Principles of Engineering Mechanics (2nd Edition) Episode 4 pot

uUA = --o rl erl + hrl eel Figure 5.14 2 If AB is of fixed length, then only two depends on the angular velocity, which is known from the velocity diagram, and the other term depends

In Fig 5.10 the instantaneous centre for member BC is found to be the intersection of AB and DC, since the velocity of B is perpendicular

to AB and the velocity of C is perpendicular to

CD

If the velocity of B is known then

(5 * 8)

VB OC V E

we=-=-=-

I2B 12C IZE Each point on link CB is, instantaneously, rotating about 12

Figure 5.9

Returning to Fig 5.6 and assuming that OAB is

anticlockwise and of given magnitude, we can 5.6 Velocityimage

place the points a, b and d on the diagram

(Fig 5.9) Note that U and d a r e the same point as

there is no relative velocity between A and D

T O construct the Point C we must view the

motion of c from two vantage Points, namely D

and B Since DC is of fixed length, the only

motion of C relative to D is perpendicular to DC;

hence we draw dc perpendicular to DC Similarly

the velocity of c relative to B is perpendicular to

CB; hence we draw bc perpendicular to BC The

intersection of these two lines locates c

The angular velocity of CB is obtained from

v~/B/CB (clockwise) The direction of rotation is

determined by observing the sense of the velocity

of C relative to B and remembering that the

relative velocity is due only to the rotation of CB

Note again that angular velocity is measured

with respect to a plane and not to any particular

point on the plane

5.5 Instantaneous centre of rotation

Another graphical technique is the use of

instantaneous centres of rotation The axes of

rotation of D C and AB are easily seen, but BC is

in general plane motion and has no fixed centre of

rotation However, at any instant a point of zero

velocity may be found by noting that the line

joining the centre to a given point is perpendicu-

lar to the velocity of that point

If the velocity diagram has been constructed for two points on a rigid body in plane motion, then the point on the velocity diagram for a third point

on the link is found by constructing a triangle on the vector diagram similar to that on the space diagram Hence in our previous example a point

E situated at, say, one third of the length of BC from c will be represented on the velocity diagram by a point e such that ce/cb = 5 , as shown

in Fig 5.9

More generally, see Fig 5.11, since ab is

perpendicular to AB, ac is perpendicular to AC and bc is perpendicular to BC, triangle abc is

similar to triangle ABC

Figure 5.10

Figure 5.1 1

Problems with sliding joints

In the mechanism shown in Fig 5.12, the block or slider B is free to move in a slot in member AO

In order to construct a velocity diagram as shown

in Fig 5.13, we designate a point B’ fixed on the link A 0 coincident in space with B The velocity

of B relative to C is perpendicular to CB, the velocity of B’ relative to 0 is perpendicular to

OB ’ and the velocity of B relative to B ’ is parallel

to the tangent of the slot at B

Figure 5.12

5.9 Simple spur gears 57

Figure 5.13

The two mechanisms used as examples, namely

the four-bar chain and the slidercrank chain,

employ just two methods of connection which are

known as turning pairs and sliding pairs It is

remarkable how many mechanisms are con-

structed using just these simple arrangements

Figure 5.17 5.7 Acceleration diagrams The complete acceleration diagram for the Having constructed the velocity diagram, it is now mechanism can now be constructed as shown in possible to draw the relevant acceleration Fig 5.17 (see also example 5.1) The acceleration diagram The relative acceleration between two of C is given by the line ac and the angular points is shown in polar co-ordinates in Fig 5.14 acceleration of CB is given by cc'/CB (clockwise),

since cc ' = hcB CB

5.8 Acceleration image

In the same way that the velocity of a point on a rigid body may be constructed once the velocities

of any two other points are known, the

accelerations of two other points

uUA = o rl erl + hrl eel

Figure 5.14

2

If AB is of fixed length, then only two

depends on the angular velocity, which is known

from the velocity diagram, and the other term

depends on the angular acceleration, which is

unknown in magnitude but is in a direction

perpendicular to AB

components remain (see Fig 5.15) One term

am = o 2 r2er2 + hrzee2

_ Figure 5.18

From Fig 5.18, the angle between (IUA and

Referring to the four-bar chain shown in

Fig 5.6 and given the angular acceleration of link

AB, the acceleration vector of B relative to A

may be drawn (Fig-5.16)- Note carefully the

directions of the accelerations: B is accelerating

centripetally towards A

arctan (2) = arctan (3)

which is independent of r l The angle between

QUA and aB,C is therefore the Same as the angle

between rl and r 2 ; hence the triangle abc in the

acceleration diagram is similar to triangle ABC

5.9 Simple spur gears

When two spur gears, shown in Fig 5.19, mesh together, the velocity ratio between the gears will

be a ratio of integers if the axes of rotation are Figure 5.16

(5.11)

