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Isolated square footing, 600 kN and 50 kNm

Short answer

Square footing 2.4 m × 2.4 m, depth 450 mm (effective 390 mm). Reinforcement along moment direction: 12 mm @ 170 mm c/c, other direction: 12 mm @ 195 mm c/c; shear checks satisfactory.

Design an isolated square footing foundation of uniform thickness for a 400 mm x 400 mm column subjected to an axial load of 600 kN and a moment of 50 kNm at service state. Consider bearing capacity of soil as 150 kN/m2^2 and concrete grade M20 and steel grade Fe415.

Data and assumptions

  • Column 400 mm × 400 mm (a=c=400a = c = 400 mm); footing and column concrete M20, steel Fe415; column bars assumed 8-16 mm.
  • Service load P=600P = 600 kN, service moment M=50M = 50 kN·m.
  • Allowable (safe) bearing capacity of soil = 150 kN/m².
  • Square footing, as required by the question.
  • 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

The weight of the footing and backfill is taken as 10% of the column load.

Areq=1.1 Pqa=1.1×600.0150=4.40 m2A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 600.0}{150} = 4.40\ \text{m}^2

Eccentricity e=M/P=50.0/600.0=0.083e = M/P = 50.0/600.0 = 0.083 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=2.4×2.4L \times B = 2.4 \times 2.4 m (A=5.76A = 5.76 m²): qmax=138.5q_{max} = 138.5 kN/m² ≤150.0\le 150.0 and qmin=90.7q_{min} = 90.7 kN/m² >0> 0. Safe.

Factored soil pressure

Pu=1.5×600.0=900.0P_u = 1.5 \times 600.0 = 900.0 kN, Mu=75.0M_u = 75.0 kN·m. pu=PuBL(1±6eL)p_u = \dfrac{P_u}{BL}\left(1 \pm \dfrac{6e}{L}\right): pu,max=188.8p_{u,max} = 188.8 kN/m², pu,min=123.7p_{u,min} = 123.7 kN/m² (average 156.2 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: (2400−400)/2=1000(2400 - 400)/2 = 1000 mm; along BB: (2400−400)/2=1000(2400 - 400)/2 = 1000 mm.

Along LL (critical section at the column face on the side of maximum pressure, linearly varying pressure): pressure at the column face =188.8−188.8−123.72.4×1.0=161.7= 188.8 - \dfrac{188.8 - 123.7}{2.4}\times 1.0 = 161.7 kN/m², so

Mu,L=B[161.7×1.022+(188.8−161.7)×1.023]=2.4 (80.85+9.03)=215.7 kN⋅mM_{u,L} = B\left[\frac{161.7 \times 1.0^2}{2} + \frac{(188.8 - 161.7)\times 1.0^2}{3}\right] = 2.4\,(80.85 + 9.03) = 215.7\ \text{kN·m}

Along BB (average pressure): Mu,B=156.2×2.4×1.02/2=187.5M_{u,B} = 156.2 \times 2.4 \times 1.0^2/2 = 187.5 kN·m.

Required depth for flexure: Mu,lim=0.138fckbd2=1007M_{u,lim} = 0.138 f_{ck} b d^2 = 1007 kN·m for d=390d = 390 mm, far larger than the moments, so the depth is governed by shear (one-way or punching). Trial D=450D = 450 mm, d=390d = 390 mm, confirmed 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 = 1296 mm².

DirectionMuM_u (kN·m)AstA_{st} req (mm²)ProvideAstA_{st} prov (mm²)
Along LL (width BB)215.7158912 mm @ 170 c/c1597
Along BB (width LL)187.5137412 mm @ 195 c/c1392

Check for one-way shear (at distance dd from the column face)

  • Along LL side: Vu=264.3V_u = 264.3 kN, τv=VuBd=0.282\tau_v = \dfrac{V_u}{B d} = 0.282 MPa; pt=0.171%p_t = 0.171\%, τc=0.297\tau_c = 0.297 MPa (Table 19). τv<τc\tau_v < \tau_c, safe.
  • Along BB side: Vu=228.8V_u = 228.8 kN, τv=0.244\tau_v = 0.244 MPa; pt=0.149%p_t = 0.149\%, τc=0.28\tau_c = 0.28 MPa (lowest tabulated value, for pt≤0.15%p_t \le 0.15\%). Safe.

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

  • Perimeter bo=2[(a+d)+(c+d)]=2[(400+390)+(400+390)]=3160b_o = 2[(a + d) + (c + d)] = 2[(400+390) + (400+390)] = 3160 mm
  • Vu=Pu−pu(a+d)(c+d)=900.0−156.2×0.790×0.790=802.5V_u = P_u - p_u (a+d)(c+d) = 900.0 - 156.2 \times 0.790 \times 0.790 = 802.5 kN (average pressure; the effect of moment transfer is neglected).
  • τv=Vubod=0.651\tau_v = \dfrac{V_u}{b_o d} = 0.651 MPa; ks=0.5+βc=1.5k_s = 0.5 + \beta_c = 1.5, limited to 1.01.0; 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): for 12 mm bars in both directions, Ld=564L_d = 564 mm. Available length from the column face (less 50 mm end cover): 950 mm along LL and 950 mm along BB. Satisfied in both directions.

Bearing stress and transfer of load from column

Column area A2=400×400=0.160A_2 = 400\times400 = 0.160 m². Bearing stress =Pu/A2=5.62= P_u/A_2 = 5.62 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. Bearing capacity = 1440 kN ≥\ge Pu=900P_u = 900 kN, so the whole load can be transferred by bearing. Provide minimum dowels of 0.5% of the column area = 800 mm²: use 8 bars of 16 mm (1608 mm²) as dowels, the same as the column bars. Dowel length in compression Ldc=0.87fyϕ4×1.25 τbd=602L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 602 mm (and the same lap with the column bars). Available depth in the footing = 450 - 50 - 24 = 376 mm <Ldc< L_{dc}; the dowels are bent at the bottom with 90° bends. (Bends do not count in compression, but this is acceptable here because the dowels are nominal: bearing alone transfers the load.)

Detailing

PLAN                      ELEVATION
+------------------+      col
| #  #  #  #  #  # |      | |
| #  +------+  #  # |     _|_|__
| #  | col  |  #  # |    |_____|  <- D = 450 mm
| #  +------+  #  # |    bars both ways
| #  #  #  #  #  # |
+------------------+
 L = 2.4 m, B = 2.4 m
  • Bottom mesh: 12 mm @ 170 c/c along LL (lower layer), 12 mm @ 195 c/c along BB (upper layer), 50 mm clear cover; dowels 8-16 mm.

Answer: Footing 2.4 m × 2.4 m × 450 mm thick; 12 mm @ 170 c/c along the length and 12 mm @ 195 c/c along the width.

✓ Answer checked by Claude Opus 5.5 (high effort)

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