CELE Geotechnical Engineering — Stresses in Soil MassStudy Notes
Thorough study notes for Stresses in Soil Mass — the fastest path from zero to ready for CELE Geotechnical Engineering. Structured for self-study reviewers who cannot attend a review centre, these notes cover the full concept library plus the CELE-specific twists Professional Regulation Commission (PRC) — Board of Civil Engineering adds to its questions.
Exam context
On the CELE 2026, the Geotechnical Engineering subtest carries a "Core" weight in Professional Regulation Commission (PRC) — Board of Civil Engineering's pattern. Stresses in Soil Mass lands at position 4th out of 11 in the standard review order. Target score is 70% weighted average, no sub-test below 50%, and roughly a meaningful share of items come from Geotechnical Engineering on a typical CELE paper.
Stresses in Soil Mass - Study Notes
Understanding stress distribution in soil is fundamental to geotechnical engineering practice. The behavior of soil—including settlement, bearing capacity, slope stability, and strength—depends critically on the stresses acting within the soil mass. Unlike most structural materials, soil behavior is governed not by total stress alone but by effective stress, which represents the stress carried by the soil skeleton itself. This distinction, formalized by Terzaghi's effective-stress principle, is perhaps the most important concept in soil mechanics. This chapter covers the effective-stress principle, pore pressure mechanics, and methods for calculating stress increases from surface loads using both rigorous (Boussinesq) and practical (2:1 approximation) approaches. Mastery of these concepts is essential for the PRC Civil Engineer Licensure Examination and professional practice.
Summary
Understanding stress in soil masses is the foundation of geotechnical engineering practice. The effective-stress principle—that σ' = σ - u governs soil behavior—is the single most important concept in soil mechanics. Total stress is calculated by summing the weight of overlying material, using the appropriate unit weight (γ above the water table, γ_sat below). Pore pressure is hydrostatic below the water table (u = γ_w × z_w) and zero above (except in capillary zones where it becomes negative, increasing effective stress). For stress increases from surface loads, the Boussinesq solution provides a rigorous method for point loads, giving Δσ = 3Q/(2πz²) directly below the load and decreasing with horizontal offset and depth. For footing loads, the 2:1 approximation (Δσ = Q/[(B+z)(L+z)]) offers a simpler alternative. These stress increases, when added to in-situ stresses and adjusted for pore pressure, determine settlement, bearing capacity, slope stability, and excavation behavior. Correct effective-stress calculations are essential for NSCP 2015 compliance and professional practice under RA 544. Common exam pitfalls include forgetting pore pressure, using incorrect unit weights, misapplying stress formulas, and ignoring the impacts of rising water tables and seepage. In the Philippine context, high water tables in coastal areas (Manila, Cebu) and seasonal monsoon flooding create significant challenges that require careful stress analysis and effective design of drainage systems.
Sections
The effective-stress principle is the cornerstone of soil mechanics. Terzaghi's principle states that the total vertical stress at any depth in a soil mass is shared between two components: the stress carried by the soil skeleton (effective stress σ') and the stress carried by pore water (pore pressure u). This relationship is expressed mathematically as: σ = σ' + u Rearranging: σ' = σ - u Where: - σ (sigma) = total vertical stress at a given depth (kPa) - σ' (sigma-prime) = effective stress carried by soil particles (kPa) - u = pore water pressure (kPa) PHYSICAL INTERPRETATION: Imagine a column of soil as a network of grains with water-filled spaces between them. The total weight of the soil column (total stress) is distributed across two load paths: the grain-to-grain contact points (effective stress) and the water within the pores (pore pressure). The water, being fluid, cannot sustain shear stress and distributes load equally in all directions (hydrostatic pressure). Only the grain-to-grain contacts can develop friction and resistance to shear. Therefore, effective stress governs all mechanical behavior: strength, settlement, stiffness, and bearing capacity. WHY EFFECTIVE STRESS MATTERS: 1. Soil strength (friction angle φ and cohesion c) depends only on effective stress, not total stress. 2. Compressibility and settlement calculations use effective stress. 3. Bearing capacity formulas are derived in terms of effective stress. 4. Consolidation theory (Terzaghi) is based entirely on effective stress. 5. Two deposits with the same total stress profile may behave very differently if their pore pressures differ. CRITICAL INSIGHT FOR BOARD EXAMS: A common exam trap is to assume that total stress alone determines soil behavior. For example, if you are asked "Will a saturated clay deposit settle more than a dry sand deposit at the same depth?" the answer depends on the pore pressure and effective stress, not just the total stress. A thick layer of water above the soil can significantly reduce effective stress and settlement potential.
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1. Terzaghi's Effective-Stress Principle
Examples
Example 1.1 — Effective Stress at Different Depths
Problem
A soil deposit consists of 2 m of sandy loam (γ = 17.5 kN/m³) above a clay layer (γsat = 19.5 kN/m³). The water table is at 2 m depth (at the interface). Calculate the total stress, pore pressure, and effective stress at 5 m depth.
Solution
At 5 m depth: - Above water table (0–2 m): Total stress from sandy loam = 17.5 × 2 = 35 kPa - Below water table (2–5 m): Depth below WT = 5 - 2 = 3 m - Total stress from clay = 19.5 × 3 = 58.5 kPa - Total stress at 5 m: σ = 35 + 58.5 = 93.5 kPa - Pore pressure (3 m below WT): u = γw × zw = 9.81 × 3 = 29.43 kPa - Effective stress: σ' = σ - u = 93.5 - 29.43 = 64.07 kPa ≈ 64 kPa Note: Use γw = 9.81 kN/m³ for water. In practical contexts, γw = 10 kN/m³ is sometimes used for quick estimates, but 9.81 is preferred for licensure exams.
Example 1.2 — Water Table at Surface
Problem
If the water table in Example 1.1 rises to the ground surface (due to flooding or high groundwater), recalculate the effective stress at 5 m depth.
Solution
With WT at surface: - Total stress from clay (0–5 m): σ = 19.5 × 5 = 97.5 kPa - Pore pressure (5 m below WT): u = 9.81 × 5 = 49.05 kPa - Effective stress: σ' = 97.5 - 49.05 = 48.45 kPa ≈ 48.5 kPa Comparison: When the water table rose from 2 m to the surface, the effective stress at 5 m decreased from 64 kPa to 48.5 kPa—a reduction of about 24%. This explains why high water tables are dangerous: they reduce effective stress and thus soil strength. A slope stable in dry conditions can fail when the water table rises.