@ A - wC r B

% - @ C rA

-

or

Figure 5.19

fixed If the two wheels are to mesh then they

must have the same circular pitch, that is the

distance between successive teeth measured along

the pitch circle must be the same for both wheels

If T i s the number of teeth on a wheel then the

circular pitchpc is r D l T , where D is the diameter

of the pitch circle The term ‘diametral pitch’ is

still used and this is defined as P = TID Another

quantity used is the module, m = DIT

The number of teeth passing the pitch point in

unit time is 27rwT, so for two wheels A and B in

mesh

l @ A T A l = I w ~ T B I

(5.9)

=

-

or - -

the minus sign indicating that the direction of

rotation is reversed

Figure 5.21 that is the motion relative to the arm or carrier is independent of the speed of the arm For example, if oc = 0 we have the case of a simple gear train where

(5.12)

O A r B

% rA

_ - -

Figure 5.22

U W Figure 5.22 shows a typical arrangement for an

epicyclic gear in which the planet is free to rotate

on a bearing on the carrier, which is itself free to rotate about the central axis of the gear If the carrier is fixed, the gear is a simple gear train so that the velocity ratio

Figure 5.20

Figure 5.20 shows a compound gear train in

which wheel B is rigidly connected to wheel C;

thus % = wc The velocity ratio for the gear is

OD % %

_ - - - .-

TS

= (3)( -2) = - T c TA

TD TB

(5.10) Note that the direction of rotation of the

annulus is the same as that of the planet, since the annulus is an internal gear Also, we see that the number of teeth on the planet wheel does not affect the velocity ratio - in this case the planet is said to act an an idler

If the carrier is not fixed, then the above velocity ratio is still valid provided the angular speeds are relative to the carrier; thus

5.10 Epicyclic motion

If the axle of a wheel is itself moving on a circular

path, then the motion is said to be epicyclic

Figure 5.21 shows the simplest type of epicyclic

motion If no slip occurs at P, the contact point,

then the velocity of P is given as

V P lO l = V O 2 / 0 1 + V P l 0 2

(5.13)

hence W A r A = @ C ( r A + r B ) - ( L ) S r B @S @C TA

5.1 1 Compound epicyclic gears 59

If two of the speeds are known then the third

may be calculated In practice it is common to fix

one of the elements (i.e sun, carrier or annulus)

and use the other two elements as input and

output Thus we see that it is possible to obtain

three different gear ratios from the same

mechanism

5,ll Compound epicyclic gears

In order to obtain a compact arrangement, and

also to enable a gearbox to have a wider choice of

selectable gear ratios, two epicyclic gears are

often coupled together The ways in which this

coupling can occur are numerous so only two

arrangements will be discussed The two chosen

are common in the automotive industry and

between them form the basis of the majority of

automatic gearboxes

Simpson gear train

In the arrangement shown in Fig 5.23(a), the two

sun wheels are on a common shaft and the carrier

of the first epicyclic drives the annulus of the

second This second annulus is the output whilst

the input is either the sun wheel or the annulus of

the first epicyclic

This design, used in a General Motors 3-speed

automatic transmission, provides three forward

gears and a reverse gear These are achieved as

follows

First gear employs the first annulus as input and

locks the carrier of the second Second gear again

uses the first annulus as input but fixes the sun

wheel shaft Third is obtained by locking the first

annulus and the sun wheel together so that the

whole assembly rotates as a solid unit Reverse

gear again locks the second carrier, as for the first

gear, but in this case the drive is via the sun

wheel

Figure 5.23(b) shows a practical layout with

three clutches and one band brake which carry

out the tasks of switching the drive shafts and

locking the second carrier or the sun wheel shaft

To engage first gear drive is applied to the

forward clutch and the second carrier is fixed In

normal drive mode this is achieved by means of

the one-way Sprag clutch This prevents the

carrier from rotating in the negative sense,

relative to the drive shaft, but allows it to

free-wheel in the positive sense This means that

no engine braking is provided during over-run

To provide engine braking the reverse/low clutch

is engaged in the lock-down mode For second

gear the reverseAow clutch (if applied) is released and the intermediate band brake is applied, thus locking the sun wheel For third gear the intermediate band is released and the direct clutch activated hence locking the whole gear to rotate in unison For reverse gear the forward clutch is released, then the direct clutch and the reverse/low clutch are both engaged thus only the second epicyclic gear is in use

