Notes on Homework #2 (2015) GEOL 4880
1
a) bulk density is the dry
density plus the water mass: rsoil
+ n rwater
= rbulk,
where n is the porosity (assuming all the porosity is filled).
1600
+ .4*1000 = 2000
b) water
pressure Pwater = rwater
* g * d , where d is the slope normal depth. (you
don’t have to use the “shallow-soil” approx. … you can just assume the water
isn’t flowing,
1000*10*2
= 20,000Pa
c) the
effective normal stress. :
total
normal stress: sn
= rbulk
* g * h
effective
normal stress: seff
= sn -
Pwater = 20,000Pa
d) (most
people got this incorrectly) It has to be zero, since no water is flowing
perpendicular to a line joining the surface to the bed. You also get the same result by calculating
the head at the surface (calling z = 0 at the bed for convenience): H = Pw/( rwater
* g) + z = z, and the head at the bed: H = Pw/( rwater
* g) + 0 = z, which gives a head
difference of 0.
2
Wading through the question, we realize that
the answer is the amount of rock needed to “tile”, or place a single layer of
rock, over a 1/10 of a square meter: i.e. put a layer of grains, 1 grain thick,
over 10% of a square meter. Fine silt is
about 0.01 mm in diameter, coarse sand is 1mm diameter
Simple solution: (ignoring
the fact that grains are more spherical than cubic)
Volume
of a silt layer,
0.01mm
thick * 0.1m2
= 0.00001 * 0.1 = 10-6m3 (or a mass of about 3gm)
Volume
of sand layer,
1mm
thick * 0.1m2 = 0.001 *0.1 =
10-4m3
(or a mass of about 0.3kg, or 300gm)
Therefore we need 100 times more sand to
make the water 10% opaque than we need silt.
3 Divide
the length L into two parts, the part under the slope (L1), and the part that
the slide mass will slide horizontally on (L2), also call the slope length down
the failure slope to be L3 (L3 = H/(sin a)
). L1 is, by trig, H/(tan
a). L2 is the length needed to get rid of the
K.E. of the slide. De-acceleration on L2
will be at -g* tan f,
so if we know the incoming K.E. (or velocity) we can use (incoming velocity)^2 = 2*a*L2 to calculate L2. The incoming velocity is given by 2*a*L3, where a is the acceleration down the
failure slope. And a is given by g*( sin a - cos a *
tan f. If you cancel all the terms you end up with
tan f = H/(L1+L2).
4 This
question is difficult , but if you understand the
forces involved with a block on a slope, and the effect of friction, then it is a really good example of a relevant
real world process.
a each day it expands and contracts by 1m x 10 degrees x
0.00001 per degree ~ 0.0001m/day
Either
the upper edge moves up, or the bottom edge moves down. Since the bottom edge will “fail” first, then this is how far the bottom edge moves
down slope each day, so the motion is
365
x 0.0001m/day ~ 0.036m/yr. Note the speed is not
twice this, even though it expands and contracts.
b this is independent of slope
e If it is touching all along the slope, then the points of
contact no longer locate the points that don’t slide. We have to solve for the points of no
motion. During expansion, some of the
upper half of the rock will move up hill while most will move downhill, on the
other hand during contraction, some of the lower end of the rock will contract
uphill, but most will contract down. The
problem can be solved by balancing forces, ie
balancing the downslope weight against friction forces. Because the block is lying at less than the
angle of friction, the friction forces generated by sliding the entire block in
one direction are more than the downslope weight. Therefore only part of the block slides down
while the over part slides up(!). It is a little lengthy to set up the algebra
to solve this but it is not difficult.
The solution gives a strong slope dependence to the speed.
For those that are interested, or those
that actually tried to do the problem:
the non-sliding point or static point in any half thermal cycle is given
as:
j
= ( tan a
+ tan f ) / (2*tan f ),
where a is
the slope angle and f
is the angle of friction, and j
is the fractional distance from the fastest sliding end of the block to the
non-sliding point (the static point is always closer to the slowest end,
therefore j
is always between ˝ and 1). The slope
dependence of the sliding speed is given by:
V = Vabove
* (2j - 1)
,
where
Vabove is the sliding speed from part a.