CELE Geotechnical Engineering — Stresses in Soil MassDetailed Explanation
Want to really understand Stresses in Soil Mass before tackling CELE Geotechnical Engineering questions? This detailed explanation breaks down every key concept, shows you why it matters for the CELE 2026, and walks through the reasoning Professional Regulation Commission (PRC) — Board of Civil Engineering expects on high-difficulty questions.
Exam context
Professional Regulation Commission (PRC) — Board of Civil Engineering runs the Civil Engineer Licensure Examination on May and November 2026. Its Geotechnical Engineering section sits under a "Core" weighting, and Stresses in Soil Mass is the 4th chapter in the 11-chapter CELE Geotechnical Engineering rotation. The CELE passing mark is 70% weighted average, no sub-test below 50%, and the most recent 2026 paper drew about a meaningful share of questions from Geotechnical Engineering.
Stresses in Soil Mass - Detailed Explanation
Understanding stresses in soil is the cornerstone of geotechnical engineering. Whether you are designing a footing, analyzing slope stability, or predicting settlement, you must know how stress distributes through the ground. Terzaghi's effective-stress principle — arguably the single most important concept in soil mechanics — tells us that soil behavior is not controlled by the total overburden pressure but by the net stress carried by the soil skeleton after pore water pressure is subtracted. On top of that, surface loads from footings, embankments, and fills add stress increments that decrease with depth. This chapter covers both concepts in the depth and format you will encounter on the PRC Civil Engineer Licensure Examination.
Concepts
Total Stress (σ)
Total stress at any point in a soil mass is the total force per unit area acting on that plane, caused by the weight of everything above it — soil, water, and any surface loads. It is computed by summing the product of unit weight and thickness for every layer above the point of interest. Below the water table, use the saturated unit weight (γ_sat); above it, use the moist or dry unit weight as appropriate. Total stress is independent of drainage conditions; it is purely a function of the overburden geometry and unit weights. Mathematically: σ = Σ(γᵢ × zᵢ). For a single homogeneous layer of thickness z: σ = γz. For multiple layers: σ = γ₁z₁ + γ₂z₂ + γ₃z₃ + … Total stress cannot be measured directly in the field; it is always calculated. It is expressed in kPa (kN/m²) in SI units.
Examples
Layer 1 is above the water table so use γ_dry = 16 kN/m³. Layer 2 is fully saturated so use γ_sat = 19 kN/m³. The total stress at z = 5 m is simply the sum of the two pressure contributions — straightforward layer-by-layer accumulation.
Scenario
A soil profile consists of: Layer 1 — 2 m dry sand, γ_d = 16 kN/m³; Layer 2 — 3 m saturated clay, γ_sat = 19 kN/m³. The water table is at the top of Layer 2. Compute total stress at the bottom of Layer 2 (z = 5 m from surface).
Solution
σ = γ₁z₁ + γ₂z₂ = 16(2) + 19(3) = 32 + 57 = 89 kPa
Even though the soil is submerged, total stress still uses γ_sat (not γ' or γ_sub). The pore water pressure will be handled separately when computing effective stress.
Scenario
A 4 m uniform deposit of saturated soil (γ_sat = 20 kN/m³) is fully submerged — the water table is at the ground surface. Compute total stress at z = 4 m.
Solution
σ = γ_sat × z = 20 × 4 = 80 kPa
Applications
- Foundation design — total stress is the starting point before subtracting pore pressure
- Retaining wall lateral earth pressure calculations
- Stability analysis of embankments
- Input into effective stress and consolidation calculations
Misconceptions
- Using γ_sub (buoyant unit weight) instead of γ_sat when computing total stress below the water table — γ_sub is only used when computing effective stress directly via σ' = Σγ'z below WT
- Forgetting to switch unit weights at the water table boundary
- Assuming total stress equals effective stress when there is no water — this is only true when u = 0 (water table far below the point)
Related Concepts
- Effective stress
- Pore water pressure
- Unit weight relationships
- Overburden pressure
Common Exam Questions
Example
A profile has 1.5 m of fill (γ = 17 kN/m³), 2 m of sand (γ_sat = 19 kN/m³), and 3 m of clay (γ_sat = 18 kN/m³). WT at top of sand. Find σ at bottom of clay: σ = 17(1.5) + 19(2) + 18(3) = 25.5 + 38 + 54 = 117.5 kPa.
Approach
Identify all layers from surface to the point. Multiply γᵢ × zᵢ for each layer. Sum all values. Always check which layers are above and which are below the water table.
