--- title: "Isolated footing 500x500, 1600 kN" subject: "Design of RCC Structure" chapter: 11 key: "semester_7/design_of_rcc_structure:isolated-footing-500x500-1600-kn" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/isolated-footing-500x500-1600-kn site: BCT Engineering --- # Isolated footing 500x500, 1600 kN Design of RCC Structure · Chapter 11: Design of Footings - Asked in: 2082 Bhadra, 10+2+2 marks - Video: [RCC 1 - Design of Isolated Column Footing](https://www.youtube.com/watch?v=epm-i1SXxIg) — Learn with Shruti - 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-500x500-1600-kn - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-11 ## Question Design an isolated footing for a column of size 500 mm x 500 mm, and carrying a service axial load of 1600 kN. Assume soil with an allowable bearing capacity of 125 kN/m$^2$ at a depth of 1.25 m below ground. Assume Fe500 TMT steel for both column and footing, and M25 grade concrete for the footing and M25 grade concrete for the column. Make construction drawings in plan and elevation. ## Short answer Provide a 3.8 m × 3.8 m square footing, depth 700 mm (d = 640 mm). Factored upward pressure 166.2 kN/m² yields bending moment 859.7 kN·m per direction; use 12 mm bars at 130 mm c/c both ways. One‑way and punching shear are safe, and 8‑20 mm dowels transfer column load. ## Answer #### Data and assumptions - Column 500 mm × 500 mm (long side $a=500$ mm along the footing length); footing concrete M25, steel Fe500; column concrete M25; column bars 8-20 mm. - Service load $P = 1600$ kN. - Allowable (safe) bearing capacity of soil = 125 kN/m² at 1.25 m depth. - The foundation depth of 1.25 m does not enter the calculation directly; the weight of the footing and backfill is taken as 10% of the column load. - Column bars assumed 8-20 mm (not given). - 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. $$ A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 1600.0}{125} = 14.08\ \text{m}^2 $$ Provide a square footing $L \times B = 3.8 \times 3.8$ m, area = 14.44 m² $\ge$ 14.08 m². #### Factored soil pressure $P_u = 1.5 \times 1600.0 = 2400.0$ kN. $p_u = \dfrac{P_u}{BL} = \dfrac{2400.0}{3.8 \times 3.8} = 166.2$ kN/m². #### Bending moments and reinforcement Cantilever projections from the column face: along $L$: $(3800 - 500)/2 = 1650$ mm; along $B$: $(3800 - 500)/2 = 1650$ mm. $M_u = p_u \times (\text{width}) \times \dfrac{(\text{projection})^2}{2}$: $M_{u,L} = 859.7$ kN·m, $M_{u,B} = 859.7$ kN·m. Required depth for flexure: $M_{u,lim} = 0.133 f_{ck} b d^2$ is far larger than the moment for $d = 640$ mm, so the depth is governed by shear (one-way or punching). Trial **$D = 700$ mm, $d = 640$ 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 $B$) | 859.7 | 3192 | 12 mm @ 130 c/c | 3306 | | Along $B$ (width $L$) | 859.7 | 3192 | 12 mm @ 130 c/c | 3306 | #### Check for one-way shear (at distance $d$ from the column face) - Along $L$ side: $V_u = 637.9$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.262$ MPa; $p_t = 0.136\%$, $\tau_c = 0.291$ MPa (Table 19). $\tau_v < \tau_c$, safe. - Along $B$ side: $V_u = 637.9$ kN, $\tau_v = 0.262$ MPa; $p_t = 0.136\%$, $\tau_c = 0.291$ MPa. Safe. #### Check for two-way (punching) shear (at $d/2$ from the column face) - Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(500+640) + (500+640)] = 4560$ mm - $V_u = P_u - p_u (a+d)(c+d) = 2400.0 - 166.2 \times 1.140 \times 1.140 = 2184.0$ kN (average pressure; the effect of moment transfer is neglected). - $\tau_v = \dfrac{V_u}{b_o d} = 0.748$ MPa; $k_s = 0.5 + \beta_c = 1.00$, $\tau_c' = k_s \times 0.25\sqrt{f_{ck}} = 1.250$ MPa. $\tau_v < \tau_c'$, **safe**. #### Development