A square aluminum (shear modulus is 25 × 10⁹ Nm⁻²) slab of side 60 cm and thickness 15 cm is subjected to a shearing force (on its narrow face) of 18.0 × 10⁴ N. The lower edge is riveted to the floor. The displacement of the upper edge is ____________ μ m. Numerical answer type. Work it out, then open the answer.

Answer: 48. FL = 60 cm15 cm thickA = 60 cm x 15 cm Force acts on the narrow face: A = 0.60×0.15 = 0.09 m². Shear stress = 18×10⁴0.09 = 2×10⁶ N m⁻² Shear strain = 2×10⁶25×10⁹ = 8×10⁻⁵ Δ x = strain× L = 8×10⁻⁵×0.60 = 4.8×10⁻⁵ m Δ x = 48

A square aluminum (shear modulus is 25 × 10⁹ Nm⁻²) slab of side 60 cm… | JEE Main 2022 (27 Jul, Shift 1)

Q6JEE Main 2022 · 27 JUL · SHIFT 1Easy

Shear Modulus & Poisson's Ratio

A square aluminum (shear modulus is $25 \times 10^{9}\text{ Nm}^{-2}$ ) slab of side $60\text{ cm}$ and thickness $15\text{ cm}$ is subjected to a shearing force (on its narrow face) of $18.0 \times 10^{4}\text{ N}$ . The lower edge is riveted to the floor. The displacement of the upper edge is ____________ $\mu$ m.

Numerical answer type. Work it out, then open the answer.

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Answer: 48

Force acts on the narrow face: $A = 0.60\times0.15 = 0.09\ \text{m}^2$ .
Shear stress $= \frac{18\times10^{4}}{0.09} = 2\times10^{6}\ \text{N m}^{-2}$
Shear strain $= \frac{2\times10^{6}}{25\times10^{9}} = 8\times10^{-5}$
$\Delta x = \text{strain}\times L = 8\times10^{-5}\times0.60 = 4.8\times10^{-5}\ \text{m}$
$\Delta x = 48\ \mu\text{m}$

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