Bearing Capacity and Settlement of Shallow Foundations

Bearing capacity of a footing depends on the soil strength and on the geometry of the footing, and it is reduced when the soil is submerged. Terzaghi's equation for a strip footing is $q_u = cN_c + \gamma D_f N_q + 0.5\gamma B N_\gamma$.

Factors affecting bearing capacity

• Size (width B): in cohesionless soil the third term $0.5\gamma BN_\gamma$ grows with $B$, so a wider footing has a larger ultimate capacity. In clay ($\phi = 0$, $N_\gamma = 0$) the capacity does not depend on $B$. A wider footing, however, stresses a deeper zone and settles more, so settlement often governs the design.
• Shape: a square or circular footing mobilises three-dimensional resistance. Terzaghi uses shape factors: strip $1.0 / 0.5$, square $1.3 / 0.4$, circular $1.3 / 0.3$ (for the $cN_c$ and $\gamma BN_\gamma$ terms). A rectangle lies between a strip and a square.
• Depth of foundation: a deeper base increases the surcharge $q = \gamma D_f$ and so the $N_q$ term.
• Soil properties: $c$, $\phi$ (which fix $N_c, N_q, N_\gamma$) and unit weight $\gamma$. Loose or soft soil fails in local shear and gives a lower capacity.
• Water table: it reduces the effective unit weight of soil (see below).
• Eccentricity and inclination of load: reduce the effective width and the capacity.

Effect of water table position

Submergence reduces the effective stress, so the soil's shear strength falls. Terzaghi's equation with correction factors is

$$q_u = cN_c + \gamma D_f N_q R_{w1} + 0.5\gamma B N_\gamma R_{w2}$$

with $R_{w1} = 0.5\left(1 + \dfrac{Z_{w1}}{D_f}\right)$ and $R_{w2} = 0.5\left(1 + \dfrac{Z_{w2}}{B}\right)$.

 Case 1: WT above base      Case 2: WT below base
 GL ______________          GL ______________
 WT ~~~~~~~~~~~~~~ Zw1       |  gamma
    |  gamma'               Base ====== 
 Base =========            WT ~~~~~~~~~~~ Zw2 < B
    |  gamma'                  |  gamma'

1. WT above the base ($Z_{w1} < D_f$): the soil on both sides of the footing and below it is submerged. Use $\gamma'$ in the $N_\gamma$ term, and the surcharge becomes $q = \gamma Z_{w1} + \gamma'(D_f - Z_{w1})$. At the ground surface $R_{w1} = R_{w2} = 0.5$, so the capacity of a cohesionless soil falls by about half.
2. WT at base level: $R_{w1} = 1$ and $R_{w2} = 0.5$, only the $N_\gamma$ term is halved.
3. WT below the base (at depth $Z_{w2}$ below it): $R_{w2}$ lies between $0.5$ and $1$. The effect lies only in the $\gamma B N_\gamma$ term, as the wedge below the base is partly submerged.
4. WT deeper than $B$ below the base: $Z_{w2} \ge B$, so $R_{w2} = 1$ and the water table has no effect.

In clay with $\phi = 0$, the water table has little effect on $q_u$ because $N_q = 1$ and $N_\gamma = 0$, except through the change in surcharge.

The plate load test (PLT) loads a rigid steel plate (30 to 75 cm) placed at foundation level and records load against settlement. Its results are not safely extrapolated to the real footing in the following circumstances.

1. Depth of influence (scale effect). The stressed zone below a plate is only about $1.5$ to $2B_p$ deep, while below the real footing it is $1.5$ to $2B$. A weak or compressible layer below the plate's influence zone is not tested at all, yet it may control the settlement of the full-size footing.
2. Clay: time-dependent settlement. The test lasts a few hours or days, so only immediate settlement is recorded. The consolidation settlement of clay, which takes months or years, is not shown. Settlement in clay also depends on the width of the footing, not in simple proportion.
3. Size effect on bearing capacity. In sand, $q_u$ increases with width (the $N_\gamma$ term), so a plate result underestimates the footing's capacity and is corrected by $q_u \propto B$. In clay $q_u$ is independent of size. A single rule cannot be applied to all soils.
4. Non-homogeneous or layered soil. Properties at the test point may not represent the whole site, and one test represents only a small area.
5. Water table. If the water table lies below the plate but within the depth of influence of the footing, the test does not reflect its effect.
6. Test conditions. Moisture and disturbance at the test pit differ from field conditions, and the pit must be dug to foundation level, so deep foundations cannot be tested.
7. Non-uniform loading and rigidity. A rigid plate and a flexible footing do not distribute pressure in the same way.
8. Cost and time. The test needs heavy reaction loading and is expensive for large footings.

Because of these limits, the PLT is used as a check and is supported by borehole data, SPT/CPT and laboratory tests.

Terzaghi and Vesic described three ways in which a soil fails below a footing. They differ in the shape of the failure surface and the load–settlement curve.

