--- title: "Built-up Sal column, 50x250 planks, 925 kN" subject: "Design of Steel and Timber Structure" chapter: 11 key: "semester_6/design_of_steel_and_timber_structure:timber-builtup-sal-925kn-3-5m" source: https://www.bctengineer.com/q/semester_6/design_of_steel_and_timber_structure/timber-builtup-sal-925kn-3-5m site: BCT Engineering --- # Built-up Sal column, 50x250 planks, 925 kN Design of Steel and Timber Structure · Chapter 11: Design of Compression Members - Asked in: 2072 Asoj, 8 marks; 2072 Magh, 6 marks (asked 2 times) - 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-builtup-sal-925kn-3-5m - Whole chapter: https://www.bctengineer.com/md/semester_6/design_of_steel_and_timber_structure/chapter-11 ## Question Design a built up salwood column fabricated with 50 mm thick and 250 mm width planks to carry an axial load of 925 kN. The effective length of the column is 3.5 m. Take E = 12700 N/mm², $f_{cp}$ = 10.6 N/mm² constant U = 0.6 and q = 1. ## Short answer Built-up box column of 50mm thick Sal planks per IS 883:1994 cl. 7.6.2. Trial 500×500mm overall, 400×400mm inside (A=90,000mm²). S/√(d1²+d2²)=5.47 <8, short column; fc=q·fcp=10.6 N/mm². Capacity P=fc·A=954 kN > 925 kN. Safe. Use 250mm wide planks, staggered nails/bolts ≤200mm pitch, solid end blocks. ## 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; 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 $S = 3.50$ m, axial load $N = 925$ kN, plank thickness $t = 50$ mm. Pin ends, effective length $= S$. **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 $E$ | 12700 | 12700 | The question gives $E = 12700$ N/mm², $f_{cp} = 10.6$ N/mm² (the Table 1 value for Sal, Grade I, inside), $U = 0.6$ and $q = 1$; these are used. For plank thickness 50 mm (cl. 7.6.2.5): $U = 0.6$, $q = 1.0$ (values given for 25 mm and 50 mm planks). #### Trial section Square box, overall side $d_1 = 500$ mm, inside side $d_2 = 500 - 2\times50 = 400$ mm. $$ A = d_1^2 - d_2^2 = 500^2 - 400^2 = 90,000\ \text{mm}^2 $$ $$ \begin{aligned} \frac{S}{\sqrt{d_1^2 + d_2^2}} &= \frac{3500}{\sqrt{500^2 + 400^2}} = \frac{3500}{640.3} = 5.47 \\ K_9 &= \frac{\pi}{2}\sqrt{\frac{UE}{5qf_{cp}}} = \frac{\pi}{2}\sqrt{\frac{0.6\times12700}{5\times1.0\times10.60}} = 18.83 \end{aligned} $$ The ratio 5.47 is less than 8, so it is a **short column** (cl. 7.6.2.1): $f_c = q f_{cp} = 1.0\times10.60 = 10.60$ N/mm². #### Check $$ P = f_cA = 10.60\times90,000/10^3 = 954\ \text{kN} > N = 925\ \text{kN} $$ Safe. #### Arrangement of planks (250 mm × 50 mm planks) ```text +------------------------------+ | plank 50 mm thick | | +----------------------+ | | | hollow | | outside 500 x 500 mm | | 400 x 400 | | | +----------------------+ | +------------------------------+ ``` Each 500 mm face is made of two 250 mm planks placed side by side; each 400 mm face is made of one 250 mm and one 150 mm plank (a plank ripped to width), so all planks are 50 mm thick. They 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 **500 mm × 500 mm** formed with 50 mm thick Sal planks (inside 400 mm). Capacity 954 kN $>$ 925 kN. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a built-up box column, which slenderness ratio is used to classify it? A. S/d, with d the overall side B. S/sqrt(d1^2 + d2^2), with d1 overall and d2 inside side C. S/r using the radius of gyration of one plank D. S/(d1 - d2) Answer: B. IS 883 cl. 7.6.2 judges a hollow box by S/sqrt(d1^2 + d2^2), an equivalent form of slenderness for the hollow section. (Key point: Slenderness of a box column) 2. S/sqrt(d1^2 + d2^2) = 5.47 for the 500 mm box. What does this tell us and what is fc? A. Intermediate column; fc is reduced by the fourth-power formula B. Long column; fc = 0.329E/(ratio)^2 C. Short column since the ratio is below 8; fc = q fcp = 10.6 N/mm^2 D. Short column; fc = fcp/2 Answer: C. A ratio below 8 means the box crushes rather than buckles, so fc = q*fcp = 1.0 x 10.6 = 10.6 N/mm^2. (Key point: Short column check) 3. How is the area of the 500 mm box with 50 mm planks found? A. A = d1^2 - d2^2 = 500^2 - 400^2 = 90,000 mm^2 B. A = d1^2 = 250,000 mm^2 C. A = 4 x 250 x 50 = 50,000 mm^2 D. A = d1^2 - (d1 - 50)^2 with d2 = 450 mm Answer: A. Inside side is d2 = 500 - 2x50 = 400 mm, so the timber area is the outer square minus the hollow. (Key point: Net area of the box) 4. Which statement matches the final design? A. Capacity 954 kN > 925 kN; each 400 mm face uses one 250 mm and one ripped 150 mm plank B. Capacity 954 kN > 925 kN; every face uses two full 250 mm planks C. Capacity 925 kN equals the load exactly, so no plank ripping is needed D. Capacity 900 kN < 925 kN, so the section is increased Answer: A. P = 10.6 x 90,000 = 954 kN exceeds 925 kN. The 400 mm faces are built from a 250 mm and a 150 mm ripped plank. 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