Module: Introduction to Geotechnical Engineering o Gravel → larger than 4.75 mm (think of small pebbles).

Lesson 1: Introduction to Geotechnical Engineering o Sand → 0.075 mm to 4.75 mm (like beach sand).
Think of every structure you see around you — houses, schools, bridges, even o Silt → 0.002 mm to 0.075 mm (powdery, smooth when dry).
roads. All of them stand on soil or rock. If the ground is weak, the structure
o Clay → smaller than 0.002 mm (sticky when wet, hard when dry).
becomes unsafe, no matter how strong the concrete or steel is. That’s where
geotechnical engineering comes in. • Engineering impact:
• In simple terms: Geotechnical engineering is the branch of civil o Gravel & Sand (coarse soils): Drain water well, strong foundation
engineering that deals with the ground — studying soil and rock, testing materials.
their properties, and designing foundations that can safely carry loads.
o Silt & Clay (fine soils): Poor drainage, can expand/shrink, may
• Why it matters: Imagine building a house on soft clay without proper cause foundation problems.
testing. The soil could compress, and the house may suddenly tilt or crack.
On the other hand, if we build on strong, well-drained gravel, the house will Think about it: Why do rice fields use clayey soils in paddies? Because clay holds
remain stable for decades. water. Why do we use gravel for road bases? Because gravel drains water quickly
and provides stability.
• What geotechnical engineers do:
o Investigate soil at project sites (through boring, sampling, and lab
tests). Lesson 3: Sieve Analysis

o Decide what type of foundation is safe (shallow or deep). When engineers need to know what kind of soil they’re dealing with, one of the
simplest tests is the sieve analysis.
o Analyze slope stability for embankments and hills.
• What it does: Separates soil particles by size using a stack of sieves.
o Help design earthworks like tunnels, retaining walls, and dams.
• How it works (simplified steps):
Key takeaway: A safe structure is not only about the materials above the ground,
but also about the soil beneath it. 1. Dry the soil sample.
2. Place it in a set of sieves (largest opening on top, smallest at the
bottom).
Lesson 2: Particle Size Classification
3. Shake the sieves so particles fall through.
Soil is not just “dirt.” It is made up of particles of different sizes. The size of the
particles controls how the soil behaves — whether it drains water quickly, holds 4. Weigh the soil retained in each sieve.
moisture, or swells when wet. 5. Compute percentages and plot the grain size distribution curve.
• Soil particle sizes: • Why it’s useful:

Prepared by: ENGR JERRY JAKE JUNE B. HAYAGAN

 o Helps classify soil (gravel, sand, silt, clay).
o Determines whether soil is well-graded (variety of sizes = stronger) Step 2: Compute % Finer
or poorly graded (uniform size = weaker).
Sieve Size (mm) Mass Retained (g) Cumulative Passing (g) Percent Finer (%)
o Important for decisions in drainage, compaction, and foundation
design. 4.75 (Gravel) 50 50 90

2.00 120 170 66

Example Problem: Particle Size Distribution 0.425 200 370 26

A soil sample weighing 500 g was subjected to sieve analysis. The following results 0.075 100 470 6
were obtained:
Pan (fines) 30 500 0
Sieve Size (mm) Mass Retained (g)

4.75 (Gravel) 50
𝑇𝑜𝑡𝑎𝑙 𝑀𝑎𝑠𝑠 − 𝐶𝑜𝑚𝑚𝑢𝑙𝑎𝑡𝑖𝑣𝑒 𝑃𝑎𝑠𝑠𝑖𝑛𝑔
𝑃𝑒𝑟𝑐𝑒𝑛𝑡 𝐹𝑖𝑛𝑒𝑟 =
2.00 120 𝑇𝑜𝑡𝑎𝑙 𝑀𝑎𝑠𝑠

0.425 200

0.075 100 Step 3: Interpret Results

Pan (fines) 30 • Most particles pass through 2 mm and 0.425 mm → so the soil is mainly
sand.
• The presence of fines (6%) shows a small amount of silt/clay.
Step 1: Compute Cumulative passing
• With a range of sizes, this soil is well-graded sand with some fines.
Sieve Size (mm) Mass Retained (g) Cumulative Passing (g)
Conclusion: This soil is suitable as a foundation material if properly compacted,
4.75 (Gravel) 50 50 since well-graded sands are stable.

