BE Civil Engineering (IOE, TU) Foundation Engineering (IOE, CE 701) Question Paper 2076 Nepal

Assumptions of Terzaghi's theory

1. The footing is a strip (plane-strain, length $\gg$ width) placed at shallow depth ($D_f \le B$).
2. The soil is homogeneous, isotropic and general shear failure occurs.
3. The soil above the base level is treated only as a surcharge $q = \gamma D_f$ (its shear strength is neglected).
4. The base of the footing is rough; the soil wedge directly beneath it (Zone I) stays in elastic equilibrium and moves as a rigid body.
5. The shear strength obeys the Mohr–Coulomb criterion $\tau = c + \sigma\tan\phi$ and the principle of superposition is valid.

Failure zones (sketch in text)

        Df  --> surcharge q = gamma*Df
   ====[ FOOTING B ]====
        \  Zone I /          Zone I  : elastic wedge (angle phi with horizontal)
         \  (a-b-c)/          Zone II : radial shear (log-spiral) fans
   Zone II \______/ Zone II   Zone III: Rankine passive zones (angle 45-phi/2)
   Zone III          Zone III

Given

$B = 2.0\ \text{m}$, $D_f = 1.5\ \text{m}$, $c = 15\ \text{kN/m}^2$, $\gamma = 18\ \text{kN/m}^3$, $N_c = 17.69$, $N_q = 7.44$, $N_\gamma = 4.97$.

Surcharge $q = \gamma D_f = 18 \times 1.5 = 27\ \text{kN/m}^2$.

(i) Ultimate bearing capacity — strip footing

$$q_u = c N_c + q N_q + 0.5\,\gamma B N_\gamma$$ $$q_u = (15)(17.69) + (27)(7.44) + 0.5(18)(2.0)(4.97)$$ $$q_u = 265.35 + 200.88 + 89.46 = \textbf{555.69 kN/m}^2$$

(ii) Net ultimate bearing capacity

$$q_{nu} = q_u - \gamma D_f = 555.69 - 27 = \textbf{528.69 kN/m}^2$$

(iii) Net safe bearing capacity ($FS = 3$)

$$q_{ns} = \frac{q_{nu}}{FS} = \frac{528.69}{3} = \textbf{176.23 kN/m}^2$$

Square footing $2.0\,\text{m} \times 2.0\,\text{m}$

Use Terzaghi shape factors ($1.3$ on $cN_c$ term, $0.4$ on the $\gamma$ term):

$$q_u = 1.3\,cN_c + qN_q + 0.4\,\gamma B N_\gamma$$ $$q_u = 1.3(265.35) + 200.88 + 0.4(18)(2.0)(4.97)$$ $$q_u = 344.96 + 200.88 + 71.57 = \textbf{617.41 kN/m}^2$$

The square footing carries a higher ultimate pressure because of the favourable shape factor on the cohesion term.

Given

Normally consolidated clay: $H = 4.0\ \text{m}$, $p_0 = 80\ \text{kPa}$, $\Delta p = 60\ \text{kPa}$, $C_c = 0.25$, $e_0 = 0.90$, $C_v = 3.0\ \text{m}^2/\text{yr}$.

(a) Primary consolidation settlement

For a normally consolidated clay:

$$S_c = \frac{C_c\,H}{1 + e_0}\,\log_{10}\!\frac{p_0 + \Delta p}{p_0}$$ $$S_c = \frac{0.25 \times 4.0}{1 + 0.90}\,\log_{10}\!\frac{80 + 60}{80}$$ $$S_c = \frac{1.0}{1.90}\times \log_{10}(1.75) = 0.5263 \times 0.2430 = 0.1279\ \text{m}$$ $$\boxed{S_c \approx \textbf{127.9 mm}}$$

(b) Time for 50 % and 90 % consolidation

The clay is double-drained (sand above and below), so the drainage path is

$$H_{dr} = \frac{H}{2} = \frac{4.0}{2} = 2.0\ \text{m}$$

Using $t = \dfrac{T_v\,H_{dr}^2}{C_v}$:

$$t_{50} = \frac{0.197 \times (2.0)^2}{3.0} = \frac{0.788}{3.0} = 0.263\ \text{yr} \approx \textbf{3.15 months}$$ $$t_{90} = \frac{0.848 \times (2.0)^2}{3.0} = \frac{3.392}{3.0} = 1.131\ \text{yr} \approx \textbf{13.6 months}$$

