BE Civil Engineering (IOE, TU) Foundation Engineering (IOE, CE 701) Question Paper 2078 Nepal

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

Water table deep, so no buoyancy correction.

Step 1 — Terzaghi's 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:

• $1.3\,c\,N_c = 1.3 \times 18 \times 25.13 = 587.9\text{ kPa}$
• $\gamma D_f N_q = 18.5 \times 1.5 \times 12.72 = 352.98\text{ kPa}$
• $0.4\,\gamma B N_\gamma = 0.4 \times 18.5 \times 2.5 \times 8.34 = 154.29\text{ kPa}$

Step 3 — Ultimate bearing capacity:

$$q_u = 587.9 + 352.98 + 154.29 = \mathbf{1095.2\ kPa}$$

Step 4 — Net ultimate bearing capacity (subtract the overburden $\gamma D_f$ that existed before loading):

$$q_{nu} = q_u - \gamma D_f = 1095.2 - (18.5 \times 1.5) = 1095.2 - 27.75 = 1067.4\text{ kPa}$$

Step 5 — Net safe bearing capacity:

$$q_{ns} = \frac{q_{nu}}{FOS} = \frac{1067.4}{3} = \mathbf{355.8\ kPa}$$

Step 6 — Net safe load:

$$Q_{ns} = q_{ns} \times A = 355.8 \times (2.5 \times 2.5) = 355.8 \times 6.25 = \mathbf{2223.8\ kN}$$

Results: $q_u = 1095.2\text{ kPa}$, net safe bearing capacity $= 355.8\text{ kPa}$, net safe load $\approx 2224\text{ kN}$.

Given: loaded area $B \times L = 3 \times 3\text{ m}$, net pressure $q = 120\text{ kPa}$. Clay from $2$ to $6\text{ m}$ below base, thickness $H = 4\text{ m}$, mid-depth at $z = 2 + 2 = 4\text{ m}$ below the footing base. $e_0 = 0.85$, $C_c = 0.30$, NC clay, $\sigma'_0 = 75\text{ kPa}$.

Step 1 — Stress increase by 2:1 method at $z = 4\text{ m}$:

$$\Delta\sigma = \frac{q\,(B \times L)}{(B+z)(L+z)} = \frac{120 \times (3 \times 3)}{(3+4)(3+4)} = \frac{120 \times 9}{7 \times 7} = \frac{1080}{49} = \mathbf{22.04\ kPa}$$

Step 2 — Settlement of a normally consolidated clay:

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

Step 3 — Substitute:

$$\frac{\sigma'_0 + \Delta\sigma}{\sigma'_0} = \frac{75 + 22.04}{75} = \frac{97.04}{75} = 1.2939$$ $$\log_{10}(1.2939) = 0.11186$$ $$S_c = \frac{0.30 \times 4}{1 + 0.85} \times 0.11186 = \frac{1.20}{1.85} \times 0.11186 = 0.6486 \times 0.11186$$ $$S_c = 0.07256\text{ m} = \mathbf{72.6\ mm}$$

Step 4 — Time for 50% consolidation (Part c): Double drainage (sand above and below), so the drainage path is half the layer thickness:

$$H_{dr} = \frac{H}{2} = \frac{4}{2} = 2\text{ m}$$ $$t_{50} = \frac{T_{v,50}\,H_{dr}^2}{c_v} = \frac{0.197 \times (2)^2}{2.5} = \frac{0.197 \times 4}{2.5} = \frac{0.788}{2.5} = 0.3152\text{ year}$$ $$t_{50} = 0.3152 \times 12 = \mathbf{3.78\ months\ }(\approx 0.315\text{ year})$$

Results: $\Delta\sigma \approx 22.0\text{ kPa}$, primary consolidation settlement $\approx \mathbf{72.6\ mm}$, and $t_{50} \approx 0.32\text{ year} \approx 3.8\text{ months}$.

Given: $d = 0.5\text{ m}$, $L = 12\text{ m}$, $c_u = 40\text{ kPa}$, $\alpha = 0.55$, $N_c = 9$, $FOS = 2.5$.

