--- title: "Strain energy due to shear in rectangular beam" subject: "Theory of Structures I" chapter: 2 key: "semester_4/theory_of_structures_i:strain-energy-shear-rect-beam" source: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/strain-energy-shear-rect-beam site: BCT Engineering --- # Strain energy due to shear in rectangular beam Theory of Structures I · Chapter 2: Analysis by the Strain Energy Method - Asked in: 2080 Chaitra, 4 marks; 2076 Bhadra, 6 marks; 2068 Bhadra, 8 marks (asked 3 times) - Video: [Strain Energy due to shear stress | Strain Energy | Strength of Materials](https://www.youtube.com/watch?v=oTz4sME0PB4) — Sampurna Engineering - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/theory_of_structures_i/strain-energy-shear-rect-beam - Whole chapter: https://www.bctengineer.com/md/semester_4/theory_of_structures_i/chapter-2 ## Question For a beam having a rectangular cross-section and subjected to lateral loads, derive an expression for the strain energy due to shear deformation only. ## Short answer Derive by integrating shear stress energy density over the section. For rectangular section, shear stress τ = V/(2I)(d^2/4 - y^2). The shear strain energy per length dU = (6/5)(V^2/(2GA))dx, so total Us = ∫ (6/5)(V^2/(2GA)) dx = ∫ (3V^2/(5GA)) dx. ## Answer Strain energy due to shear is the energy stored because of the shear deformation of the beam. For a rectangular section the shear stress varies parabolically, so the energy is found by integrating over the section. #### Derivation Take a beam of rectangular section (width $b$, depth $d$) carrying shear force $V$ at a section. Shear stress at a distance $y$ from the neutral axis: $$ \tau = \frac{V A\bar y}{I b} = \frac{V}{2I}\left(\frac{d^2}{4} - y^2\right) $$ Strain energy per unit volume $= \dfrac{\tau^2}{2G}$. For an element of length $dx$, with area element $dA = b\,dy$: $$ dU = \int_{-d/2}^{d/2} \frac{\tau^2}{2G}\, b\,dy\,dx $$ Substituting $\tau$: $$ \begin{aligned} dU &= \frac{V^2 b\,dx}{8GI^2}\int_{-d/2}^{d/2}\left(\frac{d^2}{4}-y^2\right)^2 dy \\ &= \frac{V^2 b\,dx}{8GI^2}\cdot\frac{d^5}{30} \end{aligned} $$ Using $I = \dfrac{bd^3}{12}$, so $I^2 = \dfrac{b^2d^6}{144}$: $$ dU = \frac{V^2 b\, d^5 \times 144}{8G\,b^2 d^6\times 30}\,dx = \frac{3V^2}{5\,G\,bd}\,dx = \frac{6}{5}\cdot\frac{V^2}{2GA}\,dx $$ where $A = bd$. Hence the shear strain energy for the whole beam is $$ U_s = \int_0^L \frac{6}{5}\,\frac{V^2}{2GA}\,dx = \int_0^L \frac{3V^2}{5GA}\,dx $$ The factor $\dfrac{6}{5} = 1.2$ is the **form factor (shape factor) for shear** $\alpha$ of a rectangle. If shear were uniform, $dU = \dfrac{V^2}{2GA}dx$. **Example (check):** simply supported beam with central load $P$: $V = \pm P/2$, $$ U_s = \frac{6}{5}\times\frac{(P/2)^2}{2GA}\times L = \frac{3P^2L}{20\,GA} $$ The deflection under the load (from $\delta = \partial U/\partial P$) is $\delta_s = \dfrac{3PL}{10GA}$. **Result:** $U_s = \dfrac{6}{5}\displaystyle\int_0^L \dfrac{V^2}{2GA}\,dx$ for a rectangular section. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Strain energy per unit volume due to shear stress tau is: A. tau^2/(2E) B. tau/(2G) C. tau^2/(2G) D. 2 tau^2/G Answer: C. Shear energy density uses G. (Key point: Energy density) 2. For a rectangular section the shear stress varies: A. Uniformly B. Linearly, zero at neutral axis C. Maximum at extreme fibres D. Parabolically, maximum at neutral axis Answer: D. tau = V/(2I)(d^2/4 - y^2). (Key point: Stress distribution) 3. The shear form factor for a rectangle is: A. 6/5 = 1.2 B. 10/9 C. 1 D. 3/2 Answer: A. The integration gives dU = (6/5) V^2/(2GA) dx. (Key point: Form factor) 4. Strain energy due to shear in a rectangular beam is: A. integral of V^2/(2EI) dx B. (6/5) integral of V^2/(2GA) dx C. integral of V^2/(GA) dx D. (10/9) integral of V^2/(2GA) dx Answer: B. Uniform-shear value multiplied by 1.2. 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