Hillslope morphology
Ch 8
Hillslope Types
• Soil-mantled
•
– Transport-limited
– Shape: smooth, rounded
Bare bedrock or thin soils
– Weathering-limited
– Shape: steep, irregular, jagged
= f(weathering mode, relief)
lithology, climate
structure
of Hillslope Evolution
Hillslope Process
• The transportation of regolith drives hillslope
shape
– Regolith: generated by the weathering of bedrock
• Transportation is a function of
– Slope
– Position on the hillslope (how far from crest)
– Boundary condition at the toe
• Continuous vs Episodic
• Climate
Continuous and Episodic Processes
• Creep (gravity - continuous)
• Landslides/Rockfall/Debris Flows (gravity episodic)
• Solifluction/gelifluction (gravity –
continuous)
• Gophers (biological – continuous)
• Tree throw/downfall (biological - episodic)
Model of hillslope evolution (Kirkby, 1985)
sequence of change on a fault scarp, Nevada
(Wallace, 1977)
Why are hillslopes convex?
Regolith produced between D and A is transported by creep
All regolith produced between D and B must be transported past B
If this motion is linearly related to slope,
the resulting hillslope form is convex and parabolic
The amount of regolith that must be passed increases linearly
with distance from the crest
So slope must increase linearly as well
Mass flux, Qx
Regolith thickness, R
Weathering rate, W dot
The rate of change of regolith mass = rate of regolith inputs – rate of regolith outputs
units of mass per unit time
Mass of regolith = density x volume
= r x Rdxdy
INPUTS
=
Weathering rate (production) +
Mass flux from the x and y directions
OUTPUTS
- Mass flux out of the x and y directions
Conservation of mass on a hillslope
in 1 dimension
mass flux downslope
Weathering rate
Regolith thickness as a function of time
(x-direction)
R r r . 1 Qx
 W
r b x
t r b
In steady state,the timederivative 0
dQx
rr W 
dx
.
What drives the rate of
regolith production?
What processes are involved
in moving regolith downslope?
.
Qx  r r W x
Conservation of mass
on a hillslope in 1d
R r r . 1 Qx
 W
t rb
rb x
If Qx is a function of distance and slope,
z
Q  kx
x
n
m
.
Q  rr W x
When diffusion dominates, m = 0, then
 

.
  k z
x
rr W 
x


R  r r  .
2z
   W   2
t  r b 
x
R 1

t r b
  kr
b
Landscape diffusivity
constant

z
Q  k
x


Diffusion equation
for a hillslope
In words, the rate of change
of regolith thickness
is a function of
the rate of change of slope
(curvature)
Diffusion and hillslopes
• Landscape dominated by diffusive erosion
processes have rounded edges
– Diffusion attacks sharp corners
• Diffusion combines
– Conservation of mass
– How mass flux responds to driving variables
Diffusion smooths topography
• In regions of high curvature, regolith will
either be thickened or thinned the fastest
– Creating smooth slopes
• Only works because we chose a transport
rule where mass flux (Qx) was a function
of slope only
– Flowing water is not a diffusive process
The shape of a parabola
• If you solve the
diffusion eqn, you get
 
.
  k z .
R 1 
x

r
W

r
t2 r b 
x

r
2
R  r r  .
 z
2  W   2
t  r b 
x

d z
r W 

dx
rb
  kr
b
Integrating twice results in the eqn for a parabola
This hillslope shape must arise if regolith is produced everywhere
at the same rate.
This is why hillslopes are convex.
diffusive (slope dependent) transport
leads to convex hillslopes
advective transport depends on
slope and water flow
transports sediment
creates valleys
Diffusive Roundup
•
•
•
•
Creep
Frost heave
Solifluction
Rain splash
They all round the landscape
They all have a linear dependence
on local slope
z
Q  k
x
Hillslope Question 1
d z
rr W

2
dx
rb
2
• On a hillslope with a weathering
rate of 20 microns/yr and an
effective hillslope diffusivity of
0.02 m2/yr, what is the expected
curvature of the hillslope? The
channels at the edge of the
hillslope are 200 m from the
hillcrest.
.
d2z
rr W
2700 2x105


