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?
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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.
ajitajain7:
L is not side it is the displacement of top of block
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