Ultimate consolidation settlement in soil layer

Geotechnical · FE Reference Handbook section

Geotechnical

18 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 Ultimate consolidation settlement in soil layer 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 ultimate consolidation settlement in soil layer describes physically and when it applies.
State every one of the 18 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. Ultimate consolidation settlement in soil layer 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: ultimate consolidation settlement in soil layer.
Capstone Studio instructional photograph

Geotechnical — Ultimate consolidation settlement in soil layer: 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 18 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

SULT	Quantity produced by "SULT = εvHS" — read its definition and unit from the handbook line directly above the equation.
where HS	Quantity produced by "where HS = thickness of soil layer" — read its definition and unit from the handbook line directly above the equation.
εv	Quantity produced by "εv = ∆eTOT/(1 + e0)" — read its definition and unit from the handbook line directly above the equation.
where ∆eTOT	Quantity produced by "where ∆eTOT = total change in void ratio due to recompression and virgin compression" — read its definition and unit from the handbook line directly above the equation.
ST	Quantity produced by "ST = UAVSULT" — read its definition and unit from the handbook line directly above the equation.
UAV	Quantity produced by "UAV = average degree of consolidation" — read its definition and unit from the handbook line directly above the equation.
tC	Quantity produced by "tC = elapsed time since application of consolidation load" — read its definition and unit from the handbook line directly above the equation.
τ	Quantity produced by "τ= c + σN" — read its definition and unit from the handbook line directly above the equation.
STRESS, σN	Quantity produced by "STRESS, σN" — read its definition and unit from the handbook line directly above the equation.
s	Quantity produced by "s = mean normal stress" — read its definition and unit from the handbook line directly above the equation.
t	Quantity produced by "t = maximum shear stress" — read its definition and unit from the handbook line directly above the equation.
σ1	Quantity produced by "σ1 = major principal stress" — 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.

where
Varia�on of �me factor with degree of consolida�on*
U (%) Tv U (%) Tv U (%) Tv
0 0 34 0.0907 68 0.377
1 0.00008 35 0.0962 69 0.390
2 0.0003 36 0.102 70 0.403
3 0.00071 37 0.107 71 0.417
4 0.00126 38 0.113 72 0.431
5 0.00196 39 0.119 73 0.446
Two-way
6 0.00283 40 0.126 74 0.461
drainage
7 0.00385 41 0.132 75 0.477 2Hdr
8 0.00502 42 0.138 76 0.493
9 0.00636 43 0.145 77 0.511
10 0.00785 44 0.152 78 0.529
11 0.0095 45 0.159 79 0.547
12 0.0113 46 0.166 80 0.567
13 0.0133 47 0.173 81 0.588
14 0.0154 48 0.181 82 0.610
One-way
drainage
15 0.0177 49 0.188 83 0.633 Hdr
16 0.0201 50 0.197 84 0.658

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

Primary consolidation settlement of a clay layer

A 4.0 m normally consolidated clay has e₀ = 0.90 and Cc = 0.28. Effective stress at mid-depth rises from 95 kPa to 150 kPa. Estimate the settlement.

Given

H = 4.0 m
e₀ = 0.90
Cc = 0.28
σ′₀ = 95 kPa, σ′f = 150 kPa

Find

Settlement S_c

Start with the thinking

Normally consolidated — use Cc for the whole increment.
The log is base 10.

Step-by-step solution

Stress ratio
$σ^{'} f / σ_{0}^{'} = 150/95 = 1.579$
Log term
$l o g_{10} (1.579) = 0.1984$
Settlement — S_c = (CcH/(1 + e₀))·log(σ′f/σ′₀)
Coefficient
$0.28 (4.0) /1.90 = 0.5895 m$
Substitute
$S_{c} = 0.5895 (0.1984)$
Result
$S_{c} = 0.117 m \approx 117 mm$

Answer: S_c ≈ 117 mm

Why the other options are there

269 mm (natural log used)
222 mm (1 + e₀ omitted)

Reference: FE Reference Handbook — Geotechnical — Consolidation settlement

Example 2

Primary consolidation settlement of a clay layer — Ultimate consolidation settlement in soil layer

A normally consolidated clay layer is 12 ft thick with e₀ = 0.72 and Cc = 0.26. The initial effective stress at mid-depth is 2,354 psf and a fill adds Δσ = 1,797 psf. Find the primary settlement.

