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PART I
KINEMATICS AND MECHANISMS
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iii
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1
Chapter 1
The World of Mechanisms
1.1
Sketch at least six different examples of the use of a planar four-bar linkage in practice.
They can be found in the workshop, in domestic appliances, on vehicles, on agricultural
machines, and so on.
1.2
The link lengths of a planar four-bar linkage are 25, 75, 125, and 125 mm. Assemble the
links in all possible combinations and sketch the four inversions of each. Do these
linkages satisfy Grashof's law? Describe each inversion by name--for example, a crankrocker mechanism or a drag-link mechanism.
s = 1, l = 5, p = 3, q = 5 ; these linkages all satisfy Grashof’s law since 1 + 5 < 3 + 5 .
Drag-link mechanism
Drag-link mechanism
Crank-rocker mechanism
Crank-rocker mechanism
Double-rocker mechanism
Crank-rocker mechanism
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Since the variety is unbounded no standard solutions are shown here.
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2
A crank-rocker linkage has a 100-mm frame, a 25-mm crank, a 90-mm coupler, and a
75-mm rocker. Draw the linkage and find the maximum and minimum values of the
transmission angle. Locate both toggle positions and record the corresponding crank
angles and transmission angles.
Extremum transmission angles: γ min = γ 1 = 53.1°; γ max = γ 3 = 98.1°
Toggle positions: ψ 2 = 40.1°; γ 2 = 59.1°; ψ 4 = 228.6°; γ 4 = 90.9°
1.4
In the figure, point C is attached to the coupler; plot its complete path.
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1.3
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1.5
Find the mobility of each mechanism shown in the figure.
m = 3 ( 6 − 1) − 2 ( 7 ) − 1( 0 ) = 1
Ans.
(b) n = 8, j1 = 10, j2 = 0; m = 3 ( 8 − 1) − 2 (10 ) − 1( 0 ) = 1
Ans.
(c) n = 7, j1 = 9, j2 = 0; m = 3 ( 7 − 1) − 2 ( 9 ) − 1( 0 ) = 0
Ans.
(a) n = 6, j1 = 7, j2 = 0;
Note that the Kutzbach criterion fails in this case; the true mobility is m=1. The
exception is due to a redundant constraint. The assumption that the rolling contact
joint does not allow links 2 and 3 to separate duplicates the constraint of the fixed
link length O2O3 .
(d) n = 4, j1 = 3, j2 = 2; m = 3 ( 4 − 1) − 2 ( 3) − 1( 2 ) = 1
Ans.
Note that each coaxial set of sliding joints is counted as only a single prismatic pair.
Use the mobility criterion to find a planar mechanism containing a moving quaternary
link. How many distinct variations of this mechanism can you find?
To have at least one quatenary link, a planar mechanism must have at least eight links.
The Grübler criterion then indicates that ten single-freedom joints are required for
mobility of m=1. According to H. Alt, “Die Analyse und Synthese der achtgleidrigen
Gelenkgetriebe”, VDI-Berichte, vol. 5, 1955, pp. 81-93, there are a total of sixteen
distinct eight-link planar linkages having ten revolute joints, seven of which contain a
quatenary link. These seven are shown below:
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1.6
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4
Find the time ratio of the linkage of Problem 1.3.
From the values of ψ 2 and ψ 4 we find α = 188.5° and β = 171.5° .
Then, from Eq. (1.5), Q = α β = 1.099 .
Ans.
1.8
Devise a practical working model of the drag-link mechanism.
1.9
Plot the complete coupler curve of the Roberts' mechanism of Fig. 1.19b. Use AB = CD
= AD = 60 mm and BC = 30 mm.
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1.7
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1.10
If the crank of Fig. 1.8 is turned 10 revolutions clockwise, how far and in what direction
will the carriage move?
Screw and carriage move by (10 rev ) (1.5 rev/ mm ) = 6.67mm to the left.
Carraige moves (10 rev ) ( 2 rev/mm ) = 5mm to the right with respect to the screw.
