Chapter 2 Fundamentals of the Mechanical Behavior of Materials

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Chapter 2Fundamentals of the Mechanical Behavior
of Materials
Fetweb.ju.edu.jo/staff/ie/mbarghash
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Types of Strain
FIGURE 2.1 Types of strain. (a) Tensile, (b)
compressive, and (c) shear. All deformation
processes in manufacturing involve strains of
these types. Tensile strains are involved in
stretching sheet metal to make car bodies,
compressive strains in forging metals to make
turbine disks, and shear strains in making holes
by punching.
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Universal testing machine
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Tension Test
Figure 2.2 (a) Original and final shape of a
standard tensile-test specimen. (b) Outline of a
tensile-test sequence showing stages in the
elongation of the specimen.
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Understanding shear
Not required
Definition
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Imperfections in engineering strain
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First case ln(2Lo/Lo)ln(2) Second case
ln(Lo/2Lo)ln(1/2)-ln(2)
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ln(2)0.6931 ln(1.5)ln(2/1.5)0.6931
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Stess in three dimensions
n is a vector, so tn is the stress on a plan
with n as orthogonal to it
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Special case n x, There are other cases where
ny, nz Thus we have 9 stress vectors
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Hydraustaic and deviatoric stress
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Homework prove the above relation
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The plastic deformation, the plastic deformation
Causes zero volume change
Prove the above relation for A unit cube
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Mechanical Properties of Materials
Table 2.1 Typical mechanical properties of
various materials at room temperature.
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Loading and Unloading
FIGURE 2.3 Schematic illustration of loading
and unloading of a tensile-test specimen. Note
that during unloading, the curve follows a path
parallel to the original elastic slope.
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True Stress-True-Strain Curves in Tension
FIGURE 2.5 (a) True stress-true-strain curve in
tension. Note that, unlike in an engineering
stress-strain curve, the slope is always
positive, and the slope decreases with increasing
strain. Although stress and strain are
proportional in the elastic range, the total
curve can be approximated by the power expression
shown. On this curve, Y is the yield stress and
Yf is the flow stress. (b) True-stress
true-strain curve plotted on a log-log scale. (c)
True stress-true-strain curve in tension for
1100-O aluminum plotted on a log-log scale. Note
the large difference in the slopes in the elastic
and plastic ranges. Source After R. M. Caddell
and R. Sowerby.
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Power Law Material Behavior
K strength coefficient n strain hardening
coefficient
Table 2.3 Typical values of K and n in Eq. (2.11)
at room temperature.
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True Stress - True Strain Curves for Various
Metals
FIGURE 2.6 True-stress-true-strain curves in
tension at room temperature for various metals.
The point of intersection of each curve at the
ordinate is the yield stress Y thus, the elastic
portions of the curves are not indicated. When
the K and n values are determined from these
curves, they may not agree with those given in
Table 2.3, because of the different sources from
which they were collected. Source S. Kalpakjian.
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Strain Rate Effects

• Table 2.5 Approximate range of values for C and
  m in Eq. (2.16) for various annealed materials at
  true strains ranging from 0.2 to 1.0.

C strength coefficient M strain rate
sensitivity exponent
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Effect of temperature
Power Law Creep One of the most common forms of
plastic flow is Power-Law Creep, given by the
formula Strain Rate C (Stress)n exp(-Q/RT)
Let's take each part of the formula in turn C
is a scaling constant. n means that the strain
rate increases much faster than stress. Typically
n is about 3 but can range from a bit less than 2
to 8. Recall that with viscous deformation stress
is proportional to strain rate (n1). With
power-law creep it's faster the effective
viscosity drops with stress. Q is the activation
energy required to get crystal dislocations
moving. It's typically 100-300 kilojoules per
mole, sometimes up to 500. R is the Universal
Gas Constant that turns up everywhere in physical
chemistry. In SI units it equals 8.3144
joules/mole-degree Kelvin. T is the temperature
in degrees Kelvin. As T increases, Q/RT decreases
and thus exp(-Q/RT) increases, though much more
slowly than exp(T). At very large T, Q/RT
approaches zero and the exponential term
approaches 1. This does not happen, though, at
geologically realistic temperatures.
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Barreling In Compression
Plane Strain Compression

• FIGURE 2.15 Barreling in compression of a round
  solid cylindrical specimen (7075-O aluminum)
  between flat dies. Barreling is caused by
  interfaces, which retards the free flow of the
  material. See also Figs. 6.1 and 6.2. Source K.
  M. Kulkarni and S. Kalpakjian.

