Question

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

Hint:

For isotropic materials: \[ E = 2G(1+\nu), E = 3K(1-2\nu), G = \frac{3E}{2(1+\nu)(1+\nu)} \] If any two constants among \(E, G, K, \nu\) are known, the others can be found.

Answer 1

Correct Answer: 0.4

Step 1: Recall the relationship between elastic constants.
For an isotropic material, the relation between Young’s modulus \(E\), shear modulus \(G\), and Poisson’s ratio \(\nu\) is: \[ G = \frac{E}{2(1+\nu)} \]

Step 2: Rearrange for \(\nu\).
\[ \nu = \frac{E}{2G} - 1 \]

Step 3: Substitute values.
\[ \nu = \frac{140}{2 \times 50} - 1 \] \[ \nu = \frac{140}{100} - 1 \] \[ \nu = 1.40 - 1 = 0.40 \]

Step 4: Interpretation.
- A Poisson’s ratio of 0.40 means the material is relatively ductile (metals typically have \(\nu\) in the range 0.25–0.35, rubbers near 0.50). - This result is within the theoretical limit \(0 \leq \nu \leq 0.5\), so it is valid. Final Answer:
\[ \boxed{0.40} \]