 0.00135m1
2
2
dx
rb 
2000 2x10
–
–
–
–
–
–
What are we given?
L = 200m.
=0.02 m2/yr,
W-dot = 20(10-6)m/yr
Density of rock = ~2700 kg/m3
Bulk density of regolith =~ 2000
kg/m3
– Eqn for curvature
What is the hillslope profile if the summit of the hillslope,
at x=0, is 1000 m.?
rr W 2
z
x  c2
rb 2
z(x  0)  1000
1000  c2
rr W 2
z  1000 
x
rb 2
Hillslope Profile
1005
Elevation (m.)
1000
995
990
985
980
975
970
-300
-200
-100
0
100
Horiz. Distance (m.)
200
300
Non-diffusive Hillslope Processes
aka Mass wasting
• Landslides
–
–
–
–
–
–
Fast or slow
Wet or dry
Set the location of channel heads
Define the tips of the channel network
Stochastic
A threshold system
• Elements of a landslide
– Slope, material properties, degree of saturation,
presence/absence of trees/vegetation
also flowing water (advective)
Analysis of slope stability
Case 1: Forces acting on a particle resting
on a slope surface
For a single boulder,
Resisting force (resists downslope motion) = Wcos
Driving force (promotes downslope motion) = Wsin
Wx  W sin 

F
m a rhdxdyg


sin   rgh sin 
A dxdy
dxdy
debris overlying
a slide(failure) plane
Analysis of slope stability
• What forces are acting on
•
the slab at time of failure?
All forces generated are
weight
– exactly balanced, no accel.
F  m g  r b hdxdy
Fd  r b (hdyL) g sin 
Fd  r b (hdydxcos ) g sin 
Fr  Fn 
Fr  r b (hdydxcos ) g cos t an
slab of regolith over area, A
For a block of soil,
Pressure acting perpendicular to plane = normal stress
Pressure acting in the downslope direction = shear stress
shear
normal
drivingstresses  resistingstresses
r b gh sin  cos  r b gh cos cos tan
sin 
 tan , angle of internalfriction
cos
at failure, internal friction = slope angle
Selby, 1993
Shear Strength, S
• Properties of material that resist gravity forces
• Driving force = resisting force
– Resisting = Strength + Friction
– S = c + σ’(tan)
Coulomb Equation
• Strength = cohesion + effective normal stress times angle of
internal friction
• Cohesion varies with material and condition
– Quartzite, clay, wet sand, dry sand…
• Effective normal stress varies with saturation
– Dry, moist(?), saturated
'

 rb gh  rw gd
– Normal stress-water pressure
  varies with material from 90° (quartzite) to ~33° (granular
materials) to <10° (clay).
Factor of Safety
• FS = Resisting/Driving
= Shear Strength/Shear Stress
– If strength exceeds stress (FS>>1), the slope
is “stable”.
– If strength ≈ stress (FS>1), the slope is
metastable.
– If stress minutely exceeds strength (FS≤1),
the slope is unstable and failure occurs.
drivingstresses resistingstresses
rb ghsin  cos  rb gh cos cos tan DRY
WET
rb ghsin  cos  rb gh cos cos  r w gd  tan  C
resisting rb gh cos cos  r w gd  tan  C
FS 

driving
rb ghsin  cos
rb  r g (1  )  r w
Slope stability question 1
• Consider a long linear hillslope with a slope angle of 25°.
•
It has soil that is 1 m thick, and a shear test on the soil
reveals its internal angle of friction is 31°, a cohesion
(including roots) of 5000 Pa, and a porosity of 35%.
Perform a stability analysis on this slope. Is it stable?
r gh cos cos  r gz tan( )  C '

F 
b
w
resisting
cos
 cos
 
r
 )) 
C
''
tan(
s r
b gh
w gh
r
gh
cos
cos

r
gh
tan(

C

driving
F

b
w
r
ghsin

cos

b
Fss 
r
rbb gh
gh sin
sin 
 cos
cos

2
g  9.8m / s 2
g  9.8m / s
r
 1000kg / m 33
w
rw  1000kg / m
r
 (1   )rs  rw  (0.65)2700  (0.35)1000  2105kg / m 33
b
rb  (1   )rs  rw  (0.65)2700  (0.35)1000  2105kg / m
  31
25o
  31
h  1m
h  1m
2105(9.8)(1)(0.906)(0.906)  (1000)(9.8)(1)0.6  5000 4279  5000
Fs  2105(9.8)(1)(0.906)(0.906)  (1000)(9.8)(1)0.6  5000  4279  5000  1.18
Fs 

 1.18
2105(9.8)(1)(0.422)(0.906)
7887
2105(9.8)(1)(0.422)(0.906)
7887