IV Energy Methods

Strain Energy: definition

 Strain energy is a form of potential energy.
 Strain energy is defined as the increase in energy
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.

11 - 2
 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 load-
deformation 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 P1 x1
0

4-3
 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 strain-
energy per unit volume,
x1
U P dx
V
= 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.

• Remain of the energy spent in deforming the material is

4- 4 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.
Resilience is the ability of a material to absorb
energy when it is deformed elastically
 Y2
uY = = modulus of resilience (elasticity)
4- 5 2E
 Strain Energy under Axial Loading
• In an element with a nonuniform stress distribution,
U dU
u = lim = U =  u dV = total strain energy
V →0 V dV

• For values of u < uY , i.e., below the proportional

limit,
 x2
U = dV = elastic strain energy
2E

• Under axial loading,  x = P A dV = A dx

L
P2
U = dx
2 AE
0

• For a rod of uniform cross-section,

P2L
U=
2 AE
4-6
 Comparison of Energy Stored in
Straight and Stepped bars
b
L a L/2 L/2

A P nA A P

2 P2L / 2 P2L / 2
(a) U = P L (b) U= +
2 AE 2 AE 2nAE
P2L  1+ n 
=  
2 AE  2n 
3 P2 L
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,
 x2 M 2 y2
U = dV =  2
dV
2E 2 EI

• Setting dV = dA dx,

M 2  2 
L L
M 2 y2
My U =  dA dx =  2
y dA dx
x = 2 EI 2
2 EI  A 
I 0 A 0 
L
M2
= dx
2 EI
0

➢ For an end-loaded cantilever beam,

M = − Px
L
P2 x2 P 2 L3
U = dx =
2 EI 6 EI
0
4 - 10
 Problem 1
SOLUTION:
1. Determine the reactions at A and
B from a free-body diagram of
the complete beam.

2. Develop a diagram of the

bending moment distribution.

a) Taking into account only the normal 3. Integrate over the volume of the
stresses due to bending, determine the beam to find the strain energy.
strain energy of the beam for the
loading shown. 4. Apply the particular given
conditions to evaluate the strain
b) Evaluate the strain energy knowing energy.
that the beam is a W250x67, P =
160kN, L = 3.6m, a = 0.9m, b = 2.7m,
and E = 200GPa.
44-- 11
11
 Problem 1 cont.
SOLUTION:
1. Determine the reactions at A and
B from a free-body diagram of
the complete beam.
Pb Pa
RA = RB =
L L

2. Develop a diagram of the

bending moment distribution.
Pb Pa
M1 = x M2 = v
L L

4 - 12
 Problem 1 cont.
3. Integrate over the volume of the beam to
find the strain energy.
a b
M12 M 22
U = dx +  dv
2 EI 2 EI
0 0
a 2 b 2
1  Pb  1  Pa 
Over the portion AD, =   x
2 EI  L 
 dx +   x  dx
2 EI  L 
0 0
Pb
M1 = x
1 P 2  b 2a3 a 2b3  P 2a 2b 2
L
= + = (a + b )
Over the portion BD, 2 EI L2  3 3  6 EIL2
Pa
M2 = v P 2a 2b 2
L U=
6 EIL
P = 160kN L = 3.6m
U=
(160  10 N ) (0.9m ) (2.7m )
3 2 2 2

a = 0.9m b = 2.7m 6(200  10 Pa )(104  10 m )(3.6m )

9 −6 4

E = 200 GPa I = 104  10 6 mm 4

U = 336 Nm
4 - 13
 Problem 2

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 Solution:

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Problem 4

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 Solution:
 Finding the support reactions

 Finding the internal moment

4 - 17
 Solution (Contd.)

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 Problem 5

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 Solution:

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 Problem 6

4 - 21
 Solution:

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 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,

1 G 2
 xy
2
u= 2 xy = 12  xy  xy =
2G

• The total strain energy is found from

U =  u dV

 xy
2
= dV
2G

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 Strain Energy in Torsion
• For a shaft subjected to a torsional load,
 xy
2
T 2 2
U = dV =  2
dV
2G 2GJ

• Setting dV = dA dx,
T 2  2 
L L
T 2 2
U =  dA dx =  2
 dA dx
2GJ 2
2GJ  A 
0A 0 
T
 xy = L
T2
J = dx
2GJ
0

• In the case of a uniform shaft,

T 2L
U=
2GJ

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 Strain Energy for Transverse Shear

