Fourth Edition
CHAPTER
4
MECHANICS OF
MATERIALS
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf
Pure Bending
Lecture Notes:
J. Walt Oler
Texas Tech University
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Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Pure Bending
Pure Bending
Other Loading Types
Symmetric Member in Pure Bending
Bending Deformations
Strain Due to Bending
Beam Section Properties
Properties of American Standard Shapes
Deformations in a Transverse Cross Section
Sample Problem 4.2
Bending of Members Made of Several
Materials
Example 4.03
Reinforced Concrete Beams
Sample Problem 4.4
Stress Concentrations
Plastic Deformations
Members Made of an Elastoplastic Material
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
Example 4.03
Reinforced Concrete Beams
Sample Problem 4.4
Stress Concentrations
Plastic Deformations
Members Made of an Elastoplastic Material
Plastic Deformations of Members With a Single
Plane of S...
Residual Stresses
Example 4.05, 4.06
Eccentric Axial Loading in a Plane of Symmetry
Example 4.07
Sample Problem 4.8
Unsymmetric Bending
Example 4.08
General Case of Eccentric Axial Loading
4-2
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Pure Bending
Pure Bending: Prismatic members
subjected to equal and opposite couples
acting in the same longitudinal plane
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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

y
y
x = = −
=−
(strain varies linearly)
L


c
c
m =
or ρ =

m
y
c
 x = − m
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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
Substituti ng  x = −  m
c
My
x = −
I
4-8
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Beam Section Properties
• The maximum normal stress due to bending,
Mc M
=
I
S
I = section moment of inertia
I
S = = section modulus
c
m =
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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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Properties of American Standard Shapes
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Fourth
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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 
= = anticlastic curvature
 
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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
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
1 20  90 = 1800
2 40  30 = 1200
 A = 3000
y , mm
50
20
yA, mm3
90 103
24 103
3
 yA = 114 10
3
 yA 114 10
Y =
=
= 38 mm
3000
A
(
) (121 bh3 + A d 2 )
1 90  203 + 1800  122 ) + ( 1 30  403 + 1200  182 )
= (12
12
I x =  I + A d 2 = 
I = 868  103 mm 4 = 868  10-9 m 4
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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 = +76.0 MPa
A =
=
−
9
4
I
868  10 m
M cB
3 kN  m  0.038 m  = −131.3 MPa
B
B = −
=−
−
9
4
I
868  10 m
m =
• 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
4 - 14
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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.
1 = E1 x = −
E1 y

 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 =  2dA = − 2 dA

x = −
My
I
1 =  x

• Define a transformed section such that
 2 = n x
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dF2 = −
(nE1 ) y dA = − E1 y (n dA)


E
n= 2
E1
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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 = 29x106 psi) and brass
(Eb = 15x106 psi). Determine the
maximum stress in the steel and
brass when a moment of 40 kip*in
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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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 4.03
SOLUTION:
• Transform the bar to an equivalent cross section
made entirely of brass.
Es 29 106 psi
n=
=
= 1.933
Eb 15 106 psi
bT = 0.4 in + 1.933  0.75 in + 0.4 in = 2.25 in
• Evaluate the transformed cross sectional properties
1 b h3 = 1 (2.25 in.)(3 in.)3
I = 12
T
12
= 5.063 in.4
• Calculate the maximum stresses
m =
Mc (40 kip  in.)(1.5 in.)
=
= 11.85 ksi
4
I
5.063 in.
( b )max =  m
( s )max = n m = 1.933 11.85 ksi
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( b )max = 11.85 ksi
( s )max = 22.9 ksi
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Reinforced Concrete Beams
• Concrete beams subjected to bending moments are
reinforced by steel rods.
• The steel rods carry the entire tensile load below
the neutral surface. The upper part of the
concrete beam carries the compressive load.
• In the transformed section, the cross sectional area
of the steel, As, is replaced by the equivalent area
nAs where n = Es/Ec.
• To determine the location of the neutral axis,
(bx ) x − n As (d − x ) = 0
2
1 b x2
2
+ n As x − n As d = 0
• The normal stress in the concrete and steel
x = −
My
I
c =  x
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 s = n x
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 4.4
SOLUTION:
• Transform to a section made entirely
of concrete.
• Evaluate geometric properties of
transformed section.
• Calculate the maximum stresses
in the concrete and steel.
A concrete floor slab is reinforced with
5/8-in-diameter steel rods. The modulus
of elasticity is 29x106psi for steel and
3.6x106psi for concrete. With an applied
bending moment of 40 kip*in for 1-ft
width of the slab, determine the maximum
stress in the concrete and steel.
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Sample Problem 4.4
SOLUTION:
• Transform to a section made entirely of concrete.
Es 29 106 psi
n=
=
= 8.06
Ec 3.6 106 psi
( )
2
nAs = 8.06  24 85 in  = 4.95 in 2


