Mecanica Solidos Beer-11

8/20/2019 Mecanica Solidos Beer-11

1/31

MECHANICS OFMATERIALS

Third Edition

Ferdinand P. Beer

E. Russell Johnston, Jr.

CHAPTER

11Energy Methods

.

Lecture Notes:

J. Walt Oler

Texas Tech University

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


2/31

T h i r d

E d i t i on

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

Sample Problem 11.4

Work and Energy Under Several Loads

Castigliano’s Theorem

Deflections by Castigliano’s TheoremSample Problem 11.5

© 2002 The McGraw-Hill Companies, Inc. All rights reserved. 11 - 2

xamp e .

Example 11.07

Design for Impact Loads

Work and Energy Under a Single Load

Deflection Under a Single Load


3/31

T h i r d

E d i t i on


• 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.

work elementarydxPdU ==

Strain Energy


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

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

energystrainwork totaldxPU

x

=== ∫1

0

11

2

121

2

1

0

1

xPkxdxkxU

x

=== ∫

• In the case of a linear elastic deformation,


4/31

T h i r d

E d i t i on


Strain Energy Density

• To eliminate the effects of size, evaluate the strain-

energy per unit volume,

densityenergystraind u

L

dx

A

P

V

U

x

x

==

=

∫

∫

1

1

0

0

ε

ε σ


• As the material is unloaded, the stress returns to zero

but there is a permanent deformation. Only the strainenergy 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 ε 1.


5/31

T h i r d

E d i t i on


Strain-Energy Density

• The strain energy density resulting from

setting ε 1 = ε 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.

•


limit,

E

E d E u x

22

21

21

0

1

1σ ε

ε ε

ε

=== ∫

• The strain energy density resulting from

setting σ 1 = σ Y is the modulus of resilience.

resilienceof modulus E

u Y Y ==2

2σ


6/31

T h i r d

E d i t i on



• In an element with a nonuniform stress distribution,

energystraintotallim0

===∆

∆= ∫

→∆dV uU

dV

dU

V

U u

V

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

energystrainelasticdV U x 2

∫ == σ


• Under axial loading, dx AdV AP x ==σ

∫= L

dx AE

PU

0

2

2

AE

LPU

2

2

=

• For a rod of uniform cross-section,


7/31

T h i r d

E d i t i on



y M x =σ

• For a beam subjected to a bending load,

∫∫ == dV EI

y M dV

E U x

2

222

22

σ

• Setting dV = dA dx,

dxdA y EI

M dxdA

EI

y M U

L L

∫ ∫∫ ∫

== 22

2

2

22

22


dx EI

M L

∫=0

2

2

• For an end-loaded cantilever beam,

EI

LPdx

EI

xPU

Px M

L

62

32

0

22

==

−=

∫


8/31

T h i r d

E d i t i on



• For a material subjected to plane shearing

stresses,

∫= xy

xy xy d u

γ

γ τ

0

• For values of τ xy within the proportional limit,

2


GGu y

xy xy xy 2212

21 γ τ γ ===

• The total strain energy is found from

∫∫

=

=

dV G

dV uU

xy

2

2τ


9/31

T h i r d

E d i t i on



∫∫ == dV GJ

T dV

GU

xy

2

222

22

ρ τ

• For a shaft subjected to a torsional load,

• Setting dV = dA dx,

∫ ∫∫ ∫

==

L L

dxdAT

dxdAT

U 2

2

2

2

22

ρ ρ


J

T xy

ρ τ =

∫= L

dxGJ

T

0

2

2

• In the case of a uniform shaft,

GJ

LT U

2

2

=


10/31

T h i r d

E d i t i on


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.

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

• Apply the particular given

conditions to evaluate the strainenergy.


11/31

T h i r d

E d i t i on


Sample Problem 11.2

SOLUTION:

• Determine the reactions at A and B

from a free-body diagram of the

complete beam.

