Transcript Pure Bending - Adem KARACA Kişisel Web Sayfası

Third Edition
CHAPTER
4
MECHANICS OF
MATERIALS
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf
Lecture Notes:
J. Walt Oler
Texas Tech University
Pure Bending
Basit Eğilme
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
Third
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
© 2002 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
Photo 4.3 Wide-flange steel beams form the frame of many buildings.
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4-3
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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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4-4
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MECHANICS
OF MATERIALS
Other Loading Types
Beer • Johnston • DeWolf
Photo 4.2 Clamp used to glue lumber pieces together.
• 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
Other Loading Types
• 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
Symmetric Member
in Pure Bending
Beer • Johnston • DeWolf
• 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.
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Edition
MECHANICS
OF MATERIALS
Symmetric Member
in Pure Bending
Beer • Johnston • DeWolf
• 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.
 F    dA  0
 M   z dA  0
 M    y dA  M
x
y
z
x
x
x
z
Mz  M
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MECHANICS
OF MATERIALS
Bending Deformations
Beer • Johnston • DeWolf
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
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MECHANICS
OF MATERIALS
Bending Deformations
Beer • Johnston • DeWolf
• 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
Strain Due to Bending
Beer • Johnston • DeWolf
Consider a beam segment of length L. After
deformation, the length of the neutral surface
remains L. At other sections,
DE  L  
JK  L    y 
  L  L    y       y
Strain varies linearly
x 
m 
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.

L
c


 y


or ρ 
y

c
m


y





m

x
c



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MECHANICS
OF MATERIALS
Stress Due to Bending
• For a linearly elastic material,
 x  E x  E
Beer • Johnston • DeWolf
Mz  M
y

 y 
 c 
(stress varieslinearly)
y
c
 x  E x  E   m     m
• For static equilibrium,
y
F


dA


 x  x
 c  m dA  0

m
c
 y dA  0
First moment with respect to neutral
plane is zero. Therefore, the neutral
surface must pass through the section
centroid.
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
• For static equilibrium,
M
z
0 
 y

M    y x dA    y   m  dA
 c


 I
M  m  y 2 dA  m
c
c
Mc M
m 

I
S
y
Subst it ut ing  x    m
c
My
x  
where I  I z
I
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MECHANICS
Bending RadiusOF
DueMATERIALS
to Bending
• For a linearly elastic material,
 x  E x  E
x  
E
1

y



My
I

M
EI
EI
M
EI :
Beer • Johnston • DeWolf
Mz  M
y

where I  I z
My
I
Curvature
Bending Radius
Bending Rigidity
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MECHANICS
OF MATERIALS
Beam Section Properties
Beer • Johnston • DeWolf
• 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,
I
S 
c
1
12
bh3
1
2h

1
1
S  bh3  Ah
6
6
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Edition
MECHANICS
OF MATERIALS
Beam Section Properties
Beer • Johnston • DeWolf
• 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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Edition
MECHANICS
OF MATERIALS
Properties of American
Standard Shapes
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Beer • Johnston • DeWolf
4 - 16
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MECHANICS
MATERIALS
Deformations in OF
a Transverse
Cross Section
Beer • Johnston • DeWolf
• Deformation due to bending moment M is
quantified by the curvature of the neutral surface
1
m
m
1 Mc



 c Ec Ec I

1
M

 EI
• 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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Edition
MECHANICS
Sample ProblemOF
4.2MATERIALS
Beer • Johnston • DeWolf
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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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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.
z   
Mx y
My

I x
I
• Calculate the curvature
1


M
EI
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MECHANICS
Sample ProblemOF
4.2MATERIALS
Beer • Johnston • DeWolf
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, mm 3
90 103
24 103
3
 yA  114 10
3
 yA 114 10
Y 

 38 mm
3000
A


I x   I  A d y2  


1
12

1
12
bh3  A d y2
 
90 203  1800122 
1
12

30 403  1200182
I  I x  868103 mm4  86810-9 m 4
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4 - 20
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MECHANICS
Sample ProblemOF
4.2MATERIALS
Beer • Johnston • DeWolf
• Apply the elastic flexural formula to find the
maximum tensile and compressive stresses.
M  M x  3 kNm  3106 Nm m
My
I
M cA
 3106 Nm m  22 m m
 A  76 .0 MPa
A  

3
4
I
86810 m m
M cB
 3106 Nm m   38 m m   131 .3 MPa
B  

B
I
868103 m m4
 




• Calculate the curvature
M
 3 106 N  mm


 EI 165103 MPa 868103 mm4
1



1
 20.95106 m m-1    47740m m  47.74 m
ρ
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
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MECHANICS OF MATERIALS
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Beer • Johnston • DeWolf
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MECHANICS OF MATERIALS
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Beer • Johnston • DeWolf
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MECHANICS OF MATERIALS
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
Beer • Johnston • DeWolf
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
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MECHANICS OF MATERIALS
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
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MECHANICS OF MATERIALS
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Beer • Johnston • DeWolf
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MECHANICS
OF MATERIALS
Example: Yükleme durumu
ve kesiti görülen kiriş için “n”;
Beer • Johnston • DeWolf
a) Mesnet tepkilerini bulunuz.
b) KKD ve EMD diyagramını çiziniz.
c) Kesitin ağırlık merkezinin yerini bulunuz.
d) Ağırlık merkezinden geçen yatay eksene göre kesitin atalet momentini bulunuz.
e) Kesitin en üst ve altındaki gerilmeleri hesaplayınız.
f) σak =150 MPa olduğuna göre EK=3 alarak kirişin emniyetli olup olmadığını
irdeleyiniz.
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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Beer • Johnston • DeWolf
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MECHANICS OF MATERIALS
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
Beer • Johnston • DeWolf
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MECHANICS
OF MATERIALS
Bending of Members
Made of Several Materials
Beer • Johnston • DeWolf
• 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
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
dF2  
nE1  y dA   E1 y n dA


