Third Edition MECHANICS OF MATERIALS - Kişisel Sayfalar

CHAPTER MECHA
MATER

Ferdinand P. Beer
E. Russell Johnsto
John T. DeWolf

Lecture Notes:
J. Walt Oler
Texas Tech Univers

Third Edition

ANICS OF
RIALS

on, Jr. Energy Methods

sity

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

ThirdMECHANICS OF MATE

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

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

ERIALS Beer • Johnston • DeWolf

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

ed. 11 - 2

ThirdMECHANICS OF MATE

Strain Energy

• A uniform rod

• The elementar
elongates by a

dU = P dx =

which is equa
deformation d

• The total wor

x1

U = ∫ P dx =

0

which results

• In the case of

x1

U = ∫ kx dx

0

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ERIALS Beer • Johnston • DeWolf

d is subjected to a slowly increasing load

ry work done by the load P as the rod
a small dx is

= elementary work

al to the area of width dx under the load-
diagram.

rk done by the load for a deformation x1,

= total work = strain energy

in an increase of strain energy in the rod.

a linear elastic deformation,

= 1 kx12 = 1 P1x1
2 2

ed. 11 - 3

ThirdMECHANICS OF MATE

Strain Energy Density

• To eliminate
energy per un

∫U = x1 P d

V AL

0
ε1

u = ∫σ x dε

0

• The total strai
deformation i

• As the materi
but there is a
energy repres

• Remainder of
is dissipated a

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ERIALS Beer • Johnston • DeWolf

the effects of size, evaluate the strain-
nit volume,

dx
L

ε = strain energy density

in energy density resulting from the

is equal to the area under the curve to ε1.

ial is unloaded, the stress returns to zero
permanent deformation. Only the strain
sented by the triangular area is recovered.

f the energy spent in deforming the material
as heat.

ed. 11 - 4

ThirdMECHANICS OF MATE

Strain-Energy Density

• The
settin

• The
the m
as w

• If the
limit

u

• The
settin

uY

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ERIALS Beer • Johnston • DeWolf

strain energy density resulting from

ng ε1 = εR is the modulus of toughness.

energy per unit volume required to cause
material to rupture is related to its ductility
well as its ultimate strength.

e stress remains within the proportional

t,

=∫u ε1 dε x = Eε12 = σ12

Eε1 2 2E

0

strain energy density resulting from

ng σ1 = σY is the modulus of resilience.

uY = σ 2 = modulus of resilience
Y

2E

ed. 11 - 5

ThirdMECHANICS OF MATE

Elastic Strain Energy for N

• In an elem

u = lim

∆V →

• For values

limit,

∫U = σ 2
x

2E

• Under axi

U = L P
2A
∫

0

• For a rod

U = P2L
2 AE

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

ERIALS Beer • Johnston • DeWolf

Normal Stresses

ment with a nonuniform stress distribution,

m ∆U = dU U = u dV = total strain energy

∫→0 ∆V dV

s of u < uY , i.e., below the proportional

2

x dV = elastic strain energy
E

ial loading, σ x = P A dV = A dx

P2 dx
AE

of uniform cross-section, 11 - 6

L
E

ed.

ThirdMECHANICS OF MATE

Elastic Strain Energy for N

•F

•S

σ = M y
I
x

•F

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ERIALS Beer • Johnston • DeWolf

Normal Stresses

For a beam subjected to a bending load,

σ 2 M 2y2
∫ ∫U = x dV = 2EI 2 dV

2E

Setting dV = dA dx,

L M 2y2 L M2 ⎜⎛ y2dA⎞⎟⎟ dx
0 2EI 2 0 2EI 2 ⎜
=∫ ∫ ∫ ∫U dA dx =

A ⎝A ⎠

∫= L M 2 dx
2EI
0

For an end-loaded cantilever beam,

M = −Px

∫U = L P2x2 dx = P2L3
2EI 6EI
0

ed. 11 - 7

ThirdMECHANICS OF MATE

Strain Energy For Shearing

• For
stres

u

• For

u

• The

U

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ERIALS Beer • Johnston • DeWolf

g Stresses

a material subjected to plane shearing
sses,

γ xy

∫= τ xy dγ xy

0

values of τxy within the proportional limit,

= 1 Gγ 2 = 1τ γ = τ 2
2 xy xy
2
xy xy 2G

e total strain energy is found from

U = ∫u dV

∫= τ 2 dV
xy

2G

ed. 11 - 8

ThirdMECHANICS OF MATE

Strain Energy For Shearing

•F

τ xy = Tρ •S
•I
J

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

ERIALS Beer • Johnston • DeWolf

g Stresses

For a shaft subjected to a torsional load,

τ 2 T 2ρ 2
xy 2
∫ ∫U = dV = 2GJ dV
2G

Setting dV = dA dx,

L T 2ρ 2 L T2 ⎜⎛ ρ 2dA⎟⎞⎟dx
0 2GJ 2 ⎜
= A 2GJ 2 ⎝ A⎠
∫ ∫ ∫ ∫U =
0 dA dx

∫L T2 dx

= 2GJ

0

In the case of a uniform shaft,

U = T2L
2GJ

ed. 11 - 9

ThirdMECHANICS OF MATE

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.