(c) Effect of single drainage

With one impermeable boundary the water can escape from only one face, so the drainage path becomes the full thickness $H_{dr} = H = 4.0\ \text{m}$ instead of $2.0\ \text{m}$. Because $t \propto H_{dr}^2$, doubling the path quadruples the time. Thus $t_{50}$ and $t_{90}$ would each be four times larger (e.g. $t_{50} \approx 1.05\ \text{yr}$). The ultimate settlement $S_c$ is unchanged — only the rate slows.

Given

$D = 0.4\ \text{m}$, $L = 12\ \text{m}$, $c_u = 60\ \text{kPa}$, $\alpha = 0.55$, $N_c = 9$, $FS = 2.5$.

(a) Single-pile capacity ($\alpha$-method)

Base area $A_p = \dfrac{\pi}{4}D^2 = \dfrac{\pi}{4}(0.4)^2 = 0.1257\ \text{m}^2$

Shaft area $A_s = \pi D L = \pi(0.4)(12) = 15.08\ \text{m}^2$

End-bearing resistance

$$Q_p = c_u N_c A_p = 60 \times 9 \times 0.1257 = 67.86\ \text{kN}$$

Shaft (skin) resistance

$$Q_s = \alpha\,c_u\,A_s = 0.55 \times 60 \times 15.08 = 497.63\ \text{kN}$$

Ultimate capacity

$$Q_u = Q_p + Q_s = 67.86 + 497.63 = \textbf{565.49 kN}$$

Allowable load

$$Q_{all} = \frac{Q_u}{FS} = \frac{565.49}{2.5} = \textbf{226.2 kN}$$

(b) Group efficiency — Converse–Labarre

$$\eta = 1 - \frac{\theta}{90}\cdot\frac{(n-1)m + (m-1)n}{m\,n}$$

where $\theta = \tan^{-1}(D/s)$, $m = n = 3$.

$$\theta = \tan^{-1}\!\frac{0.4}{0.9} = \tan^{-1}(0.444) = 23.96^\circ$$ $$\frac{(3-1)(3) + (3-1)(3)}{3\times3} = \frac{6+6}{9} = 1.333$$ $$\eta = 1 - \frac{23.96}{90}\times 1.333 = 1 - 0.355 = 0.645 = \textbf{64.5\%}$$

(Note: using shaft-diameter-based $\theta = \tan^{-1}(D/s)$; if the convention $\theta=\tan^{-1}(d/s)$ with pile width were used the value differs slightly. Here $D=0.4$, $s=0.9$.)

Allowable group capacity

$$Q_{g,all} = \eta \times (\text{number of piles}) \times Q_{all}$$ $$Q_{g,all} = 0.645 \times 9 \times 226.2 = \textbf{1313 kN}$$

The design group capacity is taken as the lesser of (i) the efficiency-reduced sum above and (ii) the block-failure capacity; the engineer should check both, but for this widely spaced group the efficiency value governs the estimate.

Given

$\gamma = 18\ \text{kN/m}^3$, $\phi = 30^\circ$, $H = 6.0\ \text{m}$, $q = 20\ \text{kPa}$.

Rankine active coefficient

$$K_a = \frac{1 - \sin\phi}{1 + \sin\phi} = \frac{1 - 0.5}{1 + 0.5} = \frac{0.5}{1.5} = 0.333$$

(a) Pressure distribution and thrust

Component 1 — self weight of backfill (triangular): Pressure at base $= K_a\gamma H = 0.333 \times 18 \times 6.0 = 36.0\ \text{kPa}$.

$$P_{a1} = \tfrac12 K_a \gamma H^2 = \tfrac12 (0.333)(18)(6.0)^2 = 108.0\ \text{kN/m}$$

acting at $H/3 = 2.0\ \text{m}$ above the base.

Component 2 — surcharge (rectangular): Uniform pressure $= K_a q = 0.333 \times 20 = 6.67\ \text{kPa}$.

$$P_{a2} = K_a q H = 0.333 \times 20 \times 6.0 = 40.0\ \text{kN/m}$$

acting at $H/2 = 3.0\ \text{m}$ above the base.