E
n 2
E1
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MECHANICS
Example 4.03 OF MATERIALS
Beer • Johnston • DeWolf
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.
© 2002 The McGraw-Hill Companies, Inc. All rights reserved.
Third
Edition
MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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.
• 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
Example 4.03 OF MATERIALS
Beer • Johnston • DeWolf
• Transform the bar to an equivalent cross
section made entirely of brass.
Es 29 10 6 psi
n

 1.933
6
Eb 15 10 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
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MECHANICS
Example 4.03 OF MATERIALS
Beer • Johnston • DeWolf
• 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
 b max
 s max
 11.85 ksi
 22.9 ksi
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MECHANICS
OF MATERIALS
Reinforced Concrete
Beams
Beer • Johnston • DeWolf
• 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
My
x  
I
c   x
 s  n x
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MECHANICS
Sample ProblemOF
4.4MATERIALS
Beer • Johnston • DeWolf
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
SOLUTION:
• Transform to a section made entirely of concrete.
• Evaluate geometric properties of transformed section.
• Calculate the maximum stresses in the concrete and steel.
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
• Transform to a section made entirely of concrete.
Es 29 10 6 psi
n

 8.06
6
Ec 3.6 10 psi
 
2


5
nAs  8.06  2 4 8 in
 4.95 in 2


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MECHANICS
OF MATERIALS
Solution
Beer • Johnston • DeWolf
• 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
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
• Calculate the maximum stresses.
Mc1 40 kip  in 1.45in

  c  1.306 ksi
4
I
44.4in
Mc
40 kip  in  2.55in
 s  n 2  8.06
  s  18 .52 ksi
4
I
44.4in
c 
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MECHANICS OF MATERIALS
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Beer • Johnston • DeWolf
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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MECHANICS
OF MATERIALS
Eccentric Axial Loading
in a Plane of Symmetry
Beer • Johnston • DeWolf
• 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
Example 4.07 OF MATERIALS
Beer • Johnston • DeWolf
An open-link chain is obtained by bending
low-carbon steel rods into the shape shown.
For 160 lb load, determine (a) maximum
tensile and compressive stresses, (b) distance
between section centroid and neutral axis
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf
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.
• Find the neutral axis by determining the location where the
normal stress is zero.
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MECHANICS
Example 4.07 OF MATERIALS
Beer • Johnston • DeWolf
• 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 0.6 in 
 104 lb  in
• Normal stress due to
bending moment
I  14 c 4  14  0.25 4
 3.068  10 3 in 4
Mc 104 lb  in 0.25 in 
m 

I
.068  10 3 in 4
 8475 psi
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MECHANICS
Example 4.07 OF MATERIALS
• Maximum tensile and compressive
stresses
t  0 m
 815  8475
c  0  m
 815  8475
 t  9260 psi
 c  7660 psi
Beer • Johnston • DeWolf
• 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
Sample ProblemOF
4.8MATERIALS
Beer • Johnston • DeWolf
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.
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MECHANICS OF MATERIALS
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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 tensile and
compressive stresses.
• The largest allowable load is the smallest of the two critical
loads.
From Sample Problem 2.4,
A  3 10 3 m 2
Y  0.038 m
I  868 10 9 m 4
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MECHANICS
Sample ProblemOF
4.8MATERIALS
Beer • Johnston • DeWolf
• Determine an equivalent centric and bending loads.
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MECHANICS
OF MATERIALS
Çözüm
Beer • Johnston • DeWolf
• 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
0.028 P 0.022   377 P
P McA
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  79.6 kN
• The largest allowable load
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P  77.0 kN
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MECHANICS OF MATERIALS
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MECHANICS OF MATERIALS
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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:



I 2 2 2
3
3 2
 bc  3 50  10 m 60  10 m
c 3
 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 10 6 Pa
200 109 Pa
 1.2 10 3
Y 
yY

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 10 6 m3
 306 .7 MPa  2 Y
 
m
• M=0
At the edge of the elastic core,
x 
x
E

 35 .5 10 6 Pa
200 10 9 Pa
 177 .5 10  6

yY
x

40 10 3 m
177 .5 10  6
  225 m
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MECHANICS OF MATERIALS
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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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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
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 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
or
y
 c

0   yz dA  I yz  product of inertia
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
x  
Mzy M yy

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
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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.
x  
Mzy M yy

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.
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MECHANICS OF MATERIALS
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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
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 max  1062 psi
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MECHANICS OF MATERIALS
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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
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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 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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