BE Civil Engineering (IOE, TU) Foundation Engineering (IOE, CE 701) Question Paper 2080 Nepal

Given: Square footing $B = 2.5\ \text{m}$, $D_f = 1.5\ \text{m}$, $c = 18\ \text{kPa}$, $\phi = 25^\circ$, $\gamma = 18.5\ \text{kN/m}^3$, FOS $= 3.0$.

Step 1 — Terzaghi equation for a square footing (general shear):

$$q_u = 1.3\,c\,N_c + \gamma D_f N_q + 0.4\,\gamma B N_\gamma$$

Step 2 — Substitute values:

• Cohesion term: $1.3 \times 18 \times 25.13 = 587.9\ \text{kPa}$
• Surcharge term: $18.5 \times 1.5 \times 12.72 = 352.98\ \text{kPa}$
• Self-weight term: $0.4 \times 18.5 \times 2.5 \times 8.34 = 154.29\ \text{kPa}$

$$q_u = 587.90 + 352.98 + 154.29 = 1095.17\ \text{kPa}$$

Step 3 — Net ultimate bearing capacity: The overburden (surcharge) pressure at founding level is $q = \gamma D_f = 18.5 \times 1.5 = 27.75\ \text{kPa}$.

$$q_{nu} = q_u - q = 1095.17 - 27.75 = 1067.42\ \text{kPa}$$

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

$$q_{ns} = \frac{q_{nu}}{3} = \frac{1067.42}{3} = 355.81\ \text{kPa}$$

Step 5 — Gross safe bearing capacity:

$$q_s = q_{ns} + q = 355.81 + 27.75 = 383.56\ \text{kPa}$$

Step 6 — Maximum safe gross column load: Footing area $A = 2.5 \times 2.5 = 6.25\ \text{m}^2$.

$$Q_{safe} = q_s \times A = 383.56 \times 6.25 = 2397.3\ \text{kN}$$

Final Answers:

• Ultimate bearing capacity $q_u \approx \mathbf{1095.2\ kPa}$
• Net safe bearing capacity $q_{ns} \approx \mathbf{355.8\ kPa}$
• Maximum safe gross column load $Q_{safe} \approx \mathbf{2397\ kN}$

Step 1 — Initial effective overburden stress at centre of clay layer. The water table is at the base of the raft, so all soil below is submerged. Centre of clay is at depth $z = 3\ \text{m (sand)} + 2\ \text{m (half of clay)} = 5\ \text{m}$ below raft base.

• Sand (3 m, submerged): $(20 - 9.81) \times 3 = 10.19 \times 3 = 30.57\ \text{kPa}$
• Clay (2 m to mid-layer, submerged): $(19 - 9.81) \times 2 = 9.19 \times 2 = 18.38\ \text{kPa}$

$$\sigma'_0 = 30.57 + 18.38 = 48.95\ \text{kPa}$$

Step 2 — Stress increase by 2:1 method at $z = 5\ \text{m}$.

$$\Delta\sigma = \frac{q \cdot B \cdot L}{(B+z)(L+z)} = \frac{90 \times 6 \times 10}{(6+5)(10+5)} = \frac{5400}{11 \times 15} = \frac{5400}{165} = 32.73\ \text{kPa}$$

Step 3 — Primary consolidation settlement (normally consolidated):

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

where $H = 4\ \text{m} = 4000\ \text{mm}$.

$$\frac{\sigma'_0 + \Delta\sigma}{\sigma'_0} = \frac{48.95 + 32.73}{48.95} = \frac{81.68}{48.95} = 1.6686$$ $$\log_{10}(1.6686) = 0.2223$$ $$S_c = \frac{0.32 \times 4000}{1+0.85} \times 0.2223 = \frac{1280}{1.85} \times 0.2223 = 691.89 \times 0.2223 = 153.8\ \text{mm}$$

Final Answers:

• Stress increase at clay centre $\Delta\sigma \approx \mathbf{32.7\ kPa}$
• Primary consolidation settlement $S_c \approx \mathbf{153.8\ mm \approx 0.154\ m}$

Given: $d = 0.6\ \text{m}$, $L = 15\ \text{m}$, shaft $c_u = 60\ \text{kPa}$, base $c_u = 75\ \text{kPa}$, $\alpha = 0.45$, $N_c = 9$, FOS $= 2.5$.

