Question

Problem 3: Soil Classification using the AASHTO and USCS Systems

(you will need to use the charts provided within the classification lecture)
A. Classify the soil according to the USCS classification system:
%passing sieve No. 4 - 86%,
%passing sieve No. 200 - 12%
PL =26%, PI = 10%
Dio (mm) = 0.1, Duo (mm) = 0.32, D. (mm) 0.9
B. Classify the soil samples A, B, and C below using the AASHTO and USCS Systems.
Soil Sample.% Passing
Sieve No.
B
10
68.5
79.5
20
40
36.1
69.0
69.3
59,1
48.3
38.5
28.4
19.8
4.5
60
100
200
LL
21.9
34.1
16.5
54.3
53.5
31.6
Non-plastic (NP)
PL
1​

Answer 1

`Given\\ \(\quad W=3000 Ib , \quad m=(W)/(g)=(3000)/(322) \ slug =93.1677 slug\)\\K_(e q)=2160 lbs / wp =2100 ( lbs )/(10) ( x 12)/(1 ft )=(2160 * 12) lb / ft$$`

a) The natural frequency

`\begin{aligned}&\left(\omega_(n)\right)=\sqrt{(K_(e q))/(m)}\\&=\sqrt{(2160 * 12)/(93.1677)}\\&\omega_(n)=16.68 \text { rad } | s\\&\omega_(n)=(2 \pi)/(T)\\&16.68=(2 \pi)/(T)\\&T=0.3767 s\end{aligned}`

b)

`Given, \(t=10 s , \quad y(t)=6 in = A\)\\\(y(t)=A \cos \left(\omega_(n) t+\phi\right) \rightarrow 0\)\\\(6=6 \cos (16.68 * 10+\phi)\)\\\(1=\cos (166.8+\phi)\)\\\(166.8+\phi=0\)\\\phi=-166.8\)\\At \(t=0, \quad y(0)=6 \cos (16.68 * 0-166.8)\) {y(0)}=-5.74 in`