IV
Energy Methods
Strain Energy: definition
 Strain energy is a form of potential energy.
 Strain energy is defined as the increase in energy




11 - 2
associated with the deformation of the member.
Strain energy is equal to the work done by slowly
increasing load applied to the member.
Work done to distort an elastic member is stored as
strain energy.
Some energy may be lost in plastic deformation of the
member and some may be converted into heat instead
of stored as strain energy, but the rest is recoverable.
A spring is an example of a storage device for strain
energy.
Strain Energy
• A uniform rod is subjected to a slowly increasing load
• The elementary work done by the load P as the rod
elongates by a small dx is
dU  P dx  elementary work
which is equal to the area of width dx under the loaddeformation diagram.
• The total work done by the load for a deformation x1,
x1
U   P dx  total work  strain energy
0
which results in an increase of strain energy in the rod.
• In the case of a linear elastic deformation,
x1
U   kx dx  12 kx12  12 P1x1
4-3
0
Strain Energy Density
• The strain-energy density of a material is defined as
the strain energy per unit volume.
• To eliminate the effects of size, evaluate the strainenergy per unit volume,
U

V
x1
P dx
A L
0
1
u    x d  strain energy density
0
• The total strain energy density resulting from the
deformation is equal to the area under the curve to 1.
• As the material is unloaded, the stress returns to zero
but there is a permanent deformation. Only the strain
energy represented by the triangular area is recovered.
4- 4
• Remainder of the energy spent in deforming the material
is dissipated as heat.
Strain-Energy Density
• The strain energy density resulting from
setting 1  R is the modulus of toughness.
• The energy per unit volume required to cause
the material to rupture is related to its ductility
as well as its ultimate strength.
• If the stress remains within the proportional
limit,
1
E12 12
u   E1 d x 

2
2E
0
• The strain energy density resulting from
setting 1  Y is the modulus of resilience.
uY 
4- 5
 Y2
2E
 modulus of resilience
Strain Energy under Axial Loading
• In an element with a nonuniform stress distribution,
U dU

dV
V 0 V
u  lim
U   u dV  total strain energy
• For values of u < uY , i.e., below the proportional
limit,
U 
 x2
2E
dV  elastic strain energy
• Under axial loading,  x  P A
L
P2
U 
dx
2 AE
0
• For a rod of uniform cross-section,
P2L
U
2 AE
4-6
dV  A dx
Comparison of Energy Stored in
Straight and Stepped bars
L/2
a
L
A
nA
P
2
(a) U  P L
2 AE
(b)
L/2
b
A
P
P2 L / 2 P2 L / 2
U

2 AE
2nAE
P2L  1 n 



2 AE  2n 
3 P2L
Note for n=2; case (b) has U=
which is 3/4 of case (a)
4 2 AE
4-7
Problem 1
One of the two bolts need to support a sudden tensile loading. To choose it is
necessary to determine the greatest amount of strain energy each bolt can
absorb. Bolt A has a diameter of 20 mm for 50 mm length and a root diameter
of 18 mm for 6 mm length. Bolt B has 18 mm diameter throughout the length.
Take E = 210 GPa and y= 310 Mpa.
4- 8
Solution
4- 9
Strain Energy in Bending
• For a beam subjected to a bending load,
U 
 x2
2E
dV  
M 2 y2
2 EI
2
dV
• Setting dV = dA dx,
M 2  2 
U  
dA dx  
y dA dx
2
2

2 EI
2 EI  A

0 A
0
L
x 
My
I
L

0
M 2 y2
L
M2
dx
2 EI
• For an end-loaded cantilever beam,
M   Px
L
P2 x2
P 2 L3
U 
dx 
2 EI
6 EI
0
4 - 10
Sample Problem 1
SOLUTION:
• Determine the reactions at A and B
from a free-body diagram of the
complete beam.
• Develop a diagram of the bending
moment distribution.
a) Taking into account only the normal
stresses due to bending, determine the
strain energy of the beam for the
loading shown.
b) Evaluate the strain energy knowing
that the beam is a W250x67, P =
160kN, L = 3.6m, a = 0.9m, b = 2.7m,
and E = 200GPa.
44-- 11
11
• Integrate over the volume of the
beam to find the strain energy.
• Apply the particular given
conditions to evaluate the strain
energy.
Sample Problem 1
SOLUTION:
• Determine the reactions at A and B
from a free-body diagram of the
complete beam.
RA 
Pb
L
RB 
Pa
L
• Develop a diagram of the bending
moment distribution.
M1 
4 - 12
Pb
x
L
M2 
Pa
v
L
Sample Problem 1
• Integrate over the volume of the beam to find
the strain energy.
a
b
0
0
M12
M 22
U 
dx  
dv
2 EI
2 EI
a
Over the portion AD,
0
Pb
M1 
x
L
Over the portion BD,
M2 
4 - 13
b
2
Pa
v
L
2
1  Pb 
1  Pa 

