:

1- Relative Stiffness and D.F.

Joint C:

CA : CD : CF


MOMENT DISTRIBUTION

METHOD

4

1 :

9

2 :

3

1

4

3

.25 : 0.22 : .25

.35 : 0.30 : .35 D.F

Joint D:

DC : DB : DE

8

1

4

3:

4

1

4

3:

9

2

0.22 : 0.188 : 0.09

0.44 : 0.37 : 0.19 D.F

2- Fixed end moments

Member CD:

MCD = 12

2WL

= 12

93 2

= - 20.25

MDC = + 20.25

Member DE

Case (1)


MOMENT DISTRIBUTION

METHOD

M1DE = 16

3PL

= 16

853

= 7.5 t.m

M1DE = Zero

Case (2)

M2ED = + 2

2WL

= 2

22 2

= 4 t.m


MOMENT DISTRIBUTION

METHOD

M2DE = 2

2EDM

= 2

4

= 2 t.m

Then:

MDE = - 7.5 + 2 = -5.5 t.m

MDE = + 4 t.m

Member DF, CA and DB are Zero F.E.M.

Member C-C

MCC = + 62

23 2

3- Frame without side sway:

joint C C

Member C – C C – A C – F C – D D – C D – B D – E

D.F

F.E.M

D.M

C.O.M

+ 6

0

0

.35

0

4.98

0

.35

0

4.98

0

0.30

-20.25

4.29

-3.25

.44

+20.25

-6.49

2.15

.37

0

-5.46

0

.19

-5.5

-2.80

0


MOMENT DISTRIBUTION

METHOD

D.M

C.O.M

D.M

0

0

1.14

0

.17

1.14

0

.17

0.97

-.48

.14

-.95

+.48

-021

-.79

0

-.18

-0.41

0

-.09

F.M. +6 6.29 6.29 -18.58 15.23 -6.43 -8.8

4- Horizontal Reactions:

XA = 4

15.329.6

= 2.36 t


MOMENT DISTRIBUTION

METHOD

XB = 3

29.6

= 2.09

ΣX = 0

Fo = - (2.36 + 2.0 – 1.6 – 2.09)

Fo = 0.67

5- Moment due to side sway:

Assume arbitrary F.E.M. as follows:

MAC = MCA = .6

2L

EI

=

.16

6EI

MCF =

.3

2L

EI

= .9

3EI


MOMENT DISTRIBUTION

METHOD

=

.16

3EI

If we assume any value for EI∆ , say 144 then

MAC = MCA = - 54 t.m

MCF = + 48 t.m

MDB = - 27 t.m

Joint A c D

member AC CA CF CD DC DB DE

D.F

F.E.M

D.M

C.O.M

D.M

C.O.M

D.M

C.O.M

-

-54

0

1.05

0

-1.04

0

.04

0.35

-54

0

2.1

-2.08

0

.07

.35

+48

0

2.1

-2.08

0

.07

.30

0

1.8

5.94

-1.78

-.2

.06

.44

0

0.9

11.88

-.4

-.89

.39

.37

-27

9.99

0

-.33

0

.33

.19

0

5.13

0

-.17

0

.17

F.M. -53.95 -53.91 48.09 6 11.88 -17.01 5.13


MOMENT DISTRIBUTION

METHOD

6- Horizontal load for sway:

XA = 4

95.5391.53

= 26.97 t

XB = 4

17.01

= 4.25


MOMENT DISTRIBUTION

METHOD

XF = 3

09.48

= 16.03

ΣX = 0

F1 = XA + XB + XF = 47.25 t

7- Final B.M.D. (MF)

MF = Mo + 1

1

MF

F o

= Mo + 125.47

67.0M

= Mo + 0.014 M1

MAC = + 3.15 + .014 (-53.95) = + 2.38 t.m

MCA = - 6.29 + .014 (-53.91) = - 5.53 t.m

MCF = - 6.29 + 0.014 (-48.09) = - 6.96 t.m

MCD = - 18.58 + 0.014 (+6.0) = - 18.50 t.m


MOMENT DISTRIBUTION

METHOD

3.098.73

RE= 5.91 t

1.91

4t.m1.66t.m

4t.m16.25

6.67t.m

15.418.52

5.53t.m

20.65

19.6

1.66t

16.25

6.96

1.66

16.25

4t

13.1613.84

1.98

XA=1.98

YA=19.64

2.38t.m

6

Xf=2.32

1.71

1.661.66 1.66t.m2t/m2t/m 3t/m

MDC = - 15.23 + .014 (- 11.88) = - 15.40 t.m

MDB = - 6.43 + 0.014 (+17.01) = - 6.67 t.m

MDE = - 8.8 + .014 ( 5.13) = - 8.73 t.m


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD

Example 32:

Draw B.M.D. for the given frame under the shown loading.

Solution:

Neglect the side sway and evaluate the B.M.D. (Mo.D) and

get Fo at point E which prevent the side sway, and solve again the

frame due side sway only (correction of side sway) and get M1D

and F1.Then the final B.M.D. = Mo + 1.1

MF

Fo

1- Moment without side sway

a- Relative stiffness (S or S and D.Fs.)

Joint C

CA : CB

S 5

1 :

8

2

4

3

D.F 0.52 : 0.48


MOMENT DISTRIBUTION

METHOD

b- Fixed End Moments

MAC = MCA = 0

M1CB = 64

5

WL

2

6464

25

= -10 t.m

M1CB = 0

Effect of cantilever

M2BC = + 4 t.m

Hence:

MCB = -10 + 2 = -8 t.m

MBC = 4 t.m

c- The Mo.D can be obtained by using the following table.

Joint A C B

Member AC CA CB CD BC BE


MOMENT DISTRIBUTION

METHOD

D.F.

F.E.M.

D.M.

C.O.M.

0

1.04

0.52

0

2.08

0.48

-8

1.92

+4

0

+4

-4.0

F.Mo 1.04 2.08 -6.08 +4 +4 -4

d- Horizontal force at E

From free body diagram


MOMENT DISTRIBUTION

METHOD

ΣX = 0

Fo – 4 – 12.05 = 0

Fo = 16.05 t

e- Moment due to Side Sway

Tan Y

4

3

Y = 1.33 ∆


MOMENT DISTRIBUTION

METHOD

f- Fixed End Moment

MAC = MCA = 2

6

L

EI(1.67 ∆) = -0.40 (EI. ∆)

MCB = + 2

)2(3

L

IE = (1.33 ∆) = 0.125 (EI. ∆)

Assume EI ∆ = 20

Then Assume MAC = MCA = - 8.0 t.m

And = + 2.50 t.m

Joint A C B

Member AC CA CD CB BC BE

D.F.

F.E.M.

D.M.

C.O.M.

-

-8

-

1.43

.52

-8

2.86

0

-

-

-

0.48

+2.5

2.64

- -

F.M. -6.57 -5.14 +5.14


MOMENT DISTRIBUTION

METHOD

Final bending Moments:

MA = 1.04 + )76.4

05.16( (- 6.57)

= - 21.11 t.m

MCA = - 2.08 + )76.4

05.16( (+ 5.14)

= 15.25 t.m

MCB = - 6.08 + )76.4

05.16( 5.14

= 11.25 t.m

Final Reactions:

YA = 8.26 + )76.4

05.16( (- .64)

= + 6.10 t

YA = 12.05 + )76.4

05.16( (-4.76)

= 4


MOMENT DISTRIBUTION

METHOD

YB = 3.74 + )76.4

05.16( (0.64)

= 5.90 ton

Check Σ Y = 0 YA + YB = 8 + 4 0.k

Σ x = 0 YA – 4 = 0 0.k

Example 33:

Draw B.M.D. for the given frame

Solution:

1- S and D.F.

Joint B: BA : BC

S 5

1 :

6

1

D.F .55 : 0.45

Joint C: CB : CD : CE


MOMENT DISTRIBUTION

METHOD

S 6

1 :

8

1

4

3:

3

1

4

3

B.F .33 : .49 : .18

2- Fixed End Moments:

MBC = - 4.5 t.m, MCB = + 4.5 t.m

MCE = - 24 t.m

Other moments are equal to zero.

3- Elastic Curve and Sway shape:

Fixed End Moment due to Sway

4- Moments without side sway :

The distribution of moments is according to the following table.

