More than 50%

Geotechnical · FE Reference Handbook section

Geotechnical

11 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 More than 50% 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 more than 50% describes physically and when it applies.
State every one of the 11 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. More than 50% 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: more than 50%.
Capstone Studio instructional photograph

Geotechnical — More than 50%: 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 11 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

Cu	Quantity produced by "Cu = D60/D10 Cc = D" — read its definition and unit from the handbook line directly above the equation.
Horizontal at PI	Quantity produced by "Horizontal at PI = 4 to LL = 25.5, "U "A"" — read its definition and unit from the handbook line directly above the equation.
then PI	Quantity produced by "then PI = 0.73 (LL−20) OH" — read its definition and unit from the handbook line directly above the equation.
Vertical at LL	Quantity produced by "Vertical at LL = 16 to PI = 7," — read its definition and unit from the handbook line directly above the equation.
30 then PI	Quantity produced by "30 then PI = 0.9 (LL−8)" — 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.

(50% or more of coarse (Less than 5% finesH) Cu < 6 and/or SP Poorly graded sandI
fraction passes [Cc < 1 or Cc > 3]D
No. 4 sieve)
Sands with Fines Fines classify as ML or SM Silty sandF, G, I
(More than 12% finesH) MH
Fines classify as CL or SC Clayey sandF, G, I
FINE-GRAINED SOILS Silts and Clays inorganic PI > 7 and plots on or CL Lean clayK, L, M
above "A" lineJ
Liquid limit PI < 4 or plots below "A" ML SiltK,L, M
less than 50 lineJ
organic Liquid limit − oven dried
/Liquid&#10 OL Organic clayK, L, M, N
50% or more < 0.75 Organic siltK, L, M, O
passes the No. 200 sieve Silts and Clays inorganic PI plots on or above "A" CH Fat clayK, L, M
line
Liquid limit
50 or more PI plots below "A" line MH Elastic siltK, L, M
organic Liquid limit − oven dried
/Liquid&#10 OH Organic clayK, L, M, P
< 0.75 Organic siltK, L, M, Q
HIGHLY ORGANIC SOILS Primarily organic matter, dark in color, and organic odor PT Peat
A HSands with 5 to 12% fines require dual symbols:
Based on the material passing the 3-in. (75-mm) sieve.
If field sample contained cobbles or boulders, or both, SW-SM well-graded sand with silt

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

Plasticity index, liquidity index and activity of a clay — More than 50%

A clay has a liquid limit of 30, a plastic limit of 14 and a natural water content of 28%. Compute the plasticity index and liquidity index, and estimate the clay fraction if the activity is 0.85.

Given

LL = 30
PL = 14
w = 28%
Activity A = 0.85

Find

PI, LI and the clay-size fraction

Start with the thinking

The plasticity index is the water-content band over which the soil behaves plastically.
A liquidity index above 1 means the natural water content exceeds the liquid limit — the soil is essentially a slurry.

Step-by-step solution

Formula
$P I = LL - P L$
Substituting
$P I = 30 - 14 = 16$
Formula
$L I = (w - P L) / P I$
Substituting
$L I = (28 - 14) /16 = 0.875$
Formula
$A = P I / ($
Rearranged
$c l a y f r a c t i o n = P I / A = 16/0.85 = 18.8$
Interpretation
$L I = 0.88 in d i c a t es a so f t co n s i s t e n cy$

Answer: PI = 16, LI = 0.88, clay fraction ≈ 19%

Why the other options are there

PI = 44 (limits added)
LI = -0.125 (used LL in place of PL)

Reference: FE Reference Handbook — Geotechnical → More than 50%

Example 2

Plasticity index, liquidity index and activity of a clay — More than 50% (2)

A clay has a liquid limit of 34, a plastic limit of 28 and a natural water content of 60%. Compute the plasticity index and liquidity index, and estimate the clay fraction if the activity is 1.10.

