--- title: "Timber column under axial load and bending" subject: "Design of Timber and Masonry Structures" chapter: 3 key: "semester_5/design_of_timber_and_masonry_structures:timber-column-combined-bending-axial" source: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/timber-column-combined-bending-axial site: BCT Engineering --- # Timber column under axial load and bending Design of Timber and Masonry Structures · Chapter 3: Structural Elements of Timber Structures - Asked in: year not given, 8 marks - Video: [Timber column subject to Axial load and bending moment](https://www.youtube.com/watch?v=6Bv8lOCil5g) — DAVID OLUMASAI - Answer checked by Claude Sonnet 5.5 on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_5/design_of_timber_and_masonry_structures/timber-column-combined-bending-axial - Whole chapter: https://www.bctengineer.com/md/semester_5/design_of_timber_and_masonry_structures/chapter-3 ## Question A timber column of 150 mm × 200 mm section (150 mm is the smaller dimension) is 3.0 m long and hinged at both ends. It carries an axial load of 80 kN and a bending moment of 5.0 kN·m about the axis parallel to the 150 mm side (so bending takes place over the 200 mm depth). Take $\sigma_{cc} = 8.0$ N/mm², $\sigma_{bc} = 10.0$ N/mm² and $E = 9000$ N/mm². Check the adequacy of the column with the interaction formula. ## Short answer Check interaction formula: (σc,cal/σc′) + (σbc,cal/σbc) ≤ 1. Actual stresses 2.67 N/mm² axial, 5.0 N/mm² bending. Permissible axial stress 6.61 N/mm² for le/d=20. Interaction = 0.90 < 1, column safe. ## Answer #### Interaction rule For members under combined direct compression and bending (IS 883), the condition is $$ \frac{\sigma_{c,cal}}{\sigma_c'} + \frac{\sigma_{bc,cal}}{\sigma_{bc}} \le 1 $$ where $\sigma_c'$ is the permissible axial compressive stress allowing for slenderness. (Magnification of moment due to deflection is ignored here for simplicity.) #### Step 1: Actual stresses $$ \begin{aligned} A &= 150 \times 200 = 30\,000\ \text{mm}^2 \\ \sigma_{c,cal} &= \frac{80\,000}{30\,000} = 2.67\ \text{N/mm}^2 \\ Z &= \frac{150 \times 200^2}{6} = 1.0 \times 10^{6}\ \text{mm}^3 \\ \sigma_{bc,cal} &= \frac{5.0 \times 10^6}{1.0 \times 10^6} = 5.0\ \text{N/mm}^2 \end{aligned} $$ #### Step 2: Permissible axial stress Least dimension $d = 150$ mm, so $l_e/d = 3000/150 = 20$. $$ K_8 = 0.702\sqrt{\frac{9000}{8}} = 0.702 \times 33.54 = 23.55 $$ Since $11 < 20 < 23.55$, the column is intermediate: $$ \sigma_c' = 8\left[1 - \frac{1}{3}\left(\frac{20}{23.55}\right)^4\right] = 8\,[1 - 0.1735] = 6.61\ \text{N/mm}^2 $$ #### Step 3: Interaction check $$ \frac{2.67}{6.61} + \frac{5.0}{10.0} = 0.403 + 0.500 = 0.903 < 1 $$ The column is safe; about 10 percent reserve remains. **Answer: Interaction value = 0.90 < 1.0, so the 150 mm × 200 mm column is safe.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which condition checks a column under combined compression and bending? A. sigma_c,cal/sigma_c' + sigma_bc,cal/sigma_bc <= 1 B. sigma_c,cal + sigma_bc,cal <= sigma_cc C. sigma_c,cal x sigma_bc,cal <= 1 D. sigma_c,cal/sigma_bc + sigma_bc,cal/sigma_c' <= 1 Answer: A. The sum of the ratio of actual to permissible axial stress and actual to permissible bending stress must not exceed 1. (Key point: Interaction formula) 2. What slenderness ratio is used for the 150 x 200 mm column, 3000 mm long, hinged at both ends? A. 3000/200 = 15 B. 3000/150 = 20 C. 3000/175 = 17.1 D. 3000 x 2/150 = 40 Answer: B. Slenderness uses the least dimension (150 mm) with l_e = 3000 mm, so l_e/d = 20. (Key point: Slenderness uses least dimension) 3. With K8 = 23.55, how is the column classified and what is sigma_c'? A. Short column, 8.0 N/mm^2 B. Long column, 0.329E/(l_e/d)^2 C. Long column, 3.5 N/mm^2 D. Intermediate column, about 6.61 N/mm^2 Answer: D. 11 < 20 < 23.55, so it is intermediate: sigma_c' = 8[1 - (1/3)(20/23.55)^4] = 6.61 N/mm^2. (Key point: Permissible axial stress) 4. With axial stress 2.67 N/mm^2 and bending stress 5.0 N/mm^2, what is the interaction value? A. 0.90, so the column is safe B. 1.10, so it is unsafe C. 0.50, so it is safe D. 1.50, so it is unsafe Answer: A. 2.67/6.61 + 5.0/10.0 = 0.403 + 0.500 = 0.903 < 1; Z = 150 x 200^2/6 = 1.0 x 10^6 mm^3 gives 5.0 N/mm^2. 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