Compression index

Geotechnical · FE Reference Handbook section

Geotechnical

2 formulas

10 exam-style examples

~49 min

All Geotechnical lectures

Learning objectives

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

This chapter section covers Compression index 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 compression index describes physically and when it applies.
State every one of the 2 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. Compression index 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: compression index.
Capstone Studio instructional photograph

Geotechnical — Compression index: 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 2 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.

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 — Compression index

A normally consolidated clay layer is 23 ft thick with e₀ = 0.87 and Cc = 0.37. The initial effective stress at mid-depth is 2,327 psf and a fill adds Δσ = 1,415 psf. Find the primary settlement.

Given

H = 23 ft
e₀ = 0.87
Cc = 0.37
σ′₀ = 2,327 psf
Δσ = 1,415 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 — Compression index

Step-by-step solution

Settlement
$S = (C cH) / (1 + e_{0}) \cdot l o g_{10} [(σ_{0}^{'} + Δ σ) / σ_{0}^{'}]$
Stress ratio
$(2, 327 + 1, 415) /2, 327 = 1.608$
Log term
$l o g_{10} (1.608) = 0.2063$
Coefficient
$C cH / (1 + e_{0}) = 0.37 (23) / (1 + 0.87) = 4.551 f t$
Settlement
$S = 4.551 (0.2063) = 0.939 f t = 11.27 in$

Answer: S ≈ 11.3 in. (0.94 ft)

Why the other options are there

25.9 in. (natural log used)
21.1 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Compression index

Example 3

Primary consolidation settlement of a clay layer — Compression index (2)

A normally consolidated clay layer is 19 ft thick with e₀ = 0.97 and Cc = 0.21. The initial effective stress at mid-depth is 2,616 psf and a fill adds Δσ = 1,946 psf. Find the primary settlement.

Given

H = 19 ft
e₀ = 0.97
Cc = 0.21
σ′₀ = 2,616 psf
Δσ = 1,946 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 — Compression index (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
$(2, 616 + 1, 946) /2, 616 = 1.744$
Log term
$l o g_{10} (1.744) = 0.2415$
Coefficient
$C cH / (1 + e_{0}) = 0.21 (19) / (1 + 0.97) = 2.025 f t$
Settlement
$S = 2.025 (0.2415) = 0.489 f t = 5.87 in$

Answer: S ≈ 5.9 in. (0.49 ft)

Why the other options are there

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

Reference: FE Reference Handbook — Geotechnical → Compression index

Example 4

Primary consolidation settlement of a clay layer — Compression index (3)

A normally consolidated clay layer is 21 ft thick with e₀ = 1.03 and Cc = 0.34. The initial effective stress at mid-depth is 2,927 psf and a fill adds Δσ = 922.0 psf. Find the primary settlement.

Given

H = 21 ft
e₀ = 1.03
Cc = 0.34
σ′₀ = 2,927 psf
Δσ = 922.0 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 — Compression index (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
$(2, 927 + 922.0) /2, 927 = 1.315$
Log term
$l o g_{10} (1.315) = 0.1189$
Coefficient
$C cH / (1 + e_{0}) = 0.34 (21) / (1 + 1.03) = 3.517 f t$
Settlement
$S = 3.517 (0.1189) = 0.418 f t = 5.02 in$

Answer: S ≈ 5.0 in. (0.42 ft)

Why the other options are there

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

Reference: FE Reference Handbook — Geotechnical → Compression index

Example 5

Primary consolidation settlement of a clay layer — Compression index (4)

A normally consolidated clay layer is 17 ft thick with e₀ = 0.80 and Cc = 0.45. The initial effective stress at mid-depth is 2,261 psf and a fill adds Δσ = 827.0 psf. Find the primary settlement.

Given

H = 17 ft
e₀ = 0.80
Cc = 0.45
σ′₀ = 2,261 psf
Δσ = 827.0 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 — Compression index (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
$(2, 261 + 827.0) /2, 261 = 1.366$
Log term
$l o g_{10} (1.366) = 0.1354$
Coefficient
$C cH / (1 + e_{0}) = 0.45 (17) / (1 + 0.80) = 4.250 f t$
Settlement
$S = 4.250 (0.1354) = 0.575 f t = 6.90 in$

Answer: S ≈ 6.9 in. (0.58 ft)

Why the other options are there

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

Reference: FE Reference Handbook — Geotechnical → Compression index

Example 6

Primary consolidation settlement of a clay layer — Compression index (5)

A normally consolidated clay layer is 25 ft thick with e₀ = 0.74 and Cc = 0.45. The initial effective stress at mid-depth is 2,405 psf and a fill adds Δσ = 1,084 psf. Find the primary settlement.

