BE Civil Engineering (IOE, TU) Foundation Engineering (IOE, CE 701) Question Paper 2079 Nepal

Objectives of Subsurface Exploration

• Determine the nature, sequence and thickness of soil/rock strata.
• Obtain disturbed and undisturbed samples for index and engineering tests.
• Locate the groundwater table and assess seepage/permeability.
• Determine strength and compressibility for bearing capacity and settlement.
• Detect problem soils (collapsible, expansive, organic, liquefiable).

Planning

Depth of boring should extend to where the net stress increase from the foundation is < 10% of the foundation pressure (roughly $1.5B$–$2B$ below a footing, or to bedrock). Spacing of boreholes is typically 10–30 m for buildings; at least one borehole per major column line for important structures.

SPT and Corrections

The SPT drives a split-spoon sampler $450\,\text{mm}$ using a $63.5\,\text{kg}$ hammer falling $760\,\text{mm}$; the blows for the last $300\,\text{mm}$ give $N$.

Corrections applied:

1. Overburden $C_N$
2. Hammer energy $C_E$ (to 60%)
3. Borehole diameter $C_B$
4. Rod length $C_R$
5. Sampler liner $C_S$
6. Dilatancy correction for fine saturated sand/silt.

$$N_{60}=N\cdot C_E\,C_B\,C_R\,C_S, \qquad (N_1)_{60}=C_N\,N_{60}$$

Numerical Solution

Step 1 — Effective overburden at 8 m (WT at surface):

$$\sigma'_v=(\gamma_{sat}-\gamma_w)\,z=(18-9.81)\times 8=8.19\times 8=65.52\,\text{kPa}$$

Step 2 — Energy-corrected N (all equipment factors = 1):

$$N_{60}=26\times1.0\times1.0\times1.0\times1.0=26$$

Step 3 — Dilatancy correction (fine saturated sand, $N_{60}>15$):

$$N'=15+0.5\,(N_{60}-15)=15+0.5(26-15)=15+5.5=20.5$$

Step 4 — Overburden correction (Liao & Whitman):

$$C_N=\sqrt{\frac{95.76}{65.52}}=\sqrt{1.4616}=1.209\;(\le 1.7\;\text{OK})$$

Step 5 — Corrected blow count:

$$(N_1)_{60}=C_N\times N'=1.209\times 20.5=24.8$$

Final answer: $(N_1)_{60}\approx \mathbf{25}$ blows/300 mm.

(Note: the dilatancy correction is applied to the energy-corrected value before overburden correction, per IS 2131 practice.)

Assumptions of Terzaghi's Theory

• Soil is homogeneous, isotropic and semi-infinite.
• The footing is a strip (plane-strain) of infinite length; base is rough.
• Failure is general shear; the failure surface is the Prandtl mechanism (active wedge, radial shear zone, passive zone).
• The soil above the footing base acts as a surcharge $q=\gamma D_f$ only (its shear strength is neglected).
• Loading is vertical, concentric; $D_f \le B$ (shallow foundation).

General (strip) Equation

$$q_u = cN_c + qN_q + 0.5\,\gamma B N_\gamma$$

Numerical Solution (Square footing)

For a square footing the modified equation is:

$$q_u = 1.3\,cN_c + qN_q + 0.4\,\gamma B N_\gamma$$

where the $\gamma$-term coefficient is $0.8\times0.5 = 0.4$.

Step 1 — Surcharge:

$$q=\gamma D_f = 19\times1.5 = 28.5\,\text{kPa}$$

Step 2 — Ultimate (gross) bearing capacity:

$$q_u = 1.3(15)(25.13) + (28.5)(12.72) + 0.4(19)(2.5)(8.34)$$

• $c$-term: $1.3\times15\times25.13 = 490.04\,\text{kPa}$
• $q$-term: $28.5\times12.72 = 362.52\,\text{kPa}$
• $\gamma$-term: $0.4\times19\times2.5\times8.34 = 158.46\,\text{kPa}$

$$q_u = 490.04+362.52+158.46 = 1011.02\,\text{kPa}$$

Step 3 — Net ultimate bearing capacity:

$$q_{nu}=q_u-q = 1011.02-28.5 = 982.52\,\text{kPa}$$

Step 4 — Net safe bearing capacity (FS = 3):

$$q_{ns}=\frac{q_{nu}}{FS}=\frac{982.52}{3}=327.5\,\text{kPa}$$

Final answer: net safe bearing capacity $\approx \mathbf{327.5\,kPa}$.

