Chapter 11 Energy Method
-- Utilize the Energy Method to solve
engineering mechanics problems.
-- Set aside the Equations of quilibrium
1. Introduction
The relations between forces and deformation :
Stress -- Ch 1
Fundamental concept of
Strain – Ch 2
Strain Energy – Ch 11
We will learn:
1. Modulus of Toughness
2. Modulus of resilience
3. Castigliano Theorem
11.2 Strain Energy
dU  Pdx
(11.1)
U   Pdx
x1
0
Strain energy  U   Pdx
x1
(11.2)
0
If the material response is elastic:
U
x1
0
1 2
kxdx  kx1
2
P  kx
1
U  P1 x1
2
(11.3)
11.3 Strain-Energy Density
x1 P dx
U

0 A L
V
1
U
   xd x
0
V
1
Strain energy density  u    x d  x
0
(11.4)
Modulus of Toughness = Toughness
= area under the - curve.
 x  E x
u
1
0
u
E 12
E x d  x 
2
 12
(11.5)
(11.6)
(11.7)
2E
uY 
 Y2
2E
Modulus of Resilience:
(11.8)
11.4 Elastic strain Energy for Normal Stresses
U
u  lim
V  0  V
dU
u
dV
1
u    xd x
0
1 2 1
1  x2
u  E x   x  x 
2
2
2E
(11.9)
 x  E x
1 2 1
1  x2
u  E x   x  x 
2
2
2E
U
 x2
2E
dV
P2
U
dV
2
2 EA
(11.9)
(11.11)
(11.12)
P2
U
dV
2
2 EA
U
L
0
P2
dx
2 AE
dV  Adx
P2L
U
2 AE
(11.13)
11.5 Elastic Strain Energy for Shearing
Stresses
 xy
u    xy d  xy
0
(11.18)
 xy2
1 2 1
u  G xy   xy xy 
2
2
2G
dU
u
dV
U
 xy2
2G
or
dV
U   udV
(11.19)
(11.20)
(11.21)
Strain Energy in Torsion
 xy2
T 2 2
U   dV  
dV
2
2G
2GJ
(11.19)
dV  dAdx
U
L
0
U
L
0
T2
2
(

dA)dx
2 
2GJ
T2
dx
2GJ
T 2L
U
2GJ
(11.21)
(11.22)
11.6 Strain Energy for a General State of Stress
1
u  ( x x   y y   z z   xy xy   yz yz   zx zx )
2
From Eq. (2.38)
x  
y  
z  
 xy 
u
X
E

 X
E
 X
E
 xy
G
 y


E
y
E

 z

 z
 y
E
 yz 
E

 yz
G
(2.38)
E
z
E
 zx 
 zx
G
1
1 2
( x2   y2   z2  2 ( x y   y z   z x )] 
( xy   yz2   zx2 )
2E
2G
(11.26)
(11.25)
1
1 2
2
2
2
u
( x   y   z  2 ( x y   y z   z x )] 
( xy   yz2   zx2 )
2E
2G
(11.26)
If the principal stresses are used:
1
u
( a2   b2   c2  2 ( a b   b c   c a )]
2E
Where a, b, c = the principal stresses
(11.27)
u  uv  ud
(11.28)
Where uv = the part of energy leading to volume change
= hydrostatic stress
ud = deviatoric energy = the part of energy leading to
shape change.
Defining Mean Stress or Average Stress

a b c
3

(11.29)
And set:
a   
b   
'
a

where
= mean stress
c   
'
b

'
i
 + +  0
'
b
'
c
(11.30)
= deviatoric stress
Combining Eqs. (11-29) and (11.30)
'
a
'
c
(11.31)
 a' + b' + c'  0
(11.31)
-- They only change the shape, but do not lead to
the change of the volume of the material.
The dilatation, V/V, caused by the state of stress can be
obtained, via Eq. (2.31)
 X  y  z
2 ( X   y   z )
e 

