Answer:
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Answer:
i) 796.18 N/mm^2
ii) 1111.11 N/mm^2
Explanation:
Initial diameter ( D ) = 12 mm
Gage Length = 50 mm
maximum load ( P ) = 90 KN
Fractures at = 70 KN
minimum diameter at fracture = 10mm
<u>Calculate the engineering stress at Maximum load and the True fracture stress</u>
<em>i) Engineering stress at maximum load = P/ A </em>
= P /
= 90 * 10^3 / ( 3.14 * 12^2 ) / 4
= 90,000 / 113.04 = 796.18 N/mm^2
<em>ii) True Fracture stress = P/A </em>
= 90 * 10^3 / ( 3.24 * 10^2) / 4
= 90000 / 81 = 1111.11 N/mm^2
Answer:
Option (B) increases only
Explanation:
An isolated system, in thermodynamics is defined as the system which does not allow the exchange of both the energy and mass in between the system and the surrounding.
In any practical thermodynamic process, the entropy of an isolated system increases only as the system can neither exchange energy nor mass but can generate the randomness in the molecules inside the system thus increasing its entropy up to the equilibrium point at which it reaches its maximum point.
Whiteness is often used as a derogatory term to describe Caucasian culture. This influences our society by promoting anti-White hatred.
Answer:
Explanation:
Given that:
f = 250 Hz
= 2%
= 600 Hz
= 0.5 to 1.5 increment by 0.05
![F = A sin (Xt)](https://tex.z-dn.net/?f=F%20%3D%20A%20sin%20%28Xt%29)
For 250 Hz = 250 cycle/sec
![X = 2 \pi t 250](https://tex.z-dn.net/?f=X%20%3D%202%20%5Cpi%20t%20250)
![X = 500 \pi t](https://tex.z-dn.net/?f=X%20%3D%20500%20%5Cpi%20t)
![X = Asin (500 \pi t)](https://tex.z-dn.net/?f=X%20%3D%20Asin%20%28500%20%5Cpi%20t%29)
![\omega = 250 \\ \\ \omega_n = 600](https://tex.z-dn.net/?f=%5Comega%20%3D%20250%20%20%5C%5C%20%5C%5C%20%5Comega_n%20%3D%20600)
M = 0.98 , 1.02
![M_{(w)}} = \sqrt{[1-(\frac{w}{w_n})^2 + ( 2\zeta \frac{w}{w_n})^2}](https://tex.z-dn.net/?f=M_%7B%28w%29%7D%7D%20%3D%20%5Csqrt%7B%5B1-%28%5Cfrac%7Bw%7D%7Bw_n%7D%29%5E2%20%2B%20%28%202%5Czeta%20%5Cfrac%7Bw%7D%7Bw_n%7D%29%5E2%7D)
![\frac{1}{M} = [1-(\frac{w}{w_n})^2]+(2 \zeta \frac{w}{w_n})^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BM%7D%20%3D%20%5B1-%28%5Cfrac%7Bw%7D%7Bw_n%7D%29%5E2%5D%2B%282%20%5Czeta%20%5Cfrac%7Bw%7D%7Bw_n%7D%29%5E2)
![\zeta = \frac{w_n}{2w}\sqrt{\frac{1}{M^2}-(1-(\frac{w}{w_n})^2)^2}](https://tex.z-dn.net/?f=%5Czeta%20%3D%20%5Cfrac%7Bw_n%7D%7B2w%7D%5Csqrt%7B%5Cfrac%7B1%7D%7BM%5E2%7D-%281-%28%5Cfrac%7Bw%7D%7Bw_n%7D%29%5E2%29%5E2%7D)
![\zeta = \frac{600}{2(250)}\sqrt{\frac{1}{0.98^2}-(1-(\frac{250}{600})^2)^2}](https://tex.z-dn.net/?f=%5Czeta%20%3D%20%5Cfrac%7B600%7D%7B2%28250%29%7D%5Csqrt%7B%5Cfrac%7B1%7D%7B0.98%5E2%7D-%281-%28%5Cfrac%7B250%7D%7B600%7D%29%5E2%29%5E2%7D)
At 0.7183 value of damping ratio the error value was 2% at 0.98 value of M
![\frac{1}{M} = [1-(\frac{w}{w_n})^2]+(2 \zeta \frac{w}{w_n})^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BM%7D%20%3D%20%5B1-%28%5Cfrac%7Bw%7D%7Bw_n%7D%29%5E2%5D%2B%282%20%5Czeta%20%5Cfrac%7Bw%7D%7Bw_n%7D%29%5E2)
![\zeta = \frac{w_n}{2w}\sqrt{\frac{1}{M^2}-(1-(\frac{w}{w_n})^2)^2}](https://tex.z-dn.net/?f=%5Czeta%20%3D%20%5Cfrac%7Bw_n%7D%7B2w%7D%5Csqrt%7B%5Cfrac%7B1%7D%7BM%5E2%7D-%281-%28%5Cfrac%7Bw%7D%7Bw_n%7D%29%5E2%29%5E2%7D)
![\zeta = \frac{600}{2(250)}\sqrt{\frac{1}{1.02^2}-(1-(\frac{250}{600})^2)^2}](https://tex.z-dn.net/?f=%5Czeta%20%3D%20%5Cfrac%7B600%7D%7B2%28250%29%7D%5Csqrt%7B%5Cfrac%7B1%7D%7B1.02%5E2%7D-%281-%28%5Cfrac%7B250%7D%7B600%7D%29%5E2%29%5E2%7D)
![\zeta = 0.6330](https://tex.z-dn.net/?f=%5Czeta%20%3D%200.6330)
At 0.6330 value of damping ratio the error value was 2% at 1.02 value of M.
Hence, the damping ratio
of the transducer must be placed between 0.6330 to 0.7183