Net motion of carriage = 6.67 – 5 = 1.67 mm to the left.
More in depth study of such devices is covered in Chap. 10.
1.11
Show how the mechanism of Fig. 1.12b can be used to generate a sine wave.
With the length and angle of crank 2 designated as R and ψ2, respectively, the horizontal
motion of link 4 is x4 = R cosψ 2 = R sin (ψ 2 + 90° ) .
Devise a crank-and-rocker linkage, as in Fig. 1.11c, having a rocker angle of 60°. The
rocker length is to be 0.50 m.
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1.12
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Chapter 2
Position and Displacement
2.1
Describe and sketch the locus of a point A which moves according to the equations
2.2
Find the position difference from point P to point Q on the curve y = x 2 + x − 16 , where
RPx = 2 and RQx = 4 .
2
RPy = ( 2 ) + 2 − 16 = −10 ; R P = 2ˆi − 10ˆj
2
R Q = 4ˆi + 4ˆj
RQy = ( 4 ) + 4 − 16 = 4 ;
R = R − R = 2ˆi + 14ˆj = 14.142∠81.9°
QP
2.3
Q
P
Ans.
The path of a moving point is defined by the
equation y = 2 x 2 − 28 .
Find the position
difference from point P to point Q if RPx = 4 and
RQx = −3 .
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RAx = atcos ( 2πt ) , R Ay = atsin ( 2π t ) , RAz = 0 .
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RPy = 2 ( 4 ) − 28 = 4 ;
2
R P = 4ˆi + 4ˆj
2
RQy = 2 ( −3) − 28 = −10 ; R Q = −3ˆi − 10ˆj
R = R − R = −7ˆi − 14ˆj = 15.652∠243.4°
QP
P
Ans.
The path of a moving point P is defined by the equation y = 60 − x3 / 3 . What is the
displacement of the point if its motion begins when RPx = 0 and ends when RPx = 3 ?
3
RPy ( 0 ) = 60 − ( 0 ) / 3 = 60 ; R P ( 0 ) = 60ˆj
3
RPy ( 3) = 60 − ( 3) / 3 = 51 ; R P ( 3) = 3ˆi + 51ˆj
ΔR = R ( 3) − R ( 0 ) = 3ˆi − 9ˆj = 9.487∠ − 71.6°
P
2.5
P
P
Ans.
If point A moves on the locus of Problem 2.1, find its displacement from t = 2 to t =2.5.
R A ( 2.0 ) = 2.0a cos 4π ˆi + 2.0a sin 4π ˆj = 2.0aˆi
R ( 2.5 ) = 2.5a cos 5π ˆi + 2.5a sin 5π ˆj = −2.5aˆi
A
ΔR A = R A ( 2.5 ) − R A ( 2.0 ) = −4.5aˆi
2.6
Ans.
The position of a point is given by the equation R = 100e j 2πt . What is the path of the
point? Determine the displacement of the point from t = 0.10 to t = 0.40.
The point moves in a circle of radius 100 with its center at the origin.
Ans.
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2.4
Q
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R ( 0.10 ) = 100e j 0.628 = 80.902ˆi + 58.779ˆj
R ( 0.40 ) = 100e j 2.513 = −80.902ˆi + 58.779ˆj
ΔR = R ( 0.40 ) − R ( 0.10 ) = −161.803ˆi = 161.803∠180°
2.7
(
Ans.
)
The equation R = t 2 + 4 e− j πt /10 defines the position of a point. In which direction is
the position vector rotating? Where is the point located when t = 0? What is the next
value t can have if the direction of the position vector is to be the same as it is when t =
0? What is the displacement from the first position of the point to the second?
Since the polar angle for the position vector is
θ = − j πt /10 , then dθ / dt is negative and therefore
the position vector is rotating clockwise.
Ans.
2
−
j
0
R ( 0) = 0 + 4 e
= 4∠0°
Ans.
)
The position vector will next have the same direction
when π t /10 = 2π , that is, when t=20.
Ans.
