Question

3.) Şekilde verilen sistemde aşağıdaki değerlere uygun olarak sol ve sağ düzlemde 50 mm yarıçapa yerleştirilecek dengeleme kütlelerinin değerini ve açısal pozisyonlarını bulunuz. Şaft hızı 500 ddk (30 puan) m = 2 kg, m2 = 1.5 kg, m3 = 3 kg. 11 = 25 mm, 2 = 35 mm, 13 = 40 mm R mi mi 150 m2 13 750 m m3 m3 800 200 RASAD

Şekilde verilen sistemde aşağıdaki değerlere uygun olarak sol ve sağ düzlemde 50 mm yarıçapa yerleştirilecek dengeleme kütlelerinin değerini ve açısal pozisyonlarını bulunuz. Şaft hızı 500 d/dk.

M1= 1 kg
M2= 1.5 kg
M3= 3 kg
r1= 25 mm
r2= 35 mm
r3= 40 mm

In the system given in the figu
re , find the value and angular positions of the balancing masses to be placed in a 50 mm radius in the left and right plane in accordance with the following values. Shaft speed 500 rpm.

Answer 1

In thi problem, we have to find out the value of balancing masses and there angular positions.

Given data:

With refrence to x axis

в = 90° 0,2 — 195° 03 = 285°

Balncing mass is at radius = 50 mm

Concept: Resultant force must be balanced and Resultan couple about R plane must be balanced for complete balancing.

R mi m = 1 kg. m₂ = 1.5kg. m₂ = 3 kg 2 B 75 G ome, omz. 800 mm 1 200 mm m₂ m3 3=40mm 8 = 25 mm, 82 = 35mm > 1 Let at m₂ = mass L-Plane on O Angular Position at OR Angular Position R- Plane on maa mars &= somm. for Complete balancing of system, Resultent force and Resultant couple both should be zero. wa Same {m.s.w² = 0; Em.s=0 for al mais. R-Bas Reference Plame. - Let Taking coupled at then. {m.r.l. - 0 (M = = dy = 200.). m, s, t, & mare Xdz + m₂ 313 tm. % (1000) = -O . This be above Equation will resolved in Y Component, from there angular Position x 4

becomes. The above equation Emrl coso =0 and {m.r.l.Sino=o. mix so X 1000 X COS O2 + Ix 26 x 200 cos 90' + los x35x 200 x Cos195° + 3 x 40 x 200 Cos 285 and mex 50 x 1000 X Sin + 1x25 x 200 Singo' + 115 X 35X200x sin 195* + 3x 40 x 200 Sin 285 =0 me. Cosa = + 0.0786) m, sino = kero + 0.41799 ..ME mesinoget +(m_c050112 (0.07961) 4+ (0.41943) me=0.425 kg and fana = 0.41799 0.07861 .. 10 = 79.348 Bry Now from force Balancny . (in Component form). marcoso + mar Cos OR & m, r, loso, & M2₂ Coso, fm rcoso =0. o Mpx SD x los OR =- 7-(* x 28 x Cos 90' + 105x35 x cos 195+ 3x40x Cos 288° + 0.425 x 50 x cos 79.348)

MR. COSOR- & 31449 and similarly for sin component, me.so x sin op =- (1x25 x singo + 1.5 x 35 x 58 195 + 3840 x 884265 + 0.425 x 50 x 88 79.34) mesinoR² + 1.6723 o. mes Jmo cos OR)2 + (mosinor)? - ma 1 (0.31449)2+ (106723)2 MR= 1.701 kg and OR= fani! To 6722 • 31449 OR = 79.3491 Any