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 g_m = 122 lb/ft₃, w = 14.7%, and G_s = 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

s_A = 0, u_A = 0, s’ = 0

Point B

s_B = s’ =

U_B = 0

Point C

s_C =856.69 +

= 1386.25 lb/ft²

U_C = g_w*h = 62.4*4 = 249.6 lb/ft²

s_C’ = s_C – u_C = 1386.25-249.6 = 1136.65 lb/ft²

Point D

s_C =1386.25 +

= 3156.25 lb/ft²

U_C = g_w*h = 62.4*19 = 1185.6 lb/ft²

s_C’ = s_C – u_C = 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, s_{1 (failure)} = 32 lb/in²

u_f = 5.5 lb/in²

C_cu for normally consolidated clay » 0

s₁ = s₃ tan² (45+f_cu/2)

32 = 13 tan² (45+f_cu/2)

f_cu = 25^o

s_1’ = s₁’ – u_f = 32 – 5.5 = 26.5 lb/in²

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

s₁’ = s₃’ tan² (45+f/2)

26.5 = 7.5 tan² (45+f/2)

f = 34^o