Horizontal Stress Profiles and Forces

Geotechnical · FE Reference Handbook section

Geotechnical

12 formulas

10 exam-style examples

~60 min

All Geotechnical lectures

Learning objectives

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

This chapter section covers Horizontal Stress Profiles and Forces 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 horizontal stress profiles and forces describes physically and when it applies.
State every one of the 12 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. Horizontal Stress Profiles and Forces 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: horizontal stress profiles and forces.
Capstone Studio instructional photograph

Geotechnical — Horizontal Stress Profiles and Forces: 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 12 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'1	Quantity produced by "σ'1 = γ1 × H1 0" — read its definition and unit from the handbook line directly above the equation.
γ2	Quantity produced by "γ2 = saturated σ2' = (γ1 × H1) + (γ2 × H2)" — read its definition and unit from the handbook line directly above the equation.
KA	Quantity produced by "KA = Rankine active earth pressure coefficient (smooth wall, c = 0, level backfill) = tan2 (45° - φ/2)" — read its definition and unit from the handbook line directly above the equation.
KP	Quantity produced by "KP = Rankine passive earth pressure coefficient (smooth wall, c = 0, level backfill) = tan2 (45° + φ/2)" — read its definition and unit from the handbook line directly above the equation.
K0	Quantity produced by "K0 = at rest earth pressure coefficient (smooth wall, c = 0, level backfill)" — read its definition and unit from the handbook line directly above the equation.
OCR	Quantity produced by "OCR = overconsolidation ratio" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

Effective Pore Effective
Vertical Water Horizontal
Stress Pressure Force
0 0 0
1 1 2
H1 2 2
Active forces on retaining wall per unit wall length (as shown):
Passive forces on retaining wall per unit wall length (similar to the active forces shown):
At rest forces on wall per unit length of wall

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

Vertical stress profile with surcharge and the shear stress at failure — Horizontal Stress Profiles and Forces

A profile has 3.0 m of moist sand (γ = 20.0 kN/m³) over 8.5 m of saturated sand (γ_sat = 20.0 kN/m³) with the water table at the interface. A surcharge of 33 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′ = 33°.

Given

Layer 1: 3.0 m, γ = 20.0 kN/m³
Layer 2: 8.5 m, γ_sat = 20.0 kN/m³
q = 33 kPa
K₀ = 0.45, 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
$σ_{v} = 33 + 20.0 (3.0) + 20.0 (8.5) = 263.0 k P a$
Formula
$u = γ_{w} h_{w}$
Substituting
$u = 9.81 (8.5) = 83.39 k P a$
Formula
$σ_{v}^{'} = σ_{v} - u$
Substituting
$σ_{v}^{'} = 263.0 - 83.39 = 179.6 k P a$
Formula
$σ_{h}^{'} = K_{0} σ_{v}^{'}$
Substituting
$σ_{h}^{'} = 0.45 (179.6) = 80.83 k P a$
Formula
$τ_{f} = σ^{'} tan ϕ^{'}$
Substituting
$τ_{f} = 179.6 tan 3 3^{\circ} = 116.6 k P a$

Answer: σ_v = 263.0 kPa, u = 83.4 kPa, σ′_v = 179.6 kPa, σ′_h = 80.8 kPa, τ_f = 116.6 kPa

Why the other options are there

τ_f = 170.8 kPa (total stress used)
σ′_h = 118.4 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Horizontal Stress Profiles and Forces

Example 2

Vertical stress profile with surcharge and the shear stress at failure — Horizontal Stress Profiles and Forces (2)

A profile has 2.5 m of moist sand (γ = 19.5 kN/m³) over 3.0 m of saturated sand (γ_sat = 21.0 kN/m³) with the water table at the interface. A surcharge of 15 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′ = 28°.

Given

Layer 1: 2.5 m, γ = 19.5 kN/m³
Layer 2: 3.0 m, γ_sat = 21.0 kN/m³
q = 15 kPa
K₀ = 0.45, φ′ = 28°

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
$σ_{v} = 15 + 19.5 (2.5) + 21.0 (3.0) = 126.8 k P a$
Formula
$u = γ_{w} h_{w}$
Substituting
$u = 9.81 (3.0) = 29.43 k P a$
Formula
$σ_{v}^{'} = σ_{v} - u$
Substituting
$σ_{v}^{'} = 126.8 - 29.43 = 97.32 k P a$
Formula
$σ_{h}^{'} = K_{0} σ_{v}^{'}$
Substituting
$σ_{h}^{'} = 0.45 (97.32) = 43.79 k P a$
Formula
$τ_{f} = σ^{'} tan ϕ^{'}$
Substituting
$τ_{f} = 97.32 tan 2 8^{\circ} = 51.75 k P a$

Answer: σ_v = 126.8 kPa, u = 29.4 kPa, σ′_v = 97.3 kPa, σ′_h = 43.8 kPa, τ_f = 51.7 kPa

Why the other options are there

τ_f = 67.4 kPa (total stress used)
σ′_h = 57.0 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Horizontal Stress Profiles and Forces

Example 3

Vertical stress profile with surcharge and the shear stress at failure — Horizontal Stress Profiles and Forces (3)

A profile has 3.0 m of moist sand (γ = 19.5 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 0 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′ = 35°.

