SME
Soil Mechanics
Compaction
Use this for “A compacted fill has γ = 19.8 kN/m³ at w = 12% and Gₛ = 2.68. The lab’s γd,max is 18.5 kN/m³. Find γd, γzav and the relative compaction.”
Chart
γzav(12) ≈ 19.89
Level γd,max = 18.5 meets the curve at w = 15.71
This is an example. Type your own values to start a new problem.
γd (kN/m³)
=
γ (kN/m³)
÷
(1
+
w
%)
γzav (kN/m³)
=
Gₛ
×
9.81
kN/m³
÷
(1
+
w
%
×
Gₛ
)
RC
%
=
γd (kN/m³)
÷
γd,max (kN/m³)
×
100%
Formulas
γd
=
γ
1 +
w
÷ 100
17.6786
=
19.8
1 + 12 ÷ 100
γzav
=
Gₛ
×
9.81
1 +
w
÷ 100 × Gₛ
19.8932
=
2.68
×
9.81
1 + 12 ÷ 100 × 2.68
RC
=
γd
γd,max
×
100
95.5598
=
17.6786
18.5
×
100
Assumptions
•
γd,max is the peak of the lab compaction (Proctor) curve, reached at the optimum water content.
•
No point can lie above the zero-air-voids curve, where water fills the voids and no air is left.
•
Relative compaction RC is the field γd over γd,max, in %; γw = 9.81 kN/m³.
Step-by-step
WE KNOW
γ = 19.8 kN/m³, w = 12%, γd,max = 18.5 kN/m³, Gₛ = 2.68
FIND
dry unit weight (γd), zero-air-voids unit weight (γzav), relative compaction (RC)
Step 1 · Find dry unit weight (γd)
Use γd = γ ÷ (1 + w ÷ 100)
Take
the
water
out:
the
moist
soil
weighs
1
+
w
times
its
solids.
γd
=
19.8
1 + 12 ÷ 100
γd
=
19.8
1 + 0.12
γd
=
19.8
1.12
γd
=
17.67857
kN/
m
3
Step 2 · Find zero-air-voids unit weight (γzav)
Use γzav = Gₛ × 9.81 ÷ (1 + w ÷ 100 × Gₛ)
Saturated,
the
void
ratio
is
wGₛ,
so
γd
=
Gₛγw
1 + e
becomes
this.
γzav
=
2.68
×
9.81
1 + 12 ÷ 100 × 2.68
γzav
=
2.68
×
9.81
1 + 0.12 × 2.68
γzav
=
2.68
×
9.81
1 + 0.3216
γzav
=
2.68
×
9.81
1.3216
γzav
=
26.2908
1.3216
γzav
=
19.8932
kN/
m
3
Step 3 · Find relative compaction (RC)
Use RC = γd ÷ γd,max × 100
The field’s dry unit weight as a share of the lab’s best, in %.
RC
=
17.67857
18.5
×
100
RC = 0.9556 × 100
RC = 95.5598%
Check
Put the numbers back in. Both sides should match.
17.67857
=
19.8
1 + 12 ÷ 100
✓
19.8932
=
2.68
×
9.81
1 + 12 ÷ 100 × 2.68
✓
95.5598
=
17.67857
18.5
×
100
✓
Topic check
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