Step 1 — Geometry.

• Base area: $A_b = \dfrac{\pi}{4}d^2 = \dfrac{\pi}{4}(0.6)^2 = 0.2827\ \text{m}^2$
• Shaft surface area: $A_s = \pi d L = \pi \times 0.6 \times 15 = 28.274\ \text{m}^2$

Step 2 — Ultimate skin friction (α-method):

$$Q_s = \alpha\,c_u\,A_s = 0.45 \times 60 \times 28.274 = 763.4\ \text{kN}$$

Step 3 — Ultimate end bearing:

$$Q_b = c_u\,N_c\,A_b = 75 \times 9 \times 0.2827 = 190.8\ \text{kN}$$

Step 4 — Ultimate pile capacity:

$$Q_u = Q_s + Q_b = 763.4 + 190.8 = 954.2\ \text{kN}$$

Step 5 — Safe (allowable) load:

$$Q_{safe} = \frac{Q_u}{2.5} = \frac{954.2}{2.5} = 381.7\ \text{kN}$$

Final Answers:

• Skin friction $Q_s \approx \mathbf{763.4\ kN}$
• End bearing $Q_b \approx \mathbf{190.8\ kN}$
• Safe pile load $Q_{safe} \approx \mathbf{381.7\ kN}$

Step 1 — Rankine active earth pressure coefficient:

$$K_a = \frac{1-\sin\phi}{1+\sin\phi} = \tan^2\!\left(45^\circ - \frac{\phi}{2}\right)$$

For $\phi = 32^\circ$: $\sin 32^\circ = 0.5299$.

$$K_a = \frac{1-0.5299}{1+0.5299} = \frac{0.4701}{1.5299} = 0.3073$$

Step 2 — Pressure components.

Due to soil self-weight (triangular): at base ($z = 5\ \text{m}$),

$$p_{soil} = K_a\,\gamma\,H = 0.3073 \times 17.5 \times 5 = 26.89\ \text{kPa}$$

Thrust: $P_1 = \tfrac{1}{2} \times 26.89 \times 5 = 67.23\ \text{kN/m}$, acting at $H/3 = 1.667\ \text{m}$ above base.

Due to surcharge (rectangular): uniform with depth,

$$p_{surcharge} = K_a\,q = 0.3073 \times 12 = 3.688\ \text{kPa}$$

Thrust: $P_2 = 3.688 \times 5 = 18.44\ \text{kN/m}$, acting at $H/2 = 2.5\ \text{m}$ above base.

Step 3 — Total active thrust:

$$P_a = P_1 + P_2 = 67.23 + 18.44 = 85.67\ \text{kN/m}$$

Step 4 — Location of resultant (take moments about base):

$$\bar{y} = \frac{P_1\,y_1 + P_2\,y_2}{P_a} = \frac{67.23 \times 1.667 + 18.44 \times 2.5}{85.67}$$ $$= \frac{112.07 + 46.10}{85.67} = \frac{158.17}{85.67} = 1.846\ \text{m}$$

Pressure diagram (at base):

  surcharge   soil (triangular)
  3.69 kPa  +  26.89 kPa  = 30.58 kPa at base
  |||||||||      /|
  |||||||||     / |
  |||||||||    /  |
  (uniform)   (triangle)

Final Answers:

• $K_a \approx 0.307$; total active thrust $P_a \approx \mathbf{85.7\ kN/m}$
• Resultant acts at $\bar{y} \approx \mathbf{1.85\ m}$ above the base.

(a) Subsurface site investigation.

Purpose: to determine the nature, sequence and thickness of soil/rock strata; groundwater conditions; engineering properties (strength, compressibility, permeability) for safe and economical foundation design; and to identify hazards (collapsible/expansive soils, liquefaction potential).

