Deformation of structures

Deformation of Structures
Prof. Dr. Ahmed Hasan Zubydan

Deformation of Structures
Prof. Dr. Ahmed Hasan Zubydan
Professor of Structural Engineering
Civil Engineering Department, Faculty of Engineering
PortSaid University.

Contents
Chapter 1 Double Integration
1.1 Introduction
1.2 Sign Convension
Chapter 2 Moment Area
2.1 Introduction
Chapter 3 Elastic Load
3.1 Introduction
3.2 Sign Convension
3.3 Relationship between Load, Shear, and Bending Moment
Chapter 4 Conjugate Beam
4.1 Introduction
Chapter 5 Virtual Work
5.1 Work
5.2 Complementary Work
5.3 Aplication of Complementary Virtual Work
5.4 Forms of Virtual Internal Complementary Work due to Virtual Forces
5.5 Displacement by Virtual Work
Chapter 6 Castigliano’s Second Theorem
6.1 Introduction
6.2 Deformation due to Axial Force
6.3 Deformation due to Bending Moment
6.4 Deformation due to Torsion
Chapter 7 Method of Consistent Deformations
7.1 Introduction
1
1
3
21
21
53
53
54
125
63
63
83
63
64
86
87
89
119
119
120
120
121
147
147

1
Double Integration
1.1 Introduction
Consider a beam element AB, of length dx, which is subjected to positive moment M as shown in
Fig. 1. As the element bends, the top fibers are compressed while the bottom fibers are elongated.
In between, there is longitudinal fiber remains under fixed length. This fiber is so-called neutral
fiber of the member. It is assumed that plane cross sections before deformation remain plane after
deformation. For the beam element AB, extensions of lines through end cross sections intersect at
the center of curvature O forming angle d. The line eB is constructed parallel to the cross section
at the other end A constructing triangle Bde, which is similar to triangle OAB. Then, for small
angles,
t
dx dl
d
R y
 = = (1.1)
where R is the radius of curvature of the beam element, yt is the distance from the neutral fiber to
the top fiber, and dl is the shortening of that top fiber. Equation (1.1) can be rewritten as follow:
tydl
dx R
= (1.2)
Fig. 1.1 Beam element subjected to bending moment.
yt
d
dx
dl
O
R
A B
d
e
d
M M
d

2 Deformation os Structurs, by Prof. Dr. Ahmed Zubydan.
The strain  at the top fiber is given by the change in length per unit length and can be written as
dl
dx
 = (1.3)
For linearly elastic material, the stress-strain relationship is given as
E = (1.4)
where,  is the normal stress, and E is the material modulus of elasticity. From Eqs. (1.2), (1.3),
and (1.4), the following relationship is deduced:
ty
E
R
 = − (1.5)
The minus sign is introduced to indicate that the element is being compressed. The
top fiber stress could also be expressed as
tM y
I
 = − (1.6)
where M is the applied bending moment and I is the second moment of area. The negative sign is
also to indicate the state of compressive stress. Eliminate of stress from Eqs. (1.5) and (1.6) leads
to
1 M
R EI
= (1.7)
The transverse displacement y of the beam element given in Fig. 1.1 is related to the radius of
curvature by the following relationship:
2
2
3 22
1
1
d y
dx
R dy
dx
=
  
+  
   
(1.8)
The value
dy
dx
is small compared to unity. So, Eq. (1.8) can be written as
2
2
1 d y
R dx
= (1.9)
Using Eqs. (1.7) and (1.9), the following relationship is deduced:

Double Integration 3
2
2
d y M
dx EI
= (1.10)
1.2 Sign Convention
The sign convention shown in Fig. 1.2 for displacement, rotation and moment is considered. In
this convention, the deflection is positive upward for the x direction shown in Fig.1.2a (positive x
to right). Unticlockwise rotation is considered positive (Fig. 1.2a) and the moment shown in Fig.
1.2b, with respect to x-axis, is considered positive.
(a) Positive deflection and rotation (b) Positive Moment
Fig. 1.2 Sign convention.
Example 1.1
Calculate the slope and deflection at end b for
the beam shown in figure.
Solution
Elastic curve
Free body diagram
x
+
Y +
M


Y +
x
+

2m
30 kN
a b
EI
a
b
b
b
30 kN
a
b
30 kN
60 kN.m
x
y
M(x)

Considering an origin at point a as shown in the free body diagram, the expression for bending
moment at distance x from a is
( ) 30 60M x x= −
The moment-curvature equation can be written as
2
2
( )d y M x
dx EI
=
or, ( )
2
2
1
30 60
d y
x
dx EI
= −
Integrating once with respect to x,
( )2
1
1
15 60
dy
x x c
dx EI
= − + (a)
where c1 is a constant of integration. Integrating once again with respect to x yields
( )3 2
1 2
1
5 30y x x c x c
EI
= − + + (b)
where c2 is another constant of integration. Integration constants c1 and c2 are evaluated by
substituting the two boundary conditions at support a:
• x = 0 , 0
dy
dx
=  from Eq. (a), we get c1 = 0
• x = 0 , y = 0  from Eq. (b), we get c2 =0
Slope and deflection at node b can be calculated from Eqs. (a) and (b) by substituting x = 2m as
follow:
( )
2
21 60 .
15 2 60 2b
kN m
EI EI

−
=  −  = rad (clockwise)
The negative sign indicates clockwise rotation
( )
3
3 21 80 .
5 2 30 2b
kN m
EI EI
−
 =  −  = m ()
The negative sign indicates downward deflection

Example 1.2
Calculate the slope at nodes a and b, and the maximum
deflection of the beam shown.
Solution
Elastic curve
Free body diagram
Considering an origin at point a, the expression for bending moment at distance x from a is
( ) 3
25M x x x= −
The moment-curvature equation can be written as:
2
2
( )d y M x
dx EI
=
( )
2
3
2
1
25
d y
x x
dx EI
= −
Integrating once with respect to x,
( )2 4
1
1
12.5 0.25
dy
x x c
dx EI
= − + (a)
where c1 is a constant of integration. Integrating once again with respect to x yields
( )3 5
1 2
1
0.05y x x c x c
EI
= − + + (b)
where c2 is another constant of integration. Integration constants c1 and c2 are evaluated by
substituting the boundary conditions at supports a and b:
5 m
30 kN/m
a
EI
b
a b
max.
a b
Xm
30 kN/m
25 kN 50 kN
x
a b
y M(x)

• At support a: x = 0 , 0y =  we get c2 = 0
• At support b: x = 5 m , y = 0  we get c1 from Eq. (b) as follow:
( )3 5
1
1
0 4.167(5) 0.05(5) (5)c
EI
= − +
Then, c1 = -72.93
Slope and at nodes a and b can be calculated from Eq. (a) by substituting x = 0 and x=5m,
respectively
( )
2
2 41 72.93 .
12.5(0) 0.25(0) 72.93a
kN m
EI EI

−
= − − = (clockwise)
( )
2
2 41 83.3 .
12.5 5 0.25 5 72.93b
kN m
EI EI
 =  −  − = (unticlockwise)
Determination of maximum deflection ( axm ):
The maximum deflection occurs at zero slope, then the substitution of  = 0 in Eq. (a) yields
( )2 41
0 12.5 0.25 72.93m mx x
EI
= − −
The Gauss-Seidel method is selected to solve this equation. The equation is rearranged as follows:
( )
1
2
41
72.9
12.5
m mx x
 
= + 
 
Stating with x = 0 and substituting it in the right side, then
( )
1
2
41
0 72.9 2.41495
12.5
mx
 
= + = 
 
The solution is improved by putting xm =2.41495 in the right side, then
( )
1
2
41
(2.41495) 72.9 2.55191
12.5
mx
 
= + = 
 
The previous step is repeated until the solution is conversed. Reasonable solution can be obtained
at x = 2.5963 m. The maximum deflection is resulted by substituting x = 2.5963 m. in Eq.(b)

3 5
max
3
1
4.167(2.5963) 0.05(2.5963) 72.93(2.5963)
122.3 .
( )
EI
kN m
m
EI
  = − − 
= − 
Example 1.3
Calculate the slope and deflection at point b.
Solution
Elastic curve
Since the equation of moment is different for each of part ab and bc, then the beam is divided into
two parts
Part ab (0 4x  m)
Free body diagram
Consider the origin at a and x is positive to right, the expression of bending moment at distance x
from point a is:
( ) 21.43M x x=
2
2
( )d y M x
dx EI
=
( )
2
2
1
21.43
d y
x
dx EI
=
( )2
1
1
10.7
dy
x c
dx EI
= + (a)
( )3
1 2
1
3.57y x c x c
EI
= + + (b)
Substituting the boundary condition at supports a,
4 m 3 m
50 kN
a
b
c
bbEI
a c
b
b
b
50 kN
a
b
c
bb
21.43 kN 28.57 kN
x
y M(x)

when x = 0 , y = 0  we get c2 = 0
Slope and deflection at node b can be calculated from Eqs. (a) and (b) by substituting x = 4m
( )2
1
1
10.7 4b c
EI
 =  + (c)
( )3
1
1
3.57 4 4b c
EI
 =  +  (d)
Part bc (4 7x  m)
Free body diagram
Consider the origin at a and x is positive to right, the expression of bending moment at distance x
from point a is:
( )( ) 21.43 50 4M x x x= − −
2
2
( )d y M x
dx EI
=
( )( )
2
2
1
21.43 50 4
d y
x x
dx EI
= − −
( )( )22
3
1
10.7 25 4
dy
x x c
dx EI
= − − + (e)
( )( )33
3 4
1
3.57 8.33 4y x x c x c
EI
= − − + + (f)
Substituting the boundary condition at support c,
when x = 7 , y = 0  we get
( )( )33
3 4
1
0 3.57 7 8.33 7 4 7c c
EI
=  − − +  + (g)
Slope and deflection at node b can be calculated from Eqs. (b) and (c) by substituting x = 4m
( )( )22
3
1
10.7 4 25 4 4b c
EI
 =  − − + (h)
( )( )33
3 4
1
3.57 4 8.33 4 4 4b c c
EI
 =  − − +  + (i)
50 kN
a
b
c
21.43 kN 28.57 kN
x
y M(x)

Equating of b from Eqs (c) and (h), equating b from Eqs. (d) and (i) leads to
1 3 0c c− = (j)
3 4 14 4 0c c c+ − = (k)
Solving Eqs. (g), (j) and (k), the constants c1, c3, and c4 are given as
1 3 142.8c c= = − , c4 = 0
The rotation and deflection at point b can be calculated from Eqs. (c) and (d) substituting x = 4m
as follow:
( )
2
21 28.4 .
10.7 4 142.8b
kN m
EI EI
 =  − = rad (unticlockwise)
( )
3
31 342 .
3.57 4 142.8 4b
kN m
EI EI
−
 =  −  = m ( )
Example 1.4
For the beam shown, it is required to calculate the slope
and the deflection at point a.
Solution
Elastic curve
It is convenient to start with part bc because there are two known boundary conditions. These
conditions are the deflections at support b and c which are equal zeros.
Free body diagram
2 m 6 m
50 kN
a
b c
EI2EI
a
b c
a
a
50 kN
66.7 kN 16.7 kN
xa
b c
y M(x)

( ) 16.67 100M x x= −
2
2
( )d y M x
dx EI
=
( )
2
2
1
16.67 100
d y
x
dx EI
= −
( )2
1
1
8.33 100
dy
x x c
dx EI
= − + (a)
( )3 2
1 2
1
2.78 50y x x c x c
EI
= − + + (b)
Substituting the boundary conditions,
• At support b: x = 0 , y = 0  we get c2 = 0
• At support c: x = 6 m , y = 0  we get c1 = 200
Slope at b can be calculated from Eq. (a) by substituting x = 0
( )
2
21 200 .
8.33 0 100 0 200b
kN m
EI EI
 =  −  + = rad (unticlockwise) (c)
Free body diagram
( ) 50M x x= −
2
2
( )d y M x
dx EI
=
( )
2
2
1
50
2
d y
x
dx EI
= −
( )2
3
1
25
2
dy
x c
dx EI
= − + (d)
( )3
3 4
1
8.33
2
y x c x c
EI
= − + + (e)
Substituting the boundary conditions at support b
• x = 2 ,
200
b
dy
dx EI
= =  we get c3 = 500
50 kN
66.7 kN 16.7 kN
x
a
b c
y
M(x)

• x = 2 , y = 0  we get c4 = -933.3
Slope and deflection at node a can be calculated from Eqs. (d) and (e) by substituting x = 0
( )
2
21 250 .
25 0 500
2
a
kN m
EI EI
 = −  + = rad (unticlockwise)
( )
3
31 466.7 .
16.67 0 500 0 933.3
2
a
kN m
EI EI
−
 =  +  − = m ()
Example 1.5
Calculate the slope and the deflection at point d
for the beam shown.
Solution
Elastic curve
Free body diagram
2
( ) 20 63.3 46.67M x x x= − + −
2
2
( )d y M x
dx EI
=
( )
2
2
2
1
20 63.3 46.67
2
d y
x x
dx EI
= − + −
2 m 3 m 1 m
50 kN40 kN/m
a
b c
d2EI EI EI
a
b
c
d
d
d
50 kN
40 kN/m
63.3 kN
46.7 kN.m 66.7 kN
16.7 kN
a b
b
c
d
40kN/m
63.3 kN
46.7 kN.m 16.7 kN
a b
M(x)
y

( )3 2
1
1
6.67 31.65 46.67
2
dy
x x x c
dx EI
= − + − + (a)
( )4 3 2
1 2
1
1.665 10.35 23.33
2
y x x x c x c
EI
= − + − + + (b)
Substituting the boundary conditions at support a
• At a; x = 0 , 0
dy
dx
=  we get c1=0
• At b; x = 0 , y = 0  we get c2=0
The deflection at b can be calculated from Eqs. (a) and (b) by substituting x = 2m
( )
3
4 3 21 17.8 .
1.665 2 10.35 2 23.33 2
2
b
kN m
EI EI
−
 = −  +  −  = m ()
Since there is an internal hinge at b, the elastic curve is not continuous at b and so,
( ) ( )b bleft right  .
( ) 16.67M x x= −
2
2
( )d y M x
dx EI
=
( )
2
2
1
16.67
d y
x
dx EI
= −
( )2
3
1
8.33
dy
x c
dx EI
= − + (c)
( )3
3 4
1
2.78y x c x c
EI
= − + + (d)
Substituting the boundary conditions at b and c
• At b; x = 0 ,
17.8
by y
EI
−
= =  we get c4 = -17.8
• At c; x = 3 , y = yc = 0  we get c3 = 30.95
Slope at support c can be calculated from Eq. (c) by substituting x = 3m
( )
2
21 44.1 .
8.33 3 30.95c
kN m
EI EI

