Strain Energy Part 1
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Transcript of Strain Energy Part 1
8/12/2019 Strain Energy Part 1
http://slidepdf.com/reader/full/strain-energy-part-1 1/20
ENERGY METHOD
SOLID MECHANICS II
BMCS 3333
Nadlene Razali
8/12/2019 Strain Energy Part 1
http://slidepdf.com/reader/full/strain-energy-part-1 3/20
Previous lessons : relation between force and
deformation using concept of,
Stress
Strain
Concept of Strain energy : increase in energy
associated with the deformation of the member.
Introduction
8/12/2019 Strain Energy Part 1
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Apply energy methods to solve
problems involving deflection
and deformations of the
structures.
Use principle of conservation of
energy to determine stress and
deflection of a member
subjected to impact.
Develop the method of virtual work and Castigliano’s
theorem to determine displacement and slope at points
on structural members and mechanical elements.
Introduction
8/12/2019 Strain Energy Part 1
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• A uniform rod is subjected to a slowly increasing load
• The total work done by the load for a deformation x1,
work total dx P U
x
1
0
11212
121
0
1
x P kxdxkxU
x
• In the case of a linear elastic deformation,
• The elementary work done by the load P as the rod
elongates by a small dx is
which is equal to the area of width dx under the load-
deformation diagram.
work elementarydx P dU
External Load
8/12/2019 Strain Energy Part 1
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Work of a force:
• When a force P is applied to the bar,
followed by the force P ’, total work done by
both forces is represented by the area of
the entire triangle in graph shown.
Strain energy:
When loads are applied to a body and causes deformation, the
external work done by the loads will be converted into internal
work called strain energy. This is provided no energy is
converted into other forms.
TOTAL WORK = STRAIN ENERGY
Strain Energy
8/12/2019 Strain Energy Part 1
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• To eliminate the effects of size, evaluate the strain-
energy per unit volume,
densityenergy straind u
L
dx
A
P
V
U
x
x
1
1
0
0
• 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.
• The total strain energy density resulting from the
deformation is equal to the area under the curve to 1.
Strain Energy Density
8/12/2019 Strain Energy Part 1
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• 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,
E
E d E u x
22
21
21
0
1
1
• The strain energy density resulting from
setting 1 Y is the modulus of resilience.
resilienceof modulus E
u Y Y
2
2
Strain Energy Density
8/12/2019 Strain Energy Part 1
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Example:
The stress strain diagram shown has been drawn from data obtained
during a tensile test of an aluminum alloy. Using E=72 GPa,
determine:
(a) The modulus of resilience
(b) The modulus of tougnness
Strain Energy Density
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Solution:
Strain Energy Density
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• In an element with a nonuniform stress distribution,energystraintotallim
0
dV uU
dV
dU
V
U u
V
• For values of u < uY , i.e., below the proportional
limit,
energy strainelasticdV E
U x 2
2
• Under axial loading, dx AdV A P x
L
dx AE
P U
0
2
2
AE
L P
U 2
2
• For a rod of uniform cross-section,
Elastic Strain Energy for Normal Stresses
8/12/2019 Strain Energy Part 1
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Example:
Using E = 200 GPa, determine:
(a) Strain energy of steel rod ABC when P = 25 kN
(b) The corresponding strain energy density in portion AB and BC
Elastic Strain Energy for Normal Stresses
8/12/2019 Strain Energy Part 1
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Solution:
Elastic Strain Energy for Normal Stresses
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Solution:
Elastic Strain Energy for Normal Stresses
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11 - 15
I
y M x
• For a beam subjected to a bendingload,
dV EI
y M dV
E U x
2
222
22
• Setting dV = dAdx
dx
EI
M
dxdA y EI
M dxdA
EI
y M U
L
L
A
L
A
0
2
0
2
2
2
02
22
2
22
• For an end-loaded cantilever beam,
EI
L P dx
EI
x P U
Px M
L
62
32
0
22
Elastic Strain Energy in Bending
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• For a material subjected to plane shearing
stresses,
xy
xy xy d u
0
• For values of xy within the proportional limit,
GGu
xy xy xy xy
2
2
212
21
• The total strain energy is found from,
dV G
dV uU
xy
2
2
Elastic Strain Energy for Shearing Stresses
8/12/2019 Strain Energy Part 1
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J
T xy
dV GJ
T dV
GU
xy
2
222
22
• For a shaft subjected to a torsional load,
• Setting dV = dA dx ,
L
L
A
L
A
dx
GJ
T
dxdAGJ
T dxdA
GJ
T U
0
2
0
2
2
2
02
22
2
22
• In the case of a uniform shaft,
GJ
LT U
2
2
Elastic Strain Energy for Shearing Stresses
8/12/2019 Strain Energy Part 1
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• Previously found strain energy due to uniaxial stress and plane
shearing stress. For a general state of stress,
zx zx yz yz xy xy z z y y x xu 21
• With respect to the principal axes for an elastic, isotropic body,
distortiontodue12
1
changevolumetodue6
21
221
222
2
222
accbbad
cbav
d v
accbbacba
Gu
E
vu
uu
E u
• Basis for the maximum distortion energy failure criteria,
specimentesttensileafor6
2
Guu Y
Y d d
Elastic Strain Energy for General Stresses
8/12/2019 Strain Energy Part 1
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Example:
A torque, TA = 300 N.m exerted on the disk. Using G = 73
GPa, determine the total strain energy required by the
assembly.
Elastic Strain Energy for Shearing Stresses