τ2
U = dV
v 2G

• For transverse shear ,  = VQ and dV = dAdx

It
 
2
 
L 2 2
1 VQ V Q
hence, U =
v 2G  It  dAdx = 0 2GI2  A t 2 dx
 dA

• The Integral inside the parenthesis can be replaced with

f, the form factor, which is dimension less and unique
for a specific cross section

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 Substituting the form factor, fs Strain Energy for
A Q2
f s = 2  2 dA Transverse Shear b
I A t
Q2 fs I 2
(cont)
A t 2 dA = A h 
 −y
2 
A' h
2
y
V  Q 
L 2 2 L 2
f sV
U = 2  2
 dA 
 =  dx N
2GI  A t  0 2GA h
0
A 2
 Form factor for rectangular Cross Section:
t=b  h 
  − y 
h  bh 
2
A=bh 
Q = y A' = y +  2  b − y  =  − y 2 
 2  2  2 2 
 
 
I=bh3 /12
• Substituting in the form factor equation and simplify , f=6/5

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 Problem 1

4 - 27
 Solution:

4 - 28
 Problem 2

4 - 29
 Problem 3
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.

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 Solution

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 Impact Loading
 Impact loading is a dynamic loading i.e. vary with time
 It occurs when one object strikes the other (collision)
 Consider a block released from rest falling h distance and strikes the spring to
compress it max before it comes to rest W
 Energy of falling is transformed into
strain energy of spring.
 Work done on falling of weight over a h
distance (h+ max ) is work done on spring max
to displace it max . k
 Therefore Ue=Ui

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 Impact Loading
If the weight W is applied statically (or gradually to the spring,
the end displacement of the spring is ∆st = W/k.

Ue = Ui

1
W (h + Δmax ) = (kΔmax )Δmax = 1 kΔmax 2
2 2

 Solving the above quadratic equation, the maximum root is

2
W W  W 
Δmax = +   + 2 h
k k  k 

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IMPACT LOADINGS (cont) Impact Loading
 Using this simplification, the above equation becomes
 max =  st + ( st )2 + 2 st h
  h 
 max =  st 1 + 1 + 2  
   st  
 Once ∆max is computed, the maximum force applied to the spring can be
determined from

Fmax = k max
 However if the weight is applied DYNAMICALLY, h = 0; then ∆max = 2 ∆st

Copyright © 2011 Pearson Education South Asia Pte Ltd

Example 1 Impact Loading
The aluminum pipe( OD= 60 mm, t=10 mm) shown is used to support a
load of 600 kN. Determine the maximum displacement at the top of the
pipe when the load is applied (a) gradually and (b) suddenly from h =0.
Take E =70x103N/mm2
Ue = Ui

600 1
WΔst =
W 2L
kN 2 2 AE
WL 600( 240 )
Δst = = = 0.5953
AE π( 60 − 50 )70
2 2

h when h=0, then

  0 
Δmax = Δst 1 + 1 + 2   = 2 Δst
240   Δst  
mm =1.1906 mm
 Impact Loading
• To determine the maximum stress m
- Assume that the kinetic energy is
transferred entirely to the
structure,
U m = 12 mv02

- Assume that the stress-strain

diagram obtained from a static test
is also valid under impact loading.
• Consider a rod which is hit at its
• Maximum value of the strain energy,
end with a body of mass m moving
m
2
with a velocity v0. Um =  dV
2E
• Rod deforms under impact. Stresses • For the case of a uniform rod,
reach a maximum value m and then
2U m E mv02 E
disappear. m = =
V V
4 - 36
 Impact Loading
Example 2 SOLUTION:
• Due to the change in diameter, the
normal stress distribution is nonuniform.

• Find the static load Pm which produces

the same strain energy as the impact.

• Evaluate the maximum stress

resulting from the static load Pm
Body of mass m with velocity v0 hits
the end of the nonuniform rod BCD.
Knowing that the diameter of the
portion BC is twice the diameter of
portion CD, determine the maximum
value of the normal stress in the rod.