Step 1 — Skin friction (single pile):

$$Q_s = \alpha\,c_u\,(\pi d L) = 0.55 \times 40 \times (\pi \times 0.5 \times 12)$$

Perimeter surface area $= \pi \times 0.5 \times 12 = 18.850\text{ m}^2$

$$Q_s = 0.55 \times 40 \times 18.850 = 414.7\text{ kN}$$

Step 2 — End bearing (single pile): Base area $A_b = \frac{\pi}{4}d^2 = \frac{\pi}{4}(0.5)^2 = 0.19635\text{ m}^2$

$$Q_b = c_u N_c A_b = 40 \times 9 \times 0.19635 = 70.69\text{ kN}$$

Step 3 — Ultimate load capacity (single pile):

$$Q_u = Q_s + Q_b = 414.7 + 70.69 = \mathbf{485.4\ kN}$$

Step 4 — Safe load (single pile):

$$Q_{safe} = \frac{Q_u}{FOS} = \frac{485.4}{2.5} = \mathbf{194.2\ kN}$$

Step 5 — Pile group, block failure ($3\times3$, spacing $s = 1.5\text{ m}$): Plan dimensions of the block (centre-to-centre over $2$ spaces, plus one diameter for edge cover):

$$B_g = L_g = (3-1)\times 1.5 + d = 3.0 + 0.5 = 3.5\text{ m}$$

Block perimeter $= 4 \times 3.5 = 14.0\text{ m}$; block base area $= 3.5 \times 3.5 = 12.25\text{ m}^2$.

Block skin friction (full $c_u$ on the block perimeter, $\alpha$ not applied to a soil-soil block surface):

$$Q_{s,block} = c_u \times (\text{perimeter} \times L) = 40 \times (14.0 \times 12) = 40 \times 168 = 6720\text{ kN}$$

Block end bearing:

$$Q_{b,block} = c_u N_c A_{base} = 40 \times 9 \times 12.25 = 4410\text{ kN}$$

Block (group) ultimate capacity:

$$Q_{u,block} = 6720 + 4410 = \mathbf{11130\ kN}$$

Step 6 — Compare with sum of individual piles:

$$\sum Q_u = 9 \times 485.4 = 4368.6\text{ kN}$$

Since the sum of individual capacities ($4369\text{ kN}$) is less than the block capacity ($11130\text{ kN}$), block failure does not govern; the group capacity is controlled by the individual-pile sum:

$$\boxed{Q_{group} \approx 4369\text{ kN (governed by individual piles)}}$$

Group efficiency $= \dfrac{Q_{group}}{n \, Q_u} = \dfrac{4369}{4369} = 1.0$ (i.e. taken as $\approx 100\%$ since individual action governs; safe group load $= 4369/2.5 \approx 1748\text{ kN}$).

Given: $H = 5\text{ m}$, $\gamma = 17\text{ kN/m}^3$, $\phi = 32^{\circ}$, smooth vertical back, horizontal backfill.

Step 1 — Active pressure coefficient (Rankine):

$$K_a = \frac{1-\sin\phi}{1+\sin\phi} = \frac{1-\sin 32^{\circ}}{1+\sin 32^{\circ}} = \frac{1-0.5299}{1+0.5299} = \frac{0.4701}{1.5299} = \mathbf{0.3073}$$

Part (a) — Dry backfill:

Step 2 — Total active thrust:

$$P_a = \tfrac{1}{2} K_a \gamma H^2 = \tfrac{1}{2} \times 0.3073 \times 17 \times 5^2 = 0.5 \times 0.3073 \times 17 \times 25 = \mathbf{65.3\ kN/m}$$

Step 3 — Point of application: triangular pressure distribution, acts at $H/3$ above the base:

$$\bar{z} = \frac{H}{3} = \frac{5}{3} = \mathbf{1.67\ m\ above\ the\ base}$$

Part (b) — Water table $2\text{ m}$ above base (bottom $2\text{ m}$ submerged):

Let upper dry layer thickness $= 3\text{ m}$, lower submerged layer $= 2\text{ m}$.