Pressure diagram (kPa) at base:
  surcharge ->  6.67 (rectangle, full height)
  backfill  -> 36.00 (triangle, zero at top -> 36.0 at base)
  total at base = 42.67 kPa

Total active thrust

$$P_a = P_{a1} + P_{a2} = 108.0 + 40.0 = \textbf{148.0 kN/m}$$

Line of action (take moments about the base):

$$\bar{y} = \frac{P_{a1}(H/3) + P_{a2}(H/2)}{P_a} = \frac{108.0(2.0) + 40.0(3.0)}{148.0} = \frac{216 + 120}{148.0} = \frac{336}{148.0} = \textbf{2.27 m above base}$$

(b) Stability checks

In addition, overall (deep-seated) slope stability should be verified for tall walls on weak soils.

(a) Objectives, depth and spacing of exploration

Objectives: to determine the soil/rock stratification, the engineering properties (strength, compressibility, permeability), the groundwater table, and to detect problem soils (collapsible, expansive, liquefiable) so that a safe and economical foundation can be designed.

Depth of boreholes — explore to the depth at which the net imposed stress falls to about 10 % of the surface (footing) contact pressure (or 20 % of the in-situ effective overburden). Practical guides:

• Isolated footings: $\approx 1.5\,B$ to $2\,B$ below base.
• Raft of width $B$: $\approx 1.5\,B$.
• Pile foundations: at least 5 m (or 5 pile-diameters) below the anticipated toe, and through any soft layer to a firm stratum.

Spacing of boreholes depends on the variability of the subsoil and the importance/size of the structure — typically 15–30 m for buildings, closer where strata are erratic. A minimum of one borehole per significant structural unit; for large buildings a grid is used.

(b) Standard Penetration Test (SPT) — IS 2131 / ASTM D1586

Procedure: a split-spoon sampler is driven into the bottom of a clean borehole by a $63.5\ \text{kg}$ hammer falling $760\ \text{mm}$. The blows needed to drive each of three successive $150\ \text{mm}$ increments are recorded. The first $150\ \text{mm}$ (seating drive) is discarded; the sum of the blows for the second and third increments is the field value $N$.

Corrections applied:

1. Overburden correction — normalises $N$ to a reference effective stress (e.g. Liao–Whitman $C_N = \sqrt{95.76/\sigma'_v}$, capped near 2).
2. Dilatancy / water-table correction for fine saturated sands and silts below the water table when $N > 15$: $N_c = 15 + 0.5(N - 15)$.
3. Energy / equipment corrections — hammer efficiency to $N_{60}$, plus borehole-diameter, rod-length and sampler corrections.

The corrected value is used to estimate relative density, friction angle, and allowable bearing pressure of cohesionless soils.

(c) Sampling methods

1. Thin-walled (Shelby) tube sampler — pushed (not driven) into soft to medium clays; low area ratio ($<10\%$) gives nearly undisturbed samples for strength and consolidation testing.
2. Piston sampler — a stationary piston prevents soil from entering until pushed, giving high-quality undisturbed samples in very soft clays.

(The split-spoon SPT sampler itself yields only disturbed representative samples for classification/index tests.)

Given

$B = 2.5\ \text{m}$, $D_f = 1.2\ \text{m}$, $c = 0$, $\phi = 25^\circ$, $\gamma = 19\ \text{kN/m}^3$, $N_q = 10.66$, $N_\gamma = 6.77$. Water table at the footing base.

Water-table effect

• Surcharge term (soil above base): water table is at the base, so the surcharge uses bulk unit weight.

$$q = \gamma D_f = 19 \times 1.2 = 22.8\ \text{kPa}$$

• $\gamma$-term (soil below base, in the failure wedge): fully submerged, so use effective (buoyant) unit weight.

$$\gamma' = \gamma_{sat} - \gamma_w = 19 - 9.81 = 9.19\ \text{kN/m}^3$$

Ultimate bearing capacity (c = 0)

$$q_u = q N_q + 0.5\,\gamma' B N_\gamma$$ $$q_u = (22.8)(10.66) + 0.5(9.19)(2.5)(6.77)$$ $$q_u = 243.05 + 77.77 = 320.82\ \text{kPa}$$

Net ultimate bearing capacity

$$q_{nu} = q_u - \gamma D_f = 320.82 - 22.8 = \boxed{\textbf{298.0 kPa}}$$

Cantilever vs anchored sheet pile wall

Cantilever sheet pile wall — derives all of its support from the passive resistance of the soil in which it is embedded; it acts as a vertical cantilever fixed by the soil below the dredge line. Economical only for low retained heights (≤ ~6 m) because deflections and moments grow rapidly with height.

