GATE-2019

56. The Young’s modulus(E) is related to the Lame’s parameter ‘λ’ for a Poisson solid as
(Thanks to Neeraja AU)

(A) E=2.5λ

(B) E=1.5λ

(D) E=0.5λ

Solution:

The relation between Young’s modulus(E), Lame’s parameter (λ) and Poisson’s solid(ν) is

$\lambda=\frac{\nu E}{(1+\nu)(1-2\nu)}$

Ideal Poisson Solid value is (ν) = 0.25

$\lambda=\frac{0.25 E}{(1+0.25)(1-2\times 0.25)}$

$\lambda=\frac{0.25 E}{(1.25)(1-0.5)}=\frac{0.25E}{0.625}$

$\lambda=0.4E$

$E=(\frac{1}{0.4})\lambda$

$E=2.5\lambda$

GEOPHYSICS