Page 2
Now horizontal deflection at B, may be calculated as
B
u
( ) ( )
?
= ×
D
A
v B
H
EI
dx x M x M
u
d
) 1 ( (1)
() () ( ) ( ) ( ) ( )
? ? ?
+ + =
D
C
v
C
B
v
B
A
v
EI
dx x M x M
EI
dx x M x M
EI
dx x M x M d d d
()( ) () ( )
0
5 . 2 10 5 . 2 2 5
5 . 2
0
5
0
+
- -
+ =
? ?
EI
dx x x
EI
dx x x
() ()
? ?
-
+ =
5 . 2
0
2 5
0
2
5 . 2 20 5
EI
dx x
EI
dx x
EI EI EI 3
5 . 937
3
5 . 312
3
625
= + =
Hence,
EI
u
A
3
5 . 937
= ( ?) (2)
Example 5.3
Find the rotations of joint B and C of the frame shown in Fig. 5.4a. Assume EI to
be constant for all members.
Page 3
Now horizontal deflection at B, may be calculated as
B
u
( ) ( )
?
= ×
D
A
v B
H
EI
dx x M x M
u
d
) 1 ( (1)
() () ( ) ( ) ( ) ( )
? ? ?
+ + =
D
C
v
C
B
v
B
A
v
EI
dx x M x M
EI
dx x M x M
EI
dx x M x M d d d
()( ) () ( )
0
5 . 2 10 5 . 2 2 5
5 . 2
0
5
0
+
- -
+ =
? ?
EI
dx x x
EI
dx x x
() ()
? ?
-
+ =
5 . 2
0
2 5
0
2
5 . 2 20 5
EI
dx x
EI
dx x
EI EI EI 3
5 . 937
3
5 . 312
3
625
= + =
Hence,
EI
u
A
3
5 . 937
= ( ?) (2)
Example 5.3
Find the rotations of joint B and C of the frame shown in Fig. 5.4a. Assume EI to
be constant for all members.
Rotation at B
Apply unit virtual moment at B as shown in Fig 5.5a. The resulting bending
moment diagram is also shown in the same diagram. For the unit load method, the
relevant equation is,
( ) ( )
?
= ×
D
A
v
B
EI
dx x M x M d
? ) 1 ( (1)
wherein,
B
? is the actual rotation at B, ()
v
M x d is the virtual stress resultant in the
frame due to the virtual load and
()
D
A
M x
dx
EI
?
is the actual deformation of the frame
due to real forces.
Page 4
Now horizontal deflection at B, may be calculated as
B
u
( ) ( )
?
= ×
D
A
v B
H
EI
dx x M x M
u
d
) 1 ( (1)
() () ( ) ( ) ( ) ( )
? ? ?
+ + =
D
C
v
C
B
v
B
A
v
EI
dx x M x M
EI
dx x M x M
EI
dx x M x M d d d
()( ) () ( )
0
5 . 2 10 5 . 2 2 5
5 . 2
0
5
0
+
- -
+ =
? ?
EI
dx x x
EI
dx x x
() ()
? ?
-
+ =
5 . 2
0
2 5
0
2
5 . 2 20 5
EI
dx x
EI
dx x
EI EI EI 3
5 . 937
3
5 . 312
3
625
= + =
Hence,
EI
u
A
3
5 . 937
= ( ?) (2)
Example 5.3
Find the rotations of joint B and C of the frame shown in Fig. 5.4a. Assume EI to
be constant for all members.
Rotation at B
Apply unit virtual moment at B as shown in Fig 5.5a. The resulting bending
moment diagram is also shown in the same diagram. For the unit load method, the
relevant equation is,
( ) ( )
?
= ×
D
A
v
B
EI
dx x M x M d
? ) 1 ( (1)
wherein,
B
? is the actual rotation at B, ()
v
M x d is the virtual stress resultant in the
frame due to the virtual load and
()
D
A
M x
dx
EI
?
is the actual deformation of the frame
due to real forces.
Now, and () ( ) x x M - = 5 . 2 10 ( ) ( ) x x M
v
- = 5 . 2 4 . 0 d
Substituting the values of () M x and ()
v
M x d in the equation (1),
()
?
- =
5 . 2
0
2
5 . 2
4
dx x
EI
B
?
EI
x x
x
EI 3
5 . 62
3 2
5
25 . 6
4
5 . 2
0
3 2
=
?
?
?
?
?
?
+ - = (2)
Rotation at C
For evaluating rotation at C by unit load method, apply unit virtual moment at C as
shown in Fig 5.5b. Hence,
( ) ( )
?
= ×
D
A
v
C
EI
dx x M x M d
? ) 1 ( (3)
() ( )
?
-
=
5 . 2
0
4 . 0 5 . 2 10
dx
EI
x x
C
?
EI
x x
EI 3
25 . 31
3 2
5 . 2 4
5 . 2
0
3 2
=
?
?
?
?
?
?
- = (4)
5.6 Unit Displacement Method
Consider a cantilever beam, which is in equilibrium under the action of a system of
forces . Let be the corresponding displacements and
and be the stress resultants at section of the beam. Consider a second
system of forces (virtual)
n
F F F ,....., ,
2 1 n
u u u ,....., ,
2 1
M P, V
n
F F F d d d ,....., ,
2 1
causing virtual
displacements
n
u u u d d d ,....., ,
2 1
. Let
v v
M P d d , and
v
V d be the virtual axial force,
bending moment and shear force respectively at any section of the beam.
