1

Modern University For Information and Technology

Civil Engineering Department

Lectures Notes of

Structure Design 1

CENG 216

Prepared By

Dr: Mohamed Osman Zakaria

(First Edition 2021)

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Structure Design 1

Civil Engineering Department

By

Prof. Ass. Dr. Mohamed Osman Zakaria

Faculty of Engineering

MTI University

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CONTENTS

CHAPTER ONE

Influence Lines 2

CHAPTER TWO

Deformations 19

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Chapter One

Influence Lines

1. Influence lines of Beams:

Influence lines are drawn for reactions and all internal forces for a given section

when a unit load is moving all over the beam. This is a helpful method to solve

problems of moving loads; for bridges and similar structures.

1.1 Influence line of reactions

X Unit load (1 KN)

A B

α

L Figure 1

Draw the influence line for the reaction at A when the unit load moves from A

to B. In above figure x is a variable indicating the position of the section in the

beam, while α indicates position of moving unit load.

Vertical reaction at A (VA) = 1 (L – α)/L = (1 –α/L) …….. (1)

This is a linear equation, it may be drawn (figure 2) when knowing two points;

For α = 0 , VA = 1 and for α = L , VA = 0

1

+

A Influence line of VA (I.L.VA) B Figure 2 (a)

Similarly influence line of vertical reaction at B may be obtained, where;

VB= α/L …….. (2)

For α = 0 , VB = 0 , and for α= L , VB = 1

1

A B

Influence line of VB (I.L.VB) Figure2 (b)

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1.2 Influence line of internal forces

a- Bending Moment

Draw the influence line of the bending moment for the section at distance x

shown in figure 1. (Say section S)

Since the unit load is moving across the beam than 2 cases may be obtained;

- Unit load to the right of the studied section ,that is:

X ≤ α ≤ L

Bending moment at section S (M) is:

MS (α,X) = VA . X = (1 – α/L) X ……… (3)

Above moment equation is linear and may be drawn by calculating the two

boundary conditions;

For α = X (load is over studied section in this case);

M (α,X) = (1 – X/L)X ……… (4)

For α = L ; MS (α,X) = 0

- Unit load to the left of the studied section ,that is:

0 ≤ α≤ X

MS (α,X) = VB . (L – X) = (L – X) α/L …….. (5)

As before For α = X ; MS (α,X) =(1 – X/L)X

And for α =0 , MS (α,X) = 0

Above results are drawn in figure 5

X

S

+

(1 – X/L) X

Influence line of bending moment MS (I.L.MS) Figure 3

b- Shearing Force

Draw the influence line of the section at distance α shown in figure 1.

- Unit load to the right of the studied section ,that is:

X ≤ α ≤ L

Shearing force at section S (QS):

QS (α,X) = VA = (1 – α/L)

Again this is a linear equation, and limits are;

For α = X (load is over studied section in this case);

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Q (α,X) = (1 – X/L) ……… (6)

And for α = L , QS (α,X) = 0

- Unit load to the left of the studied section ,that is:

0 ≤ α ≤ X

QS (α,X) = - VB = - α/L

For α = X (load is over studied section in this case);

Q (α,X) = - X/L ……….. (7)

And for α = 0, QS (α,X) = 0

1- X/L

+

-

X – X/L

Influence line of Shearing Force QS (I.L. QS)

Figure 4

In above drawing for shear diagram positive sign is drawn upward

Solved Example 1;

For the beam shown in figure 5; draw the influence lines of both reactions,

bending moment and shearing forces for a given section, say the S-S section

shown in figure 5.

S 3m

A 1m B S 5m C 2m D

Figure 5

Solution:

For the reaction at B, we apply equation 1, to obtain:

VB = (1- α/L), for point B → α = 0, VB = 1

And for C → α = 5m (L = 5m), VC = 0. Figure 6 a show the I.L.VB.

7

Similarly VC is obtained by applying equation 2. Figure 6 b show I.L.VC

6/5 1

(+) D

A B I.L.VB(1KN factor) C -2/5

Figure 6 a

For parts over AB and CD they are obtained simply by extending the part over

BC to the right and left. All values are calculating using simple geometric theory.

1 7/5

A (+)

-1/5 B I.L.VC(1KN factor) C D

Figure 6 b

Influence lines of Internal Forces

- Bending Moment

This may be obtained by applying equations 4 and 5; where:

X= 5 – 3 = 2m (X measured from left support);

MS = (1 – X/L) X = (1 – 2/5)2 = 6/5

As before portion over parts AB and CD are simply obtained by the extension of

parts over BC to both directions right and left as done above in influence lines of

reactions. The reader should note the negative parts obtained by the extension.

See figure 7.

-3/5 -4/5

A (-) B S C (-) D

(+)

6/5 I.L.MS(1KN.m factor) Figure 7

- Shearing Force

From equations 6 & 7 along with figure 4 the I.L.QS is drawn as shown in figure 8.

Note that X = 2m and L = 5m. Similarly parts AB & CD are obtained by extension of

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part BC as done previously. Change of signs should be noted.(negative shear is

drawn above datum which is possible).

A - -2/5 - -2/5

1/5 B 3/5 + I.L.QS C D

(1KN factor) Figure 8

Comment:

Equations from 1 to 7 may be applied for all influence line problems in beams. It

should be noted that X is measured from the left support till the studied section,

and L is the distance between the two supports. Above equations are not

applied for cantilever.

Solved Example 2;

For the same beam of figure 5 obtain the maximum positive bending moment and

minimum negative shearing force at section S when a moving uniformly

distributed load (UDL) of intensity 5KN/m cross the beam. Length of UDL is 2m.

Solution:

a- Maximum positive Bending Moment

The problem may be solved by studying the drawing in figure 9. The position of

the UDL (2m length) lies between two points 1m to the left and one meter to the

right of section S. Conveniently the drawing is shown hereafter where the

required area is hachured. The area between lower and upper ordinates is

calculated by using the theory of similar triangle as follows;

Ordinate at E = (6/5)1/2 = 3/5

Ordinate at F = (6/5)2/3 = 4/5

Area = [(3/5 + 6/5)/2]1 + [(6/5 + 4/5)2]1 = 1.9 m2

Mmax =1.9 (5) = 9.5 KN.m

-3/5 1m 2m -4/5

A - B E S F C - D

+

6/5 I.L.MS Figure 9

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b- Minimum negative shearing force

This is obtained by studying figure 8. It is clear that required position of loading is

between B and S or the part CD, both give the same result.

Qmin = (½)(-2/5)2(5) = -2 KN.

Qmin means can also be called the maximum negative.

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Tutorial Sheet of

Influence lines for Beams

Problem (1) Draw the influence lines of:

- reactions at supports,

- bending moment and shearing forces at sections shown in the figures 1 & 2:

x ∑

A B

L

Figure 1 x1 ∑1 x2 ∑2

A B

2L 4L L

Figure 2

Problem (2) Draw the influence lines of all the internal forces for the beam shown in figure 2 for the

following data:

L = 2m , x1 = 2m and x2 = 3m

Problem (3) 4m

20 15 20KN 10KN

A ∑ B

2m 1 3m 1 2m

Figure 3

Draw the influnce line for bending and shear for section ∑ in the beam of figure 3.