 General shear          Local shear         Punching shear
 ___ ___ ___            __|____|__           _|____|_
 /\ |  | /\ /           \  \  /  /           |    |
 \_\|__|/_/ (reaches    (fades in soil)      |    | (no heave)
    surface)

Ascertaining the mode of failure

• From relative density (sand): $D_r > 70\%$ gives general shear, $D_r < 20\%$ gives punching and in between local shear. SPT: $N > 30$ general, $N < 10$ loose.
• From stress–strain behaviour (clay): unconfined or triaxial test. A well-defined peak at small strain (below about 5%) means general shear. A strain at failure of 20% or more, with no clear peak, indicates local or punching shear.
• From the plate load test: a defined breaking point in the load–settlement curve means general shear.
• If the soil is likely to fail in local shear, Terzaghi's design reduces the strength: $c' = \tfrac{2}{3}c$ and $\tan\phi' = \tfrac{2}{3}\tan\phi$, and modified factors $N_c', N_q', N_\gamma'$ are used.

Proportioning for uniform settlement means choosing the plan size of each footing so that all footings of a structure settle by about the same amount, even though their loads differ. This keeps the differential settlement and the angular distortion within the permitted limits.

Procedure

1. Collect loads. Find the load at every column: dead load plus the part of the live load that acts permanently (about 25% to 50%, as per usual practice). This is the load that causes long-term settlement.
2. Fix the allowable total settlement $S_{all}$ (for example 25 mm in sand, 40 mm in clay) and the allowable differential settlement.
3. Soil data. From the borehole log, SPT or plate load test, find the settlement characteristics of the soil (modulus of compressibility, $C_c$, $e_0$, $\sigma'_{vo}$).
4. Trial sizes. Assume sizes with equal soil pressure $q = Q/A$ (equal pressure is the first trial for sand) and check bearing capacity with $FS \ge 3$.
5. Compute settlement of each footing.
   • In sand: from the Terzaghi–Peck or Meyerhof curves, find the pressure that gives $S_{all}$ for each width $B$. The settlement increases with $B$, so for a given settlement the larger footing must carry a lower pressure than a smaller one.
   • In clay: find the stress increase $\Delta\sigma$ at mid-depth of the compressible layer for each footing and compute $S_c = \dfrac{C_c H}{1+e_0}\log\dfrac{\sigma'_{vo}+\Delta\sigma}{\sigma'_{vo}}$.
6. Revise. Increase the width of footings that settle more and reduce the width of those that settle less. Because a wide footing stresses a deeper zone, heavily loaded footings need a lower contact pressure to settle equally.
7. Repeat until all footings give nearly the same settlement, and finally check shear (bearing capacity) failure for each.

The final design gives different contact pressures for different footings, but equal settlements.

Given: strip footing, $B = 1.5$ m, $D_f = 1.2$ m, $c = 0$, $N_q = 81.3$, $N_\gamma = 100.4$, $G = 2.70$, $\gamma_d = 16$ kN/m³, $\gamma_w = 9.81$ kN/m³. For a strip footing $q_u = \gamma D_f N_q + 0.5\gamma B N_\gamma$ (the $cN_c$ term is zero).

Case 1: dry sand

$$\begin{aligned} q &= 16 \times 1.2 = 19.2\ \text{kN/m}^2 \\ q_{u1} &= 19.2 \times 81.3 + 0.5 \times 16 \times 1.5 \times 100.4 \\ &= 1560.96 + 1204.8 = 2765.76\ \text{kN/m}^2 \end{aligned}$$

Case 2: water table at 0.5 m below ground level

The sand below the water table is saturated. Find its saturated unit weight from $\gamma_d = \dfrac{G\gamma_w}{1+e}$:

$$\begin{aligned} 1 + e &= \frac{2.70 \times 9.81}{16} = 1.655 \Rightarrow e = 0.655 \\ \gamma_{sat} &= \frac{(G+e)\gamma_w}{1+e} = \frac{(2.70 + 0.655) \times 9.81}{1.655} = 19.88\ \text{kN/m}^3 \\ \gamma' &= 19.88 - 9.81 = 10.07\ \text{kN/m}^3 \end{aligned}$$

The water table is above the base, so the soil from 0.5 m to 1.2 m is submerged (0.7 m thick), and the soil below the base is also submerged, so $\gamma'$ is used in the $N_\gamma$ term:

$$\begin{aligned} q &= 16 \times 0.5 + 10.07 \times 0.7 = 15.05\ \text{kN/m}^2 \\ q_{u2} &= 15.05 \times 81.3 + 0.5 \times 10.07 \times 1.5 \times 100.4 \\ &= 1223.7 + 758.6 = 1982.3\ \text{kN/m}^2 \end{aligned}$$

Change in capacity

$$\Delta q_u = 2765.76 - 1982.29 = 783.5\ \text{kN/m}^2$$

Answer: the ultimate bearing capacity falls from 2765.8 kN/m² to 1982.3 kN/m², a reduction of about 783 kN/m² (28.3%).

