SME
Physical Geology
Surface processes
Use this for “A soil slope of 30° has a slip surface 2 m down, c = 5 kPa, φ = 35°, γ = 20 kN/m³. What is its factor of safety?”
Diagram
σ = γz cos²θ = 20 × 2 × cos²30°
σ = 30 kPa
τ = γz sin θ cos θ
τ = 20 × 2 × sin 30° × cos 30°
τ = 17.32 kPa
s = c + σ tan φ = 5 + 30 × tan 35°
s = 26.01 kPa
FS
=
s
τ
=
26.01
17.32
=
1.501
FS ≥ 1: the strength holds the driving stress, so the slab stays.
The slab is long next to its depth; the depth is not drawn to scale.
This is an example. Type your own values to start a new problem.
θ
Slope angle
example · °
z
Depth to slip surface
example
c
Cohesion
example · kPa
φ
Friction angle
example · °
γ
Unit weight
example · kN/m³
σ
Normal stress
calculated · kPa
τ
Driving shear stress
calculated · kPa
s
Shear strength
calculated · kPa
FS
Factor of safety
calculated
Formulas
σ
=
γ
×
z
×
cos
(
θ
°)
2
30
=
20
×
2
×
cos
(30°)
2
τ
=
γ
×
z
×
sin(
θ
°)
×
cos(
θ
°)
17.3 = 20 × 2 × sin(30°) × cos(30°)
s
=
c
+
σ
×
tan(
φ
°)
26 = 5 + 30 × tan(35°)
FS
=
s
τ
1.5
=
26
17.3
Assumptions
•
The slab is long next to its depth, so its ends don’t matter (an infinite slope).
•
The soil is dry: water in its pores would cut σ and so the strength.
•
FS below 1 means the slab slides; FS above 1 means it holds.
Step-by-step
WE KNOW
θ = 30°, z = 2 m, c = 5 kPa, φ = 35°, γ = 20 kN/m³
FIND
normal stress (σ), driving shear stress (τ), shear strength (s), factor of safety (FS)
Step 1 · Find normal stress (σ)
Use σ = γ × z × cos(θ°)²
The
column’
s
weight
γz
rests
on
a
base
tilted
by
θ
;
the
part
pressing
square
onto
it,
per
area
of
base,
is
γz
cos²
θ
.
σ
=
20
×
2
×
cos
(30°)
2
σ
=
20
×
2
×
0.866
2
σ
=
20
×
2
×
0.75
σ
=
40
×
0.75
σ
=
30
kPa
Step 2 · Find driving shear stress (τ)
Use τ = γ × z × sin(θ°) × cos(θ°)
The
weight’
s
part
along
the
slope,
spread
over
the
same
tilted
base.
τ
=
20
×
2
×
sin(30°)
×
cos(30°)
τ
=
20
×
2
×
0.5
×
0.866
τ
=
40
×
0.5
×
0.866
τ
=
20
×
0.866
τ
=
17.3
kPa
Step 3 · Find shear strength (s)
Use s = c + σ × tan(φ°)
Mohr–Coulomb: the cohesion plus the friction on the normal stress.
s
=
5
+
30
×
tan(35°)
s
=
5
+
30
×
0.7002
s
=
5
+
21.006
s
=
26
kPa
Step 4 · Find factor of safety (FS)
Use FS = s ÷ τ
How many times the strength covers the stress pushing the slab downhill.
FS
=
26
17.3
FS = 1.5
Check
Put the numbers back in. Both sides should match.
30
=
20
×
2
×
cos
(30°)
2
✓
17.3 = 20 × 2 × sin(30°) × cos(30°) ✓
26 = 5 + 30 × tan(35°) ✓
1.5
=
26
17.3
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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