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MECHANICS OF
MATERIALS
Fourth Edition
Ferdinand P. Beer
E. Russell Johnston, Jr.
John T. DeWolf
Lecture Notes:
J. Walt Oler
Texas Tech University
CHAPTER
2002 The McGraw-Hill Companies, Inc. All rights reserved.
2Stress and Strain Axial Loading
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Contents
Stress & Strain: Axial Loading
Normal Strain
Stress-Strain Test
Stress-Strain Diagram: Ductile Materials
Stress-Strain Diagram: Brittle Materials
Hookes Law: Modulus of Elasticity
Elastic vs. Plastic Behavior
Fatigue
Deformations Under Axial Loading
Example 2.01
Sample Problem 2.1Static Indeterminacy
Example 2.04
Thermal Stresses
Poissons Ratio
Generalized Hookes Law
Dilatation: Bulk Modulus
Shearing Strain
Example 2.10
Relation Among E, n, and G
Sample Problem 2.5
Composite Materials
Saint-Venants Principle
Stress Concentration: Hole
Stress Concentration: Fillet
Example 2.12Elastoplastic Materials
Plastic Deformations
Residual Stresses
Example 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 inthe 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 underaxial loading. Later chapters will deal with torsional and pure bendingloads.
MECHANICS OF MATERIA S
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Normal Strain
strainnormal
stress
L
A
P
L
A
P
A
P
2
2
LL
A
P
2
2
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Stress-Strain Test
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Stress-Strain Diagram: Ductile Materials
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Stress-Strain Diagram: Brittle Materials
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Hookes Law: Modulus of Elasticity
Below the yield stress
ElasticityofModulus
orModulusYoungs
E
E
Strength is affected by alloying,
heat treating, and manufacturing
process but stiffness (Modulus ofElasticity) is not.
(i.e. the slope of the linear part of
the stress-strain diagram)
Isotropic Materials: Material properties are independent of direction considered.
Unisotropic Materials: Material properties are direction dependent, e.g. wood.
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Elastic vs. Plastic Behavior
If the strain disappears when the
stress is removed, the material is
said to behave elastically.
When the strain does not return
to zero after the stress is
removed, the material is said tobehaveplastically.
The largest stress for which this
occurs is called the elastic limit.
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Deformations Under Axial Loading
AE
P
EE
From Hookes Law:
From the definition of strain:
L
Equating and solving for the deformation,
AE
PL
With variations in loading, cross-section ormaterial properties,
i ii
ii
EA
LP
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Example 2.01
Determine the deformation of
the steel rod shown under the
given loads.
mm56.15dmm18.27DGPa200E
SOLUTION:
Divide the rod into components at
the load application points.
Apply a free-body analysis on each
component to determine theinternal force
Evaluate the total of the component
deflections.
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SOLUTION:
Divide the rod into three
components:
2
21
21
mm580AA
mm300LL
2
3
3
mm190A
mm.400L
Apply free-body analysis to each
component to determine internal forces,
kN120P
kN60P
kN240P
3
2
1
Evaluate total deflection,
mm.1.73m1073.110190
4.010120
10580
3.01060
10580
3.010240
10200
1
A
LP
A
LP
A
LP
E
1
EA
LP
3
6
3
6
3
6
3
9
3
33
2
22
1
11
i ii
ii
mm.73.1
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Sample Problem 2.1
The rigid barBDEis supported by two linksAB and CD.
LinkAB is made of aluminum (E= 70 GPa)
and has a cross-sectional area of 500 mm2.
LinkCD 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) ofB, b) ofD, and c) ofE.
SOLUTION:
Apply a free-body analysis to the bar
BDEto find the forces exerted by
linksAB andDC.
Evaluate the deformation of linksAB
andDCor the displacements ofBandD.
Work out the geometry to find the
deflection at E given the deflections
at B and D.
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E
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Displacement ofB:
m10514
Pa1070m10500
m3.0N1060
6
926-
3
AEPL
B
mm514.0B
Displacement ofD:
m10300
Pa10200m10600m4.0N1090
6
926-
3
AE
PLD
mm300.0D
Free body: BarBDE
ncompressioF
F
tensionF
F
M
AB
AB
CD
CD
B
kN60
m2.0m4.0kN3000M
kN90
m2.0m6.0kN300
0
D
SOLUTION:
Sample Problem 2.1
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Displacement of D:
mm7.73
mm200
mm0.300
mm514.0
x
x
x
HD
BH
DD
BB
mm928.1E
mm928.1mm7.73
mm7.73400
mm300.0
E
E
HD
HE
DD
EE
Sample Problem 2.1
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Static Indeterminacy
Structures for which internal forces and reactions
cannot be determined from statics alone are saidto bestatically indeterminate.
