Question

The top and bottom surfaces of a metal block each have an area of A = 0.030 m 2, and the height of the block is d = 0.11 m. At the top surface of the block, a force F1 is applied to the right, while at the bottom surface of the block, a force F2 is applied to the left, causing a shear in the metal block. If F1 = F2 = 30 ⨯ 106 N and the displacement between the two edges due to the shear is 1.12 10-3 m, what is the shear modulus of the metal

Answer 1

Answer:

Shear modulus is equal to $9.82\times 10^{10}N/m^2$

Explanation:

We have given area $A=0.030m^2$

Force is given $F_1=F_2=30\times 10^6N$

Height of the block d = 0.11 m

Change in height of the block $\Delta d=1.12\times 10^{-3}m$

Stress is given by

$stress=\frac{force}{area}$

$stress=\frac{30\times 10^6}{0.030}=10^9N/m^2$

Strain is equal to

$strain=\frac{\Delta d}{d}$

$strain=\frac{1.12\times 10^{-3}}{0.11}=10.18\times 10^{-3}$

Shear modulus is equal to

Shear modulus $=\frac{stress}{strain}$

$=\frac{10^9}{10.18\times 10^{-3}}=9.82\times 10^{10}N/m^2$