--- title: "Rectangular Sal column, 150 kN + 15 kNm" subject: "Design of Steel and Timber Structure" chapter: 11 key: "semester_6/design_of_steel_and_timber_structure:timber-column-sal-4m-150kn-15knm" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/timber-column-sal-4m-150kn-15knm site: BCT Engineering --- # Rectangular Sal column, 150 kN + 15 kNm Design of Steel and Timber Structure · Chapter 11: Design of Compression Members - Asked in: 2076 Bhadra, 12 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/timber-column-sal-4m-150kn-15knm - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-11 ## Question Design a Sal timber rectangular column having length 4.0 m, subjected to axial compressive force 150 kN and bending moment 15 kNm. ## Short answer Provide solid Sal column 175 mm × 250 mm (Grade I, inside). Long column, permissible axial stress 7.98 N/mm², bending 16.90 N/mm²; combined ratio 0.92 < 1. ## Answer **Approach.** Working stress design to IS 883:1994 (cl. 7.6 for columns, cl. 7.7 for combined axial compression and bending). Ends are taken as pin-ended, so the effective length equals the unsupported length $S$ (cl. 7.6.1.5). #### Data and permissible stresses Length $S = 4.00$ m = 4000 mm. Axial load $N = 150$ kN, bending moment $M = 15$ kN·m. **Species:** Sal (Shorea robusta), Group A. **Location:** inside. **Grade:** Grade I (standard) (Table 1 values are for Grade I). Permissible stresses (IS 883:1994 Table 1, inside location): | Property | Table 1 value (N/mm²) | Used (N/mm²) | |---|---|---| | Bending / tension along grain $f_b$ | 16.9 | 16.90 | | Compression parallel to grain $f_{cp}$ | 10.6 | 10.60 | | Compression perpendicular to grain $f_{cn}$ | 4.6 | 4.60 | | Horizontal shear $H$ | 0.94 | 0.94 | | Modulus of elasticity $E$ | 12670 | 12670 | Location and grade not stated: **inside**, Grade I (standard) assumed. Bending about the major axis. #### Trial section Try **175 mm × 250 mm**. Area $A = 43,750$ mm²; least dimension $d = 175$ mm. #### Slenderness and permissible axial stress (cl. 7.6.1) $$ \begin{aligned} S/d &= \frac{4000}{175} = 22.9 \\ K_8 &= 0.584\sqrt{\frac{E}{f_{cp}}} = 0.584\sqrt{\frac{12670}{10.60}} = 20.19 \end{aligned} $$ $S/d = 22.9 > K_8 = 20.2$, so it is a **long column**: $$ f_c = \frac{0.329E}{(S/d)^2} = \frac{0.329\times12670}{22.9^2} = 7.98\ \text{N/mm}^2 $$ $S/d = 22.9 < 50$ (cl. 7.6.1.4). OK. #### Combined axial compression and bending (cl. 7.7.1) $$ \begin{aligned} f_{ac} &= \frac{N}{A} = \frac{150000}{43,750} = 3.429\ \text{N/mm}^2 \\ f_{ab} &= \frac{6M}{bd^2}\ \text{(bending about the strong axis)} = \frac{6\times15000000}{175\times250^2} = 8.229\ \text{N/mm}^2 \end{aligned} $$ Permissible bending stress $f_b = 16.90$ N/mm²; the form factor is not applied for depths up to 300 mm (cl. 7.5.4). $$ \frac{f_{ac}}{f_c} + \frac{f_{ab}}{f_b} = \frac{3.429}{7.98} + \frac{8.229}{16.90} = 0.917 \le 1.0 $$ Safe. #### Answer Provide a solid Sal column of **175 mm × 250 mm** cross-section (Grade I (standard), inside location). Combined stress ratio $= 0.92 < 1$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. S/d = 22.9 and K8 = 20.19. Which class and formula apply? A. Intermediate; fc = fcp[1 - (1/3)(S/d/K8)^4] B. Short; fc = fcp C. Long; fc = 0.329E/(S/d)^2 D. Long; fc = fcp/(S/d) Answer: C. S/d exceeds K8, so the column buckles elastically: fc = 0.329 x 12670/22.9^2 = 7.98 N/mm^2. (Key point: Column class) 2. For the 175 x 250 mm section, which dimension is used in S/d? A. 250 mm B. 175 mm C. 212.5 mm D. The diagonal 305 mm Answer: B. Slenderness uses the least lateral dimension: 4000/175 = 22.9. (Key point: Least dimension) 3. How is the bending stress about the strong axis computed? A. fab = N/A B. fab = 6M/(bd^2) with b = 175, d = 250, giving 8.23 N/mm^2 C. fab = 6M/(db^2) with d = 250, b = 175 D. fab = M/(bd) Answer: B. Bending about the major axis uses d = 250 mm: 6 x 15 x 10^6/(175 x 250^2) = 8.229 N/mm^2. (Key point: Bending stress) 4. What is the interaction value and no form factor reason? A. 0.92; depth 250 mm is up to 300 mm so no form factor is applied B. 1.15; section fails C. 0.92; form factor 0.95 is applied D. 0.50; bending is ignored Answer: A. 3.429/7.98 + 8.229/16.9 = 0.917 <= 1; the form factor applies only to depths over 300 mm. 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