0RL
Deformations due to actual loads and redundant
reactions are determined separately and then added
orsuperposed.
Redundant reactions are replaced with
unknown loads which along with the other
loads must produce compatible deformations.
A structure will be statically indeterminate
whenever it is held by more supports than are
required to maintain its equilibrium.
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Example 2.04
Determine the reactions atA andB for the steel
bar and loading shown, assuming a close fit at
both supports before the loads are applied.
Solve for the reaction atA due to applied loads
and the reaction found atB.
Require that the displacements due to the loadsand due to the redundant reaction be compatible,
i.e., require that their sum be zero.
Solve for the displacement atB due to the
redundant reaction atB.
SOLUTION:
Consider the reaction atB as redundant, release
the bar from that support, and solve for the
displacement atB due to the applied loads.
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SOLUTION:
Solve for the displacement atB due to the applied
loads with the redundant constraint released,
EEA
LP
LLLL
AAAA
PPPP
i ii
ii9
L
4321
2643
2621
34
3321
10125.1
m150.0
m10250m10400
N10900N106000
Solve for the displacement atB due to the redundant
constraint,
i
B
ii
iiR
B
E
R
EA
LP
LL
AA
RPP
3
21
262
261
21
1095.1
m300.0
m10250m10400
Example 2.04
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Require that the displacements due to the loads and due to
the redundant reaction be compatible,
kN577N10577
01095.110125.1
0
3
39
B
B
RL
R
E
R
E
Find the reaction atA due to the loads and the reaction atB
kN323
kN577kN600kN3000
A
Ay
R
RF
kN577
kN323
B
A
R
R
Example 2.04
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Thermal Stresses
A temperature change results in a change in length or
thermal strain. There is no stress associated with thethermal strain unless the elongation is restrained by
the supports.
tcoefficienexpansionthermal
AE
PLLT PT
Treat the additional support as redundant and apply
the principle of superposition.
0
0
AE
PLLT
PT
The thermal deformation and the deformation from
the redundant support must be compatible.
TEA
P
TAEPPT
0
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Poissons Ratio
For a slender bar subjected to axial loading:
0 zyx
xE
The elongation in the x-direction is
accompanied by a contraction in the other
directions. Assuming that the material isisotropic (no directional dependence),
0 zy
Poissons ratio is defined as
x
z
x
y
n
strainaxial
strainlateral
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Generalized Hookes Law
For an element subjected to multi-axial loading,
the normal strain components resulting from thestress components may be determined from the
principle of superposition. This requires:
1) strain is linearly related to stress
2) deformations are small
EEE
EEE
EEE
zyxz
zyxy
zyxx
nn
n
n
nn
With these restrictions:
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Dilatation: Bulk Modulus
Relative to the unstressed state, the change in volume is
e)unit volumperin volume(changedilatation
21
111111
zyx
zyx
zyxzyx
E
e
n
For element subjected to uniform hydrostatic pressure,
modulusbulk
213
213
n
n
Ek
k
p
Epe
Subjected to uniform pressure, dilatation must be
negative, therefore
210 n
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Shearing Strain
Consider the effect of the presence of shearing
stresses on the faces of a small element.
In the case of isotropic materials, these stresses come
in pairs of equal and opposite vectors applied to the
opposite sides of the given element and have no
effect on normal strains.
A cubic element subjected to a shear stress will deform into a rhomboid.
The correspondingshearstrain is quantified in terms of the change in angle
between the sides,
xyxy f
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Shearing Strain
A plot of shear stress vs. shear strain is similar the
previous plots of normal stress vs. normal strain
except that the strength values are approximately
half. For small strains,
zxzxyzyzxyxy GGG
where G is the modulus of rigidity or shear modulus.
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Example
A rectangular block of material with
modulus of rigidity G = 620 MPa is
bonded to two rigid horizontal plates.
The lower plate is fixed, while the
upper plate is subjected to a horizontal
forceP. Knowing that the upper plate
moves through 1 mm under the action
of the force, determine a) the average
shearing strain in the material, and b)
the forcePexerted on the plate.
SOLUTION:
Determine the average angulardeformation or shearing strain of
the block.
