Consider a cubic element ABCD fixed at the bottom face and subjected to shearing force at the top face. The block experiences the following effects due to this shearing load:
(a) shearing stress t is induced at the faces DC and AB.
(b) complimentary shearing stress of the same magnitude is set up on the faces AD and BC.
(c) The block distorts to a new configuration ABC’D’.
(d) The diagonal AC elongates (tension) and diagonal BD shortens (compression). Longitudinal strain in diagonal AC
= (AC' - AC)/ AC = (AC' - AE)/ AC = EC′/AC ...(i)
where CE is perpendicular from C onto AC′
Since extension CC′ is small, ∠ACB can be assumed to be equal ∠ACB which is 45°. Therefore
EC′ = CC′ cos 45° = CC′/ √2
Longitudinal strain =
...(ii)
Where, Φ = CC’/BC represents the shear strain
In terms of shear stress t and modulus of rigidity C, shear strain = τ/C
longitudinal strain of diagonal AC = τ/2C ...(iii)
The strain in diagonal AC is also given by
= strain due to tensile stress in AC – strain due to compressive stress in BD