Key Points
- Total stress σ = effective stress σ' + pore pressure u
- Effective stress alone governs strength, settlement, and bearing capacity
- Pore water pressure u cannot sustain shear; it acts hydrostatic
- Two soil profiles with same total stress but different pore pressures will behave differently
- Below the water table, pore pressure is always positive; above it, u = 0 (or capillary tension if present)
- Terzaghi's principle applies to all saturated soils regardless of stress level
Total stress at any depth is simply the cumulative weight of all material above that point, divided by the area. For a vertical column (which is the standard assumption), we calculate by summing the weight contributions layer by layer. FORMULA: σ = Σ(γ_i × z_i) Where: - γ_i = unit weight of the i-th layer (kN/m³) - z_i = thickness of the i-th layer (m) CRITICAL RULE FOR UNIT WEIGHT SELECTION: 1. ABOVE the water table: Use γ (moist unit weight) or γd (dry unit weight). Typical values: γ = 16–18 kN/m³. 2. BELOW the water table: Always use γsat (saturated unit weight). Typical values: γsat = 19–22 kN/m³. 3. For submerged soil (below WT), some engineers use γ' = γsat - γw (buoyant/submerged unit weight) instead of γsat. This is mathematically equivalent: σ = γsat × z = (γ' + γw) × z, and when you subtract pore pressure u = γw × z, you get σ' = γ' × z directly. However, the standard approach (use σ = γsat × z, then subtract u = γw × z) is preferred in exams because it clearly shows the effective-stress calculation. STEP-BY-STEP PROCEDURE: 1. Identify the water table location. 2. Divide the soil profile into layers with different unit weights. 3. For each layer above the WT, use the in-situ (or moist) unit weight. 4. For each layer below the WT, use the saturated unit weight. 5. Sum the stress contributions: σ = γ₁z₁ + γ₂z₂ + ... + γₙzₙ. 6. Subtract pore pressure to get effective stress. COMMON EXAM ERRORS: - Using γsat above the water table (wrong—use γ or γd). - Using dry unit weight γd below the water table (wrong—must use γsat). - Forgetting that pore pressure acts everywhere below the WT, even in fine-grained soils. - Not recognizing that capillary tension above the WT can create negative pore pressure and increase effective stress.
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2. Total Stress Calculation (σ)
Examples
Example 2.1 — Layered Soil Profile with Water Table
Problem
A site profile consists of: - Layer 1: 3 m of dry sand, γ = 16 kN/m³ - Layer 2: 4 m of moist clay, γ = 18 kN/m³ - Layer 3: Saturated sandy clay, γsat = 20 kN/m³ The water table is at 4 m depth (at the boundary between layers 2 and 3). Calculate the total stress at 8 m depth.
Solution
At 8 m depth: - Stress from Layer 1 (0–3 m): σ₁ = 16 × 3 = 48 kPa - Stress from Layer 2 (3–4 m): σ₂ = 18 × 1 = 18 kPa (Note: Layer 2 is only 1 m thick in this calculation, from 3 m to 4 m) - Stress from Layer 3 (4–8 m): This is 4 m below the WT. σ₃ = 20 × 4 = 80 kPa - Total stress: σ = 48 + 18 + 80 = 146 kPa If the question had asked for effective stress at 8 m: - Pore pressure: u = 9.81 × (8 - 4) = 9.81 × 4 = 39.24 kPa - Effective stress: σ' = 146 - 39.24 = 106.76 kPa ≈ 107 kPa
Example 2.2 — Uniform Soil with Simple Water Table
Problem
A uniform clay deposit has γsat = 19 kN/m³ and the water table is at the surface (z_w = 0). What is the total stress at 10 m depth, and what is the effective stress?
Solution
This is the simplest case—all stress is from saturated soil. - Total stress: σ = 19 × 10 = 190 kPa - Pore pressure: u = 9.81 × 10 = 98.1 kPa - Effective stress: σ' = 190 - 98.1 = 91.9 kPa Alternatively, using submerged unit weight γ' = γsat - γw = 19 - 9.81 = 9.19 kN/m³: - σ' = 9.19 × 10 = 91.9 kPa (same result) This second method is sometimes taught but can be confusing if you forget that you must use submerged unit weight consistently throughout the calculation. The first method is more transparent for exams.
Key Points
- Total stress σ is the sum of weight contributions from all layers above the point
- Use γ (moist) or γd (dry) above the water table; use γsat below the water table
- Total stress increases linearly with depth (σ = γ × z for uniform soil)
- Total stress is independent of soil stiffness or strength; it depends only on weight and depth
- At any given depth, total stress is the same whether the soil is dense or loose, strong or weak
Pore water pressure, also called pore pressure or neutral stress, is the pressure of water in the void spaces of soil. Understanding and calculating pore pressure correctly is essential because it directly affects effective stress and therefore all aspects of soil behavior. BASIC HYDROSTATIC PORE PRESSURE: In static conditions (no groundwater flow), pore pressure at any depth is simply the weight of the water column above that point: u = γ_w × z_w Where: - γ_w = unit weight of water = 9.81 kN/m³ (or 10 kN/m³ in quick estimates) - z_w = depth below the water table (m) IMPORTANT POINTS: 1. Pore pressure is zero above the water table (in dry soil). 2. Pore pressure increases linearly with depth below the water table at a rate of ~10 kPa per meter. 3. Pore pressure acts equally in all directions (hydrostatic); it does not depend on soil type or stiffness. 4. Pore pressure cannot sustain shear stress—only hydrostatic (normal) stress. 5. Pore pressure is independent of total stress once you are below the water table; it depends only on the water column height. CAPPLARY RISE AND NEGATIVE PORE PRESSURE: Above the water table, in fine-grained soils (clay, silt), capillary action can draw water upward, creating a capillary zone. In this zone, pore pressure is negative (suction), which increases effective stress. Heights of capillary rise vary widely: - Sand: 0.3–1 m - Silt: 1–5 m - Clay: >5 m (sometimes very high) In the capillary zone: u = -γ_w × h_c (negative, where h_c is height above the water table). Example: If capillary rise in a clay is 4 m and you are 2 m above the water table, then u = -9.81 × 2 = -19.62 kPa (suction). This increases effective stress: σ' = σ - (-19.62) = σ + 19.62 kPa. This is why clay can be very hard and stiff above the water table but becomes soft and weak when saturated. SEEPAGE AND SEEPAGE PRESSURE: When groundwater flows (seepage), additional pressure develops in the direction of flow. The seepage pressure (or seepage force) can increase pore pressure in the direction of flow and decrease it in the opposite direction. This is critical for: - Slope stability analysis (upward seepage beneath a slope) - Piping failure in dams (critical gradient) - Bearing capacity under flowing groundwater For downward seepage with hydraulic gradient i: u_seepage = u_hydrostatic + ρ_w × g × i × distance_in_flow_direction For upward seepage (most dangerous): u = γ_w × (z_w + i × distance) where i is the hydraulic gradient When upward seepage is significant, the "quick condition" (also called "boiling") can occur when the seepage force equals the weight of the soil, making it behave like a fluid. This is critical in excavation and dam analysis.