Classical bearing capacity theories were developed step by step. Each author added a term that the previous theory neglected.

1. Rankine (1857) and Bell. Based on Rankine earth pressure. The soil below the footing is in active state and the soil beside it in passive state. For cohesionless soil $q_u = \gamma D_f\left(\dfrac{1+\sin\phi}{1-\sin\phi}\right)^2$. Bell (1915) extended it to $c$–$\phi$ soil: $q_u = \gamma D_f N_\phi^2 + 2c\sqrt{N_\phi}(N_\phi + 1)$ with $N_\phi = \tan^2(45^\circ + \phi/2)$. Pauker's method is a similar approach for cohesionless soil. These ignore footing width and base friction.
2. Prandtl (1921). Studied punching of a rigid die into a weightless $c$–$\phi$ medium. Found a failure surface with a wedge, a log-spiral fan and a passive zone, giving $q_u = cN_c$ with $N_c = (N_q - 1)\cot\phi$ and $N_q = e^{\pi\tan\phi}\tan^2(45^\circ + \phi/2)$. The footing is on the surface and soil is weightless.
3. Reissner (1924). Extended Prandtl to include the surcharge $q = \gamma D_f$, adding the $N_q$ term.
4. Terzaghi (1943). Added the weight of soil ($N_\gamma$ term), a rough base and the effect of shape (square, circular), and gave $q_u = cN_c + qN_q + 0.5\gamma BN_\gamma$.
5. Meyerhof (1951, 1963). Included the shear strength of soil above the base, and gave shape, depth and inclination factors.
6. Hansen (1970) and Vesic (1973). Further refined the factors for shape, depth, inclination of load, base and ground tilt, with improved values of $N_\gamma$.

So the progression is: Rankine (weight and surcharge, approximate) → Prandtl (cohesion) → Reissner (surcharge) → Terzaghi (soil weight, practical form) → Meyerhof, Hansen, Vesic (general equation with correction factors).

Classical bearing capacity theories (Rankine, Prandtl, Reissner, Terzaghi) have several limitations.

• Strip footing assumed. Plane strain is assumed, so shape effects for square, circular and rectangular footings are put in by empirical factors rather than being derived.
• Homogeneous, isotropic soil. Real soil is layered, variable and anisotropic, and the theories cannot deal with it directly.
• Rankine's and Bell's theories ignore the width of the footing and the friction on the base, so they are approximate and conservative.
• Prandtl's theory assumes weightless soil and no surcharge; Reissner's still neglects soil weight.
• Terzaghi's theory ignores the shear strength of the soil above the base (treated only as surcharge) and assumes a rough base and $D_f \le B$, so it is not suited to deep footings.
• General shear assumed. Local and punching shear are handled only by an empirical two-thirds reduction of $c$ and $\tan\phi$.
• Load conditions. Central, vertical, static load only. Eccentric, inclined and dynamic loads are not included in the original form.
• Nγ values differ between authors (Terzaghi, Meyerhof, Hansen, Vesic) because of the uncertainty in the failure surface, giving different results for the same soil.
• Compressibility is ignored. The soil is treated as rigid–plastic, so settlement is not obtained. Settlement may govern the design.
• Water table effect is accounted for only approximately by correction factors.

The plate load test is an in-situ test used to find the ultimate bearing capacity of soil and the settlement of a footing at a given load. A rigid steel plate is loaded in stages and the settlement is recorded at each stage.

Procedure

1. A pit is dug to the foundation level, with width at least 5 times the plate size. A square or circular plate of 30 to 75 cm (about 25 mm thick) is bedded on the levelled soil.
2. A load is applied through a hydraulic jack acting against a reaction (a loaded platform or a kentledge, or an anchored truss). The jack load is measured by a pressure gauge or proving ring.
3. Load is applied in increments of about one-fifth of the estimated safe load. Settlement is read on dial gauges (usually 2 to 4, supported on an independent datum beam) at set time intervals, until the rate of settlement falls to $0.02$ mm/min (or 24 h in clay) before the next increment.
4. The test continues until failure or until the settlement is 25 mm or more (about 1.5 times the proposed design load).

        Jack + gauge
           |   Dial gauge
  ====Reaction beam====
           |
        [Plate]    <- test pit
  ///////////////

Results

• The load–settlement curve is plotted. The ultimate load is the point of failure, or the intersection of tangents on the initial and final parts. For sand, $q_u(\text{footing}) = q_u(\text{plate}) \times B/B_p$ and settlement $S = S_p\left[\dfrac{B(B_p+0.3)}{B_p(B+0.3)}\right]^2$. For clay, $q_u$ is the same for plate and footing.