The operation of the various clutches and band brakes is conventionally achieved by a hydraulic circuit which senses throttle position and road speed The system is designed to change down at

a lower speed than it changes up at a given throttle position to prevent hunting Electronic control is now used to give more flexibility in changing parameters to optimise for economy or for performance

To determine the gear ratios two equations of the same type as equation 5.13 are required and they are solved by applying the constraints dictated by the gear selected A more convenient

set of symbols will be used to represent rotational speed We shall use the letter A to refer to the

annulus, C for the carrier and S for the sun, also

we shall use 1 to refer to the first simple epicyclic gear and 2 for the second In this notation, for example, the speed of the second carrier will be referred to as C2

For the first epicyclic gear

and for the second epicyclic gear

(5.14)

(5.15)

Where R is the ratio of teeth on the annulus to

teeth on the sun In all cases S2 = S1 and C1 = A2 = wo , the output

With the first gear selected C2 = 0 and Al = o i ,

the input

From equation 5.14 S1 = -wi X R 1 + wo(l+ R1 )

and from equation 5.15 S1 = -wo x R2

wo (1 + R1+ R2 1

R1

Eliminating S1 wi =

thus the first gear ratio = wi/wo = (1 + R 1 + R t ) / R 1

With second gear selected S1 = 0 and wi is still A I

Figure 5.23(a)

Figure 5.23(b)

From equation 5.14 0 = wo(l + R1 ) - wi x R1

thus the second gear ratio q l w g = (1 + R 1 ) / R 1

Summarising we have GEAR

2nd

GEAR RATIO The third gear is, of course, unity

For the reverse gear C2 = 0 and wi = S1 so from (1 + R1 )lRl

m i l o g = - R2 Reverse -R2

5.1 1 Compound epicyclic gears 61

Figure 5.23(c)

Figure 5.23(d)

Ravigneaux gearbox

The general arrangement of the Ravigneaux gear

is shown in Fig 5.23(c) This gear is used in the

Borg Warner automatic transmission which is to

be found in many Ford vehicles

In this design there is a common planet carrier

Discussion examples

Example 5.1

The four-bar chain mechanism will now be analysed in greater detail We shall consider the mechanism in the configuration shown in Fig 5.24 and determine v c , z+, oz, w 3 , a B , a c ,

aE , ;2 and h, , and the suffices 1, 2, 3 and 4 will refer throughout to links AB, BC, C D and D A respectively

and the annulus is rigidly connected to the output

shaft The second epicyclic has two planets to

effect a change in the direction of rotation

compared with a normal set In the actual design,

shown in Fig 5.23(d), the first planet wheel

doubles as the idler for the second epicyclic gear

When first gear is selected, the front clutch

provides the drive to the forward sun wheel and

the common carrier is locked, either by the rear

band brake in lock-down mode or by the free-

wheel in normal drive For second gear the drive

is still to the forward sun wheel but the reverse

sun wheel is fixed by means of the front band

brake For top gear drive both suns are driven by

the drive shaft thereby causing the whole gear

train to rotate as a unit For the reverse gear the

rear clutch applies the drive to the reverse sun

wheel and the carrier is locked by the rear band

brake

For the first gear the input w j = S2 and

C1 = C2 = 0, the output wo = A l = A Z So, from

equation 5.15,

Figure 5.24

Velocities

In general, for any link PQ of length R and rotating with angular velocity w (see Fig 5.25(a))

we have, from equation 2.17,

S Z = R 2 X A

therefore wi/wo = S21A = R2

For second gear S2 is the input but SI = 0

From equation 5.14 0 = -AX R1 + (1 + R I ) C

and from equation 5.15 S2 = R2 X A + C ( l - R2)

Elimination of C gives

S2 = R2 + A X RI X (1 - R2)/(1+ R1)

