--- title: "Check 25 cm x 25 cm Sal column, 200 kN + 20 kNm" subject: "Design of Steel and Timber Structure" chapter: 11 key: "semester_6/design_of_steel_and_timber_structure:timber-column-sal-25x25-200kn-20knm" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/timber-column-sal-25x25-200kn-20knm site: BCT Engineering --- # Check 25 cm x 25 cm Sal column, 200 kN + 20 kNm Design of Steel and Timber Structure · Chapter 11: Design of Compression Members - Asked in: 2068 Magh (old course), 10 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-25x25-200kn-20knm - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-11 ## Question If a sal-wood column of 25cm × 25cm has a length of 4m, determine whether the column can carry 200KN axial load and 20KNm bending moment. Assume suitable data if necessary. ## Short answer Check combined axial and bending per IS 883. Actual stresses: f_ac=3.20, f_ab=7.68 N/mm²; permissible: f_c=9.207, f_b=16.9 N/mm². Ratio 0.80<1, column safe. ## Answer **Approach.** Combined axial compression and bending, IS 883:1994 cl. 7.7.1: safe if $\dfrac{f_{ac}}{f_c} + \dfrac{f_{ab}}{f_b} \le 1$. #### Data and permissible stresses Section 250 × 250 mm, $S = 4000$ mm (pin ends), $N = 200$ kN, $M = 20$ 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 are not stated: **inside**, Grade I (standard) assumed. #### Permissible axial stress $$ \begin{aligned} S/d &= 4000/250 = 16.0 \\ K_8 &= 0.584\sqrt{12670/10.6} = 20.19 \end{aligned} $$ $11 < S/d < K_8$: intermediate column. $$ f_c = 10.6\left[1 - \frac{1}{3}\left(\frac{16.0}{20.19}\right)^4\right] = 9.207\ \text{N/mm}^2 $$ #### Actual stresses $$ \begin{aligned} f_{ac} &= \frac{200\times10^3}{250\times250} = 3.200\ \text{N/mm}^2 \\ f_{ab} &= \frac{6M}{bd^2} = \frac{6\times20\times10^6}{250^3} = 7.680\ \text{N/mm}^2 \end{aligned} $$ (The depth is 250 mm, which is less than 300 mm, so no form factor: $f_b = 16.9$ N/mm².) #### Check $$ \frac{3.200}{9.207} + \frac{7.680}{16.9} = 0.348 + 0.454 = 0.802 $$ **Answer: the value is 0.80, which is less than 1: the column can carry the 200 kN load and 20 kN·m moment safely.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a timber column under axial load and bending, which condition must be satisfied for safety? A. f_ac/f_c + f_ab/f_b <= 1 B. f_ac + f_ab <= f_c C. f_ac/f_c <= 1 and f_ab/f_b <= 1 separately only, with no interaction D. f_ac/f_b + f_ab/f_c <= 1 Answer: A. IS 883 cl. 7.7.1 uses an interaction equation: actual axial stress over permissible axial stress plus actual bending stress over permissible bending stress must not exceed 1. (Key point: Combined stress check) 2. The column has S = 4000 mm and d = 250 mm, so S/d = 16, and K8 = 20.19. How is it classified? A. Short column, since S/d is above 11 B. Long column, since S/d is above 11 C. Intermediate column, since 11 < S/d < K8 D. Intermediate column, since S/d > K8 Answer: C. S/d = 16 lies between 11 and K8 = 20.19, so the intermediate-column formula for f_c is used. (Key point: Slenderness and column type) 3. Which expression gives the permissible compressive stress for this intermediate column? A. f_c = f_cp (no reduction) B. f_c = f_cp[1 - (1/3)(S/d / K8)^4] C. f_c = 0.329 E / (S/d)^2 D. f_c = f_cp x (S/d) / K8 Answer: B. For intermediate columns the parallel-to-grain stress is reduced by the factor [1 - (1/3)(S/(d K8))^4], giving 9.207 N/mm^2. (Key point: Permissible axial stress) 4. With f_ac = 3.2, f_c = 9.207, f_ab = 7.68 and f_b = 16.9 N/mm^2, what is the conclusion? A. Ratio sum = 0.80 < 1, so the column is safe B. Ratio sum = 1.20 > 1, so the column is unsafe C. Ratio sum = 0.80 but the column is unsafe because bending is large D. Ratio sum = 0.35, because bending is ignored below 300 mm depth Answer: A. 3.2/9.207 + 7.68/16.9 = 0.348 + 0.454 = 0.802, which is below 1, so the column carries the loads safely. 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