Given

H = 12 ft
e₀ = 0.72
Cc = 0.26
σ′₀ = 2,354 psf
Δσ = 1,797 psf

Find

Primary consolidation settlement S

Start with the thinking

Normally consolidated → use the virgin compression branch only.
Stresses are evaluated at mid-depth of the compressible layer.

Figure for Primary consolidation settlement of a clay layer — Ultimate consolidation settlement in soil layer

Step-by-step solution

Settlement
$S = (C cH) / (1 + e_{0}) \cdot l o g_{10} [(σ_{0}^{'} + Δ σ) / σ_{0}^{'}]$
Stress ratio
$(2, 354 + 1, 797) /2, 354 = 1.763$
Log term
$l o g_{10} (1.763) = 0.2463$
Coefficient
$C cH / (1 + e_{0}) = 0.26 (12) / (1 + 0.72) = 1.814 f t$
Settlement
$S = 1.814 (0.2463) = 0.447 f t = 5.36 in$

Answer: S ≈ 5.4 in. (0.45 ft)

Why the other options are there

12.3 in. (natural log used)
9.2 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Ultimate consolidation settlement in soil layer

Example 3

Primary consolidation settlement of a clay layer — Ultimate consolidation settlement in soil layer (2)

A normally consolidated clay layer is 14 ft thick with e₀ = 0.73 and Cc = 0.24. The initial effective stress at mid-depth is 1,654 psf and a fill adds Δσ = 1,581 psf. Find the primary settlement.

Given

H = 14 ft
e₀ = 0.73
Cc = 0.24
σ′₀ = 1,654 psf
Δσ = 1,581 psf

Find

Primary consolidation settlement S

Start with the thinking

Normally consolidated → use the virgin compression branch only.
Stresses are evaluated at mid-depth of the compressible layer.

Figure for Primary consolidation settlement of a clay layer — Ultimate consolidation settlement in soil layer (2)

Step-by-step solution

Settlement
$S = (C cH) / (1 + e_{0}) \cdot l o g_{10} [(σ_{0}^{'} + Δ σ) / σ_{0}^{'}]$
Stress ratio
$(1, 654 + 1, 581) /1, 654 = 1.956$
Log term
$l o g_{10} (1.956) = 0.2913$
Coefficient
$C cH / (1 + e_{0}) = 0.24 (14) / (1 + 0.73) = 1.942 f t$
Settlement
$S = 1.942 (0.2913) = 0.566 f t = 6.79 in$

Answer: S ≈ 6.8 in. (0.57 ft)

Why the other options are there

15.6 in. (natural log used)
11.7 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Ultimate consolidation settlement in soil layer

Example 4

Primary consolidation settlement of a clay layer — Ultimate consolidation settlement in soil layer (3)

A normally consolidated clay layer is 16 ft thick with e₀ = 1.17 and Cc = 0.22. The initial effective stress at mid-depth is 1,407 psf and a fill adds Δσ = 1,696 psf. Find the primary settlement.

Given

H = 16 ft
e₀ = 1.17
Cc = 0.22
σ′₀ = 1,407 psf
Δσ = 1,696 psf

Find

Primary consolidation settlement S

Start with the thinking

Normally consolidated → use the virgin compression branch only.
Stresses are evaluated at mid-depth of the compressible layer.

Figure for Primary consolidation settlement of a clay layer — Ultimate consolidation settlement in soil layer (3)

Step-by-step solution

Settlement
$S = (C cH) / (1 + e_{0}) \cdot l o g_{10} [(σ_{0}^{'} + Δ σ) / σ_{0}^{'}]$
Stress ratio
$(1, 407 + 1, 696) /1, 407 = 2.205$
Log term
$l o g_{10} (2.205) = 0.3435$
Coefficient
$C cH / (1 + e_{0}) = 0.22 (16) / (1 + 1.17) = 1.622 f t$
Settlement
$S = 1.622 (0.3435) = 0.557 f t = 6.69 in$

Answer: S ≈ 6.7 in. (0.56 ft)

Why the other options are there

15.4 in. (natural log used)
14.5 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Ultimate consolidation settlement in soil layer

Example 5

Primary consolidation settlement of a clay layer — Ultimate consolidation settlement in soil layer (4)

A normally consolidated clay layer is 19 ft thick with e₀ = 0.85 and Cc = 0.36. The initial effective stress at mid-depth is 2,015 psf and a fill adds Δσ = 1,034 psf. Find the primary settlement.

