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Isolated footing 450x450, 1600 kN service

Short answer

Provide a 2.7 m × 2.7 m footing, 700 mm deep, with 12 mm bars at 130 mm c/c both ways. Development length 680 mm is satisfied. Bearing stress requires dowels; column bars extended into footing are adequate.

Design an isolated footing for a square column 450 mm x 450 mm, reinforcement with 8-20 dia bars and carrying a service load of 1600 kN. Assume bearing capacity of soil as 250 kN/m2^2 and depth of foundation as 1.5 m. Adopt M20 concrete and Fe 500 steel. Also check the development length and bearing stress in concrete.

Data and assumptions

  • Column 450 mm × 450 mm (long side a=450a=450 mm along the footing length); footing concrete M20, steel Fe500; column concrete M20; column bars 8-20 mm.
  • Service load P=1600P = 1600 kN.
  • Allowable (safe) bearing capacity of soil = 250 kN/m² at 1.5 m depth.
  • 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=1.1×1600.0250=7.04 m2A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 1600.0}{250} = 7.04\ \text{m}^2

Provide a square footing L×B=2.7×2.7L \times B = 2.7 \times 2.7 m, area = 7.29 m² ≥\ge 7.04 m².

Factored soil pressure

Pu=1.5×1600.0=2400.0P_u = 1.5 \times 1600.0 = 2400.0 kN. pu=PuBL=2400.02.7×2.7=329.2p_u = \dfrac{P_u}{BL} = \dfrac{2400.0}{2.7 \times 2.7} = 329.2 kN/m².

Bending moments and reinforcement

Cantilever projections from the column face: along LL: (2700−450)/2=1125(2700 - 450)/2 = 1125 mm; along BB: (2700−450)/2=1125(2700 - 450)/2 = 1125 mm. Mu=pu×(width)×(projection)22M_u = p_u \times (\text{width}) \times \dfrac{(\text{projection})^2}{2}: Mu,L=562.5M_{u,L} = 562.5 kN·m, Mu,B=562.5M_{u,B} = 562.5 kN·m. Required depth for flexure: Mu,lim=0.133fckbd2M_{u,lim} = 0.133 f_{ck} b d^2 is far larger than the moment for d=640d = 640 mm, so the depth is governed by shear (one-way or punching). Trial D=700D = 700 mm, d=640d = 640 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)562.5226812 mm @ 130 c/c2349
Along BB (width LL)562.5226812 mm @ 130 c/c2349

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

  • Along LL side: Vu=431.1V_u = 431.1 kN, τv=VuBd=0.249\tau_v = \dfrac{V_u}{B d} = 0.249 MPa; pt=0.136%p_t = 0.136\%, τc=0.288\tau_c = 0.288 MPa (IS 456 Table 19, SP 16 formula). τv<τc\tau_v < \tau_c, safe.
  • Along BB side: Vu=431.1V_u = 431.1 kN, τv=0.249\tau_v = 0.249 MPa; pt=0.136%p_t = 0.136\%, 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+640)+(450+640)]=4360b_o = 2[(a + d) + (c + d)] = 2[(450+640) + (450+640)] = 4360 mm
  • Vu=Pu−pu(a+d)(c+d)=2400.0−329.2×1.090×1.090=2008.9V_u = P_u - p_u (a+d)(c+d) = 2400.0 - 329.2 \times 1.090 \times 1.090 = 2008.9 kN (average pressure; the effect of moment transfer is neglected).
  • τv=Vubod=0.720\tau_v = \dfrac{V_u}{b_o d} = 0.720 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): 12 mm bars (both directions) → Ld=680L_d = 680 mm. Available length from column face (less 50 mm end cover): 1075 mm along LL, 1075 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=11.85= P_u/A_2 = 11.85 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=2400P_u = 2400 kN, so the excess 578 kN must be carried by dowels: As,dow=5775000.87fy=1328A_{s,dow} = \dfrac{577500}{0.87 f_y} = 1328 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=906L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 906 mm (and the same lap with the column bars). Available depth in footing = 700 - 50 - 24 = 626 mm <Ldc< L_{dc}; dowels are bent at the bottom with 90° bends.

Detailing

PLAN                      ELEVATION
+------------------+      col
| #  #  #  #  #  # |      | |
| #  +------+  #  # |     _|_|__
| #  | col  |  #  # |    |_____|  <- D = 700 mm
| #  +------+  #  # |    bars both ways
| #  #  #  #  #  # |
+------------------+
 L = 2.7 m, B = 2.7 m
  • Bottom mesh: 12 mm @ 130 c/c along LL (lower layer), 12 mm @ 130 c/c along BB (upper layer), 50 mm clear cover; dowels 8-20 mm.

Answer: Footing 2.7 m × 2.7 m × 700 mm thick; 12 mm @ 130 c/c along the length and 12 mm @ 130 c/c along the width.

✓ Answer checked by Claude Opus 5.5 (high effort)

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