Anchored sheet pile wall — has an additional support: a tie rod connected near the top to an anchorage (deadman, anchor pile, or grout anchor). The anchor reduces the embedment depth, bending moment and deflection, making it suitable for greater heights.

  CANTILEVER                 ANCHORED
   ____                       __o====  anchor/tie rod
  |    | active                |    | active
  |    |---- dredge line       |    |---- dredge line
  | // | passive (front)       | // | passive
  | \\ | passive (back, lower) | \\ |
   pivot point near toe         lower moment

Assumed pressure distribution (cantilever, cohesionless soil)

The wall rotates about a point O a little above its toe. Above O the front (dredge) side mobilises passive pressure and the back side active. Below O the situation reverses — passive develops on the back face. The classical analysis replaces the curved net-pressure diagram below the pivot by a simplified linear net-pressure diagram, giving a characteristic distribution that is active down to the dredge line, then a net pressure that crosses zero and reverses near the toe.

Depth of embedment and maximum moment

1. Net pressure diagram is drawn using $K_a$ and $K_p$ (Rankine/Coulomb) for the soil above and below the dredge line.
2. Depth of embedment $D$: apply the two equilibrium conditions —
   • $\sum F_H = 0$ (net horizontal force = 0), and
   • $\sum M = 0$ about the base/toe. These yield a cubic (4th-degree) equation in $D$; solving gives the theoretical depth $D_0$. A safety factor of 20–40 % is added ($D = 1.2\text{–}1.4\,D_0$) for the design embedment.
3. Maximum bending moment occurs at the depth where the net shear force is zero; setting the area of the pressure diagram above that point equal to zero locates it, and the moment of the pressure diagram about that level gives $M_{max}$, which sizes the section modulus of the pile.

Well (caisson) foundation

A well foundation is a large-diameter, hollow, monolithic deep foundation that is sunk to the required founding level by excavating ('grabbing') the soil from inside while the well sinks under its own weight (and kentledge). It is the standard foundation for major river bridges, because it can be taken below the maximum scour depth, resists large horizontal and vertical loads, and avoids the deep open excavation/dewatering needed for spread footings.

Components (sketch in text)

   ___________________   <- top plug / well cap
  | |             | |
  | |   STEINING  | |    <- thick masonry/RCC wall (load transfer + weight)
  | |  (dredge   | |
  | |   hole)    | |
  |__\         /__|       <- WELL CURB (transition, RCC)
      \ cutting/          <- CUTTING EDGE (sharp steel-shod tip)
       \ edge/
   ====BOTTOM PLUG (concrete)====

Functions:

• Cutting edge — the sharp, steel-shod lowest part that cuts/penetrates the soil and offers least resistance as the well sinks.
• Well curb — the wedge-shaped RCC ring above the cutting edge that transmits the steining load to the cutting edge and houses it.
• Steining — the main thick wall of the well; provides the self-weight needed for sinking and transfers superstructure loads to the founding stratum; its thickness governs stability.
• Bottom plug — concrete placed (often by tremie) at the base after the founding level is reached; it seals the bottom and distributes the base load like an inverted dome.

Difficulties during sinking and remedy

1. Tilt — the well leans from the vertical (uneven soil resistance). Remedy: apply eccentric kentledge / eccentric grabbing on the higher side, or use water jetting on the high side.
2. Shift — horizontal displacement of the well from its design position (often accompanies tilt). Remedy: combined with tilt correction, apply lateral pulling/strutting and eccentric loading to bring it back; provide a small permissible allowance in design.

Ground improvement

Ground improvement is the deliberate modification of the engineering behaviour of in-situ soil — increasing its strength and bearing capacity, reducing its compressibility/settlement, controlling permeability, or mitigating liquefaction — so that a site otherwise unsuitable for a given structure can be used economically.

