Sal timber column, 50 kN, 4 m

Approach. Working stress design to IS 883:1994 (cl. 7.6 for columns, cl. 7.7 for combined axial compression and bending). Ends are taken as pin-ended, so the effective length equals the unsupported length $S$ (cl. 7.6.1.5).

Data and permissible stresses

Length $S = 4.00$ m = 4000 mm. Axial load $N = 50$ kN.

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):

Location and grade not stated: inside, Grade I assumed. A square section is chosen.

Trial section

Try 125 mm × 125 mm (square). Area $A = 15,625$ mm²; least dimension $d = 125$ mm.

Slenderness and permissible axial stress (cl. 7.6.1)

$$\begin{aligned} S/d &= \frac{4000}{125} = 32.0 \\ K_8 &= 0.584\sqrt{\frac{E}{f_{cp}}} = 0.584\sqrt{\frac{12670}{10.60}} = 20.19 \end{aligned}$$

$S/d = 32.0 > K_8 = 20.2$, so it is a long column:

$$f_c = \frac{0.329E}{(S/d)^2} = \frac{0.329\times12670}{32.0^2} = 4.07\ \text{N/mm}^2$$

$S/d = 32.0 < 50$ (cl. 7.6.1.4). OK.

Check

$$P = f_cA = 4.07\times15,625/10^3 = 63.6\ \text{kN} \ge N = 50\ \text{kN}$$

Safe.

Answer

Provide a solid Sal column of 125 mm × 125 mm cross-section (Grade I (standard), inside location). Capacity $64$ kN $> 50$ kN.