Mat Foundations

Compensated (floating) foundation

A compensated foundation is a raft or mat placed at a depth such that the weight of soil excavated is equal to (fully compensated) or a large part of (partially compensated) the weight of the structure. The net pressure on the soil is then zero or small, so settlement is very small, like a ship floating in water.

Net pressure: $q_{net} = \dfrac{Q}{A} - \gamma D_f$. When $q_{net} = 0$, $D_f = \dfrac{Q}{\gamma A}$. It is used on soft, compressible clays and for buildings with basements.

Types of mat foundation

 (a) Flat plate       (b) Thickened under columns
 _____|____|____       _____|____|_____
 ===============      ===/==\==/==\====
 
 (c) Beam and slab     (d) Cellular (box)
   |    |    |         |  _____  _____ |
 ==|====|====|==       |_|_____|_____|_|

1. Flat plate (uniform thickness) mat: a plain slab of uniform thickness. Suitable for light loads and closely spaced columns.
2. Flat plate thickened under columns: pedestals or drop panels increase the punching shear resistance. For heavier column loads with uniform spacing.
3. Beam and slab mat: beams in both directions with a slab below or above. Stiff, economical for heavy and unequal column loads and large spans.
4. Slab with basement walls (rigid frame): the basement walls act as stiffeners. Used where a basement is needed.
5. Cellular (box) raft: a thick slab, a top slab and walls forming hollow cells. Very stiff, reduces differential settlement, and the hollow space reduces net pressure. For high-rise buildings on compressible soils.
6. Piled raft: the mat is supported on piles, used when settlement is large or the soil is very weak.

Suitability

Mats are used where the allowable pressure is low (spread footings would cover more than about 50% of the plan area), where the soil is erratic or compressible, for uplift due to a high water table, and for tall or heavy buildings needing uniform settlement.

The conventional (rigid) method treats the mat as an infinitely rigid body, so the soil pressure follows a straight line (planar) distribution.

Assumptions

• The mat is rigid, so its deflection is small compared with the soil's.
• The soil pressure is linearly distributed, with its centroid on the line of action of the resultant of the column loads.

Steps

1. Total load: $Q = Q_1 + Q_2 + \dots + Q_n$ (column loads, plus mat weight if not included).
2. Eccentricities: the resultant lies at $(\bar x, \bar y)$ from the centre; $e_x = \dfrac{\sum Q_ix_i}{Q}$ and $e_y = \dfrac{\sum Q_iy_i}{Q}$. So $M_x = Qe_y$ and $M_y = Qe_x$.
3. Soil pressure at any point $(x, y)$:

$$q = \frac{Q}{A} \pm \frac{M_y\,x}{I_y} \pm \frac{M_x\,y}{I_x}$$

with $A = BL$, $I_x = \dfrac{BL^3}{12}$ and $I_y = \dfrac{LB^3}{12}$. Check that $q_{max} \le q_{all}$ and $q_{min} > 0$ (no tension). 4. Divide into strips: separate the mat into strips of width $B_1$ in each direction, each strip bounded by the centre lines between columns. 5. Strip loads: for a strip, average soil pressure $q_{av}$ and total soil reaction $q_{av}B_1L_1$; the sum of column loads on the strip $\Sigma Q$ is generally not equal to it. 6. Modify the loads (average of the two) so the strip is in equilibrium: soil load factor $F = \dfrac{Q_{av}}{\Sigma Q}$, etc. 7. Shear force and bending moment diagrams of the strip loaded with the modified column loads and the soil pressure, treated as a continuous beam. 8. Design: reinforcement for the moments, and check for punching (two-way) shear at columns and beam shear, as in IS 456.

 Column loads Q1  Q2  Q3
              v   v   v
        ______________________
        ^^^^^^^^^^^^^^^^^^^^^^   <- q (soil pressure)

Cohesive soil (clay, $\phi = 0$)

1. Find undrained cohesion $c_u$ (from $q_u/2$ of an unconfined test or a vane shear test).
2. Use Skempton's bearing capacity factor: for a square mat $N_c = 6\left(1+0.2\dfrac{D_f}{B}\right)\le 9$, and for a rectangle $N_c = \left(0.84+0.16\dfrac{B}{L}\right)N_{c(sq)}$.
3. Net ultimate capacity $q_{nu} = c_uN_c$; net safe capacity $q_{ns} = q_{nu}/F$ ($F \ge 3$).
4. Check that the net pressure $\left(\dfrac{Q}{A} - \gamma D_f\right) \le q_{ns}$ and then check consolidation settlement.

