Elasticity & Stress-Strain: Edge Cases and Subtle Traps (3)

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Tags Elasticity And Stress Strain

Question

A uniform steel wire of length L=2L = 2 m, cross-sectional area A=2×10−6A = 2 \times 10^{-6} m² and density ρ=7800\rho = 7800 kg/m³ hangs vertically from a rigid support, with a mass M=100M = 100 kg attached at the bottom. Young’s modulus of steel Y=2×1011Y = 2 \times 10^{11} N/m². Find (a) the elongation of the wire ignoring its weight, (b) the additional elongation due to the wire’s own weight and (c) the total elongation. Take g=10g = 10 m/s².

Solution — Step by Step

For uniform tension, ΔL=FLAY\Delta L = \frac{F L}{A Y}.

ΔL1=Mg⋅LAY=100×10×22×10−6×2×1011=20004×105=5×10−3 m=5 mm\Delta L_1 = \frac{Mg \cdot L}{A Y} = \frac{100 \times 10 \times 2}{2 \times 10^{-6} \times 2 \times 10^{11}} = \frac{2000}{4 \times 10^5} = 5 \times 10^{-3} \text{ m} = 5 \text{ mm}

The tension in the wire varies along its length — at the top, the wire supports its full weight plus MM; at the bottom, only MM. So the strain isn’t uniform. We integrate.

Consider an element dxdx at distance xx from the bottom. Tension at this point due to wire below it is T(x)=(ρAx)gT(x) = (\rho A x) g (wire weight only — we already counted MM in step 1).

ΔL2=∫0LT(x) dxAY=∫0LρgxYdx=ρgL22Y\Delta L_2 = \int_0^L \frac{T(x)\, dx}{A Y} = \int_0^L \frac{\rho g x}{Y} dx = \frac{\rho g L^2}{2 Y} ΔL2=7800×10×42×2×1011=3.12×1054×1011=7.8×10−7 m≈0.78 μm\Delta L_2 = \frac{7800 \times 10 \times 4}{2 \times 2 \times 10^{11}} = \frac{3.12 \times 10^5}{4 \times 10^{11}} = 7.8 \times 10^{-7} \text{ m} \approx 0.78~\mu\text{m} ΔL=ΔL1+ΔL2≈5 mm+0.00078 mm≈5.00078 mm\Delta L = \Delta L_1 + \Delta L_2 \approx 5 \text{ mm} + 0.00078 \text{ mm} \approx 5.00078 \text{ mm}

Total ≈5.0008\approx 5.0008 mm. The self-weight contribution is six orders of magnitude smaller than the load contribution.

Why This Works

Young’s modulus relates stress to strain at a point. When tension varies, strain varies, and total elongation is the integral of strain over length. The constant-tension formula ΔL=FL/(AY)\Delta L = FL/(AY) is just the special case where FF doesn’t depend on xx.

The physically important takeaway: for a wire of effective length LL under self-weight only, the total elongation is ρgL2/(2Y)\rho g L^2 / (2Y), not ρgL2/Y\rho g L^2 / Y. The factor of 12\tfrac{1}{2} comes from the integration — the average tension is half the maximum.

Alternative Method

Use the centroid trick: replace the distributed weight with a point load equal to the wire’s total weight W=ρALgW = \rho A L g acting at the centre of mass of the wire. Then ΔL2=W⋅(L/2)/(AY)=ρgL2/(2Y)\Delta L_2 = W \cdot (L/2)/(AY) = \rho g L^2 /(2Y). Same answer through statics.

Common Mistake

Two traps in this problem. First, students forget the 12\tfrac{1}{2} in ΔL2=ρgL2/(2Y)\Delta L_2 = \rho g L^2 / (2Y) — they treat the wire’s weight as if it acted at the bottom. Second, they double-count by including the wire’s weight in step 1 and in step 2. The mass MM is separate from the wire’s weight; treat them as two independent contributions.

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