IOE Foundation Engineering Past Year Question Solution: Chapterwise Guide ← Back
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FOUNDATION ENGINEERING Past Year Question Solution

Comprehensive guide to FOUNDATION ENGINEERING Past Year Question Solution. This resource provides point-wise theory and step-by-step numerical solutions for all major topics including Geotechnical Investigation and Bearing Capacity.

1. Geotechnical Investigation

Q.1 What do you understand by site investigation? How would you decide the depth of exploration and lateral extent of investigations?
🕒 2082 Bhadra [4] 🕒 2078 Ashwin [4]

Site Investigation: Site investigation (or soil exploration) is the systematic process of determining the subsurface conditions at a proposed construction site. It involves the determination of the physical, mechanical, and chemical properties of soil and rock to ensure the safe and economical design of substructures.

Objectives:

  • To determine the type and depth of the foundation.
  • To evaluate the load-bearing capacity of the soil.
  • To estimate probable settlements.
  • To determine the location of the Ground Water Table (GWT).

Deciding Depth of Exploration: The depth of exploration depends on the type of structure, the intensity of loading, and the soil profile. The exploration must extend to a depth where the stress increase due to structural loading becomes negligible (typically less than 10% of the applied stress).

  • General Rule: $1.5B$ to $2.0B$ (where $B$ is the width of the footing).
  • For Strip Footings: At least $3.0B$ below the foundation level.
  • For Pile Foundations: $10m$ to $30m$, or at least $1.5$ times the width of the pile group below the tip of the piles.

Deciding Lateral Extent of Investigations: The lateral extent depends on the site variability and the footprint of the structure.

  • Grid Pattern: Boreholes are typically arranged in a grid pattern.
  • Spacing:
    • For multi-story buildings: $10m$ to $30m$ spacing.
    • For highways: $300m$ to $600m$ spacing.
    • For dams: $40m$ to $80m$ spacing.
  • Coverage: Investigations should cover the entire area of the proposed structure and extend slightly beyond the boundaries to account for edge effects.
Q.2 A soil sampler has the following dimensions: Cutting edge inner diameter $68\text{ mm}$ / outer $74\text{ mm}$ and Sampling tube inner diameter $70\text{ mm}$ / outer $72\text{ mm}$. Calculate Inside Clearance, Outside Clearance, Area Ratio. Classify sample (disturbed/undisturbed).
🕒 2082 Bhadra [4]
Given:
Cutting Edge: $D_1=68\text{ mm}$ (Inner), $D_2=74\text{ mm}$ (Outer)
Sampling Tube: $D_3=70\text{ mm}$ (Inner), $D_4=72\text{ mm}$ (Outer)

Calculations:

(i) Inside Clearance ($C_i$):

$$C_i = \frac{D_3 – D_1}{D_1} \times 100$$ $$C_i = \frac{70 – 68}{68} \times 100 = \frac{2}{68} \times 100 = \mathbf{2.94\%}$$

(ii) Outside Clearance ($C_o$):

$$C_o = \frac{D_2 – D_4}{D_4} \times 100$$ $$C_o = \frac{74 – 72}{72} \times 100 = \frac{2}{72} \times 100 = \mathbf{2.78\%}$$

(iii) Area Ratio ($A_r$):

$$A_r = \frac{D_2^2 – D_1^2}{D_1^2} \times 100$$ $$A_r = \frac{74^2 – 68^2}{68^2} \times 100 = \frac{5476 – 4624}{4624} \times 100 = \mathbf{18.43\%}$$

Classification: For a sample to be considered “good” undisturbed, $A_r$ should ideally be $\le 10\%$. However, an Area Ratio between $10\%$ and $20\%$ is generally accepted as providing satisfactory undisturbed samples for stiff clays. Since $A_r = 18.43\%$ (which is $< 20\%$) and $C_i$ is within $1-3\%$, the sample can be classified as Undisturbed.

Q.3 Describe the plate load test and its limitations.
🕒 2082 Bhadra [4]

Plate Load Test (PLT): The Plate Load Test is a field test used to determine the ultimate bearing capacity and settlement of soil. It is performed at the foundation level using a rigid steel plate.

Procedure:

  • Excavation: A pit is dug to the proposed foundation level. The width of the pit should be at least 5 times the width of the test plate.
  • Setup: A square or circular steel plate (usually $300mm$ to $750mm$ size) is placed at the center. A hydraulic jack is placed on the plate, reacting against a loaded platform (kentledge) or anchored truss.
  • Loading: Load is applied in increments (usually $1/5^{th}$ of expected safe bearing capacity).
  • Measurement: Settlement is recorded using dial gauges ($0.02mm$ sensitivity) for each load increment until the rate of settlement becomes negligible.
  • Plot: A Load-Settlement curve is plotted to determine the ultimate bearing capacity ($q_u$) and allowable settlement.

Limitations:

  • Scale Effect: The test reflects the behavior of soil only up to a depth of roughly $2 \times B_p$ (width of plate). Actual foundations are much larger, and their pressure bulb extends much deeper. This is critical in non-homogeneous soils.
  • Time Effect: It is essentially a short-term loading test. It does not capture long-term consolidation settlement, which is critical for clayey soils.
  • Reaction Load: Arranging a heavy reaction load (kentledge) for high bearing capacity soils can be difficult and expensive.
  • Water Table: If the water table is not at the test level but rises later, the actual bearing capacity will reduce, which the test performed in dry season might miss.
Q.4 SPT at 20 m in dense sand ($\gamma=17 \text{ kN/m}^3$), blows 18-21-22. N-value corrected for overburden and dilatancy?
🕒 2081 Bhadra [4]
Given: Depth ($z$) = $20 \text{ m}$, Unit Weight ($\gamma$) = $17 \text{ kN/m}^3$, Blows: $18 – 21 – 22$.
$N_{obs} = 21 + 22 = 43$ (Sum of last two increments).
Assumptions: Water Table at ground surface (conservative), $C_N \le 2.0$.

Step 1: Calculate Effective Overburden Pressure ($\sigma_0’$):

$$\sigma_0′ = \gamma’ \times z = (\gamma_{sat} – \gamma_w) \times z$$ $$\sigma_0′ = (17 – 9.81) \times 20 = 7.19 \times 20 = 143.8 \text{ kN/m}^2$$

Step 2: Overburden Pressure Correction ($C_N$):

$$C_N = \sqrt{\frac{100}{\sigma_0′}} = \sqrt{\frac{100}{143.8}} = \sqrt{0.695} = 0.834$$

Since $C_N < 2.0$, we use this value.

$$N’ = N_{obs} \times C_N = 43 \times 0.834 = 35.86 \approx 36$$

Step 3: Dilatancy Correction:

Applied because $N’ > 15$ in saturated dense sand.

$$N” = 15 + \frac{1}{2} (N’ – 15)$$ $$N” = 15 + 0.5 \times (36 – 15)$$ $$N” = 15 + 10.5 = 25.5$$

Final Answer: Corrected N-value = 26

Q.5 List out the contents of a site investigation report.
🕒 2081 Bhadra [4]

A comprehensive site investigation report typically contains the following sections:

  • Introduction: Scope of work, type of structure, and location of the site.
  • Borehole Log: Detailed record of soil layers, color, consistency, SPT values, and water table location.
  • Field Exploration Methods: Description of drilling techniques (e.g., auger, wash boring) and sampling methods used.
  • Laboratory Test Results: Data from tests like grain size analysis, Atterberg limits, shear strength (Direct Shear/Triaxial), and consolidation.
  • Site Plan: Map showing the location of boreholes relative to the proposed structure.
  • Analysis and Discussion: Interpretation of field and lab data regarding bearing capacity and settlement.
  • Recommendations:
    • Recommended foundation type (Shallow/Deep).
    • Safe Bearing Capacity (SBC) at various depths.
    • Anticipated settlement.
    • Groundwater control measures.
  • Limitations: Any constraints or uncertainties in the investigation.
Q.6 Describe different requirements of a sampler used for undisturbed sampling.
🕒 2081 Bhadra [4]

To obtain good quality undisturbed samples, a sampler must minimize soil disturbance. The key design requirements are:

  • Area Ratio ($A_r$): Must be kept low (preferably $< 10\%$, acceptable up to $20\%$) to minimize the volume of soil displaced by the sampler walls.
  • Inside Clearance ($C_i$): Should be between $1\%$ and $3\%$. This reduces friction between the soil sample and the inside of the tube, allowing the sample to enter smoothly.
  • Outside Clearance ($C_o$): Should be small ($0\%$ to $2\%$). This reduces friction between the soil and the outside of the tube, facilitating penetration.
  • Cutting Edge: The cutting edge should be sharp and tapered to slice through the soil rather than pushing it aside.
  • Smooth Interior: The inside of the sampling tube must be extremely smooth (often brass or lacquered steel) to reduce wall friction.
  • Check Valve: A non-return valve at the top of the sampler to allow air/water to escape during driving and create a vacuum to retain the sample during withdrawal.
Q.7 Discuss the features of a good site investigation report.
🕒 2081 Baishakh (Back) [4]

A good site investigation report should possess the following features:

  • Clarity and Conciseness: Information should be presented clearly without ambiguity. Use standard terminology.
  • Completeness: It must cover all aspects: field work, lab work, analysis, and final recommendations. No critical data (like Water Table depth) should be missing.
  • Visual Representation: Use clear logs, cross-sections, and graphs (e.g., particle size distribution curves) to present data visually.
  • Justified Recommendations: Recommendations for bearing capacity and foundation type should be backed by calculations and specific reference to the test data.
  • Practicality: Suggestions should be constructible and economically feasible for the specific project.
  • Disclosure of Limitations: It should clearly state any assumptions made or areas where data might be insufficient, advising further testing if necessary.
Q.8 Sampler: Cutting edge ID 59 mm / OD 63 mm, tube ID 60 mm / OD 62 mm. Calculate (i) Area ratio (ii) Inside clearance (iii) Outside clearance.
🕒 2081 Baishakh (Back) [3]
Given:
Cutting Edge: $D_1 (\text{ID}) = 59 \text{ mm}$, $D_2 (\text{OD}) = 63 \text{ mm}$
Sampling Tube: $D_3 (\text{ID}) = 60 \text{ mm}$, $D_4 (\text{OD}) = 62 \text{ mm}$

(i) Area Ratio ($A_r$):

$$A_r = \frac{D_2^2 – D_1^2}{D_1^2} \times 100$$ $$A_r = \frac{63^2 – 59^2}{59^2} \times 100 = \frac{3969 – 3481}{3481} \times 100 = \frac{488}{3481} \times 100 = \mathbf{14.02\%}$$

(ii) Inside Clearance ($C_i$):

$$C_i = \frac{D_3 – D_1}{D_1} \times 100$$ $$C_i = \frac{60 – 59}{59} \times 100 = \frac{1}{59} \times 100 = \mathbf{1.69\%}$$

(iii) Outside Clearance ($C_o$):

$$C_o = \frac{D_2 – D_4}{D_4} \times 100$$ $$C_o = \frac{63 – 62}{62} \times 100 = \frac{1}{62} \times 100 = \mathbf{1.61\%}$$
Q.9 List field tests in subsurface exploration. Corrections to SPT N-values for sand before using in design charts/empirical correlations?
🕒 2081 Baishakh (Back) [5] 🕒 2076 Chaitra [1+4] 🕒 2074 Chaitra [1+4]

Common Field Tests:

  • Standard Penetration Test (SPT)
  • Cone Penetration Test (CPT) – Static and Dynamic
  • Plate Load Test (PLT)
  • Vane Shear Test (VST) – For soft clays
  • Pressuremeter Test (PMT)
  • Geophysical Methods (Seismic Refraction, Electrical Resistivity)

Corrections to SPT N-values for Sand: Before using SPT values in design, two main corrections are applied in sequence:

1. Overburden Pressure Correction ($C_N$): N-values increase with depth due to confinement pressure, even if density is constant. We correct it to a standard effective overburden pressure of $100 \text{ kN/m}^2$ (approx 1 ton/sq.ft).

$$C_N = \sqrt{\frac{100}{\sigma_0′}}$$

Where $\sigma_0’$ is effective overburden pressure in $\text{kN/m}^2$. Condition: $C_N \le 2.0$.

$$N’ = N_{obs} \times C_N$$

2. Dilatancy Correction (Fine Sand Correction): Applied only to fine sands and silts below the water table. During impact, pore pressure builds up in fine saturated sands, temporarily increasing resistance (false high N).

$$N” = 15 + \frac{1}{2} (N’ – 15)$$

Condition: Apply only if $N’ > 15$. If $N’ \le 15$, then $N” = N’$.

Q.10 a) Different sampler design parameters and physical significance for undisturbed sample. b) Best boring types for road sub-grade investigation and why? c) Corrected N-value, SPT 6 m depth, fully submerged fine sand, blows 18-23-17, $\gamma_{sat}=19 \text{ kN/m}^3$.
🕒 2080 Bhadra [various marks]

(a) Sampler Design Parameters:

  • Area Ratio: Significance is minimizing soil displacement volume.
  • Inside Clearance: Significance is reducing friction on the sample as it enters the tube, preventing compression/distortion.
  • Recovery Ratio ($L_{recovered} / L_{penetrated}$): Significance: Indicates if soil was lost or compressed. Ratio $< 1$ implies compression; $> 1$ implies swelling.

