--- title: "Footing for 400x400 column, 1500 kN" subject: "Design of RCC Structure" chapter: 11 key: "semester_7/design_of_rcc_structure:footing-for-400x400-column-1500-kn" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/footing-for-400x400-column-1500-kn site: BCT Engineering --- # Footing for 400x400 column, 1500 kN Design of RCC Structure · Chapter 11: Design of Footings - Asked in: 2081 Baisakh, 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/footing-for-400x400-column-1500-kn - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-11 ## Question Design a footing for supporting a square column of 400 mm x 400 mm. The column carries a service load of 1500 kN. The safe bearing capacity of soil is 150 kN/m$^2$. Use M20 concrete and Fe415 grade steel. ## Short answer Provide a 3.4 m x 3.4 m square footing, overall depth 650 mm (d=590 mm). Reinforcement: 16 mm bars @185 mm c/c both ways. One-way shear and punching shear are within limits; bearing stress check requires dowels, satisfied by continuing the column’s 8–20 mm bars. ## 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 $P = 1500$ kN. - Allowable (safe) bearing capacity of soil = 150 kN/m². - Column bars not given; 8-20 mm assumed. Loads are service loads. - 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. $$ A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 1500.0}{150} = 11.00\ \text{m}^2 $$ Provide a square footing $L \times B = 3.4 \times 3.4$ m, area = 11.56 m² $\ge$ 11.00 m². #### Factored soil pressure $P_u = 1.5 \times 1500.0 = 2250.0$ kN. $p_u = \dfrac{P_u}{BL} = \dfrac{2250.0}{3.4 \times 3.4} = 194.6$ kN/m². #### Bending moments and reinforcement Cantilever projections from the column face: along $L$: $(3400 - 400)/2 = 1500$ mm; along $B$: $(3400 - 400)/2 = 1500$ mm. $M_u = p_u \times (\text{width}) \times \dfrac{(\text{projection})^2}{2}$: $M_{u,L} = 744.5$ kN·m, $M_{u,B} = 744.5$ kN·m. Required depth for flexure: $M_{u,lim} = 0.138 f_{ck} b d^2$ is far larger than the moment for $d = 590$ mm, so the depth is governed by shear (one-way or punching). Trial **$D = 650$ mm, $d = 590$ 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 $B$) | 744.5 | 3633 | 16 mm @ 185 c/c | 3695 | | Along $B$ (width $L$) | 744.5 | 3633 | 16 mm @ 185 c/c | 3695 | #### Check for one-way shear (at distance $d$ from the column face) - Along $L$ side: $V_u = 602.2$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.300$ MPa; $p_t = 0.184\%$, $\tau_c = 0.315$ MPa (Table 19). $\tau_v < \tau_c$, safe. - Along $B$ side: $V_u = 602.2$ kN, $\tau_v = 0.300$ MPa; $p_t = 0.184\%$, $\tau_c = 0.315$ MPa. Safe. #### Check for two-way (punching) shear (at $d/2$ from the column face) - Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(400+590) + (400+590)] = 3960$ mm - $V_u = P_u - p_u (a+d)(c+d) = 2250.0 - 194.6 \times 0.990 \times 0.990 = 2059.2$ kN (average pressure; the effect of moment transfer is neglected). - $\tau_v = \dfrac{V_u}{b_o d} = 0.881$ 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): 1450 mm along $L$, 1450 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 $= P_u/A_2 = 14.06$ 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 $<$ $P_u = 2250$ kN, so the excess 810 kN must be carried by dowels: $A_{s,dow} = \dfrac{810000}{0.87 f_y} = 2243$ mm²; the column bars (8-20 mm, 2513 mm²) are continued into the footing as dowels - 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 = 650 - 50 - 32 = 568 mm $< L_{dc}$; dowels are bent at the bottom with 90° bends. #### Detailing ```text PLAN (3.4 m x 3.4 m) +------------------------------------+ | | | | | | | | | | | |--+---+---+---+---+---+---+---+---+-| | | | | | | | | | | | |--+---+---+---+---+---+---+---+---+-| | | | | +----------+ | | | | |--+---+---+-| column |--+---+---+-| | | | | +----------+ | | | | |--+---+---+---+---+---+---+---+---+-| | | | | | | | | | | | |--+---+---+---+---+---+---+---+---+-| | | | | | | | | | | | +------------------------------------+ | = bars along B, - = bars along L: 16 mm @ 185 c/c both ways ELEVATION | column | | | +---------------+----------+---------------+ | (dowels) | D = 650 mm | o o o o o o o o o o o o o | bottom mesh, 50 mm cover +------------------------------------------+ <---------------- 3400 mm -----------------> ``` - Bottom mesh: 16 mm @ 185 c/c along $L$ (lower layer), 16 mm @ 185 c/c along $B$ (upper layer), 50 mm clear cover; dowels 8-20 mm. **Answer:** Footing 3.4 m × 3.4 m × 650 mm thick; 16 mm @ 185 c/c along the length and 16 mm @ 185 c/c along the width. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the plan size of the 400 x 400 column footing found? A. Area = Pu / qa with factored load B. Area = 1.1P/qa = 11.00 m^2, then a square side giving at least that area (3.4 m) C. Area = P / (1.1 qa) D. Area equals 3 times the column area Answer: B. Service load plus 10% for weight is divided by the safe bearing capacity; 3.4 x 3.4 = 11.56 m^2 is at least 11.00 m^2. (Key point: Footing size) 2. Which pressure is used to compute bending moments and shears? A. Factored pressure pu = Pu/(BL) = 194.6 kN/m^2 B. Safe bearing capacity of 150 kN/m^2 C. Service pressure including footing weight D. Pressure with factor 0.9 Answer: A. Limit state design uses factored load Pu = 1.5P divided by the base area. (Key point: Design pressure) 3. Where is the critical section for two-way (punching) shear? A. At the column face B. At distance d from the column face C. At d/2 from the column face all round, perimeter 2[(a+d)+(c+d)] D. At the edge of the footing Answer: C. Punching shear is checked on a perimeter at d/2 from the column face, and tau_v is compared with ks x 0.25 sqrt(fck). (Key point: Punching shear section) 4. Bearing capacity is 1440 kN but Pu = 2250 kN. What follows? A. The footing fails and must be redesigned B. The bearing area is doubled C. Only minimum dowels are needed D. The excess 810 kN is carried by dowels (2243 mm^2 required); the 8-20 mm column bars (2513 mm^2) suffice Answer: D. Dowel area = excess load / (0.87 fy) = 2243 mm^2, which the continued column bars exceed. 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