As the similar triangles OA'Q' and OB'P' stand on common line OB' we conclude that pointsO,Q' and P' lie on the same straight line. Further, as lines OQ'P' and OA'B' cut parallellinesA'D'and B'C', we have from similar triangles OA'Q' and OB'P',

OP' OB' (5.12)= --OQ' OA

OP' = OB = k (5.13)orOQ' OA

As point 0 is a position-fixed pivoVit follows from equations (5.8) and (5.13) thatinstantaneous radii of curvatures at corresponding points P and Q, on the two respective curves,always bear a constant ratio k. Hence, point Q describes a curve similar to the one described bypoint P, to a scale 11k. Similarly, if point Q is made to trace a given curve, point P will trace asimilar curve to a scale k. It may be noted that the scaling factor k is given by

k = OB = BP (5.14)OA AQ

Prior to the development of milling machines, in the late seventeenth century, it was extremelydifficult to machine straight, flat surfaces. Production of good prismatic (sliding) pairs, free of'backlash was therefore considered to be a difficult proposition. It was during this period that theproblem of obtaining approximate straight line motion, using only turning pairs, attracted greatattention of kinematicians. In a majority of cases, a reference was made to the flat segments ofcoupler curves, generated by mechanisms with turning pairs only. Such mechanisms are knownas "approximate straight line motion mechanisms" and they are capable of tracing straight lineduring a finite range of traverse only.

An obvious way to have constrained motion of a point in a mechanism along a straight lineis to use a sliding pair. Sliding pairs, however, have a limitation in that they are bulky and arealways susceptible to comparatively rapid wear. In certain situations, therefore, it is necessaryto obtain straight line motion using turning pairs only. In rest of the cases, the modem practice~.is to produce straight line motions using sliders.

Besides self-recording instruments in indicator mechanism, requiring a point to move alonga straight line, a few other cases requiring a point to move along approximate straight path .illustrated in Fig. 5.3. The figures are self-explanatory.

Constraining a point in a plane to follow a straight line was one of the stipulations which ledL. Burmster to lay the foundations of modem mechanism synthesis. The design of straight linetrajectory crank mechanisms has always attracted great attention of kinematicians and Kempdedicated a whole book to this topic by the year 1877. More recent work in this direction hasled to the establishment of coupler curves and tables from which the best four-bar linkages forstraight line guidance can be extracted.

~,/

I,.r-:Coupler ~

curve "

FIg.5.3 (a) The level luffing crane. The path tracer point P traces approximately a horizontal straight line,(b) A four-bar mechanism used for advancing film of a movie camera. The coupler point P generatesa straight line for a small portion.

Mechanisms for exact and approximate straight line motion may either use all turning pairs,or turning pairs with one sliding pair. The straight line motion mechanisms can be classified intwo important groups of paths:

(a) Mechanisms in which coincidence of the coupler curve arcs with a straight line istheoretically exact (called Exact straight line motion mechanisms), and

(b) Mechanisms in which coincidence of the coupler curve arcs with a straight line is onlyapproximate (called Approximate straight line motion mechanisms).

Compared to exact straight line motion mechanisms, approximate straight line motionmechanismshave fewer links and this turns out to be an advantage of major importance. A largenumberof 4-bar mechanisms exist in this category through which remarkable accuracy can beachieved.As against this, most 'exact straight-line trajectories' are obtainable using mechanisms

six or more links. It must be noted that increase in the number of links and joints increasesthe error due to production tolerances and clearances. This necessitates calculation of tolerancesin cases where extreme accuracy is desired. Chebyshev proposed and carried through design ofmechanisms, for straight line guidance, such that the segment of coupler curve, used forgeneratingapproximate straight line, lies between two selected tolerance limits on linearity.

5..5 EXACT STRAIGHT LINE MOTION MECHANISMS~l

Statement:As illustrated in Fig. 5.4, let 0, Band e represent three distinct points of a mechanismwhichare always constrained to lie along a common straight line as the line oe turns about 081\ centre. As the line oe turns, let the point B move along the circumference of a circle withOA as diameter and, at the same time, let the position of point e be such that the product (0 B)(OC) is constant. Then the coupler point e will move along a straight line ex perpendicular tothe diameter OA.

Proof: Let X be the foot of perpendicular dropped fromCon OA produced. Then triangles OAB and OCX are similarbecause

LOBA = LOXC = 90° and angle LAOB is common tothe two triangles.

Thus, from similar triangles,

OBOA

OXOC

OX = (OB) (OC)(OA)

Clearly, as the diameter of circle OA remains constant,OX is also constant.

In other words, for all positions of point B and line OC, tracing/coupler point C moves alongstraight line in a direction perpendicular to OA produced.

Several mechanisms have been proposed to connect points 0, C and B in such a way thatfor all the configurations of mechanism, condition (OB)(OC)/(OA) = constant, is satisfied. Thisis explained further through Peaucellier and Hart mechanisms.

Noting thatand

(OR2 - CR2) = OF2 - CF2

= (OF + CF)(OF - FC)

OF + CF = OCOF- FC = OB

In order to satisfy conditions stipulated in Section 5.5.1,an eight-link mechanism was proposed by PeilUcellier.The frame link OP also constitutes the radius of circle,along the circumference of which point B is constrainedto move through crank PB. The fixed link PO equals theradius of crank PB. Points B and C represent pins atopposite comers of a rhombus BQCR whose pins Q andR are connected through equal links OQ and OR to fixedpivot O. 0

The tracing point C will describe a straight line CXperpendicular to the line of centres OP extended, if andonly if,

To prove this, join the diagonal RQ of the rhombus, which will bisect the other diagonal Be.From the right angled triangles ORF and CRF,

OR2 = OF2 + RF2

CR2 = CF2 + RF2

(5.15)(5.16) .

