Design of Footings

A mat (raft) foundation is a large, continuous slab that supports the whole building (or several columns) and spreads the load over the full plan area. It is used when soil is weak, when isolated footings would cover more than about half the plan area, or to reduce differential settlement.

Design steps

1. Collect the data: column loads and positions (dead + live, and moments), safe bearing capacity of the soil, grades of concrete and steel, depth of foundation.
2. Find the resultant of all column loads, $\sum P$, and its position $(\bar x, \bar y)$. Choose the plan size of the mat (with edge projections) so that its centroid coincides, or nearly coincides, with the resultant. Eccentricity $e =$ resultant - centroid should be small.
3. Area: $A = \dfrac{\sum P (1 + 0.1)}{q_{a}}$ (10% allowance for self-weight), and check that net upward pressure $$q = \frac{\sum P}{A} \pm \frac{\sum P\, e_x}{Z_y} \pm \frac{\sum P\, e_y}{Z_x} \le q_a, \quad q_{min} \ge 0.$$
4. Choose the method of analysis:
   • Rigid (conventional) method: if the column spacing is less than $1.75/\lambda$ (where $\lambda = \sqrt[4]{k_sB/4E_cI}$), the mat is taken as rigid. Contact pressure is linear. The mat is divided into strips between the column lines, each strip is analysed as a continuous beam under the pressure and the column loads, and the shear and bending moment diagrams are drawn.
   • Flexible method (beam on elastic foundation, or finite element/ plate on springs), when the mat is not rigid.
5. Thickness: determine the depth from punching shear around the heaviest columns (at $d/2$) and one-way shear at $d$ from the face of the column or wall (shear reinforcement is avoided when possible). Check also the bending requirement.
6. Bending moments and reinforcement: calculate positive and negative moments in each strip in both directions; design steel at top and bottom in each direction (top steel at the supports (columns), bottom steel at mid-span). Provide minimum steel of 0.12% each way.
7. Check the development length, the bearing at the column base and the settlement.
8. Detailing: top and bottom mesh in both directions, extra bars at columns, cover 50 mm (75 mm if cast against the soil without blinding), dowels for the column bars, construction joints and waterproofing if below ground water.

A mat can be a plain slab, a slab with beams (ribbed), or a cellular/ box type.

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, linear pressure): $M_{u,L} = 630.5$ 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 (22 bars).
• 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).

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; $p_t = 0.196\%$, $\tau_c = 0.324$ MPa (Table 19). $\tau_v < \tau_c$, safe.
• Along $B$ side: $V_u = 185.9$ kN, $\tau_v = 0.073$ MPa; $p_t = 0.136\%$, $\tau_c = 0.288$ 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                      ELEVATION
+------------------+      col
| #  #  #  #  #  # |      | |
| #  +------+  #  # |     _|_|__
| #  | col  |  #  # |    |_____|  <- D = 700 mm
| #  +------+  #  # |    bars both ways
| #  #  #  #  #  # |
+------------------+
 L = 4.0 m, B = 2.3 m

• Bottom mesh: 16 mm @ 160 c/c along $L$ (lower layer), 12 mm @ 130 c/c along $B$ (upper layer), 50 mm clear cover; dowels 8-20 mm.

Answer: Footing 4.0 m × 2.3 m × 700 mm thick; 16 mm @ 160 c/c along the length and 12 mm @ 130 c/c along the width.

Data and assumptions

• Column 400 mm × 400 mm (long side $a=400$ mm along the footing length); footing concrete M20, steel Fe415; column concrete M20; column bars 8-25 mm.
• Service load $P = 1200$ kN, service moment $M = 120$ kN·m.
• Allowable (safe) bearing capacity of soil = 150 kN/m².
• 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.

$$A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 1200.0}{150} = 8.80\ \text{m}^2$$

Eccentricity $e = M/P = 120.0/1200.0 = 0.100$ 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 $L \times B = 4.2 \times 2.4$ m ($A = 10.08$ m²): $q_{max} = 149.7$ kN/m² $\le 150.0$ and $q_{min} = 112.2$ kN/m² $> 0$. Safe.