Using the obtained bending and shear obtain the bending and shear of the shown system of

loading shown in figure 3 at section ∑

Problem (4) In the beam of figure 2, obtain the bending and shear for the sections of problem 2 when

subjected to UDL of intensity 10KN/m

Problem (5) 20 25KN 25 KN

2m 3m Figure 4

The moving load of figure 4 will be applied for the beams of figure 2&3:

Obtain the maximum and minimum bending and shear for the sections ∑ shown in each

figure.

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2. Influence lines of Trusses:

Member of trusses carry only axial forces. Therefore influence lines are drawn for

axial forces when moving load of one ton travel across the truss (in the upper or

lower chord).

Let us study the N truss shown in figure 10.

1 3 5 7 9 11

2m

2 4 6 8 10 12

3m 3m 3m 3m 3m

Figure 10

It is required to draw first the influence lines of the reactions at support 2 and 10

due to 1KN moving on the upper chord from node 1 to 11.

Influence lines for truss reactions are similar to what have been done in beams.

Therefore we can draw the influence lines of a beam having the same span (12m)

and a cantilever part of 2m, as shown in figure 11.

V2 = (1 –α/L) from equation 1

1 0.75

+ 0.5 0,25

-1/4

Influence line of reaction at support 2 Figure 11 a

V10 = α/L from equation 2 1 1.25

0.25 0.5 + +

Influence line of reaction at support 10 Figure 11 b

Now let us draw the influence line of the member 2-4.

Simple analysis of node 2 indicates that force in member 2-4 is always zero.

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For influence line of member 1-3, we must study 2 cases;

1- Unit load to the right of node 3

In this case force in member 1-3 is easily obtained by studying the following left

hand side, see figure 12. F1-3

Figure 12 F1-4 2m

V2 F2-4

Moment about node 4 → F1-3 (2) = V2 (3) 3 m

F1-3= - 1.5 V2.

This means draw influence line of reaction V2 but in opposite side (because the

sign is negative). All values are multiplied by 1.5

2- Unit load on member 1-3

In this case it is easier to study the right hand side. Taking moment about node 4;

F1-3 (2) =- 9 V10 → F1-3 = - V10 (4.5)

Thus V10 is multiplied by 4.5 and drawn in opposite side;

Influence line of member 1-3 is shown in figure 13

0.375

-4.5V10 - - -1.5 V2

-1.125 I.L.F1-3

Figure 13

For influence line of member 4-6, we should study the following cases;

1. Unit load on, and right of node 5

Figure 14 shows the left hand side section of member 4-6;

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F3-5

β

2 m

F3-6

F4-6

V2 3 m

Figure 14

Moment about node 3;

M3 = F4-6 (2) – V2(3) = 0 → F4-6 = 1.5 V2

2. Unit load on, and left of node 3

Similar approach may be done as above but with the right hand side to obtain:

F4-6 = 4.5 V10; figure 15 shows the influence line of member 4-6.

1.125

+ +

I.L.F4-6 -0.375

Figure 15

To draw the influence line for any inclined member such as 3-6, we proceed as

follow;

Again we should study different cases regarding the position of the unit moving

load;

1. Unit load to the right of node 6 (5 of upper chord)

To this end let us consider the left hand side of a section cutting member 3-6, see

figure 14. The vertical components of forces for the above section are:

V2 = F3-6 sin β → F3-6 = V2 / sin β = V2(√13/2).

2. Unit load to the left of node 3

In this case we should study the right hand side, see figure 16.

14

3 5 7 9 11

F3-6 2m

4 6 8 10 12

3m 3m 3m V10

Figure 16

The vertical projection of forces, gave:

F3-6 = - V10 (√13/2); figure 17 shows the influence line of member 3-6.

V2 (√13/2)

+

- V10 (√13/2) - I.L.F3-6 Figure 17 -

As a demonstration of influence line in a vertical member, let us study the

member 3-4. Again we have two cases to study.

1. Unit load on nodes 1, 5, 7, 9 and 11

This member is easier to study by the equilibrium of joint 3, figure 18.

F3-1 β F3-5

F3-6 Node 3

F3-4 Figure 18

Vertical equilibrium of node 3, gave;

F3-6 sin β + F3-4 = 0 → F3-4 = - F3-6 (2/√13)

As we know the influence line of F3-6 , see figure 17, then F3-4 is known.

2. Unit load on node 3

This particular case should be studied alone, figure 19.

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Unit load

F3-1 β F3-5

F3-6 Node 3

F3-4 Figure 19

F3-4 = -(F3-6 (2/√13) + 1) = V10 - 1 = - 0.75

Figure 20 shows the influence line of member 3-4.

- V2

- -

-0.75 I.L.F3-4 Figure 20

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Tutorial Sheet of

Influence lines for Trusses

Problem (1)

Draw the influence lines for:

- reactions

1 3 5 7 9

L

2 4 6 8 10

L L L L

Figure 1

- vertical members 1-2 and 5-6

- top members 1-3 and 7-9

- bottom members 2-4, 6-8 and 8-10

- diagonals 1-4 and 7-10

Problem (2) If the following moving load (shown in figure 2) cross the truss of figure 1; obtain:

- maximum reation at support 8

- maximum normal forces in members 1-2, 7-9, 2-4 and 6-8

Assume L = 3m

40 KN 50 KN

3m

Figure 2

Problem (3) Draw the influence lines for all the members of the truss shown in figure 3:

1 3 5

2m

2 4 6

3m 2m 2m 2m 2m

Figure 3

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3. Problems to be solved by the reader:

1) Draw the influence lines of:

- reactions at supports,

- bending moment and shearing forces at sections shown in the following

figures:

x

A Σ B figure 1

L x1 x2 Σ 2

Figure 2 2L Σ1 A 4L B L

2) Draw the influence lines of all the internal forces for the beam shown in figure

2 using the following data:

L = 2m, x1 = 2m and x2 = 3m

3) 20 4m Σ 20KN 10KN

A 2 4m 2m B 2m figure 3

Draw the influnce line for bending and shear for section ∑ in the beam of figure

3.

Using the obtained bending and shear obtain the bending and shear of the shown

system of loading shown in figure 3 at section ∑

4) In the beam of figure 2, obtain the bending and shear for the sections of

problem 2 when subjected to UDL of intensity 10KN/m

5) The moving load of figure 4 will be applied for the beams of figure 2&3:

Obtain the maximum and minimum bending and shear for the sections ∑ shown

in each figure.

20 25KN 25 KN

Figure 4 2 m 3 m

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6) Draw the influence lines for:

- reactions

1 3 5 7 9

L

2 4 6 8 10

L L L L

Figure 5

- vertical members 1-2 and 5-6

- top members 1-3 and 7-9

- bottom members 2-4, 6-8 and 8-10

- diagonals 1-4 and 7-10

7) If the following moving load (shown in figure 6 cross the truss of figure 5;

obtain:

- maximum reation at support 8

- maximum normal forces in members 1-2, 7-9, 2-4 and 6-8

Assume L = 3m

40 KN 50 KN

3m

Figure 6

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Chapter Two

Deformations

1. Introduction

In this chapter we are dealing with the deflection and angle of slope for beams,

those two are known as deformations. This deflection (linear displacement) and

angle of slope (angular displacement) may be calculated by several methods.

However in order to understand those methods of calculation, it is important to

demonstrate some fundamental relations in the simple (pure) bending of beams.

To this end let us consider two sections SS’ & S1S1’ as shown in figure 1, where SS’

is taken as a reference section (will not move).

