Effective stress

Geotechnical · FE Reference Handbook section

Geotechnical

3 formulas

10 exam-style examples

~51 min

All Geotechnical lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Effective stress within Geotechnical. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what effective stress describes physically and when it applies.
State every one of the 3 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: unit weight in pcf with depth in ft gives stress in psf, not psi.

Lecture

Why this section exists. Effective stress is the part of Geotechnical that lets you connect a layered soil profile beneath a footing or wall to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as a phase-diagram quantity, an effective stress, or a bearing capacity. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. unit weight in pcf with depth in ft gives stress in psf, not psi. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Truck-mounted drill rig taking a soil boring beside a bridge, with sample jars in the foreground. — Photo 1. Where this shows up in practice: effective stress.
Capstone Studio instructional photograph

Geotechnical — Effective stress: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a layered soil profile beneath a footing or wall. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 3 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

σ'	Quantity produced by "σ' = σ - u" — read its definition and unit from the handbook line directly above the equation.
u	Quantity produced by "u = huγw" — read its definition and unit from the handbook line directly above the equation.
where hu	Quantity produced by "where hu = uplift or pressure head" — read its definition and unit from the handbook line directly above the equation.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Effective stress below a water table — Effective stress

A profile has 7 ft of moist sand (γ = 106.0 pcf) over 10 ft of saturated sand (γsat = 137.0 pcf). The water table sits at the top of the saturated layer. Find σ, u and σ′ at the base.

Given

H₁ = 7 ft, γ = 106.0 pcf
H₂ = 10 ft, γsat = 137.0 pcf
γw = 62.4 pcf

Find

Total stress, pore pressure and effective stress at depth

Start with the thinking

Total stress accumulates the actual unit weight of every layer above.
Pore pressure uses only the head of water above the point.

Figure for Effective stress below a water table — Effective stress

Step-by-step solution

Total stress
$σ = 106.0 (7) + 137.0 (10) = 2, 112 p s f$
Pore pressure — u = γw h = 62.4(10) = 624.0 psf
Effective stress
$σ^{'} = σ - u = 2, 112 - 624.0 = 1, 488 p s f$
In tsf
$σ^{'} = 0.74 t s f$

Answer: σ ≈ 2,112 psf, u ≈ 624.0 psf, σ′ ≈ 1,488 psf

Why the other options are there

σ′ = 2,112 psf (pore pressure not subtracted)
u = 1,061 psf (water table taken at ground surface)

Reference: FE Reference Handbook — Geotechnical → Effective stress

Example 2

Vertical stress profile with surcharge and the shear stress at failure — Effective stress

A profile has 3.5 m of moist sand (γ = 19.0 kN/m³) over 9.0 m of saturated sand (γ_sat = 20.5 kN/m³) with the water table at the interface. A surcharge of 13 kPa acts at the surface. At the base compute the total normal stress, pore pressure, effective stress, horizontal stress (K₀ = 0.45) and the shear stress at failure for φ′ = 35°.

Given

Layer 1: 3.5 m, γ = 19.0 kN/m³
Layer 2: 9.0 m, γ_sat = 20.5 kN/m³
q = 13 kPa
K₀ = 0.45, φ′ = 35°

Find

σ_v, u, σ′_v, σ′_h and τ_f at the base

Start with the thinking

Total normal stress accumulates every overburden layer plus any surcharge.
Only effective stress governs strength — the horizontal stress and shear stress at failure both use σ′.

Step-by-step solution

Formula — σ_v = q + Σγh
Substituting
$σ_{v} = 13 + 19.0 (3.5) + 20.5 (9.0) = 264.0 k P a$
Formula
$u = γ_{w} h_{w}$
Substituting
$u = 9.81 (9.0) = 88.29 k P a$
Formula
$σ_{v}^{'} = σ_{v} - u$
Substituting
$σ_{v}^{'} = 264.0 - 88.29 = 175.7 k P a$
Formula
$σ_{h}^{'} = K_{0} σ_{v}^{'}$
Substituting
$σ_{h}^{'} = 0.45 (175.7) = 79.07 k P a$
Formula
$τ_{f} = σ^{'} tan ϕ^{'}$
Substituting
$τ_{f} = 175.7 tan 3 5^{\circ} = 123.0 k P a$

Answer: σ_v = 264.0 kPa, u = 88.3 kPa, σ′_v = 175.7 kPa, σ′_h = 79.1 kPa, τ_f = 123.0 kPa

Why the other options are there

τ_f = 184.9 kPa (total stress used)
σ′_h = 118.8 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Effective stress

Example 3

Effective stress below a water table — Effective stress (2)

A profile has 10 ft of moist sand (γ = 107.0 pcf) over 7 ft of saturated sand (γsat = 131.0 pcf). The water table sits at the top of the saturated layer. Find σ, u and σ′ at the base.