Factored soil pressure

$P_u = 1.5 \times 1200.0 = 1800.0$ kN, $M_u = 180.0$ kN·m. $p_u = \dfrac{P_u}{BL}\left(1 \pm \dfrac{6e}{L}\right)$: $p_{u,max} = 204.1$ kN/m², $p_{u,min} = 153.1$ kN/m² (average 178.6 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 $L$: $(4200 - 400)/2 = 1900$ mm; along $B$: $(2400 - 400)/2 = 1000$ mm. Along $L$ (critical section at the column face on the side of maximum pressure, linear pressure): $M_{u,L} = 817.4$ kN·m. Along $B$ (average pressure): $M_{u,B} = 375.0$ kN·m. Required depth for flexure: $M_{u,lim} = 0.138 f_{ck} b d^2$ is far larger than the moment for $d = 740$ mm, so the depth is governed by shear (one-way or punching). Trial $D = 800$ mm, $d = 740$ 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 $B = 2400$ mm in the proportion $\dfrac{2}{\beta+1} = \dfrac{2}{1.75+1} = 0.727$ (IS 456 cl. 34.3.1), where $\beta = L/B$.

• Central band: $0.727 \times 4032 = 2932$ mm² → 16 mm @ 160 mm c/c (15 bars).
• Each outer strip (900 mm wide): remaining steel 550 mm² or minimum 0.12% = 864 mm², whichever is more → 16 mm @ 180 mm c/c (5 bars).

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

• Along $L$ side: $V_u = 548.5$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.309$ MPa; $p_t = 0.181\%$, $\tau_c = 0.312$ MPa (Table 19). $\tau_v < \tau_c$, safe.
• Along $B$ side: $V_u = 195.0$ kN, $\tau_v = 0.063$ MPa; $p_t = 0.133\%$, $\tau_c = 0.288$ MPa. Safe.

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

• Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(400+740) + (400+740)] = 4560$ mm
• $V_u = P_u - p_u (a+d)(c+d) = 1800.0 - 178.6 \times 1.140 \times 1.140 = 1567.9$ kN (average pressure; the effect of moment transfer is neglected).
• $\tau_v = \dfrac{V_u}{b_o d} = 0.465$ 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; 16 mm bars → $752$ mm. Available length from column face (less 50 mm end cover): 1850 mm along $L$, 950 mm along $B$. Satisfied in both directions.

Bearing stress and transfer of load from column

Column area $A_2 = 400\times400 = 0.160$ m². Bearing stress $= P_u/A_2 = 11.25$ 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 = 1440 kN $<$ $P_u = 1800$ kN, so the excess 360 kN must be carried by dowels: $A_{s,dow} = \dfrac{360000}{0.87 f_y} = 997$ mm²; the column bars (8-25 mm, 3927 mm²) are continued into the footing as dowels - adequate. Dowel length in compression $L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 940$ mm (and the same lap with the column bars). Available depth in footing = 800 - 50 - 32 = 718 mm $< L_{dc}$; dowels are bent at the bottom with 90° bends.

Detailing

PLAN                      ELEVATION
+------------------+      col
| #  #  #  #  #  # |      | |
| #  +------+  #  # |     _|_|__
| #  | col  |  #  # |    |_____|  <- D = 800 mm
| #  +------+  #  # |    bars both ways
| #  #  #  #  #  # |
+------------------+
 L = 4.2 m, B = 2.4 m

• Bottom mesh: 16 mm @ 150 c/c along $L$ (lower layer), 16 mm @ 205 c/c along $B$ (upper layer), 50 mm clear cover; dowels 8-25 mm.

Answer: Footing 4.2 m × 2.4 m × 800 mm thick; 16 mm @ 150 c/c along the length and 16 mm @ 205 c/c along the width.