Joint A B C

Member AB BA BC CB CE CD

D.F .55 .45 .44 .18 .49


MOMENT DISTRIBUTION

METHOD

F.E.M

D.M

C.O.M

D.M

C.O.M

D.M

C.O.M

0

0

1.24

0

-0.89

0

+.05

0

2.48

0

-1.77

0

+0.09

-4.5

2.02

3.22

-1.45

-0.17

+.08

+4.5

6.44

1.01

-.33

-.73

0.24

-24

3.51

0

-0.18

0

.13

0

9.55

0

-0.5

0

.36

F.M. 0.04 0.80 -0.80 11.13 -20.54 9.41

Determination of Fo

From Σ x = 0 , 4 + YA + XA + XD +Fo = 0

XA = 4

48 = 3 , XD =

3

41.9 = 3.13 t,

Fo = 4 + 3 + 3.13 = 10.13 ton

Correction of Side sway:


MOMENT DISTRIBUTION

METHOD

The correction of side sway moments can be correct

according to following table:

Joint A B C

Member AB BA BC CB CE CD

D.F

F.E.M

D.M

C.O.M

D.M

C.O.M

D.M

C.O.M

-

-30

0

4.82

0

-.8

0

0.18

.55

-30

9.63

0

-1.59

0

0.36

.45

12.5

7.87

2.89

-1.30

-.65

0.29

.33

12.5

5.78

3.94

-1.3

-.65

.21

.18

0

3.15

0

-.71

0

.12

.49

-30

8.57

0

-1.93

0

.32

F.M. -25.8 -21.6 21.6 20.48 2.56 -23.04

The final B.M.D. can be drawn as follows

M = Mo + 1

1

MF

Fo


MOMENT DISTRIBUTION

METHOD

MA = 0.4 + 52.19

13.10 (-25.8) = - 13.02

MBA = -.8 + .52 ( 21.6) = - 10.43

MBC = -.8 + .52 ( 21.6) = - 10.43

MCB = +11.13 + .52 (+ 20.48) = + 21.78

MCD = -20.54 + .52 ( 2.54) = - 19.22


MOMENT DISTRIBUTION

METHOD

Temperature Effect for Frames without sway:

a- Uniform rise in temp.

In some cases, the temperature variation resulting in an

additional moments at supports. Due to increase of temperature

∆t degree in the shown frame, the different fixed end moments

for each member can be obtain from the geometry of deformed

shape.

MAB = MBA = ABL

EI2

6. ∆1

MBC = MCB = BCL

EI2

6 . ∆2

∆1 = . ∆t. LBC


MOMENT DISTRIBUTION

METHOD

∆2 = . ∆t. LBB

For the shown frame

∆1 = . ∆t. L1

∆2 = . ∆t. h1

∆3 = . ∆t. L2

∆4 = . ∆t. (L1+L2)

b- Non uniform rise in tempre. For beams:

The effect can be divided into two stages;


MOMENT DISTRIBUTION

METHOD

1-Elongation at centre line ∆ = . )2

21(

tt . L

The effect of this elongation can be solved as above in

uniform rise in temperature.

2- Rotation of member for member for free member:

d = EI

dLM .

= dLh

tt.

12


MOMENT DISTRIBUTION

METHOD

M = EI h

tt 12

For fixed member AB the restrain of rotation at A and B to

satisfy the fixations create a fixed end moment M equal to;

MAB = + EI. h

tt 12

.

and the bending moment diagram a shown;

IF t1 > t2 MAB = - EI. h

tt 2.

1

and the B.M.D will be as shown;

c- Temperature Effect for Frames with Side sway:

The solution of this type consist of two steps, first step

solve the problem without side – sway, and get Fo. Hence, the

second step; as before, solve the due to sway and get F1.

M = Mo + 1

0

F

F. M1

Example 34:

For the shown frame draw B.M.D due to rise in temp.

∆t = 20O. , EI = 20000 t.m

2


MOMENT DISTRIBUTION

METHOD

1- Without Side sway

∆1 = . ∆t. (6+10)

∆2 = . ∆t. (10)

∆3 = . ∆t. (6-4)

then get fixed end moment, MO, and F0.

2- Side–Sway, assume the fixed end moment (as before) and

get M1, and F1.

M Final = M0 + 11

1M

F

F

Example 35:

Sketch the elastic Line for the following frames due to uniform

rise in temperature. Sketch B.M.D.

a- Uniform rise in whole frame.


MOMENT DISTRIBUTION

METHOD

b- Uniform rise in A B C D E only.

c- Non uniform rise ∆t = 200 frame in t2 – t1 = 20

0


MOMENT DISTRIBUTION

METHOD

Example 36 :

Draw B.M. Diagram due to rise in temperature as shown in

figure.


MOMENT DISTRIBUTION

METHOD

∆1 = 10–5

2

4020 (4.0) = 1.2 × 10

–3 m

∆2 = 10–5

2

4030 (5.0) = 1.75 × 10

–3 m

∆3 = 10–5

2

3020 (5.0) = 1.25 × 10

–3 m

∆4 = 10–5

2

3020 (10.0) = 2.5 × 10

–3 m

∆5 = ∆4 + 10–5

2

4020 (10.0) = 5.5 × 10

–3 m

MAD = 24

)10000(6 (1.2 × 10

–3) + 10

–5

8.0

2040 (10000) = 2 m.t

MDA = 24

)10000(6 (1.2 × 10

–3) + 10

–5

8.0

2040 (10000) = 7 m.t

MEB=25

)10000(3 (2.5 × 10

–3) - 1.5×0

–5

8.0

2040 (10000) = 1.125m.t

MFC = - 1.5 × 10–5

) - 8.0

2030 (10000) = 1.125m.t

MDE = + 6210

)15000((1.75×10

–3)- 10

–5

0.1

2040 (15000) = -2.505 m.t

MED = 6210

)15000((1.75×10

–3)-1.2×10

–5

0.1

2040 (15000)=-3.495 m.t

MEF = 62)10(

)10000(6(1.75×10

–3)-1.25×10

–3

0.1

2030 (15000)=-1.59 m.t

MFE=62)10(

)10000((1.75×10

–3-1.25×10

–3)+10

–5

0.1

2030 (15000)=-1.05 m.t


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD

Frames with Several Degrees of Freedom of Sways:

in all cases of multi – story building frames, vierendeel

girders and any structures, sway take place at more than one

level. For these cases the final solution may be obtained by

superimposing the results of the following cases:

1- Case 1 of no sways get Mo.D and F10, F20, ….. etc F1n.

Where F10,… Dn0 are the horizontal (or vertical) external

force necessary to act at the storey levels as shown in the

following figure ( case of two level of sway or two degree

of freedom).


MOMENT DISTRIBUTION

METHOD

2- Case 2 of an arbitrary sway in the first storey case of

M1.D and calculate the horizontal forces at levels (1) and

(2) F11 and F21

3- Case 3 of an arbitrary sway in the second storey (sway2)

as shown in the following figure.


MOMENT DISTRIBUTION

METHOD

For this case calculate F12 and F22 from the equation

express the condition that the sum of the horizontal forces at each

level mist be zero.

Now at each level Σx = 0, then:

F10 + X1 F11 + X2 F12 = 0 )

) Get X1 and X2

F20 + X1 F21 + X2 F22 = 0 )

Where X1 is the ratio of the actual Sway at level (1) to the

arbitrary sway in case 2 while X2 is the ratio of the actual sway at

level (2) to arbitrary sway in case(3).

The final bending moment can be obtained from the

following equation:

M = Mo + X1 . M1 + X2. M2

In cases of n degrees of freedom , it is generally required n

sway corrections. For example, the following figure shows case

of three degrees of freedom.


MOMENT DISTRIBUTION

METHOD

For this case:

F10 + X1.F11 + X2 F12 + X3 F13 = 0

F20 + X1.F21 + X2 F22 + X3 F23 = 0

F30 + X1.F31 + X2 F32 + X3 F33 = 0

From these equations the values of X1 , X2 , can be

obtained, then, th final moments can be computed as follow.


MOMENT DISTRIBUTION

METHOD

M = Mo + X1.M1 + X2.M2 + X3.M3.

Example 37:

For the given frame Draw B.M.D. due to given loading.

Solution:

The frame has two degree of freedom in translation as

follow:

1- Translation of level B E G

2- Translation of level C F

The solution is consist of three cases:

a- Case (1): Fine the B.M.D. (Mo.D) due to given lods

without sway.


MOMENT DISTRIBUTION

METHOD

b- Case (2): Find the B.M.D. (M1.D) due to sway of

level B E G, while level C F is prevented from

sway.

c- Case (3): Find the B.M.D. (M2.D) due to sway of

level C F while level B E G is prevented from sway.

