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Footing, 1250 kN column load, depth 1.8 m

Short answer

Net safe bearing capacity 107.6 kN/m² gives required area 11.62 m², so 3.5 m square footing. Factored pressure 153.1 kN/m²; depth D=550 mm (d=490 mm) governed by shear. Reinforcement 16 mm @ 180 c/c both ways; shear safe. Bearing stress 11.72 MPa exceeds 9 MPa, excess 435 kN carried by column dowels.

Design a RC footing to carry a column load of 1250 kN from 400 x 400 mm square column having 20 mm diameter bar as longitudinal steel. The bearing capacity of soil is 140 kN/m2^2. Consider the depth of foundation as 1.8 m. Take unit weight of earth as 18 kN/m3^3. Use M20 concrete mix and Fe415 grade steel. Also sketch the reinforcements in plan and section.

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 6-20 mm.
  • Service load P=1250P = 1250 kN.
  • Allowable (safe) bearing capacity of soil = 140 kN/m² at 1.8 m depth, soil unit weight 18 kN/m³.
  • Bearing capacity is taken as the gross value at the foundation level.
  • 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

Net safe bearing capacity = q−γDf=140−18×1.8=107.60q - \gamma D_f = 140 - 18 \times 1.8 = 107.60 kN/m² (this takes care of the weight of footing and backfill).

Areq=Pqnet=1250.0107.60=11.62 m2A_{req} = \frac{P}{q_{net}} = \frac{1250.0}{107.60} = 11.62\ \text{m}^2

Provide a square footing L×B=3.5×3.5L \times B = 3.5 \times 3.5 m, area = 12.25 m² ≥\ge 11.62 m².

Factored soil pressure

Pu=1.5×1250.0=1875.0P_u = 1.5 \times 1250.0 = 1875.0 kN. pu=PuBL=1875.03.5×3.5=153.1p_u = \dfrac{P_u}{BL} = \dfrac{1875.0}{3.5 \times 3.5} = 153.1 kN/m².

Bending moments and reinforcement

Cantilever projections from the column face: along LL: (3500−400)/2=1550(3500 - 400)/2 = 1550 mm; along BB: (3500−400)/2=1550(3500 - 400)/2 = 1550 mm. Mu=pu×(width)×(projection)22M_u = p_u \times (\text{width}) \times \dfrac{(\text{projection})^2}{2}: Mu,L=643.5M_{u,L} = 643.5 kN·m, Mu,B=643.5M_{u,B} = 643.5 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=490d = 490 mm, so the depth is governed by shear (one-way or punching). Trial D=550D = 550 mm, d=490d = 490 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)643.5381516 mm @ 180 c/c3910
Along BB (width LL)643.5381516 mm @ 180 c/c3910

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

  • Along LL side: Vu=567.9V_u = 567.9 kN, τv=VuBd=0.331\tau_v = \dfrac{V_u}{B d} = 0.331 MPa; pt=0.228%p_t = 0.228\%, τc=0.345\tau_c = 0.345 MPa (IS 456 Table 19, SP 16 formula). τv<τc\tau_v < \tau_c, safe.
  • Along BB side: Vu=567.9V_u = 567.9 kN, τv=0.331\tau_v = 0.331 MPa; pt=0.228%p_t = 0.228\%, τc=0.345\tau_c = 0.345 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+490)+(400+490)]=3560b_o = 2[(a + d) + (c + d)] = 2[(400+490) + (400+490)] = 3560 mm
  • Vu=Pu−pu(a+d)(c+d)=1875.0−153.1×0.890×0.890=1753.8V_u = P_u - p_u (a+d)(c+d) = 1875.0 - 153.1 \times 0.890 \times 0.890 = 1753.8 kN (average pressure; the effect of moment transfer is neglected).
  • τv=Vubod=1.005\tau_v = \dfrac{V_u}{b_o d} = 1.005 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): 1500 mm along LL, 1500 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=11.72= P_u/A_2 = 11.72 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=1875P_u = 1875 kN, so the excess 435 kN must be carried by dowels: As,dow=4350000.87fy=1205A_{s,dow} = \dfrac{435000}{0.87 f_y} = 1205 mm²; the column bars (6-20 mm, 1885 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 = 550 - 50 - 32 = 468 mm <Ldc< L_{dc}; dowels are bent at the bottom with 90° bends.

Detailing

PLAN                      ELEVATION
+------------------+      col
| #  #  #  #  #  # |      | |
| #  +------+  #  # |     _|_|__
| #  | col  |  #  # |    |_____|  <- D = 550 mm
| #  +------+  #  # |    bars both ways
| #  #  #  #  #  # |
+------------------+
 L = 3.5 m, B = 3.5 m
  • Bottom mesh: 16 mm @ 180 c/c along LL (lower layer), 16 mm @ 180 c/c along BB (upper layer), 50 mm clear cover; dowels 6-20 mm.

Answer: Footing 3.5 m × 3.5 m × 550 mm thick; 16 mm @ 180 c/c along the length and 16 mm @ 180 c/c along the width.

✓ Answer checked by Claude Opus 5.5 (high effort)

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