(OR2 - CR2) = (OC)(OB)

Since OR and CR are both constant link lengths, the R.H.S. is also constant.Hence, (OC)( OB) = constant for all configurations.

As against eight links in Peaucellier's mechanism, this mechanism consists of six links only. Asseenin Fig. 5.6, link OP is the frame 1 while link PB ( = OP) is the driving crank, through whichit transmits motion to link DG. Links DE, EF, FG and DG constitutes a crossed parallelogram,so that, links DE = FG and DG = EF. Under this situation, for any configuration, lines joiningDF and EG are parallel. Lines joining D, E, G and F thus forms a trapezium.

Link DE is pivoted at 0 to the frame link. Clearly, if points B and C on link EF are selectedon a straight line from 0, parallel to DF, the line will remain parallel to DF and EG in all otherconfigurations. Note that crank PB is connected at B to link DG through a turning pair, whileC is a tracing point on link EF.

Being the end point of crank PB,.point B is constrained to move along circle of radius PB.Draw FM perpendicular to EG. It remains to be shown next that for this mechanism too,

(0 B) (OC) = constant.

From triangle EFG,

EF2 = FG2 + EG2 - 2(FG)(EG)cosLEGF

Now, DEGF being a trapezium MG = (EF - DF)/2, and by symmetry

cosLEGF = (EG- DF)2(FG)

Substituting in equation (5.19),

EF2 = FG2 + EG2 _ 2(FG)(EG) (EG- DF)2(FG)

or EF2 = FG2 + EG2 - [(EG)2 - (EG)(DF)]

Substituting for EG and DF from above in equation (5.20), whose R.H.S. is constant,

(OB)(DE) (OC)(DE)

(OD) (OE)

(OB)(OC) DE2(OD)(OE)

or EF2 = FG2 + (EG)(DF)

or (EG)(DF) = (EF2 - FG2)

EF and FG being link lengths, the R.H.S. of equation (5.20) is constant. It remains to beshown now that left hand side of equation (5.20) equals (OB)(OC). Triangles DEF and OECare similar since all the three sides are parallel. Hence,

DF OC-=-DE OE ~

DF = (OC)(DE)(OE)

Also, triangles DEG and ODB are similar, as all the three sides are parallel. Hence,

EG OB=DE OD

EG = (OB)(DE)(OD)

As link lengths DE, OD and OE are constant it follows from Eq. (5.21) that (OB)(OC)constant. Thus, as all the conditions of Section 5.4.1 are fulfilled, point C traces a straight r .perpendicular to diameter OA. .

Though Hart's mechanism appears to be preferable to Peaucellier mechanism due to smaU(number of links, it suffers on account of large space requirement. Even for a short path ofthe Hart's mechanism requires larger space and this constitutes a great practical disadvantag

This exact straight line motion mechanism differs from earlier two cases in that it incorporaa slider also. As will be appreciated by readers during the course of discussions, this mechanisdoes not generate but merely copies straight line motion.

The mechanism consists of a slider-crank mechanism with crank OB equal in length toof connecting rod BS (Fig. 5.7) and the connecting rod length SB extended to include couplpoint C such that

BC = BS = OB.

The end S of the connecting rod is constrained to move along straight path OS using a slidelement and guides. It can be shown that with this arrangement, C moves along a straight .through 0 perpendicular to OS. For this purpose it is enough to show that underconfigurations, line joining C with 0 makes right angle with OS.

As BC = BO = BS, a circle drawn with B ascentre and CS as diameter will necessarily passthrough point 0 in all the configurations of themechanism. Further, 0 being a point on thesemicircle,LCOS = 90° always.

Hence point C always moves along a straightpathperpendicular to OS through O.

This can also be proved by considering isoscelestrianglesaBS and OBC, in which,

LOBC = LBOS + LBSO,LOBC = 28

LOBS = LOCB + LCOB = 2<1>,

and LOBC + LOBS = 180°, it follows that 28 + 2<1> = 180° or (8 + <1» = 90°

This shows that angle

LCOS = 90° in all configurations.

The main problem with Scott-Russel mechanism is that exactness with which point Cfollowsa straight line obviously depends upon exactness with which slider S is guided along thestraightpath. Needless to say that friction and wear of slider pair is much larger than the turningpairand as such the resulting straight line motion of point C is of little practical significance.

Thisapproximate straight line motion mechanism was used by J. Watt to guide piston rod alongstraightpath in many of his early steam engines. In the absence of any means to machine straightflat surfaces, he had to rely on approximate straight line motion mechanism.

As shown in Fig. 5.8 it is a crossed four-bar mechanism OABQ with cranks OA and QBofunequal lengths and pivoted to frame at 0 and Q respectively. The coupler (tracing) point isso chosen on link AB that it traces a figure of eight as shown in dotted line. The length of linkAB is such that when links QB and OA are parallel, AB is perpendicular to them.

