3. INTRODUCTION
ο±Energy methods are widely used to
obtain solutions to elasticity problems
and determine deflection of structures.
3
β’ Like deflections of joint on a truss or
points on a beam or shaft.
ο±In energy method, strain energy is
associated with it. Hence it is also
known as Strain energy method.
4. P
ο± Strain energy is the energy stored in the material due to
deformation under external load.
STRAIN ENERGY
10. 1 0
STRAIN
ENERGY
CALCULATION
UNDER AXIAL LOADING
Strain Energy [ U ] = Γ βl Γ P
Since , βl =
Strain Energy [ U ] = Γ Γ P
Strain Energy [ U ] =
P = Force applied
l = Length of the
bar
E = Young's
Modulus
A = Area of the
bar
Where ,
J or N-m
13. 1 3
STRAIN
ENERGY
CALCULATION
UNDER TORSION
Strain Energy [ U ] = Γ ΞΈ Γ Ξ€
Since , ΞΈ =
Strain Energy [ U ] = Γ Γ Ξ€
Strain Energy [ U ] = J or N-m
Ξ€ = Applied Torque
l = Length of the
bar
G = Modulus of
Rigidity
J = Polar moment of
inertia
Where ,
17. 1 7
STRAIN
ENERGY
CALCULATION
UNDER BENDING
Strain Energy [ U ] = Γ
π
π
Γ M
Since ,
π
πΌ
=
πΈ
π
Strain Energy [ U ] = Γ Γ M
Strain Energy [ U ] =
Since , π = π π
ππ
πΈπΌ
π2
π
2πΈπΌ
This Equation is valid only for β PURE BENDING
J or N-m
18. 1 8
STRAIN
ENERGY
CALCULATION
UNDER BENDING
dx
Suppose a bending stress W is acting on the bar.
Let , on the dx portion bending moment is M.
Strain Energy [ dU ] =
π2
ππ₯
2πΈπΌ
l
W
Strain Energy [ U ] =
0
π
π2ππ₯
2πΈπΌ
For the whole bar
19. 1 9
EXERCISE
3KN
5KN
2 m 1 m
Find the strain energy (in N-m ) stored in the beam.
EI = πππ
.
x
O B
A
ππ΄π΅ = β3π₯
Moment for the section AB will be __
Where x = 0 to 1
Similarly , Moment for the section OA will be __
πππ΄= β3(π₯ + 1) β 5π₯
πππ΄= β8π₯ β 3
Where x = 0 to 2
x
20. 2 0
EXERCISE
3KN
5KN
2 m 1 m
Find the strain energy (in N-m ) stored in the beam.
EI = πππ
.
x
O B
A
x
Here the total strain energy U will be ___
π = πππ΄+ ππ΄π΅
π =
πππ΄
2
ππ₯
2πΈπΌ
+
ππ΄π΅
2
ππ₯
2πΈπΌ
π =
0
2
(β8π₯ β 3)2 ππ₯
2πΈπΌ
+
0
1
(β3π₯)2 ππ₯
2πΈπΌ
21. 2 1
EXERCISE
3KN
5KN
2 m 1 m
Find the strain energy (in N-m ) stored in the beam.
EI = πππ
.
x
O B
A
x
π =
1
2πΈπΌ
[ 0
2
64π₯2 + 9 + 48π₯ ππ₯ + 0
1
9π₯2 dx ]
π =
1
2Γ104[
64Γ8
3
+ 9 Γ 2 + (48 Γ 2)]
π = 14.38 π β π