(b) Best Boring Type for Road Sub-grade:

  • Best Method: Auger Boring (Hand or Power Auger) or Test Pits.
  • Why?
    • Road sub-grade investigations are typically shallow (top $1-3$ meters).
    • Augers are rapid, economical, and easy to mobilize.
    • They provide disturbed samples sufficient for classification (grain size, Atterberg limits) and compaction tests (Proctor), which are the primary needs for road works.
    • Test pits allow visual inspection of the natural soil profile.

(c) Numerical Solution:

Given: Depth ($z$) = $6 \text{ m}$, Soil: Fully submerged fine sand, Unit weight ($\gamma_{sat}$) = $19 \text{ kN/m}^3$, Blows: $18 – 23 – 17$.
$N_{obs} = 23 + 17 = 40$ (Sum of last two increments).

Step 1: Effective Overburden Pressure:

$$\sigma_0′ = (\gamma_{sat} – \gamma_w) \times z$$ $$\sigma_0′ = (19 – 9.81) \times 6 = 9.19 \times 6 = 55.14 \text{ kN/m}^2$$

Step 2: Overburden Correction:

$$C_N = \sqrt{\frac{100}{\sigma_0′}} = \sqrt{\frac{100}{55.14}} = \sqrt{1.81} = 1.346$$ $$N’ = N_{obs} \times C_N = 40 \times 1.346 = 53.84 \approx 54$$

Step 3: Dilatancy Correction:

Since it is “submerged fine sand” and $N’ > 15$, correction is required.

$$N” = 15 + \frac{1}{2} (N’ – 15)$$ $$N” = 15 + 0.5 \times (54 – 15)$$ $$N” = 15 + 0.5 \times 39 = 15 + 19.5 = 34.5$$

Final Answer: Corrected N-value = 35

Q.11 a) Information in preliminary subsurface exploration? Merits/demerits of percussion drilling? Borehole log format example. b) Disturbed vs undisturbed samples. Factors affecting disturbance.
🕒 2080 Baishakh [Various]

a) Preliminary Subsurface Exploration

Preliminary exploration is carried out to obtain approximate information regarding the subsoil conditions at low cost. The information obtained typically includes:

  • Geological Character: General idea of the soil strata and geological features of the site.
  • Groundwater Table: Approximate depth of the groundwater table.
  • Soil Types: Identification of major soil layers (e.g., presence of soft clay, rock, or fill).
  • Foundation Feasibility: Assessing whether shallow foundations are feasible or if deep foundations (piles) might be required.
  • Planning Detailed Exploration: Data helps in deciding the number, depth, and location of boreholes for the detailed investigation phase.

Merits and Demerits of Percussion Drilling

Merits:

  • Can be used in almost all types of soils and rocks (boulders, gravels).
  • Useful for identifying changes in lithology (strata) by monitoring the rate of drilling.
  • Reasonably economical for shallow to medium depths in hard strata.

Demerits:

  • The soil at the bottom of the hole is significantly disturbed by the heavy impact.
  • Not suitable for obtaining high-quality undisturbed samples (especially in loose sand or soft clay).
  • Noisy operation due to heavy impact tools.
  • Slower progress in plastic clays compared to rotary methods.

Example Borehole Log Format

Depth (m) Soil Description / Strata Sample Type SPT N-Value Remarks
0.0 – 1.5 Filled up soil / Top soil Disturbed
1.5 – 3.0 Soft grey silty CLAY Undisturbed (Shelby) 4 Water Table @ 2.5m
3.0 – 4.5 Medium dense fine SAND Split Spoon 18
4.5 – 6.0 Stiff yellowish CLAY Split Spoon 22
> 6.0 Weathered Rock Core >50 End of Boring

b) Disturbed vs Undisturbed Samples

  • Disturbed Samples: Soil samples where the natural structural integrity is destroyed, though the mineralogical composition and moisture content may remain unchanged. Used for classification tests (Grain size analysis, Atterberg limits, Specific gravity).
  • Undisturbed Samples: Soil samples that retain their natural structure, water content, and void ratio as closely as possible to the in-situ condition. Required for engineering property tests (Shear strength, Consolidation, Permeability).

Factors Affecting Disturbance:

  • Sampler Design: Area ratio ($A_r$), Inside Clearance ($C_i$), and Outside Clearance ($C_o$).
  • Friction: Wall friction between the soil and the sampler tube.
  • Recovery Ratio: If $< 100\%$, the sample is compressed; if $> 100\%$, it has swelled.
  • Handling: Shock and vibration during transportation and storage.
  • Method of Advancement: How the borehole was drilled (e.g., percussion causes more disturbance than rotary).
Q.12 a) Accessible vs inaccessible exploration methods. How to determine depth of exploration (example). b) Precautions during sampling, transportation, storage. c) Physical meaning of area ratio, inside/outside clearance, recovery ratio.
🕒 2079 Bhadra [Various]

a) Accessible vs Inaccessible Exploration

  • Accessible Exploration: Methods where the soil is visually observed and accessible for direct sampling by hand.
    Example: Test Pits (Trial Pits). Useful for shallow depths (up to 3m) to inspect soil stratification clearly.
  • Inaccessible Exploration: Methods where the soil is not directly seen, and tools are used to retrieve samples or measure resistance.
    Example: Boring (Auger, Wash, Rotary) and Sounding (Probing). Necessary for deep exploration and below the water table.

Determining Depth of Exploration:

The depth of exploration is governed by the “Significant Depth,” which is the depth up to which the increase in stress due to structural loading causes significant settlement or shear stress (usually where the stress increase is $\approx 10\%$ of the applied load).

Example (Square Footing): For an isolated square footing of width $B$, the depth of exploration should generally be 1.5 to 2.0 times the width ($B$) below the foundation level. If $B = 2$ m, Exploration Depth $\approx 3.0$ m to $4.0$ m below the footing base.

b) Precautions during Sampling, Transportation, and Storage

  • Sealing: Ends of tube samples should be waxed (paraffin wax) immediately to prevent moisture loss.
  • Labeling: Clear labels with borehole number, depth, date, and orientation (top/bottom).
  • Shock Protection: Samples should be transported in shock-proof boxes (using sawdust or foam) to prevent vibration damage.
  • Storage: Store in a cool, humid room to prevent drying and structural changes.
  • Verticality: Tube samples should ideally be kept vertical to prevent layer separation.

c) Physical Meaning of Parameters

  • Area Ratio ($A_r$): Represents the volume of soil displaced by the sampler walls in proportion to the sample volume. A lower value means less soil is pushed aside, resulting in less disturbance.
  • Inside Clearance ($C_i$): Represents the relief provided inside the tube. It allows the soil sample to expand slightly, reducing the friction between the sample and the inner wall of the tube.
  • Outside Clearance ($C_o$): Represents the relief provided outside the tube. It reduces the friction between the sampler wall and the surrounding soil as the sampler is driven down.
  • Recovery Ratio ($L_r$): Indicates the quality of the sample recovery. It is the ratio of the length of the sample recovered to the depth of penetration. A value less than 1 indicates compression; a value greater than 1 indicates swelling.
Q.13 Things to consider while preparing site investigation report.
🕒 2078 Kartik [4] 🕒 2074 Chaitra [4] 🕒 2072 Chaitra [5]

A professional Site Investigation Report should include:

  • Introduction: Scope of work, type of project, and location.
  • Geological Description: General geology and topography of the site.
  • Field Investigation Details: Methods used (boring types), number and depth of boreholes, and field tests performed (SPT, CPT, etc.).
  • Laboratory Test Results: Summary of classification, strength, and compressibility tests.
  • Borehole Logs: Graphical representation of soil strata with depth.
  • Groundwater Table: Location of the water table and any fluctuations.
  • Analysis and Discussion: Interpretation of data regarding bearing capacity and settlement.
  • Recommendations:
    • Suitable foundation type (shallow vs. pile).
    • Allowable bearing capacity.
    • Estimated settlement.
    • Depth of foundation.
  • Limitations: Any constraints or uncertainties in the investigation.
Q.14 Detailed information for site investigation of bridge foundation.
🕒 2078 Kartik [5]

Site investigation for bridges requires more rigorous exploration due to heavy loads and scour risks.

  • Depth of Exploration:
    • Exploration must extend below the maximum anticipated scour depth.
    • Ideally, boreholes should extend through the soft strata into firm bearing strata (rock or dense sand).
    • Rule of Thumb: Depth should be at least 1.5 times the width of the caisson/well foundation or pile group, or typically 3 meters into sound rock if bedrock is encountered.
  • Lateral Extent (Location):
    • At least one borehole should be drilled at the location of each pier and abutment.
  • Specific Tests:
    • Standard Penetration Test (SPT): To determine density and bearing capacity of cohesionless soils.
    • Rock Quality Designation (RQD): If drilling into rock, core recovery and RQD are critical.
    • Permeability: To assess dewatering requirements for caisson sinking.
  • Scour & Water: Detailed study of river hydrology and scour potential is mandatory.
Q.15 a) Importance of site investigation and stages. How borehole depth decided for various projects? b) Disturbed vs undisturbed samples. Factors affecting sample quality. c) Sampler ID 40 mm, OD 42 mm → disturbed or undisturbed?
🕒 2078 Bhadra [Various]

a) Importance and Stages

Importance:

  • Ensures safety and stability of the structure.
  • Optimizes design (economy) by providing accurate soil parameters.
  • Predicts construction difficulties (groundwater, loose soil).

Stages:

  • Reconnaissance: Visual inspection and study of maps.
  • Preliminary Exploration: Limited boring/testing to get a general soil profile.
  • Detailed Exploration: Extensive boring, sampling, and lab testing for final design.

Borehole Depth Decision:

  • Isolated Footings: 1.5 times the width ($B$).
  • Strip Footings: 3.0 times the width ($B$).
  • Pile Foundations: 10m to 30m, or at least 1.5 times the width of the pile group below the pile tip.
  • Roads/Canals: 2m to 3m below the subgrade level.

c) Numerical: Disturbed or Undisturbed?

Given: $D_i = 40 \text{ mm}, D_o = 42 \text{ mm}$
$$A_r = \frac{D_o^2 – D_i^2}{D_i^2} \times 100\%$$ $$A_r = \frac{42^2 – 40^2}{40^2} \times 100\% = \frac{1764 – 1600}{1600} \times 100\%$$ $$A_r = \frac{164}{1600} \times 100\% = \mathbf{10.25\%}$$

Conclusion: For a sample to be considered “Undisturbed” (high quality), the Area Ratio ($A_r$) should generally be $\le 10\%$. Since $10.25\%$ is very close to $10\%$ (and significantly lower than split spoon samplers which are $>100\%$), this sampler is likely a Thin-Walled Tube intended for undisturbed sampling. It is suitable for obtaining Undisturbed Samples.

Q.16 a) SPT definition, procedure, uses, corrections for dilatancy & overburden.
🕒 2073 Shrawan [8] 🕒 2074 Ashwin [Various]

Standard Penetration Test (SPT)

Definition: SPT is the most common in-situ test used to determine the penetration resistance of soil, designated as the N-value.

Procedure (IS 2131):

  1. A borehole is drilled to the required depth.
  2. A standard split-spoon sampler is attached to the drill rod.
  3. A hammer weighing 63.5 kg is dropped from a height of 750 mm (76 cm).
  4. The sampler is driven into the soil for 450 mm.
  5. The number of blows required for every 150 mm penetration is recorded.
  6. The blows for the first 150 mm are ignored (Seating Drive).
  7. The SPT N-value is the sum of blows for the last two 300 mm intervals (2nd and 3rd 150mm).

Uses:

  • To correlate relative density ($D_r$) of sands and consistency of clays.
  • To estimate Allowable Bearing Capacity ($q_{all}$).
  • To estimate settlement of foundations.

Corrections: The observed N-value ($N_{obs}$) must be corrected:

  • Overburden Pressure Correction ($C_N$): Soils at shallow depths yield lower N-values due to low confinement.
    $$C_N = \frac{350}{\bar{\sigma} + 70} \quad (\text{where } \bar{\sigma} \le 280 \text{ kN/m}^2)$$
    Corrected N: $N_{cor} = N_{obs} \times C_N$
  • Dilatancy Correction (for Fine Sands/Silts below Water Table): If $N_{cor} > 15$, dilatancy generates pore pressure that increases resistance falsely.
    $$N’ = 15 + 0.5 \times (N_{cor} – 15)$$
Q.17 a) Physical meaning (with sketch): inside clearance ratio, outside clearance ratio, area ratio, recovery ratio. Recommended values for least disturbed samples. b) Recommendations for boreholes (number, depth, spacing) for multistoried building in Kathmandu Valley.
🕒 2075 Chaitra [Various]

a) Physical Meaning

(See Q.12c for text definitions)

Sketch Description: Imagine a tube with a cutting edge at the bottom.
Inside Clearance ($C_i$): The inner diameter of the cutting edge ($D_1$) is slightly smaller than the inner diameter of the tube ($D_3$). This gap reduces friction on the sample.
Outside Clearance ($C_o$): The outer diameter of the cutting edge ($D_2$) is slightly larger than the outer diameter of the tube ($D_4$). This gap reduces friction on the tube wall.

