Question

A cube of aluminium of sides 0.1 m is subjected to a
shearing force of 100 N. The top face of the cube is
displaced through 0.02 cm with respect to the bottom
face. The shearing strain would be
(1) 0.02 (2) 0.1 (3) 0.005 (4) 0.002​

Answer 1

Answer:

strain = change in length / original length

strain = displaced length/ side length

strain = 0.02/0.1

strain = 0.002

Answer 2

Answer:

(4)0.002

Explanation:

As we know, shearing strain =

Length of deformation at its maximum

------------------------------------

Perpendicular length in the plane of force

Here, Length of deformation at its maximum = 0.02 ×10power-2

& side of cube (Perpendicular length in the plane of force) = 0.1

On solving.

= 0.02×10power-2÷0.1

= 0.2×10power-2

=0.2÷100 = 0.002.... @ns....

Hence, shearing strain = 0.002