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Determine the pressure on each rail allowing for centrifugal force and gyroscopic couple action.

Total mass of four wheeled trolley car is 1800 kg. The car runs on rails of 1.6 m gauge and rounds a curve of 24m radius at 36 kmph. The track is banked at 100 .The external diameter of wheel is 600 mm and each pair with the axel has 2 mass of 180 kg, with radius of gyration of 240 mm. The height of the center of the mass of the mass of the car above the wheel base is 950 mm. Determine the pressure on each rail allowing for centrifugal force and gyroscopic couple action.

Mumbai University > Mechanical Engineering > Sem 5 > Theory of Machines-II

Marks: 10 Marks

Year: May 2016

1 Answer
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M= 1800 kg

X= 1.6 m

R= 24 m

V$= 36 kmph \\ = 36× \dfrac{5}{18}$

∴V = 10 m/sec.

dw= 600 mm = 0.6 m

rw= 0.3 m

h= 0.95 m

Mass of the wheel= 180 kg

Radius of gyration $= 240 mm \\ = 0.24 m$

enter image description here

Considering

First conndering the effect of (w=mg) of the car and of the centrifugal force, on it determining the Reactions RA and RB Ref. above fig. 1

Resolving the 1hr forces to the trag.

RA+RB $= Mg \cos ѳ +\dfrac{ Mv^2}{R} \sin θ \\ = 1800×9.81×\cos 10°+\dfrac{1800(10)^2}{24}×\sin 10°$

$\boxed{RA+RB= 18692 N}$

Now taking moment @ B.

$RA×W= Mg \cos θ×\dfrac{W}{2} + Fe \sin θ× \dfrac{W}{2} + Mg \sin θ×h ─ Fc \cos θ×h$

$RA= [Mg \cos ѳ+\dfrac{Mr^2}{R} \sin ѳ] × ½ + [Mg \sin ѳ– \dfrac{Mr^2}{R} \cos ѳ]×\dfrac{h}{w}$

$RA = [1800×9.81×\cos 10+1800×\dfrac{10^2}{24}× \sin 10]×\dfrac{1}{2} +[1800×9.81×\sin 10 ─(1800×\dfrac{10^2}{24}× \cos 10]×\dfrac{0.95}{1.6}$

RA= 6781N

RB= 11911N

Reactions due to gyroscopic couples.

Cw$= 2 Iw×Wω× \cosθ×wp \\ = 2×mk^2×\dfrac{V^2}{rR} × \cosθ \\ = 2×180×(0.24)^2× \dfrac{10^2}{(0.3×24)}× \cos10 \\ = 283.6 N─M$

Reactions on outer wheel,

$R_{G \ \ outer}= \dfrac{C_W}{2_W} = \dfrac{283.6}{(2×1.6 )}= 88.6 N$ upwards(↑)

$R_{G \ \ inner} = 88.6 N downwards(↓)$

Pressure on each outer rails= RB + RG outer

= 11911+88.6=11999.6 (downwards)

Similarly on each inner wheel $= R_A─ R_{G \ \ inner} \\ = 6781─ 88.6= 6692.4 N (upwards)$

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