Question

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

a) We can find the stress from the formula:

$$\sigma=\dfrac{F}{A}=\dfrac{200\ lb\cdot (\dfrac{1\ N}{0.2248\ lb})}{\dfrac{\pi(1.5\cdot10^{-3}\ m)^2}{4}}=503\cdot10^6\ Pa=503\ MPa.$$

b) We can find strain from the formula:

$$\epsilon=\dfrac{\Delta L}{L_0}=\dfrac{100.01\ cm-100\ cm}{100\ cm}=1.0\cdot10^{-4}.$$

c) We can find Young’s Modulus of the wire from the formula:

$E=\dfrac{\sigma}{\epsilon}=\dfrac{503\cdot10^6\ Pa}{1.0\cdot10^{-4}}=5030\cdot10^9\ Pa=5030\ GPa.$

Answer:

a) $\sigma=503\cdot10^6\ Pa=503\ MPa.$

b) $\epsilon=1.0\cdot10^{-4}.$

c) $E=5030\cdot10^9\ Pa=5030\ GPa.$