Types of Settlement

• Immediate (elastic) settlement $S_i$: occurs rapidly on loading, at constant volume in saturated clay; computed from elasticity theory.
• Primary consolidation settlement $S_c$: time-dependent volume change as excess pore water pressure dissipates; dominant in saturated clays.
• Secondary settlement $S_s$: creep of the soil skeleton at constant effective stress after primary consolidation ends; governed by the secondary compression index $C_\alpha$.

Numerical Solution

Step 1 — Initial effective overburden at clay mid-depth (6 m, WT at surface): From 0–4 m sand, 4–6 m clay (mid-depth is at 6 m below ground since footing is at surface):

$$\sigma'_0=(18-9.81)(4)+(19-9.81)(2)$$ $$=8.19\times4 + 9.19\times2 = 32.76 + 18.38 = 51.14\,\text{kPa}$$

Step 2 — Stress increase (given): $\Delta\sigma = 28\,\text{kPa}$

$$\sigma'_0+\Delta\sigma = 51.14+28 = 79.14\,\text{kPa}$$

Step 3 — Primary consolidation (normally consolidated):

$$S_c=\frac{C_c\,H}{1+e_0}\log_{10}\!\left(\frac{\sigma'_0+\Delta\sigma}{\sigma'_0}\right)$$ $$=\frac{0.30\times4}{1+0.90}\log_{10}\!\left(\frac{79.14}{51.14}\right)$$

• $\dfrac{0.30\times4}{1.90}=\dfrac{1.2}{1.90}=0.6316\,\text{m}$
• $\dfrac{79.14}{51.14}=1.5475;\quad \log_{10}(1.5475)=0.18965$

$$S_c = 0.6316\times0.18965 = 0.1198\,\text{m}$$

Final answer: primary consolidation settlement $\approx \mathbf{0.120\,m = 120\,mm}$.

(For higher accuracy the layer could be subdivided, but a single mid-depth computation is acceptable for a 4 m layer.)

Load Transfer Mechanism

A pile transfers structural load to the soil by two mechanisms: skin (shaft) friction along the embedded length, and end (base/point) bearing at the tip. As load is applied, the upper shaft mobilises friction first; with increasing settlement the base resistance develops. Total capacity:

$$Q_u = Q_s + Q_b$$

Classification by Load Transfer

• Friction (floating) piles: load carried mainly by shaft friction (soft/medium clays).
• End-bearing piles: load transferred mainly to a firm stratum at the tip.
• Combined friction + end-bearing piles: both contribute significantly.

Numerical Solution ($\alpha$-method)

Geometry:

• Perimeter $p=\pi d=\pi\times0.5=1.5708\,\text{m}$
• Base area $A_b=\dfrac{\pi}{4}d^2=\dfrac{\pi}{4}(0.5)^2=0.19635\,\text{m}^2$

Step 1 — Shaft resistance:

$$Q_s=\alpha\,c_u\,(p\,L)=0.55\times60\times(1.5708\times12)$$ $$=0.55\times60\times18.850 = 33\times18.850 = 622.04\,\text{kN}$$

Step 2 — Base resistance:

$$Q_b=c_u\,N_c\,A_b=60\times9\times0.19635 = 540\times0.19635 = 106.03\,\text{kN}$$

Step 3 — Ultimate capacity:

$$Q_u=Q_s+Q_b=622.04+106.03 = 728.07\,\text{kN}$$

Step 4 — Allowable load (FS = 2.5):

$$Q_a=\frac{Q_u}{FS}=\frac{728.07}{2.5}=291.2\,\text{kN}$$

Final answer: allowable pile load $\approx \mathbf{291\,kN}$.

Modes of Failure Checked

1. Sliding along the base.
2. Overturning about the toe.
3. Bearing capacity failure of the foundation soil (and limiting base pressure / no tension).
4. Overall (deep-seated) slope failure.