x
dx

x  dx





2 EI  L 
2 EI  L 
0
1 P 2  b 2a3 a 2b3  P 2a 2b 2
a  b 



2 EI L2  3
3  6 EIL2
P 2a 2b2
U
6 EIL
P  160kN
L  3.6m
a  0.9m
b  2.7m
E  200GPa
I  104  10 6 mm 4
160  10 N  0.9m 2.7m
U
6200  10 Pa 104  10 m 3.6m 
3
9
U  336Nm
2
2
6
2
4
Problem 2
4 - 14
Solution:
4 - 15
Problem 4
11 - 16
Solution:
 Finding the support reactions
 Finding the internal moment
4 - 17
Solution (Contd.)
4 - 18
Problem 5
4 - 19
Solution:
4 - 20
Problem 6
4 - 21
Solution:
4 - 22
Strain Energy For Shearing Stresses
• For a material subjected to plane shearing
stresses,
 xy
u
 xy d xy
0
• For values of xy within the proportional limit,
2

xy
2
u  12 G xy  12  xy  xy 
2G
• The total strain energy is found from
U   u dV

4 - 23
2
 xy
2G
dV
Strain Energy in Torsion
• For a shaft subjected to a torsional load,
2
 xy
T 2 2
U 
2G
dV  
2GJ
2
dV
• Setting dV = dA dx,
T 2  2 
U 
dA dx  
 dA dx
2
2

2GJ
2GJ  A

0A
0
L
 xy 
T
J
T 2 2
L
L
T2

dx
2GJ
0
• In the case of a uniform shaft,
T 2L
U
2GJ
4 - 24
Problem 1
4 - 25
Solution:
4 - 26
Problem 2
Rod AC is made of aluminum (G = 73 GPa) and is subjected to torque T applied at end
C. Knowing that portion BC of the rod is hollow and has an inside diameter of 16 mm,
determine the strain energy of the rod for a maximum shearing stress of 120 MPa.
4 - 27
Solution
4 - 28
Castigliano’s Theorem
• Strain energy for any elastic structure
subjected to two concentrated loads,

U  12 11P12  212 P1P2   22 P22

• Differentiating with respect to the loads,
* Castigliano’s Theorem is named
after the Italian engineer, Alberto
Castigliano (1847-1884)
U
 11P1  12 P2  x1
P1
U
 12 P1   22 P2  x2
P2
• Castigliano’s theorem: For an elastic structure
subjected to n loads, the deflection xj of the
point of application of Pj, measured along the
line of action of Pj, can be expressed as
4- 29
xj 
U
Pj
and  j 
U
M j
j 
U
T j
Deflections by Castigliano’s Theorem
• Application of Castigliano’s theorem is
simplified if the differentiation with respect to
the load Pj is performed before the integration
or summation to obtain the strain energy U.
• In the case of a beam,
L
M2
U 
dx
2 EI
0
L
U
M M
xj 

dx
Pj
EI Pj
0
• For a truss,
n
Fi2 Li
U 
2A E
i 1 i
4- 30
n
U
F L F
xj 
 i i i
Pj i 1 Ai E Pj
Problem 1
 Determine the displacement of point B on the beam shown
 Apply an external force at B
 Using method of section the internal moment and partial derivative are to be
w
determined
B
A
L
P
wx
x
VM
x
  M NA  0; M  wx    px   0
2
wx 2
M
 px
2
M
 x
P
wx 2
M
IfP  0, thenM  
and
 x
wx 2
M 
p
2
Solution
(Contd.)
M
 x
P
wx 2
M
If P  0, then M  
and
 x
2
P
L
L
2



M
dx
wx
dx



 x 
  M
 




2 
EI
 P  EI
0
0 
wL4

8 EI
Problem 2
(
)
Solution (Contd.)
Problem 3
Solution (Contd.)
Solution (Contd.)