−
= −  + = rad (clockwise)
Part cd (0 1x  m)
50 kN
66.67 kN16.67 kN
b
c
d
y M(x)
x

( ) 50 50M x x= −
2
2
( )d y M x
dx EI
=
( )
2
2
1
50 50
d y
x
dx EI
= −
( )2
5
1
25 50
dy
x x c
dx EI
= − + (e)
( )3 2
5 6
1
8.33 25y x x c x c
EI
= − + + (f)
Substituting the boundary conditions at support c
• x = 0 ,
44.1
c
dy
dx EI

−
= =  we get c5 = -44.1
• x = 0 , y = 0  we get c6 = 0
Slope and deflection at node d can be calculated from Eqs. (e) and (f) by substituting x = 1m
( )
2
21 69.1 .
25 1 50 1 44.1d
kN m
EI EI

−
=  −  − = rad (clockwise)
( )
3
3 21 60.7 .
8.33 1 25 1 44.1 1d
kN m
EI EI
−
 =  −  −  = m ()
Example 1.6
Using the Double Integration method, calculate the slope and the
deflection at point c for the shown frame.
EI is the same for each member.
Solution
50 kN
66.7 kN16.7 kN
b c
d
y
M(x)
x
3 m
3 m
20 kN
40 kN
a
b
c

Free body diagram Elastic curve
Considering the origin at node a and x is positive
upward
( ) 20 60M x x= − −
2
2
( )d y M x
dx EI
=
( )
2
2
1
20 60
d y
x
dx EI
= − −
( )2
1
1
10 60
dy
x x c
dx EI
= − − + (a)
( )3 2
1 2
1
3.33 30y x x c x c
EI
= − − + + (b)
• x = 0 , 0
dy
dx
=  we get c1 = 0
• x = 0 , y = 0  we get c2 = 0
Slope and deflection at node b can be calculated from Eqs (a) and (b) by substituting x = 3 m
( )
2
21 270 .
10 3 60 3b
kN m
EI EI

−
= −  −  = rad (clockwise)
( )
3
3 21 360 .
( ) 3.33 3 30 3b
kN m
Hal
EI EI
−
 = −  −  = m (→)
20 kN
40 kN
20 kN
40 kN
60 kN.m
a
b
c
a
b
c
c(Val)
c
c(Hal)b(Hal)
20 kN
40 kN
20 kN
40 kN
60 kN.m
a
b
c
M(x)
x
y

It is noted that b is perpendicular to member ab, so, b is horizontal.
Considering the origin at node b and x is positive to
the right.
( ) 40 120M x x= −
2
2
( )d y M x
dx EI
=
( )
2
2
1
40 120
d y
x
dx EI
= −
( )2
3
1
20 120
dy
x x c
dx EI
= − +
(c)
( )3 2
3 4
1
6.67 60y x x c x c
EI
= − + + (d)
Substituting the boundary conditions at node b
• x = 0 ,
270dy
dx EI
−
=  we get c3 = -270
• x = 0 , y = ∆b(Val) = 0 (neglecting the axial deformation in member ab)  we get
c4 = 0
Slope and deflection at node c can be calculated from Eqs. (b) and (c) substituting x = 3 m
( )
2
21 450 .
20 3 120 3 270c
kN m
EI EI

−
=  −  − = rad (clockwise)
( )
3
3 21 11760 .
( ) 6.67 3 60 3 270 3c
kN m
Val
EI EI
−
 =  −  −  = m ()
Since the axial deformation is neglected in member bc, the horizontal displacements at c and b are
the same, or
3
360 .
( ) ( )c b
kN m
Hal Hal
EI
−
 =  = m (→)
20 kN
40 kN
20 kN
40 kN
60 kN.ma
b c
x
M(x)y

Example 1.7
Calculate the horizontal displacement of roller
support at d for the shown frame.
Solution
Free body diagram Elastic curve
Considering the origin at node a and x is positive upward
( ) 25 125M x x= −
2
2
( )d y M x
dx EI
=
( )
2
2
1
25 125
d y
x
dx EI
= −
( )2
1
1
12.5 125
dy
x x c
dx EI
= − + (a)
( )3 2
1 2
1
4.167 62.5y x x c x c
EI
= − + + (b)
8 m
5 m
150 kN
50 kN/m
25kN/m
2EI
EI EI
a
b
d
c
150 kN
50 kN/m 25kN/m
25 kN
160.9 kN
125 kN.m
239.1 kN
c
d
b
a
c
d
b
a
b(Hal)
c
c(Hal)
d(Hal)
c
150 kN
25 kN
160.9 kN
125 kN.m
b
a
x
50 kN/m
y
M(x)

• x = 0 , 0
dy
dx
=  we get c1 = 0
• x = 0 , y = 0  we get c2 = 0
The deflection at node b can be calculated from Eq. (b) by substituting x = 5 m
( )
3
3 21 1042 .
( ) 4.167 5 62.5 5b
kN m
Hal
EI EI
−
 =  −  = m (→)
Free body diagram of member bc
Considering the origin at node b and x is positive to left
2
( ) 25 160.9M x x x= − +
2
2
( )d y M x
dx EI
=
( )
2
2
2
1
25 160.9
2
d y
x x
dx EI
= − +
( )3 2
3
1
8.33 80.45
2
dy
x x c
dx EI
= − + + (c)
( )4 3
3 4
1
2.08 26.82
2
y x x c x c
EI
= − + + + (d)
Substituting the boundary conditions at b and c
• At b; x = 0 , y = ∆b(Val) = 0  we get c4 = 0
• At c; x = 8 m , y = ∆c(Val) =0  we get c3 = -651.5
The rotation at node b can be calculated from Eq. (c) substituting x = 8 m
( )3 21
8.33 8 80.45 8 651.5
2
c
EI
 = −  +  − =
2
116.5 .kN m
EI
rad (unticlockwise)
Neglecting the axial deformation of member bc, then
125 kN
50 kN/m
25kN/m
160.9 kN
c
b
x
M(x)y

( ) ( )
3
1042 .
c b
kN m
Hal Hal
EI
 =  = m (→)
Part dc (0 5x  m)
Considering the origin at node c and x is positive to bottom
2
( ) 12.5 125 312.5M x x x= − + −
2
2
( )d y M x
dx EI
=
( )
2
2
2
1
12.5 125 312.5
d y
x x
dx EI
= − + −
( )3 2
5
1
4.167 62.5 312.5
dy
x x x c
dx EI
= − + − + …….(e)
( )4 3 2
5 6
1
1.0416 20.83 156.25y x x x c x c
EI
= − + − + + ….(f)
Substituting the boundary conditions at c
• x = 0 ,
116.5
c
dy
dx EI
= = we get c5 = 116.5
• x = 0,
1042
( )cy Hal
EI
=  = we get c6 = 1042
The deflection at end d can be calculated from Eq. (f) by substituting x = 5 m
( )4 3 21
( ) 1.0416 5 20.83 5 156.25 5 116.5 5 1042d Hal
EI
 = −  +  −  +  +
=
3
328 .kN m
EI
−
m ()


25 kN/m
239.1 kN
c
d
x
125 kN
239.1 kN
312.5 kN.m
M(x)
y

Problems
1) Calculate θa and ∆a for the shown beam.
2) Calculate θb and ∆b for the shown beam.
3) Calculate θd and ∆d for the shown beam.
4) Calculate θd and ∆a for the shown beam.
6) Calculate θc and ∆c for the shown frame.
1.0 m 1.0 m
50 kN/m
a b c
4 m 3 m
50 kN15 kN/m
a
b
c
6 m 2 m
80 kN
20 kN/m
a c
d
1.5 m 7 m 2 m
30 kN 50 kN
a
b c
d
2 m 3 m 1 m
50 kN 20 kN
a
b
c
d
2EI EI EI
3 m
3 m
40 kN
b
c
a
1 m
2EI
EI

7) Calculate ∆b and θa for the shown frame.
5 m
50 kN
30 kN/m
a
b
c
EI
2EI
8 m

2
Moment Area
2.1 Intoduction
The moment area method is based on a consideration of the geometry of the elastic curve and the
relation between the rate of change of slope and the bending moment at a point on the elastic curve.
Two theorems are associated with the moment area method.
First moment-area theorem:
Provided that the elastic curve is continuous between two points a and b (i.e., there are no internal
hinges between a and b). The angle between points a and b on the deflected structure, or the slope
at point b relative to the slope at point a, is given by the area under M/EI diagram between these
two points.
Proof:
Let ab be a continuous portion of the elastic curve as shown in Fig. 2.1.a. Within region ab, an
element of length dx is shown with tangents to the deflected member constructed at each end of
the element.
(a) Elastic curve (b) Elastic load
Fig.2.1 The elastic curve and the corresponding M/EI diagram
The angel between these end tangents, which represents the angle change that occurs over the
length dx, is denoted d. According to Eq. (1.10), this angle change is given by
y
d ba
ba
d
a b
x
+
M/EI a b
x dx x

M
d dx
EI
 = (2.1)
where M and I are the bending moment and the second moment of area at distance x, respectively,
and E is the modulus of elasticity of the beam element material. With reference to Fig. 2.1a, it is
clear that d is given by the shaded area.
The total angle change that occurs between tangents constructed at points a and b (
a
b ) results
from the summation of the incremental angle changes between a and b and is given by
b
a
b
a
d =  (2.2)
Substituting of Eq. (2.1) into Eq. (2.2) leads to
b
a
b
a
M
dx
EI
 =  (2.3)
M
area of diagrambetweena and b
EI
=
Second moment-area theorem:
Provided that the elastic curve is continuous between two points a and b (i.e., there are no internal
hinges between a and b). The deflection of point b on the deflected structure with respect to a line
drawn tangent to point a on the structure is given by the static moment of area under M/EI diagram
between points a and b taken about an axis through point b.
Proof:
As shown in Fig. 2.1a, if the tangents to the element of length dx are extended up to the vertical
line through point b. For small angels, the vertical distance between these tangents (d) is given
by
d x d = (2.4)
Substituting of Eq. (2.1) into Eq. (2.4), we obtain
M
d x dx
EI
 = (2.5)
which shows that the intercept d is given by the static moment of the shaded area of the M/EI
diagram (Fig. 2.1b) taken about an axis through point b. The accumulation of these intercepts for
all increments between points a and b gives

Moment Area 23
b
a
b
a
d =  (2.6)
where a
b is the vertical displacement of point b on the deflected structure with respect to a line
drawn tangent to the beam element at point a. Substituting of Eq. (2.5) into Eq. (2.6) leads to
b
a
b
a
M
x dx
EI
 =  (2.7)
= Moment of M/EI diagram between a and b
Example 2.1
Use the Moment Area method to determine the vertical
deflection and the slope at point b for the cantilever
shown.
Solution
Bending moment diagram (kN.m) M / EI diagram
Elastic curve
Rotation at b:
From the first moment-area theorem,
 
2
1
2
1 80 .
80 2a
b
kN m
EI EI
 =   =
and therefore,
2m
40 kN
a b
EI
80 80/EI
a b
a
b
b=b
ab =b
a

a
b a b  = −
Since a is a fixed end, a = 0. Therefore,
2
80 80 .
0b
kN m
EI EI

−
= − = rad (clockwise)
Negative sign indicates a clockwise rotation
Rotation at b:
From the second moment-area theorem,
1 2
2 3
1 106.7
80 2 2a
b
EI EI
 =     =  
Since the tangent of elastic line at a is horizontal. 2, then
3
106.7 .a
b b
kN m
EI
 =  = m ( )
Example 2.2
Calculate the rotation and the displacement at b and c for
the beam shown.
Solution
Bending moment diagram M/EI diagram
Elastic curve
1.5 m 1.5 m
30 kN/m
a b c
101.3
33.7
a b c
8.44
101.3/EI
33.7/EI
8.44/EI
a b c
a
b
c
c
c

Moment Area 25
Rotation at b:
 
2
1 1
2 2
1 101 .
101.3 1.5 33.7 1.5a
b
kN m
EI EI

−
= −   −   =
and therefore,
a
b a b  = −
Since a is a fixed end, then b = 0. Therefore,
2
101 101 .
0b
kN m
EI EI

−
Rotation at c:
 1 1
2 2
1
101.3 1.5 33.7 1.5a
c
EI
 = −   −   1 2
2 3
1
33.7 1.5 8.44 1.5
EI
+ −   +    
2
118 .kN m
EI
−
=
Since a is a fixed end, b = 0. Therefore,
a
c a c  = −
2
118 118 .
0
kN m
EI EI
= − = − rad (clockwise)
Deflection at b:
From the second moment area theorem,
 
3
1 1
2 2
1 88.6 .
101.3 1.5 1 33.7 1.5 0.5a
b
kN m
EI EI
 =    +    =
From geometry,
3
88.6 .a
b b
kN m
EI
 =  = m ()

Deflection at c:
From the second moment area theorem,
 1 1
2 2
1
101.3 1.5 2.5 33.7 1.5 2a
c
EI
 =    +   
1 2
2 3
1
33.7 1.5 1 8.44 1.5 0.75
EI
+    −     
3
259 .kN m
EI
=
From geometry,
3
259 .a
c c
kN m
EI
 =  = m ()
Example 2.3
Use the Moment Area method to determine the slope at
point b and the vertical deflection at mid-span for the
simple beam shown.
Solution
Elastic curve
Rotation at a:
4 m
10 kN/m
a c
bEI
4 m
a b c
80 80/EI
a b c
a
b
c
c
a
b
b
a
a

Moment Area 27
3
2
3
1 1706.7 .
80 8 4a
c
kN m
EI EI
 =    =  
From geometry,
a
c
a
acL