Applying a negative moment (-M) to the section S’1S’1 which becomes after

deformation S’2S’2 by rotating through point n. the following relations are

obtained;

S2 S1 ΔX S

m2 m1 m

M - y X

α

S1’ S2’ S’

α

Y

O Figure 1

m1m2˃ 0 is a strain elongation, thus the corresponding σ ˃ 0 is a tension stress.

From the geometry we get;

m1m2 = (-y) tan (-α) = y.α (where α is a very small angle). From Hook’s law of

stress-strain relationship we obtain:

m1m2 / mm1 = y.α / ΔX = σ / E

Gives: σ = E. y. α / ΔX ……. (1)

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The beam shown in figure 1 becomes an arc of a circle, figure 2, in which the

upper side is in tension while the lower is in compression as shown below.

(+) tension side

(-) comp. side

Ρ Ρ

Figure 2

In between the two extremes we obtain a neutral fiber, where both stresses and

strains are zero. Point n of figure 1 is known as the neutral point, and the axis

passing through n in the lateral cross section S’1S’1 is the neutral axis.

In case of simple bending the neutral axis and the center of gravity are the same.

From figures 1&2, we obtain the following relations:

Sin α = α = ΔX / Ρ → α / ΔX = 1 / Ρ

Where Ρ, is the radius of the neutral circle. However Ρ may be taken as the radius

of the inner or outer fibers, since the depth of the beam is negligible with respect

to the radius. We can easily write for any element (i) of the beam:

Fi = σiwi = E yiwi α / ΔX

Where wi = area of element (i)

Equilibrium in the x-axis direction gives;

N + (E α / ΔX) Σ yiwi = 0

But N (normal force) is assumed zero because we are studying the case of simple

bending, and the quantity Σ yiwi is the statically moment of area. Thus yG A = 0

→ yG = 0

This means that neutral axis is the same as the axis GZ of the lateral cross section

in case of simple bending.

The second equation of equilibrium in the y-axis direction is:

Q + τiwi = 0

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But Q = 0 (as a first assumption), therefore shear stresses τ = 0.

The third and last equation of equilibrium will be for the moment:

M + (E yiwi α / ΔX) yi = 0

M = - (E α / ΔX)yi2wi

The quantity yi2wi =IGZ = I

M = - (E α / ΔX) I ……. (2)

From equations (1) & (2), we obtain:

α /ΔX = σ / Ey = - M / EI = 1 / Ρ

And from the fundamental mathematics we have:

Ρ = [(1 + y’2)

3/2] / y’’

Where, as before, Ρ is the radius of curvature and y’ = first derivative of y with

respect to x. y’’ = second derivative of y with respect to x.

The small quantity y’2 in the above equation use to be neglected, thus:

1 / Ρ = y’’ = - M(x) / EI ……… (3)

Which is the general differential equation of any deformed bar (beam), see

figure3.

P w/m’

Deformed curve deformed curve

Figure 3

2. Double Integration Method

To obtain the deformation in any beam many methods may be applied. The

double integration method is one of them. This method is based on the

differential equation of the deformed curve (elastic curve), equation 3. Clearly

integrating equation 3 twice will give the algebraic equation of the elastic line

from which any deflection may be obtained.

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Let us study the cantilever beam of figure 4 in order to obtain the deflection and

angle of rotation at point B.

A P X

L B

Y

S.F.D. P

-PL

B.M.D.

Elastic curve Ө

Figure 4

Reactions at support A are:

VA = P ↑ and MA = PL anti Clockwise direction

Bending moment equation is ;

M(x) = - MA + P X = P (X – L)

Q(x) = P = constant

Assuming constant EI, Maximum deflection will occur at point B.

Applying equation 3; y’’ = -P(X – L) / EI

Y’ = -(P/EI)(X2/2 – LX + c1)

For x =0, y’ = 0 → c1 = 0

Y’ = -(P/EI)(X2/2 – XL)

Y = -(P/EI)(X3/6 – LX

2/2 + c2)

For X = 0 , Y = 0 → c2 = 0

Y = -(P/EI)(X3/6 – LX

2/2)

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YB = maximum deflection = - (P/EI)(L3/6 – L

3/2) = PL

3/3EI

Rotation at B is obtained from equation of y’ at X = L:

Y’ (X = L) = tan Ө = PL2/2EI = Ө for our case of small angles

Solved example 1:

Determine the mathematical expression of the deformed (elastic) curve of the

beam AB shown in figure 5. Obtain the maximum deflection and the rotation at

support A. Assume constant EI.

w/m’

A L B x

Figure 5 y

Solution

The general differential equation of the deformed shape is:

Y’’ = -M(x) / EI

M(x) = wx L/2 – wx2/2

EI y’’ = - wx L/2 + wx2/2

EI y’ = - wx2 L/4 + wx

3/6 + c1

EI y = - wx3L/12 + wx

4/24 +c1x + c2

For x = 0, y = 0 → c2 = 0

For x = L, y = 0 → 0 = - wL4/12 + wL

4/24 + L c1

C1=w L3 / 24

Y = (w/EI)(-Lx3/12 +x

4 /24+ xL

3 / 24)

The maximum deflection is in the middle of the span AB; that is for x = L/2

Ymax = (w/EI)(-L4 /96 + L

4 /384 + L

4/48) = 5wL

4/384EI

Rotation at point A, may be obtained from the equation of y’ for x = 0;

24

Y’A =wL3/24EI

Solved example 2:

Determine the rotation at points A and B of the beam shown in figure 6,

subjected to a concentrated load P. the beam have constant EI.

a P

A C B

L Figure 6

Solution

0 ≤ x ≤ a

M(x) = (P/L)(L – a) x

EI y1’’ = - P[(L – a) x/L]

EI y1’ = -P [(L – a) x2 / 2L + c1]

EI y1 = - P [(L – a) x3/6 L + c1x + c2]

a ≤ x ≤ L

M(x) = (P/L)(L – a) x – P (x – a)

EI y2“ =- P [(L – a) x/L – (x – a)]

EI y2’= - P[x2/2 – ax

2/2L – x

2/2 – ax + c3]

EI y2 = - P[x3/6 – ax

3/6L – x

3/6 – ax

2/2 + c3 x + c4]

1) From the first condition x = 0, y1 = 0 → c2 = 0

2) X = L , y2 = 0 →

0 = L3/6 –aL

2/6 – L

3/6 + aL

2/2 + c3L + c4

3) For x = a, y1 = y2

C1 a = - a3/6 – a

2/2 + c3 a + c4

4) For x = a, y1’ = y2’ → c1 = - a2/2 + a

2 + c3

From the above three conditions, we obtain:

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C1 = a2/2 – (a

3/6 + aL

2/3)/L

C3 = - (a3/6 + aL

2/3)/L

C4 = a3/6

To obtain rotation at A, we use the equation of y1’;

Y1’ = - (P/EI)[(L – a) x2 / 2L + a

2/2 – (a

3/6 + aL

2/3)/L]

YA’ = - (P/EI)[a2/2 – (a

3/6 + aL

2/3)/L] = Pa(2L – a)(L – a)/6EIL

For the rotation at B, we use the expression of y2’;

YB’ = - (P/EI)[(L – a)L/2 – L2/2 + aL – (a

3/6 + aL

2/3)/L]

= - Pa(L – a)(L + a)/6EIL

For the special case of a = L/2

YA’ = - Yb’ = PL3 /16EI

It is useful to calculate the maximum deflection in case of a = L/2, from any of the

y equations above, to obtain;

Ymax = Y(x=L/2) = PL3/48EI

This example may be solved by changing the origin for the second part only to be

from right to left (origin at B). In this case solution will be much easier (see next

solved example 3), this is should be done by the reader.

