Castigliano’s Theorems - 2 | Structural Analysis - Civil Engineering (CE) PDF Download

 Page 1


Example 3.4 
Find the vertical deflection at A of the structure shown Fig. 3.5. Assume the 
flexural rigidity EI and torsional rigidity GJ to be constant for the structure. 
 
 
 
The beam segment is subjected to bending moment   ( ; x is 
measured from C )and the beam element 
BC Px a x < < 0
AB is subjected to torsional moment of 
magnitude and a bending moment of Pa ) b x ( Px B from measured is x ; 0 = = . The 
strain energy stored in the beam  is,  ABC
 
    dx
EI
Px
dx
GJ
Pa
dx
EI
M
U
b
b a
? ? ?
+ + =
0
2
0
2
0
2
2
) (
2
) (
2
      (1) 
After simplifications, 
     
    
GJ
b a P
EI
a P
U
2 6
2 2 3 2
+ = +
EI
b P
6
3 2
        (2) 
 
Vertical deflection  at 
A
u A is, 
 
    
GJ
b Pa
EI
Pa
u
P
U
A
2 3
3
+ = =
?
?
EI
Pb
3
3
+       (3) 
 
 
 
 
Page 2


Example 3.4 
Find the vertical deflection at A of the structure shown Fig. 3.5. Assume the 
flexural rigidity EI and torsional rigidity GJ to be constant for the structure. 
 
 
 
The beam segment is subjected to bending moment   ( ; x is 
measured from C )and the beam element 
BC Px a x < < 0
AB is subjected to torsional moment of 
magnitude and a bending moment of Pa ) b x ( Px B from measured is x ; 0 = = . The 
strain energy stored in the beam  is,  ABC
 
    dx
EI
Px
dx
GJ
Pa
dx
EI
M
U
b
b a
? ? ?
+ + =
0
2
0
2
0
2
2
) (
2
) (
2
      (1) 
After simplifications, 
     
    
GJ
b a P
EI
a P
U
2 6
2 2 3 2
+ = +
EI
b P
6
3 2
        (2) 
 
Vertical deflection  at 
A
u A is, 
 
    
GJ
b Pa
EI
Pa
u
P
U
A
2 3
3
+ = =
?
?
EI
Pb
3
3
+       (3) 
 
 
 
 
Example 3.5 
Find vertical deflection at C of the beam shown in Fig. 3.6. Assume the flexural 
rigidity EI to be constant for the structure. 
 
 
 
The beam segment CB is subjected to bending moment  ( ) and 
beam element 
Px a x < < 0
AB is subjected to moment of magnitude .  Pa
To find the vertical deflection at , introduce a imaginary vertical force Q at . 
Now, the strain energy stored in the structure is, 
C C
 
      dy
EI
Qy Pa
dx
EI
Px
U
b a
? ?
+
+ =
0
2
0
2
2
) (
2
) (
         (1) 
 
Differentiating strain energy with respect toQ , vertical deflection atC is obtained. 
 
    dy
EI
y Qy Pa
u
Q
U
b
C
?
+
= =
?
?
0
2
) ( 2
    (2) 
     
          dy Qy Pay
EI
u
b
C
?
+ =
0
2
1
    (3) 
 
 
Page 3


Example 3.4 
Find the vertical deflection at A of the structure shown Fig. 3.5. Assume the 
flexural rigidity EI and torsional rigidity GJ to be constant for the structure. 
 
