Four Bar Linkages in Two Dimensions. A link has fixed length and is joined to other links and also possibly to a fixed point.

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1 Four bar lnkages 1 Four Bar Lnkages n Two Dmensons lnk has fed length and s oned to other lnks and also possbly to a fed pont. The relatve velocty of end B wth regard to s gven by V B = ω r y v B B = +y = ω B k( + y) = ω B ( y + ) 1 Vector r = + y n two dmensons θ ω B Note ω B s the counter clockwse drecton angular velocty. The relatve velocty s always perpendcular to the lne onng the lnkage ends, Formally ω = k ω and ω r = 0 0 ω y 0 k but Equaton 1 can be seen by nspecton. Or note V B = ω B r. The component s V B snθ = V B y r = ω B y. Ths we have n Equaton 1. The component arses n a smlar way. The relatve acceleraton has two components. One component s centrpetal and the other azmuthal or, one mght say, crcumferental. There s no Corols acceleraton snce the "radus" or barlnk length s constant. The centrpetal acceleraton of B w. r. t. a Brad = ω (ω r) = ω B 2 ( + y) 2 The azmuthal acceleraton a Bazm = ρ = B ( y + ) 3 a B = a Brad + a Bazm a Bazm = B 2 k(+y) = B 2 (-y+) y a Brad = - ω B 2 (+y) B = +y B, angular acceleraton ω B, angular velocty = ω B 2 ( + y) + B ( y + ) = ( ω B 2 B y ) + ( ω B 2 y + B )

2 Four bar lnkages 2 Eample 260 mm B 240 mm 100 mm 100 mm y ω ο O 1 C 60 mm O s rotatng at constant angular velocty ω o counterclockwse. Fnd all veloctes and acceleratons. Veloctes Take O. In eq. 1 60, y 80 The velocty of w. r. t. O s v O = ω o ( 80 +( 60) ) = ω o ( ) 4 Smlarly v B = ω B ( ) 5 where ω B s the ( at present unknown) counterclockwse angular velocty of B nd v CB = ω CB [ ( 180) + 0 )] = ω CB where ω CB s the ( at present unknown) counterclockwse angular velocty of CB. Thus we can epress the velocty of C whch s zero as 0 = v CO = v O + v B + v CB = ω o ( ) + ω B ( ) + ω CB Compare components 0 = ω o 60 + ω B 240 ω B = ω o 4 Compare components; 0 = ω o 80 ω B ω CB 180 ω CB = 7 ω o Fnally v B = ω CB ( 180) = 105 ω o n -drecton as n the fgure 10

3 Four bar lnkages 3 cceleratons 260 mm B 240 mm 100 mm 100 mm y ω ο O C 1 60 mm Usng the formulae: Frst a O = ω o 2 ( ) 11 Then usng eqq. 2 & 3 a B = ω B 2 ( ) + B ( ) 12 and a BC = ω B 2 ( + y) + B ( y + ) Note that we have ω B = ω o 4 ; B s yet to be determned. nd a CB = ω CB 2 ( 180) + CB [ 0 (- 180)] = 180 CB ω CB 2 13 Thus the acceleraton of C, whch s zero s gven by 0 = a CO = a O + a B + a CB = ω o 2 ( 60 80) ω B 2 ( ) + B ( ) ω CB CB Compare coeffcents: 0 = 80 ω o 2 100ω B B ω CB 2 14 whence 240 B = 80 ω o ω B 2 80 ω CB 2 = ω o and B = 5 48 ω o 2 15 Compare coeffcents: 0 = 60ω o 2 240ω B B CB 0 = ω o CB

4 Four bar lnkages 4 Thus CB = ω o 2 16 From the angular acceleratons the lnear acceleratons follow farly easly. To recap the data: ω B = ω o 4 ; ω CB = 7 ω o 12 B = 5 48 ω o 2 ; CB = ω o 2 From eq. 11 a O = ( 60 80)ω o 2 and we may wrte a BC = a BO = 180 CB 180 ω CB 2 = = mm B 100 mm β ω ο O y 60 mm 1 C

5 Four bar lnkages 5 lternatve approach We may work drectly wth the dagrams for veloctes and acceleratons eamnng the geometry of each. Velocty Dagram -125 g v B (v) v CB O () -10 v B -20 () v β e () d -60 The angles and β are taken from the geometry of the system. sn = 5 12 ; cos = ;tan = 5 12 snβ = 4 5 ; cosβ = 3 5 () We may enter v at the angle β as shown and wth ω o = 1 usng mm s 1 as unts. Or take v wth component 80 vertcally and 60 horzontally. () v B s at an angle as shown and snce v B has only a horzontal component the length c g = 60 () Now tan = 5 12 then c-d = = 25 (v) Whence v CB = 105 We have v B = 60 cos = = 65 mm s 1 Snce the length of B s 260 mm ω B = = 0 25 rad s 1 Smlarly the angular velocty of BC ω BC = = 7 12 rad s 1.