2
− j 2π
R ( 20 ) = ( 20 + 4 ) e
= 404∠0°
ΔR = R ( 20 ) − R ( 0 ) = 400∠0°
2.8
Ans.
2
The location of a point is defined by the equation R = ( 4t + 2 ) e j πt / 30 , where t is time in
seconds. Motion of the point is initiated when t = 0. What is the displacement during the
first 3 s? Find the change in angular orientation of the position vector during the same
time interval.
R ( 0 ) = ( 0 + 2 ) e j 0 = 2∠0° = 2ˆi
R ( 3) = (12 + 2 ) e j π9 / 30 = 14∠54° = 8.229ˆi + 11.326ˆj
ΔR = R ( 3) − R ( 0 ) = 6.229ˆi + 11.326ˆj = 12.926∠61.2°
Ans.
Δθ = 54° − 0° = 54° ccw
Ans.
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(
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2.9
Link 2 in the figure rotates according to the equation θ = πt / 4 . Block 3 slides outward
on link 2 according to the equation r = t 2 + 2 . What is the absolute displacement
ΔR P from t = 1 to t = 2? What is the apparent displacement ΔR P ?
3
3/ 2
R P3 = re jθ = ( t 2 + 2 ) e jπ t / 4
R P3 (1) = 3∠45° = 2.121ˆi + 2.121ˆj
R ( 2 ) = 6∠90° = 6ˆj
P3
ΔR P3 = R P3 ( 2 ) − R P3 (1) = −2.121ˆi + 3.879ˆj = 4.421∠118.7°
Ans.
R P3 / 2 = re j 0 = ( t 2 + 2 ) ˆi 2
R P3 / 2 (1) = 3ˆi 2
R ( 2 ) = 6ˆi
2
ΔR P3 / 2 = R P3 / 2 ( 2 ) − R P3 / 2 (1) = 3ˆi 2
2.10
Ans.
A wheel with center at O rolls without slipping so that its center is displaced 10 in to the
right. What is the displacement of point P on the periphery during this interval?
Since the wheel rolls without slipping,
ΔRO = −Δθ RPO .
Δθ = −ΔRO / RPO
= −10 in / 6 in = −1.667 rad = −95.5°
For R PO ,
θ ′ = θ + Δθ = 270° − 95.5° = 174.5°
R ′ = 6∠174.5° = −5.972ˆi + 0.574ˆj in
PO
(
ΔR P = ΔR O + R PO′ − R PO
)
= 10ˆi − 5.972ˆi + 0.574ˆj + 6ˆj
ΔR P = 4.028ˆi + 6.574ˆj = 7.710∠58.5° in
2.11
Ans.
A point Q moves from A to B along link 3 while link 2 rotates from θ 2 = 30° to
θ 2′ = 120° . Find the absolute displacement of Q.
R = 75∠30° = 64.95ˆi + 37.5ˆj
Q3
R Q3 ′ = 75∠120° = −37.5ˆi + 64.95ˆj
ΔR Q3 = R Q3 ′ − R Q3 = −102.45ˆi + 27.45ˆj
ΔR Q5 / 3 = R BA = 150ˆi
ΔR Q5 = ΔR Q3 + ΔR Q5 / 3
ΔR Q5 = 47.5ˆi + 27.5ˆj = 55∠30° mm
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P3 / 2
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2.12
The linkage shown is driven by moving the sliding block 2. Write the loop-closure
equation. Solve analytically for the position of sliding block 4. Check the result
graphically for the position where φ = −45° .
The loop-closure
equation is
R A = R B + R AB .
RAe jπ /12 = RB + RAB e j (π +φ )
= RB − RAB e jφ
Taking the imaginary components of this, we get
RA sin15° = − RAB sin φ
sin φ
sin − 45°
RA = − RAB
= −200
= 546 mm
sin15°
sin15°
The offset slider-crank mechanism is driven by rotating crank 2. Write the loop-closure
equation. Solve for the position of the slider 4 as a function of θ 2 .