Numerical Solution

Step 1 — Rankine active coefficient:

$$K_a=\frac{1-\sin\phi}{1+\sin\phi}=\frac{1-0.5}{1+0.5}=\frac{0.5}{1.5}=0.3333$$

Step 2 — Active thrust (per metre run):

$$P_a=\tfrac{1}{2}K_a\gamma H^2=0.5\times0.3333\times18\times6^2$$ $$=0.5\times0.3333\times18\times36 = 108.0\,\text{kN/m}$$

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

Step 3 — Sliding check:

$$F_{slide}=\frac{\mu W}{P_a}=\frac{0.5\times420}{108.0}=\frac{210}{108.0}=1.94$$

FS$_{slide}=1.94 \ge 1.5$ → Safe in sliding.

Step 4 — Overturning check (moments about toe):

• Resisting moment $M_R = W\times \bar{x} = 420\times2.0 = 840\,\text{kN·m}$
• Overturning moment $M_O = P_a\times(H/3) = 108.0\times2.0 = 216\,\text{kN·m}$

$$F_{ot}=\frac{M_R}{M_O}=\frac{840}{216}=3.89$$

FS$_{ot}=3.89 \ge 2.0$ → Safe against overturning.

Final answer: $P_a=\mathbf{108\,kN/m}$; FS$_{sliding}=\mathbf{1.94}$ (safe); FS$_{overturning}=\mathbf{3.89}$ (safe).

Comparison (any three)

Numerical Solution ($\phi=0$, undrained)

Step 1 — Surcharge: $q=\gamma D_f = 18\times1 = 18\,\text{kPa}$

Step 2 — Gross ultimate bearing capacity:

$$q_u=c_uN_c+qN_q+0.5\gamma BN_\gamma = 40(5.14)+18(1.0)+0$$ $$=205.6+18 = 223.6\,\text{kPa}$$

Step 3 — Net ultimate bearing capacity:

$$q_{nu}=q_u-q = 223.6-18 = 205.6\,\text{kPa}$$

Final answer: net ultimate bearing capacity $=\mathbf{205.6\,kPa}$ (equals $c_uN_c$, as expected for the $\phi=0$ case).

Pressure Distribution (cantilever sheet pile, cohesionless soil)

        Retained side        Dredge side
  ____   active (Ka)
 |    |\
 |    | \  active increases with depth
 |    |  \____ dredge line
 |    |  /|   below this, net pressure reverses:
 |    | / |   passive on front (left) + active behind
 |    |/  |   near the toe pressure reverses again
  ----   pivot point O

Above the dredge line, active pressure acts behind the wall and increases with depth. Below the dredge line, the wall tends to rotate about a point near its base: passive resistance develops in front and active behind down to the pivot, then the senses reverse near the toe.

Principle for Embedment Depth

The required depth is found from statics: $\sum F_H = 0$ and $\sum M = 0$ about the toe (or about the assumed pivot), using the net pressure diagram. The driving (active) moment is balanced by the resisting (passive) moment with a factor of safety (typically by reducing $K_p$ or increasing computed depth by ~20%).

Numerical Solution

Step 1 — Earth pressure coefficients:

$$K_a=\frac{1-\sin32^\circ}{1+\sin32^\circ}=\frac{1-0.5299}{1+0.5299}=\frac{0.4701}{1.5299}=0.307$$ $$K_p=\frac{1}{K_a}=\frac{1+\sin32^\circ}{1-\sin32^\circ}=\frac{1.5299}{0.4701}=3.255$$

Step 2 — Active thrust on retained height (H = 4 m):

$$P_a=\tfrac12 K_a\gamma H^2 = 0.5\times0.307\times17\times4^2$$ $$=0.5\times0.307\times17\times16 = 41.75\,\text{kN/m}$$

Acting at $H/3 = 1.33\,\text{m}$ above the dredge line.

Final answer: $K_a=\mathbf{0.307}$, $K_p=\mathbf{3.255}$, active thrust $P_a=\mathbf{41.75\,kN/m}$.

Five Ground Improvement Techniques

1. Compaction (dynamic/vibro/rolling): densifies loose granular soils; best for sands and gravels. Reduces voids, increases strength and reduces settlement.
2. Preloading with vertical (sand/wick) drains: accelerates consolidation of soft saturated clays before construction.
3. Stone columns (vibro-replacement): install compacted granular columns in soft clays/silts; provide reinforcement, drainage and increased bearing.
4. Grouting (permeation/compaction/jet): injection of cement or chemical grout to fill voids, reduce permeability, increase strength; suited to fissured rock, sands, fills.
5. Soil stabilisation with lime/cement: chemical modification of clayey/silty soils to reduce plasticity and increase strength; used for subgrades and expansive soils.