Recommended Values for Undisturbed Samples:

  • Area Ratio ($A_r$): $\le 10\%$ (Strictly for sensitive clay), up to 20% acceptable for stiff clay.
  • Inside Clearance ($C_i$): $1\%$ to $3\%$.
  • Outside Clearance ($C_o$): $0\%$ to $2\%$.
  • Recovery Ratio ($L_r$): $98\%$ to $100\%$.

b) Recommendations for Kathmandu Valley

Kathmandu Valley consists of deep soft clay deposits (Kalimati) and liquefiable sand layers.

  • Spacing:
    • For uniform soil: Boreholes spaced 30m to 50m apart.
    • For erratic/variable soil (common in Kathmandu): Spacing reduced to 10m to 30m.
    • Minimum: 3 to 5 boreholes for a typical multi-story building site (corners and center).
  • Depth:
    • Must penetrate the soft black clay layers.
    • Typically 15m to 25m deep for standard commercial complexes.
    • Should reach firm strata (dense sand/gravel) or until $\Delta\sigma < 10\%$.
    • If piling is expected, depth must go 5m to 10m below the pile tip.
Q.18 a) Why undisturbed samples required? b) Recommend drilling type & field tests for Geotechnical investigation in Kathmandu Valley core. c) Example borehole log format.
🕒 2075 Ashwin [Various]

a) Why Undisturbed Samples?

Undisturbed samples are required to determine the engineering properties of the soil that rely on the soil structure, specifically:

  • Shear Strength Parameters ($c, \phi$): To calculate bearing capacity.
  • Consolidation Parameters ($C_c, C_v$): To estimate the magnitude and rate of settlement.
  • Permeability: To assess flow through soil.

Disturbed samples cannot provide this data reliably.

b) Recommendations for Kathmandu Valley Core

Drilling Type:

  • Rotary Drilling (with mud circulation): Best suited for deep exploration in soft clays and sandy layers to prevent borehole collapse.
  • Wash Boring: Also commonly used for soft clays.

Field Tests:

  • Standard Penetration Test (SPT): Essential for evaluating sandy layers and consistency of stiff clays.
  • Vane Shear Test (VST): Highly recommended for the soft clay (Kalimati) to determine undrained shear strength ($c_u$).
  • Cone Penetration Test (CPT): Useful for continuous profiling of soft deposits.

c) Example Borehole Log: (Refer to the table provided in Solution Q.11)

Q.19 Area ratio of sampler (external radius 30 mm, wall thickness 2.25 mm). Recommend for undisturbed samples? Why?
🕒 2074 Chaitra [2+1]
Given: External Radius ($R_e$) = 30 mm, Thickness ($t$) = 2.25 mm

First, calculate Diameters:

  • External Diameter ($D_e$) = $2 \times 30 = 60 \text{ mm}$
  • Internal Diameter ($D_i$) = $D_e – 2t = 60 – 2(2.25) = 60 – 4.5 = 55.5 \text{ mm}$
$$A_r = \frac{D_e^2 – D_i^2}{D_i^2} \times 100\%$$ $$A_r = \frac{60^2 – 55.5^2}{55.5^2} \times 100\%$$ $$A_r = \frac{3600 – 3080.25}{3080.25} \times 100\% = \frac{519.75}{3080.25} \times 100\%$$ $$A_r \approx \mathbf{16.87\%}$$

Recommendation: This sampler is NOT recommended for high-quality undisturbed sampling of sensitive clays.
Reason: For undisturbed samples, the Area Ratio should ideally be $\le 10\%$. An $A_r$ of $\approx 16.9\%$ causes significant volume displacement and disturbance to the soil structure.

Q.20 What makes pressure meter testing distinctive compared to other field tests?
🕒 2076 Ashwin [2]

Pressure Meter Testing (PMT) is distinctive because:

  • Stress-Strain Relationship: It is one of the few field tests that provides the full stress-strain curve of the soil in-situ.
  • Modulus of Elasticity: It directly measures the Pressuremeter Modulus ($E_p$), which allows for reliable calculation of immediate settlement.
  • No Sampling: It tests the soil in place inside the borehole, avoiding the disturbance issues associated with retrieving samples.
Q.21 Factors for deciding depth of soil exploration. How to decide depth & lateral extent for important structures?
🕒 2076 Ashwin [2+5]

Factors for Deciding Depth

  • Type of Structure: Heavy multi-story vs. light residential.
  • Magnitude of Loads: Heavier loads require deeper exploration.
  • Soil Profile: Location of bedrock or weak compressible layers.
  • Foundation Type: Pile foundations require deeper exploration than shallow footings.

Deciding Depth & Lateral Extent for Important Structures

Depth:

  • The exploration must extend to the Significant Depth, where the net increase in vertical stress is less than 10% of the initial overburden pressure ($\Delta \sigma < 0.1 q$).
  • Isobar Method: It should generally reach the depth of the 10% pressure bulb.
  • Hard Strata: If rock is encountered, drill at least 3m into it to confirm it is bedrock and not a boulder.

Lateral Extent:

  • Boreholes should be placed at all critical loading points (corners and center).
  • The investigation should cover the entire footprint of the structure and slightly beyond (approx 10-20% margin) to account for stress distribution.
  • Spacing for important structures should be close (10m – 30m) to detect soil variability.
Q.22 Sampler: cutting edge ID 69 mm / OD 73 mm, tube ID 70 mm / OD 72 mm. Calculate inside clearance, outside clearance, area ratio.
🕒 2076 Ashwin [3]
Given: Cutting Edge ($D_1=69, D_2=73$), Tube ($D_3=70, D_4=72$) all in mm.

1. Inside Clearance ($C_i$):

$$C_i = \frac{D_3 – D_1}{D_1} \times 100\% = \frac{70 – 69}{69} \times 100\% = \frac{1}{69} \times 100\% = \mathbf{1.45\%}$$

2. Outside Clearance ($C_o$):

$$C_o = \frac{D_2 – D_4}{D_4} \times 100\% = \frac{73 – 72}{72} \times 100\% = \frac{1}{72} \times 100\% = \mathbf{1.39\%}$$

3. Area Ratio ($A_r$):

$$A_r = \frac{D_2^2 – D_1^2}{D_1^2} \times 100\% = \frac{73^2 – 69^2}{69^2} \times 100\%$$ $$A_r = \frac{5329 – 4761}{4761} \times 100\% = \frac{568}{4761} \times 100\% = \mathbf{11.93\%}$$

Chapter 2: Slope Stability Analysis – Exam Solutions

Q.1 Describe different types of slope failures. Derive an expression for the factor of safety of an infinite slope of cohesive soil.
🕒 2079 Regular [3+5]

Part 1: Types of Slope Failures

Slope failures are generally classified into the following types based on the shape of the failure surface and the extent of the soil mass involved:

  • Translational Failure: Occurs in infinite slopes where the failure surface is parallel to the slope surface. A thin layer of soil slides over an underlying hard stratum.
  • Rotational Failure: Common in finite slopes. The soil mass slides along a curved surface.
    • Toe Failure: The failure surface passes through the toe of the slope. This occurs when the slope is steep and the soil is relatively homogeneous.
    • Face Failure: The failure surface passes above the toe, intersecting the slope face. This typically occurs when a weaker soil layer exists above the toe or the slope angle is very high.
    • Base Failure: The failure surface passes below the toe. This occurs when the soil beneath the toe is soft and weak.
  • Compound Failure: A combination of rotational and translational failure. The slip surface is curved at the ends but flat in the middle, often due to the presence of a hard soil layer at a shallow depth.
  • Wedge Failure: Also known as plane failure or block failure, where a wedge of soil slides along a distinct plane of weakness (interface).

Part 2: Factor of Safety for Infinite Slope (Cohesive Soil)

Consider an infinite slope of inclination $\beta$ with the horizontal. The soil is cohesive ($c > 0$) and has an angle of internal friction $\phi$. We consider a prism of soil of width $b$ and vertical depth $z$.

1. Forces acting on the soil element:
Weight of the soil element ($W$): $W = \gamma \cdot z \cdot b \cdot 1$ (considering unit length perpendicular to the page). The weight acts vertically downwards.

2. Resolving the weight on the failure plane:
The failure plane is parallel to the slope surface at depth $z$.

$$N = W \cos \beta = (\gamma z b) \cos \beta$$ $$T = W \sin \beta = (\gamma z b) \sin \beta$$

3. Stresses on the failure plane:
The area of the base of the prism on the failure plane ($A$) is $A = \frac{b}{\cos \beta}$.

$$\text{Normal Stress }(\sigma) = \frac{N}{A} = \frac{\gamma z b \cos \beta}{b / \cos \beta} = \gamma z \cos^2 \beta$$ $$\text{Shear Stress }(\tau) = \frac{T}{A} = \frac{\gamma z b \sin \beta}{b / \cos \beta} = \gamma z \sin \beta \cos \beta$$

4. Shear Strength ($S$):
According to Mohr-Coulomb failure criterion: $S = c + \sigma \tan \phi$. Substituting $\sigma$:

$$S = c + \gamma z \cos^2 \beta \tan \phi$$

5. Factor of Safety ($F$):
The factor of safety is defined as the ratio of shear strength ($S$) to mobilized shear stress ($\tau$).

$$F = \frac{S}{\tau} = \frac{c + \gamma z \cos^2 \beta \tan \phi}{\gamma z \sin \beta \cos \beta}$$ $$\mathbf{F = \frac{c}{\gamma z \sin \beta \cos \beta} + \frac{\tan \phi}{\tan \beta}}$$
Q.2 Define the factor of safety. Explain the procedure of the Method of Slices for the analysis of the stability of finite slopes.
🕒 2079 Back [2+6]

Part 1: Factor of Safety

The Factor of Safety ($F$) in slope stability analysis is defined as the ratio of the available shear strength of the soil to the mobilized shear stress required to maintain equilibrium along a potential failure surface.

$$F = \frac{\text{Shear Strength (Resisting Force)}}{\text{Mobilized Shear Stress (Driving Force)}} = \frac{s}{\tau}$$
  • If $F > 1$, the slope is considered stable.
  • If $F = 1$, the slope is in a state of critical equilibrium (failure is imminent).
  • If $F < 1$, the slope is unstable.

Part 2: Method of Slices (Swedish Circle Method)

This method is used for the stability analysis of finite slopes assuming a circular slip surface. It is applicable to heterogeneous soils ($c-\phi$ soils) and conditions with seepage.

Procedure:

  1. Assume a Slip Surface: Select a trial circular failure surface with center $O$ and radius $R$.
  2. Divide into Slices: Divide the soil mass above the slip surface into a number of vertical slices (usually 6 to 12).
  3. Forces on a Slice: Consider a typical slice. The forces acting on it are Weight ($W$), Cohesion ($c$), Friction, and inter-slice forces (neglected in the simplified method).
  4. Resolve Weight: For each slice, calculate the weight $W$. Resolve $W$ into two components relative to the base of the slice:
    • Normal component ($N$): $N = W \cos \alpha$ (where $\alpha$ is the inclination of the base of the slice).
    • Tangential component ($T$): $T = W \sin \alpha$ (Driving force).
  5. Calculate Resisting and Driving Forces:
    • Total Driving Moment ($M_d$) = $\sum T \times R = R \sum (W \sin \alpha)$.
    • Total Resisting Moment ($M_r$) = $R [c \cdot L + \sum N \tan \phi]$.
  6. Calculate Factor of Safety: The ratio of resisting moments to driving moments.
    $$\mathbf{F = \frac{c L + \tan \phi \sum (W \cos \alpha)}{\sum (W \sin \alpha)}}$$

Note: The analysis is repeated for several trial slip circles to find the minimum factor of safety.

Q.3 What are the different types of slope failures? Explain the friction circle method for the stability analysis of finite slopes.
🕒 2078 Regular [2+6]

Part 1: Types of Slope Failures

(Refer to Q1 Part 1 for the list: Translational, Rotational (Toe, Face, Base), Compound, and Wedge failures.)

Part 2: Friction Circle Method

The Friction Circle Method is a graphical technique for analyzing the stability of finite slopes in homogeneous soils. It assumes that the resultant reaction between the slipping mass and the stable soil is tangent to a “Friction Circle.”

Concept of Friction Circle: A small circle is drawn around the center of rotation ($O$) with a radius $r = R \sin \phi_m$, where $R$ is the radius of the slip circle and $\phi_m$ is the mobilized friction angle.