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3. Pore Water Pressure (u)
Examples
Example 3.1 — Hydrostatic Pore Pressure
Problem
A clay layer lies at depths 5–10 m. The water table is at 3 m depth. Calculate the pore pressure at 5 m depth (top of clay) and at 10 m depth (bottom of clay).
Solution
At 5 m depth: - Depth below WT: z_w = 5 - 3 = 2 m - Pore pressure: u = 9.81 × 2 = 19.62 kPa At 10 m depth: - Depth below WT: z_w = 10 - 3 = 7 m - Pore pressure: u = 9.81 × 7 = 68.67 kPa Note: The pore pressure increases by 9.81 × 5 = 49.05 kPa over the 5 m thickness of the clay, but this increase is due to the added water column height, not the presence of clay.
Example 3.2 — Quick Condition and Upward Seepage
Problem
A sandy soil deposit is 8 m thick. Water table is at the top (z = 0). The average hydraulic gradient due to upward seepage is i = 0.5. The unit weight of the sand is γsat = 20 kN/m³. At what depth will the quick condition occur? (Quick condition occurs when effective stress becomes zero.)
Solution
With upward seepage, the pore pressure at depth z includes both hydrostatic and seepage components: u = γ_w × (z_w + i × z) = 9.81 × (0 + 0.5 × z) = 4.905 × z kPa (since z_w = z for WT at surface) Total stress at depth z: σ = 20 × z Effective stress: σ' = σ - u = 20z - 4.905z = 15.095z The quick condition occurs when σ' = 0: 15.095z = 0, which gives z → ∞ (never, in this scenario) Wait, let me reconsider. If the hydraulic gradient is measured as the head drop divided by distance, and upward flow occurs, the pore pressure increase due to flow is: u_flow = γ_w × i × L (where L is the distance over which the head drops) For a simpler case: assume the seepage pressure is u_seep = γ_w × i = 9.81 × 0.5 = 4.905 kPa/m of depth. Then: σ' = 20z - (9.81z + 4.905z) ... (this gets complex with flow interpretation) More practically: The critical hydraulic gradient at which quick condition occurs is: i_critical = γ'sat / γ_w = (γ_sat - γ_w) / γ_w = (20 - 9.81) / 9.81 = 10.19 / 9.81 ≈ 1.04 Since i = 0.5 < 1.04, the quick condition does not occur at any depth. If i had been > 1.04, the quick condition would occur throughout the deposit.
Key Points
- Pore pressure u = γw × zw below the water table (hydrostatic)
- Pore pressure is zero above the water table in dry conditions
- Pore pressure acts equally in all directions; it cannot sustain shear
- Capillary rise above the WT creates negative pore pressure (suction), increasing effective stress
- Seepage pressure from groundwater flow adds to (or subtracts from) hydrostatic pore pressure
- Upward seepage is dangerous because it reduces effective stress and can cause the quick condition
Now that we understand total stress, pore pressure, and the effective-stress principle, we can systematically calculate effective stress at any depth and construct an effective-stress profile. STEP-BY-STEP PROCEDURE: 1. Determine the soil layers and water table location. 2. Calculate total stress at the depth of interest: σ = Σ(γ_i × z_i). 3. Calculate pore pressure at the same depth: u = γ_w × z_w (or u = 0 if above WT). 4. Compute effective stress: σ' = σ - u. 5. Repeat for multiple depths to construct a profile. PROPERTIES OF EFFECTIVE-STRESS PROFILES: - Above the water table: σ' = σ (since u = 0). The effective-stress line coincides with the total-stress line. - Below the water table: σ' < σ because u > 0. The effective-stress line lags behind the total-stress line. - At the water table: σ' = σ and u = 0 (transition point). - Below the WT, as depth increases, both σ and u increase, but σ increases faster (because of soil weight). The gap between σ and σ' remains constant at γ_w × (z - z_w), which grows linearly with depth below the WT. - In a uniform soil below the WT, effective stress increases at a rate of (γ_sat - γ_w) = γ' per meter of depth. EFFECTIVE STRESS UNDER SLOPING GROUNDWATER TABLE: If the water table is not horizontal (e.g., near a river or coastal area), the vertical distance from any point to the water table varies. The effective-stress calculation remains the same—u = γ_w × (vertical distance to WT)—but the shape of the pore-pressure profile becomes non-linear when plotted against absolute depth. DRAWING EFFECTIVE-STRESS DIAGRAMS: In professional practice and on exams, effective-stress profiles are often plotted with depth on the vertical axis and stress on the horizontal axis. Two diagrams are typically drawn side by side: 1. Total stress (σ) vs. depth: A single line increasing with depth. 2. Pore pressure (u) vs. depth: u = 0 above WT, then increasing linearly below WT. 3. Effective stress (σ') vs. depth: Follows σ above WT, then diverges below WT. Alternatively, all three are plotted on the same axis (with different scales) for comparison.
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4. Effective Stress Calculation and Profiles
Examples
Example 4.1 — Complete Effective-Stress Profile
Problem
Construct a complete effective-stress profile for the following soil deposit: - 0–2 m: Sand, γ = 17 kN/m³ - 2–6 m: Clay, γ = 18 kN/m³ - 6–10 m: Saturated sandy clay, γsat = 20 kN/m³ - Water table at 6 m depth Calculate σ, u, and σ' at depths 0, 2, 6, and 10 m.
Solution
At z = 0 m (ground surface): - σ = 0 kPa - u = 0 kPa (above WT) - σ' = 0 kPa At z = 2 m (sand–clay interface): - σ = 17 × 2 = 34 kPa - u = 0 kPa (above WT) - σ' = 34 kPa At z = 6 m (WT at this depth): - σ = 17 × 2 + 18 × 4 = 34 + 72 = 106 kPa - u = 0 kPa (exactly at WT, no depth below) - σ' = 106 kPa At z = 10 m (4 m below WT): - σ = 17 × 2 + 18 × 4 + 20 × 4 = 34 + 72 + 80 = 186 kPa - u = 9.81 × 4 = 39.24 kPa - σ' = 186 - 39.24 = 146.76 kPa ≈ 147 kPa Summary table: | Depth (m) | σ (kPa) | u (kPa) | σ' (kPa) | |-----------|---------|---------|----------| | 0 | 0 | 0 | 0 | | 2 | 34 | 0 | 34 | | 6 | 106 | 0 | 106 | | 10 | 186 | 39.24 | 146.76 | Note: The effective stress at 10 m (147 kPa) is significantly less than the total stress (186 kPa) because of pore pressure. If this soil is used as a foundation, the bearing capacity depends on σ', not σ.
Example 4.2 — Rising Water Table
Problem
For the same soil deposit as Example 4.1, what happens to the effective stress at 10 m if the water table rises from 6 m to 4 m (perhaps due to seasonal flooding or dam construction)?
