Chapter 2 Mechanics of materials

baconossifiedMechanics

Oct 29, 2013 (3 years and 10 months ago)

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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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9

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




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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14



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