# Chapter 2 Mechanics of materials

Mechanics

Oct 29, 2013 (4 years and 7 months ago)

89 views

1

Normal

stress

=

Chapter 2 Mechanics of Materials

Example: Estimate the normal stress on a shin bone (

)

A

Tensile stress (+)

Compressive stress (
-
)

At a point:

2

Shear stress (

) =

= F tangential to the area /
A

A

At a point,

3

Normal strain (

)

=
fractional change of length=

x

F

F

fixed

Shear strain (?) = d
eformation under shear stress =

l

x

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
.

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

6

Bulk modulus

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

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

9

y

z

Joint effect of three normal stress components

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

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

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

13

Point C moves further along x
-

and y
-
direction by distances

/2) and

/2) respectively.

nn

/2)
2

/2)
2
]
1/2

2

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’

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

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

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

17

Elastic Strain Energy

The

energy

stored

in

a

small

volume
:

The

energy

stored

:

Energy

density

in

the

material

:

e=extension

x