Gépészet | Géprajz » Procedure for Constructing Velocity Diagrams

Source: http://www.doksinet 5 Procedure for constructing veloci& diagrams 5.1 Summary In this Unit you have been drawing velocity diagrams for many forms of mechanisms. You will have noticed that the velocity diagram is built up steadily, one point at a time. Any mechanism that you will be faced with on this course can be broken down into a succession of simple parts which can be solved by the methods that you have l e a d . You may be surprised to see that the key ideas that you need, to solve even a complicated mechanism, can be summarized into only three problem elements shown in Figure 58. U you arc comfortable with these thnc problem-solving elements then you will have no trouble with velocity diagrams. Link (position diagram) TVps name TAN b-ll SLI isIWiw1 - Existing polnm on miocHy diagram ConstrucHon of ~locitv diwnm Ic F / b,c PROP (pmponlonl Notg pD bc parallel to sliding surface bc- (W). if known cb = c0 ipmponion) bed imaw of BC0 B b b , ~ B bc

bd cd BD 1 CD bed image of BCD Source: http://www.doksinet TAN (-1i.l) The velocity of B on the rigid link BC is known. The velocity of C relative to B has to be simply tangential (TAN)so c must be on a Line through b perpendicular to BC. If the angular velocity o is known, then be is given by or, and therefore the position of c is known directly. sn (m Two links slide relative to each other, and the velocity of the point B on one is known. The point C on the other link, which is at this instant coincident with B,must have a velocity relative to B which is pure sliding. Therefore c must be on a line through b parallel to the sliding surface. If the magnitude of the sliding velocity (8,), is known, then bc = (v,),, and c is known directly. PROP W-) The velocities of two points B and C on a body arc known; find the third. If the thrce points are in a straight lie, then usc proportion. If they are not, then construct perpendiculars to the lines betwan unknown (D) and known (B,C)

points. In both PROP cases the velocity diagram is image of the position diagram, producing a useful check. The mechanism shown in Figure S9 is more complicated than those you have analysed so far. However, although it may be daunting at flrst glana, it still contains only the samc three problem elements. Figure 59 8 K G 00 Figure M ) In Figure 60 the mechanism has b m exploded to show the individual links more clearly. Drawing such an exploded diagram is o k n a useful step in the analysis of more complex mechanisms. It can help with the identification of the basic chain and is particularly usdul for ensuring that you dont confusc the lettering of sliders and their coincident points. If you are in a i y doubt about how to start or procad with a velocity analysis, or about the labelliig of slidcrs and coincident points, I recommend that you sketch such a diagram. Source: http://www.doksinet obgk fixed Input to OC Basic chain OCSB Points fad Velocity d-am points lnholved oc CS

sb bad ed eg h cfs hf hk Constructton rYPe Bxed TAN SLI TAN PROP TAN TAN PROP TAN SLI Flgutr 61 Figure 61 tabulates tha step taken in the construction of thc vdocity diagram for thk mecbanirun and Figure 62 shows the velocity diagram sketch I suggest that you look through the tabulation and satisfy yourself that the velocity components are correctly listed. The remainder of this Unit is a series of SAQs for you to gain expcriena. In tackling these questions I suggest that you work through the following steps - a systematic approach will help you to keep in mind a dear picture of the solution, If in any doubt, dnw an exploded d i a g ~ ~ ".:: -. , , ,,., ?B Label the position dininsertin# anyjnaaaestJrlpdindda:i{ll pointc at tk slidern. Identify aW axed point& Identify link with input d Identify b& chain. Plot flxcd Looate flmt whbb point. Plot around b* chain * SA0 19 Refer back to Figure 42. Identify: (a) the fixed points, (b) the link with input data, (c)

the basic chain, and (d) the h t solvable point. What problem element type (TAN,SLI or PROP)is involved in determhh~the following velocities Figure 63 shows a mechanism driven through the flywheel BA. Figure 64 shows its velocity diagram. (a) What are the fixed points? (b) To which l i is the input data? (C) What is the first solvable point? Source: http://www.doksinet Figure 63 Position diagram S d e I mm : l0 mm Figure M Velocity d i w a m Scale 1 mm : 5 mm S- (d) What is the basic chain and what is the appendage? (e) Which of the three problem element types arc involved in this question? (I) Label the velocity diagram. (g) Determine the angular velocity of the link El? SA0 2 i (v8) Sheet V8 shows a slider linkage with BC making an angle of 30" with the vertical. BC is rotating clockwise at 5 rad S- and also has an angular acceleration of l rad S- in the same direction. Determine the velocity of the slider at F and the angular velocity of the link FD at the instant