Given

Layer 1: 3.0 m, γ = 19.5 kN/m³
Layer 2: 6.0 m, γ_sat = 20.0 kN/m³
q = 0 kPa
K₀ = 0.55, 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
$σ_{v} = 0 + 19.5 (3.0) + 20.0 (6.0) = 178.5 k P a$
Formula
$u = γ_{w} h_{w}$
Substituting
$u = 9.81 (6.0) = 58.86 k P a$
Formula
$σ_{v}^{'} = σ_{v} - u$
Substituting
$σ_{v}^{'} = 178.5 - 58.86 = 119.6 k P a$
Formula
$σ_{h}^{'} = K_{0} σ_{v}^{'}$
Substituting
$σ_{h}^{'} = 0.55 (119.6) = 65.80 k P a$
Formula
$τ_{f} = σ^{'} tan ϕ^{'}$
Substituting
$τ_{f} = 119.6 tan 3 5^{\circ} = 83.77 k P a$

Answer: σ_v = 178.5 kPa, u = 58.9 kPa, σ′_v = 119.6 kPa, σ′_h = 65.8 kPa, τ_f = 83.8 kPa

Why the other options are there

τ_f = 125.0 kPa (total stress used)
σ′_h = 98.2 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Horizontal Stress Profiles and Forces

Example 4

Vertical stress profile with surcharge and the shear stress at failure — Horizontal Stress Profiles and Forces (4)

A profile has 2.5 m of moist sand (γ = 19.5 kN/m³) over 7.5 m of saturated sand (γ_sat = 19.5 kN/m³) with the water table at the interface. A surcharge of 15 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′ = 26°.

Given

Layer 1: 2.5 m, γ = 19.5 kN/m³
Layer 2: 7.5 m, γ_sat = 19.5 kN/m³
q = 15 kPa
K₀ = 0.55, φ′ = 26°

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
$σ_{v} = 15 + 19.5 (2.5) + 19.5 (7.5) = 210.0 k P a$
Formula
$u = γ_{w} h_{w}$
Substituting
$u = 9.81 (7.5) = 73.58 k P a$
Formula
$σ_{v}^{'} = σ_{v} - u$
Substituting
$σ_{v}^{'} = 210.0 - 73.58 = 136.4 k P a$
Formula
$σ_{h}^{'} = K_{0} σ_{v}^{'}$
Substituting
$σ_{h}^{'} = 0.55 (136.4) = 75.03 k P a$
Formula
$τ_{f} = σ^{'} tan ϕ^{'}$
Substituting
$τ_{f} = 136.4 tan 2 6^{\circ} = 66.54 k P a$

Answer: σ_v = 210.0 kPa, u = 73.6 kPa, σ′_v = 136.4 kPa, σ′_h = 75.0 kPa, τ_f = 66.5 kPa

Why the other options are there

τ_f = 102.4 kPa (total stress used)
σ′_h = 115.5 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Horizontal Stress Profiles and Forces

Example 5

Vertical stress profile with surcharge and the shear stress at failure — Horizontal Stress Profiles and Forces (5)

A profile has 2.5 m of moist sand (γ = 18.5 kN/m³) over 4.5 m of saturated sand (γ_sat = 18.5 kN/m³) with the water table at the interface. A surcharge of 42 kPa acts at the surface. At the base compute the total normal stress, pore pressure, effective stress, horizontal stress (K₀ = 0.60) and the shear stress at failure for φ′ = 29°.