= −
2
1706.7/ 213.3 .
8
a
EI kN m
EI

−
Deflection at b:
32
3 8
1 320
80 4 4a
b
EI EI
 =     =  
From geometry,
3
213.3 320 533.3 .
. 4a
b a ab b
kN m
L
EI EI EI
 = −  =  − = m ( )
Example 2.4
Determine the slope at d and c and the vertical
deflection at b for the simple beam shown.
Solution
Bending moment Diagram M/EI diagram
2 m 2 m 2 m
30 kN 60 kN
a
b c
d
EI
80
100
a b c d
80/EI
100/EI
a b c d

Elastic curve
Rotation at d:
 1 1
2 2
1
80 4 2 100 4 4d
a
EI
 =    +   
3
1120 .kN m
EI
=
From geometry,
d
a
d
adL


=
1120/
6
EI
=
2
186.7 .kN m
EI
= rad (unticlockwise)
Rotation at c:
 
2
1
2
1 100 .
100 2d
c
kN m
EI EI

−
= −   =
Then,
d
c d c  = −
2
186.7 100 86.7 .
c
kN m
EI EI EI
 = − = rad (unticlockwise)
Deflection at b:
From the second moment area,
3
1 1
2 2
1 2 453.3 .
80 2 100 4 2
3
d
b
kN m
EI EI
 
 =    +    = 
 
From geometry,
a
b
c
d
a
d
b
d
c
d
b
c
c
d
d

Moment Area 29
d
b d bd bL =  − 
3
186.7 453.3 293.5 .
4b
kN m
EI EI EI
 =  − =
Example 2.5
Using the Moment Area method, determine the rotation
at a, c and d, and the vertical deflection at b and d. EI
is the same for each member.
Solution
Elastic curve
Rotation at c:
To determine the sign of rotation at c, each of positive and negative rotation is determined
individually as follows:
Positive moment on span ac Negative moment on span ac
3 m 3 m 2 m
50 kN
20 kN/m
a
b
c
d
100
a b
c d
90
100/EI
a b
c
90/EI
d
50/EI
a
b
c
d
b
b
a
d
c d
ca
c
a
a
90/EI
cb
100/EI
a
c

Rotation at c due to positive moment Rotation at c due to negative moment
From second moment-area theorem,
3
2
3
1 1080 .
90 6 3c
a
kN m
EI EI
+
 =    =  
3
1 2
2 3
1 1200 .
100 6 6c
a
kN m
EI EI
−
 =     =  
It is observed that c c
a a
− +
   , then rotation at c is negative and can be calculated as follows:
( ) 2
1200 1080 / 20 .
6
c
a
c
ac
EI kN m
L EI

 − −
= − = − = rad. (clockwise)
Rotation at a:
From first moment-area theorem,
2
2 1
3 2
1 60 .
90 6 100 6c
a
kN m
EI EI
 =   −   =  
and therefore,
2
20 60 80 .c
a c a
kN m
EI EI EI
  
− −
= − = − = rad. (clockwise)
Deflection at b:
3
32 1 1
3 8 2 3
1 127.5 .
90 3 3 50 3 3a
b
kN m
EI EI
 =     −     =  
a c
a
+
b
a c
c
-
a
c-

Moment Area 31
From geometry,
3
80 127.5 112.5 .
. 3a
b a ab b
kN m
L
EI EI EI
 = −  =  − = m ( )
Rotation at d:
 
2
1
2
1 100 .
100 2c
d
kN m
EI EI

−
= −   =
and then,
2
20 100 120 .d
d c c
kN m
EI EI EI
  
− −
= − = − = rad (clockwise)
Deflection at d:
3
1 2
2 3
1 133.3 .
100 2 2c
d
kN m
EI EI
 =     =  
and then,
3
133.3 20 173.3 .
. 2c
d d c cd
kN m
L
EI EI EI
 =  + = +  = m ( )
Example 2.6
Calculate the rotations at b, c and d, and the deflections
at a, c and e for the beam shown.
Solution
1.5 m 3 m 4 m 2 m
30 kN 100 kN 50 kN
a
b
c
d
e

Elastic curve
Rotation at d:
To determine the sign of rotation at d, each of positive and negative rotation is determined
Positive moment on span bd Negative moment on span bd
Rotation at d due to positive moment Rotation at d due to negative moment
45
102.9
100
a b
c
d e
45/EI
102.9/EI
100/EI
a b
c
d e
b
c
d
e
a
a
b
e
b
a
c
102.9/EI
b c d
45/EI
100/EI
b c d
b dd
+
b
d+
b d
d
-b
d-

Moment Area 33
 
3
1 1
2 2
1 1200 .
102.9 3 2 102.9 4 4.33d
b
kN m
EI EI
+
 =    +    =
 
3
1 1
2 2
1 1200 .
45 3 1 100 4 5.67d
b
kN m
EI EI
−
 =    +    =
It is observed that d d
b b
− +
 =  , then the tangent of elastic curve at d is horizontal or the rotation at
d is equal zero
0d =
Deflection at e:
3
1
2
1 4 134 .
100 2
3
d
e
kN m
EI EI
 
 =    = 
 
Since 0d = , then,
3
134 .d
e e
kN m
EI
 =  = m ()
Deflection at c:
3
1 1
2 2
1 8 4 259 .
100 2 102.9 4
3 3
d
c
kN m
EI EI
 
 =    −    = 
 
Since 0d = , then,
3
259 .d
c c
kN m
EI
 =  = m ()
Rotation at c:
From the first moment area theorem,
 
2
1 1
2 2
1 5.8 .
100 4 102.9 4d
c
kN m
EI EI

−
=   −   =

Then,
d
c d c  = −
2
5.8 5.8 .
0c
kN m
EI EI
 = − = − rad (clockwise)
Rotation at b:
 
2
1 1 1
2 2 2
1 92 .
102.9 7 45 3 100 4d
b
kN m
EI EI

−
= −   +   +   =
Then,
d
b d b  = −
2
92 92 .
0b
kN m
EI EI
Deflection at a:
From the second moment area theorem
 
3
1
2
1 33.7 .
45 1.5 1b
a
kN m
EI EI
 =    =
From geometry
b
a b ab aL =  − 
3
92 33.7 105 .
1.5a
kN m
EI EI EI
 =  − =
Example 2.7
Calculate the deflections at b and d for the beam shown.
Solution
2 m 3 m 2 m
30 kN 20 kN
a
b
c
d

Moment Area 35
Bending moment diagram M / EI diagram
Elastic curve
Deflection at b:
3
1
2
1 4 44.44 .
33.3 2
3
a
b
kN m
EI EI
 
 =    = 
 
Since 0a = , then
3
44.44 .a
b b
kN m
EI
 =  = m ()
Deflection at d:
From the second moment-area theorem
 
3
1
2
1 120 .
40 3 2c
b
kN m
EI EI
 =    =
From geometry
( ) 2
120/ 44.4/ 25.2 .
3
c
b b
c
bc
EI EI kN m
L EI

− −  −
From the second moment-area theorem
3
1
2
1 4 53.3 .
40 2
3
c
d
kN m
EI EI
 
 =    = 
 
From geometry
33.3 40
a b c d
33.3/EI
40/EI
a b c d
a
b
c
d
b
c
d
c
b d
c

c
d c cd dL =  + 
3
25.2 53.3 104 .
2d
kN m
EI EI EI
 =  + = m ()
Example 2.8
For the frame shown in figure, it s required to calculate the
rotation at b and c, the horizontal displacement at b, and the
deflection at node c.
Solution
3 m
4 m
40 kN20 kN
a
b cEI
2EI
200
120
a
b c
200/2EI
120/EI
a
b c120/2EI

Moment Area 37
Elastic curve
Rotation at b:
Since the rotation at fixed end a is zero, then the application of the first moment-area theorem
leads to
 
2
1 1
2 2
1 320 .
60 4 100 4a
b b
kN m
EI EI
 
−
= = −   +   = rad (clockwise)
Horizontal displacement at b:
Since the rotation at fixed end a is zero, then the application of the second moment-area theorem
lead to
3
1 1 1 2
2 3 2 3
1 693.3 .
( ) 60 4 4 100 4 4a
b b
kN m
Hal
EI EI
 =  =     +     =   m ( )→
Rotation at c:
 
2
1
2
1 180 .
120 3b
c
kN m
EI EI

−
= −   =
From geometry,
2
320 180 500 .b
c b c
kN m
EI EI EI
  
− −
Vertical deflection at c:
b(Hal)
b
c(Val)
c
b
b
c
a
b

3
1 2
2 3
1 360 .
120 3 3b
c
kN m
EI EI
 =     =  
and then, from geometry,
3
320 360 1320 .
( ) . 3b
c bc cb
kN m
Val L
EI EI EI
 = +  =  + = m ( )
Example 2.9
For the frame shown in figure, calculate the
rotation at a, b, and d , and the horizontal
displacement at d and e.
Solution
Elastic curve
4 m 4 m
4 m
30 kN
100 kN
EI
2EI 2EI
EI
a
b
c
d
e
120
120
260
a
b c d
e
120/EI
60/EI
130/EI
a
b c d
e
b(Hal) b
a
b
d
b d
a
d(Hal)
ea
b
d
e
d

Moment Area 39
It is conventional to start with member bd because there are two known boundary conditions
(∆b(Val) = ∆d(Val) = 0).
Rotation at d:
3
1 1 1 2 1
2 3 2 3 2
1 2240 .
60 4 4 130 4 4 130 4 5.33d
b
kN m
EI EI
 =     +     +    =  
For small displacements,
2
2240/ 280 .
8
d
b
d
bd
EI kN m
L EI


= = = rad. (unticlockwise)
Rotation at b:
 
2
1 1 1
2 2 2
1 640 .
60 4 130 4 130 4d
b
kN m
EI EI

−
= −   −   −   =
and therefore,
2
280 640 360 .d
b d b
kN m
EI EI EI
  
−
Rotation at a:
From first moment-area theorem
 
2
1
2
1 240 .
120 4b
a
kN m
EI EI

−
= −   =
and therefore,
2
360 240 600 .b
a b a
kN m
EI EI EI
  
− −
Horizontal displacement at d:

From second moment-area theorem
3
1 1
2 3
1 320 .
120 4 4a
b
kN m
EI EI
 =     =  
From geometry
3
600 320 2080 .
( ) . 4a
b a ab b
kN m
Hal L
EI EI EI
 = −  =  − = m ( )→
Neglecting the axial deformation in member bd, then
3
2080 .
( ) ( )d b
kN m
Hal Hal
EI
 =  = m ( )→
Horizontal displacement at e:
From geometry,
3
2080 280 3200 .
( ) ( ) 4e d d de
kN m
Hal Hal L
EI EI EI
 =  +  = +  = m ( )→
Example 2.10
Calculate the slope at c and the horizontal displacement
at a for the frame shown.
Solution
Bending momentdiagram M / EI diagram
8 m 2 m
4 m
60 kN
40 kN
20 kN/m
a
b
c
d
80
80
120
a
b
c d
160
80/EI
80/EI
120/EI
a
b
c d
160/EI

Moment Area 41
Elastic curve
It is convenient to start with member bc because there are two known boundary conditions (b =
0 and c = 0)
Rotation at c:
To determine the sign of rotation at d, each of positive and negative rotation is determined
Positive moment on span bc Negative moment on span bd
Rotation at d due to positive moment Rotation at d due to negative moment
3
2
3
1 3413.3 .
160 8 4c
b
kN m
EI EI
+
 =    =  
a
b
c
d
a
b
a
b
d
b
b c
160/EI
80/EI
120/EI
b c
b c
c
+
d
c+
b cc
-
b
c-

3
1 1
2 2
1 8 16 3413.3 .
80 8 120 8
3 3
c
b
kN m
EI EI
−  
 =    +    = 
 
It is observed that c c
b b
− +
 =  , then the tangent of elastic curve at c is horizontal or the rotation at
d is equal zero
0c =
Horizontal displacement at a:
From the first moment-area theorem
( )
2
1 2
2 3
1 53.3 .
80 120 8 160 8c
b
kN m
EI EI

−
 = +  −   = 
Then
c
b c b  = −
2
53.3 53.3 .
0b
kN m
EI EI
 
3
1
2
1 266.67 .
80 2 3.33b
a
kN m
EI EI
 =    =
From geometry
( ) b
a a b abHal L =  − 
3
266.67 53.3 53.5 .
( ) 4a
kN m
Hal
EI EI EI
 = −  = m (→)
Example 2.11
For the frame shown in figure, it is required to calculate
the rotation at b, d and e, and the vertical deflection at
c and e.
Solution 2 m 4 m 4 m 3 m
4 m
100 kN 50 kN20 kN
a
b
c
d
ef
EI
EI 2EI 2EI 2EI

Moment Area 43
Elastic curve
Rotation at b:
3
1 2
2 3
1 1400 .
210 4.47 4.47b
a
kN m
EI EI
 =     =  
For small displacements,
2
1400/ 313 .
4.47
b
a
b
ab
EI kN m
L EI


= − = − = − rad. (clockwise)
Deflection at c:
3
1 2
2 3
1 666.67 .
125 4 4b
c
kN m
EI EI
 =     =  
and from geometry,
210
250
15040
a
b c d e
f
210/EI
125/EI 75/EI
40/EI
a
b c d e
f
a
b
c
c
c
b
a
b
b
b
d
e
f
a
b
c
c
c
d
d
d ef e
e
d

3
313 666.67 1918.7 .
. 4b
c b bc c
kN m
L
EI EI EI
 = +  =  + = m ( )
Rotation at d:
It should be noticed that, the first moment area theorem couldn’t be directly applied between b and
d due to the presence of internal hinge at c. This hinge causes a discontinuity in the elastic curve
at c (i.e. ( ) ( )c cleft right  ).
The application of the second moment-area theorem leads to
3
1 2
2 3
1 400 .
75 4 4d
c
kN m
EI EI
 =     =  
From geometry
2
1918.7/ 400/ 379.7 .
4
d
c c
d
cd
EI EI kN m
L EI

 −  −
= = = rad (unticlockwise)
Deflection at e:
3
1 2
2 3
1 225 .
75 3 3d
e
kN m
EI EI
 =     =  
From geometry,
3
379.7 225 914 .
. 3e
e d de d
kN m
L
EI EI EI
 = −  =  − = m ( )
Rotation at e:
From first moment-area theorem
 
