Check 200x200 Debdar column, 30 kN + 1.5 kNm

Approach. Combined axial compression and bending, IS 883:1994 cl. 7.7.1: the column is safe if $\dfrac{f_{ac}}{f_c} + \dfrac{f_{ab}}{f_b} \le 1$.

Data and permissible stresses

Section 200 × 200 mm, effective length $S = 2500$ mm, $N = 30$ kN, $M = 1.5$ kN·m.

Species: Deodar (Cedrus deodara), Group C. Location: outside. Grade: Grade I (standard) (Table 1 values are for Grade I).

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

The column is located outside the building, so the outside values are taken; the grade is not stated, so Grade I (standard).

Permissible axial stress

$$\begin{aligned} S/d &= 2500/200 = 12.5 \\ K_8 &= 0.584\sqrt{E/f_{cp}} = 0.584\sqrt{9480/6.9} = 21.65 \end{aligned}$$

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

$$f_c = f_{cp}\left[1 - \frac{1}{3}\left(\frac{S/d}{K_8}\right)^4\right] = 6.9\left[1 - \frac{1}{3}\left(\frac{12.5}{21.65}\right)^4\right] = 6.644\ \text{N/mm}^2$$

Actual stresses

$$\begin{aligned} f_{ac} &= \frac{30\times10^3}{200\times200} = 0.750\ \text{N/mm}^2 \\ f_{ab} &= \frac{6M}{bd^2} = \frac{6\times1.5\times10^6}{200^3} = 1.125\ \text{N/mm}^2 \end{aligned}$$

Check

$$\frac{f_{ac}}{f_c} + \frac{f_{ab}}{f_b} = \frac{0.750}{6.644} + \frac{1.125}{8.7} = 0.113 + 0.129 = 0.242$$

Answer: the interaction value is 0.24 < 1, so the column is safe. (Depth is 200 mm, so no form factor is applied to the bending stress.)