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ERIALS Beer • Johnston • DeWolf

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.

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

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

E

ed. 11 - 10

ThirdMECHANICS OF MATE

Sample Problem 11.2

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ERIALS Beer • Johnston • DeWolf

SOLUTION:

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

RA = Pb RB = Pa
L L

• Develop a diagram of the bending
moment distribution.

M1 = Pb x M2 = Pa v
L L

ed. 11 - 11

ThirdMECHANICS OF MATE

Sample Problem 11.2

• Inte
the

U

Over the portion AD,

M1 = Pb x
L

Over the portion BD,

M2 = Pa v U
L

P = 45kips L = 144 in. U
U
a = 36 in. b = 108 in.

E = 29 ×103ksi I = 248 in4

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

ERIALS Beer • Johnston • DeWolf

egrate over the volume of the beam to find
strain energy.

a M12 b M 2
2
U = ∫ ∫2EI 2EIdx+ dv

00

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

= 1 P2 ⎜⎛ b2a3 + a2b3 ⎞⎟ = P2a2b2 (a + b)
2EI L2 ⎜⎝ 3 3 ⎠⎟ 6EIL2

U = P2a2b2
6EIL

= (40kips)2(36 in)2(108 in)2
29 ×103 ksi 248 in4 (144 in)
( )( )U6

U = 3.89 in ⋅ kips

ed. 11 - 12

ThirdMECHANICS OF MATE

Strain Energy for a Genera

• Previously found strain energy

shearing stress. For a general

(u= 1 σ xε x +σ yε y + σ zε z +τ x
2

• With respect to the principal a

[u = 1 σ 2 + σ 2 + σ 2 − 2ν (σ
2E a b c

= uv + ud

uv = 1 − 2v (σ a + σb +σc )2 = d
6E

[ud
=1 (σ a − σ b )2 + (σ b −σ
12G

• Basis for the maximum distort

ud < (ud )Y = σ 2 for a tensile
Y

6G

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ERIALS Beer • Johnston • DeWolf

al State of Stress

y due to uniaxial stress and plane
state of stress,

)xyγ xy + τ yzγ yz + τ zxγ zx

axes for an elastic, isotropic body,

]σ aσb + σbσ c + σ cσ a )

due to volume change

]σ c )2 + (σ c − σ a )2 = due to distortion

tion energy failure criteria,

test specimen

ed. 11 - 13

ThirdMECHANICS OF MATE

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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ERIALS Beer • Johnston • DeWolf

• To determine the maximum stress σm

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

Um = 1 mv02
2

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

• Maximum value of the strain energy,

∫Um = σ 2 dV
m

2E

• For the case of a uniform rod,

σm = 2UmE = mv02E
V V

ed. 11 - 14

ThirdMECHANICS OF MATE

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.

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ERIALS Beer • Johnston • DeWolf

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

ed. 11 - 15

ThirdMECHANICS OF MATE

Example 11.06

SOLUTION:

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

Um = 1 mv02
2

∫= σ 2 dV ≠ σ m2V
m

2E 2E

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ERIALS Beer • Johnston • DeWolf

• Find the static load Pm which produces
the same strain energy as the impact.

Um = Pm2 (L 2) + Pm2(L 2) = 5 Pm2L
16 AE
AE 4 AE

Pm = 16 Um AE
5L

• Evaluate the maximum stress resulting

from the static load Pm

σ m = Pm
A

= 16 UmE
5 AL

= 8 mv02E
5 AL

ed. 11 - 16

ThirdMECHANICS OF MATE

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.

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

ERIALS Beer • Johnston • DeWolf

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
resulting from the static load Pm

ed. 11 - 17

ThirdMECHANICS OF MATE

Example 11.07

SOLUTION:

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

Um = Wh

∫= σ 2 dV ≠ σ m2V
m

2E 2E

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

ERIALS Beer • Johnston • DeWolf

• Find the static load Pm which produces
the same strain energy as the impact.

