Impact
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MEC 3351 Impact Lecture
1
Resilience – Strain energy stored in a specimen when strained within elastic
limit.
Proof Resilience – The maximum energy stored at the elastic limit.
Modulus of Resilience – Proof resilience per unit volume.
Stress/strain
developed
Magnitude of load
applied
Manner of load
application
Gradual
Sudden
Impact or
Shock
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MEC 3351 Impact Lecture
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Gradually Applied Loads
Let a vertical bar of length l and x-sectional area A be rigidly held at one end
and carry a gradually applied load P.
If
Resultant extension – l
l
Resultant stress in the bar - σ
Therefore axial load P
P  A
l
Now energy stored in the bar is
P
U  Work done by gradually applied axial load P
 * Average load x elongation
P
1
Pl
 l  A
2
2
AE
1
l  2
 A 
Al
2
E 2E
08:34
MEC 3351 Impact Lecture
Note
* Load is gradually increased
from 0 to P, hence average.
3
Or
U
2
2E
x Volume of bar
... (1)
If the value of stress σ at the elastic limit is σe then proof resilience is
UP 
 e2
2E
x Volume of bar
Modulus of resilience

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 e2
2E
... (2)
MEC 3351 Impact Lecture
4
Sudden Applied Loads
Let
Instantaneous Elongation – l
Instantaneous Stress
–σ
Then equating the strain energy in the bar to the work done by the applied
load we get:
U  Work done by load

or
 
* * A l  Pl
2 
2P

A
** Stress increases with
increase in stretch from 0 to
l , hence average stress
Instantaneous stress developed in a bar subjected to suddenly applied load is
thus twice the stress produced by the same load applied gradually.
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MEC 3351 Impact Lecture
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Therefore, Instantaneous elongation:
l 

E
l
2 Pl
AE
Instantaneous elongation produced in a bar subjected to sudden applied load
is thus twice that produced by the load applied gradually.
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MEC 3351 Impact Lecture
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Stresses due to Impact Loading
Let
W – Dropping Weight
A – Bar cross-sectional area
l – Maximum elongation of bar
σ – Corresponding stress
P – Equivalent static or gradually applied load
which could produce same elongation l.
W
l
h
l
Pl
Strain energy in the bar at this instant 
2
Also work done by the weight  W h  l 
But work done by the weight = Strain energy stored in the bar
therefore
Pl
W h  l  
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MEC 3351 Impact Lecture
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But
l 
Therefore
Pl
AE
Pl  P Pl

Wh 

AE  2 AE

WPl
P 2l
Wh 

AE 2 AE
or
P 2  2WP 
Therefore
2 AEWh
0
l
2W  4W 2 
P
8 AEWh
l
2
(neglectin g negative value)
Therefore
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
 2 AEh  
P  W 1  1 

Wl

 

... (4)
MEC 3351 Impact Lecture
8
If, however the value of l is very small as compared to h, then
Pl P Pl
P 2l
Wh 


2
2 AE 2 AE
Therefore
P2 
2WAEh
l
or
P 2 2WEh

2
A
Al
or
P
2WEh
 
A
Al
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... (5)
MEC 3351 Impact Lecture
9