SIMPLE STRESSES AND STRAINS

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29 Νοε 2013 (πριν από 3 χρόνια και 7 μήνες)

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SIMPLE STRESSES AND STRAINS


Elasticity:

It is the property of certain materials of returning back to their original position after
removing the external force, is known as elasticity.

Stress
:

It is the resistance of unit area to deformation. Mathematical
ly, stress may be defined as
the force per unit area, i.e,

σ= P/A

Where:


σ = Intensity of stress (written as stress)


P = Load or force acting on the body, and


A = Cross sectional area of the boy

If the load or force is in kg and area is in cm
2
, then the unit of stress will be kg
/cm
2
.

Strain;

Is the deformation per unit length. Mathematically, strain may be defined as deformation
per unit length, i.e.,



=

L/L

Where


= Strain



L = Change of length of the body, and


L = Original length

Types of str
esses:

There are many types of stresses, yet the following two types are important from the
subject point of view:

1)

Tensile stress, and 2) Compressive stress

Tensile stress:



P
P

When a section is subjected to two equal and opposite pul
ls, as a result of which the body
tends to lengthen, as shown in the figure, the stress included is called
tensile stress
. The
corresponding strain is called
tensile strain.

Compressive stress:


P
P


When a section is subjected to two
equal and opposite pushes, as shown in the above
figure, as a result of which the body tends to shorten its length, the stress included is
called
compressive stress
and the corresponding strain is called
compressive strain.

Elastic limit:

It has been found

that, for a given section, there is a limiting value of force up to and
within which, the deformation disappears on removal of the force. The value of stress
corresponding to thi
s

limiting force is called
elastic limit
of the material.

Hooke

猠污s:

It sta
tes that when a material is loaded, within its elastic limit, the stress is proportional to
the strain. Mathematically:

Stress/strain = E =a constant

It may be noted that the
Hooke

猠污s

equally well for tension as well as compression.

Modulus of elasticit
y:

As already mentioned whenever a material is loaded within its elastic limit, the stress is
proportional to strain:

σ



σ = E


or E= σ/


Where:


E = A constant of proportionality known as modulus of elasticity.


Example 1:

A rod 100 mm long and 2cmx2cm c
ross section is subjected to a pull of 100 kg force. If
the modulus of elasticity of the material is 2.0x10
6

kg/cm
2
, determine the elongation of
the rod.

Solution:

L = 100cm, P = 1000 kg , A = 2x2= 4 cm
2

, E = 2.0x10
6

kg/cm
2

σ =P/A =1000/4 =250

kg/cm
2



=σ /E = 250/2.0x10
6

=0.000125


L =

L =0.000125x100 = 0.0125 cm


Principle of superposition:


Sometimes, a body is subjected to a number of forces acting on its outer edges as well as
at some other sections, along the length of the body. In
such case, the forces are split up,
and their effects are considered on individual sections. The resulting deformation of the
body is equal to the algebraic sum of the deformation of the individual sections. Such a
principle of finding out the resultant de
formation is called the principle of superposition.


The relation for the resulting deformation is modified as:


L = PL/EA
=1/EA
(P
1
L
1

+ P
2
L
2

+ P
3
L
3

+……….)


Example 2:

A brass bar, having cross sectional area of 10 cm
2

is subjected to axial forces as shown i
n
figure. Find the total elongation of the bar. Take E = 0.8x10
6

kg/cm
2
.

A
B
C
D
1000 kg
2000 kg
8000 kg
5000 kg
120 cm
100 cm
60 cm

Solution:


5000 kg
5000 kg
3000 kg
3000 kg
1000 kg
1000 kg


For section A
-
B:

Tension force of 5000 kg
=
P
1

For section B
-
C:

Compression force of 300 kg =P
2

For section C
-
D:

Compression

force of 1000 kg =P
3




L

=
1/ (10x0.8x10
6
)[ 5000x60


3000x100


1000x120]


= 1/(8x10
6
)[300000


300000


120000]


=
-
12/800 =
-

0.015 cm


Example

3:

A reinforced concrete column is 50cmx50cm in section. The column is reinforced with 4
ste
el bars of 2.5 cm diameter, one in each corner and is carrying a load of 200 tones. Find
the stress in the concrete and steel bars. Take E for steel = 2.1x10
6

kg/cm
2

and E for
concrete = 0.14x10
6

kg/cm
2
.

Solution:

Area of column = 50x50 = 2500 cm
2


Area of

steel =4(

/4) (2.5)
2

= 19.63 cm
2

Area of concrete = 2500


19.63 = 2480.37 cm
2

P = 200000 kg

σ
s

=stress in steel , σ
c

= stress in concrete


L
s
=

L
c

σ
s
/E
s
= σ
c
/E
c

σ
s

= σ
c
(E
s
/E
c
) = σ
c
(2.1x10
6
/0.14x10
6
) = 15 σ
c

(1)

P = P
s

+P
c


P = σ
s
xA
s

+ σ
c
xA
c

= σ
s
x19.63 + σ
c
x2480.37 (2)

From (1) & (2)

σ
s

= 1080 kg/cm
2

σ
c

=

72 kg/cm
2