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Question: A rod is fixed between two points at \(20^{o}C\). The coefficient of linear expansion of material of...

A rod is fixed between two points at 20oC20^{o}C. The coefficient of linear expansion of material of rod is 1.1×10−5/oC1.1 \times 10^{- 5}/^{o}C and Young’s modulus is 1.2×1011N/m21.2 \times 10^{11}N/m^{2}. Find the stress developed in the rod if temperature of rod becomes 10oC10^{o}C

A

1.32×107N/m21.32 \times 10^{7}N/m^{2}

B

1.10×1015N/m2\mathbf{1.10}\mathbf{\times}\mathbf{1}\mathbf{0}^{\mathbf{15}}\mathbf{N/}\mathbf{m}^{\mathbf{2}}

C

1.32×108N/m2\mathbf{1.32 \times 1}\mathbf{0}^{\mathbf{8}}\mathbf{N/}\mathbf{m}^{\mathbf{2}}

D

1.10×106N/m2\mathbf{1.10}\mathbf{\times}\mathbf{1}\mathbf{0}^{\mathbf{6}}\mathbf{N/}\mathbf{m}^{\mathbf{2}}

Answer

1.32×107N/m21.32 \times 10^{7}N/m^{2}

Explanation

Solution

Thermal stress FA=YαΔθ=1.2×1011×1.1×10−5×(20−10)=1.32×107N/m2\frac{F}{A} = Y\alpha\Delta\theta = 1.2 \times 10^{11} \times 1.1 \times 10^{- 5} \times (20 - 10) = 1.32 \times 10^{7}N/m^{2}