1-Stiffness and D.F.

Joint B: BA : BC : BE

S 6

1 :

4

1 :

12

2

D.F. .29 : .42 : .29

Joint C CB : CF

S 4

1 :

12

2

D.F. 0.60 : 0.40

Joint E EB : ED : EG : EF

S 6

1 :

6

1 :

6

1 :

4

1

D.F .4 : .6

Joint G GE : GH

S 6

1

6

1

D.F .5 : .5


MOMENT DISTRIBUTION

METHOD

2- case (1): Assume that the frame is prevented from side sway

with the two forces F10 and F20 at the joint G and F. then get the

values of F10 and F20.

Fixed End Moments:

MBE = -15, MEB = + 15

MCF = - 24, MFC = + 24

MEG = - 48, MGE = + 48

The moment distribution is shown in the following tables


MOMENT DISTRIBUTION

METHOD

Case 1 (Mo. D)

Joint C B E G F

Member CF CB BC BA BE EF ED ED EG GE GH FE FC

D.F

F.E.M.

D.M.

C.O.M.

D.M.

C.O.M

D.M.

C.O.M

D.M.

.40

-24

9.6

-4.8

.66

-2.08

1.74

-.65

.46

.60

0

14.4

3.15

.99

-2.27

2.61

-.49

.68

.42

0

6.3

7.2

-4.55

.5

-.98

1.30

-.80

.29

0

4.35

0

-3.14

0

-.70

0

-.56

.29

-15

4.35

363

-3.14

1.88

-.70

.62

-.56

.22

15

7.26

2.17

3.75

-1.57

1.23

-.35

.39

.34

0

11.22

-7.2

5.78

-3.12

1.90

-.97

.62

.22

0

7.26

0

3.75

0

1.23

0

.39

.22

-48

7.26

-12

3.75

-.90

1.23

-.47

.39

.5

48

-24

3.63

-1.815

1.88

-.49

0.62

-.31

.5

0

-24

0

1.815

0

-.94

0

-.31

0.6

0

-14.4

5.61

-6.25

2.89

-1.93

.95

-1.09

0.4

24

-9.6

4.8

-4.16

.33

1.29

.87

-.73

F.M. -19.07 19.07 8.97 -.05 -8.92 27.88 8.23 12.63 -48.74 27.0 -27.0 -14.22 14.22

Notes: F.M C.O.M.

MAB = 2

MBA +

2

56. = -.30 t.m


MOMENT DISTRIBUTION

METHOD

MDE = 2

63.12 +

2

39. = 6.51 t.m

MHG = 2

27 +

2

31. = -13.67 t.m


MOMENT DISTRIBUTION

METHOD

From Mo. D, we can calculate the F10 and F20 as follows:

From Σx at Level C F:

3-Sway Correction:

Case (2) the displacement ∆1 is due to acting of F11 a point H.

Fixed End Moments:

Assume EI ∆1 = 48

MAB = MAB = 26

6EI.∆1 =

6

1EI ∆1 = 8


MOMENT DISTRIBUTION

METHOD

MBC = MBC = - 24

6EI ∆1 = -

16

)48(6 = - 18

MEF = MEF = MBC – MCB = - 18

MED = MDE = MGH = MH6 = 6

1EI ∆1 = 8

The moment distribution of Case 2 is according to the

following table:


MOMENT DISTRIBUTION

METHOD

Case 2 (M1.D)

Joint C B E G F

Member CF CB BC BA BE EB EF ED EG GE GH FE FC

D.F

F.E.M.

D.M.

C.O.M.

D.M.

C.O.M

D.M.

0.4

0

7.2

3.6

-2.28

-1.08

-.98

.6

-18

10.8

2.1

-3.42

-1.365

1.47

.42

-18

4.2

5.4

-2.73

-1.71

0.94

.29

8

2.90

0

-1.89

0

0.65

.29

0

2.90

1.1

-1.89

-0.53

0.65

.22

0

2.2

1.45

-1.07

-0.94

0.62

.34

-18

3.4

5.40

-1.65

-1.59

0.96

.22

8

2.2

0

-1.07

0

0.62

.22

0

2.2

-2

-1.07

-0.28

0.62

.5

0

-4

1.1

-0.55

-0.54

0.27

.5

8

-4

0

-0.55

0

0.27

.60

-18

10.8

1.7

-3.18

-0.82

1.18

.4

0

7.20

3.6

-2.12

-1.14

0.78

F.M. 8.42 -8.42 -11.90 9.66 2.23 2.26 -11.48 9.75 -.053 3.72 3.72 -8.32 8.32

F.E.M. C.O.M.

MAB = 8 + 2

65.089.19.2 = 8.83 t.m

MDE = 8 + 2

62.007.12.2 = 8.88 t.m


MOMENT DISTRIBUTION

METHOD

MHG = 8 + 2

27.055.04 = 5.86 t.m


MOMENT DISTRIBUTION

METHOD

To get f12 and f11 from F12 and f11 from Σx = 0 at levels of and

DEH


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD

Case (3)

The displacement ∆2 is due to acting of F22 at Joint F

Fixed End Moment

Assume EI ∆2 = 48

MBC = MCB = MEF = MFE = 24

6EI 2 = -

8

3EI ∆2 = -18 t.m

The moment distribution of case (3) is according to the following

table:

The horizontal forces F22 and F12 can be obtained as follows:

Σx = 0 at level CF

F22 = - 8.7

F12 = -10.76


MOMENT DISTRIBUTION

METHOD

Case (3) (M2.D)

Joint C B E G F

Member CF CB BC BA BE EB EF ED EG GE GH FE FC

D.F

F.E.M.

D.M.

C.O.M.

D.M.

C.O.M

D.M.

.4

0

7.2

3.6

-2.95

-1.26

1.12

.6

-18

10.8

3.78

-4.43

-1.55

1.69

.42

-18

7.56

5.4

-3.10

-2.21

1.27

0.29

0

5.22

0

-2.14

0

.88

.29

0

5.22

1.98

-2.14

-.83

.88

.22

0

3.96

2.61

-1.76

-1.07

.7

.34

-18

5.4

5.4

-2.73

-1.84

.81

.22

0

3.96

0

-1.76

0

.7

.22

0

3.96

0

-1.76

-.25

.7

.5

0

0

1.98

-.5

-.88

.44

.5

0

0

0

-.5

0

.44

.6

-18

10.8

2.7

-3.78

-1.36

1.69

.4

0

7.2

3.6

-2.52

-1.45

1.12

F.M. 7.71 -7.71 -9.1 3.96 5.14 4.84 -10.40 2.90 2.66 1.04 1.04 -7.95 7.95

MAB = 2

96.3 = 1.92 t.m

MDE = 2

90.2 = 1.45 t.m

MHG = 2

04.1 = 0.52 t.m


MOMENT DISTRIBUTION

METHOD

From the previous 3 cases we get at the two levels X = O as follows:

F10 + X1 F11 + X2 F12 = 0 (at level BEG)

F20 + X1 F21 + X2 F22 = 0 (at level CF)

6.16 + X1 (17.815) + X2 = (- 10.03) = 0

7.0 + X1 (10.033) + X2 (- 8.07) = 0

0.61 – 1.776 X1 + X2 = 0

1.41 – 2.929 X1 = 0

X1 = 0.48

X2 = 0.246

The Final Moment MF = M0 + X1.M1 + X2.M2

MF = M0 + 0.48M1 + 0.246M2

MAB = - .30 + 0.48 (8.83) + 0.246 (1.92) = 4.425 t.m

MBA = - .05 + 0.48 (-9.66) + 0.246 (3.96) = 5.56 t.m

MBC = + 8.97 + 0.48 (-11.90) + 0.246 (-9.1) = 1.0194 t.m

MCB = + 19.07 + 0.48 (-8.42) + 0.246 (-7.71) = 13.13 t.m

MBE = - 8.92 + 0.48 (-2.33) + 0.246 (5.14) = - 6.58 t.m

MEB =+27.88 - 0.48(-2.26)+0.246(4.84) = + 30.155 t.m

MFC = + 14.22 - 0.48(8.32) + 0.246(7.95) = 20.17 t.m

MDE = 6.51 - 0.48 (8.88) + 0.246(7.95) = 20.17 t.m

MED = 12.63 - 0.48 (9.75) + 0.246 (2.90) = 18.02 t.m

MEF =+8.23 - 0.48 (-11.48) + 0.246 (-10.40) = 0.161 t.m

MEG =- 48.74 - 0.48 (-0.53) + 0.246 (2.66) = -48.34 t.m

MGE = 27.0 - 0.48 (-3.72) + 0.246 (1.04) = 25.33 t.m

MHG =- 13.67 - 0.48 (5.86) + 0.246 (0.52) = -10.7 t.m


MOMENT DISTRIBUTION

METHOD

Frame with n – degree of freedom of Side sway:

Boundary condition:

The sum of the forces at each level will equal to zero, hence;

F10 + X1.F11 + X2F12 + …. + XnF1n = 0


MOMENT DISTRIBUTION

METHOD

F20 + X1.F21 + X2F22 + …. + XnF2n = 0

: :

Fn0 + X1Fn1 + X2Fn2 + …. + XnFnn = 0

Solving the above equations to et , X1, X2, …… Xn.