• Evaluate the geometric properties of the
transformed section.
 x
12 x  − 4.95(4 − x ) = 0
 2
(
x = 1.450 in
)
I = 13 (12 in )(1.45 in )3 + 4.95 in 2 (2.55 in )2 = 44.4 in 4
• Calculate the maximum stresses.
Mc1 40 kip  in 1.45 in
=
I
44.4 in 4
Mc
40 kip  in  2.55 in
 s = n 2 = 8.06
I
44.4 in 4
c =
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
 c = 1.306 ksi
 s = 18.52 ksi
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Plastic Deformations
• For any member subjected to pure bending
y
c
 x = − m
strain varies linearly across the section
• If the member is made of a linearly elastic material,
the neutral axis passes through the section centroid
and
x = −
My
I
• For a material with a nonlinear stress-strain curve,
the neutral axis location is found by satisfying
Fx =   x dA = 0
M =  − y x dA
• For a member with vertical and horizontal planes of
symmetry and a material with the same tensile and
compressive stress-strain relationship, the neutral
axis is located at the section centroid and the stressstrain relationship may be used to map the strain
distribution from the stress distribution.
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Plastic Deformations
• When the maximum stress is equal to the ultimate
strength of the material, failure occurs and the
corresponding moment MU is referred to as the
ultimate bending moment.
• The modulus of rupture in bending, RB, is found
from an experimentally determined value of MU
and a fictitious linear stress distribution.
RB =
MU c
I
• RB may be used to determine MU of any
member made of the same material and with the
same cross sectional shape but different
dimensions.
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Members Made of an Elastoplastic Material
• Rectangular beam made of an elastoplastic material
Mc
I
 x  Y
m =
 m = Y
I
M Y =  Y = maximum elastic moment
c
• If the moment is increased beyond the maximum
elastic moment, plastic zones develop around an
elastic core.
M=
2

3 M 1 − 1 yY
2 Y
3 2



c 
yY = elastic core half - thickness
• In the limit as the moment is increased further, the
elastic core thickness goes to zero, corresponding to a
fully plastic deformation.
M p = 32 M Y = plastic moment
Mp
k=
= shape factor (depends only on cross section shape)
MY
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Plastic Deformations of Members With a
Single Plane of Symmetry
• Fully plastic deformation of a beam with only a
vertical plane of symmetry.
• The neutral axis cannot be assumed to pass
through the section centroid.
• Resultants R1 and R2 of the elementary
compressive and tensile forces form a couple.
R1 = R2
A1 Y = A2 Y
The neutral axis divides the section into equal
areas.
• The plastic moment for the member,
(
)
M p = 12 A Y d
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Residual Stresses
• Plastic zones develop in a member made of an
elastoplastic material if the bending moment is
large enough.
• Since the linear relation between normal stress
and strain applies at all points during the
unloading phase, it may be handled by assuming
the member to be fully elastic.
• Residual stresses are obtained by applying the
principle of superposition to combine the stresses
due to loading with a moment M (elastoplastic
deformation) and unloading with a moment -M
(elastic deformation).
• The final value of stress at a point will not, in
general, be zero.
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 4.05, 4.06
A member of uniform rectangular cross section is
subjected to a bending moment M = 36.8 kN-m.
The member is made of an elastoplastic material
with a yield strength of 240 MPa and a modulus
of elasticity of 200 GPa.
Determine (a) the thickness of the elastic core, (b)
the radius of curvature of the neutral surface.
After the loading has been reduced back to zero,
determine (c) the distribution of residual stresses,
(d) radius of curvature.
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 4.05, 4.06
• Thickness of elastic core:
M =
2

3 M 1 − 1 yY
2 Y
3 2


c 

36.8 kN  m =
2

3 (28.8 kN  m )1 − 1 yY
2
 3 2
yY
yY
=
= 0.666
c
60 mm



c 
2 yY = 80 mm
• Radius of curvature:
• Maximum elastic moment:
(
)(
)
2
I 2 2 2
= 3 bc = 3 50  10−3 m 60  10−3 m
c
= 120  10− 6 m3
I
M Y =  Y = 120  10− 6 m3 (240 MPa )
c
= 28.8 kN  m
(
)
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Y =
Y
E
=
240 106 Pa
200 109 Pa
= 1.2 10−3
y
Y = Y

