Definition of strain Energy and Modulus Of Resilience. Development Of Strain Energy Formulae. Computation of the Strain Energy and Modulus of Resilience Of Engineering Materials.

1. STRAIN ENERGY
BY
L.E EZE

2. Strength of Material
Module 1: Strain Energy
Lecture: 03
Topic: Strain Energy and Modulus Of Resilience
Learning Outcomes
 Definition of strain Energy and Modulus Of Resilience
 Development Of Strain Energy Formulae
 Determine the Strain Energy and Modulus of Resilience Of
Engineering Materials

3. STRAIN ENERGY(U)
Strain energy is the internal energy which is stored in any
material that is loaded within its elastic limit. It is also the
work done by a loaded material.
it is also called RESILIENCE.
Strain Energy (U) =
δ2
2ε
. 𝑽
Resilience is the energy stored within Elastic limit. It is given by
‘U’
Mathematically,

4. Development of strain energy formula
Consider the bar rod below;
Axially loaded member
Stress stored in the bar δ =
p
𝐴
Strain(ε )=
extension(𝑒)
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ(𝐿)
Young modulus(E)=
𝑠𝑡𝑟𝑒𝑠𝑠 (δ)
Strain(ε)
Young modulus(E) =
δ
extension(𝑒)
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔𝑡ℎ(𝐿)
Re-arranging equation 1;
(E) =
δ
𝑒
×L 1
e =
δ
𝐸
×L

5. But,
Hence;
Stress stored in the bar δ =
p
𝐴
If we plot a graph of load against extension for the
axially loaded rod, we shall obtain a straight line
graph.
e =
PL
𝐴𝐸 2

6. Area of the graph = Energy stored
Area of shaded portion =
1
2
𝑃 × 𝑒
Area =
1
2
× δA .
PL
𝐴𝐸
Area =
1
2
× δA . δ.
L
𝐸
.
But, δ =
p
𝐴
Recall:
P = δA and e =
PL
𝐴𝐸
Area of the shaded portion =
Area of a triangle
Hence;
Area
Load
extension

7. Area of the graph = Energy stored
Alternatively, Substitute e =
PL
𝐴𝐸
To obtain;
Therefore;
Area Of graph =
1
2
×
δ2A .L
𝐸
.
A × L = Volume
Area Of graph =
1
2
×
δ2
𝐸
.V
Energy stored = Strain Energy(U)
Area Of graph =
1
2
×
P2L
𝐴𝐸
.
Strain Energy(U) =
1
2
δ2
𝐸
.V 3
Strain Energy(U) =
1
2
P2L
𝐴𝐸
. 4

8. Proof Resilience (Umax) =
1
2
δ2
max
𝐸
.V
mathematically;
Proof Resilience: it is the
maximum strain energy stored
in a material. Proof resilience
simply means (maximum
strain-energy).
Modulus of Resilience =
1
2
δ2
𝐸
.V
𝑉
Modulus of resilience is
calculated as area under the
stress- strain graph
Modulus of Resilience: it is
defined as the strain energy
per unit volume.
Prove Of M.O.D :
M.O.D =
δ2
2E
5

9. Area of the graph =
1
2
δ×ε
Area =
1
2
×
e
L
×
𝑃
𝐴
Area of the graph = M.O.R
1
2
×P.e = Area under the load-
extension graph = Strain
Energy
ε =
e
L
δ =
𝑃
𝐴
But:
stress
Area
strain
A× L = Volume
M.O.R =
1
2
×
𝑃𝑒
𝐴𝐿
6

10. Hence;
Area of bar = 500mm2 =
0.0005m
Length of bar = 600mm = 0.6m
P = 50kN =50,000N
E = 200GN = 200×109 N/m2
Solution
Invoking equation 2
Example 1:
A prismatic bar of cross –
section 500mm2 length 600mm
is acted upon by a 50kN force.
Determine;
 The total elongation of the
bar
 The strain Energy
Consider E= 200GN/m2
M.O.R =
strain Energy
volume
7 600mm
50kN
50kN
e =
PL
𝐴𝐸 2

11. e =
50,000×0.6m
0.0005×200×109
U =
1
2
50,0002×0.6
0.0005× 200×109
U = 7.5J
e =
30,000
100,000,000
e = 0.0003m
Strain Energy(U) =
1
2
P2L
𝐴𝐸
Applying equation 4
ANSWERS
Total elongation bar =
0.0003m
Strain Energy in the bar = 7.5J

12. Area of bar = 0.25m × 0.45m
Length of bar = 3.5m
P = 2000N
E = 3MN/m2 = 3×106 N/m2
Solution
Example 2:
A prismatic bar 3.5m long is
subjected to axial compressive
force 2000N. If the cross-
sectional area of the bar is
0.25m × 0.45m, and modulus
of Elasticity is 3MN/m2
calculate;
 The strain Energy in the bar
 The modulus of Resilience
3.5m
2000N
2000N
Invoking equation 4
4
Strain Energy(U) =
1
2
P2L
𝐴𝐸

13. U =
1
2
20002×3.5
0.25m × 0.45m ×3×106
U = 20.741J
ANSWERS
Applying equation 7
Strain Energy in bar = 20.74J
Modulus Of Resilience = 52.67N/m2
M.O.R =
strain Energy
volume
Strain energy = 20.741J
Volume = area × length of bar
Volume = 3.5 × 0.25 × 0.45
Volume = 0.39375m2
M.O.R =
20.741J
0.39375m2
M.O.R = 52.67N/m2

14. Area of bar = 4m2
Length of bar = 10m
P = 4000N
E = 3000N/m2
Solution
Invoking equation 4
Example 3:
A prismatic bar 10m long is
subjected to axial compressive
forces P = 4000N. Compute
the amount of U stored in the
bar if the cross-sectional is
4m2, and M.O.E is 3,000N/m2
10m
4000N
4000N
4
Strain Energy(U) =
1
2
P2L
𝐴𝐸

15. U =
1
2
40002×10
4×3000
U = 6666.67J
ANSWERS
Strain Energy in bar = 20.74J

16. Reference Books
1. Robert L .Mott & Joseph A Untener –Applied Strength Of
Material
2. R.k. Bansal – Strength Of Material
3. R.k Rajput – Strength of material

17. L.E Eze
Teacher Scientist
Sismo Academy
ezelotachukwu91@gmail.com
Thank you