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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf • Mazurek
Chapter 4
Pure Bending
Dr. Mohammed Eldosoky
Associate prof.
FAHAD BIN SULTAN UNIVERSITY
College of Engineering
Department of Mechanical Engineering
FAHAD BIN SULTAN UNIVERSITY
College of Engineering
Department of Mechanical Engineering
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Pure Bending
Pure Bending:
Prismatic members
subjected to equal
and opposite
couples acting in
the same
longitudinal plane
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Other Loading Types
• Eccentric Loading: Axial loading which
does not pass through section centroid
produces internal forces equivalent to an
axial force and a couple
• Transverse Loading: Concentrated or
distributed transverse load produces
internal forces equivalent to a shear
force and a couple
• Principle of Superposition: The normal
stress due to pure bending may be
combined with the normal stress due to
axial loading and shear stress due to
shear loading to find the complete state
of stress.
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Symmetric Member in Pure Bending
• Internal forces in any cross section are equivalent
to a couple. The moment of the couple is the
section bending moment.
• From statics, a couple M consists of two equal
and opposite forces.
• The sum of the components of the forces in any
direction is zero.
• The moment is the same about any axis
perpendicular to the plane of the couple and
zero about any axis contained in the plane.
• These requirements may be applied to the sums
of the components and moments of the statically
indeterminate elementary internal forces.
Fx    x dA  0
M y   z x dA  0
M z    y x dA  M
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Beer • Johnston • DeWolf • Mazurek
Bending Deformations
Beam with a plane of symmetry in pure
bending:
• member remains symmetric
• bends uniformly to form a circular arc
• cross-sectional plane passes through arc center
and remains planar
• length of top decreases and length of bottom
increases
• a neutral surface must exist that is parallel to the
upper and lower surfaces and for which the length
does not change
• stresses and strains are negative (compressive)
above the neutral plane and positive (tension)
below it
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Strain Due to Bending
Consider a beam segment of length L.
After deformation, the length of the neutral
surface remains L. At other sections,
L     y 
  L  L     y      y
x 
m 

L
c


y

or

ρ
y

(strain varies linearly)
c
m
y
c
 x   m
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Stress Due to Bending
• For a linearly elastic material,
y
c
 x  E x   E m
y
   m (stress varies linearly)
c
• For static equilibrium,
y
Fx  0    x dA     m dA
c

0   m  y dA
c
First moment with respect to neutral
plane is zero. Therefore, the neutral
surface must pass through the
section centroid.
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• For static equilibrium,
 y

M    y x dA    y    m  dA
 c


 I
M  m  y 2 dA  m
c
c
Mc M
m 

I
S
y
Substituting  x    m
c
My
x  
I
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MECHANICS OF MATERIALS
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Beam Section Properties
• The maximum normal stress due to bending,
Mc M

I
S
I  section moment of inertia
m 
S
I
 section modulus
c
A beam section with a larger section modulus
will have a lower maximum stress
• Consider a rectangular beam cross section,
3
1
I 12 bh
S 
 16 bh3  16 Ah
c
h2
Between two beams with the same cross
sectional area, the beam with the greater depth
will be more effective in resisting bending.
• Structural steel beams are designed to have a
large section modulus.
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Properties of American Standard Shapes
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Deformations in a Transverse Cross Section
• Deformation due to bending moment M is
quantified by the curvature of the neutral surface


1 Mc
 m  m 

c
Ec Ec I
M

EI
1
• Although cross sectional planes remain planar
when subjected to bending moments, in-plane
deformations are nonzero,
 y   x 
y

 z   x 
y

• Expansion above the neutral surface and
contraction below it cause an in-plane curvature,
1 
  anticlasti c curvature
 
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Sample Problem 4.2
SOLUTION:
Based on the cross section geometry,
calculate the location of the section
centroid and moment of inertia.
Y 
 yA
A

I x   I  A d 2

Apply the elastic flexural formula to
find the maximum tensile and
compressive stresses.
m 
A cast-iron machine part is acted upon
by a 3 kN-m couple. Knowing E = 165
GPa and neglecting the effects of
fillets, determine (a) the maximum
tensile and compressive stresses, (b)
the radius of curvature.
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Mc
I
Calculate the curvature
1


M
EI
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf • Mazurek
Sample Problem 4.2
SOLUTION:
Based on the cross section geometry, calculate
the location of the section centroid and
moment of inertia.
Area, mm 2
y , mm
yA, mm 3
1 20  90  1800
50
90  103
2 40  30  1200
20
24  103
3
 A  3000
 yA  114  10
3
 yA 114 10
Y 

 38 mm
3000
A



1 bh3  A d 2
I x   I  A d 2   12

 