We select a position for which the links QB and OA are parallel and on opposite sides of thelinkAB. In this position the instantaneous centre of link AB is at infinity and accordingly if linkQBbe given a small rotation <1>, the link OA will rotate through e such that

arcBB' = arcAA'

Such a rotation will shift the instantaneous centre to position I as shown. Best position fortracing point C is obtained by drawing a horizontal line IC' to meet corresponding position A'B'of link at C. I being the instantaneous centre (13), point C' will move in vertical direction alonga straight line. For the position QBAO, all the points in link AB move up in vertical direction. Forsmall angles 9 and <1>,

QB 9:= -

OA <1>

Again A'C' =(IC')9 and B'C' =(/C')<1>

A'C' 9Therefore --= -

B'C' <1>

Substituting for 9/<1> from equation (5.22)

(A'C'/B'C/) = QB/OA

Thus, the tracing point C divides the link AB in two parts whose lengths vary inversely asthe lengths of adjacent links.

The Grasshopper mechanism is also known as modified Scott Russel mechanism. This namefollows because the slider of Scott Russel mechanism in Fig. 5.7 is replaced by a lever QB whichoscillates about centre Q with radius QB. In Grasshopper mechanism 0 and Q are the fixedpivots and points B and A are constrained to move along arcs of circle with radii QB and OArespectively. Tracing point C, which is located on extension of coupler link BA, then describesa straight path perpendicular to line OB. For this, necessary condition is that the instantaneouscentre I (= 13) of link 3 w.r. to 1 be such that IC is parallel to OB as shown in Fig. 5.9

There is one more deviation which this mechanism has compared to the Scott and Russelmechanism. Unlike Scott Russel mechanism, the path of coupler point C in Grasshoppermechanism does not pass through the fixed pivot O.

The point C will describe an approximate straight line if,

AC = ( ~~) (5.23)

13C _._._.- ._._.-. _._._._._._._._._._._._._._._.- ._._._.;. Ii . I.'I I.'I I

I 1I I

I I

I I

I 34iB!-._._._._.-._._._._._._._. _._._._._._._ .. -. B"I ....- ,

I ~ B' :I I

I

: .... A"''''-, 4 :I .•.•- .....:

I •••. I'. ,

~ ..._,.... ........J' :C,.",:·~_·_._._._._.- ._._._._._._. _._._._. _._.-' _._._._.-': Q

14

This is again an approximate straight line motion mechanismconsisting of a 4-bar rocker-rocker mechanism in which thecrossed links OA and QB are of equal lengths (see Fig. 5.10).The tracing point C is located at the middle of the coupler linkAB.

The link-lengths are so proportioned that in the extremepositions of mechanism (namely, configurations OA'B'e' andOA''B'' C"), the coupler point C" is directly above A" when ABlies along link QB and coupler point C' is directly above B'whenlink AB lies along link OA. The proportions of the linkscan be established as follows.

Let us consider coupler AB to be of unit length and findout proportionate lengths of other links. Thus, Let,

AB = IOA = QB = x

and OQ=y

Now from right angled triangle OA"Q of Fig. 5.10,

(OA")2 = (QA")2 + (OQ)2

or ~ = (x - 1)2 + lor 2x=1+l

12oI 14

Q

It follows therefore that the proportion of link lengths should be

AB : OQ : OA : AP = 1 : 2 : 2.5 : 2.0

For the configuration OABQ, the location of I.c. 13, which is just below C, justifies the'on of point C in horizontal direction.

's is again a 4-bar rocker-rocker mechanism, with linksectedthrough turning pairs only. As seen in Fig. 5.11,OA and QB are of equal lengths while the coupler

. s the tracing point C on an extension of coupler in thetion of perpendicular bisector of coupler AB. Thec:locationof tracing point on the perpendicular bisector

< canbe decided by considering instantaneous centre Itmationof coupler w.r. to frame link.

I,n Fig.5.11 the I.e. 13 of link 3 w.r. to 1 is located at I., , if the tracing point C is to describe a horizontal,.ghtlinethe point C must also lie on a vertical line from I.is shown in Fig. 5.11. It can be verified that for this

.sm good dimensions can be OQ :AB : OA :PC =3: 0.584: 1.112.

VJENGINE INDICATORS

e mechanisms used in engine indicators demonstrate one of the most interesting'onsof straight line motion. A small piston is acted upon by steam or gas pressure in the. Variations in steam/gas pressure cause displacement of piston against a spring force.ount of displacement of piston is directly proportional to the steam/gas pressure. The;ment of piston is communicated to the pencil which traces variation of pressure in, on a graph paper (or sheet of paper) wrapped around indicator drum which rotatesl!1OOutits axis so as to help plotting graph along x-direction.

,-<Simplex Indicator

< ~e simplest of all engine indicator mechanisms and consists of a pantograph with fixedk;:;rched to the body of indicator and the point Q attached to the piston. The links, CQ and AQ constitute a parallelogram with link BC extended upto the tracing point P.. in pantograph, points 0, Q and P lie on a straight line. This is shown in Fig. 5.12.

n in Section 5.3 point P will reproduce motion of point Q to an enlarged scale givenfactor,

OP = OB = BP = kOQ OA BC

'Steeringgear of an automobile is a mechanism which is used in automobile for turning axes ofrotationof two front wheels, with reference to the chasis, so as to cause the automobile to followsomecurved path. In the steering mechanism of a motor vehicle the front wheels are mounted

'onshortaxles (called stul)-axles), pivoted on separate axis called side-pivot steering. The pivot'pointsare fixed to the chasis. The rear wheels are placed over the back axle, at the two ends,ofthedifferential tube. As against this, in certain types of vehicles such as trailers or steamtractionengines, the forward wheels are mounted on a turntable, free to swivel, so that theiforwardaxle is directed towards the instantaneous centre I which lies on the line of the rear axle.

