Page 177 - WCS - Electrical

π

Force
π 2
d =
Stress =
4
4
sectional

Original

area
Load

500
=
Stress =
=

N
Load
(or)
forced
22
16π
Area
16x
σ =
Area
2
mm
500x7
=
16x22

stress
E =
strein
Applied

stress
Deflection
G =
strain

shear
10
newtons
600

30

mm
30

stress
volumetric

K =
volumetric
strain

Ultimate
stress

Safety =

of

Factor
stress

strain

Lateral
Poisson'
ratio =
s
strain
Linear

stress

Ultimate
Safe

stress =
stress
Safe

E
μ =
1+
2.5
=
4
E

−
μ =
1
stress
Ultimate

2G
stress =

Safe
of
safety
Factor

π
2.5
2
×
(4.2)
=
 E

 

4
4
 − 21
−1
= 3K
1
 

 −


 2G
G

 

2.5
22
4.2×
4.2
×
=

4
4
E
 3G

-


 
3K shear 2G cross = applied E E + 2    Spring stiffness = 7× 500 × Safe load 7
3
 −
= 3K

 



  G    G  Working stress = Load = 30× 1000
Area 600
 9KG  WORKSHOP CALCULATION & SCIENCE - CITS
Ultimate
stress

E =   Factor of safety =
 G + 3K  Working stress

450 M.Pa
=
50 M.Pa
= 9 Ans.
Assignment
1 When two equal and opposite forces act on a body they have the same line of action.If they tend to increase
the length of the body, the applied forces are called .
a Compressive force
b Tensile force
c Shear force
d Parallel force
2 The ratio between ultimate stress to working stress is called as .
a Bulk modulus
b Young’s modulus
c Factor of safety
d Modulus of rigidity
3 The side of the square rod is 10 mm which is subjected to the tensile load of 5000 newtons. Find the tensile
stress?
a 5 Nw/cm2
b 50 Nw/mm2
c 500 Nw/cm2
d 5000 Nw/mm 2
4 A steel brake rod is subjected to a tensile force of 3000 newton.If the diameter of the rod is 9 mm calculate the
stress developed in the rod ?
a 67.09 N/mm2
b 63.78 N/mm2
c 52.65 N/mm2
e 47.09 N/mm 2
5 A copper rod of 45 mm diameter is subjected to tensile load of 4500 newtons. Calculate the stress.
a 263 N/cm2
b 273 N/cm2
c 283 N/cm2
d 293 N/cm 2
164
CITS : WCS - Electrical - Exercise 16