Theorem of Least Work - 2 | Structural Analysis - Civil Engineering (CE) PDF Download

 Page 2


 
 
The free body diagram of the ring is as shown in Fig. 4.4. Due to symmetry, the 
slopes at  is zero.  The value of redundant moment is such as to make 
slopes at zero.  The bending moment at any section 
D C and 
0
M
D C and ? of the beam is, 
 
    ) cos 1 (
2
0
? - - =
PR
M M    (1) 
 
Now strain energy stored in the ring due to bending deformations is, 
 
          
?
=
p
?
2
0
2
2
d
EI
R M
U     (2) 
 
Due to symmetry, one could consider one quarter of the ring. According to 
theorem of least work, 
 
    ?
p
Rd
M
M
EI
M
M
U
0
2
0
0
0
?
?
= =
?
?
?
    (3) 
 
     1
0
=
?
?
M
M
 
 
          ?
p
Rd
EI
M
M
U
?
=
?
?
2
0 0
        (4) 
     
?
- - =
2
0
0
 )] cos 1 (
2
[
4
0
p
? ? d
PR
M
EI
R
    (5) 
 
Integrating and solving for 
0
M , 
 
     
0
11
2
MPR
p
?
=-
?
??
?
?
     (6) 
 
     PR M 182 . 0
0
= 
 
Now, increase in diameter , may be obtained by taking the first partial derivative 
of strain energy with respect to . Thus,   
?
P
 
         
U
P
?
?=
?
 
 
 
Page 3


 
 
The free body diagram of the ring is as shown in Fig. 4.4. Due to symmetry, the 
slopes at  is zero.  The value of redundant moment is such as to make 
slopes at zero.  The bending moment at any section 
D C and 
0
M
D C and ? of the beam is, 
 
    ) cos 1 (
2
0
? - - =
PR
M M    (1) 
 
Now strain energy stored in the ring due to bending deformations is, 
 
          
?
=
p
?
2
0
2
2
d
EI
R M
U     (2) 
 
Due to symmetry, one could consider one quarter of the ring. According to 
theorem of least work, 
 
    ?
p
Rd
M
M
EI
M
M
U
0
2
0
0
0
?
?
= =
?
?
?
    (3) 
 
     1
0
=
?
?
M
M
 
 
          ?
p
Rd
EI
M
M
U
?
=
?
?
2
0 0
        (4) 
     
?
- - =
2
0
0
 )] cos 1 (
2
[
4
0
p
? ? d
PR
M
EI
R
    (5) 
 
Integrating and solving for 
0
M , 
 
     
0
11
2
MPR
p
?
=-
?
??
?
?
     (6) 
 
     PR M 182 . 0
0
= 
 
Now, increase in diameter , may be obtained by taking the first partial derivative 
of strain energy with respect to . Thus,   
?
P
 
         
U
P
?
?=
?
 
 
 
Now strain energy stored in the ring is given by equation (2). Substituting the value 
of  and equation (1) in (2), we get, 
0
M
  
   
?
- - - =
2 /
0
2
)} cos 1 (
2
) 1
2
(
2
{
2
p
? ?
p
d
PR PR
EI
R
U   (7) 
 
Now the increase in length of the diameter is, 
 
 
?
- - - - - - =
?
?
2 /
0
)} cos 1 (
2
) 1
2
(
2
)}{ cos 1 (
2
) 1
2
(
2
{ 2
2
p
? ?
p
?
p
d
R R PR PR
EI
R
P
U
     (8) 
 
After integrating, 
 
    
33
2
{ ) 0.149
4
PR PR
EI EI
p
p
?= - =   (9) 
 
 
4.3     Maxwell–Betti Reciprocal theorem   
Consider a simply supported beam of span L as shown in Fig. 4.5. Let this beam 
be loaded by two systems of forces 
 
and 
  
separately 
 
 as shown in the figure. 
Let be the deflection below the load point when only load is acting. 
Similarly let be the deflection below load , when only load is acting on the 
beam. 
1
P
2
P
21
u
2
P
1
P
12
u
1
P
2
P
 
