Question

The relation of three elastic constants of a metal in terms of Young's Modulus (E), Modulus of rigidity (G) and Bulk Modulus (K) is expressed as

• A. \( E = \frac{3\text{KG}}{(3\text{K}+\text{G})} \)
• B. \( E = \frac{6\text{KG}}{(3\text{K}+\text{G})} \)
• C. \( E = \frac{8\text{KG}}{(3\text{K}+\text{G})} \)
• D. \( E = \frac{9\text{KG}}{(3\text{K}+\text{G})} \)

Hint:

An easy way to remember this formula is to memorize its reciprocal form: \( \frac{9}{E} = \frac{3}{G} + \frac{1}{K} \). Finding a common denominator and flipping this reciprocal equation gives the standard formula directly!

Answer 1

The Correct Option is D

Concept: In the mechanics of materials and isotropic elasticity theory, the deformation behavior of a material is defined by three primary elastic constants alongside Poisson's ratio (\( \nu \)):

• Young's Modulus (\( E \)): A measure of tensile or compressive stiffness.

• Modulus of Rigidity / Shear Modulus (\( G \)): A measure of shear stiffness.

• Bulk Modulus (\( K \)): A measure of volumetric stiffness under uniform hydrostatic pressure.
These values are interconnected, and we can derive the relationship between all three constants by mathematically eliminating Poisson's ratio (\( \nu \)) from the standard constitutive equations.

Step 1: Listing the individual formulas involving Poisson's ratio.
We use two fundamental equations from elasticity theory that express Young's Modulus in terms of \( K \) and \( G \) separately using Poisson's ratio \( \nu \): \[ E = 3K(1 - 2\nu) \quad \cdots (1) \] \[ E = 2G(1 + \nu) \quad \cdots (2) \]

Step 2: Rearranging the equations to isolate the Poisson's ratio terms.
Let us rearrange equation (1) to isolate the term \( 2\nu \): \[ \frac{E}{3K} = 1 - 2\nu \quad \Rightarrow \quad 2\nu = 1 - \frac{E}{3K} \quad \Rightarrow \quad \nu = \frac{1}{2} - \frac{E}{6K} \quad \cdots (3) \] Now, let us rearrange equation (2) to isolate \( \nu \): \[ \frac{E}{2G} = 1 + \nu \quad \Rightarrow \quad \nu = \frac{E}{2G} - 1 \quad \cdots (4) \]

Step 3: Equating the two expressions to eliminate \( \nu \).
Since equations (3) and (4) both equal \( \nu \), we set them equal to each other: \[ \frac{E}{2G} - 1 = \frac{1}{2} - \frac{E}{6K} \]

Step 4: Grouping all terms containing Young's Modulus \( E \) onto one side.
Add \( \frac{E}{6K} \) to both sides and add 1 to both sides of the equation: \[ \frac{E}{2G} + \frac{E}{6K} = 1 + \frac{1}{2} \] Simplify the right side constant sum: \[ \frac{E}{2G} + \frac{E}{6K} = \frac{3}{2} \] Factor out the common variable \( E \) on the left-hand side: \[ E \left( \frac{1}{2G} + \frac{1}{6K} \right) = \frac{3}{2} \]

Step 5: Finding a common denominator and isolating \( E \).
Find a common denominator for the terms inside the brackets, which is \( 6\text{KG} \): \[ E \left( \frac{3K + G}{6\text{KG}} \right) = \frac{3}{2} \] Isolate Young's Modulus \( E \) by multiplying both sides by the reciprocal fraction: \[ E = \frac{3}{2} \times \left( \frac{6\text{KG}}{3K + G} \right) \] Simplifying the constant multipliers: \[ E = 3 \times \left( \frac{3\text{KG}}{3K + G} \right) = \frac{9\text{KG}}{3K + G} \] This derived expression matches Option (D).