Question

A bar of circular cross-section is loaded by a compressive force P = 100 kN. The bar has length L = 2.0 m and diameter d = 30 mm. It is made of aluminum alloy with modulus of elasticity E = 73 GPa. What is the strain of the bar? Express the answer in mm/m.

Answer 1

Answer on Question#50402 - Physics - Mechanics - Kinematics - Dynamics

A bar of circular cross-section is loaded by a compressive force $P = 100 \mathrm{kN}$. The bar has length $L = 2.0 \mathrm{~m}$ and diameter $d = 30 \mathrm{~mm}$. It is made of aluminum alloy with modulus of elasticity $E = 73 \mathrm{GPa}$. What is the strain of the bar? Express the answer in mm/m.

Solution:

The stress of cylinder is given by the Hooke's law

$$\sigma = E \varepsilon,$$

where $\sigma = \frac{P}{A}$ is a stress ($F$ is the compressive force acting on the bar, and $A$ is the cross-sectional area of the bar), $E$ is the modulus of elasticity, and $\varepsilon$ is the strain of the cylinder.

The cross-sectional area of the bar is given by

$$A = \frac{\pi}{4} d^{2}$$

The stress of the cylinder

$$\sigma = \frac{P}{A} = \frac{4P}{\pi \cdot d^{2}}$$

The strain of the cylinder

$$\varepsilon = \frac{\sigma}{E} = \frac{4P}{\pi \cdot E \cdot d^{2}} = \frac{100 \mathrm{kN}}{\pi \cdot 73 \mathrm{GPa} \cdot (0.03)^{2} \mathrm{m}^{2}} = 4.8 \cdot 10^{-4} = 0.48 \frac{\mathrm{mm}}{\mathrm{m}}$$

Answer: $\varepsilon = \frac{4P}{\pi \cdot E \cdot d^{2}} = 0.48 \frac{\mathrm{mm}}{\mathrm{m}}$.

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