Module 6 · Lesson 6.2

The compaction curve and the zero-air-voids line

Optimum water content, maximum dry unit weight, and the ceiling no curve can cross.

Why this matters

Compact the same soil at several water contents and plot dry unit weight against water content: the points fall on a single humped curve. Its peak defines the two numbers every specification is written around — the maximum dry unit weight and the optimum water content — and the curve always sits below a hard theoretical ceiling: the zero-air-voids line.

The laboratory compaction test (Proctor test, Module 12) compacts a soil at several water contents using a standard effort, and plots $γ_{d}$ against $w$ . The result is a compaction curve: rising to a peak, then falling.

The peak dry unit weight is the maximum dry unit weight $γ_{d, m a x}$ .
The water content at the peak is the optimum water content $w_{o pt}$ .

The rising (dry) side is limited by friction between grains; the falling (wet) side is limited by water occupying the voids. The optimum is where these two effects balance for that compactive effort.

From first principles

The zero-air-voids line

We want to show: Find the dry unit weight a soil would have at a given water content if all air were removed — the ceiling every compaction curve lies below.

There is a hard limit to compaction: you can remove air, but you cannot remove the solids or the water present. If every last air void were squeezed out, the soil would be saturated, and its dry unit weight would depend only on how much water is in it. That saturated limit is the ceiling.

Try it

Compaction-curve explorer

Dry unit weight against water content. Find the optimum, and see the zero-air-voids ceiling every curve sits below.

Soil

Compactive effort

Optimum water content w_opt: 16.0 %
Max dry unit weight γ_d,max: 17.6 kN/m³
ZAV ceiling at w_opt: 18.5 kN/m³
Air voids at optimum: 4.8 %

Read the diagram

Orange = the compaction curve; its peak is (w_opt, γd,max).
Dark dashed = zero-air-voids ceiling (S = 100%); faint = 10% air voids.
More effort raises the peak and moves it left (drier).

At the optimum this soil reaches 17.6 kN/m³ at 16.0% water content, about 4.8% air voids below the zero-air-voids ceiling. Adding water beyond the optimum pushes the point down the wet side — γd falls.

Compactive effort shifts the whole curve. Rolling harder — heavier plant, more passes, thinner layers — raises the peak (a higher $γ_{d, m a x}$ ) and moves it to the left (a lower $w_{o pt}$ ). More energy packs the grains tighter and needs less lubrication to do it. But every curve, at every effort, still hugs the same zero-air-voids ceiling on its wet side, because that side is limited by saturation, not by effort.

Worked example

Air voids at the optimum

Given

A silty clay compacts to γd,max = 18.2 kN/m³ at w_opt = 14%
Gs = 2.70
γw = 9.81 kN/m³

Find

The air-voids content at the optimum, and the zero-air-voids dry unit weight at that water content.

Assumptions

Solids + water + air three-phase soil; Gs as given.

Practice

A soil has Gs = 2.65 and is compacted at w = 12%. What is the zero-air-voids dry unit weight at this water content? (γw = 9.81 kN/m³, answer in kN/m³.)

Summary

The compaction curve plots γd against w; its peak gives γd,max and the optimum water content w_opt.
The zero-air-voids line γd,zav = Gs·γw/(1 + w·Gs) is the saturated ceiling no curve can cross.
Real curves run a few per cent of air voids below the ceiling on the wet side.
More compactive effort raises the peak and shifts it left (drier optimum).

Progress is kept in this browser only.

This is educational material. It uses simplified examples to teach principles, and must not be relied on for real design or safety-critical decisions. Module overview and checkpoint