Isolated footing 450x450, service moment 110 kNm

Data and assumptions

• Column 450 mm × 450 mm (long side $a=450$ mm along the footing length); footing concrete M20, steel Fe415; column concrete M20; column bars 8-20 mm.
• Service load $P = 1000$ kN, service moment $M = 110$ kN·m.
• Allowable (safe) bearing capacity of soil = 140 kN/m².
• The service moment acts about one axis.
• 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.

$$A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 1000.0}{140} = 7.86\ \text{m}^2$$

Eccentricity $e = M/P = 110.0/1000.0 = 0.110$ m, which is less than $L/6$, so there is no tension under the base. The size is chosen so that

$$q_{max} = 1.1\frac{P}{BL}\left(1 + \frac{6e}{L}\right) \le q_a$$

Try $L \times B = 4.0 \times 2.3$ m ($A = 9.20$ m²): $q_{max} = 139.3$ kN/m² $\le 140.0$ and $q_{min} = 99.8$ kN/m² $> 0$. Safe.

Factored soil pressure

$P_u = 1.5 \times 1000.0 = 1500.0$ kN, $M_u = 165.0$ kN·m. $p_u = \dfrac{P_u}{BL}\left(1 \pm \dfrac{6e}{L}\right)$: $p_{u,max} = 189.9$ kN/m², $p_{u,min} = 136.1$ kN/m² (average 163.0 kN/m²). Self-weight does not cause bending in the footing, so it is excluded.

Bending moments and reinforcement

Cantilever projections from the column face: along $L$: $(4000 - 450)/2 = 1775$ mm; along $B$: $(2300 - 450)/2 = 925$ mm. Along $L$ (critical section at the column face on the side of maximum pressure): the pressure is 166.1 kN/m² at the face and 189.9 kN/m² at the edge, so per metre width $M = 166.1 \times \dfrac{1.775^2}{2} + (189.9 - 166.1) \times \dfrac{1.775^2}{3} = 261.6 + 25.0 = 286.6$ kN·m/m, and $M_{u,L} = 286.6 \times 2.3 = 659.3$ kN·m. Along $B$ (average pressure): $M_{u,B} = 279.0$ kN·m. Required depth for flexure: $M_{u,lim} = 0.138 f_{ck} b d^2$ is far larger than the moment for $d = 640$ mm, so the depth is governed by shear (one-way or punching). Trial $D = 700$ mm, $d = 640$ 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 $bD$.

For the rectangular footing, the short-direction steel is placed in a central band of width $B = 2300$ mm in the proportion $\dfrac{2}{\beta+1} = \dfrac{2}{1.74+1} = 0.730$ (IS 456 cl. 34.3.1), where $\beta = L/B$.

• Central band: $0.730 \times 3360 = 2453$ mm² → 12 mm @ 100 mm c/c (23 bars, 2601 mm²).
• Each outer strip (850 mm wide): remaining steel 453 mm² or minimum 0.12% = 714 mm², whichever is more → 12 mm @ 120 mm c/c (7 bars, 792 mm²).

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

• Along $L$ side: $V_u = 475.9$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.323$ MPa; $p_t = 0.209\%$, $\tau_c = 0.327$ MPa (Table 19). $\tau_v < \tau_c$, safe.
• Along $B$ side: $V_u = 185.9$ kN, $\tau_v = 0.073$ MPa; $p_t = 0.16\%$, $\tau_c \approx 0.29$ MPa. Safe.

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

• Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(450+640) + (450+640)] = 4360$ mm
• $V_u = P_u - p_u (a+d)(c+d) = 1500.0 - 163.0 \times 1.090 \times 1.090 = 1306.3$ kN (average pressure; the effect of moment transfer is neglected).
• $\tau_v = \dfrac{V_u}{b_o d} = 0.468$ 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 → $L_d = 752$ mm; 12 mm bars → $564$ mm. Available length from column face (less 50 mm end cover): 1725 mm along $L$, 875 mm along $B$. Satisfied in both directions.

Bearing stress and transfer of load from column

Column area $A_2 = 450\times450 = 0.203$ m². Bearing stress $= P_u/A_2 = 7.41$ 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 = 1822 kN $\ge$ $P_u = 1500$ kN, so the whole load can be transferred by bearing. Provide minimum dowels of 0.5% of column area = 1012 mm²: use 8 bars of 20 mm (2513 mm²) as dowels, same as column bars. 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 = 700 - 50 - 28 = 622 mm $< L_{dc}$; dowels are bent at the bottom with 90° bends.

Detailing

PLAN (4.0 m x 2.3 m; L horizontal)
+--------------------------------------------+
|  |   |   |   |   |   |   |   |   |   |   | |
|--+---+---+---+---+---+---+---+---+---+---+-|
|  |   |   |   |   |   |   |   |   |   |   | |
|--+---+---+---+-+----------+--+---+---+---+-|
|  |   |   |   | |  column  |  |   |   |   | |
|--+---+---+---+-+----------+--+---+---+---+-|
|  |   |   |   |   |   |   |   |   |   |   | |
|--+---+---+---+---+---+---+---+---+---+---+-|
|  |   |   |   |   |   |   |   |   |   |   | |
+--------------------------------------------+
 <-0.85 m><---central band 2.3 m---><-0.85 m>
- = 16 mm @ 150 c/c (along L)
| = 12 mm along B: @ 100 c/c in central band, @ 120 c/c in outer strips

ELEVATION
                 |  column  |
                 |          |
+----------------+----------+----------------+
|                  (dowels)                  |  D = 700 mm
|  o  o  o  o  o  o  o  o  o  o  o  o  o  o  |  bottom mesh, 50 mm cover
+--------------------------------------------+
<----------------- 4000 mm ------------------>

• Bottom mesh: 16 mm @ 150 c/c along $L$ (lower layer); along $B$ (upper layer), 12 mm @ 100 c/c in the central 2.3 m band and 12 mm @ 120 c/c in each 0.85 m outer strip; 50 mm clear cover; dowels 8-20 mm.

Answer: Footing 4.0 m × 2.3 m × 700 mm thick; 16 mm @ 150 c/c along the length; along the width, 12 mm @ 100 c/c in the central 2.3 m band and 12 mm @ 120 c/c in the outer strips.