R C = R A + R BA + R CB
RC = RAe jπ / 2 + RBAe jθ 2 + RCB e jθ3
Taking real and imaginary parts,
0 = RA + RBA sin θ 2 + RCB sin θ 3
RC = RBA cosθ 2 + RCB cosθ 3 and
and, solving simultaneously, we get
⎛ − R − RBA sin θ 2 ⎞
θ 3 = sin −1 ⎜ A
⎟ with −90° < θ 3 < 90°
RCB
⎝
⎠
2
− ( RA + RBA sin θ 2 ) = 63cos θ 2 + 1043.75 − 125sin θ 2 + 156.25cos 2 θ 2
RC = RBA cos θ 2 + RCB
2
Ans.
2.14
Write a calculator program to find the sum of any number of two-dimensional vectors
expressed in mixed rectangular or polar forms. The result should be obtainable in either
form with the magnitude and angle of the polar form having only positive values.
Because the variety of makes and models of calculators is vast and no standards exist for
programming them, no standard solution is shown here.
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2.13
Ans.
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12
Write a computer program to plot the coupler curve of any crank-rocker or double-crank
form of the four-bar linkage. The program should accept four link lengths and either
rectangular or polar coordinates of the coupler point relative to the coupler.
Again the variety of programming languages makes it difficult to provide a standard
solution. However, one version, written in FORTRAN IV, is supplied here as an
example. There are no accepted standards for programming graphics. Therefore the
Tektronix PLOT-10 subroutine library, for display on Tektronix 4010 series displays, is
chosen as an older but somewhat recognized alternative. The symbols in the program
correspond to the notation shown in Figure 2.19 of the text. The required input data are:
⎧ X5, Y5, -1
R1, R2, R3, R4, ⎨
⎩R5, θ5, 1
The program can be verified using the data of Example 2.7 and checking the results
against those of Table 2.3.
PROGRAM CCURVE
C
C
C
C
C
C
C
C
C
C
C
C
C
A FORTRAN IV PROGRAM TO PLOT THE COUPLER CURVE OF ANY CRANK-ROCKER
OR DOUBLE-CRANK FOUR-BAR LINKAGE, GIVEN ITS DIMESNIONS.
WRITTEN FOR A DEC 11/70 COMPUTER SYSTEM, USING SUBROUTINES FROM
TEKTRONIX PLOT-10 FOR DISPLAY ON 4010 SERIES DISPLAYS.
REF:J.J.UICKER,JR, G.R.PENNOCK, & J.E.SHIGLEY, ‘THEORY OF MACHINES
AND MECHANISMS,’ THIRD EDITION, OXFORD UNIVERSITY PRESS, 2003.
EXAMPLE 2.6
WRITTEN BY: JOHN J. UICKER, JR.
ON:
01 JANUARY 1980
READ IN THE DIMENSIONS OF THE LINKAGE.
READ(5,1000)R1,R2,R3,R4,X5,Y5,IFORM
1000 FORMAT(6F10.0,I2)
C
C
FIND R5 AND ALPHA.
IF(IFORM.LE.0)THEN
R5=SQRT(X5*X5+Y5*Y5)
ALPHA=ATAN2(Y5,X5)
ELSE
R5=X5
ALPHA=Y5/57.29578
END IF
Y5=AMAX1(0.0,R5*SIN(ALPHA))
C
C
INITIALIZE FOR PLOTTING AT 120 CHARACTERS PER SECOND.
CALL INITT(1200)
C
C
SET THE WINDOW FOR THE PLOTTING AREA.
CALL DWINDO(-R2,R1+R2+R4,-R4,R4+R4+Y5)
C
C
CYCLE THROUGH ONE CRANK ROTATION IN FIVE DEGREE INCREMENTS.
TH2=0.0
DTH2=5.0/57.29578
IPEN=-1
DO 2 I=1,73
CTH2=COS(TH2)
STH2=SIN(TH2)
C
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2.15
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CALCULATE THE TRANSMISSION ANGLE.
