11 energy methods

MECHANICS OF MATERIALS

Third Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

John T. DeWolf

Lecture Notes:

J. Walt Oler

Texas Tech University

CHAPTER

© 2002 The McGraw-Hill Companies, Inc. All rights reserved.

11Energy Methods



Th

irdE

ditio

n

Beer • Johnston • DeWolf

11 - 2

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

Sample Problem 11.4

Work and Energy Under Several Loads

Castigliano’s Theorem

Deflections by Castigliano’s Theorem

Sample Problem 11.5



Th

irdE

ditio

n


11 - 3

• 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

which is equal to the area of width dx under the load-deformation diagram.

workelementarydxPdU

• The total work done by the load for a deformation x1,

which results in an increase of strain energy in the rod.

energystrainworktotaldxPUx

1

0

11212

121

0

1

xPkxdxkxUx

• In the case of a linear elastic deformation,

Strain Energy



Th

irdE

ditio

n


11 - 4

Strain Energy Density• To eliminate the effects of size, evaluate the strain-

energy per unit volume,

densityenergy straindu

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



Th

irdE

ditio

n


11 - 5

Strain-Energy Density

• The strain energy density resulting from setting 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

EdEu x 22

21

21

01

1

• The strain energy density resulting from setting Y is the modulus of resilience.

resilience of modulusE

u YY

2

2



Th

irdE

ditio

n


11 - 6


• In an element with a nonuniform stress distribution,

energystrain totallim0

dVuU

dV

dU

V

Uu

V

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

energy strainelasticdVE

U x 2

2

• Under axial loading, dxAdVAPx

L

dxAE

PU

0

2

2

AE

LPU

2

2

• For a rod of uniform cross-section,



Th

irdE

ditio

n


11 - 7


I

yMx

• For a beam subjected to a bending load,

dVEI

yMdV

EU x

2

222

22

• Setting dV = dA dx,

dxEI

M

dxdAyEI

MdxdA

EI

yMU

L

L

A

L

A

0

2

0

22

2

02

22

2

22

• For an end-loaded cantilever beam,

EI

LPdx

EI

xPU

PxM

L

62

32

0

22



Th

irdE

ditio

n


11 - 8


• For a material subjected to plane shearing stresses,

xy

xyxy du

0

• For values of xy within the proportional limit,

GGu xy

xyxyxy 2

2

212

21

• The total strain energy is found from

dVG

dVuU

xy

2

2



Th

irdE

ditio

n


11 - 9


J

Txy

dVGJ

TdV

GU xy

2

222

22

• For a shaft subjected to a torsional load,

• Setting dV = dA dx,

L

L

A

L

A

dxGJ

T

dxdAGJ

TdxdA

GJ

TU

0

2

0

22

2

02

22

2

22

• In the case of a uniform shaft,

GJ

LTU

2

2



Th

irdE

ditio

n


11 - 10

Sample Problem 11.2

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.

SOLUTION:

• Determine the reactions at A and B from a free-body diagram of the complete beam.

• Integrate over the volume of the beam to find the strain energy.

• Apply the particular given conditions to evaluate the strain energy.

• Develop a diagram of the bending moment distribution.



Th

irdE

ditio

n


11 - 11

Sample Problem 11.2

SOLUTION:

• Determine the reactions at A and B from a free-body diagram of the complete beam.

L

PaR

L

PbR BA

• Develop a diagram of the bending moment distribution.

vL

PaMx

L

PbM 21



Th

irdE

ditio

n


11 - 12

Sample Problem 11.2

vL

PaM

xL

PbM

2

1

BD,portion Over the

AD,portion Over the

43 in 248ksi1029

in. 108in. 36a

in. 144kips45

IE

b

LP

• Integrate over the volume of the beam to find the strain energy.

baEIL

baPbaab

L

P

EI

dxxL

Pa

EIdxx

L

Pb

EI

dvEI

Mdx

EI

MU

ba

ba

2

2223232

2

2

0

2

0

2

0

22

0

21

6332

1

2

1

2

1

22

EIL

baPU

6

222

in 144in 248ksi 10296

in 108in 36kips4043

222

U

kipsin 89.3 U



Th

irdE

ditio

n


11 - 13

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,

zxzxyzyzxyxyzzyyxxu 21

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

distortion todue 12

1

change volume todue 6

21

22

1

222

2

222

accbbad

cbav

dv

accbbacba

Gu

E

vu

uu

Eu

• Basis for the maximum distortion energy failure criteria,

specimen test tensileafor 6

2

Guu Y

Ydd



Th

irdE

ditio

n


11 - 14

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 m and then disappear.

• To determine the maximum stress m

- Assume that the kinetic energy is transferred entirely to the structure,

202

1 mvUm

- Assume that the stress-strain diagram obtained from a static test is also valid under impact loading.

dVE

U mm 2

2

• Maximum value of the strain energy,

• For the case of a uniform rod,

V

Emv

V

EUmm

202



Th

irdE

ditio

n


11 - 15

Example 11.06

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.