Sand Drains and Acceleration of Consolidation

Sand drains provide a short horizontal drainage path so that excess pore water can flow radially to the drains instead of only vertically over the full clay thickness. Because drainage path length is drastically reduced, and consolidation time varies with the square of the drainage path length ($t \propto d^2$), dissipation of pore pressure (and hence settlement) occurs much faster.

The primary controlling parameter is the drain spacing $s$ (which sets the diameter of influence $d_e$ of each drain): smaller spacing → shorter radial drainage path → faster consolidation. (The coefficient of consolidation for radial flow $c_h$ and the time factor $T_r$ also govern the rate, but spacing is the main design lever.)

Key relation: $U = 1-(1-U_v)(1-U_r)$ — combined vertical + radial degree of consolidation.

Well Foundation

A well foundation is a large-diameter, hollow, box-type deep foundation sunk into the ground (often through water and soil to a firm stratum). It is widely used for bridge piers and abutments across rivers because it can carry heavy vertical and lateral loads, resist scour, and be sunk to large depths.

Components

• Cutting edge — sharp lower edge that eases sinking.
• Well curb — wedge-shaped RCC ring above the cutting edge.
• Steining — the main thick wall of the well (provides weight for sinking and carries load).
• Bottom plug — concrete seal at the base after sinking.
• Dredge hole — central opening through which soil is excavated.
• Top plug and well cap — close the top and transfer pier load to the steining.

Major Forces in Design

• Self-weight (dead load) and superimposed live loads.
• Lateral forces: water current pressure, wind, earthquake, braking/tractive forces.
• Earth pressure and buoyancy.
• Resistance: base reaction and side (passive) soil resistance.

Tilt and Shift

• Tilt: the well sinks out of vertical (rotation). Corrected by eccentric excavation (dredging more on the higher side), eccentric loading/kentledge, water jetting on the high side, or pulling with hooks.
• Shift: horizontal displacement of the well from its design position, usually accompanying tilt. Controlled by careful guiding during early sinking and corrected together with the tilt. Tilt should generally be kept within 1 in 60 and shift within about 1% of depth sunk.

Pile Group Efficiency

Group efficiency $E_g$ is the ratio of the actual capacity of the pile group to the sum of the individual capacities of the same number of isolated piles:

$$E_g=\frac{Q_{g(\text{group})}}{n\,Q_{u(\text{single})}}$$

Converse–Labarre Formula

$$E_g=1-\frac{\theta}{90}\left[\frac{(m-1)n+(n-1)m}{m\,n}\right]$$

where $\theta=\tan^{-1}(d/s)$ in degrees, $m=$ rows, $n=$ piles per row.

Numerical Solution

Step 1 — Angle $\theta$:

$$\theta=\tan^{-1}\!\left(\frac{d}{s}\right)=\tan^{-1}\!\left(\frac{0.4}{1.2}\right)=\tan^{-1}(0.3333)=18.435^\circ$$

Step 2 — Bracket term ($m=n=3$):

$$\frac{(m-1)n+(n-1)m}{mn}=\frac{(2)(3)+(2)(3)}{9}=\frac{6+6}{9}=\frac{12}{9}=1.3333$$

Step 3 — Efficiency:

$$E_g=1-\frac{18.435}{90}\times1.3333 = 1-0.20483\times1.3333$$ $$=1-0.2731 = 0.7269$$

Final answer: group efficiency $E_g \approx \mathbf{0.727 = 72.7\%}$.

Shallow vs Deep Foundations

When a Raft Foundation is Preferred

• The soil has low bearing capacity and individual footings would need very large areas (footings would cover > ~50% of the plan area).
• To reduce/control differential settlement by tying the whole structure together on one slab.
• Where the soil is non-uniform or contains soft pockets/lenses.
• For structures with basements below the water table (raft acts as a water-resisting slab) and to use the floating (compensated) foundation principle.
• For heavily loaded structures such as silos, tall buildings and water tanks on compressible soil.