strain energy part 1

8/12/2019 Strain Energy Part 1

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ENERGY METHOD

SOLID MECHANICS II

BMCS 3333

Nadlene Razali



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



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



• 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



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



• 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



• 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




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




Solution:




• 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



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




Solution:




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



• 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



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




• 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



Example:

A torque, TA = 300 N.m exerted on the disk. Using G = 73

GPa, determine the total strain energy required by the

assembly.




Solution:

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strain energy part 1

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