Solution
New condition: Water table at 4 m depth. At z = 10 m (6 m below new WT): - σ = 17 × 2 + 18 × 4 + 20 × 4 = 186 kPa (unchanged—weight above doesn't change) - u_new = 9.81 × 6 = 58.86 kPa (pore pressure increased) - σ'_new = 186 - 58.86 = 127.14 kPa Comparison: - Original σ' (WT at 6 m): 146.76 kPa - New σ' (WT at 4 m): 127.14 kPa - Reduction: 146.76 - 127.14 = 19.62 kPa This reduction in effective stress is significant. Depending on the soil type, it could: - Reduce bearing capacity of a foundation - Increase settlement potential - Reduce slope stability (increase slope failure risk) - Increase lateral earth pressure on retaining walls This example illustrates why managing groundwater is critical in civil engineering projects.
Key Points
- σ' = σ - u at all depths
- Above the water table: u = 0, so σ' = σ
- Below the water table: u > 0, so σ' < σ
- Effective stress depends on both the soil weight (σ) and the water column height (u)
- In uniform soil below the WT, σ' increases at rate γ' = γsat - γw per meter depth
- Effective-stress profiles are essential tools for understanding soil settlement and strength
- The vertical distance to the water table (not absolute elevation) determines pore pressure
When a load is applied at the ground surface (e.g., a building footing, a wheel load, an embankment), it creates stresses that propagate downward and outward through the soil. These stress increases, superimposed on the in-situ stress from the soil's own weight, determine settlement and influence stability. The most rigorous method for calculating these stress increases is the Boussinesq solution, derived for an elastic half-space under a point load. BOUSSINESQ FORMULA FOR POINT LOAD: For a concentrated vertical load Q (in Newtons or kN) applied at the ground surface, the vertical stress increase Δσ at a depth z and horizontal distance r from the load is: Δσ = (3Q / 2πz²) × [1 / (1 + (r/z)²)]^(5/2) Special case — directly below the load (r = 0): Δσ_max = 3Q / (2πz²) Where: - Q = concentrated load (kN) - z = depth below surface (m) - r = horizontal distance from the vertical line through the load (m) - Δσ = vertical stress increase (kPa) CHARACTERISTICS OF THE BOUSSINESQ SOLUTION: 1. At r = 0 (directly below the load), Δσ is maximum: Δσ_max = 1.5Q / (πz²). 2. As r increases (moving horizontally away from the load), Δσ decreases rapidly. 3. As z increases (going deeper), Δσ decreases, proportional to 1/z². 4. The stress spreads with depth—at twice the depth (2z), the stress is one-quarter as large (for r = 0). 5. The Boussinesq solution assumes the soil is elastic, homogeneous, isotropic, and semi-infinite (no bottom boundary). It is an upper bound for most natural soils. LIMITATIONS OF BOUSSINESQ: - Assumes elastic material (not plastic deformation). - Assumes homogeneous soil (same properties everywhere). - Assumes no bottom boundary (very deep deposits). - Does not account for soil layering or variations. - Does not account for time-dependent consolidation. DESPITE LIMITATIONS, Boussinesq is widely used because: - It provides a closed-form solution (no numerical integration needed). - Results are reasonably accurate for design purposes. - It is conservative (often overestimates stress increases). - It is the standard in most geotechnical textbooks and codes. FOR PRACTICE AND EXAMS: The Boussinesq formula is usually applied in two contexts: 1. Directly below a point load: Δσ = 3Q / (2πz²). 2. At an offset from the load, using the full formula with the ratio (r/z). Many reference tables and charts (influence factor charts) simplify the calculation by expressing Δσ in terms of an influence factor I_z: Δσ = I_z × (Q / z²) Where I_z is tabulated as a function of r/z ratio.
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5. Stress Increase from Surface Loads — Boussinesq Solution
Examples
Example 5.1 — Point Load Directly Below
Problem
A concentrated load of Q = 150 kN is applied at the ground surface. What is the vertical stress increase directly below the load (r = 0) at depths z = 1 m, 2 m, and 3 m?
Solution
Using Δσ = 3Q / (2πz²): At z = 1 m: Δσ = (3 × 150) / (2 × π × 1²) = 450 / (2π) = 450 / 6.283 = 71.62 kPa At z = 2 m: Δσ = (3 × 150) / (2 × π × 2²) = 450 / (2 × π × 4) = 450 / 25.133 = 17.91 kPa At z = 3 m: Δσ = (3 × 150) / (2 × π × 3²) = 450 / (2 × π × 9) = 450 / 56.549 = 7.96 kPa Observation: As depth doubles from 1 m to 2 m, the stress decreases by a factor of 4 (from 71.62 to 17.91). This illustrates the 1/z² dependence. At 3 m, the stress is less than 8 kPa, which is often negligible for practical purposes.
Example 5.2 — Point Load at Horizontal Offset
Problem
For the same 150 kN load, find the vertical stress increase at z = 2 m and r = 2 m (i.e., at a point 2 m horizontally away from the load and 2 m deep).
Solution
Using the full Boussinesq formula: Δσ = (3Q / 2πz²) × [1 / (1 + (r/z)²)]^(5/2) Calculate the ratio: r/z = 2/2 = 1 Calculate the bracket term: [1 / (1 + 1²)]^(5/2) = [1 / 2]^(5/2) = [0.5]^2.5 = 0.1768 Calculate the base Boussinesq term (from Example 5.1, at z = 2 m, r = 0): 3Q / (2πz²) = 450 / (2π × 4) = 17.91 kPa Final result: Δσ = 17.91 × 0.1768 = 3.17 kPa Note: Compared to the stress directly below (17.91 kPa at z = 2 m, r = 0), the stress at the same depth but 2 m away is only 3.17 kPa—about 18% of the maximum. This rapid fall-off with horizontal distance is why stresses from distant loads usually don't affect a foundation.
Example 5.3 — Multiple Point Loads (Superposition)
Problem
Two building columns, each carrying 200 kN, are 4 m apart on the ground surface. Find the vertical stress increase at a depth z = 3 m, at a point equidistant from both loads (i.e., 2 m from each column).
Solution
By superposition, the total stress is the sum of stresses from each load. For each load (Q = 200 kN), at z = 3 m, r = 2 m: r/z = 2/3 ≈ 0.667 Boussinesq base term: 3Q / (2πz²) = (3 × 200) / (2π × 9) = 600 / 56.549 = 10.61 kPa Influence factor: [1 / (1 + (2/3)²)]^(5/2) = [1 / (1 + 0.444)]^(5/2) = [1 / 1.444]^(5/2) = [0.6923]^2.5 = 0.3849 Stress from one load: Δσ_one = 10.61 × 0.3849 = 4.08 kPa Stress from both loads (same contribution due to symmetry): Δσ_total = 2 × 4.08 = 8.16 kPa This example demonstrates the superposition principle: the total stress increase is the sum of stress increases from individual loads. For multiple loads, this can be calculated element by element.
