Castigliano’s Theorems - 1 | Structural Analysis - Civil Engineering (CE) PDF Download

 Page 1


Instructional Objectives 
After reading this lesson, the reader will be able to; 
1. State and prove first theorem of Castigliano. 
2. Calculate deflections along the direction of applied load of a statically 
determinate structure at the point of application of load. 
3. Calculate deflections of a statically determinate structure in any direction at a 
point where the load is not acting by fictious (imaginary) load method. 
4. State and prove Castigliano’s second theorem. 
  
  
3.1 Introduction  
In the previous chapter concepts of strain energy and complementary strain 
energy were discussed. Castigliano’s first theorem is being used in structural 
analysis for finding deflection of an elastic structure based on strain energy of the 
structure. The Castigliano’s theorem can be applied when the supports of the 
structure are unyielding and the temperature of the structure is constant. 
 
 
3.2 Castigliano’s First Theorem  
For linearly elastic structure, where external forces only cause deformations, the 
complementary energy is equal to the strain energy. For such structures, the 
Castigliano’s first theorem may be stated as the first partial derivative of the 
strain energy of the structure with respect to any particular force gives the 
displacement of the point of application of that force in the direction of its line of 
action.  
 
 
Page 2


Instructional Objectives 
After reading this lesson, the reader will be able to; 
1. State and prove first theorem of Castigliano. 
2. Calculate deflections along the direction of applied load of a statically 
determinate structure at the point of application of load. 
3. Calculate deflections of a statically determinate structure in any direction at a 
point where the load is not acting by fictious (imaginary) load method. 
4. State and prove Castigliano’s second theorem. 
  
  
3.1 Introduction  
In the previous chapter concepts of strain energy and complementary strain 
energy were discussed. Castigliano’s first theorem is being used in structural 
analysis for finding deflection of an elastic structure based on strain energy of the 
structure. The Castigliano’s theorem can be applied when the supports of the 
structure are unyielding and the temperature of the structure is constant. 
 
 
3.2 Castigliano’s First Theorem  
For linearly elastic structure, where external forces only cause deformations, the 
complementary energy is equal to the strain energy. For such structures, the 
Castigliano’s first theorem may be stated as the first partial derivative of the 
strain energy of the structure with respect to any particular force gives the 
displacement of the point of application of that force in the direction of its line of 
action.  
 
 
 
 
Let 
 
be the forces acting at  from the left end on a simply 
supported beam of span
n
P P P ,...., ,
2 1 n
x x x ,......, ,
2 1
L . Let 
 
be the displacements at the loading 
points  respectively as shown in Fig. 3.1. Now, assume that the 
material obeys Hooke’s law and invoking the principle of superposition, the work 
done by the external forces is given by (vide eqn. 1.8 of lesson 1) 
n
u u u ,..., ,
2 1
n
P P P ,...., ,
2 1
 
    
n n
u P u P u P W
2
1
..........
2
1
2
1
2 2 1 1
+ + + =    (3.1) 
 
 
Page 3


Instructional Objectives 
After reading this lesson, the reader will be able to; 
1. State and prove first theorem of Castigliano. 
2. Calculate deflections along the direction of applied load of a statically 
determinate structure at the point of application of load. 
3. Calculate deflections of a statically determinate structure in any direction at a 
point where the load is not acting by fictious (imaginary) load method. 
4. State and prove Castigliano’s second theorem. 
  
  
3.1 Introduction  
In the previous chapter concepts of strain energy and complementary strain 
energy were discussed. Castigliano’s first theorem is being used in structural 
analysis for finding deflection of an elastic structure based on strain energy of the 
structure. The Castigliano’s theorem can be applied when the supports of the 
structure are unyielding and the temperature of the structure is constant. 
 
 
3.2 Castigliano’s First Theorem  
For linearly elastic structure, where external forces only cause deformations, the 
complementary energy is equal to the strain energy. For such structures, the 
Castigliano’s first theorem may be stated as the first partial derivative of the 
strain energy of the structure with respect to any particular force gives the 
displacement of the point of application of that force in the direction of its line of 
action.  
 
