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Chapter 7: Problem 896

A rod is fixed between two points at \(20^{\circ} \mathrm{C}\). The Coefficient of linear expansion of material of rad is $1.1 \times 10^{-5} /{ }^{\circ} \mathrm{C}\( and Young's modulus is \)1.2 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)$. Find the stress developed in the rod if temperature of rod becomes \(10^{\circ} \mathrm{C}\). (A) \(1.32 \times 10^{7}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (B) \(1.10 \times 10^{15}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (C) \(1.32 \times 10^{8}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (D) \(1.10 \times 10^{6}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\)

Short Answer

Expert verified
The short answer is: The stress developed in the rod is \(-1.32 \times 10^{7}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (compressive stress). The closest given answer choice is (A) \(1.32 \times 10^{7}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (not considering the negative sign).

Step by step solution

01

Find the change in temperature

Determine the change in temperature by calculating the difference between the final and initial temperatures. ΔT = T_final - T_initial ΔT = \(10^{\circ} \mathrm{C}\) - \(20^{\circ} \mathrm{C}\) ΔT = \(-10^{\circ} \mathrm{C}\)
02

Calculate the thermal strain

Use the coefficient of linear expansion (α) and the change in temperature (ΔT) to find the thermal strain denoted as (ε). ε = α * ΔT ε = \(1.1 \times 10^{-5} /{ }^{\circ} \mathrm{C}\) * \(-10^{\circ} \mathrm{C}\) ε = \(-1.1 \times 10^{-4}\)
03

Find the stress developed in the rod

Use Young's modulus (Y) and the thermal strain (ε) to calculate the stress (σ) developed in the rod. σ = Y * ε σ = \(1.2 \times 10^{11}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) * \(-1.1 \times 10^{-4}\) σ = \(-1.32 \times 10^{7}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) The stress developed in the rod is \(-1.32 \times 10^{7}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\). Since the stress is negative, it indicates a compressive stress developed in the rod. None of the given answer choices match the calculated stress, but the closest one is: (A) \(1.32 \times 10^{7}\left(\mathrm{~N} / \mathrm{m}^{2}\right)\) (not considering the negative sign)

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