Given

H₁ = 10 ft, γ = 107.0 pcf
H₂ = 7 ft, γsat = 131.0 pcf
γw = 62.4 pcf

Find

Total stress, pore pressure and effective stress at depth

Start with the thinking

Total stress accumulates the actual unit weight of every layer above.
Pore pressure uses only the head of water above the point.

Figure for Effective stress below a water table — Effective stress (2)

Step-by-step solution

Total stress
$σ = 107.0 (10) + 131.0 (7) = 1, 987 p s f$
Pore pressure — u = γw h = 62.4(7) = 436.8 psf
Effective stress
$σ^{'} = σ - u = 1, 987 - 436.8 = 1, 550 p s f$
In tsf
$σ^{'} = 0.78 t s f$

Answer: σ ≈ 1,987 psf, u ≈ 436.8 psf, σ′ ≈ 1,550 psf

Why the other options are there

σ′ = 1,987 psf (pore pressure not subtracted)
u = 1,061 psf (water table taken at ground surface)

Reference: FE Reference Handbook — Geotechnical → Effective stress

Example 4

Vertical stress profile with surcharge and the shear stress at failure — Effective stress (2)

A profile has 4.5 m of moist sand (γ = 19.5 kN/m³) over 3.5 m of saturated sand (γ_sat = 19.5 kN/m³) with the water table at the interface. A surcharge of 59 kPa acts at the surface. At the base compute the total normal stress, pore pressure, effective stress, horizontal stress (K₀ = 0.40) and the shear stress at failure for φ′ = 30°.

Given

Layer 1: 4.5 m, γ = 19.5 kN/m³
Layer 2: 3.5 m, γ_sat = 19.5 kN/m³
q = 59 kPa
K₀ = 0.40, φ′ = 30°

Find

σ_v, u, σ′_v, σ′_h and τ_f at the base

Start with the thinking

Total normal stress accumulates every overburden layer plus any surcharge.
Only effective stress governs strength — the horizontal stress and shear stress at failure both use σ′.

Step-by-step solution

Formula — σ_v = q + Σγh
Substituting
$σ_{v} = 59 + 19.5 (4.5) + 19.5 (3.5) = 215.0 k P a$
Formula
$u = γ_{w} h_{w}$
Substituting
$u = 9.81 (3.5) = 34.34 k P a$
Formula
$σ_{v}^{'} = σ_{v} - u$
Substituting
$σ_{v}^{'} = 215.0 - 34.34 = 180.7 k P a$
Formula
$σ_{h}^{'} = K_{0} σ_{v}^{'}$
Substituting
$σ_{h}^{'} = 0.40 (180.7) = 72.27 k P a$
Formula
$τ_{f} = σ^{'} tan ϕ^{'}$
Substituting
$τ_{f} = 180.7 tan 3 0^{\circ} = 104.3 k P a$

Answer: σ_v = 215.0 kPa, u = 34.3 kPa, σ′_v = 180.7 kPa, σ′_h = 72.3 kPa, τ_f = 104.3 kPa

Why the other options are there

τ_f = 124.1 kPa (total stress used)
σ′_h = 86.0 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Effective stress

Example 5

Effective stress below a water table — Effective stress (3)

A profile has 5 ft of moist sand (γ = 118.0 pcf) over 9 ft of saturated sand (γsat = 138.0 pcf). The water table sits at the top of the saturated layer. Find σ, u and σ′ at the base.

Given

H₁ = 5 ft, γ = 118.0 pcf
H₂ = 9 ft, γsat = 138.0 pcf
γw = 62.4 pcf

Find

Total stress, pore pressure and effective stress at depth

Start with the thinking

Total stress accumulates the actual unit weight of every layer above.
Pore pressure uses only the head of water above the point.

Figure for Effective stress below a water table — Effective stress (3)

Step-by-step solution

Total stress
$σ = 118.0 (5) + 138.0 (9) = 1, 832 p s f$
Pore pressure — u = γw h = 62.4(9) = 561.6 psf
Effective stress
$σ^{'} = σ - u = 1, 832 - 561.6 = 1, 270 p s f$
In tsf
$σ^{'} = 0.64 t s f$

Answer: σ ≈ 1,832 psf, u ≈ 561.6 psf, σ′ ≈ 1,270 psf

Why the other options are there

σ′ = 1,832 psf (pore pressure not subtracted)
u = 873.6 psf (water table taken at ground surface)

Reference: FE Reference Handbook — Geotechnical → Effective stress

Example 6

Vertical stress profile with surcharge and the shear stress at failure — Effective stress (3)

A profile has 2.0 m of moist sand (γ = 17.5 kN/m³) over 3.5 m of saturated sand (γ_sat = 20.0 kN/m³) with the water table at the interface. A surcharge of 55 kPa acts at the surface. At the base compute the total normal stress, pore pressure, effective stress, horizontal stress (K₀ = 0.45) and the shear stress at failure for φ′ = 34°.