Ingeneral, when the vehicle takes a turn the front wheels, along with their respective axles,"turnabout the respective pivot points. The rear wheels remain straight and do not turn. Thus,'thesteeringis done by means of front wheels only.

Conditionof correct steering requires that the relative motion between the wheels and the roadsurfaceshould be that of pure rolling while taking a turn, avoiding lateral slip (skidding). This

:ispossiblewhen the wheels move along concentric arcs of circle about an instantaneous centre.(l) of rotation. Under this condition the motion of the entire vehicle can be conceived to be a.'motionof body rotation about a vertical axis through I.

Letr be the wheel radius while RjOli and RoOlo be the radial distances and angular speeds of'rotationof inner and outer wheels respectively. If Ol' be the angular speed of body rotation ofthevehicleabout I, then

, rOlo rOljOl =--=

Ro Rj

The condition of correct steering therefore dictates that, while turning, the wheels should. be in planes which are tangential to the respective concentric paths. As seen above, this isPossibleonly when the axes of two front stub-axles, when produced, intersect the common axisofrear wheels at the same point which is the instantaneous centre I.

. CF AEFrom the geometry of FIg. 5.16, tane = -'; and tan<jl= -FI EI

EI FIcot<jl- cote = - - - (5.44)AE CF

AE = CF = H, the wheel base

Substitutingin equation (5.44),

(cot<jl_ cote) = (EI - FI) = EFH H

EF = W, the distance between pivots

Hence,for correct steering, the inner and outer stub-axles should rotate through angles 6 and~respectively such that

W(cot<jl- cot6) = - (5.45)H

W = Distance between pivotsH = Wheel base

'--' '--'~ Whe~/rack ~

Wheel tread width

Flg.5.16 Automobile steering gear.

5.8.2 Davis Steering GearSpecial features of this steering mechanism (Fig. 5.17) are: t

1. The bell-crank levers BAK and DCL, which consists of slotted arms AK and Crespectively, while the other arms constitute stub-axles for front wheels.

2. The angles of the bell-crank levers namely, LBAK and LDCL, are equal.3. The die-blocks M and N, pin-connected at the ends of sliding link MN, provides a sliding

pair between die-blocks and the slotted arms.4. The link MN is constrained to move in a direction parallel to link AC by two guides

and G2 as shown. .5. The bell-crank levers BAK and DCL are pivoted to the front axle at A and C called king:

pins.

The sliding is effected by sliding link MN either to the left or to the right. When the steeringgear is in the mid-position and the car is moving straight, each of the slotted arms AK and Cis inclined at an angle a to the longitudinal axis of the car. If the link MN is now moved to theright through a distance x, relative to the chasis, the bell-crank levers BAK and DCL are movto the position shown by dotted lines. The stub axle lines BA and CD in new position B'ACD' when produced, intersect at P.

Let cl> and e be the angles through which the arms AK and CL are turned as a result ofdisplacement x of link MN to the right. Let the new position of sliding link MN be M'N', ancorresponding position of bell-crank levers be K'AB' and [:CD' as shown in Fig. 5.17. Let h bethe distance of MN from wheel axis AC and 2b be the difference between lengths MN and AC:When the vehicle moves straight LKAR = LLCR' = a and

tana = (*)

·c.. Dr-'-'--. !- e

-'-. iJ- -'-.-.--'-.-.-. -'--'-. -'-. '-.-'-. -'--.-. '-.'-'-'-'- -'- p'- - - ..::-.::-.:::..:,.,. .•...-

With link MN shifted to next position M'N', angles of inclination of arms KA and LC with'tudinal direction change to (a. + ep)and (a. - a) respectively. Thus,

LK'AR = (a. + ep) and LL'CR' = (a. - a)

From the geometry of figure,

tan(a. + ep)= (_b_:_X)

tan (a. _ a) = (_b _~_X)

tan (a. + ep)= _tan_a._+_tan_ep_1- tan a. tanep

tan (a. _ a) = _tan_a._-_tan_a_1+ tan a. tan a(

bstituting for tana., we have

tan (a. + ep)= _(_bl_h)_+_ta_n_ep_1- (blh) tanep

tan(a. _ a) =' (plh) - tan a1+ (blh) tan a

stituting for L.R.S. from equations (5.47) and (5.48), in above equations,

(b+X) = (blh)+tanep

h 1- (blh) tanep

(b - x) = (blh) - tan a

h 1+(blh) tan a

W2(tana) = -H tana = (2:)

Simplifying (5.51) through cross-multiplication,

(b + x) - b(b + x) tan<\>= b + h tan<\>h

or [ h + b2: bx )tan<\>= x or tan<\>= (h2 +:~ + bx)

Also, simplifying (5.52) through cross-multiplication,

(b - x) + b(b-x) tane = b - htaneh

[ h + b2

~ bx ] tane = x

tane = hx(h2 + b2 - bx)

From equations (5.53) and (5.54), therefore

(h2 +b2 +bx) (h2 +b2 - bx)cot<\>- cote = ------hx hx

Therefore, for Davis steering gear, cot<\>- cote = 2(tana)

However for correct steering condition, from (5.45), cot<\>- cot e = WH

Hence, correct steering with Davis steering gear is obtained when

IIn this case, the point of intersection P of front stub axles in new position will lie on the

extension of rear axle line. With this condition, namely, tana = W/2H, Davis steering gear wibe equally effective at all the angles of steering in either direction.