Step 4 — Lateral effective stresses (use $K_a$ on effective vertical stress):

• At depth $3\text{ m}$ (top of water): $\sigma'_v = 17 \times 3 = 51\text{ kPa}$; $\;\sigma'_h = K_a \sigma'_v = 0.3073 \times 51 = 15.67\text{ kPa}$
• At base ($5\text{ m}$): $\sigma'_v = 51 + 9.5 \times 2 = 51 + 19 = 70\text{ kPa}$; $\;\sigma'_h = 0.3073 \times 70 = 21.51\text{ kPa}$

Step 5 — Active earth thrust (split the diagram):

• Triangle, upper $3\text{ m}$: $\tfrac12 \times 15.67 \times 3 = 23.51\text{ kN/m}$
• Rectangle over lower $2\text{ m}$: $15.67 \times 2 = 31.34\text{ kN/m}$
• Triangle over lower $2\text{ m}$: $\tfrac12 \times (21.51-15.67) \times 2 = \tfrac12 \times 5.84 \times 2 = 5.84\text{ kN/m}$

Earth thrust $P_{a,earth} = 23.51 + 31.34 + 5.84 = 60.69\text{ kN/m}$

Step 6 — Water thrust over bottom $2\text{ m}$:

$$P_w = \tfrac12 \gamma_w h_w^2 = \tfrac12 \times 9.81 \times 2^2 = \tfrac12 \times 9.81 \times 4 = 19.62\text{ kN/m}$$

Step 7 — Total horizontal thrust:

$$P_{total} = 60.69 + 19.62 = \mathbf{80.3\ kN/m}$$

The rise of water table increases the total horizontal thrust from $65.3\text{ kN/m}$ to about $\mathbf{80.3\ kN/m}$.

Part (a) — Subsoil investigation programme:

Purpose: to determine the nature, sequence and thickness of subsurface strata; the engineering properties (strength, compressibility, permeability) of each stratum; groundwater conditions; and to provide data for selecting foundation type, depth and allowable bearing pressure, and for estimating settlement.

Stages:

1. Reconnaissance / desk study — geological maps, existing records, site visit.
2. Preliminary (exploratory) investigation — a few boreholes/trial pits, in-situ tests (SPT/CPT), to define a general picture.
3. Detailed investigation — adequate boreholes, undisturbed sampling, lab testing, groundwater observation.
4. Supplementary / construction-stage investigation if needed.

Methods of boring: trial pits (shallow), auger boring (cohesive soils above WT), wash boring, percussion boring (gravels/boulders), rotary drilling (rock & deep holes).

Sampling: Disturbed samples (from auger/SPT split spoon) — preserve grain-size and Atterberg limits but not structure; used for classification. Undisturbed samples (thin-walled Shelby tubes, area ratio $\le 10\%$) — preserve in-situ structure and water content; used for strength and consolidation tests.

Depth of boring: generally taken to a depth where the net stress increase from the foundation is small (commonly $\approx 10\%$ of applied net pressure, roughly $1.5$–$2$ times the width of the loaded area below founding level), or to firm/incompressible stratum/bedrock.

Number/spacing of boreholes: depends on building size and soil variability; for a multi-storey building a grid spacing of about $15$–$30\text{ m}$ is common, with at least one borehole per major column-load region and a minimum of $3$–$4$ for a reasonable building footprint.

Part (b) — SPT corrections:

Step 1 — Overburden correction (Liao & Whitman):

$$C_N = \sqrt{\frac{95.76}{\sigma'_v}} = \sqrt{\frac{95.76}{95}} = \sqrt{1.0080} = 1.0040$$ $$N_1 = C_N \times N = 1.0040 \times 28 = 28.11 \approx 28$$

Step 2 — Dilatancy correction (applies for fine/silty saturated sand when $N_1 > 15$):

$$N_{corr} = 15 + \tfrac12 (N_1 - 15) = 15 + 0.5\,(28.11 - 15) = 15 + 0.5 \times 13.11 = 15 + 6.55$$ $$N_{corr} = 21.56 \approx \mathbf{22}$$

Corrected SPT value $\approx 22$ (overburden-corrected $N_1 \approx 28$, then reduced for dilatancy to $\approx 22$).

Effect of water table: The presence of water reduces the effective unit weight of soil (from $\gamma$ to $\gamma' = \gamma_{sat} - \gamma_w$), which reduces both the surcharge term ($q = \gamma D_f N_q$) and the self-weight term ($0.5\gamma B N_\gamma$) of the bearing capacity equation. Hence a high water table lowers the bearing capacity (by up to roughly $50\%$ when the soil is fully submerged). The cohesion term is essentially unaffected.

Two correction factors are introduced:

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

• $R_{w1}$ corrects the surcharge (overburden) term for the water table located between the ground surface and the base of the footing.
• $R_{w2}$ corrects the self-weight term for the water table located between the base and a depth $B$ below the base (within the failure zone beneath the footing).

