Strain Energy-Definition and Related Formulas, Strain Energy due to Shear Loading, Strain Energy due to Bending, Strain Energy due to Torsion and Examples

1. STUDY OF STRAIN
ENERGY DUE TO SHEAR,
BENDING AND TORSION
SHANTILAL SHAH ENGINEERING
COLLEGE, BHAVNAGAR
Affiliated to
Gujarat Technological University,
Gandhinagar
CIVILENGINEERINGDEPARTMENT
SUB: STRUCTURAL ANALYSIS- 1
SUB CODE: 2140306
PREPARED BY:
SINGHANIA JAY SANJAY
(170433106034)(1131)(D2D)
VASAVA ASHISHKUMAR
RAMSINGBHAI
(170433106035)(1132)(D2D)
VYAS NAMAN
PRAKASHKUMAR
(170433106036)(1133)(D2D)
ZAPADIYA YOGESHKUMAR
RANCHODBHAI
(170433106037)(1134)(D2D)
GUIDED BY :
PROF. K.A.MEHTA
PROF. D.P. ADVANI
GROUP NO. = 30

2. Strain Energy
What is Strain Energy?
 When a body is subjected to gradual, sudden
or impact load, the body deforms and work is
done upon it. If the elastic limit is not
exceed, this work is stored in the body. This
work done or energy stored in the body is
called strain energy.
 Energy is stored in the body during
deformation process and this energy is called
“Strain Energy”.
Strain energy = Work done

3. Resilience :
Total strain energy stored in a body is called resilience.
Where σ = stress
V = volume of the body
Proof Resilience :
Maximum strain energy which can be stored in a body is
called proof resilience.
Where σE = stress at Elastic
Limit
2
*
2
u V
E


2
( )
*
2
E
pu V
E



4. Modulus of Resilience :
Maximum strain energy which can be stored in a body
per unit volume, at elastic limit is called modulus of
resilience.
2
( )
2
E
mu
E



5. Strain Energy Due to Shear Loading
If t is the uniform shear stress produce in the material by
external forces applied within elastic limit, the energy
stored due to Shear Loading is given by,
Where, τ = shear stress
G = Modulus
of
Rigidity
2
*
2
u V
G



6.  Consider a square block ABCD of length l , faces BC and AD are
subjected to shear stress τ , Let face AD is fixed.
 The section ABCD will deform to AB1C1D through the angle ∅.
∅ = Shear strain
∅ is very small
∴ tan ∅ = ∅
….. Shear Strain
1
tan
BB
l
 
1BB
l
 

7.  Force P on Face BC, P=τ * BC * l
 When P is applied gradually, Gradual Load is:
In Case of Average Force
 The elastic energy stored due to shear loading is known as Shear
Resilience.
1
1
2
*
2
1
*( * * )*
2
1
*( * * )* .
2
1
*( * )* .
2
1
* * *
2
P
u BB
u BC l BB
u BC l l
u A l
G
u A l
G

 








1
0
2 2
.
*
P P
BB l
G
BC l A




 



2
*
2
u V
G



8. Strain Energy Due To Bending (Flexure)
 Consider two transverse section ‘1-1’ and ‘2-2’ of a beam, distance
‘dx’ apart as shown in fig.
 Consider a small strip of area ‘da’ at distance ‘y’ from the neutral
axis.
 B.M. in small portion
‘dx’ will be constant.
*
M
I y
M
y
I



 

9.  Strain energy stored in small strip of area da.
 Strain Energy stored in entire section of a beam.
2
2
2
2 2
2
*
2
*( . )
2
1
*( . ) *( . )
2
1
*( )*( . )
2
u V
E
u da dx
E
M
u y da dx
E I
M y
u da dx
E I






2 2
2
1 .
( )*( . )
2
c
t
y y
total
y y
M y
u da dx
E I


 

10.  Now, for Strain Energy in Entire Beam, integrate between
limits 0 to l.
…Strain Energy due to Bending.
2
2
2
2
2
2
1 .
. . .
2
1 .
. .
2
.
2
c
t
y
total y
total
total
M dx
u y da
E I
M dx
u I
E I
M
u dx
EI




2
2
0
.
l
M
EIu dx  

11. Strain Energy due to Torsion
 We have seen that, when a member is subjected to a uniform shear
stress 𝛕, the strain energy stored in the member is:
 Consider a small
elemental ring of
thickness ‘dr’, at
radius ‘r’.
2
*
2
V
G


12.  Strain energy due to torsion for uniform shear stress, is the
...Strain Energy for One Ring
2
2
2 2
2
2 2
2
2 3
2
*
2
( )
*
2
.
*
.2
.
*(2 * )*
.2
.
* . .
.
r
r
r
r
r
r
u V
G
r
Ru V
G
r
u V
R G
r
u r dr l
R G
r
u l dr
R G













13.  Total strain energy for whole section, is obtained by
integrating over a range from r = 0 to r = D/2 for a solid shaft.
2 2 3
2
0
22
3
2
0
22 4
2
0
4
2
2
.
. . .
.
. .
.
.
. .
.
. 4
( )
. . 2.
. 4
D
D
D
r
u l dr
R G
l
u r dr
R G
l r
u
R G
D
l
u
R G


 
 
 


 
  
 
 
 
  
 
 



14. …Strain Energy due to Torsion
2 4
2
2 4
2
2 2
2 2
2
. .
.
. 64
. .
.
64
.
4
. . .
16
. . .
4* 4*
. .
4
l D
u
R G
l D
u
D
G
l D
u
G
l D
u
G
l A
u
G
 
 
 
 






2
*
4
u V
G



15. Example 1:
 An axial pull of 50 kN is suddenly applied to a steel bar 2m
long and 1000 mm2 in cross section. If modulus of elasticity of
steel is 200 kN/mm2.
Find, (i) Maximum Instantaneous Stress
(ii) Maximum Instantaneous Extension
(iii) Strain Energy
(iv) Modulus of Resilience.
 Here, P = 50 kN (Sudden load)
A = 1000 mm2
l = 2m = 2000 mm
E = 200 kN/mm2 = 200× 103 N/mm2

16. i. Maximum Instantaneous Stress:
ii. Maximum Instantaneous Extension:
3
22 2*50*10
100 /
1000
P
N mm
A
   
4
3
4
100
5*10
200*10
.
5*10 *2000
1
E
E
l
l
l l
l
l mm






 





  





17. iii. Strain Energy (u):
iv. Modulus of Resilience (um):
2
2
3
*
2
100
*(1000*2000)
2*200*10
50000 .
u V
E
u
u N mm




2
2
3
3
2
100
2*200*10
0.025 .
m
m
m
u
E
u
u N mm mm