Stages:

1. Reconnaissance / desk study — review geological maps, aerial photos, existing records, site visit.
2. Preliminary (exploratory) investigation — a few boreholes/test pits, simple field tests to outline the profile.
3. Detailed investigation — full grid of boreholes, in-situ tests (SPT, CPT, vane shear), undisturbed sampling, groundwater monitoring.
4. Laboratory testing & reporting — classification, consolidation, shear tests; interpretation and geotechnical report.

Depth of boreholes: generally taken to a depth where the net stress increase due to the foundation is small (commonly $\le 10\%$ of net applied pressure, i.e. about $1.5\,B$ to $2\,B$ below the footing for isolated footings, deeper for rafts) or to firm bearing stratum/bedrock.

Spacing of boreholes: depends on uniformity of strata and structure importance — typically $10$–$30\ \text{m}$ for buildings; closer where strata are erratic.

(b) SPT corrections.

Step 1 — Overburden correction (Liao & Whitman):

$$C_N = \sqrt{\frac{95.76}{\sigma'_v}} = \sqrt{\frac{95.76}{120}} = \sqrt{0.798} = 0.8933$$ $$N' = C_N \times N_{field} = 0.8933 \times 28 = 25.01 \approx 25$$

Step 2 — Dilatancy (water-table) correction. For fine/silty saturated sand, the correction applies only when $N' > 15$:

$$N_{corr} = 15 + \tfrac{1}{2}(N' - 15) = 15 + 0.5(25 - 15) = 15 + 5 = 20$$

Final Answer: Corrected SPT value $N \approx \mathbf{20}$.

Terzaghi strip footing (c = 0):

$$q_u = \gamma D_f N_q + 0.5\,\gamma' B N_\gamma$$

Step 1 — Surcharge term. Water table is at the base, so soil above founding level is moist:

$$\gamma D_f N_q = 18 \times 1.2 \times 22.46 = 485.14\ \text{kPa}$$

Step 2 — Self-weight term. Soil below the base is submerged, so use effective (submerged) unit weight:

$$\gamma' = \gamma_{sat} - \gamma_w = 20 - 9.81 = 10.19\ \text{kN/m}^3$$ $$0.5\,\gamma' B N_\gamma = 0.5 \times 10.19 \times 2.0 \times 19.13 = 194.94\ \text{kPa}$$

Step 3 — Ultimate bearing capacity:

$$q_u = 485.14 + 194.94 = 680.08\ \text{kPa}$$

Step 4 — Net ultimate bearing capacity. Overburden at base $q = \gamma D_f = 18 \times 1.2 = 21.6\ \text{kPa}$:

$$q_{nu} = q_u - q = 680.08 - 21.6 = 658.5\ \text{kPa}$$

Final Answer: Net ultimate bearing capacity $q_{nu} \approx \mathbf{658.5\ kPa}$.

Step 1 — Parameters. $m = 3$ rows, $n = 3$ columns, $d = 0.4\ \text{m}$, $s = 1.2\ \text{m}$.

Step 2 — 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 3 — Converse-Labarre group efficiency:

$$\eta = 1 - \frac{\theta}{90}\left[\frac{(m-1)n + (n-1)m}{m\,n}\right]$$ $$= 1 - \frac{18.435}{90}\left[\frac{(2)(3) + (2)(3)}{9}\right] = 1 - 0.2048 \times \frac{12}{9}$$ $$= 1 - 0.2048 \times 1.3333 = 1 - 0.2731 = 0.7269 \approx 72.7\%$$

Step 4 — Safe group capacity:

$$Q_g = \eta \times n_{piles} \times Q_{single} = 0.7269 \times 9 \times 250 = 1635.5\ \text{kN}$$

Final Answers:

• Group efficiency $\eta \approx \mathbf{72.7\%}$
• Safe group capacity $Q_g \approx \mathbf{1635\ kN}$

Cantilever sheet pile wall — principle. A cantilever sheet pile derives its stability solely from the passive resistance of the soil mobilised over the embedded depth below the dredge (excavation) line. The wall behaves like a vertical cantilever fixed by the soil. Under the active pressure from the retained side, the wall tends to rotate about a point near its toe; soil resists this rotation through passive pressure acting on alternate faces above and below the rotation point.

Pressure distribution and rotation point. The wall rotates about a point O located a short distance above the toe.

