--- title: "Rectangular footing, 450x450, uniaxial moment 240 kNm" subject: "Design of RCC Structure" chapter: 11 key: "semester_7/design_of_rcc_structure:rectangular-footing-450x450-uniaxial-moment-240" source: https://www.bctengineer.com/q/semester_7/design_of_rcc_structure/rectangular-footing-450x450-uniaxial-moment-240 site: BCT Engineering --- # Rectangular footing, 450x450, uniaxial moment 240 kNm Design of RCC Structure · Chapter 11: Design of Footings - Asked in: 2068 Baisakh (old course), 15 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/rectangular-footing-450x450-uniaxial-moment-240 - Whole chapter: https://www.bctengineer.com/md/semester_7/design_of_rcc_structure/chapter-11 ## Question Design a reinforced concrete rectangular footing for a square column of size 450 mm x 450 mm, which is subjected to an axial load of 1650 kN and uni-axial moment of 240 kNm at service state. Consider allowable bearing capacity of soil as 120 kN/m$^2$. Show design summary and reinforcement detailing with neat sketch. ## Short answer Rectangular footing 5.5 m × 3.2 m, depth 850 mm (effective 790 mm). Longitudinal reinforcement: 20 mm @ 170 mm c/c; transverse: 16 mm @ 195 mm c/c with central band of 16 mm @ 150 mm c/c per IS 456. ## Answer #### Data and assumptions - Column 450 mm × 450 mm (long side $a=450$ mm along the footing length); footing concrete M20, steel Fe415; column concrete M20; column bars 8-20 mm. - Service load $P = 1650$ kN, service moment $M = 240$ kN·m. - Allowable (safe) bearing capacity of soil = 120 kN/m². - Materials are not given; M20 and Fe415 are assumed, with 8-20 mm column bars. - 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. $$ A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 1650.0}{120} = 15.13\ \text{m}^2 $$ Eccentricity $e = M/P = 240.0/1650.0 = 0.145$ 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 $L \times B = 5.5 \times 3.2$ m ($A = 17.60$ m²): $q_{max} = 119.5$ kN/m² $\le 120.0$ and $q_{min} = 86.8$ kN/m² $> 0$. **Safe.** #### Factored soil pressure $P_u = 1.5 \times 1650.0 = 2475.0$ kN, $M_u = 360.0$ kN·m. $p_u = \dfrac{P_u}{BL}\left(1 \pm \dfrac{6e}{L}\right)$: $p_{u,max} = 162.9$ kN/m², $p_{u,min} = 118.3$ kN/m² (average 140.6 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 $L$: $(5500 - 450)/2 = 2525$ mm; along $B$: $(3200 - 450)/2 = 1375$ 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 $p_f = 142.5$ kN/m² at the face and $x = 2.525$ m: $M_{u,L} = 1592.5$ kN·m. Along $B$ (average pressure): $M_{u,B} = 731.1$ kN·m. Required depth for flexure: $M_{u,lim} = 0.138 f_{ck} b d^2$ is far larger than the moment for $d = 790$ mm, so the depth is governed by shear (one-way or punching). Trial **$D = 850$ mm, $d = 790$ 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 $B$) | 1592.5 | 5869 | 20 mm @ 170 c/c | 5914 | | Along $B$ (width $L$) | 731.1 | 5610 | 16 mm @ 195 c/c | 5671 | For the rectangular footing, the short-direction steel is placed in a central band of width $B = 3200$ mm in the proportion $\dfrac{2}{\beta+1} = \dfrac{2}{1.72+1} = 0.736$ (IS 456 cl. 34.3.1), where $\beta = L/B$. - Central band: $0.736 \times 5610 = 4127$ mm² → 16 mm @ 150 mm c/c (21 bars). - Each outer strip (1150 mm wide): remaining steel 742 mm² or minimum 0.12% = 1173 mm², whichever is more → 16 mm @ 190 mm c/c (6 bars). #### Check for one-way shear (at distance $d$ from the column face) - Along $L$ side: $V_u = 865.6$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.342$ MPa; $p_t = 0.234\%$, $\tau_c = 0.349$ MPa (IS 456 Table 19, SP 16 formula). $\tau_v < \tau_c$, safe. - Along $B$ side: $V_u = 452.5$ kN, $\tau_v = 0.104$ MPa; $p_t = 0.131\%$, $\tau_c = 0.288$ MPa. Safe. #### Check for two-way (punching) shear (at $d/2$ from the column face) - Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(450+790) + (450+790)] = 4960$ mm - $V_u = P_u - p_u (a+d)(c+d) = 2475.0 - 140.6 \times 1.240 \times 1.240 = 2258.8$ kN (average pressure; the effect of moment transfer is neglected). - $\tau_v = \dfrac{V_u}{b_o d} = 0.576$ 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): 20 mm bars → $L_d = 940$ mm; 16 mm bars → $752$ mm. Available length from column face (less 50 mm end cover): 2475 mm along $L$, 1325 mm along $B$. **Satisfied** in both directions. #### Bearing stress and transfer of load from column Column area $A_2 = 450\times450 = 0.203$ m². Bearing stress $= P_u/A_2 = 12.22$ 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 = 1822 kN $<$ $P_u = 2475$ kN, so the excess 652 kN must be carried by dowels: $A_{s,dow} = \dfrac{652500}{0.87 f_y} = 1807$ 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 = 850 - 50 - 36 = 764 mm $\ge L_{dc}$, **adequate**. #### Detailing ```text PLAN ELEVATION +------------------+ col | # # # # # # | | | | # +------+ # # | _|_|__ | # | col | # # | |_____| <- D = 850 mm | # +------+ # # | bars both ways | # # # # # # | +------------------+ L = 5.5 m, B = 3.2 m ``` - Bottom mesh: 20 mm @ 170 c/c along $L$ (lower layer); along $B$ (upper layer), 16 mm @ 150 c/c in the central 3.2 m band and 16 mm @ 190 c/c in each outer strip, 50 mm clear cover; dowels 8-20 mm. **Answer:** Footing 5.5 m × 3.2 m × 850 mm thick; 20 mm @ 170 c/c along the length; along the width, 16 mm @ 150 c/c in the central 3.2 m band and 16 mm @ 190 c/c in the outer strips. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. After A_req = 1.1P/q_a = 15.13 m^2, why is the trial size 5.5 x 3.2 m (17.6 m^2) taken rather than 15.13 m^2? A. Sizes are always rounded up by 50% B. Moment makes pressure non-uniform, so q_max must be limited to q_a; the area needs to be larger C. The column is large D. To avoid shear reinforcement Answer: B. Moment raises the edge pressure above the average, so size is increased until q_max = 1.1(P/BL)(1+6e/L) <= q_a. (Key point: Footing size selection) 2. Which moment is used for the long-direction steel? A. Moment at the column face on the side of maximum pressure, using linear pressure B. Moment at d from the face using average pressure C. The service moment 240 kN-m directly D. Moment at the footing centre Answer: A. The long-direction cantilever is designed at the column face using the linearly varying pressure, giving the larger value of 1522.8 kN-m. (Key point: Critical section for moment) 3. For one-way shear, where is the section checked and against what? A. At the column face against 0.25 sqrt(f_ck) B. At d/2 from the face against tau_c max C. At the footing edge against tau_c D. At distance d from the column face; tau_v = V_u/(B d) is compared with tau_c from Table 19 Answer: D. One-way shear is checked at d from the face, and tau_v must be less than tau_c, which depends on p_t. (Key point: One-way shear check) 4. Which permissible bearing stress governs the load transfer from the column, and why? A. Footing concrete value (18 MPa), because it is larger B. Column concrete 0.45 f_ck = 9 MPa, because it is the lesser of the two C. Steel yield stress D. 0.45 f_ck sqrt(A_1/A_2) always Answer: B. The lesser of the column value (0.45 f_ck) and the footing value (up to twice that) governs, so 9 MPa. 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