FIGURE 2.16 Schematic illustration of the
plane-strain compression test. The dimensional
relationships shown should by satisfied for this
test to be useful and reproducible. This test
give the yield stress of the material in plane
strain, Y. Source After A. Nadai and H. Ford.
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Toughness It is the Area under the stress strain
curve
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Derive the instability point for plain stress
case with equal stresses
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Principle stresses
Since x, y, z are optional coordinates, then we
can obtain Certain coordinates where the shear
stresses disappear
Example, the normal directions in simple tension
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Two dimensional-plain stress case
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Homework, (dont submitted it, just solve
it) Find the maximum stress case for the simple
tension And Find the principle stresses for the
simple shear case
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Not required
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Plain strain transformation
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Not required
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Plane Strain

• A state of plane strain exits when the strains
  are confined to a single plane, such as the x-y
  plane.
• This generally means that the stresses in the
  other direction eg., the z direction, are
  non-zero.
• Plane Strain occurs in thick sections that
  constrain out of plane deformations

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Transformations in plane strain
e
e
,
g
,
Determine the
.
The state of plane strain at a point p is given
by
x
y
xy
principal strains and the maximum in-plane shear
strain and show the
orientations of the elements subjected to these
strains. Also determine the
absolute maximum shear strain.
e
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Solution
g
xy
tan
2
q
p
e
e
x
y
g
-0.006
1
xy
1
.
.
q
atan
atan
p
(0.003-0.001)
e
e
2
2
x
y
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Principal Strains
e
e
e
e
g
x
y
x
y
xy
.
.
.
.
e
'
cos
2
q
sin
2
q
x
p
p
2
2
2
e
e
e
e
g
x
y
x
y
xy
.
.
.
.
e
'
cos
2
q
sin
2
q
y
p
p
2
2
2
-0.006
0.004
0.002
-79.6º
.
sin
-79.6º
e
'
cos
x
2
2
2
0.005162
ey-0.001162
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Maximum Shear Strain
e
'
e
'
x
y

e
'
g
x
test
e
'
y
gmax 0.006325
Use diameters of circles!, Mohrs circle plots
g/2, so g is the diameter (2x radius).
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Max In-Plane Shear Strain
9.2º
-0.006325
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Mohrs Circle

• Plot normal strain on the x-axis
• Plot ½ the shear strain on the y-axis
• Solve as you would for plane stress problem

g/2
e
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Strain energy
For the elastic region
For 3D case
For principle stress case
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Plane Strain

• A state of plane strain exits when the strains
  are confined to a single plane, such as the x-y
  plane.
• This generally means that the stresses in the
  other direction eg., the z direction, are
  non-zero.
• Plane Strain occurs in thick sections that
  constrain out of plane deformations

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Transformations in plane strain
e
e
,
g
,
Determine the
.
The state of plane strain at a point p is given
by
x
y
xy
principal strains and the maximum in-plane shear
strain and show the
orientations of the elements subjected to these
strains. Also determine the
absolute maximum shear strain.
e
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Solution
g
xy
tan
2
q
p
e
e
x
y
g
-0.006
1
xy
1
.
.
q
atan
atan
p
(0.003-0.001)
e
e
2
2
x
y
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Principal Strains
e
e
e
e
g
x
y
x
y
xy
.
.
.
.
e
'
cos
2
q
sin
2
q
x
p
p
2
2
2
e
e
e
e
g
x
y
x
y
xy
.
.
.
.
e
'
cos
2
q
sin
2
q
y
p
p
2
2
2
-0.006
0.004
0.002
-79.6º
.
sin
-79.6º
e
'
cos
x
2
2
2
0.005162
ey-0.001162
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Maximum Shear Strain
e
'
e
'
x
y

e
'
g
x
test
e
'
y
gmax 0.006325
Use diameters of circles!, Mohrs circle plots
g/2, so g is the diameter (2x radius).
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Max In-Plane Shear Strain
9.2º
-0.006325
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Mohrs Circle

• Plot normal strain on the x-axis
• Plot ½ the shear strain on the y-axis
• Solve as you would for plane stress problem

