--- title: "Sal timber column, 50 kN, 4 m" subject: "Design of Steel and Timber Structure" chapter: 11 key: "semester_6/design_of_steel_and_timber_structure:timber-column-sal-50kn-4m" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/timber-column-sal-50kn-4m site: BCT Engineering --- # Sal timber column, 50 kN, 4 m Design of Steel and Timber Structure · Chapter 11: Design of Compression Members - Asked in: 2075 Baisakh, 10 marks - Video: [Steel and timber numerical 2081/sal timber numerical](https://www.youtube.com/watch?v=BGZ65AQOerk) — Rabin silwal - 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-50kn-4m - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-11 ## Question Design a timber column of Sal species to carry the axial load of 50 KN. Unsupported length of column is 4 m. ## Short answer Provide solid Sal column 125 mm × 125 mm (Grade I, inside). Long column, permissible axial stress 4.07 N/mm², capacity 64 kN > 50 kN. ## 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 $N = 50$ kN. **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 assumed. A square section is chosen. #### Trial section Try **125 mm × 125 mm** (square). Area $A = 15,625$ mm²; least dimension $d = 125$ mm. #### Slenderness and permissible axial stress (cl. 7.6.1) $$ \begin{aligned} S/d &= \frac{4000}{125} = 32.0 \\ K_8 &= 0.584\sqrt{\frac{E}{f_{cp}}} = 0.584\sqrt{\frac{12670}{10.60}} = 20.19 \end{aligned} $$ $S/d = 32.0 > K_8 = 20.2$, so it is a **long column**: $$ f_c = \frac{0.329E}{(S/d)^2} = \frac{0.329\times12670}{32.0^2} = 4.07\ \text{N/mm}^2 $$ $S/d = 32.0 < 50$ (cl. 7.6.1.4). OK. #### Check $$ P = f_cA = 4.07\times15,625/10^3 = 63.6\ \text{kN} \ge N = 50\ \text{kN} $$ Safe. #### Answer Provide a solid Sal column of **125 mm × 125 mm** cross-section (Grade I (standard), inside location). Capacity $64$ kN $> 50$ kN. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. S/d = 32 against K8 = 20.19. What kind of column is this? A. Short B. Intermediate C. Long D. Spaced Answer: C. S/d exceeds K8, so it is a long column that fails by elastic buckling. (Key point: Long column) 2. What formula gives fc for the 125 mm square? A. fc = fcp B. fc = 0.329E/(S/d)^2 = 4.07 N/mm^2 C. fc = fcp[1 - (1/3)(S/d/K8)^4] D. fc = 0.329E x (S/d)^2 Answer: B. Euler-type formula: 0.329 x 12670/32^2 = 4.07 N/mm^2. (Key point: Long column stress) 3. What is the upper limit of S/d for solid columns, and is it met? A. 50; 32 < 50 so OK B. 11; 32 > 11 so it fails C. 80; 32 < 80 so OK D. 20; 32 > 20 so it fails Answer: A. IS 883 cl. 7.6.1.4 limits solid columns to S/d <= 50. (Key point: Slenderness limit) 4. What is the capacity of the 125 x 125 mm section? A. P = 4.07 x 15,625 = 63.6 kN > 50 kN B. P = 10.6 x 15,625 = 165 kN C. P = 4.07 x 125 = 0.5 kN D. P = 40 kN < 50 kN Answer: A. P = fc x A = 4.07 x 15,625 = 63.6 kN, above 50 kN. 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