 
 
The beam segment is subjected to bending moment   ( ; x is 
measured from C )and the beam element 
BC Px a x < < 0
AB is subjected to torsional moment of 
magnitude and a bending moment of Pa ) b x ( Px B from measured is x ; 0 = = . The 
strain energy stored in the beam  is,  ABC
 
    dx
EI
Px
dx
GJ
Pa
dx
EI
M
U
b
b a
? ? ?
+ + =
0
2
0
2
0
2
2
) (
2
) (
2
      (1) 
After simplifications, 
     
    
GJ
b a P
EI
a P
U
2 6
2 2 3 2
+ = +
EI
b P
6
3 2
        (2) 
 
Vertical deflection  at 
A
u A is, 
 
    
GJ
b Pa
EI
Pa
u
P
U
A
2 3
3
+ = =
?
?
EI
Pb
3
3
+       (3) 
 
 
 
 
Example 3.5 
Find vertical deflection at C of the beam shown in Fig. 3.6. Assume the flexural 
rigidity EI to be constant for the structure. 
 
 
 
The beam segment CB is subjected to bending moment  ( ) and 
beam element 
Px a x < < 0
AB is subjected to moment of magnitude .  Pa
To find the vertical deflection at , introduce a imaginary vertical force Q at . 
Now, the strain energy stored in the structure is, 
C C
 
      dy
EI
Qy Pa
dx
EI
Px
U
b a
? ?
+
+ =
0
2
0
2
2
) (
2
) (
         (1) 
 
Differentiating strain energy with respect toQ , vertical deflection atC is obtained. 
 
    dy
EI
y Qy Pa
u
Q
U
b
C
?
+
= =
?
?
0
2
) ( 2
    (2) 
     
          dy Qy Pay
EI
u
b
C
?
+ =
0
2
1
    (3) 
 
 
    
?
?
?
?
?
?
+ =
3 2
1
3 2
Qb Pab
EI
u
C
    (4)       
 
 But the force  is fictitious force and hence equal to zero. Hence, vertical 
deflection is, 
Q
 
     
EI
Pab
u
C
2
2
=     (5)  
 
 
3.3 Castigliano’s Second Theorem  
In any elastic structure having independent displacements  
corresponding to external forces along their lines of action, if strain 
energy is expressed in terms of displacements then equilibrium equations may 
be written as follows. 
n
n
u u u ,..., ,
2 1
n
P P P ,...., ,
2 1
n
 
    ,     1,2,...,
j
j
U
Pj
u
?
==
?
n     (3.9) 
 
This may be proved as follows. The strain energy of an elastic body may be 
written as 
 
    
n n
u P u P u P U
2
1
..........
2
1
2
1
2 2 1 1
+ + + =  (3.10) 
 
We know from Lesson 1 (equation 1.5) that  
  
       (3.11)  
11 2 2
..... ,       1,2,..,
ii i inn
Pku ku ku i n =+ + + =
 
where is the stiffness coefficient and is defined as the force at  due to unit 
displacement applied at 
ij
k i
j . Hence, strain energy may be written as, 
 
1111 12 2 2 211 22 2 11 2 2
11 1
[ ...] [ ...] ....... [ ...]
22 2
nn n
Uukuku ukuku ukuku =+ ++ + +++ + + (3.12)  
 
We know from reciprocal theorem
ij ji
kk = . Hence, equation (3.12) may be 
simplified as,  
 
[]
22 2
11 1 22 2 12 1 2 13 1 3 1 1
1
.... .... ...
2
nn n n n
Ukuku ku kuukuu kuu ?? = + ++ + + ++ +
??
 (3.13)
 
 
Page 4


Example 3.4 
Find the vertical deflection at A of the structure shown Fig. 3.5. Assume the 
flexural rigidity EI and torsional rigidity GJ to be constant for the structure. 
 
 
 
The beam segment is subjected to bending moment   ( ; x is 
measured from C )and the beam element 
BC Px a x < < 0
AB is subjected to torsional moment of 
magnitude and a bending moment of Pa ) b x ( Px B from measured is x ; 0 = = . The 
strain energy stored in the beam  is,  ABC
 
    dx
EI
Px
dx
GJ
Pa
dx
EI
M
U
b
b a
? ? ?
+ + =
0
2
0
2
0
2
2
) (
2
) (
2
      (1) 
After simplifications, 
     
    
GJ
b a P
EI
a P
U
2 6
2 2 3 2
+ = +
EI
b P
6
3 2
        (2) 
 
Vertical deflection  at 
A
u A is, 
 
    
GJ
b Pa
EI
Pa
u
P
U
A
2 3
3
+ = =
?
?
EI
Pb
3
3
+       (3) 
 
 
 
 
Example 3.5 
Find vertical deflection at C of the beam shown in Fig. 3.6. Assume the flexural 
rigidity EI to be constant for the structure. 
 
