--- title: "Isolated square footing, 650 kN service" subject: "Design of RCC Structure" chapter: 11 key: "semester_7/design_of_rcc_structure:isolated-square-footing-650-kn-service" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/isolated-square-footing-650-kn-service site: BCT Engineering --- # Isolated square footing, 650 kN service Design of RCC Structure · Chapter 11: Design of Footings - Asked in: 2078 Bhadra, 12 marks - Video: [Design of a Square Footing](https://www.youtube.com/watch?v=jYCc0QFKwR4) — Stan Academy - 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/isolated-square-footing-650-kn-service - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-11 ## Question Design an isolated square footing foundation of uniform thickness for a 400 mm x 400 mm column subjected to an axial load of 650 kN at service state. Consider safe bearing capacity of soil as 170 kN/m$^2$ and concrete of M20 and steel grade of Fe500. Show the reinforcements in plan and in section of footing. ## Short answer Provide a 2.1 m square footing, 450 mm thick. Reinforcement: 12 mm bars @ 205 mm c/c both ways. Provide 8 dowels of 16 mm diameter. ## Answer #### Data and assumptions - Column 400 mm × 400 mm (long side $a=400$ mm along the footing length); footing concrete M20, steel Fe500; column concrete M20; column bars 8-16 mm. - Service load $P = 650$ kN. - Allowable (safe) bearing capacity of soil = 170 kN/m². - Column bars assumed 8-16 mm. - Clear cover 50 mm for the footing (IS 456 cl. 26.4.2.2); effective depth $d = D - 60$ mm (cover + half bar). Load factor 1.5. #### Size of footing Weight of footing and backfill is taken as 10% of the column load. $$ A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 650.0}{170} = 4.21\ \text{m}^2 $$ Provide a square footing $L \times B = 2.1 \times 2.1$ m, area = 4.41 m² $\ge$ 4.21 m². #### Factored soil pressure $P_u = 1.5 \times 650.0 = 975.0$ kN. $p_u = \dfrac{P_u}{BL} = \dfrac{975.0}{2.1 \times 2.1} = 221.1$ kN/m². #### Bending moments and reinforcement Cantilever projections from the column face: along $L$: $(2100 - 400)/2 = 850$ mm; along $B$: $(2100 - 400)/2 = 850$ mm. $M_u = p_u \times (\text{width}) \times \dfrac{(\text{projection})^2}{2}$: $M_{u,L} = 167.7$ kN·m, $M_{u,B} = 167.7$ kN·m. Required depth for flexure: $M_{u,lim} = 0.133 f_{ck} b d^2$ is far larger than the moment for $d = 390$ mm, so the depth is governed by shear (one-way or punching). Trial **$D = 450$ mm, $d = 390$ mm**, found by the shear checks below. $A_{st} = \dfrac{0.5 f_{ck}}{f_y}\left[1 - \sqrt{1 - \dfrac{4.6 M_u}{f_{ck} b d^2}}\right] b d$; minimum steel = 0.12% of $bD$. | Direction | $M_u$ (kN·m) | $A_{st}$ req (mm²) | Provide | $A_{st}$ prov (mm²) | |---|---|---|---|---| | Along $L$ (width $B$) | 167.7 | 1134 | 12 mm @ 205 c/c | 1159 | | Along $B$ (width $L$) | 167.7 | 1134 | 12 mm @ 205 c/c | 1159 | #### Check for one-way shear (at distance $d$ from the column face) - Along $L$ side: $V_u = 213.6$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.261$ MPa; $p_t = 0.141\%$, $\tau_c = 0.288$ MPa (IS 456 Table 19, SP 16 formula). $\tau_v < \tau_c$, safe. - Along $B$ side: $V_u = 213.6$ kN, $\tau_v = 0.261$ MPa; $p_t = 0.141\%$, $\tau_c = 0.288$ MPa. Safe. #### Check for two-way (punching) shear (at $d/2$ from the column face) - Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(400+390) + (400+390)] = 3160$ mm - $V_u = P_u - p_u (a+d)(c+d) = 975.0 - 221.1 \times 0.790 \times 0.790 = 837.0$ kN (average pressure; the effect of moment transfer is neglected). - $\tau_v = \dfrac{V_u}{b_o d} = 0.679$ MPa; $k_s = 0.5 + \beta_c = 1.00$, $\tau_c' = k_s \times 0.25\sqrt{f_{ck}} = 1.118$ MPa. $\tau_v < \tau_c'$, **safe**. #### Development length $L_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}$, $\tau_{bd} = 1.92$ MPa (M20, deformed bars): 12 mm bars (both directions) → $L_d = 680$ mm. Available length from column face (less 50 mm end cover): 800 mm along $L$, 800 mm along $B$. **Satisfied** in both directions. #### Bearing stress and transfer of load from column Column area $A_2 = 400\times400 = 0.160$ m². Bearing stress $= P_u/A_2 = 6.09$ MPa. Permissible bearing stress: column concrete $0.45 f_{ck} = 9.00$ MPa; footing concrete $0.45 f_{ck}\sqrt{A_1/A_2}$ with $\sqrt{A_1/A_2} \le 2$ gives up to 18.00 MPa. The lesser, 9.00 MPa, governs. Capacity of bearing = 1440 kN $\ge$ $P_u = 975$ kN, so the whole load can be transferred by bearing. Provide minimum dowels of 0.5% of column area = 800 