---
title: "RC wall footing, 1000 kN total load"
subject: "Design of RCC Structure"
chapter: 11
key: "semester_7/design_of_rcc_structure:rc-wall-footing-1000-kn-total"
source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/rc-wall-footing-1000-kn-total
site: BCT Engineering
---

# RC wall footing, 1000 kN total load

Design of RCC Structure · Chapter 11: Design of Footings

- Asked in: 2065 Kartik (old course), 12 marks
- Video: [Limit State Design of Wall Footing | Design of RC Element | Lec - 25 | Civil Tutor](https://www.youtube.com/watch?v=Cjyt8XyEuOQ) — CIVIL TUTOR
- 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/rc-wall-footing-1000-kn-total
- Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-11

## Question

Design a RC concrete spread footing of an RC wall having 4 m length and 300 mm width. Total load on wall is equal to 1000 kN at service state. Take M20 and Fe415 grades of concrete and steel respectively, and safe bearing capacity of soil = 150 kN/m$^2$.

## Short answer

For 1 m length of wall, required width 1.85 m including 10% footing weight. Factored pressure 202.7 kN/m2 gives Mu=60.87 kNm/m; provide depth 375 mm, main steel 10 mm @ 140 mm c/c and distribution 8 mm @ 110 mm c/c.

## Answer

#### Data and assumptions
- Wall 300 mm thick and 4 m long carrying a total service load of 1000 kN, so the load per metre run is $w = 1000/4 = 250$ kN/m. M20 concrete, Fe415 steel, $q_a = 150$ kN/m². The footing is designed for a 1 m length of wall. Cover 50 mm.

#### Width of footing
$$
B = \frac{1.1\,w}{q_a} = \frac{1.1 \times 250}{150} = 1.833\ \text{m} \Rightarrow \text{provide } B = 1.85\ \text{m}
$$
(10% is added for the weight of the footing and backfill.)

#### Factored pressure and bending moment
$w_u = 1.5 \times 250 = 375$ kN/m, $p_u = \dfrac{w_u}{B} = \dfrac{375}{1.85} = 202.70$ kN/m²
Projection beyond the wall face $= (1850 - 300)/2$ $= 775$ mm.
$$
M_u = \frac{p_u\, a^2}{2} = \frac{202.70 \times 0.775^2}{2} = 60.87\ \text{kN·m per metre}
$$

#### Depth
The depth is governed by one-way shear at $d$ from the wall face (checked below); trial **$D = 375$ mm**, $d = 375 - 50 - 6 = 319$ mm (conservative). Flexure check: $M_{u,lim} = 0.138 f_{ck} b d^2 = 280.9$ kN·m $> 60.87$ kN·m, so the depth is adequate and the section is under-reinforced.

#### Reinforcement
$A_{st} = 548$ mm²/m from $M_u$ (minimum 0.12% of $bD$ = 450 mm²/m). Adopt 548 mm²/m → provide **10 mm @ 140 mm c/c** across the width (561 mm²/m, $p_t = 0.176\%$).
Distribution steel along the wall (0.12% of $bD$) = 450 mm²/m → 8 mm @ 110 mm c/c (457 mm²/m).

#### Check for shear
$V_u = p_u (a - d) = 202.70 \times (0.775 - 0.319) = 92.43$ kN/m; $\tau_v = \dfrac{92432}{1000 \times 319} = 0.290$ MPa < $\tau_c = 0.30$ MPa (M20, $p_t = 0.176\%$, Table 19). Safe.

#### Development length
$L_d = \dfrac{0.87 \times 415 \times 10}{4 \times 1.92} = 470$ mm; available length $= 775 - 50 = 725$ mm. Satisfied.

#### Detailing
```text
SECTION ACROSS WALL
        |  wall 300  |
        |            |
   _____|            |_____
  |                       |  D = 375 mm
  |_______________________|
   o o o o o o o o o o o o   <- 10 mm @ 140 c/c
  <---------- 1.85 m ------->
```
Distribution bars of 8 mm @ 110 mm c/c are tied above the main bars, parallel to the wall.

**Answer:** Strip footing 1.85 m wide and 375 mm thick; main bars 10 mm @ 140 mm c/c across the wall, distribution bars 8 mm @ 110 mm c/c.

## Check yourself

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

1. A 4 m wall carries 1000 kN in total. How is the strip footing designed?
   A. For 1000 kN as a single block
   B. For a 1 m length with w = 1000/4 = 250 kN/m
   C. For 1000 kN per metre
   D. For 4 m width
   Answer: B. A wall footing is designed per metre length, using the load per metre run. (Key point: Design per metre run)

2. How is the footing width B = 1.85 m obtained?
   A. B = w / q_a = 1.67 m
   B. B = 1.5 w / q_a
   C. B = 1.1 w / q_a = 1.833 m, rounded up to 1.85 m
   D. B = wall thickness + 2d
   Answer: C. 10% is added for the weight of footing and backfill, giving B = 1.1 x 250 / 150, then rounded up. (Key point: Width of footing)

3. How is the design moment found?
   A. Factored upward pressure p_u acting on the cantilever projection: M_u = p_u a^2/2
   B. M_u = w L^2/8 over the wall length
   C. Using service load p x a^2/2
   D. M_u = p_u a^2/8
   Answer: A. The projection beyond the wall face acts as a cantilever under the factored soil pressure, so M_u = p_u a^2/2 = 60.87 kN-m/m. (Key point: Bending moment on cantilever)

4. What is the purpose of the 8 mm @ 110 mm distribution bars along the wall?
   A. To resist the main bending moment
   B. To resist shear
   C. To provide minimum steel (0.12% of bD) and control shrinkage/temperature cracking while holding the main bars
   D. To increase development length
   Answer: C. Bars parallel to the wall provide the minimum 0.12% steel and keep the main bars in position. (Key point: Distribution steel)

## Related in Design of RCC Structure

- [Footing with load transfer check, 1000 kN](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Afooting-with-load-transfer-check-1000) — Design of RCC Structure, chapter 11
- [Isolated footing 400x400, 1200 kN service](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aisolated-footing-400x400-1200-kn-service) — Design of RCC Structure, chapter 11
- [Isolated square footing, 650 kN service](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aisolated-square-footing-650-kn-service) — Design of RCC Structure, chapter 11, asked 2×
- [Footing, 1250 kN column load, depth 1.8 m](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Afooting-1250-kn-column-load-depth) — Design of RCC Structure, chapter 11
- [Isolated footing 450x450, 1600 kN service](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aisolated-footing-450x450-1600-kn-service) — Design of RCC Structure, chapter 11
- [Isolated footing 450x450, service moment 110 kNm](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aisolated-footing-450x450-service-moment-110) — Design of RCC Structure, chapter 11, asked 2×

## Related in other subjects

- [Friction pile length, 300 mm, 250 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Afriction-pile-300mm-250kn) — Foundation Engineering, chapter 8
- [Safe load P on bolted bracket, M20 4.6](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abracket-bolted-safe-load-p-m20-120) — Design of Steel and Timber Structure, chapter 4
- [Circular footing 1 m dia, 1000 kN](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Acircular-footing-1m-1000kn) — Foundation Engineering, chapter 6
- [Shop welded bracket, 5 mm fillet, safe load](https://www.bctengineer.com/question.md?key=semester_6%2Fdesign_of_steel_and_timber_structure%3Abracket-weld-safe-load-5mm) — Design of Steel and Timber Structure, chapter 4