6 Four bar lnkages 6 cceleraton Dagram s before the angles and β are taken from the geometry of the system. sn = 5 12 ; cos = (v) 415/12 f g snβ = 4 5 ; cosβ = 3 5 The order of steps s shown by roman numerals on the dagram. () We may put n a straght off and enter the 80 dmenson g a ( settng ω o = 1) () & ()a e s the radal acceleraton of B towards and ths we get knowng ω B. a e = 260 (ω o /4) 2. Hence the dmensons a b and b c. (v) The length f-e s the radal acceleraton of B w. r. t. C We have ω CB = 7/12 and so f-e = 245/4 (v) So, knowng the dmensons on the rght we fnd d e = 25. (v)snce c-e s the transverse acceleraton to B w. r. t. and we have the angle and can compute the dmenson d-c (v) β e a B d 125/12 (v) 245/4 (v) c a b 15 () 80 () 25/4 () (v) Whence f O follows. The dagram s solved. To get the angular acceleratons, BC = f- O BC = = rad s 2 clockwse : see dagram. B = c e 260 = = 5 48 rad s 2 counter clockwse : see dagram. Ths process could be done appromately wth a ruler and pencl. It does not always happen that the numbers come n as convenent ntegers as they do here.

7 Four bar lnkages 7 Slder and Crank & Connectng Rod ω 0 c r o θ B The slder B moves only horzontally. It s as f B were on the end on an nfntely long rod whose fed end was at ±. Veloctes. v = ω o ( c snθ + c cosθ) = ω o c ( snθ + cosθ ) 1 v B = ω B ( r ( sn) + r cos ) = ω B r ( sn+ cos) 2 Thus v B = v + v B = ( ω ο c snθ + ω ΒΑ r sn) + ( ω o c cosθ + ω B r sn) 3 Snce the vertcal or component of v B = 0 we get ω B = ω o c cosθ r cos 4 and v B = ω o c( snθ cosθ tan) 5 Note tan = c snθ c cosθ so v B = ω o c snθ + c snθ cosθ c snθ cosθ c cosθ = ω o c snθ c cosθ 6 We could have arrved at the last epresson by usng the cosne rule and the geometry of the connectons: r 2 = c c cosθ 7 Take the dervatve w. r. t. tme: or snce ω o = θ ' 0 = ' 2 c ' cosθ + 2 c snθ θ ' ' = ω o c snθ c cosθ = v B 8 The velocty dagram s shown below.

8 Four bar lnkages 8 v v B = r ω B c ω 0 θ v B Usng the sne rule we get: v B sn(θ+) = cω o cos ; v b = c ω osn(θ + ) cos o 9 whch agrees wth equaton 5 above. Note that there are varous other ways of approachng ths analyss: look carefully at the problem. For eample we could use the sne rule on the orgnal geometry sn c = snθ r : sn = r c snθ Take the dervatve w. r. t. : cos c ' = cosθ r θ ' ; ' = c cosθ r cos θ' so ω B = ' = c cosθ r cos θ' = ω o c cosθ r cos : whch s eq. 4

9 Four bar lnkages 9 cceleraton ω 0 c r o θ B We suppose that the crank O s gong at constant speed so O = 0 a = ω o 2 ( c cosθ + c snθ) = ω o 2 c ( cosθ + snθ) nd a B = ω B 2 ( r cos r sn) + B ( ( r sn) + r cos) = ω B 2 r ( cos + sn) + B r ( sn + cos) and a B = a + a B = ( ω o 2 c cosθ ω B 2 r cos+ B r sn) + ( ω o 2 c snθ+ ω B 2 r sn + B r cos) Snce B moves only horzontally the component of acceleraton s zero and so B = 0 = ω o 2 c snθ+ ω B 2 r sn + B r cos 1 r cos ( ω o 2 c snθ ω B 2 r sn) Snce ω B = ω o c cosθ r cos B = ω o 2 c r 2 cos 3 ( r snθ cos2 c sn cos 2 θ) and ω a o 2 c B = r r (snθ tan cosθ) c cos2 θ cos 3 Ths can hardly be counted as a smple epresson. For partcular cases the solutons may be smpler. In any event the complety of the formula suggests ether to use a computer or to nsert numercal values early n the analyss.

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