Hence the final bending moment is

M = MO + X1. M1 + X2M2 + …. + XnMn.

Influence Lines by Moment Distribution:

The principle of unit displacement is valid to all stru – ctures either

determinate or indeterminate. The influence lines for inderminate beams

and frames can be solved very easy by the method of moment

distribution. In case of internal forces; M,Q,N; we apply by a unit internal

relative displacement.While for external forces; as reactions, we apply a

unit external displacement. Considering, as an example. The shown fixed

beam;

a- To draw I.L.MA

1- Produce a unit rotation = 1 rad. At A and direction of MA

(-ve moment), keeping all other joints fixes.


MOMENT DISTRIBUTION

METHOD

2- Obtain fixed end moment at A.

at A

MA = L

EI4.1

And at B

MB = L

EI2.1

Hence the bending moment diagram can be obtained for this

simple case, and the B.M.D will be used to determine the elastic curve by

using the conjugate beam method. Similarly the I.L.MB can be drawn.

The influence line of any of the internal forces at any section can be

obtained in terms of I.L.MA & I.L.MB.


MOMENT DISTRIBUTION

METHOD

Case of Continuous Beam:

To draw the influence line of bending moment at B, as an example,

for the given continuous beam.


MOMENT DISTRIBUTION

METHOD

procedures (For beams or Frames)

Produce a unit relative rotation at B just to right of B as shown in

the previous fig. , and keeping all other joints fixed. The obtained

moment will be F.E.Ms; as follows;

MBC = L

4EI

MCB =

bcL

2EI

Notice that the unit relative rotation can be produce just to left of

B, in which case F.E.Ms will be.

MBA =

BAL

3EI

The unit relative rotation can also be divided in any arbitrary

suitable ratio between BA, to the levt; and BC; to the right;

2- Get the final B.M.D for all the beam by moment distribution

method as shown in the previous fig (b).

3- To obtain the elastic line, we use the conjugate beam method as

shown in fig . (c), then draw the elastic curve.

4- The above procedures are used in case of frames without sway. In

case of frames having side sway we must obtain the final B.M.D

due to assumed init displacement after making the sway

correction.


MOMENT DISTRIBUTION

METHOD

5- To obtain I.L.XA produce a unit horizontal translation at support A in

the direction of XA (assumed). Support A translate without rotation,

while all other joints and supports will remain fixed ; (except support c

wil remain always hinged;). The developing F.E.Ms will be.

Mad = Mda = + ad2L

6EI

Similarly , complete the above procedures.

5- To obtain I.L.YA in the shown frame ,

produce a unit vertical upward translation

at support A without rotation.

This causes a unit translation upward at D and G, in the some time.

For this case F.E.Ms will be;

Mgh = Mhg = gh2L

6EI.1

Similarly;

Mgh = Mhg = de2L

6EI.1

7- To obtain I.L. for N.G.M. at any section;

I.L.for normal force N, required to produce a unit relative axial;

horizontal or vertical; translation. Due to this unit translation at a certain

joint calculate the dived end moment. All other joint are kept fixed. For

an example,

I.L.NDE,


MOMENT DISTRIBUTION

METHOD

Produce unit horizontal translation at D to right , then, F.E.Ms as

follows;

Mad = Mda = 12L

6EI.1

Mdg = Mgd = s2L

6EI.1

Other method by produce a unit translation at E to right , then joint

F translate unit displacement also , while D is fixed.

I.L.Q

The influence line for shearing force at any section ( say) I.L.Qn,

can be obtained by produce a unit relative vertical translation at the

section. Then calculate the F.E.Ms. notice that at a section a unit vertical

drop in the deflection line.

I.L.M

Produce a unit relative rotation at the section.


MOMENT DISTRIBUTION

METHOD

Example 1:

For the shown continuous beam Draw I.L. for the followings;

1- I.L.MB

2- I.L.Mn

3- I.L.YA

4- I.L.YB

5- I.L.Qn

Solution

1- I.L.MB

a- Produce a unit angle of rotation at BA Just to left of B.

F.E.MB4

assume EI = 1

F.E.Ms = 0.25 Multiplied

by 100 equals to 25 use the

conjufate beam method to draw

elastic curve as follow;

θ A = rad(100)EI

EI30

= 100

30= 0.3 RAD

= - 1.067


MOMENT DISTRIBUTION

METHOD

y2 = (-30 × 8 + 38.40 )

100

1

= - 1.33

θ Br = 10040

= 0.40 rad

θ c = 100

20= 0.20 rad

y3 = (- 0.20×4 + 15× 5

4) =

100

1

The I.L.MB can be obtained as shown;

Nots:

YA = MB , Mn = YA (6)

I.L.YA = I.L.YA0 + I.L. 12

MB

I.L.Mn = I.L.Mn0 + 2

1I.LMB

I.L.YB = I.L.YB0 + 12

1I.LMB

For beam BA and =

I.L.YBO -8

1 I.L.MB for beam Bc.

Where I.L.YAO is the influence line of YA at Simple beam AB and

I.Lmn0 is the influence line for Mn at simple beam Ab as follows;

I.L.Mn

The influence line of Mn can be obtained by knowing I.L.YB and

I.L.MnO as follows.


MOMENT DISTRIBUTION

METHOD

I.L.Mn = I.L.Mn0 + 2

1I.LMB

I.L.YA = I.L.YA0 + 12

1I.LMB

Similarly:


MOMENT DISTRIBUTION

METHOD

Example 2:

For the given frame, draw the influence lines for;

1- I.L.YA

2- I.L.YB

3- I.L.Me-d

4- I.L.Qn

5- I.L.Md

6- I.L.Mn

Solution:

Solution

Producing a unit vertical

Translation at support A.

MFD. =MFED = -6EI / 100

Put EI 1000


MOMENT DISTRIBUTION

METHOD

MFDE = MFED = - 60

1000y1 = - 94.91(2.5)

+ (2/6) (8.295)

(2.5)2 + (2.5)

2/2

(24.89)

= - 39.19

1000y2 = - 94.91(5) -

(18.12/6) (5)2+

(2/6) 16.59* (5)2+

+ (52/2)16.59

= 20.20

1000y3 = - 94.91(7.5) -

(27.31/6) (7.5)2+

(2/6)(24.89)(7.5)2

(7.52/2)(8.295)

= 69.63

1000y4 = 33.92(2.5) +2.52/6

(1.70) - (2/6) (2.5)2

(3.39)-(2.52/2)*10.18

= 47.70

1000y5 = 33.92(5) + (52/6)*3.39

(2/6)*52 (6.79)-(5

2/2)

(6.79)

= 42.27

1000y6 = 33.92(7.5) +7.52/6

(5.09)- (2/6)*7.52

(10.18)-(7.52/2)3.39

= 15.90

I. L. YB

MF = 6EI / L2

EI = 1000

MF = + 60

1000y1 = 43.82(2.5) + (2.52)/6

(12.94)- (2/6)*2.52

(9.76)-(2.52/2)*29.27

=11.23

1000y2 = 43.82(5) +52/6


MOMENT DISTRIBUTION

METHOD

(25.88)- (2/6)*52

(19.51)- (52/2)*19.51

- (52/2)*19.51

= -79.53

1000y3 = 43.82(7.5) +7.52/6

(38.82)- (2/6)7.52

(29.27)-7.52/2

(9.76)

= -130.73

1000y4 = - 92.40(2.5) -2.52/6

(14.62)+ (2/6)14.24

(2.5)2+2.5

2/2

(42.72)

