Mechanical Engineering - Theory of machines - Discussion
Discussion Forum : Theory of machines - Section 6 (Q.No. 35)
35.
A shaft has two heavy rotors mounted on it. The transverse natural frequencies, considering each of the rotor separately, are 100 Hz and 200 Hz respectively. The lowest critical speed is
Discussion:
4 comments Page 1 of 1.
Hemant patel said:
1 decade ago
Dunkerley's empirical formula:
1/(Fn)^2 = (1/ (Fn1)^2) + (1/ (Fn2)^2).
Fn = 200/5^(1/2).
= Critical speed = Whirling speed = Unit is r.p.s.
= (200/5^(1/2))*60 r.p.m.
= 5367 r.p.m.
1/(Fn)^2 = (1/ (Fn1)^2) + (1/ (Fn2)^2).
Fn = 200/5^(1/2).
= Critical speed = Whirling speed = Unit is r.p.s.
= (200/5^(1/2))*60 r.p.m.
= 5367 r.p.m.
(1)
PAATEL said:
6 years ago
When two shafts 100+200 hz = 60 rpm * 60 (2).
rpm = 7200.
rpm the transverse natural frequency 1.35 to 1.5,
7200/1.3 =5367 rpm.
rpm = 7200.
rpm the transverse natural frequency 1.35 to 1.5,
7200/1.3 =5367 rpm.
Sheetal said:
9 years ago
Thanks for the Explanation @Hemant.
Gaurav said:
1 decade ago
How is it calculated?
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