Question

A rubber cube of side 5 cm has one side fixed while a tangential force equal to 1800 N is applied to opposite face, find the shearing strain and the lateral displaceement of the strained face, modulus of rigidity for rubber is `2.4xx10^(6)N//m^(2)`

Answer 1

Stress `=(F)/(A)=eta(x)/(L)`
strain `=theta=(F)/(Aeta)=(1800)/(25xx10^(-4)xx2.4xx10^(6))=(180)/(25xx24)=(3)/(10)=0.3rad`
as `(x)/(L)=0.3impliesx=0.3xx5xx10^(-2)m=1.5mm`