--- title: "Square deodar column 4 m, 350 kN, outside" subject: "Design of Steel and Timber Structure" chapter: 11 key: "semester_6/design_of_steel_and_timber_structure:timber-column-deodar-square-4m-350kn" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/timber-column-deodar-square-4m-350kn site: BCT Engineering --- # Square deodar column 4 m, 350 kN, outside Design of Steel and Timber Structure · Chapter 11: Design of Compression Members - Asked in: 2069 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-deodar-square-4m-350kn - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-11 ## Question Design a 4m long square column of deodar planks to carry an axial load of 350KN. Take outside location and select grade of timber. ## Short answer Provide a solid Deodar column of 250 mm × 250 mm (Select grade, outside). S/d = 16 (intermediate), fc = 6.93 N/mm², capacity = 433 kN > 350 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 $N = 350$ kN. **Species:** Deodar (Cedrus deodara), Group C. **Location:** outside. **Grade:** Select grade (Grade I values multiplied by 1.16, IS 883:1994 cl. 6.3). Permissible stresses (IS 883:1994 Table 1, outside location): | Property | Table 1 value (N/mm²) | Used (N/mm²) | |---|---|---| | Bending / tension along grain $f_b$ | 8.7 | 10.09 | | Compression parallel to grain $f_{cp}$ | 6.9 | 8.00 | | Compression perpendicular to grain $f_{cn}$ | 2.1 | 2.44 | | Horizontal shear $H$ | 0.7 | 0.81 | | Modulus of elasticity $E$ | 9480 | 9480 | The question states **outside** location and **Select** grade. #### Trial section Try **250 mm × 250 mm** (square). Area $A = 62,500$ mm²; least dimension $d = 250$ mm. #### Slenderness and permissible axial stress (cl. 7.6.1) $$ \begin{aligned} S/d &= \frac{4000}{250} = 16.0 \\ K_8 &= 0.584\sqrt{\frac{E}{f_{cp}}} = 0.584\sqrt{\frac{9480}{8.00}} = 20.10 \end{aligned} $$ $11 < S/d = 16.0 < K_8 = 20.1$, so it is an **intermediate column**: $$ f_c = f_{cp}\left[1 - \frac{1}{3}\left(\frac{S/d}{K_8}\right)^4\right] = 8.00\left[1 - \frac{1}{3}\left(\frac{16.0}{20.1}\right)^4\right] = 6.93\ \text{N/mm}^2 $$ $S/d = 16.0 < 50$ (cl. 7.6.1.4). OK. #### Check $$ P = f_cA = 6.93\times62,500/10^3 = 433.3\ \text{kN} \ge N = 350\ \text{kN} $$ Safe. #### Answer Provide a solid Deodar column of **250 mm × 250 mm** cross-section (Select grade, outside location). Capacity $433$ kN $> 350$ kN. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What fcp is used for Select grade Deodar outside? A. 6.9 N/mm^2 B. 8.00 N/mm^2 (6.9 x 1.16) C. 7.8 N/mm^2 D. 10.09 N/mm^2 Answer: B. Select grade is Grade I x 1.16; outside fcp 6.9 gives about 8.0 N/mm^2. (Key point: Select grade outside values) 2. How is K8 found and what is its value? A. 0.584 sqrt(E/fcp) = 20.10 using the Select fcp B. 0.329 sqrt(E/fcp) C. pi/2 sqrt(UE/5qfcp) D. 0.584 sqrt(fcp/E) Answer: A. K8 = 0.584 sqrt(9480/8.0) = 20.10. (Key point: K8) 3. S/d = 16 and K8 = 20.1. What formula? A. fc = fcp B. Intermediate: fc = fcp[1 - (1/3)(S/d / K8)^4] = 6.93 C. Long: 0.329E/(S/d)^2 D. Box column formula Answer: B. 11 < 16 < 20.1 so the intermediate formula is used. (Key point: Classification) 4. What is the conclusion for the 250 x 250 mm section? A. 433 kN > 350 kN, safe B. 433 kN < 350 kN C. 350 kN D. 500 kN, unsafe Answer: A. P = 6.93 x 62,500 = 433 kN. 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