Given

LL = 34
PL = 28
w = 60%
Activity A = 1.10

Find

PI, LI and the clay-size fraction

Start with the thinking

The plasticity index is the water-content band over which the soil behaves plastically.
A liquidity index above 1 means the natural water content exceeds the liquid limit — the soil is essentially a slurry.

Step-by-step solution

Formula
$P I = LL - P L$
Substituting
$P I = 34 - 28 = 6$
Formula
$L I = (w - P L) / P I$
Substituting
$L I = (60 - 28) /6 = 5.333$
Formula
$A = P I / ($
Rearranged
$c l a y f r a c t i o n = P I / A = 6/1.10 = 5.5$
Interpretation
$L I = 5.33 in d i c a t es a v er y so f t, n e a r - l i q u i d co n s i s t e n cy$

Answer: PI = 6, LI = 5.33, clay fraction ≈ 5%

Why the other options are there

PI = 62 (limits added)
LI = 4.333 (used LL in place of PL)

Reference: FE Reference Handbook — Geotechnical → More than 50%

Example 3

Plasticity index, liquidity index and activity of a clay — More than 50% (3)

A clay has a liquid limit of 58, a plastic limit of 19 and a natural water content of 34%. Compute the plasticity index and liquidity index, and estimate the clay fraction if the activity is 1.00.

Given

LL = 58
PL = 19
w = 34%
Activity A = 1.00

Find

PI, LI and the clay-size fraction

Start with the thinking

The plasticity index is the water-content band over which the soil behaves plastically.
A liquidity index above 1 means the natural water content exceeds the liquid limit — the soil is essentially a slurry.

Step-by-step solution

Formula
$P I = LL - P L$
Substituting
$P I = 58 - 19 = 39$
Formula
$L I = (w - P L) / P I$
Substituting
$L I = (34 - 19) /39 = 0.385$
Formula
$A = P I / ($
Rearranged
$c l a y f r a c t i o n = P I / A = 39/1.00 = 39.0$
Interpretation
$L I = 0.38 in d i c a t es a s t i f f co n s i s t e n cy$

Answer: PI = 39, LI = 0.38, clay fraction ≈ 39%

Why the other options are there

PI = 77 (limits added)
LI = -0.615 (used LL in place of PL)

Reference: FE Reference Handbook — Geotechnical → More than 50%

Example 4

Plasticity index, liquidity index and activity of a clay — More than 50% (4)

A clay has a liquid limit of 36, a plastic limit of 17 and a natural water content of 27%. Compute the plasticity index and liquidity index, and estimate the clay fraction if the activity is 0.95.

Given

LL = 36
PL = 17
w = 27%
Activity A = 0.95

Find

PI, LI and the clay-size fraction

Start with the thinking

The plasticity index is the water-content band over which the soil behaves plastically.
A liquidity index above 1 means the natural water content exceeds the liquid limit — the soil is essentially a slurry.

Step-by-step solution

Formula
$P I = LL - P L$
Substituting
$P I = 36 - 17 = 19$
Formula
$L I = (w - P L) / P I$
Substituting
$L I = (27 - 17) /19 = 0.526$
Formula
$A = P I / ($
Rearranged
$c l a y f r a c t i o n = P I / A = 19/0.95 = 20.0$
Interpretation
$L I = 0.53 in d i c a t es a so f t co n s i s t e n cy$

Answer: PI = 19, LI = 0.53, clay fraction ≈ 20%

Why the other options are there

PI = 53 (limits added)
LI = -0.474 (used LL in place of PL)

Reference: FE Reference Handbook — Geotechnical → More than 50%

Example 5

Plasticity index, liquidity index and activity of a clay — More than 50% (5)

A clay has a liquid limit of 53, a plastic limit of 22 and a natural water content of 35%. Compute the plasticity index and liquidity index, and estimate the clay fraction if the activity is 0.60.