Davis steering gear, however, has an inherent weakness. It can, at best, be described as atheoretically correct steering gear. The presence of sliding pair involves more of friction. Wearat the sliding surfaces produce slackness and, as a result, original accuracy is soon lost. For this.reason, even though Davis steering gear gives correct steering at all angles of turn, it is rarelyused in practice.

Letus now consider Ackerman steering gear in which sliding pairs are eliminated and are replacedby turning pairs. The gear is more commonly used and gives an approximation good enough forpractical purposes.

This steering mechanism consists of a four-bar mechanism involving turning pairs only. Thetwoopposite linksAC and KL, of unequal lengths, are parallel in normal position when the vehiclemoves straight. The other two links AK and CL, which are the arms of bell-crank levers, are ofequallengths. These arms are inclined at an angle of a to the longitudinal axis when the vehicleis moving straight. Under this condition, the four-bar mechanism AKCL constitutes a trapeziumwithparallel and unequal sides AC and KL.

In order to steer the car to, say right, the link CL is turned clockwise so as to increase anglea. This results in c.c.w. rotation of arm AK, reducing angle a at the other end. For a given valueofratio (AK/AC) and angle a, a unique value of angle S may be obtained for a given value <peithergraphicallyor analytically.

From the geometry of Fig. 5.18,AQ-CQ

cot~ - cotS = ---------PQ

(cot~ - cotS) = (;~)

-. _'. :..-.---.; ..•..."..:.

whereP is the point of intersection of stub-axles B'A and CD' produced. Clearly, for correctsteering,the point of intersection P of front stub axle lines should lie on rear axle line producedif necessary.In other words, equation (5.56) should conform with

W(cot~ - cotS) = -H

CL [sin (a + e)sina] = AK[sina - sin(a - <\»]AK = CL, simplifying further,

(sinacos e + cosasine) - sina = sina - (sinacos<\>- cosasin<\»sina(cose + cos<\>- 2) = cosa(sin<\>- sine)

sin<\>-sine(cose + cos<\>- 2)

For private cars with AC: AK = 8.5: 1 and a = 18°, if the distance AC is 0.4 times thewheel-base, Ackerman steering mechanism would give correct steering for e z 24°. Thus wi~above proportions, Ackerman steering gear provides correct steering in the following thretpositions onlyJ

(i) When moving straight, i.e., when e = 0°.(ii) When turning to the left at e z 24°.(iii) When turning to the right at e z 24°.

At all other values of angle e, therefore, wheels tend to follow path...along circular arcs that.do not have a common centre. In all such cases, the instantaneous centre I does not lie on theaxis of rear wheels but lies on a line parallel to the rear axle towards the front side. It followSthat some amount of skidding is bound to occur at steering angles other than e = 0 and 24°'Hence, angle a and the proportions of the mechanism should be so chosen as to reduceinherent tendency to skid to a minimum, especially for turning radii that could be followed athispeeds, where the centrifugal action also contributes to skidding.

Despite the limitation stated above, Ackerman steering gear is still the most commonlY'steering gear of the two. The advantage of the Ackerman steering gear lies in the use of revo\u(pin) joint rather than the sliding pairs. Resulting simplicity facilitates in upkeeping. Further,will be shown later, resulting deviation from condition of correct steering is small in the caseAckerman steering mechanism. \

The position of cross-arm (i.e., the arm of bell-crank lever) corresponding to correct anec of steering, relative to the maximum angle of steering are governed by the design requiremenSmaller values of angle ec are most frequently used while large angles occur infrequently. It .usual therefore to select correct steering angle as equal to one-half to two-thirds of the maximuangle of steering, i.e., between 20° and 25° approximately. Angle a, needed for providingspecified values angle ec of correct steering, is established below. Note that expression tan a =WI2H of equation (5.55) is valid only for Davi's Gear.

Required Value of a, Neglecting ObliqUity of Cross-LinkLet e and <\>denote angles of steering corresponding to correct steering position. Now, forsatisfactory operation of Ackerman steering gear, movement in a direction parallel to AC(see Fig. 5.18) of ends K and L of cross-bar KL must always be same. Thus for displacementsshown in Fig. 5.18,

projection of arc LL' on AC = projection of arc KK' on AC.Treating arcs LL' and KK' to be equal to chords, for small angles e and <\>,we have

Unlike Davis' steering gear, thus, the expression for a is not quite simple (compare withequation5.55). Thus to obtain desired value of a, one needs to know about the dimensions, likewheelbase H and distance between pivots W of a 4-wheeler. With (WIH) ratio thus known, valueofcorrect angle of steering </)for outer wheel, corresponding to each arbitrarily assumed valuesof angle e -for inner wheel, can then be calculated from,

Wcot</)- cote = -

HWith the set of values e and </)for correct steering established, corresponding value of~

requiredangle a can then be calculated from equation (5.57). These calculated values of </)andIX for a set of values, e = 10°, 15°, 20°, 25°, and 30°, are listed in Table 5.1.

