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2 years ago

Compute the solution to x + 2x + 2x = 0 for Xo = 0 mm, vo = 1 mm/s and write down the closed-form expression for the response.

Engineering

1 answer:

Nutka1998 [239]2 years ago

6 0

Answer:

β = $\frac{c}{\sqrt{km} }$ = 0.7071 ≈ 1 ( damping condition )

closed-form expression for the response is attached below

Explanation:

Given : x + 2x + 2x = 0 for Xo = 0 mm and Vo = 1 mm/s

computing a solution :

M = 1,

c = 2,

k = 2,

Wn = $\sqrt{\frac{k}{m} }$ = $\sqrt{2}$

next we determine the damping condition using the damping formula

β = $\frac{c}{\sqrt{km} }$ = 0.7071 ≈ 1

from the condition above it can be said that the damping condition indicates underdamping

attached below is the closed form expression for the response

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Answer:

The correct option is;

1) Test the model and analyze the results of the test

Explanation:

Based on the flowchart, a model improvement process involves the implementation of a process or model improvement cycle such as the Plan Do Study Act, PDSA cycle, however feedback to the process will be be gotten from testing the model and analyzing the results of the tests. When grey areas or aspects of the model are found that cause the system to malfunction are determined, steps should then be taken to improve the performance of the model.

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3 years ago

A cast-iron tube is used to support a compressive load. Knowing that E 5 10 3 106 psi and that the maximum allowable change in l

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Answer:

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Explanation:

(a)

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Making $\sigma$ the subject then

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$\sigma=(10\times 10^{6} psi)\times 0.00025=2500 si= 2.5 ksi$

(b)

Stress, $\sigma= \frac {F}{A}$ making A the subject then

$A=\frac {F}{\sigma}$

$A=\frac {\pi(d_o^{2}-d_i^{2})}{4}$

where d is the diameter and subscripts o and i denote outer and inner respectively.

We know that $2t=d_o - d_i$ where t is thickness

Now substituting

$\frac {\pi(d_o^{2}-d_i^{2})}{4}=\frac {1600}{2500}$

$\pi(d_o^{2}-d_i^{2})=\frac {1600}{2500}\times 4$

$(d_o^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4$

But the outer diameter is given as 2 in hence

$(2^{2}-d_i^{2})=\frac {1600}{2500\times \pi}\times 4$

$2^{2}-(\frac {1600}{2500\times \pi}\times 4)=d_i^{2}$

$d_i=\sqrt {2^{2}-(\frac {1600}{2500\times \pi}\times 4)}=1.784692324 in\approx 1.785 in$

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$t=0.5(2-1.785)=0.1075 in$

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3 years ago

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A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur

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Question in order:

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0.603 225

0.203 368

0.162 442

Answer:

191 MPa

Explanation:

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