Lecture Mechanics of materials (Third edition) - Chapter 11: Energy methods

Third Edition

CHAPTER

MECHANICS OF
MATERIALS
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf

Energy Methods

Lecture Notes:
J. Walt Oler
Texas Tech University

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Third
Edition

MECHANICS OF MATERIALS

Beer • Johnston • DeWolf

Energy Methods
Strain Energy
Strain Energy Density
Elastic Strain Energy for Normal Stresses
Strain Energy For Shearing Stresses

Sample Problem 11.2
Strain Energy for a General State of Stress
Impact Loading
Example 11.06
Example 11.07
Design for Impact Loads
Work and Energy Under a Single Load
Deflection Under a Single Load

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Sample Problem 11.4
Work and Energy Under Several Loads
Castigliano’s Theorem
Deflections by Castigliano’s Theorem
Sample Problem 11.5

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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
0
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Strain Energy Density
• To eliminate the effects of size, evaluate the strainenergy per unit volume,

U
=
V

x1

P dx
∫A L
0

ε1

u = ∫ σ x dε = strain energy density
0

• The total strain energy density resulting from the
deformation is equal to the area under the curve to ε1.
• 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.
• Remainder of the energy spent in deforming the material
is dissipated as heat.
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MECHANICS OF MATERIALS

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Strain-Energy Density
• The strain energy density resulting from
setting ε1 = εR 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,
ε1

Eε12 σ 12
=
u = ∫ Eε1 dε x =
2
2E
0

• The strain energy density resulting from
setting σ1 = σY is the modulus of resilience.
uY =
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σ Y2
2E

= modulus of resilience

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Elastic Strain Energy for Normal Stresses
• In an element with a nonuniform stress distribution,
∆U dU
=
V
∆
dV
∆V →0

u = lim

U = ∫ u dV = total strain energy

• For values of u < uY , i.e., below the proportional
limit,
σ x2
U =∫

2E

dV = elastic strain energy

• Under axial loading, σ x = P A

dV = A dx

L

P2
U =∫
dx
2 AE
0

• For a rod of uniform cross-section,
P2L
U=
2 AE
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Elastic Strain Energy for Normal Stresses
• For a beam subjected to a bending load,
σ x2
M 2 y2
U =∫

2E

dV = ∫

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
0 A
L

σx =

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

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Strain Energy For Shearing Stresses
• For a material subjected to plane shearing
stresses,
γ xy

u=

∫τ xy dγ xy
0

• For values of τxy within the proportional limit,
u=

1 Gγ 2
xy
2

= 12 τ xy γ xy =

2
τ xy

2G

• The total strain energy is found from
U = ∫ u dV
=∫

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2
τ xy

2G

dV

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Strain Energy For Shearing Stresses
• For a shaft subjected to a torsional load,
2
τ xy
T 2ρ 2
U =∫

2G

dV = ∫

2GJ

2

dV

• Setting dV = dA dx,
T 2 ⎛⎜ 2 ⎞⎟
U =∫∫
dA dx = ∫
ρ dA dx
2
2⎜∫
⎟
2GJ
2GJ ⎝ A
⎠
0A
0

L

τ xy =

Tρ
J

T 2ρ 2

L

L

T2
=∫
dx
2GJ
0

• In the case of a uniform shaft,
T 2L
U=
2GJ

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Sample Problem 11.2
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 W10x45, P = 40
kips, L = 12 ft, a = 3 ft, b = 9 ft, and E
= 29x106 psi.
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• Integrate over the volume of the
beam to find the strain energy.
• Apply the particular given
conditions to evaluate the strain
energy.

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Sample Problem 11.2
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 =

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Pb
x
L

M2 =

Pa
v
L

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Sample Problem 11.2
• Integrate over the volume of the beam to find
the strain energy.
a

b

0

0

M 22
M12
dv
dx + ∫
U =∫
2 EI
2 EI
a

Over the portion AD,

0

Pb
M1 =
x
L
Over the portion BD,
M2 =

Pa
v
L

2

b

2

1 ⎛ Pb ⎞
1 ⎛ Pa ⎞
=
x
dx
x ⎟ dx
+
⎜
⎟
⎜
∫
∫
2 EI ⎝ L ⎠
2 EI ⎝ L ⎠
0

1 P 2 ⎜⎛ b 2a3 a 2b3 ⎞⎟ P 2 a 2b 2
(a + b )
=
=
+
2 EI L2 ⎜⎝ 3
3 ⎠⎟ 6 EIL2

P 2a 2b 2
U=
6 EIL

P = 45 kips

L = 144 in.

a = 36 in.

b = 108 in.

E = 29 × 103 ksi I = 248 in 4

(
40 kips )2 (36 in )2 (108 in )2
U=
6(29 × 103 ksi )(248 in 4 )(144 in )
U = 3.89 in ⋅ kips

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MECHANICS OF MATERIALS

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Strain Energy for a General State of Stress
• Previously found strain energy due to uniaxial stress and plane
shearing stress. For a general state of stress,

(

u = 12 σ xε x + σ yε y + σ zε z + τ xyγ xy + τ yzγ yz + τ zxγ zx

)

• With respect to the principal axes for an elastic, isotropic body,
u=

[

]

1 2
σ a + σ b2 + σ c2 − 2ν (σ aσ b + σ bσ c + σ cσ a )
2E

= uv + ud
uv =

1 − 2v
(σ a + σ b + σ c )2 = due to volume change
6E

ud =

1
(σ a − σ b )2 + (σ b − σ c )2 + (σ c − σ a )2 = due to distortion
12G

[

]

• Basis for the maximum distortion energy failure criteria,
ud < (ud )Y =

σ Y2
6G

for a tensile test specimen

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Impact Loading
• To determine the maximum stress σm
- 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
end with a body of mass m moving
with a velocity v0.
• Rod deforms under impact. Stresses
reach a maximum value σm and then
disappear.
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• Maximum value of the strain energy,
Um = ∫

σ m2
2E

dV

• For the case of a uniform rod,
mv02 E
2U m E
=
σm =
V
V
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Example 11.06
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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Example 11.06
• Find the static load Pm which produces
the same strain energy as the impact.
Pm2 (L 2 )
+
Um =
AE
Pm =

16 U m AE
5 L

• Evaluate the maximum stress resulting
from the static load Pm

SOLUTION:
• Due to the change in diameter,
the normal stress distribution is
nonuniform.