Solved example 3:

Using the double integration method, obtain the maximum deflection and the

rotation at left support for the beam of variable moment of inertia shown in

figure 7. Take EI = 40000 KN.m2

10KN/m

A B

4m(EI) C 4m (2EI)

Figure 7

Solution:

VA = 10x4x2/8 = 10 KN

VB = 10x4x6/8 = 30 KN

For part AC 0≤ x1 ≤ 4m

26

M(x) = 10 x1

Y’’1 = - 10x1/EI

Y’1 = - (10/EI)(x2

1/2 + C1) ..... (a)

Y1 = - (10/EI)(x3

1/6 + C1x1 + C2) ...... (b)

for x1 = 0 , Y1 = 0 → C2 = 0

For part CB 0 ≤ x2≤ 4m

M(x) = 30x2 – 10(x2

2/2)

Y’’2 = - (10/EI)(3x2- x2

2/2)/ 2

Y’2 = -(10/EI)(3x2

2/4 - x3

2/12 + C3) ........ (c)

Y2 = - (10/EI)(x3

2/4 - x4

2/48+ C3x2 + C4) ........(d)

For x2 = 0, y2 = 0 → C4 = 0

We now have two remaining unknown constants. They may be obtained from the

following two conditions;

1) For x1 = x2 = 4m , y1 = y2

2) For x1 = x2 = 4m , y’1 = - y’2 (note the negative sign in this case)

Substituting above values of x1& x2 in equations a, b,c & d;

(4)2/2 + C1 = - [3(4)

2/4 -(4)

3/12 + C3]

(4)3/6 + C14 =(4)

3/4 - (4)

4/48 + C34)

From which C3 = -22/3 = C1

For the rotation at left support we use equation (a) for x1 = 0

Y’A = ӨA =-(10/EI)(C1)= -(10/40000)(-22/3) = 1.833 x10-3

rad

The position of maximum deflection may be obtained by equating equation (a) to

zero;

0 = x2

1/2 – 22/3 → x1 = 3.83 m ≈ 4 m

27

And from equation (b), the maximum deflection is;

Y(x = 4m) = - (10/40000)(3.833/6 – (22/3)3.83) = 4.667/1000 m = 4.667 mm

Solved example 4:

For the cantilever shown in figure 8, obtain the deflection and the rotation at the

free end A. assume EI = 30000 KN.m2

20KN.m 15KN/m x

A B

3 m Figure 8

Solution:

VB = 15x3 = 45 KN ↑

MB = 20 + 15x3x1.5 = 87.5 KN.m clockwise direction

M(x) = VB x – MB – 15x2/2 (note the direction of x in figure 8)

= 45x – 87.5 – 7.5x2

EI y’’= - (45x – 87.5 – 7.5x2)

EI y’ = -(45x2/2 – 87.5x – 2.5x

3 + c1)

EI y = - (45x3/6 – 87.5x

2/2 – 2.5x

4/4 + c1x + c2)

For x = 0, y = 0 → c2 = 0

For x = 0, y’ = 0 → c1 = 0

To obtain the rotation at A we substitute for x = 3m in the equation of y’;

y’A = ӨA (x = 3m) = - (1/EI)[45(3)2/2 – 87.5(3) – 2.5(3)

3 + 0] = 127.5/30000 =

0.00425 rad.

28

Since x is taken from left to right (figure 8), therefore positive sign mean

anticlockwise direction.

To obtain the deflection at A we substitute for x = 3m in the equation of y;

yA = - 1/EI [45(3)3/6 – 87.5 (3)

2/2 – 2.5(3)

4/4 +0 + 0] = 241.875/30000 = 0.008063m

= 8.063 mm downward.

29

Tutorial sheet of

Double Integration Method

Problem (1) Using the double integration method obtain the rotation at A and the deflection at C for

the following beams (consider EI = constant). Draw the elastic curve

a

A P B

C

L Figure 1

A w/m B C

4L L

Figure 2

Problem (2) Obtain the maximun deflection in problem of figure 1. Assume a = 4m, L = 6m, P = 30 KN

and take EI = 20000 KN m2

Compare the maximum deflection when a = L/2 = 3m with above results.

Problem (3) Draw and Derive the equation of the elastic curve for the beams shown in figures 3 &4.

Determine the slope at supports and the maximum deflection using the double integration

method. Assume EI = 30000 KN m2

10KN/m

20KN.m 24KN/m

A B A B

3m 6m

Figure 3 Figure 4

Problem (4) Using the double integration method, obtain the maximum deflection and the rotation at

left support for the beam of variable moment of inertia shown in figure 5. Take EI = 40000

KN.m2

10KN/m

A B

EI C 2EI

4m 4m Figure 5

Problem (5) Using the double integration method, obtain the maximum deflection and the rotation at

the free end of the cantilever of figure 6. Assume constant EI

w/m

A B

Figure 6 L

30

3. Elastic Load Method

As shown in the double integration method the deflection may be calculated by;

Y = - ∫∫ (M(x) / EI) dx ……. (1)

And the slope is obtained by;

Y’ = - ∫ (M(x) / EI) dx …….. (2)

The analogy between the two above equations and the fundamental equations of

bending and shear, namely;

M = ∫∫P(x)dx ………. (3)

Q = ∫P(x) dx ………. (4)

Where P(x) is the load function

Let us now replace the load function in (3) and (4) by the quantity M(x)/EI, thus the

resulting bending moment and shearing force will be the deflection and the angle

of slope respectively. The quantity M(x)/EI is referred to as elastic load, from

which the name of the method. The resulting reaction, shear and bending are also

known as the elastic reaction, elastic shearing force (angle of slope) and elastic

bending moment (deflection). The following statements summarize this method;

“The slope of the elastic curve at any point along an end supported beam is

numerically equal to value of the elastic shearing force at this point of a

corresponding simply supported beam”

“The deflection at any point along an end supported beam is numerically equal to

value of the elastic bending moment at this point of a corresponding simply

supported beam”.

Let us illustrate this method in detail by solving the following applications;

Solved example 1:

Obtain the deflection at point C and slope at support A of the beam shown in

figure 9.

31

10KN

2KN/m

A 4m C 4m B

Figure 9

Solution:

Bending moment diagram is conveniently split into two parts M1 and M2for each

case of loading as shown in figure 10. EI = 24000 KN.m2

Wl2/8 = 16KN.m B.M1

W1 W2 W2’ W1’

4m 4m

PL/4 = 20 KN.m B.M2

W3 W3’ Figure 10

8/3 8/3 8/3m

The elastic weights are

W1 = W’1 = 2/3 (2x42/8) 4 = 32/3EI KN.m

2 at 2 m from C

W2 = W2’ = ½ (16 x 4) = 32/EI KN.m2 at 4/3 m from C

W3= W3’ = ½ (20 x 4) = 30/EI KN.m2 at 4/3 m from C

To obtain the rotation at A, we simply calculate the shearing force at A which is

equal to the reaction VA.

VA = ӨA= [(32/3)6 +(32/3)2 + 32x16/3 + 32x8/3 + 30(16/3) + 30 (8/3)]/8EI =154/3EI

KN.m2 = 2.1388

-3 = 0.1225 degree

Deflection at C is obtained by calculating the bending moment of elastic weights

at point C.