Table 5.1 Table for optimum value of a for each value steering angle Q.; for WIH= 0.4 satisfying condition ofcorrect steering

Set of values for a and <I>to satisfy condition ofcorrect steering: cot <I>- cot a = WIH = 0.4

Angle a required for providing correct steeringat a given set of values a and <I>

a _{ 1 ) tana =sin <I>- sina aO<I>= tan 0.4 + cota

cosa + cos<l>- 2

1. 10° 9.35° 0.3927 21.44°2. 15° 13.605° 0.3797 20.79°3. 20° 17.626° 0.3658 20.09°4. 25° 21.455° 0.3488 19.228°5. 30° 25.128° 0.3296 18.244°

-EXAMPLE 5.1 In a Davis steering gear, the distance between the pivots of front axle is 1 metreandthe wheel base is 2.5 metre. Find the inclination of the track arm to the longitudinal axis of

Uhecar when it is moving along a straight path. (DAV Indore, Nov.-Dec. 1986)

Refer to Figs. 5.16 and 5.17.

w= 1 m = 100 emH = wheel base = 2.5 m = 250 em

As shown in the discussions on Davis steering gear, for correct steeringWcot</)- cote = -H

which leads to the condition for Davis gear asWtana =-2H

= 100 = ~ = 0.2(250)2 5

a = 11.309° = 11° - 18.54'

5 10 15 20 25 30 35 40 45Graph: Variation of II>V s 9 -? 9°

Hooke's joint is used to connect two intersecting shafts. The axes of the shafts usually make,smallangle <X with each other which may vary during operation. The Hooke's joint finds its useinautomobiles for power transmission from the gear box to the back axle. The joint permit some

unt of angular misalignment in connected shafts. This becomes a very desirable feature oncount of possible misalignment arising out of elastic frame and also due to spring in

;mspension.Sometimes two such joints are used in the same drive system, as in automobiles,whereone joint is used at the power transmission and the second joint at the rear axle. Hook jointlalso finds application in the transmission of the drive to the spindles of multi-spindle drilling, chine.

Verticalplane

Fig.5.19 Hooke's (cardan) joint.

Essentially a Hooke's joint consists of two semi-circular yokes (forks) 2 and 4 formed attheends of driving and driven shafts (see Fig. 5.19). Bearings are provided in the yokes to receive

the arms of right angled cross 3. This arrangement ensures pin connection between the croarms and yokes.

Although both shafts must complete one revolution in the same time-interval, it is easyshow that the angular velocity ratio of the two shafts is not constant all throughout the revolutiIt will be shown that the angular velocity ratio varies as a function of the angle a betweenshafts and the angle of rotation e of the driver.

Essentially a Hooke's joint belongs to the category of space-mechanism and represents·spherical four-bar linkage. Here, the axes of the four revolute pairs are not parallel, but havecommon point of intersection at the centre of an imagin!U'Ysphere on whose surface all the £..revolute pairs move. The links are not straight but have a curvature same as that of great circof the imaginary sphere. In the case of Hooke's joint, three of the links subtend angles ofand the fourth link 1 subtends an angle of more than 90° (see Fig. 5.20).

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·_·_·-r-·.i.-'·~.\.·""O "',.,

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5.9.1 Transmission Characteristics

With reference to Figs. 5.19 and 5.20, to an observer looking along the axis of driving shaftshown) point B on the input crank will appear to move along arc of a circle with DB as rThis is obviously because the plane in which arm DB is rotating is normal to the directionviewing. However, true length of arm DC will be seen only in the position shown in Fig. 5as also in position obtained by rotating it through 180°. In all other positions of rotation,apparent (shorter) length of arm DC will be seen. This is obvious as only in position DC theis parallel to the plane of projection (normal to direction of viewing). In Fig. 5.21(a) therepath of point B, as seen by a viewer looking in the direction of arrowhead, is indicated by a c'As against this, path of point C will be seen as an ellipse with major axis along arm OC ofcross. Figure 5.21(b) shows the traces of planes of rotation of the two arms of the cross.arm DC of the cross revolves in a plane at an angle of a to the plane in which arm OB ofcross revolves about driving shaft.

Let the arm DB of the cross rotate through an angle e. Then, in the plane of projec .represented by trace PP in (Fig. 5.21 b), the arm DC will also appear to have rotatedangle e (this is because DB and DC are the arms of the same rigid cross), and point C willalong the elliptical path to occupy new position DC,. Actual angle of rotation (true angle) ofDC, however, has to be obtained in its own plane of rotation represented by trace QQ. Since

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I,i-- Axis of driving

" I shaft", ~ I 1

". '0' M::N'p -'-'-'-'-'-'-'-'-'-'-'-,~f':-']a'-'-'T:-'-'-'-'-'- p

Axis of driven ,/ , : :h ft· I Isa............... .' , "

~.' •••• 1:',. N ' ......-,

arms OB and OC are of the same length, actual path of point C is identical to the path of pointB, which is shown as circle in Fig. 5.21(a). A line C1Czdrawn parallel to the line OB to meetthecircle at Cz then identifies OCz as the actual position of line OC. Angle LCOCz = $ therefore,gives actual angle of rotation of ~ OC in its own plane of rotation.

In order to establish relation between input angle of rotation a and the output angle $, let usproject points C1and Cz on the traces PP and QQ as also on vertical diameter CC'. Let M andN' be the foots of perpendicular on trace PP from points C1and Cz. Point N is obtained on traceQQ by drawing an arc with 0' as centre and O'N' as radius. Clearly NM is at right angles tothe trace PP. Also drop perpendicular CIL from point Cion vertical diameter CC'.

