--- title: "Solid Sal column, 75 kN + 12 kNm, 2 m" subject: "Design of Steel and Timber Structure" chapter: 11 key: "semester_6/design_of_steel_and_timber_structure:timber-solid-column-75kn-12knm" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/timber-solid-column-75kn-12knm site: BCT Engineering --- # Solid Sal column, 75 kN + 12 kNm, 2 m Design of Steel and Timber Structure · Chapter 11: Design of Compression Members - Asked in: 2074 Bhadra, 6 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-solid-column-75kn-12knm - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-11 ## Question Design a solid wood column to resist a factored axial load of 75 KN and Factored moment of 12 KNm. The column is made of Sal wood and is 2m long. ## Short answer Provide a solid Sal column of 125 mm × 225 mm. S/d=16 (intermediate), fc=9.21 N/mm²; combined stress ratio (fac/fc + fab/fb) = 0.96 < 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 $S = 2.00$ m = 2000 mm. Axial load $N = 75$ kN, bending moment $M = 12$ 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 | The loads are used as given (IS 883 is a working-stress code, so they are treated as the design loads); location and grade are not stated: **inside**, Grade I. Bending about the major axis. #### Trial section Try **125 mm × 225 mm**. Area $A = 28,125$ mm²; least dimension $d = 125$ mm. #### Slenderness and permissible axial stress (cl. 7.6.1) $$ \begin{aligned} S/d &= \frac{2000}{125} = 16.0 \\ K_8 &= 0.584\sqrt{\frac{E}{f_{cp}}} = 0.584\sqrt{\frac{12670}{10.60}} = 20.19 \end{aligned} $$ $11 < S/d = 16.0 < K_8 = 20.2$, so it is an **intermediate column**: $$ f_c = f_{cp}\left[1 - \frac{1}{3}\left(\frac{S/d}{K_8}\right)^4\right] = 10.60\left[1 - \frac{1}{3}\left(\frac{16.0}{20.2}\right)^4\right] = 9.21\ \text{N/mm}^2 $$ $S/d = 16.0 < 50$ (cl. 7.6.1.4). OK. #### Combined axial compression and bending (cl. 7.7.1) $$ \begin{aligned} f_{ac} &= \frac{N}{A} = \frac{75000}{28,125} = 2.667\ \text{N/mm}^2 \\ f_{ab} &= \frac{6M}{bd^2}\ \text{(bending about the strong axis)} = \frac{6\times12000000}{125\times225^2} = 11.378\ \text{N/mm}^2 \end{aligned} $$ 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). $$ \frac{f_{ac}}{f_c} + \frac{f_{ab}}{f_b} = \frac{2.667}{9.21} + \frac{11.378}{16.90} = 0.963 \le 1.0 $$ Safe. #### Answer Provide a solid Sal column of **125 mm × 225 mm** cross-section (Grade I (standard), inside location). Combined stress ratio $= 0.96 < 1$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For 125 x 225 mm, S/d = 16 and K8 = 20.19. What class? A. Short B. Intermediate C. Long D. Not allowed Answer: B. 11 < 16 < K8, so the intermediate formula is used to get fc = 9.21 N/mm^2. (Key point: Classification) 2. How is fab calculated? A. 6M/(bd^2) with d = 225 mm, 11.38 N/mm^2 B. 6M/(db^2) with b = 125 mm C. M/A = 0.43 N/mm^2 D. 6M/(bd^2) with d = 125 mm Answer: A. Bending is about the major axis, so depth d = 225 mm: 6 x 12 x 10^6/(125 x 225^2) = 11.378 N/mm^2. (Key point: Bending stress) 3. What is the combined check and result? A. 2.667/9.21 + 11.378/16.9 = 0.96 <= 1, safe B. 2.667/16.9 + 11.378/9.21 = 1.39, unsafe C. 2.667 + 11.378 = 14.0 < 16.9, safe D. 11.378/16.9 = 0.67 only Answer: A. Each stress is divided by its own permissible value; 0.289 + 0.673 = 0.963. 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