Question Type
Multi-layer profile — compute total stress at a given depth
Key Points To Remember
- σ = Σ(γᵢ × zᵢ) — sum layer by layer from the surface down to the point
- Use γ_sat (saturated unit weight) for soil below the water table
- Use γ_moist or γ_dry for soil above the water table
- Total stress INCREASES monotonically with depth
- Total stress does NOT control soil strength or consolidation — effective stress does
- Typical γ_sat values: 18–22 kN/m³; γ_w = 9.81 kN/m³ (use 10 kN/m³ only when the problem specifies)
Pore Water Pressure (u)
Pore water pressure is the pressure carried by the water filling the voids between soil particles. Under hydrostatic (no-flow) conditions, it equals γ_w × z_w, where z_w is the depth of the point below the water table (the piezometric head). Pore pressure acts equally in all directions and does NOT contribute to intergranular contact forces; it pushes soil grains apart. Under seepage conditions, pore pressure is modified by the seepage gradient: upward seepage increases pore pressure (reduces effective stress and may cause a 'quick' or 'boiling' condition), while downward seepage decreases pore pressure (increases effective stress). In consolidation problems, excess pore pressure (Δu) develops when a load is applied rapidly and dissipates over time as water drains — this is the mechanism behind time-dependent settlement. Key formula: u = γ_w × z_w (hydrostatic, below the water table). Above the water table: u = 0 (for simplicity unless capillarity is considered; capillary suction gives negative u, increasing σ', but this is rarely tested in board exams without explicit data).
Examples
z_w = 7 − 3 = 4 m (depth below the water table). Pore pressure is strictly a function of the height of water above the point, not total depth from surface.
Scenario
The water table is at 3 m depth. Find pore pressure at z = 7 m (i.e., 4 m below WT).
Solution
u = γ_w × z_w = 9.81 × 4 = 39.24 kPa
The critical hydraulic gradient (i_cr) at which effective stress becomes zero (quick condition) is i_cr = γ'/γ_w = (γ_sat − γ_w)/γ_w. This is a classic board exam topic — if i ≥ i_cr, the soil liquefies or 'boils'.
Scenario
Upward seepage occurs with a hydraulic gradient i = 0.3 in a saturated sand layer. The water table is at the ground surface. Find effective pore pressure gradient.
Solution
Under upward seepage: u at depth z = γ_w(z + iz) = γ_w × z(1 + i). The seepage pressure per unit volume adds γ_w × i × z to the normal hydrostatic pressure.
Applications
- Effective stress computation in multi-layer profiles
- Quick condition and piping analysis in dams and excavations
- Consolidation settlement time-rate analysis
- Triaxial test interpretation (total vs. effective stress paths)
Misconceptions
- Using depth from the surface (z) instead of depth from the water table (z_w) when computing u
- Forgetting artesian (confined) conditions where piezometric head is above the ground — z_w must be measured from the piezometric surface, not the WT
- Assuming pore pressure always increases effective stress — it reduces effective stress
- Ignoring negative pore pressure (suction) above the water table — conservative to ignore, but it means effective stress above WT may be higher than calculated
Related Concepts
- Total stress
- Effective stress
- Seepage and hydraulic gradient
- Consolidation
- Quick condition
Common Exam Questions
Example
Artesian pressure head is 5 m above the ground surface. At z = 3 m below ground in the confined aquifer: z_w = 3 + 5 = 8 m (measure from piezometric head down to point). u = 9.81 × 8 = 78.48 kPa.
Approach
Identify z_w = depth of point below water table. Multiply by γ_w = 9.81 kN/m³. If the problem says 'artesian' conditions, z_w is measured from the piezometric level (which may be above the ground surface), not the water table.
Question Type
Compute pore pressure at a given depth given water table location
Key Points To Remember
- u = γ_w × z_w (hydrostatic); z_w = depth below the phreatic surface
- γ_w = 9.81 kN/m³ (use the exact value given in the problem)
- u = 0 above the water table (unless capillary data is given)
- Upward seepage: u increases → σ' decreases (dangerous for stability)
- Downward seepage: u decreases → σ' increases (conservative, safer)
- Excess pore pressure from loading dissipates during consolidation
- Pore pressure does NOT contribute to shear strength
Effective Stress (σ') — Terzaghi's Principle
Terzaghi's effective stress principle (1936) is the foundation of modern soil mechanics: σ' = σ − u. Effective stress is the portion of total stress transmitted through the soil skeleton at grain-to-grain contacts. It is the stress that directly governs: (1) shear strength via the Mohr-Coulomb criterion τ = c' + σ'tan φ'; (2) volumetric compression and consolidation settlement; and (3) permeability-driven drainage behavior. The principle means that if pore pressure increases (e.g., due to rainfall raising the water table, artesian pressure, or rapid loading), the effective stress decreases and the soil weakens — this is the mechanism behind many landslides and foundation failures in the Philippines during typhoon season. Conversely, lowering the water table (e.g., by pumping) increases effective stress and can improve bearing capacity. An alternative computation: σ' = Σγ'ᵢzᵢ (using submerged unit weight γ' = γ_sat − γ_w below WT, and γ above WT) — this gives the same result as σ − u and is often faster for fully submerged profiles.
Examples
Build σ layer by layer using γ_sat for the submerged clay. Compute u from depth below WT (z_w = 2 m). Subtract: σ' = σ − u. Alternatively: σ' = γ(3) + γ'(2) = 18(3) + (20−9.81)(2) = 54 + 20.38 = 74.38 kPa — same answer, confirming both approaches.