From right angled triangle O'MN of Fig. 5.21(b),

O'M-- = cosa.O'N

O'N = O'N', and hence,O'M = cos<XO'N'

Again, from Fig. 5.21(a) and (b),

O'M = OM = LC1 = OL (tana)

And (fN' = ON = LC2 = OL(tan<1»

Substituting for O'M and O'N' in equation (A),

(OL) tane----= cosa(OL) tan <1>

Equation (5.58) relates angle of rotation <1> of the driven shaft to the angle of rotation e ofdriving shaft through the angle a between the two shaft axes.

Differentiating equation (5.58) with respect to time t,

d<1> de(sec2

<1» - = (sec2e) - secadt dt

Angle of inclination a between the two shaft axes is assumed constant for the purpose of

differentiation above. Remembering that d<1>== 0)' is the angular speed of rotation of driven shaftdt

and de == 0) is the angular speed of rotation of driving shaft, from equation (5.59) we havedt

, [sec2e seca)0) = 0)

sec2<1>

sec2<1> == I + tan2

<1>

sec2<1> = I + (taneseca)2

2m cos2e cos2a + sin2esec 'I' = cos2ecos2a

2m (1- sin2a) cos2e + sin2esec 'I' = ---------

cos2e cos2a

But

or, using Eq. (5.58),

2m [1-sin2acos

2e)sec 'I' = cos2a cos2e

Substituting for sec2<1> in equation (5.60),

, [sec2eseca) cos2acos2e0) = ---- 0)

1 1- sin2a cos2e

0)' = 0)(-1-_-S-i:-~-:-ac-o-s-2e-)

It follows from equation (5.61) that for a constant input speed 0), output speed 00' chwith angle of rotation e of driving crank.

Polar Angular Velocity DiagramFor a maximum value of 0)', it follows from equation (5.61) that the denominator shou(

minimum, which requires that for given a, cos28 be maximum. Hence maximum value of ro'occurs when 8 = 0, and 1t, for which .

ro' = ro C ~::~a)Thus (ro')max= (~) (5.62)

cosa

and occurs at 8 = 0 and 1t.

Again, for minimum value of ro', cos28 should be minimum. This ~ccurs when a = ~ or 31t,2 2

(ro')min = ro(co;a) = (ro)cosa (5.63)

Further ro' = ro, when the ratio ro' in equation (5.61) is equal to 1. This requires that,ro

ro' = 1 = cosaro 1- sin2 cos2a

cosa = 1 - sin2a cos28

cos2a = 1- cosa = (1- cosa)sin2a (1- cosa) (1 + cosa)

cos28 = 1 _(1+ cosa)

sinza = 1 - C+:osa) = C::::a)tanza = cosa or, tana = ± .J cosa

a = 30°, and ro = 3 radlsro = ro' requires a = tan-\±.J cos30)

a = 42.94, (180 + 42.94) and a = - 42.94 and (180 - 42.94)8 = 42.94; 222.94; -42.94 and 137.058°

And, therefore,

Taking

or, wheni.e., atFurther

(ro')min = ro cos30

(ro')min = 3 x 0.866 = 2.598 radls which occurs at a = 1t and 31t2 2

ro(ro')max = --- = 3.46 radls which occurs at 8 = 0 and 1t

oos30°

- (oo2cosa) sin2a sin2e(1- sin2acos2e)2

The elliptical curve, shown in broken line in Fig. 5.22, is the polar diagram representing variableoutput velocity 00'. The circle in full line represents input velocity 00 which is constant at all e.

Fig. 5.22 Polar velocity diagram for a = 30°.

Angular Accelerations in Hooke's joint

From equation (5.61),

00' = oo( 1- Si:~:~Os2e )

The angular acceleration, for constant a, is given by

doo' doo' de doo'-- = ---- = 00--dt de dt de

doo' = oo[-cosa {O + 2cose sin2a Sinel]de (1- sin2a cos2e)2,

- (oocosa) sin2a sin2e=---------

(1- sin2a cos2e)2

doo'

dtThus

Angular acceleration of driven shaft,

doo' - (oo2cosa) sin2a sin2e-- = 2 2 2 (5.~d t (1 - sin a cos 9)

The value of 8 at which the acceleration is maximum, is obtained by differentiating equa .(5.64) w.r. to e and equating the same to zero. The resulting equation will not be simple.'A simplified e~pression for e at which acceleration is maximum is- .

cos28 "" ( 2sin2

ex. ) (5.65)2-sin2ex.

Even for values of ex. as high as 30°, the expression (5.65) is sufficiently accurate within afew minutes of angular measure. It may be of interest to have a look into the effect of angle ex.,included between shaft axes, on the percentage variation of angular speed of driven shaft. Fromequations (5.62) and (5.63), it follows that

Percentage of speed variation

= oo:ru.x - oo~ x 10000

Effects of Large Angular AccelerationHooke's joint involves a rigid connection be-tween the connected shafts and as such fromthe point of view of torque transmission, mo-ment of inertia of connected parts should besmall.In case this is not so, very high alternatingtorques, and therefore very high alternating·stressesmay be set up in the parts of the joint.For instance, the angular acceleration expres-sion (5.64) for ex. = 20° modifies as under. Fora = 20°,

x~_.~:3I :3 40

J~ 20

30 40a in degrees

Effect of angle aon % variation in outputspeed.

doo' - 002 (0.1099)sin28=-------dt (1- 0.117cos28)2

Thus, the angular accelerations at different values of 8 are:

6= 0° 30° 60° 90° 120° 150° 180° 210° 240°

dro' 0 -{).114ci -{),101ro20 O,101ro2 0.114ro2 0 -{).114ro2-{).101ro2d(=

9= 270° 300° 330° 360°

dro' 0 O,101ro2 O,114ci 0d(=

It is concluded that the angular acceleration varies cyclically between positive and negativemaximum value through zero. Thus frequency of fluctuation is twice during one revolution.Further, angular acceleration is directly proportional to the square of angular speed 00. Thus athigh speeds, the problem becomes much more severe.

