Test: Stress-Strain Relations & Elastic Constants - 2 - Mechanical Engineering MCQ

As we know E = 2G(1 + μ) = 3K(1 – 2μ) = 9KG / (3K + G).

We can use E = 2G(1 + μ) = 3K(1 – 2μ) = 9KG / (3K + G) to get the relation between E, K and G.

Under plane stress condition, the strain in the direction perpendicular to the plane is not zero. It has been found experimentally that when a body is stressed within the elastic limit, the lateral strain bears a constant ratio to the linear strain.

On using E = 2G(1 + μ) we can put the values of E and G to get the Poissons value.

As E = 2G(1 + μ) = 3K(1 – 2μ) = 9KG / (3K + G)
The value of μ must be between 0 to 0.5, so as E never equal to G but if μ = 1/3, then E=K.

As E = 2G(1 + μ)
1.25 x 10² = 2G(1 + 0.34)
G = 0.4664 x 10² MPa.

As E = 2G(1 + μ) = 3K(1 – 2μ) = 9KG / (3K + G), if any two of these four are known, the other two can be calculated by the relations between them.

For longitudinal strain we need Youngs modulus and for calculating transverse strain we need Poisson’s ratio. We may calculate Poissons ratio from E = 2G(1 + μ) for that we need shear modulus.

As E = 2G(1 + μ) putting μ=1 we get E = -3K.

As we know G = E / 2(1 +μ) so this gives the ratio of E to G = 2(1 + μ).