CGAM=(R3*R3+R4*R4-R1*R1-R2*R2+2.0*R1*R2*CTH2)/(2.0*R3*R4)
IF(ABS(CGAM).GT.0.99)THEN
CALL MOVABS(100,100)
CALL ANMODE
WRITE(7,1001)
1001
FORMAT(//’ *** THE TRANSMISSION ANGLE IS TOO SMALL. ***’)
GO TO 1
END IF
SGAM=SQRT(1.0-CGAM*CGAM)
GAM=ATAN2(SGAM,CGAM)
C
C
CALCULATE THETA 3.
STH3=-R2*STH2+R4*SIN(GAM)
CTH3=R3+R1-R2*CTH2-R4*COS(GAM)
TH3=2.0*ATAN2(STH3,CTH3)
C
C
CALCULATE THE COUPLER POINT POSITION.
TH6=TH3+ALPHA
XP=R2*CTH2+R5*COS(TH6)
YP=R2*STH2+R5*SIN(TH6)
C
C
PLOT THIS SEGMENT OF THE COUPLER CURVE.
IF(IPEN.LT.0)THEN
IPEN=1
CALL MOVEA(XP,YP)
ELSE
IPEN=-1
CALL DRAWA(XP,YP)
END IF
TH2=TH2+DTH2
2 CONTINUE
C
C
DRAW THE LINKAGE.
CALL MOVEA(0.0,0.0)
CALL DRAWA(R2,0.0)
XC=R2+R3*COS(TH3)
YC=R3*SIN(TH3)
CALL DRAWA(XC,YC)
CALL DRAWA(XP,YP)
CALL DRAWA(R2,0.0)
CALL MOVEA(XC,YC)
CALL DRAWA(R1,0.0)
1 CALL FINITT(0,0)
CALL EXIT
STOP
END
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C
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14
For each linkage shown in the figure, find the path of point P: (a) inverted slider-crank
mechanism; (b) second inversion of the slider-crank mechanism; (c) straight-line
mechanism; (d) drag-link mechanism.
(a)
(c)
(b)
(d)
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2.16
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2.17
Using the offset slider-crank mechanism of Fig. 2.15, find the crank angles corresponding
to the extreme values of the transmission angle.
As shown, γ = 90° − θ 3 .
Also from the figure
e + r2 sin θ 2 = r3 cos γ .
Differentiating with
respect to θ 2 ,
dγ
r2 cos θ 2 = −r3 sin γ
;
dθ 2
r cos θ 2
dγ
=− 2
.
dθ 2
r3 sin γ
Now, setting dγ / dθ 2 = 0 , we get cos θ 2 = 0 .
2.18
In Section 1.10 it is pointed out that the transmission angle reaches an extreme value for
the four-bar linkage when the crank lies on the line between the fixed pivots. Referring
to Fig. 2.19, this means that γ reaches a maximum or minimum when crank 2 is
coincident with the line O2 O4 . Show, analytically, that this statement is true.
From ΔO4O2 A :
s 2 = r12 + r22 − 2r1r2 cos θ 2 .
Also, from ΔABO4 :
s 2 = r32 + r42 − 2r3 r4 cos γ .
Equating these we differentiate
with respect to θ 2 to obtain
dγ
or
2r1r2 sin θ 2 = 2r3 r4 sin γ
dθ 2
r r sin θ 2
dγ
= 12
.
dθ 2 r3 r4 sin γ
Now, for
dγ
= 0 , we have sin θ 2 = 0 . Thus, θ 2 = 0, ± 180°, ± 360°,…
dθ 2
Q.E.D.
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Therefore, we conclude that θ 2 = ± ( 2k + 1) π / 2 = ±90°, ±270°,… Ans.
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2.19
The figure illustrates a crank-and-rocker linkage in the first of its two limit positions. In
a limit position, points O2 , A, and B lie on a straight line; that is, links 2 and 3 form a
straight line. The two limit positions of a crank-rocker describe the extreme positions of
the rocking angle. Suppose that such a linkage has r1 = 400 mm , r2 = 200 mm ,
r3 = 500 mm , and r4 = 400 mm .
(b)
(c)
Find θ 2 and θ 4 corresponding to each limit position.
