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# Theorem of Least Work - 2 Civil Engineering (CE) Notes | EduRev

## Civil Engineering (CE) : Theorem of Least Work - 2 Civil Engineering (CE) Notes | EduRev

``` 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

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## Structural Analysis

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