Given

LL = 53
PL = 22
w = 35%
Activity A = 0.60

Find

PI, LI and the clay-size fraction

Start with the thinking

The plasticity index is the water-content band over which the soil behaves plastically.
A liquidity index above 1 means the natural water content exceeds the liquid limit — the soil is essentially a slurry.

Step-by-step solution

Formula
$P I = LL - P L$
Substituting
$P I = 53 - 22 = 31$
Formula
$L I = (w - P L) / P I$
Substituting
$L I = (35 - 22) /31 = 0.419$
Formula
$A = P I / ($
Rearranged
$c l a y f r a c t i o n = P I / A = 31/0.60 = 51.7$
Interpretation
$L I = 0.42 in d i c a t es a s t i f f co n s i s t e n cy$

Answer: PI = 31, LI = 0.42, clay fraction ≈ 52%

Why the other options are there

PI = 75 (limits added)
LI = -0.581 (used LL in place of PL)

Reference: FE Reference Handbook — Geotechnical → More than 50%

Example 6

Plasticity index, liquidity index and activity of a clay — More than 50% (6)

A clay has a liquid limit of 66, a plastic limit of 15 and a natural water content of 51%. Compute the plasticity index and liquidity index, and estimate the clay fraction if the activity is 0.75.

Given

LL = 66
PL = 15
w = 51%
Activity A = 0.75

Find

PI, LI and the clay-size fraction

Start with the thinking

The plasticity index is the water-content band over which the soil behaves plastically.
A liquidity index above 1 means the natural water content exceeds the liquid limit — the soil is essentially a slurry.

Step-by-step solution

Formula
$P I = LL - P L$
Substituting
$P I = 66 - 15 = 51$
Formula
$L I = (w - P L) / P I$
Substituting
$L I = (51 - 15) /51 = 0.706$
Formula
$A = P I / ($
Rearranged
$c l a y f r a c t i o n = P I / A = 51/0.75 = 68.0$
Interpretation
$L I = 0.71 in d i c a t es a so f t co n s i s t e n cy$

Answer: PI = 51, LI = 0.71, clay fraction ≈ 68%

Why the other options are there

PI = 81 (limits added)
LI = -0.294 (used LL in place of PL)

Reference: FE Reference Handbook — Geotechnical → More than 50%

Example 7

Plasticity index, liquidity index and activity of a clay — More than 50% (7)

A clay has a liquid limit of 73, a plastic limit of 16 and a natural water content of 56%. Compute the plasticity index and liquidity index, and estimate the clay fraction if the activity is 1.35.

Given

LL = 73
PL = 16
w = 56%
Activity A = 1.35

Find

PI, LI and the clay-size fraction

Start with the thinking

The plasticity index is the water-content band over which the soil behaves plastically.
A liquidity index above 1 means the natural water content exceeds the liquid limit — the soil is essentially a slurry.

Step-by-step solution

Formula
$P I = LL - P L$
Substituting
$P I = 73 - 16 = 57$
Formula
$L I = (w - P L) / P I$
Substituting
$L I = (56 - 16) /57 = 0.702$
Formula
$A = P I / ($
Rearranged
$c l a y f r a c t i o n = P I / A = 57/1.35 = 42.2$
Interpretation
$L I = 0.70 in d i c a t es a so f t co n s i s t e n cy$

Answer: PI = 57, LI = 0.70, clay fraction ≈ 42%

Why the other options are there

PI = 89 (limits added)
LI = -0.298 (used LL in place of PL)

Reference: FE Reference Handbook — Geotechnical → More than 50%

Example 8

Plasticity index, liquidity index and activity of a clay — More than 50% (8)

A clay has a liquid limit of 64, a plastic limit of 27 and a natural water content of 55%. Compute the plasticity index and liquidity index, and estimate the clay fraction if the activity is 1.10.

Given

LL = 64
PL = 27
w = 55%
Activity A = 1.10

Find

PI, LI and the clay-size fraction

Start with the thinking

The plasticity index is the water-content band over which the soil behaves plastically.
A liquidity index above 1 means the natural water content exceeds the liquid limit — the soil is essentially a slurry.