 
 
 
Let 
 
be the forces acting at  from the left end on a simply 
supported beam of span
n
P P P ,...., ,
2 1 n
x x x ,......, ,
2 1
L . Let 
 
be the displacements at the loading 
points  respectively as shown in Fig. 3.1. Now, assume that the 
material obeys Hooke’s law and invoking the principle of superposition, the work 
done by the external forces is given by (vide eqn. 1.8 of lesson 1) 
n
u u u ,..., ,
2 1
n
P P P ,...., ,
2 1
 
    
n n
u P u P u P W
2
1
..........
2
1
2
1
2 2 1 1
+ + + =    (3.1) 
 
 
Work done by the external forces is stored in the structure as strain energy in a 
conservative system. Hence, the strain energy of the structure is, 
 
n n
u P u P u P U
2
1
..........
2
1
2
1
2 2 1 1
+ + + =   (3.2) 
 
Displacement 
 
below point  is due to the action of  acting at 
distances respectively from left support
. 
Hence,  may be expressed 
as, 
1
u
1
P
n
P P P ,...., ,
2 1
n
x x x ,......, ,
2 1 1
u
 
n n
P a P a P a u
1 2 12 1 11 1
.......... + + + =    (3.3) 
 
In general,  
 
   n i P a P a P a u
n in i i i
,... 2 , 1             ..........
2 2 1 1
= + + + =  (3.4) 
 
where  is the flexibility coefficient at  due to unit force applied at 
ij
a i j . 
Substituting the values of  in equation (3.2) from equation (3.4), we 
get, 
n
u u u ,..., ,
2 1
 
...] [
2
1
....... ...] [
2
1
...] [
2
1
2 2 1 1 2 22 1 21 2 2 12 1 11 1
+ + + + + + + + + = P a P a P P a P a P P a P a P U
n n n
(3.5) 
 
We know from Maxwell-Betti’s reciprocal theorem
ji ij
a a = . Hence, equation (3.5) 
may be simplified as,  
 
[]
22 2
11 1 22 2 12 1 2 13 1 3 1 1
1
.... .... ...
2
nn n n n
U a P a P a P a PP a PP a PP ?? =+ ++ + + ++
??
+ (3.6) 
 
Now, differentiating the strain energy with any force  gives,  
1
P
 
n n
P a P a P a
P
U
1 2 12 1 11
1
.......... + + + =
?
?
    (3.7) 
 
It may be observed that equation (3.7) is nothing but displacement  at the 
loading point. 
1
u
In general,  
n
n
u
P
U
=
?
?
     (3.8) 
 
Hence, for determinate structure within linear elastic range the partial derivative 
of the total strain energy with respect to any external load is equal to the 
 
Page 4


Instructional Objectives 
After reading this lesson, the reader will be able to; 
1. State and prove first theorem of Castigliano. 
2. Calculate deflections along the direction of applied load of a statically 
determinate structure at the point of application of load. 
3. Calculate deflections of a statically determinate structure in any direction at a 
point where the load is not acting by fictious (imaginary) load method. 
4. State and prove Castigliano’s second theorem. 
  
  
3.1 Introduction  
In the previous chapter concepts of strain energy and complementary strain 
energy were discussed. Castigliano’s first theorem is being used in structural 
analysis for finding deflection of an elastic structure based on strain energy of the 
structure. The Castigliano’s theorem can be applied when the supports of the 
structure are unyielding and the temperature of the structure is constant. 
 
 
3.2 Castigliano’s First Theorem  
For linearly elastic structure, where external forces only cause deformations, the 
complementary energy is equal to the strain energy. For such structures, the 
Castigliano’s first theorem may be stated as the first partial derivative of the 
strain energy of the structure with respect to any particular force gives the 
displacement of the point of application of that force in the direction of its line of 
action.  
 
 
 
 
Let 
 
be the forces acting at  from the left end on a simply 
supported beam of span
n
P P P ,...., ,
2 1 n
x x x ,......, ,
2 1
L . Let 
 
be the displacements at the loading 
points  respectively as shown in Fig. 3.1. Now, assume that the 
material obeys Hooke’s law and invoking the principle of superposition, the work 
done by the external forces is given by (vide eqn. 1.8 of lesson 1) 
n
u u u ,..., ,
2 1
n
P P P ,...., ,
2 1
 
    
n n
u P u P u P W
2
1
..........
2
1
2
1
2 2 1 1
+ + + =    (3.1) 
 
 
Work done by the external forces is stored in the structure as strain energy in a 
conservative system. Hence, the strain energy of the structure is, 
 
n n
u P u P u P U
2
1
..........
2
1
2
1
2 2 1 1
+ + + =   (3.2) 
 