shown. SA0 11 (v#) Determine the angular velocity of the link CG and the velocity of sliding of the link CG in the trunnion at F for the position diagram shown on V9. Determine the velocity of the piston at R and the angular velocity of the link RQ for the mechanism shown on V10. SA0 24 (V1 1) Sheet V1 1 shows a quick-return mechanism. Determine the velocity of the slider at E and the angular velocity of the link DE. OB is rotating at 5 rad S- clockwise SA0 28 (v12) Sheet V12 shows a valve mechanism. The crank OA rotates at a constant velocity of S00 rad S-, clockwise. Determine the angular velocity of arm BCD and the velocity of the slider at F. Source: http://www.doksinet Glossary Text -h refumce Appendage Basic chain . . C~lnndentpoint Compound link Fmt aolvabk point Fued link Fixed point Grounded link Input data Inversion Trunnion Velocity image Vclocity origin Parts of a mechaniim attached to the basic chain The part of a mechanism which is analysed Brat. Always a

fow-link chain containing the link with the input data The point on a link which is instantanwusly coincident with either (1) a s*t moving along that link or (2) a trunnion through which thc link slides A link which fonis pairs with three or more other links The Brat moving point to be analysed in the h i c chain The link which is considered to be stationary Any point on the frame. In particular, the pivot of a grounded link The statiowy component of a mechanism. Modclled as thc 6xed link A link pivoted on the fixed link The known dam, desnibing the motion of one link of a mechanism A mechanism derived from another mechanism by a change of the fixed link A sliding link, c.g, in a slider-crank linkage The velocity of either (1) a slidcr. rclative to the coincident point on it8 adjacent link, or (2) the coinddrnt point on a link, relative to the trunnion through which it slides A grounded rotating link through which another link slides The velocity diagram of a compound lid,which is

always the same shape as the link The point in the velocity diagram which represents pro velocity Source: http://www.doksinet Answers to Self-Assessment Questions SA0 1 (h),is tbe velocity of 0 relative to the point 8; tbat is (#o),=b=5ms-1 as the observation point (or point of rrlcma)is now B. 4 5 6 Wrong - shape wrong (wmparc 5) correct Wrong - wrong shape SA0 8 (a) f 6 or 7 (h) 9 (c) 5 100 o=--100Iads-I 1 From Figure 65, o is anticlcckwire. . F w r c 66 showa a akctcb of tbe velocity diagun. F ~ o n 0 , B n g e x w o ; d M i n dirsdioniromo. Tnnsla:totbccaddtbclinkOD;thisgivoado*wisc rotation d OBCD. a 08 F i g u ~68~shorn tbc velocity diagram. - SA0 5 (a) (&)o-aZ-4ms-1J 0 (b) ( & ) , = E = S ~ S - ~d (C) (u,)c=a=5ms-1 r c Flgure 66 Velocity diagram /m SAQ 8 Scalelmm:100mms- (d) ( B o ) B - b = 3 m s - 1 c Tbe input date is for link BC.Heace C is the BRt mlvablc Remember, velocity is a vector quantity and must ahvays point. haw a magnitude and

dirsdion. SA0 4 (a) 5(7i(o=O). (b) l . Z 4 ( 7 i f o - 0 ) 3and6bawlcttnsoutofaqucnawitbtboreoltbc position diagml. dA0 6 1 Wrong - becaun it M not rotated through W 2 Wrong - wrong shape 3 wrong - lencrs not in mrmpondiag positions Veloclty m l y s b : 0 is a Bxed point. ( ~ ) ~ - 6 r n s - d-82 (h), -7 m S- from c to b (Lto BC) (~)o-?m~~~motob(~toOB) (t~,)~-(100X 0.1) me- (h),- 10ms-l AM" Source: http://www.doksinet BA0 9 On Figure 68 dresswe bc and halve it to locate m on the line bc. By proportion Sa Figure 69. 5 Mark m on your velocity diagram. Join om 6 (&),=mi-49~0.2 -9.8 m 8 - 4 13" BA0 l 1 On the complete velocity diagram of Figure 31(b) join o to r and measure its length and direction relative to the vcrtid (q)O=H=(65.5 Figure 69 Vcloclty diagram for SAQ 9 The velocity of M relative to the fixed point 0 is ( E M ) O - ~ 7.7 = m S-4 17" 910 10 Ficre 70 shows the velocity diagram. X 0.05) 1= 33 m S-T82 BA0 12 The analysis has to be

