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A square aluminium (shear modulus is 25 × 10 9 Nm - 2 ) slab of side 60   cm and thickness 15   cm is subjected to a shearing force (on its narrow face) of 18 . 0 × 10 4   N . The lower edge is riveted to the floor. The displacement of the upper edge is _____ μ m .

Correct answer

0

Step-by-step solution

Shearing stress is given by F A = η x l Thus, displacement is x = F l A η ⇒ x = 18 × 10 4 × 60 × 10 - 2 60 × 10 - 2 × 15 × 10 - 2 × 25 × 10 9 = 48 × 10 - 6   m = 48   μ m

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