Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

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

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
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.

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.

Grade:

1 Answers

AKASH GOYAL AskiitiansExpert-IITD
419 Points
11 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..

 

 


Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free