Use the definition of shearing stress to
find the forceP.
Apply Hookes law for shearing stress
and strain to find the corresponding
shearing stress.
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2 - 27
Determine the average angular deformation
or shearing strain of the block.
rad020.0mm50
mm1tan xyxyxy
Apply Hookes law for shearing stress and
strain to find the corresponding shearing
stress.
MPa6.13rad020.0MPa680G xyxy
Use the definition of shearing stress to find
the forceP.
kN170105.62200MPa6.13 -3 mmmmAP xy
kN170P
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Relation Among E, n, and G
An axially loaded slender bar will
elongate in the axial direction andcontract in the transverse directions.
n 12G
E
Components of normal and shear strain are
related,
If the cubic element is oriented as in the
bottom figure, it will deform into a
rhombus. Axial load also results in a shear
strain.
An initially cubic element oriented as in
top figure will deform into a rectangular
parallelepiped. The axial load produces a
normal strain.
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Sample Problem
A circle of diameterd= 225 mm is scribed on
an unstressed aluminum plate of thickness, t,
20 mm. Forces acting in the plane of the plate
later cause normal stresses x = 80 MPa and
z = 140 MPa.
ForE= 70 GPa and n = 1/3, determine thechange in:
a) the length of diameterAB,
b) the length of diameterCD,
c) the thickness of the plate, and
d) the volume of the plate.
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2 - 30
SOLUTION:
Apply the generalized Hookes Law to
find the three components of normal
strain.
mm/mm10619.1
EEE
mm/mm10047.1
EEE
mm/mm10476.0
MPa1403
10MPa80
MPa000,70
1
EEE
3
zyx
z
3
zyxy
3
zyxx
n
n
n
n
n
n
Evaluate the deformation components.
mm225mm/mm10476.0d 3xAB
mm225mm/mm10619.1d 3zDC
mm20mm/mm10047.1t 3yt
mm11.0AB
mm36.0DC
mm02.0t
Find the change in volume
33
333
mm2038038010048.1
/mmmm10048.1
eVV
ezyx
3mm6.026,3V
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2 - 31
Saint-Venants Principle
Loads transmitted through rigid
plates result in uniform distribution
of stress and strain.
Saint-Venants Principle:
Stress distribution may be assumed
independent of the mode of load
application except in the immediate
vicinity of load application points.
Stress and strain distributions
become uniform at a relatively short
distance from the load application
points.
Concentrated loads result in large
stresses in the vicinity of the load
application point.
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Example 2.14, 2.15, 2.16
A cylindrical rod is placed inside a tube
of the same length. The ends of the rodand 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
c) determine the permanent set
d) calculate the residual stresses in
the rod and tube.
ksi36
psi1030
in.075.0
,
6
2
rY
r
r
E
A
ksi45
psi1015
in.100.0
,
6
2
tY
t
t
E
A
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a) draw a load-deflection diagram for the rod-
tube assembly
in.1036in.30psi1030
psi1036
kips7.2in075.0ksi36
3-6
3
,
,,
2,,
LE
L
AP
rY
rYrYY,r
rrYrY
in.1009in.30psi1015
psi1045
kips5.4in100.0ksi45
3-6
3
,
,,
2,,
LE
L
AP
tY
tYtYY,t
ttYtY
tr
tr PPP
Example 2.14, 2.15, 2.16
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b,c) determine the maximum elongation and permanent set
at a load ofP= 5.7 kips, the rod has reached the
plastic range while the tube is still in the elastic range
in.30psi1015
psi1030
ksi30in0.1
kips0.3
kips0.3kips7.27.5
kips7.2
6
3
t
2t
,
LE
L
A
P
PPP
PP
t
tt
t
t
rt
rYr
in.1060 3max t
the rod-tube assembly unloads along a line parallel to
0Yr
in.106.4560
in.106.45in.kips125
kips7.5
slopein.kips125in.1036
kips5.4
3maxp
3max
3-
m
P
m
in.104.14 3p
Example 2.14, 2.15, 2.16
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calculate the residual stresses in the rod and tube.
calculate the reverse stresses in the rod and tubecaused by unloading and add them to the maximum
stresses.
ksi2.7ksi8.2230
ksi69ksi6.4536
ksi8.22psi10151052.1
ksi6.45psi10301052.1
in.in.1052.1
in.30
in.106.45
,
,
63
63
33
tttresidual
rrrresidual
tt
rr
.
E
E
L
Example 2.14, 2.15, 2.16
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