Transcript Copyright © 2015 McGraw-Hill Education. Permission

Seventh Edition
CHAPTER
2
MECHANICS OF
MATERIALS
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf
David F. Mazurek
Stress and Strain
– Axial Loading
Lecture Notes:
Brock E. Barry
U.S. Military Academy
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Contents
Stress & Strain: Axial Loading
Normal Strain
Stress-Strain Test
Stress-Strain Diagram: Ductile Materials
Stress-Strain Diagram: Brittle Materials
Hooke’s Law: Modulus of Elasticity
Elastic vs. Plastic Behavior
Fatigue
Deformations Under Axial Loading
Concept Application 2.1
Sample Problem 2.1
Static Indeterminate Problems
Concept Application 2.4
Problems Involving Temperature Change
Poisson’s Ratio
Multiaxial Loading: Generalized
Hooke’s Law
Dilatation: Bulk Modulus
Shearing Strain
Concept Application 2.10
Relation Among E, n, and G
Composite Materials
Sample Problem 2.5
Saint-Venant’s Principle
Stress Concentration: Hole
Stress Concentration: Fillet
Concept Application 2.12
Elastoplastic Materials
Plastic Deformations
Residual Stresses
Concept Applications 2.14, 2.15, 2.16
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Stress & Strain: Axial Loading
• Suitability of a structure or machine may depend on the deformations in
the structure as well as the stresses induced under loading. Statics
analyses alone are not sufficient.
• Considering structures as deformable allows determination of member
forces and reactions which are statically indeterminate.
• Determination of the stress distribution within a member also requires
consideration of deformations in the member.
• Chapter 2 is concerned with deformation of a structural member under
axial loading. Later chapters will deal with torsional and pure bending
loads.
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Normal Strain
Fig. 2.1 Undeformed
and deformed axially
loaded rod.
Fig. 2.3 Twice the load is
required to obtain the same
deformation  when the
cross-sectional area is
doubled.
Fig. 2.4 The deformation is
doubled when the rod length
is doubled while keeping the
load P and cross-sectional
area A.


P
 stress
A

L
 normal strain
2P P


2A A


L
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P

A
2 


2L L
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Stress-Strain Test
Photo 2.2 Universal test machine used to test
tensile specimens.
Photo 2.3 Elongated tensile test
specimen having load P and deformed
length L > L0.
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Stress-Strain Diagram: Ductile Materials
Fig. 2.6 Stress-strain diagrams of two typical ductile materials.
Photo 2.4 Ductile material tested
specimens: (a) with cross-section
necking, (b) ruptured.
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Stress-Strain Diagram: Brittle Materials
Fig 2.7 Stress-strain diagram for a typical brittle material.
Photo 2.5 Ruptured brittle materials specimen.
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Hooke’s Law: Modulus of Elasticity
• Below the yield stress
  E
E  Youngs Modulus or
Modulus of Elasticity
• Strength is affected by alloying,
heat treating, and manufacturing
process but stiffness (Modulus of
Elasticity) is not.
Fig 2.11 Stress-strain diagrams for iron and
different grades of steel.
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Elastic vs. Plastic Behavior
• If the strain disappears when the
stress is removed, the material is
said to behave elastically.
• The largest stress for which this
occurs is called the elastic limit.
Fig. 2.13 Stress-strain response of ductile
material load beyond yield and unloaded.
• When the strain does not return
to zero after the stress is
removed, plastic deformation
of the material has taken place.
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Fatigue
• Fatigue properties are shown on
-N diagrams.
• A member may fail due to fatigue
at stress levels significantly below
the ultimate strength if subjected
to many loading cycles.
Fig. 2.16 Typical -n curves.
• When the stress is reduced below
the endurance limit, fatigue
failures do not occur for any
number of cycles.
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Deformations Under Axial Loading
• From Hooke’s Law:
  E


E

P
AE
• From the definition of strain:


