STRENGTH OF MATERIALS NOTES

http://www.eduresourcecollection.com/civil_sm.php

THE EXCERPTS OF THE LINK ARE EASY TO FIND HERE BUT YOU FIND IT IN A FULL WHEN YOU FOLLOW THE LINK.

EXCERPTS:

Relationship between modulus of elasticity and modulus of rigidity

Consider a square element ABCD of side ‘a’ subjected to simple shear of intensity q as shown in figure.
It is deformed to the shape AB’C’D under the shear stress. Drop perpendicular BE to the diagonal DB’. Let ? be the shear strain induced and let N be the modulus of rigidity.
The diagonal DB gets elongated to DB’. Hence there is tensile strain et in the diagonal.
et = (DB’ – DB) / DB = EB’ / DB
since this deformation is very small we can take L BB’E = 45º
EB’ = BB’ / v2 = AB tan ? / v2 = a tan ? / v2
DB = v2 a
et = (a tan ? / v2)/ v2 a = tan ? /2 = = ? / 2 since ? is small
ie et = ½ X q/N —————— (1)
We know that stress along the diagonal DB is a pure tensile stress pt = q and that along the diagonal AC is a pure compressive stress pc also equal to q. hence the strain along the diagonal DB is et = q/E + 1/m X q/E
Ie et = q/E (1+1/m) —————— (2)
From (1) and (2) we have,
E = 2N(1+1/m)
This is the required relationship between E and N……………………………………..

FOR A FULL KNOWLEDGE FOLLOW THE LINK SAID ABOVE/////////////////////SHAMSUL HUSAIN

I AM REALY VERY THANKFUL TO THE BLOGGERS WHO ARE ACTIVE IN SUCH ACTIVITIES………….