--- title: "Box deodar column, 60 mm planks, 350 kN" subject: "Design of Steel and Timber Structure" chapter: 11 key: "semester_6/design_of_steel_and_timber_structure:timber-box-column-deodar-5m-350kn" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/timber-box-column-deodar-5m-350kn site: BCT Engineering --- # Box deodar column, 60 mm planks, 350 kN Design of Steel and Timber Structure · Chapter 11: Design of Compression Members - Asked in: 2071 Magh, 10 marks - Video: [Find axial load on on rectangular box column|timber numerical |numerical from timber|5th semester](https://www.youtube.com/watch?v=Db12TE4o8cM) — Hamro Engineering - 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-box-column-deodar-5m-350kn - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-11 ## Question Design a 5 m long rectangular box column built by 60 mm thick deodar planks to carry an axial load of 350 kN. ## Short answer Provide a square box column of overall 300 mm × 300 mm with 60 mm thick Deodar planks (inside 180 mm). Slenderness ratio S/√(300²+180²) = 14.29 (intermediate), fc = 6.96 N/mm², capacity = 401 kN > 350 kN. ## Answer **Approach.** Box (built-up) column designed to IS 883:1994 cl. 7.6.2: the planks are nailed or bolted to form a hollow square section (a square is a special case of the rectangular box); the strength depends on $S/\sqrt{d_1^2 + d_2^2}$, where $d_1$ is the overall side and $d_2$ the inside side of the box. #### Data and permissible stresses Length $S = 5.00$ m, axial load $N = 350$ kN, plank thickness $t = 60$ mm. Pin ends, effective length $= S$. **Species:** Deodar (Cedrus deodara), Group C. **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 $f_b$ | 10.2 | 10.20 | | Compression parallel to grain $f_{cp}$ | 7.8 | 7.80 | | Compression perpendicular to grain $f_{cn}$ | 2.7 | 2.70 | | Horizontal shear $H$ | 0.7 | 0.70 | | Modulus of elasticity $E$ | 9480 | 9480 | Location and grade are not stated, so **inside** and Grade I (standard) are assumed. For plank thickness 60 mm (cl. 7.6.2.5): $U = 0.6$, $q = 1.0$ (the code gives values for 25 mm and 50 mm planks; the 50 mm values are used). #### Trial section Square box, overall side $d_1 = 300$ mm, inside side $d_2 = 300 - 2\times60 = 180$ mm. $$ A = d_1^2 - d_2^2 = 300^2 - 180^2 = 57,600\ \text{mm}^2 $$ $$ \begin{aligned} \frac{S}{\sqrt{d_1^2 + d_2^2}} &= \frac{5000}{\sqrt{300^2 + 180^2}} = \frac{5000}{349.9} = 14.29 \\ K_9 &= \frac{\pi}{2}\sqrt{\frac{UE}{5qf_{cp}}} = \frac{\pi}{2}\sqrt{\frac{0.6\times9480}{5\times1.0\times7.80}} = 18.97 \end{aligned} $$ $8 < 14.29 < K_9 = 18.97$, so it is an **intermediate column**: $$ f_c = q f_{cp}\left[1 - \frac{1}{3}\left(\frac{S}{K_9\sqrt{d_1^2 + d_2^2}}\right)^4\right] = 1.0\times7.80\left[1 - \frac{1}{3}\left(\frac{14.29}{18.97}\right)^4\right] = 6.96\ \text{N/mm}^2 $$ #### Check $$ P = f_cA = 6.96\times57,600/10^3 = 401\ \text{kN} > N = 350\ \text{kN} $$ Safe. #### Arrangement ```text +------------------------------+ | plank 60 mm thick | | +----------------------+ | | | hollow | | outside 300 x 300 mm | | 180 x 180 | | | +----------------------+ | +------------------------------+ ``` The 60 mm thick planks are fixed with staggered nails or bolts at a pitch not exceeding about 150 mm to 200 mm so that they act as one unit; the ends are fitted with solid end blocks. #### Answer Provide a box column of overall size **300 mm × 300 mm** formed with 60 mm thick Deodar planks (inside 180 mm). Capacity 401 kN $>$ 350 kN. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Planks are 60 mm, but code gives U and q for 25 and 50 mm. What is done? A. The 50 mm values U = 0.6, q = 1.0 are used B. Linearly extrapolate U C. Use U = 0.8 of 25 mm planks D. Reject 60 mm planks Answer: A. The answer states the 50 mm values are used. (Key point: Plank thickness constants) 2. With 300 mm overall and 60 mm planks, what is d2 and A? A. d2 = 180 mm, A = 57,600 mm^2 B. d2 = 240 mm, A = 32,400 mm^2 C. d2 = 200 mm, A = 50,000 mm^2 D. d2 = 180 mm, A = 90,000 mm^2 Answer: A. d2 = 300 - 2x60 = 180; A = 90,000 - 32,400 = 57,600 mm^2. (Key point: Inside side) 3. Ratio 14.29, K9 = 18.97. What class? A. Short B. Intermediate C. Long D. Spaced Answer: B. 8 < 14.29 < K9, so the intermediate formula gives fc = 6.96 N/mm^2. (Key point: Classification) 4. What is the final capacity? A. 401 kN > 350 kN, safe B. 401 kN < 350 kN C. 350 kN exactly D. 250 kN, unsafe Answer: A. P = 6.96 x 57,600 = 401 kN. 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