Scenario
Board-exam classic: A soil profile has 3 m of sand (γ = 18 kN/m³) above the water table and 2 m of saturated clay (γ_sat = 20 kN/m³) below. Find σ, u, and σ' at z = 5 m (2 m below WT).
Solution
σ = 18(3) + 20(2) = 54 + 40 = 94 kPa; u = 9.81(2) = 19.62 kPa; σ' = 94 − 19.62 = 74.38 kPa ≈ 74.4 kPa
When the WT rises to the surface, pore pressure at z = 5 m is 9.81 × 5 = 49.05 kPa (full column of water above the point). Effective stress drops from 74.4 kPa to ~45 kPa — a 40% reduction that severely impacts bearing capacity and slope stability. This scenario is common during Philippine monsoon season.
Scenario
The water table rises to the ground surface. Recompute σ' at z = 5 m for the same profile (now both layers are saturated: γ_sat1 = 18 kN/m³ [assume it saturates], γ_sat2 = 20 kN/m³).
Solution
σ = 18(3) + 20(2) = 94 kPa (same); u = 9.81(5) = 49.05 kPa; σ' = 94 − 49.05 = 44.95 kPa
Notice: σ' increases above WT at the same rate as σ (since u=0). Below the WT, σ increases at rate γ_sat but σ' increases at rate γ' = γ_sat − γ_w = 9.19 kN/m³ only — the increase is slower because pore pressure also builds up.
Scenario
Compute the effective stress profile at z = 0, 2, 4, and 6 m for a 6 m deposit: WT at 2 m, γ_moist = 17 kN/m³ (0–2 m), γ_sat = 19 kN/m³ (2–6 m). Use γ_w = 9.81 kN/m³.
Solution
z=0: σ=0, u=0, σ'=0 kPa | z=2m: σ=17×2=34 kPa, u=0, σ'=34 kPa | z=4m: σ=34+19×2=72 kPa, u=9.81×2=19.62 kPa, σ'=52.38 kPa | z=6m: σ=34+19×4=110 kPa, u=9.81×4=39.24 kPa, σ'=70.76 kPa
Applications
- Bearing capacity: q_ult uses φ' and c' (effective parameters)
- Consolidation settlement: settlement is driven by increase in σ'
- Slope stability: Factor of Safety uses τ = c' + σ'tan φ'
- Retaining walls: active/passive pressures (Rankine, Coulomb) in effective stress terms
- Pile design: skin friction uses σ' (β-method: f_s = β × σ'_v)
Misconceptions
- Thinking σ' = σ when the soil is 'dry' above the WT — correct (u=0 so σ'=σ), but students sometimes apply this rule below the WT
- Using γ' in σ computation instead of σ' computation — γ' is for effective stress direct calculation, NOT for total stress
- Forgetting that effective stress CANNOT be negative in a granular soil (negative means tension, which soil cannot sustain — sand 'boils' or crack forms)
Related Concepts
- Shear strength (Mohr-Coulomb)
- Consolidation and settlement
- Total stress
- Pore water pressure
- Bearing capacity
Common Exam Questions
Example
Profile: 2m fill (γ=16), 3m sand (γ_sat=19), WT at top of sand. Find σ' at base of sand: σ=16(2)+19(3)=89 kPa; u=9.81(3)=29.43 kPa; σ'=59.57 kPa.
Approach
Step 1: Draw the profile and mark WT. Step 2: Compute σ = Σγᵢzᵢ (γ_sat below WT). Step 3: Compute u = γ_w × z_w. Step 4: σ' = σ − u. Verify with σ' = Σγ'ᵢzᵢ if time permits.
Question Type
Three-variable computation: find σ, u, and σ' at a specified depth
Example
WT drops 1 m: new z_w decreases by 1 m → u decreases by 9.81 kPa → σ' increases by ~9.81 kPa (net effect depends on which zone changes unit weight).
Approach
Compute σ' at original WT, then recompute with new WT. Compare. Raising WT decreases σ'; lowering WT increases σ'.
Question Type
Effect of water table change on effective stress
Key Points To Remember
- σ' = σ − u (ALWAYS — this is Terzaghi's principle)
- σ' controls strength, settlement, and consolidation — not σ
- Equivalent formula: σ' = Σγᵢzᵢ (γ above WT) + Σγ'ᵢzᵢ (γ' below WT), where γ' = γ_sat − γ_w
- σ' is always ≥ 0 for stable soil (if σ' = 0, soil is at the verge of 'quick' condition)
- Raising WT → u increases → σ' decreases → weaker soil
- Lowering WT → u decreases → σ' increases → stronger soil
- In an over-consolidated clay, OCR = σ'_p / σ'_v0 (past effective stress / current effective stress)
Stress Increase from Surface Loads — Boussinesq Point Load
When a concentrated point load Q acts on the surface of a semi-infinite, homogeneous, isotropic, elastic half-space, the vertical stress increase at any point inside the mass is given by Boussinesq's (1885) solution: Δσ_z = (3Q / 2π z²) × [1 / (1 + (r/z)²)]^(5/2), where z is the depth below the surface and r is the horizontal distance from the load to the point. Directly below the load (r = 0): Δσ_z = 3Q / (2π z²). The stress bulb concept: Δσ diminishes rapidly with depth (proportional to 1/z²) and with horizontal distance. At the same depth, stress is maximum directly below the load and decreases as r increases. The Boussinesq solution assumes: linear elastic, homogeneous, isotropic, semi-infinite soil — assumptions that are approximations for real soil but give results accurate enough for engineering practice. For board exams, the most commonly tested case is r = 0 (directly below the load).