2
1
2
1 112.5 .
75 3d
e
kN m
EI EI

−
= −   =
and therefore,
2
379.7 112.5 267.2 .d
e d e
kN m
EI EI EI
  = − = − = rad (unticlockwise)

Moment Area 45
Example 2.12
For the frame shown, calculate the rotation at b, d and e,
the horizontal deflection at b, and the vertical deflection
at c.
Solution
Elastic curve
Horizontal displacement at b:
Since the rotation at fixed end a is equal zero, then the application of the second moment-area
theorem leads to
2 m 3 m 2 m
5 m
50 kN 150 kN
a
b c d
e
2EI
EI EIEI
2 m
125
125
225
150
100
a
b
c de
125/2EI
125/2EI
225/EI
150/EI
100/EI
a
b
c de
e
b
b
d
b
e(Val)
c
c
b
c
d
a
b
e
c
d
b
a

 
3
1 781.25 .
( ) 125 5 2.5
2
a
b b
kN m
Hal
EI EI
−
 =  = −   = m ( )→
Rotation at b:
 
2
1 312.5 .
125 5
2
a
b
kN m
EI EI

−
= −  =
Because the rotation at a is equal zero, then,
2
312.5 .
0 a
b b
kN m
EI
 
−
Deflection at e:
3
1 2
2 3
1 133.3 .
100 2 2b
e
kN m
EI EI
 =     =  
From geometry,
( )
3
312.5 133.3 491.7 .
. 2b
e b be e
kN m
Val L
EI EI EI
 = −  =  − = m ( )
Rotation at e:
 
2
1
2
1 100 .
100 2b
e
kN m
EI EI
 =   =
and then,
2
312.5 100 212.5 .b
e b e
kN m
EI EI EI
  
− −
= + = + = rad (clockwise)
Deflection at c:

Moment Area 47
3
1 2
2 3
1 675 .
225 3 3b
c
kN m
EI EI
 =     =  
From geometry,
3
312.5 675 1612.5 .
. 3b
c b bc c
kN m
L
EI EI EI
 = +  =  + = m ( )
Rotation at d:
 
3
1
2
1 600 .
150 4 2d
c
kN m
EI EI
 =    =
From geometry,
2
1612.5/ 600/ 553.1 .
4
d
c c
d
cd
EI EI kN m
L EI

 +  +
Example 2.13
Calculate the horizontal and the vertical displacements
at f for the shown structure.
Solution
3 m 3 m 2 m
2 m
80 kN
40 kN
a
b
c
d
e
f
120
80
80
80
80
a b c d
e
f
120/EI
80/EI
80/EI
80/EI
80/EI
a b c d
ef

Elastic curve
Rotation at c:
 
3
1
2
1 1080 .
120 6 3c
a
kN m
EI EI
 =    =
From geometry
2
1080/ 180 .
6
c
a
c
ac
EI kN m
L EI


Displacement at d:
Part cdef
3
1
2
1 2 53.3 .
80 2
3
c
d
kN m
EI EI
 
 =    = 
 
From geometry
( ) c
d c cd dVal L =  + 
a
b
c
d
e
f
ca
c
c
d
e
f
d
c
d(Val)
e(Val)
e(Hal)
e
d
f(Hal)
f(Val)
f
e
c
d
e

Moment Area 49
3
180 53.3 413 .
( ) 2d
kN m
Val
EI EI EI
 =  + = m ()
Displacements at e:
 
2
1
2
1 80 .
80 2c
d
kN m
EI EI
 =   =
Then,
c
d c d  = +
2
80 180 260 .
d
kN m
EI EI EI
 = + =
 
3
1 160 .
80 2 1d
e
kN m
EI EI
 =   =
From geometry
( ) d
e d de eHal L =  + 
3
260 160 680 .
( ) 2e
kN m
Hal
EI EI EI
 =  + = m ()
Neglecting the axial deformation in member de, then
3
413 .
( ) ( )e d
kN m
Val Val
EI
 =  = m ()
Displacements at f:
 
2
1 160 .
80 2d
e
kN m
EI EI
 =  =
Then
d
e d e  = +

2
260 160 420 .
e
kN m
EI EI EI
 = + = rad (clockwise)
3
1
2
1 4 106.67 .
80 2
3
e
f
kN m
EI EI
 
 =    = 
 
From geometry
( ) ( )e
f e ef f eVal L Val =  +  − 
3
420 106.67 413 533 .
2
kN m
EI EI EI EI
=  + − = m ()
Neglecting the axial deformation in member ef, then
3
680 .
( ) ( )f e
kN m
Hal Hal
EI
 =  = m ()

Moment Area 51
Problems
1) Calculate θc and ∆c for the shown beam.
4) Calculate θa and ∆a for the shown beam.
5) Calculate θd and ∆d for the shown structure.
6) Calculate θa and ∆b for the shown structure.
1.5 m 1 m
50 kN/m
a b c
4 m 2 m 1.5 m
50 kN20 kN/m
EI 2EI
a
b
c
d
2EI
2 m 3 m 1 m
50 kN40 kN/m
a
b c
d2EI EI EI
2 m 6 m
60 kN 30 kN/m
a
b
c d
1.5 m
8 m 1.5
4 m
1.5 m
40 kN
20 kN/m
a
b
c
d
1.5
60 kN
5 m
8 m
150 kN
EI
2EI
a
b
c
d
5 m
60 kN

6) Calculate θa and ∆b for the shown structure.
8) Calculate θd and ∆d for the shown structure.
9) Calculate θe and ∆e for the shown structure.
5 m 5 m
5 m
100 kN
a
b c
d
e
8 m
5 m
100 kN
50 kN/m
20kN/m
2EI
EI EI
a
b
d
c
3 m 6 m 3 m
2 m
40 kN/m
a
b
c
d
e
2EI
EI EI
EI
2 m
50 kN

3
Elastic Load
3.1 Introduction
The moment area method can be expressed in the form of an analogy known as the elastic load
method. Consider the simply supported beam shown in Fig. 3.1a along with the resulting M/EI
diagram (Fig. 3.1b). The systematic application of the moment area theorems is outlined in Fig.
3.1d, and the expressions for θa ,θc, and ∆c are given by
b
a
b a
a
M EI xdx
l l


= =

(3.1)
c
a
c a c a
a
M
dx
EI
   = − = −  (3.2)
c
a
c a c a
a
M
s s xdx
EI
  = −  = −  (3.3)
The quantities given by Eqs. (3.1) to (3.3) can be determined by considering an analogous problem.
Consider an imaginary simply supported beam of the same length as the actual beam, which is
acted on by a loading formed by the M/EI diagram (elastic load) as illustrated in Fig.31b. By
taking moments about point b, we determine the reaction at a to be
b
a
a
M EI xdx
R
l
=

(3.4)
This reaction is shown in Fig. 3.1b. A free-body diagram of section ac (Fig.3.1c) permits one to
determine the shear and moment at point c. With the aid of Fig. 3.1c, we have
c
c a
a
M
Q R dx
EI
= −  (3.5)
and

54 Deformation of Structurs, by Prof. Dr. Ahmed Zubydan.
c
c a
a
M
M R s xdx
EI
= −  (3.6)
Comparing Eqs. (3.1) and (3.4), we see that Ra on the imaginary beam is equivalent to θa on the
real beam. Furthermore, by comparing Eqs. (3.2) and (3.3) with Eqs. (3.5) and (3.6), we see that
Qc and Mc on the imaginary beam are equivalent to θc and ∆c on the real beam, respectively. From
this discussion on can say that:
The slope and deflection at any point on a simply supported beam segment are given by the shear
and moment that resulted from applying the M/EI diagram as loading on an imaginary simply
supported beam of the same length as the given real beam.
(a) Loaded beam(b) Bending moment (elastic load)
(c) Free body of segment ac (d) Elastic curve
Fig. 3.1 Elastic load method
3.2 Sign Convention
The sign conventions shown in Fig.3.3 for displacement, rotation, and elastic shear and moment
are considered. In these conventions, the deflection is positive upward and the rotation is positive
unticlockwise rotation as shown in Fig. 3.3a. The elastic shear and moment on conjugate beam
is considered positive for the directions shown in Fig. 3.3b.
c
s
a b
w(x)
l
x dx
ca
b
s
x dx
ca
s
Ra
Mc (c)
Qc (c)
ca
c
ba
a
.s
a
ca
a bc

Elastic Load 55
(a) Positive deflection and rotation
(b) Positive elastic shear (c) Positive elastic moment.
Fig. 3.3 Sign convention.
Example 3.1
For loaded beam shown, determine θa, θb, θc, and ∆b.
Solution
Bending moment Elastic load M/EI
Elastic forces on the imaginary beam Elastic curve
Q(left) Q(right)
M M


Y +
x
+

4 m 3 m
50 kN20 kN/m
a
b
c
154.3
a b c
154.3/EI
40/EI
a b c
a b c
f2
f1
f3
Ra
Rc
a
b
c
b
a c
b

Considering the imaginary beam, the elastic forces on the beam can be calculated as follows:
2
1 3
1 106.67
40 4f
EI EI
=   =  
 1
2 2
1 308.6
154.3 4f
EI EI
=   =
 1
3 2
1 231.45
154.3 3f
EI EI
=   =
The elastic reaction Rc can calculated considering the moment of elastic load about node c for part
ac as follow:
0cM = ;
 
1 333.3
106.67 5 308.6 4.33 231.45 2
7
aR
EI EI
=  +  +  =
The elastic shear Qc can be calculated as follow:
333.3
a aQ R
EI
= =
Then,
2
333.3 .
( ) ( )a a
kN m
realbeam Q imaginarybeam
EI
 = = rad (unticlockwise)
The elastic reaction Rc can be calculated by considering the summation of elastic forces, so,
0Y = ;
 
1 313.3
106.67 308.6 231.45 333.33cR
EI EI
= + + − =
313.3
c cQ R
EI
= − = −
Then,
2
313.3 .
( ) ( )c c
kN m
EI
 = = − rad (clockwise)
The elastic moment at b is given as
 
1 709
231.45 1 313.3 3bM
EI EI
=  −  = −
Then,

Elastic Load 57
3
709 .
( ) ( )b b
kN m
realbeam M imaginarybeam
EI
 = = − m ()
The elastic shear at b is given as
 
1 81.9
313.3 231.45bQ
EI EI
= − =
2
81.9 .
( ) ( )b b
kN m
EI
Example 3.2
For loaded beam shown, determine θa, θc, θd, and
∆b.
Solution
Bending moment Elastic load
Elastic forces on the imaginary beam Elastic curve
Considering the imaginary beam ad, the elastic forces on the beam can be calculated as follows:
 1
1 2
1 133.3
133.3 2f
EI EI
=   =
2 m 2 m 2 m 2 m
80 kN 80 kN 40 kN
a
b c
d
e
133.3
106.7
80
a b c
d e
133.3/EI
106.7/EI
80/EI
a b c d e
a b c d
f2
f1 f3
f4
f5
Ra
Rd
a
c
e
d
b
a
c
d
b

 1
2 2
1 133.3
133.3 2f
EI EI
=   =
 1
3 2
1 106.7
106.7 2f
EI EI
=   =
 1
4 2
1 106.7
106.7 2f
EI EI
=   =
 1
5 2
1 80
80 2f
EI EI
=   =
The elastic reaction Ra can calculated by considering the moment of elastic forces about node d as
follow:
0dM = ;
1 4 2 240
133.3 4.67 133.3 3.33 106.7 2.67 106.7 80
6 3 3
aR
EI EI
 
=  +  +  +  −  = 
 
The elastic shear at a is given by
240
a aQ R
EI
= − = −
Then,
2
240 .
( ) ( )a a
kN m
EI
The elastic reaction Rd can be calculated by considering the summation of elastic forces, so,
0Y = ;
 
1 160
133.3 133.3 106.7 106.7 80 240dR
EI EI
= + + + − − =
160
d dQ R
EI
= =
2
160 .
( ) ( )d d
kN m
EI
The elastic moment at b is given as
1 2 391
133.3 240 2
3
bM
EI EI
 
=  −  = − 
 
Then,
3
391 .
( ) ( )b b
kN m
EI
 = = − m ()

Elastic Load 59
The elastic shear at c is given by
 
1 133
160 80 106.7cQ
EI EI
= + − =
Then,
2
133 .
( ) ( )c c
kN m
EI
Example 3.3
For loaded beam shown, determine θb, θc and ∆ at
midspan bc
Solution
Bending moment Elastic load
Elastic forces on the imaginary beam. Elastic curve
Applying the elastic load method on the simply supported beam bc, the elastic forces on the
imaginary beam can be calculated as follow:
1 m 6 m 2 m
40 kN 50 kN
30 kN/m
a b c d
40
100
a b c d
135
40/EI
100/EI
a
b c
d
135/EI
b c
f2
f1
f3
Rb Rc
a
b c
dm
b c
m

 1
1 2
1 120
40 6f
EI EI
=   =
2
2 3
1 540
135 6f
EI EI
=   =  
 1
3 2
1 300
100 6f
EI EI
=   =
The intermediate elastic reaction Rc can calculated considering the moment about node a for part
ac as follow:
 
1 90
120 4 540 3 300 2
6
bR
EI EI
= −  +  −  =
90
b bQ R
EI
= − = −
Then,
2
90 .
( ) ( )b b
kN m
EI
The elastic reaction Rc is calculated by considering the summation of elastic forces, or,
0Y = ;
 
1 30
120 300 540 90cR
EI EI
= − − + − =
30
c cQ R
EI
= =
Then,
2
30 .
( ) ( )c c
kN m
EI
To evaluate the deflection at midspan bc (point m) consider the following section at m:
Free body of section
at midspan bc
Then,
b
270/EI
120/EI
45/EI
Rb =90/EI
m
1.125 m
1.5 m
1 m
Mm
Qm
3 m

Elastic Load 61
 
3
1 191.25 .
270 1.125 45 1 120 1.5 90 3m
kN m
M
EI EI
=  −  −  −  = −
and,
2
191.25 .
( ) ( )m m
kN m
EI
 = = − m ()