g/2
e
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Prove the above relations prove that equation
21 and 22 are the same (for the plain stress case)
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Principle stresses in three direction
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We want to determine the normal and shear
stresses on the plane. The normal stresses are
easiest. Unfortunately, we can't add stresses,
only forces, so we have to determine the forces
the stresses exert, add them up, then convert
back to stress. Consider stress S1. It acts along
the X1 axis, .but the stress "sees" only the area
of the plane visible along the X1 axis, which is
c1. So F1 S1c1. Similarly, F2 S2c2 and F3
S3c3. The force normal to the plane exerted by F1
is F1c1, and the total force normal to the plane
is F1c1 F2c2 F3c3. Since F1 S1c1, we
find Fn S1c12 S2c22 S3c32 Furthermore,
stress force/area, but the area of the plane is
one, so we have Sn S1c12 S2c22
S3c32 Determining shear stress can be a lot
messier, if we do things the brute force way. Or
we can do it the easy way.
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Here we are looking in the plane of the normal
and shear forces. It's obvious from the vector
diagram that F2 Fn2 Fs2. Since the plane has
an area of one and stress force per unit area,
we have F2 Sn2 Ss2. Note that it's only the
magnitudes of the stresses that we are adding.
Stresses do not add vectorially! The total force
F can be found from the three vectors F1, F2 and
F3 above. Since these three components are
mutually perpendicular, we have F2 F12 F22
F32 or Sn2 Ss2 S12c12 S22c22 S32c32 (this
will be very useful a bit later)
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So we have Ss2 F2 - Sn2  F12 F22 F32 -
(S1c12 S2c22 S3c32)2   S12c12 S22c22
S32c32 - (S1c12 S2c22 S3c32)2   S12c12
S22c22 S32c32 - S12c14 - S22c24 - S32c34 -
2S1S2c12c22 - S2S3c32c22 - S3S1c12c32 We regroup
terms to get Ss2 S12c12(1 - c12) S22c22(1 -
c22) S32c32(1 - c32) - 2S1S2c12c22 -
S2S3c32c22 - S3S1c12c32 Now, since 1 - c12 c22
c32, we can rewrite the above as Ss2
S12c12(c22 c32) S22c22(c32 c12)
S32c32(c22 c12) - 2S1S2c12c22 - S2S3c32c22 -
S3S1c12c32 Gathering terms, we get (S12 -
2S1S2 S22)c12c22 (S22 - 2S2S3 S32)c32c22
(S32 - 2S3S1 S12)c12c32 Ss2 (S1 - S2)2c12c22
(S2 - S3)2c32c22 (S1 - S3)2c12c32
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Triaxial Stress State
?
?
(ve sense shown)
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3D Principal Triaxial Stress
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3D Stress Principal Stresses
The three principal stresses are obtained as the
three real roots of the following equation
where
I1, I2, and I3 are known as stress invariants as
they do not change in value when the axes are
rotated to new positions.
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Stress Invariants for Principal Stress
Zero shear stress on principal planes
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Derive the two dimensional principle stress case
using the Equations for the tri-axial stress case
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Mohrs Circle?

• There is no Mohrs circle solution for problems
  of triaxial stress state
• Solution for maximum principal stresses and
  maximum shear stress is analytical
• Either closed form solution or numerical solution
  (or computer program) are used to solve the
  eigenvalue problem.

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Maximum Shear Stresses
Absolute max shear stress is the numerically
larger of
tyz, tabs max
txy
tyz
s3
s1
s2
Normal Stress, s
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3D Mohrs Circle Plane Stress
A Case Study The two principal stresses are of
the same sign
s3
s1
s2
-s
t
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3D Mohrs Circle Plane Stress
A Case Study The two principal stresses are of
opposite sign
s3
s1
s2
s
t
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Example
For the following state of stress, find the
principal and critical values.
y
Tensor shows that sz 0 and t xz t yz 0
x
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The other 2 faces
x
y
z
z
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3-D Mohrs Circles
t max 77 MPa
Shear Stress, MPa
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Example triaxial stress state, not plane stress

• Determine the maximum principal stresses and the
  maximum shear stress for the following triaxial
  stress state. (ve values as defined in slide 1)

s
MPa
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Solution
s

MPa
20 30 10 40 MPa
-3025 MPa
89500 MPa
Solve
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Results
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Mohrs circles
Shear (MPa)
tyz, tabs max58.5
s226.5
s363.5
s1 -51.8
Normal Stress, s (MPa)
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Safety Factor?
Not Required

• If the stress state was determined on a steel
  crankshaft, made of forged SAE1045 steel with a
  yield strength of 300 MPa, what is the factor of
  safety against yield?
• Tresca Criterion tmax 58.5 MPa

2. Max Principal Stress Criterion smax 63.5 MPa
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NOT REQUIRED
3. Von Mises Criterion
103.31 MPa
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Yield criterion
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For simple tension
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Tresca yield locus
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Flow rules
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Prove it Ignoring the Shear strains
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Efficiency ideal work / (total work)
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Next lecture problems and solutions for two
lectures And tutorials
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Using abacus, the program is a finite element
software for Modelling and analysis of
Manufacturing processes We will use it to
understand the operations and its analysis
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