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MECHANICS OF MATERIALS Third Edition Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Texas Tech University CHAPTER

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Published by , 2017-02-19 08:05:03

Third Edition MECHANICS OF MATERIALS - Kişisel Sayfalar

MECHANICS OF MATERIALS Third Edition Ferdinand P. Beer E. Russell Johnston, Jr. John T. DeWolf Lecture Notes: J. Walt Oler Texas Tech University CHAPTER

ThirdMECHANICS OF MATE

Work and Energy Under S

• Defle
conce

x1
x2

• Com
evalu
P1 fo

U

• Reve

U

• Strain
It fol
theor

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

ERIALS Beer • Johnston • DeWolf

Several Loads

ections of an elastic beam subjected to two
entrated loads,
1 = x11 + x12 = α11P1 + α12P2

2 = x21 + x22 = α21P1 + α22P2

mpute the strain energy in the beam by

uating the work done by slowly applying

ollowed by P2,
( )=
1 α11P12 + 2α12P1P2 + α 22 P22
2

ersing the application sequence yields

( )=
1 α 22 P22 + 2α 21P2 P1 + α11P12
2

n energy expressions must be equivalent.

llows that α12=α21 (Maxwell’s reciprocal

rem).

ed. 11 - 26

ThirdMECHANICS OF MATE

Castigliano’s Theorem

• Stra
subj

U

• Diff






• Cast
subje
poin

x

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

ERIALS Beer • Johnston • DeWolf

ain energy for any elastic structure

jected to two concentrated loads,

( )U
= 1 α11P12 + 2α12P1P2 + α22P22
2

ferentiating with respect to the loads,

∂U = α11P1 + α12P2 = x1
∂P1

∂U = α12P1 + α22P2 = x2
∂P2

tigliano’s theorem: For an elastic structure

ected to n loads, the deflection xj of the
nt of application of Pj can be expressed as

j = ∂U and θ j = ∂U φj = ∂U
∂Pj ∂M j ∂T j

ed. 11 - 27

ThirdMECHANICS OF MATE

Deflections by Castigliano’

• Applic
simplif
the loa
or sum

• In the c

L

U=∫

0

• For a tr

n

U =∑

i=

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

ERIALS Beer • Johnston • DeWolf

’s Theorem

cation of Castigliano’s theorem is

fied if the differentiation with respect to

ad Pj is performed before the integration
mmation to obtain the strain energy U.

case of a beam,

M 2 dx ∫x j= ∂U = LM ∂M dx
2EI ∂Pj EI ∂Pj

0

russ, =∑x j ∂U = n FiLi ∂Fi
∂Pj i=1 AiE ∂Pj
∑n Fi2Li

=1 2 AiE

ed. 11 - 28

ThirdMECHANICS OF MATE

Sample Problem 11.5

S


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.

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

ERIALS Beer • Johnston • DeWolf

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.

• 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.

ed. 11 - 29

ThirdMECHANICS OF MATE

Sample Problem 11.5

SOLUTION:
• Find the re

at C from a

Ax =

• Apply the m
force in eac

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

ERIALS Beer • Johnston • DeWolf

:

eactions at A and B due to a dummy load Q
a free-body diagram of the entire truss.

− 3 Q Ay = Q B = 3 Q
4 4

method of joints to determine the axial
ch member due to Q.

FCE = FDE = 0

FAC = 0; FCD = −Q

FAB = 0; FBD = − 3 Q
4

ed. 11 - 30

ThirdMECHANICS OF MATE

Sample Problem 11.5

• Combine with the results of Sample P
with respect to Q of the strain energy

∑yC = ⎝⎛⎜⎜ Fi Li ⎟⎟⎠⎞ ∂Fi = 1 (4306P + 42
Ai E ∂Q E

• Setting Q = 0, evaluate the derivative

displacement at C.

( )yC
= 4306 40 ×103 N yC =
73×109 Pa

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

ERIALS Beer • Johnston • DeWolf

Problem 11.4 to evaluate the derivative
y of the truss due to the loads P and Q.

263Q)

e which is equivalent to the desired

2.36 mm ↓ 11 - 31

ed.


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