= -83.06

1000y5 = 92.40(5) -52/6

(29.24)+ (2/6)2 *8.48

(5)2+ (5

2/2)*28.48

=9.50

1000y6 = - 92.40(7.5) -7.52/6

(43.86)+ (2/6)*42.72

(7.5)2+ (7.5

2/2)*14.24

= 97.31

I. L. Y E-D

MED = 4EI / L

MDE = 2EI / L

EI = 100

MED = 60

MDE = 20

100y1 = - 9.63(2.5) -2/6

(2.41)(2.5)2

-7.22(2.52/2)

6.87(2.52/6)

= -44.4812

100y2 = - 9.63(5) -2/6

(4.82)(5)2

-4.82(52/2)

+13.74(52/6)

= -91.316


MOMENT DISTRIBUTION

METHOD

100y3 = - 9.63(7.5) – (3/6)*7.22

(7.5)2-2.41(7.5

2/2)

+ (20.61/6)*7.52

= -82.1625

100y4 = -24.47(2.5)-(1.25/6)*

(2.5)2+ (2/6)*2.45

(2.5)2+ (2.5

2/2)

(7.34)

= -34.435

100y5 = - 24.47(5) – (2.45/6)*

(5)2+4.9(2/6) (5)

2

+4.9(52/2)

= 31.475

100y6 = - 24.47(7.5) -3.68/6

(7.5)2+7.34(2/6) (7.5)

2

+2.45(7.5)2/2

= -11.4937

I. L. Q N

From n-c

Qn = ya

I. L.Qn = L.L.ya

From d-n

Qn = ya – 1

I.L.Qn = I.L.ya – 1

I. L. d-e

MDE = -4EI / L

MED = -2EI / L

EI = 100

MDE = -40

MED = -20

100y1 = - 63.73(2.5) -

(2.38/6)*2.52+

2/6(5.96) (2.5)2

+17.90(2.5)2/2

= -93.458


MOMENT DISTRIBUTION

METHOD

1000y2 = - 63.7333(5) -

(4.75/6)*52+11.94

(2/6)*52+11.94

+ (5)2/2

= -89.705

100y3 = -63.733(7.5) – (7.13/6)

(7.5)2+17.90(2/6)

(7.5)+(5.97/2)

(7.5)2

= -41.310

100y4 = 8.88(2.5) + (0.42/6)

(2.5)2-(2/6)*1.14

(2.5)2-3.42(2.5)

2/2

= 9.575

100y5 = 8.88(5) + (.85/6)*

(5)2-2.28(2/6) (5)

2

-2.28(52/2)

= 0.442

100y6 = 8.88(7.5) (1.26/6)

(7.5)2-(1.14/2)(5)

2

= 0.038


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD

Problems

1- Using Moment distribution Method to draw B.M.D. and S.F.D. for

the following indeterminate beams due to given lodes, settlement,

and rise in temperature sketch elastic curve.

2- Draw B.M.D., S.F.D., and N.F.D. for the following frames, sketch

elastic curve.

3- Draw B.M.D., S.F.D., N.F.D. for the following frames due to given

loads and settlements. Sketch the elastic curve for each case.


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD

4- Draw B.M., S.F. and N.F.Ds for the following frames.

4- Draw B.M.D for the following structure due to shown loads

and rise in temperature. Sketch the elastic curve for each

case.


MOMENT DISTRIBUTION

METHOD

5- Draw B.M.D. for the following structures due to given loads and

sketch the elastic curve.


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD


MOMENT DISTRIBUTION

METHOD

8 Approximate Analysis of

Statically

Indeterminate Structures

1. Introduction

In this chapter we will present some of the approximate methods

used to analyze statically indeterminate trusses and frames. These

methods were developed on the basis of structural behavior and their

accuracy in most cases compares favorably with more exact methods of

analysis. Although not all types of structural forms will be discussed here,

it is hoped that enough insight is gained from the study of these methods

so that one can judge what would be the best approximations to make

when performing an approximate force analysis of a statically

indeterminate structure.

If a model used to represent a structure is statically indeterminate,

then the analysis of it must satisfy both the conditions of equilibrium and

compatibility of displacement at the joints. As will be shown in later

chapters of this text, the compatibility conditions can be related to the

loads, provided we know the material's modulus of elasticity and the size

and shape of the members. For design, however, we will not know a

member's size, and so an analysis that provides further simplifying

assumptions for modeling the structure must be made. This analysis is

called an approximate analysis, and throughout this chapter, we will use it

to simplify the model of a statically indeterminate structure to one that is

statically determinate. By performing an approximate analysis, a

preliminary design of the members of a structure can be made, and once

this is complete, the more exact indeterminate analysis can then be

543

Approximate

performed and the design refined. An approximate analysis also provides

insight as to a structure's behavior under load and is beneficial when

checking an exact analysis, or when time, money, or capabilities are not

available for performing a more exact analysis.

Realize that in a general sense, all methods of structural analysis are

approximate, simply because the conditions of loading, geometry, material

behavior, and joint resistance at the supports are never known exactly,

2. Trusses

A common type of truss often used for lateral bracing of a building or

for the top and bottom cords of a bridge is shown in Fig. l-a. When used for

this purpose, this truss is not considered a primary element for the support

of the structure, and as a result it is often analyzed by approximate methods.

In the case shown, it will be noticed that if a diagonal is removed from each

of the three panels, it will render the truss statically determinate. Hence, the

truss is statically indeterminate to the third degree,* and therefore we must

make three assumptions regarding the bar forces in order to reduce the truss

to one that is statically determinate. These assumptions can be made with

regard to the cross diagonals,

Realizing that when one diagonal in a panel is in tension the

P0P0a

b

R1 R2R1

R1V=

FIG.1 (a) , (b)

F1

FaFb

F2

544

Approximate

responding cross diagonal will be in compression. This is evident from i.g.

7-6, where the "panel shear" V is carried by the vertical component of

tensile force in member (a) and the vertical component of compressive

force in member b. Two methods of analysis are generally acceptable.

Method 1: If the diagonals are intentionally designed to

be long and sender, it is reasonable to assume

that they cannot support a compressive force,

otherwise, they may easily buckle the tension

diagonal, whereas the compressive diagonal is

assumed to be a zero-force member.

Method 2: If the diagonal members are intended to be constructed from

large rolled sections such as angles or channels, they may be

equally capable of supporting a tensile and compressive

force. Hence, we can assume that the tension and

compression diagonals each carry half the panel shear.

Both of these methods of approximate analysis are illustrated

numerically in the following examples.

Example 1:

Determine (approximately) the forces in the members of the truss

shown in Fig. Assume the diagonals are slender and there force will not

support a compressive force.

545

Approximate

2t2t2t

3t3t

1 3 5 7 9

642 BA

4*3=12m

Fig.(2-a)

3m

Solution

Reactions;

From symmetry yA = yB = 3 t

The truss is statically to the forth degree thus we can assume the

compression diagonals sustains zero forces, i.e. Can be omitted from the

truss and the remaining truss can easily analyzed.

The zero members are (A-2 , 4-5, 6-13} The remaining member

forces are easily found by method of joints, and are shown in Fig. 2-a-b

2t2t2t

3t3t

0 -3t4.243-1t1.4141.4141t

4.243t

-3t

-3t1 3 -4t 5 -4t 7 -3t 9

642 BA

4*3=12m

Fig.(2-b)

3m

546

Approximate

Example 2:

Determine (approximately ) the forces m the members of the truss

shown in Fig. 3-a. the diagonals are to be designed to support both

tension and compression forces, and there fore each is assumed to

carry half the panel shear. The support reactions have been computed.

