Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping

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.

9 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.                                                           


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..



9 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Course Features

  • 101 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution

Course Features

  • 110 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details