Apply the first system of forces on the beam, which has been
previously bent by virtual forces
n
F F F ,....., ,
2 1
n
F F F d d d ,....., ,
2 1
. From the principle of virtual
displacements we have,
( ) ( )
?
?
=
=
n
j
v
j j
EI
ds x M x M
u F
1
d
d
(5.11)
?
=
V
T
v d de s
Page 5
Now horizontal deflection at B, may be calculated as
B
u
( ) ( )
?
= ×
D
A
v B
H
EI
dx x M x M
u
d
) 1 ( (1)
() () ( ) ( ) ( ) ( )
? ? ?
+ + =
D
C
v
C
B
v
B
A
v
EI
dx x M x M
EI
dx x M x M
EI
dx x M x M d d d
()( ) () ( )
0
5 . 2 10 5 . 2 2 5
5 . 2
0
5
0
+
- -
+ =
? ?
EI
dx x x
EI
dx x x
() ()
? ?
-
+ =
5 . 2
0
2 5
0
2
5 . 2 20 5
EI
dx x
EI
dx x
EI EI EI 3
5 . 937
3
5 . 312
3
625
= + =
Hence,
EI
u
A
3
5 . 937
= ( ?) (2)
Example 5.3
Find the rotations of joint B and C of the frame shown in Fig. 5.4a. Assume EI to
be constant for all members.
Rotation at B
Apply unit virtual moment at B as shown in Fig 5.5a. The resulting bending
moment diagram is also shown in the same diagram. For the unit load method, the
relevant equation is,
( ) ( )
?
= ×
D
A
v
B
EI
dx x M x M d
? ) 1 ( (1)
wherein,
B
? is the actual rotation at B, ()
v
M x d is the virtual stress resultant in the
frame due to the virtual load and
()
D
A
M x
dx
EI
?
is the actual deformation of the frame
due to real forces.
Now, and () ( ) x x M - = 5 . 2 10 ( ) ( ) x x M
v
- = 5 . 2 4 . 0 d
Substituting the values of () M x and ()
v
M x d in the equation (1),
()
?
- =
5 . 2
0
2
5 . 2
4
dx x
EI
B
?
EI
x x
x
EI 3
5 . 62
3 2
5
25 . 6
4
5 . 2
0
3 2
=
?
?
?
?
?
?
+ - = (2)
Rotation at C
For evaluating rotation at C by unit load method, apply unit virtual moment at C as
shown in Fig 5.5b. Hence,
( ) ( )
?
= ×
D
A
v
C
EI
dx x M x M d
? ) 1 ( (3)
() ( )
?
-
=
5 . 2
0
4 . 0 5 . 2 10
dx
EI
x x
C
?
EI
x x
EI 3
25 . 31
3 2
5 . 2 4
5 . 2
0
3 2
=
?
?
?
?
?
?
- = (4)
5.6 Unit Displacement Method
Consider a cantilever beam, which is in equilibrium under the action of a system of
forces . Let be the corresponding displacements and
and be the stress resultants at section of the beam. Consider a second
system of forces (virtual)
n
F F F ,....., ,
2 1 n
u u u ,....., ,
2 1
M P, V
n
F F F d d d ,....., ,
2 1
causing virtual
displacements
n
u u u d d d ,....., ,
2 1
. Let
v v
M P d d , and
v
V d be the virtual axial force,
bending moment and shear force respectively at any section of the beam.
Apply the first system of forces on the beam, which has been
previously bent by virtual forces
n
F F F ,....., ,
2 1
n
F F F d d d ,....., ,
2 1
. From the principle of virtual
displacements we have,
( ) ( )
?
?
=
=
n
j
v
j j
EI
ds x M x M
u F
1
d
d
(5.11)
?
=
V
T
v d de s
The left hand side of equation (5.11) refers to the external virtual work done by the
system of true/real forces moving through the corresponding virtual displacements
of the system. The right hand side of equation (5.8) refers to internal virtual work
done. The principle of virtual displacement states that the external virtual work of
the real forces multiplied by virtual displacement is equal to the real stresses
multiplied by virtual strains integrated over volume. If the value of a particular force
element is required then choose corresponding virtual displacement as unity. Let
us say, it is required to evaluate , then choose
1
F 1
1
= u d and n i u
i
,....., 3 , 2 0 = = d .
From equation (5.11), one could write,
()
?
=
EI
ds M M
F
v 1
1
) (
1
d
(5.12)
where, (
1 v
M) d is the internal virtual stress resultant for 1
1
= u d . Transposing the
above equation, we get
?
=
EI
Mds M
F
v 1
1
) ( d
(5.13)
The above equation is the statement of unit displacement method. The above
equation is more commonly used in the evaluation of stiffness co-efficient .
ij
k
Apply real displacements in the structure. In that set and the other
all displacements . For such a case the quantity in
equation (5.11) becomes i.e. force at 1 due to displacement at 2. Apply virtual
displacement
n
u u ,.....,
1
1
2
= u
) ,......, 3 , 1 ( 0 n i u
i
= =
j
F
ij
k
1
1
= u d . Now according to unit displacement method,
()
?
=
EI
ds M M
k
v 2 1
12
) (
1
d
(5.14)
Summary
In this chapter the concept of virtual work is introduced and the principle of virtual
work is discussed. The terms internal virtual work and external virtual work has
been explained and relevant expressions are also derived. Principle of virtual
forces has been stated. It has been shown how the principle of virtual load leads to
unit load method. An expression for calculating deflections at any point of a
structure (both statically determinate and indeterminate structure) is derived. Few
problems have been solved to show the application of unit load method for
calculating deflections in a structure.
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