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Sergeeva-Olga [200]
2 years ago
10

Problem 3: Soil Classification using the AASHTO and USCS Systems

Engineering
1 answer:
nataly862011 [7]2 years ago
5 0

<u>Solution:</u>

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$$

a) The natural frequency

\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}

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 \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

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