Deodar column 4 m, 300 kN

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 $N = 300$ kN.

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):

Location and grade not stated: inside, Grade I (standard) assumed; a square section is chosen.

Trial section

Try 225 mm × 225 mm (square). Area $A = 50,625$ mm²; least dimension $d = 225$ mm.

Slenderness and permissible axial stress (cl. 7.6.1)

$$\begin{aligned} S/d &= \frac{4000}{225} = 17.8 \\ K_8 &= 0.584\sqrt{\frac{E}{f_{cp}}} = 0.584\sqrt{\frac{9480}{7.80}} = 20.36 \end{aligned}$$

$11 < S/d = 17.8 < K_8 = 20.4$, so it is an intermediate column:

$$f_c = f_{cp}\left[1 - \frac{1}{3}\left(\frac{S/d}{K_8}\right)^4\right] = 7.80\left[1 - \frac{1}{3}\left(\frac{17.8}{20.4}\right)^4\right] = 6.29\ \text{N/mm}^2$$

$S/d = 17.8 < 50$ (cl. 7.6.1.4). OK.

Check

$$P = f_cA = 6.29\times50,625/10^3 = 318.4\ \text{kN} \ge N = 300\ \text{kN}$$

Safe.

Answer

Provide a solid Deodar column of 225 mm × 225 mm cross-section (Grade I (standard), inside location). Capacity $318$ kN $> 300$ kN.