Cohesionless soil (sand, gravel)

1. Find $\phi$ from SPT $N$ (corrected for overburden) or direct tests.
2. Use Terzaghi's equation with shape factors, $q_u = qN_q + 0.5\gamma BN_\gamma$ (use $\gamma'$ and correction factors if the water table is near).
3. Obtain the safe pressure: $q_s = q_u/F$ with $F = 3$.
4. As $q_u$ is very large for a wide mat, the allowable settlement governs. Use the SPT-based formula $q_{na} = 19.16\,N\,F_d\,\dfrac{S_a}{25.4}$ (Meyerhof, for mats) with $F_d = 1 + 0.33D_f/B \le 1.33$.

Basic difference

Given: mat $B = 18$ m, $L = 22$ m, $D_f = 3$ m, $\gamma = 20$ kN/m³, $q_u$ (unconfined) $= 75$ kN/m², so $c_u = 37.5$ kN/m², FS = 3.

Use Skempton's equation for clay ($\phi = 0$).

Bearing capacity factor

$$\begin{aligned} \frac{D_f}{B} &= \frac{3}{18} = 0.167 < 2.5 \\ N_{c(sq)} &= 6\left(1 + 0.2 \times 0.167\right) = 6.20 \\ N_{c(rect)} &= \left(0.84 + 0.16\frac{B}{L}\right)N_{c(sq)} = (0.84 + 0.1309)(6.20) = 6.02 \end{aligned}$$

Safe pressure

$$\begin{aligned} q_{nu} &= c_uN_c = 37.5 \times 6.02 = 225.7\ \text{kN/m}^2 \\ q_{ns} &= \frac{225.7}{3} = 75.2\ \text{kN/m}^2 \\ q_s &= q_{ns} + \gamma D_f = 75.2 + 20 \times 3 = 135.2\ \text{kN/m}^2 \end{aligned}$$

Safe total weight

$$Q_{safe} = q_s \times A = 135.2 \times (18 \times 22) = 135.2 \times 396 = 53\,557\ \text{kN}$$

Answer: the raft can safely support a total weight (building plus foundation) of about 53 600 kN (53.6 MN).

A mat (raft) foundation is a large slab covering the whole building area and supporting all columns. It is designed (chosen) in the following situations.

• Low bearing capacity: if spread footings would cover more than about 50% of the plan area, or would overlap, a single mat is cheaper and simpler.
• Reduce settlement: a mat spreads the load over a wide area and reduces the net pressure, and its stiffness reduces differential settlement on erratic or compressible soil.
• Basement and water table: a mat acts as a floor of the basement, resists uplift and water pressure, and is waterproofed easily.
• Compensated foundation: removing the soil weight reduces net pressure on soft clay.
• Heavy, unequal column loads or loads from tall buildings and silos, chimneys, towers.
• Earthquake areas: a rigid mat ties the structure into one unit.

The basic assumptions of the conventional (rigid) method of mat analysis are:

1. The mat is perfectly rigid compared with the soil, so it settles as a plane (rigid body) and does not bend appreciably.
2. The soil pressure is linear (planar), given by $q = \dfrac{Q}{A} \pm \dfrac{M_xy}{I_x} \pm \dfrac{M_yx}{I_y}$.
3. The centroid of soil pressure coincides with the line of action of the resultant of the column loads (static equilibrium).
4. The mat is divided into independent strips, with no shear transfer between adjacent strips.
5. The soil is homogeneous, and column spacing and loads vary little (not more than about 20%), and spacing is less than $1.75/\lambda$.

Fully compensated raft: a raft placed at such depth that the weight of the soil excavated is equal to the total weight of the structure, including the raft. The net pressure on the soil is then zero, and, in principle, no extra settlement or shear failure occurs.

Derivation of depth

Let $Q$ = total load of the structure (including the raft), $A$ = area of the raft, $\gamma$ = unit weight of the excavated soil, $D_f$ = depth of the raft.

Gross pressure at foundation level: $q = \dfrac{Q}{A}$

Weight of soil removed per unit area (overburden pressure): $\gamma D_f$

Net pressure on the soil: $q_{net} = \dfrac{Q}{A} - \gamma D_f$

For a fully compensated raft $q_{net} = 0$:

$$\frac{Q}{A} = \gamma D_f$$ $$D_f = \frac{Q}{\gamma A} = \frac{q}{\gamma}$$

If the water table is above the raft, the buoyancy reduces the soil weight, so $\gamma D_f$ is replaced by the vertical effective stress at base, or the buoyant uplift is subtracted from $Q$.

The spacings and the points $A, B, C, D$ are not legible, so these assumptions are made: the mat is $13\ \text{m}$ (x) $\times$ $15\ \text{m}$ (y); columns are at $x = -5, 0, +5$ m and $y = +6, 0, -6$ m (top row at $y = +6$); and $A, B, C, D$ are the corners $A(-6.5, +7.5)$, $B(+6.5, +7.5)$, $C(+6.5, -7.5)$, $D(-6.5, -7.5)$. The column loads include the mat weight.