MC = yC = (154/3EI)4 – (32/3EI)2 - (32/EI)4/3 – (30/EI)(4/3))

= 304/3EI KN.m3 = 4.222 mm

32

Solved example 2:

Calculate the rotation at C and the deflection at the free end D of the overhanging

beam shown in figure 11. Assume EI = 25000 KN.m2

2 KN/m 9KN

A 4 m B 2m C 2m D Figure 11

The bending moment diagram is drawn for each case of loading separately as

shown in figure 12.

8KN.m

w1 w2 w3 w4

12KN.m

w5 w6 Figure 12

w1 =2/3(2x42/8)4/EI = 32/3EI KN.m

2 at 2m from A

w2 = ½ (8x4)/EI= 16/EI KN.m2 at 8/3 m from A

w3 = ½(8x2)/EI = 8/EI KN.m2at 14/3 m from A

w4 = 2/3(2x22

/8)2/EI = 4/3EI KN.m2 at 5 m from A

w5 = ½(12x4)/EI = 24/EI KN.m2 at 8/3 m from A

w6 = ½(12x2)/EI = 12/EI KN.m2 at 14/3 m from A

To obtain the rotation at C, we simply calculate the shear at C;

ӨC = -[(32/3)2 + 16(8/3) + 8(14/3) + (4/3)5 + 24(8/3) + 12(14/3)]/EI =-228/EI =-9.2-

3radian =-0.53 degree

From figure 13 where the elastic curve is drawn,

2m yD

ӨC ӨC Figure 13

33

YD = 2 tan ӨC = 2 tan(-0.53) = -0.0184 m =- 1.84 cm

4. Conjugate Beam

It may be noticed that elastic weight method is only applicable for the part

between the two supports. In order to overcome this problem, the conjugate

beam is introduced.

Let us consider the cantilever shown in figure 4, this problem has been solved by

the double integration method. It is clear that position of maximum deflection

(point B), have zero moment, and position of zero deflection have maximum

bending moment (at support A). Thus the elastic weight method is not applicable

in this case. Therefore the conjugate beam method is established, in which the

kind of supports are changed. The following table (table 1) gave the relation

between the original beam and the conjugate beam. The elastic weights are then

applied to the conjugate beam, from which we can calculate any required

deformation.

End of Original Beam End of Conjugate Beam

1 A A

2 B B

3 A A

4 B B

5

6

Table 1

Shear of conjugate beam is equal to the rotation at a given point, while bending

of conjugate beam is equivalent to the deflection.

Solved example 1:

For the beam shown in figure 14, obtain the deflection at the middle of the span,

and the rotation at the free end C using the conjugate beam method. Assume EI =

30000 KN.m2.

34

4KN/m 9 KN

A B C

6 m 2 m Figure 14

Reaction at A VA = (4x6x3 – 9x2)/6 = 9 KN

Reaction at B VB = (4x6x3 + 9x8)/6 = 24 KN

The bending moment diagram is drawn in figure 15 conveniently for each case of

loading.

wL2/8 = 18KN.m W1

+ B.M1

W2 -18 KN.m W3

- B.M2

4 m 4/3 Figure 15

W1 = 2/3(18)(6)/EI = 72/ EI

W2 = ½(18)(6)/EI = 54/EI

W3 = ½ (18)(2)/EI = 18/EI

It is important to underline that negative moment produce upward

elastic weight, while positive moment gave downward elastic weight,

as shown in figure 15. Also the positive shearing force indicates clockwise

direction and positive bending moment means downward deflection.

Above weights are applied on conjugate beam using table 1, see figure 16.

W1 W2 W3

A x B C

3 m 4/3m

4m Figure 16

35

First we calculate the reaction of the conjugate beam.

VA may be calculated by taking moment about B for the left part

VA =(3 W1 - 2 W2)/6 = 18/EI upward

VC = W1 - W2 - W3 – VA = - 18/EI downward

MC =4/3(W3 ) = 24/EI clockwise

The rotation at the free end is directly equal to the above reaction at C, that is :ѳC

= VC = 18/EI = 18/30000 = 0.0006radian = 0.0344 degree

To obtain the deflection at the middle of the span (D), we need to write the

equation of bending moment for each case of loading in order to take moment

about point D for the left part;

M1 (x) = wL/2 x – wx2 /2 = [4(6)/2] x – 4x

2/2 = 12x – 2x

2

M2 (x) = -3x

Moment about D for the conjugate beam will produce the deflection Y;

MD = Y= [18x – 2/3(12x – 2x2)x(3/8 x) + ½ (3x)x(x/3)]/EI

Substituting in above equation of Y for x = 3m:

Y = [18(3) – 2/3(12x3 – 2(3)2

)3(3x3/8) + ½(3)3]/EI = [27/30000]1000

= 0.90 mm.

Solved example 2:

The cantilever shown in figure 17 has a variable moment of inertia as indicated.

Determine the slope and deflection at the free end.

P

A 2EI B EI C

L/2 L/2

Elastic curve Ө yc Figure 17

36

Solution

This example was solved with constant EI, however the bending moment

equation still valid. Conveniently we rewrite it again;

M(x)= P(X – L), the bending moment is drawn in figure 18.

-PL

- B.M.

Figure 18

Since we have a variation of inertia, the above B.M. is modified by dividing each

value by the corresponding inertia and applies the whole to the conjugate beam

as shown in figure 19.

-PL/2 W1 W2 W3

A - B modified B.M.

Conjugate Beam Figure 19

W1 = ½[(PL/4)L/2] = PL2/16EI at 5L/6 from B

W2 = (PL/4)(L/2) = PL2/8EI at 3L/4 from B

W3 = ½[(PL/2)(L/2)] = PL2/8EI at L/3 from B

The vertical elastic reaction at B, gave the angle of slope ӨB;

ӨB = W1 + W2+ W3 = -5PL2/16EI which is a positive shear (clockwise)

YB is the moment at B;

YB = (PL2/16EI)5L/6 + (PL

2/8EI)3L/4 + (PL

2/8EI)(L/3) = 3PL

3/16EI downward.

37

Solved example 3:

For the beam shown in figure 20, obtain the rotation at supports A&B. calculate

also the deflection at C, and the maximum deflection. Assume constant EI = 25000

KN.m2.

30KN 20KN

A 4m C 2m 2m B Figure 20

Solution:

VA = (30x4 + 20x2)/8 = 20 KN A B

VB = (30x4 + 20x6)/8 = 30 KN 80 60

W1 W2 W3 W4

Figure 21

Bending moment and elastic weights are shown in figure 21

W1 = (80x4)/2 = 160 KN.m2 at 8/3 m from A

W2 = ½x20x2 = 20 KN.m2 at 14/3 m from A

W3 = 60x2 = 120 KN.m2 at 5m from A

W4 = ½ x 60x2 = 60 KN.m2 at 20/3 m from A

From figure 19, the elastic reactions are;

VA = 170 KN.m2 and VB = 190 KN.m

2.

From VA& VB we could obtain the rotations at both supports by dividing those

values by EI;

ӨA = 170/25000 =0.0068 rad. = 0.39 degree

ӨB= 190/25000 = 0.0076 rad. = 0.435 degree

38

EI YC = 170x4 – 160x4/3 = 1400/3

YC = 1400/(3x25000) = 0.01867 m = 18.67 mm

Let us assume the maximum deflection at point D somewhere between the two

loads shown in figure 22. At this point elastic shearing force must equal zero.