Problems
1) Calculate ∆c and θa for the shown beam.
2) Calculate ∆ at mid-span bc and θc for the
shown beam.
3) Calculate ∆c and θd for the shown beam.
3 m 3 m 3 m
30 kN 80 kN
a
b c
d
EI
2 m 6 m
60 kN 40 kN/m
a
b c
1 m 3 m 4 m 2 m
30 kN 100 kN 60 kN
a
b
c
d
e

4
Conjugate Beam
4.1 Introduction
The method of elastic loads can be applied conveniently to simply supported beams, but the
analogies developed can be extended to beams with various support conditions. The conjugate
beam method uses the analogies between internal elastic generalized forces and actual deflections,
but it avoids the difficulties of interpreting the computed quantities for beams that are not simply
supported. To predict true deflection behavior using this method, it is required to make additional
comparisons between real beam behavior and that of an imaginary conjugate beam.
Since the problem of beam statics is governed by the following equation:
2
2
d M dQ
w
dxdx
= = (4.1)
where M, Q, and w are the bending moment, shearing force, and the applied load, respectively.
From the previous deferential equation, the integration of load gives the shear, and the bending
moment can be calculated by integration the shear.
Similarly, the beam deflection problem is governed by the following equation:
2
2
d y d M
dx EIdx

= = (4.2)
This is also a second order linear differential equation. Then, starting with the curvature, M/EI; the
first integration gives the slope, and the second integration yields the deflection. Then, if M/EI is
taken as loading on the conjugate beam, the resulting shear and moment will be the slope and the
deflections on the real beam.
The equations of of beam equilibrium and deformation can be summarized as follow:
Deflection y elastic bending moment

Slope
dy
dx
elastic shear
Curvature
2
2
d y M
EIdx
= bending moment (elastic load)
3
3
d y Q
EIdx
= shear force
4
4
d y w
EIdx
= loading
The supports for the conjugate beam can be readily determined from the following two
relationships:
Slope in actual beam  shear in conjugate beam
and
Deflection in actual beam  bending moment in conjugate beam
Depending on these relationships, the boundary conditions of Q and M on the conjugate beam are
forced to match the boundary conditions of  and y on the real beam.
Following these rules, we have
1. At a hinged or rollered end support of a real beam, there is slope and no deflection. So, the
corresponding conjugate beam support must have shear and no moment. Therefor, the
corresponding conjugate beam support must also be hinged or rollered end support.
2. At a fixed-end support of an actual beam, there is neither slope nor deflection. So, there must
be neither shear nor moment at the corresponding end of the conjugate beam; that is, the conjugate
beam end must be free end.
3. At a free end of a real beam, there is slope as well as deflection. Thus, the corresponding
conjugate-beam end must have both shear and moment; that is, the conjugate-beam support must
be fixed-end support.
4. At an interior support of an actual beam, there is slope and no deflection; thus, the
corresponding conjugate beam must have shear but no moment at that point.; that is, the conjugate-
beam must have an internal hinge at that location.
5. At an internal hinge of the real beam, there is sudden change of slope but no sudden change in
deflection. Therefore, the conjugate-beam must have a sudden change in shear but no sudden
change in moment at the same section. Therefor, the conjugate beam must have an interior
support to satisfy these conditions.
The various real support conditions and their corresponding conjugate substitutes are summarized
in Fig. 4.1. Figure 4.2 shows some examples of real beams and the corresponding conjugate beams,
based on the above principles.

Conjugate Beam 65
To use the conjugate beam method, an imaginary beam (conjugate beam) is conceived that has the
same length as the real beam and has the previous boundary condition transformation. The final
conjugate beam is then subjected to loading that corresponds to the M/EI diagram of the real beam.
The resulting shear and moment diagrams of the conjugate beam gives the slope and deflection
diagrams, respectively, of the real beam.
Note:
The sign convention of the elastic load method is followed in the conjugate beam method.
Real beam Transforms to Conjugate beam
Hinged support
y = 0,   0
 Hinged support
M = 0, Q  0
Roller support
y = 0,   0
 Roller support
M = 0, Q  0
Fixed support
y = 0,  = 0
 Free end
M = 0, Q = 0
Free end
y  0,   0
 Fixed end
M  0, Q  0
Interior support
y = 0,   0
 Interior hinge
M = 0, Q  0
Interior hinge
y  0,   0
 Interior support
M  0, Q  0
Fig. 4.1 Transformation of real beam to conjugate beam.

Real beam Conjugate beam
Fig. 4.2 Some real beams and the corresponding conjugate beams.
a b
a b
a b c
a b c
a b c d
a b c d
a b c d e
a b
a b
a b c
a b c
a b c d
a b c d
a b c d e

Conjugate Beam 67
Example 4.1
Calculate the slopes and deflections at points b and c
for the structure shown under indicated loading.
Solution
Bending moment diagram (kN.m). Conjugate beam and loading.
Free body diagram of conjugate beam. Deflected shape.
The elastic forces on the conjugate beam can be calculated as follow:
 =   =  
21 675 .1 450 3
1 2
kN m
f
EI EI
 =  =
2
2
1 450 .
150 3
kN m
f
EI EI
 
=   = 
 
2
3
1 1 150 .
150 2
2
kN m
f
EI EI
 
=   = 
 
2
4
1 2 50 .
37.5 2
3
kN m
f
EI EI
The equilibrium of conjugate beam leads to
 
−
= − − = − − =
2
1 2
1 1125 .
675 450b
kN m
Q f f
EI EI
3 m 2 m
75 kN/m
a b c
600
150
a b c
37.5
600/EI
150/EI
37.5/EIa b
c
f1
f2
f3
f4
a b
c
Rc
Mc
a
b
c
c
c
b
b

 
−
= −  −  = −  −  =
2
1 2
1 2025 .
2 1.5 675 2 450 1.5b
kN m
M f f
EI EI
Then,

−
= =
2
1125 .
(Re ) ( )b b
kN m
al Beam Q Conjugate Beam
EI
rad (clockwise)
and,
−
 = =
3
2025 .
(Re ) ( )b b
kN m
al Beam M Conjugate Beam
EI
m ( )
 
−
= − − − = − − − =4 3 2 1
1 1225
50 150 450 675cQ f f f f
EI EI
=  −  −  − 4 3 2 1
4
1 3.5 4
3
cM f f f f
− 
=  −  −  −  = 
 
3
1 4 4425 .
50 1 150 450 3.5 675 4
3
kN m
EI EI
Then,

−
= =
2
1225 .
(Re ) ( )c c
kN m
al Beam Q Conjugate Beam
EI
rad (clockwise)
and,
−
 = =
3
4425 .
(Re ) ( )c c
kN m
al Beam M Conjugate Beam
EI
m ( )
Example 4.2
Determine the rotation at c and the deflection at a for
the beam shown under indicated loading. consider EI
is the same for all members.
Solution
2 m 6 m
60 kN 30 kN/m
a
b c
120
a b
c135
120/EI
135/EI
a
b
c

Conjugate Beam 69
The elastic forces on the conjugate beam are calculated as follow:
 =   =
2
1
1 2
1 120 .
120 2
kN m
f
EI EI
 =   =
2
1
2 2
1 360 .
120 6
kN m
f
EI EI
=   =  
2
2
3 3
1 540 .
135 6
kN m
f
EI EI
Considering part bc of the conjugate beam,
= 0cM ;
   = −  +  = −  +  =
2
2 3
1 1 30 .
4 3 360 4 540 3
6 6
b
kN m
R f f
EI EI
= 0YF ;
 = − + − = − + − =
2
2 3
1 150 .
360 540 30c b
kN m
R f f R
EI EI
= =
2
150 .
c c
kN m
Q R
EI
 = =
2
150 .
(Real ) ( )c c
kN m
Beam Q ConjugateBeam
EI
rad (unticlockwise)
Positive sign indicates unticlockwise rotation.
 
= −  +  = − 
 
3
1 4 100 .
120 30 2
3
a
kN m
M
EI EI
 = = −
3
100 .
(Re ) ( )a a
kN m
alBeam M ConjugateBeam
EI
m ( )
a b
c
f2
f1
f3
Ra
b
Rb
Rc
Ma a
b c
a c

Negative sign indicates downward displacement.
Example 4.3
Calculate the slopes at point c and d, and
deflection at point d for the shown beam.
Solution
 
=   = 
 
2
1
1 2 208.3 .
62.5 5
3
kN m
f
EI EI
 
=   = 
 
2
2
1 1 250 .
100 5
2
kN m
f
EI EI
 
=   = 
 
2
3
1 1 89.3 .
35.7 5
2
kN m
f
EI EI
 
=   = 
 
2
4
1 1 50 .
100 2
2 2
kN m
f
EI EI
5 m 2 m 1 m
50 kN
20 kN/m
EI 2EI
a
b
c
d
2EI
50
a b
c d
35.7/EI 50/2EI
100/2EI
62.5/EI
10/2EI
100/EI
35.7/2EI
a
b c d
f1
f2
f3
f4
f5
f6
f7
f8
Rc
a
b
c
c
d
Ra
Rd
Md
a
b
c
d
d
b
b

Conjugate Beam 71
 
=   = 
 
2
5
1 2 6.7 .
10 2
2 3
kN m
f
EI EI
 =  =
2
6
1 35.7 .
35.7 2
2
kN m
f
EI EI
 
=   = 
 
2
7
1 1 7.1 .
14.3 2
2 2
kN m
f
EI EI
 
=   = 
 
2
8
1 1 12.5 .
50 1
2 2
kN m
f
EI EI
The intermediate elastic reaction Rc can be calculated considering the moment about node a for
part ac as follow:
=  +   +  +   
2
1 2 4 53
1
2.5 5 5.67 6
7
cR f f f f
EI
−   +  +   
2
3 6 73
1
5 6 6.33
7
f f f
EI
=  +   +  +   
2
3
1
208.3 2.5 250 5 50 5.67 6.7 6
7EI
−   +  +  =  
2
2
3
1 160 .
89.3 5 35.8 6 7.1 6.33
7
kN m
EI EI
( )
Then,
= =
2
160 .
c c
kN m
Q R
EI
 = =
2
160 .
(Re ) ( )c c
kN m
alBeam Q ConjugateBeam
EI
rad (unticlockwise)
For part cd of conjugate beam,
 = − = −  =
2
8
1 147.5 .
160 12.5 1d c
kN m
Q R f
EI EI
 = =
2
147.5 .
(Re ) ( )d d
kN m
EI
rad (unticlockwise)
 
=  −  =  −  = 
 
3
8
2 1 2 151.7 .
1 160 1 12.5
3 3
d c
kN m
M R f
EI EI
 = =
3
151.7 .
(Re ) ( )d d
kN m
EI
m ( )

Example 4.4
Calculate the slope and deflection at points b and c
for the beam shown under indicated loading.
Solution
 
=   = 
 
2
1
1 1 5 .
40 1
4 2
kN m
f
EI EI
 
=   = 
 
2
2
1 1 20 .
40 1
2
kN m
f
EI EI
 
=   = 
 
2
3
1 1 40 .
40 2
2
kN m
f
EI EI
For conjugate beam,
= 0bM ;
 
= −  −  −  
 
1 2 3
1 2 2
1.67
3 3 3
dR f f f
1 m 1 m 2 m
60 kN
a b c
d
4EI EI
40
40
a b
c d
40/4EI
40/EI
a
b c d
f1
f2
f3
Rb
Rd
a
b c d a
b
d
c
c
d
b
c

Conjugate Beam 73
 
= −  −  −  = 
 
2
1 2 2 27.78 .
5 20 40 1.67
3 3 3
kN m
EI EI
( )
= 0YF ;
 = + − −2 3 1b dR f f f R
   = + − − = + − − =
2
2 3 1
1 27.22 .
20 40 5 27.78b d
kN m
R f f f R
EI EI
( )
= =
2
27.78 .
d d
kN m
Q R
EI
 = =
2
27.78 .
(Real ) ( )d d
kN m
Beam Q Conjugate Beam
EI
rad (unticlockwise)
 = =
2
27.22 .
(Real ) ( )b b
kN m
Beam R Conjugate Beam
EI
rad
− 
= −  = −  = 
 
3
1
2 1 2 3.33 .
5
3 3
b
kN m
M f
EI EI
−
 = =
3
3.33 .
(Real ) ( )b b
kN m
Beam M Conjugate Beam
EI
m ( )
 
−
= − − = − − =
2
2 1
1 12.22 .
20 5 27.22c b
kN m
Q f f R
EI EI

−
= =
2
12.22 .
(Real ) ( )c c
kN m
Beam Q Conjugate Beam
EI
rad (clockwise)
=  −  − 2 1
1
1 1.67
3
c bM f R f
− 
=  −  −  = 
 
3
1 1 28.88 .
20 27.22 1 5 1.67
3
kN m
EI EI
−
 = =
3
28.88 .
(Real ) ( )c c
kN m
Beam M Conjugate Beam
EI
m ( )
Example 4.5
Calculate the slope at point c and d, and the deflection at
points a and d for the beam shown under indicated loading.
Consider EI is the same for all members.
Solution
2 m 4 m 3 m
25 kN/m
a b c d

Free body diagram of conjugate beam. Deflected shape
=   =  
2
2
1 3
1 16.65 .
12.5 2
kN m
f
EI EI
 =   =
2
1
2 2
1 50 .
50 2
kN m
f
EI EI
 =  =
2
3
1 200 .
50 4
kN m
f
EI EI
=   =  
2
2
4 3
1 133.3 .
50 4
kN m
f
EI EI
 =   =
2
1
5 2
1 125 .
62.5 4
kN m
f
EI EI
 =   =
2
1
6 2
1 168.8 .
112.5 3
kN m
f
EI EI
=   =  
2
2
7 3
1 56.25 .
28.13 3
kN m
f
EI EI
Considering part bc for the conjugate beam,
Σ Mc = 0;
 
=  −  +  =  −  +  = 
 
2
3 4 5
1 1 1 1 1 1 1 75 .
200 133.3 125
2 2 3 2 2 3
b
kN m
R f f f
EI EI
( )
ΣMb = 0;
50
112.5
a b c d
28.13
5012.5
50/EI
112.5/EI
12.5/EI
50/EI 28.13/EI
a
b c
d
f1
f2
f4
f3
f5
RC
f6
f7
Rb
a
b c
d
b cRa Rd
Md
Ma
a
b c
d
d
c
b
a