• Above the dredge line: active pressure on the back.
• Below the dredge line and above O: net pressure = passive (front) − active (back).
• Below O: the pressures reverse — passive acts on the back face, giving a sharp reversal near the toe.

  Retained side                Dredge line
  active (triangular)  -----------------
        \|                  | net passive
         \|                 |  (front)
          \|        O  -----+----  point of rotation
           |       reversed |
           |  passive(back) | (toe)

(a) Depth of embedment: found from equilibrium of horizontal forces and moments (sum of horizontal forces = 0 and sum of moments about the toe / rotation point = 0). The theoretical depth $d$ is then increased by about $20\text{–}40\%$ (factor $\approx 1.2$–$1.4$, or design $K$ values with FOS) for safety.

(b) Maximum bending moment: occurs at the depth where the shear force is zero (net lateral pressure changes sign). It is obtained by taking moments of all forces above that section.

Contrast with anchored sheet pile wall. An anchored wall has an additional support — an anchor/tie rod near the top. This reduces the required embedment depth, lowers the maximum bending moment, and makes the wall economical for greater retained heights. It is analysed by the free-earth or fixed-earth support methods, whereas a cantilever wall relies only on toe fixity.

(a) Well (caisson) foundation. A well foundation is a large-diameter, hollow, monolithic deep foundation that is sunk to the required depth by excavating the soil from within. It is widely used for bridge piers and abutments across rivers because it can carry very large vertical and lateral (scour, water current, impact) loads, can be founded below the deepest scour level, and provides a large bearing area. Preferred over piles where: heavy lateral loads exist, deep scour is expected, boulders/obstructions make pile driving difficult, or a massive, rigid foundation is required.

(b) Main components of a well foundation:

1. Cutting edge — sharp steel-shod edge at the bottom that facilitates sinking by cutting into the soil.
2. Well curb — wedge-shaped RCC ring above the cutting edge transferring load to the cutting edge.
3. Steining — the main vertical wall (RCC/brick) of the well; provides weight for sinking and structural strength.
4. Bottom plug — concrete seal at the base after sinking, transferring load to the founding stratum.
5. Sand filling — fills the dredge hole to add stability and reduce stresses.
6. Top plug — concrete seal over the sand fill.
7. Well cap — RCC slab on top transferring pier load to the steining.

(c) Tilts and shifts during sinking — problems & remedies:

1. Unequal sinking causing tilt → apply eccentric loading / kentledge on the higher side to correct.
2. Horizontal shift from vertical line → excavate more on the side toward which the well must move (dredge unevenly).
3. Tilt due to a hard stratum or obstruction on one side → dewater and remove obstruction, or push with water jetting/struts.
4. Excessive friction holding the well → apply additional kentledge or water jetting along the outer steining to reduce skin friction and resume sinking.

Other remedial aids: hooking/strutting, pulling with steel wire ropes, and using a heavier curb.

Three ground improvement techniques for loose granular soils:

1. Vibro-compaction (vibroflotation): A vibrating probe is inserted to rearrange and densify loose clean sands/gravels by vibration (sometimes with water jetting), reducing void ratio and increasing relative density and bearing capacity. Best for free-draining, low-fines sands.

2. Dynamic compaction: A heavy tamper (10–40 t) is repeatedly dropped from a height onto the surface, sending shock waves that densify loose sands and granular fills to moderate depths. Suited to large open areas with granular soils.

3. Stone columns (vibro-replacement): Columns of compacted gravel/stone are installed to reinforce, drain and densify the surrounding ground; effective in loose silty sands and soft soils with some fines, improving bearing capacity and reducing settlement, and accelerating drainage.

(Other valid options: compaction grouting, blasting for deep loose sands, preloading with vertical drains.)

Shallow vs deep foundations:

Two situations favouring a raft (mat) foundation:

1. When the soil has low bearing capacity and the combined area of isolated footings would exceed about $50\%$ of the building footprint — a raft spreads the load over the whole area and reduces contact pressure.
2. Where differential settlement must be minimised (e.g. on compressible/non-uniform soils or for structures sensitive to settlement), as the rigid raft bridges weak pockets and equalises settlement. Also useful where a basement/water-tightness is required.