Data and assumptions

• Column 400 mm × 400 mm (long side $a=400$ mm along the footing length); footing concrete M20, steel Fe500; column concrete M20; column bars 8-16 mm.
• Service load $P = 650$ kN.
• Allowable (safe) bearing capacity of soil = 170 kN/m².
• Column bars assumed 8-16 mm.
• 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.

$$A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 650.0}{170} = 4.21\ \text{m}^2$$

Provide a square footing $L \times B = 2.1 \times 2.1$ m, area = 4.41 m² $\ge$ 4.21 m².

Factored soil pressure

$P_u = 1.5 \times 650.0 = 975.0$ kN. $p_u = \dfrac{P_u}{BL} = \dfrac{975.0}{2.1 \times 2.1} = 221.1$ kN/m².

Bending moments and reinforcement

Cantilever projections from the column face: along $L$: $(2100 - 400)/2 = 850$ mm; along $B$: $(2100 - 400)/2 = 850$ mm. $M_u = p_u \times (\text{width}) \times \dfrac{(\text{projection})^2}{2}$: $M_{u,L} = 167.7$ kN·m, $M_{u,B} = 167.7$ kN·m. Required depth for flexure: $M_{u,lim} = 0.133 f_{ck} b d^2$ is far larger than the moment for $d = 390$ mm, so the depth is governed by shear (one-way or punching). Trial $D = 450$ mm, $d = 390$ 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: $V_u = 213.6$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.261$ MPa; $p_t = 0.141\%$, $\tau_c = 0.288$ MPa (Table 19). $\tau_v < \tau_c$, safe.
• Along $B$ side: $V_u = 213.6$ kN, $\tau_v = 0.261$ MPa; $p_t = 0.141\%$, $\tau_c = 0.288$ MPa. Safe.

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

• Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(400+390) + (400+390)] = 3160$ mm
• $V_u = P_u - p_u (a+d)(c+d) = 975.0 - 221.1 \times 0.790 \times 0.790 = 837.0$ kN (average pressure; the effect of moment transfer is neglected).
• $\tau_v = \dfrac{V_u}{b_o d} = 0.679$ 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): 12 mm bars → $L_d = 680$ mm; 12 mm bars → $680$ mm. Available length from column face (less 50 mm end cover): 800 mm along $L$, 800 mm along $B$. Satisfied in both directions.

Bearing stress and transfer of load from column

Column area $A_2 = 400\times400 = 0.160$ m². Bearing stress $= P_u/A_2 = 6.09$ 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 = 1440 kN $\ge$ $P_u = 975$ kN, so the whole load can be transferred by bearing. Provide minimum dowels of 0.5% of column area = 800 mm²: use 8 bars of 16 mm (1608 mm²) as dowels, same as column bars. Dowel length in compression $L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 725$ mm (and the same lap with the column bars). Available depth in footing = 450 - 50 - 24 = 376 mm $< L_{dc}$; dowels are bent at the bottom with 90° bends.

Detailing

PLAN                      ELEVATION
+------------------+      col
| #  #  #  #  #  # |      | |
| #  +------+  #  # |     _|_|__
| #  | col  |  #  # |    |_____|  <- D = 450 mm
| #  +------+  #  # |    bars both ways
| #  #  #  #  #  # |
+------------------+
 L = 2.1 m, B = 2.1 m

• Bottom mesh: 12 mm @ 205 c/c along $L$ (lower layer), 12 mm @ 205 c/c along $B$ (upper layer), 50 mm clear cover; dowels 8-16 mm.

Answer: Footing 2.1 m × 2.1 m × 450 mm thick; 12 mm @ 205 c/c along the length and 12 mm @ 205 c/c along the width.

Data and assumptions

• Column 400 mm × 400 mm (long side $a=400$ mm along the footing length); footing concrete M20, steel Fe415; column concrete M20; column bars 8-16 mm.
• Service load $P = 600$ kN, service moment $M = 50$ kN·m.
• Allowable (safe) bearing capacity of soil = 150 kN/m².
• Square footing as required by the question; column bars assumed 8-16 mm.
• 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.

$$A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 600.0}{150} = 4.40\ \text{m}^2$$

Eccentricity $e = M/P = 50.0/600.0 = 0.083$ 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 $L \times B = 2.4 \times 2.4$ m ($A = 5.76$ m²): $q_{max} = 138.5$ kN/m² $\le 150.0$ and $q_{min} = 90.7$ kN/m² $> 0$. Safe.