Total load and centroid

$$\begin{aligned} Q &= 600+700+600+1100+1300+1400+700+800+600 = 7800\ \text{kN} \\ \sum Qx &= (-5)(600+1100+700) + (5)(600+1400+600) = +1000\ \text{kN·m} \\ \sum Qy &= 6(1900) - 6(2100) = -1200\ \text{kN·m} \\ e_x &= \frac{1000}{7800} = 0.128\ \text{m},\quad e_y = \frac{-1200}{7800} = -0.154\ \text{m} \end{aligned}$$

So $M_y = Qe_x = 1000$ kN·m and $M_x = Qe_y = -1200$ kN·m.

Section properties

$$\begin{aligned} A &= 13 \times 15 = 195\ \text{m}^2 \\ I_x &= \frac{13 \times 15^3}{12} = 3656.25\ \text{m}^4,\quad I_y = \frac{15 \times 13^3}{12} = 2746.25\ \text{m}^4 \\ \frac{Q}{A} &= \frac{7800}{195} = 40.0\ \text{kN/m}^2 \end{aligned}$$

Soil pressure

$$q = \frac{Q}{A} + \frac{M_x\,y}{I_x} + \frac{M_y\,x}{I_y} = 40 - 0.3282\,y + 0.3641\,x$$

Answer: $q_A = 35.2$, $q_B = 39.9$, $q_C = 44.8$, $q_D = 40.1$ kN/m² (all compressive, so no tension under the mat; maximum at C).

Given: raft $B = 3$ m, $L = 12$ m, $D_f = 2$ m, corrected $N = 22$, cohesionless soil, water table 4 m below GL.

Method (Meyerhof/Bowles, SPT-based, for raft): for a raft the settlement is nearly independent of width, so the net allowable pressure for a settlement $S_a$ is

$$q_{na} = 19.16\,N\,F_d\,\frac{S_a}{25.4}\ \ \text{(kN/m}^2\text{, } S_a \text{ in mm)},\qquad F_d = 1 + 0.33\frac{D_f}{B} \le 1.33$$

Allowable settlement: the differential settlement is limited to 20 mm. As the total settlement is at least equal to the differential settlement, the conservative value $S_a = 20$ mm is used.

Calculation

$$\begin{aligned} F_d &= 1 + 0.33\times\frac{2}{3} = 1.22 \\ q_{na} &= 19.16 \times 22 \times 1.22 \times \frac{20}{25.4} = 404.9\ \text{kN/m}^2 \end{aligned}$$

Water table correction

The water table lies at 4 m, which is within a depth $B$ below the base ($D_f + B = 5$ m), so

$$C_w = 0.5 + 0.5\frac{D_w}{D_f + B} = 0.5 + 0.5\times\frac{4}{5} = 0.90$$ $$q_{na}' = 404.9 \times 0.90 = 364.4\ \text{kN/m}^2$$

Answer: net allowable bearing pressure of the raft $\approx 365$ kN/m² (settlement criterion governs; the shear failure check will have a large margin in sand).

Given: raft $20\ \text{m} \times 10\ \text{m}$, gross pressure $q = 200$ kN/m², $D_f = 2.5$ m, clay $c_u = 80$ kN/m², $\gamma = 19$ kN/m³.

Skempton's equation: $q_{nu} = c_uN_c$, with $B = 10$ m, $L = 20$ m.

$$\begin{aligned} \frac{D_f}{B} &= \frac{2.5}{10} = 0.25 < 2.5 \\ N_{c(sq)} &= 6\left(1 + 0.2 \times 0.25\right) = 6.30 \\ N_{c(rect)} &= \left(0.84 + 0.16 \times \frac{10}{20}\right)(6.30) = 0.92 \times 6.30 = 5.80 \\ q_{nu} &= 80 \times 5.796 = 463.7\ \text{kN/m}^2 \end{aligned}$$

Net applied pressure:

$$q_{net} = q - \gamma D_f = 200 - 19 \times 2.5 = 152.5\ \text{kN/m}^2$$ $$FS = \frac{q_{nu}}{q_{net}} = \frac{463.7}{152.5} = 3.04$$

Answer: factor of safety $\approx 3.04$.

Given: raft $B = 20$ m, $L = 30$ m, $A = 600$ m², $c_u = 10$ kN/m², $\gamma = 19$ kN/m³, total load $Q = 100$ MN $= 100\,000$ kN.

Gross pressure: $q = \dfrac{100\,000}{600} = 166.67$ kN/m².