Limitations

• The stressed depth is small (about $2B_p$), so deeper weak layers are not tested.
• Only immediate settlement is measured, not long-term consolidation in clay.
• Size effects: results cannot be extrapolated directly to large footings.
• Represents only a small area, and is expensive and slow.

Terzaghi's theory (1943) gives the ultimate bearing capacity of a shallow footing by considering the equilibrium of a failure wedge. It is the most widely used classical theory and extends Prandtl's and Reissner's work by including soil weight and a rough base.

Assumptions

1. The footing is a long strip (plane strain), and $D_f \le B$ (shallow foundation).
2. The base is rough, so the soil wedge directly below it stays elastic and moves down with the footing.
3. The soil is homogeneous and isotropic, and obeys Mohr–Coulomb's law $\tau = c + \sigma\tan\phi$.
4. The shear strength of the soil above the base level is ignored. It is replaced by a uniform surcharge $q = \gamma D_f$.
5. Failure occurs in general shear (for local shear, strength is reduced to two-thirds).
6. The load is vertical and central.

Failure mechanism

   q_u  ____________
       |  I  |
  II  /\    /\  II
 III /  \  /  \ III

Under the footing, zone I is an elastic (triangular) wedge with base angles $\phi$. Zone II is a radial shear zone bounded by a logarithmic spiral. Zone III is a passive Rankine zone.

Derivation (outline)

Considering vertical equilibrium of the wedge: $q_u B + W = 2P_p + 2C\sin\phi$, where $C = \dfrac{cB}{2\cos\phi}$ is the cohesive force on each inclined face, $W$ the weight of the wedge, and $P_p$ the passive resistance from zones II and III (made of three parts, due to $c$, $q$ and $\gamma$). This gives

$$q_u = cN_c + \gamma D_f N_q + 0.5\gamma B N_\gamma \quad (\text{strip})$$

For other shapes: square $q_u = 1.3cN_c + \gamma D_fN_q + 0.4\gamma BN_\gamma$; circular $q_u = 1.3cN_c + \gamma D_fN_q + 0.3\gamma BN_\gamma$. $N_c, N_q, N_\gamma$ depend only on $\phi$ (e.g. for $\phi = 35^\circ$: $57.8, 41.4, 42.4$). For local shear use $c' = \tfrac{2}{3}c$, $\tan\phi' = \tfrac{2}{3}\tan\phi$.

Skempton (1951) studied the bearing capacity of saturated clay ($\phi_u = 0$) and showed that the bearing capacity factor $N_c$ is not constant. It increases with depth of foundation relative to width.

• For a strip footing: $N_c = 5\left(1 + 0.2\dfrac{D_f}{B}\right)$ for $D_f/B \le 2.5$, and $N_c = 7.5$ for larger depths.
• For a square or circular footing: $N_c = 6\left(1 + 0.2\dfrac{D_f}{B}\right)$ up to $D_f/B = 2.5$, and $N_c = 9$ beyond that.
• For a rectangular footing: $N_c(\text{rect}) = \left(0.84 + 0.16\dfrac{B}{L}\right)N_c(\text{square})$.

The net ultimate capacity is $q_{nu} = cN_c$ and the net safe bearing capacity is

$$q_{ns} = \frac{q_{nu}}{F} = \frac{cN_c}{F}$$

with $F$ usually 3, where $c = q_u/2$ is the undrained cohesion. The gross safe pressure is $q_{ns} + \gamma D_f$.

More differential settlement is allowed in clay than in sand because the nature of settlement differs.

• Time: settlement in clay occurs slowly through consolidation, over months or years. The structure can gradually adjust, creep and redistribute stresses, so cracking is reduced. In sand settlement is rapid, mostly during construction, and the structure cannot adjust.
• Pattern: settlement in clay is smoother and bowl-shaped, and the angular distortion between adjacent footings is small even if the total settlement is large. In sand, local variations in density cause an erratic pattern, and differential settlement may be as high as 75% of the total.
• Magnitude: clay settles by more, so the same ratio of differential to total settlement gives larger values, with limits of about 40 mm in clay against 25 mm in sand (Terzaghi and Peck).

So the allowable values are higher in clay without greater risk of damage.

Settlement of a foundation, if excessive or uneven, affects both appearance and safety of the structure.

• Cracks in walls, floors, beams and plaster from differential settlement.
• Tilting of tall structures such as towers and chimneys, as in the Leaning Tower of Pisa.
• Extra stresses in members of statically indeterminate frames, which can cause structural distress or even failure.
• Damage to services: pipes, drains, sewer lines and cables connected to the building may break.
• Loss of serviceability: doors and windows jam, floors slope, lifts and crane rails go out of line, and machinery loses alignment.
• Drainage problems and ponding around the structure.
• Aesthetic and economic loss, with costly repairs and a lower property value.

For this reason settlement is checked against permissible limits in addition to the bearing capacity check.

Total settlement is $S = S_i + S_c + S_s$.

Settlement of a structure can take three forms, depending on how the footings settle relative to each other.