2.00 120 170

0.425 200 370

0.075 100 470

Pan (fines) 30 500

Just add the mass retained from the top sieve downward.
Prepared by: ENGR JERRY JAKE JUNE B. HAYAGAN
 Interpretation:
• If Cu > 4 → soil is well-graded (good mix of sizes).
Lesson 4: Key Terms in Grain Size Distribution • If Cu ≈ 1–3 → soil is poorly graded (uniform size, weaker).
1. D10 (Effective Size)
• The particle size for which 10% of the soil sample (by weight) is finer. 5. Coefficient of Curvature (Cc)
• Found by reading the grain size distribution curve at 10% passing. (𝑫𝟑𝟎 )𝟐
𝑪𝒄 =
• Purpose: 𝑫𝟔𝟎 × 𝑫𝟏𝟎

o Represents the size of the smaller particles in the soil. • Purpose: Checks the shape of the grain size distribution curve.

o Used to estimate permeability (drainage capacity) and filter Interpretation:

design. • If 1 ≤ Cc ≤ 3 → soil is considered well-graded.
o Example: If D10 = 0.2 mm, it means 10% of the soil particles are • If outside this range → soil may be gap-graded (missing certain particle
smaller than 0.2 mm. sizes).
2. D30
• Particle size at which 30% of the soil sample is finer. Why Do Engineers Care?
• Used in the calculation of Cc (Coefficient of Curvature). • Permeability (Drainage): D10 helps predict how water flows through soil.
• Foundation Strength: Well-graded soils (Cu > 4 and 1 ≤ Cc ≤ 3) compact
3. D60 better and are more stable for construction.

• Particle size at which 60% of the soil sample is finer. • Filter Design: D10 is used in designing filters in dams or drainage systems.

• Represents the coarser fraction of the soil. • Soil Classification: Cu and Cc help classify soil as well-graded sand (SW),
poorly graded sand (SP), etc., following USCS (Unified Soil Classification
System).
4. Coefficient of Uniformity (Cu)
𝑫𝟔𝟎
𝑪𝒖 =
𝑫𝟏𝟎
• Purpose: Tells how well-graded (wide variety of particle sizes) or poorly
graded (uniform particle size) a soil is.
Prepared by: ENGR JERRY JAKE JUNE B. HAYAGAN
Lesson 5: Determination of diameters of particles by ENGR JEK 𝑫𝟏𝟎 is in between the 6% and 26% finer or in between 0.075 mm and 0.425 mm
diameter.
Using the example problem above in Particle size distribution
Let (0.425mm, 26%) be the 1st coordinate or (𝑥1 , 𝑦1 ), (0.075 mm, 6%) be the 2nd
coordinate or (𝑥2 , 𝑦2 ), and (𝐷10 , 10%) be the 3rd coordinate or (𝑥3 , 𝑦3 ),.
Using similar triangles:
26 − 6 26 − 10
=
𝑙𝑜𝑔(0.425) − 𝑙𝑜𝑔(0.075) 𝑙𝑜𝑔(0.425) − 𝑙𝑜𝑔(𝐷10 )
16
26.54887227 =
𝑙𝑜𝑔(0.425) − 𝑙𝑜𝑔(𝐷10 )
16
𝑙𝑜𝑔(0.425) − 𝑙𝑜𝑔(𝐷10 ) = = 0.6026621333
26.54887227
𝑙𝑜𝑔(𝐷10 ) = 𝑙𝑜𝑔(0.425) − 0.6026621333 = −0.9742732033

Review principles of logarithms

𝑙𝑜𝑔10 𝑥 = 𝑦 𝑜𝑟 10𝑦 = 𝑥
By Similar triangles:
𝒚𝟏 − 𝒚𝟐 𝒚 𝟏 − 𝒚𝟑
=
𝒍𝒐𝒈(𝒙𝟏 ) − 𝒍𝒐𝒈(𝒙𝟐 ) 𝒍𝒐𝒈(𝒙𝟏 ) − 𝒍𝒐𝒈(𝒙𝟑 ) 𝐷10 = 10−0.9742732033 = 0.1061027882 𝑚𝑚

For example: Finding 𝑫𝟏𝟎 Using calculator technique: checking only

Sieve Size (mm) Mass Retained (g) Cumulative Passing (g) Percent Finer (%) Mode 3-4
4.75 (Gravel) 50 50 90 X (diameter) Y (% finer)

2.00 120 170 66 0.075 6

0.425 26
0.425 200 370 26
Then 10𝑥̂ = 0.1061027882 𝑚𝑚
0.075 100 470 6

Pan (fines) 30 500 0

Prepared by: ENGR JERRY JAKE JUNE B. HAYAGAN