Strain and Hooke's Law

This section provides practical examples and step-by-step solutions for problems involving Strain and Hooke's Law. It covers the calculation of normal and shear strain, the application of Hooke's Law in one and multiple dimensions, and the use of Poisson's ratio to determine transverse deformations.

Example 1: Basic Normal Strain

Problem: A steel rod has an original length of 2 m2 \text{ m}. When subjected to a tensile force, it elongates by 1.5 mm1.5 \text{ mm}. Calculate the normal strain in the rod.

1. Identify given values and goal
  • Original length, L=2 m=2000 mmL = 2 \text{ m} = 2000 \text{ mm}
  • Change in length, δ=1.5 mm\delta = 1.5 \text{ mm}
  • Goal: Find the normal strain, ϵ\epsilon
2. Apply the normal strain formula

The formula for normal strain is:

ϵ=δL\epsilon = \frac{\delta}{L}
3. Substitute the values and calculate
ϵ=1.5 mm2000 mm=0.00075 mm/mm\epsilon = \frac{1.5 \text{ mm}}{2000 \text{ mm}} = 0.00075 \text{ mm/mm}

Or, expressed in microstrain:

ϵ=750 μϵ\epsilon = 750 \text{ } \mu\epsilon

Example 2: Hooke's Law and Axial Deformation

Problem: An aluminum alloy rod (E=70 GPaE = 70 \text{ GPa}) has a diameter of 10 mm10 \text{ mm} and a length of 500 mm500 \text{ mm}. If it is subjected to a tensile force of 5 kN5 \text{ kN}, determine the normal stress and the total elongation of the rod.

1. Identify given values and goal
  • Modulus of Elasticity, E=70 GPa=70000 MPaE = 70 \text{ GPa} = 70000 \text{ MPa} (N/mm2\text{N/mm}^2)
  • Diameter, d=10 mmd = 10 \text{ mm}
  • Length, L=500 mmL = 500 \text{ mm}
  • Force, P=5 kN=5000 NP = 5 \text{ kN} = 5000 \text{ N}
  • Goal: Find normal stress σ\sigma and total elongation δ\delta
2. Calculate the cross-sectional area
A=π4d2=π4(10)2≈78.54 mm2A = \frac{\pi}{4} d^2 = \frac{\pi}{4} (10)^2 \approx 78.54 \text{ mm}^2
3. Calculate the normal stress
σ=PA=5000 N78.54 mm2≈63.66 MPa\sigma = \frac{P}{A} = \frac{5000 \text{ N}}{78.54 \text{ mm}^2} \approx 63.66 \text{ MPa}
4. Calculate the total elongation using Hooke's Law

First, find the strain using Hooke's Law (σ=Eϵ\sigma = E \epsilon):

ϵ=σE=63.66 MPa70000 MPa≈0.000909 mm/mm\epsilon = \frac{\sigma}{E} = \frac{63.66 \text{ MPa}}{70000 \text{ MPa}} \approx 0.000909 \text{ mm/mm}

Then, find the elongation (δ=ϵL\delta = \epsilon L):

δ=0.000909×500 mm≈0.455 mm\delta = 0.000909 \times 500 \text{ mm} \approx 0.455 \text{ mm}

Alternatively, use the combined formula δ=PLAE\delta = \frac{PL}{AE}:

δ=5000 N×500 mm78.54 mm2×70000 MPa≈0.455 mm\delta = \frac{5000 \text{ N} \times 500 \text{ mm}}{78.54 \text{ mm}^2 \times 70000 \text{ MPa}} \approx 0.455 \text{ mm}

Example 3: Poisson's Ratio

Problem: A standard structural steel rod (E=200 GPaE = 200 \text{ GPa}, ν=0.3\nu = 0.3) is 2 m2 \text{ m} long and has a diameter of 20 mm20 \text{ mm}. It is subjected to an axial tensile load of 60 kN60 \text{ kN}. Calculate the change in its length and the change in its diameter.

1. Identify given values and goal
  • Modulus of Elasticity, E=200 GPa=200000 MPaE = 200 \text{ GPa} = 200000 \text{ MPa}
  • Poisson's ratio, ν=0.3\nu = 0.3
  • Original length, L=2 m=2000 mmL = 2 \text{ m} = 2000 \text{ mm}
  • Original diameter, d=20 mmd = 20 \text{ mm}
  • Force, P=60 kN=60000 NP = 60 \text{ kN} = 60000 \text{ N}
  • Goal: Find change in length δL\delta_L and change in diameter δd\delta_d
2. Calculate the longitudinal strain and change in length

Cross-sectional area:

A=π4(20)2≈314.16 mm2A = \frac{\pi}{4} (20)^2 \approx 314.16 \text{ mm}^2

Normal stress:

σx=60000 N314.16 mm2≈190.99 MPa\sigma_x = \frac{60000 \text{ N}}{314.16 \text{ mm}^2} \approx 190.99 \text{ MPa}

Longitudinal strain:

ϵx=σxE=190.99200000≈0.000955\epsilon_x = \frac{\sigma_x}{E} = \frac{190.99}{200000} \approx 0.000955

Change in length:

δL=ϵxL=0.000955×2000=1.91 mm\delta_L = \epsilon_x L = 0.000955 \times 2000 = 1.91 \text{ mm}
3. Calculate the lateral strain and change in diameter

Using Poisson's ratio:

ν=−ϵyϵx  ⟹  ϵy=−νϵx=−0.3×0.000955≈−0.0002865\nu = -\frac{\epsilon_y}{\epsilon_x} \implies \epsilon_y = -\nu \epsilon_x = -0.3 \times 0.000955 \approx -0.0002865

Change in diameter:

δd=ϵyd=−0.0002865×20=−0.00573 mm\delta_d = \epsilon_y d = -0.0002865 \times 20 = -0.00573 \text{ mm}

The negative sign indicates a contraction in diameter.

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