=
yY
Y
=
40 10−3 m
1.2 10−3
 = 33.3 m
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MECHANICS OF MATERIALS
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Example 4.05, 4.06
• M = 36.8 kN-m
yY = 40 mm
 Y = 240 MPa
• M = -36.8 kN-m
Mc 36.8 kN  m
=
I
120 106 m3
= 306.7 MPa  2 Y
 =
m
• M=0
At the edge of the elastic core,
x =
x
E
=
− 35.5 106 Pa
200 109 Pa
= −177.5 10− 6
 =−
yY
x
=
40 10−3 m
177.5 10− 6
 = 225 m
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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
=
• 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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MECHANICS OF MATERIALS
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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 160 lb 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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MECHANICS OF MATERIALS
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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
and bending moment
P = 160 lb
M = Pd = (160 lb )(0.65 in )
= 104 lb  in
• Normal stress due to
bending moment
I = 14 c 4 = 14  (0.25)4
m
= 3.068  10−3 in 4
Mc (104 lb  in )(0.25 in )
=
=
I
3.068  10−3 in 4
= 8475 psi
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 4.07
• Maximum tensile and compressive
stresses
t = 0 +m
= 815 + 8475
c = 0 − m
= 815 − 8475
 t = 9260 psi
 c = −7660 psi
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• Neutral axis location
0=
P My0
−
A
I
P I
3.068 10−3 in 4
y0 =
= (815 psi )
AM
105 lb  in
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 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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MECHANICS OF MATERIALS
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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
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
P = 77.0 kN
4 - 35
Fourth
Edition
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.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
4 - 36
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Unsymmetric Bending
• 0 = Fx =   x dA =   −  m dA
y
 c

or 0 =  y dA
neutral axis passes through centroid
Wish to determine the conditions under
which the neutral axis of a cross section
of arbitrary shape coincides with the
axis of the couple as shown.
• The resultant force and moment
from the distribution of
elementary forces in the section
must satisfy
Fx = 0 = M y M z = M = applied couple
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
 y

M
=
M
=
−
y
  −  m dA
•
z
 c

σ I
or M = m
I = I z = moment of inertia
c
defines stress distribution
• 0 = M y =  z x dA =  z −  m dA
y
 c

or 0 =  yz dA = I yz = product of inertia
couple vector must be directed along
a principal centroidal axis
4 - 37
Fourth
Edition
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
x = −
Mzy Myy
+
Iz
Iy
• Along the neutral axis,
x = 0 = −
tan  =
(M cos ) y + (M sin  ) y
Mzy Myy
+
=−
Iz
Iy
Iz
Iy
y Iz
= tan 
z Iy
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
4 - 38
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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.
Mz y M yz
x =
+
Iz
Iy
A 1600 lb-in couple is applied to a
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
tan

=
=
tan 
stress in the beam, (b) the angle that the
z Iy
neutral axis forms with the horizontal
plane.
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
4 - 39
Fourth
Edition
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
1 (1.5 in )(3.5 in )3 = 5.359 in 4
I z = 12
1 (3.5 in )(1.5 in )3 = 0.9844 in 4
I y = 12
The largest tensile stress due to M z occurs along AB
1 =
M z y (1386 lb  in )(1.75 in )
= 452.6 psi
=
4
Iz
5.359 in
The largest tensile stress due to M z occurs along AD
2 =
M yz
Iy
=
(800 lb  in )(0.75 in ) = 609.5 psi
0.9844 in 4
• The largest tensile stress due to the combined loading
occurs at A.
 max = 1 +  2 = 452.6 + 609.5
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
 max = 1062 psi
4 - 40
Fourth
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
Example 4.08
• Determine the angle of the neutral axis.
Iz
5.359 in 4
tan  = tan  =
tan 30
4
Iy
0.9844 in
= 3.143
 = 72.4o
© 2006 The McGraw-Hill Companies, Inc. All rights reserved.
4 - 41
Fourth
Edition
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 Mz y M yz
x = −
+
A
Iz
Iy
• If the neutral axis lies on the section, it may
be found from
My
Mz
P
y−
z=
Iz
Iy
A
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4 - 42