Question

P4.2: The four-bar mechanism in Figure P4.2 has the link lengths: LAB = 4", LED = 24", Loc = 14", LA = 30". The location of COM is as follows: LAG = 2.120"; LEG2 = 11.863"; LDG = 6.722". The crank 1 rotates counterclockwise with the angular displacement 8. = 27t-t [rad), angular velocity 0. = , = 21 [rad/s], and angular accelera- tion &= 7= 0 [rad/s). The mass and moment of inertia relative to each link's COM are as fol- lows: m = 0.596 [lbm]; I. = 1.047 [lbm - in.?]; m2 = 3.939 [lbm]; Ig2 = 197.744 [lbm-in.); m3 = 2.389 [lbm); Ic3 = 40.887 [Ibm-in.). At t= 0.25s, the crank is in vertical position 0 = 1/2 (rad). For this position, determine

Answer 1

The following figure shows the mechanism scaled. Q3 ² 20° 04 600 B 03 Qadc

The following figure shows the mechanism scaled. Q3 ² 20° 04 600 B 03 Qadc

a: let be at (0:0) and we get Bat (0,4) Cat (30,0) and for D, a Component is 24 Cosas 24 Cos (20) 2 22:55 and y component is 4+245in (03) 24+24 Sinzo 12-208 .. location of Dis at (22:55, 12:208) and location of Gi is (0, 2-12), that G2 is (LBarcos O3, 44LBG25 in O3) (11.863 Cos 20, 4711.863 S1n2o (11-14) 8.057) of 2

and location of G3 is (2460503 + L083 65 04 , (14-2043) Sin 04 - (246520 +6.722 60560 (14-6-722) Sin60) (25.913, 6-302) 2

bi To find angular velocities of link 2 and 3, we need to draw the velocity diagram we have, Vb 4 x 27" 87 ins ABWI Velocity of d wit b. Valb to to Bo and Vd to to DC with this information, the velouty diagram is given by,

d Na Vab Vb Ca 5

from the diagram, Vab 2 14.25 in/s and we Vab 14.252 0:593 rad's DB 24 also vd 24.25 in/s and W3 z vd 24.25 1-732 rad/s 14 le angular velocities link 2 and 3 2 DC of C: of Now to find the absolute angulor accelarations link 2 and 3, we need to draw the acceleration diagram of mechanism for that we have, al AB W12 4 xen) 2 2 157.9 in/s²

arz 2 arz 2 radial accelerations, 2 BD W22 2 24 x 0.5932 28.43 in/s² DC W3² 14x1.7322 42-14 in/s² and arall to BD, ar3 ll to CD, tangential accelerations, Atz to to BD and ats to to CD. with this information we could able to locate d in the acceleration diagram

ac arz A3 atz d al αν atzn १ 1982

And now absolute angular acceleralions of links is given by ar 92-91-25 in/s from diagram ар Во 91-252 224 3.802 rad/s² and as аз D C A3 285 in/s² 2 85 26.071 rad/s² 14 ie absolute angulor accelerations link 2 and 3 is 3.802 rad/s and 6.071 rad/srespectively. of

di 2 0 , aly Now to find a and y components of linear accelerations, from diagram air 157.9 in/s² Now drawing to of ab at a meeting at e, we get ed 40 in /3 , azy zeb: 77.5 in/s2. and that of link 3, A3xzed 4011/s, azy soin/8² Aza 2 cez

ac larz ED atz • al ☆ e ar atz 9 eeb

ei N x and y linear 1 No to find x Components of acceleration of Gi, G2, G3, we need to locate 91 92 and 93 in the acceleration diagram for that we have the ratios, n. agi ab and bg2 2 LAGI 2.12 0.53 LAB 4 LBG2 11.863 0.494 LBD 24 Loa3 2 0.48 Loc 14 and bd d93 de 2 6.722

а) с ags I arz 93 at3 Az ag, d aga gi Az H2 a, 1 9 aga 2 atz b arz

Now drawing trs at gz and 93 to ab meeting at f and g respectively as shown in the above diagram Now, agix zo agiu ag, 283.68 in/s² and that of gz , agzx 929 19.25 in/s agzyzgb 2 42.5 in/5² and that of 93, agzz ²93f 2 20.75 in/3², 9q3y2 af 240 in/s² 2 Note: The reason we are taking to to ab is because its ll to y-axis