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Sadasivan N
Registrace 14. 04. 2009
Velocity in mechanisms Solution to Example problem 7.2 in R S Khurmi book
Velocity in Mechanism - Relative Velocity Method
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Video
4.5.2 Problem 2
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Problems on Epicyclic Gear Train Reference: R.S. Khurmi - Theory of Machines Text Book
4.5.1 Problem 1
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Problems on Epicyclic Gear Train Reference: R.S. Khurmi - Theory of Machines Text Book
5.8.5 Problem 5
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Problems on Differential Band Brakes Reference: R.S. Khurmi Theory of Machines Text Book
5.8.4 Problem 4
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Problems on Differential Band Brakes Reference: R.S. Khurmi - Theory of Machines Text Book
5.8.3 Problem 3
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Problems on Band Brakes Reference: R.S. Khurmi - Theory of Machines Text Book
5.8.2 Problem 2
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Problems on Band Brakes Reference: R.S. Khurmi - Theory of Machines Text Book
5.8.1 Problem 1
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Problems on Band Brakes Reference: R.S. Khurmi - Theory of Machines Text Book
5.8 Introduction to Band Brakes
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Reference: R.S. Khurmi - Theory of Machines Text Book
5.7.5 Problem 5
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Problems on Brakes Reference: R.S. Khurmi - Theory of Machines Text Book
5.7.4 Problem 4
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Problems on Brakes Reference: R.S. Khurmi - Theory of Machines Text Book
5.7.3 Porblem 3
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Problems on Brakes Reference: R.S. Khurmi - Theory of Machines Text Book
5.7.2 Problem 2
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Problems on Brakes Reference: R.S. Khurmi - Theory of Machines Text Book
5.7.1 Problem 1
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Problems on Brakes Reference: R.S. Khurmi - Theory of Machines Text Book
5.7 Introduction to Brakes
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Reference: R.S. Khurmi - Theory of Machines Text Book
5.5. Introduction to thrust collar bearings
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5.5. Introduction to thrust collar bearings
del = Wl/AE is same for all longitudinal vibrations?
Useful the videos
Sir delta 2 value shown in my Casio is 0.0108 how to convert 0.0108 into 10.86×10 to the power of minus 3 ???
@vijayvijju6993 10^-3 simply means divided by 10^3 which is nothing but divided by 1000. It's very simple. 10.86/1000 = 0.01086 10.86/10^3 = 0.01086 Therefore 10.86*10^-3 = 0.01086 There is no real calculation involved Just understand 1/10^3 = 10^-3
Sir BE length is 1.5, how do we get 0.5
Sir try 2/3 times but i got magnitude 9 and degree 210 😢
Make sure you are moving the scale properly while doing perpendiculars, use a drafter for better results
Thanks sir
Sir need ans for these two question
A vibrating system having mass 1 kg is suspended by a spring of stiffness 1000 N/m and it is put to harmonic excitation of 10 N. Assuming viscous damping, determine (i) the resonant frequency (ii) the phase angle at resonance (iii) the amplitude at resonance (iv) the frequency corresponding to the peak amplitude and(v) damped frequency. Take C = 40 N-sec/m.
A rotor of mass 10 kg is mounted mid-way on a 20 mm diameter horizontal shaft supported at the ends by two bearings. The bearing span is 800 mm. Because of certain manufacturing defect, the centre of gravity of the disc is 0.1 mm away from the geometric centre of the rotor. If the system rotates at 50 r.p.s., determine the amplitude of steady state vibrations and the dynamic load transmitted to the bearings Take E = 200 GN/m2
Brilliant work sir
Nice explanation sir..
Thank you sir...
Very good explanation and helpful
Hi sir tq you sir it is very helpful in lockdown period so plse upload more University solved problems sir🙏🙏🙏🙏
My pleasure aravind... Feel free to post your doubts..
Ok sir give me a mail id
@Aravind K send it to this mail ID.. sadasivan88@gmail.com from which I'm hosting this channel
Sir It is a important derivation it will ask any previous question paper ?
SUPER Sir
GREAT SIR
Spr sir