--- title: "Isolated footing 350x500, 2000 kN" subject: "Design of RCC Structure" chapter: 11 key: "semester_7/design_of_rcc_structure:isolated-footing-350x500-2000-kn" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/isolated-footing-350x500-2000-kn site: BCT Engineering --- # Isolated footing 350x500, 2000 kN Design of RCC Structure · Chapter 11: Design of Footings - Asked in: 2069 Chaitra, 14 marks - 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-350x500-2000-kn - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-11 ## Question Design the isolated footing of a column of 350 mm x 500 mm. Column is subjected to design axial load of 2000 kN and design BM of 80 kN-m. Allowable bearing capacity of soil is equal to 175 kN/m$^2$. ## Short answer Rectangular footing 3.9 m × 2.3 m, depth 800 mm (effective 740 mm). Longitudinal steel: 16 mm @ 150 mm c/c; transverse: 16 mm @ 205 mm c/c with central band of 14 bars @ 160 mm c/c per IS 456. ## Answer #### Data and assumptions - Column 500 mm × 350 mm (long side $a=500$ mm along the footing length); footing concrete M20, steel Fe415; column concrete M20; column bars 8-20 mm. - Given loads are factored: $P_u = 2000$ kN, $M_u = 80$ kN·m. Service loads = factored / 1.5: $P = 1333.3$ kN, $M = 53.3$ kN·m. - Allowable (safe) bearing capacity of soil = 175 kN/m². - The question gives design (factored) loads; M20 concrete and Fe415 steel are assumed, with 8-20 mm column bars. - Moment acts about the minor axis, varying the pressure along the 500 mm side. - 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. $$ A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 1333.3}{175} = 8.38\ \text{m}^2 $$ Eccentricity $e = M/P = 53.3/1333.3 = 0.040$ m, which is less than $L/6$, so there is no tension under the base. The size is chosen so that $$ q_{max} = 1.1\frac{P}{BL}\left(1 + \frac{6e}{L}\right) \le q_a $$ Try $L \times B = 3.9 \times 2.3$ m ($A = 8.97$ m²): $q_{max} = 173.6$ kN/m² $\le 175.0$ and $q_{min} = 153.4$ kN/m² $> 0$. **Safe.** #### Factored soil pressure $P_u = 1.5 \times 1333.3 = 2000.0$ kN, $M_u = 80.0$ kN·m. $p_u = \dfrac{P_u}{BL}\left(1 \pm \dfrac{6e}{L}\right)$: $p_{u,max} = 236.7$ kN/m², $p_{u,min} = 209.2$ kN/m² (average 223.0 kN/m²). Self-weight does not cause bending in the footing, so it is excluded. #### Bending moments and reinforcement Cantilever projections from the column face: along $L$: $(3900 - 500)/2 = 1700$ mm; along $B$: $(2300 - 350)/2 = 975$ mm. Along $L$ (critical section at the column face on the side of maximum pressure, linear pressure): $M_{u,L} = B\left[p_f\dfrac{x^2}{2} + (p_{max}-p_f)\dfrac{x^2}{3}\right]$, with $p_f = 224.7$ kN/m² at the face and $x = 1.7$ m: $M_{u,L} = 773.4$ kN·m. Along $B$ (average pressure): $M_{u,B} = 413.3$ kN·m. Required depth for flexure: $M_{u,lim} = 0.138 f_{ck} b d^2$ is far larger than the moment for $d = 740$ mm, so the depth is governed by shear (one-way or punching). Trial **$D = 800$ mm, $d = 740$ 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 $B$) | 773.4 | 3006 | 16 mm @ 150 c/c | 3083 | | Along $B$ (width $L$) | 413.3 | 3744 | 16 mm @ 205 c/c | 3825 | For the rectangular footing, the short-direction steel is placed in a central band of width $B = 2300$ mm in the proportion $\dfrac{2}{\beta+1} = \dfrac{2}{1.70+1} = 0.742$ (IS 456 cl. 34.3.1), where $\beta = L/B$. - Central band: $0.742 \times 3744 = 2778$ mm² → 16 mm @ 160 mm c/c (14 bars). - Each outer strip (800 mm wide): remaining steel 483 mm² or minimum 0.12% = 768 mm², whichever is more → 16 mm @ 200 mm c/c (4 bars). #### Check for one-way shear (at distance $d$ from the column face) - Along $L$ side: $V_u = 