R1 +R2

w ~ I w O = S2IA = _

1 + R l thus

The top gear ratio is again unity

Reverse has C = 0 with input S1 so from

equation 5.14

S l = - R l X A

giving the gear ratio

witwo = S11A = -R1

Summarising we have

G E A R G E A R RATIO

Figure 5.25 VQfp = Rer + Roee

If PQ is of fixed length then R = 0 and VQ/P has

a magnitude Rw and a direction perpendicular to the link and in a sense according the the direction

of 0

Velocity diagram (section 5.4) Since II is constant, the magnitude of vBIA is wllI and its direction is perpendicular to A B in the sense indicated in Fig 5.25(b), so we can draw to a

suitable scale the vector ab- which represents

Z ) B / ~ The velocity of C is determined by considering the known directions of v U B and VUD

Discussionexamples 63

Link Velocity Direction Sense Magnitude ( d s ) Line

AB %/A LAB \ (AB)wl = (0.15)12 = 1.8 ab

From the concept of the velocity image we can

find the position of e on bc from

be BE

_ - - Thus

be = 1.28 - = 0 3 3 7 d s

t:0) and by noting that (see equation 2.24) The magnitude of + is ae and this is found

(i)

ve1ocity There are sufficient data to draw the Znstantaneous centre (section 5.5) In Fig 5.28, ve1ocity triang1e representing equation (i) I, the instantaneous centre of rotation of BC, is at

From this figure it can be Seen that the location rotates instantaneously about I From the known

of point C on the velocity diagram is the direction of v B , the angular velocity of the

intersection of a line drawn through b perpen- triangle is clearly Seen to be clockwise

dicular to BC and a line drawn through a, d

perpendicular to DC By scaling we find that the

magnitude of dc is 1.50 d s and thus

v U A = 'uC = 1.50 d s 14"

from the diagram to be 1.63 d s Thus

V U A = %/A -k V U B

and vUA = vUD since A and D each have zero V E = 1.63 d~ 20"

The magnitude of 02 is

V B wl(AB) 12(0.15)

- = 6.7 r a d s

Cr);!=-=

and q = -6.7 k r a d s

0.27 The magnitude of w;? is

bc 1.28

BC 0.19

w;?=- =- =6.7rad/s

To determine the direction, we note that vuB,

the velocity of C relative to B is the sense from b

to c (and that %IC is in the opposite sense) so that

BC is rotating clockwise (see Fig 5.27) Thus

% = -6.7 k r a d s

The magnitude of q is

where k is the unit vector coming out of the page

cd 1.5

CD 0.15

0 3 = - = - - - 10 r a d s

and the direction is clearly anticlockwise, so that

o3 = 10 k r a d s

e - - - - - - The magnitude of vc is

VC = %(IC) = 6.7(0.225) = 1.50 d s

and the sense is in the direction shown

The magnitude of q is

V C 1.47

= 9.8 r a d s

w 3 = - = -

CD 0.15 and the sense is clearly anticlockwise so that

w3 = 9.8k r a d s

Point E lies on link BC so that the instant-

aneous centre for E is also I The magnitude of %

is

and the sense is in the direction shown

are obviously due to inaccuracies in drawing

Accelerations

For any link PQ of length R, angular velocity w

and angular acceleration h (see Fig 5.29) we

have, from equation 2.18,

+ = %(IE) = 6.7(0.245) = 1.64 m / s

The discrepancies between the two methods

The magnitude of 4 is

C’C 4.7

BC 0.19

4 = - = - = 24.7 rads2

To determine the sense of 4 we note that the normal component of urn is c’c in the sense of c’

to c; thus BC has a clockwise angular accelera- tion

uQIP = (R - Ro2) e, + (Rh + 2Ro) ee & = -24.7k rads2

If PQ is of fixed length then R = R = 0 and uQIP

has one component of magnitude Rw2 always in

the sense of Q to P and another of magnitude Rh,

perpendicular to PQ and directed according to CD 0.15

the sense of h

Acceleration diagram (section 5.7) See Fig 5.30

The radial and normal components of U B / A are

both known, and summing these gives the total

acceleration uB since A is a fixed point can find the position of e on bc from

(ab’ + b’b = ab in the diagram) The radial

directed from C to B The normal component of

uUB is perpendicular to BC but is as yet unknown

in magnitude or sense Similar reasoning applies

to uUD However we have enough data to locate

point c on the acceleration diagram shown in

Fig 5.30

The magnitudes and directions of UB and uc are

Similarly we find that the magnitude of ;3 is

28 187rads2

o 3 = - = - =

C”C

and the sense is anticlockwise,

;3 = 187krad/s2 From the concept of the acceleration image we

- + - + - - 9

be BE

bc BC

_ - component of uuB has a magnitude of l2 %2 and is

Thus

be = 0.99 (;io) - =0.260m/s2 The magnitude and direction of uE are taken from the diagram and we find

taken directly from the diagram UE = auA = ae - = 24.2 m / s 2 45”

