Chapter Eight
Energy Method
8.1 Introduction
8.2 Strain energy expressions
8.3 Principal of stationary potential energy; several degrees of
freedom ------ Castigliano’s first theorem ---- Examples
8.4 Principal of stationary complementary energy -----Castigliano;s second theorem ----Examples
8.5 Statically indeterminate problems ----Examples
Review and Summary
8.1 Introduction

Two methods in solid mechanics ---- vectorial methods
and energy methods or variational methods
(1) vectorial methods ---- emphasized in elementary courses
and are formulated in terms of vector quantities such as
forces and displacement
(2) Energy methods---- formulated in terms of scalar quantities
such as work and energy. Advantages o f energy methods ----- ability to avoid some extraneous detail and to yield
approximate solution for complicated problems.

Degree of freedom (d.o.f.) ------ the number of independent
quantities needed to define uniquely the configuration of a
system ----generalized coordinates.
8.2 Strain energy expression

Work is done by a force as it moves through a distance,
and by a moment as it turns through an angle.
B
W   f cos  du
dW  ( f cos  )du
A
(the work done by a moment : mcosd )

m
d
If the effect of force is to distort an elastic body (such as a linear
spring), work done by f id stored as strain energy U (expressed
in terms of displacement). Complementary strain energy U *
(expressed in terms of force). For a linear elastic material U and
U* are numerically equal.
(1) The strain energy U=U()
B

k2
U   fdu   kudu 
A
0
2
f=ku
(2)
The complementary strain energy U*=U*(F)
B
F f
F2
*
U   udf  
df 
A
0 k
2k
U=f/k
 A unit column of linear elastic material can be viewed as a
linear spring, so the complementary strain energy U* is
 The complementary strain energy U* (expressed in terms
of stress or force) for a bar of length L under general complex
loading is (by volume integral)
 Expression for the strain energy U (in terms of displacements)
2
2
of a slender straight bar is (replacing My by  EI y (d w / dx ), )
M z  EI z (d 2 w / dx 2 ),
T  GJ (d / dx)
 Slender circular rings
U 
2
0
Moment-curvature relations for the
ring
M
1 d 2v
( 2  v)
2
2
2
EI
R
d
EI  1  d v 
 Rd
 
2  R 2  d 2  v 
U  strain energy for straight bar)
( R  ,
 Energy of transverse shear
6V z h 2
E zx 
(  z 2 ),
A  bh
2
Ah 4
L h/2 1
U*   
 zx2 bdzdx
0  h / 2 2G
L
 V 
  1.2 z dx
0
 2GA 
8.3 Principal of stationary potential energy; several degrees
of freedom ---- Castigliano’s first theorem ---- Examples
 Admissible or kinematically admissible configurations
 Potential energy  of a structure ---- =U+
U ---- the strain energy
 ---- the potential of the loads
For linear spring, d.o.f.=
U  k2 / 2,
   F
  U    k2 / 2  F
 The principal of stationary potential energy: Among all
admissible configuration (that satisfied static equilibrium
conditions) makes the potential energy stationary with
respect to small admissible variations of displacement. If
the stationary condition is a minimum, the equilibrium
state is stable.
d
 0  k  F  0  F  k
d
d  (k  F )d  0
( 
F
)
k
 Several degrees of freedom ---- Castigliano’s first theorem
(1) the potential energy  for n degree of freedom Di
 U 
U  U ( D1 , D2 , Dn )
n
   Pi  i
1
(2)
then principal of stationary potential energy gives
d  0 
(3)



dD1 
dD2   
dDn
D1
D2
Dn
For any and all of these dDi, d must vanish, this is
possible only if

0
(i  1,2, , n)
Di
Castigilano’s first theorem: If strain energy U is
expressed in terms of independent displacement d.o.f.,
then the load Pi that corresponds to d.o.f. Di is given by
the partial derivative of U with respect to Di
From above
We have