Given

Layer 1: 2.0 m, γ = 17.5 kN/m³
Layer 2: 3.5 m, γ_sat = 20.0 kN/m³
q = 55 kPa
K₀ = 0.45, φ′ = 34°

Find

σ_v, u, σ′_v, σ′_h and τ_f at the base

Start with the thinking

Total normal stress accumulates every overburden layer plus any surcharge.
Only effective stress governs strength — the horizontal stress and shear stress at failure both use σ′.

Step-by-step solution

Formula — σ_v = q + Σγh
Substituting
$σ_{v} = 55 + 17.5 (2.0) + 20.0 (3.5) = 160.0 k P a$
Formula
$u = γ_{w} h_{w}$
Substituting
$u = 9.81 (3.5) = 34.34 k P a$
Formula
$σ_{v}^{'} = σ_{v} - u$
Substituting
$σ_{v}^{'} = 160.0 - 34.34 = 125.7 k P a$
Formula
$σ_{h}^{'} = K_{0} σ_{v}^{'}$
Substituting
$σ_{h}^{'} = 0.45 (125.7) = 56.55 k P a$
Formula
$τ_{f} = σ^{'} tan ϕ^{'}$
Substituting
$τ_{f} = 125.7 tan 3 4^{\circ} = 84.76 k P a$

Answer: σ_v = 160.0 kPa, u = 34.3 kPa, σ′_v = 125.7 kPa, σ′_h = 56.5 kPa, τ_f = 84.8 kPa

Why the other options are there

τ_f = 107.9 kPa (total stress used)
σ′_h = 72.0 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Effective stress

Example 7

Effective stress below a water table — Effective stress (4)

A profile has 12 ft of moist sand (γ = 124.0 pcf) over 17 ft of saturated sand (γsat = 135.0 pcf). The water table sits at the top of the saturated layer. Find σ, u and σ′ at the base.

Given

H₁ = 12 ft, γ = 124.0 pcf
H₂ = 17 ft, γsat = 135.0 pcf
γw = 62.4 pcf

Find

Total stress, pore pressure and effective stress at depth

Start with the thinking

Total stress accumulates the actual unit weight of every layer above.
Pore pressure uses only the head of water above the point.

Figure for Effective stress below a water table — Effective stress (4)

Step-by-step solution

Total stress
$σ = 124.0 (12) + 135.0 (17) = 3, 783 p s f$
Pore pressure — u = γw h = 62.4(17) = 1,061 psf
Effective stress
$σ^{'} = σ - u = 3, 783 - 1, 061 = 2, 722 p s f$
In tsf
$σ^{'} = 1.36 t s f$

Answer: σ ≈ 3,783 psf, u ≈ 1,061 psf, σ′ ≈ 2,722 psf

Why the other options are there

σ′ = 3,783 psf (pore pressure not subtracted)
u = 1,810 psf (water table taken at ground surface)

Reference: FE Reference Handbook — Geotechnical → Effective stress

Example 8

Vertical stress profile with surcharge and the shear stress at failure — Effective stress (4)

A profile has 5.5 m of moist sand (γ = 18.5 kN/m³) over 6.0 m of saturated sand (γ_sat = 20.0 kN/m³) with the water table at the interface. A surcharge of 28 kPa acts at the surface. At the base compute the total normal stress, pore pressure, effective stress, horizontal stress (K₀ = 0.55) and the shear stress at failure for φ′ = 33°.

Given

Layer 1: 5.5 m, γ = 18.5 kN/m³
Layer 2: 6.0 m, γ_sat = 20.0 kN/m³
q = 28 kPa
K₀ = 0.55, φ′ = 33°

Find

σ_v, u, σ′_v, σ′_h and τ_f at the base

Start with the thinking

Total normal stress accumulates every overburden layer plus any surcharge.
Only effective stress governs strength — the horizontal stress and shear stress at failure both use σ′.

