--- title: "Tubular column load and eccentric loading" subject: "Strength of Materials" chapter: 10 key: "semester_4/strength_of_materials_mechanical:tube-column-and-eccentric-short-column" source: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/tube-column-and-eccentric-short-column site: BCT Engineering --- # Tubular column load and eccentric loading Strength of Materials · Chapter 10: Columns - Asked in: year not given, 4+4 marks - Video: [TT27-a Critical Buckling Load for a Hollow Cylindrical Column](https://www.youtube.com/watch?v=9CQefrNbY_o) — Seeing Structures - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/strength_of_materials_mechanical/tube-column-and-eccentric-short-column - Whole chapter: https://www.bctengineer.com/md/semester_4/strength_of_materials_mechanical/chapter-10 ## Question (a) A hollow circular steel tube of 80 mm outside diameter and 64 mm inside diameter, 3 m long, is used as a column. E = 200 GPa. Find the Euler crippling load and the slenderness ratio when (i) both ends are hinged and (ii) one end is fixed and the other free. (b) A short column of rectangular section 150 mm x 250 mm carries a compressive load of 200 kN with an eccentricity of 40 mm measured along the 250 mm side. Find the maximum and minimum stresses at the base, and state the limit of eccentricity for no tension. ## Short answer Using I=1.187×10^6 mm^4, k=25.6 mm: hinged-hinged λ=117, Pcr=π²EI/L²=260.4 kN; fixed-free λ=234, Pcr=65.1 kN. Eccentric short column: σmax=P/A+Pe/Z=10.45 MPa, σmin=0.21 MPa; no-tension limit e≤h/6=41.7 mm. ## Answer #### (a) Euler load for the tube $$ I = \frac{\pi}{64}(80^4 - 64^4) = 1.187\times10^6\ \text{mm}^4 $$ $$ A = \frac{\pi}{4}(80^2 - 64^2) = 1809.6\ \text{mm}^2, \qquad k = \sqrt{\frac IA} = 25.6\ \text{mm} $$ | End condition | $L_e$ (mm) | $\lambda = L_e/k$ | $P_{cr} = \dfrac{\pi^2EI}{L_e^2}$ | |---|---|---|---| | (i) Hinged-hinged | 3000 | 117 | $\dfrac{\pi^2\times200000\times1.187\times10^6}{3000^2} = 260.4$ kN | | (ii) Fixed-free | 6000 | 234 | $\dfrac{\pi^2\times200000\times1.187\times10^6}{6000^2} = 65.1$ kN | The crippling load of the fixed-free column is one quarter of that of the hinged column. The Euler stresses are $143.9$ MPa and $36.0$ MPa, both below the yield stress and consistent with $\lambda>100$, so Euler's formula is applicable. #### (b) Eccentrically loaded short column Eccentricity is along the 250 mm side ($h = 250$ mm, $b = 150$ mm). $$ A = 150\times250 = 37\,500\ \text{mm}^2, \qquad Z = \frac{bh^2}{6} = \frac{150\times250^2}{6} = 1.5625\times10^6\ \text{mm}^3 $$ $$ \sigma_0 = \frac PA = \frac{200\times10^3}{37500} = 5.33\ \text{MPa} $$ $$ \sigma_b = \frac{Pe}{Z} = \frac{200\times10^3\times40}{1.5625\times10^6} = 5.12\ \text{MPa} $$ $$ \sigma_{max} = 5.33 + 5.12 = 10.45\ \text{MPa (compressive)}, \qquad \sigma_{min} = 5.33 - 5.12 = 0.21\ \text{MPa (compressive)} $$ No tension if $\sigma_{min} \ge 0$: $$ \frac{P}{A}\ge\frac{Pe}{Z} \;\Rightarrow\; e\le\frac ZA = \frac h6 = 41.7\ \text{mm} $$ The load must lie within the **middle third** of the depth (the core, kernel). Here $e = 40 < 41.7$ mm, so there is no tension. **Answer: (a) $P_{cr} = 260.4$ kN ($\lambda=117$) and $65.1$ kN ($\lambda=234$); (b) $\sigma_{max} = 10.45$ MPa, $\sigma_{min} = 0.21$ MPa, limit $e = h/6 = 41.7$ mm.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does Pcr of the fixed-free tube compare with the hinged one? A. One quarter, since Le = 2L B. Half, since Le = √2 L C. Double, since Le = L/2 D. The same Answer: A. Pcr varies as 1/Le², so Le = 2L gives 1/4 (65.1 vs 260.4 kN). (Key point: Fixed-free comparison) 2. What is the slenderness ratio for the hinged tube? A. 234 B. 117, from Le/k = 3000/25.6 C. 75 D. 25.6 Answer: B. k = sqrt(I/A) = 25.6 mm, so λ = 3000/25.6 = 117; the fixed-free value is 234. (Key point: Slenderness ratio) 3. What are the maximum and minimum stresses at the base? A. 5.33 MPa and 5.12 MPa B. 10.45 MPa compression and 0.21 MPa tension C. 10.45 MPa and 0.21 MPa, both compressive D. 5.12 MPa and 0 Answer: C. P/A = 5.33 plus and minus Pe/Z = 5.12. (Key point: Eccentric load stresses) 4. What is the eccentricity limit for no tension? A. e ≤ h/2 = 125 mm B. e ≤ h/3 = 83 mm C. e ≤ h/12 = 20.8 mm D. e ≤ h/6 = 41.7 mm (middle third) Answer: D. No tension needs P/A ≥ Pe/Z, i.e. e ≤ Z/A = h/6. 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