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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

Respuesta :

Answer:

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

Explanation:

We have given area [tex]A=0.030m^2[/tex]

Force is given [tex]F_1=F_2=30\times 10^6N[/tex]

Height of the block d = 0.11 m

Change in height of the block [tex]\Delta d=1.12\times 10^{-3}m[/tex]

Stress is given by

[tex]stress=\frac{force}{area}[/tex]

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

Strain is equal to

[tex]strain=\frac{\Delta d}{d}[/tex]

[tex]strain=\frac{1.12\times 10^{-3}}{0.11}=10.18\times 10^{-3}[/tex]

Shear modulus is equal to

Shear modulus [tex]=\frac{stress}{strain}[/tex]

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