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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​

Problem 3 Soil Classification using the AASHTO and USCS Systemsyou will need to use the charts provided within the classification lectureA Classify the soil acc class=

Respuesta :

Solution:

[tex]Given\\ \(\quad W=3000 Ib , \quad m=\frac{W}{g}=\frac{3000}{322} \ slug =93.1677 slug\)\\K_{e q}=2160 lbs / wp =2100 \frac{ lbs }{10} \frac{ x 12}{1 ft }=(2160 \times 12) lb / ft$[/tex]

a) The natural frequency

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

b)

[tex]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 \times 10+\phi)\)\\\(1=\cos (166.8+\phi)\)\\\(166.8+\phi=0\)\\\phi=-166.8\)\\At \(t=0, \quad y(0)=6 \cos (16.68 \times 0-166.8)\) {y(0)}=-5.74 in[/tex]

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