mm²: use 8 bars of 16 mm (1608 mm²) as dowels, same as column bars. Dowel length in compression $L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 725$ mm (and the same lap with the column bars). Available depth in footing = 450 - 50 - 24 = 376 mm $< L_{dc}$; dowels are bent at the bottom with 90° bends. #### Detailing ```text PLAN ELEVATION +------------------+ col | # # # # # # | | | | # +------+ # # | _|_|__ | # | col | # # | |_____| <- D = 450 mm | # +------+ # # | bars both ways | # # # # # # | +------------------+ L = 2.1 m, B = 2.1 m ``` - Bottom mesh: 12 mm @ 205 c/c along $L$ (lower layer), 12 mm @ 205 c/c along $B$ (upper layer), 50 mm clear cover; dowels 8-16 mm. **Answer:** Footing 2.1 m × 2.1 m × 450 mm thick; 12 mm @ 205 c/c along the length and 12 mm @ 205 c/c along the width. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the required footing area found as 1.1P/q_a (= 4.21 m^2) in this answer? A. The 1.1 allows for self-weight of the footing and backfill, taken as 10% of the column load B. The 1.1 is the partial safety factor for soil C. The 1.1 converts service load into factored load D. The 1.1 accounts for moment transferred from the column Answer: A. The soil must also carry the footing and backfill weight, assumed as 10% of P, so the load on soil is 1.1P and the area is 1.1P/q_a. (Key point: Footing area from service load) 2. Which load is used to get the soil pressure p_u = 221.1 kN/m^2 for bending and shear design? A. Service load 650 kN divided by the area B. 1.1 x 650 kN divided by the area C. Factored load 1.5 x 650 = 975 kN divided by the footing area D. Safe bearing capacity 170 kN/m^2 directly Answer: C. Sizing uses service load, but structural design (limit state) uses the factored load P_u = 1.5P over the area, giving 975/(2.1x2.1) = 221.1 kN/m^2. (Key point: Factored soil pressure) 3. Why is the footing depth D = 450 mm fixed by the shear checks and not by flexure? A. Because Fe500 steel cannot resist bending B. Because M_u,lim for the section is far larger than the 167.7 kN.m moment, so shear controls the depth C. Because the minimum depth is always 450 mm by code D. Because punching shear is never checked in flexure Answer: B. The section is hugely adequate in bending (M_u,lim >> 167.7 kN.m), so depth is chosen to satisfy one-way and two-way shear. (Key point: Depth governed by shear) 4. Bearing capacity of the column section is 1440 kN against P_u = 975 kN. What does this lead to? A. The whole load transfers by bearing, so only minimum dowels (0.5% of column area = 800 mm^2) are needed B. Dowels must carry the entire load C. The footing depth must be increased D. The column must be enlarged Answer: A. Since bearing capacity (0.45 f_ck x A_2 = 1440 kN) is at least P_u, no extra dowel area is needed beyond the 0.5% minimum; 8-16 mm bars are provided. (Key point: Load transfer by bearing) ## Close variants set in other exams - [Isolated square footing, 600 kN and 50 kNm](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aisolated-square-footing-600-kn-and) — Design of RCC Structure, chapter 11, asked 2× ## Related in Design of RCC Structure - [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 400x400, service moment 120 kNm](https://www.bctengineer.com/question.md?key=semester_7%2Fdesign_of_rcc_structure%3Aisolated-footing-400x400-service-moment-120) — Design of RCC Structure, chapter 11, asked 2× - [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 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× - [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 - [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 ## Related in other subjects - [Brick column design for 80 kN axial load](https://www.bctengineer.com/question.md?key=semester_5%2Fconcrete_technology_and_masonry_structure%3Abrick-column-80kn-roof) — Concrete Technology and Masonry Structure, chapter 8 - [Safe axial load on a timber column](https://www.bctengineer.com/question.md?key=semester_5%2Fdesign_of_timber_and_masonry_structures%3Aaxial-capacity-timber-column) — Design of Timber and Masonry Structures, chapter 3 - [Soil exploration, safe bearing capacity, improvement](https://www.bctengineer.com/question.md?key=semester_6%2Fbuilding_technology%3Aexploration-bc-065s) — Building Technology, chapter 2 - [Allowable load of 400 mm pile, layered soil](https://www.bctengineer.com/question.md?key=semester_5%2Ffoundation_engineering%3Apile-400mm-layered-soil-chart) — Foundation Engineering, chapter 8