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Box Sal column 5 m, 400 kN

Short answer

Provide box column 275 mm × 275 mm overall built from 50 mm thick Sal planks (inside 175 mm × 175 mm). Permissible compressive stress 9.04 N/mm², area 45000 mm², capacity 407 kN > 400 kN.

Design a 5m long rectangular box columns built up by 5cm thick Sal wood planks to carry on axial load 400KN.

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/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=5.00S = 5.00 m, axial load N=400N = 400 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 EE1267012670

Location is not stated, so inside and Grade I are assumed.

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=275d_1 = 275 mm, inside side d2=275−2×50=175d_2 = 275 - 2\times50 = 175 mm.

A=d12−d22=2752−1752=45,000 mm2A = d_1^2 - d_2^2 = 275^2 - 175^2 = 45,000\ \text{mm}^2 Sd12+d22=50002752+1752=5000326.0=15.34K9=π2UE5qfcp=π20.6×126705×1.0×10.60=18.81\begin{aligned} \frac{S}{\sqrt{d_1^2 + d_2^2}} &= \frac{5000}{\sqrt{275^2 + 175^2}} = \frac{5000}{326.0} = 15.34 \\ K_9 &= \frac{\pi}{2}\sqrt{\frac{UE}{5qf_{cp}}} = \frac{\pi}{2}\sqrt{\frac{0.6\times12670}{5\times1.0\times10.60}} = 18.81 \end{aligned}

8<15.34<K9=18.818 < 15.34 < K_9 = 18.81, so it is an intermediate column:

fc=qfcp[1−13(SK9d12+d22)4]=1.0×10.60[1−13(15.3418.81)4]=9.04 N/mm2f_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\times10.60\left[1 - \frac{1}{3}\left(\frac{15.34}{18.81}\right)^4\right] = 9.04\ \text{N/mm}^2

Check

P=fcA=9.04×45,000/103=407 kN>N=400 kNP = f_cA = 9.04\times45,000/10^3 = 407\ \text{kN} > N = 400\ \text{kN}

Safe.

Arrangement

   +------------------------------+
   |  plank 50 mm thick           |
   |   +----------------------+   |
   |   |        hollow        |   |   outside 275 x 275 mm
   |   |      175 x 175       |   |
   |   +----------------------+   |
   +------------------------------+

The 50 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 275 mm × 275 mm formed with 50 mm thick Sal planks (inside 175 mm). Capacity 407 kN >> 400 kN.

✓ Answer checked by Claude Opus 5.5 (high effort)

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