SME
Soil Mechanics · Problem type
Settlement of an overconsolidated clay
Use this for “The 4 m clay (Cc = 0.3, e₀ = 0.9, σ′₀ = 80 kPa) once carried σ′ₚ = 120 kPa; Cₛ = 0.05. How far does it settle under Δσ = 60 kPa?”
Diagram
S
S
=
H
1 + e₀
×
(Cₛ
log(
σ′ₚ
σ′₀
)
+
Cc
log(
σ′₀ + Δσ
σ′ₚ
))
S
=
4
1.9
×
(0.05
log(
120
80
)
+
0.3
log(
140
120
))
S = 0.06082 m
The dashed line is the surface after settling, drawn to the same scale as H.
This is an example. Type your own values to start a new problem.
H
Clay thickness
example · m
Cc
Compression index
example
e₀
Initial void ratio
example
σ′₀
Effective stress now
example · kPa
Δσ
Added stress
example · kPa
S
Settlement
calculated · m
Cₛ
Swell index
example
σ′ₚ
Preconsolidation stress
example · kPa
Formulas
S
=
H
1 +
e₀
×
(Cₛ
×
log(
σ′ₚ
σ′₀
)
+
Cc
×
log(
σ′₀ + Δσ
σ′ₚ
))
0.06082
=
4
1 + 0.9
×
(0.05
×
log(
120
80
)
+
0.3
×
log(
80 + 60
120
))
Assumptions
•
Overconsolidated: the clay once carried σ′ₚ, more than today’s σ′₀.
•
Up to σ′ₚ it recompresses along the swell index Cₛ; past σ′ₚ, along Cc.
•
σ′₀ and Δσ are taken at the layer’s middle; one-dimensional compression.
Step-by-step
WE KNOW
H = 4 m, Cc = 0.3, e₀ = 0.9, σ′₀ = 80 kPa, Δσ = 60 kPa, Cₛ = 0.05, σ′ₚ = 120 kPa
FIND
settlement (S)
Step 1 · Find settlement (S)
Use S = H ÷ (1 + e₀) × (Cₛ × log(σ′ₚ ÷ σ′₀) + Cc × log((σ′₀ + Δσ) ÷ σ′ₚ))
Recompress along Cₛ up to σ′ₚ, then compress along Cc from σ′ₚ to σ′₀ + Δσ.
80 + 60 > 120: past σ′ₚ
S
=
4
1 + 0.9
×
(0.05
×
log(
120
80
)
+
0.3
×
log(
80 + 60
120
))
S
=
2.105
×
(0.05
×
0.1761
+
0.3
×
0.06695)
S
=
2.105
×
(0.008805
+
0.02008)
S
=
2.105
×
0.02889
S
=
0.06082
m
Check
Put the numbers back in. Both sides should match.
0.06082
=
4
1 + 0.9
×
(0.05
×
log(
120
80
)
+
0.3
×
log(
80 + 60
120
))
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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