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



Th

irdE

ditio

n


11 - 16

Example 11.06

SOLUTION:

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

E

VdV

E

mvU

mm

m

22

22

202

1


L

AEUP

AE

LP

AE

LP

AE

LPU

mm

mmmm

5

16

16

5

4

22 222


AL

Emv

AL

EU

A

P

m

mm

20

5

8

5

16



Th

irdE

ditio

n


11 - 17

Example 11.07

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

SOLUTION:

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





Th

irdE

ditio

n


11 - 18

Example 11.07

SOLUTION:

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

E

VdV

E

WhU

mm

m

22

22


For an end-loaded cantilever beam,

3

32

6

6

L

EIUP

EI

LPU

mm

mm


2266

cIL

WhE

cIL

EU

I

LcP

I

cM

m

mmm



Th

irdE

ditio

n


11 - 19

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

V

EUmm

2

V

EU

VLcccLcIL

cIL

EU

mm

mm

24

//

6

412

4124

412

2

• For the case of the cantilever beam

Maximum stress reduced by:

• uniformity of stress• low modulus of elasticity with

high yield strength• high volume

• For the case of the nonuniform rod,

V

EU

ALLALAV

AL

EU

mm

mm

8

2/52/2/4

5

16



Th

irdE

ditio

n


11 - 20


• Previously, we found the strain energy by integrating the energy density over the volume. For a uniform rod,

AE

LPdxA

E

AP

dVE

dVuU

L

22

2

21

0

21

2

• Strain energy may also be found from the work of the single load P1,

1

0

x

dxPU

• For an elastic deformation,

11212

121

00

11

xPxkdxkxdxPUxx

• Knowing the relationship between force and displacement,

AE

LP

AE

LPPU

AE

LPx

2

211

121

11



Th

irdE

ditio

n


11 - 21


• Strain energy may be found from the work of other types of single concentrated loads.

EI

LP

EI

LPP

yPdyPUy

63

321

31

121

1121

0

1

• Transverse load

EI

LM

EI

LMM

MdMU

2

211

121

1121

0

1

• Bending couple

JG

LT

JG

LTT

TdTU

2

211

121

1121

0

1

• Torsional couple



Th

irdE

ditio

n


11 - 22

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.

lLlL BDBC 8.06.0

From statics,PFPF BDBC 8.06.0

From the given geometry,

• Strain energy of the structure,

AE

lP

AE

lP

AE

LF

AE

LFU BDBDBCBC

2332

22

364.02

8.06.0

22

• Equating work and strain energy,

AE

Ply

yPAE

LPU

B

B

728.0

364.021

2



Th

irdE

ditio

n


11 - 23

Sample Problem 11.4

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.

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.

• 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.



Th

irdE

ditio

n


11 - 24

Sample Problem 11.4SOLUTION:

• Find the reactions at A and B from a free-body diagram of the entire truss.

821821 PBPAPA yx

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

PF

PF

CE

DE

815

817

0

815

CD

AC

F

PF

PF

PF

CE

DE

821

45

0ABF



Th

irdE

ditio

n


11 - 25

Sample Problem 11.4

• Evaluate the strain energy of the truss due to the load P.

2

22

297002

1

2

1

2

PE

A

LF

EEA

LFU

i

ii

i

ii

• Equate the strain energy to the work by P and solve for the displacement.

9

33

2

21

1073

1040107.29

2

2970022

E

E

E

y

E

P

PP

Uy

UPy

mm27.16Ey



Th

irdE

ditio

n


11 - 26

Work and Energy Under Several Loads• Deflections of an elastic beam subjected to

two concentrated loads,

22212122212

21211112111

PPxxx

PPxxx

• Reversing the application sequence yields

21111221

22222

1 2 PPPPU

• Strain energy expressions must be equivalent. It follows that (Maxwell’s reciprocal theorem).

22222112

21112

1 2 PPPPU

• Compute the strain energy in the beam by evaluating the work done by slowly applying P1 followed by P2,



Th

irdE

ditio

n


11 - 27

Castigliano’s Theorem

22222112

21112

1 2 PPPPU

• Strain energy for any elastic structure subjected to two concentrated loads,

• Differentiating with respect to the loads,

22221122

12121111

xPPP

U

xPPP

U

• Castigliano’s theorem: For an elastic structure subjected to n loads, the deflection xj of the point of application of Pj can be expressed as

and j

jj

jj

j T

U

M

U

P

Ux



Th

irdE

ditio

n


11 - 28

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

jjj

L

dxP

M

EI

M

P

Uxdx

EI

MU

00

2

2

• For a truss,

j

in

i i

ii

jj

n

i i

iiP

F

EA

LF

P

Ux

EA

LFU

11

2

2



Th

irdE

ditio

n


11 - 29

Sample Problem 11.5

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 joint C caused by the load P.

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

• 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.

• Setting Q = 0, evaluate the derivative which is equivalent to the desired displacement at C.

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.



Th

irdE

ditio

n


11 - 30

Sample Problem 11.5

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.

QBQAQA yx 43

43

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

QFF

QFF

FF

BDAB

CDAC

DECE

43;0

;0

0



Th

irdE

ditio

n


11 - 31

Sample Problem 11.5

• 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.

QPEQ

F

EA

LFy i

i

iiC 42634306

1

• Setting Q = 0, evaluate the derivative which is equivalent to the desired displacement at C.

Pa1073

104043069

3

NyC mm 36.2Cy

Date post:	12-Nov-2014
Category:	Documents
Upload:	tigin
View:	720 times
Download:	3 times

11 energy methods

Documents