Step-by-step solution

Formula — σ_v = q + Σγh
Substituting
$σ_{v} = 28 + 18.5 (5.5) + 20.0 (6.0) = 249.8 k P a$
Formula
$u = γ_{w} h_{w}$
Substituting
$u = 9.81 (6.0) = 58.86 k P a$
Formula
$σ_{v}^{'} = σ_{v} - u$
Substituting
$σ_{v}^{'} = 249.8 - 58.86 = 190.9 k P a$
Formula
$σ_{h}^{'} = K_{0} σ_{v}^{'}$
Substituting
$σ_{h}^{'} = 0.55 (190.9) = 105.0 k P a$
Formula
$τ_{f} = σ^{'} tan ϕ^{'}$
Substituting
$τ_{f} = 190.9 tan 3 3^{\circ} = 124.0 k P a$

Answer: σ_v = 249.8 kPa, u = 58.9 kPa, σ′_v = 190.9 kPa, σ′_h = 105.0 kPa, τ_f = 124.0 kPa

Why the other options are there

τ_f = 162.2 kPa (total stress used)
σ′_h = 137.4 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Effective stress

Example 9

Effective stress below a water table — Effective stress (5)

A profile has 10 ft of moist sand (γ = 114.0 pcf) over 18 ft of saturated sand (γsat = 124.0 pcf). The water table sits at the top of the saturated layer. Find σ, u and σ′ at the base.

Given

H₁ = 10 ft, γ = 114.0 pcf
H₂ = 18 ft, γsat = 124.0 pcf
γw = 62.4 pcf

Find

Total stress, pore pressure and effective stress at depth

Start with the thinking

Total stress accumulates the actual unit weight of every layer above.
Pore pressure uses only the head of water above the point.

Figure for Effective stress below a water table — Effective stress (5)

Step-by-step solution

Total stress
$σ = 114.0 (10) + 124.0 (18) = 3, 372 p s f$
Pore pressure — u = γw h = 62.4(18) = 1,123 psf
Effective stress
$σ^{'} = σ - u = 3, 372 - 1, 123 = 2, 249 p s f$
In tsf
$σ^{'} = 1.12 t s f$

Answer: σ ≈ 3,372 psf, u ≈ 1,123 psf, σ′ ≈ 2,249 psf

Why the other options are there

σ′ = 3,372 psf (pore pressure not subtracted)
u = 1,747 psf (water table taken at ground surface)

Reference: FE Reference Handbook — Geotechnical → Effective stress

Example 10

Vertical stress profile with surcharge and the shear stress at failure — Effective stress (5)

A profile has 4.5 m of moist sand (γ = 18.0 kN/m³) over 8.0 m of saturated sand (γ_sat = 19.5 kN/m³) with the water table at the interface. A surcharge of 30 kPa acts at the surface. At the base compute the total normal stress, pore pressure, effective stress, horizontal stress (K₀ = 0.45) and the shear stress at failure for φ′ = 32°.

Given

Layer 1: 4.5 m, γ = 18.0 kN/m³
Layer 2: 8.0 m, γ_sat = 19.5 kN/m³
q = 30 kPa
K₀ = 0.45, φ′ = 32°

Find

σ_v, u, σ′_v, σ′_h and τ_f at the base

Start with the thinking

Total normal stress accumulates every overburden layer plus any surcharge.
Only effective stress governs strength — the horizontal stress and shear stress at failure both use σ′.

Step-by-step solution

Formula — σ_v = q + Σγh
Substituting
$σ_{v} = 30 + 18.0 (4.5) + 19.5 (8.0) = 267.0 k P a$
Formula
$u = γ_{w} h_{w}$
Substituting
$u = 9.81 (8.0) = 78.48 k P a$
Formula
$σ_{v}^{'} = σ_{v} - u$
Substituting
$σ_{v}^{'} = 267.0 - 78.48 = 188.5 k P a$
Formula
$σ_{h}^{'} = K_{0} σ_{v}^{'}$
Substituting
$σ_{h}^{'} = 0.45 (188.5) = 84.83 k P a$
Formula
$τ_{f} = σ^{'} tan ϕ^{'}$
Substituting
$τ_{f} = 188.5 tan 3 2^{\circ} = 117.8 k P a$

Answer: σ_v = 267.0 kPa, u = 78.5 kPa, σ′_v = 188.5 kPa, σ′_h = 84.8 kPa, τ_f = 117.8 kPa

Why the other options are there

τ_f = 166.8 kPa (total stress used)
σ′_h = 120.2 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Effective stress

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a layered soil profile beneath a footing or wall, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Effective stress contains 3 relations; you must be able to find this page in under 15 seconds.
Exam style: a phase-diagram quantity, an effective stress, or a bearing capacity.
Unit rule: unit weight in pcf with depth in ft gives stress in psf, not psi.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

unit weight in pcf with depth in ft gives stress in psf, not psi
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.