Factored soil pressure

$P_u = 1.5 \times 600.0 = 900.0$ kN, $M_u = 75.0$ kN·m. $p_u = \dfrac{P_u}{BL}\left(1 \pm \dfrac{6e}{L}\right)$: $p_{u,max} = 188.8$ kN/m², $p_{u,min} = 123.7$ kN/m² (average 156.2 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 $L$: $(2400 - 400)/2 = 1000$ mm; along $B$: $(2400 - 400)/2 = 1000$ mm. Along $L$ (critical section at the column face on the side of maximum pressure, linear pressure): $M_{u,L} = 204.9$ kN·m. Along $B$ (average pressure): $M_{u,B} = 187.5$ kN·m. Required depth for flexure: $M_{u,lim} = 0.138 f_{ck} b d^2$ is far larger than the moment for $d = 390$ mm, so the depth is governed by shear (one-way or punching). Trial $D = 450$ mm, $d = 390$ 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: $V_u = 264.3$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.282$ MPa; $p_t = 0.161\%$, $\tau_c = 0.297$ MPa (Table 19). $\tau_v < \tau_c$, safe.
• Along $B$ side: $V_u = 228.8$ kN, $\tau_v = 0.244$ MPa; $p_t = 0.149\%$, $\tau_c = 0.288$ MPa. Safe.

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

• Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(400+390) + (400+390)] = 3160$ mm
• $V_u = P_u - p_u (a+d)(c+d) = 900.0 - 156.2 \times 0.790 \times 0.790 = 802.5$ kN (average pressure; the effect of moment transfer is neglected).
• $\tau_v = \dfrac{V_u}{b_o d} = 0.651$ 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): 12 mm bars → $L_d = 564$ mm; 12 mm bars → $564$ mm. Available length from column face (less 50 mm end cover): 950 mm along $L$, 950 mm along $B$. Satisfied in both directions.

Bearing stress and transfer of load from column

Column area $A_2 = 400\times400 = 0.160$ m². Bearing stress $= P_u/A_2 = 5.62$ 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 = 1440 kN $\ge$ $P_u = 900$ kN, so the whole load can be transferred by bearing. Provide minimum dowels of 0.5% of column area = 800 mm²: use 8 bars of 16 mm (1608 mm²) as dowels, same as column bars. Dowel length in compression $L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 602$ mm (and the same lap with the column bars). Available depth in footing = 450 - 50 - 24 = 376 mm $< L_{dc}$; dowels are bent at the bottom with 90° bends.

Detailing

PLAN                      ELEVATION
+------------------+      col
| #  #  #  #  #  # |      | |
| #  +------+  #  # |     _|_|__
| #  | col  |  #  # |    |_____|  <- D = 450 mm
| #  +------+  #  # |    bars both ways
| #  #  #  #  #  # |
+------------------+
 L = 2.4 m, B = 2.4 m

• Bottom mesh: 12 mm @ 180 c/c along $L$ (lower layer), 12 mm @ 195 c/c along $B$ (upper layer), 50 mm clear cover; dowels 8-16 mm.

Answer: Footing 2.4 m × 2.4 m × 450 mm thick; 12 mm @ 180 c/c along the length and 12 mm @ 195 c/c along the width.

Steps for design of a rectangular RC footing (isolated)

A rectangular footing is used when the column is rectangular, when there is a moment on the column, or when there is a space restriction on one side.

