--- title: "Isolated footing 350x350, 1500 kN" subject: "Design of RCC Structure" chapter: 11 key: "semester_7/design_of_rcc_structure:isolated-footing-350x350-1500-kn" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/isolated-footing-350x350-1500-kn site: BCT Engineering --- # Isolated footing 350x350, 1500 kN Design of RCC Structure · Chapter 11: Design of Footings - Asked in: 2075 Chaitra, 10 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-footing-350x350-1500-kn - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-11 ## Question Design a R.C.C isolated footing to carry an axial load of 1500 kN. The column is 350 mm x 350 mm in size with 20 mm diameter, 8 Nos longitudinal bars. The bearing capacity of soil is 175 kN/m$^2$. Use M20 grade concrete and Fe415 grade steel. Assume missing datas. ## Short answer Area = 1.1*1500/175 = 9.43 m², provide 3.1 m square. Factored pressure 234.1 kN/m²; depth D=650 mm (d=590 mm). Reinforcement 16 mm @ 185 c/c both ways; shear safe. Bearing stress 18.37 MPa > 9 MPa, excess 1148 kN requires dowels; column bars plus 4 extra 16 mm bars provide 3317 mm². ## Answer #### Data and assumptions - Column 350 mm × 350 mm (long side $a=350$ 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 = 175 kN/m². - Axial load taken as service load; column concrete M20 assumed. - 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. $$ A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 1500.0}{175} = 9.43\ \text{m}^2 $$ Provide a square footing $L \times B = 3.1 \times 3.1$ m, area = 9.61 m² $\ge$ 9.43 m². #### Factored soil pressure $P_u = 1.5 \times 1500.0 = 2250.0$ kN. $p_u = \dfrac{P_u}{BL} = \dfrac{2250.0}{3.1 \times 3.1} = 234.1$ kN/m². #### Bending moments and reinforcement Cantilever projections from the column face: along $L$: $(3100 - 350)/2 = 1375$ mm; along $B$: $(3100 - 350)/2 = 1375$ mm. $M_u = p_u \times (\text{width}) \times \dfrac{(\text{projection})^2}{2}$: $M_{u,L} = 686.1$ kN·m, $M_{u,B} = 686.1$ 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 $B$) | 686.1 | 3350 | 16 mm @ 185 c/c | 3369 | | Along $B$ (width $L$) | 686.1 | 3350 | 16 mm @ 185 c/c | 3369 | #### Check for one-way shear (at distance $d$ from the column face) - Along $L$ side: $V_u = 569.8$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.312$ MPa; $p_t = 0.184\%$, $\tau_c = 0.315$ MPa (IS 456 Table 19, SP 16 formula). $\tau_v < \tau_c$, safe. - Along $B$ side: $V_u = 569.8$ kN, $\tau_v = 0.312$ 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 $b_o = 2[(a + d) + (c + d)] = 2[(350+590) + (350+590)] = 3760$ mm - $V_u = P_u - p_u (a+d)(c+d) = 2250.0 - 234.1 \times 0.940 \times 0.940 = 2043.1$ kN (average pressure; the effect of moment transfer is neglected). - $\tau_v = \dfrac{V_u}{b_o d} = 0.921$ 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): 1325 mm along $L$, 1325 mm along $B$. **Satisfied** in both directions. #### Bearing stress and transfer of load from column Column area $A_2 = 350\times350 = 0.122$ m². Bearing stress $= P_u/A_2 = 18.37$ 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 = 1102 kN $<$ $P_u = 2250$ kN, so the excess 1148 kN must be carried by dowels: $A_{s,dow} = \dfrac{1147500}{0.87 f_y} = 3178$ mm²; the column bars (8-20 mm, 2513 mm²) are continued into the footing as dowels, and 4 extra bars of 16 mm (804 mm²) are added, giving 3317 mm² $\ge$ 3178 mm². 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 ELEVATION +------------------+ col | # # # # # # | | | | # +------+ # # | _|_|__ | # | col | # # | |_____| <- D = 650 mm | # +------+ # # | bars both ways | # # # # # # | +------------------+ L = 3.1 m, B = 3.1 m ``` - 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 3.1 m × 3.1 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 area found for a 1500 kN service load on soil of 175 kN/m^2? A. A = P/q = 8.57 m^2 B. A = 1.1P/q_a = 9.43 m^2, with 10% added for footing and backfill weight C. A = 1.5P/q_a D. A = P/(1.1 q_a) Answer: B. Adding 10% of P for self-weight and backfill gives 1.1 x 1500/175 = 9.43 m^2, so 3.1 x 3.1 m (9.61 m^2) is provided. (Key point: Size of footing) 2. What is the punching shear result for d = 590 mm? A. tau_v = 0.921 MPa < allowable k_s x 0.25 sqrt(f_ck) = 1.118 MPa, safe B. tau_v = 0.921 MPa exceeds 0.25 sqrt(f_ck), unsafe C. tau_v is compared with tau_c from Table 19 D. Punching shear is not needed for square columns Answer: A. Two-way shear is checked at d/2 from the column face against k_s x 0.25 sqrt(f_ck) = 1.118 MPa; 0.921 is less, so safe. (Key point: Punching shear) 3. Bearing capacity is 1102 kN, P_u = 2250 kN. What follows? A. Nothing, column bars are enough B. The excess 1148 kN needs A_s = 1147500/(0.87 f_y) = 3178 mm^2, more than the column bars 2513 mm^2, so extra dowels are required C. The footing must be made 2.1 m deep D. Bearing stress is raised to 18 MPa Answer: B. Dowels must carry the load beyond bearing capacity; 3178 mm^2 exceeds the 8-20 mm bars (2513 mm^2), so extra dowel bars are added. 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