Procedure:

  1. Trial Surface: Select a trial slip circle (Center $O$, Radius $R$).
  2. Forces: Consider the failure wedge as a rigid body in equilibrium under the action of three forces:
    • Weight ($W$): The total weight of the soil wedge acting vertically through its centroid.
    • Cohesion ($C$): The total cohesive force ($C_m = c_m \times L_{chord}$) acting parallel to the chord of the slip arc at a distance $a$ from the center $O$.
    • Reaction ($P$): The resultant of the normal reaction and frictional resistance along the slip surface.
  3. Direction of Reaction ($P$): The reaction force $P$ must be tangent to the friction circle (radius $r = R \sin \phi$).
  4. Force Triangle:
    • Draw the vector representing weight $W$ (magnitude and direction known).
    • Draw the line of action of the Cohesive Force ($C_m$).
    • Draw the line of action of the Reaction ($P$) passing through the intersection of $W$ and $C_m$ and tangent to the friction circle.
    • Construct the force triangle to determine the magnitude of the mobilized cohesion ($C_m$) required for stability.
  5. Factor of Safety: Usually, a Factor of Safety with respect to friction ($F_\phi$) is assumed, and the Factor of Safety with respect to cohesion ($F_c$) is calculated. The process is repeated until $F_\phi \approx F_c$.
Q.4 An infinite slope is made of clay having a specific gravity of 2.70, a liquid limit of 40%, and a liquidity index of 0.7. The slope angle is $25^\circ$. If the mobilized friction angle is $20^\circ$, determine the mobilized cohesion for a factor of safety of 1.5.
🕒 2078 Back [8]
Given: $G = 2.70$, $LL = 40\%$, $LI = 0.7$, $\beta = 25^\circ$, $\phi_{mob} = 20^\circ$, $F = 1.5$

Step 1: Estimate Soil Unit Weight ($\gamma$)

Formula for Liquidity Index: $LI = \frac{w – PL}{LL – PL}$. Assuming $PI \approx 20\%$, then $PL = 20\%$.

$$0.7 = \frac{w – 20}{40 – 20} \Rightarrow w = 34\%$$ $$e = wG = 0.34 \times 2.70 = 0.918$$ $$\gamma_{sat} = \frac{G + e}{1 + e} \gamma_w = \frac{2.70 + 0.918}{1 + 0.918} \times 9.81 \approx 18.5 \text{ kN/m}^3$$

Step 2: Analysis of Stability

For an infinite slope, the mobilized shear stress must equal the mobilized shear strength.

$$\tau = c_{mob} + \sigma \tan \phi_{mob}$$ $$\gamma z \sin \beta \cos \beta = c_{mob} + (\gamma z \cos^2 \beta) \tan \phi_{mob}$$ $$\Rightarrow c_{mob} = \gamma z \cos^2 \beta (\tan \beta – \tan \phi_{mob})$$

Step 3: Calculation

Using $\cos(25^\circ) = 0.9063$, $\tan(25^\circ) = 0.4663$, $\tan(20^\circ) = 0.3640$:

$$c_{mob} = \gamma z (0.9063)^2 (0.4663 – 0.3640)$$ $$c_{mob} = 0.084 \cdot \gamma \cdot z$$

Assuming critical depth $z = 5 \text{ m}$ (standard assumption when not given):

$$c_{mob} = 0.084 \times 18.5 \times 5 = \mathbf{7.77 \text{ kN/m}^2}$$

Answer: The mobilized cohesion is approx $7.8 \text{ kN/m}^2$.

Q.5 What are the causes of slope failure? Explain the Swedish Circle Method (Method of Slices) for the stability analysis of finite slopes.
🕒 2076 Regular [3+5]

Part 1: Causes of Slope Failure

Slope failures occur when the shear stress exceeds the shear strength of the soil. The causes are categorized as:

  • Factors increasing Shear Stress:
    • Gravity: Weight of the soil mass.
    • External Loading: Surcharge loads from buildings, vehicles, or embankments.
    • Steepening of Slope: Erosion at the toe or excavation.
    • Seepage Pressure: Force exerted by flowing water.
  • Factors decreasing Shear Strength:
    • Increase in Water Content: Softens clay and reduces cohesion.
    • Pore Water Pressure: Reduces effective stress ($\sigma’ = \sigma – u$), thereby reducing frictional strength.
    • Weathering: Degradation of soil structure.
    • Vibration/Earthquakes: Can cause liquefaction or immediate strength loss.

Part 2: Swedish Circle Method

(Refer to the solution in Q2 Part 2 for the detailed explanation of the Method of Slices).

Q.6 Explain the uses of stability charts in slope stability analysis. How are they helpful in the preliminary design of slopes?
🕒 2076 Back [4+4]

Uses of Stability Charts (Taylor’s Charts)

Stability charts are graphical representations relating the Factor of Safety to the slope properties ($c, \phi, \gamma$), slope height ($H$), and slope angle ($\beta$).

  • Determination of Factor of Safety: For a slope with known geometry and soil properties, the Stability Number ($S_n$) can be read from the chart to calculate $F$.
  • Calculation of Critical Height: They allow finding the maximum stable height ($H_c$) for a specific slope angle.
  • Determination of Safe Slope Angle: For a fixed height and desired FOS, the maximum safe angle $\beta$ can be determined.

Helpfulness in Preliminary Design

  • Speed: They provide a very quick estimate of stability without the need for complex iterative calculations or computer software.
  • Feasibility: They help engineers quickly assess whether a proposed slope geometry is feasible or requires modification early in the project.
  • Parametric Study: They allow designers to easily see the effect of changing the slope angle or height on the factor of safety.
Q.7 A cutting of depth 10 m is to be made in a soil having $c = 30 \text{ kN/m}^2$, $\gamma = 19 \text{ kN/m}^3$, and $\phi = 0$. The slope angle is $30^\circ$. Find the factor of safety with respect to cohesion if the stability number ($S_n$) for a $30^\circ$ slope is 0.172.
🕒 2075 Regular [8]
Given: $H = 10 \text{ m}$, $c = 30 \text{ kN/m}^2$, $\gamma = 19 \text{ kN/m}^3$, $\phi = 0$, $\beta = 30^\circ$, $S_n = 0.172$

Formula:

$$S_n = \frac{c_{mob}}{\gamma H} = \frac{c}{F_c \cdot \gamma \cdot H}$$

Calculation:
Rearranging the formula to solve for $F_c$:

$$F_c = \frac{c}{S_n \cdot \gamma \cdot H}$$ $$F_c = \frac{30}{0.172 \times 19 \times 10} = \frac{30}{32.68}$$ $$\mathbf{F_c = 0.918}$$

(Note: Since $F_c < 1$, the slope is theoretically unstable).

Q.8 Differentiate between infinite and finite slopes. Describe the step-by-step procedure for the stability analysis of a finite slope using the Friction Circle Method.
🕒 2075 Back [3+5]

Part 1: Difference between Infinite and Finite Slopes

Feature Infinite Slope Finite Slope
Extent Extends indefinitely in the lateral direction; the surface is constant. Limited in extent; bounded by a top surface (crest) and bottom surface (toe).
Analysis Stability is analyzed for a single vertical prism. The soil properties are constant at any specific depth. Stability is analyzed for the entire soil mass above a specific failure surface.
Failure Type Usually translational failure (parallel to slope). Usually rotational failure (curved slip surface).
Example Long natural hillsides. Embankments, earth dams, highway cuttings.

Part 2: Friction Circle Method Procedure

(Refer to the solution in Q3 Part 2 for the detailed step-by-step procedure).

Q.9 Derive an expression for the factor of safety against sliding for an infinite slope in cohesionless soil ($c=0$) when the slope is submerged.
🕒 2074 Regular [6]

Condition: Infinite slope, Cohesionless soil ($c=0$), Submerged.

1. Assumptions:

  • The slope is submerged under static water.
  • The effective stress analysis uses the submerged unit weight $\gamma’$.

2. Forces on the Element:

Consider a soil element of depth $z$ on a slope of angle $\beta$.

$$\text{Effective Normal Stress } (\sigma’) = \gamma’ z \cos^2 \beta$$ $$\text{Shear Stress } (\tau) = \gamma’ z \sin \beta \cos \beta$$

(Where $\gamma’ = \gamma_{sat} – \gamma_w$).

3. Shear Strength ($s$):

For cohesionless soil ($c=0$), strength is purely frictional: $s = \sigma’ \tan \phi$.

$$s = (\gamma’ z \cos^2 \beta) \tan \phi$$

4. Factor of Safety ($F$):

$$F = \frac{\text{Resisting Strength ($s$)}}{\text{Driving Stress ($\tau$)}} = \frac{\gamma’ z \cos^2 \beta \tan \phi}{\gamma’ z \sin \beta \cos \beta}$$ $$F = \frac{\cos \beta \tan \phi}{\sin \beta}$$ $$\mathbf{F = \frac{\tan \phi}{\tan \beta}}$$

(Note: For a fully submerged slope without seepage flow, the factor of safety is the same as for a dry slope).

Q.10 Explain Taylor’s Stability Number. How is it used to determine the stability of slopes?
🕒 2074 Back [4+4]

Taylor’s Stability Number ($S_n$)

Taylor (1937) analyzed the stability of friction circles and introduced a dimensionless parameter called the Stability Number ($S_n$). It relates the cohesion ($c$) required to maintain stability to the unit weight ($\gamma$) and height ($H$) of the slope.

$$S_n = \frac{c}{F_c \cdot \gamma \cdot H}$$

If the slope is just stable ($F_c = 1$): $S_n = \frac{c}{\gamma H_c}$.

$S_n$ is primarily a function of the slope angle ($\beta$) and the mobilized friction angle ($\phi_m$).

Usage in Determining Stability

  1. Using Stability Charts: Taylor provided charts (nomograms) where $S_n$ is plotted against the slope angle $\beta$ for various values of $\phi$.
  2. Calculating Factor of Safety:
    • Determine $\beta$ and $\phi$ for the given slope.
    • Read the value of $S_n$ from Taylor’s chart.
    • Calculate the Factor of Safety using the formula: $F = \frac{c}{S_n \cdot \gamma \cdot H}$.
  3. Design: It is used to calculate the required cohesion or the maximum permissible height for a slope to ensure a desired factor of safety.
Q.11 A canal is to be excavated through a soil with $c = 15 \text{ kN/m}^2$, $\phi = 20^\circ$, $e = 0.9$, and $G = 2.67$. The side slope is 1 in 1. The depth of the canal is 6 m. Determine the factor of safety with respect to cohesion when the canal runs full. (Assume Taylor’s stability number $S_n$ for these conditions).
🕒 2073 Regular [8]

Given Data:

  • Cohesion ($c$) = $15 \text{ kN/m}^2$
  • Angle of internal friction ($\phi$) = $20^\circ$
  • Void ratio ($e$) = $0.9$
  • Specific gravity ($G$) = $2.67$
  • Side slope = 1:1 (Horizontal:Vertical), therefore Slope angle $\beta = 45^\circ$
  • Depth of canal ($H$) = $6 \text{ m}$
  • Condition: Canal runs full (Submerged slope)

Formulae Used:

Saturated Unit Weight:

$$\gamma_{sat} = \left( \frac{G + e}{1 + e} \right) \gamma_w$$

Submerged Unit Weight:

$$\gamma’ = \gamma_{sat} – \gamma_w$$

Taylor’s Stability Number Equation:

$$S_n = \frac{c}{F_c \cdot \gamma’ \cdot H}$$

(Note: For submerged slopes, the submerged unit weight $\gamma’$ is used).

Calculations:

Step 1: Calculate Unit Weights

Assuming unit weight of water $\gamma_w = 9.81 \text{ kN/m}^3$.

$$\gamma_{sat} = \left( \frac{2.67 + 0.9}{1 + 0.9} \right) \times 9.81$$ $$\gamma_{sat} = \left( \frac{3.57}{1.9} \right) \times 9.81$$ $$\gamma_{sat} = 1.8789 \times 9.81 = 18.43 \text{ kN/m}^3$$ $$\gamma’ = 18.43 – 9.81 = 8.62 \text{ kN/m}^3$$

Step 2: Determine Taylor’s Stability Number ($S_n$)

From Taylor’s Stability Charts for $\phi = 20^\circ$ and Slope angle $\beta = 45^\circ$:

The approximate value of $S_n$ is 0.062.

Step 3: Calculate Factor of Safety ($F_c$)

Rearranging the stability equation to solve for $F_c$:

$$F_c = \frac{c}{S_n \cdot \gamma’ \cdot H}$$ $$F_c = \frac{15}{0.062 \times 8.62 \times 6}$$ $$F_c = \frac{15}{3.2066}$$ $$F_c = 4.677$$

Answer:

$$\boxed{F_c \approx 4.68}$$
Q.12 Discuss the stability analysis of an infinite slope in cohesive soil with steady seepage parallel to the slope.
🕒 2073 Back [6]

Stability Analysis of Infinite Slope in Cohesive Soil (with Seepage):

An infinite slope is one that extends indefinitely in steady boundaries. In this analysis, the failure plane is assumed to be parallel to the ground surface at a depth $z$.

Conditions:

  • Soil is cohesive ($c – \phi$ soil).
  • Seepage is steady and parallel to the slope.
  • The water table is at the ground surface (worst-case scenario for seepage parallel to slope).

Forces Acting on a Soil Element:

Consider a prismatic slice of soil of width $b$ and depth $z$ on a slope inclined at angle $\beta$.

Total Weight ($W$): The weight of the saturated soil slice acts vertically downwards.

$$W = \gamma_{sat} \cdot z \cdot b$$

Normal Stress ($\sigma$):

$$\sigma = \gamma_{sat} \cdot z \cdot \cos^2 \beta$$

Shear Stress ($\tau$): The mobilised shear stress due to gravity.

$$\tau = \gamma_{sat} \cdot z \cdot \cos \beta \sin \beta$$

Pore Water Pressure ($u$): Due to seepage parallel to the slope, the flow net equipotential lines are perpendicular to the slope. The pore pressure at depth $z$ is:

$$u = \gamma_w \cdot z \cdot \cos^2 \beta$$

Effective Stress Analysis:

The shear strength is governed by effective stress parameters ($c’$ and $\phi’$).

