HOMEWORK SOLUTION

HW #1

CE 455

FOUNDATION AND EARTH STRUCTURE

(Question no. 1.4, 1.8, 1.12, 1.16, & 1.27 ..Braja M. Das textbook)

1.4 gm = 122 lb/ft3, w = 14.7%, and Gs = 2.68

a. Void ratio

b. Porosity

c. Degree of saturation

d. Dry unit weight

1.8 Soil B; Passing #4 = 100%, passing #10 = 80%, passing #40 = 61%, passing #200 = 55%, LL = 38, PL = 25.

Classification according  to USCS (group symbol and group name)

R4 = 0, R200 =100-55 = 45%, PI = 38-25 = 13 from table 1.9 the soil is Fine-grained (inorganic) and group symbol is CL

Sand fraction = R200-R4 = 45% ³ 15%

Gravel fraction = R4 = 0% < 15%

Sand fraction/gravel fraction = ~ > 1

From table 1.11 the group name is Sandy Lean Clay

 A

1.12

Determine the total stress (s), pore water pressure (u) and effective stress (s’)

Point A

sA = 0, uA = 0, s’ = 0

Point B

sB = s’ =

UB = 0

Point C

sC =856.69 +

= 1386.25 lb/ft2

UC = gw*h = 62.4*4 = 249.6 lb/ft2

sC’ = sC uC = 1386.25-249.6 = 1136.65 lb/ft2

Point D

sC =1386.25 +

= 3156.25 lb/ft2

UC = gw*h = 62.4*19 = 1185.6 lb/ft2

sC’ = sC uC = 3156.25-1185.6 = 1970.65 lb/ft2

1.16 Normally consolidated clay; thickness (Hc) = 3.7 m, e = 0.82, LL = 42,

Po = 110 kN/m2 and (Po+DP) = 155 kN/m2

Cc = 0.009*(LL-10)=0.009*32 = 0.288

Settlement =

1.27 Saturated normally consolidated clay; s3 = 13 lb/in2, s1 (failure) = 32 lb/in2

uf = 5.5 lb/in2

Ccu for normally consolidated clay » 0

s1 = s3 tan2 (45+fcu/2)

32 = 13 tan2 (45+fcu/2)

fcu = 25o

s1’ = s1’ – uf = 32 – 5.5 = 26.5 lb/in2

s3’ = s3’ – uf = 13 – 5.5 = 7.5 lb/in2

s1’ = s3’ tan2 (45+f/2)

26.5 = 7.5 tan2 (45+f/2)

f = 34o