Third Edition

MECHANICS OF
MATERIALS
Ferdinand P. Beer
E. Russell Johnston
Combined Stresses:
Examples

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Third

MECHANICS OF MATERIALS Beer • Johnston • DeWolf

Combined stresses

The combined loadings which will be studied

are:

1.Axial load + Bending moment

2.Axial load + Torsional loads
3.Bending moment + Torsional load
4.Shear load + Bending moment

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Combined Stresses: Axial load + Bending moment

• Eccentric Loading: Axial loading which

does not pass through section centroid
produces internal forces equivalent to an
axial force and a couple

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MECHANICS OF MATERIALS 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
load and linear stress distribution due a pure
bending moment
 x   x centric   x bending
P My
 
A I
• Eccentric loading
• Validity requires stresses below proportional
FP limit, deformations have negligible effect on
M  Pd geometry, and stresses not evaluated near points
of load application.

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Example 4.07
SOLUTION:
• Find the equivalent centric load and
bending moment

• Superpose 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
and outer edges, respectively, of the
An open-link chain is obtained by superposed stress distribution.
bending low-carbon steel rods into the
shape shown. For 160 lb load, determine • Find the neutral axis by determining
(a) maximum tensile and compressive the location where the normal stress
stresses, (b) distance between section is zero.
centroid and neutral axis
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MECHANICS OF MATERIALS Beer • Johnston • DeWolf

Example 4.07
• Normal stress due to a
centric load
A  c 2   0.25 in 2
 0.1963 in 2
P 160 lb
0  
A 0.1963 in 2
 815 psi

• Equivalent centric load • Normal stress due to

and bending moment bending moment
P  160 lb I  14 c 4  14  0.254
M  Pd  160 lb  0.65in   3.068  103 in 4
 104 lb  in Mc 104 lb  in 0.25 in 
m  
I .068  103 in 4
 8475 psi

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Example 4.07

• Maximum tensile and compressive • Neutral axis location

stresses P My0
t  0 m 0 
A I
 815  8475  t  9260 psi
P I 3.068  103 in 4
y0   815 psi 
c  0  m AM 105 lb  in
 815  8475  c  7660 psi
y0  0.0240 in

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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
MPa in compression. Determine the largest
force P which can be applied to the link.

SOLUTION:
• Determine an equivalent centric load and
bending moment.

• Superpose the stress due to a centric

load and the stress due to bending.

• Evaluate the critical loads for the allowable

From Sample Problem 2.4, tensile and compressive stresses.
A  3 103 m 2
• The largest allowable load is the smallest
Y  0.038 m of the two critical loads.
I  868 109 m 4

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Sample Problem 4.8

• Determine an equivalent centric and bending loads.
d  0.038  0.010  0.028 m
P  centric load
M  Pd  0.028 P  bending moment

• Superpose stresses due to centric and bending loads

P McA P (0.028 P)(0.022)
A     3
  377 P
A I 3 10 868 109
P Mc P (0.028 P)(0.038)
B    B     1559 P
A I 3 103 868 109
• Evaluate critical loads for allowable stresses.
 A  377 P  30 MPa P  79.58kN
 B  1559 P   120 MPa P  76.97 kN

• The largest allowable load P  76.9 kN

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MECHANICS OF MATERIALS Beer • Johnston • DeWolf

Unsymmetric Bending
• Analysis of pure bending has been limited
to members subjected to bending couples
acting in a plane of symmetry.

• Members remain symmetric and bend in

the plane of symmetry.

• The neutral axis of the cross section

coincides with the axis of the couple

• Will now consider situations in which the

bending couples do not act in a plane of
symmetry.

• Cannot assume that the member will bend

in the plane of the couples.

• In general, the neutral axis of the section will

not coincide with the axis of the couple.
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Unsymmetric Bending
• 0  Fx    x dA      m dA
y
 c 
or 0   y dA

neutral axis passes through centroid

 y 
Wish to determine the conditions under • M  M z    y   m dA
 c 
which the neutral axis of a cross section σ I
of arbitrary shape coincides with the or M  m I  I z  moment of inertia
c
axis of the couple as shown. defines stress distribution

• The resultant force and moment

• 0  M y   z x dA   z   m dA
from the distribution of y
elementary forces in the section  c 
must satisfy or 0   yz dA  I yz  product of inertia

Fx  0  M y M z  M  applied couple couple vector must be directed along

a principal centroidal axis

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MECHANICS OF MATERIALS Beer • Johnston • DeWolf

Unsymmetric Bending
Superposition is applied to determine stresses in
the most general case of unsymmetric bending.
• Resolve the couple vector into components along
the principle centroidal axes.
M z  M cos  M y  M sin 

• Superpose the component stress distributions

Mz y Myz
x   
Iz Iy
• Along the neutral axis,
x  0  
Mz y M yz
 
 M cos   y   M sin   z
Iz Iy Iz Iy
y Iz
tan    tan 
z Iy

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Example 4.08
SOLUTION:
• Resolve the couple vector into
components along the principle
centroidal axes and calculate the
corresponding maximum stresses.
M z  M cos  M y  M sin 

• Combine the stresses from the

component stress distributions.
M y M z
x   z  y
A 1600 lb-in couple is applied to a Iz Iy
rectangular wooden beam in a plane
• Determine the angle of the neutral
forming an angle of 30 deg. with the
axis.
vertical. Determine (a) the maximum y Iz
stress in the beam, (b) the angle that the tan    tan 
z Iy
neutral axis forms with the horizontal
plane.
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MECHANICS OF MATERIALS Beer • Johnston • DeWolf

Example 4.08
• Resolve the couple vector into components and calculate
the corresponding maximum stresses.
M z  1600 lb  in  cos 30  1386 lb  in
M y  1600 lb  in  sin 30  800 lb  in
I z  121 1.5in  3.5in   5.359in 4
3

I y  121  3.5in 1.5in   0.9844in 4

3

The largest tensile stress due to M z occurs along AB

M z y 1386 lb  in 1.75in 
1    452.6 psi
Iz 5.359in 4
The largest tensile stress due to M y occurs along AD

2 
Myz

 800 lb  in  0.75in   609.5 psi
Iy 0.9844in 4

• The largest tensile stress due to the combined loading

occurs at A.
 max   1   2  452.6  609.5  max  1062 psi

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Example 4.08

• Determine the angle of the neutral axis.

I 5.359 in 4
tan   z tan   tan 30
Iy 0.9844 in 4
 3.143

  72.4o

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MECHANICS OF MATERIALS Beer • Johnston • DeWolf

General Case of Eccentric Axial Loading

• Consider a straight member subject to equal
and opposite eccentric forces.

• The eccentric force is equivalent to the system

of a centric force and two couples.
P  centric force
M y  Pa M z  Pb

• By the principle of superposition, the

combined stress distribution is
P M z y M yz
x   
A Iz Iy

• If the neutral axis lies on the section, it may

be found from
Mz My P
y z
Iz Iy A

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Combined Stresses: Axial and Torsional Loads

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MECHANICS OF MATERIALS Beer • Johnston • DeWolf

Combined Stresses: Axial and Torsional Loads

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Example 4.09

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Example 4.09
SOLUTION

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Combined Stresses: Bending Moment&Torsional Load

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MECHANICS OF MATERIALS Beer • Johnston • DeWolf

Combined Stresses: Bending Moment&Torsional Load

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Example 4.10

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Example 4.10
SOLUTION

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Example 4.10
SOLUTION

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