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Box deodar column, 60 mm planks, 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.

Design a 5 m long rectangular box column built by 60 mm thick deodar planks to carry an axial load of 350 kN.

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=350N = 350 kN, plank thickness t=60t = 60 mm. Pin ends, effective length =S= 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):

PropertyTable 1 value (N/mm²)Used (N/mm²)
Bending / tension along grain fbf_b10.210.20
Compression parallel to grain fcpf_{cp}7.87.80
Compression perpendicular to grain fcnf_{cn}2.72.70
Horizontal shear HH0.70.70
Modulus of elasticity EE94809480

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=0.6U = 0.6, q=1.0q = 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 d1=300d_1 = 300 mm, inside side d2=300−2×60=180d_2 = 300 - 2\times60 = 180 mm.

A=d12−d22=3002−1802=57,600 mm2A = d_1^2 - d_2^2 = 300^2 - 180^2 = 57,600\ \text{mm}^2 Sd12+d22=50003002+1802=5000349.9=14.29K9=π2UE5qfcp=π20.6×94805×1.0×7.80=18.97\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<K9=18.978 < 14.29 < K_9 = 18.97, so it is an intermediate column:

fc=qfcp[1−13(SK9d12+d22)4]=1.0×7.80[1−13(14.2918.97)4]=6.96 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\times7.80\left[1 - \frac{1}{3}\left(\frac{14.29}{18.97}\right)^4\right] = 6.96\ \text{N/mm}^2

Check

P=fcA=6.96×57,600/103=401 kN>N=350 kNP = f_cA = 6.96\times57,600/10^3 = 401\ \text{kN} > N = 350\ \text{kN}

Safe.

Arrangement

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

✓ Answer checked by Claude Opus 5.5 (high effort)

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