5t10t

F E D

CBA3m3m

4m

5t

Fig.(3-a)

10t

Solution

By inspection the truss is statically indeterminate to the second

degree. The two assumptions require the tensile and compressive

diagonals to carry equal forces, that is , FFB=FAE = F.for vertical

section through the left panel, Fig.3-b we have

Fig.(3-c)Fig.(3-b)

F

F

F

F

D

FCD

C

ED

DB

EC

BC

=F

=F

F

F

F

F

AE

FB

AE

AB

=F

=F

F

F

A

AF

5t

10t

0

0

0

0

5t

547

Approximate

0fY )5/4(F2510

t125.3F)5

8(F5

So that

FFB )T(125.3

FAE )C(125.3

At joint F

0FΧ FFE sin125.3

)C(275.1)5

3(125.3

at joint A

F X= 0 FAB sinFAE

)T(87.1)5

3(125.3

F Y=0 )T(5.7)5

4(125.35FAF

A vertical section throught the right panel is shown in Fig. 3-c

F y=0

5)5

4(F2

FDB )T(125.3

FDB )C(125.3

548

Approximate

At joints D

F X=0 FDB = )C(t875.1)5

3(125.3

F y=0 FCD = )C(t5.2)5

4(125.3

At joint C

F X=0 FBC = 3.125 ( )T(t875.1)5

3

3. Building Frames Subjected to Vertical Loads

Building frames often consist of girders that are rigidly connected to

columns so that the entire structure is better able to resist the effects of

lateral forces due to wind and earthquake. An example of such a rigid

framework, often called a building bent, is shown in Fig. 4. In this section

we will establish a method for analyzing (approximately) the forces in

building frames due to vertical loads, and in Sees. 5 and 6 an approximate

analysis for frames subjected to lateral loads will be presented. In all these

cases it should be noticed that most of the simplifying assumptions made to

reduce a frame from a statically indeterminate structure to one that is

statically determinate are based on the way the structure deforms under load.

3.1 Assumptions for Approximate Analysis

Consider a typical girder located within a building bent and

subjected to a uniform vertical load, as shown in Fig. 5-a. The column

supports at A and B will each exert three reactions on the girder, and

therefore the girder will be statically indeterminate to the third degree.

549

Approximate

To make the girder statically determinate, an approximate analysis

will therefore require three assumptions. If the columns are extremely

stiff, no

Fig.4 Typical Building Frame

L

W

(c)

W

0.21L0.21L

LL

momentsPoints of zero Girder

W stiffstiffcolumn column

(a)(b)

A B

550

Approximate

rotation at A and B will occur, and the deflection curve for the girder will

look like that shown in Fig. 5-b. Using one of the known methods an

exact analysis reveals that for this case inflection points, or zero moment

occur at 0.211 from each support. If, however, the column connections

at A and B are very flexible, then like a simply supported beam, zero

moment will occur at the supports, Fig. 5-c. In reality, however, the

columns will provide some flexibility at the supports, and therefore we

will assume that zero moment occurs at the average point between the

above two extremes, i .e., at (0.211+0)/2= 0.11 from each support, Fig.

5-d. Furthermore, an exact analysis of frames supporting vertical loads

indicate- that the axial forces in the girder are negligible.

As a result of the above discussion, each girder of length/may

be modeled by a simply-supported span of length 0.81 resting on

two cantilevered ends, each having a length of 0.11, Fig. 5-e. The i

olio wing three assumptions are incorporated in this model:

1 - There is zero moment in the girder,o. 1 Lfrom the left support.

0.8L

L

momentsPoints of zero Assume

WW

0.1L0.1L

0.1L0.1L

Approximate case Model

A B

(e)(d)

551

Approximate

There is zero moment in the girder,o. 1 Lfrom the right support.

The girder does not support an axial force.

By using static's, the internal loadings in the girders can now

be obtained. The following example illustrates this numerically.

Example 3

Determine (approximately) and exactly the bending moment

diagram for the following frame.

12m

W

DC

A B

Fig. (6-a)

552

Approximate

Fig. (6-c)

Approximate Solution

Assume intermediate hinges as

Shown in Fig.6-b

1.29.61.2

FE

Fig. (6-b)

553

Approximate

19.419.4

19.4

9.72 9.72

19.4

34.56

+

-

-

-

-

Fig. (6-d) 4.

Portal Frames

4.1. Frames

Portal frames are frequently used over the entrance of a bridge and

as a main element in building design in order to transfer horizontal forces

applied at their top joints to the foundation. On bridges, these forces are

caused by wind, earthquake, and unbalanced traffic loading on the bridge

deck. Portals can be pin-supported, fixed-supported, or supported by

partial fixity. The approximate analysis of each case will now be discussed

for a simple three-member portal,

4.2. Pin-Supported

A pin-supported type of portal consists of pin-supported vertical

members having equal length and size and a rigidly connected horizontal

girder, Fig. 7-a. Since four unknowns exist at the supports but only three

equilibrium equations are available for solution, this structure is statically

indeterminate to the first degree. Consequently, only one assumption must

be made to reduce the frame to one that is statically determinate.

The elastic deflection of the portal is shown in Fig. 7-b. This

554

Approximate

diagram indicates that a point of inflection, that is where the moment

changes from positive bending to negative bending, is located

approximately at the girder's midpoint. Since the moment in the girder is

zero at this point, we can assume a hinge exists there and then proceed to

determine the reactions at the supports using statics. If this is done, it is

found that the horizontal reactions at the base of each column are equal

and the other reactions are those indicated in Fig. 7-c. Furthermore , the

moment diagrams for this frame are indicated in Fig. 7-d.

h

L

h

assumed hinge

PP

Fig. (7-b)Fig. (7-a)

hh

L/2 L/2

Ph/2

Ph/2Ph/2

P/2P/2

P/2P/2P

Fig. (7-c)

555

Approximate

Ph/2

Ph/2

Ph/2

Ph/2

B. M. D

+

+

-

-

Fig. (7-c)

4.3. Fixed-Supported

Portals with two fixed supports, Fig. 8-a are statically

indeterminate to the third degree since there is a total of six unknown at

the supports. If the vertical members have equal lengths and cross

sectional areas, the frame will deflect as shown in Fig. 8-b. For this

case we will assume points of inflection occur at the midpoints of all

three members, and therefore hinges are placed at these points. The

reactions and moment diagrams for each member can therefore be

determined by dismembering the frame at the hinges and applying the

equations of equilibrium to each of the four parts. The results are shown

in Fig. 8-c. Note that, as in the case of the pin connected portal, the

horizontal reactions at the base of each column are equal. The moment

diagrams for this frame are indicated in Fig. 8-d.

4.4. Partial Fixity

556

Approximate

Since it is both difficult and costly to construct a perfectly fixed

support or foundation for a portal frame, it is conservative and somewhat

realistic to assume a slight rotation occurs at the supports, Fig. 9-a. As a

result, the points of inflection on the columns lie somewhere between the

case of having a pin supported portal, Fig. 7-a9 where the "inflection

points" are at the supports (base of columns), and a fixed supported portal,

Fig. 8-a, where the inflection points are at the center of the columns. Many

engineers arbitrarily define the location at h/3, Fig. 9, and therefore place

hinges at these points, and also at the center of the girder. Fig. 9-b shows

B.M.D.

h

L

h

assumed hinge

PP

Fig. (7-b) Fig. (8-a)

557

Approximate

h/2

L/2 L/2

Ph/2L

P/2

P/2

P/2P

Ph/2L

Ph/2LPh/2L

P/2

Ph/2L

P/2

Ph/2L

P/2

Ph/2L

Ph/4

P/2

Ph/4

Ph/2L

h/2

Fig. (7-c)

Ph/4

Ph/4

Ph/4

Ph/4

B. M. D

+

+

-

-

+

-

Ph/4 Ph/4

Fig. (7-d)

558

Approximate

h

L

assumed hinge

P

h/3h/3

00

partial fixation

Fig. (9-a)

M

B. M. D

+

+

-

-

+

-

M/2 M/2

MM

M

Fig. (9-b) 4.5.

Trusses

When a portal is used to span large distances, a truss may be

used in place of the horizontal girder. Such a structure is used on large

bridges and as transverse bents auditoriums . A typical example is

shown in fig 10-a.in all cases, the suspended truss is assumed to be

pin connected at its points of attachment to the columns. Furthermore,

the truss keeps the columns straight within the region of attachment

when the portal is subjected to the side sway A, Fig. 10-b.

Consequently, we can analyze trussed portals using the same

assumptions as those used for portal frames, for pin supported columns,

assume the horizontal reactions are equal as in Fig. For fixed

559

Approximate

supported columns, assume the horizontal reactions are equal and an

inflection point (or hinge) occurs on each column, measured midway

between the base of the column and the lowest point of truss member

connection to the column. See Fig. S-c and Fig. 10-b

The following example illustrates how to determine the forces in

the members of a trussed portal using the approximate method of analysis

described above

P/2

Fig. (10-a)

L

P

Fig. (10-b)

L

P

assumed hinge

h

h/2

P/2

5. Lateral Loads on Building Frames

5.1. Portal Method

In Sec. 4 we discussed the action of lateral loads on portal frames

and found that for a frame fixed supported at its base, points of

inflection occur at approximately the center of each girder and column

and the columns carry equal shear loads, fig.8. A building bent deflects

in the same way as a portal frame, fig. 11-a, and therefore it would be

appropriate to assume inflection points occur at the center of the

columns and girders. If we consider each bay of the frame to be

composed of a series. If we consider each bay of the frame to be

composed of a series of portals, fig. 11-b, then as further assumption the

interior columns would represent the effect of two portal columns

560

Approximate

P

Fig. (11-a)

X3X2X1

=inflection point

X3=VX2/2=V

Fig. (11-b)

X2/2=VX1=V

and

would therefore carry twice the shear V as the two exterior columns . In

summary, then the portal method for analyzing fixed supported

building frames requires the following assumptions:

1. A hinge is placed at the center of each girder since

this is assumed u = he a point o* zero moment

2. A hinge is placed at the center of each column s ince

this is assumed to be a point of zero moment.

