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Footing with load transfer check, 1000 kN

Short answer

Use a 3.4 m x 3.4 m square footing, overall depth 500 mm (d=440 mm). Reinforcement: 16 mm bars @205 mm c/c both ways. One-way and two-way shear are safe; load transfer check shows that column bars as dowels can carry the excess 60 kN, so the connection is adequate.

Design an isolated footing supporting a square column 400 mm x 400 mm with 8-20 ϕ\phi longitudinal reinforcement. The column is subjected to axial service load of 1000 kN. Consider base of footing is 1 m below the ground level. Take allowable bearing capacity of soil 100 kN/m2^2, M15 concrete and Fe 415 steel. Also check transfer of load from column to footing.

Data and assumptions

  • Column 400 mm × 400 mm (long side a=400a=400 mm along the footing length); footing concrete M15, steel Fe415; column concrete M20; column bars 8-20 mm.
  • Service load P=1000P = 1000 kN.
  • Allowable (safe) bearing capacity of soil = 100 kN/m² at 1.0 m depth.
  • Column concrete grade is not given; M20 is assumed for the column (footing is M15).
  • 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=1.1×1000.0100=11.00 m2A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 1000.0}{100} = 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=1.5×1000.0=1500.0P_u = 1.5 \times 1000.0 = 1500.0 kN. pu=PuBL=1500.03.4×3.4=129.8p_u = \dfrac{P_u}{BL} = \dfrac{1500.0}{3.4 \times 3.4} = 129.8 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=496.3M_{u,L} = 496.3 kN·m, Mu,B=496.3M_{u,B} = 496.3 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=440d = 440 mm, so the depth is governed by shear (one-way or punching). Trial D=500D = 500 mm, d=440d = 440 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)496.3333116 mm @ 205 c/c3335
Along BB (width LL)496.3333116 mm @ 205 c/c3335

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

  • Along LL side: Vu=467.6V_u = 467.6 kN, τv=VuBd=0.313\tau_v = \dfrac{V_u}{B d} = 0.313 MPa; pt=0.223%p_t = 0.223\%, τc=0.335\tau_c = 0.335 MPa (Table 19). τv<τc\tau_v < \tau_c, safe.
  • Along BB side: Vu=467.6V_u = 467.6 kN, τv=0.313\tau_v = 0.313 MPa; pt=0.223%p_t = 0.223\%, τc=0.335\tau_c = 0.335 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+440)+(400+440)]=3360b_o = 2[(a + d) + (c + d)] = 2[(400+440) + (400+440)] = 3360 mm
  • Vu=Pu−pu(a+d)(c+d)=1500.0−129.8×0.840×0.840=1408.4V_u = P_u - p_u (a+d)(c+d) = 1500.0 - 129.8 \times 0.840 \times 0.840 = 1408.4 kN (average pressure; the effect of moment transfer is neglected).
  • τv=Vubod=0.953\tau_v = \dfrac{V_u}{b_o d} = 0.953 MPa; ks=0.5+βc=1.00k_s = 0.5 + \beta_c = 1.00, τc′=ks×0.25fck=0.968\tau_c' = k_s \times 0.25\sqrt{f_{ck}} = 0.968 MPa. τv<τc′\tau_v < \tau_c', safe.

Development length

Ld=0.87fyϕ4τbdL_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}, τbd=1.60\tau_{bd} = 1.60 MPa (M15, deformed bars): 16 mm bars (both directions) → Ld=903L_d = 903 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=9.38= P_u/A_2 = 9.38 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 13.50 MPa. The lesser, 9.00 MPa, governs. Capacity of bearing = 1440 kN << Pu=1500P_u = 1500 kN, so the excess 60 kN must be carried by dowels: As,dow=600000.87fy=166A_{s,dow} = \dfrac{60000}{0.87 f_y} = 166 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=903L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 903 mm (and the same lap with the column bars). Available depth in footing = 500 - 50 - 32 = 418 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 @ 205 c/c both ways

ELEVATION
                |  column  |
                |          |
+---------------+----------+---------------+
|                 (dowels)                 |  D = 500 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 @ 205 c/c along LL (lower layer), 16 mm @ 205 c/c along BB (upper layer), 50 mm clear cover; dowels 8-20 mm.

Answer: Footing 3.4 m × 3.4 m × 500 mm thick; 16 mm @ 205 c/c along the length and 16 mm @ 205 c/c along the width.

✓ Answer checked by Claude Opus 5.5 (high effort)

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