Examples
Since r = 0, the simplified formula applies. Note 2π × 2² = 2π × 4 = 25.13. This is a board exam staple — memorize 3Q/(2πz²).
Scenario
A concentrated load Q = 100 kN acts at the surface. Find Δσ directly below at z = 2 m (r = 0).
Solution
Δσ = 3Q / (2πz²) = 3(100) / (2π × 4) = 300 / 25.133 = 11.94 kPa
(0.5)^2.5 = (0.5)^2 × (0.5)^0.5 = 0.25 × 0.7071 = 0.1768. The stress at r = 2 m is only 17.6% of the stress directly below — showing rapid attenuation with horizontal distance.
Scenario
For the same Q = 100 kN at the surface, find Δσ at z = 2 m, r = 2 m.
Solution
r/z = 2/2 = 1; Δσ = (3×100)/(2π×4) × [1/(1+1²)]^(5/2) = 11.94 × [1/2]^(5/2) = 11.94 × (0.5)^2.5 = 11.94 × 0.1768 = 2.11 kPa
This is Exercise 3 from the reference. The key steps: compute r/z, evaluate the bracket term, multiply by 3Q/(2πz²). Always compute the bracket exponent carefully — (5/2) = 2.5, so split into square and square root for mental calculation.
Scenario
Find Δσ at z = 3 m, r = 2 m for Q = 200 kN.
Solution
r/z = 2/3 = 0.667; [1/(1+(0.667)²)]^(5/2) = [1/(1+0.444)]^(5/2) = [1/1.444]^(5/2) = (0.6926)^(5/2); (0.6926)^2 = 0.4797; (0.6926)^0.5 = 0.8322; product = 0.3992; Δσ = (3×200)/(2π×9) × 0.3992 = (600/56.55) × 0.3992 = 10.61 × 0.3992 = 4.24 kPa
Applications
- Estimating stress increase below isolated column footings (conservative upper bound)
- Settlement computation when combined with consolidation theory
- Superposition of multiple point loads (sum individual Δσ from each load)
- Stress under a pile point load
Misconceptions
- Using z as total depth from surface instead of depth from load application point (if load is at depth, adjust accordingly)
- Forgetting the (5/2) power — some students use (3/2) which gives a completely wrong answer
- Confusing Boussinesq (elastic theory) with the 2:1 method — they give different answers; Boussinesq is more accurate for point loads
- Using the full formula when r=0 and simplifying incorrectly — at r=0, [1/(1+0)]^2.5 = 1, so Δσ = 3Q/(2πz²) cleanly
Related Concepts
- 2:1 stress distribution method
- Stress bulb
- Settlement computation
- Influence factor charts (Newmark)
Common Exam Questions
Example
Q = 150 kN at z = 3 m, r = 0: Δσ = 3(150)/(2π×9) = 450/56.55 = 7.96 kPa.
Approach
Use Δσ = 3Q/(2πz²). Substitute Q (kN) and z (m). Answer in kPa. Always confirm r = 0 is stated or implied.
Question Type
Compute Δσ directly below a point load at given depth
Example
Q=100 kN, z=2 m, r=1 m: r/z=0.5; bracket=[1/1.25]^2.5=(0.8)^2.5=0.8^2×0.8^0.5=0.64×0.894=0.572; Δσ=11.94×0.572=6.83 kPa.
Approach
Compute r/z. Evaluate [1/(1+(r/z)²)]^2.5. Multiply by 3Q/(2πz²). Take care with the exponent calculation.