L

Pa R

L

Pb R B A ==


• eve op a agram o t e en ng

moment distribution.

v L

Pa M x

L

Pb M == 21


12/31

T h i r d

E d i t i on


Sample Problem 11.2

Pb

AD,portionOver the

• Integrate over the volume of the beam to find

the strain energy.

dx x L

Pa

EI dx x

L

Pb

EI

dv EI

M dx

EI

M U

ba

ba

+

=

+=

∫∫

∫∫

0

2

0

2

0

22

0

21

2

1

2

1

22


v L

Pa M

x L

M

=

=

2

1

BD,portionOver the

43

in248ksi1029

in.108in.36a

in.144kips45

=×=

==

==

I E

b

LP

( )ba EIL

baPbaab

L

P EI

+=

+=

22223232

22

63321

EIL

baPU

6

222

=

( ) ( ) ( )

( )( )( )in144in248ksi10296in108in36kips40

43

222

×=U

kipsin89.3 ⋅=

U


13/31

T h i r d

E d i t i on


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,

( ) zx zx yz yz xy xy z z y y x xu γ τ γ τ γ τ ε σ ε σ ε σ +++++= 21

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

( )[ ]22

1 222 ++−++= accbbacba E

u σ σ σ σ σ σ ν σ σ σ


( )

( ) ( ) ( )[ ] distortiontodue12

1

changevolumetodue6

21

222

2

=−+−+−=

=++−

=

+=

accbbad

cbav

d v

G

u

E

vu

uu

σ σ σ σ σ σ

σ σ σ

• Basis for the maximum distortion energy failure criteria,

( ) specimentesttensileafor

6

2

G

uu Y Y d d

σ =<


14/31

T h i r d

E d i t i on


Impact Loading

• To determine the maximum stress σ m

- Assume that the kinetic energy is

transferred entirely to the

structure,202

1 mvU m =

- Assume that the stress-strain


• 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 σ m and then

disappear.

is also valid under impact loading.

∫= dV

E U m

m 2

2σ

• Maximum value of the strain energy,

• For the case of a uniform rod,

V

E mv

V

E U mm

202 ==σ


15/31

T h i r d

E d i t i on


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.

•


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 ofportion CD, determine the maximum

value of the normal stress in the rod.

resulting from the static load Pm


16/31

T h i r d

E d i t i on


Example 11.06



( ) ( )

L

AE U P

AE

LP

AE

LP

AE

LPU

mm

mmmm

5

16

16

5

4

22 222

=

=+=


SOLUTION:

• Due to the change in diameter,

the normal stress distribution is

nonuniform.

E

V dV

E

mvU

mm

m

22

22

202

1

σ σ ≠=

=

∫

• va ua e e max mum s ress resu ng

from the static load Pm

AL

E mv

AL

E U

A

P

m

mm

20

5

8

5

16

=

=

=σ


17/31

T h i r d

E d i t i on


Example 11.07

SOLUTION:

• The normal stress varies linearly along

the length of the beam as across a

transverse section.




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.

• Evaluate the maximum stressresulting from the static load Pm


18/31

T h i r d

E d i t i on


Example 11.07



For an end-loaded cantilever beam,

32

6

6

EI U P

EI

LPU

m

mm

=

=


SOLUTION:

• The normal stress varies linearly

along the length of the beam as

across a transverse section.

E

V dV

E

WhU

mm

m

22

22 σ σ ≠=

=

∫

L

• Evaluate the maximum stress

resulting from the static load Pm

( ) ( )2266

c I L

WhE

c I L

E U

I

LcP

I

c M

m

mmm

==

==σ


19/31

T h i r d

E d i t i on


Design for Impact Loads

• For the case of a uniform rod,

V

E U mm

2=σ

• For the case of the nonuniform rod,

( ) ( ) AL L A L AV

AL

E U mm

2 / 52 / 2 / 4

5

16

=+=

=σ


( )( ) ( ) ( )

V

E U

V Lccc Lc I L

c I L

E U

m

m

m

m

24

/ /

6

412

4124

412

2

=

===

=

σ

π π

σ

• For the case of the cantilever beam

Maximum stress reduced by:

• uniformity of stress

• low modulus of elasticity with

high yield strength

• high volume

V E U mm 8=σ


20/31

T h i r d

E d i t i on



• Strain energy may also be found from

the work of the single load P1,

∫=1

0

x

dxPU

• For an elastic deformation,

11 x x


,

energy by integrating the energydensity over the volume.

For a uniform rod,

( )

AE

LPdx A

E

AP

dV E

dV uU

L

22

2

21

0

21

2

==

==

∫

∫ ∫

σ

1121200

xP xk dxkxdxPU ====

• Knowing the relationship between

force and displacement,

AE

LP

AE

LPPU

AE

LP x

2

211

121

11

=

=

=

T E


21/31

Th i r d

E d i t i on



• Strain energy may be found from the work of other types

of single concentrated loads.