L
• Equating and solving for the deformation,
PL

AE
Fig. 2.17 Undeformed and
deformed axially-loaded rod.
• With variations in loading, cross-section or
material properties,
PL
  i i
i Ai Ei
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Concept Application 2.1
SOLUTION:
• Divide the rod into components at
the load application points.
E  29 106 psi
Determine the deformation of
the steel rod shown under the
given loads.
• Apply a free-body analysis on each
component to determine the
internal force
• Evaluate the total of the component
deflections.
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Concept Application 2.1
SOLUTION:
• Divide the rod into three
components:
• Apply free-body analysis to each
component to determine internal forces,
P1  60  10 3 lb
P2  15  10 3 lb
P3  30  10 3 lb
• Evaluate total deflection,
Pi Li 1  P1L1 P2 L2 P3 L3 

 


A
E
E
A
A
A
i i i
 1
2
3 
 

 
 

 60 103 12  15 103 12 30 103 16 




6
0
.
9
0
.
9
0
.
3
29 10 

1
L1  L2  12 in.
L3  16 in.
A1  A2  0.9 in 2
A3  0.3 in 2
 75.9 10 3 in.
  75.9 10 3 in.
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Sample Problem 2.1
SOLUTION:
The rigid bar BDE is supported by two
links AB and CD.
• Apply a free-body analysis to the bar
BDE to find the forces exerted by
links AB and DC.
• Evaluate the deformation of links AB
and DC or the displacements of B
and D.
• Work out the geometry to find the
Link AB is made of aluminum (E = 70
deflection at E given the deflections
GPa) and has a cross-sectional area of 500
at B and D.
mm2. Link CD is made of steel (E = 200
GPa) and has a cross-sectional area of (600
mm2).
For the 30-kN force shown, determine the
deflection (a) of B, (b) of D, and (c) of E.
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Sample Problem 2.1
SOLUTION:
Displacement of B:
B 
Free body: Bar BDE
PL
AE

 60 103 N 0.3 m 

500 10-6 m2 70 109 Pa
 514 10  6 m
M  0
B
0  30 kN)(0.6 m   FCD (0.2 m)
 B  0.514 mm 
Displacement of D:
FCD  90 kN FCD  90 kN tension
M  0
D
0  30 kN  0.4 m   FAB  0.2 m
FAB  60 kN FAB  60 kN compressio n
D 
PL
AE

90 10 3 N 0.4 m 

600 10-6 m2 200 109 Pa
 300 10  6 m
 D  0.300 mm 
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Sample Problem 2.1
Displacement of D:
BB BH

DD HD
0.514 mm 200 mm   x

0.300 mm
x
x  73.7 mm
EE  HE

DD HD
E
0.300 mm

400  73.7 mm
73 .7 mm
 E  1.928 mm
 E  1.928 mm 
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Static Indeterminate Problems
• Structures for which internal forces and reactions
cannot be determined from statics alone are said
to be statically indeterminate.
• A structure will be statically indeterminate
whenever it is held by more supports than are
required to maintain its equilibrium.
• Redundant reactions are replaced with
unknown loads which along with the other
loads must produce compatible deformations.
• Deformations due to actual loads and redundant
reactions are determined separately and then
added.
  L R  0
Fig. 2.23
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Concept Application 2.4
Determine the reactions at A and B for the steel
bar and loading shown, assuming a close fit at
both supports before the loads are applied.
SOLUTION:
• Consider the reaction at B as redundant, release
the bar from that support, and solve for the
displacement at B due to the applied loads.
• Solve for the displacement at B due to the
redundant reaction at RB.
• Require that the displacements due to the loads
and due to the redundant reaction be
compatible, i.e., require that their sum be zero.
• Solve for the reaction at RA due to applied loads
and the reaction found at RB.
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Concept Application 2.4
SOLUTION:
• Solve for the displacement at B due to the applied
loads with the redundant constraint released,
P1  0 P2  P3  600 103 N P4  900 103 N
A1  A2  400 10 6 m 2
A3  A4  250 10 6 m 2
L1  L2  L3  L4  0.150 m
Pi Li 1.125 109
L  