Question Type
Compute Δσ at offset point (r ≠ 0)
Key Points To Remember
- Boussinesq: Δσ = (3Q/2πz²) × [1/(1+(r/z)²)]^(5/2)
- At r = 0 (directly below): Δσ = 3Q / (2πz²) — memorize this
- Δσ decreases as z increases (1/z² relationship directly below)
- Δσ decreases as r increases (stress bulb spreads out)
- Units: Q in kN, z and r in m → Δσ in kPa
- Assumption: homogeneous, isotropic, linear elastic semi-infinite mass
- For r = 0: Δσ is maximum at the surface (theoretically infinite at z=0) and decreases with depth
Stress Increase from Surface Loads — 2:1 Approximation Method
The 2:1 (two-to-one) method is a practical, simplified approach to estimate the average stress increase Δσ below a uniformly loaded rectangular footing of dimensions B × L carrying a total load Q. The load is assumed to spread at a 2-vertical : 1-horizontal ratio in both plan dimensions, so at depth z below the footing, the stressed area becomes (B + z)(L + z). The formula is: Δσ = Q / [(B + z)(L + z)]. This gives the average stress increase at depth z, which is suitable for settlement computations. Note: Q here is the total applied load (kN), not a pressure — be careful about units. If given a contact pressure q₀ (kPa) instead: Δσ = q₀ × B × L / [(B + z)(L + z)]. The 2:1 method is simpler and less accurate than Boussinesq theory but is widely used in preliminary design and is a board exam favorite for footing problems. It is conservative (overestimates stress) near the edges of the loaded area but underestimates at the center for shallow depths. For a square footing (B = L): Δσ = Q / (B + z)².
Examples
This is the textbook example. B = L = 2 m, z = 3 m: each dimension becomes 2+3=5 m. Area = 5×5 = 25 m². Δσ = 800/25 = 32 kPa. Clean and fast — exactly the type of problem that appears on board exams.
Scenario
A 2 m × 2 m footing carries Q = 800 kN. Find Δσ at z = 3 m below the footing.
Solution
Δσ = Q/[(B+z)(L+z)] = 800/[(2+3)(2+3)] = 800/(5×5) = 800/25 = 32 kPa
This is Exercise 2 from the reference. At greater depth, stress decreases. Notice that doubling the depth from 2m to roughly 5m reduces Δσ from 50 to 20.8 kPa. This attenuation is important for multi-layer settlement calculations.
Scenario
A 3 m × 4 m footing carries 1500 kN. Find Δσ at z = 2 m and z = 5 m below the footing.
Solution
At z=2m: Δσ = 1500/[(3+2)(4+2)] = 1500/(5×6) = 1500/30 = 50 kPa. At z=5m: Δσ = 1500/[(3+5)(4+5)] = 1500/(8×9) = 1500/72 = 20.83 kPa
For strip footings, use L = 1 m (per unit length) and Q = load per unit length. Apply the same formula. Some texts use Δσ = q₀B/(B+z) for strip footings in plane strain — clarify the form used in the problem.
Scenario
A strip footing 0.6 m wide carries a load of 90 kN/m. Find the stress increase at z = 1.2 m.
Solution
For strip footing (per unit length, L=1): Δσ = (90×1)/[(0.6+1.2)(1+1.2)] = 90/[1.8×2.2] = 90/3.96 = 22.73 kPa
Applications
- Settlement computation for individual footings (Terzaghi-Peck method)
- Stress check at the center of a clay layer below a footing
- Quick comparison of stress distribution from different footing sizes
- Preliminary foundation design and footing sizing
Misconceptions
- Writing (B+z)(L+z) as B×L + z² — algebraically wrong; it must be expanded as BL + Bz + Lz + z²
- Measuring z from the ground surface instead of the footing base — for deeply embedded footings, this gives significantly different results
- Confusing total load Q (kN) with contact pressure q₀ (kPa) — always check units
- Using the 2:1 formula for flexible strip loads or circular footings without appropriate modification
Related Concepts
- Boussinesq point load
- Settlement computation
- Bearing capacity of shallow foundations
- Stress bulb under footings
Common Exam Questions
Example
1.5m × 1.5m footing, q₀ = 200 kPa, find Δσ at z=1.5m: Q=200×1.5×1.5=450 kN; Δσ=450/[(1.5+1.5)²]=450/9=50 kPa.
Approach
Identify B, L, Q, and z. Apply Δσ = Q/[(B+z)(L+z)]. Check if z is from footing base or ground surface — usually from footing base. If q₀ (pressure) is given, convert: Q = q₀ × B × L.
Question Type
Compute Δσ at a given depth using 2:1 method
Example
2m×2m footing, Q=500kN, find z where Δσ=20kPa: (2+z)²=500/20=25; 2+z=5; z=3m.
Approach
Set Δσ_target = Q/[(B+z)(L+z)]. Solve for z algebraically. For square footing: (B+z)² = Q/Δσ → z = √(Q/Δσ) − B.
