Isolated footing 350x350, 1500 kN

Data and assumptions

• Column 350 mm × 350 mm (long side $a=350$ mm along the footing length); footing concrete M20, steel Fe415; column concrete M20; column bars 8-20 mm.
• Service load $P = 1500$ kN.
• Allowable (safe) bearing capacity of soil = 175 kN/m².
• Axial load taken as service load; column concrete M20 assumed.
• Clear cover 50 mm for the footing (IS 456 cl. 26.4.2.2); effective depth $d = D - 60$ mm (cover + half bar). Load factor 1.5.

Size of footing

Weight of footing and backfill is taken as 10% of the column load.

$$A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 1500.0}{175} = 9.43\ \text{m}^2$$

Provide a square footing $L \times B = 3.1 \times 3.1$ m, area = 9.61 m² $\ge$ 9.43 m².

Factored soil pressure

$P_u = 1.5 \times 1500.0 = 2250.0$ kN. $p_u = \dfrac{P_u}{BL} = \dfrac{2250.0}{3.1 \times 3.1} = 234.1$ kN/m².

Bending moments and reinforcement

Cantilever projections from the column face: along $L$: $(3100 - 350)/2 = 1375$ mm; along $B$: $(3100 - 350)/2 = 1375$ mm. $M_u = p_u \times (\text{width}) \times \dfrac{(\text{projection})^2}{2}$: $M_{u,L} = 686.1$ kN·m, $M_{u,B} = 686.1$ kN·m. Required depth for flexure: $M_{u,lim} = 0.138 f_{ck} b d^2$ is far larger than the moment for $d = 590$ mm, so the depth is governed by shear (one-way or punching). Trial $D = 650$ mm, $d = 590$ mm, found by the shear checks below.

$A_{st} = \dfrac{0.5 f_{ck}}{f_y}\left[1 - \sqrt{1 - \dfrac{4.6 M_u}{f_{ck} b d^2}}\right] b d$; minimum steel = 0.12% of $bD$.

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

• Along $L$ side: $V_u = 569.8$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.312$ MPa; $p_t = 0.184\%$, $\tau_c = 0.315$ MPa (IS 456 Table 19, SP 16 formula). $\tau_v < \tau_c$, safe.
• Along $B$ side: $V_u = 569.8$ kN, $\tau_v = 0.312$ MPa; $p_t = 0.184\%$, $\tau_c = 0.315$ MPa. Safe.

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

• Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(350+590) + (350+590)] = 3760$ mm
• $V_u = P_u - p_u (a+d)(c+d) = 2250.0 - 234.1 \times 0.940 \times 0.940 = 2043.1$ kN (average pressure; the effect of moment transfer is neglected).
• $\tau_v = \dfrac{V_u}{b_o d} = 0.921$ MPa; $k_s = 0.5 + \beta_c = 1.00$, $\tau_c' = k_s \times 0.25\sqrt{f_{ck}} = 1.118$ MPa. $\tau_v < \tau_c'$, safe.

Development length

$L_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}$, $\tau_{bd} = 1.92$ MPa (M20, deformed bars): 16 mm bars (both directions) → $L_d = 752$ mm. Available length from column face (less 50 mm end cover): 1325 mm along $L$, 1325 mm along $B$. Satisfied in both directions.

Bearing stress and transfer of load from column

Column area $A_2 = 350\times350 = 0.122$ m². Bearing stress $= P_u/A_2 = 18.37$ MPa. Permissible bearing stress: column concrete $0.45 f_{ck} = 9.00$ MPa; footing concrete $0.45 f_{ck}\sqrt{A_1/A_2}$ with $\sqrt{A_1/A_2} \le 2$ gives up to 18.00 MPa. The lesser, 9.00 MPa, governs. Capacity of bearing = 1102 kN $<$ $P_u = 2250$ kN, so the excess 1148 kN must be carried by dowels: $A_{s,dow} = \dfrac{1147500}{0.87 f_y} = 3178$ mm²; the column bars (8-20 mm, 2513 mm²) are continued into the footing as dowels, and 4 extra bars of 16 mm (804 mm²) are added, giving 3317 mm² $\ge$ 3178 mm². Dowel length in compression $L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 752$ mm (and the same lap with the column bars). Available depth in footing = 650 - 50 - 32 = 568 mm $< L_{dc}$; dowels are bent at the bottom with 90° bends.

Detailing

PLAN                      ELEVATION
+------------------+      col
| #  #  #  #  #  # |      | |
| #  +------+  #  # |     _|_|__
| #  | col  |  #  # |    |_____|  <- D = 650 mm
| #  +------+  #  # |    bars both ways
| #  #  #  #  #  # |
+------------------+
 L = 3.1 m, B = 3.1 m

• Bottom mesh: 16 mm @ 185 c/c along $L$ (lower layer), 16 mm @ 185 c/c along $B$ (upper layer), 50 mm clear cover; dowels 8-20 mm.

Answer: Footing 3.1 m × 3.1 m × 650 mm thick; 16 mm @ 185 c/c along the length and 16 mm @ 185 c/c along the width.