MECHANICS OF MATERIALS

FourthEdition

Beer • Johnston • DeWolf

Eccentric Axial Loading in a Plane of Symmetry• Stress due to eccentric loading found by

superposing the uniform stress due to a centric l d d li t di t ib ti dload and linear stress distribution due a pure bending moment

( ) ( )xxx += bendingcentric σσσ ( ) ( )

IMy

AP−=

bendingcentric

E t i l di• Eccentric loading

PdMPF

==

• Validity requires stresses below proportional limit, deformations have negligible effect on geometry and stresses not evaluated near pointsPdM geometry, and stresses not evaluated near points of load application.

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MECHANICS OF MATERIALS

FourthEdition

Beer • Johnston • DeWolf

Example 4.07SOLUTION:

• Find the equivalent centric load and qbending moment

• Superpose the uniform stress due toSuperpose the uniform stress due to the centric load and the linear stress due to the bending moment.

• Evaluate the maximum tensile and compressive stresses at the inner

d d i l f hAn open-link chain is obtained by bending low-carbon steel rods into the

and outer edges, respectively, of the superposed stress distribution.

gshape shown. For 160 lb load, determine (a) maximum tensile and compressive stresses (b) distance between section

• Find the neutral axis by determining the location where the normal stress is zero.

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stresses, (b) distance between section centroid and neutral axis

MECHANICS OF MATERIALS

FourthEdition

Beer • Johnston • DeWolf

Example 4.07

( )22

• Normal stress due to a centric load

( )

lb160in1963.0

in25.02

22

=

==

P

cA ππ

psi815in1963.0

lb16020

=

==APσ

• Equivalent centric load and bending moment

• Normal stress due to bending moment

( )( )ilb104

in65.0lb160lb160==

=PdM

P ( )

in10068.3

25.043

4414

41

×=

==

−

cI ππg

inlb104 ⋅= ( )( )

i8475in10068.3

in25.0inlb10443×

⋅== −I

Mcmσ

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psi8475=

MECHANICS OF MATERIALS

FourthEdition

Beer • Johnston • DeWolf

Example 4.07

• Maximum tensile and compressive stresses

0 +=t σσσ

• Neutral axis location

0 0−=I

MyAP

8475815

0

0

−=+=

+=

mc

mt

σσσ

σσσpsi9260=tσ

i7660

( )inlb105

in10068.3psi81543

0 ⋅×

==−

MI

APy

IA

8475815−= psi7660−=cσin0240.00 =y

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MECHANICS OF MATERIALS

FourthEdition

Beer • Johnston • DeWolf

Sample Problem 4.8The largest allowable stresses for the cast iron link are 30 MPa in tension and 120 MP i i D t i th l tMPa in compression. Determine the largest force P which can be applied to the link.

SOLUTION:

• Determine equivalent centric load and b di tbending moment.

• Superpose the stress due to a centric load and the stress due to bending

• Evaluate the critical loads for the allowable tensile and compressive stressesFrom Sample Problem 4 2

load and the stress due to bending.

tensile and compressive stresses.

• The largest allowable load is the smallest of the two critical loads.

From Sample Problem 4.2,23

m038.0m103 −

=

×=

YA

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of the two critical loads.49 m10868 −×=I

MECHANICS OF MATERIALS

FourthEdition

Beer • Johnston • DeWolf

Sample Problem 4.8• Determine equivalent centric and bending loads.

loadcentricm028.0010.0038.0 =−=

Pd

moment bending 028.0loadcentric

====

PPdMP

• Superpose stresses due to centric and bending loads( )( ) PPP

IMc

AP A

A 37710868

022.0028.0103 93 +=

×+

×−=+−= −−σ

( )( ) PPPI

McAP A

B 155910868

022.0028.0103

10868103

93 −=×

−×

−=−−=

××

−−σ

• Evaluate critical loads for allowable stresses.kN6.79MPa30377 ==+= PPAσkN0.77MPa1201559 =−=−= PPBσ

kN0.77=P• The largest allowable load

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kN0.77PThe largest allowable load