Lesson 8: Bulk, dry, saturated or submerged: which density should you use?
One soil, four densities · about 40 minutesYou will learn
Why one number is not enough
The grains in a soil do not change, but the water around them does. Rain fills the voids, a dry summer empties them, and below the water table the soil is surrounded by water that pushes up on it. Each situation gives the same soil a different density. Engineers name four of them, and choosing the wrong one is a common source of error.
- Explain bulk, dry, saturated and submerged density, and when each one applies.
- Calculate all four, and the matching unit weights, from Gs, e and w.
You will use the phase diagram and S e = w Gs from Lesson 6, and γ = ρ g from Lesson 7.
The idea, in plain words
Four conditions
| Density | What it describes |
|---|---|
| Bulk (ρbulk) | The soil as it is in the ground, with whatever water it holds: total mass divided by total volume. A dry soil’s bulk density equals its dry density; a saturated soil’s equals its saturated density. |
| Dry (ρdry) | Only the mass of the solids, divided by the total volume. Tells you how tightly the grains are packed, whatever the water content. |
| Saturated (ρsat) | The density when every void is full of water. |
| Submerged (ρ′) | A saturated soil under water feels lighter, because the water around it pushes up (buoyancy). Also called the effective or buoyant density. |
Where the formulas come from
One phase diagram gives them all
Use the trick from Lesson 6: draw a phase diagram with 1 unit volume of solids. Then:
| Part | Volume | Mass |
|---|---|---|
| Solids | 1 | Gs ρw |
| Water | S e | S e ρw |
| Air | (1 − S) e | 0 |
| Total | 1 + e | (Gs + S e) ρw |
Bulk density is total mass divided by total volume:
The other three come from the same expression:
For unit weights, replace ρ with γ and ρw with γw = 9.81 kN/m³ in every formula.
Fractions, not percentages
In all these formulas, use w and S as decimals: a water content of 18% goes in as 0.18. Putting in 18 is one of the most common mistakes in this topic.
Why submerged soil is lighter
Buoyancy
Anything under water is pushed up by a force equal to the weight of water it displaces. That is why a stone feels lighter in a river. A cubic metre of saturated soil under water displaces a cubic metre of water, so it is pushed up by 9.81 kN. What remains, γsat − γw, is roughly half of its saturated unit weight. This “lighter” weight is what presses the grains together below the water table, which is why it matters so much in Lesson 11, on effective stress.
Wherever you are
Names and symbols vary. You will see “submerged”, “buoyant” and “effective” unit weight, written γ′, γsub or γb. Bulk density is sometimes called “wet” or “moist” density. They all mean the same things described here, so check the definitions in whichever report or textbook you are reading.
Worked example
All four for one soil
A soil has Gs = 2.68, e = 0.62 and water content w = 18%. Find all four densities and unit weights.
| Density | Working | kg/m³ | kN/m³ |
|---|---|---|---|
| Bulk | (1 + 0.18) × 2.68 ÷ 1.62 × 1000 | 1952 | 19.15 |
| Dry | 2.68 ÷ 1.62 × 1000, or 1952 ÷ 1.18 | 1654 | 16.23 |
| Saturated | (2.68 + 0.62) ÷ 1.62 × 1000 | 2037 | 19.98 |
| Submerged | 2037 − 1000, or (2.68 − 1) ÷ 1.62 × 1000 | 1037 | 10.17 |
Unit weights are each density × 9.81 ÷ 1000. Notice the order, which always holds: submerged < dry < bulk < saturated (bulk can equal dry or saturated at the extremes).
Using it
Which one should you use?
| Situation | Use |
|---|---|
| Weight of soil above the water table | Bulk unit weight |
| Weight of soil below the water table (total stress) | Saturated unit weight |
| Stress between the grains below the water table | Submerged (effective) unit weight |
| How well a fill has been compacted | Dry density, because it ignores the water |
State which unit weight you used
In design checks I regularly find calculations that use the bulk unit weight below the water table, or a single unit weight for every layer. The answers can be noticeably wrong. A good report states, for each layer, the bulk and saturated unit weights and where the water table was assumed, so anyone checking can follow the numbers.
Your turn
From a lab sample
Task (15 minutes)
A sample of volume 1.000 × 10−3 m³ has a mass of 2.050 kg. Its water content is 15% and Gs = 2.70.
- Find its bulk and dry density.
- Find its void ratio and degree of saturation.
- Find its saturated and submerged unit weights.
Show the answers
1. ρbulk = 2.050 ÷ 0.001 = 2050 kg/m³. ρdry = 2050 ÷ 1.15 = 1783 kg/m³.