3. At a given floor level the shear at the interior column

hinges is twice that at the exterior column hinges since

the frame is considered to be a superposition of portals.

561

Approximate

Fig. (12-a)

A C E H

16.0m16.0m16.0m

120 ton B D F G

KJ

O

L

Fig. (12-b)

120 ton

KJ

V L 2V K2V J

LyKyJyIy

These assumptions provide an adequate reduction of the frame to one

that is statically determinate yet stable under loading.

By comparison with the more exact statically determinate

analysis, the portal method is most suitable for buildings having low

elevation and uniform framing. The reason for this has to do with the

structure's action under load In this regard, consider the frame as

acting l ike cantilevered beam that is fixed to the ground. Recall from

mechanics o f materials that shear resistance becomes more important in

the design of short beams whereas bedding is more important if the

beam is long (See. 6. )The portal method accounts for this effect, listed

as assumption 3 above.

The following examples illustrate how to apply the portal method

to analyze a building bent.

Example 5

Determine (approximately) the reactions at the base of the

columns of the frame shown in Fig. 13-a, use the portal method of

analysis.

Solution

Applying the two assumptions of the portal method, we place

562

Approximate

2t

12 ton

Iy=1.5t

6m

MMx=10t8m 10 ton

6m

Jy=0

Iy=1.5t

8m 8m

N

Nx=6t

Ny=1.5t

4t

Fig. (13-c) Fig. (13-d)

6t

6m

Ky=0

1.5t

8m 8m

O

Ox=2t

Ox=1.5t

4t

Fig. (13-e)

8m

6m

Ly 1.5t

2t

2t

1.5t

Fig. (13-f)

hinges at the centers of the girders and columns of the frame. The

locations of these points are indicated by the letters I through O in Fig.

13-a. A section through the column hinges at I, J. K, L yields the free

body diagram shown in Fig. 13-b. Here the third assumption regarding

the column shears. We require

;0Fx 1200 – 6V = 0 v=2t

Using this result, we can now proceed to dismember the frame at

the hinges and determine their reactions. As general rule, always start this

analysis at the corner where the horizontal load is applied. Hence, the

free-body diagram of segment IBM is shown in Fig. 13-c. The three

reaction components at the hinges Iy, Mx and My are determined by

applying ,0Fy .0Fy

The adjacent MJN is analyzed next Fig 13-d followed by segment NKO,

Fig. 13-e, finally segment OL, Fig. 13-f. Using these results, the free

body diagrams of the columns with their support reactions are shown in

Fig. 13-g.

Ax=2t

Ay=1.5t

6m

Fig. (13-g)

2 t

1.5t

Mx=12t.m

Cx=2t

6m

4 t

Mc=24t.m

Cx=2t

6m

4 t

ME=24t.m

Hx=2t

6m

4 t

MH=12t.m

1.5t

Hy

563

Approximate

B CA

20 t

30t

D E F

G H IJ S

M N

PQ

O

J K L

5m

5m

8m8m

Fig. (14-a)

20 t G H IJ S

V

PyOyQy

2VV

2.5m

V

KyJyLy

2VV

3m

5m

30t

20t

Fig. (14-b)

Example 6

Determine (approximately) the reactions at the base of the columns

of the frame shown in fig. 14-a. Use the portal method of analysis.

Solution

First hinges are placed at the centers of the girders and columns of

the frame. The locations of these points are indicated by the letters J

through S in Fig. 14-a. A section through the hinges at O, P, Q and J, K,

L yield the free body diagrams in fig 14-b . The column shears are

calculated as follows:

Using these results, we can now proceed to analyze each part the

frame. The analysis starts with the corner segment OGR, 14-c. The three

564

Approximate

4m4m

S

Sx=5t

Sy=3.125t12.5t

12.5t

12.5t

12.5t

12.5t

12.5t

12.5t

12.5t

12.5t12.5t

12.5t

12.5t

12.5t

12.5t

12.5t

12.5t

12.5t

12.5t

12.5t

12.5t

12.5t

12.5t12.5t

12.5t

12.5t

12.5t 12.5t12.5t

12.5t

12.5t

12.5t

12.5t 12.5t

12.5t12.5t

12.5t

12.5t12.5t

12.5t

12.5t

12.5t12.5t

12.5t

12.5t

12.5t12.5t

12.5t

12.5t

12.5t

12.5t

unknowns Oy. Rx and Rx have been calculated using the equations of

equilibrium. With these results segment OJM is analyzed next, fig. 14-d

then segment JA Fig. 14-e ; RPS Fig . 14-h PMNK , Fig 14-g; and KB,

Fig 14-h complete this example and analysze segments SIQ then QNL,

and finally LC, and show that Cx = 12.5 t , Cy = 15.625 t , and Mc = 37.5

t.m.

B.M.D

12.5t12.5t

12.5t12.5t

12.5t 12.5t

12.5t

50t50t

50t

37.5

37.5

7525

50t37.5

37.5 75

565

Approximate

(b)

(C)(T)

beam

(a)

M

building frame

Fig. (15.a,b)

6- Lateral Loads on Building Frames

6-1 Cantilever Method

The cantilever method assumes that the stress in a column is

proportional to its distance from the centered of all the column areas at a

given floor level. This assumption is based on the same action as a

cantilevered beam subjected to a transverse load. It may be recalled from

mechanics of materials that such a loading causes a bending stress in the

in summary , then using the cantilever method , the following

566

Approximate

assumptions apply to a fixed supported frame.

1- A hinge is place at the center of each girder sing this is

assumed to be a point of Zero moment.

2- A hinge is placed at the center of each column since this is

assumed to be a point of Zero moment.

3- The axial stress in a column is proportional to its distance

from the centroid of the cross sectional area of the columns

at a given floor level. Since stress equals per area, then in

the special beam that vanes linearly from the beam's

neutral axis, Fig. 15-a. In a similar manner the lateral loads

on a frame tend to tip the frame over, or cause a rotation of

the frame about a "neutral axis" lying in a horizontal plan

that passes through the columns at each floor level. To

counteract this tipping, the axial forces (or stress) in the

columns will be tensile on one side of the neutral axis and

compressive on the other side, Fig. 15-b. Like the

cantilevered beam, it therefore seems reasonable to assume

this axial stress has a linear variation from the centered of the

column areas or neutral axis. The cantilever method is

therefore appropriate if the frame is tall and slender, or has

columns with different cross-sectional areas. Case of the

columns having equal cross sectional areas, the force in

a column is also proportional to its distance from the

centroid of the column areas.

These three assumptions reduce the frame to one that is both stable

and statically determinate

567

Approximate

The following examples illustrate how to apply the cantilever

method to analyze a building bent.

Example 7

Determine (approximately) the reactions at the base of the columns

of the frame shown in Fig. 16-a. The columns are assumed to have equal

cross-sectional areas. Use the cantilever method of analysis.

Solution

First hinges placed at the midpoints of the columns and girders.

The locations of these points are indicated by letters G through L in Fig.

16-a.