Given

Layer 1: 2.5 m, γ = 18.5 kN/m³
Layer 2: 4.5 m, γ_sat = 18.5 kN/m³
q = 42 kPa
K₀ = 0.60, φ′ = 29°

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
$σ_{v} = 42 + 18.5 (2.5) + 18.5 (4.5) = 171.5 k P a$
Formula
$u = γ_{w} h_{w}$
Substituting
$u = 9.81 (4.5) = 44.15 k P a$
Formula
$σ_{v}^{'} = σ_{v} - u$
Substituting
$σ_{v}^{'} = 171.5 - 44.15 = 127.4 k P a$
Formula
$σ_{h}^{'} = K_{0} σ_{v}^{'}$
Substituting
$σ_{h}^{'} = 0.60 (127.4) = 76.41 k P a$
Formula
$τ_{f} = σ^{'} tan ϕ^{'}$
Substituting
$τ_{f} = 127.4 tan 2 9^{\circ} = 70.59 k P a$

Answer: σ_v = 171.5 kPa, u = 44.1 kPa, σ′_v = 127.4 kPa, σ′_h = 76.4 kPa, τ_f = 70.6 kPa

Why the other options are there

τ_f = 95.1 kPa (total stress used)
σ′_h = 102.9 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Horizontal Stress Profiles and Forces

Example 6

Vertical stress profile with surcharge and the shear stress at failure — Horizontal Stress Profiles and Forces (6)

A profile has 6.0 m of moist sand (γ = 19.0 kN/m³) over 6.5 m of saturated sand (γ_sat = 19.0 kN/m³) with the water table at the interface. A surcharge of 31 kPa acts at the surface. At the base compute the total normal stress, pore pressure, effective stress, horizontal stress (K₀ = 0.50) and the shear stress at failure for φ′ = 30°.

Given

Layer 1: 6.0 m, γ = 19.0 kN/m³
Layer 2: 6.5 m, γ_sat = 19.0 kN/m³
q = 31 kPa
K₀ = 0.50, 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
$σ_{v} = 31 + 19.0 (6.0) + 19.0 (6.5) = 268.5 k P a$
Formula
$u = γ_{w} h_{w}$
Substituting
$u = 9.81 (6.5) = 63.77 k P a$
Formula
$σ_{v}^{'} = σ_{v} - u$
Substituting
$σ_{v}^{'} = 268.5 - 63.77 = 204.7 k P a$
Formula
$σ_{h}^{'} = K_{0} σ_{v}^{'}$
Substituting
$σ_{h}^{'} = 0.50 (204.7) = 102.4 k P a$
Formula
$τ_{f} = σ^{'} tan ϕ^{'}$
Substituting
$τ_{f} = 204.7 tan 3 0^{\circ} = 118.2 k P a$

Answer: σ_v = 268.5 kPa, u = 63.8 kPa, σ′_v = 204.7 kPa, σ′_h = 102.4 kPa, τ_f = 118.2 kPa

Why the other options are there

τ_f = 155.0 kPa (total stress used)
σ′_h = 134.3 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Horizontal Stress Profiles and Forces

Example 7

Vertical stress profile with surcharge and the shear stress at failure — Horizontal Stress Profiles and Forces (7)

A profile has 3.0 m of moist sand (γ = 18.0 kN/m³) over 7.5 m of saturated sand (γ_sat = 20.5 kN/m³) with the water table at the interface. A surcharge of 58 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: 3.0 m, γ = 18.0 kN/m³
Layer 2: 7.5 m, γ_sat = 20.5 kN/m³
q = 58 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
$σ_{v} = 58 + 18.0 (3.0) + 20.5 (7.5) = 265.8 k P a$
Formula
$u = γ_{w} h_{w}$
Substituting
$u = 9.81 (7.5) = 73.58 k P a$
Formula
$σ_{v}^{'} = σ_{v} - u$
Substituting
$σ_{v}^{'} = 265.8 - 73.58 = 192.2 k P a$
Formula
$σ_{h}^{'} = K_{0} σ_{v}^{'}$
Substituting
$σ_{h}^{'} = 0.45 (192.2) = 86.48 k P a$
Formula
$τ_{f} = σ^{'} tan ϕ^{'}$
Substituting
$τ_{f} = 192.2 tan 3 4^{\circ} = 129.6 k P a$

Answer: σ_v = 265.8 kPa, u = 73.6 kPa, σ′_v = 192.2 kPa, σ′_h = 86.5 kPa, τ_f = 129.6 kPa

Why the other options are there

τ_f = 179.3 kPa (total stress used)
σ′_h = 119.6 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Horizontal Stress Profiles and Forces

Example 8

Vertical stress profile with surcharge and the shear stress at failure — Horizontal Stress Profiles and Forces (8)

A profile has 2.5 m of moist sand (γ = 17.5 kN/m³) over 4.5 m of saturated sand (γ_sat = 19.5 kN/m³) with the water table at the interface. A surcharge of 18 kPa acts at the surface. At the base compute the total normal stress, pore pressure, effective stress, horizontal stress (K₀ = 0.50) and the shear stress at failure for φ′ = 36°.

