---
title: "Load combinations for column per IS 456"
subject: "Design of RCC Structure"
chapter: 3
key: "semester_7/design_of_rcc_structure:load-combinations-for-column-per-is"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/load-combinations-for-column-per-is
site: BCT Engineering
---

# Load combinations for column per IS 456

Design of RCC Structure · Chapter 3: Limit State Method of design

- Asked in: 2067 Asar (old course), 8 marks
- Video: [Load combinations as per IS 456-2000 code | civil engineering | Building design | online course |](https://www.youtube.com/watch?v=TwgeH1DwwnY) — Civil Engineering by Shravan
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/load-combinations-for-column-per-is
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-3

## Question

A reinforced concrete column of a moment resisting frame has its cross section 400 mm x 600 mm and effective height 3.6 m. The column has to carry loading combination of dead load, live load and moment due to wind load. The computed dead load and live load on column are 250 kN and 100 kN respectively, whereas the induced horizontal load on column due to wind load is 5 kN/m. Calculate loading combination values for all possible loading combinations as per IS 456.

## Short answer

Column 400×600mm, leff=3.6m, DL=250kN, LL=100kN, wind UDL=5kN/m. Wind force=18kN, fixed-end moment=5.4kNm. IS 456 load combinations: 1) 1.5(DL+LL)=525kN, M=0; 2) 1.2(DL+LL±WL)=420kN, M=±6.48kNm, H=±21.6kN; 3) 1.5(DL±WL)=375kN, M=±8.10kNm, H=±27kN; 4) 0.9DL±1.5WL=225kN, M=±8.10kNm, H=±27kN. Check all four.

## Answer

**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)**

| No. | Combination | Axial $P_u$ (kN) | Moment $M_u$ (kN·m) | Shear $H_u$ (kN) |
|---|---|---|---|---|
| 1 | $1.5\,(DL+LL)$ | 525 | 0 | 0 |
| 2 | $1.2\,(DL+LL\pm WL)$ | 420 | $\pm$6.48 | $\pm$21.6 |
| 3 | $1.5\,(DL\pm WL)$ | 375 | $\pm$8.10 | $\pm$27 |
| 4 | $0.9\,DL\pm1.5\,WL$ | 225 | $\pm$8.10 | $\pm$27 |

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.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. What axial load and moment result from 1.2(DL + LL ± WL)?
   A. Pu = 420 kN, Mu = ±6.48 kN·m
   B. Pu = 525 kN, Mu = 0
   C. Pu = 375 kN, Mu = ±8.10 kN·m
   D. Pu = 420 kN, Mu = ±8.10 kN·m
   Answer: A. 1.2 x 350 = 420 kN and 1.2 x 5.4 = 6.48 kN·m. (Key point: Combination 2)

2. What does 0.9 DL ± 1.5 WL give for axial load, and why is it considered?
   A. 225 kN; minimum axial load with maximum moment, critical for stability or reversal
   B. 375 kN; maximum axial load
   C. 525 kN; the governing load
   D. 225 kN; because live load is ignored by mistake
   Answer: A. 0.9 x 250 = 225 kN; the reduced dead load with full wind gives the worst case for overturning and stress reversal. (Key point: Combination 4)

3. What is the combination for DL + LL only?
   A. 1.2(DL+LL) = 420 kN
   B. 1.5(DL+LL) = 525 kN
   C. 1.5 DL = 375 kN
   D. 0.9 DL = 225 kN
   Answer: B. 1.5 x (250 + 100) = 525 kN with no wind moment. (Key point: Combination 1)

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