Q1. A load of 4000 N is applied to a wire of area 2 mm². Its length changes from 2 m by 1 mm. Find Young's modulus.
Stress = F/A = 4000 / (2×10⁻⁶) = 2×10⁹ Pa
Strain = ΔL/L = 1×10⁻³ / 2 = 5×10⁻⁴
Y = stress/strain = 2×10⁹ / 5×10⁻⁴ = 4×10¹² Pa
Why does a wire stretch and a bridge sag? Learn stress, strain and elasticity, and compute stress and Young's modulus.
Enter the force and cross-sectional area to get the stress (in MPa, where 1 N/mm² = 1 MPa).
Beyond the elastic limit a material no longer returns to its original shape — it deforms permanently.
Stress, strain and Young's-modulus problems with steps.
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Q1. A load of 4000 N is applied to a wire of area 2 mm². Its length changes from 2 m by 1 mm. Find Young's modulus.
Stress = F/A = 4000 / (2×10⁻⁶) = 2×10⁹ Pa
Strain = ΔL/L = 1×10⁻³ / 2 = 5×10⁻⁴
Y = stress/strain = 2×10⁹ / 5×10⁻⁴ = 4×10¹² Pa
Q2. Why is steel more elastic than rubber?
For the same stress, steel produces a much smaller strain than rubber.
So steel has a much larger Young's modulus.
A larger modulus means greater elasticity — steel returns to shape more strongly.
| Quantity | Formula | SI unit |
|---|---|---|
| Stress | σ = F / A | pascal (Pa) |
| Strain | ε = ΔL / L | — (no unit) |
| Young's modulus | Y = (F/A) / (ΔL/L) | Pa |
| Elastic potential energy | U = ½ × stress × strain × volume | J |
Stress is the internal restoring force per unit area (Pa); strain is the fractional deformation produced (no unit).
The ratio of longitudinal stress to longitudinal strain, Y = (F/A)/(ΔL/L), a measure of a material's stiffness.
Within the elastic limit, stress is directly proportional to strain.
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