Notes on Homework #2 (2012) GEOL 4880

1 a) bulk density is the dry density plus the water mass: r_soil + n r_water = r_bulk, where n is the porosity (assuming all the porosity is filled).

1600 + .4*1000 = 2000

b) water pressure P_water = r_water * 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: s_n = r_bulk * g * h {note: recognizing that cos a is 1 since a is 0} = 40,000Pa

effective normal stress: s_eff = s_n - P_water = 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 = P_w/( r_water * g) + z = z, and the head at the bed: H = P_w/( r_water * g) + 0 = z, which gives a head difference of 0.

2 a) P_water = r_water * g * d * cos a , where d is the slope normal depth (shallow soil approx).

(1000 * 10 * 1 * cos a ) = 8,600Pa

b) t_s or s_driving = r_bulk * g * d * sin a ; (r_bulk see 1a above)

(1600 * 10 * 1 + 0.4 * 1000 * 10 * 1) * sin a = 10,000Pa

c) s_resisting = ( r_bulk * g * d * cos a - P_water ) * tan q

((1600 * 10 * 1 + 0.4 * 1000 * 10 * 1) * .866 – 8,600 ) * .65 = 5,600Pa

d) therefore the F.S. for the wet slide is:

F.S. = 5,600/10,000 = .56 ; Failure

e) As compared to the dry slide: (as above but use dry density and 0 water pressure)

F.S. = [(1600 *10 * 1* .86 ) * 0.65]/(1600 * 10 * 1 * .5) = 1.12 ; Stable

f) The hydraulic gradient is just the slope, so:

q = K_sat * d * tan a = 0.003m²/sec

g) Apply a cosine correction, vertical depth = slope perpendicular depth/cos a = 1/.86 = 1.2m

3 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.

4 Discuss in class

5 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 ½ and 1). The slope dependence of the sliding speed is given by:

V = V_above * (2j - 1) ,

where V_above is the sliding speed from part a.