1. Module
1
Energy Methods in
Structural Analysis
Version 2 CE IIT, Kharagpur

2. Lesson
4
Theorem of Least Work
Version 2 CE IIT, Kharagpur

3. Instructional Objectives
After reading this lesson, the reader will be able to:
1. State and prove theorem of Least Work.
2. Analyse statically indeterminate structure.
3. State and prove Maxwell-Betti’s Reciprocal theorem.
4.1 Introduction
In the last chapter the Castigliano’s theorems were discussed. In this chapter
theorem of least work and reciprocal theorems are presented along with few
selected problems. We know that for the 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 load. This
theorem when applied to the statically indeterminate structure results in the
theorem of least work.
4.2 Theorem of Least Work
According to this theorem, the partial derivative of strain energy of a statically
indeterminate structure with respect to statically indeterminate action should
vanish as it is the function of such redundant forces to prevent any displacement
at its point of application. The forces developed in a redundant framework are
such that the total internal strain energy is a minimum. This can be proved as
follows. Consider a beam that is fixed at left end and roller supported at right end
as shown in Fig. 4.1a. Let P1 , P2 ,...., Pn be the forces acting at distances
x1 , x 2 ,......, x n from the left end of the beam of span L . Let u1 , u 2 ,..., u n be the
displacements at the loading points P1 , P2 ,...., Pn respectively as shown in Fig. 4.1a.
This is a statically indeterminate structure and choosing Ra as the redundant
reaction, we obtain a simple cantilever beam as shown in Fig. 4.1b. Invoking the
principle of superposition, this may be treated as the superposition of two cases,
viz, a cantilever beam with loads P1 , P2 ,...., Pn and a cantilever beam with redundant
force Ra (see Fig. 4.2a and Fig. 4.2b)
Version 2 CE IIT, Kharagpur

4. Version 2 CE IIT, Kharagpur

5. In the first case (4.2a), obtain deflection below A due to applied loads P1 , P2 ,...., Pn .
This can be easily accomplished through Castigliano’s first theorem as discussed
in Lesson 3. Since there is no load applied at A , apply a fictitious load Q at A as in
Fig. 4.2. Let u a be the deflection below A .
Now the strain energy U s stored in the determinate structure (i.e. the support A
removed) is given by,
1 1 1 1
US = P1u1 + P2 u 2 + .......... + Pn u n + Qu a (4.1)
2 2 2 2
It is known that the displacement u1 below point P1 is due to action of P , P2 ,...., Pn
1
acting at x1 , x 2 ,......, x n respectively and due to Q at A . Hence, u1 may be
expressed as,
Version 2 CE IIT, Kharagpur

6. u1 = a11 P + a12 P2 + .......... + a1n Pn + a1a Q
1 (4.2)
where, aij is the flexibility coefficient at i due to unit force applied at j . Similar
equations may be written for u2 , u3 ,...., un and ua . Substituting for u2 , u3 ,...., un and ua
in equation (4.1) from equation (4.2), we get,
1 1
US = P [a11 P + a12 P2 + ... + a1n Pn + a1a Q ] + P2 [a21 P + a22 P2 + ...a2 n Pn + a2 a Q ] + .......
1 1 1
2 2 (4.3)
1 1
+ Pn [an1 P + an 2 P2 + ...ann Pn + ana Q] + Q[aa1 P + aa 2 P2 + .... + aan Pn + aaa Q]
1 1
2 2
Taking partial derivative of strain energy U s with respect to Q , we get deflection
at A .
∂U s
= aa1 P + aa 2 P2 + ........ + aan Pn + aaa Q (4.4)
∂Q
1
Substitute Q = 0 as it is fictitious in the above equation,
∂U s
= ua = aa1 P + aa 2 P2 + ........ + aan Pn (4.5)
∂Q
1
Now the strain energy stored in the beam due to redundant reaction RA is,
Ra L3
2
Ur = (4.6)
6 EI
Now deflection at A due to Ra is
∂U r R L3
= −ua = a (4.7)
∂Ra 3EI
The deflection due to Ra should be in the opposite direction to one caused by
superposed loads P , P2 ,...., Pn , so that the net deflection at A is zero. From
1
equation (4.5) and (4.7) one could write,
∂Us ∂U
= ua = − r (4.8)
∂Q ∂Ra
Since Q is fictitious, one could as well replace it by Ra . Hence,
Version 2 CE IIT, Kharagpur

7. ∂
(U s + U r ) = 0 (4.9)
∂Ra
or,
∂U
=0 (4.10)
∂Ra
This is the statement of theorem of least work. Where U is the total strain energy
of the beam due to superimposed loads P , P2 ,...., Pn and redundant reaction Ra .
1
Example 4.1
Find the reactions of a propped cantilever beam uniformly loaded as shown in Fig.
4.3a. Assume the flexural rigidity of the beam EI to be constant throughout its
length.
Version 2 CE IIT, Kharagpur

