STRAIN ENERGY AND IMPACT LOADING FREE NOTES

1. What is Strain Energy ?

Strain energy is the energy stored in a material or body when it is deformed by an external force, provided the deformation is within the elastic limit. When the force is removed, this stored energy is released, allowing the material to return to its original shape.

When you stretch a spring or a rubber band, work is done on it. This work is stored as strain energy. Once released, the spring or rubber band returns to its original shape by releasing the stored energy.

In simple words:

Strain energy is the work done by external forces in deforming a body, which is stored as internal energy.

Unit of strain energy is Joule (J)

2. Basic Formula

For a gradually applied load: $U = \frac{1}{2} P \delta$ Where

U = Strain energy (Joule)
P = Applied load (N)
δ = Deformation or extension (m)

3. Strain Energy in Terms of Stress and Strain

$U = \frac{\sigma^2}{2E} \times V$ Where

σ = Stress
E = Young’s Modulus
V = Volume of material

Strain Energy per Unit Volume (Strain Energy Density)

$u = \frac{\sigma^2}{2E}$ This represents energy stored per unit volume of material.

Strain Energy in a Bar

For a bar subjected to axial load: $U = \frac{P^2 L}{2AE}$ Where

$L$ = Length of bar
$A$ = Cross-sectional area
$E$ = Young’s modulus

4. Strain Energy in Different Types of Loading

Type of Loading	Strain Energy Formula
Axial loading	$U = \frac{P^2L}{2AE}$ U=2AEP2L
Bending	$U = \int \frac{M^2}{2EI} dx$ U=∫2EIM2dx
Torsion	$U = \frac{T^2L}{2GJ}$ U=2GJT2L

Where

$M$ = Bending moment
$T$ = Torque
$I$ = Moment of inertia
$G$ = Shear modulus
$J$ = Polar moment of inertia

5. Applications of Strain Energy

Design of springs
Shock absorbing systems
Impact load calculations
Structural analysis
Energy methods like Castigliano’s theorem

6. Impact Loading

Impact loading is a type of loading in which a force is applied suddenly or due to a collision, causing a much higher stress than the same load applied gradually.

Unlike static loading, where the load is applied slowly, impact loading produces dynamic effects because the body experiences rapid acceleration and deceleration.

Examples:

Hammer striking metal
Drop test in structures
Railway wheel hitting rail joints
Punching and forging operations

Case 1: Suddenly Applied Load

If a load P is suddenly applied to a bar:

Maximum stress produced is: $\sigma_{max} = 2\sigma$ This means stress becomes twice the static stress.

Maximum deflection: $\delta_{max} = 2\delta$ Thus, sudden loading causes double deformation compared to gradual loading.

Case 2: Falling Load (Impact of Falling Weight)

Consider a weight W falling from height h on a bar.

Energy balance principle:

Potential Energy = Strain Energy $W(h + \delta) = \frac{\sigma^2}{2E} \times V$ From this relation, the maximum stress in the bar can be calculated.

Maximum Stress Formula

$\sigma_{max} = \frac{W}{A} \left(1 + \sqrt{1 + \frac{2AEh}{WL}} \right)$ Where

$W$ = Falling load
$h$ = Height of fall
$A$ = Area of bar
$L$ = Length of bar
$E$ = Young’s modulus

7. Comparison of Loading Types

Loading Type	Stress Produced
Gradual load	$\sigma = P/A$
Suddenly applied load	$2\sigma$
Impact load	Much greater than $2\sigma$

8. Practical Engineering Examples

Drop forging
Pile driving
Railway track loading
Machine hammering
Crash and collision design

9. Numerical Problems

Problem

A steel bar has the following dimensions:

Length, $L = 2 \, \text{m}$ L=2m
Cross-sectional area, $A = 600 \, \text{mm}^2$ A=600mm2
Young’s modulus, $E = 200 \, \text{GPa}$ E=200GPa
Axial tensile load, $P = 60 \, \text{kN}$ P=60kN

Find:

Extension of the bar
Strain energy stored in the bar

Solution

$L = 2000$ L=2000 mm
$A = 600$ A=600 mm²
$E = 200000$ E=200000 N/mm²
$P = 60000$ P=60000 N

Step 1: Calculate Extension

$\delta=\frac{PL}{AE}$ $=\frac{60000\times2000}{600\times200000}$ $=\frac{120000000}{120000000} =1\text{ mm}$

Extension = 1 mm

Step 2: Calculate Strain Energy

$U=\frac{1}{2}P\delta$ $=\frac{1}{2}\times60000\times1 =30000\text{ N-mm}$

Convert to Joules $30000\text{ N-mm}=30\text{ J}$

Answer

Extension = 1 mm
Strain Energy = 30 J

Frequently Asked Questions (FAQ) on Strain Energy and Impact Loading

1. What is resilience?

Resilience is the ability of a material to absorb and store energy within its elastic limit and recover its original shape after the load is removed.

2. What is proof resilience?

Proof resilience is the maximum strain energy that a material can store without undergoing permanent deformation.

3. What is modulus of resilience?

Modulus of resilience is the strain energy stored per unit volume of a material up to the elastic limit. $\text{Modulus of Resilience}=\frac{\sigma_y^2}{2E}$ where:

$\sigma_y$ = Yield stress
$E$ = Young’s modulus

4. What is impact loading?

Impact loading is a dynamic load that acts suddenly due to a falling object or collision, producing higher stresses than a gradually applied load.

5. Why is impact loading more dangerous than static loading?

Impact loading creates high stresses because the load is applied suddenly, generating dynamic effects and higher strain energy in the material.

6. What is the impact stress formula?

$\sigma=\frac{W}{A}\left[1+\sqrt{1+\frac{2AEh}{WL}}\right]$ where:

$W$ = Falling load
$h$ = Height of fall
$A$ = Cross-sectional area
$E$ = Young’s modulus
$L$ = Length of the member

7. What happens if the load is applied suddenly without any fall?

When the height of fall is zero ( $h = 0$ h=0), the stress becomes: $\sigma=\frac{2W}{A}$ This is twice the stress produced by the same load applied gradually.

8. What is the stress due to a gradually applied load?

$\sigma=\frac{W}{A}$ This is the basic static stress formula.

9. What is the principle used in impact loading problems?

Impact loading problems are solved using the principle of conservation of energy, where the potential energy of the falling weight equals the strain energy stored in the member.

10. Which factors affect impact stress?

Impact stress depends on:

Weight of the falling object
Height of fall
Cross-sectional area
Length of the member
Young’s modulus of the material

11. What is the difference between gradual, sudden, and impact loading?

Loading Type	Load Application	Stress Produced
Gradual Loading	Applied slowly	Lowest
Sudden Loading	Applied instantly without fall	About twice the gradual stress
Impact Loading	Falling load or collision	Highest

12. Why is Young’s modulus important in impact loading?

Young’s modulus determines the stiffness of a material. A stiffer material (higher $E$ E) deforms less under load and influences the stress developed during impact.

13. Can strain energy exist during plastic deformation?

Strain energy is mainly recoverable only during elastic deformation. During plastic deformation, part of the applied energy is permanently dissipated and cannot be fully recovered.