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Safe load, 225x225 select Sal column, 3 m

Short answer

Slenderness S/d = 13.33 (intermediate). Permissible stress fc = 11.25 N/mm², safe axial load = 569 kN.

A timber column 225 × 225 in cross section having an unsupported length of 3 m. Assuming the column to be of sal wood of selected grade, find the safe axial load.

Approach. Safe axial load of a solid column, IS 883:1994 cl. 7.6.1: P=fc×AP = f_c\times A, where fcf_c depends on the slenderness S/dS/d.

Data and permissible stresses

Section 225 × 225 mm, unsupported length S=3000S = 3000 mm (pin ends), Select grade Sal.

Species: Sal (Shorea robusta), Group A. Location: inside. Grade: Select grade (Grade I values multiplied by 1.16, IS 883:1994 cl. 6.3).

Permissible stresses (IS 883:1994 Table 1, inside location):

PropertyTable 1 value (N/mm²)Used (N/mm²)
Bending / tension along grain fbf_b16.919.60
Compression parallel to grain fcpf_{cp}10.612.30
Compression perpendicular to grain fcnf_{cn}4.65.34
Horizontal shear HH0.941.09
Modulus of elasticity EE1267012670

Location not stated: inside.

Slenderness

S/d=3000225=13.33K8=0.584Efcp=0.5841267012.30=18.75\begin{aligned} S/d &= \frac{3000}{225} = 13.33 \\ K_8 &= 0.584\sqrt{\frac{E}{f_{cp}}} = 0.584\sqrt{\frac{12670}{12.30}} = 18.75 \end{aligned}

11<S/d=13.33<K8=18.7511 < S/d = 13.33 < K_8 = 18.75, so it is an intermediate column.

fc=fcp[1−13(S/dK8)4]=12.30[1−13(13.3318.75)4]=11.25 N/mm2f_c = f_{cp}\left[1 - \frac{1}{3}\left(\frac{S/d}{K_8}\right)^4\right] = 12.30\left[1 - \frac{1}{3}\left(\frac{13.33}{18.75}\right)^4\right] = 11.25\ \text{N/mm}^2

Safe load

P=fcA=11.25×50,625/103=569 kNP = f_cA = 11.25\times50,625/10^3 = 569\ \text{kN}

Answer: the safe axial load is about 569 kN (permissible stress fc=11.25f_c = 11.25 N/mm²; S/d=13.3<50S/d = 13.3 < 50).

✓ Answer checked by Claude Opus 5.5 (high effort)

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