Again at a = 30°, the expression 5.64 reduces to

doo' OOZ (0.2165) sin2e---;jf = - (1- 0.117cosz e)z

A comparison with the corresponding expression for a = 20° reveals that larger the shaftangle a, higher is the acceleration and, therefore, this angle should' be kept as small as possible..

tan<1>= (tane seca)

and at the second universal joint, considering", as the input angle and <1>as output angles

tan<1>= (tan", seca)tan", = tan<1>cosa

It is possible to connect two shafts by two Hooke's couplings through an intermediate shaft suchthat the uneven velocity ratio of the first coupling will be cancelled out by the other one.

Thus, two parallel or intersecting shafts may be connected by a double universal joint andhave uniform output motions, provided that the intermediate shaft makes equal angles with theconnected shafts and that the forks on the intermediate shaft are in the same plane.

Let e be the angle of rotation of driving shaft and let the intermediate shaft rotate througllangle <1>.Again, at the other end of intermediate shaft, let '" be the angle of rotation of ~.intermediate shaft for the angle of rotation <1>of the intermediate shaft. Then at the first universaljoint,

(d<1» secze

de = secz<1> cosa

and differentiating (5.68) w.r. to <1>,

d", = secz<1> cosad<1> secz",

Multiplying (5.69) by (5.70) on respective sides,

d", = secze = ( 1+ tanze)de secz", 1+ tanZ

",

Substituting for tane from (5.67) and for tan", from (5.68),

d", = 1+ (tan <1>cosa)z = 1de 1+ (tan <1>cosa)z

Thus the angular speed of input shaft is equal to the angular speed of output shaft and thatsecondHooke's joint exactly neutralises the effect of the first one. Necessary conditions tofulfilledby a double Hooke's joint of this type are as under.

Twoshafts that are neither parallel nor intersecting may be connected by a double Hooke'st and will have uniform angular speed ratio, provided that1. intermediate shaft makes equal angles with the connected shafts (the input and output),

and2: forks on the intermediate shaft are so arranged that they simultaneously lie in the plane

of shaft axes at the respective ends.

Clearly,the intermediate shaft is the only shaft in a double Hooke's joint which is subjectedmaximumaccelerations and retardations and as such, in order to cut down high alternatingrtiatorques to and high values of alternating stresses, the intermediate shaft should be as lightpossible.

PLE 5.3 Two shafts are to be connected by a Hooke's joint. The driving shaft rotate atuniformspeed of 500 r.p.m. and the speed of driven shaft must lie between 475 and 525 r.p.m .. tenninethe maximum permissible angle between the shafts (Univ. of Indore, April 1976, May

2; DAVV,March 1990).

lution: When the driving shaft rotates uniformly at N r.p.m., the maximum and minimumd of driven shaft are given by,

~i L~a) and (Ncosa)n;lince,N = 500 r.p.m., it follows that maximum variation of speed is given by

N (_1_ - cosa) = (525 - 475) = 50cosa

Ch81lt1r:a

MECHANISMS WITH

PAIRS

As discussed in section 2.4, a linear motion lower pair (e.g., turning pairs, prismatic pairs andscrew pairs) has a degree offreedom of one, as each point on one element of the pair can move

lonly along a single line/curve relative to the other element. Unless otherwise mentioned, the term'~lower pair will be used to describe linear motion lower pairs. Since lower pairs involve surfacecontact rather than line or point contact, it follows that lower pairs can be more heavily loaded forthe same unit pressure. They are considerably more wear-resistant. For this reason, developmentin kinematics has centred more and more around lower pairs. It may be pertinent to make apassing reference to another common term Linkage which denotes a mechanism consisting onlyof lower pairs.

5.2 OFFSET SLIDER-CRANK MECHANISM AS A QUICK RETURNMECHANISM

In many situations, mechanisms are required to perform repetitive operations such as pushingparts along an assembly line and folding cardboard boxes in an automated packaging machine.Besides above there are applications like a shaper machine and punching/rivetting press in which·working stroke is completed under load and must be executed slowly compared to the returnstroke. This results in smaller work done per unit time. A quick return motion mechanism isuseful in all such applications.

A quick return motion mechanism is essentially a slider-crank mechanism in which the sliderhas different average velocities in forward and return stroke. Thus even if crank rotatesuniformly, the slider completes one stroke quickly compared to the other stroke. An offset slider·crank mechanism, shown in Fig. 5.1, can be used conveniently to achieve the above objective.

In Fig. 5.1, the crank centre 0 is offset by an amount e with respect to the line of strokeof slider. If r be the radius of crank OC and I the length of connecting rod CP, the extremeright-hand position PR and extreme left-hand position PL of slider P occur when the crank and