Page 4


 
 
The free body diagram of the ring is as shown in Fig. 4.4. Due to symmetry, the 
slopes at  is zero.  The value of redundant moment is such as to make 
slopes at zero.  The bending moment at any section 
D C and 
0
M
D C and ? of the beam is, 
 
    ) cos 1 (
2
0
? - - =
PR
M M    (1) 
 
Now strain energy stored in the ring due to bending deformations is, 
 
          
?
=
p
?
2
0
2
2
d
EI
R M
U     (2) 
 
Due to symmetry, one could consider one quarter of the ring. According to 
theorem of least work, 
 
    ?
p
Rd
M
M
EI
M
M
U
0
2
0
0
0
?
?
= =
?
?
?
    (3) 
 
     1
0
=
?
?
M
M
 
 
          ?
p
Rd
EI
M
M
U
?
=
?
?
2
0 0
        (4) 
     
?
- - =
2
0
0
 )] cos 1 (
2
[
4
0
p
? ? d
PR
M
EI
R
    (5) 
 
Integrating and solving for 
0
M , 
 
     
0
11
2
MPR
p
?
=-
?
??
?
?
     (6) 
 
     PR M 182 . 0
0
= 
 
Now, increase in diameter , may be obtained by taking the first partial derivative 
of strain energy with respect to . Thus,   
?
P
 
         
U
P
?
?=
?
 
 
 
Now strain energy stored in the ring is given by equation (2). Substituting the value 
of  and equation (1) in (2), we get, 
0
M
  
   
?
- - - =
2 /
0
2
)} cos 1 (
2
) 1
2
(
2
{
2
p
? ?
p
d
PR PR
EI
R
U   (7) 
 
Now the increase in length of the diameter is, 
 
 
?
- - - - - - =
?
?
2 /
0
)} cos 1 (
2
) 1
2
(
2
)}{ cos 1 (
2
) 1
2
(
2
{ 2
2
p
? ?
p
?
p
d
R R PR PR
EI
R
P
U
     (8) 
 
After integrating, 
 
    
33
2
{ ) 0.149
4
PR PR
EI EI
p
p
?= - =   (9) 
 
 
4.3     Maxwell–Betti Reciprocal theorem   
Consider a simply supported beam of span L as shown in Fig. 4.5. Let this beam 
be loaded by two systems of forces 
 
and 
  
separately 
 
 as shown in the figure. 
Let be the deflection below the load point when only load is acting. 
Similarly let be the deflection below load , when only load is acting on the 
beam. 
1
P
2
P
21
u
2
P
1
P
12
u
1
P
2
P
 
 
 
The reciprocal theorem states that the work done by forces acting through 
displacement of the second system is the same as the work done by the second 
system of forces acting through the displacements of the first system. Hence, 
according to reciprocal theorem, 
 
21 2 12 1
u P u P × = ×     (4.11) 
 
Now, can be calculated using Castiglinao’s first theorem. Substituting 
the values of  in equation (4.27) we get, 
21 12
 and u u
21 12
 and u u
 
EI
L P
P
EI
L P
P
48
5
48
5
3
1
2
3
2
1
× = ×    (4.12) 
  
Hence it is proved. This is also valid even when the first system of forces is 
 
and the second system of forces is given by . Let 
 be the displacements caused by the forces  only and 
n
P P P ,...., ,
2 1 n
Q Q Q ,...., ,
2 1
n
u u u ,...., ,
2 1 n
P P P ,...., ,
2 1
n
d d d ,...., ,
2 1
be the displacements due to system of forces  only acting 
on the beam as shown in Fig. 4.6.  
n
Q Q Q ,...., ,
2 1
 
Page 5


 
 
The free body diagram of the ring is as shown in Fig. 4.4. Due to symmetry, the 
slopes at  is zero.  The value of redundant moment is such as to make 
slopes at zero.  The bending moment at any section 
D C and 
0
M
D C and ? of the beam is, 
 
