--- title: "Tapering rod: total extension under load and self weight" subject: "Strength of Materials" chapter: 3 key: "semester_3/strength_of_materials:tapered-rod-45kn-extension" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/tapered-rod-45kn-extension site: BCT Engineering --- # Tapering rod: total extension under load and self weight Strength of Materials · Chapter 3: Simple Stress and Strain - Asked in: 2069 Chaitra, 6 marks - Video: [Deformation of a uniformily tapering circular bar](https://www.youtube.com/watch?v=viRLTivxhQY) — Make Mechanical Simple - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/strength_of_materials/tapered-rod-45kn-extension - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3 ## Question A vertical rod of length 3m tapers uniformly from a diameter of 80mm at the top to 40mm at the bottom. If it is rigidly fixed at the upper end and is subjected to an axial load of 45KN, determine the total extension in the bar. Take density of material = $2\times10^5\ \text{kg/m}^3$ and young's modulus = $210\ \text{GN/m}^2$. ## Short answer For a frustum of a cone fixed at top, extension due to 45 kN load is 0.2558 mm (using δ = 4PL/(π E d1 d2)). Self-weight of rod (density 200000 kg/m³) adds 0.028 mm, giving total extension 0.284 mm. ## Answer The rod is a frustum of a cone, fixed at the top (d₁ = 80 mm) with the 45 kN load at the bottom (d₂ = 40 mm). The extension has two parts: that due to the axial load and that due to the rod's own weight. **Data:** $L = 3000$ mm, $E = 210\times10^3\ \text{N/mm}^2$, $P = 45\,000$ N, $\rho = 2\times10^5\ \text{kg/m}^3$ (as given), $g = 9.81\ \text{m/s}^2$. #### Extension due to the load For a uniformly tapering circular bar: $$ \delta_P = \frac{4PL}{\pi E d_1 d_2} = \frac{4(45\,000)(3000)}{\pi(210\,000)(80)(40)} = 0.2558\ \text{mm} $$ #### Extension due to self weight Volume of the frustum $V = \dfrac{\pi L}{12}(d_1^2 + d_1d_2 + d_2^2) = 0.008796\ \text{m}^3$, so the weight is $$ W = \rho g V = 2\times10^5(9.81)(0.008796) = 17.26\ \text{kN} $$ Let $y$ be measured up from the bottom, with diameter $d = d_2 + ky$ where $k = \dfrac{d_1-d_2}{L} = 0.01333$. The weight below that section is $\dfrac{\rho g\pi}{12k}(d^3-d_2^3)$, and the self-weight extension is $$ \delta_w = \int_0^L \frac{\text{(weight below)}}{E\,\frac{\pi}{4}d^2}\,dy = \frac{\rho g}{3k^2E}\left[\frac{d_1^2-d_2^2}{2} + d_2^3\left(\frac{1}{d_1}-\frac{1}{d_2}\right)\right] = 0.0280\ \text{mm} $$ #### Total extension $$ \delta = \delta_P + \delta_w = 0.2558 + 0.0280 = 0.2838\ \text{mm} $$ **Answer: total extension $\approx 0.284\ \text{mm}$.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which two parts make up the total extension? A. Extension due to the axial load and due to self weight B. Extension due to temperature and load C. Load only D. Self weight only Answer: A. The answer sums 0.2558 and 0.0280 mm. (Key point: Components) 2. Which formula gives extension of a tapering circular bar under load? A. PL/(A E) with mean diameter B. 4PL/(pi E d1 d2) C. 4PL/(pi E d1^2) D. PL d1 d2/E Answer: B. For uniform taper the exact result is 4PL/(pi E d1 d2). (Key point: Load formula) 3. Why is the self-weight integral needed? A. Axial force at each section varies with the weight below it, and area varies B. Weight is negligible C. Density is constant D. The load is zero Answer: A. Both force and area vary with position. (Key point: Self weight) 4. What is the total extension? A. 0.256 mm B. 0.028 mm C. 0.284 mm D. 2.84 mm Answer: C. 0.2558 + 0.0280 = 0.2838 mm. 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