Given

H = 19 ft
e₀ = 0.85
Cc = 0.36
σ′₀ = 2,015 psf
Δσ = 1,034 psf

Find

Primary consolidation settlement S

Start with the thinking

Normally consolidated → use the virgin compression branch only.
Stresses are evaluated at mid-depth of the compressible layer.

Figure for Primary consolidation settlement of a clay layer — Ultimate consolidation settlement in soil layer (4)

Step-by-step solution

Settlement
$S = (C cH) / (1 + e_{0}) \cdot l o g_{10} [(σ_{0}^{'} + Δ σ) / σ_{0}^{'}]$
Stress ratio
$(2, 015 + 1, 034) /2, 015 = 1.513$
Log term
$l o g_{10} (1.513) = 0.1799$
Coefficient
$C cH / (1 + e_{0}) = 0.36 (19) / (1 + 0.85) = 3.697 f t$
Settlement
$S = 3.697 (0.1799) = 0.665 f t = 7.98 in$

Answer: S ≈ 8.0 in. (0.67 ft)

Why the other options are there

18.4 in. (natural log used)
14.8 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Ultimate consolidation settlement in soil layer

Example 6

Primary consolidation settlement of a clay layer — Ultimate consolidation settlement in soil layer (5)

A normally consolidated clay layer is 15 ft thick with e₀ = 0.84 and Cc = 0.24. The initial effective stress at mid-depth is 2,749 psf and a fill adds Δσ = 1,024 psf. Find the primary settlement.

Given

H = 15 ft
e₀ = 0.84
Cc = 0.24
σ′₀ = 2,749 psf
Δσ = 1,024 psf

Find

Primary consolidation settlement S

Start with the thinking

Normally consolidated → use the virgin compression branch only.
Stresses are evaluated at mid-depth of the compressible layer.

Figure for Primary consolidation settlement of a clay layer — Ultimate consolidation settlement in soil layer (5)

Step-by-step solution

Settlement
$S = (C cH) / (1 + e_{0}) \cdot l o g_{10} [(σ_{0}^{'} + Δ σ) / σ_{0}^{'}]$
Stress ratio
$(2, 749 + 1, 024) /2, 749 = 1.372$
Log term
$l o g_{10} (1.372) = 0.1375$
Coefficient
$C cH / (1 + e_{0}) = 0.24 (15) / (1 + 0.84) = 1.957 f t$
Settlement
$S = 1.957 (0.1375) = 0.269 f t = 3.23 in$

Answer: S ≈ 3.2 in. (0.27 ft)

Why the other options are there

7.4 in. (natural log used)
5.9 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Ultimate consolidation settlement in soil layer

Example 7

Primary consolidation settlement of a clay layer — Ultimate consolidation settlement in soil layer (6)

A normally consolidated clay layer is 11 ft thick with e₀ = 0.90 and Cc = 0.30. The initial effective stress at mid-depth is 1,342 psf and a fill adds Δσ = 1,763 psf. Find the primary settlement.

Given

H = 11 ft
e₀ = 0.90
Cc = 0.30
σ′₀ = 1,342 psf
Δσ = 1,763 psf

Find

Primary consolidation settlement S

Start with the thinking

Normally consolidated → use the virgin compression branch only.
Stresses are evaluated at mid-depth of the compressible layer.

Figure for Primary consolidation settlement of a clay layer — Ultimate consolidation settlement in soil layer (6)

Step-by-step solution

Settlement
$S = (C cH) / (1 + e_{0}) \cdot l o g_{10} [(σ_{0}^{'} + Δ σ) / σ_{0}^{'}]$
Stress ratio
$(1, 342 + 1, 763) /1, 342 = 2.314$
Log term
$l o g_{10} (2.314) = 0.3643$
Coefficient
$C cH / (1 + e_{0}) = 0.30 (11) / (1 + 0.90) = 1.737 f t$
Settlement
$S = 1.737 (0.3643) = 0.633 f t = 7.59 in$

Answer: S ≈ 7.6 in. (0.63 ft)

Why the other options are there

17.5 in. (natural log used)
14.4 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Ultimate consolidation settlement in soil layer

Example 8

Primary consolidation settlement of a clay layer — Ultimate consolidation settlement in soil layer (7)

A normally consolidated clay layer is 12 ft thick with e₀ = 0.95 and Cc = 0.38. The initial effective stress at mid-depth is 2,507 psf and a fill adds Δσ = 1,050 psf. Find the primary settlement.