Effective Normal Stress ($\sigma’$):

$$\sigma’ = \sigma – u$$ $$\sigma’ = (\gamma_{sat} \cdot z \cdot \cos^2 \beta) – (\gamma_w \cdot z \cdot \cos^2 \beta)$$ $$\sigma’ = (\gamma_{sat} – \gamma_w) z \cos^2 \beta = \gamma’ z \cos^2 \beta$$

Shear Strength ($S$):

$$S = c’ + \sigma’ \tan \phi’$$ $$S = c’ + \gamma’ z \cos^2 \beta \tan \phi’$$

Factor of Safety ($F$):

The factor of safety is the ratio of resisting shear strength to the mobilizing shear stress.

$$F = \frac{S}{\tau} = \frac{c’ + \gamma’ z \cos^2 \beta \tan \phi’}{\gamma_{sat} z \cos \beta \sin \beta}$$

Conclusion: For a cohesive soil with seepage parallel to the slope, the stability depends on the effective cohesion, the submerged unit weight (contributing to friction), and the saturated unit weight (contributing to the driving force).

Q.13 Explain the $\phi_u = 0$ analysis for finite slopes. In which conditions is this analysis applicable?
🕒 2072 Regular [4+2]

Explanation of $\phi_u = 0$ Analysis:

The $\phi_u = 0$ analysis (or “Phi-Zero” analysis) is a total stress method used to determine the stability of slopes in saturated clayey soils under undrained conditions.

  • Concept: In this analysis, it is assumed that the angle of shearing resistance with respect to total stress is zero ($\phi_u = 0$). The shear strength of the soil is purely due to undrained cohesion ($c_u$). $$\tau_f = c_u$$
  • Failure Surface: The failure surface is typically assumed to be a circular arc.

Method:

Consider a trial slip circle of radius $R$.

Driving Moment ($M_d$): Caused by the weight of the sliding soil mass ($W$) acting at a lever arm ($\bar{x}$) from the center of rotation.

$$M_d = W \cdot \bar{x}$$

Resisting Moment ($M_r$): Provided by the cohesion acting along the slip arc length ($L$). Since $\phi_u = 0$, friction does not contribute to resistance.

$$M_r = c_u \cdot L \cdot R$$

(Where $L = R \cdot \theta$, $\theta$ in radians).

Factor of Safety ($F$):

$$F = \frac{M_r}{M_d} = \frac{c_u \cdot L \cdot R}{W \cdot \bar{x}}$$

Applicability Conditions:

The $\phi_u = 0$ analysis is applicable in the following specific conditions:

  • Immediately after Construction: When an embankment is constructed rapidly over a saturated clay foundation, or a cut is made in saturated clay.
  • Saturated Clays: Specifically for fully saturated clays where the application of load leads to an immediate increase in pore pressure, preventing any immediate volume change or drainage.
  • Short-term Stability: It assesses the “End-of-Construction” stability before pore water pressures have had time to dissipate.
Q.14 What are the assumptions made in the analysis of slope stability? Distinguish between face failure, toe failure, and base failure with neat sketches.
🕒 2072 Back [3+5]

Assumptions in Slope Stability Analysis:

  • Two-Dimensional Analysis: The problem is treated as two-dimensional; stresses and strains parallel to the slope length are neglected (Plane strain condition).
  • Shear Strength: The shear strength of the soil follows Coulomb’s Law: $\tau = c + \sigma \tan \phi$.
  • Homogeneity: The soil mass is assumed to be homogeneous and isotropic (unless specific layers are defined).
  • Limit Equilibrium: The factor of safety is assumed to be constant along the entire failure surface. The soil mass is on the verge of failure.
  • Failure Surface: The slip surface is generally assumed to be a circular arc (or a specific shape like a logarithmic spiral or plane).

Distinction between Failure Types:

  • 1. Face Failure:
    • Description: The failure surface intersects the slope face above the toe.
    • Condition: Occurs when the slope angle is very high, or a weak soil layer exists in the upper part of the slope.
    • Sketch: Draw a slope. Draw a curved line starting from the top surface and exiting on the slope face (above the bottom corner).
  • 2. Toe Failure:
    • Description: The failure surface passes exactly through the toe of the slope.
    • Condition: Most common mode of failure for steep slopes in homogeneous soil.
    • Sketch: Draw a slope. Draw a curved line starting from the top surface and exiting exactly at the bottom corner (toe).
  • 3. Base Failure:
    • Description: The failure surface passes below the toe into the foundation soil.
    • Condition: Occurs when the slope angle is low ($\beta < 53^\circ$) or the foundation soil is weaker than the embankment soil.
    • Sketch: Draw a slope. Draw a curved line starting from the top surface, going deep under the ground, and exiting on the ground surface far beyond the toe.
Q.15 An embankment is to be constructed with a slope of $30^\circ$ and a height of 10 m. The soil parameters are $c’ = 25 \text{ kN/m}^2$, $\phi’ = 25^\circ$, and $\gamma = 18 \text{ kN/m}^3$. Using the Method of Slices, determine the factor of safety along a given trial slip circle.
🕒 2071 Regular [8]

Note: The specific geometry of the “given trial slip circle” (radius, center coordinates, or slice details) is required to perform the numerical summation. In an exam context, this would be provided via a diagram or a table of slice data. Below is the solution setup using the standard Fellenius (Swedish) Method of Slices which you must follow.

Given Data:

  • Slope Angle ($\beta$) = $30^\circ$
  • Height ($H$) = $10 \text{ m}$
  • Cohesion ($c’$) = $25 \text{ kN/m}^2$
  • Friction Angle ($\phi’$) = $25^\circ$
  • Unit Weight ($\gamma$) = $18 \text{ kN/m}^3$

Methodology:

The soil mass above the slip circle is divided into vertical slices. The Factor of Safety ($F$) is given by:

$$F = \frac{\sum [c’ \Delta L + (W \cos \alpha – u \Delta L) \tan \phi’]}{\sum W \sin \alpha}$$

Where:

  • $c’, \phi’$ = Effective shear strength parameters.
  • $\Delta L$ = Length of the curved arc for a slice ($b / \cos \alpha$).
  • $W$ = Weight of the slice ($\gamma \times \text{Height} \times \text{Width}$).
  • $\alpha$ = Angle of inclination of the tangent to the slip circle at the base of the slice.
  • $u$ = Pore water pressure at the base of the slice (if explicitly given; if dry/moist without seepage, $u=0$).

Step-by-Step Calculation Procedure:

  1. Draw the Slope: To scale, using $\beta=30^\circ$ and $H=10m$.
  2. Draw Slip Circle: Using the given center and radius.
  3. Divide into Slices: Divide the soil mass into vertical slices (usually 6-10 slices).
  4. Tabulate Data: Create a table with columns for Slice No., Width ($b$), Avg Height ($h$), Weight ($W$), Base Angle ($\alpha$), $\Delta L$, $N = W \cos \alpha$, and $T = W \sin \alpha$.

Calculate FOS: (Assuming no pore pressure $u=0$ for simplicity unless specified)

$$F = \frac{c’ \sum \Delta L + \tan \phi’ \sum N}{\sum T}$$

Substituting given values:

$$F = \frac{25 \cdot (\sum \Delta L) + \tan(25^\circ) \cdot (\sum W \cos \alpha)}{\sum W \sin \alpha}$$

To solve this completely, one would need to measure $b$, $h$, and $\alpha$ from the diagram. Ensure you show this table in the exam.

Q.16 Describe the procedure for locating the critical slip circle for a slope in a generic $c-\phi$ soil.
🕒 2070 Regular [5]

For a $c-\phi$ soil, the critical slip circle (the one yielding the minimum Factor of Safety) does not necessarily pass through the toe. The Fellenius guideline is commonly used to locate the center of the critical circle.

Procedure:

  1. Locate Center $O_1$: Determine the directional angles $\alpha_1$ and $\beta_1$ based on the slope angle $i$ (or $\beta$). Fellenius provides a table or chart where these angles are defined relative to the toe and the top of the slope. Plot the point defined by these angles to find the first trial center, $O_1$.
  2. Locate Center $P$ (Directional Line): Draw a line from the toe of the slope making an angle with the horizontal (often denoted as $\psi$ or derived from empirical charts). The centers of critical circles often lie on a specific locus.
  3. Establish the Locus of Centers: Connect $O_1$ with the theoretical point required for the given slope geometry. This line serves as the locus for potential centers of critical circles.
  4. Trial and Error (Grid Search): Select multiple trial centers along this locus or in a grid pattern near the theoretical critical center. For each center, draw a slip circle passing through the toe (or below it for base failure).
  5. Calculate the Factor of Safety ($F$): For each circle using the Method of Slices or Taylor’s method.
  6. Identify Minimum FOS: Plot the calculated $F$ values as contours (iso-safety lines) around the trial centers. The center point corresponding to the lowest value of $F$ represents the Critical Slip Circle.
Q.17 Derive the expression for the factor of safety of an infinite slope for a $c-\phi$ soil when the water table is at the ground surface and seepage is taking place parallel to the slope.
🕒 2070 Back [6]

Derivation:

Consider an infinite slope with inclination $\beta$. The water table is at the ground surface, and seepage occurs parallel to the slope.

Element Definition:

Consider a soil element of depth $z$ and unit area on the slope.

  • Saturated Unit Weight = $\gamma_{sat}$
  • Unit Weight of Water = $\gamma_w$
  • Submerged Unit Weight = $\gamma’ = \gamma_{sat} – \gamma_w$

Mobilizing Stress ($\tau$):

The shear stress causing instability is due to the component of the total saturated weight parallel to the slope.

$$\tau = \gamma_{sat} \cdot z \cdot \sin \beta \cos \beta$$

Resisting Strength ($S$):

The soil strength is governed by the Mohr-Coulomb equation in terms of effective stress:

$$S = c’ + \sigma’ \tan \phi’$$

Total Normal Stress ($\sigma$):

$$\sigma = \gamma_{sat} \cdot z \cdot \cos^2 \beta$$

Pore Water Pressure ($u$):

From the flow net for flow parallel to the surface, the pore pressure head at depth $z$ corresponds to the vertical depth adjusted for the slope:

$$u = \gamma_w \cdot z \cdot \cos^2 \beta$$

Effective Normal Stress ($\sigma’$):

$$\sigma’ = \sigma – u$$ $$\sigma’ = (\gamma_{sat} \cdot z \cdot \cos^2 \beta) – (\gamma_w \cdot z \cdot \cos^2 \beta)$$ $$\sigma’ = (\gamma_{sat} – \gamma_w) z \cos^2 \beta$$ $$\sigma’ = \gamma’ z \cos^2 \beta$$

Substituting $\sigma’$ back into the strength equation:

$$S = c’ + \gamma’ z \cos^2 \beta \tan \phi’$$

Factor of Safety ($F$):

The Factor of Safety is the ratio of Resisting Strength to Mobilizing Stress.

$$F = \frac{S}{\tau}$$ $$F = \frac{c’ + \gamma’ z \cos^2 \beta \tan \phi’}{\gamma_{sat} z \sin \beta \cos \beta}$$

Dividing the numerator and denominator by the term $\gamma_{sat} z \sin \beta \cos \beta$, or simply simplifying the frictional component:

$$\boxed{F = \frac{c’}{\gamma_{sat} z \sin \beta \cos \beta} + \frac{\gamma’}{\gamma_{sat}} \frac{\tan \phi’}{\tan \beta}}$$

(This is the standard form of the expression).

Chapter 3: Earth Pressure Theories – Solutions

Q.1 Differences between Rankine’s and Coulomb’s theory.
🕒 2082 Bhadra [2]
Feature Coulomb’s Theory Rankine’s Theory
Wall Friction Friction between the wall and the backfill is considered ($\delta \neq 0$). Friction between the wall and the backfill is not considered ($\delta = 0$), assuming the wall is smooth.
Wall Inclination Applicable for any inclination of the back of the wall. The back of the wall must be vertical (though modifications exist, basic theory assumes vertical).
Methodology More rational/versatile. Deals with total thrust (wedge theory). Relatively simple. Deals with stress at a point (stress analysis).
Result Accuracy More accurate for active earth pressure. Often overestimates active pressure if wall friction is ignored.
Solution Type Graphical methods (Culmann’s, Rebhann’s) available. Primarily analytical.
Q.2 Retaining wall 6 m high, two layers: 3 m sand ($\gamma=17.5$ kN/m³, $\phi=30^\circ$) over saturated sandy clay ($\gamma=19$ kN/m³, $\phi=30^\circ$, $c=10$ kN/m²), GWT at sandy clay surface. Calculate Active pressure distribution, total thrust/m, and point of application.
🕒 2082 Bhadra [6]

Given Data:

  • Total Height $H = 6$ m.
  • Layer 1 (0-3m) Sand: $\gamma_1 = 17.5$ kN/m³, $\phi_1 = 30^\circ \Rightarrow K_{a1} = \frac{1}{3}$, $c_1=0$.
  • Layer 2 (3-6m) Sandy Clay: $\gamma_{sat} = 19$ kN/m³, $\phi_2 = 30^\circ \Rightarrow K_{a2} = \frac{1}{3}$, $c_2=10$ kN/m².
  • GWT: At depth 3 m (Interface).

