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Isolated footing 500x500, 1600 kN

Short answer

Provide a 3.8 m × 3.8 m square footing, depth 700 mm (d = 640 mm). Factored upward pressure 166.2 kN/m² yields bending moment 859.7 kN·m per direction; use 12 mm bars at 130 mm c/c both ways. One‑way and punching shear are safe, and 8‑20 mm dowels transfer column load.

Design an isolated footing for a column of size 500 mm x 500 mm, and carrying a service axial load of 1600 kN. Assume soil with an allowable bearing capacity of 125 kN/m2^2 at a depth of 1.25 m below ground. Assume Fe500 TMT steel for both column and footing, and M25 grade concrete for the footing and M25 grade concrete for the column. Make construction drawings in plan and elevation.

Data and assumptions

  • Column 500 mm × 500 mm (long side a=500a=500 mm along the footing length); footing concrete M25, steel Fe500; column concrete M25; column bars 8-20 mm.
  • Service load P=1600P = 1600 kN.
  • Allowable (safe) bearing capacity of soil = 125 kN/m² at 1.25 m depth.
  • The foundation depth of 1.25 m does not enter the calculation directly; the weight of the footing and backfill is taken as 10% of the column load.
  • Column bars assumed 8-20 mm (not given).
  • 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=1.1×1600.0125=14.08 m2A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 1600.0}{125} = 14.08\ \text{m}^2

Provide a square footing L×B=3.8×3.8L \times B = 3.8 \times 3.8 m, area = 14.44 m² ≥\ge 14.08 m².

Factored soil pressure

Pu=1.5×1600.0=2400.0P_u = 1.5 \times 1600.0 = 2400.0 kN. pu=PuBL=2400.03.8×3.8=166.2p_u = \dfrac{P_u}{BL} = \dfrac{2400.0}{3.8 \times 3.8} = 166.2 kN/m².

Bending moments and reinforcement

Cantilever projections from the column face: along LL: (3800−500)/2=1650(3800 - 500)/2 = 1650 mm; along BB: (3800−500)/2=1650(3800 - 500)/2 = 1650 mm. Mu=pu×(width)×(projection)22M_u = p_u \times (\text{width}) \times \dfrac{(\text{projection})^2}{2}: Mu,L=859.7M_{u,L} = 859.7 kN·m, Mu,B=859.7M_{u,B} = 859.7 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)859.7319212 mm @ 130 c/c3306
Along BB (width LL)859.7319212 mm @ 130 c/c3306

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

  • Along LL side: Vu=637.9V_u = 637.9 kN, τv=VuBd=0.262\tau_v = \dfrac{V_u}{B d} = 0.262 MPa; pt=0.136%p_t = 0.136\%, τc=0.291\tau_c = 0.291 MPa (Table 19). τv<τc\tau_v < \tau_c, safe.
  • Along BB side: Vu=637.9V_u = 637.9 kN, τv=0.262\tau_v = 0.262 MPa; pt=0.136%p_t = 0.136\%, τc=0.291\tau_c = 0.291 MPa. Safe.

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

  • Perimeter bo=2[(a+d)+(c+d)]=2[(500+640)+(500+640)]=4560b_o = 2[(a + d) + (c + d)] = 2[(500+640) + (500+640)] = 4560 mm
  • Vu=Pu−pu(a+d)(c+d)=2400.0−166.2×1.140×1.140=2184.0V_u = P_u - p_u (a+d)(c+d) = 2400.0 - 166.2 \times 1.140 \times 1.140 = 2184.0 kN (average pressure; the effect of moment transfer is neglected).
  • τv=Vubod=0.748\tau_v = \dfrac{V_u}{b_o d} = 0.748 MPa; ks=0.5+βc=1.00k_s = 0.5 + \beta_c = 1.00, τc′=ks×0.25fck=1.250\tau_c' = k_s \times 0.25\sqrt{f_{ck}} = 1.250 MPa. τv<τc′\tau_v < \tau_c', safe.

Development length

Ld=0.87fyϕ4τbdL_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}, τbd=2.24\tau_{bd} = 2.24 MPa (M25, deformed bars): 12 mm bars (both directions) → Ld=583L_d = 583 mm. Available length from column face (less 50 mm end cover): 1600 mm along LL, 1600 mm along BB. Satisfied in both directions.

Bearing stress and transfer of load from column

Column area A2=500×500=0.250A_2 = 500\times500 = 0.250 m². Bearing stress =Pu/A2=9.60= P_u/A_2 = 9.60 MPa. Permissible bearing stress: column concrete 0.45fck=11.250.45 f_{ck} = 11.25 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 22.50 MPa. The lesser, 11.25 MPa, governs. Capacity of bearing = 2812 kN ≥\ge Pu=2400P_u = 2400 kN, so the whole load can be transferred by bearing. Provide minimum dowels of 0.5% of column area = 1250 mm²: use 8 bars of 20 mm (2513 mm²) as dowels, same as column bars. Dowel length in compression Ldc=0.87fyϕ4×1.25 τbd=777L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 777 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 (3.8 m x 3.8 m)
+------------------------------------+
|  |   |   |   |   |   |   |   |   | |
|--+---+---+---+---+---+---+---+---+-|
|  |   |   |   |   |   |   |   |   | |
|--+---+---+---+---+---+---+---+---+-|
|  |   |   | +----------+  |   |   | |
|--+---+---+-|  column  |--+---+---+-|
|  |   |   | +----------+  |   |   | |
|--+---+---+---+---+---+---+---+---+-|
|  |   |   |   |   |   |   |   |   | |
|--+---+---+---+---+---+---+---+---+-|
|  |   |   |   |   |   |   |   |   | |
+------------------------------------+
| = bars along B, - = bars along L: 12 mm @ 130 c/c both ways

ELEVATION
                |  column  |
                |          |
+---------------+----------+---------------+
|                 (dowels)                 |  D = 700 mm
|  o  o  o  o  o  o  o  o  o  o  o  o  o   |  bottom mesh, 50 mm cover
+------------------------------------------+
<---------------- 3800 mm ----------------->
  • 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 3.8 m × 3.8 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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