Key Points
- Boussinesq solution gives vertical stress increase from a point load on elastic half-space
- Δσ directly below load: Δσ = 3Q / (2πz²)
- Stress decreases with depth (proportional to 1/z²) and with horizontal offset (proportional to 1/(1+(r/z)²)^2.5)
- Boussinesq is most accurate at moderate depths; less accurate near the surface
- Solution assumes homogeneous, elastic, semi-infinite soil—rarely true in nature
- Influence factor charts simplify calculations for non-vertical stresses and lateral distances
While Boussinesq is rigorous and widely used, many practical problems involve distributed loads over finite areas (footings, slabs, embankments) rather than point loads. For such cases, rigorous solutions require integration over the loaded area (very tedious) or use of influence charts. A simpler and widely accepted approximation is the 2:1 (or 45°) method, which assumes the load spreads at a slope of 2 vertical to 1 horizontal (or equivalently, at a 45° angle to the vertical). 2:1 APPROXIMATION FORMULA: For a uniformly loaded rectangular footing of dimensions B × L carrying a total load Q: Δσ = Q / [(B + z) × (L + z)] Where: - Q = total load on footing (kN) - B = width of footing (m) - L = length of footing (m) - z = depth below the footing base (m) - Δσ = average vertical stress increase at depth z (kPa) PHYSICAL INTERPRETATION: The method assumes that the loaded area, when projected to depth z, has expanded by z in all directions (spreading at 1:1 slope in plan), so the width increases from B to (B + z) and the length from L to (L + z). The stress "spreads out" over this larger area, so the intensity decreases. WHEN TO USE 2:1 VS. BOUSSINESQ: 1. Point loads (concentrated, non-distributed): Use Boussinesq. 2. Small footings (B, L are small compared to z): Use Boussinesq or influence charts. 3. Large footings (B, L are significant compared to z): Use 2:1 method or influence charts. 4. For exams, if the problem doesn't specify which method, use 2:1 for footings and Boussinesq for concentrated loads. ACCURACY OF 2:1 METHOD: - Generally gives reasonable estimates of average stress at depth z directly below the footing. - Most conservative (slightly overestimates) in the near field (z < B). - Less accurate for points far from the footing or at shallow depths. - Underestimates stresses at points far laterally from the footing. VARIATION: STRESS AT POINTS AWAY FROM CENTER The 2:1 method as written gives the average stress directly below the footing. For points away from the footing centerline, more sophisticated methods (influence charts, numerical integration) are needed. Some approximate extensions exist but are beyond the scope of licensure exams. COMPARISON WITH BOUSSINESQ: For a distributed load, the 2:1 method and Boussinesq give similar results at moderate depths (z > B/2 or so) but diverge at shallow depths. For practical foundation design, both are acceptable. The 2:1 method is preferred because it is simpler and requires no complex calculations or tables.
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6. Stress Increase from Footing Loads — 2:1 Approximation Method
Examples
Example 6.1 — Square Footing with 2:1 Method
Problem
A 2.5 m × 2.5 m square footing carries a total load of 1000 kN. Using the 2:1 method, calculate the stress increase at depths z = 1 m, 2 m, and 3 m below the footing base.
Solution
Using Δσ = Q / [(B + z)(L + z)] with B = L = 2.5 m, Q = 1000 kN: At z = 1 m: Δσ = 1000 / [(2.5 + 1) × (2.5 + 1)] = 1000 / (3.5 × 3.5) = 1000 / 12.25 = 81.63 kPa At z = 2 m: Δσ = 1000 / [(2.5 + 2) × (2.5 + 2)] = 1000 / (4.5 × 4.5) = 1000 / 20.25 = 49.38 kPa At z = 3 m: Δσ = 1000 / [(2.5 + 3) × (2.5 + 3)] = 1000 / (5.5 × 5.5) = 1000 / 30.25 = 33.06 kPa Summary: The stress decreases significantly with depth, from 82 kPa at 1 m to 33 kPa at 3 m. At 3 m (depth = 1.2 × footing width), the stress is about 40% of the maximum.
Example 6.2 — Rectangular Footing (Non-Square)
Problem
A rectangular footing 3 m × 5 m carries 1500 kN. Find the stress increase at z = 2.5 m depth.
Solution
Using Δσ = Q / [(B + z)(L + z)] with B = 3 m, L = 5 m, Q = 1500 kN, z = 2.5 m: Δσ = 1500 / [(3 + 2.5) × (5 + 2.5)] = 1500 / (5.5 × 7.5) = 1500 / 41.25 = 36.36 kPa Note: For a rectangular footing, the formula still treats the footing as a simple rectangle. The longer dimension (5 m) spreads faster (becomes 7.5 m), so the stress is lower than for a square footing of similar size at the same depth.
Example 6.3 — Comparison: 2:1 Method vs. Boussinesq
Problem
For a 2 m × 2 m footing carrying 500 kN, compare the stress increase at z = 2 m using (a) 2:1 method and (b) Boussinesq (assuming equal distribution to 4 point loads at footing corners).
Solution
(a) 2:1 method: Δσ_2:1 = 500 / [(2 + 2) × (2 + 2)] = 500 / 16 = 31.25 kPa (b) Boussinesq—equivalent point load approach: Assume the 500 kN is distributed to 4 point loads at the corners, each 125 kN. The distance from the footing center to a corner is √(1² + 1²) = 1.414 m. At z = 2 m, each corner load contributes: Δσ_corner = (3 × 125) / (2π × 2²) × [1 / (1 + (1.414/2)²)]^(5/2) = (375 / 25.133) × [1 / (1 + 0.5)]^(5/2) = 14.92 × [1 / 1.5]^(5/2) = 14.92 × 0.272 = 4.06 kPa (per corner) Total from 4 corners: 4 × 4.06 ≈ 16.24 kPa This is an oversimplified Boussinesq estimate and lower than the 2:1 method (31.25 kPa). The discrepancy is due to the approximation of a distributed load as 4 point loads and the geometry. In practice, rigorous Boussinesq for distributed loads requires integration, and the 2:1 method is a practical compromise.