Step 1: Pressure Calculations ($p_a = K_a \sigma’_v – 2c\sqrt{K_a} + u$)

  • At $z=0$ m: $\sigma’_v = 0 \Rightarrow p_a = 0$.
  • At $z=3$ m (In Sand): $$p_a = \frac{1}{3}(17.5 \times 3) = 17.5 \text{ kN/m}^2$$
  • At $z=3$ m (In Clay): $$p_a = K_{a2}\sigma’_v – 2c_2\sqrt{K_{a2}} = \frac{1}{3}(52.5) – 2(10)\sqrt{0.333}$$ $$p_a = 17.5 – 11.54 = 5.96 \text{ kN/m}^2$$
  • At $z=6$ m (Base): $$\sigma’_v = 52.5 + (19 – 9.81) \times 3 = 80.07 \text{ kN/m}^2$$ $$u = 9.81 \times 3 = 29.43 \text{ kN/m}^2$$ $$p_{soil} = \frac{1}{3}(80.07) – 11.54 = 15.15 \text{ kN/m}^2$$ $$p_{total} = 15.15 + 29.43 = 44.58 \text{ kN/m}^2$$

Step 2: Thrust (P) and Moment (M) Calculation

1. Sand Triangle: $P_1 = 0.5 \times 3 \times 17.5 = 26.25$ kN (at 4.0m)
2. Clay Surcharge Rect: $P_2 = 3 \times 5.96 = 17.88$ kN (at 1.5m)
3. Clay Soil Triangle: $P_3 = 0.5 \times 3 \times (15.15 – 5.96) = 13.79$ kN (at 1.0m)
4. Water Triangle: $P_4 = 0.5 \times 3 \times 29.43 = 44.15$ kN (at 1.0m)

Totals:

$$P_{total} = 26.25 + 17.88 + 13.79 + 44.15 = \mathbf{102.07 \text{ kN/m}}$$ $$M_{total} = 105.0 + 26.82 + 13.79 + 44.15 = 189.76 \text{ kNm/m}$$

Point of Application:

$$\bar{z} = \frac{189.76}{102.07} = \mathbf{1.86 \text{ m}} \text{ from base}$$
Q.3 Step-by-step procedure (sketch) for Culmann’s graphical method – active earth pressure, no surcharge, cohesionless soil.
🕒 2081 Bhadra [4]

Culmann’s method is based on Coulomb’s wedge theory.

  1. Draw Wall: Draw retaining wall $AB$ to scale.
  2. $\phi$-line: From heel $B$, draw a line $BD$ making angle $\phi$ with horizontal.
  3. Ground Line: Draw the backfill surface profile.
  4. E-Line (Earth Pressure Line): Draw line $BL$ from $B$ making angle $\psi = \beta – \delta$ with the $\phi$-line (where $\beta$ is backfill slope, $\delta$ is wall friction).
  5. Failure Planes: Draw arbitrary planes $BC_1, BC_2, \dots$ cutting ground at $C_1, C_2 \dots$.
  6. Plot Weights: Calculate wedge weights ($W_1, W_2 \dots$) and plot them along the $\phi$-line from $B$ (e.g., $Bd_1 = W_1$).
  7. Parallels: From points $d_1, d_2 \dots$, draw lines parallel to the E-line ($BL$) intersecting failure planes at $e_1, e_2 \dots$.
  8. Culmann’s Curve: Join $e_1, e_2 \dots$ with a smooth curve.
  9. Max Pressure: Draw a tangent to the curve parallel to the $\phi$-line. Let point of tangency be $F$.
  10. Result: Draw $FG$ parallel to E-line intersecting $\phi$-line. Length $FG$ represents Active Earth Pressure ($P_a$).
Q.4 Derive relation for maximum height of unsupported excavation in clayey soil.
🕒 2081 Bhadra [4]

Objective: Find critical height ($H_c$) where vertical cut stands without support.

Bell’s equation for $\phi=0, K_a=1$: $$p_a = \gamma z – 2c$$

1. At surface ($z=0$), $p_a = -2c$ (Tension).
2. Depth of tension crack ($z_0$): $p_a = 0 \Rightarrow \gamma z_0 = 2c \Rightarrow z_0 = \frac{2c}{\gamma}$.

Condition for Unsupported Excavation: Total net lateral force must be zero (Tensile zone balances Compressive zone).

$$P_a = \int_{0}^{H_c} (\gamma z – 2c) dz = 0$$ $$\left[ \frac{\gamma z^2}{2} – 2cz \right]_{0}^{H_c} = 0$$ $$\frac{\gamma H_c^2}{2} – 2cH_c = 0$$

Dividing by $H_c$:

$$\frac{\gamma H_c}{2} = 2c \Rightarrow \mathbf{H_c = \frac{4c}{\gamma}}$$
Q.5 Earth pressure coefficient and its types.
🕒 2081 Baishakh (Back) [2]

Definition: The ratio of horizontal effective stress ($\sigma’_h$) to vertical effective stress ($\sigma’_v$). $$K = \frac{\sigma’_h}{\sigma’_v}$$

Types:

  • At Rest ($K_0$): Rigid wall, no movement. $K_0 \approx 1 – \sin\phi$.
  • Active ($K_a$): Wall moves away from soil. $K_a = \frac{1 – \sin\phi}{1 + \sin\phi}$.
  • Passive ($K_p$): Wall moves towards soil. $K_p = \frac{1 + \sin\phi}{1 – \sin\phi}$.
Q.6 Earth pressure diagram for purely cohesive soil ($\gamma_t, \phi=0, c$), rigid wall height H, active condition. Magnitude & line of action of force/unit length from base.
🕒 2081 Baishakh (Back) [6]

1. Pressure Diagram:
Equation: $p_a = \gamma z – 2c$.
Top ($z=0$): $-2c$ (Tension).
Zero pressure at $z_0 = 2c/\gamma$.
Base ($z=H$): $\gamma H – 2c$ (Compression).

2. Magnitude ($P_a$):
Ignorning tensile cracks, we calculate force for the positive pressure triangle only (below $z_0$).
Height of comp. zone = $H – \frac{2c}{\gamma}$. Base width = $\gamma H – 2c$.

$$P_a = \frac{1}{2} \times \text{Base} \times \text{Height}$$ $$P_a = \frac{1}{2} (\gamma H – 2c)(H – \frac{2c}{\gamma}) = \mathbf{\frac{1}{2}\gamma\left(H – \frac{2c}{\gamma}\right)^2}$$

3. Line of Action:
Acts at the centroid of the positive triangle.

$$\bar{y} = \frac{1}{3}\left(H – \frac{2c}{\gamma}\right) \text{ from the base}$$
Q.7 Culmann’s graphical method – passive earth pressure (sketch).
🕒 2080 Bhadra [6]

Procedure is similar to Active but with geometric differences reflecting passive resistance:

  1. Draw wall $AB$ and ground line.
  2. $\phi$-line: Drawn below the horizontal from heel $B$ at angle $\phi$.
  3. E-line: Drawn below $\phi$-line based on wall friction angle $\delta$.
  4. Draw failure wedges $BC_1, BC_2 \dots$ and plot weights $W$ on the $\phi$-line.
  5. Draw lines parallel to E-line from weight points to intersect failure planes.
  6. Join intersections to form the Passive Curve.
  7. Result: Find the point on the curve that gives the minimum value (lowest ordinate) relative to the $\phi$-line tangent. This represents $P_p$.
Q.8 a) Why retaining walls designed for active pressure? b) How Coulomb’s theory improves on Rankine’s.
🕒 2080 Baishakh [2+2]

a) Design for Active Pressure:

  • Yielding: Walls (gravity/cantilever) usually yield/tilt slightly, determining the mobilization of shear strength.
  • Economy: Active pressure ($K_a \approx 0.33$) is the minimum pressure compared to At-Rest or Passive ($K_p \approx 3$). Designing for this lower value makes structures economical.

b) Coulomb’s Improvement over Rankine:

  • Wall Friction: Coulomb accounts for friction ($\delta$) between wall and soil, leading to more realistic/economical designs (Rankine assumes smooth wall).
  • Versatility: Can easily handle inclined walls, irregular backfill slopes, and surcharge loads via the wedge theory.
Q.9 a) Reasons for not considering passive pressure in rigid retaining structures design. b) Culmann’s method – active thrust, inclined backfill + line load (sketch). c) 6 m wall, vertical back, horizontal cohesionless backfill $\gamma=17$ kN/m³, $\phi=30^\circ$, surcharge 25 kN/m², WT 3.5 m from base. Total active thrust & direction.
🕒 2079 Bhadra [Various]

Part (a): Reasons for not considering passive pressure

Passive earth pressure is generally not relied upon in the design of rigid retaining structures for the following reasons:

  • Magnitude of Wall Movement: The wall movement required to mobilize full passive resistance is very large (typically 2% to 10% of height). Rigid walls often do not undergo such displacement.
  • Uncertainty of Soil Presence: Soil at the toe may be removed by scouring, erosion, or future excavation (e.g., utility trenches).
  • Safety Conservative Approach: Neglecting passive pressure yields a safer design.
  • Weathering: Toe soil is shallow and susceptible to softening from environmental factors.

Part (b): Culmann’s Method (Line Load)

Procedure:

  1. Draw wall $AB$ and backfill surface $AC$. Draw $\phi$-line $AD$ at angle $\phi$ to horizontal.
  2. Draw $\theta$-line $AE$ at angle $\psi = \alpha – \delta$ to the $\phi$-line.
  3. For a trial wedge $ABC_1$, plot weight $W_1$ on $\phi$-line as vector $Ad_1$.
  4. Line Load Effect: Add the line load $q_L$ to the wedge weight ($W_1 + q_L$). Plot modified weight $Ad_1’$.
  5. Draw a line from $d_1’$ parallel to the $\theta$-line to intersect the failure ray $AC_1$.
  6. The locus of intersections gives the Culmann curve; the maximum intercept represents $P_a$.

Part (c): Numerical Solution

Given: $H = 6\text{ m}$, $\gamma = 17\text{ kN/m}^3$, $\phi = 30^\circ$, $q = 25\text{ kN/m}^2$. WT at $3.5\text{ m}$ from base ($H_2$). Dry depth $H_1 = 2.5\text{ m}$.
$$K_a = \frac{1 – \sin 30^\circ}{1 + \sin 30^\circ} = \frac{1}{3} = 0.333$$ $$\gamma’ = 17 – 9.81 = 7.19 \text{ kN/m}^3$$

Pressure Intensities:

At Top ($z=0$): $p_a = K_a q = 0.333 \times 25 = 8.33 \text{ kN/m}^2$
At WT ($z=2.5$): $p_a = 8.33 + (0.333 \times 17 \times 2.5) = 8.33 + 14.17 = 22.50 \text{ kN/m}^2$
At Base ($z=6$): $p_{soil} = 22.50 + (0.333 \times 7.19 \times 3.5) = 30.88 \text{ kN/m}^2$ (Soil+Surcharge only)
Water Pressure at Base: $u = 9.81 \times 3.5 = 34.34 \text{ kN/m}^2$

Total Thrust Calculation ($P_{Total}$):

$P_1$ (Surcharge Rect): $8.33 \times 6 = 49.98$ kN/m (at 3.0m)
$P_2$ (Soil Top Tri): $0.5 \times 14.17 \times 2.5 = 17.71$ kN/m (at $3.5 + 2.5/3 = 4.33$m)
$P_3$ (Soil Top Rect acting on bottom): $14.17 \times 3.5 = 49.60$ kN/m (at $1.75$m)
$P_4$ (Soil Bottom Tri): $0.5 \times (0.333 \times 7.19 \times 3.5) \times 3.5 = 14.66$ kN/m (at $1.17$m)
$P_5$ (Water Tri): $0.5 \times 34.34 \times 3.5 = 60.09$ kN/m (at $1.17$m)
$P_{Total} = 49.98 + 17.71 + 49.60 + 14.66 + 60.09 = 192.04 \text{ kN/m}$

Point of Application ($\bar{z}$):

$$192.04 \times \bar{z} = (49.98 \times 3) + (17.71 \times 4.33) + (49.60 \times 1.75) + (14.66 \times 1.17) + (60.09 \times 1.17)$$ $$192.04 \times \bar{z} = 149.94 + 76.68 + 86.8 + 17.15 + 70.30 = 400.87$$ $$\bar{z} = \frac{400.87}{192.04} = \mathbf{2.09 \text{ m from base}}$$
Q.10 Retaining wall 7.5 m, two layers: 4.5 m sand ($\gamma=18$, $\phi=35^\circ$) over saturated clayey ($\gamma=19.5$, $\phi=30^\circ$, $c=16$), GWT at clay surface. Active pressure distribution, total thrust/m, point of application.
🕒 2078 Kartik [8] 🕒 2074 Chaitra [10]
Layer 1: $H_1=4.5$m, $\gamma=18$, $\phi=35^\circ$. Layer 2: $H_2=3.0$m, $\gamma_{sat}=19.5$, $\phi=30^\circ$, $c=16$.
$K_{a1} = \frac{1-\sin 35}{1+\sin 35} = 0.271$
$K_{a2} = \frac{1-\sin 30}{1+\sin 30} = 0.333$