E
i Ai Ei
• Solve for the displacement at B due to the redundant
constraint,
P1  P2   RB
A1  400  10  6 m 2
L1  L2  0.300 m

A2  250  10  6 m 2

Pi Li
1.95  10 3 RB
δR  

E
i Ai Ei
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Concept Application 2.4
• Require that the displacements due to the loads and due to
the redundant reaction be compatible,
  L R  0


1.125 109 1.95 103 RB


0
E
E
RB  577 103 N  577 kN
• Find the reaction at A due to the loads and the reaction at B
 Fy  0  R A  300 kN  600 kN  577 kN
R A  323 kN
RA  323 kN
RB  577 kN
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Problems Involving Temperature Change
Fig. 2.26 (partial)
• A temperature change results in a change in length or
thermal strain. There is no stress associated with the
thermal strain unless the elongation is restrained by
the supports.
• Treat the additional support as redundant and apply
the principle of superposition.
 T   T L
P 
  coefficient of thermal expansion
PL
AE
• The thermal deformation and the deformation from
the redundant support must be compatible.
  T   P  0
Fig. 2.27 Superposition method to find force at point B of restrained
rod AB undergoing thermal expansion. (a) Initial rod length; (b)
thermally expanded rod length; (c) force P pushes point B back to
zero deformation.
PL
0
AE
P   AE T 
P
    E T 
A
 T L 
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Poisson’s Ratio
• For a slender bar subjected to axial loading:

x  x  y z  0
E
• The elongation in the x-direction is
accompanied by a contraction in the other
directions. Assuming that the material is
homogeneous and isotropic (no directional
Fig. 2.29 A bar in uniaxial tension and a
dependence),
representative stress element.
 y  z  0
• Poisson’s ratio is defined as
n 

lateral strain

 y  z
axial strain
x
x
Fig. 2.30 Materials undergo transverse
contraction when elongated under axial load.
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Multiaxial Loading: Generalized Hooke’s Law
• For an element subjected to multi-axial loading,
the normal strain components resulting from the
stress components may be determined from the
principle of superposition. This requires:
1) Each effect is linearly related to the load that
produces it.
2) The deformation resulting form any given
load is small and does not affect the conditions
of application of the other loads.
• With these restrictions:
 x n y n z
x  
E
y  
z  
Fig. 2.33 Deformation of unit cube under
multiaxial loading: (a) unloaded; (b) deformed.

n x
E

E

 y n z
E
n x n y
E

E

E

 z
E
E
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Dilatation: Bulk Modulus
• Relative to the unstressed state, the change in volume is
e  n  1  1   x   y   z   1
 x y z
1  2n
 x   y   z 
E
 dilatation (change in volume per unit volum e)

• For element subjected to uniform hydrostatic pressure,
31  2n 
p
p
E
k
E
k
 bulk modulus or modulus of compression
31  2n 
e
• Subjected to uniform pressure, dilatation must be
negative, therefore
Fig. 2.33 Deformation of unit
cube under multiaxial loading:
(a) unloaded; (b) deformed.
0  n  12
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Shearing Strain
• A cubic element subjected to only shearing stress will
deform into a rhomboid. The corresponding
shearing strain is quantified in terms of the change in
angle between the sides,
 xy  f  xy 
Fig. 2.36 Unit cubic element
subjected to shearing stress.
• A plot of shearing stress vs. shearing strain is similar
to the previous plots of normal stress vs. normal
strain except that the strength values are
approximately half. For values of shearing strain
that do not exceed the proportional limit,
 xy  G  xy  yz  G  yz  zx  G  zx
where G is the modulus of rigidity or shear modulus.
Fig. 2.37 Deformation of unit cubic
element due to shearing stress.
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Concept Application 2.10
SOLUTION:
• Determine the average angular
deformation or shearing strain of
the block.
Fig. 2.41(a) Rectangular block loaded in shear.
A rectangular block of material with
modulus of rigidity G = 90 ksi is
bonded to two rigid horizontal plates.
The lower plate is fixed, while the
upper plate is subjected to a horizontal
force P. Knowing that the upper plate
moves through 0.04 in. under the action
of the force, determine (a) the average
shearing strain in the material, and (b)
the force P exerted on the plate.
• Apply Hooke’s law for shearing stress
and strain to find the corresponding
shearing stress.
• Use the definition of shearing stress to
find the force P.
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Concept Application 2.10
• Determine the average angular deformation
or shearing strain of the block.
 xy  tan  xy 
Fig. 2.41(b) Deformed block showing
the shear strain.
0.04 in.
2 in.
 xy  0.020 rad
• Apply Hooke’s law for shearing stress and
strain to find the corresponding shearing
stress.