Limiting values:

• When water table is at or above the ground surface: $R_{w1} = 0.5$.
• When water table is at or below depth $D_f$ (i.e. at the base): $R_{w1} = 1.0$.
• When water table is at the base level: $R_{w2} = 0.5$.
• When water table is at depth $\ge B$ below the base: $R_{w2} = 1.0$.

Linear interpolation is used for intermediate positions.

Three cases (sketch in text):

Case 1: WT at/above GL
  GL ===WT===========    R_w1 = 0.5, R_w2 = 0.5
     |          |
  ---+---footing+---  (base)

Case 2: WT between GL and base
  GL ================
     | ===WT===      |  R_w1 = 0.5..1.0 (interp.), R_w2 = 0.5
  ---+---footing+---  (base)

Case 3: WT between base and depth B below base
  GL ================
     |          |
  ---+---footing+--- (base)  R_w1 = 1.0
        ===WT===            R_w2 = 0.5..1.0 (interp. to depth B)

When the water table is deeper than $D_f + B$, both factors equal $1.0$ and there is no reduction.

Given: $\phi = 30^{\circ}$, $H = 4\text{ m}$, $\gamma = 18\text{ kN/m}^3$ (uniform, dry).

Step 1 — Earth pressure coefficients (Rankine):

$$K_a = \frac{1-\sin 30^{\circ}}{1+\sin 30^{\circ}} = \frac{1-0.5}{1+0.5} = \frac{0.5}{1.5} = \mathbf{0.333}$$ $$K_p = \frac{1+\sin 30^{\circ}}{1-\sin 30^{\circ}} = \frac{1.5}{0.5} = \mathbf{3.0}$$

Step 2 — Active pressure at the dredge line (depth $H$):

$$p_a = K_a \gamma H = 0.333 \times 18 \times 4 = 24.0\text{ kPa}$$

Step 3 — Active thrust above dredge line (triangle):

$$P_a = \tfrac12 p_a H = \tfrac12 \times 24.0 \times 4 = \mathbf{48.0\ kN/m}$$

Step 4 — Point of zero net pressure below the dredge line, depth $z_0$: Below the dredge line, the active side pressure continues to grow while passive pressure develops on the front. Net pressure is zero where active = passive:

$$K_a \gamma (H + z_0) = K_p \gamma z_0$$ $$K_a (H + z_0) = K_p z_0 \;\Rightarrow\; K_a H = (K_p - K_a) z_0$$ $$z_0 = \frac{K_a H}{K_p - K_a} = \frac{0.333 \times 4}{3.0 - 0.333} = \frac{1.333}{2.667} = \mathbf{0.50\ m}$$

Equivalently $z_0 = \dfrac{H}{(K_p/K_a) - 1} = \dfrac{4}{9-1} = 0.50\text{ m}$.

Results: $K_a = 0.333$, $K_p = 3.0$, active thrust above dredge line $= 48.0\text{ kN/m}$, point of zero net pressure at $z_0 = 0.50\text{ m}$ below the dredge line.

1. Vibro-compaction / vibroflotation — A vibrating probe densifies loose cohesionless (sand/gravel) deposits by rearranging particles into a denser state, increasing relative density, bearing capacity and liquefaction resistance. Best for clean sands with low fines.

2. Stone columns (vibro-replacement) — Columns of compacted crushed stone are installed in soft soil; they act as stiff, free-draining inclusions that carry load, accelerate consolidation and reduce settlement. Suitable for soft clays and silts and loose silty sands.

3. Preloading with vertical drains (PVDs/sand drains) — A surcharge (preload) is placed to consolidate the soil in advance; prefabricated vertical drains shorten the drainage path and speed up primary consolidation. Most suitable for soft, saturated, compressible clays/silts.

4. Grouting (cement / chemical / jet grouting) — Injecting grout fills voids/fissures, cements particles together, and increases strength and reduces permeability. Permeation grouting suits coarse sands/gravels; jet grouting and chemical grouting can be used in finer soils and to form columns/cut-offs.

(Other acceptable methods: dynamic compaction for loose granular fills; deep soil mixing for soft clays; soil reinforcement/geosynthetics; compaction piles.)

Use: Well (caisson) foundations are deep, large-section foundations sunk through water and soft strata to a firm stratum. They are widely used for bridge piers and abutments across rivers, for heavy structures, and where large lateral loads and scour must be resisted. They are preferred where heavy concentrated loads and deep scour make piles uneconomical.

