--- title: "Poisson's ratio and E from rod and G" subject: "Strength of Materials" chapter: 3 key: "semester_3/strength_of_materials:modulus-rigidity-rod-poisson-modulus" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/modulus-rigidity-rod-poisson-modulus site: BCT Engineering --- # Poisson's ratio and E from rod and G Strength of Materials · Chapter 3: Simple Stress and Strain - Asked in: 2068 Baisakh (old course), 8 marks - Video: [Poisson’s ratio, Modulus of elasticity , Modulus of rigidity ( Shear Modulus), Bulk Modulus](https://www.youtube.com/watch?v=53cUUpDSr5w) — Prof.Dr.Pravin Patil - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/modulus-rigidity-rod-poisson-modulus - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3 ## Question The modulus of rigidity for a material is $0.5\times10^5\ \text{N/mm}^2$. A 10mm diameter rod of the material was subjected to an axial pull of 10kN and the change in diameter was observed to be $3\times10^{-3}\ \text{mm}$. Calculate the Poisson's ratio and the modulus of elasticity. ## Short answer From axial stress 127.32 N/mm², lateral strain 3×10⁻⁴, and G=0.5×10⁵ N/mm², solve ν = 0.308 using ε_lat = νσ/[2G(1+ν)]. Then E = 2G(1+ν) = 1.308×10⁵ N/mm². ## Answer **Method:** use the lateral strain from the change in diameter, the axial stress, and the relation $E = 2G(1+\nu)$. #### Stress and lateral strain $$ \sigma = \frac{P}{A} = \frac{10\,000}{\frac{\pi}{4}(10)^2} = 127.32\ \text{N/mm}^2, \qquad \varepsilon_{lat} = \frac{\Delta d}{d} = \frac{3\times10^{-3}}{10} = 3\times10^{-4}\ \text{(contraction)} $$ #### Poisson's ratio Lateral strain $= \nu\,\dfrac{\sigma}{E}$ and $E = 2G(1+\nu)$, so $$ \varepsilon_{lat} = \frac{\nu\sigma}{2G(1+\nu)} \Rightarrow 3\times10^{-4} = \frac{\nu(127.32)}{2(0.5\times10^5)(1+\nu)} $$ $$ 127.32\,\nu = 30(1+\nu) \Rightarrow \nu = \frac{30}{127.32-30} = 0.3082 $$ #### Modulus of elasticity $$ E = 2G(1+\nu) = 2(0.5\times10^5)(1+0.3082) = 130825\ \text{N/mm}^2 $$ **Answer:** $\nu = 0.308$ and $E \approx 1.308\times10^5\ \text{N/mm}^2$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the lateral strain computed from the observed change in diameter? A. Change in diameter divided by the rod length B. Change in diameter multiplied by the area C. Change in diameter divided by original diameter = 3x10^-4 D. Change in diameter divided by the axial stress Answer: C. Lateral strain is the change in diameter over the original diameter: 0.003/10 = 3x10^-4. (Key point: Lateral strain from diameter change) 2. What is the axial stress in the rod? A. 127.32 N/mm^2 (10 000 N over pi/4 x 10^2 mm^2) B. 31.83 N/mm^2 C. 1000 N/mm^2 D. 63.66 N/mm^2 Answer: A. Stress = P/A = 10 000 / 78.54 = 127.32 N/mm^2. (Key point: Axial stress) 3. Why can nu be found without knowing E beforehand? A. Because E equals G for any material B. Because lateral strain equals axial strain C. Because E = 2G(1+nu) is substituted into lateral strain = nu x sigma/E, leaving nu as the only unknown D. Because Poisson's ratio is always 0.3 Answer: C. Substituting E = 2G(1+nu) gives 3x10^-4 = nu(127.32)/(2G(1+nu)), which is solved for nu. (Key point: Eliminating E using G) 4. What are the final values of Poisson's ratio and E? A. nu = 0.308, E = 1.308x10^5 N/mm^2 B. nu = 0.308, E = 1.0x10^5 N/mm^2 C. nu = 0.5, E = 1.5x10^5 N/mm^2 D. nu = 0.692, E = 0.846x10^5 N/mm^2 Answer: A. nu = 30/(127.32-30) = 0.3082 and E = 2(0.5x10^5)(1.3082) = 1.308x10^5 N/mm^2. 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