 xy  G xy  90 10 3 psi 0.020 rad   1800 psi
• Use the definition of shearing stress to find
the force P.
P   xy A  1800 psi 8 in. 2.5 in.   36 10 3 lb
P  36.0 kips
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Relation Between E, n, and G
• An axially loaded slender bar will
elongate in the x direction and contract in
both of the transverse y and z directions.
• An initially cubic element oriented as in
Figure 2.42(a) will deform into a
rectangular parallelepiped. The axial load
produces a normal strain.
• If the cubic element is oriented as in
Figure 2.42(b), it will deform into a
rhombus. Axial load also results in a
shearing strain.
• Components of normal and shearing strain
are related,
Fig. 2.42 Representations of strain in an axially
loaded bar: (a) cubicstrain element faces
aligned with coordinate axes; (b) cubicstrain
element faces rotated 45º about z-axis.
E
 1  n 
2G
or
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G
E
21  n 
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Composite Materials
• Fiber-reinforced composite materials are fabricated
by embedding fibers of a strong, still material into a
weaker, softer material called a matrix.
• Normal stresses and strains are related by Hooke’s
Law but with directionally dependent moduli of
elasticity,

Ex  x
x
Ey 
y
y

Ez  z
z
• Transverse contractions are related by directionally
dependent values of Poisson’s ratio, e.g.,
n xy
Fig. 2.44 Orthotropic Fiberreinforced composite material
under uniaxial tensile load.
y


n xz   z
x
x
• The three components of strain x, y, and z for
orthotropic materials can be expressed in terms of
normal stress only and do not depend upon any
shearing stresses.
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Sample Problem 2.5
A circle of diameter d = 9 in. is scribed on an
unstressed aluminum plate of thickness t = 3/4
in. Forces acting in the plane of the plate later
cause normal stresses x = 12 ksi and z = 20
ksi.
For E = 10x106 psi and n = 1/3, determine the
change in:
a) the length of diameter AB,
b) the length of diameter CD,
c) the thickness of the plate, and
d) the volume of the plate.
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Sample Problem 2.5
• Evaluate the deformation components.
SOLUTION:
• Apply the generalized Hooke’s Law to
find the three components of normal
strain.
x  

 x n y n z
E

E

1
 0.533 10 3 in./in.
n x  y n z
E

E
 1.067 10
z  

3
E
in./in.
n x n y
E

C
A
D


 z
E
E
 1.600 10 3 in./in.

  x d   0.533 10 3 in./in. 9 in. 