• Transverse load • Bending couple • Torsional couple


EI

LP

EI

LPP

yPdyPU

y

63

321

31

121

1121

0

1

=

=

== ∫

EI

L M

EI

L M M

M d M U

2

211

121

1121

0

1

=

=

== ∫ θ θ θ

JG

LT

JG

LT T

T d T U

2

211

121

1121

0

1

=

=

== ∫ φ φ φ

T E


22/31

Th i r d

E d i t i on


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,

LF LF U BD BD BC BC

22

+=


l Ll L BD BC 8.06.0 ==

From statics,

PF PF BD BC 8.06.0 −=+=

From the given geometry,

( ) ( )[ ] AE

lP

AE

lP 2332364.0

2

8.06.0=

+=

• Equating work and strain energy,

AE

Pl y

yP AE

LPU

B

B

728.0

364.021

2

=

==

T E


23/31

Th i r d

E d i t i on


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.

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

• Equate the strain energy to the work

of P and solve for the displacement.

T E


24/31

Th i r d

E d i t i on


Sample Problem 11.4

SOLUTION:

• Find the reactions at A and B from a free-body

diagram of the entire truss.

821821 P BP AP A y x

==−=

• 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

−=

= 0= ABF

T

E


25/31

hi r d

E d i t i on


Sample Problem 11.4


• va uate t e stra n energy o t e

truss due to the load P.

( )2

22

297002

1

2

1

2

P E

A

LF

E E A

LF U

i

ii

i

ii

=

== ∑∑

• quate t e stra n energy to t e wor y

and solve for the displacement.

( )( )9

33

2

21

1073

1040107.29

2

2970022

×

××=

==

=

E

E

E

y

E

P

PP

U

y

U Py

↓= mm27.16 E y

T h

E d


26/31

hi r d

di t i on


Work and Energy Under Several Loads

• Deflections of an elastic beam subjected to twoconcentrated loads,

22212122212

21211112111

PP x x x

PP x x x

α α

α α

+=+=

+=+=

• Compute the strain energy in the beam by

evaluating the work done by slowly applying


• Reversing the application sequence yields

21111221

22222

1

2 PPPPU α α α ++=

• Strain energy expressions must be equivalent.

It follows that α 12=α

21( Maxwell’s reciprocal

theorem).

22222112

21112

1 2 PPPPU α α α ++=

1 2,

T h

E d


27/31

hi r d

di t i on


Castigliano’s Theorem

22222112

21112

1 2 PPPPU α α α ++=

• Strain energy for any elastic structure

subjected to two concentrated loads,

• Differentiating with respect to the loads,

1212111 xPPU

=+=∂

α α


22221122

xPPP

U =+=

∂

∂α α

• Castigliano’s theorem: For an elastic structure

subjected to n loads, the deflection x j of the

point of application of P j can be expressed as

and j

j j

j j

jT

U

M

U

P

U x

∂

∂=

∂

∂=

∂

∂= φ θ

T h

E d


28/31

hi r d

di t i on


Deflections by Castigliano’s Theorem

• Application of Castigliano’s theorem issimplified if the differentiation with respect to

the load P j is performed before the integration

or summation to obtain the strain energy U .

• In the case of a beam,

∫∫ ∂∂

=∂

∂==

L

j

L

dxP

M

EI

M

P

U xdx

EI

M U

2

2


• For a truss,

j

in

i i

ii

j j

n

i i

ii

P

F

E A

LF

P

U x

E A

LF U

∂

∂=

∂

∂== ∑∑

== 11

2

2

T h

E d


29/31

hi r d

di t i on


Sample Problem 11.5

• A l the method of oints to determine

SOLUTION:

• For application of Castigliano’s theorem,

introduce a dummy vertical load Q at C .

Find the reactions at A and B due to thedummy load from a free-body diagram of

the entire truss.


Members of the truss shownconsist of sections of aluminum

pipe with the cross-sectional areas

indicated. Using E = 73 GPa,

determine the vertical deflection ofthe joint C caused by the load P.

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 .

T h

E d

B J h t D W lf


30/31

ir d i t i on


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.

Q BQ AQ A y x 43

43

==−=

• Apply the method of joints to determine the axial


.

QF F

QF F

F F

BD AB

CD AC

DE CE

43;0

;0

0

−==

−==

==

T h i

E d i

Beer Johnston DeWolf


31/31

ir d i t i on


Sample Problem 11.5

• Combine with the results of Sam le 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.

( )QP E Q

F

E A

LF y i

i

iiC 42634306

1+=

∂

∂

= ∑

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

Pa1073

104043069

3

×

×=

N yC ↓= mm36.2C y

Documents

Mecanica Solidos Beer-11