Given

Layer 1: 2.5 m, γ = 17.5 kN/m³
Layer 2: 4.5 m, γ_sat = 19.5 kN/m³
q = 18 kPa
K₀ = 0.50, φ′ = 36°

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
$σ_{v} = 18 + 17.5 (2.5) + 19.5 (4.5) = 149.5 k P a$
Formula
$u = γ_{w} h_{w}$
Substituting
$u = 9.81 (4.5) = 44.15 k P a$
Formula
$σ_{v}^{'} = σ_{v} - u$
Substituting
$σ_{v}^{'} = 149.5 - 44.15 = 105.4 k P a$
Formula
$σ_{h}^{'} = K_{0} σ_{v}^{'}$
Substituting
$σ_{h}^{'} = 0.50 (105.4) = 52.68 k P a$
Formula
$τ_{f} = σ^{'} tan ϕ^{'}$
Substituting
$τ_{f} = 105.4 tan 3 6^{\circ} = 76.54 k P a$

Answer: σ_v = 149.5 kPa, u = 44.1 kPa, σ′_v = 105.4 kPa, σ′_h = 52.7 kPa, τ_f = 76.5 kPa

Why the other options are there

τ_f = 108.6 kPa (total stress used)
σ′_h = 74.8 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Horizontal Stress Profiles and Forces

Example 9

Vertical stress profile with surcharge and the shear stress at failure — Horizontal Stress Profiles and Forces (9)

A profile has 5.5 m of moist sand (γ = 17.0 kN/m³) over 4.5 m of saturated sand (γ_sat = 20.5 kN/m³) with the water table at the interface. A surcharge of 51 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: 5.5 m, γ = 17.0 kN/m³
Layer 2: 4.5 m, γ_sat = 20.5 kN/m³
q = 51 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
$σ_{v} = 51 + 17.0 (5.5) + 20.5 (4.5) = 236.8 k P a$
Formula
$u = γ_{w} h_{w}$
Substituting
$u = 9.81 (4.5) = 44.15 k P a$
Formula
$σ_{v}^{'} = σ_{v} - u$
Substituting
$σ_{v}^{'} = 236.8 - 44.15 = 192.6 k P a$
Formula
$σ_{h}^{'} = K_{0} σ_{v}^{'}$
Substituting
$σ_{h}^{'} = 0.45 (192.6) = 86.67 k P a$
Formula
$τ_{f} = σ^{'} tan ϕ^{'}$
Substituting
$τ_{f} = 192.6 tan 3 4^{\circ} = 129.9 k P a$

Answer: σ_v = 236.8 kPa, u = 44.1 kPa, σ′_v = 192.6 kPa, σ′_h = 86.7 kPa, τ_f = 129.9 kPa

Why the other options are there

τ_f = 159.7 kPa (total stress used)
σ′_h = 106.5 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Horizontal Stress Profiles and Forces

Example 10

Vertical stress profile with surcharge and the shear stress at failure — Horizontal Stress Profiles and Forces (10)

A profile has 4.0 m of moist sand (γ = 19.0 kN/m³) over 3.5 m of saturated sand (γ_sat = 21.0 kN/m³) with the water table at the interface. A surcharge of 4 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′ = 31°.

Given

Layer 1: 4.0 m, γ = 19.0 kN/m³
Layer 2: 3.5 m, γ_sat = 21.0 kN/m³
q = 4 kPa
K₀ = 0.55, φ′ = 31°

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
$σ_{v} = 4 + 19.0 (4.0) + 21.0 (3.5) = 153.5 k P a$
Formula
$u = γ_{w} h_{w}$
Substituting
$u = 9.81 (3.5) = 34.34 k P a$
Formula
$σ_{v}^{'} = σ_{v} - u$
Substituting
$σ_{v}^{'} = 153.5 - 34.34 = 119.2 k P a$
Formula
$σ_{h}^{'} = K_{0} σ_{v}^{'}$
Substituting
$σ_{h}^{'} = 0.55 (119.2) = 65.54 k P a$
Formula
$τ_{f} = σ^{'} tan ϕ^{'}$
Substituting
$τ_{f} = 119.2 tan 3 1^{\circ} = 71.60 k P a$

Answer: σ_v = 153.5 kPa, u = 34.3 kPa, σ′_v = 119.2 kPa, σ′_h = 65.5 kPa, τ_f = 71.6 kPa

Why the other options are there

τ_f = 92.2 kPa (total stress used)
σ′_h = 84.4 kPa (pore pressure ignored)

Reference: FE Reference Handbook — Geotechnical → Horizontal Stress Profiles and Forces

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

Horizontal Stress Profiles and Forces contains 12 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.