B
A
 4.8  10 3 in.

  z d   1.600 10 3 in./in. 9 in. 
E
1






12
ksi

0

20
ksi

3
10 10 6 psi 
y  
B

C
D
 14.4 10 3 in.

 t   y t   1.067 10 3 in./in. 0.75 in. 
 t  0.800 10 3 in.
• Find the change in volume
e   x   y   z  1.067 10 3 in 3/in 3
V  eV  1.067 10 3 15 15  0.75 in 3
V  0.187 in 3
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Beer • Johnston • DeWolf • Mazurek
Saint-Venant’s Principle
• Loads transmitted through rigid
plates result in uniform distribution
of stress and strain.
• Concentrated loads result in large
stresses in the vicinity of the load
application point.
Fig. 2.47 Axial load
applied by rigid plates to
rubber model.
Fig. 2.48 Concentrated
axial load applied to
rubber model.
Fig. 2.49 Stress distributions in a
plate under concentrated axial loads.
• Stress and strain distributions
become uniform at a relatively short
distance from the load application
points.
• Saint-Venant’s Principle:
Stress distribution may be assumed
independent of the mode of load
application except in the immediate
vicinity of load application points.
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Stress Concentration: Hole
Fig. 2.50 Stress distribution near
circular hole in flat bar under axial
loading.
Stress concentration factor

K  max
Fig. 2.52(a) Stress concentration
factors for flat bars under axial loading.
Discontinuities of cross section may result in
high localized or concentrated stresses.
 ave
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Beer • Johnston • DeWolf • Mazurek
Stress Concentration: Fillet
Fig. 2.52(b) Stress concentration
factors for flat bars under axial loading.
Fig. 2.51 Stress distribution near
fillets in flat bar under axial loading.
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Concept Application 2.12
SOLUTION:
• Determine the geometric ratios and
find the stress concentration factor
from Figure 2.52.
Determine the largest axial load P
that can be safely supported by a
flat steel bar consisting of two
portions, both 10 mm thick, and
respectively 40 and 60 mm wide,
connected by fillets of radius r = 8
mm. Assume an allowable normal
stress of 165 MPa.
• Find the allowable average normal
stress using the material allowable
normal stress and the stress
concentration factor.
• Apply the definition of normal stress to
find the allowable load.
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MECHANICS OF MATERIALS
Concept Application 2.12
Beer • Johnston • DeWolf • Mazurek
• Determine the geometric ratios and
find the stress concentration factor
from Figure 2.52.
D 60 mm

 1.50
d 40 mm
r
8 mm

 0.20
d 40 mm
K  1.82
• Find the allowable average normal
stress using the material allowable
normal stress and the stress
concentration factor.
 ave 
Fig. 2.52(b) Stress concentration
factors for flat bars under axial loading.
 max
K