Given

H = 12 ft
e₀ = 0.95
Cc = 0.38
σ′₀ = 2,507 psf
Δσ = 1,050 psf

Find

Primary consolidation settlement S

Start with the thinking

Normally consolidated → use the virgin compression branch only.
Stresses are evaluated at mid-depth of the compressible layer.

Figure for Primary consolidation settlement of a clay layer — Ultimate consolidation settlement in soil layer (7)

Step-by-step solution

Settlement
$S = (C cH) / (1 + e_{0}) \cdot l o g_{10} [(σ_{0}^{'} + Δ σ) / σ_{0}^{'}]$
Stress ratio
$(2, 507 + 1, 050) /2, 507 = 1.419$
Log term
$l o g_{10} (1.419) = 0.1519$
Coefficient
$C cH / (1 + e_{0}) = 0.38 (12) / (1 + 0.95) = 2.338 f t$
Settlement
$S = 2.338 (0.1519) = 0.355 f t = 4.26 in$

Answer: S ≈ 4.3 in. (0.36 ft)

Why the other options are there

9.8 in. (natural log used)
8.3 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Ultimate consolidation settlement in soil layer

Example 9

Primary consolidation settlement of a clay layer — Ultimate consolidation settlement in soil layer (8)

A normally consolidated clay layer is 13 ft thick with e₀ = 1.16 and Cc = 0.40. The initial effective stress at mid-depth is 2,368 psf and a fill adds Δσ = 921.0 psf. Find the primary settlement.

Given

H = 13 ft
e₀ = 1.16
Cc = 0.40
σ′₀ = 2,368 psf
Δσ = 921.0 psf

Find

Primary consolidation settlement S

Start with the thinking

Normally consolidated → use the virgin compression branch only.
Stresses are evaluated at mid-depth of the compressible layer.

Figure for Primary consolidation settlement of a clay layer — Ultimate consolidation settlement in soil layer (8)

Step-by-step solution

Settlement
$S = (C cH) / (1 + e_{0}) \cdot l o g_{10} [(σ_{0}^{'} + Δ σ) / σ_{0}^{'}]$
Stress ratio
$(2, 368 + 921.0) /2, 368 = 1.389$
Log term
$l o g_{10} (1.389) = 0.1427$
Coefficient
$C cH / (1 + e_{0}) = 0.40 (13) / (1 + 1.16) = 2.407 f t$
Settlement
$S = 2.407 (0.1427) = 0.343 f t = 4.12 in$

Answer: S ≈ 4.1 in. (0.34 ft)

Why the other options are there

9.5 in. (natural log used)
8.9 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Ultimate consolidation settlement in soil layer

Example 10

Primary consolidation settlement of a clay layer — Ultimate consolidation settlement in soil layer (9)

A normally consolidated clay layer is 19 ft thick with e₀ = 1.15 and Cc = 0.44. The initial effective stress at mid-depth is 2,772 psf and a fill adds Δσ = 844.0 psf. Find the primary settlement.

Given

H = 19 ft
e₀ = 1.15
Cc = 0.44
σ′₀ = 2,772 psf
Δσ = 844.0 psf

Find

Primary consolidation settlement S

Start with the thinking

Normally consolidated → use the virgin compression branch only.
Stresses are evaluated at mid-depth of the compressible layer.

Figure for Primary consolidation settlement of a clay layer — Ultimate consolidation settlement in soil layer (9)

Step-by-step solution

Settlement
$S = (C cH) / (1 + e_{0}) \cdot l o g_{10} [(σ_{0}^{'} + Δ σ) / σ_{0}^{'}]$
Stress ratio
$(2, 772 + 844.0) /2, 772 = 1.304$
Log term
$l o g_{10} (1.304) = 0.1154$
Coefficient
$C cH / (1 + e_{0}) = 0.44 (19) / (1 + 1.15) = 3.888 f t$
Settlement
$S = 3.888 (0.1154) = 0.449 f t = 5.39 in$

Answer: S ≈ 5.4 in. (0.45 ft)

Why the other options are there

12.4 in. (natural log used)
11.6 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Ultimate consolidation settlement in soil layer

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

Ultimate consolidation settlement in soil layer contains 18 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.