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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.
TVps name
TAN b-ll
SLI isIWiw
Link (position diagram)
Existing polnm on miocHy diagram
cb c
bed imaw of BC
bd BD cd 1 CD bed image of BCD
PROP (pmponlonl
B
b , c
pD b , ~
B
\ b
bc
bc parallel to sliding surface
bc- (W). if known
= ipmponion)
ConstrucHon of ~ l o c i t v d i w n m
Ic
F
Notg
\
TAN (- 1 i.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 don't 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.
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?
S A 0 2i (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.
S A 0 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.
S A 0 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.
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
BA 0 9
On Figure 68 dresswe bc and halve it to locate m on the line bc.
Sa Figure 69.
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"
Ficre 70 shows the velocity diagram.
P i p e 70 Velocity diagrmn for SAQ 10 Scalelmm:mmms-'
(8) No.
(b) P.
(c)
l Velocity m l y J l a. 0 is the hxed point. (@,),-lOms-L+ =@ (h)p=?ma-'/(LtoBP) (h)O=? m S-'\ (L to BO) 2 Draw^ the velocity^ diagram^ to the^ d^ e^ given, Figurc 70. ( Y ) ~ ob (59 X 0.2) 3 OB 0.5 0. rad 8-l
h0- 23.6 rad S-'>
By proportion
(^5) Mark m on your velocity diagram. Join om 6 ( & ) , = m i - 4 9 ~ 0. 2
-9.8 m 8 - ' 4 13"
BA 0 l 1 On the complete velocity diagram of Figure 31(b) join o to r and measure its length and direction relative to the v c r t i d (q)O=H=(65.5 X 0.05) 1 = 3.3 m S-'T82'
BA 0 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).
Figure 71 Velocity diagram of h i e cludn, SAQ I (hUactucrl size)
= 8.6 rad S-'
&==8.6rads-'> 4 Proportion to determine m:
qm = 44 X 0. = 13.5 mm (Scc Figure 72).
Figure 72 Veloclty dfqgramfor SAQ 12 ( W a c t u a l size)
BA 0 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.
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. ( ~ ~ = 1 0 ~ 0. 3 ~ - 3 r n r - ~ ~ - i P ( ~ ) , =? / ( I t o T U ) u t o t l i n e (h),-?(+toTS)atorlinc (^2) Draw the velocity diagram for the basic chain
SA0 14
Fbat Input to Basic ~01vcrblc Flsure llnk c h i n Appendages point
41 OB OBCG - B
42 UR URQP RSTO* R
43 sliderats WTSO - S
44 Slidant E DCEG OBC E
45 08 OBCD CEO B
SA0 l
The bssic chain is OBCG. OBE is a compound link
appendage driviug 6nal appsadapc EHE Veloflty analysb: 0. G and H uo Gxcd points. 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).
Flgure 74 Velocity dlagmmfor SAQ I S Scak l mm: 1 0 0 m m ~ - ~
(E,),,=w=3X1ms-~J
I
Figure 78 Velocity diagram for SAQ 21 Scalelmm:5mms-'
I -
MO n Figure 78 shows the velocity diagram.
Dimensions from pmition digram 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.
The basic chain is BCDE; the appendage is DFG. From the position diagram,
BC = 0.1 m; CD = 0.38 m;
ED - 0.15 m; DF = 0.5 m For BCDE (U,), = 5 X 0.1 I - 0.5 m S-' I = E (tan) (h), =? m S-' (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) (h)D=?ms-'/(h.toFD,tm) ( h ) O - 3 = 9 6 x o.mst -0.48 m S-'+
SA0 22
See Fium 79.
Figure 79 Velocity diagram for SAQ 22 ScaleImm:25mms-'
BC=CD=0.6m CF=DE-0.6m The basic chain is OBDE. Note: OBCTF is not a four-link chain (remcmk: tmnnion is a Link). Velocity anulysls: 0. F. E am fixcd pointa.
&),-l x0.25k =0.25ms-' -
( G ). =? r n ~ - ~ f(LtolinkDB) (%),-?m~-~(LtolinkDE) c is half-way along db; so it is casily located. T is on the link CG whcident with E (h), =? m S-'\ (A to link CG) (h)p =? m S-' /(along link CO) Draw the velocity diagram to scale (Figure 79). -8 =(23.5 X 0.0029 d
= 0.34 rad S-' 3
SA0 23
, r h Figure 80. VeIodty adysls: 0 , M and G arc fixed points. The basic chain is OPSG. ( 4 ) 0 = 1 0 x 0. Z ~ = Z O m s ~ ' ~ = g ( t a n )
(&.),-?m S-'/(L to link SP, tan)
(h), =? m S-' /(along slot, sli)
For QRM
pq P Q 0.
ps PS 0.
pq = 0.357 X 88 - 31.5 mm (h),=? m S-'(tan) ( g ) , =? m s - ' ~ ( s l i )
Solution
(h),, = W =(26 X 0.025) c -0.65ms-'+
(".)~3-89~0.M53 =4,Srads-'>
%,=QR- 0.
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. (^) Figure 80 Velocity diagram for SAQ 23
(%),=S x0.2 ,= =l.Oms-l %=&(tan) Scale 1^ mm^ :^25 mm^ S-' (q). =? m S-' /(along link FTC, sli)
(h), = 7 m S-' \ (L to link TC, tan)
Draw the basii chain octb.
For CDEG
cd CD 0.
et CT 0.
cd-0.67 X 95=64mm
(E,), = l m S-' /(L to link DE, tan)
Complete the velocity diagram (Figure 81).
-P= 63 X 0.01 + =0.63 ms-' +
=^ - - = ed>-^ 3 1 ~ 0. 0 1^ Figure81 Velocity^ diagram^ for^ SAQ^^24 (Mf^ actual^ size)
ED ED 0. rad S-'>
=0.39 rad S-'> The velocity diagram for the basic chain can now be drawn: o h.
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.
(Ec),= 147 X 0.05 =, = 7.35 m B-' I=&(tan)
(B,), = 7 m S-' f (sli) Velocity analysis: 0. B and G are fixed points. The input (F&=? m s - ' / ( L to CF, tan) data is to link OA. The basic chain is OAEB. The velocity diagram can now be completed (Figure 82). (E,), = 500 X 0.03 r = l5 m S-' r = m(tan) (^) ( ~ , ) ~ = a = 4 5. 5~ 0. 1 5 ~~ 6. 8m 5-1 T
(E*)e= 7 m S' l ". (along line CED, sli)
Note that the velocity image of the cranked link appears
(Ee). -? m S-' (L to line BE, tan) in the velocity diagram; it is ahown as a dashed line.
Index to Block 3
BCQllsCY
angular ~cceluation
angular velocity
appendage
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
gear ratio
gem wheels
lpoundod link
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