8. There three reactions Ra , Rb and M b as shown in the figure. We have only two
equation of equilibrium viz., ∑F y = 0 and ∑ M = 0 . This is a statically
indeterminate structure and choosing Rb as the redundant reaction, we obtain a
simple cantilever beam as shown in Fig. 4.3b.
Now, the internal strain energy of the beam due to applied loads and redundant
reaction, considering only bending deformations is,
L
M2
U =∫ dx (1)
0
2 EI
According to theorem of least work we have,
Version 2 CE IIT, Kharagpur

9. ∂U M ∂M
L
=0=∫ (2)
∂Rb 0
EI ∂Rb
wx 2
Bending moment at a distance x from B , M = Rb x − (3)
2
∂M
=x (4)
∂Rb
Hence,
∂U ( R x − wx 2 / 2) x
L
=∫ b dx (5)
∂Rb 0 EI
∂U ⎡ RB L3 wL4 ⎤ 1
=⎢ − ⎥ =0 (6)
∂Rb ⎣ 3 8 ⎦ EI
Solving for Rb , we get,
3
RB = wL
8
5 wL2
Ra = wL − Rb = wL and M a = − (7)
8 8
Example 4.2
A ring of radius R is loaded as shown in figure. Determine increase in the
diameter AB of the ring. Young’s modulus of the material is E and second
moment of the area is I about an axis perpendicular to the page through the
centroid of the cross section.
Version 2 CE IIT, Kharagpur

10. Version 2 CE IIT, Kharagpur

11. The free body diagram of the ring is as shown in Fig. 4.4. Due to symmetry, the
slopes at C and D is zero. The value of redundant moment M 0 is such as to make
slopes at C and D zero. The bending moment at any section θ of the beam is,
PR
M = M0 − (1 − cos θ ) (1)
2
Now strain energy stored in the ring due to bending deformations is,
2π
M 2R
U = ∫ 2 EI dθ
0
(2)
Due to symmetry, one could consider one quarter of the ring. According to
theorem of least work,
∂U 2π M ∂M
=0=∫ Rdθ (3)
∂M 0 0 EI ∂M
0
∂M
=1
∂M 0
2π
∂U M
∂M 0
= ∫ EI Rdθ
0
(4)
π
4R 2 PR
0= ∫ [M 0 − 2 (1 − cosθ )] dθ
EI 0
(5)
Integrating and solving for M 0 ,
⎛1 1⎞
M 0 = PR ⎜ − ⎟ (6)
⎝2 π ⎠
M 0 = 0.182 PR
Now, increase in diameter Δ , may be obtained by taking the first partial derivative
of strain energy with respect to P . Thus,
∂U
Δ=
∂P
Version 2 CE IIT, Kharagpur

12. Now strain energy stored in the ring is given by equation (2). Substituting the value
of M 0 and equation (1) in (2), we get,
π /2
2R PR 2 PR
U=
EI ∫{ 2
0
( − 1) −
π 2
(1 − cosθ )}2 dθ (7)
Now the increase in length of the diameter is,
π /2
∂U 2 R PR 2 PR R 2 R
=
∂P EI ∫ 2{ 2
0
( − 1) −
π 2
(1 − cosθ )}{ ( − 1) − (1 − cosθ )}dθ
2 π 2
(8)
After integrating,
PR 3 π 2 PR 3
Δ= { − ) = 0.149 (9)
EI 4 π EI
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 P1 and P2 separately as shown in the figure.
Let u 21 be the deflection below the load point P2 when only load P1 is acting.
Similarly let u12 be the deflection below load P1 , when only load P2 is acting on the
beam.
Version 2 CE IIT, Kharagpur

13. 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,
P1 × u12 = P2 × u 21 (4.11)
Now, u12 and u 21 can be calculated using Castiglinao’s first theorem. Substituting
the values of u12 and u 21 in equation (4.27) we get,
5 P2 L3 5 P L3
P1 × = P2 × 1 (4.12)
48 EI 48 EI
Hence it is proved. This is also valid even when the first system of forces is
P1 , P2 ,...., Pn and the second system of forces is given by Q1 , Q2 ,...., Qn . Let
u1 , u 2 ,...., u n be the displacements caused by the forces P1 , P2 ,...., Pn only and
δ 1 , δ 2 ,...., δ n be the displacements due to system of forces Q1 , Q2 ,...., Qn only acting
on the beam as shown in Fig. 4.6.
Version 2 CE IIT, Kharagpur

14. Now the reciprocal theorem may be stated as,
Pi δ i = Qi u i i = 1,2,...., n (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 the
Version 2 CE IIT, Kharagpur

15. end, the celebrated theorem of Maxwell-Betti’s reciprocal theorem has been sated
and proved.
Version 2 CE IIT, Kharagpur