f a shear force of 300 N is applied on a cube of side 40 cm, find the displacement of the top surface of the cube. The rigidity modulus of the material is 5xx10^(10)N//m^(2)

f a shear force of 300 N is applied on a cube of side 40 cm, find the

1.	f a shear force of 300 N is applied on a cube of side 40 cm, find the displacement of the top surface of the cube. The rigidity modulus of the material is 5xx10^(10)N//m^(2)
Answer» SOLUTION :`eta=("shearing stress")/("shearing strain")=(300//0.4xx0.4)/(theta)`, since area of the face of the cube`=(40xx40) cm^(2)` `5xx10^(10)=(30000)/( 16 theta) :. theta=(3)/(80xx10^(6))` displacement of the surface x=,But `(x)/(L) tan theta= theta` since `theta` is small `:. (x)/(0.4)=(3)/(80xx16^(6)) or x=(12)/(8xx10^(7))=0.15xx10^(-7)m ` So displacement of surface is `0..15xx10^(-7)m`