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1. 1
Recap
In the last class we have learnt about
• Hook's law
• Young’s modulus
• Modulus of rigidity
• Bulk modulus

2. 2
Objective
On the completion of this class, you will
be able to solve
•Problems related to Simple Stress and
Strain in uniform cross section bars

3. 3
1. A steel rod 20mm diameter and 600mm long
is subjected to an axial pull of 40KN.
Determine the elongation of rod if E=2x10
5

N/mm
2
Given Data:
Diameter, d = 20mm
Length, l = 600mm
Load, P = 40KN = 40x1000N
Young's modulus,

E=2x10
5
N/mm
2

Simple problem 1

4. 4
Simple problem 1
Pl
l
AE

2
4
A d


2
20
4


3
5
40 10 600
314.16 2 10
l
 
 
 

Solution:
0.382mm
Elongation,
Area,
Elongation,
2
314.16mm

5. 5
1. A steel rod 24mm diameter and 500mm
long is
subjected to an axial pull of 30KN. Determine
the stress & elongation of rod if E=2x10
5

N/mm
2

Exercise problem 1

6. 6
2. A mild steel specimen is tested in tension and
the following results were found
Diameter of specimen : 20mm
Length of specimen : 200mm
Extension under load of 10KN : 0.032mm
Load at yield point : 82KN
Maximum load : 133KN
Length of specimen after fracture : 252mm
Diameter of neck : 12.6mm
Simple Problem 2

7. 7
Calculate:
(a) Young’s modulus
(b) Working stress if factor of safety is 2
(c) Percentage of elongation
(d) Percentage of reduction in area
(e) Stress at yield point
(f) Ultimate stress
Simple Problem 2 (Contd..)

8. 8
Given data:
Diameter of specimen : 20mm
Length of specimen : 200mm
Extension under load of 10KN : 0.032mm
Load at yield point : 82KN
Maximum load : 133KN
Length of specimen after fracture :
252mm
Diameter of neck : 12.6mm
Simple Problem 2 (Contd..)

9. 9
Solution:
E
e


Simple Problem 2 (Contd..)
P
A

2
4
A d


2
20
4


2
314.16mm
3
10 10
314.16



2
31.83 /N mm
Young'sModulus,
Stress,
Area,

10. 10
Simple Problem 2 (Contd..)
30.032
0.00016 0.16 10
200
e

   
3
31.83
0.16 10e


 

5 2
1.989 10 /N mm 
Strainat10kN ,load
'Young s ,ModulusE
Working ,Stress
Stress
Factor

at
of
yield
safety
Stressatyield
3
82 10
314.16


2
261 /N mm
WorkingStress
2261
130.5 /
2
N mm 

11. 11
Simple Problem 2 (Contd..)
%of ,elongation
( )
100
o
o
l l
l


(252 200)
100
200

 
26%
%
ofreductionin,area
( )
100
o
o
a a
a


(314.06 124.689)
100 60.31%
314.06

  

12. 12
Simple Problem 2 (Contd..)
Ultimatestress
max.load
area
3
133 10
314.16


2
423.35N/mm

13. 13
A mild steel specimen is tested in tension and the
following results were found
diameter of specimen : 20mm
length of specimen : 100mm
extension under load of 80KN : 0.125mm
load at yield point : 110KN
maximum load : 185KN
final elongation : 30mm
diameter of neck : 12.6mm

Exercise 2

14. 14
Calculate:
(a) Young’s modulus
(b) Working stress if factor of safety is 2
(c) Percentage of elongation
(d) Percentage of reduction in area
(e) Stress at yield point
(f) Ultimate stress
Exercise 2

15. 15
Summary
In this class, you have learnt solving
• Problems related to stress and
strain
• Problems related to stress-strain
diagram