Key Points
- 2:1 method assumes load spreads at 2:1 slope (or 45° angle)
- Δσ = Q / [(B + z)(L + z)] for rectangular footing
- Simple and quick calculation compared to Boussinesq
- Reasonably accurate for depths z > B (width of footing)
- Gives average stress directly below footing, not stress at arbitrary points
- Most conservative (safe) for design purposes
- Assumes uniform load distribution over footing base
For more accurate calculations of stress increase under distributed loads—especially at points away from the footing center—engineers use influence factors (also called influence coefficients) or Newmark's chart. These are graphical or tabular tools that account for the geometry of the loaded area and the position of the point of interest. INFLUENCE FACTOR METHOD: The stress increase under a loaded area is expressed as: Δσ = I_z × q Where: - I_z = influence factor (dimensionless), typically in the range 0 to ~0.5 - q = uniform pressure on the loaded area (σ_footing = Q/[B × L], in kPa) The influence factor depends on: - The depth z relative to footing dimensions B and L - The horizontal position relative to the footing corners - The shape of the loaded area (rectangle, circle, etc.) INFLUENCE FACTORS FOR RECTANGULAR AREAS: For a rectangular loaded area B × L with uniform pressure q, the maximum influence factor (directly below the center) is given by Boussinesq-based solutions or tables. Common values: - At z = 0 (footing base): I_z ≈ 0.5 to 1.0 (depending on geometry) - At z = B (depth = footing width): I_z ≈ 0.1 to 0.3 - At z = 2B: I_z ≈ 0.05 to 0.1 - At z → ∞: I_z → 0 NEWMARK'S CHART: Newmark's chart is a graphical tool (influence chart) that simplifies the calculation of stress increase at any point below a uniformly loaded rectangular area. The chart is typically divided into small squares, each representing an influence factor of 0.001 to 0.01 (depending on the scale). To use the chart: 1. The loaded area is drawn to scale. 2. The point of interest (depth z) is marked. 3. A transparent overlay (with a grid scaled to depth z) is placed on the chart. 4. The number of grid squares within the loaded area is counted. 5. The stress increase is calculated as: Δσ = (number of squares / total squares) × q × influence factor per square. IN PRACTICE: Newmark's chart is less commonly used now because: - Digital calculators and spreadsheet programs have made tabular integration easy. - Boussinesq-based formulas are available in most software. - The chart requires careful drafting and is prone to reading errors. For PRC licensure exams, you should be familiar with the concept but may not need to use Newmark's chart in detail. However, understanding that stress is non-uniform below a footing and varies with lateral distance is important. FOR BOARD EXAMS: Questions on influence factors typically ask you to: 1. Look up or calculate I_z for given B, L, z dimensions. 2. Apply Δσ = I_z × q to find stress increase. 3. Understand that I_z decreases with depth and with distance from the footing center. 4. Compare results with the 2:1 method (which gives average stress).
Heading
7. Influence Factors and Newmark's Chart
Examples
Example 7.1 — Influence Factor Lookup and Application
Problem
A 3 m × 4 m footing applies a uniform pressure of 200 kPa. Using an influence factor table (typical geotechnical references), find the vertical stress increase at z = 2 m depth directly below the footing center.
Solution
From Boussinesq-based influence factor tables (or calculation): For B/z = 3/2 = 1.5 and L/z = 4/2 = 2.0, the influence factor I_z ≈ 0.215 (typical value; actual value depends on the reference table used). Stress increase: Δσ = I_z × q = 0.215 × 200 = 43 kPa Comparison with 2:1 method (Example 6): 2:1: Δσ = Q / [(B+z)(L+z)] = 1200 / [(3+2)(4+2)] = 1200 / (5 × 6) = 40 kPa Note: q = Q / (B × L) = 1200 / 12 = 100 kPa is the footing pressure. Using this in the influence factor method: Δσ = 0.215 × 100 = 21.5 kPa (if I_z = 0.215 for this geometry—check your reference). The exact values vary by reference, but this illustrates how influence factors are applied.
Key Points
- Influence factors I_z simplify stress calculation: Δσ = I_z × q
- I_z is dimensionless and depends on footing geometry and depth
- I_z decreases with depth and lateral distance from footing center
- Newmark's chart is a graphical tool to determine I_z without calculation
- Influence factors account for non-uniform stress distribution below footing
- For centers of rectangular footings, I_z can be tabulated or looked up in handbooks
- 2:1 method gives average stress; influence factors give local stress
Real soil deposits often consist of multiple layers with different properties (sand, clay, silt) and different unit weights. When calculating effective stress in a layered profile subject to surface loads, we must: 1. Calculate the total stress: σ = Σ(γ_i × z_i) + Δσ_load, where Δσ_load is the stress increase from surface loads. 2. Calculate pore pressure: u = γ_w × z_w (depends only on depth below WT). 3. Calculate effective stress: σ' = σ - u. MULTI-LAYER STRESS CALCULATION: Procedure: - At each layer boundary and within each layer, calculate σ by summing weights down to that depth, then add the Δσ from the applied load (using Boussinesq or 2:1 method). - Calculate u based on the depth below the water table (same calculation regardless of layers). - Compute σ' = σ - u. IMPORTANCE FOR FOUNDATION DESIGN: The stress distribution through layers determines: - Settlement potential: If a compressible layer (clay) has high σ', it will settle; if another clay at the same depth has low σ' (high water table), it may not. - Bearing capacity: Different layers have different strengths; the critical layer (weakest or most stressed) governs. - Lateral stress: Horizontal stress in each layer depends on K_0 (coefficient of lateral stress at rest), which varies by layer type. DRAWING PROFILES FOR LAYERED SOILS: For a complete analysis, plot three diagrams for each layer: 1. Total stress σ (increases in steps at layer boundaries) 2. Pore pressure u (increases linearly with depth below WT) 3. Effective stress σ' = σ - u (step changes at layer boundaries, adjusted by u) 4. Stress increase Δσ from applied load (decreases with depth) 5. Total stress including load: σ_total = σ_initial + Δσ 6. Effective stress with load: σ'_total = σ_total - u
Heading
8. Multi-Layer Soil with Stress Increase
Examples
Example 8.1 — Multi-Layer Profile with Foundation Load
Problem
A 2 m × 2 m footing carries 800 kN and is embedded 1 m below the ground surface. The soil profile is: - 0–1 m: Sand, γ = 17 kN/m³ (above footing base) - 1–5 m: Clay, γ = 18 kN/m³ - 5–10 m: Sandy clay, γsat = 20 kN/m³ - Water table at 5 m depth Calculate σ, u, and σ' at 5 m depth (at WT) and at 7 m depth (2 m into sandy clay). Note: At the footing base (1 m depth), assume a 2:1 stress increase based on load.
