SME
Soil Mechanics · Problem type
At-rest pressure
Use this for “A basement wall 6 m high holds back sand with φ′ = 32°, OCR = 2 and γ = 18.5 kN/m³. Find K₀ and the at-rest thrust.”
Chart
σ′ₕ(6) ≈ 75.34
Shaded: ∫ from 0 to 6 of σ′ₕ(z) dz = 226
This is an example. Type your own values to start a new problem.
K₀
=
(1
−
sin(
φ′
°))
×
OCR
sin(
°)
σ′₀ (kPa)
=
K₀
×
γ (kN/m³)
×
H (m)
P₀ (kN/m)
=
½
×
σ′₀ (kPa)
×
H (m)
Formulas
K₀
=
(1
−
sin(φ′))
×
OC
R
sin(φ′)
0.6787
=
(1
−
sin(32°))
×
2
sin(32°)
σ′₀
=
K₀
×
γ
×
H
75.3385 = 0.6787 × 18.5 × 6
P₀
=
1
2
×
σ′₀
×
H
226.0155
=
1
2
×
75.3385
×
6
Assumptions
•
The wall does not move, like a basement wall or a braced wall.
•
Jaky’s K₀ = 1 − sin φ′ for a normally consolidated soil (OCR = 1), times OCR to the power sin φ′.
•
A level, dry backfill; P₀ is per metre of wall.
Step-by-step
WE KNOW
φ′ = 32°, OCR = 2, γ = 18.5 kN/m³, H = 6 m
FIND
at-rest coefficient (K₀), pressure at the base (σ′₀), at-rest thrust (P₀)
Step 1 · Find at-rest coefficient (K₀)
Use K₀ = (1 − sin(φ′)) × OCRsin(φ′)
Jaky’s 1 − sin φ′ for the soil as first laid down, raised by the overconsolidation (Mayne and Kulhawy).
K₀
=
(1
−
sin(32°))
×
2
sin(32°)
K₀
=
(1
−
0.5299)
×
2
0.5299
K₀
=
0.4701
×
1.4438
K₀
=
0.678725
Step 2 · Find pressure at the base (σ′₀)
Use σ′₀ = K₀ × γ × H
The vertical stress γz at the base, times the coefficient, pushes sideways on the wall.
σ′₀ = 0.678725 × 18.5 × 6
σ′₀ = 12.556 × 6
σ′₀ = 75.3385 kPa
Step 3 · Find at-rest thrust (P₀)
Use P₀ = ½ × σ′₀ × H
The pressure grows from 0 at the top to its base value: the thrust is the triangle’s area.
P₀
=
1
2
×
75.3385
×
6
P₀
=
37.67
×
6
P₀
=
226.0155
kN/m
Check
Put the numbers back in. Both sides should match.
0.678725
=
(1
−
sin(32°))
×
2
sin(32°)
✓
75.3385 = 0.678725 × 18.5 × 6 ✓
226.0155
=
1
2
×
75.3385
×
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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Lateral earth pressure
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