1. Load and area: take the column service load $P$ (add about 10% for the weight of footing and backfill) and the safe bearing capacity $q_a$. $$A_{req} = \frac{1.1P}{q_a}$$
2. Dimensions: choose $L \times B$ so that the projections on both sides of the column are about equal, i.e. $L/B \approx$ ratio of column sides, with $LB \ge A_{req}$. If the column carries a moment $M$, check $q_{max} = \dfrac{1.1P}{LB}\left(1 + \dfrac{6e}{L}\right) \le q_a$ and $q_{min} \ge 0$ with $e = M/P$.
3. Factored upward soil pressure: $p_u = \dfrac{P_u}{LB}$ ($P_u = 1.5P$) (or the maximum pressure for a footing with moment). The self-weight is not included in bending and shear design.
4. Depth from the greater of: (a) one-way shear at a distance $d$ from the column face, and (b) punching shear at $d/2$ from the column face (permissible stress $k_s \cdot 0.25\sqrt{f_{ck}}$), and the bending requirement. Adopt $D$ and $d$.
5. Bending moment at the face of the column in each direction: $$M_{u,L} = p_u B\frac{(L-a)^2}{8}, \qquad M_{u,B} = p_u L\frac{(B-c)^2}{8}$$ Find $A_{st}$ in each direction (minimum 0.12% of $bD$).
6. Distribution of steel: long-direction bars are spread uniformly over the width $B$. For short-direction bars, a fraction $\dfrac{2}{\beta + 1}$ of the total ($\beta = L/B$) is concentrated in a central band of width $B$; the rest is spread equally in the two outer portions (IS 456 cl. 34.3.1).
7. Checks: one-way shear $\tau_v \le \tau_c$ in both directions, punching shear, development length of the bars ($L_d \le$ available length from the face of the column), and bearing at the column-footing interface; provide dowels.
8. Detail: bars at the bottom, with 50 mm cover, plan and sections.

Why shear reinforcement is not provided in a footing

• A footing is a thick member, and its depth is made large enough that the shear stress is less than the permissible shear stress of concrete (for both one-way and punching shear). The concrete alone carries the shear, so stirrups are not needed.
• Stirrups in a thick, heavy slab cast against soil are difficult to place and hold, and would cause congestion.
• Bent-up bars or shear reinforcement are not efficient against punching in a thick footing, and the depth is easily increased at little cost.
• Increasing the depth also gives rigidity, so the assumption of a uniform soil pressure is better satisfied.

Data and assumptions

• Column 500 mm × 500 mm (long side $a=500$ mm along the footing length); footing concrete M25, steel Fe500; column concrete M25; column bars 8-20 mm.
• Service load $P = 1600$ kN.
• Allowable (safe) bearing capacity of soil = 125 kN/m² at 1.25 m depth.
• Depth of foundation 1.25 m is used only for the cover to the base; weight of footing and soil 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 $d = D - 60$ mm (cover + half bar). Load factor 1.5.

Size of footing

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

$$A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 1600.0}{125} = 14.08\ \text{m}^2$$

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

Factored soil pressure

$P_u = 1.5 \times 1600.0 = 2400.0$ kN. $p_u = \dfrac{P_u}{BL} = \dfrac{2400.0}{3.8 \times 3.8} = 166.2$ kN/m².

Bending moments and reinforcement

Cantilever projections from the column face: along $L$: $(3800 - 500)/2 = 1650$ mm; along $B$: $(3800 - 500)/2 = 1650$ mm. $M_u = p_u \times (\text{width}) \times \dfrac{(\text{projection})^2}{2}$: $M_{u,L} = 859.7$ kN·m, $M_{u,B} = 859.7$ kN·m. Required depth for flexure: $M_{u,lim} = 0.133 f_{ck} b d^2$ 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 $bD$.

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

• Along $L$ side: $V_u = 637.9$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.262$ MPa; $p_t = 0.136\%$, $\tau_c = 0.291$ MPa (Table 19). $\tau_v < \tau_c$, safe.
• Along $B$ side: $V_u = 637.9$ kN, $\tau_v = 0.262$ MPa; $p_t = 0.136\%$, $\tau_c = 0.291$ MPa. Safe.