1. Uniform (total) settlement

All footings settle by the same amount. The structure moves down without distortion and there is little damage. It affects only the connections to services and the level of the building.

 Before:  |======|      After:  |======|
          ########             ########
                               (all down)

2. Tilting (rotation)

One side settles more than the other, so the building rotates as a rigid body. It occurs with eccentric loads, non-uniform soil, or adjacent heavy loads. Tall structures such as chimneys and towers are most affected, and the Leaning Tower of Pisa is an example.

     /|
    / |  <- tilted rigid body
   /__|
  ######

3. Differential settlement

Different footings settle by different amounts, giving angular distortion $\delta/L$ between points. It distorts the frame and causes cracks and damage to beams, walls and floors. Two shapes are common:

• Sagging (dishing): centre settles more than the ends. Cracks appear in the lower part of the wall.
• Hogging: ends settle more than the centre. Cracks appear in the upper part.

 Sagging:  ______      Hogging:       ____
          /      \                   /    \
 ========        ======     ========        ========

The differential settlement is limited to about 1/300 to 1/500 of span to avoid cracking (about 20 mm for footings in sand, 40 mm in clay).

The load–settlement curve of a plate load test on sand is plotted as pressure $q$ against settlement $s$.

1. Ultimate bearing capacity of the plate $q_{up}$: take the pressure at the point of failure (peak). If the curve has no clear peak, use the intersection of tangents drawn to the initial and the final straight parts of the curve (or the break point on a log–log plot).
2. Footing capacity: in sand, $q_u$ increases with width, so $q_{u}(\text{footing}) = q_{up}\dfrac{B}{B_p}$.
3. Safe bearing capacity (shear): $q_s = q_u/F$ with $F = 2.5$ to $3$.
4. Settlement criterion: for the permitted settlement of the footing (25 mm), the equivalent plate settlement is found from $\dfrac{S_f}{S_p} = \left[\dfrac{B(B_p+0.3)}{B_p(B+0.3)}\right]^2$, and the pressure for that plate settlement is read from the curve.
5. The allowable bearing pressure is the smaller of the safe pressure by shear failure and the pressure by settlement.

Given: $L/B = 1.5$, $D_f = 1.5$ m, $Q = 6000$ kN, $\phi = 34^\circ$, $N_q = 41.4$, $N_\gamma = 42.4$, $c = 0$, $\gamma_{sat} = 2.0\ \text{g/cc} = 19.62$ kN/m³, FS = 3, water table at ground surface (worst case). Use $\gamma_w = 9.81$ kN/m³.

Assumptions: Terzaghi's equation for a rectangular footing with $B/L = 1/1.5 = 0.667$. The load of 6000 kN includes the weights of column and footing, so the gross allowable pressure is $q_u/F$.

$$\begin{aligned} \gamma' &= 19.62 - 9.81 = 9.81\ \text{kN/m}^3 \\ q &= \gamma' D_f = 9.81 \times 1.5 = 14.715\ \text{kN/m}^2 \\ q_u &= qN_q + 0.5\left(1 - 0.2\tfrac{B}{L}\right)\gamma' B N_\gamma \\ &= 14.715 \times 41.4 + 0.5(1 - 0.133)(9.81)(42.4)B \\ &= 609.2 + 180.3\,B\ \text{kN/m}^2 \end{aligned}$$

Load capacity: $Q = \dfrac{q_u}{3} \times B \times L = \dfrac{609.2 + 180.3B}{3} \times 1.5B^2 = 6000$

Solving for $B$ (trial and error, checked in Python): $B = 3.185$ m.

Answer: provide a rectangular footing of 3.2 m × 4.8 m (safe load 6072 kN > 6000 kN).

Given: strip footing $B = 2$ m, $D_f = 1.2$ m, $c = 0$ (cohesionless), $N_q = 41$, $N_\gamma = 42$, applied pressure $q = 400$ kN/m², $\gamma = 16.8$ kN/m³, $\gamma_{sat} = 19.5$ kN/m³, $\gamma_w = 9.81$ kN/m³. FOS $= q_u/q$. Strip footing: $q_u = \gamma D_f N_q + 0.5\gamma B N_\gamma$.

(i) WT at 4 m below GL

The water table is $4 - 1.2 = 2.8$ m below the base, greater than $B = 2$ m, so it has no effect.

$$\begin{aligned} q_u &= 16.8 \times 1.2 \times 41 + 0.5 \times 16.8 \times 2 \times 42 \\ &= 826.6 + 705.6 = 1532.2\ \text{kN/m}^2 \\ FOS &= 1532.2/400 = 3.83 \end{aligned}$$

(ii) WT at 2.5 m below GL

$Z_{w2} = 2.5 - 1.2 = 1.3$ m below the base ($< B$), so $R_{w2} = 0.5(1 + 1.3/2) = 0.825$. The surcharge is unchanged ($R_{w1} = 1$).

