Skip to main content

Box column of teak planks, 400 kN

Short answer

Built-up box column of 50mm thick Teak planks per IS 883:1994 cl. 7.6.2. Trial 300×300mm overall, 200×200mm inside (A=50,000mm²). S/√(d1²+d2²)=12.48, K9=17.72, intermediate column; fc=q·fcp[1-⅓(S/K9√(d1²+d2²))⁴]=8.63 N/mm². Capacity P=fc·A=431 kN > 400 kN. Safe. Inside location, Grade I, staggered fasteners ≤200mm pitch, solid end blocks.

Design a built-up box column to carry axial load of 400 kN having length 4.5 m. Use 50 mm thick Teak planks.

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=4.50S = 4.50 m, axial load N=400N = 400 kN, plank thickness t=50t = 50 mm. Pin ends, effective length =S= S.

Species: Teak (Tectona grandis), Group B. 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_b15.515.50
Compression parallel to grain fcpf_{cp}9.49.40
Compression perpendicular to grain fcnf_{cn}4.54.50
Horizontal shear HH1.151.15
Modulus of elasticity EE99709970

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=300d_1 = 300 mm, inside side d2=300−2×50=200d_2 = 300 - 2\times50 = 200 mm.

A=d12−d22=3002−2002=50,000 mm2A = d_1^2 - d_2^2 = 300^2 - 200^2 = 50,000\ \text{mm}^2 Sd12+d22=45003002+2002=4500360.6=12.48K9=π2UE5qfcp=π20.6×99705×1.0×9.40=17.72\begin{aligned} \frac{S}{\sqrt{d_1^2 + d_2^2}} &= \frac{4500}{\sqrt{300^2 + 200^2}} = \frac{4500}{360.6} = 12.48 \\ K_9 &= \frac{\pi}{2}\sqrt{\frac{UE}{5qf_{cp}}} = \frac{\pi}{2}\sqrt{\frac{0.6\times9970}{5\times1.0\times9.40}} = 17.72 \end{aligned}

8<12.48<K9=17.728 < 12.48 < K_9 = 17.72, so it is an intermediate column:

fc=qfcp[1−13(SK9d12+d22)4]=1.0×9.40[1−13(12.4817.72)4]=8.63 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\times9.40\left[1 - \frac{1}{3}\left(\frac{12.48}{17.72}\right)^4\right] = 8.63\ \text{N/mm}^2

Check

P=fcA=8.63×50,000/103=431 kN>N=400 kNP = f_cA = 8.63\times50,000/10^3 = 431\ \text{kN} > N = 400\ \text{kN}

Safe.

Arrangement

   +------------------------------+
   |  plank 50 mm thick           |
   |   +----------------------+   |
   |   |        hollow        |   |   outside 300 x 300 mm
   |   |      200 x 200       |   |
   |   +----------------------+   |
   +------------------------------+

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 300 mm × 300 mm formed with 50 mm thick Teak planks (inside 200 mm). Capacity 431 kN >> 400 kN.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Design of Steel and Timber Structure chapter 11 questions →