The centroid of the columns' cross-sectional area can be determined by

inspection, Fig. 16-b, ur analytically as follows;

m3AA

)A(6)A(0

A

AXX

The axial force in each column is thus proportional to its distance

6m

3

144,46

4m

4m

15t

30tI

J

KH

LG

EB

DC

(b)

(a)

FA

568

Approximate

from this point. Hence, a section through the hinges H and K at the top

story yields the free-body diagram shown in Fig. 16-c. Note how the

column to the left of the centered must be subjected to tension and the

one of the right is subjected to compression. This is necessary in order to

counteract the tipping caused by the 30 ton force. Summing moments

about the neutral axis (NA) we have :

0M NA 0Hy3)2(30

The unknowns can be related by proportional triangles, Fig. 16-c

that is ,

3

K

3

H yy or Hy = Ky

Thus,

Hy = Ky = 10 ton

3m3m

4m

2m

3m3m

2m

30t

30t

15t

KxHx

LxGx

Hy Ky

NA

NA

(c)

(d)

Gy Ly

569

Approximate

In a similar manner, using a section of the frame through the hinges at G

and Fig. 16-d, we have

0M NA 30(6) + 15(2)-3Gy – 3 Ly =0

Since

3

L

3

G yy or Gy = Ly

Gy = Ly 35 =ton

Each part of the frame can now be analyzed using the above

results. As Examples 5 to 6 , we begin at the upper corner where the

applying loading occurs, i.e ., segment HCI, Fig 16-a . Applying of

equilibrium 0F,0M xI , 0Fy , yields the results for HX , IX

and Iy shown in the free-body diagram in fig, 16-f, followed by HJG ,

Fig. 16-g, then KJL, Fig . 16-h and bottom portions of the columns, Fig.

16-i

570

Approximate

30

30

30

30

30

45

30

45 45

45

75

30

3m

2m

Kx=15t

Ky=10t

10t

15t

10

15

3m

2.5t

7.5t

Lx=22.5t

Ly=35t

35t

22.5t

Fx=22.5t

Mf=45t.m

Fy=35tAy=35t

MA=45t.m.

Ax=22.5t

2m

22.5t

35t

Gy=35t

G=1.5tJy=25t

Jx=7.5t3m

2m

15t10t

I

Ix=15t3m

Iy=10t

30t

Hx=15t

Hy=10t

2m

2m

2m

2m

571

Approximate

Example 8

Show how to determine (approximately) the reactions at base of

the columns of the frame shown in fig h. 17-a . The columns have the

cross-sectional area shown in Fig. 17-b Use the cantilever method of

analysis.

Solution

First hinges are assumed to exist at the centers of the girders and

columns of the frame. The locations of these points are indicated by the

letters E through R in Fig. 17-a . The centroid of the columns' cross-

sectional area is determined from Fig. 17-b as follows:

X = m53.28106810

)10(60)8(20)10(0

A

AX

DCBA

QP

12m L 10in2

M 8 in

I J

N 6 in

K

O 10 in

H 10 in6 inG8in10 in16m E

(a)

8t

10t

2

2 2

22

22

25m15m20m

(b)

25m15m20m

X

10 in6 in8 in10 in2222

572

Approximate

Here the columns have different cross- sectional area , so the axial

stress in a column is proportional to its distance from the neutral axis,

located at X = 28.53m . Hence, a section through the hinges at L, M, N,

O yields the free=body diagram shown in Fig .17-c. Note how the

columns to the left of the centroid are subjected to tension and those on

the right are

Subjected to compression. Why? Summing moments about the neutral

axis, we have

;0M NA

0)47.31(O)47.6(N)53.8(M)53.28(L)6(8 yyyy (1)

8t

8t

10t

Fy=.867t

Oy

OxNx

Ny

Mx

MyLy

Lx

Hx

Hy=4.001tGy=.493tEy=.867t

Ex Fx G

6.478.53

31.47m28.53m

6m

12m

8m

31.47m28.53m

6.478.53

(C)

(d)

Fig. (17-c,d)

573

Approximate

Since any columns stress is proportional to its distance the

neutral axis, we can relate the column stresses by proportional triangles .

Expressing the relations in terms of the force Ly and take the columns

area into consideration, we have

;53.28

53.8LM

)10(

L

53.28

53.8

)8(

M yy My = 0.239LY (

;53.28

47.6LN

)10(

L

53.28

47.6

)6(

N yy Ny = 0.136Ly (3)

;LO53.28

47.31

)10(

L

53.28

47.31

)10(

O yy Oy = 1.103Ly (4)

Solving Equation (1) to (4) yields

Ly = 0.726 t My = 0.173t Ny 35t Oy =35 t

Using this same method, show that one obtains the results in Fig. 17 –d

for the columns at E, F G and H.

We can proceed to analyze each part of frame. As in the previous

examples, we begin with the upper corner segment LP, Fig. 17-e. Using

the calculated results, segment LEI is analyzed next, Fig . 17-f, followed

by segment EA, Fig. 17-g One can continue to analyze the other

segments in sequence, i.e., PQM, then MJFI, then FB, and so on.

3.628t

2.720t

Ax=2.720t

MA=21.760t.m

Ay=3.628t

Fy=3.628t

Fx=2.720t

Iy=2.902t

Ix=8.490t10m

6m

1.210t

0.726tPy=.726t

Px=6.79t10m

P

8t

Lx=1.21t

Ly=.726t

6m

8m

8m

A

(f)

(e)

(g)

Fig.(17-e.f and g)

574

Approximate

PROBLEMS

For the given statically indeterminate structures shown in figs. Draw

N.F. , S.F. & B.M.Ds . by using approximation methods.

10t 10t

5t

10t

1t/m

2t/m

3.06.06.0

6.0

6.0

3.0

4.0

4.0

6.0

8.0

4.0

(b)(a)

(c)

(d)

575

Approximate

5.0

3.0

3.0

6.0

5.05.0

3.0

3.0

4.0

6.06.06.06.0

3.0

4.0

(e)

(f)

2t/

m1t/

m2t/

m1t/

m

10t

10t

10t

5t

10t

10t

15t

576

Approximate

6.0

6.0

3.0

3.0

6.06.0

3.0

3.0

4.0

3.0

3.0

4.0

6.0

(g)

(h)

10t

10t

10t

10t

5t

10t

10t

15t

REFERENCES

Abdel-Rohman, M. “structural Analysis – A transition from

classical to Matrix methods”, Publishing House for University,

second edition, 1998.

Bazaraa & A.Akl “structural Analysis II”, First edition, 1999.

R.C. Hibbeler, “structural Analysis”, second edition, Macmillan

publishing company, New York, 1990.

C.H. Norris, J.B. Wilbur, S. Utku, ” Elementary structural

analysis”, Mc Graw-Hill Book comp., 1976.

Emam, H., “Fundamental Theory of Structures”, faculty of

Engineering, Cairo University.

Mouner Badir, ” Statically Indeterminate Plane Structures",

faculty of Engineering, University of Alexandria, 1980.

Chu-Kai Wang, “Statically Indeterminate Structures”, McGraw-

Hill Book Comp.; 1953.

A.S. Diwan, A.F.A. ElRahman, Y.M. Mansour, “Statically

Indeterminate Structures”, Vol. II, Alex. Univ., 1981

W.M. El-Dakhakhni, “Theory of Structures, Part 2”, Dar El-

Maaref, Cairo, 1974.

Shaker, “Plane Analysis of Indeterminate structures”, Ain Shams

Univ., Cairo, 1968.

R.C. Coates, M.G. Coutie, F.K. Kong, “Structural Analysis”,

E.L. Book Society And Nelson, 1980.

N. Makhlouf, G. Sherif, “Theory Of Structures, Statically Ind",

Al Azhar University, 1984.

. Bazaraa, “Structural Analysis and Mechanics II”, 1989.

. S.N. Mickhaiel, “Notes on structural Analysis and Mechanics for

Third Year Civil”, 1996

C.H. NORRIS, J. B WILBUR , & S. UTKU " Elementary Structural

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Eman, H. , " Structural Analysis and Mechanics " 3 rd year civil.

Facly. Of Engg. Cairo univ. 1991 – 1992.

Chu- Kai Wang "Statically Indeterminate Structures " Mcgraw – Hill

Book Comp. , 1953.

A.S. Diwan, A.F.A. EL RAHMAN, Y.M. Mansour "Statically

Indeterminate Structures" vil. Il. Alex . Univ., 1981

W.M. EL – DAKHAKHNI " Theory of Structures, part 2 " Dar El –

Maaref, Cairo, 1974

FHAKER "Plane Analysis of Indeterminate Structures" Aim Shams

Univ. , Cairo , 1986 .

R.C. COATES, M.G COUTIE, F.K.KONG "Structural Analysis" E.L.

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N. Makhlouf, G. Sherif "Theory of Structures, statically Ind." , AL

SAZHAR UNIVERSITY, 1984 .

A.Bazaraa , " Structural analysis and Mechanics II " , 299 P, 1989 .

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Alex. University. 1980.

Documents

Contents - ZU · PDF filecontents 2- buckling 1 ... 13-maxwell's law of reciprocal deflections; 93 betti' s law 14-influence line for deflection 95 15-problems 97