The area of the upper face of a rectangular block is 0.5 m × 0.5 m and the lower face is fixed. The height of the block is 1 cm.
The area of the upper face of a rectangular block is 0.5 m × 0.5 m and the lower face is fixed. The height of the block is 1 cm. A shearing force applied at the top face produces a displacement of 0.0 15 mm. Find the strain and shearing force.
(Modulus of rigidity: n = 4.5 × 1010 N/m2)
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Given: A = 0.5 m × 0.5 m = 0.25 m2,
h = 1 cm = 10-2m,
x = 0.015 mm = 15 × 10-6m
n = 4.5 × 1010 N/m2
To find: Strain (θ). Shearing force (F)
Formulae:
i. θ = \(\frac{\text{x}}{h}\)
ii. F = nAθ
Calculations:
Using formula (i),
θ = \(\frac{15\times10^{-6}}{10^{-2}}\) = 1.5 × 10-3
Using formula (ii),
F = 4.5 × 1010 × 0.25 × 1.5 × 10-3
= 1.688 × 107 N
Shearing force is 1.688 × 107 N and strain is 1.5 × 10-3
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