$$\begin{aligned} q_u &= 826.6 + 705.6 \times 0.825 = 826.6 + 582.1 = 1408.7\ \text{kN/m}^2 \\ FOS &= 1408.7/400 = 3.52 \end{aligned}$$

(iii) WT at 0.5 m below GL

The water table is above the base. $\gamma' = 19.5 - 9.81 = 9.69$ kN/m³.

$$\begin{aligned} q &= 16.8 \times 0.5 + 9.69 \times 0.7 = 15.18\ \text{kN/m}^2 \\ q_u &= 15.18 \times 41 + 0.5 \times 9.69 \times 2 \times 42 \\ &= 622.5 + 407.0 = 1029.5\ \text{kN/m}^2 \\ FOS &= 1029.5/400 = 2.57 \end{aligned}$$

Answer: FOS = 3.83 (WT at 4 m), 3.52 (WT at 2.5 m) and 2.57 (WT at 0.5 m). A rising water table lowers the factor of safety.

Given: $B = L = 4$ m, $\phi = 32^\circ$, $c = 35$ kN/m², $\gamma = 18$ kN/m³, $Q = 2000$ kN, $e = 0.3$ m in one direction, FS = 2.5, $N_c = 55$, $N_q = 40$, $N_\gamma = 45$.

Method: use the effective width $B' = B - 2e$ (Meyerhof) and Terzaghi's square footing equation $q_u = 1.3cN_c + \gamma D_f N_q + 0.4\gamma B' N_\gamma$. The load must satisfy $\dfrac{q_u}{FS} A' \ge Q$.

$$\begin{aligned} B' &= 4 - 2(0.3) = 3.4\ \text{m},\quad A' = 3.4 \times 4 = 13.6\ \text{m}^2 \\ \text{Required } q_u &= \frac{2.5 \times 2000}{13.6} = 367.6\ \text{kN/m}^2 \end{aligned}$$

Terms of $q_u$ (independent of depth):

$$\begin{aligned} 1.3cN_c &= 1.3 \times 35 \times 55 = 2502.5\ \text{kN/m}^2 \\ 0.4\gamma B' N_\gamma &= 0.4 \times 18 \times 3.4 \times 45 = 1101.6\ \text{kN/m}^2 \end{aligned}$$

So even at the ground surface ($D_f = 0$): $q_u = 2502.5 + 1101.6 = 3604.1\ \text{kN/m}^2$, which is far more than the required $367.6$ kN/m². The factor of safety at $D_f = 0$ is $3604.1 \times 13.6/2000 = 24.5$.

Answer: from the shear failure condition, no minimum depth is required (the requirement is met even at $D_f \approx 0$, FS = 24.5 against 2.5). The depth is therefore decided by other needs: a minimum of about 1.0 m below the ground (to get below topsoil, frost and shrinkage zones) is adopted in practice, which gives FS = 29.4.

Note: with the data given, the load is small compared with the capacity of the soil, so shear failure does not control the depth.

Given: square footing, $Q = 750$ kN, clay with $c = 35$ kN/m², $\gamma = 17$ kN/m³, FS = 3. The depth is not given.

Assumptions: the clay is purely cohesive ($\phi = 0$), so Terzaghi's factors are $N_c = 5.7$, $N_q = 1$, $N_\gamma = 0$. Because the footing is backfilled, the weight of footing plus backfill is taken as about equal to the soil it replaces, so the net load intensity on the soil is $Q/B^2$ and the net ultimate capacity is used.

$$q_u = 1.3cN_c + \gamma D_f N_q \Rightarrow q_{nu} = q_u - \gamma D_f = 1.3cN_c$$ $$\begin{aligned} q_{nu} &= 1.3 \times 35 \times 5.7 = 259.35\ \text{kN/m}^2 \\ q_{ns} &= \frac{259.35}{3} = 86.45\ \text{kN/m}^2 \\ B^2 &= \frac{750}{86.45} = 8.68\ \text{m}^2 \Rightarrow B = 2.95\ \text{m} \end{aligned}$$

Since the depth cancels out ($N_q = 1$), the answer does not depend on $D_f$.

Answer: provide a square footing of 3.0 m × 3.0 m (net pressure $750/9 = 83.3$ kN/m² < 86.45 kN/m²; FS $= 259.35/83.3 = 3.11$).

Given: $B = 2$ m, $L = 3$ m, $D_f = 1.5$ m, $e = 0.26$ m along the width, $c = 20$ kN/m², $\phi = 35^\circ$, $\gamma = 18$ kN/m³ (taken as the saturated unit weight), water table 0.5 m below GL, Terzaghi's theory.

Bearing capacity factors for $\phi = 35^\circ$ (Terzaghi chart, general shear): $N_c = 57.8$, $N_q = 41.4$, $N_\gamma = 42.4$. Factor of safety assumed $F = 3$, $\gamma_w = 9.81$ kN/m³.

