Homework No. 2 2.7. w/c = 16% 2.12. e = 0.6 2.13. w/c = 24%, T = 110.7 pcf, d = 89.3 pcf, e = 0.842 2.17. VTNeeded = 124,277 m3 2.19. T = 78.4 pcf, w/c = 65.4%, d = 0.797 g/cm3 or 0.797 Mg/m3, d = 7.81 kN/m3 2.35. e = 1.4, = 0.583, T = 1694 kg/m3 or 1.694 g/cm3, d = 1114 kg/m3 or 1.114 g/cm3, = 694 kg/m3

Homework No. 3 1. a.

Graph Curve Symbol Group Name 2 2 OH Organic Silt 3 1 SC Clayey Sand 4 3 SP-SM Poorly Graded Silty

Sand 5 1 GW Well-Graded Gravel 6 1 ML Silt w/ sand

b. A-4: Fair to Poor 2. Grain size for grain size distribution curves is plotted on log-scale because of the wide range in particle size. 3. a. PI = 30%, LI = -0.1% b. Group Symbol: CH

Homework No. 4 1. a.

Borrow Site Volume Needed (cu. yd) 1 112,853 2 108,485 3 100,676

b.

Borrow Site Estimated Cost ($) 1 1,184,857 2 1,301,820 3 1,107,436

c. T = 110 pcf 2. 7.4.)

k m/s 1.51E-6 cm/s 1.51E-4 Ft/d 4.28E-1

Soil Type: Very Fine sands, organic and inorganic silts, mixtures of sand, silt, and clay, glacial till, stratified clay deposits, etc 7.16.)

Case Datum Location

Enter Exit Point A

1 10 m from top hp = 4 m he = 6 m ht = 10 m

hp = -2 m he = 2 m ht = 0 m

hp = 0.5 m he = 3 m

ht = 2.5 m 2 9 m from top hp = 9 m

he = 0 m ht = 9 m

hp = 1 m he = 4 m ht = 5 m

hp = 7 m he = 1 m ht = 8 m

3 9 m from top (through center

of soil)

hp = 9 m he = 0 m ht = 9 m

hp = 5 m he = 0 m ht = 5 m

hp = 6 m he = 0 m ht = 6 m

Most Economical

Homework No. 5

Flow Net Rules

a. any surface of constant head (e.g., bottom of a flat-bottomed reservoir) is by definition an equipotential, and flow lines must meet it at right angles.

b. since flow cannot cross impermeable boundaries, the flow at such a boundary must be parallel to it, i.e., impermeable boundaries are flow lines, and equipotentials must meet them at right angles.

c. the water table is, by definition, the surface where P = 0; it can thus be an equipotential only if it is horizontal. At any point on the water table (no matter whether it is flat or sloping) h = z, where z is the elevation of the water table above the datum. If there is no seepage percolating down to the water table, it can be considered a flow line. In the general case however (sloping water table, seepage across it), the water table is neither a flow line nor an equipotential, and flow lines will intersect it at an angle.

1. a. q = 492.5 ft3/day b.

Point Uplift Pressures along base of dam

(psf) 1 5179 2 4618 3 4056 4 3494 5 2933 6 2371 7 1810

Resultant 419,340 lb/ft of wall or 3495 lb

2.

Point Uplift Pressures along base of dam

(psf) 1 4056 2 3588 3 3120 4 2652 5 2184 6 1719

Resultant 346,320 lb/ft of wall or 2886 lb

The resultant of the uplift pressures for the alternative design is less than the design in problem 1. Problem 2 has a smaller resultant force due to the additional head loss through the soil.

Homework No. 6 6.28.)

Depth (ft) Total Stress (psf) Pore Pressure (psf) Effective Stress (psf) 0 2000 0 2000 12 3392 0 3392 25 4939 811.2 4238 38 6486 1622 4864 48 7696 2246 5450

6.29.) a.)

Depth (ft) Total Stress (psf) Pore Pressure (psf) Effective Stress (psf) 0 0 0 0 5 580 0 580 10 1175 312 863 40 4745 2184 2561

b.) Water table rises 4 ft above ground surface in spring @ 25 ft: Total Stress = 3210 psf Pore Pressure = 1810 psf Effective Stress = 1400 psf 6.32.)

Depth (m) Total Stress (kPa) Pore Pressure (kPa) Effective Stress (kPa)

0 0 0 0 3 54 0 54 7 133 39 94 12 227 88 139 16 302 127 175

6.35.) a.) k= 0.45

Depth (m) oh' (kPa) oh (kPa) 0 0 0 3 24.3 24.3 7 42.3 81.3 12 62.55 151 16 79 206

b.) k= 1.6

Depth (m) oh' (kPa) oh (kPa) 0 0 0 3 86.4 86.4 7 150 189 12 222 310 16 280 407

Homework No. 7 1. a.)

Preconsolidation Stress (psf): 4900 to 5050

b.) Sc = 0.384 ft to 0.400 ft 2. a.) Sc = 0.202 m b.) Sc = 0.0202 m

0.5 0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 1

10 100 1000 10000 100000

e

Vertical Effective Stress (psf)