W’3 and W’2 will not equal the previous W3 and W2 since the distance is now

different; they are calculated as follow;

D x

W’3 W4 190

Y W’2 Figure 22

W4 = 60 KN.m2 as before

W’3 = (x – 2) 60 = 60x – 120

W’2 = ½(x – 2) y the value of y may be obtained from the geometry of the 2

triangles shown in figure 23;

2 m

Y / 20 = (x – 2)/2 from which;

20 x – 2 Figure 23

Y = 10(x – 2)

Thus W’2 = ½(x – 2)2 10

Elastic shear at D is now calculated from the right hand side as;

190 – 60 – 60x + 120 – 5x2 + 20x – 20 = 0

From which x = 3.874m. yD is now calculated by taking the moment of elastic

loads about D for the right hand side;

39

YD = Ymax = Y(x = 3.874) = (190x3.874 – 60x2.541 – W’2 x1.874/2 – W’3 x 1.874/3)/EI =

18.71 mm

Notes;

- From the value of maximum deflection above, one can note the

little difference between it and the deflection at the middle of

the span that is why most engineers calculate the maximum

deflection at the middle.

- W’3 and W’2 are calculated after knowing the value of x, as;

W’3 = (x – 2)60 = (3.874x60 – 2) 60 = 112.44 KN.m2

W’2 = ½(x – 2)2 10 = ½(3.874 – 2)

2 10 = 17.56 KN.m

2

Solved example 4;

Obtain the rotation in A & B, and the deflection at the middle of the beam AB

subjected to couple C. assume constant EI.

x C

A a L – a B (a)

W1 -Ca/L

-

C(1 – a/L) B.M.D. Figure 24 (b)

W2

Solution ;

0 ≤ x ≤ a

M(x) = - C x / L

a ≤ x ≤ L

M(x) = - C x / L + C

40

Bending moment diagram is shown in figure 24 (b).

W1 = ½(Ca / L)a = C a2/ 2L upward and at 2a/3 from A

W2 = ½[C(1 – a/L)(L – a)] downward and at 2(L – a)/3

Elastic shear at A = rotation at A = ӨA = reaction at A

ӨA = 1/EIL[- Ca=2 (L – 2a/3)/2L + C (L – a)

3/3L]

= - (C/6EIL)[2L2 – 6La + 3a

2]

ӨB = (1/EIL)[(Ca2/2L)2a/3 – (C/2L)(L – a)

2(a + L/3 – a/3)]

= (C/6EI L)[- L2 + 3a

2]

The elastic moment at the middle of the span gave the required deflection. For

the case of a ˂ L/2 and for the moment from the right hand side;

Y = (C/6EIL)(- L2 + 3a

2) L/2 – ½(CL/2x2)(L/3x2EI) = C /16EI(- L

2 + 4a

2)

Special Cases

1) a = 0

ӨA = CL/3EI , ӨB = - CL/6EI and y = - CL2/16EI

2) a = L

ӨA = - CL/3EI , ӨB = CL/6EI and y = CL2/16EI

3) a = L/2

ӨA = - CL/24EI = ӨB and y = 0.

Properties of Area;

It is clear that calculation of area and position of CG is important for above

methods. For this reason we introduce hear-after some important properties of

area in table 2.

41

Shape Area C.G.

1

b

L

bL

L/2

2

b

L

bL/2

L/3 from base

3

b

L

2bL/3

3L/8 from base

4

b 2nd

degree

L

bL/3

L/4 from base

5 3rd

degree

b

L

bL/4 L/5 from base

Table 2

42

Tutorial Sheet of

Conjugate Beam and elastic weight

Problem (1) Using the conjugate beam method obtain the rotation at A and the deflection at C for the

following beams (consider EI = constant). Draw the elastic curve

a

A P B

C

L Figure 1

A w/m B C

4L L

Figure 2

Problem (2) Determine the slope at supports and the maximum deflection for the beams shown in

figures 3 & 4 using the conjugate beam method. Draw the elastic curve.

Assume EI = 30000KNm2

10KN/m

10KN.m 21KN/m

A B A B

3m 6m

Figure 3 Figure 4

Problem (3) Using the conjugate beam method, obtain the deflection at C and the rotation at left

support for the beam of variable moment of inertia shown in figure 5. Take EI = 40000

KN.m2

10KN/m

A B

EI C 2EI

4m 4m Figure 5

Problem (4) Using the conjugate beam and the moment area method, obtain the maximum deflection

and the rotation at the free end of the cantilever of figure 6. Assume constant EI

w/m

A B

Figure 6 L

43

5. Virtual Work

This method (theory) is considered as the most important one used in the

calculation of deformation. It is important to note that calculation of deformation

is only one application of this method.

Let us consider the beam shown in figure 25(a), which is known as the original

system, where we can obtain the bending moment Mo. It is required to calculate

the deflection at point C.

P1 P2 P3

A C B

a) Original system of load produce bending moment Mo

1KN

A C B

b) Unit load system produce bending moment M1

dα

Figure 25 M M

Dx

c) Section after deformation

It is well known that external work We is equal to internal Work Wi;

We = Wi ……. (a)

We also have dα = (M/EI) dx

Where M is the bending moment of the original system (Mo); thus the

deformation of the original system is;

dα = (Mo/EI) dx …….. (b)

Let us now assume that all loading in the original system are removed, and only a

unit load is applied at point C (figure 25 b) producing the deflection δC. . The total

external work (virtual work) = 1 x δC , and the internal work is;

Wi = ʃ M1 dα ……… (c)

44

Where M1, is the bending moment of the unit load system.

From equations a, b&c, we get;

1 x δC = ʃ M1 dα ……… (d)

Substituting for dα from equation b into (d);

δC = ʃ(MOM1/EI) dx ……….. (1)

Equation 1, is the fundamental equation of the virtual work theory.

Let us now solve the problem shown in figure 26. It is required to obtain the

rotation and deflection at the free end using the virtual work theory. Assume

constant EI.

M A x B

Figure 26 P L

Solution :

The bending moment is conveniently drawn twice for the total load and for each

case of loading separately, as shown in figure 27.

-M - PL -M -(M + PL)

a)B.M for (M) b) B.M for (P) c) total B.M.D.

Figure 27

The equation of bending moment for each load is;

M(x)1 = - M ……. (a)

M(x)2 = - Px ………. (b)

The total bending moment is the sum of a & b as ;

M(x)0 = - (M + Px) …….. (c)

45

If in this problem we replace the applied force P and moment M by a unit load

and unit moment respectively, equations a & b became;

M(x)1 = -1 ….…. (d)

M(x)2 = -x ……… (e)

ӨA =1/EI ∫0L MoM1 dx = 1/EI ∫0

L-(M + Px) (-1) dx = 1/EI[(Mx + (P x

2)/2]0

L

ӨA = 1/EI [ML + PL2/2]

δA = 1/EI∫0L MoM2 dx = 1/EI ∫0

L - (M + Px) (-x) dx = 1/EI[Mx

2/2 + Px

3/3]0

L

δA = 1/EI [ML2/2 + PL

3/3]

From the above analysis we notice that M1 in equation 1 is replaced by M2 for

calculation of deflection, therefore it is conveniently to write equation 1 in the

following form;

δ = ʃ(MO Mi/EI) dx ……….. (2)

Where i is an index changing from 1 to n, and n is the number of the required

calculated displacements.