Effective dimensions

$B' = B - 2e = 2 - 2(0.26) = 1.48$ m; $L = 3$ m.

Overburden and submerged weight

The water table is above the base, so the soil below the base is submerged: $\gamma' = 18 - 9.81 = 8.19$ kN/m³.

$$q = 18 \times 0.5 + 8.19 \times 1.0 = 17.19\ \text{kN/m}^2$$

Ultimate bearing capacity (rectangular footing)

$$\begin{aligned} q_u &= \left(1 + 0.3\tfrac{B'}{L}\right)cN_c + qN_q + 0.5\left(1 - 0.2\tfrac{B'}{L}\right)\gamma' B' N_\gamma \\ &= (1.148)(20)(57.8) + 17.19 \times 41.4 + 0.5(0.9013)(8.19)(1.48)(42.4) \\ &= 1327.1 + 711.7 + 231.6 \\ &= 2270.4\ \text{kN/m}^2 \end{aligned}$$

Safe bearing capacity

$$\begin{aligned} q_{nu} &= q_u - q = 2270.4 - 17.19 = 2253.2\ \text{kN/m}^2 \\ q_{ns} &= \frac{q_{nu}}{3} + q = 751.1 + 17.19 = 768.2\ \text{kN/m}^2 \end{aligned}$$

Safe load on the footing $= q_{ns} \times B'L = 768.2 \times 1.48 \times 3 = 3411\ \text{kN}$.

Answer: safe bearing capacity $q_s \approx 768$ kN/m² (net ultimate 2253 kN/m², FS = 3), corresponding to a safe load of about 3410 kN.

Given: $B = 2$ m (X-direction), $L = 3$ m (Y-direction), $D_f = 2$ m, $\phi = 35^\circ$, $c = 25$ kPa, $e_x = 0.2$ m, $e_y = 0.3$ m, WT 0.8 m below GL, $\gamma = 16.5$ kN/m³ (above WT), $\gamma_{sat} = 19.6$ kN/m³, $N_c = 57.8$, $N_q = 41.4$, $N_\gamma = 42.4$, FOS = 3, $\gamma_w = 9.81$ kN/m³.

Method: Meyerhof's effective area for double eccentricity, with Terzaghi's equation for a rectangular footing.

Effective dimensions

$$B' = 2 - 2(0.2) = 1.6\ \text{m},\quad L' = 3 - 2(0.3) = 2.4\ \text{m},\quad A' = 3.84\ \text{m}^2$$

Overburden pressure

The water table is above the base, so $\gamma' = 19.6 - 9.81 = 9.79$ kN/m³.

$$q = 16.5 \times 0.8 + 9.79 \times 1.2 = 13.2 + 11.75 = 24.95\ \text{kN/m}^2$$

Ultimate bearing capacity

$$\begin{aligned} q_u &= \left(1 + 0.3\tfrac{B'}{L'}\right)cN_c + qN_q + 0.5\left(1 - 0.2\tfrac{B'}{L'}\right)\gamma' B' N_\gamma \\ &= (1.2)(25)(57.8) + 24.95 \times 41.4 + 0.5(0.8667)(9.79)(1.6)(42.4) \\ &= 1734.0 + 1032.8 + 287.8 \\ &= 3054.6\ \text{kN/m}^2 \end{aligned}$$

Allowable load

$$Q_{all} = \frac{q_u}{FOS} A' = \frac{3054.6}{3} \times 3.84 = 3909.9\ \text{kN}$$

Answer: allowable load $\approx 3910$ kN (allowable pressure $q_u/3 = 1018$ kN/m² on the effective area; gross basis).

Given: column load $= 150$ t $= 150 \times 9.81 = 1471.5$ kN, $D_f = 1.5$ m, $\gamma = 20$ kN/m³, sand ($c = 0$), $N_q = 39.51$, $N_\gamma = 40.13$ (and $N_c = 55.2$, not needed). The water table is 3 m below the footing, which is more than $B$, so it has no effect.

Assumptions: square footing, FS = 3 (not given), Terzaghi's equation, general shear, weight of footing included in the load.

$$\begin{aligned} q_u &= \gamma D_f N_q + 0.4\gamma B N_\gamma \\ &= 20 \times 1.5 \times 39.51 + 0.4 \times 20 \times 40.13\,B \\ &= 1185.3 + 321.0\,B\ \text{kN/m}^2 \end{aligned}$$

Safe load: $\dfrac{q_u}{3}B^2 = 1471.5$, giving (by trial) $B = 1.61$ m.

Answer: provide a square footing of 1.7 m × 1.7 m (safe load 1668 kN > 1471.5 kN).

Given: circular footing $B = 2.5$ m, gross load $Q = 2000$ kN, $\phi = 30^\circ$, $c = 0$, $\gamma = 19$ kN/m³, FS = 3, $N_q = 22.5$, $N_\gamma = 19.7$.

