--- title: "Elongation of tapering circular bar (derivation)" subject: "Strength of Materials" chapter: 3 key: "semester_3/strength_of_materials:tapered-circular-bar-elongation-derive" source: https://www.bctengineer.com/q/semester_3/strength_of_materials/tapered-circular-bar-elongation-derive site: BCT Engineering --- # Elongation of tapering circular bar (derivation) Strength of Materials · Chapter 3: Simple Stress and Strain - Asked in: 2079 Bhadra, 6 marks; 2075 Chaitra, 6 marks; 2074 Chaitra, 6 marks (asked 3 times) - 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-circular-bar-elongation-derive - Whole chapter: https://www.bctengineer.com/md/semester_3/strength_of_materials/chapter-3 ## Question Derive the expression for the elongation of a circular bar of tapering section due to axial load. ## Short answer For a circular bar tapering uniformly from d1 to d2 over length L under axial load P, elongation δ = 4PL/(πEd1d2). Derived by integrating dδ = (4P)/(πE(d1+kx)^2) dx with k = (d2-d1)/L. ## Answer Consider a circular bar of length $L$ whose diameter varies uniformly from $d_1$ at one end to $d_2$ at the other, subjected to an axial pull $P$. Young's modulus is $E$. ```text ____ ______----'''' | _____----'''' | P <--- |d1 x --> d2| ---> P '''''----______ | ''''''----____ | '''' |<------------- L ------------>| ``` #### Derivation Measure $x$ from the small end (diameter $d_1$, taking $d_2 > d_1$). The diameter at distance $x$ is $$ d_x = d_1 + \frac{d_2 - d_1}{L}\,x = d_1 + kx,\qquad k = \frac{d_2 - d_1}{L} $$ Cross-sectional area: $A_x = \dfrac{\pi}{4}d_x^2$, so the stress is $\sigma_x = \dfrac{4P}{\pi d_x^2}$. Elongation of a small length $dx$: $$ d\delta = \frac{\sigma_x}{E}dx = \frac{4P}{\pi E d_x^2}dx $$ Total elongation: $$ \delta = \frac{4P}{\pi E}\int_0^L\frac{dx}{(d_1 + kx)^2} = \frac{4P}{\pi E}\left[-\frac{1}{k(d_1 + kx)}\right]_0^L = \frac{4P}{\pi E k}\left[\frac{1}{d_1} - \frac{1}{d_2}\right] $$ Since $d_1 + kL = d_2$, and $\dfrac{1}{d_1} - \dfrac{1}{d_2} = \dfrac{d_2 - d_1}{d_1d_2} = \dfrac{kL}{d_1d_2}$: $$ \boxed{\delta = \frac{4PL}{\pi E\,d_1\,d_2}} $$ #### Checks - If $d_1 = d_2 = d$, then $\delta = \dfrac{4PL}{\pi Ed^2} = \dfrac{PL}{AE}$, the formula for a prismatic bar. - The formula holds only for a gradual taper, away from the ends, where the stress is assumed uniform over each section (Saint-Venant's principle). **Example:** $P = 50$ kN, $L = 1000$ mm, $d_1 = 20$ mm, $d_2 = 40$ mm, $E = 200$ GPa gives $\delta = \dfrac{4(50\times10^3)(1000)}{\pi(200\times10^3)(20)(40)} = 0.398$ mm. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the elongation of the tapering bar found? A. Use PL/AE with the average area B. Use PL/AE with the smaller diameter C. Integrate dδ = σx dx/E over the length with area varying with x D. Use the average of the two stresses Answer: C. Area changes along the length so the strain must be integrated over elements. (Key point: Approach) 2. What is the diameter at x from the small end? A. d1 + x B. d1 + (d2 - d1)x/L C. (d1 + d2)/2 D. d1 L/x Answer: B. The taper is uniform, so diameter varies linearly with x. (Key point: Diameter law) 3. What is the final elongation? A. 4PL/(π E d1 d2) B. 4PL/(π E d1^2) C. PL/(π E (d1+d2)^2/4) D. 2PL/(π E d1 d2) Answer: A. The integral gives (4P/πEk)(1/d1 - 1/d2) with k = (d2-d1)/L, which simplifies to 4PL/(πE d1 d2). (Key point: Result) 4. What happens to the formula if d1 = d2 = d? A. It becomes zero B. It becomes infinite C. It halves D. It reduces to PL/AE Answer: D. 4PL/(πEd^2) = PL/(AE), the prismatic formula. 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