SME
Soil Mechanics · Problem type
Active thrust with the backfill under water
Use this for “A 6 m wall holds back saturated sand (φ′ = 32°, γₛₐₜ = 20 kN/m³) with water up to its top. Find the total thrust.”
Chart
σₕ(6) ≈ 77.65
Shaded: ∫ from 0 to 6 of σₕ(z) dz = 232.9
This is an example. Type your own values to start a new problem.
Kₐ
=
(1
−
sin(
φ′
°))
÷
(1
+
sin(
φ′
°))
σ′ₐ (kPa)
=
Kₐ
×
γ′ (kN/m³)
×
H (m)
P (kN/m)
=
½
×
(
σ′ₐ (kPa)
+
u (kPa)
)
×
H (m)
γ′ = γₛₐₜ − 9.81 kN/m³ is the soil’s weight under water; σ′ₐ is the soil’s own pressure at the base and u the water’s.
γₛₐₜ
Saturated unit weight
example · kN/m³
Formulas
Kₐ
=
1 − sin(φ′)
1 + sin(φ′)
0.3073
=
1 − sin(32°)
1 + sin(32°)
γ′ = γₛₐₜ − 9.81
10.19 = 20 − 9.81
σ′ₐ
=
Kₐ
×
γ′
×
H
18.7858 = 0.3073 × 10.19 × 6
u
=
9.81
×
H
58.86 = 9.81 × 6
P
=
1
2
×
(σ′ₐ
+
u
)
×
H
232.9374
=
1
2
×
(18.7858
+
58.86)
×
6
Assumptions
•
The water table is at the surface of the backfill and the wall is not drained.
•
Kₐ acts on the effective stress only; water pushes equally every way (K = 1).
•
γw = 9.81 kN/m³; P is per metre of wall.
Step-by-step
WE KNOW
φ′ = 32°, γₛₐₜ = 20 kN/m³, H = 6 m
FIND
active coefficient (Kₐ), buoyant unit weight (γ′), effective pressure at the base (σ′ₐ), water pressure at the base (u), total thrust (P)
Step 1 · Find active coefficient (Kₐ)
Use Kₐ = (1 − sin(φ′)) ÷ (1 + sin(φ′))
Rankine’s
active
wedge:
tan²(45°
−
φ′
2
)
=
1 − sin φ′
1 + sin φ′
.
Kₐ
=
1 − sin(32°)
1 + sin(32°)
Kₐ
=
1 − 0.5299
1 + 0.5299
Kₐ
=
0.4701
1.5299
Kₐ = 0.307259
Step 2 · Find buoyant unit weight (γ′)
Use γ′ = γₛₐₜ − 9.81
Below the water table the soil is buoyed up: take off the unit weight of water.
γ′ = 20 − 9.81
γ′
=
10.19
kN/
m
3
Step 3 · Find effective pressure at the base (σ′ₐ)
Use σ′ₐ = Kₐ × γ′ × H
The vertical stress γz at the base, times the coefficient, pushes sideways on the wall.
σ′ₐ = 0.307259 × 10.19 × 6
σ′ₐ = 3.131 × 6
σ′ₐ = 18.7858 kPa
Step 4 · Find water pressure at the base (u)
Use u = 9.81 × H
The water pressure at the base of the wall, from the full height of water.
u
=
9.81
×
6
u
=
58.86
kPa
Step 5 · Find total thrust (P)
Use P = ½ × (σ′ₐ + u) × H
Two triangles, soil and water, each growing from 0 at the top: add their areas.
P
=
1
2
×
(18.7858
+
58.86)
×
6
P
=
1
2
×
77.6458
×
6
P
=
38.8229
×
6
P
=
232.9374
kN/m
Check
Put the numbers back in. Both sides should match.
0.307259
=
1 − sin(32°)
1 + sin(32°)
✓
10.19 = 20 − 9.81 ✓
18.7858 = 0.307259 × 10.19 × 6 ✓
58.86 = 9.81 × 6 ✓
232.9374
=
1
2
×
(18.7858
+
58.86)
×
6
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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