Answer to Question #146307 in Physics for Chan

Question

A piece of steel plate is of 1.22 x 1.22 meters in dimension and 20 mm thick. The plate is rigidly supported to its lower end so as to position vertically. Stress is applied to the upper end uniformly and it was observed to deformed at 0.05 mm. Find the strain of the material.

A wire 100 cm long with a diameter of 1.5 mm. The material is stretched to the other end with a 200 Ib force. After the application of the force, the length of wire becomes 100.01 cm. Find the stress, strain, and the Young’s Modulus of the wire.

Answer 1

1)

$$\epsilon=\frac{0.05}{20}=0.0025$$

2)

Strain:

$$\epsilon=\frac{100.01-100}{100}=0.0001$$

Stress:

$$\sigma=\frac{F}{0.25\pi d^2}=\frac{(4.44822)200}{0.25\pi (1.5\cdot10^{-3})^2}=5\cdot10^{8}\ Pa$$

Young’s Modulus of the wire

$$E=\frac{\sigma}{\epsilon}=\frac{5\cdot10^{8}}{1\cdot10^{-4}}=5\cdot10^{12}\ Pa$$