Solution
First, calculate the footing pressure and stress increase at various depths. Footing pressure: q = 800 / (2 × 2) = 200 kPa at 1 m depth. Using 2:1 method, at depth z below the footing base: Δσ = 800 / [(2 + z) × (2 + z)] At z = 4 m below footing (i.e., at 5 m absolute depth): Δσ = 800 / [(2 + 4) × (2 + 4)] = 800 / 36 = 22.22 kPa Total stress at 5 m: σ = 17 × 1 + 18 × 4 + 0 = 17 + 72 = 89 kPa (in-situ, no sandy clay yet) σ_with_load = 89 + 22.22 = 111.22 kPa Pore pressure at 5 m (at WT): u = 0 (exactly at WT by convention, no depth below) Effective stress at 5 m: σ'_initial = 89 kPa σ'_with_load = 111.22 kPa At z = 7 m (2 m below WT): Δσ = 800 / [(2 + 6) × (2 + 6)] = 800 / 64 = 12.5 kPa Total stress at 7 m: σ = 17 × 1 + 18 × 4 + 20 × 2 = 17 + 72 + 40 = 129 kPa (in-situ) σ_with_load = 129 + 12.5 = 141.5 kPa Pore pressure at 7 m (2 m below WT): u = 9.81 × 2 = 19.62 kPa Effective stress at 7 m: σ'_initial = 129 - 19.62 = 109.38 kPa σ'_with_load = 141.5 - 19.62 = 121.88 kPa Summary table (with load): | Depth (m) | σ (kPa) | Δσ (kPa) | σ_total (kPa) | u (kPa) | σ'_total (kPa) | |-----------|---------|---------|---|---------|--------| | 5 (WT) | 89 | 22.22 | 111.22 | 0 | 111.22 | | 7 | 129 | 12.5 | 141.5 | 19.62 | 121.88 | Interpretation: The load increases total stress at 5 m by 22 kPa and at 7 m by 12.5 kPa. Below the water table (7 m), the pore pressure (19.62 kPa) reduces effective stress but does not eliminate it. The effective stress is still high enough to support foundation stability.
Key Points
- In layered soil, total stress has step changes at layer boundaries
- Pore pressure increases linearly with depth below WT (independent of layers)
- Effective stress reflects both layer changes and pore pressure
- Stress increase from loads diminishes with depth and is independent of existing soil layers
- For settlement and bearing capacity, effective stress in critical layers is most important
- Plotting profiles helps visualize stress distribution and identify critical depths
PITFALL 1: FORGETTING PORE PRESSURE BELOW THE WATER TABLE One of the most common errors is calculating total stress correctly but then either: - Not subtracting pore pressure (incorrectly stating that σ' = σ below the WT). - Using pore pressure as if it acts like soil stress (it doesn't—water has no shear strength). Reminder: σ' = σ - u always applies. Below the WT, u is always positive and reduces σ'. PITFALL 2: USING THE WRONG UNIT WEIGHT - Using γ_moist above the WT is correct. - Using γ_sat above the WT is wrong—leads to overstated total stress. - Using γ_d (dry) below the WT is wrong—gives zero stress. - Always use γ_sat below the WT. Example error: "A layer is 4 m thick and has γ = 20 kN/m³. Below the WT, I'll use γ = 20..." Wrong! If 20 is the in-situ (moist) value, and the soil is saturated below the WT, the actual γ_sat is likely 19–21 kN/m³, and this should be used. PITFALL 3: MISAPPLYING THE 2:1 METHOD - The formula is Δσ = Q / [(B + z)(L + z)], not Q / (B × L) (which gives the footing pressure). - Some students write Δσ = Q / [(B + z)(L - z)] or other variants—these are wrong. - The 2:1 method is an approximation; it doesn't work well at very shallow depths (z < B/2) or for highly non-uniform loads. PITFALL 4: CONFUSING BOUSSINESQ FORMULAS - For a point load directly below: Δσ = 3Q / (2πz²). This is correct. - At an offset: Use the full formula with (r/z) term. Don't forget the fifth power in the denominator. - Don't apply the direct formula to distributed loads without integration. PITFALL 5: NOT ACCOUNTING FOR WATER TABLE LOCATION CORRECTLY - Pore pressure depends on depth below the WT, not absolute elevation or depth below the surface. - If the WT is at 3 m and you are at 7 m depth, then z_w = 7 - 3 = 4 m, and u = 9.81 × 4. - If the WT is above the surface (e.g., in flooded conditions), treat the water layer as having γ = γ_w. PITFALL 6: IGNORING CAPILLARY EFFECTS - Above the WT, in fine-grained soils, capillary rise creates negative pore pressure (suction). - This suction increases effective stress and can make clay very strong and stiff. - When the capillary water is removed (e.g., by construction drainage), the soil loses this suction and settles or becomes weaker. - For exams, if the problem doesn't explicitly mention capillary rise, assume u = 0 above the WT. But be aware of this concept. PITFALL 7: SEEPAGE AND THE QUICK CONDITION - Upward seepage increases pore pressure, reducing effective stress and creating instability. - The critical hydraulic gradient i_c ≈ 1.0 for most sands (varies from 0.8 to 1.2). - If the applied hydraulic gradient exceeds i_c, the quick condition develops, and the soil behaves like a fluid. - This is critical for excavations, dams, and any situation with upward groundwater flow. BOARD EXAM STRATEGY: 1. Always draw a clear sketch of the problem showing depths, WT location, layers, and loads. 2. List known values: γ values, depths, WT location, applied loads. 3. Calculate in order: total stress → pore pressure → effective stress (→ stress increases if loads are present). 4. Check units (kPa, kN, m) consistently. 5. For multiple-choice, check if the answer makes physical sense: effective stress should decrease with rising WT, increase with depth, and be reduced by pore pressure. 6. For free-response, show all steps and unit conversions. 7. If using Boussinesq or 2:1, state the method clearly and justify (e.g., "For this concentrated load, Boussinesq applies...").
Heading
9. Common Pitfalls and Board Exam Tips
Examples
Example 9.1 — Common Error: Wrong Unit Weight
Problem
A 5 m thick clay layer has γ_sat = 19.5 kN/m³ and γ_d = 16 kN/m³. The WT is at the bottom of the layer (5 m depth). A student calculates: Student's solution: σ = 16 × 5 = 80 kPa (WRONG) Student's statement: "I used γ_d because the soil is not saturated at depths 0–5 m." Correct solution: If the soil is above the WT from 0–5 m and then saturated below 5 m, we need more information about the moist γ. If the problem states γ = 18 kN/m³ (in-situ, moist): σ = 18 × 5 = 90 kPa If the WT is exactly at 5 m (at the layer boundary), u = 0 at that depth, so σ' = 90 kPa. Error explanation: The student used dry unit weight (16 kN/m³), which gives too low a stress. Dry unit weight is used only in very specific contexts (e.g., calculating internal friction angle from triaxial tests). For stress calculations, use the in-situ (moist) unit weight above the WT and saturated unit weight below.
Example 9.2 — Quick Condition Verification
Problem
In an excavation with upward seepage, the average hydraulic gradient is measured as i = 1.2. The sand has γ_sat = 20.5 kN/m³. Will the quick condition occur?
Solution
Critical hydraulic gradient: i_c = (γ_sat - γ_w) / γ_w = (20.5 - 9.81) / 9.81 = 10.69 / 9.81 = 1.09 Applied gradient: i = 1.2 Comparison: i (1.2) > i_c (1.09), so the quick condition WILL occur. Consequence: At the point where the quick condition develops, effective stress becomes zero (σ' = 0), and the sand loses all strength. It behaves like a liquid. This is extremely dangerous in excavations and can lead to collapse. Mitigation: reduce seepage (dewatering, pumping) or stabilize with sheet piling or grouting.
