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Worked Examples · Example 8.1

Q.A structural steel rod has a radius of 10 mm and a length of 1.0 m. A 100 kN force stretches it along its length. Calculate

(a) stress,
(b) elongation, and
(c) strain on the rod. Young's modulus of structural steel is 2.0×1011 N m−22.0 \times 10^{11}\ \text{N m}^{-2}.
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This problem applies Young’s modulus to find stress, strain, and elongation in a steel rod under axial tension. Stress is force per area, strain is the fractional change in length, and Young’s modulus links them. The answers are: stress = 3.18×108 N/m23.18 \times 10^8\ \text{N/m}^2, elongation = 1.59 mm1.59\ \text{mm}, strain = 1.59×10−31.59 \times 10^{-3}.

The key idea here is that Young’s modulus YY is a material property that tells you how stiff a material is — how much it resists being stretched. For a rod under tension, the relationship is:

Y=stressstrain=F/AΔL/LY = \frac{\text{stress}}{\text{strain}} = \frac{F/A}{\Delta L / L}

So if you know the force, the geometry, and YY, you can find everything else. Let’s go step by step.


  1. Find the cross-sectional area of the rod.

    The rod is circular with radius r=10 mm=0.010 mr = 10\ \text{mm} = 0.010\ \text{m}.

    Area A=πr2=π(0.010)2=π×10−4 m2A = \pi r^2 = \pi (0.010)^2 = \pi \times 10^{-4}\ \text{m}^2.

    Numerically: A=3.1416×10−4 m2A = 3.1416 \times 10^{-4}\ \text{m}^2.

  2. Calculate stress.

    Stress σ=FA\sigma = \frac{F}{A}, where F=100 kN=100×103 N=1.0×105 NF = 100\ \text{kN} = 100 \times 10^3\ \text{N} = 1.0 \times 10^5\ \text{N}.

    So

σ=1.0×1053.1416×10−4=3.183×108 N/m2.\sigma = \frac{1.0 \times 10^5}{3.1416 \times 10^{-4}} = 3.183 \times 10^8\ \text{N/m}^2.

Rounding to three significant figures: σ=3.18×108 N/m2\sigma = 3.18 \times 10^8\ \text{N/m}^2.

Tip

Always convert units to SI before plugging into formulas. A common slip is leaving the radius in mm — that would give an area off by a factor of 10610^6.

  1. Find strain using Young’s modulus. Young’s modulus Y=2.0×1011 N/m2Y = 2.0 \times 10^{11}\ \text{N/m}^2. From Y=σεY = \frac{\sigma}{\varepsilon}, we get …

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