Conjugate Beam 75
( )
=  −  + 
 
=  −  +  =  
 
3 4 5
2
1 1 2
2 2 3
1 1 1 2 116.65 .
200 133.3 125
2 2 3
cR f f f
kN m
EI EI
= =
2
75 .
b b
kN m
Q R
EI
 = =
2
75 .
(Real ) ( )b b
kN m
Beam Q ConjugateBeam
EI
rad (unticlockwise)
−
= − =
116.6
c cQ R
EI

−
= =
2
116.6 .
(Re ) ( )c c
kN m
EI
rad (clockwise)
Considering part ab of the conjugate beam,
− 
=  −  −  =  −  −  = 
 
3
1 2
4 1 4 200 .
1 2 16.6 1 50 75 2
3 3
a b
kN m
M f f R
EI EI
−
 = =
3
200 .
(Re ) ( )a a
kN m
EI
m ( )
Considering part cd of the conjugate beam,
 
=  −  − 
−
=  −  −  =
7 6
3
1.5 2 3
1 603.45 .
56.25 1.5 168.75 2 116.65 3
d cM f f R
kN m
EI EI
−
 = =
3
603.45 .
(Re ) ( )d d
kN m
EI
m ( )
Maximum positive deflection in part bc:
The maximum positive deflection occurs on part bc.
Assume that the maximum deflection occurs at
distance x from node b. At maximum deflection, the
slope of the elastic line is equal zero, then,
116.65/EI75/EI
b c
x
50/EI
62.5/EI62.5x/4EI
25x 2/8EI
25.x.(4-x) /2
s
s

0YF = (+) for the portion left to section s-s;
( )−
+  +  − −  =
2
25 475 2 25 1 50 1 62.5
. 0
3 8 2 2 2 4
x xx x x
x x x
EI EI EI EI EI
− + − + =3 2
4.165 17.187 50 75 0x x x
2.3x m
( )max
3
1 2 2.3 1 2.3
. 75 2.3 16.53 2.3 48.9 2.3
3 2 2 3
1 2.3 1 2.3 80.8 .
50 2.3 35.95 2.3
2 2 3
M at Sec s s
EI
kN m
EI EI
 
− =  +    +    
 
 
+ −   −    = 
 
 = =
3
max max
80.8 .
(Re ) ( )
kN m
EI
m ( )
Example 4.6
Calculate the deflections at b and c, and the rotation
at c for the beam shown.
Solution
75/EI
b
2.3 m
50/EI
35.95/EI
16.53
48.9
s
s
Mmax
2 m 2 m 4 m 2 m
100 kN
a b
c
d e
EI2EI 2EI
133.3
133.3
66.7
a
b c d
e
133.3/2EI
133.3/EI
a
b c d
e
66.7/2EI

Conjugate Beam 77
The elastic forces on the conjugate beam are determined as follows:
 
=   = 
 
2
1
1 1 66.65 .
133.3 2
2 2
kN m
f
EI EI
 
=   = 
 
2
2
1 1 133.3 .
133.3 2
2
kN m
f
EI EI
 
=   = 
 
2
3
1 1 266.6 .
133.3 4
2
kN m
f
EI EI
 
=   = 
 
2
4
1 1 33.3 .
66.7 2
2 2
kN m
f
EI EI
For the conjugate beam,
= 0dM ;
 
=  −  −  −  
 
1 2 3 4
1 8 4
7.33 4.67
6 3 3
bR f f f f
 
=  −  −  −  
 
= −
2
1 8 4
66.65 7.33 133.3 4.67 266.6 33.33
6 3 3
148.16 .
EI
kN m
EI
 
= −  = −  = − 
 
3
1
4 1 4 88.9 .
66.65
3 3
b
kN m
M f
EI EI
 = = −
3
88.9 .
(Re ) ( )b b
kN m
EI
m ( )
= −  −  + 1 2
2
3.33 2
3
c bM f R f
 
= −  −  +  = − 
 
3
1 2 403 .
66.65 3.33 148.16 2 133.3
3
kN m
EI EI
 = = −
3
403 .
(Re ) ( )c c
kN m
EI
a
b c d
e
f2
f1
f3
f4
Rb Rd
a
b
c
d
e
c
c
b

 = − − + = − − + = −
2
1 2
1 81.5 .
66.65 148.16 133.3c b
kN m
Q f R f
EI EI
 = = −
2
81.5 .
(Re ) ( )c c
kN m
EI
Example 4.7
Calculate the slope at b and d, and the deflection at c
for the beam shown under indicated loading.
Solution
The elastic forces are determined as follow:
 
=   = 
 
2
1
1 2 360 .
180 6
2 3
kN m
f
EI EI
 
=   = 
 
2
2
1 1 105 .
70 6
2 2
kN m
f
EI EI
6 m 1 m 3 m
40 kN
40 kN/m 20 kN/m
a
c
d
2EI EI
EI
b
70
22.5
180
70/2EI
180/2EI
22.5/EI
70/EI
a
b
c d
a
b
c d
b
Ra
Rb
f1
f2
f3
f4
Rc
Rd
a b
c
d

Conjugate Beam 79
 
=   = 
 
2
3
1 1 35 .
70 1
2
kN m
f
EI EI
 
=   = 
 
2
4
1 2 45 .
22.5 3
3
kN m
f
EI EI
Considering part ab of the conjugate beam,
= 0aM ;
   =  −  =  −  =
2
1 2
1 1 110 .
3 4 360 3 105 4
6 6
b
kN m
R f f
EI EI
= =
2
110 .
b b
kN m
Q R
EI
 = =
2
110 .
(Re ) ( )b b
kN m
EI
rad (unticlockwise)
Considering part bcd of the conjugate beam,
 
=  −  =  −  = 
 
3
3
2 1 2 86.67 .
1 110 1 35
3 3
c b
kN m
M R f
EI EI
 = =
3
86.67 .
(Re ) ( )c c
kN m
EI
= 0cM ;
 
=  −  −  
 
 
=  −  −  = 
 
3 4
2
1 2 3
1
3 3 2
1 2 3 6.4 .
110 1 35 45
3 3 2
d bR R f f
kN m
EI EI
= − = −
2
6.4 .
d d
kN m
Q R
EI
 = = −
2
6.4 .
(Re ) ( )d d
kN m
EI
rad (clockwise)
Example 4.8
Calculate the slope difference at c, slope at d, and the
deflection at points b, c and g for the beam shown
under indicated loading.
2 m 2 m 2 m 1 m 6 m
30 kN 60 kN
12 kN/m
a b c
d
e
f
EIEI2EI g

 =   =
2
1
1 2
1 54 .
108 2
2
kN m
f
EI EI
=   =  
2
2
2 3
1 64 .
24 4
kN m
f
EI EI
 =   =
2
1
3 2
1 12 .
24 1
kN m
f
EI EI
 =   =
2
1
4 2
1 36 .
24 3
kN m
f
EI EI
 =   =
2
1
5 2
1 117 .
78 3
kN m
f
EI EI
 =   =
2
1
6 2
1 117 .
78 3
kN m
f
EI EI
Considering part df of the conjugate beam,,
= 0fM ;
( )   =   −  =   −  = 
2
5 6 4
1 1 87 .
3 5 117 2 3 36 5
6 6
d
kN m
R f f f
EI EI
( )
−
= − =
2
87 .
d d
kN m
Q R
EI
108
24
24
78
a b c d
e fg
108/2EI
24/EI
24/EI
78/EI
a
b c
d
e fg
f1
f2
f3
Rd f5 f6
f4
a
b c
d
d
e f
Rc
g
Rb
Rf
a
b
g
c
d
e
fc
d
g
b

Conjugate Beam 81

−
= =
2
87 .
(Re ) ( )d d
kN m
EI
rad (clockwise)
Considering part abcd of the conjugate beam,,
 
=  −  =  −  = 
 
2
3
2 1 2 79 .
1 87 1 12
3 3
c d
kN m
M R f
EI EI
 = =
3
79 .
(Re ) ( )c c
kN m
EI
m ( )
= 0bM ;
 
=  +  +  −  
 
1 2 3
1 4
2 5 4.67
4 3
c cR f f R f
 
=  +  +  −  = 
 
2
1 4 144.75 .
54 64 2 87 5 12 4.67
4 3
kN m
EI EI
( )
 = =
2
144.75 .
(Re ) ( )c c
kN m
alBeam R ConjugateBeam
EI
rad
− 
= −  = −  = 
 
3
1
4 1 4 72 .
54
3 3
b
kN m
M f
EI EI
−
 = =
3
72 .
(Re ) ( )b b
kN m
EI
m ( )
 
=  +   −  −  
 
2 3
18
1
3 2 2.67 2
2
g d c
f
M R f R
EI
−
=  +   −  −  =  
3
3
8
1 36.5 .
87 3 32 2 12 2.67 144.75 2
kN m
EI EI
−
 = =
3
36.5 .
(Re ) ( )g g
kN m
EI
m ( )

Problems
1) Calculate ∆c and θa for the shown beam.
2) Calculate ∆d and θd for the shown beam.
3) Calculate ∆d and θd for the shown beam.
4) Calculate ∆d and θa for the shown beam.
5) Calculate ∆a and θf for the shown beam.
6) Calculate ∆b for the shown beam.
3 m 2 m 2 m
50 kN
a
b c
d
2EI 2EI EI
12 m 3 m 3 m
200 kN
50 kN/m
a
b c d
EI2EI 2EI
2 m 3 m 1 m
50 kN
40 kN/m
a
b c
d2EI EI EI
1 m 6 m 2 m
40 kN 50 kN
30 kN/m
a b c d
2 m 4 m 4 m 2 m 6 m
40 kN 80 kN
15 kN/m
a
b
c
d e
f
2EI 2EI EIEI
2 m 3 m
80 kN
a
b
c

5
Virtial Work
5.1 Work
When a force acting on a body and moves its point of application a certain distance, the product
of force and that distance is defined as work. Consider a nonrigid body is subjected a force of
magnitude P as shown in Fig. 5.1a. The displacement of the body along the line of action of P is
. If an incremental displacement d occurs along the line of action of force P, the incremental
work, dW, done by the force is defined as
(a) Nonrigid body (b) General force-displacement relationship.
(c) Constant force-displacement
relationship.
(d) Linear force-displacement
relationship.
Fig. 5.1 Work.
P
d
Undeformed
body
Deformed
body
P
d 
Displacement
Force
P
d 
Displacement
Force
P
d 
Displacement
Force

.dW P d=  (5.1)
which is the small shaded area in Fig. 5.1b. Thus, the total work done over the full displacement
 is
0
.W P d

=  (5.2)
which is the general case of material behavior. If the  occurs under a constant force P as shown
in Fig. 5.1c, the external work done by P is
.W P=  (5.3)
Another case of interest is when the force is proportional to the displacement. This case
represented in Fig. 5.1d and the work done is
1
2
.W P=  (5.4)
This case occurs when the force is causing the displacement that produces the work, and the
structure is linearly elastic.
For the system of n displacements and forces, the total work for the n components of general case
of behavior is given by
1 0
.
in
i i
i
W P d

=
=  (5.5)
For constant load values over the full range of the displacements, the total work done becomes
1
.
n
i
i
W P d
=
=  (5.6)
and for linear force-displacement relations, the total work done is
1
2
1
.
n
i
i
W P d
=
=  (5.7)
5.2 Complementary work
If the work is done by an incremental force dP being swept through a displacement of , as shown
in Fig. 5.2a, then the resulting work done is named complementary work. The incremental
complementary work, dW*
, is given by

Virtual Work 85
(a) Nonrigid body
(b) General force-displacement relationship.
(c) Constant force-displacement
relationship.
(d) Linear force-displacement
relationship.
Fig. 5.2 Complementary work.
*
.dW dP=  (5.8)
The total complementary work, W*
, done over the full magnitude of the force P is given by
*
0
.
P
W dP=  (5.9)
When the total complementary work results from the case of constant displacement value over the
full range of the force P as shown in Fig. 5.2c, the complementary work is
*
.W P=  (5.10)
If the structure is linearly elastic as shown in Fig. 5.2d, then the total complementary work is given
by
* 1
2
.W P=  (5.11)
For a system with n forces and n displacements, the total complementary work is given by
dP

Undeformed
body
Deformed
body
P

Displacement
Force
dP
P

Displacement
Force
dP
P

Displacement
Force
dP

*
1 0
.
iPn
i i
i
W dP
=
=  (5.12)
For the case of constant displacements over the full range of the forces, the total complementary
work is given by
*
1
.
n
i i
i
W dP
=
=  (5.13)
For linearly elastic case, the integration of Eq. (5.12) leads to
* 1
2
1
.
n
i i
i
W dP
=
=  (5.14)
A comparison between work and the complementary work shows that W = W*
for only the cases
of linear systems.
5.3 Application of complementary virtual work
Deflections can be determined by using the complementary virtual work method which uses virtual
forces. This method has the disadvantage that only one displacement quantity can be obtained at
a time for a designated loading condition. In this method, a fictitious loading system is introduced,
and the displacements of the actual system are used as displacements for this loading. This
fictitious loading system is a set of virtual forces, P, that are used to determine an actual
displacement ∆. This procedure thus requires two systems as shown in Fig. 5.3. These systems
are virtual force system in equilibrium, and actual deformation pattern that is geometrically
compatible.
Using these systems, we equate the external complementary virtual work to the internal
complementary virtual work of deformation as follow
(a) D system (actual deformation system). (b) P system (virtual force system).
Fig. 5.3 Virtual forces
D
P
P

Virtual Work 87
* *
e iW W= (5.15)
where W*
e and W*
i are the complementary virtual external and internal works, respectively.
Introducing the components of these works leads to
( ) ( )
1
n
i P Di
i Vol
P dVol  
=
 =  (5.16)
where σP is the stress due to virtual force δP, εD is the actual strain due to actual displacement ∆i
and Vol is the volume of the element.
5.4 Forms of virtual internal complementary work due to virtual forces
To evaluate the displacement in a structure, an expression for the complementary virtual internal
work is required. Consider a beam ab of length L and subjected to arbitrary loading (Fig. 5.4a) so
that an elemental length dx is subjected to an axial force ND, moment MD, and torque TD. While
the deformations due to shearing forces are neglected. The internal virtual work developed due to
each of the virtual forces is as follows:
Due to Virtual Axial Force NP
Since the element dx experiences a real displacement dLD due to actual force ND, the virtual work
performed by the virtual force NP on the element dx is
Beam under general case of loading.
(b) Beam element under
axial force.
(c) Beam element under
bending moment.
(d) Beam element under
torsional moment.
Fig. 5.4 Forms of internal work.
a bdx
L
NP
dx dLD
NP
dD
dx
MP
MP
dx
TP TP
D

