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Rectangular footing, 450x450, uniaxial moment 240 kNm

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.

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/m2^2. Show design summary and reinforcement detailing with neat sketch.

Data and assumptions

  • Column 450 mm × 450 mm (long side a=450a=450 mm along the footing length); footing concrete M20, steel Fe415; column concrete M20; column bars 8-20 mm.
  • Service load P=1650P = 1650 kN, service moment M=240M = 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−60d = 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.

Areq=1.1 Pqa=1.1×1650.0120=15.13 m2A_{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=0.145e = M/P = 240.0/1650.0 = 0.145 m, which is less than L/6L/6, so there is no tension under the base. The size is chosen so that

qmax=1.1PBL(1+6eL)≤qaq_{max} = 1.1\frac{P}{BL}\left(1 + \frac{6e}{L}\right) \le q_a

Try L×B=5.5×3.2L \times B = 5.5 \times 3.2 m (A=17.60A = 17.60 m²): qmax=119.5q_{max} = 119.5 kN/m² ≤120.0\le 120.0 and qmin=86.8q_{min} = 86.8 kN/m² >0> 0. Safe.

Factored soil pressure

Pu=1.5×1650.0=2475.0P_u = 1.5 \times 1650.0 = 2475.0 kN, Mu=360.0M_u = 360.0 kN·m. pu=PuBL(1±6eL)p_u = \dfrac{P_u}{BL}\left(1 \pm \dfrac{6e}{L}\right): pu,max=162.9p_{u,max} = 162.9 kN/m², pu,min=118.3p_{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 LL: (5500−450)/2=2525(5500 - 450)/2 = 2525 mm; along BB: (3200−450)/2=1375(3200 - 450)/2 = 1375 mm. Along LL (critical section at the column face on the side of maximum pressure, linear pressure): Mu,L=B[pfx22+(pmax−pf)x23]M_{u,L} = B\left[p_f\dfrac{x^2}{2} + (p_{max}-p_f)\dfrac{x^2}{3}\right], with pf=142.5p_f = 142.5 kN/m² at the face and x=2.525x = 2.525 m: Mu,L=1592.5M_{u,L} = 1592.5 kN·m. Along BB (average pressure): Mu,B=731.1M_{u,B} = 731.1 kN·m. Required depth for flexure: Mu,lim=0.138fckbd2M_{u,lim} = 0.138 f_{ck} b d^2 is far larger than the moment for d=790d = 790 mm, so the depth is governed by shear (one-way or punching). Trial D=850D = 850 mm, d=790d = 790 mm, found by the shear checks below.

Ast=0.5fckfy[1−1−4.6Mufckbd2]bdA_{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 bDbD.

DirectionMuM_u (kN·m)AstA_{st} req (mm²)ProvideAstA_{st} prov (mm²)
Along LL (width BB)1592.5586920 mm @ 170 c/c5914
Along BB (width LL)731.1561016 mm @ 195 c/c5671

For the rectangular footing, the short-direction steel is placed in a central band of width B=3200B = 3200 mm in the proportion 2β+1=21.72+1=0.736\dfrac{2}{\beta+1} = \dfrac{2}{1.72+1} = 0.736 (IS 456 cl. 34.3.1), where β=L/B\beta = L/B.

  • Central band: 0.736×5610=41270.736 \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 dd from the column face)

  • Along LL side: Vu=865.6V_u = 865.6 kN, τv=VuBd=0.342\tau_v = \dfrac{V_u}{B d} = 0.342 MPa; pt=0.234%p_t = 0.234\%, τc=0.349\tau_c = 0.349 MPa (IS 456 Table 19, SP 16 formula). τv<τc\tau_v < \tau_c, safe.
  • Along BB side: Vu=452.5V_u = 452.5 kN, τv=0.104\tau_v = 0.104 MPa; pt=0.131%p_t = 0.131\%, τc=0.288\tau_c = 0.288 MPa. Safe.

Check for two-way (punching) shear (at d/2d/2 from the column face)

  • Perimeter bo=2[(a+d)+(c+d)]=2[(450+790)+(450+790)]=4960b_o = 2[(a + d) + (c + d)] = 2[(450+790) + (450+790)] = 4960 mm
  • Vu=Pu−pu(a+d)(c+d)=2475.0−140.6×1.240×1.240=2258.8V_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).
  • τv=Vubod=0.576\tau_v = \dfrac{V_u}{b_o d} = 0.576 MPa; ks=0.5+βc=1.00k_s = 0.5 + \beta_c = 1.00, τc′=ks×0.25fck=1.118\tau_c' = k_s \times 0.25\sqrt{f_{ck}} = 1.118 MPa. τv<τc′\tau_v < \tau_c', safe.

Development length

Ld=0.87fyϕ4τbdL_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}, τbd=1.92\tau_{bd} = 1.92 MPa (M20, deformed bars): 20 mm bars → Ld=940L_d = 940 mm; 16 mm bars → 752752 mm. Available length from column face (less 50 mm end cover): 2475 mm along LL, 1325 mm along BB. Satisfied in both directions.

Bearing stress and transfer of load from column

Column area A2=450×450=0.203A_2 = 450\times450 = 0.203 m². Bearing stress =Pu/A2=12.22= P_u/A_2 = 12.22 MPa. Permissible bearing stress: column concrete 0.45fck=9.000.45 f_{ck} = 9.00 MPa; footing concrete 0.45fckA1/A20.45 f_{ck}\sqrt{A_1/A_2} with A1/A2≤2\sqrt{A_1/A_2} \le 2 gives up to 18.00 MPa. The lesser, 9.00 MPa, governs. Capacity of bearing = 1822 kN << Pu=2475P_u = 2475 kN, so the excess 652 kN must be carried by dowels: As,dow=6525000.87fy=1807A_{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 Ldc=0.87fyϕ4×1.25 τbd=752L_{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 ≥Ldc\ge L_{dc}, adequate.

Detailing

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 LL (lower layer); along BB (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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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