--- title: "Isolated footing 400x400, 1200 kN service" subject: "Design of RCC Structure" chapter: 11 key: "semester_7/design_of_rcc_structure:isolated-footing-400x400-1200-kn-service" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/isolated-footing-400x400-1200-kn-service site: BCT Engineering --- # Isolated footing 400x400, 1200 kN service Design of RCC Structure · Chapter 11: Design of Footings - Asked in: 2074 Chaitra, 10 marks - Video: [Design of Isolated Footing | Pad Footing for Axial Load as per IS 456:2000.](https://www.youtube.com/watch?v=Wm_EzAtud3s) — Visual Civil - 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-footing-400x400-1200-kn-service - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-11 ## Question Design an isolated footing to support a square column of 400 x 400 mm. The column (400 x 400 mm) carries a service load of 1200 kN. The allowable soil pressure is 150 kN/m$^2$. Use M20 concrete and Fe415 grade steel. Unit weight of soil above footing base = 18 kN/m$^3$. Necessary missing data assume suitably. ## Short answer Net safe bearing capacity = 150 - 18*1.5 = 123 kN/m²; area = 1200/123 = 9.76 m², provide 3.2 m square. Factored pressure 175.8 kN/m²; depth D=550 mm (d=490 mm). Reinforcement 16 mm @ 195 c/c both ways; shear safe. Bearing stress 11.25 MPa > 9 MPa, excess 360 kN carried by column dowels. ## Answer #### Data and assumptions - Column 400 mm × 400 mm (long side $a=400$ mm along the footing length); footing concrete M20, steel Fe415; column concrete M20; column bars 8-20 mm. - Service load $P = 1200$ kN. - Allowable (safe) bearing capacity of soil = 150 kN/m² at 1.5 m depth, soil unit weight 18 kN/m³. - Depth of foundation is not given; 1.5 m is assumed. Column bars assumed 8-20 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 Net safe bearing capacity = $q - \gamma D_f = 150 - 18 \times 1.5 = 123.00$ kN/m² (this takes care of the weight of footing and backfill). $$ A_{req} = \frac{P}{q_{net}} = \frac{1200.0}{123.00} = 9.76\ \text{m}^2 $$ Provide a square footing $L \times B = 3.2 \times 3.2$ m, area = 10.24 m² $\ge$ 9.76 m². #### Factored soil pressure $P_u = 1.5 \times 1200.0 = 1800.0$ kN. $p_u = \dfrac{P_u}{BL} = \dfrac{1800.0}{3.2 \times 3.2} = 175.8$ kN/m². #### Bending moments and reinforcement Cantilever projections from the column face: along $L$: $(3200 - 400)/2 = 1400$ mm; along $B$: $(3200 - 400)/2 = 1400$ mm. $M_u = p_u \times (\text{width}) \times \dfrac{(\text{projection})^2}{2}$: $M_{u,L} = 551.2$ kN·m, $M_{u,B} = 551.2$ kN·m. Required depth for flexure: $M_{u,lim} = 0.138 f_{ck} b d^2$ is far larger than the moment for $d = 490$ mm, so the depth is governed by shear (one-way or punching). Trial **$D = 550$ mm, $d = 490$ 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 $B$) | 551.2 | 3258 | 16 mm @ 195 c/c | 3299 | | Along $B$ (width $L$) | 551.2 | 3258 | 16 mm @ 195 c/c | 3299 | #### Check for one-way shear (at distance $d$ from the column face) - Along $L$ side: $V_u = 511.9$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.326$ MPa; $p_t = 0.210\%$, $\tau_c = 0.334$ MPa (IS 456 Table 19, SP 16 formula). $\tau_v < \tau_c$, safe. - Along $B$ side: $V_u = 511.9$ kN, $\tau_v = 0.326$ MPa; $p_t = 0.210\%$, $\tau_c = 0.334$ MPa. Safe. #### Check for two-way (punching) shear (at $d/2$ from the column face) - Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(400+490) + (400+490)] = 3560$ mm - $V_u = P_u - p_u (a+d)(c+d) = 1800.0 - 175.8 \times 0.890 \times 0.890 = 1660.8$ kN (average pressure; the effect of moment transfer is neglected). - $\tau_v = \dfrac{V_u}{b_o d} = 0.952$ 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): 16 mm bars (both directions) → $L_d = 752$ mm. Available length from column face (less 50 mm end cover): 1350 mm along $L$, 1350 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 $= P_u/A_2 = 11.25$ 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 $<$ $P_u = 1800$ kN, so the excess 360 kN must be carried by dowels: $A_{s,dow} = \dfrac{360000}{0.87 f_y} = 997$ mm²; the column bars (8-20 mm, 2513 mm²) are continued into the footing as dowels, which is adequate. Dowel length in compression $L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 752$ mm (and the same lap with the column bars). Available depth in footing = 550 - 50 - 32 = 468 mm $< L_{dc}$; dowels are bent at the bottom with 90° bends. #### Detailing ```text PLAN ELEVATION +------------------+ col | # # # # # # | | | | # +------+ # # | _|_|__ | # | col | # # | |_____| <- D = 550 mm | # +------+ # # | bars both ways | # # # # # # | +------------------+ L = 3.2 m, B = 3.2 m ``` - Bottom mesh: 16 mm @ 195 c/c along $L$ (lower layer), 16 mm @ 195 c/c along $B$ (upper layer), 50 mm clear cover; dowels 8-20 mm. **Answer:** Footing 3.2 m × 3.2 m × 550 mm thick; 16 mm @ 195 c/c along the length and 16 mm @ 195 c/c along the width. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does this answer use A = P/q_net with q_net = 150 - 18 x 1.5 = 123 kN/m^2 instead of 1.1P/q_a? A. Because depth (1.5 m assumed) and soil unit weight are given, the overburden weight is subtracted from the capacity B. Because the column is 400 x 400 C. Because the load is factored D. Because Fe415 steel is used Answer: A. With foundation depth and soil weight known, the weight of footing and backfill is accounted for by reducing the bearing capacity by gamma x D_f. (Key point: Net bearing capacity method) 2. Where is the bending moment taken for the square footing, and how is it computed? A. At the footing centre, using total load B. At the column face, as a cantilever projection of 1400 mm under p_u = 175.8 kN/m^2 C. At d from the column face as a simply supported beam D. At the footing edge with zero pressure Answer: B. The footing is treated as a cantilever from the column face: M_u = p_u x width x projection^2/2 = 551.2 kN.m. (Key point: Critical section for bending) 3. Excess load beyond bearing is 360 kN. How is dowel adequacy shown? A. A_s,dow = 360000/(0.87 f_y) = 997 mm^2, less than the continued column bars (2513 mm^2), so they are adequate B. Dowels are not required C. Dowel area equals 1% of footing area D. Dowels must be 997 mm long Answer: A. Steel required to take the excess load is 997 mm^2; the 8-20 mm column bars provide 2513 mm^2 when continued into the footing. 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