Question

The Poisson’s ratio ($\mu$) is of order

• A. 1 $<$ $\mu$ $<$ 0
• B. 0 $<$ $\mu$ $<$ 1/2
• C. $\infty$ $<$ $\mu$ $<$ -$\infty$
• D. 1 $<$ $\mu$ $<$ -1

Hint:

For standard isotropic materials, the physical range of Poisson's ratio is $0 < \mu < 0.5$. A value of 0.5 represents a completely incompressible material (like rubber).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Poisson's ratio ($\mu$) is a dimensionless elastic constant that measures the lateral contraction of a material when stretched longitudinally.
For stable, isotropic, and elastic materials, this ratio is governed by the laws of thermodynamics and the relationships between elastic moduli.

Step 2: Detailed Explanation:
For a three-dimensional isotropic material, the relationship between the Bulk Modulus ($K$), Young's Modulus ($E$), and Poisson's ratio ($\mu$) is given by:
\[ K = \frac{E}{3(1 - 2\mu)} \]
For the bulk modulus ($K$) to remain positive (meaning the material does not expand when compressed), the denominator must be positive:
\[ 1 - 2\mu > 0 \implies \mu < \frac{1}{2} \]
Similarly, the relationship between the Shear Modulus ($G$), Young's Modulus ($E$), and Poisson's ratio ($\mu$) is:
\[ G = \frac{E}{2(1 + \mu)} \]
For the shear modulus ($G$) to remain positive, the denominator must be positive:
\[ 1 + \mu > 0 \implies \mu > -1 \]
While theoretical limits allow Poisson's ratio to range from -1 to 0.5, almost all common physical and engineering materials (such as metals, concrete, and polymers) exhibit positive lateral contraction.
Thus, for standard engineering materials, the value of Poisson's ratio falls within the range:
\[ 0 < \mu < 0.5 \text{ (or } 1/2) \]
An ideal, completely incompressible material (such as rubber or water) has a Poisson's ratio of exactly 0.5.

Step 3: Final Answer
The Poisson's ratio ($\mu$) of standard engineering materials is of the order $0 < \mu < 1/2$.