Pressure Calculation ($p_a = K_a \sigma_v – 2c\sqrt{K_a}$):

At $z=0$: $p_a = 0$
At $z=4.5$ (Just above): $p_a = 0.271 \times (18 \times 4.5) = 21.95 \text{ kN/m}^2$
At $z=4.5$ (Just below): $\sigma_v = 81$. $p_a = 0.333(81) – 2(16)\sqrt{0.333} = 26.97 – 18.46 = 8.51 \text{ kN/m}^2$
At $z=7.5$ (Base): $\sigma_v’ = 81 + (19.5-9.81)\times 3 = 110.07$.
$p_{soil} = 0.333(110.07) – 18.46 = 18.19 \text{ kN/m}^2$
Water Pressure: $u = 9.81 \times 3 = 29.43 \text{ kN/m}^2$

Total Thrust & Moments:

$P_1$ (Top Tri): $0.5 \times 21.95 \times 4.5 = 49.39$ kN (Arm: $3+1.5 = 4.5$m)
$P_2$ (Bottom Rect): $8.51 \times 3 = 25.53$ kN (Arm: $1.5$m)
$P_3$ (Bottom Soil Tri): $0.5 \times (18.19-8.51) \times 3 = 14.52$ kN (Arm: $1.0$m)
$P_4$ (Water Tri): $0.5 \times 29.43 \times 3 = 44.15$ kN (Arm: $1.0$m)
$$P_{Total} = 49.39 + 25.53 + 14.52 + 44.15 = \mathbf{133.59 \text{ kN/m}}$$

Point of Application:

$$\bar{z} = \frac{(49.39 \times 4.5) + (25.53 \times 1.5) + (14.52 \times 1) + (44.15 \times 1)}{133.59}$$ $$\bar{z} = \frac{319.22}{133.59} = \mathbf{2.39 \text{ m from base}}$$
Q.11 a) Lateral earth pressure force vs wall movement plot. b) Four basic assumption differences Rankine vs Coulomb.
🕒 2078 Bhadra [2+4]

Part (a): Pressure vs Movement

The plot depicts Earth Pressure (Y-axis) vs. Wall Strain (X-axis):

  • At Rest ($K_0$): Zero movement.
  • Active ($P_a$): Wall moves away. Pressure decreases to minimum value ($P_a$) at small strain.
  • Passive ($P_p$): Wall moves towards. Pressure increases to maximum ($P_p$) at large strain. $P_p \gg P_a$.

Part (b): Rankine vs Coulomb

Feature Rankine Coulomb
Wall Friction Neglected (Smooth wall). Considered ($\delta$).
Failure Surface Planar (derived from stress element). Planar Wedge (Trial wedge).
Analysis Type Plastic Equilibrium (Stress at a point). Wedge Equilibrium (Force balance).
Versatility Limited (Simple boundaries). Versatile (Inclined backs, slopes).
Q.12 a) Reasons for not considering passive pressure in stability analysis. Culmann’s method – active thrust, inclined backfill + surcharge. b) Total passive thrust & point of application procedure.
🕒 2076 Chaitra [Various]

Part (a): Culmann’s Method with Surcharge

For uniform surcharge $q$: Convert it to equivalent soil height $h_e = q/\gamma$. The weight of any trial wedge is increased by the surcharge load acting on top of it ($W_{total} = W_{soil} + q \times \text{width}$). Plot these total weights on the $\phi$-line and proceed as usual to find the max intercept.

Part (b): Passive Thrust Procedure

  • Use $K_p = (1+\sin\phi)/(1-\sin\phi)$.
  • Calculate effective vertical stress $\sigma_v’$.
  • Formula: $p_p = K_p \sigma_v’ + 2c\sqrt{K_p} + u$. (Note: Cohesion adds to resistance).
  • Integrate the area of the pressure diagram to find Total Thrust $P_p$.
Q.13 a) “General State of Plastic Equilibrium” meaning. b) 6 m vertical wall, saturated cohesive backfill (top 3 m $\gamma=18, c=18$; bottom 3 m $\gamma=20, c=25$). Tension crack depth, active pressure if crack develops, distribution & resultant point.
🕒 2076 Ashwin [Various]

Part (a): Plastic Equilibrium

A soil mass is in a general state of plastic equilibrium when shear stress at every point on the potential failure plane has reached the soil’s shear strength ($\tau_f$). The entire mass is on the verge of yielding.

Part (b): Numerical Solution

Assumed $\phi=0$ (Saturated Cohesive). $K_a=1$.
Top Layer (0-3m): $\gamma=18, c=18$.
Bottom Layer (3-6m): $\gamma=20, c=25$.

Pressure & Tension Crack:

At $z=0$: $p_a = -2c = -36 \text{ kN/m}^2$.
Tension Crack Depth $z_c$: $p_a = \gamma z_c – 2c = 0 \Rightarrow z_c = \frac{2(18)}{18} = \mathbf{2.0 \text{ m}}$
At $z=3$ (Top Interface): $p_a = (18 \times 3) – 36 = 18 \text{ kN/m}^2$
At $z=3$ (Bottom Interface): $p_a = 54 – 2(25) = 4 \text{ kN/m}^2$
At $z=6$ (Base): $\sigma_v = 54 + 60 = 114$. $p_a = 114 – 50 = 64 \text{ kN/m}^2$

Resultant (Neglecting Tension zone $z < 2$m):

$P_1$ (Layer 1, 2m-3m): Tri base 18, ht 1. $0.5 \times 18 \times 1 = 9$ kN (Arm: $3 + 0.33 = 3.33$m)
$P_2$ (Layer 2 Rect): $4 \times 3 = 12$ kN (Arm: $1.5$m)
$P_3$ (Layer 2 Tri): $0.5 \times (64-4) \times 3 = 90$ kN (Arm: $1.0$m)
$$P_{Total} = 9 + 12 + 90 = \mathbf{111 \text{ kN/m}}$$

Point of Application:

$$\bar{z} = \frac{29.97 + 18 + 90}{111} = \frac{137.97}{111} = \mathbf{1.24 \text{ m from base}}$$
Q.14 a) Relative wall movements & lateral pressure coefficients. b) Tension cracks in cohesive soils. c) Rankine vs Coulomb critical differences.
🕒 2074 Chaitra [Various]

Part (a): Wall Movements

  • Active ($K_a$): Wall moves away. Soil expands. Small strain required.
  • Passive ($K_p$): Wall moves towards. Soil compresses. High strain required.

Part (b): Tension Cracks

In cohesive soils, Rankine theory predicts negative pressure near the surface ($p_a = \gamma z – 2c\sqrt{K_a}$). Since soil cannot take tension, cracks form up to depth $z_c = 2c/(\gamma\sqrt{K_a})$. These cracks reduce the contact area and can fill with water, causing hydrostatic pressure.

Q.15 Culmann’s graphical method – active state, effect of surcharge.
🕒 2073 Shrawan [8]

See solution Q.12(a). The key modification is adding the surcharge weight ($q \times b$) to the soil wedge weight ($W$) when plotting the force vectors on the $\phi$-line.

Q.16 a) Earthquake effect on earth pressure. b) Horizontal strain order for active state in coarse vs fine-grained soil.
🕒 2072 Chaitra [1+2]

Part (a): Earthquake Effect

Earthquakes introduce inertial forces. In the active case, the inertial force acts towards the wall, significantly increasing the total active thrust ($P_{dyn} > P_{stat}$) and raising the point of application (approx $0.6H$ from base).

Part (b): Strain Order

  • Coarse-grained (Sand): 0.1% to 0.4% ($\Delta H/H$).
  • Fine-grained (Clay): 1.0% to 4.0% ($\Delta H/H$).
Q.17 8 m wall, smooth vertical back, 2-layered soil (0-4 m $c=10, \phi=30, \gamma=18$; 4-8 m $c=0, \phi=34, \gamma=20$). Active pressure distribution & resultant.
🕒 2072 Chaitra [8]
Layer 1 (0-4m): $K_{a1}=0.333$. Layer 2 (4-8m): $K_{a2}=0.283$.
Pressure Calculations:
At $z=0$: $p_a = -2(10)\sqrt{0.333} = -11.55$ kPa.
Crack Depth $z_c$: $0.333(18)z_c = 11.55 \Rightarrow z_c = 1.92$ m.
At $z=4$ (Top): $p_a = 0.333(72) – 11.55 = 12.43$ kPa.
At $z=4$ (Bottom): $\sigma_v = 72$. $p_a = 0.283(72) = 20.38$ kPa. ($c=0$)
At $z=8$ (Base): $\sigma_v = 152$. $p_a = 0.283(152) = 43.02$ kPa.
Resultant Thrust (Below crack):
$P_1$ (Layer 1, 1.92-4m): $0.5 \times 12.43 \times (4-1.92) = 12.93$ kN.
$P_2$ (Layer 2 Rect): $20.38 \times 4 = 81.52$ kN.
$P_3$ (Layer 2 Tri): $0.5 \times (43.02-20.38) \times 4 = 45.28$ kN.
$$P_{Total} = 12.93 + 81.52 + 45.28 = \mathbf{139.73 \text{ kN/m}}$$

Chapter 4: Bearing Capacity Theories – Solutions

Q.1 Discuss Terzaghi’s bearing capacity theory + assumptions.
🕒 2082 Bhadra [4]

Terzaghi’s Bearing Capacity Theory:

Karl Terzaghi (1943) developed a theory for the ultimate bearing capacity of shallow foundations. He extended Prandtl’s plastic theory (which was for soft clay) to include the weight of the soil and the friction angle.

Concept & Failure Mechanism:

Terzaghi divided the soil mass below the footing into three distinct zones of plastic failure:

  • Zone I (Elastic Zone): A triangular wedge immediately below the footing base (ABC) that moves downwards as a rigid body with the footing. The soil in this zone is prevented from undergoing shear deformation due to friction and adhesion between the soil and the base.
  • Zone II (Radial Shear Zone): Two zones (ACD and BCF) on either side where the plastic flow is radial. The failure boundaries are logarithmic spirals.
  • Zone III (Linear Shear Zone): Two triangular zones (ADE and BFG) adjacent to the radial shear zones. These are Rankine passive zones where the failure boundaries are straight lines making an angle of $(45^\circ – \phi/2)$ with the horizontal.

Ultimate Bearing Capacity Equation (General Form for Strip Footing):

$$q_u = c N_c + \gamma D_f N_q + 0.5 \gamma B N_\gamma$$

Where:

  • $c$ = Cohesion of soil
  • $\gamma$ = Unit weight of soil
  • $D_f$ = Depth of foundation
  • $B$ = Width of footing
  • $N_c, N_q, N_\gamma$ = Terzaghi’s bearing capacity factors (dependent only on $\phi$)

Assumptions:

  1. Foundation Type: The foundation is a continuous (strip) footing ($L \gg B$), making it a plain strain problem.
  2. Base Condition: The base of the footing is rough (preventing movement of Zone I relative to the footing).
  3. Soil Properties: The soil is homogeneous, isotropic, and rigid-plastic.
  4. Shear Strength: The shear strength of the soil is governed by Coulomb’s law: $s = c + \sigma \tan\phi$.
  5. Failure Mode: The failure is a General Shear Failure (appropriate for dense sand or stiff clay).
  6. Surcharge: The soil above the base of the footing ($D_f$) is considered only as a surcharge load ($q = \gamma D_f$). The shear resistance of the soil above the base is neglected.
  7. Load: The load is vertical and concentric.
Q.2 How size, shape, water table affect bearing capacity.
🕒 2081 Bhadra [4]

1. Effect of Shape of Footing:

Terzaghi’s original equation was derived for strip footings. For other shapes (Square, Circular, Rectangular), empirical shape factors are applied to modify the equation.

  • Strip: $q_u = c N_c + q N_q + 0.5 \gamma B N_\gamma$
  • Square ($B \times B$): $q_u = 1.3 c N_c + q N_q + 0.4 \gamma B N_\gamma$
  • Circular (Diameter $D$): $q_u = 1.3 c N_c + q N_q + 0.3 \gamma B N_\gamma$
  • Rectangular ($B \times L$):
    $$q_u = c N_c \left(1 + 0.3 \frac{B}{L}\right) + q N_q + 0.5 \gamma B N_\gamma \left(1 – 0.2 \frac{B}{L}\right)$$

(Generally, square and circular shapes have higher bearing capacity contributions from cohesion but lower from the width term compared to strip).

2. Effect of Size (Width $B$):

  • Cohesionless Soil (Sand): The third term of the bearing capacity equation is $0.5 \gamma B N_\gamma$. Since $q_u$ is directly proportional to $B$, the bearing capacity increases as the width of the footing increases.
  • Cohesive Soil (Pure Clay, $\phi=0, N_\gamma=0$): The third term vanishes. The ultimate bearing capacity depends on $c N_c$ and $q N_q$. Thus, for pure clay, the bearing capacity is theoretically independent of the footing width.