0
Di
(for i  1,2, , n )
U
 Pi (for i  1,2, , n )
Di
For example : D1  , D2   then
U
U
 F,
M


 Examples (Determine the displacement d.o.f. that define the static
equilibrium configuration)
(1) Two-Bar Linkage
The only d.o.f. is , that =0 as reference state,
we have (neglecting strain energy U)
L

    2W  (1  cos  )  F (2 L sin  )
2

d
2F
 0  tan  
d
W
( is the value for static equilibrium.)
(2) Rigid Bar
The d.o.f. is B (C=2B), the rotation of the rigid bar is
=B/b
AE 2 AE
 
B 
(2 B ) 2
  M C  B 
2L
2L
 b 
AE 2
 

5 B  M C  B 
2L
 b 
d
L
 0  B 
MC
d B
5 AEb
the forces in the wires
M
2M C
AE
AE
PB 
B  C ,
PC 
( 2 B ) 
L
5b
L
5b
U
(3) Spring in Series
Two d.o.f. are needed to define the configuration:
A, B. Initially , A=B=0.
U
1
1
k12A  k 2 ( B   A ) 2
2
2
Using Castigliano’s First theorem, we have
U
 FA  k1  A  k 2  B   A   FA
 A
U
 FB 
 B
A 
k 2  B   A   FB
FA  FB
F  FB FB
,B  A

k1
k1
k2
8.4 Principal of stationary complementary energy --------castigliano’s second theorem ---- Examples
 Complementary energy of a structure ---- U*+
For linear spring
U   F 2 / 2k   F 2
 F
 
2k
   F 
 The principal of stationary complementary energy: among
all statically admissible stress fields, the actual stress field
(that yields kinematically admissible displacements) makes
the complementary energy stationary with respect small
statically variations of stress.
d 
F
 0     0,
F  k
(1)
dF
k
(2) If there are several forces, we have
 *
 0 for i=1,2,……,n
Pi
In previous example of two springs in series, we have
  FA  FB

 A  0
FA
k1
  FA  FB FB


 B  0
FB
k1
k2

The complementary energy of a structure loaded by
concentrated forces and/or moments is
n
  U   Pi Di


1


U 
0
 Di
Pi
Pi
(i  1,2,......,n)
 Castigliaon’s second theorem: the partial derivative of
complementary energy U* with respect to a load yields the
displacement component of the loaded point in the direction
of that load.
U *
 Di
Pi
(i  1,2,n)
Pi can be a force F or moment M.
U *
 i
Fi
,
U *
 i
M i
 Unit load method (A convient format of Castigliano’s
second theorem)
Using complementary strain energy expression U * for curved
U 
bar and Di  P , we find
i
L  M y M y
M M z
k V V z 
Di   
 z
 z z
dx

0  EI

P
EI

P
GA

P
i
z
i
i 
 y
introducing
my
mz
Vz
 M y m y M z m z Tt





 dx
 EI

EI z
GJ
y


M y
my 
Pu is a moment produced by a unit load Pi (a unit
force or a unit moment)
 Examples
(1) Cantilever beam
(a) First determine the vertical deflection of B.
The bending moment is (neglect shearing)
2
qx
M ( x)   Fx 
2
using unit load methods, M F   x  m y
1
qx 2
FL3 qL4
B  
( Fx 
)(  x)dx 

EI
2
3EI 8EI
(b) Find the deflection at B due to q along (F=0), two method:
(1) simply set F=0 in above expression for B,
(2) temporarily apply a load (such as unit load) in the
desired dorection, after using unit load method, then
set this load to zero: ((case (b))

1  qx 2
qL4

B  
 1( x) ( x)dx 
EI  2
8EI

Set this load=0
(c) the horizontal deflection at C
h 1
L 1 
qx 2 
qL3 h

hdx  
H  
(0)(s)ds  
0 EI
0 EI 
2
6 EI


(2) Split Ring
(a)

