Question

A square lead slab of side 50cm and thickness 10cm is subjected to a shearing force of 9×10 4 .The lower edge is riveted to the floor. How much will the upper edge be displaced?

Answer 1

Thank you for asking for this question, here is your answer:

Area of the face on which force is applied:

A = 50 x 10

= 500 cm²

= 500 x 10 ^ -4 m²

Length

L = 0.5 m

Rigidity:

n = 5.6 x 10⁹ Pa

Force:

F = 9 x 10⁴ N

Expression for Rigidity modulus:

n = FL/AΔL

F is force of shearing

L is side

A is the area of the face on which force is applied

Displacement  of upper edge is

ΔL = FL/An

= (9x10^4) (0.5)/ (500 x 10^ - 4) ( 5.6 x 10^9)

= 0.00016 m

= 0.16 mm

If there is any confusion please leave a comment below.