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 Impact Loading
Example 1 • Find the static load Pm which produces
the same strain energy as the impact.
Pm2 (L 2 ) Pm2 (L 2 ) 5 Pm2 L
Um = + =
AE 4 AE 16 AE
16 U m AE
Pm =
5 L

• Evaluate the maximum stress resulting

SOLUTION:
from the static load Pm
• Due to the change in diameter, Pm
the normal stress distribution is m =
A
nonuniform.
16 U m E
=
U m = 12 mv02 5 AL

m
2
m
2 8 mv02 E
V =
= dV  5 AL
2E 2E
4 - 38
 Example 2 Impact Loading
SOLUTION:
• The normal stress varies linearly along
the length of the beam as across a
transverse section.

• Find the static load Pm which produces

the same strain energy as the impact.

• Evaluate the maximum stress

A block of weight W is dropped from a resulting from the static load Pm
height h onto the free end of the
cantilever beam. Determine the
maximum value of the stresses in the
beam.

4 - 39
 Example 2 Impact Loading
• Find the static load Pm which produces
the same strain energy as the impact.

For an end-loaded cantilever beam,

Pm2 L3
Um =
6 EI
6U m EI
Pm =
L3
SOLUTION:
• The normal stress varies linearly • Evaluate the maximum stress
along the length of the beam as resulting from the static load Pm
across a transverse section. M m c Pm Lc
m = =
U m = Wh I I
6U m E 6WhE
m
2
m
2
=
2E
dV 
2E
V =
( )
LI c 2
=
( )
L I c2

4- 40
 Problem 5
Impact Loading
The cylindrical block E has a speed o= 5 m/s when it strikes squarely the
yoke BD that is attached to the 22 mm diameter rods AB and CD. Knowing that
the rods are made of a steel for which Y = 345 MPa and E = 200 GPa,
determine the weight of the block E for which the factor of safety is five with
respect to permanent deformation of the rods.
Solution
 Solution (Contd.)

Substituting the data

 Impact Loading
Problem 6
Solution

l
 Impact Loading
Problem 7
The composite aluminium bar is made from two segments having diameters of 5
mm and 10 mm. Determine the maximum axial stress developed in the bar if the
5kg collar is dropped from a height of h= 100 mm. E al= 70 GPa and Y=410 MPa

P = k max
W = k st

4 - 46
 Problem 8
Impact Loading
The composite aluminium bar is made from two segments having diameters of
5 mm and 10 mm. Determine the maximum h from which the 5kg collar is
dropped so that it produces a maximum axial stress in the bar of max=300
MPa. Eal= 70 GPa and Y=410 MPa

P = k max
W = k st
4 - 47
 Impact Loading
Problem 9
The composite aluminium 2014-T6 bar is made from two segments having
diameters of 7.5 mm and 15 mm. Determine the maximum axial stress
developed in the bar if the 10 kg collar is dropped from a height of h=100 mm.
Eal= 73.1 GPa and Y=414 MPa

n = impact factor

n = max
 st

4 - 48
 Impact Loading
Problem 10
The composite aluminium 2014-T6 bar is made from two segments having diameters of
7.5 mm and 15 mm. Determine the maximum height h from which the 10 kg collar
should be dropped so that it produces a maximum axial stress in the bar of max=300
MPa.
Eal= 73.1 GPa and Y=414 MPa
n = impact factor

n = max
 st

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 Tutorial in Class

4 - 50
4 - 51
4 - 52
 ENERGY UNDER SEVERAL LOADS
 The strain energy of a structure subjected to several loads
will be considered in terms of loads and the resulting
deflections.
 Consider an elastic beam AB subjected to two concentrated
loads P1 and P2. The strain energy of the beam is equal to the
work of P1 and P2 as they are slowly applied at C1 and C2.

4 - 53
 Energy under several loads
 Let assume ONLY P1 is applied to the beam. C1 and C2 are
deflected and the deflection are proportional to the load P1.
Results in deflection x11 and x21.
 x11=11P1 x21=21P1 (eq.1)
 11 and 21 are constants called influence coefficients which
represent the deflection of C1 and C2.

4 - 54
 Energy under several loads
 Let assume ONLY P2 is applied to the beam. C1 and C2 are
deflected and the deflection are proportional to the load P2.
Results in deflection x12 and x22.
 x12=12P2 x22=22P2 (eq.2)
 12 and 22 are constants called influence coefficients which
represent the deflection of C1 and C2.

 The deflections of x1 and x2 of C1 and C2 when both loads

are applied as:
 x1=x11+x12=11P1+ 12P2 (eq.3)
 x2=x21+x22=21P1+ 22P2 (eq.4)
4 - 55
 Energy under several loads
 To compute the work done by P1 and P2.
 Assume P1 is applied first at C1. From eqn.1, expression of
the work P1 given as:
 0.5P1(x11)=0.5P1(11P1)= 0.511P12 (eq.5)
 *note that P2 does not work while C2 moves through x21
because its does not been applied.