 
 
The beam segment CB is subjected to bending moment  ( ) and 
beam element 
Px a x < < 0
AB is subjected to moment of magnitude .  Pa
To find the vertical deflection at , introduce a imaginary vertical force Q at . 
Now, the strain energy stored in the structure is, 
C C
 
      dy
EI
Qy Pa
dx
EI
Px
U
b a
? ?
+
+ =
0
2
0
2
2
) (
2
) (
         (1) 
 
Differentiating strain energy with respect toQ , vertical deflection atC is obtained. 
 
    dy
EI
y Qy Pa
u
Q
U
b
C
?
+
= =
?
?
0
2
) ( 2
    (2) 
     
          dy Qy Pay
EI
u
b
C
?
+ =
0
2
1
    (3) 
 
 
    
?
?
?
?
?
?
+ =
3 2
1
3 2
Qb Pab
EI
u
C
    (4)       
 
 But the force  is fictitious force and hence equal to zero. Hence, vertical 
deflection is, 
Q
 
     
EI
Pab
u
C
2
2
=     (5)  
 
 
3.3 Castigliano’s Second Theorem  
In any elastic structure having independent displacements  
corresponding to external forces along their lines of action, if strain 
energy is expressed in terms of displacements then equilibrium equations may 
be written as follows. 
n
n
u u u ,..., ,
2 1
n
P P P ,...., ,
2 1
n
 
    ,     1,2,...,
j
j
U
Pj
u
?
==
?
n     (3.9) 
 
This may be proved as follows. The strain energy of an elastic body may be 
written as 
 
    
n n
u P u P u P U
2
1
..........
2
1
2
1
2 2 1 1
+ + + =  (3.10) 
 
We know from Lesson 1 (equation 1.5) that  
  
       (3.11)  
11 2 2
..... ,       1,2,..,
ii i inn
Pku ku ku i n =+ + + =
 
where is the stiffness coefficient and is defined as the force at  due to unit 
displacement applied at 
ij
k i
j . Hence, strain energy may be written as, 
 
1111 12 2 2 211 22 2 11 2 2
11 1
[ ...] [ ...] ....... [ ...]
22 2
nn n
Uukuku ukuku ukuku =+ ++ + +++ + + (3.12)  
 
We know from reciprocal theorem
ij ji
kk = . Hence, equation (3.12) may be 
simplified as,  
 
[]
22 2
11 1 22 2 12 1 2 13 1 3 1 1
1
.... .... ...
2
nn n n n
Ukuku ku kuukuu kuu ?? = + ++ + + ++ +
??
 (3.13)
 
 
Now, differentiating the strain energy with respect to any displacement  gives 
the applied force  at that point, Hence,   
1
u
1
P
 
   
12
11 1 2 1
1
........
nn
U
ku ku k u
u
?
= +++
?
    (3.14) 
 
Or, 
 
   ,            1,2,...,
j
j
U
Pj
u
?
==
?
n     (3.15) 
  
     
Summary 
In this lesson, Castigliano’s first theorem has been stated and proved for linearly 
elastic structure with unyielding supports. The procedure to calculate deflections 
of a statically determinate structure at the point of application of load is illustrated 
with examples. Also, the procedure to calculate deflections in a statically 
determinate structure at a point where load is applied is illustrated with examples. 
The Castigliano’s second theorem is stated for elastic structure and proved in 
section 3.4.