*
i P DdW N dL= (5.17)
in which the displacement dLD is given by
D
D
N dx
dL
EA
= (5.18)
where, E and A is the modulus of elasticity and the cross sectional area of the element, respectively.
Substituting the value of dLD in Eq. (5.17) yields
* D
i P
N dx
dW N
EA
= (5.19)
For the entire member ab, the complementary virtual internal work becomes
*
0
L
D
i P
N dx
W N
EA
=  (5.20)
When analyzing a structure consists of m members and the axial force is constant along each
member, as in the case of trusses, Eq. (5.20) can be written as
*
1
m
D
i P
j
N L
W N
EA=
=  (5.21)
Due to Virtual Moment MP
Figure 5.4c shows a virtual moment MP acting on the element dx which is under a real displacement
dD caused by the applied external real moment MD. The virtual work performed by MP on the
element dx is
*
.i P DdW M d= (5.22)
in which, the real displacement dD is given by
D
D
M dx
d
EI
 = (5.23)
Therefore,
* D
i P
M dx
dW M
EI
= (5.24)

Virtual Work 89
The virtual work, for the entire beam, becomes
*
0
L
D
i P
M dx
W M
EI
=  (5.25)
Due to Virtual Torque TP
Applying the above procedure for twist, the following expressions for the internal virtual work for
twist:
*
0
L
D
i P
T dx
W T
GJ
=  (5.26)
where, TP and TD are the virtual and the real torque, respectively, G is the element shear modulus,
and J is cross sectional polar moment of inertia.
5.5 Displacemet by virtual work
To determine the real displacements in a structure by the method of virtual work, we require an
expression for the complementary external virtual work given by the left-hand side of Eq. (5.16),
that is,
( )*
1
n
e ii
i
W P D
=
=  (5.27)
If a displacement D at only one point in the deformed structure is required, we can use a unit value
for the external virtual force P and then, Eq. (5.27) becomes
*
1.eW =  (5.28)
The substitution of the external and the internal complementary virtual work into Eq. (5.15) leads
to
1 0 0
1.
L Lm
D D D
P P P
j
N L M dx T dx
N M T
EA EI GJ=
 = + +   (5.29)
where ∆ is the required displacement.
For convenience, values of 1 2
0
( ) ( )
L
f x f x dx in Eq. (5.25) for general loading shown in Fig. 5.5a is
given as follow:
1 2 1 2
0
( ) ( )
L
f x f x dx A Y=  (5.30)

where A1 is the area enclosed by the function f1(x) and Y2 is the coordinated value of function f2(x)
corresponding to the center of area of the function f1(x). As a special case if both the functions are
linear as shown in Fig. 5.5b the previous integration is given as follow:
( )1 2 1 1 2 2 1 2 2 1
0
( ) ( ) 2 2
6
L
L
f x f x dx a b a b a b a b= + + + (5.31)
(a) General integration (a) Special integration
Fig. 5.5 Integration of different functions.
Example 5.1
Determine the vertical deflection at b and the
rotation at a for the simple beam shown in
figure.
Solution
Bending moment MD (kN.m). Elastic curve.
f2 (x)
f1(x)
C.G.
A1
L
linear
Y2
f2(x)
f1(x)
L
linear
linear
a1
a2
b1
b2
4 m 3 m
50 kN
a
b
c
bb
85.7
+ +
a b c
a c
b
b
a

Virtual Work 91
Vertical displacement at b
Apply unit load at b in the required direction (Vertical) as shown in the virtual system.
Virtual load system
MP (kN.m).
D P
b
L
M M
dx
EI
 = 
2 2
3 3
1 1 1
85.7 4 1.71 85.7 3 1.71
2 2
b
EI
 
 =     +     
 
3
342 .kN m
EI
= m ()
Positive sign indicates that the displacement is in the same direction of virtual force (downward).
Rotation at a
Apply unit moment at a as shown in the virtual system.
Virtual load system
MP (kN.m).
D P
a
L
M M
dx
EI
 = 
1 2
2 3
1 4 3 3
2 85.7 85.7 1 85.7 3
6 7 7
a
EI

  
= −   +  −      
  
2
143 .kN m
EI
−
= rad (clockwise)
Negative sign indicates that the rotation is opposite to the assumed moment.
Example 5.2
Determine the vertical deflection and the rotation at d for
the beam shown in figure.
Solution
x =1.0 kN
a cb
+ +
1.71
x =1.0 kN.m
a c
b
1.0
- -
3/7
12 m 3 m 3 m
200 kN
50 kN/m
a b c d
EI2EI 2EI

Vertical displacement at d
Virtual load system
MP (kN.m).
1 2 2 1
2 3 3 2
1
1200 12 6 900 12 6
2
d
EI
 =     −      
( )
1 3
2 1200 6 2 600 3 1200 3 600 6
2 6EI
 
+   +   +  +  
 
3
1 2
2 3
1 11700 .
600 3 3
kN m
EI EI
+     =  
m ()
Rotation at d
Virtual load system
MP (kN.m).
1 2 2 1
2 3 3 2
1
1200 12 1 900 12 1
2
d
EI
 =     −      
( )1
2
1
1200 600 3 1
2EI
 + +   
 
2
1
2
1 2856 .
600 3 1
kN m
EI EI
+    = rad (clockwise)
1200
600
900
-
+
a
b c d
a b
c
d
c
d
6
3
-
x =1.0 kN
a b
c
d
1.0
-
1.0
x1=1.0 kN.m
a
b c d

Virtual Work 93
Example 5.3
Calculate the vertical deflection at a and the rotation at
f for the shown beam.
Solution
Rotation at f
Virtual load system
MP (kN.m).
1 4 1 1 4 1 1 1 1
2 130 120 2 130 2 60 130 60
2 6 6 6 6 6 3 3 6
f
EI

    
=   −  +   −   +  −     
    
1
2
1 2
60 2
9EI
 
+ −    
 
2
3
1 1
45 6
2EI
 
+ −    
 
2
83.3 .kN m
EI
= −
Vertical displacement at a
Virtual load system
MP (kN.m).
 1
2
1
80 2 2
2
a
EI
 =   
( ) ( )
1 4 4
2 80 2 2 130 1 80 1 130 2 2 130 1 60 1
2 6 6EI
 
+   −   +  −  + −   +  
 
2 m 4 m 4 m 2 m 6 m
40 kN 100 kN
10 kN/m
a
b
c
d e
f
2EI 2EI EIEI
80
130
60
-
+
-
45
a b
c
d e
f
a
b
c
d
e
f
a
f
x=1.0 kN.m
1.0
1/3
-
+
a b c d e
f
1/6
x=1.0 kN
2.0
-
a b c
d e f
1.0

3
53.3 .kN m
EI
= − m ()
Example5.4
Determine the horizontal displacement at b, the vertical
displacement at c, and the rotation at d for the shown frame.
Solution
Horizontal displacement at b
Virtual load system
MP (kN.m).
10 m
10 m
100 kN 300 kN
EI
2EI
a
b c
d
1000
1000
1250
++
+
a
b c d
a
b c
dc(Val)
d
b(Hal)
x =1.0 kN
+
+
10
10
1.0
1.0
1.0
a
b c d
+
5

Virtual Work 95
1 2
2 3
1
( ) 1000 10 10b Hal
EI
 =      
( ) 1 2
2 3
1 5
2 100 10 2 1250 5 1000 5 1250 10 1250 5 5
2 6EI
 
+   +   +  +  +     
 
3
51875 .kN m
EI
= m( )→
Vertical displacement at c
Virtual load system
MP (kN.m).
( ) 1 2
2 3
1 5
2 1250 2.5 1000 2.5 1250 5 2.5
2 6
c
EI
 
 =   +  +     
 
3
6250 .kN m
EI
= m( )
Rotation at d
Virtual load system
MP (kN.m).
( ) ( )
1 5 5
2 1250 0.5 1000 0.5 2 1250 0.5 1250 1
2 6 6
d
EI

 
= −   +  −   +  
 
x =1.0 kN
+
2.5 0.5
0.5
+
a
b c d
x =1.0 kN.m0.5
1/10
-
1.0
1/10
a
b
c
d

2
1770.8 .kN m
EI
−
Example 5.5
Calculate the vertical displacement at d, the
horizontal displacement at c, and the rotation at d for
the shown frame.
Solution
Virtual load system
MP (kN.m).
 1
2
1
180 3.354 2d
EI
 =     1
2
1
375.1 5 2
EI
+   
3
2480 .kN m
EI
= m ()
8 m 3 m
4 m
1.5 m
60 kN
20 kN/m
a
b
c
d
195.1
375.1
180
+
+
-
a
b
c
d
178.9
a
b
c
d
d(Val)
c(Hal)
x=1.0kN
3.0
3.0 -
+
a
b
c
d

Virtual Work 97
Rotation at d
Virtual load system
MP (kN.m).
1 2
2 3
1 16 4
195.1 8.94 178.9 8.94
33 11
d
EI

 
= −    −    
 
1
2
1 2
375.1 5
11EI
 
+    
 
 1
2
1
180 3.354 1
EI
+   
2
339 .kN m
EI
= − rad (unticlockwise)
Horizontal displacement at c
Virtual load system
MP (kN.m).
1 2
2 3
1 8
195.1 8.94 178.9 8.94 2
3
c
EI
 
 = −    −    
 
1
2
1 8
375.1 5
3EI
 
+ −    
 
3
6962 .kN m
EI
= − m (→)
Example 5.6
Determine the vertical deflection at c, the rotation at
e, and the rotation difference at c for the shown frame
solution
x=1.0 kN.m
3/11
8/11
-
+
1.0
1.0
-
a
b
c
d
x=1.0kN
4.04.0
- -
a
b
c
d
5 m 5 m
5 m
100 kN
200 kN
a
b c
d
e
EI EI
2EI2EI

Virtual load system.
MP (kN.m).
1 2
2 3
1
250 5 2.5c
EI
 =      
1 2
2 3
1
250 5 2.5
2EI
+      
1 2
2 3
1
750 5 2.5
2EI
+      
1 2
2 3
1
750 5 2.5
EI
+      
3
6250 .kN m
EI
= m( )
Rotation at e
Virtual load system
MP (kN.m).
250
750
-
--
-
a
b c d
e
750
250
a
b
c
d
e
c
c
e
2.5
2.5 2.5
2.5
x = 1.0 kN
0.50.5
0.50.5
a
b c d
e
+
-
+
-
0.5
0.5
0.5
0.5
1.0
1/10
1/10 1/10
1/10
x1=1.0kN.m
a
b
c d
e

Virtual Work 99
1 2
2 3
1
250 5 0.5e
EI
 = −      
1 2
2 3
1
250 5 0.5
2EI
+ −      
1 2
2 3
1
750 5 0.5
2EI
+       ( )
1 5
2 750 0.5 750 1
6EI
 
+   +  
 
2
1250 .kN m
EI
= rad (clockwise(
Rotation difference at c
Virtual load system
MP (kN.m).
1 2
2 3
1
250 5 1c
EI
 =        1
2
1
250 5 1
2EI
+     1
2
1
750 5 1
2EI
+   
2
1 2
2 3
1 2916 .
750 5 750 1
kN m
EI EI
+   +   =   rad
Example 5.7
Calculate the vertical displacements at c and b
for the shown structure. Consider axial
deformations in link members only.
Solution
Normal forces ND (kN). Bending moment MD (kN.m).
-x =1.0 kN.m
1.0
1.0 1.0
1.0
1/5 1/5a
b c d
e
4 m 4 m 4 m 4 m
3 m
100 kN 100 kN 100 kN
-266.7
200
-333.3
200
-333.3
a b c d e
f
g 200 200
a
b c d
e- - - -

Elastic curve.
Virtual load system NP (kN). Bending moment MP (kN.m).
D pP D
c
L
N N LM M
dx
EI EA
 = + 
1 2
2 3
4
200 4 2c
EI
 =       +  
2
5 333.3 1.67 3 200 1
EA
  +   +  
1
8 266.7 1.33
EA
 
3
2133.3 . 9630 .kN m kN m
EI EA
= + m ( )
Virtual load system NP (kN). Bending moment MP (kN.m).
 
