Soil Mechanics Geothechinical

Question

An undisturbed clays sample extracted from the field was of a diameter 38 mm, and a height of 80 mm. The mass of the soil was determined to be 182g. Water content was determined as 40%. The void ratio of the sample is 0.82. Draw the phase diagram and determine the specific gravity, bulk density, and dry density of the specimen.

Answer 1

Before You Read the Solution:

This is not just an answer. This is a way of thinking.

Any calculator can plug numbers into a formula. But an engineer knows when the answer is wrong—before anyone tells them. An engineer checks. Questions. Verifies.

In this solution, I will show you how to solve this problem. But more importantly, I will show you how to think about it. I will show you the moment I knew something was wrong—and what I did about it.

Because a number is not an answer. A number is a question. The question is: What do we do about it?

Step 1: Total Volume

The specimen is cylindrical:

V = (π/4) × d² × h = (π/4) × (3.8) ² × (8.0) = 90.729 cm³

Step 2: Using Given Values Directly

Let's solve using all given values:

Vs = V / (1 + e) = 90.729 / 1.82 = 49.851 cm³

V v = e × Vs = 0.82 × 49.851 = 40.878 cm³

M s = M / (1 + w) = 182 / 1.40 = 130.0 g

Mw = M - M s = 182 - 130 = 52.0 g

V w = Mw / ρw = 52.0 / 1.0 = 52.00 cm³

Check

Vs + V w = 49.851 + 52.00 = 101.851 cm³

But V total = 90.729 cm³

Vs + Vw > Vtotal IMPOSSIBLE.

Also, calculate degree of saturation:

S = (w × Gs) / e = (0.40 × 2.61) / 0.82 = 127%

S = 127% > 100% IMPOSSIBLE.

You cannot fit 127 cm³ of water into 100 cm³ of voids. What's the solution?

Step 3: The Correction - Assume S = 100%

Since this is a saturated clay below the water table, we assume S = 100%.

This means Vw = Vv = 40.878 cm³ (capped at the void volume).

Now we recalculate:

Mw = Vw × ρw = 40.878 × 1.0 = 40.878 g

Ms = M total - Mw = 182 - 40.878 = 141.122 g

Step 4: Recalculate All Properties

Bulk Density:

ρ = Mtotal / Vtotal = 182 / 90.729 = 2.006 g/cm³ = 2006 kg/m³

Dry Density:

ρd = Ms / Vtotal = 141.122 / 90.729 = 1.555 g/cm³ = 1555 kg/m³

Unit Weight of Solids:

γs = Ms / Vs = 141.122 / 49.851 = 2.831 g/cm³

Specific Gravity:

Gs = γs / γw = 2.831 / 1.0 = 2.83

Step 5: Verification

Check
Vs + Vv	49.851 + 40.878 = 90.729 cm³ OK
Ms + Mw	141.122 + 40.878 = 182 g OK
Gs = ρd(1+e)/ρw	1.555 × 1.82 / 1.0 = 2.83 OK
S = w·Gs/e	(0.290 × 2.83) / 0.82 = 100% OK

Everything is consistent. Gs = 2.83 is reasonable for clay.

Step 6: Phase Diagram Summary

Physical Model:

1. Vs = 49.851 cm³, Ms = 141.122 g
2. Vw = 40.878 cm³, Mw = 40.878 g
3. Va = 0, Ma = 0
4. V total = 90.729 cm³, M total = 182 g

Normalized Model:

1. Vs = 1.0
2. Vv = 0.82
3. Vtotal = 1.82

Think About This I Check Your Understanding:

1. Vs and Vw are almost equal. What does this mean? This soil holds a lot of water-nearly as much water as solids. It's soft, wet, and compressible.
2. If e = 0.60 instead of 0.82? Vv decreases, density increases. This is why compaction works.
3. Settlement or frost heave? Settlement is the primary concern. The soil is soft, saturated, and compressible.
4. Is Gs = 2.83 reasonable? Yes-clays typically range from 2.60 to 2.80. 2.83 is slightly high but acceptable.
5. Porosity: n = Vv / Vtotal = 40.878 / 90.729 = 45.06%. Nearly half the soil volume is voids.

Engineering Intuition

This problem teaches a critical lesson: When the numbers don't match reality, question them. We used physical constraints (S ≤ 100%) to find the correct solution.

A number is not an answer. A number is a question. The question is: What do we do about it?

Thank You for Learning With Me

This is how I teach. Not just formulas—understanding. If this helped you, reach out. I'm here to help you master the "why" behind the "what."

— Louay Awad, PMP, CCM, EIT

The Structural Expert

Understand the "Why" — Master the "What"