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

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

Development length

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

Bearing stress and transfer of load from column

Column area $A_2 = 500\times500 = 0.250$ m². Bearing stress $= P_u/A_2 = 9.60$ MPa. Permissible bearing stress: column concrete $0.45 f_{ck} = 11.25$ MPa; footing concrete $0.45 f_{ck}\sqrt{A_1/A_2}$ with $\sqrt{A_1/A_2} \le 2$ gives up to 22.50 MPa. The lesser, 11.25 MPa, governs. Capacity of bearing = 2812 kN $\ge$ $P_u = 2400$ 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 $L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 777$ mm (and the same lap with the column bars). Available depth in footing = 700 - 50 - 24 = 626 mm $< L_{dc}$; dowels are bent at the bottom with 90° bends.

Detailing

PLAN                      ELEVATION
+------------------+      col
| #  #  #  #  #  # |      | |
| #  +------+  #  # |     _|_|__
| #  | col  |  #  # |    |_____|  <- D = 700 mm
| #  +------+  #  # |    bars both ways
| #  #  #  #  #  # |
+------------------+
 L = 3.8 m, B = 3.8 m

• Bottom mesh: 12 mm @ 130 c/c along $L$ (lower layer), 12 mm @ 130 c/c along $B$ (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.

Data and assumptions

• Column 400 mm × 400 mm (long side $a=400$ 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 = 150 kN/m².
• Column bars not given; 8-20 mm assumed. Loads are service loads.
• 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.

$$A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 1500.0}{150} = 11.00\ \text{m}^2$$

Provide a square footing $L \times B = 3.4 \times 3.4$ m, area = 11.56 m² $\ge$ 11.00 m².

Factored soil pressure

$P_u = 1.5 \times 1500.0 = 2250.0$ kN. $p_u = \dfrac{P_u}{BL} = \dfrac{2250.0}{3.4 \times 3.4} = 194.6$ kN/m².

Bending moments and reinforcement

Cantilever projections from the column face: along $L$: $(3400 - 400)/2 = 1500$ mm; along $B$: $(3400 - 400)/2 = 1500$ mm. $M_u = p_u \times (\text{width}) \times \dfrac{(\text{projection})^2}{2}$: $M_{u,L} = 744.5$ kN·m, $M_{u,B} = 744.5$ 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: $V_u = 602.2$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.300$ MPa; $p_t = 0.184\%$, $\tau_c = 0.315$ MPa (Table 19). $\tau_v < \tau_c$, safe.
• Along $B$ side: $V_u = 602.2$ kN, $\tau_v = 0.300$ 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 $b_o = 2[(a + d) + (c + d)] = 2[(400+590) + (400+590)] = 3960$ mm
• $V_u = P_u - p_u (a+d)(c+d) = 2250.0 - 194.6 \times 0.990 \times 0.990 = 2059.2$ kN (average pressure; the effect of moment transfer is neglected).
• $\tau_v = \dfrac{V_u}{b_o d} = 0.881$ 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; 16 mm bars → $752$ mm. Available length from column face (less 50 mm end cover): 1450 mm along $L$, 1450 mm along $B$. Satisfied in both directions.

Bearing stress and transfer of load from column

Column area $A_2 = 400\times400 = 0.160$ m². Bearing stress $= P_u/A_2 = 14.06$ 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 = 1440 kN $<$ $P_u = 2250$ kN, so the excess 810 kN must be carried by dowels: $A_{s,dow} = \dfrac{810000}{0.87 f_y} = 2243$ mm²; the column bars (8-20 mm, 2513 mm²) are continued into the footing as dowels - adequate. 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 = 3.4 m, B = 3.4 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 3.4 m × 3.4 m × 650 mm thick; 16 mm @ 185 c/c along the length and 16 mm @ 185 c/c along the width.

Data and assumptions

• Column 400 mm × 400 mm (long side $a=400$ mm along the footing length); footing concrete M15, steel Fe415; column concrete M20; column bars 8-20 mm.
• Service load $P = 1000$ 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 $d = D - 60$ mm (cover + half bar). Load factor 1.5.