(a) Fully compensated

Net pressure is zero: $q = \gamma D_f$

$$D_f = \frac{166.67}{19} = 8.77\ \text{m}$$

(b) Depth for FS = 3

Use Skempton's equation: $q_{nu} = c_uN_c$ and $FS = \dfrac{q_{nu}}{q - \gamma D_f}$. For $D_f/B \le 2.5$:

$$N_c = \left(0.84 + 0.16\frac{20}{30}\right)\times 6\left(1 + 0.2\frac{D_f}{20}\right) = 5.68\,(1 + 0.01D_f)$$ $$\begin{aligned} 10 \times 5.68(1 + 0.01D_f) &= 3\,(166.67 - 19D_f) \\ 56.8 + 0.568D_f &= 500 - 57D_f \\ D_f &= \frac{443.2}{57.568} = 7.70\ \text{m} \end{aligned}$$

Check: $N_c = 6.12$, $q_{nu} = 61.2$ kN/m², $q_{net} = 166.67 - 19(7.70) = 20.4$ kN/m², $FS = 3.0$ (O.K.; $D_f/B = 0.385$).

Answer: (a) $D_f = 8.77$ m for a fully compensated raft; (b) $D_f = 7.70$ m for FS = 3.

Given: mat $B = 10$ m, $L = 15$ m, $D_f = 2.5$ m, saturated clay $c_u = 30$ kN/m², total load $Q = 4000$ kN. The unit weight of soil is not given.

Assumption: the relief due to excavation is ignored (conservative) because $\gamma$ is not given, so the net applied pressure is taken equal to $Q/A$.

Bearing capacity factor (Skempton)

$$\begin{aligned} N_{c(sq)} &= 6\left(1 + 0.2 \times \frac{2.5}{10}\right) = 6.30 \\ N_{c(rect)} &= \left(0.84 + 0.16\times\frac{10}{15}\right)(6.30) = 5.96 \\ q_{nu} &= c_uN_c = 30 \times 5.964 = 178.9\ \text{kN/m}^2 \end{aligned}$$

Applied pressure and factor of safety

$$\begin{aligned} q &= \frac{4000}{10 \times 15} = 26.67\ \text{kN/m}^2 \\ FS &= \frac{178.9}{26.67} = 6.71 \end{aligned}$$

Answer: factor of safety $\approx 6.7$. If the weight of the soil excavated for the basement ($\gamma D_f \approx 18 \times 2.5 = 45$ kN/m²) is taken into account, it exceeds the applied pressure of 26.7 kN/m², so the mat is over-compensated and the true factor of safety is even higher.

Given: raft $B = 14$ m, $L = 21$ m, $q_u$ (unconfined) $= 14$ kN/m² so $c_u = 7$ kN/m², gross pressure at base $q = 135$ kN/m², $\gamma = 19$ kN/m³, FS = 3.

Skempton: $FS = \dfrac{c_uN_c}{q - \gamma D_f}$ (net ultimate capacity over net pressure). For $D_f/B \le 2.5$:

$$N_c = \left(0.84 + 0.16\times\frac{14}{21}\right)\times 6\left(1 + 0.2\frac{D_f}{14}\right) = 5.68\,(1 + 0.01429D_f)$$ $$\begin{aligned} 7 \times 5.68(1 + 0.01429D_f) &= 3\,(135 - 19D_f) \\ 39.76 + 0.568D_f &= 405 - 57D_f \\ D_f &= \frac{365.24}{57.568} = 6.34\ \text{m} \end{aligned}$$

Check: $D_f/B = 0.45 < 2.5$; $N_c = 6.19$; $q_{nu} = 43.4$ kN/m²; $q_{net} = 135 - 19(6.34) = 14.5$ kN/m²; $FS = 3.0$.

Answer: the base of the raft should be placed at a depth of about 6.3 m.

Given: mat $B = 8$ m, $L = 10$ m, $D_f = 5$ m, saturated clay $c_u = 50$ kPa, $\gamma = 19$ kN/m³.

Skempton's method ($\phi = 0$):

$$\begin{aligned} \frac{D_f}{B} &= \frac{5}{8} = 0.625 < 2.5 \\ N_{c(sq)} &= 6\left(1 + 0.2 \times 0.625\right) = 6.75 \\ N_{c(rect)} &= \left(0.84 + 0.16 \times \frac{8}{10}\right)(6.75) = 0.968 \times 6.75 = 6.53 \\ q_{nu} &= c_uN_c = 50 \times 6.534 = 326.7\ \text{kPa} \end{aligned}$$

With a factor of safety $F = 3$ (assumed, usual value):

$$q_{ns} = \frac{q_{nu}}{F} = \frac{326.7}{3} = 108.9\ \text{kPa}$$

Answer: net ultimate capacity $= 326.7$ kPa and net safe bearing capacity $\approx 108.9$ kPa (gross safe pressure $= 108.9 + 19 \times 5 = 203.9$ kPa).