    ) cos 1 (
2
0
? - - =
PR
M M    (1) 
 
Now strain energy stored in the ring due to bending deformations is, 
 
          
?
=
p
?
2
0
2
2
d
EI
R M
U     (2) 
 
Due to symmetry, one could consider one quarter of the ring. According to 
theorem of least work, 
 
    ?
p
Rd
M
M
EI
M
M
U
0
2
0
0
0
?
?
= =
?
?
?
    (3) 
 
     1
0
=
?
?
M
M
 
 
          ?
p
Rd
EI
M
M
U
?
=
?
?
2
0 0
        (4) 
     
?
- - =
2
0
0
 )] cos 1 (
2
[
4
0
p
? ? d
PR
M
EI
R
    (5) 
 
Integrating and solving for 
0
M , 
 
     
0
11
2
MPR
p
?
=-
?
??
?
?
     (6) 
 
     PR M 182 . 0
0
= 
 
Now, increase in diameter , may be obtained by taking the first partial derivative 
of strain energy with respect to . Thus,   
?
P
 
         
U
P
?
?=
?
 
 
 
Now strain energy stored in the ring is given by equation (2). Substituting the value 
of  and equation (1) in (2), we get, 
0
M
  
   
?
- - - =
2 /
0
2
)} cos 1 (
2
) 1
2
(
2
{
2
p
? ?
p
d
PR PR
EI
R
U   (7) 
 
Now the increase in length of the diameter is, 
 
 
?
- - - - - - =
?
?
2 /
0
)} cos 1 (
2
) 1
2
(
2
)}{ cos 1 (
2
) 1
2
(
2
{ 2
2
p
? ?
p
?
p
d
R R PR PR
EI
R
P
U
     (8) 
 
After integrating, 
 
    
33
2
{ ) 0.149
4
PR PR
EI EI
p
p
?= - =   (9) 
 
 
4.3     Maxwell–Betti Reciprocal theorem   
Consider a simply supported beam of span L as shown in Fig. 4.5. Let this beam 
be loaded by two systems of forces 
 
and 
  
separately 
 
 as shown in the figure. 
Let be the deflection below the load point when only load is acting. 
Similarly let be the deflection below load , when only load is acting on the 
beam. 
1
P
2
P
21
u
2
P
1
P
12
u
1
P
2
P
 
 
 
The reciprocal theorem states that the work done by forces acting through 
displacement of the second system is the same as the work done by the second 
system of forces acting through the displacements of the first system. Hence, 
according to reciprocal theorem, 
 
21 2 12 1
u P u P × = ×     (4.11) 
 
Now, can be calculated using Castiglinao’s first theorem. Substituting 
the values of  in equation (4.27) we get, 
21 12
 and u u
21 12
 and u u
 
EI
L P
P
EI
L P
P
48
5
48
5
3
1
2
3
2
1
× = ×    (4.12) 
  
Hence it is proved. This is also valid even when the first system of forces is 
 
and the second system of forces is given by . Let 
 be the displacements caused by the forces  only and 
n
P P P ,...., ,
2 1 n
Q Q Q ,...., ,
2 1
n
u u u ,...., ,
2 1 n
P P P ,...., ,
2 1
n
d d d ,...., ,
2 1
be the displacements due to system of forces  only acting 
on the beam as shown in Fig. 4.6.  
n
Q Q Q ,...., ,
2 1
 
 
 
Now the reciprocal theorem may be stated as, 
 
n i u Q P
i i i i
,...., 2 , 1                        = = d   (4.13) 
 
 
Summary 
In lesson 3, the Castigliano’s first theorem has been stated and proved. For 
statically determinate structure, the partial derivative of strain energy with respect 
to external force is equal to the displacement in the direction of that load at the 
point of application of the load. This theorem when applied to the statically 
indeterminate structure results in the theorem of Least work. In this chapter the 
theorem of Least Work has been stated and proved. Couple of problems is solved 
to illustrate the procedure of analysing statically indeterminate structures. In th e