Key Points
- Always subtract pore pressure: σ' = σ - u
- Use γsat below WT; use γ (moist) above WT
- 2:1 method: Δσ = Q / [(B+z)(L+z)], not Q/(B×L)
- Pore pressure depends on depth below WT, not absolute depth
- Capillary rise above WT creates negative u (suction), increasing σ'
- Upward seepage reduces σ' and can cause quick condition
- Quick condition occurs when hydraulic gradient ≥ γ'_sat / γ_w ≈ 1.0
- Draw sketches, label everything, check units, verify answers make physical sense
FOUNDATION DESIGN (NSCP 2015 AND LOCAL PRACTICE): The National Structural Code of the Philippines (NSCP 2015) emphasizes effective-stress calculations for foundation design. When calculating bearing capacity, the designer must: 1. Determine the effective stress profile at the foundation level and at critical depths below. 2. Use effective stress in bearing-capacity formulas (e.g., Terzaghi's formula adjusted for shape and depth). 3. Account for groundwater conditions; if the water table is high, the bearing capacity is reduced. For example, in the Manila area where water tables are often high (1–3 m), effective stress is significantly reduced, limiting bearing capacity. This is why deep foundations (piles, caissons) are common in Manila and Cebu. SETTLEMENT ANALYSIS (TERZAGHI'S CONSOLIDATION THEORY): When a load is applied to clay, the soil compresses. Settlement occurs in two phases: 1. Immediate settlement (elastic, occurs instantly). 2. Consolidation settlement (time-dependent, as pore water is squeezed out and effective stress increases). The rate of consolidation depends on: - The thickness of the clay layer - The drainage conditions (is water free to escape?) - The permeability (how easily water flows through the clay) - The increase in effective stress (Δσ') from the applied load Without correct calculation of Δσ' from the footing load, consolidation estimates are useless. SLOPE STABILITY AND WATER TABLE MANAGEMENT: In the Philippines, during the monsoon season, water tables rise significantly. This is a major cause of slope failures on roads and embankments. The mechanism: 1. Rising WT increases pore pressure u. 2. Higher u reduces effective stress σ'. 3. Lower σ' reduces friction strength (tan φ decreases in terms of effective stress). 4. Slope fails. Solutions include: - Controlling drainage (interceptor drains, subsurface drainage) - Reducing slopes (flatter inclination) - Removing weight from slope top - Using pile reinforcement (soil nails, geo-grids) All these require understanding how u and σ' vary with depth and groundwater conditions. EXCAVATION AND EARTH PRESSURE: When an excavation is made, the stress conditions change: - The in-situ stress above the excavation is removed. - Lateral earth pressure on retaining walls decreases if the water table is above the wall. - But if water seeps into the excavation, upward seepage pressure acts on the base, reducing effective stress and potentially causing base heave or piping failure. Correct calculation of pore pressure is essential to: 1. Design safe retaining walls. 2. Prevent base heave (instability of the excavation bottom). 3. Prevent piping (where water seeps along a path, removing soil). DAM AND LEVEE DESIGN: For dams and levees, especially in flood-prone areas of the Philippines, effective stress in the dam material and at the foundation is critical: - Rapid drawdown (quick removal of water from the reservoir) can cause slope failure because the pore pressure inside the dam doesn't drop as fast as the outside water level, creating a hydraulic gradient that points outward. Effective stress decreases, and the slope fails. - During seepage through the dam, high pore pressures in zones where seepage is concentrated can cause piping and failure. For PRC ENGINEERS IN PRACTICE: Before submitting a design, a PE (Professional Engineer) in civil engineering must: 1. Verify that all stress calculations are correct (total, pore, effective). 2. Check that bearing capacities, settlements, and stability factors are based on correct effective stress. 3. Ensure that water-table assumptions are reasonable and conservative (usually assume worst-case, highest WT during design life). 4. Document all assumptions clearly for future reviewers and inspectors. FAILURE TO DO SO can result in: - Unsafe structures (settlements, failures) - Legal liability (RA 544 accountability) - Damage to public safety - Professional sanctions from the Board of Examiners
Heading
10. Real-World Applications and PRC Context
Examples
Example 10.1 — Foundation Settlement in Manila
Problem
A low-rise building (3-story) is being designed in Manila on a site with: - 2 m of loose sand, γ = 16.5 kN/m³ - 5 m of soft clay, γ_sat = 18 kN/m³ - Water table at 1.5 m depth (typical for Manila bay area) A 3 m × 3 m footing will be placed at 1 m depth (just above the WT). The footing carries 900 kN. Using 2:1 method, find the stress increase in the clay layer at 5 m depth (4 m below the footing base).
Solution
At 4 m below footing base (z = 4 m): Δσ = 900 / [(3 + 4) × (3 + 4)] = 900 / 49 = 18.37 kPa For consolidation settlement calculation, the effective stress increase in the clay at 5 m depth is: Δσ' = 18.37 kPa (since pore pressure doesn't change with surface load distribution—the same stress increase acts on both soil grains and water) This 18.37 kPa increase, acting on a soft Manila clay, would be analyzed using the consolidation coefficient c_v and the compression index C_c to estimate settlement. Typical Manila clays have C_c = 0.3–0.5, so a 4 m thick layer could settle 50–200 mm under this load. This is significant and would crack the 3-story building.
Example 10.2 — Slope Stability During Monsoon
Problem
A roadside slope in the Philippines is 30° incline, 10 m high. Soil is clay with φ' = 25° and c' = 10 kPa. Normally (dry season), the factor of safety is FS = 1.5. During the monsoon, the water table rises from 8 m deep (dry) to 2 m deep (wet). Estimate the new factor of safety.
Solution
Simplified analysis (not a full slope stability calculation, which is complex): In the dry season: - Effective stress is high (u = 0 or low). - Shear strength τ = c' + σ' tan φ' is high. - FS = 1.5 (stable, with margin). During monsoon: - Water table rises to 2 m. - Pore pressure increases significantly in the zone 2–10 m depth. - Effective stress decreases: σ' = σ - u becomes much smaller. - Shear strength τ = c' + σ' tan φ' decreases (because σ' is smaller). - FS decreases, perhaps to 0.8–1.0 (unstable). Typical outcome: The slope fails during heavy monsoon rains, especially in the 2–5 m depth zone where water table rise is most dramatic. Mitigation: Install subsurface drains to keep the water table at least 2–3 m below the slope surface, or reduce the slope angle to increase stability.
Key Points
- NSCP 2015 requires effective-stress-based foundation design
- Settlement calculations depend critically on Δσ' from loads
- Rising water tables during monsoon season are a major cause of slope failures in the Philippines
- Consolidation theory requires correct Δσ' calculation
- Excavation and retaining wall design depend on pore pressure distribution
- Dam and levee design requires careful analysis of seepage and pore pressure
- PEs must verify all effective-stress calculations and document assumptions
- High water tables in Manila, Cebu, and coastal areas require deep foundations and special design
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