Strain energy due to sudden load

1. Subject Name: Theory of Structures
Topic Name:Strain Energy Due Sudden
Load
Lecture No: 30
Dr.Omprakash Netula
Professor & HOD
Department of Civil Engineering
9/28/2017 Lecture Number, Unit Number 1

2. ∴(σ −
P
A
)2 =
2EPh
Al
+
p2
A2
∴(σ −
P
A
) =
2EPh
Al
+
p2
A2
… Stresses due to impact load
∴ σ =
P
A
+
2EPh
Al
+
p2
A2
If load is applied suddenly, h = 0
∴ σ =
P
A
+
p2
A2
+ 0
∴ σ =
2P
A

3.  When 𝛿𝑙 is very small as compered to h , then
Work done = P × h
σ2
2E
× A × l = P ×h
𝜎2
=
2𝐸𝑃ℎ
𝐴𝑙
𝜎 =
2𝐸𝑃ℎ
𝐴𝑙

4. Let, Work done on the bar by shock = u
Work stored in the bar =
... max instantaneous
stress
Strain Energy due to shock loading :
1
2
× R × δl
=
1
2
× σ × A ×
σl
E
=
σ2
2E
× A × l
∴ U =
σ2
2E
× A × l
∴ σ2 =
2UE
Al
∴ σ =
2UE
Al

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,
𝐮 =
𝛕 𝟐
𝟐𝐆
× 𝐕
Where, t = shear stress
G = Modulus of rigidity

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
tan ∅ =
BB1
l
∅ is very small
∴ tan ∅ = ∅
∴ ∅ =
BB1
l
….. Shear Strain

7. Force P on face BC
P = τ × BC × l
When P in applied gradually In
case of gradual load.
u =
P
2
× BB1 average
force
=
1
2
× (τ × BC × l ) × BB1 =
0+P
2
=
P
2
=
1
2
× (τ × BC × l ) × ∅ ∙ lBB1 = ∅ ∙ l
=
1
2
× (τ × A) ∙
τ
G
∙ l G =
τ
∅
=
1
× τ2
× A × l BC × l= A
u =
τ2
2G
× V
 The elastic energy stored due to shear loading is known as shear
resilience