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Footing for 400x400 column, 1500 kN

Short answer

Provide a 3.4 m x 3.4 m square footing, overall depth 650 mm (d=590 mm). Reinforcement: 16 mm bars @185 mm c/c both ways. One-way shear and punching shear are within limits; bearing stress check requires dowels, satisfied by continuing the column’s 8–20 mm bars.

Design a footing for supporting a square column of 400 mm x 400 mm. The column carries a service load of 1500 kN. The safe bearing capacity of soil is 150 kN/m2^2. Use M20 concrete and Fe415 grade steel.

Data and assumptions

  • Column 400 mm × 400 mm (long side a=400a=400 mm along the footing length); footing concrete M20, steel Fe415; column concrete M20; column bars 8-20 mm.
  • Service load P=1500P = 1500 kN.
  • Allowable (safe) bearing capacity of soil = 150 kN/m².
  • Column bars not given; 8-20 mm assumed. Loads are service loads.
  • 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=1.1×1500.0150=11.00 m2A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 1500.0}{150} = 11.00\ \text{m}^2

Provide a square footing L×B=3.4×3.4L \times B = 3.4 \times 3.4 m, area = 11.56 m² ≥\ge 11.00 m².

Factored soil pressure

Pu=1.5×1500.0=2250.0P_u = 1.5 \times 1500.0 = 2250.0 kN. pu=PuBL=2250.03.4×3.4=194.6p_u = \dfrac{P_u}{BL} = \dfrac{2250.0}{3.4 \times 3.4} = 194.6 kN/m².

Bending moments and reinforcement

Cantilever projections from the column face: along LL: (3400−400)/2=1500(3400 - 400)/2 = 1500 mm; along BB: (3400−400)/2=1500(3400 - 400)/2 = 1500 mm. Mu=pu×(width)×(projection)22M_u = p_u \times (\text{width}) \times \dfrac{(\text{projection})^2}{2}: Mu,L=744.5M_{u,L} = 744.5 kN·m, Mu,B=744.5M_{u,B} = 744.5 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=590d = 590 mm, so the depth is governed by shear (one-way or punching). Trial D=650D = 650 mm, d=590d = 590 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)744.5363316 mm @ 185 c/c3695
Along BB (width LL)744.5363316 mm @ 185 c/c3695

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

  • Along LL side: Vu=602.2V_u = 602.2 kN, τv=VuBd=0.300\tau_v = \dfrac{V_u}{B d} = 0.300 MPa; pt=0.184%p_t = 0.184\%, τc=0.315\tau_c = 0.315 MPa (Table 19). τv<τc\tau_v < \tau_c, safe.
  • Along BB side: Vu=602.2V_u = 602.2 kN, τv=0.300\tau_v = 0.300 MPa; pt=0.184%p_t = 0.184\%, τc=0.315\tau_c = 0.315 MPa. Safe.

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

  • Perimeter bo=2[(a+d)+(c+d)]=2[(400+590)+(400+590)]=3960b_o = 2[(a + d) + (c + d)] = 2[(400+590) + (400+590)] = 3960 mm
  • Vu=Pu−pu(a+d)(c+d)=2250.0−194.6×0.990×0.990=2059.2V_u = P_u - p_u (a+d)(c+d) = 2250.0 - 194.6 \times 0.990 \times 0.990 = 2059.2 kN (average pressure; the effect of moment transfer is neglected).
  • τv=Vubod=0.881\tau_v = \dfrac{V_u}{b_o d} = 0.881 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): 16 mm bars (both directions) → Ld=752L_d = 752 mm. Available length from column face (less 50 mm end cover): 1450 mm along LL, 1450 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=14.06= P_u/A_2 = 14.06 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 = 1440 kN << Pu=2250P_u = 2250 kN, so the excess 810 kN must be carried by dowels: As,dow=8100000.87fy=2243A_{s,dow} = \dfrac{810000}{0.87 f_y} = 2243 mm²; the column bars (8-20 mm, 2513 mm²) are continued into the footing as dowels - 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 = 650 - 50 - 32 = 568 mm <Ldc< L_{dc}; dowels are bent at the bottom with 90° bends.

Detailing

PLAN (3.4 m x 3.4 m)
+------------------------------------+
|  |   |   |   |   |   |   |   |   | |
|--+---+---+---+---+---+---+---+---+-|
|  |   |   |   |   |   |   |   |   | |
|--+---+---+---+---+---+---+---+---+-|
|  |   |   | +----------+  |   |   | |
|--+---+---+-|  column  |--+---+---+-|
|  |   |   | +----------+  |   |   | |
|--+---+---+---+---+---+---+---+---+-|
|  |   |   |   |   |   |   |   |   | |
|--+---+---+---+---+---+---+---+---+-|
|  |   |   |   |   |   |   |   |   | |
+------------------------------------+
| = bars along B, - = bars along L: 16 mm @ 185 c/c both ways

ELEVATION
                |  column  |
                |          |
+---------------+----------+---------------+
|                 (dowels)                 |  D = 650 mm
|  o  o  o  o  o  o  o  o  o  o  o  o  o   |  bottom mesh, 50 mm cover
+------------------------------------------+
<---------------- 3400 mm ----------------->
  • Bottom mesh: 16 mm @ 185 c/c along LL (lower layer), 16 mm @ 185 c/c along BB (upper layer), 50 mm clear cover; dowels 8-20 mm.

Answer: Footing 3.4 m × 3.4 m × 650 mm thick; 16 mm @ 185 c/c along the length and 16 mm @ 185 c/c along the width.

✓ Answer checked by Claude Opus 5.5 (high effort)

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