TSM 352 HOMEWORK ASSIGNMENT 01



TSM 352 HOMEWORK ASSIGNMENT 04 Solutions

Page 23 - 26 of text:

Questions 1, 5, 9, 10, 11, 12, 13

1. Clay

5. A: -0.7 + (-6.6) = -7.3 B: -3 + (-1) = -4

Since A has a lower energy, water moves from B to A.

9.

A) Pb = Ms/Vt

Volume = π (1.5 in)2(3) = 21.2 in3 = 0.0123ft3

Pb = 1.07lb/0.0123ft3 = 87.2lb/ft3

B) f = (Va+Vw) / Vt *Assume a particle density of 2.65 g/cm3

Vt – Vs = Va + Vw = 0.0123 ft3- (cm3/2.65g)(1.07lb)(453.6g/1lb)(3.53*10-5ft3/cm3) = 0.0058 ft3

f = 0.0058/0.0123 = 0.5

C) w = Mw/Ms = (1.32 – 1.07) / 1.07 = 0.23

D) θ = Vw/Vt = [0.25lb water (ft3/62.4lb)] / 0.0123 = 0.3

E) fa = Va/Vt = [0.0058 – (0.25/62.4)] / 0.0123 = 0.15

F) 0.15*( π0.252) = 0.0288 ft3 air in a 0.5 ft diameter, 1 ft long soil column.

Cross-sectional area of column = π0.252

0.0288 ft3 air / π0.252 = 0.15ft = 1.7”

10.

A)91 lb/ft3

B)0.45

C)0.098

D) 0.14

E) 0.31

F) 0.61 ft

11.

A) 1.6 g/cm3

B) 0.39

C) 0.6

D) 0.06

E)0.24

F) 0.15

12) See Excel sheet

13)

Find V of borrow material to fill in:

e = 0.75 = Va + Vw / Vs = 4285.7 yd3 / 5714.3 yd3

Therefore, 4285.7 yd3 of borrow material must be used to fill (assuming the fill contains no moisture).

For the borrow material:

e = 1.15 = 23/20 = 2292.4 yd3 / 1993.3 yd3

Ps = Ms/Vs = 150 lb/ yd3 = 4050 lb/ yd3 and w = Mw/Ms = 12%

Ms = (Ms/Vs)(Vs) = 4050lb/ yd3 * 1993.3 yd3 = 8072865 lbs = 4036 tons

Mw = (Mw/Ms)(Ms) = 0.12 * 8072865 lbs = 968744 lbs = 484 tons

................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download