done in two parts: determination of the velocity diagram for the basic chain and then the location of M on the velocity diagram. The batk c h i n is PQRS. The Arst solvable point is Q as the input data is for link PQ. 1 Velocity d y s l s : The Bxcd points an P and S. 2 Draw bBBic chain velocity diagram (sec Figum 71). P i p e 70 Velocity diagrmnfor SAQ 10 Scalelmm:mmms- (8) No. (b) P. Figure 71 Velocity diagram of h i e cludn, SAQ I2 (hUactucrl size) (c) l Velocity mlyJla. 0 is the hxed point (@,),-lOms-L+ =@ (h)p=?ma-/(LtoBP) (h)O=? m S- 2 3 (L to BO) 0.4 Draw the velocity diagram to the d e given, Figurc 70. - ( Y ) ~ ob OB 0.5 (59 X 0.2) rad 8 - l 0.5 &==8.6rads-> 4 Proportion to determine m: h0 23.6 rad S-> qm = 44 X 0.308 = 13.5 mm (Scc Figure 72). = 8.6 rad S- Source: http://www.doksinet SA0 14 Flsure 41 42 43 44 45 Figure 72 Veloclty dfqgramfor SAQ 12 ( W a c t u a l size) Input to llnk Basic chin OB OBCG URQP UR sliderats Slidant E 08

WTSO Fbat ~01vcrblc Appendages point RSTO* - DCEG OBC OBCD CEO SA0 l8 The bssic chain is OBCG. OBE is a compound link appendage driviug 6nal appsadapc EHE Veloflty analysb: 0. G and H uo Gxcd points BA0 15 The angular acalmtion hrs no beariq on the velocity diagram (Figure 73) and m can be ignored at thin stae. The fixed link is SR. The bmii chain is STUR which in considered first and then W is obtained by proportion later. For 0- For OBE (a compound link) (&),=?mr-f ( L t o B E ) b t o e h For EFH (E,),-7ms--(AtoFE) (E,). - 7 m S-$ sliding wJtially h a w the velocity diagram (Fiwre 74). Fl#ure 73 Veloclty dlqgrmnjbr SAQ 13 (hoyactual she) l Velocity rmolyais: The 6 x 4 pinta uo S and R. The input data is to UR.The basic chain is STUR Tbc first lolvablo point ia U. (~~=10~0.3~-3rnr-~~-iP (~),=?/(ItoTU)utotline (h),-?(+toTS)atorlinc 2 B R S E B Draw the velocity diagram for the basic chain Flgure 74 Velocity dlagmmfor SAQ I S Scak l mm: 1 0 0 m m ~ - ~

(E,),,=w=3X1ms-~J Source: http://www.doksinet M O 17 Velocity d y s i s . The coincident point is S on link CD The input data M for link OB.B U the 6mt solvable point Thc bdc chain h OBSC. From the position CS 18.5 X 002 037 m - - (e,)0-10x0.5P (U,), = 7 m S- (U,)= - 7 m S- -5ms- P / (parallsl to CD) (L to CD) C I Figure 76 Veloe1ty drPgrm for SAQ 18 . 0. d # ScslcImm:5mma- SA0 19 (3P, U, G;(b) RU (c) URQP, (d) R; (0) TAN; (I) TAN; (g) TAN (h) PROP;(i) TAN;(j) SLI. Figure 75 Velocity diagramfor SAQ 17 (Wactual&l Dmw the velocity diagam (Figure 75). (h),-3-4.8 M OW (3 F i point8 A, D and F. (b) AB. (C) B. (d) ABCD. Appendage CEE m1-L47 (e) TAN only, no SLI nor PROP (0 scc Fipm 77. M O 18 0d OBED are both four-link cbainr in their OM ri& pad contain the input date Thedore in tbh csre cithor could be treated os the h i c chain. velwily andysis: For OBCG (h)o=20x0.02% =04msr-I (C).-?mu-1 % =3 (Lto BC) (C),-Tms-u For OBED Draw the velocity diagram