Terzaghi's equation for a circular footing: $q_u = 1.3cN_c + \gamma D_f N_q + 0.3\gamma B N_\gamma$. With $c = 0$ the first term is zero.

Required ultimate capacity

$$\begin{aligned} A &= \frac{\pi}{4}(2.5)^2 = 4.909\ \text{m}^2 \\ q_{applied} &= \frac{2000}{4.909} = 407.4\ \text{kN/m}^2 \\ q_u &= FS \times q_{applied} = 3 \times 407.4 = 1222.3\ \text{kN/m}^2 \end{aligned}$$

Solve for depth

$$\begin{aligned} 0.3\gamma BN_\gamma &= 0.3 \times 19 \times 2.5 \times 19.7 = 280.7\ \text{kN/m}^2 \\ \gamma D_fN_q &= 19 \times 22.5 \times D_f = 427.5\,D_f \\ 1222.3 &= 427.5\,D_f + 280.7 \\ D_f &= \frac{941.6}{427.5} = 2.20\ \text{m} \end{aligned}$$

Answer: the footing should be placed at a depth of about 2.2 m (gross capacity basis; $D_f/B = 0.88 < 1$, so the shallow-footing equation is valid).

Given: $B = 3$ m, $L = 4$ m, $D_f = 1.5$ m, $\phi = 34^\circ$, $c = 0$, $\gamma = 20$ kN/m³ (above WT), $\gamma_{sat} = 21.5$ kN/m³, $N_q = 29$, $N_\gamma = 41$, FS = 3, $\gamma_w = 9.81$ kN/m³. As the excavation is backfilled, the overburden pressure $q$ acts in full.

Terzaghi's rectangular footing ($c = 0$): $q_u = qN_q + 0.5\left(1 - 0.2\dfrac{B}{L}\right)\gamma B N_\gamma$, with $1 - 0.2(0.75) = 0.85$. Safe load $= \dfrac{q_u}{3}\times BL$ (gross).

(i) WT at 1 m below GL (0.5 m above the base)

Sand below the base is submerged: $\gamma' = 21.5 - 9.81 = 11.69$ kN/m³.

$$\begin{aligned} q &= 20(1.0) + 11.69(0.5) = 25.85\ \text{kN/m}^2 \\ q_u &= 25.85 \times 29 + 0.5(0.85)(11.69)(3)(41) \\ &= 749.5 + 611.1 = 1360.6\ \text{kN/m}^2 \\ Q_{safe} &= \frac{1360.6}{3}\times 12 = 5442\ \text{kN} \end{aligned}$$

(ii) WT at 1 m below the base

$Z_{w2} = 1$ m $< B = 3$ m, so $R_{w2} = 0.5(1 + 1/3) = 0.667$. The overburden is dry: $q = 20 \times 1.5 = 30$ kN/m².

$$\begin{aligned} q_u &= 30 \times 29 + 0.5(0.85)(20)(3)(41)(0.667) \\ &= 870.0 + 1045.5 \times 0.667 = 870.0 + 697.0 \\ &= 1567.0\ \text{kN/m}^2 \\ Q_{safe} &= \frac{1567.0}{3}\times 12 = 6268\ \text{kN} \end{aligned}$$

Answer: (i) safe load $\approx 5440$ kN; (ii) safe load $\approx 6270$ kN. The higher water table reduces the safe load by about 13%.

Given: plate $B_p = 0.3$ m on sand. Net load $Q = 3200$ kN, allowable footing settlement $S_f = 25$ mm, square footing $B \times B$.

The figure could not be read exactly, so the plate readings below are assumed (typical for sand): pressure (kN/m²) 0, 100, 200, 300, 400, 500, 600, 700, 800 against settlement (mm) 0, 1.5, 4, 7.5, 12, 17.5, 24, 33, 45. If your curve differs, repeat the same steps with your own readings.

Method (Terzaghi–Peck, sand)

The settlement of a footing and a plate under the same pressure are related by

$$\frac{S_f}{S_p} = \left[\frac{B(B_p + 0.3)}{B_p(B + 0.3)}\right]^2 = \left[\frac{2B}{B + 0.3}\right]^2 \quad (B_p = 0.3\ \text{m})$$

So for each trial size $B$: the pressure is $q = 3200/B^2$; the plate settlement equivalent to 25 mm is $S_p = 25\left(\dfrac{B+0.3}{2B}\right)^2$; read the plate settlement from the curve at pressure $q$ and compare.

The curve settlement equals the required value at $B \approx 3.25$ m.

Safety check

The pressure is $\approx 300$ kN/m², which is well below the failure region of the curve (the curve keeps rising smoothly up to 800 kN/m²), so shear failure is not critical. Check FS $= q_u/q$ with $q_u$ (plate) scaled by $B/B_p$, which gives a large value.

Answer: provide a square footing of about 3.3 m × 3.3 m (pressure $\approx 294$ kN/m², settlement 25 mm), on the assumed curve.