2
5 333.3 0.83 3 200 0.5b
EA
 =   +   +  
1
8 266.7 0.67
EA
 
4800 .kN m
EA
= m ( )
a
b
c
d
f
e
g
b c
-1.33 1.0
-1.67
1.0
-1.67
a
b
c
d
e
f g
x = 1 kN
a
b d
- -
2 kN.m
c
2 kN.m
e- -
-0.67
0.5
-0.83
0.5
-0.83
a
b c d
e
f g
x = 1 kN
a b
d
1.0
c
1.0
e- -
+ +

Virtual Work 101
Example 5.8
Determine the displacements and the rotation at c for
the shown structure. Consider the axial deformation
in link members only.
Solution
Free body diagram. Bending moment. MD (kN.m).
Normal forces. ND (kN).
Elastic curve
Virtual load system.
MP (kN.m). NP (kN).
3 m 3 m 3 m 3 m
4 m
200 kN
a
b
c
d e
EI EI
EA
EA
200 kN
200 kN
150 kN
200 kN
150 kN
200 kN
b
c
d e
a
1200 kN.m
a b c
d e
+ +
250kN
-250kN
a
b c
d e
b(Val)
c (Hal)
2.5kN
x =1.0 kN
1 kN
2 kN
1.5 kN
2 kN
b
c
d
e
a
1.5 kN
+ +
-2.5kN
6 kN.m

1 2
2 3
2
( ) 1200 6 6c Val
EI
 =        
2
250 2.5 5
EA
+  
3
28800 . 6250 .kN m kN m
EI EA
= + m( )
Horizontal displacement at c
Virtual load system
MP (kN.m).
1 2
2 3
2
( ) 1200 6 8c Hal
EI
 =        
1
250 1.67 5 250 3.33 5
EA
+   +  
3
38400 . 3125 .kN m kN m
EI EA
= + m( )
Rotation at c
Virtual load system
MP (kN.m).
NP (kN).
1
2
1 1
1200 6
3
c
EI

 
= −    
 
+  
1
250 0.208 5
EA
 
2
1200 .kN m
EI
−
525kN
EA
+ rad (clockwise)
x = 1.0 kN
4/3 kN
2 kN
8/3 kN 4/3 kN
a b c
d
e
+ +
1 kN
8 kN.m
3.33
-3.33
x = 1.0 kN.m
1/8 kN
1/6 kN
1/8 kN
1/6 kN
a
b c
d e
1.0
-
0.208
-0.208

Virtual Work 103
Example 5.9
Calculate the vertical and the horizontal displacements at d
and the vertical displacement at h for the shown structure.
Solution
Bending moment. MD (kN.m). Normal force.ND (kN).
Elastic curve.
5 m 4 m 4 m
4 m
2.5 m
1 m
80 kN 40 kN30 kN/m
a b
c
d
e
f
g
h
265
375
375
_
_
+
+
93.75
b
c
d e
-80
160160
-164.9
-210
26.1
84.8
a b
c
d e
f
g
h
a
b
c
d
e
f
g hd(Hal)
d(Val) h(Val)

Virtual load system
MP (kN.m).
NP (kN).
1 2
2 3
1 10 5
( ) 375 5 93.75 5
3 2
d Val
EI
 
 =    −    
 
( )
1 1
375 265 3.5 5
2EI
 
+ −   
 
1
2
1 10
265 4
3EI
 
+ −    
 
 
1
84.8 1.6 6.4
EA
+  
3
1540 .kN m
EI
= − m ()
868 .kN m
EA
+ m ()
Horizontal displacement at d
Virtual system
MP (kN.m).
NP (kN).
( )
1 3.5
( ) 2 265 3.5 3.5 375
6
d Hal
EI
 
 = −   +  
 
1
2
1 7
265 4
3EI
 
+ −    
 
+  
1
84.8 2.4 6.4
EA
−  
3
1553 .kN m
EI
= −
1303 .kN m
EA
− m (→)
x=1.0 kN
-1.6 kN
5 kN.m
5 kN.m
5 kN.m-
-
-
a
b
c
d e
f
g
h
x=1.0 kN
-2.4 kN
-
3.5 kN.m
-
a b
c
d e f
g
h

Virtual Work 105
Vertical displacement at h
Virtual system
MP (kN.m).
NP (kN).
( )
1 3.5
( ) 2 265 8 375 8
6
h Val
EI
 
 =   −  
 
 1
2
1
265 4 5.333
EI
+   
 
1
84.8 2.56 6.4 160 4 4 2 210 2.676 4.72
EA
+   +    +  
 
1
164.9 4.123 4.123 26.1 1.758 4.123
EA
+   −  
3
3550 . 11775 .kN m kN m
EI EA
= + m ()
Example 5.10
Determine the horizontal and the vertical displacements at d,
and the vertical deflection at b for the shown truss.
EA is the same for each member
Solution
Member forces ND (kN). Deflected shape.
4.0 kN4.0 kN
-4.123 kN
-2.676 kN
-1.758 kN
2.561 kN
8 kN.m
x=1.0 kN
+
+
a b
c
d
e f
g
h
0
4 m 4 m
3 m
3 m225 kN
150 kN
a b
c
d
37.5kN
-187.5 kN
200 kN
-62.5 kN
-250kN
225 kN
150 kN
50 kN
75 kN
50 kN
a
b
d
c
a
b
d
c

Horizontal displacement at d
Virtual load system
NP (kN).
Member L (m) ND (kN) NP (kN) ND NP L
ab 8 200 0 0
ac 6 -37.5 0.375 -84.38
ad 5 -187.5 0.625 -585.95
bd 5 -250 0 0
cd 5 -62.5 0.625 -195.31
D PN N L -865.63
1 865630 .
( )d D P
kN m
Hal N N L
EA EA
−
 = = m ()
Virtual load system
NP (kN).
ab 8 200 0 0
ac 6 37.5 -0.5 -112.5
X=1 kN
-0.38kN
0.63 kN
0.63 kN
0.5 kN
0.5 kN
0
0
a
b
d
c
-0.5kN
-0.83 kN
0.83 kN X=1 kN
0.67 kN
1 kN
0.67 kN
0
0
a
b
d
c

Virtual Work 107
ad 5 -187.5 -0.834 781.88
bd 5 -250 0 0
cd 5 -62.5 0.834 -260.63
D PN N L 408.75
1 408750 .
( )d D P
kN m
Val N N L
EA EA
 = = m ( )
Virtual load system
NP (kN).
ab 8 200 -1.33 -2128
ac 6 37.5 -1.0 -225
ad 5 -187.5 0 0
bd 5 -250 1.67 -2087.5
cd 5 -62.5 1.67 -521.875
D PN N L -4962.375
1 4962375
( )b D PVal N N L
EA EA
−
 = = m( )
-1kN
-1.33 kN
1.67 kN
1.67 kN x = 1 kN
1.33 kN
1 kN
1.33 kN
0
a
b
d
c

Example 5.11
Calculate the horizontal and the vertical displacements at f for the
truss shown in figure.
EA is the same for each member.
Solution
Horizontal displacement at f
Virtual load system
NP (kN).
ab 7 4 0.286 8
ac 5.1 7.3 1.02 37.97
1 1 3 1 1
5 m
5 m
8 kN
8 kN
4 kN
a b
c
d
e f
8 kN
8 kN 4 kN
4
7.3
13.7
-20.4
-8
7.3
-9.2
-4.1
-0.8
16 kN
16 kN 20 kN
a
b
c
d
e f
a b
c
d
e f
x=1.0 kN
0.286
1.02
0.669
-1.457
1.02
-1.281
1.0
1.0 kN
1.429 kN 1.429 kN
0
0
a
b
c d
e
f

Virtual Work 109
ad 7.81 13.7 0.669 71.6
bd 5.1 -20.4 -1.457 151.6
cd 5 -8 0 0
ce 5.1 7.3 1.02 37.97
de 6.4 -9.2 -1.281 75.4
df 5.1 -4.1 0 0
ef 3 -0.8 1 -2.4
D PN N L 381
1 381
( )f D PHal N N L
EA EA
 = = m (→)
Vertical displacement at f
Virtual load system
NP (kN).
ab 7 4 0.143 4
ac 5.1 7.3 -0.204 -7.6
ad 7.81 13.7 -0.134 -14.34
bd 5.1 -20.4 -0.728 75.74
cd 5 -8 0 0
ce 5.1 7.3 -0.204 -7.6
de 6.4 -9.2 0.256 -15.07
df 5.1 -4.1 -1.02 21.33
ef 3 -0.8 -0.2 0.48
D PN N L 56.7
1 56.7 .
( )f D P
kN m
Val N N L
EA EA
 = = m ()
x=1.0 kN
0.143
-0.204
-0.134
-0.728
-0.204
0.256
-1.02
-0.2
0.286 kN 0.714 kN
0
a b
c
d
e f

Example 5.12
For the shown truss, it is required to
a) Calculate the vertical and the horizontal displacements at c
due to applied loads.
b) Calculate the vertical displacement at c due to temperature
change of +80 C0
in members ac and ce.(α = 1210-6
per C0
)
c) Calculate the horizontal displacement at c due lake of fit of 4
cm in member de
EA is the same for each member
solution
Vertical displacement at e due to applied loads
Member L (m) ND (kN) NP (kN)= ND /100 ND NP L
ab 11.18 -67.1 -0.671 503.4
ac 11.18 134.2 1.342 2013.5
bc 15.81 -21.1 -0.211 70.4
bd 15.81 -210.8 -2.108 7025.4
cd 10 -133.3 -1.333 1776.9
ce 22.36 223.6 2.236 11179.3
de 14.14 -282.8 -2.828 11308.6
D PN N L 33878
5 m 5 m 5 m 10 m
5 m
10 m
10 m
100 kN
a
b
c d
e
100 kN
-67.1
134.2
-21.1
-210.8
-133.3
223.6
-282.8
150 kN
250 kN
a
b
c
d
e
a
b
c d
e

Virtual Work 111
1 33878 .
( )e D P
kN m
Val N N L
EA EA
 = = m ()
Vertical displacement at e due to temperature change in members ac and ce.
D t L =
Member L (m) ∆D (m) NP (kN) ∆D NP
ab 11.18 0 -0.671 0
ac 11.18 10.7310-3 1.342 14.410-3
bc 15.81 0 -0.211 0
bd 15.81 0 -2.108 0
cd 10 0 -1.333 0
ce 22.36 21.4710-3 2.236 4810-3
de 14.14 0 -2.828 0
D PN 62.410-3
0( ) 62.4e PVal N =  = mm ()
Horizontal displacement at e due to applied loads
Virtual load system
NP (kN).
ab 11.18 -67.1 0 0
ac 11.18 134.2 2.236 3354.8
bc 15.81 -21.1 -1.054 351.6
bd 15.81 -210.8 -1.054 3512.7
cd 10 -133.3 -0.667 889
ce 22.36 223.6 2.236 11179.3
de 14.14 -282.8 -1.414 5654.3
D PN N L 24945
x=1.0 kN
2.236
-1.054
-1.054
-0.667
2.236
-1.414
1.0 kN
2.0 kN
2.0 kN
0a
b
c
d
e

1 24945 .
( )e D P
kN m
Hal N N L
EA EA
 = = m (→)
Horizontal displacement at e due to lake of fit in member de.
D t L =
Member L (m) ∆D(m) NP (kN) ND NP L
ab 11.18 0 0 0
ac 11.18 0 2.236 0
bc 15.81 0 -1.054 0
bd 15.81 0 -1.054 0
cd 10 0 -0.667 0
ce 22.36 0 2.236 0
de 14.14 -0.04 -1.414 0.05656
0 PN 0.05656
0( ) 56.56e PHal N =  = mm (→)
Example 5.13
Calculate the vertical displacement at h for shown
truss.
EA is the same for each member.
Solution
3 m 3 m 3 m 3 m
4 m
80 kN
40 kN
a
b
c
d
e
f
g
h
40 kN
15
-25
15
80
-30
-75
-50
30 30 30
20 kN 60 kN 40 kN
80 kN
0 0
a
b
c
d
e
f
g
h
a
b
c
d
e
f
g
h

Virtual Work 113
Vertical displacement at h
Virtual load system
NP (kN).
ab 3 15 -0.375 -16.88
ac 5 -25 0.625 -78.1
bc 4 80 0 0
bd 3 15 -0.375 -16.88
cd 5 -75 -0.625 234.38
ce 3 30 0.75 67.5
de 4 0 0 0
df 3 -30 -0.75 67.5
eg 3 30 0.75 67.5
fg 4 0 0 0
fh 5 -50 -1.25 312.5
gh 3 30 0.75 67.5
D PN N L 705
1 705 .
( )h D P
kN m
Val N N L
EA EA
 = = m ()
Example 5.14
Calculate the vertical displacement at d and the rotation of
member df for the truss shown.
E is the same for each member.
Solution
x=1.0 kN
-0.375
0.625
-0.375 -0.75
-0.625
-1.25
0.75 0.75 0.75
0.5 kN 0.5 kN 1.0 kN
0 00
a
b
c
d
e
f
g
h
4 m 4 m 4 m 4 m
5 m
100 kN
120 kN80 kN
a
b
c
d
e
f
g
h
2A 2A
2A
2A 2A 2A 2A
2A
A
A A A
A

Rotation of member df
Virtual load system
NP (kN).
Member L (m) A ND (kN) NP (kN) ND NP L
ab 4 2A 116 0.2 46.4/A
ac 6.4 2A -185.7 -0.32 190/A
bc 5 A 80 0 0
bd 4 2A 116 0.2 46.4/A
cd 6.4 A 83.2 0.32 170.4/A
ce 4 2A -168 -0.4 134.4/A
de 5 A 0 0 0
df 4 2A 124 -0.2 -49.6/A
dg 6.4 A 70.4 0.96 432.5/A
eg 4 2A -168 -0.4 134.4/A
fg 5 A 0 -1.0 0
fh 4 2A 124 -0.2 -49.6/A
gh 6.4 2A -198.5 0.32 -203.3/A
D PN N L
A

843/A
843 . 210.8
4D P
df df
N N L kN m kN
L
EA EA EA
 = = = rad (unticlockwise)
80 kN 120 kN
100 kN
116
-185.7
11680 124
83.2
70.4
124
-198.5
-168 -168
145 kN 155 kN
0
0
a
b
c
d
e
f
g
h
a
b
c
d
e
f
g
h
x=1.0 kN x=1.0 kN
0.2
-0.32
0.2 -0.2
0.32
0.96
-0.2
-1
0.32
-0.4 -0.4
0.25 kN 0.25 kN
0
0
a
b
c
d
e
f
g
h

Virtual Work 115
Virtual load system
NP (kN).
Member L (m) A ND (kN) NP (kN) ND NP L/A
ab 4 2A 116 0.4 92.8/A
ac 6.4 2A -185.7 -0.64 380.3/A
bc 5 A 80 0 0
bd 4 2A 116 0.4 92.8/A
cd 6.4 A 83.2 0.64 34.8/A
ce 4 2A -168 -0.8 268.8/A
de 5 A 0 0 0
df 4 2A 124 0.4 99.2/A
dg 6.4 A 70.4 0.64 288.4/A
eg 4 2A -168 -0.8 268.8/A
fg 5 A 0 0 0
fh 4 2A 124 0.4 99.2/A
gh 6.4 2A -198.5 -0.64 406.5/A
D PN N L
A

2338/A
2338 .
( ) D P
d
N N L kN m
Val
EA EA
 = = m ()
x=1.0 kN
0.4
-0.64
0.4 0.4
0.64
0.64
0.4
-0.64
-0.8 -0.8
0.5 kN 0.5 kN
0
0
0
a
b
c
d
e
f
g
h

Deformation of structures

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