(Figure 76). Volodty of piston at C - - ( 6 ~=g ) ~ 67.5 X 0005 034 m S- + Figure 77 Velocity dia$rm for SAQ 10 Scaklmm:5mms- Source: http://www.doksinet I Figure 78 Velocity diagram for SAQ 21 Scalelmm:5mms- MO n IDimensions from pmition digram Figure 78 shows the velocity diagram. OB = 0.25 m Vcloeily d y s k : Input data is for link BC. C is the fint solvable point. The Gxcd points an B, E and G BC=CD=0.6m The basic chain is BCDE; the appendage is DFG. From the position diagram, - ED 0.15 m; DF = 05 m (U,), = 5 X 0.1 I (h), = ? m S- - Velocity anulysls: 0. F E am fixcd pointa 0.5 m S- I = E (tan) (A to DC) c to d line (tan) (h). - ? m S- -(A to DE) e to d line (tan) For DFG (h)O- I m S-- (parallel to slot, sli) o.mst -048 &),-l x0.25k =025ms- m S-+ (%),-?m~-~(LtolinkDE) c is half-way along db; so it is casily located. T is on the link CG whcident with E (A to link CG) (h)p =? m S- /(along link CO) Draw the velocity diagram to scale (Figure 79).

-8 =(23.5 X 00029 d SA0 22 See Fium 79. = 0.34 rad S- Figure 79 Velocity diagram for SAQ 22 ScaleImm:25mms- -3 ( G ) . = ? r n ~ - ~ (LtolinkDB) f (h), =? m S- (h)D=?ms-/(h.toFD,tm) (h)O-3=96x The basic chain is OBDE. Note: OBCTF is not a four-link chain (remcmk: tmnnion is a Link). BC = 0.1 m; CD = 038 m; For BCDE CF=DE-0.6m 3 Source: http://www.doksinet , SA0 23 h Figure 80. r VeIodty adysls: 0 , M and G arc fixed points. The basic chain is OPSG. (4)0=10x0.Z~=ZOms~~=g(tan) to link SP, tan) (&.),-?m S-/(L (h), = ? m S- /(along slot, sli) For QRM pq P Q 0.25 = -= -= 0.357 ps PS 0.7 pq = 0.357 X 88 - 31.5 mm (h),=? m S-(tan) (g),=?ms-~(sli) Solution (h),, = W =(26 X 0.025) c -0.65ms-+ (".)~3-89~0M53 =4,Srads-> 0.5 %,=QR- SA0 14 Insert coincident points G and T.G at E, and T (on FD) at B. Velocity analysis: 0 , C and G an fixed points. The input data is to link OB. The basic chain is OBTC (%),=S x0.2 ,= =lOms-l %=&(tan) (q).=? Figure 80 Velocity

diagram for SAQ 23 Scale 1 mm :25 mm S- m S- /(along link FTC, sli) (h),= 7 m S- (Lto link TC, tan) Draw the basii chain octb. For CDEG cd CD et CT cd-0.67 0.2 0.3 X 0.67 95=64mm (E,), = l m S- /(L to link DE, tan) -7ms-~(sli) Complete the velocity diagram (Figure 81). -P=63 X 0.01 + =063 ms- =- - = ED ed> ED + 31~0.01 rad S-> 0.8 The velocity diagram for the basic chain can now be drawn: o h . =0.39 rad S-> SA0 25 It is heldul to ioin B to E and B to C on the wsition diagram; you & then imagine the cranked &on of the link nplaad by a kinematically identical solid triangular portion BCE. Velocity analysis: 0. B and G are fixed points The input data is to link OA. The basic chain is OAEB (E,), = 500 X 0.03 Figure81 Velocity diagram for SAQ 24 (Mf actual size) r = l5 m S- r = m(tan) (E*)e= 7 m S l ". (along line CED, sli) (Ee). -? m S- (L to line BE, tan) (Ec),= 147 X 0.05 =, = 735 m B- I=&(tan) (B,), = 7 m S- f (sli) (F&=? ms-/(L

to CF, tan) The velocity diagram can now be completed (Figure 82). ( ~ , ) ~ = a = 4 5 . ~5 0 1 ~5 6~ m 8 5-1 T Note that the velocity image of the cranked link appears in the velocity diagram; it is ahown as a dashed line. Source: http://www.doksinet F i g m 82 Velocltty dkgramfep SAQ 25 Scplc1mm:150mms- Source: http://www.doksinet Index to Block 3 BCQllsCY angular ~cceluation angular velocity appendage gear ratio gem wheels lpoundod link basic chain belt drive centripetal acceleration chain drive circular motion coincident point compound link constant-acceleration model consuained slidex dot notation first solvable point fixed point four-link chain friction wheels PROP rule slidu-crank mechanism SLI rule sliding velocity TAN tale IaIlgentirl ~ C Y a t i o n mpntial velocity terminal velocity trunnion velocity dhpam vebw imeee velocity o&in