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

Shear modulus is equal to
`9.82* 10^(10)N/m^2`

Step-by-step explanation:

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

Force is given
`F_1=F_2=30* 10^6N`

Height of the block d = 0.11 m

Change in height of the block
`\Delta d=1.12* 10^(-3)m`

Stress is given by

`stress=(force)/(area)`

`stress=(30* 10^6)/(0.030)=10^9N/m^2`

Strain is equal to

`strain=(\Delta d)/(d)`

`strain=(1.12* 10^(-3))/(0.11)=10.18* 10^(-3)`

Shear modulus is equal to

Shear modulus
`=(stress)/(strain)`

`=(10^9)/(10.18* 10^(-3))=9.82* 10^(10)N/m^2`