Displacement 
 
below point  is due to the action of  acting at 
distances respectively from left support
. 
Hence,  may be expressed 
as, 
1
u
1
P
n
P P P ,...., ,
2 1
n
x x x ,......, ,
2 1 1
u
 
n n
P a P a P a u
1 2 12 1 11 1
.......... + + + =    (3.3) 
 
In general,  
 
   n i P a P a P a u
n in i i i
,... 2 , 1             ..........
2 2 1 1
= + + + =  (3.4) 
 
where  is the flexibility coefficient at  due to unit force applied at 
ij
a i j . 
Substituting the values of  in equation (3.2) from equation (3.4), we 
get, 
n
u u u ,..., ,
2 1
 
...] [
2
1
....... ...] [
2
1
...] [
2
1
2 2 1 1 2 22 1 21 2 2 12 1 11 1
+ + + + + + + + + = P a P a P P a P a P P a P a P U
n n n
(3.5) 
 
We know from Maxwell-Betti’s reciprocal theorem
ji ij
a a = . Hence, equation (3.5) 
may be simplified as,  
 
[]
22 2
11 1 22 2 12 1 2 13 1 3 1 1
1
.... .... ...
2
nn n n n
U a P a P a P a PP a PP a PP ?? =+ ++ + + ++
??
+ (3.6) 
 
Now, differentiating the strain energy with any force  gives,  
1
P
 
n n
P a P a P a
P
U
1 2 12 1 11
1
.......... + + + =
?
?
    (3.7) 
 
It may be observed that equation (3.7) is nothing but displacement  at the 
loading point. 
1
u
In general,  
n
n
u
P
U
=
?
?
     (3.8) 
 
Hence, for determinate structure within linear elastic range the partial derivative 
of the total strain energy with respect to any external load is equal to the 
 
displacement of the point of application of load in the direction of the applied 
load, provided the supports are unyielding and temperature is maintained 
constant. This theorem is advantageously used for calculating deflections in 
elastic structure. The procedure for calculating the deflection is illustrated with 
few examples.  
 
Example 3.1 
Find the displacement and slope at the tip of a cantilever beam loaded as in Fig. 
3.2. Assume the flexural rigidity of the beam EI to be constant for the beam. 
 
 
 
Moment at any section at a distance x away from the free end is given by 
 
Px M - =      (1) 
 
Strain energy stored in the beam due to bending is   
?
=
L
dx
EI
M
U
0
2
2
  (2)  
 
Substituting the expression for bending moment M in equation (3.10), we get, 
 
?
= =
L
EI
L P
dx
EI
Px
U
0
3 2 2
6 2
) (
    (3) 
 
Page 5


Instructional Objectives 
After reading this lesson, the reader will be able to; 
1. State and prove first theorem of Castigliano. 
2. Calculate deflections along the direction of applied load of a statically 
determinate structure at the point of application of load. 
3. Calculate deflections of a statically determinate structure in any direction at a 
point where the load is not acting by fictious (imaginary) load method. 
4. State and prove Castigliano’s second theorem. 
  
  
3.1 Introduction  
In the previous chapter concepts of strain energy and complementary strain 
energy were discussed. Castigliano’s first theorem is being used in structural 
analysis for finding deflection of an elastic structure based on strain energy of the 
structure. The Castigliano’s theorem can be applied when the supports of the 
structure are unyielding and the temperature of the structure is constant. 
 
 
3.2 Castigliano’s First Theorem  
For linearly elastic structure, where external forces only cause deformations, the 
complementary energy is equal to the strain energy. For such structures, the 
Castigliano’s first theorem may be stated as the first partial derivative of the 
strain energy of the structure with respect to any particular force gives the 
displacement of the point of application of that force in the direction of its line of 
action.  
 