Given

H = 25 ft
e₀ = 0.74
Cc = 0.45
σ′₀ = 2,405 psf
Δσ = 1,084 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 — Compression index (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
$(2, 405 + 1, 084) /2, 405 = 1.451$
Log term
$l o g_{10} (1.451) = 0.1616$
Coefficient
$C cH / (1 + e_{0}) = 0.45 (25) / (1 + 0.74) = 6.466 f t$
Settlement
$S = 6.466 (0.1616) = 1.045 f t = 12.54 in$

Answer: S ≈ 12.5 in. (1.04 ft)

Why the other options are there

28.9 in. (natural log used)
21.8 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Compression index

Example 7

Primary consolidation settlement of a clay layer — Compression index (6)

A normally consolidated clay layer is 17 ft thick with e₀ = 0.71 and Cc = 0.43. The initial effective stress at mid-depth is 2,580 psf and a fill adds Δσ = 641.0 psf. Find the primary settlement.

Given

H = 17 ft
e₀ = 0.71
Cc = 0.43
σ′₀ = 2,580 psf
Δσ = 641.0 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 — Compression index (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
$(2, 580 + 641.0) /2, 580 = 1.248$
Log term
$l o g_{10} (1.248) = 0.0964$
Coefficient
$C cH / (1 + e_{0}) = 0.43 (17) / (1 + 0.71) = 4.275 f t$
Settlement
$S = 4.275 (0.0964) = 0.412 f t = 4.94 in$

Answer: S ≈ 4.9 in. (0.41 ft)

Why the other options are there

11.4 in. (natural log used)
8.5 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Compression index

Example 8

Primary consolidation settlement of a clay layer — Compression index (7)

A normally consolidated clay layer is 15 ft thick with e₀ = 1.13 and Cc = 0.23. The initial effective stress at mid-depth is 2,829 psf and a fill adds Δσ = 1,385 psf. Find the primary settlement.

Given

H = 15 ft
e₀ = 1.13
Cc = 0.23
σ′₀ = 2,829 psf
Δσ = 1,385 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 — Compression index (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
$(2, 829 + 1, 385) /2, 829 = 1.490$
Log term
$l o g_{10} (1.490) = 0.1731$
Coefficient
$C cH / (1 + e_{0}) = 0.23 (15) / (1 + 1.13) = 1.620 f t$
Settlement
$S = 1.620 (0.1731) = 0.280 f t = 3.36 in$

Answer: S ≈ 3.4 in. (0.28 ft)

Why the other options are there

7.7 in. (natural log used)
7.2 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Compression index

Example 9

Primary consolidation settlement of a clay layer — Compression index (8)

A normally consolidated clay layer is 9 ft thick with e₀ = 0.92 and Cc = 0.44. The initial effective stress at mid-depth is 2,547 psf and a fill adds Δσ = 1,141 psf. Find the primary settlement.

Given

H = 9 ft
e₀ = 0.92
Cc = 0.44
σ′₀ = 2,547 psf
Δσ = 1,141 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 — Compression index (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
$(2, 547 + 1, 141) /2, 547 = 1.448$
Log term
$l o g_{10} (1.448) = 0.1608$
Coefficient
$C cH / (1 + e_{0}) = 0.44 (9) / (1 + 0.92) = 2.063 f t$
Settlement
$S = 2.063 (0.1608) = 0.332 f t = 3.98 in$

Answer: S ≈ 4.0 in. (0.33 ft)

Why the other options are there

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

Reference: FE Reference Handbook — Geotechnical → Compression index

Example 10

Primary consolidation settlement of a clay layer — Compression index (9)

A normally consolidated clay layer is 16 ft thick with e₀ = 0.96 and Cc = 0.23. The initial effective stress at mid-depth is 1,316 psf and a fill adds Δσ = 1,895 psf. Find the primary settlement.

Given

H = 16 ft
e₀ = 0.96
Cc = 0.23
σ′₀ = 1,316 psf
Δσ = 1,895 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 — Compression index (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
$(1, 316 + 1, 895) /1, 316 = 2.440$
Log term
$l o g_{10} (2.440) = 0.3874$
Coefficient
$C cH / (1 + e_{0}) = 0.23 (16) / (1 + 0.96) = 1.878 f t$
Settlement
$S = 1.878 (0.3874) = 0.727 f t = 8.73 in$

Answer: S ≈ 8.7 in. (0.73 ft)

Why the other options are there

20.1 in. (natural log used)
17.1 in. (1+e₀ omitted)

Reference: FE Reference Handbook — Geotechnical → Compression index

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

Compression index contains 2 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.