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MECHANICS OF MATERIALS Fourth Edition 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 d load and linear stress distribution due a pure bending moment ( ) ( ) x x x + = bending centric σ σ σ I My A P = bending centric E ti l di Eccentric loading Pd M P F = = Validity requires stresses below proportional limit, deformations have negligible effect on geometry and stresses not evaluated near points Pd M geometry , and stresses not evaluated near points of load application. © 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 1

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Microsoft PowerPoint - 4_3_pure_bending [Compatibility Mode]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 ( ) ( )
PdM PF
© 2006 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 1
MECHANICS OF MATERIALS
• Find the equivalent centric load and q bending 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 h An open-link chain is obtained by bending low-carbon steel rods into the
and outer edges, respectively, of the superposed stress distribution.
g shape 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
( )
( )( ) ilb104
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psi8475=
0 +=t σσσ
• Neutral axis location
0 0−= I
My A P
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 4.8 The 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
m038.0 m103 −
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of the two critical loads. 49 m10868 −×=I
MECHANICS OF MATERIALS
loadcentric m028.0010.0038.0 =−=
P d
I Mc
kN0.77=P• The largest allowable load
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kN0.77PThe largest allowable load