2. From ρdry = Gsρw ÷ (1 + e): 1 + e = 2700 ÷ 1783 = 1.514, so e = 0.51. S = w Gs ÷ e = 0.15 × 2.70 ÷ 0.514 = 0.79.
3. ρsat = (2.70 + 0.514) ÷ 1.514 × 1000 = 2122 kg/m³, so γsat = 2122 × 9.81 ÷ 1000 = 20.8 kN/m³. γ′ = 20.8 − 9.81 = 11.0 kN/m³.
Quick check
Five quick questions
Choose an answer to see the explanation.
Key points
What to take away
- Bulk: the soil as found. Dry: solids only. Saturated: voids full of water. Submerged: saturated soil under water, lighter because of buoyancy.
- ρbulk = (Gs + S e) ÷ (1 + e) × ρw. Set S = 0 for dry and S = 1 for saturated.
- ρdry = ρbulk ÷ (1 + w) and ρ′ = ρsat − ρw.
- The same formulas give unit weights with γw = 9.81 kN/m³.
- Use w and S as decimals in formulas.
- Above the water table use bulk; below it use saturated for total stress and submerged for stress between the grains. Dry density is the measure of compaction.
Up next: Lesson 9
How loose or dense is a sand? You will learn relative density, which compares a sand with its loosest and densest possible states.
Assessment (optional)
Test yourself properly
Unlike the practice questions, this assessment has no answers on the page. Attempt it on your own, as if it were an exam.
Instructions
- Total: 40 marks. Suggested time: 60 minutes.
- Answer all questions. The marks for each question are shown in brackets.
- You may refer to the lesson notes, but write your answers in your own words.
- Show your working in calculations. Where a question asks you to justify or explain, marks go to your reasoning, not just the final answer.
Section A: Concepts (12 marks)
- Define bulk, dry, saturated and submerged density. [4 marks]
- State which density or unit weight you would use (a) for soil above the water table, (b) for total stress below the water table, (c) for stress between the grains below the water table, and (d) to check compaction. [4 marks]
- Explain why a soil below the water table is effectively lighter. [2 marks]
- Explain why dry density, rather than bulk density, is used to judge how well a soil has been compacted. [2 marks]
Section B: Calculations (16 marks)
A soil has Gs = 2.70, e = 0.60 and w = 15%. Take ρw = 1000 kg/m³ and γw = 9.81 kN/m³.
(a) Calculate the degree of saturation. [2 marks]
(b) Calculate the bulk density. [2 marks]
(c) Calculate the dry density, and check it with ρbulk ÷ (1 + w). [2 marks]
(d) Calculate the saturated density. [2 marks]
(e) Calculate the submerged density. [2 marks]
(f) Give the bulk, saturated and submerged unit weights in kN/m³. [3 marks]
- On site, a compacted fill has a bulk density of 2050 kg/m³ and a water content of 12%. The specification requires a dry density of at least 95% of 1900 kg/m³. Does the fill pass? Show your working. [3 marks]
Section C: Applied case (12 marks)
A designer’s calculation for a retaining wall uses a single unit weight, γ = 19.0 kN/m³, for all the soil behind the wall. The water table is 2 m below ground level. Tests on the soil give Gs = 2.68, e = 0.65 and, above the water table, w = 20%.
Write a short technical note to the designer (maximum 300 words) that:
- explains which unit weights should be used above and below the water table, and for which purpose [4 marks]
- calculates the bulk, saturated and submerged unit weights for this soil [4 marks]
- explains what goes wrong if 19.0 kN/m³ is used for the soil below the water table when working out the stress between the grains [2 marks]
Clear, well-organised communication. [2 marks]
How your work is judged
| Criterion | What a strong answer shows |
|---|---|
| Technical accuracy | Correct terms, methods and calculations |
| Use of evidence | Conclusions backed by specific information in the question (test results, descriptions, site details) |
| Engineering judgement | Results linked to what they mean for design and construction |
| Communication | Clear, concise and well organised, in your own words |
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References
- British Standards Institution (BSI) (2014) BS EN ISO 17892-2:2014 Geotechnical investigation and testing. Laboratory testing of soil. Determination of bulk density. London: BSI.
- Das, B.M. and Sobhan, K. (2018) Principles of Geotechnical Engineering. 9th edn. Boston, MA: Cengage Learning.
- Knappett, J.A. and Craig, R.F. (2019) Craig’s Soil Mechanics. 9th edn. Boca Raton, FL: CRC Press.
- Powrie, W. (2014) Soil Mechanics: Concepts and Applications. 3rd edn. Boca Raton, FL: CRC Press.
Further learning
- Practise: take the worked example and change the water content from 0% to the saturated value. Watch the bulk density move from the dry value to the saturated value.
- Read: Chapter 1 of Craig’s Soil Mechanics works through these relationships with more examples.
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