Above problem may be solved graphically by multiplying the diagram of M0 by the

two diagrams of M1 and M2 respectively as follow;

ӨA= 1/EI [(-ML)(-1) + (-PL2/2)(-1)] = ML/EI + PL

2/2EI

δA = 1/EI [(-ML)(-L/2) + (-PL2/2)(-2L/3)] = ML

2/2EI + PL

3/3EI

Same results obtained previously.

We can therefore rewrite equation 2 for the graphical solution as;

δ = A0 M1 ……… (3)

Where A0 = area of the M0 diagram divided by EI, and M1 = ordinate in M1

corresponding to the CG of the M0 diagram.

46

Solved example 1:

For the beam of constant EI shown in figure 28 (a), obtain the deflection at mid

span and the rotation at left support.

x w/m x P=1

A B A B

wL/2 L wL/2 ½ L/2 L/2 ½

(a) (b)

wL2/8 L/4

B.M.D. (M0) M1

(c) Figure 28 (d)

The equation of the bending moment of the original system M0 is;

M0(x) = wLx/2 - wx2/2

And the equation of M1 for the unit system shown in (b) is;

M1(x) = x/2 0 ≤ x ≤ L/2

M1(x) = x/2 – 1(x – L/2) = ½ (L – x) 0 ≤ x ≤ L/2

In this problem the symmetry for both M0 and M1 is clear, thus we can study half

the beam and multiply by two;

From equation 2 we can obtain the deflection at the middle of the beam;

δ = 2/EI ∫0L/2

(M0M1) dx

δ = 2/EI ∫0L/2

(wLx/2 – wx2/2) x/2 dx = 5 wL

4/ 384 EI

Alternatively equation 3 can give the same answer but graphically;

δ = 2/EI [2/3(wL2/8) L/2(5/8 x L/4)] = 5 wL

4/ 384 EI as before

47

And to calculate the rotation ӨA, we apply a unit moment at A, see figure28. The

reader may note that no symmetry in this case. Therefore integration should be

from 0 to L.

ӨA = 1/EI ∫0L (wlx/2 – wx

2) (x/L) dx = wL

3/ 24EI

1

1/L L 1/L

Figure 29

1 M2 diagram

Graphically the same result may be obtained by multiplying M0 and M2 diagrams;

ӨA = 1/EI [(2/3)wL2/8(L)1/2)] = wL

3/24EI

Solved example 2:

For the beam shown in figure 30 (a), obtain the deflections at C and D. Calculate

also the angle of rotation at B. assume EI = 40000 KN.m2.

20 KN 10KN

A C B D

3 m 3 m 1.5m (a)

-7.5 - 15KN.m

30 KN.m M0

(b)

48

1 KN

1.5 KN.m

M1 (c)

-0.75 - 1.5 1 KN

M2 (d)

1 KN.m

1KN.m M3

Figure 30 (e)

Solution:

Conveniently the bending moment diagram for the original system M0 is drawn in

figure 30 (b) separately for each case of loading. In figure 30 (c) the M1diagram is

drawn by replacing the given load by a unit load at the position of the required

deflection (point C). Similarly the M2 and M3 diagrams are drawn for unit load at d

and unit moment at B respectively.

δC = ∫M0 M1/EI dx

= 1/EI [- ½ x 7.5 x 3 x 1 – 7.5x3x0.75 – ½ x 7.5x3x1 + 2[1/2 x 30x3x1] = 50.625/EI

δC = 50.625x1000/40000 = 1.265 mm

δd = ∫M0M1 /EI dx

= 1/EI [1/2 x 15x6x1 + ½ x15x1.5x1 – ½ x30x3x0.5 – ½ x 30x3x1] = -11.25/EI = -

11.25x1000/4000 = - 0.281 mm

ӨB =1/EI [- ½ x 15x6x2/3 + ½ x 30x3x1/3 + ½ x 30x3x2/3] = 15/EI = 15/40000 =

0.000375 rad = 0.021 degree.

49

Tutorial Sheet of

Virtual Work for Beams

Problem (1) Using the virtual work theory obtain the rotation at A and the deflection at C for the

following beams (consider EI = 25000 KN.m2)

P = 40KN, a = 2m, L= 5m and w = 5KN/m

a

A P B

C

L Figure 1

A w/m B C

L L/4

Figure 2

Problem (2) Using the virtual work theory, obtain the deflection at C and the rotation at left support

for the beam of variable moment of inertia shown in figure 3. Take EI = 40000 KN.m2

10KN/m

A B

EI C 2EI

4m 4m Figure 3

Problem (3) For the frame in figure 4 obtain the vertical deflection at C and the horizontal

displacement at E. Obtain also the rotation at A. E = 20000KN/cm2

and I = 5000cm4.

B B 10 KN/m D

10 KN

C 2I

4m I I 2m

E

Figure 4

A

4 m 4 m

50

Problem (4) Obtain the the deflection and the slope at point D for the overhanging beam of variable

moment of inertia shown in figure 5. E = 20000KN/cm2

and I = 5000cm4.

20 KN/m

A B C

1 4 m 2m

2EI EI

51

5.1 Application to Frames:

Let us solve the following problem shown in figure 31 (a). It is required to

calculate the horizontal movement of support C and vertical deflection at D. All

members have constant inertia. Assume EI = 25000KN.m2.

40 KN

A D B

20 KN

3 m 3 m 4 m

C

20 KN

(a)

2m 2m

60 KN.m

M0 diagram

(b)

3 m -4 KN.m

1 KN 4/3 - 2 8/3 -

2/3 KN -

M1 diagram

1 KN

(c)Unit load 1 2/3 KN

52

1 KN

1.5KN.m

M2 diagram

(d)Unit load 2

Figure 31

Bending moment for original system (M0) and unit loads 1 & 2 (M1& M2) are

drawn in figures 31 (b), (c) and (d) respectively.

δC = 1/EIʃM0 M1 dx =1/EI[1/2x60x3x(-4/3)+/2x60x3x(-/3)]=-360/25000 =-0.0144m

= -14.4mm

Negative sign means direction is opposite to the assumed. That is from left to

right.

δD = 1/EIʃM0 M2 dx = 2/EI[1/2 x 60 x 3 x 1] = 180/25000 = 0.0072m = 7.2mm

In the above calculation of δD benefit of symmetry is applied.

Solved example 2:

For the frame shown in figure 32(a), calculate the deflection at D and rotation at

A. EI = 25000 KN.m2. 10 KN/m

B C 1m

3 m 2EI EI 10KN

EI D 1m

A 4 m (a)

53

-30KN.m

-30 -10

20KN.m

M0 diagram (b)

4/3 8/3m

-3KN.m - -2KN.m

- -

M1 diagram 5/3m

1KN

1KN 1/4KN

1/4KN Figure 32 (c)

Solution :

δD = 1/EI[1/2(-30x3x(-2)) + ½(-10x1x(-5/3)] + 1/2EI[-10x4(-1.5) + 1/2(-20x4(-8/3) +

2/3(20x4(-2.5)] = 115/EI = 115/25000 = 0.0046 m = 4.6 mm

ѳA = 1/EI[1/2(-30x3(-1)) + 1/2EI[-10x4(-1/2) + ½(-20x4(-2/3)) + 2/3(20x4(-1/2)] =

55/EI = 55/25000 = 0.0022 rad. = 0.126 degree

-1KN.m

- M2 diagram

-

1KN.m 1/4KN

1/4KN Figure 32 (d)

54

Solved example 3:

For the frame shown in figure 33 (a), calculate the deflection at point C and the

horizontal displacement of support E. Assume EI = 25000 KN.m2.

