--- title: "Safe load, 225x225 select Sal column, 3 m" subject: "Design of Steel and Timber Structure" chapter: 11 key: "semester_6/design_of_steel_and_timber_structure:timber-column-sal-select-225x225" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/timber-column-sal-select-225x225 site: BCT Engineering --- # Safe load, 225x225 select Sal column, 3 m Design of Steel and Timber Structure · Chapter 11: Design of Compression Members - Asked in: 2071 Bhadra, 8 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-select-225x225 - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-11 ## Question A timber column 225 × 225 in cross section having an unsupported length of 3 m. Assuming the column to be of sal wood of selected grade, find the safe axial load. ## Short answer Slenderness S/d = 13.33 (intermediate). Permissible stress fc = 11.25 N/mm², safe axial load = 569 kN. ## Answer **Approach.** Safe axial load of a solid column, IS 883:1994 cl. 7.6.1: $P = f_c\times A$, where $f_c$ depends on the slenderness $S/d$. #### Data and permissible stresses Section 225 × 225 mm, unsupported length $S = 3000$ mm (pin ends), Select grade Sal. **Species:** Sal (Shorea robusta), Group A. **Location:** inside. **Grade:** Select grade (Grade I values multiplied by 1.16, IS 883:1994 cl. 6.3). 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 | 19.60 | | Compression parallel to grain $f_{cp}$ | 10.6 | 12.30 | | Compression perpendicular to grain $f_{cn}$ | 4.6 | 5.34 | | Horizontal shear $H$ | 0.94 | 1.09 | | Modulus of elasticity $E$ | 12670 | 12670 | Location not stated: **inside**. #### Slenderness $$ \begin{aligned} S/d &= \frac{3000}{225} = 13.33 \\ K_8 &= 0.584\sqrt{\frac{E}{f_{cp}}} = 0.584\sqrt{\frac{12670}{12.30}} = 18.75 \end{aligned} $$ $11 < S/d = 13.33 < K_8 = 18.75$, so it is an **intermediate column**. $$ f_c = f_{cp}\left[1 - \frac{1}{3}\left(\frac{S/d}{K_8}\right)^4\right] = 12.30\left[1 - \frac{1}{3}\left(\frac{13.33}{18.75}\right)^4\right] = 11.25\ \text{N/mm}^2 $$ #### Safe load $$ P = f_cA = 11.25\times50,625/10^3 = 569\ \text{kN} $$ **Answer: the safe axial load is about 569 kN** (permissible stress $f_c = 11.25$ N/mm²; $S/d = 13.3 < 50$). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the safe axial load of a solid column found? A. P = fc x A, with fc depending on S/d B. P = fcp x A always C. P = fb x A D. P = E x A/(S/d) Answer: A. The permissible stress fc is first found from slenderness and then multiplied by the area. (Key point: Safe load formula) 2. S/d = 13.33 and K8 = 18.75. What class? A. Short B. Intermediate C. Long D. Exceeds limit Answer: B. 11 < 13.33 < 18.75, so it is intermediate. (Key point: Classification) 3. What fcp is used for Select grade Sal? A. 10.6 N/mm^2 B. 12.30 N/mm^2 = 10.6 x 1.16 C. 9.4 N/mm^2 D. 16.9 N/mm^2 Answer: B. Select grade multiplies Grade I values by 1.16. (Key point: Select grade fcp) 4. What is the safe axial load? A. About 569 kN (fc = 11.25 N/mm^2) B. About 620 kN C. About 400 kN D. About 569 kN with fc = 12.3 N/mm^2 Answer: A. P = 11.25 x 50,625 = 569 kN. 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