Size of footing

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

$$A_{req} = \frac{1.1\,P}{q_a} = \frac{1.1 \times 1000.0}{100} = 11.00\ \text{m}^2$$

Provide a square footing $L \times B = 3.4 \times 3.4$ m, area = 11.56 m² $\ge$ 11.00 m².

Factored soil pressure

$P_u = 1.5 \times 1000.0 = 1500.0$ kN. $p_u = \dfrac{P_u}{BL} = \dfrac{1500.0}{3.4 \times 3.4} = 129.8$ kN/m².

Bending moments and reinforcement

Cantilever projections from the column face: along $L$: $(3400 - 400)/2 = 1500$ mm; along $B$: $(3400 - 400)/2 = 1500$ mm. $M_u = p_u \times (\text{width}) \times \dfrac{(\text{projection})^2}{2}$: $M_{u,L} = 496.3$ kN·m, $M_{u,B} = 496.3$ kN·m. Required depth for flexure: $M_{u,lim} = 0.138 f_{ck} b d^2$ is far larger than the moment for $d = 440$ mm, so the depth is governed by shear (one-way or punching). Trial $D = 500$ mm, $d = 440$ 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: $V_u = 467.6$ kN, $\tau_v = \dfrac{V_u}{B d} = 0.313$ MPa; $p_t = 0.223\%$, $\tau_c = 0.335$ MPa (Table 19). $\tau_v < \tau_c$, safe.
• Along $B$ side: $V_u = 467.6$ kN, $\tau_v = 0.313$ MPa; $p_t = 0.223\%$, $\tau_c = 0.335$ MPa. Safe.

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

• Perimeter $b_o = 2[(a + d) + (c + d)] = 2[(400+440) + (400+440)] = 3360$ mm
• $V_u = P_u - p_u (a+d)(c+d) = 1500.0 - 129.8 \times 0.840 \times 0.840 = 1408.4$ kN (average pressure; the effect of moment transfer is neglected).
• $\tau_v = \dfrac{V_u}{b_o d} = 0.953$ MPa; $k_s = 0.5 + \beta_c = 1.00$, $\tau_c' = k_s \times 0.25\sqrt{f_{ck}} = 0.968$ MPa. $\tau_v < \tau_c'$, safe.

Development length

$L_d = \dfrac{0.87 f_y \phi}{4\tau_{bd}}$, $\tau_{bd} = 1.60$ MPa (M15, deformed bars): 16 mm bars → $L_d = 903$ mm; 16 mm bars → $903$ mm. Available length from column face (less 50 mm end cover): 1450 mm along $L$, 1450 mm along $B$. Satisfied in both directions.

Bearing stress and transfer of load from column

Column area $A_2 = 400\times400 = 0.160$ m². Bearing stress $= P_u/A_2 = 9.38$ 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 13.50 MPa. The lesser, 9.00 MPa, governs. Capacity of bearing = 1440 kN $<$ $P_u = 1500$ kN, so the excess 60 kN must be carried by dowels: $A_{s,dow} = \dfrac{60000}{0.87 f_y} = 166$ mm²; the column bars (8-20 mm, 2513 mm²) are continued into the footing as dowels - adequate. Dowel length in compression $L_{dc} = \dfrac{0.87 f_y\phi}{4 \times 1.25\,\tau_{bd}} = 903$ mm (and the same lap with the column bars). Available depth in footing = 500 - 50 - 32 = 418 mm $< L_{dc}$; dowels are bent at the bottom with 90° bends.

Detailing

PLAN                      ELEVATION
+------------------+      col
| #  #  #  #  #  # |      | |
| #  +------+  #  # |     _|_|__
| #  | col  |  #  # |    |_____|  <- D = 500 mm
| #  +------+  #  # |    bars both ways
| #  #  #  #  #  # |
+------------------+
 L = 3.4 m, B = 3.4 m

• Bottom mesh: 16 mm @ 205 c/c along $L$ (lower layer), 16 mm @ 205 c/c along $B$ (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.