Step-by-step solution

Formula
$P I = LL - P L$
Substituting
$P I = 64 - 27 = 37$
Formula
$L I = (w - P L) / P I$
Substituting
$L I = (55 - 27) /37 = 0.757$
Formula
$A = P I / ($
Rearranged
$c l a y f r a c t i o n = P I / A = 37/1.10 = 33.6$
Interpretation
$L I = 0.76 in d i c a t es a so f t co n s i s t e n cy$

Answer: PI = 37, LI = 0.76, clay fraction ≈ 34%

Why the other options are there

PI = 91 (limits added)
LI = -0.243 (used LL in place of PL)

Reference: FE Reference Handbook — Geotechnical → More than 50%

Example 9

Plasticity index, liquidity index and activity of a clay — More than 50% (9)

A clay has a liquid limit of 69, a plastic limit of 20 and a natural water content of 32%. Compute the plasticity index and liquidity index, and estimate the clay fraction if the activity is 0.55.

Given

LL = 69
PL = 20
w = 32%
Activity A = 0.55

Find

PI, LI and the clay-size fraction

Start with the thinking

The plasticity index is the water-content band over which the soil behaves plastically.
A liquidity index above 1 means the natural water content exceeds the liquid limit — the soil is essentially a slurry.

Step-by-step solution

Formula
$P I = LL - P L$
Substituting
$P I = 69 - 20 = 49$
Formula
$L I = (w - P L) / P I$
Substituting
$L I = (32 - 20) /49 = 0.245$
Formula
$A = P I / ($
Rearranged
$c l a y f r a c t i o n = P I / A = 49/0.55 = 89.1$
Interpretation
$L I = 0.24 in d i c a t es a s t i f f co n s i s t e n cy$

Answer: PI = 49, LI = 0.24, clay fraction ≈ 89%

Why the other options are there

PI = 89 (limits added)
LI = -0.755 (used LL in place of PL)

Reference: FE Reference Handbook — Geotechnical → More than 50%

Example 10

Plasticity index, liquidity index and activity of a clay — More than 50% (10)

A clay has a liquid limit of 74, a plastic limit of 21 and a natural water content of 39%. Compute the plasticity index and liquidity index, and estimate the clay fraction if the activity is 1.10.

Given

LL = 74
PL = 21
w = 39%
Activity A = 1.10

Find

PI, LI and the clay-size fraction

Start with the thinking

The plasticity index is the water-content band over which the soil behaves plastically.
A liquidity index above 1 means the natural water content exceeds the liquid limit — the soil is essentially a slurry.

Step-by-step solution

Formula
$P I = LL - P L$
Substituting
$P I = 74 - 21 = 53$
Formula
$L I = (w - P L) / P I$
Substituting
$L I = (39 - 21) /53 = 0.340$
Formula
$A = P I / ($
Rearranged
$c l a y f r a c t i o n = P I / A = 53/1.10 = 48.2$
Interpretation
$L I = 0.34 in d i c a t es a s t i f f co n s i s t e n cy$

Answer: PI = 53, LI = 0.34, clay fraction ≈ 48%

Why the other options are there

PI = 95 (limits added)
LI = -0.660 (used LL in place of PL)

Reference: FE Reference Handbook — Geotechnical → More than 50%

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

More than 50% contains 11 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.

More than 50%

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

retainedonNo.200sieveSandsCleanSandsCu≥6and1≤Cc≤3DSWWell−gradedsandI

IIfsoilcontains≥15

DLIfsoilcontains≥30

Cu=D60/D10Cc=D

10×D60add"sandy"togroupname.

EIfsoilcontains≥15

PI≥4andplotsonorabove"A"line.

HorizontalatPI=4toLL=25.5,"U"A"

thenPI=0.73(LL−20)OH

VerticalatLL=16toPI=7,

30thenPI=0.9(LL−8)

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section