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somy chaturvedi Grade:
        

a steel plate of face area 4cm2 and thickness 0.5cm is fixed rigidly at lower surface . a tangential force of 10N is applied on upper surface . find the lateral displacement of the upper surface wrt lower surface . rigidity modulus=8.4*10to the power 10N per metre square.

6 years ago

Answers : (1)

AKASH GOYAL AskiitiansExpert-IITD
419 Points
										

Dear Somy


rigidity modulus (G) = shear stress/shear strain= s/e


shear stress, s=10/4x10-4 = 2.5 x 104 N/m2


shear strain, e = s/G = 2.5 x 104/8.4 x 1010


e= 0.297 x 10-6 radians


fro small angle e=tane= relative displacement/thickness= d/0.005


d=0.005e= 0.005 x 0.297 x 10-6 m = 1.485 x 10-6 mm


 


All the best.                                                           


AKASH GOYAL


AskiitiansExpert-IIT Delhi


 


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6 years ago
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