Main components of a well:

• Cutting edge — sharp lowest edge that facilitates sinking.
• Well curb — wedge-shaped RC ring above the cutting edge that transmits load to the cutting edge.
• Steining — the main thick wall (body) of the well, usually masonry/RCC, that provides weight for sinking and resists earth pressure.
• Bottom plug — concrete seal at the base after the well reaches founding level; transfers load to the soil.
• Sand filling / dredge hole — sand inside the steining to reduce stresses and add weight.
• Top plug and well cap — top concrete plug and the RC cap that distributes the pier load to the steining.

Shifts and tilts: During sinking, a well may tilt (rotate from vertical) or shift (translate horizontally from its planned position) due to non-uniform soil resistance, obstructions (boulders/logs), uneven dredging, or sloping hard strata. Permissible tilt is usually limited to about $1$ in $80$ and shift to about $1\%$ of the depth sunk.

Remedial measures (any two):

1. Eccentric / one-sided dredging — remove more material from the higher side to let it sink faster and correct the tilt.
2. Eccentric loading (kentledge) — place extra load on the higher side to push it down.
3. Water jetting on the high side to reduce skin friction there; or applying a horizontal pull/strut to shift the well back.
4. Hooking / pulling with anchors and inserting wooden sleepers under the cutting edge on the lower side to arrest further tilting.

Given: plate width $B_p = 0.30\text{ m}$, plate settlement $S_p = 8\text{ mm}$; footing width $B_f = 2.0\text{ m}$; same load intensity, cohesionless (sandy) soil.

Step 1 — Settlement scaling relation for footings on SAND:

$$\frac{S_f}{S_p} = \left[\frac{B_f\,(B_p + 0.3)}{B_p\,(B_f + 0.3)}\right]^2$$

(with widths in metres; $0.3$ is in metres).

Step 2 — Substitute:

$$\frac{B_f(B_p+0.3)}{B_p(B_f+0.3)} = \frac{2.0\,(0.30+0.30)}{0.30\,(2.0+0.30)} = \frac{2.0 \times 0.60}{0.30 \times 2.30} = \frac{1.20}{0.69} = 1.7391$$

Step 3 — Square the ratio and multiply:

$$\frac{S_f}{S_p} = (1.7391)^2 = 3.0245$$ $$S_f = 8 \times 3.0245 = 24.2\text{ mm}$$

Settlement of the prototype footing $\approx \mathbf{24.2\ mm}$.

(Note: the simpler relation $S_f = S_p\,[2B_f/(B_f+B_p)]^2$ is sometimes quoted; the Terzaghi–Peck form used above is the standard one for sands and is preferred here.)

Shallow vs. deep foundations:

A raft (mat) foundation is preferred when:

1. The soil has low bearing capacity so that individual footings would need very large areas (commonly when the total footing area would exceed about $50\%$ of the building plan area).
2. The soil is highly compressible or erratic/variable, and a raft bridges over soft spots reducing differential settlement.
3. Heavy/concentrated column loads are closely spaced, so isolated footings would overlap.
4. To control differential settlement and provide a rigid base, e.g. for tall buildings, silos, water tanks.
5. Where a basement below the water table is required, the raft also acts as a watertight floor (buoyancy/compensated raft).

Part (b) — Floating (compensated) foundation: A floating or compensated foundation is one in which the weight of the soil excavated for the basement is made equal to (or a large part of) the structural load, so the net increase in pressure on the bearing stratum is zero or small. Because the soil at founding level was already carrying the overburden it now supports a similar load, settlement is minimised.

• Fully compensated: weight of excavated soil $=$ total structural load $\Rightarrow$ net pressure $= 0$.
• Partially compensated: excavated soil weight $<$ structural load $\Rightarrow$ a reduced (but non-zero) net pressure remains. This approach is used for heavy structures on soft, compressible clays where conventional footings would settle excessively.

Part (c) — Depth for full compensation: Plan area $A = 20 \times 15 = 300\text{ m}^2$.

$$\text{Gross applied pressure} = \frac{Q}{A} = \frac{63000}{300} = 210\text{ kPa}$$

For full compensation, removed soil pressure $= \gamma D_f$ must equal the applied pressure:

$$\gamma D_f = 210 \;\Rightarrow\; D_f = \frac{210}{18} = \mathbf{11.67\ m}$$

The basement raft must be founded at a depth of about $11.7\text{ m}$ for a fully compensated (zero net pressure) foundation.