Question

For a certain linear elastic isotropic material, the Young’s modulus is 140 GPa and the shear modulus is 50 GPa. The Poisson’s ratio for the material is ........... (rounded off to two decimal places).

Hint:

For linear elastic isotropic materials, the Poisson's ratio can be calculated using the formula \( \nu = \frac{E}{2G} - 1 \), where \( E \) is the Young's modulus and \( G \) is the shear modulus.

Answer 1

Step 1: In the case of a linear elastic isotropic material, the Poisson’s ratio \( \nu \) is related to the Young’s modulus \( E \) and the shear modulus \( G \) by the following formula: \[ \nu = \frac{E}{2G} - 1 \] This relationship is fundamental in material science and helps define the deformation characteristics of isotropic materials under stress. 
Step 2: The given values for this material are: \[ E = 140 \, {GPa}, G = 50 \, {GPa} \] We substitute these values into the formula for Poisson’s ratio: \[ \nu = \frac{140 \, {GPa}}{2 \times 50 \, {GPa}} - 1 \] Simplifying the expression: \[ \nu = \frac{140}{100} - 1 = 1.4 - 1 = 0.40 \]
Step 3: After simplifying the calculation, we find that the Poisson's ratio \( \nu \) for this material is \( 0.40 \). This value indicates how the material deforms in response to applied stress, where the lateral strain is 0.40 times the axial strain. 
Step 4: It is important to note that Poisson’s ratio for most common engineering materials typically lies between 0 and 0.5. Values closer to 0.5 indicate more incompressible materials (such as metals), while values lower than 0.5 indicate more compressible materials (such as rubber). 
Step 5: Therefore, the Poisson’s ratio for the material is \( 0.40 \), which matches the given answer.