Question

a cube of aluminium of each side 4 CM is subjected to tangential force. the top face of the cube is stretched through 0.016cm with respect to the face. find 1 shearing stress and sheering strain. given meh =2.08×10 to power 11 dyu/cm square​

Answer 1

The value of sheering strain = 0.004 and shearing stress = 8×10^8 dyne/cm^2.

Explanation:

Given data:

Length of each side "L"=4 cm

Lateral displacement "x" =0.016 cm

η=2.08 ×10^11 dyne/cm^2

Solution:

(a)

Shearing strain θ = xL= 0.016/4 = 0.004

(b) Using relation:

η = F/θa

or

Shearing stress = F/a = ηθ

=2 × 10^11 × 4 × 10^−3

= 8×10^8 dyne/cm^2

Thus the value of sheering strain = 0.004 and shearing stress = 8×10^8 dyne/cm^2.

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