515.1$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.303$ MPa; $p_t = 0.181\%$, $\tau_c = 0.312$ MPa (IS 456 Table 19, SP 16 formula). $\tau_v < \tau_c$, safe. - Along $B$ side: $V_u = 204.3$ kN, $\tau_v = 0.071$ MPa; $p_t = 0.133\%$, $\tau_c = 0.288$ MPa. Safe. #### Check for two-way (punching) shear (at $d/2$ from the column face) - Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(500+740) + (350+740)] = 4660$ mm - $V_u = P_u - p_u (a+d)(c+d) = 2000.0 - 223.0 \times 1.240 \times 1.090 = 1698.6$ kN (average pressure; the effect of moment transfer is neglected). - $\tau_v = \dfrac{V_u}{b_o d} = 0.493$ 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): 1650 mm along $L$, 925 mm along $B$. **Satisfied** in both directions. #### Bearing stress and transfer of load from column Column area $A_2 = 500\times350 = 0.175$ m². Bearing stress $= P_u/A_2 = 11.43$ 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 = 1575 kN $<$ $P_u = 2000$ kN, so the excess 425 kN must be carried by dowels: $A_{s,dow} = \dfrac{425000}{0.87 f_y} = 1177$ 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 = 800 - 50 - 32 = 718 mm $< L_{dc}$; dowels are bent at the bottom with 90° bends. #### Detailing ```text PLAN ELEVATION +------------------+ col | # # # # # # | | | | # +------+ # # | _|_|__ | # | col | # # | |_____| <- D = 800 mm | # +------+ # # | bars both ways | # # # # # # | +------------------+ L = 3.9 m, B = 2.3 m ``` - Bottom mesh: 16 mm @ 150 c/c along $L$ (lower layer); along $B$ (upper layer), 16 mm @ 160 c/c in the central 2.3 m band and 16 mm @ 200 c/c in each outer strip, 50 mm clear cover; dowels 8-20 mm. **Answer:** Footing 3.9 m × 2.3 m × 800 mm thick; 16 mm @ 150 c/c along the length; along the width, 16 mm @ 160 c/c in the central 2.3 m band and 16 mm @ 200 c/c in the outer strips. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The question gives design axial load 2000 kN and BM 80 kN-m. How are they used for soil pressure checks? A. Used directly against the allowable bearing capacity B. Divided by 1.5 to get service loads (1333.3 kN, 53.3 kN-m) C. Multiplied by 1.5 again D. Multiplied by 1.1 only Answer: B. The given loads are already factored, so service loads = factored / 1.5 are used for the allowable bearing pressure check. (Key point: Factored vs service loads) 2. No soil depth is given, so how is the footing and backfill weight allowed for when sizing? A. The area is taken as 1.1 P / q_a, i.e. 10% extra load B. It is ignored C. q_a is reduced by 50% D. The area is taken as P / (1.5 q_a) Answer: A. The weight of footing and backfill is taken as 10% of the column load, so A_req = 1.1 P / q_a. (Key point: Footing area estimate) 3. Bearing capacity of the column concrete is 0.45 f_ck x A_2 = 1575 kN, less than P_u = 2000 kN. What follows? A. The footing must be enlarged B. The footing concrete grade must be raised C. The excess 425 kN is carried by dowels/column bars continued into the footing D. The column must be redesigned with M30 Answer: C. The excess load is carried by steel: A_s = 425000/(0.87 f_y) = 1177 mm^2, which the 8-20 mm bars (2513 mm^2) provide. (Key point: Bearing at column-footing interface) 4. With L/B = 1.70, the central band takes 0.742 of the short-direction steel. What is the reason for this arrangement? A. Steel is concentrated near the column where the short-direction moment is more effective B. It reduces total steel C. It prevents punching shear D. It satisfies development length Answer: A. The central band under the column is more heavily stressed, so more of the short-direction steel is concentrated there (IS 456 cl. 34.3.1). 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