165 MPa
 90.7 MPa
1.82
• Apply the definition of normal stress
to find the allowable load.
P  A ave  40 mm 10 mm 90.7 MPa 
 36.3 103 N
P  36.3 kN
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Elastoplastic Materials
• Previous analyses based on assumption of
linear stress-strain relationship, i.e.,
stresses below the yield stress
• Assumption is good for brittle material
which rupture without yielding
• If the yield stress of ductile materials is
exceeded, then plastic deformations occur
Fig. 2.53 Stress-strain diagram for an
idealized elastoplastic material.
• Analysis of plastic deformations is
simplified by assuming an idealized
elastoplastic material
• Deformations of an elastoplastic material
are divided into elastic and plastic ranges
• Permanent deformations result from
loading beyond the yield stress
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Plastic Deformations
K
• Elastic deformation while maximum
stress is less than yield stress
 A
PY  Y
K
• Maximum stress is equal to the yield
stress at the maximum elastic
loading
P   ave A 
 max A
• At loadings above the maximum
elastic load, a region of plastic
deformations develop near the hole
PU   Y A
 K PY
• As the loading increases, the plastic
region expands until the section is at
a uniform stress equal to the yield
stress
Fig. 2.56 Distribution of stresses in
elastic-perfectly plastic material under
increasing load.
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Residual Stresses
• When a single structural element is loaded uniformly
beyond its yield stress and then unloaded, it is permanently
deformed but all stresses disappear. It should not be
assumed that this will always be the case.
• Residual stresses will remain in a structure after
loading and unloading if
- only part of the structure undergoes plastic
deformation
- different parts of the structure undergo different
plastic deformations
• Residual stresses also result from the uneven heating or
cooling of structures or structural elements
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Concept Applications 2.14, 2.15, 2.16
A cylindrical rod is placed inside a tube
of the same length. The ends of the rod
and tube are attached to a rigid support
on one side and a rigid plate on the
other. The load on the rod-tube
assembly is increased from zero to 5.7
kips and decreased back to zero.
a) draw a load-deflection diagram
for the rod-tube assembly
b) determine the maximum
elongation
Fig. 2.54(a) Concentric rod-tube
assembly axially loaded by rigid plate.
Ar  0.075 in. 2
At  0.100 in. 2
Er  30  106 psi
Et  15  106 psi
σ r Y
 36 ksi
σt Y
 45 ksi
c) determine the permanent set
d) calculate the residual stresses in
the rod and tube.
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Concept Applications 2.14, 2.15, 2.16
a) Draw a load-deflection diagram for the rod-tube
assembly
Pr Y   r Y Ar  36 ksi 0.075 in 2   2.7 kips
 r Y
36  103 psi
δr Y   r Y L 
30 in. 
L
6
Er
30  10 psi
 36  10-3 in.
Pt Y   t Y At  45 ksi 0.100 in 2   4.5 kips
 t Y
45  103 psi
δt Y   t Y L 
 30 in. 
L
6
Et
15  10 psi
 90  10-3 in.
Fig. 2.55 (a) Rod load-deflection response
with elastic unloading (red dashed line). (b)
Tube load-deflection response. Given loading
does not yield tube so unloading is along the
original elastic loading line. (c) Combined
rod-tube assembly load deflection response
with elastic unloading (red dashed line).
P  Pr  Pt
   r  t
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Beer • Johnston • DeWolf • Mazurek
b,c) determine the maximum elongation and permanent set
Concept Applications
2.14, 2.15, 2.16
• At a load of P = 5.7 kips, the rod has reached the plastic
range while the tube is still in the elastic range
Pr  Pr Y  2.7 kips
Pt  P  Pr  5.7  2.7  kips  3.0 kips
t 
Pt 3.0 kips

 30 ksi
2
At 0.1in
 t  t L 
t
Et
L
30  103 psi
15  106 psi
30 in.
 max   t  60 103in.
• The rod-tube assembly unloads along a line parallel to 0Yr
m
Fig. 2.55 (a) Rod load-deflection response
with elastic unloading (red dashed line). (b)
Tube load-deflection response. Given loading
does not yield tube so unloading is along the
original elastic loading line. (c) Combined
rod-tube assembly load deflection response
with elastic unloading (red dashed line).
4.5 kips
-3
36 10 in.
  
 125 kips in.  slope
Pmax
5.7 kips

 45.6 10 3 in.
m
125 kips in.
 p   max     60  45.6 10 3 in.
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 p  14 .4  10 3 in.
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MECHANICS OF MATERIALS
Beer • Johnston • DeWolf • Mazurek
Concept Applications 2.14, 2.15, 2.16
• Calculate the residual stresses in the rod and tube.
Calculate the reverse stresses in the rod and tube caused
by unloading and add them to the maximum stresses.

 45.6  10 3 in.
  
 1.52  10 3 in. in.
L
30 in.
 r   Er   1.52  10 3 30  10 6 psi  45.6 ksi
 t   Et   1.52  10 3 15  10 6 psi  22.8 ksi
 residual ,r   r   r  36  45.6  ksi  9.6 ksi
Fig. 2.55 (a) Rod load-deflection response
with elastic unloading (red dashed line). (b)
Tube load-deflection response. Given loading
does not yield tube so unloading is along the
original elastic loading line. (c) Combined
rod-tube assembly load deflection response
with elastic unloading (red dashed line).
 residual ,t   t   t  30  22.8 ksi  7.2 ksi
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