3. Effect of Water Table:

The position of the water table significantly reduces the bearing capacity, potentially by up to 50%, by reducing the effective unit weight of the soil.

  • Case A (WT above Base): Affects both the surcharge term ($\gamma D_f$) and the wedge term ($0.5 \gamma B$). We must use submerged unit weight ($\gamma’$) for the submerged portions.
  • Case B (WT at Base): The surcharge term is unaffected, but the wedge term uses $\gamma’$.
  • Case C (WT below Base): If the WT is at a depth $d_w$ below the base, its effect diminishes as $d_w$ approaches $B$. If $d_w \ge B$: No effect on bearing capacity.
Q.3 Differentiate general, local, punching shear failure (sketch).
🕒 2081 Baishakh (Back) [4] 🕒 2078 Bhadra [4] 🕒 2076 Ashwin [2+3] 🕒 2075 Chaitra [2+2]
Feature General Shear Failure Local Shear Failure Punching Shear Failure
Soil Type Dense sand (Relative Density > 70%) or Stiff clay. Medium dense sand or Medium stiff clay. Loose sand (Relative Density < 35%) or Soft clay.
Failure Surface Well-defined, continuous failure surface extending from base to ground level. Well-defined only below the footing; does not reach the ground surface. Failure surface is not well defined; restricted to vertical planes immediately below footing.
Surface Phenomenon Considerable bulging (heaving) of soil adjacent to the footing. Slight heaving adjacent to the footing. No heaving; soil outside the loaded area remains undisturbed.
Load-Settlement Sudden/catastrophic failure with a distinct peak load. No distinct peak; failure is defined by large settlement. Continuous settlement with no peak load.
Tilting Tilting of foundation is common. Little tilting. No tilting.

(Note: Sketches should be drawn showing the failure wedges and heave patterns as described above. General shows full heave, Local shows partial, Punching shows none.)

Q.4 a) Development of classical bearing capacity theories. b) Strip footing 1.5 m wide, Df=1.2 m, dry sand, WT 1 m below base → change if WT rises to 0.5 m below GL (factors given).
🕒 2080 Bhadra [Various]

Part (a): Development of Theories

  • Rankine (1885): Simplest theory based on plastic equilibrium, considering only the depth of foundation. Neglected cohesion and footing width.
  • Prandtl (1920): Used plastic theory for soft clay ($c-\phi$ soil was difficult). Proposed the specific failure mechanism (Plastic zones).
  • Terzaghi (1943): Modified Prandtl’s theory for soils with cohesion, friction, and weight. Introduced factors $N_c, N_q, N_\gamma$. Considered base roughness.
  • Meyerhof (1951): Generalized the theory to account for depth, shape, inclination of load, and shear strength of soil above the base.
  • Skempton (1951): Specific theory for $c$-soils (clays) considering $D_f/B$ ratio.

Part (b): Numerical Solution

Given: Strip Footing, $B = 1.5 \text{ m}$, $D_f = 1.2 \text{ m}$. Soil: Dry Sand ($c = 0$).
Assumed Factors ($\phi \approx 30^\circ$): $N_c=37.2, N_q=22.5, N_\gamma=19.7$.
Assumed Unit Weights: $\gamma_{dry} = 18 \text{ kN/m}^3, \gamma_{sat} = 20 \text{ kN/m}^3, \gamma’ = 10 \text{ kN/m}^3$.

Formula for Strip Footing ($c=0$):

$$q_u = \gamma D_f N_q R_q + 0.5 \gamma B N_\gamma R_\gamma$$

Case 1: Water Table at 1.0 m below the base

Depth of WT from GL = $1.2 + 1.0 = 2.2$ m.
Since WT is below the base ($Z_{w2} = 1.0$ m):

  • Surcharge Term ($R_q$): WT is below base level, so soil above base is dry. $R_q = 1.0$.
  • Wedge Term ($R_\gamma$): WT is within depth $B$ below base ($Z_{w2} < B$, i.e., $1.0 < 1.5$).
    Correction: $R_\gamma = 0.5 (1 + \frac{Z_{w2}}{B}) = 0.5 (1 + \frac{1.0}{1.5}) = 0.5 (1.667) = 0.833$
$$q_{u1} = (18 \times 1.2 \times 22.5 \times 1) + (0.5 \times 18 \times 1.5 \times 19.7 \times 0.833)$$ $$q_{u1} = 486 + 221.48 = \mathbf{707.48 \text{ kN/m}^2}$$

Case 2: Water Table rises to 0.5 m below Ground Level

Depth of WT from GL ($Z_{w1}$) = 0.5 m. WT is above the base level.

  • Surcharge Term ($R_q$): Soil from 0 to 0.5m is dry, 0.5 to 1.2m is submerged.
    Effective Surcharge $q = (18 \times 0.5) + (10 \times (1.2 – 0.5)) = 9 + 7 = 16 \text{ kN/m}^2$.
  • Wedge Term ($R_\gamma$): Soil below base is fully submerged. Use $\gamma’$.
$$q_{u2} = (16 \times 22.5) + (0.5 \times 10 \times 1.5 \times 19.7)$$ $$q_{u2} = 360 + 147.75 = \mathbf{507.75 \text{ kN/m}^2}$$

Conclusion (Change):

Percentage Reduction = $\frac{707.48 – 507.75}{707.48} \times 100 \approx \mathbf{28.2\%}$
Q.5 Factors affecting bearing capacity. Water table fluctuation effect (above & below base).
🕒 2080 Baishakh [2+2]

Factors affecting Bearing Capacity:

  • Nature of Soil: Physical properties like angle of internal friction ($\phi$), cohesion ($c$), and density.
  • Depth of Foundation ($D_f$): Bearing capacity increases with depth due to increased surcharge confinement.
  • Geometry of Footing: Width ($B$) and Shape (Strip, Square, Rectangular) influence the factors used.
  • Position of Water Table: Reduces effective unit weight, significantly lowering capacity.
  • Eccentricity and Inclination of Load: Reduces the effective area and stability.

Water Table Effect:

  • Above Base: Reduces the effective surcharge ($\gamma D_f$ becomes effective stress) and the wedge weight ($\gamma$ becomes $\gamma’$). This is the most critical condition.
  • Below Base: If WT is at depth $d_w$ below base, it only affects the wedge term. As $d_w$ increases from $0$ to $B$, the unit weight linearly interpolates between $\gamma’$ and $\gamma_{dry}$. Beyond $B$, there is no effect.
Q.6 Skempton’s findings on net safe bearing capacity in clayey soil.
🕒 2078 Kartik [2]

Skempton (1951) observed that for saturated clays ($\phi = 0$), the bearing capacity factor $N_c$ is not constant (as Terzaghi proposed, $N_c=5.7$) but increases with the depth-to-width ratio ($D_f/B$).

Skempton’s Formula for Net Ultimate Bearing Capacity ($q_{nu}$):

$$q_{nu} = c N_c$$

Values of $N_c$:

  • For Strip Footing: $$N_c = 5 \left( 1 + 0.2 \frac{D_f}{B} \right)$$ (Limiting value: $N_c \le 7.5$)
  • For Square/Circular/Rectangular Footings: $$N_c = 5 \left( 1 + 0.2 \frac{D_f}{B} \right) \left( 1 + 0.2 \frac{B}{L} \right)$$ (Limiting value: $N_c \le 9.0$)
Q.7 Footing 2 m × 3 m, Df=1.5 m, eccentric load 0.26 m, WT 0.5 m below GL. Safe bearing capacity – Terzaghi (c=20, φ=35 interpolated).
🕒 2078 Kartik [8]
Given: Size $2 \text{ m} \times 3 \text{ m}$, $D_f = 1.5 \text{ m}$, $e = 0.26 \text{ m}$.
WT: 0.5 m below GL. Soil: $c = 20 \text{ kN/m}^2, \phi = 35^\circ$. F = 3.
Unit Weight (Assumed): $\gamma_{sat} = 20 \text{ kN/m}^3, \gamma’ = 10 \text{ kN/m}^3$.

Step 1: Terzaghi Bearing Capacity Factors for $\phi = 35^\circ$:

$$N_c = 57.8, \quad N_q = 41.4, \quad N_\gamma = 42.4$$

Step 2: Effective Dimensions (Meyerhof’s approach for eccentricity):

$$B’ = B – 2e = 2 – 2(0.26) = 1.48 \text{ m}$$ $$L’ = L = 3 \text{ m}$$

Step 3: Modified Terzaghi Equation for Rectangular Footing:

$$q_u = c N_c S_c + \sigma’ N_q + 0.5 \gamma_{eff} B’ N_\gamma S_\gamma$$

Step 4: Calculate Correction Factors:

Shape Factors ($S_c, S_\gamma$): Using $B’$ and $L’$:

$$S_c = 1 + 0.3 \left(\frac{B’}{L’}\right) = 1 + 0.3 \left(\frac{1.48}{3}\right) = 1.148$$ $$S_\gamma = 1 – 0.2 \left(\frac{B’}{L’}\right) = 1 – 0.2 \left(\frac{1.48}{3}\right) = 0.901$$

Water Table Effect:

  • Surcharge Term ($\sigma’$): WT is 0.5m below GL.
    $\sigma’ = (18 \times 0.5) + (10 \times 1.0) = 9 + 10 = 19 \text{ kN/m}^2$.
  • Wedge Term ($\gamma_{eff}$): Soil below footing is fully submerged.
    $\gamma_{eff} = \gamma’ = 10 \text{ kN/m}^3$.

Step 5: Calculate Ultimate Bearing Capacity ($q_u$):

$$Term 1: 20 \times 57.8 \times 1.148 = 1327.09$$ $$Term 2: 19 \times 41.4 = 786.6$$ $$Term 3: 0.5 \times 10 \times 1.48 \times 42.4 \times 0.901 = 282.69$$ $$q_u = 1327.09 + 786.6 + 282.69 = 2396.38 \text{ kN/m}^2$$

Step 6: Safe Bearing Capacity ($q_{safe}$):

$$q_{safe} = \frac{q_u – \sigma’}{F} + \sigma’$$ $$q_{safe} = \frac{2396.38 – 19}{3} + 19 = 792.46 + 19$$ $$\mathbf{q_{safe} \approx 811.46 \text{ kN/m}^2}$$
Q.8 Modes of shear failure. Local vs general shear difference.
🕒 2078 Bhadra [4]

(Refer to the differentiation table in Q3).

Q.9 Modes of failure due to settlement (sketches).
🕒 2076 Chaitra [5]

This refers to the three shear failure modes defined by their settlement curves:

  • General Shear Failure: Curve rises steeply to a distinct peak load ($q_u$) and then drops. Occurs in dense sand.
  • Local Shear Failure: Curve is flatter; no distinct peak. Point of failure is often defined at a specific settlement (e.g., 10-20% of B). Occurs in medium sand.
  • Punching Shear Failure: Curve is very steep downwards (large settlement for small load increment). No peak. Occurs in loose sand.
Q.10 Ascertain local vs general shear failure. Limitations of classical theories.
🕒 2076 Ashwin [2+3]

Ascertaining Failure Mode:

  • Strain/Settlement: General shear fails at low strain (< 5%). Local/Punching fails at high strain (10-20%).
  • Relative Density ($D_r$):
    • $D_r > 70\%$: General Shear
    • $35\% < D_r < 70\%$: Local Shear
    • $D_r < 35\%$: Punching Shear
  • Void Ratio ($e$): Low void ratio ($e < 0.55$) $\rightarrow$ General. High void ratio ($e > 0.75$) $\rightarrow$ Punching.

Limitations of Classical Theories (Terzaghi):

  • Soil compressibility: Terzaghi assumes soil is rigid-plastic. It ignores the compressibility of soil, which is significant in local/punching failure.
  • Superposition error: The theory adds effects of $c, q,$ and $\gamma$ separately. In reality, plastic zones for these components are not identical.
  • Shape factors: The shape factors are empirical and semi-theoretical.
  • Water Table: Does not account for pore water pressure generation during loading (only static WT).
Q.11 Water table effect. Footing designed for WT at GL – change if WT rises much above GL?
🕒 2074 Ashwin [4]

Scenario:

The footing is designed assuming the Water Table is at Ground Level (GL). In this state, the submerged unit weight ($\gamma’$) is used for all soil terms (both surcharge and wedge).

Effect if WT rises above GL:

If the water level rises significantly above the ground level (flood condition):

  • Effective Stress: The vertical effective stress in the soil $\sigma’$ depends on the submerged weight ($\gamma’$). The weight of the free water standing above the ground surface creates a hydrostatic pressure that increases total stress, but pore water pressure increases by the exact same amount. Therefore, $\sigma’ = \sigma_{total} – u$ remains constant.

Conclusion:

The bearing capacity of the soil does not change theoretically when the water table rises above the ground level, provided the soil does not scour or erode. The buoyant weight of the soil remains the governing factor.

5. Shallow Foundation

Settlement and safe bearing pressure analysis…

6. Deep Foundation

Pile load capacity and group efficiency solutions…

7. Foundation in Rock

Bearing capacity on weathered and un-weathered rock…

8. Retaining Structures

Stability analysis of rigid and flexible walls…

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