46“

43”

2

+

U B = uB/A = ab = 22.0 m / ~

uc = a c / D = dc + = 31.6 m / ~

aB/A (radial) [(AB A/ l1oI2 = 0.15(12)2 = 21.6 ab‘

aB/A (normal) LAB 7 11 hi = 0.15(35) = 5.25 b‘b

uuB (radial) IIBC A/ 1 2 ~ 2 ~ = 0.19(6.7)2 = 8.53 be’

AB {

BC {

UQD (radial) llCD L 13w32 = 0.15(10)2 = 15.0 de’’

Vector-algebra methods

Vector algebra can be used in the solution of

mechanism problems Such methods are a

powerful tool in the solution of three-dimensional

mechanism problems but usually take much

longer than graphical methods for problems of

plane mechanisms They do, however, give a

systematic approach which is amenable to

computer programming

An outline of a vector-algebra solution to the

present problem is given below Students who are

following a course leading to the analysis of

three-dimensional mechanisms should find this a

useful introduction and are encouraged to try

these techniques on a few simple plane mechan-

isms

values of d is consistent with the links BC and C D

joining at C, and one of the values of c

corresponds with the mechanism being in the

alternative position shown dotted in Fig 5.31

The vector Z2 can then be found from equation (ii) The results are

Z1 = (0.075Oi+O.l299j) m

Z2 = (0.1893i+O.l58Oj) m

l3 = (0.0350i - 0.14571’) m Now,

vC = %+vUB

and, from equation 5.3,

v , = 0 , x l , + O , x Z ,

V c = 0 3 x ( - 1 3 )

0 1 x z1 +O, x z, + 0 3 x z3 = 0

also

(iv)

(VI Equating the two expressions for vc 7

Writing w1 = 12k, 02 = * k and o3 = u 3 k 7

and carrying out the vector products in equation (v), gives

From Fig 5.31 we note that

(-1.559-0.01580, + O.1457~3)i

+ (0.9 + 0.18930, + 0.035 65w3)j = 0

Z, + Z2 + l3 + 1, = 0 (ii)

and the vectors 1, and l3 can be determined by first

evaluating angles 13, and O3 by the methods of

The vector I 1 = I1 (cos 61 i + sin Od) is known

Equating the coeffjcients of i and j to z e r ~ and solving for O, and w 3 , we find

normal trigonometry and then writing O, = -6.634

Z2 = 12 (cos 62i + sin 6d)

Z3 = l3 (cos 63 i - sin 63j)

Alternatively we can write

Z2 = 12e2 = l2 (ai + b j )

Z3 = 13e3 = 13(ci+dj)

and w3 = 9.980

Using % = w1 X Zl and equation (iv) leads to

l%l = d[(1.559)2+(0.9)2] = 1 8 0 0 d s

% = -1.5593+0.9jm/s

and

and determine the values of a, b, c and d Noting

that

(iii)

vc = -(1.4533+0.3558j) m/s

lvcl = d[(1.453)2+(0.3558)2] = 1.497 m/s

A quicker way of finding vc, if 02 is not

d = k d ( 1 - c 2 )

and substituting in equation (ii) with z4 = -14i and

insertion of numerical values gives

required, iS to note that Since DC/B is perpendicu- lar to BC, we can write

0 190e2 = (0.225 - 0 1 8 0 ~ ) i vC/B-z2 = o

-[0.1299fO.l80d(l -c2)U or (vc-%).Z2 = 0 Taking the modulus of this equation eliminates

e2 and rearranging and squaring we find two

values for c , each with two corresponding values

of d from equation (iii) Only one of each pair of and carrying out the dot product we find

% is known and writing from equation (iv)

vc = o 3 k x (-0.035 Oli+O.l457j)