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Built-up Sal column, 50x250 planks, 925 kN

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.

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, fcpf_{cp} = 10.6 N/mm² constant U = 0.6 and q = 1.

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/d12+d22S/\sqrt{d_1^2 + d_2^2}, where d1d_1 is the overall side and d2d_2 the inside side of the box.

Data and permissible stresses

Length S=3.50S = 3.50 m, axial load N=925N = 925 kN, plank thickness t=50t = 50 mm. Pin ends, effective length =S= 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):

PropertyTable 1 value (N/mm²)Used (N/mm²)
Bending / tension along grain fbf_b16.916.90
Compression parallel to grain fcpf_{cp}10.610.60
Compression perpendicular to grain fcnf_{cn}4.64.60
Horizontal shear HH0.940.94
Modulus of elasticity EE1270012700

The question gives E=12700E = 12700 N/mm², fcp=10.6f_{cp} = 10.6 N/mm² (the Table 1 value for Sal, Grade I, inside), U=0.6U = 0.6 and q=1q = 1; these are used.

For plank thickness 50 mm (cl. 7.6.2.5): U=0.6U = 0.6, q=1.0q = 1.0 (values given for 25 mm and 50 mm planks).

Trial section

Square box, overall side d1=500d_1 = 500 mm, inside side d2=500−2×50=400d_2 = 500 - 2\times50 = 400 mm.

A=d12−d22=5002−4002=90,000 mm2A = d_1^2 - d_2^2 = 500^2 - 400^2 = 90,000\ \text{mm}^2 Sd12+d22=35005002+4002=3500640.3=5.47K9=π2UE5qfcp=π20.6×127005×1.0×10.60=18.83\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): fc=qfcp=1.0×10.60=10.60f_c = q f_{cp} = 1.0\times10.60 = 10.60 N/mm².

Check

P=fcA=10.60×90,000/103=954 kN>N=925 kNP = f_cA = 10.60\times90,000/10^3 = 954\ \text{kN} > N = 925\ \text{kN}

Safe.

Arrangement of planks (250 mm × 50 mm planks)

   +------------------------------+
   |  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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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