Question Type
Find depth at which Δσ equals a target value
Key Points To Remember
- Δσ = Q / [(B+z)(L+z)] — the load spreads over a larger area with depth
- For square footing B = L: Δσ = Q / (B+z)²
- For strip footing (L → ∞): Δσ = Q/L / (B+z) per unit length
- z is measured from the BASE of the footing, not from the ground surface
- Q is total load (kN); multiply by B×L to convert from q₀ (kPa) if needed
- The denominator is (B+z)(L+z) NOT (B×z)(L×z) — common mistake
- Δσ approaches zero as z → ∞ (stress dissipates completely at great depth)
Stress Increase Under Uniformly Loaded Areas — Influence Factors and Newmark's Chart
For uniformly loaded rectangular, circular, or irregular areas, Boussinesq's solution is integrated over the area to give the stress increase at any point. For a rectangular load of dimensions B × L with uniform pressure q₀, the stress increase at depth z below a CORNER of the rectangle is: Δσ = q₀ × I, where I is a dimensionless influence factor that depends on m = B/z and n = L/z (or from published tables/charts). To find Δσ below an interior point or a point outside the loaded area, use superposition — add or subtract rectangular contributions. The influence factor I is bounded between 0 (at infinite depth) and 0.25 (directly below the center of an infinitely large load, at z→0). Newmark's chart (integration circles) provides a graphical method for irregular load shapes: draw the footing to scale, count the influence areas covered, and multiply by the influence value (0.005 per unit for Newmark's original chart). While full chart construction is rarely required in board exams, the concept of superposition and the corner-load formula are frequently tested.
Examples
m = B/z = 4/2 = 2; n = L/z = 6/2 = 3. Look up I from Boussinesq influence factor table. Apply Δσ = q₀ × I directly. For below the center, use 4 × I for the quarter-rectangle.
Scenario
A rectangular footing 4m × 6m carries q₀ = 100 kPa. Using an influence factor I = 0.084 at m=2, n=3 (from standard tables at z=2m), find Δσ below a corner at z = 2 m.
Solution
Δσ = q₀ × I = 100 × 0.084 = 8.4 kPa
Applications
- Stress increase below embankments and mat foundations
- Settlement calculations for large raft foundations
- Analysis of adjacent footing interaction
- Design of MSE (Mechanically Stabilized Earth) walls — stress on reinforcement layers
Misconceptions
- Using m=B/z and n=L/z without ensuring B ≤ L (some tables require m ≤ n; swap if needed)
- Applying the corner formula directly to a center point without the superposition split
- Confusing Boussinesq rectangular influence factor tables with Westergaard's formula (used for rigid pavements)
Related Concepts
- Boussinesq point load
- 2:1 method
- Settlement computation
- Newmark's chart
Common Exam Questions
Example
Point P is below the center of 4m×4m footing at z=2m: split into 4 corner rectangles of 2m×2m; m=n=2/2=1; I(1,1)≈0.175 (from table); Δσ = 4×100×0.175 = 70 kPa for q₀=100 kPa.
Approach
Express the point as a corner of one or more rectangles (add/subtract). Apply Δσ = q₀ × I for each rectangle using table values for I(m,n). Sum algebraically.
Question Type
Use superposition to find stress at an interior or exterior point
Key Points To Remember
- Δσ = q₀ × I (corner of rectangular load); I = f(m=B/z, n=L/z)
- For a point below the CENTER of a rectangle B×L: split into four corners of (B/2)(L/2) areas and sum: Δσ = 4 × q₀ × I(m=B/2z, n=L/2z)
- Superposition: combine rectangular areas to cover any load pattern
- Influence factor I ranges from 0 to 0.25
- Newmark's chart: each block represents equal influence; count blocks under the loaded area
- For circular loads of radius R at depth z: Δσ = q₀[1 − (1+(R/z)²)^(−3/2)]
Practice Problems
The profile shows three distinct zones: (1) above WT — effective stress equals total stress since u=0; (2) at WT — u is still zero but transitions; (3) below WT — u builds up as γ_w × z_w. Notice σ increases continuously with depth, but σ' increases at a slower rate below the WT because pore pressure also increases. The effective stress gradient below WT equals γ' = γ_sat − γ_w = 19 − 9.81 = 9.19 kN/m³ per meter, compared to γ_sat = 19 kN/m³ per meter for total stress.
Problem
PROBLEM 1 — Effective Stress Profile (Board Exam Level): A soil deposit at a construction site in Metro Manila has the following profile: 0–2 m: Dry sandy fill, γ_d = 16 kN/m³; 2–5 m: Moist sand, γ_moist = 18 kN/m³; 5–9 m: Saturated clay, γ_sat = 19 kN/m³. The water table is at z = 5 m. Using γ_w = 9.81 kN/m³, compute the total stress σ, pore pressure u, and effective stress σ' at depths z = 2 m, z = 5 m, and z = 9 m.
Solution
AT z = 2 m (base of fill): σ = 16(2) = 32 kPa; u = 0 (above WT); σ' = 32 − 0 = 32 kPa. AT z = 5 m (top of clay / at WT): σ = 16(2) + 18(3) = 32 + 54 = 86 kPa; u = 9.81(0) = 0 kPa (exactly at WT); σ' = 86 − 0 = 86 kPa. AT z = 9 m (base of clay): σ = 16(2) + 18(3) + 19(4) = 32 + 54 + 76 = 162 kPa; u = 9.81(4) = 39.24 kPa; σ' = 162 − 39.24 = 122.76 kPa ≈ 122.8 kPa. VERIFICATION at z=9m via γ' method: σ' = 16(2) + 18(3) + (19−9.81)(4) = 32 + 54 + 36.76 = 122.76 kPa ✓
When the WT rises 3 m, total stress increases slightly (sand unit weight changed from 18 to 19 kN/m³ in the 2–5 m zone: +3 kPa), but pore pressure increases much more (from 39.24 to 68.67 kPa: +29.43 kPa). Net effect: σ' drops by about 26.4 kPa. This is the mechanism behind rainfall-induced slope failures and reduced bearing capacity during typhoon season — a critical real-world application in the Philippine context.
