Load combinations for column per IS 456

Given: column 400 x 600 mm, effective height $l=3.6$ m, $DL=250$ kN, $LL=100$ kN, wind UDL $w=5$ kN/m along the height.

Wind effect on the column

• Total horizontal force: $H_w=5\times3.6=18.0$ kN.
• The column is part of a moment resisting frame, so it is assumed fixed at both ends under the lateral UDL (assumption). Fixed-end moment: $M_w=\dfrac{wl^2}{12}=\dfrac{5\times3.6^2}{12}=5.40$ kN·m (wind can act in either direction, so the moment can be $\pm$).

Load combinations (IS 456:2000, Table 18 / cl. 36.4)

Working:

$$\begin{aligned} 1.5(250+100)&=525\ \text{kN}\\ 1.2(250+100)&=420\ \text{kN},\quad 1.2\times5.4=6.48\ \text{kN·m},\quad 1.2\times18=21.6\ \text{kN}\\ 1.5\times250&=375\ \text{kN},\quad 1.5\times5.4=8.10\ \text{kN·m},\quad 1.5\times18=27\ \text{kN}\\ 0.9\times250&=225\ \text{kN},\quad 1.5\times5.4=8.10\ \text{kN·m},\quad 27\ \text{kN} \end{aligned}$$

Combination 4 applies where stability against overturning or stress reversal is critical (minimum axial load with maximum moment).

Answer: Combination 1 gives the maximum axial load ($P_u=525$ kN, $M_u=0$ apart from the minimum eccentricity moment); combination 2 gives $P_u=420$ kN with $M_u=6.48$ kN·m; combinations 3 and 4 give the maximum moment ($M_u=8.10$ kN·m) with $P_u=375$ kN and $225$ kN. No single case can be assumed to govern, so the column is checked for all four and the worst case is adopted.