E
E
as
1  2 '
e
( a + b' + c' )
E
or
e=0
The uv can be obtained by substituting  into Eq. (11.27)
to obtain
1
3(1  2 ) 2
2
2
uv 
[3  2 (3 )] 

2E
2E
By means of Eq. (11.29) we have
1  2
uv 
( a + b + c )2
6E
(11.32)
The distortion or deviatoric energy can be obtained as
1
ud  u  uv 
[3( a2   b2   c2 )  6 ( a b   b c   c a ) 
6E
(1  2 )( a   b   c )2 ]
After simplification:
ud 
1
[( a2  2 a b   b2 )  ( b2  2 b c   c2 )  ( c2  2 c a   a2 )]
6E
Recalling Eq. (2.43)
E
1
2G
or
1 1

2G
E
(2.43)
Hence, the previous equation takes a new form
1
ud 
[( a   b )2  ( b   c )2  ( c   a )2 ]
12G
For 2-D cases,  c  0
Eq. (11.33) reduces to
1
ud 
( a2   a b   b2 )
6G
For uniaxial tension, i.e. 1-D cases, b = 0 and a = y
( ud ) y 
 y2
6G
(11.33)
(11.34)
Substituting this equation to the previous equation, it leads to
 a2   a b   b2   Y2
(7.26)
Expanding the same operation to a 3-D case, one can have
( a   b )2  ( b   c )2  ( c   a )2  2 Y2
(11.35)
Replacing “<“ by “=“, it follows
( a   b ) 2  ( b   c ) 2  ( c   a ) 2  2 Y2
This is a circular cylinder of radius 2 / 3  y
(11.36)
The 2- D Yield Locus:
   a b    
2
a
2
b
2
Y
(7.26)
The 3- D Yield Locus:
( a   b )2  ( b   c )2  ( c   a )2  2 Y2
(11.35)
11.7 Impact Loading
1 2
K. E. of the ball = K. E. = mvo
2
The strain energy in the bar:
Um  
 m2
2E
dV
(11.37)
Assuming:
1. No heat dissipation
2. The ball sticks with the rod after impact.
1 2
U m  K . E .  mv0
2
(11.38)
If the stress is uniform within the rod:
Um  
 m2
2E
dV 
 m2 V
2E
Therefore, m can be determined as:
2U m E
mv02 E
m 