4 - 56
 Energy under several loads
 Now slowly apply P2 at C2
 From eqn.2, expression of the work P2 as
 0.5P2(x22)=0.5P2(22P2)= 0.522P22 (eq.6)
 as P2 applied at C2 the point of P1 moves through x12 from
C’1 to C1 and the load P1 does its work.
 Since P1 is fully applied during displacement(fig), work is
equal to: P1x12=P1(12P2)= 12P1P2 (eq.7)

4 - 57
 Energy under several loads
 Adding eqn.5,6 and 7, the strain energy under loads P1 and
P2 given as:
 U=0.5(11P12+ 212P1P2+ 22P22
 However if load P2 had first applied to the beam and then
load P1, the strain energy of the beam given as:
 U=0.5(22P22+ 221P2P1+ 11P12
 work done is shown as:


4 - 58
 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,
U
= 11P1 + 12 P2 = x1
* Castigliano’s Theorem is named
P1
after the Italian engineer, Alberto U
Castigliano (1847-1884) = 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
U U U
xj = and  j = j =
4- 59 Pj M j 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 L
M2 U M M
U = dx xj = = dx
2 EI Pj EI Pj
0 0

• For a truss,
n n
Fi2 Li U F L F
U = xj = = i i i
2A E
i =1 i
Pj i =1 Ai E Pj

4- 60
 Problem 7
 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
determined w
B A
L
wx
P x
V
M  NA
+ M = 0; M + wx   + p(x ) = 0
2
x
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
 M  dx
L
 wx 2 
 = M  =  − 
 (− x ) dx
0  P  EI 0 
2  EI
wL4
=
8 EI
 w

B A
L
 Determine the SLOPE of point B.
 Apply an external MOMENT at B
 Using method of section the internal moment and partial derivative are to be
determined  x
+  M NA = 0; M + Mb + wx  = 0
2
 x  M
M = − Mb − wx ; = −1
wx  2  Mb
Mb dU L  M  dx L  wx 2 
− Mb (− 1)
dx
V = = M  = −
M dMb 0  Mb  EI 0  2  EI
x L
 wx 2  dx
=  
0 2  EI
L
1  wx 3 
=
EI  6  0
wL3
= ccw
6 EI
Problem 8
(
)
Solution (Contd.)
Problem 9
Solution (Contd.)
Solution (Contd.)
 Problem 1
SOLUTION:
• For application of Castigliano’s theorem,
introduce a dummy vertical load Q at C.
Find the reactions at A and B due to the
dummy load from a free-body diagram of
the entire truss.
• Apply the method of joints to determine
the axial force in each member due to Q.
Members of the truss shown
consist of sections of aluminum • evaluate the derivative with respect to Q
pipe with the cross-sectional areas of the strain energy of the truss due to
indicated. Using E = 73 GPa, the loads P and Q.
determine the vertical deflection of
the joint C caused by the load P.
• Setting Q = 0, evaluate the derivative
which is equivalent to the desired
displacement at C.
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 SOLUTION:
• Find the reactions at A and B due to a dummy load Q
at C from a free-body diagram of the entire truss.
Ax = − 34 Q Ay = Q B = 34 Q

• Apply the method of joints to determine the axial

force in each member due to Q.

FCE = FDE = 0
FAC = 0; FCD = −Q

FAB = 0; FBD = − 34 Q

4- 70
 • Combine with the results of Sample Problem 11.4 to evaluate the derivative
with respect to Q of the strain energy of the truss due to the loads P and Q.
 F L  F 1
yC =   i i  i = (4306 P + 4263Q )
 Ai E  Q E

• Setting Q = 0, evaluate the derivative which is equivalent to the desired

displacement at C.

yC =
(
4306 40  103 N ) yC = 2.36 mm 
9
73  10 Pa
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 Problem 2

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11 - 73
 Problem 3

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11 - 75
 Problem 4

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11 - 77
11 - 78
  An L shaped bar with length AB and BC of 0.5m and 1m
respectively.. Load P=25kN acting at point A. Diameter is
25mm along the rod. Given E=190GN/m^2 and G=74
GN/m^2. Calculate total strain energy of the bar members.

11 - 79