σm =
=

U m = 12 mv02

σ m2
σ m2 V
=
dV ≠

∫ 2E

Pm2 ( L 2 ) 5 Pm2 L
=
4 AE
16 AE

2E

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Pm
A
16 U m E
5 AL

8 mv02 E
=
5 AL

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MECHANICS OF MATERIALS

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Example 11.07
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.

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Example 11.07
• 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
Pm =

SOLUTION:
• The normal stress varies linearly
along the length of the beam as
across a transverse section.

=∫

2E

dV ≠

L3

• Evaluate the maximum stress
resulting from the static load Pm

U m = Wh

σ m2

6U m EI

σ m2 V
2E

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M m c Pm Lc
=
σm =
I
I
=

6U m E

( )

=
2
LI c

6WhE

( )

L I c2

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Design for Impact Loads
• For the case of a uniform rod,
σm =

2U m E
V

• For the case of the nonuniform rod,
σm =

16 U m E
5 AL

V = 4 A( L / 2 ) + A( L / 2 ) = 5 AL / 2

σm =

8U m E
V

• For the case of the cantilever beam

Maximum stress reduced by:
• uniformity of stress
• low modulus of elasticity with
high yield strength
• high volume
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σm =

6U m E

( )
L (I / c 2 ) = L (14 πc 4 / c 2 ) = 14 (πc 2 L ) = 14 V

σm =

L I c2

24U m E
V
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MECHANICS OF MATERIALS

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Work and Energy Under a Single Load
• Strain energy may also be found from
the work of the single load P1,
x1

U = ∫ P dx
0

• For an elastic deformation,
• Previously, we found the strain
energy by integrating the energy
density over the volume.
For a uniform rod,
U = ∫ u dV = ∫
L

=∫
0

σ

2

2E

dV

(P1 A)2 Adx =
2E

x1

x1

0

0

U = ∫ P dx = ∫ kx dx = 12 k x12 = 12 P1x1

• Knowing the relationship between
force and displacement,

P12 L
2 AE

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PL
x1 = 1
AE
2
P
L
P
L
⎛
⎞
1
1

U = 12 P1⎜
=
⎟
⎝ AE ⎠ 2 AE
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Work and Energy Under a Single Load
• Strain energy may be found from the work of other types
of single concentrated loads.
• Transverse load

U=

y1

1Py
P
dy
=
∫
2 1 1
0

⎛ P1L3 ⎞
1
⎟=
= 2 P1⎜
⎜ 3EI ⎟
⎝
⎠

• Bending couple

θ1

U = ∫ M dθ = 12 M1θ1
0

P12 L3
6 EI

2
M
L
M
L
⎛
⎞
= 12 M1⎜ 1 ⎟ = 1
⎝ EI ⎠ 2 EI

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• Torsional couple

φ1

U = ∫ T dφ = 12 T1φ1
0
2
T
L
T
L
⎛
⎞
= 12 T1⎜ 1 ⎟ = 1
⎝ JG ⎠ 2 JG

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Deflection Under a Single Load
• If the strain energy of a structure due to a
single concentrated load is known, then the

equality between the work of the load and
energy may be used to find the deflection.
• Strain energy of the structure,
2
2
FBC
LBC FBD
LBD
+
U=
2 AE
2 AE

[

From the given geometry,
LBC = 0.6 l

LBD = 0.8 l

From statics,
FBC = +0.6 P FBD = −0.8 P

]

P 2l (0.6 )3 + (0.8)3
P 2l
=
= 0.364
AE

2 AE

• Equating work and strain energy,
P2L 1
U = 0.364
= 2 P yB
AE
yB = 0.728

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Pl
AE
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Sample Problem 11.4
SOLUTION:
• Find the reactions at A and B from a
free-body diagram of the entire truss.
• Apply the method of joints to
determine the axial force in each
member.

Members of the truss shown consist of
sections of aluminum pipe with the
cross-sectional areas indicated. Using
E = 73 GPa, determine the vertical
deflection of the point E caused by the
load P.

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• Evaluate the strain energy of the
truss due to the load P.
• Equate the strain energy to the work
of P and solve for the displacement.

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Sample Problem 11.4
SOLUTION:
• Find the reactions at A and B from a freebody diagram of the entire truss.
Ax = −21 P 8

Ay = P

B = 21 P 8

• Apply the method of joints to determine the
axial force in each member.

FDE = − 17
P
8

FAC = + 15
P
8

FDE = 54 P

FCE = + 15
P
8

FCD = 0

FCE = − 21
P
8

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FAB = 0

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Sample Problem 11.4

• Evaluate the strain energy of the
truss due to the load P.
1
Fi2 Li
Fi2 Li
U =∑
=
∑ Ai
2 Ai E 2 E
=

(

1
29700 P 2
2E

)

• Equate the strain energy to the work by P
and solve for the displacement.
1 Py = U
E
2

2U 2 ⎛⎜ 29700 P 2 ⎞⎟
yE =
=
P P ⎜⎝ 2 E ⎟⎠

(
29.7 × 103 )(40 × 103 )
yE =
73 × 10

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9

y E = 16.27 mm ↓

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