SME
Soil Mechanics · Problem type
Effective stress at a depth
Use this for “Dry sand 3 m thick (γ = 18 kN/m³) lies on clay with γₛₐₜ = 19 kN/m³, the water table on the clay. Find σ, u and σ′ at 5 m.”
Diagram
At z = 5 m.
σ′ = σ − u = 92 − 19.6 = 72.4 kPa.
u = γw(z − zw) = 9.81 × 2
u = 19.6 kPa
That holds below the water table; above it u = 0.
This is an example. Type your own values to start a new problem.
σ (kPa)
=
γ (kN/m³)
×
zw (m)
+
γₛₐₜ (kN/m³)
×
(
z (m)
−
zw (m)
)
u (kPa)
=
9.81
kN/m³
×
(
z (m)
−
zw (m)
)
σ′ (kPa)
=
σ (kPa)
−
u (kPa)
Formulas
σ
=
γ
×
zw
+
γₛₐₜ
×
(
z
−
zw)
92 = 18 × 3 + 19 × (5 − 3)
u
=
9.81
×
(
z
−
zw)
19.62 = 9.81 × (5 − 3)
σ
′
=
σ
−
u
72.38 = 92 − 19.62
Assumptions
•
The water table is at the top of the clay: dry sand above it (γ), saturated clay below (γₛₐₜ).
•
Still pore water: u = γw × the depth below the water table, with γw = 9.81 kN/m³.
•
Effective stress σ′, carried from grain to grain, is the total stress σ less the pore pressure u.
Step-by-step
WE KNOW
zw = 3 m, γ = 18 kN/m³, γₛₐₜ = 19 kN/m³, z = 5 m
FIND
total stress (σ), pore water pressure (u), effective stress (σ′)
Step 1 · Find total stress (σ)
Use σ = γ × zw + γₛₐₜ × (z − zw)
Add the weight of each layer above the point: its unit weight times its thickness.
σ
=
18
×
3
+
19
×
(5
−
3)
σ
=
18
×
3
+
19
×
2
σ
=
54
+
38
σ
=
92
kPa
Step 2 · Find pore water pressure (u)
Use u = 9.81 × (z − zw)
Water pressure grows with the depth below the water table.
u
=
9.81
×
(5
−
3)
u
=
9.81
×
2
u
=
19.62
kPa
Step 3 · Find effective stress (σ′)
Use σ′ = σ − u
The grains carry what the water does not.
σ
′
=
92
−
19.62
σ
′
=
72.38
kPa
Check
Put the numbers back in. Both sides should match.
92 = 18 × 3 + 19 × (5 − 3) ✓
19.62 = 9.81 × (5 − 3) ✓
72.38 = 92 − 19.62 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Related lessons
Consolidation
Main lesson
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