V
V
(11.39)
11.8 Design for Impact Loads
Case A: For a Uniform-Diameter Rod:
m 
2U m E
V
(11.45a)
Case B: For a Multiple-Section Rod:
 L
 L
 L
V  4 A   A   5 A 
 2
 2
 2
m 
8U m E
V
(11.45b)
Case C: For a Circular Cantilever Rod:
m 
6U m E
L( I / c )2
(11.44)
1 4
I  c
4
However,
Hence,
1 4 2 1 2
1
L( I / c )  L(  c / c )   c L  V
4
4
4
2
Eq. (11.44) can be reduced to
24U m E
m 
V
(11.45c)
Since
m 
E
V
We conclude that:
--- in order to develop lower m in the rod, the rod
should have
1. Lower E
2. Larger V
3. Uniform m
11.9 Work and Energy under a Single Load
Case A: For an Uniaxial Load:
Strain energy  U   Pdx
x1
(11.2)
0
1
U  P1 x1
2
(11.3)
Case B: For a Cantilever Beam:
1
U  P1 y1
2
1 P1 L3
P12 L3
U  P1 (
)
2 3 EI
6 EI
(11.46)
Case C: For a Beam in Bending:
U
1
0
1
Md  M11
2
1
M1 L M12 L
U  M1 (
)
2
EI
2 EI
(11.47)
(11.48)
Case D: For a Beam in Torsion:
U
1
0
1
Td  T11
2
1 T1 L T12 L
U  T1 ( ) 
2 JG
2JG
(11.49)
11.10 Deflection under a Single Load by the
Work-Energy Method
Case A: For an Uniaxial Load:
1
U  P1 x1
2
2U
 x1 
P1
(11.3)
where x1 = deflection due to P1
Case B: For a Beam in Bending:
1
U  M11
2
2U
 1 
M1
(11.47)
where 1 = deflection due to single moment M1
11.11 Work and Energy under Several Loads
Deflection due to P1:
x11  11 P1
x21   21 P1
(11.54)
Deflection due to P2:
x12  12 P2
x22   22 P2
(11.54)
ij = influence coefficients
The combining effect of P1 and P2
x1  x11  x12  11 P1  12 P2
x2  x21  x22   21P1   22 P2
(11.54)
(11.54)
Calculating the Work Done by P1 and P2:
Case I: Assuming P1 is applied first --At Point 1, the work done by P1 is
1
1
1
P1 x11  P1 (11 P1 )  11 P12
2
2
2
(11.58)
At Point 2, the work done by P1 is zero.
P2 is applied next --At Point 2, the work done by P2 is
1
1
1
P2 x22  P2 ( 22 P2 )   22 P22
2
2
2
(11.59)
At Point 1, the work done by P1 due to additional defection
[caused by P2] is
P1 x12  P1 (12 P2 )  12 P1 P2
(11.60)
(11.58) + (11.59) + (11.60), the total strain
energy is
1
U  (11 P12  212 P2 P1   22 P22 )
2
(11.61)
Case II: Assuming P2 is applied first --1
U  ( 22 P22  212 P2 P1  11 P12 )
2
(11.62)
Equating Eqs. (11.61) and (11.62) leads
to
12  12
-- Maxwell Reciprocal Theorem
11.12 Castigliano’s Theorem
1
U  (11 P12  212 P2 P1   22 P22 )
2
U
 11 P1  12 P2  x1
P1
U
 12 P1   22 P2  x2
P2
(11.61)
(11.63)
(11.64)
Or, in general
U
xj 
Pj
(11.65)
U
j 
M j
(11.66)
[ Castigliano’s Theorem ]
General Formulation for Castigliano’s Theorem
For multiple loading, P1, P2, …., Pn
the deflection of the point of application of Pi can be expressed as
x j   jk Pk
(11.66)
k
The total strain energy of the structure is
1
U     ik Pi Pk
2i k
(11.67)
Differentiating U w.r.to Pj
U 1
1
  jk Pk   ji Pi
Pj 2 k
2 i
Since ij = ji, the above equation becomes
U 1
1
  jk Pk   ji Pi    jk Pk
k
Pj 2 k
2 i
(11.65)
xj 
U
Pj
-- for concentrated loads
j 
U
M j
-- for moment loads
U
j 
T j
-- for torsion
11.13 Deflections by Castigliano’s Theorem
Total strain energy of a beam
subjected to bending
U
L
0
M2
dx
2 EI
(11.17)
But, the differentiation
can be applied prior to
integration.
Deflection at point Pj
L M M
U
xj 

dx
0 EI P
Pj
j
(11.70)
Total strain energy of a truss member
Fi 2 Li
U 
i 1 2 Ai E
n
(11.71)
Deflection at point Pj
U n Fi Li Fi
xj 

Pj i 1 Ai E Pj
(11.72)
If no load is applied to a point, where we desire to obtain a
deflection:
-- Apply a dummy (fictitious) load Q at that point, determine
U
xj 
Q j
Then, set Q = 0.
[ Castigliano’s Theoem ]
(11.76)
11.14 Statically Indeterminate Structures
Structure indeterminate to the 1st degree:
Procedures:
1. Assume one support as redundant
2. Replace it with an unknown force
3. y = U/RA = 0  solving for RA
L M M
U
yA 

dx
0
R A
EI RA
1
M  RA x  wx 2
2
M
x
RA
1 L
1 3
2
yA 
( RA x  wx )dx

0
EI
2
1 RA L3 wL4

(

)0
EI 3
8
3
3
RA  wL RA  wL 
8
8
5
1
RB  wL  M B  wL2
8
8