30 KN

B 2EI D 9 KN

EI C EI

3 m

9KN A E

27KN 1.5m 4.5 m 3 KN (a)

-27KN.m - -20.25

- + 33.75

M0 diagram

(b)

1m 1 KN 3 m

0.75 + 0.75

1.125KN.m

M1 diagram (c)

-3 -3KN.m

-3 -3KN.m

M2 diagram

1 KN 1 KN (d) Figure 33

Solution:

55

M0 diagram has been conveniently split into positive and negative parts, see

figure 33 (b). A unit load is applied at C and the corresponding M1 diagram is

drawn (Figure 33 c). M2 diagram is drawn for unit horizontal load at E (Figure 33

d).

δC = ʃ(M0 M1 / EI) dx = A0 M¯1 / EI

δC = (1/2EI)[1/2(33.75)1.5x0.75 + ½(33.75)4.5(0.75) – ½(20.25)4.5x0.75 –

20.25x1.5(1.125/2) – ½(6.75x1.5(0.375) = 22.77/(2x25000) = 0.455x10-3

m =0.455

mm

δE = ʃ(M0 M2 / EI) dx = A0 M¯2 / EI

δE = (1/EI)[1/2(-27)3(-3)] + (1/2EI)[1/2(-27)6(-3) + ½(33.75)1.5(-3) + ½(33.75)4.5(-

3)] = 121.5/EI + (1/2EI)[243 – 76 – 227.8125] = (121.5 – 30.4)/EI = 0.0037m = 3.7

mm

56

Tutorial Sheet of

Virtual Work for Frames

Problem (1) Using the virtual work theory obtain the rotation at A and the deflection at C for the

following frame (consider EI = 25000 KN.m2)

30KN

B D

C 2I

4 m I I

Figure 1

A E

2 m 4 m

Problem (2) For the frame of figure 2 obtain the rotation at C and the displacement at A. Assume EI

= 20000 KN.m2

W = 15KN/m 20KN

A

2EI B

3 m

Figure 2 EI

6m C

Problem (3) Obtain the rotation and the displacement C of the frame shown in figure 3. Assume EI =

30000 KN.m2

20KN W = 15KN/m

C

B

2EI

2 m

A

Figure 3 1.5m 1.5 m

2EI EI

Problem (4)

57

Using the virtual work theory, obtain the deflection at the middle of BC and the rotations

at A & D for the frame shown in figure below. EI = 25000 KN.m2

10 KN/m

B C 15 KN

2 EI

EI EI 2 m

4 m D

6 m

A

58

5.2 Application to Trusses:

Calculation of deformation for trusses may be done by the virtual work method.

Equation (2) may be written for trusses as;

δ = ʃ(FOFiL/EA) dx ……….. (4)

Where; F0 = forces in all members due to original (given) system of loading, Fi =

forces in all members due to unit load, A = area of truss members and L= the

length of member.

Let us study the truss shown in figure 34 (a), where the area of each member is

shown. E = 20000 KN/cm2. It is required to calculate the vertical deflection at joint

3, and relative movement between joints 3 and 6.

2 20cm2 4 20cm

2 6

25 25 15 25 25cm2

3m

1 20cm2 3 20cm

2 5

120 4m 240KN 4m 120KN (a)

-160KN - 160KN

200 0 200

-120KN - 120KN

0 0 (b) Figure 34

Joint 1:

From the vertical equilibrium of joint 1, we obtain the force in member 1-2 as;

F1-2 = - 120 KN

Joint 2; F24

α

Figure 35 F21 F23

59

From the simple geometry sin α = 3/5 and cos α = 4/5

From vertical equilibrium;

F21 + F23 sin α = 0 → F23 = 200 KN

And from horizontal equilibrium;

F24 + F23cos α = 0 → F24 = - 160 KN

From symmetry we can obtain all other forces as shown in Figure 34 (b).

To obtain the vertical displacement of joint 3, we apply a unit load downward at

joint 3, and get the corresponding forces in all members as shown in figure 34.

The reader may note the similarity between the unit load case and the original

load system. Thus all obtained forces in figure 34(b) are divided by 240 to obtain

forces in all members due to unit load. See figure 36.

Now we apply equation 4 to obtain the deflection of joint 3, as shown in table 3.

-160/240= -2/3 KN -2/3 KN

200/240=5/6 0 5/6KN

-120/240=-1/2 -1/2KN

0 0

1 KN forces due to unit load Figure 36

Member L (m) A (cm2) F0 (KN) F1(KN) F2(KN) F0F1L/A F0F2L/A

1-2

1-3

2-3

2-4

3-4

3-5

3-6

4-6

5-6

3

4

5

4

3

4

5

4

3

25

20

25

20

15

20

25

20

25

-120

0

200

-160

0

0

200

-160

-120

-1/2

0

5/6

-2/3

0

0

5/6

-2/3

-1/2

0

0

0

0

0

0

-1

0

0

7.2

0

100/3

21.333

0

0

100/3

21.333

7.2

0

0

0

0

0

0

-40

0

0

Total 123.732 -40

Table 3

60

From above table;

δ3 = ΣF0F1L/A = 123.732 x 1000 / 20000 = 6.1866 mm

The obtained deflection is positive, that is as assumed downward.

0 0 1KN

0 0 0 -1KN 0

0 1KN 0 Figure 37

To obtain the relative movement between node 3 and 6, we apply a unit load in

this direction 3-6 as shown in figure 37. Corresponding member forces are shown

on same figure and in table 3, column 6 (F2).

Joint 6; F64 3/5

α 4/5

F63

Figure 38 F65

ΣY = 0 → F65 + 3/5 + F63 sin α = 0

0 + 3/5 + F63 (3/5) = 0 → F63 = -1KN

Σ X = 0 → 4/5 + F64 + F63cos α = 0

4/5 + F64 + (-1)4/5 = 0 → F64 = 0

The relative displacement between node 3 and 6 is:

δ36 = - 40 x 1000/20000 = - 2mm

Assumed direction is not correct and the two joints will move away from each

other.

61

Tutorial Sheet of

Virtual Work for Trusses

Problem (1) Using the virtual work theory obtain the vertical displacement at 1 and the horizontal

displacement at 2 for the truss shown in figure 1 (consider E = 25000 KN/cm2) all cross

sections area are as shown.

100KN 150KN

1 3 5 7 9

25 20 25 20 cm2

25 25 20 15 20 15 20 20 25 cm2

2m

20 25 25 25cm2 10

2 4 6 8

3m 2m 2m 2m 2m

Figure 1

Problem (2) Obtain the vertical displacement at 6 and the relative displacement between joints 2 and

3 for the truss shown in figure 2 (consider E = 25000 KN/cm2) all cross sections area are as

shown.

100 100 150 50KN

1 3 5 7 9 11 20KN

25 20 25 25 25cm2

25 25 25 20 20 20 25 20 20 25 25cm2

2m

25 20 25 25 20cm2

2 4 6 8 10 12

3m 2m 2m 2m 2m

Figure 2

Problem (3) Obtain the vertical displacement at 6, horizontal displacement at 1 and the relative

displacement between joints 4 and 7 for the truss shown in figure 3 (consider E = 25000

KN/cm2). Assume all cross sections 25 cm

2.

100KN 100 KN 100 KN

1 3 5 7 9 30KN

1 m

1 m

2 4 6 8 10

3m 3m 3m 3m Figure 3