Question icon
Mechanics

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.

Profile image of somy chaturvedi
15 Years agoGrade
Answers icon

1 Answer

Profile image of AKASH GOYAL AskiitiansExpert-IITD
15 Years ago

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

 

Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation.

Win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.

Now you score 5+15 POINTS by uploading your Pic and Downloading the Askiitians Toolbar  respectively : Click here to download the toolbar..