1 90  203  1800  122  1 30  403  1200  182
 12
12
I  868  103 mm 4  868  10-9 m 4
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MECHANICS OF MATERIALS
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Sample Problem 4.2
Apply the elastic flexural formula to find the
maximum tensile and compressive stresses.
Mc
I
M c A 3 kN  m  0.022 m
A 

I
868  109 m 4
M cB
3 kN  m  0.038 m
B  

I
868  109 m 4
m 
 A  76.0 MPa
 B  131.3 MPa
Calculate the curvature
1



M
EI
3 kN  m
165 GPa 868 10-9 m 4 
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1
 20.95  103 m -1

  47.7 m
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf • Mazurek
Bending of Members Made of Several Materials
Consider a composite beam formed from
two materials with E1 and E2.
Normal strain varies linearly.
x  
y

Piecewise linear normal stress variation.
For upper part
 1  E1 x  
E1 y

For lower part
 2  E2 x  
E2 y

Neutral axis does not pass through
section centroid of composite section.
Elemental forces on the section are
Ey
E y
dF1   1dA   1 dA dF2   2 dA   2 dA

x  
My
I
1   x

Define a transformed section such that
 2  n x
dF2  
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nE1  y dA   E1 y n dA


E
n 2
E1
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf • Mazurek
Example 4.03
SOLUTION:
Transform the bar to an equivalent cross
section made entirely of brass
Evaluate the cross sectional properties of
the transformed section
Calculate the maximum stress in the
transformed section. This is the correct
maximum stress for the brass pieces of
the bar.
Bar is made from bonded pieces of
steel (Es = 200 GPa) and brass (Eb
= 100 GPa). Determine the
maximum stress in the steel and
brass when a moment of 4.5 KNm
is applied.
Determine the maximum stress in the steel
portion of the bar by multiplying the
maximum stress for the transformed
section by the ratio of the moduli of
elasticity.
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Example 4.03
SOLUTION:
Transform the bar to an equivalent cross section
made entirely of brass.
n
Es 200GPa

 2.0
Eb 100GPa
bT  10 mm  2 18 mm  10 mm  56 mm
Evaluate the transformed cross sectional properties
I  121 bT h3  121 56 mm 75 mm 
3
 1.96875 106 mm 4
Calculate the maximum stresses
m 
Mc 4500 Nm 0.0375 m 

 85.7 MPa
-6
4
I
1.9687510 m
 b max   m
 s max  n m  2  85.7 MPa
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 b max  85.7 MPa
 s max  171.4 MPa
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf • Mazurek
Stress Concentrations
Stress concentrations may occur:
in the vicinity of points where the
loads are applied
m  K
Mc
I
in the vicinity of abrupt changes in
cross section
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Eccentric Axial Loading
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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

Eccentric loading
FP
M  Pd
P My

A I
Validity requires stresses below proportional
limit, deformations have negligible effect on
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
superposed stress distribution.
An open-link chain is obtained by
bending low-carbon steel rods into the
shape shown. For 700 N load, determine Find the neutral axis by determining
the location where the normal stress
(a) maximum tensile and compressive
is zero.
stresses, (b) distance between section
centroid and neutral axis
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Example 4.07
Normal stress due to a
centric load
A  c 2   6 mm 
2
 113.1 mm 2
0 
P
700 N

A 113.1106 m 2
 6.2 MPa
Equivalent centric load and
bending moment
P  700 N
M  Pd  700 N 0.016 m 
 11.2 Nm
Normal stress due to
bending moment
I  14 c 4  14  6 mm 
4
 1017.9 mm 4
m 
Mc 11.2 Nm 0.006 m 

I
1017.9 10-12 m 4
 66 MPa
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Example 4.07
Maximum tensile and compressive
stresses
t  0 m
 6.2  66
P My 0

A
I
P I
1017 .9  10 12 m 4
6
y0 
 6.2  10 Pa
AM
11.2 Nm
0
 t  72 .2 MPa
c  0  m
 6.2  66
Neutral axis location
 c  59 .8 MPa


y0  0.56 mm
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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 equivalent centric load and
bending moment.
Superpose the stress due to a centric load
and the stress due to bending.
From Sample Problem 4.2,
A  3 103 m 2
Y  0.038 m
I  868 109 m 4
Evaluate the critical loads for the allowable
tensile and compressive stresses.
The largest allowable load is the smallest of
the two critical loads.
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Sample Problem 4.8
Determine 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
0.028 P 0.022  377 P
P Mc A
P



A
I
3 103
868 109
0.028 P 0.022  1559 P
P Mc
P
B    A  

A
I
3 103
868 109
A  
Evaluate critical loads for allowable stresses.
 A  377 P  30 MPa
P  79.6 kN
 B  1559 P  120 MPa P  77.0 kN
The largest allowable load
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P  77.0 kN
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