Problem
PROBLEM 2 — Water Table Rise Effect (Board Exam Level): For the same profile as Problem 1, the water table rises 3 m due to heavy rainfall (new WT at z = 2 m, i.e., at the base of the fill). Assuming the sand between 2–5 m becomes fully saturated with γ_sat = 19 kN/m³, recompute σ, u, and σ' at z = 9 m. By how much does σ' change?
Solution
With WT at z = 2 m: σ = 16(2) + 19(3) + 19(4) = 32 + 57 + 76 = 165 kPa (γ_sat for sand is now 19 kN/m³ for 2–5m zone). Note: use γ_sat=19 for both sand and clay below new WT. u = 9.81(7) = 68.67 kPa (z_w = 9 − 2 = 7 m below new WT). σ' = 165 − 68.67 = 96.33 kPa. CHANGE in σ': Original σ' = 122.76 kPa (Problem 1); New σ' = 96.33 kPa; Decrease = 122.76 − 96.33 = 26.43 kPa. That is a 21.5% reduction in effective stress.
Part (a) uses the simplified r=0 formula. Part (b) shows the attenuation with horizontal distance — the offset pipe at r=3m receives only 4.89/14.92 = 32.8% of the stress experienced by the pipe directly below. This is important for assessing stress on utilities adjacent to heavy foundations. Also note: 2π × 4² = 2π × 16 = 100.53, not 50.27 — students often forget to square z.
Problem
PROBLEM 3 — Boussinesq Point Load: A water tower is idealized as a concentrated surface load Q = 500 kN. A buried pipe runs 4 m directly below the tower base. (a) Find the stress increase in the pipe at z = 4 m, r = 0. (b) Find the stress increase at z = 4 m, r = 3 m (offset pipe). Use γ_w = 9.81 kN/m³.
Solution
(a) r = 0: Δσ = 3Q/(2πz²) = 3(500)/(2π × 16) = 1500/100.53 = 14.92 kPa. (b) r = 3 m, z = 4 m: r/z = 3/4 = 0.75; [1/(1+(0.75)²)]^(5/2) = [1/(1+0.5625)]^(2.5) = [1/1.5625]^(2.5) = (0.64)^(2.5). Compute: (0.64)^2 = 0.4096; (0.64)^0.5 = 0.8; (0.64)^(2.5) = 0.4096 × 0.8 = 0.3277. Δσ = 14.92 × 0.3277 = 4.89 kPa.
The stress at the contact level (z=0⁺) is simply the contact pressure (Q/A). The 2:1 method then tracks attenuation through the clay layer. A common board exam twist is to compute the average stress increase in the clay layer as Δσ_avg ≈ (Δσ_top + 4×Δσ_mid + Δσ_bot)/6 for Simpson's rule, then use it in the consolidation settlement formula: S_c = C_c/(1+e₀) × H × log[(σ'₀ + Δσ)/σ'₀]. Note the dramatic decrease from 160 kPa at contact to 26.4 kPa at 4 m depth — foundation loads dissipate quickly with depth.
Problem
PROBLEM 4 — 2:1 Method (Settlement Analysis): A 2.5 m × 3.0 m reinforced concrete footing is subjected to a column load of 1200 kN (total, including footing weight). The footing bears on top of a 4 m thick clay layer. Using the 2:1 method, find the stress increase at the top (z = 0⁺), middle (z = 2 m), and bottom (z = 4 m) of the clay layer.
Solution
B = 2.5 m, L = 3.0 m, Q = 1200 kN. At z = 0⁺ (contact pressure): Δσ = Q/(BL) = 1200/(2.5×3.0) = 1200/7.5 = 160 kPa. At z = 2 m: Δσ = 1200/[(2.5+2)(3.0+2)] = 1200/[(4.5)(5.0)] = 1200/22.5 = 53.33 kPa. At z = 4 m: Δσ = 1200/[(2.5+4)(3.0+4)] = 1200/[(6.5)(7.0)] = 1200/45.5 = 26.37 kPa.
The factor of safety less than 1.0 means the upward seepage force exceeds the submerged weight of the soil — the soil will 'boil' or liquefy. In practice, this requires dewatering (lowering the water table by pumping) before excavation can safely proceed. The critical hydraulic gradient i_cr ≈ 1.0 for most saturated sands is a key boardable value. The quick condition is one of the most dangerous failure modes in excavations and cofferdam construction.