length $L_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}$, $\tau_{bd} = 2.24$ MPa (M25, deformed bars): 12 mm bars (both directions) → $L_d = 583$ mm. Available length from column face (less 50 mm end cover): 1600 mm along $L$, 1600 mm along $B$. **Satisfied** in both directions. #### Bearing stress and transfer of load from column Column area $A_2 = 500\times500 = 0.250$ m². Bearing stress $= P_u/A_2 = 9.60$ MPa. Permissible bearing stress: column concrete $0.45 f_{ck} = 11.25$ MPa; footing concrete $0.45 f_{ck}\sqrt{A_1/A_2}$ with $\sqrt{A_1/A_2} \le 2$ gives up to 22.50 MPa. The lesser, 11.25 MPa, governs. Capacity of bearing = 2812 kN $\ge$ $P_u = 2400$ kN, so the whole load can be transferred by bearing. Provide minimum dowels of 0.5% of column area = 1250 mm²: use 8 bars of 20 mm (2513 mm²) as dowels, same as column bars. Dowel length in compression $L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 777$ mm (and the same lap with the column bars). Available depth in footing = 700 - 50 - 24 = 626 mm $< L_{dc}$; dowels are bent at the bottom with 90° bends. #### Detailing ```text PLAN (3.8 m x 3.8 m) +------------------------------------+ | | | | | | | | | | | |--+---+---+---+---+---+---+---+---+-| | | | | | | | | | | | |--+---+---+---+---+---+---+---+---+-| | | | | +----------+ | | | | |--+---+---+-| column |--+---+---+-| | | | | +----------+ | | | | |--+---+---+---+---+---+---+---+---+-| | | | | | | | | | | | |--+---+---+---+---+---+---+---+---+-| | | | | | | | | | | | +------------------------------------+ | = bars along B, - = bars along L: 12 mm @ 130 c/c both ways ELEVATION | column | | | +---------------+----------+---------------+ | (dowels) | D = 700 mm | o o o o o o o o o o o o o | bottom mesh, 50 mm cover +------------------------------------------+ <---------------- 3800 mm -----------------> ``` - Bottom mesh: 12 mm @ 130 c/c along $L$ (lower layer), 12 mm @ 130 c/c along $B$ (upper layer), 50 mm clear cover; dowels 8-20 mm. **Answer:** Footing 3.8 m × 3.8 m × 700 mm thick; 12 mm @ 130 c/c along the length and 12 mm @ 130 c/c along the width. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With P = 1600 kN and q_a = 125 kN/m^2, the required area is A. 12.8 m^2 B. 14.08 m^2 (1.1 x 1600/125) C. 19.2 m^2 D. 20.5 m^2 Answer: B. 10% is added for the weight of footing and backfill, giving 1.1 x 1600/125 = 14.08 m^2, so 3.8 x 3.8 m is provided. (Key point: Area of footing) 2. How is p_u = 166.2 kN/m^2 obtained? A. 1600/(3.8 x 3.8) B. 1.5 x 1600/(3.8 x 3.8) = 2400/14.44 C. 125 x 1.5 D. 1600/14.08 Answer: B. The factored column load 2400 kN is divided by the footing area. (Key point: Factored soil pressure) 3. What governs the footing depth D = 700 mm? A. Flexure B. Shear (one-way or punching) C. Minimum steel D. Dowel length Answer: B. The flexural capacity is far larger than the moment for d = 640 mm, so shear checks control the depth. (Key point: Governing depth) 4. Punching shear is checked on a perimeter located A. At d from the column face with b_o = 4 x 500 B. At d/2 from the column face, b_o = 2[(a+d)+(c+d)] = 4560 mm C. At the footing edge D. At the column face Answer: B. The critical section is d/2 from the column face, with permissible stress k_s x 0.25 sqrt(fck) = 1.25 MPa. (Key point: Punching shear) 5. Bearing capacity of 2812 kN exceeds Pu = 2400 kN. What is provided? A. No connection at all B. Minimum dowels of 0.5% of the column area, here 8-20 mm bars C. Dowels of 5% of footing area D. Only extra concrete Answer: B. Even when bearing carries all the load, minimum dowels of 0.5% of column area (1250 mm^2) are provided. 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