--- title: "Relation between E, G and Poisson's ratio" subject: "Strength of Materials" chapter: 3 key: "semester_3/strength_of_materials:modulus-e-g-poisson-relation" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/modulus-e-g-poisson-relation site: BCT Engineering --- # Relation between E, G and Poisson's ratio Strength of Materials · Chapter 3: Simple Stress and Strain - Asked in: 2081 Baisakh, 8 marks; 2067 Asar (old course), 4 marks; 2066 Chaitra (old course), 6 marks (asked 3 times) - Video: [Elastic Constants and their Relationship | Lec-5](https://www.youtube.com/watch?v=sZi0W3pr8Jg) — Unacademy GATE & ESE - Civil & ES - 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-e-g-poisson-relation - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3 ## Question Derive the relationship between Young's modulus of Elasticity, Modulus of rigidity and Poisson's ratio. ## Short answer Under pure shear τ, the tensile diagonal stress is τ, compressive is -τ. Diagonal strain from Hooke's law is τ(1+μ)/E, and from shear strain φ = τ/G it is τ/(2G); equating yields E = 2G(1+μ). ## Answer **Result:** $E = 2G(1 + \mu)$, where $G$ is the modulus of rigidity. #### Setting up Consider a square element ABCD of side $a$ subjected to pure shear: shear stress $\tau$ on its faces. The diagonals then carry equal tensile and compressive stresses of magnitude $\tau$ at $45^\circ$ (principal stresses $\sigma_1 = +\tau$ and $\sigma_2 = -\tau$). ```text A────────────D Shear τ on all faces │ │ shear strain φ = τ/G │ ╲ │ diagonal BD stretches B────────────C diagonal AC shortens ``` #### Strain in the diagonal in terms of E and μ The tensile principal stress along diagonal BD is $\tau$ and the compressive one along AC is $-\tau$. The strain of diagonal BD is $$ \varepsilon_{BD} = \frac{\tau}{E} - \mu\left(\frac{-\tau}{E}\right) = \frac{\tau}{E}(1 + \mu) \qquad (1) $$ #### Strain in the diagonal in terms of the shear strain Keep face AB fixed. Under the shear strain $\phi$, face DC slides parallel to AB to the position D′C′, so every point on it moves by $DD' = CC' = a\phi$. Diagonal BD lengthens by the component of $DD'$ along BD, i.e. $$ \delta_{BD} = DD'\cos45^\circ = a\phi\cos45^\circ $$ and the original diagonal length is $BD = a\sqrt{2} = a/\cos45^\circ$. Therefore $$ \varepsilon_{BD} = \frac{a\phi\cos45^\circ}{a/\cos45^\circ} = \phi\cos^245^\circ = \frac{\phi}{2} = \frac{\tau}{2G} \qquad (2) $$ (using $\phi = \tau/G$). #### Equating (1) and (2) $$ \frac{\tau}{2G} = \frac{\tau}{E}(1 + \mu) \;\Rightarrow\; \boxed{E = 2G(1 + \mu)} $$ For steel with $\mu = 0.3$: $G = E/2.6 = 0.385E$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What principal stresses act on the diagonals of an element in pure shear τ? A. 2τ on both B. τ on both in tension C. +τ along one diagonal and -τ along the other, at 45° D. Zero Answer: C. Pure shear is equivalent to equal tension and compression on planes at 45°. (Key point: Stress state) 2. What is the diagonal strain in terms of E and μ? A. (τ/E)(1 - μ) B. (τ/E)(1 + μ) C. τ/(2E) D. τ(1 + μ)/G Answer: B. The tension τ/E is increased by the Poisson effect of the compression: τ/E + μτ/E. (Key point: Strain from E and μ) 3. What is the diagonal strain in terms of shear strain φ? A. φ/2 = τ/2G B. φ C. 2φ D. φ/4 Answer: A. Displacement component along the diagonal over length gives φ cos^2 45° = φ/2. (Key point: Strain from G) 4. Equating the two expressions gives which relation? A. E = 2G(1 - μ) B. E = G(1 + 2μ) C. E = 3G(1 + μ) D. E = 2G(1 + μ) Answer: D. τ/2G = τ(1+μ)/E gives E = 2G(1+μ). 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