For an end-loaded cantilever beam,

Um = Pm2L3
6EI

Pm = 6U m EI
L3

• Evaluate the maximum stress
resulting from the static load Pm

σm = M mc = Pm Lc
I I

6U m E 6WhE
L I c2 L I c2
( ) ( )= =

ed. 11 - 18

ThirdMECHANICS OF MATE

Design for Impact Loads

•

•

•

Maximum stress reduced by:
• uniformity of stress
• low modulus of elasticity with

high yield strength
• high volume

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

ERIALS Beer • Johnston • DeWolf

• For the case of a uniform rod,

σm = 2U m E
V

For the case of the nonuniform rod,

σm = 16 UmE
5 AL

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

σm = 8U m E
V

For the case of the cantilever beam

6U m E
L I c2
( )σm =

( ) ( ) ( )L I /c2
= L 1 πc4 / c2 = 1 πc2L = 14V
4 4

σm = 24U m E
V

ed. 11 - 19

ThirdMECHANICS OF MATE

Work and Energy Under a

•

•

• Previously, we found the strain

energy by integrating the energy

density over the volume.

For a uniform rod, •

U = ∫u dV = ∫ σ2 dV

2E

∫= L (P1 A)2 Adx = P12L
2E 2AE
0

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

ERIALS Beer • Johnston • DeWolf

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,

x1 x1 x12

∫ ∫U = = 1 = 1
P dx = kx dx 2 k 2 P1x1

00

Knowing the relationship between

force and displacement,

x1 = P1L
AE

U = 1 P1⎛⎝⎜ P1L ⎟⎞ = P12L
2 AE ⎠ 2 AE

ed. 11 - 20

ThirdMECHANICS OF MATE

Work and Energy Under a

• Strain energy may be found from the w
of single concentrated loads.

• Transverse load • Bendin

y1 θ1

∫U = P dy = 1 P1y1 U = ∫M d
2
0
0

= 1 P1⎛⎜⎜⎝ P1L3 ⎞⎟ = P12L3 = 1 M1⎜⎝⎛
2 3EI ⎠⎟ 6EI 2

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ERIALS Beer • Johnston • DeWolf

Single Load

work of other types

ng couple • Torsional couple

φ1

dθ = 1 M1θ1 ∫U = T dφ = 1 T1φ1
2 2

0

⎛⎜ M1L ⎟⎞ = M12L = 1 T1⎜⎛⎝ T1L ⎞⎟ = T12L
⎝ EI ⎠ 2EI 2 JG ⎠ 2JG

ed. 11 - 21

ThirdMECHANICS OF MATE

Deflection Under a Single

• If
sin
eq
en

• Str

From the given geometry, • Eq

LBC = 0.6l LBD = 0.8l

From statics,

FBC = +0.6P FBD = −0.8P

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

ERIALS Beer • Johnston • DeWolf

Load

the strain energy of a structure due to a
ngle concentrated load is known, then the
quality between the work of the load and
nergy may be used to find the deflection.

rain energy of the structure,

U = FB2C LBC + FB2DLBD
2AE 2AE

[ ]= P2l (0.6)3 + (0.8)3 = 0.364 P2l
2AE AE

quating work and strain energy,

U = 0.364 P2L = 1 P yB
AE 2

yB = 0.728 Pl
AE

ed. 11 - 22

ThirdMECHANICS OF MATE

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.

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

ERIALS Beer • Johnston • DeWolf

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.

ed. 11 - 23

ThirdMECHANICS OF MATE

Sample Problem 11.4

SOLUT

• Find
body

Ax = −

• Appl
axial

FDE = − 17 P FAC = + 15 P
8 8

FCE = + 15 P FCD = 0
8

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ERIALS Beer • Johnston • DeWolf

TION:

the reactions at A and B from a free-
y diagram of the entire truss.

−21P 8 Ay = P B = 21P 8

ly the method of joints to determine the
force in each member.

FDE = 5 P FAB = 0
4
11 - 24
FCE = − 21 P
8

ed.

ThirdMECHANICS OF MATE

Sample Problem 11.4

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

∑ ∑U = Fi2Li = 1 Fi2Li
2AiE 2E Ai
( )= 1 29700P2
2E

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

ERIALS Beer • Johnston • DeWolf

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

1 PyE =U
2

yE = 2U = 2 ⎛⎜ 29700P2 ⎟⎞
P P ⎜⎝ 2E ⎟⎠

( )( )yE
= 29.7 ×103 40 ×103 yE = 16.27mm ↓
73 × 109

ed. 11 - 25