 
 
 
Let 
 
be the forces acting at  from the left end on a simply 
supported beam of span
n
P P P ,...., ,
2 1 n
x x x ,......, ,
2 1
L . Let 
 
be the displacements at the loading 
points  respectively as shown in Fig. 3.1. Now, assume that the 
material obeys Hooke’s law and invoking the principle of superposition, the work 
done by the external forces is given by (vide eqn. 1.8 of lesson 1) 
n
u u u ,..., ,
2 1
n
P P P ,...., ,
2 1
 
    
n n
u P u P u P W
2
1
..........
2
1
2
1
2 2 1 1
+ + + =    (3.1) 
 
 
Work done by the external forces is stored in the structure as strain energy in a 
conservative system. Hence, the strain energy of the structure is, 
 
n n
u P u P u P U
2
1
..........
2
1
2
1
2 2 1 1
+ + + =   (3.2) 
 
Displacement 
 
below point  is due to the action of  acting at 
distances respectively from left support
. 
Hence,  may be expressed 
as, 
1
u
1
P
n
P P P ,...., ,
2 1
n
x x x ,......, ,
2 1 1
u
 
n n
P a P a P a u
1 2 12 1 11 1
.......... + + + =    (3.3) 
 
In general,  
 
   n i P a P a P a u
n in i i i
,... 2 , 1             ..........
2 2 1 1
= + + + =  (3.4) 
 
where  is the flexibility coefficient at  due to unit force applied at 
ij
a i j . 
Substituting the values of  in equation (3.2) from equation (3.4), we 
get, 
n
u u u ,..., ,
2 1
 
...] [
2
1
....... ...] [
2
1
...] [
2
1
2 2 1 1 2 22 1 21 2 2 12 1 11 1
+ + + + + + + + + = P a P a P P a P a P P a P a P U
n n n
(3.5) 
 
We know from Maxwell-Betti’s reciprocal theorem
ji ij
a a = . Hence, equation (3.5) 
may be simplified as,  
 
[]
22 2
11 1 22 2 12 1 2 13 1 3 1 1
1
.... .... ...
2
nn n n n
U a P a P a P a PP a PP a PP ?? =+ ++ + + ++
??
+ (3.6) 
 
Now, differentiating the strain energy with any force  gives,  
1
P
 
n n
P a P a P a
P
U
1 2 12 1 11
1
.......... + + + =
?
?
    (3.7) 
 
It may be observed that equation (3.7) is nothing but displacement  at the 
loading point. 
1
u
In general,  
n
n
u
P
U
=
?
?
     (3.8) 
 
Hence, for determinate structure within linear elastic range the partial derivative 
of the total strain energy with respect to any external load is equal to the 
 
displacement of the point of application of load in the direction of the applied 
load, provided the supports are unyielding and temperature is maintained 
constant. This theorem is advantageously used for calculating deflections in 
elastic structure. The procedure for calculating the deflection is illustrated with 
few examples.  
 
Example 3.1 
Find the displacement and slope at the tip of a cantilever beam loaded as in Fig. 
3.2. Assume the flexural rigidity of the beam EI to be constant for the beam. 
 
 
 
Moment at any section at a distance x away from the free end is given by 
 
Px M - =      (1) 
 
Strain energy stored in the beam due to bending is   
?
=
L
dx
EI
M
U
0
2
2
  (2)  
 
Substituting the expression for bending moment M in equation (3.10), we get, 
 
?
= =
L
EI
L P
dx
EI
Px
U
0
3 2 2
6 2
) (
    (3) 
 
Now, according to Castigliano’s theorem, the first partial derivative of strain 
energy with respect to external force P gives the deflection  at A in the 
direction of applied force. Thus,  
A
u
    
        
EI
PL
u
P
U
A
3
3
= =
?
?
     (4) 
  
To find the slope at the free end, we need to differentiate strain energy with 
respect to externally applied momentM atA . As there is no moment atA , apply 
a fictitious moment  at
0
M A. Now moment at any section at a distance x away 
from the free end is given by    
 
0
M Px M - - = 
 
Now, strain energy stored in the beam may be calculated as, 
 
    
?
+ + =
+
=
L
EI
L M
EI
PL M
EI
L P
dx
EI
M Px
U
0
2
0
2
0
3 2 2
0
2 2 6 2
) (
   (5) 
     
Taking partial derivative of strain energy with respect to , we get slope at
0
M A . 
 
2
0
0
2
A
M L UPL
M EI EI
?
?
== +
?
    (6) 
 
But actually there is no moment applied atA. Hence substitute in 
equation (3.14) we get the slope at A. 
0
0
= M
 
   
EI
PL
A
2
2
= ?      (7) 
 
Example 3.2 
A cantilever beam which is curved in the shape of a quadrant of a circle is loaded 
as shown in Fig. 3.3. The radius of curvature of curved beam isR, Young’s 
modulus of the material is E and second moment of the area is I about an axis 
perpendicular to the plane of the paper through the centroid of the cross section. 
Find the vertical displacement of point A on the curved beam.