Three techniques

1. Vibro-compaction / Vibroflotation

• Best suited: clean, loose cohesionless sands (low fines content).
• Mechanism: a vibrating probe (vibroflot) is inserted; the vibration rearranges the loose sand grains into a denser packing, often with backfill added at the surface. This increases relative density, friction angle, bearing capacity and reduces liquefaction potential.

2. Preloading with Prefabricated Vertical Drains (PVDs / sand drains)

• Best suited: soft, saturated, compressible clays and silts.
• Mechanism: a surcharge (preload) is placed to consolidate the clay before construction. Vertical drains shorten the drainage path so that pore water dissipates radially and quickly, accelerating primary consolidation; settlement that would have occurred under the structure happens in advance, and the clay gains shear strength.

3. Stone columns (granular piles)

• Best suited: soft clays and loose silty soils of medium strength.
• Mechanism: compacted columns of crushed stone are installed on a grid. They act as stiff, free-draining inclusions that carry part of the load, reduce settlement (load sharing / stress concentration), and provide radial drainage to speed consolidation.

(Other valid techniques: dynamic compaction, grouting/jet grouting, deep soil mixing, reinforced earth, lime/cement stabilisation.)

(a) Contact pressures

$$q_A = \frac{400}{1.5 \times 1.5} = \frac{400}{2.25} = 177.8\ \text{kPa}$$ $$q_B = \frac{1100}{2.5 \times 2.5} = \frac{1100}{6.25} = 176.0\ \text{kPa}$$

The two are almost equal, with footing A imposing the marginally higher contact pressure ($177.8$ vs $176.0\ \text{kPa}$).

(b) Which settles more, and why

Even though A has the slightly higher (essentially equal) contact pressure, footing B will settle more. For a given pressure on sand, the depth of the significant pressure bulb is proportional to the footing width $B$. The larger footing B stresses a much greater volume/depth of compressible sand, so its settlement is larger. This is captured by classical relations such as

$$\frac{S_2}{S_1} = \left[\frac{B_2\,(B_1 + 0.3)}{B_1\,(B_2 + 0.3)}\right]^2$$

(for footings on sand at equal pressure), which gives $S_B > S_A$. Hence wider footings settle more at equal pressure.

Total vs differential settlement

• Total settlement is the absolute downward movement of a single foundation element.
• Differential settlement is the difference in settlement between two points (e.g. between footings A and B, or across a raft).

Why differential settlement is more critical: a structure can usually tolerate large uniform total settlement (it simply moves down as a whole), but differential settlement induces angular distortion, additional bending moments, cracking of walls/finishes, jamming of doors, and in severe cases structural failure. Design therefore limits angular distortion (commonly $\le 1/500$ for framed buildings).

Negative skin friction (NSF / downdrag)

Cause: when the soil surrounding a pile settles more than the pile itself, the soil drags downward along the shaft. This typically occurs where a recently placed fill, or a soft compressible layer undergoing consolidation (e.g. due to surcharge or lowering of the water table) settles around an end-bearing pile.

Where it acts: along the upper part of the shaft — over the depth of the settling (consolidating) layer, down to the neutral point where the relative movement between pile and soil reverses.

Effect on capacity: instead of helping to support the load, this friction acts downward as an additional load (drag load) on the pile. It must be subtracted from the available capacity (or added to the structural/working load). Hence:

$$Q_{allowable,\ net} = \frac{Q_{ultimate}}{FS} - Q_{NSF}$$

NSF can also induce extra settlement and additional axial stress in the pile.

Drag load calculation

Given: $D = 0.5\ \text{m}$, fill thickness $L = 7\ \text{m}$, unit (adhesion) skin friction $f_n = \alpha c_u = 12\ \text{kPa}$.

Shaft surface area in the fill:

$$A_s = \pi D L = \pi \times 0.5 \times 7 = 10.996\ \text{m}^2$$

Drag load (negative skin friction force):

$$Q_{NSF} = f_n \times A_s = 12 \times 10.996 = 131.9\ \text{kN}$$ $$\boxed{Q_{NSF} \approx \textbf{132 kN}}$$

This $132\ \text{kN}$ acts downward and must be allowed for as an extra load when checking the pile's safe capacity.