Determine the Z-direction displacement of the
loaded end and its rotation component about the y
axis
Bending and twisting moments (in ring) MR and T are:
M R   FR sin   C sin 
T  FR (1  cos  )  C cos 
here C is the unit couple for calculation of rotation
(b)
(c)
Calculate deflection, we set C=0
2  M
M R
T T 
z    R

dx
0
GJ F 
 EI F
2  FR sin 

( R sin  ) Rd
0
EI
2  FR(1  cos )

R (1  cos ) Rd
0
GJ
FR3 3FR3


EI
GJ
Calculate the rotation, C=1 mR=-sin and T=-cos, in
M R T
the above equation, we use M R C 0 , T C 0 ,
,
, we
C
C
FR 2 FR 2

obtain  y 
EI
GJ
(3)
Truss Analysis
(a)
Calculate the vertical deflection at C by unit force method:
first calculate the internal force N in each bar due to Q;
then the force n due to unit force. Nn is nonzero only bars
DB and DC, so
 NL   NL 
 NL 
C   
n  
n  
n
 EA  DB  EA  DC
i  EA  i
 Q L  1  QL  1  0

EA
EA
(b)
Relative motion of points C. G. : we apply collinear force
(unit load) as shown. Then
QL
 NL 
 CG   
n   3.414
EA
CD , FC , FD  EA 
Rotation of bar FG: we apply couple forces 1/L (unit
moment), then
 NL 
 FG   
n 
BD , AC ,CF  EA 
(c)
8.5 Statically indeterminate problems ---- Examples
(1) propped cantilever beam

Determine the bending moment in
the statically indeterminate beam
(a)
Find the redundant and regard it as a known load on the
structure.
(b)
The bending moment is
M  Rx  qx 2 / 2
L M M
U *
 R  0
or
0 EI R dx  0
R
i.e.,
1 L
qx 2 
 Rx 
x dx  0  R  3qL / 8
EI 0 
2 
(c)
we can choose ML as redundant, then by
L M M
U *
 0 or
0 EI M L ds  0 we
M L
Obtain the same result as in (b).
(3) Elastically support cantilever beam
We use two methods to solve this problem:
(a) Consider the U* of the beam only, then
U 

L
0
M2
qx 2
dx where M  Rx 
2 EI
2
k
U 
U 

 
R
 (k)
or

L
0
M
mdx
EI
-x
when  is known, then R=k
(b) Consider the U* of whole system (beam and spring).
2
L M
R2

U 

dx where M  Rx  qx 2 2
2k 0 2 EI
R is the reaction at the base of the spring, by Castigliano’s
second theorem or unit load method:
U 
0
R
or
R LM

mdx  0
0
k
EI
M
x
R
we can get R    R k
(3) Semicircular arch

Find the support reaction of the arch
This is the statically indeterminate to the third degree. There are
several methods to solve the problem:
(a) Find U* of the arch as U*= U*(V, h, M0), then by using
U 
U 
U 
0 ,
0 ,
0
V
H
M 0
we obtain V, H, M0.
(b) By using symmetry, only have 2 unknows, H , M  as
2






U

U
H
,
M
  , by
shown in (c), then

2 

U 
U 
 0 , we solve the problem.
0 ,

M
H

2
(c)


As shown in (d), U  U ( H , M 0 ) , by
U 
 0,
H
U 
0
M 0

(4) Internally indeterminate truss Find the forces in
all bars
we elect to use forces in bars AD and CF as redundants. These
forces FAD and FCF are exposed if bars AD and CF are cut. Then
the truss is rendered as statically determinate, i.e., all forces in
bars can be written in terms of Q1, Q2, FAD and FCF. We can
write U* as
11
N i2 L

U 
i 1 2 Ai Ei
FAD and FCF can be obtain by
U 
U 
0 ,
0 ,
FCF
FAD
Since in the imagined cut in each bar (before and after loads F AD,
FCF), the relative approach or separation of the cut ends is zero.