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1
Normal
stress
=
Chapter 2 Mechanics of Materials
Example: Estimate the normal stress on a shin bone (
脛骨
)
A
Tensile stress (+)
Compressive stress (

)
At a point:
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2
Shear stress (
切應力
) =
= F tangential to the area /
A
A
At a point,
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3
Normal strain (
正應變
)
=
fractional change of length=
x
F
F
fixed
Shear strain (?) = d
eformation under shear stress =
l
x
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4
Stress

strain curve
o
Yield pt
.
Work
hardening
break
Elastic
deformation
H
ooke's law:
In elastic region,
,
or
/
=
E
E
is a constant, named as
Young’s modulus
or
modulus of
elasticity
Similarly, in elastic region,
/
= G, where G is a constant,
named as
shear modulus
or
modulus of rigidity
.
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5
Exercise set 2 (Problem 3)
Find the total
extension of the bar.
X
15
mm
W
5
mm
1.2
m
0.6
m
o
dx
Width of a cross

sectional element at
x
:
Stress in this element :
Strain of this element:
The extension of this element :
The total extension of the whole bar is :
= 2.13 x 10

4
m
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6
Bulk modulus
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7
Poisson's ratio :
For a homogeneous isotropic material
normal
strain
:
lateral
strain
:
Poisson's
ratio
:
value
of
:
0
.
2

0
.
5
F
F
x
d
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8
Double index notation
for stress and strain
1
st
index: surface, 2
nd
index: force
For normal stress components :
x
xx
,
y
yy
,
z
zz,
x
xx
y
z
zx
zy
y
z
xz
xy
yx
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y
z
Joint effect of three normal stress components
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10
Symmetry of shear stress components
Take moment a
bout the z axis, total torque = 0,
(
xy
y
z
)
x
= (
yx
x
z
)
y, hence,
xy
=
yx
.
Similarly,
yz
=
zy
and
xz
=
zx
z
y
x
xy
yx
x
y
z
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11
Define pure rotation angle
rot
and
pure
shear strain
, such that
the angular
displacements of the two surfaces are:
1
=
rot
+
def
and
2
=
rot

def
. Hence,
rot
= (
1
+
2
)/2 and
def
= (
1

2
)/2
Original shear strain is “simple” strain =
There is no real deformation during pure rotation,
but “simple” strain
0.
x
2
=

Example:
1
= 0 and
2
=

,
so
def
= (0+
)/2 =
/2 and
rot
= (0

)/2 =

/2
Pure shear strain is
⼲
x
d
y
rot
2
def
def
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12
Example: Show that
Proof:
xx
=
yy
=
zz
=
, hence
3
=
xx
+
yy
+
zz
= (1

2
v
)(
xx
+
yy
+
zz
)/
E
xx
=
yy
=
zz
=

p
(compressive stress)
For hydrostatic pressure
l
l
l
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13
Point C moves further along x

and y

direction by distances
of AD(
/2) and
AD(
/2) respectively.
nn
= [(AD
/2)
2
+ (AD
/2)
2
]
1/2
/ [(AD)
2
+ (AD)
2
]
1/2
=
/2
True shear strain:
yx
=
/2
Therefore, the normal component of strain is equal to the
shear component of strain:
nn
=
yx
and
nn
=
/2
Example : Show that
nn
=
/2
A
C’
C
D
D’
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yx
(
lW)
sin 45
o
x2 = 2 (
l
cos 45
o
)
W
nn
Example : Show that
nn
=
nn
/(2
G
)
Consider equilibrium along n

direction:
Therefore
yx
=
nn
From definition :
=
xy
/
G =
nn
/
G
= 2
nn
l
l
n
yx
xy
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15
xx
=
xx
/
E

yy
/
E

v
zz
/
E
Set
xx
=
nn
=

yy
,
zz
= 0,
xx
=
nn
nn
= (1+
)
nn
/
E =
nn
/2
G
(previous example)
Example : Show
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16
Ex. 12 kN forces are applied to the top
& bottom of a cube (20 mm edges),
E
=
60 GPa,
= 0.3. Find (i) the force
exerted by the walls, (ii)
yy
z
y
12
kN
x
(i)
xx
= 0,
yy
= 0 and
zz
=

12
3
N/(20

3
m)
2
= 3
7
Pa
xx
=
(
xx

v
yy

v
zz
) /
E
0 = [
xx

0
–
0.3
(

3
7
)]/60
9
xx
=

9
6
Pa (compressive)
Force =
A
xx
= (20

3
m)
2
(

9
6
Pa) =

3.6
3
N
(ii)
yy
=
(
yy

v
zz

v
xx
) /
E
= [
0
–
0.3
(

3
7
)
–
0.3
(

9
6
)]/60
9
= 1.95

4
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17
Elastic Strain Energy
The
energy
stored
in
a
small
volume
:
The
energy
stored
:
Energy
density
in
the
material
:
e=extension
x
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18
Similarly for shear strain :
F
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