STRAIN ENERGY AND IMPACT LOADING

For a gradually applied load:U=12PฮดU = \frac{1}{2} P \deltaWhere

  • UUU = Strain energy (Joule)
  • PPP = Applied load (N)
  • ฮด\deltaฮด = Deformation or extension (m)

Strain Energy in Terms of Stress and Strain

U=ฯƒ22Eร—VU = \frac{\sigma^2}{2E} \times VWhere

  • ฯƒ\sigmaฯƒ = Stress
  • EEE = Youngโ€™s Modulus
  • VVV = Volume of material

Strain Energy per Unit Volume (Strain Energy Density)

u=ฯƒ22Eu = \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=P2L2AEU = \frac{P^2 L}{2AE}Where

  • LLL = Length of bar
  • AAA = Cross-sectional area
  • EEE = Young’s modulus

Strain Energy in Different Types of Loading

Type of LoadingStrain Energy Formula
Axial loadingU=P2L2AEU = \frac{P^2L}{2AE}U=2AEP2Lโ€‹
BendingU=โˆซM22EIdxU = \int \frac{M^2}{2EI} dxU=โˆซ2EIM2โ€‹dx
TorsionU=T2L2GJU = \frac{T^2L}{2GJ}U=2GJT2Lโ€‹

Where

  • MMM = Bending moment
  • TTT = Torque
  • III = Moment of inertia
  • GGG = Shear modulus
  • JJJ = Polar moment of inertia

Applications of Strain Energy

  1. Design of springs
  2. Shock absorbing systems
  3. Impact load calculations
  4. Structural analysis
  5. Energy methods like Castiglianoโ€™s theorem

2. Impact Loading

Definition

Impact loading occurs when a load is applied suddenly or by collision, such as a falling weight striking a structure.

Unlike gradual loading, impact loading produces much higher stress because energy is transferred quickly.

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:ฯƒmax=2ฯƒ\sigma_{max} = 2\sigmaThis means stress becomes twice the static stress.

Maximum deflection:ฮดmax=2ฮด\delta_{max} = 2\deltaThus, 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 EnergyW(h+ฮด)=ฯƒ22Eร—VW(h + \delta) = \frac{\sigma^2}{2E} \times VFrom this relation, the maximum stress in the bar can be calculated.

Maximum Stress Formula

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

  • WWW = Falling load
  • hhh = Height of fall
  • AAA = Area of bar
  • LLL = Length of bar
  • EEE = Youngโ€™s modulus

Comparison of Loading Types

Loading TypeStress Produced
Gradual loadฯƒ=P/A\sigma = P/Aฯƒ=P/A
Suddenly applied load2ฯƒ2\sigma2ฯƒ
Impact loadMuch greater than 2ฯƒ2\sigma2ฯƒ

Practical Engineering Examples

  1. Drop forging
  2. Pile driving
  3. Railway track loading
  4. Machine hammering
  5. Crash and collision design

Leave a Reply

Your email address will not be published. Required fields are marked *