Problem
PROBLEM 5 — Quick Condition Check: A 3 m deep construction excavation is made in a saturated fine sand (γ_sat = 20 kN/m³). The water table is at the ground surface. A standpipe shows upward seepage with total head at the bottom of the excavation = 4 m above the bottom level. (a) Compute the critical hydraulic gradient i_cr. (b) Determine the factor of safety against heave (boiling). γ_w = 9.81 kN/m³.
Solution
(a) γ' = γ_sat − γ_w = 20 − 9.81 = 10.19 kN/m³. i_cr = γ'/γ_w = 10.19/9.81 = 1.039. (b) The upward hydraulic gradient in the 3 m sand column below the excavation: the total head difference is 4 m over a seepage path of 3 m (from bottom of excavation to the surface of the soil below): i = Δh/L = (4−3)/3. Wait, restate: head at bottom of excavation = 4 m above bottom level; head at top of soil-below-excavation (bottom of excavation) = 0 (datum at bottom of excavation); seepage path = 3 m upward; Δh = 4 m; i = 4/3 = 1.333. FS = i_cr / i_actual = 1.039/1.333 = 0.78 < 1.0 → HEAVE/BOILING OCCURS. Excavation is unstable.
Exam Preparation Tips
- MEMORIZE the four core formulas: σ=Σγᵢzᵢ, u=γ_w×z_w, σ'=σ−u, and Δσ=Q/[(B+z)(L+z)]. These appear in almost every board exam Geotechnical Engineering problem.
- ALWAYS draw a soil profile sketch first — mark layer boundaries, the water table, and label unit weights. A 30-second sketch prevents unit weight mix-ups that cost full marks.
- REMEMBER: γ_sat goes below the water table in total stress calculations; γ' (= γ_sat − γ_w) goes below WT only when computing effective stress directly by the γ' method. Never use γ' in the total stress formula.
- For Boussinesq, remember the exponent is 5/2 = 2.5. Split it as (value)² × √(value) for mental math. Practice with r/z = 0, 0.5, 1.0 to build intuition on attenuation.
- For 2:1 method: the denominator is (B+z)(L+z) — both dimensions expand. NOT (B×z)(L×z). The depth z is measured from the BASE of the footing, not the ground surface.
- When water table rises to the ground surface: u at depth z = 9.81z — the maximum possible hydrostatic pressure. Effective stress is then computed with full pore pressure over the entire depth.
- Artesian (confined) conditions: the piezometric surface is ABOVE the WT or even above the ground. u at the bottom of the confined layer = γ_w × (depth from piezometric surface to the point). This can result in very low or even negative effective stress in the overlying clay — a trigger for heave.
- Practice the three-column tabular approach for effective stress profiles: column 1 = σ, column 2 = u, column 3 = σ'. Check that σ' increases by γ' per meter below WT and by γ per meter above WT.
- Superposition for multiple loads: Δσ_total = Δσ₁ + Δσ₂ + Δσ₃. This applies to both Boussinesq (sum of individual point-load contributions) and the 2:1 method for adjacent footings.
- Critical hydraulic gradient: i_cr = γ'/γ_w ≈ 1.0 for typical saturated soils. If the problem gives upward seepage gradient i > i_cr, the soil is at risk of heaving — always check this in excavation problems.
- In board exams, when γ_w is not given, use 9.81 kN/m³ (not 10). Only use 10 if the problem explicitly states it. Many answers differ by this small factor.
- Link stresses to behavior: σ' controls Mohr-Coulomb shear strength (τ = c' + σ'tanφ'), drives consolidation settlement (ΔS = Cc/(1+e₀) × H × log[(σ'₀+Δσ)/σ'₀]), and governs pile skin friction (f_s = β × σ'_v). Understanding the WHY prevents formula confusion.
In summary
Stresses in soil mass is not just a theoretical chapter — it is the backbone of every geotechnical design decision you will encounter as a licensed Civil Engineer in the Philippines. Terzaghi's effective stress principle (σ' = σ − u) is perhaps the most exam-critical equation in soil mechanics: it tells you that what matters is not how heavy the soil column is, but how much of that weight is actually borne by the grains. Misapplying unit weights above and below the water table, forgetting to subtract pore pressure, or confusing z (depth from surface) with z_w (depth below water table) are the most frequent sources of error on the PRC board exam. For stress increases from loads, master the 2:1 method [Δσ = Q/((B+z)(L+z))] for quick footing problems and the Boussinesq formula [Δσ = 3Q/(2πz²) for r=0] for point loads — these two cover the vast majority of board exam items. Always set up the problem with a sketch, identify the water table, assign the correct unit weights, and apply the formulas systematically. Connect these stress calculations to their downstream uses — consolidation settlement, bearing capacity, and slope stability — and you will demonstrate the integrated understanding that the PRC licensure examination rewards. The effective stress principle underpins RA 544 (Republic Act 544, the Civil Engineering Law of the Philippines) in ensuring that engineers design structures that are safe, functional, and protective of public welfare, especially in a country as geologically active and typhoon-exposed as the Philippines.
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