Let ABCD is a square block

Shear strain = ϕ

Strain of AC = AC_{1} – AC / AC = C_{1} C_{2} / AC

= CC_{1} cos 45° / √2 = CC_{1} / 2 AB

1 /2 (CC1 / AB) = 1/2 (CC1 / BC) = 1/2 ϕ

= 1/2 x t / G

Again tensile strain of AC due to tensile stress σ = σ / E

= t / E

and the tensile strain on diagonal AC due to compressive stress on BD = µ . t /E

Combined effect of the above two stresses on

Diagonal AC = t / E + µ t /E = t /E (1+µ)

Equating (3.5) and (3.6), we get

t = 2G = t / E (1+µ)

G = E / 2 (1 +µ)

where µ = Poisson’s ratio

G = modulus of rigidity

E = modulus of elasticity.