Arectangular block of material with shear modulus g= 620 mpa is fixed to rigid plates at its top and bottom surfaces. thelower plate remains fixed while the upper plate is subjected to a horizontal force, p. if the top plate displaces 2 mm horizontally, determine:

a. the average shear strain in the material
b. the force p exerted on the upper plate.

γ =0.01, P=248 kN

Explanation:

Given Data:

displacement = 2mm ;

height = 200mm ;

l = 400mm ;

w = 100 ;

G = 620 MPa = 620 N//mm²;    1MPa = 1N//mm²

a. Average Shear Strain:

The average shear strain can be determined by dividing the total displacement of plate by height

γ = displacement / total height

     = 2/200 = 0.01

b. Force P on upper plate:

Now, as we know that force per unit area equals to stress

τ = P/A

Also,  τ = Gγ

By comapring both equations, we get

P/A = Gγ   ------------ eq(1)

First we need to calculate total area,

A = l*w = 400 * 100= 4*10^4mm²

By putting the values in equation 1, we get

P/40000 = 620 * 0.01

P = 248000 N or 2.48 *10^5 N or 248 kN

answer:

sure does

explanation: