Elasticity & Stress-Strain: Application Problems (1)

easy • 3 min read
Tags Elasticity And Stress Strain

Question

A steel wire of length 2 m2 \text{ m} and cross-sectional area 1 mm21 \text{ mm}^2 supports a load of 50 kg50 \text{ kg}. Young’s modulus of steel is Y=2×1011 N/m2Y = 2 \times 10^{11} \text{ N/m}^2. Find (a) the stress, (b) the strain, (c) the elongation of the wire, and (d) the elastic potential energy stored. Take g=10 m/s2g = 10 \text{ m/s}^2.

Solution — Step by Step

Force on the wire:

F=mg=50×10=500 NF = mg = 50 \times 10 = 500 \text{ N}

Cross-section: A=1 mm2=10−6 m2A = 1 \text{ mm}^2 = 10^{-6} \text{ m}^2.

σ=FA=50010−6=5×108 N/m2\sigma = \frac{F}{A} = \frac{500}{10^{-6}} = 5 \times 10^{8} \text{ N/m}^2

By definition Y=σ/ϵY = \sigma/\epsilon:

ϵ=σY=5×1082×1011=2.5×10−3\epsilon = \frac{\sigma}{Y} = \frac{5 \times 10^{8}}{2 \times 10^{11}} = 2.5 \times 10^{-3}

A strain of 0.25%0.25\% — tiny, as expected for steel.

ΔL=ϵ⋅L=2.5×10−3×2=5×10−3 m=5 mm\Delta L = \epsilon \cdot L = 2.5 \times 10^{-3} \times 2 = 5 \times 10^{-3} \text{ m} = 5 \text{ mm}

For a wire stretched within the elastic limit:

U=12×stress×strain×volumeU = \tfrac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume}

Volume: V=A⋅L=10−6×2=2×10−6 m3V = A \cdot L = 10^{-6} \times 2 = 2 \times 10^{-6} \text{ m}^3.

U=12×(5×108)×(2.5×10−3)×(2×10−6)=1.25 JU = \tfrac{1}{2} \times (5 \times 10^{8}) \times (2.5 \times 10^{-3}) \times (2 \times 10^{-6}) = 1.25 \text{ J}

Why This Works

Young’s modulus is the wire’s stiffness rating: how much stress per unit strain. Once YY is known, the four quantities — stress, strain, elongation, energy — are all linked by σ=Yϵ\sigma = Y\epsilon and the geometry ϵ=ΔL/L\epsilon = \Delta L/L.

The energy formula U=12σϵVU = \tfrac{1}{2}\sigma\epsilon V is the elastic analogue of 12kx2\tfrac{1}{2}kx^2 for a spring: integrate force times displacement up to the final stretch. The factor of 12\tfrac{1}{2} comes from the linear ramp-up of force during stretching.

Alternative Method

Treat the wire as a spring with effective stiffness k=YA/Lk = YA/L:

k=(2×1011)(10−6)2=105 N/mk = \frac{(2 \times 10^{11})(10^{-6})}{2} = 10^{5} \text{ N/m}

Then ΔL=F/k=500/105=5×10−3 m\Delta L = F/k = 500/10^5 = 5 \times 10^{-3} \text{ m}, and U=12k(ΔL)2=12×105×(5×10−3)2=1.25 JU = \tfrac{1}{2}k(\Delta L)^2 = \tfrac{1}{2} \times 10^5 \times (5\times 10^{-3})^2 = 1.25 \text{ J}. Identical answer.

Memorise three forms for elastic PE: U=12FΔL=12k(ΔL)2=12σϵVU = \tfrac{1}{2}F\Delta L = \tfrac{1}{2}k(\Delta L)^2 = \tfrac{1}{2}\sigma\epsilon V. Pick whichever matches the data given. JEE often gives volume + stress + strain (last form) or force + extension (first form).

Common Mistake

Unit slips dominate this topic. Cross-section in mm2\text{mm}^2 versus m2\text{m}^2 differs by a factor of 10610^6. Length in cm vs m differs by 100100. Always convert to SI base units before the first calculation — never carry mixed units through the algebra.

The other common trap: students multiply stress by strain (without the 12\tfrac{1}{2}) for energy density, giving twice the right answer. Energy density of an elastically stretched solid is 12σϵ\tfrac{1}{2}\sigma\epsilon, just like the spring.

Final answer: σ=5×108 Pa\sigma = 5 \times 10^8 \text{ Pa}, ϵ=2.5×10−3\epsilon = 2.5 \times 10^{-3}, ΔL=5 mm\Delta L = 5 \text{ mm}, U=1.25 JU = 1.25 \text{ J}.

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