What is the total rocking angle of link 4?
What are the transmission angles at the extremes?
(a)
From isosceles triangle O4O2 B we
can calculate or measure θ 2 = 29° , θ 4 = 58°
and θ 2′ = 248° , θ 4′ = 136° .
Ans.
(b)
Then Δθ 4 = θ 4′ − θ 4 = 78° .
(c)
Finally, from isosceles triangle
Ans.
O4O2 B , γ = 29° and γ ′ = 68° .
2.20
A double-rocker mechanism has a dead-center position and may also have a limit
position (see Prob. 2.19). These positions occur when links 3 and 4 in the figure lie along
a straight line. In the dead-center position the transmission angle is 180° and the
mechanism is locked. The designer must either avoid such positions or provide the
external force, such as a spring, to unlock the linkage. Suppose, for the linkage shown in
the figure, that r1 = 14 in , r2 = 5.5 in , r3 = 5 in , and r4 = 12 in . Find θ 2 and θ 4
corresponding to the dead-center position. Is there a limit position?
For the given dimensions, there are
two dead-center positions, and they
correspond to the two extreme
travel positions of crank O2 A .
From O4 AO2 using the law of
cosines, we can find θ 2 = 114.0° ,
θ 4 = 162.8° and, symmetrically,
θ 2′′′ = −114.0° , θ 4′′′ = −162.8° . There
are also two limit positions, these
occur at θ 2′ = 56.5° , θ 4′ = 133.1°
and, symmetrically, at θ 2′′ = −56.5° ,
θ 4′′ = −133.1° .
Ans.
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(a)
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2.21
The figure shows a slider-crank mechanism that has an offset e and that is placed in one
of its limiting positions. By changing the offset e, it is possible to cause the angle that
crank 2 makes in traversing between the two limiting positions to vary in such a manner
that the driving or forward stroke of the slider takes place over a larger angle than the
angle used for the return stroke. Such a linkage is then called a quick-return mechanism.
The problem here is to develop a formula for the crank angle traversed during the
forward stroke and also develop a similar formula for the angle traversed during the
return stroke. The ratio of these two angles would then constitute an time ratio of the
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drive to return strokes. Also determine which direction the crank should rotate.
From the figure we can see that e = ( r3 + r2 ) sin θ 2 = ( r3 − r2 ) sin (θ 2′ − 180° ) or
⎛
e ⎞
e ⎞
−1 ⎛
⎟ , θ 2′ = 180° + sin ⎜
⎟
⎝ r3 + r2 ⎠
⎝ r3 − r2 ⎠
θ 2 = sin −1 ⎜
⎛ e ⎞
e ⎞
−1 ⎛
Δθ drive = θ 2′ − θ 2 = 180° + sin −1 ⎜
⎟ − sin ⎜
⎟
⎝ r3 − r2 ⎠
⎝ r3 + r2 ⎠
⎛ e ⎞
e ⎞
−1 ⎛
Δθ return = θ 2 + 360° − θ 2′ = 180° + sin −1 ⎜
⎟ − sin ⎜
⎟
⎝ r3 + r2 ⎠
⎝ r3 − r2 ⎠
Assuming driving when sliding to the right, the crank should rotate clockwise.
Ans.
Ans.
Ans.
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Chapter 3
Velocity
3.1
The position vector of a point is given by the equation R = 100e j πt , where R is in inches.
Find the velocity of the point at t = 0.40 s.
R ( t ) = 2.5e jπ t
( t ) = jπ 2.5e jπ t
R
= jπ 2.5 ( cos 0.40π + j sin 0.40π )
= −2.5π sin 72° + j 2.5π cos 72°
( 0.40 ) = −7.46 + j 2.427 = 7.85∠162° m/s
R
3.2
(
Ans.
)
The equation R = t 2 + 4 e-j πt/ 10 defines the path of a particle. If R is in meters, find the
velocity of the particle at t = 20 s.
R ( t ) = ( t 2 + 4 ) e − jπ t /10
( t ) = 2te − jπ t /10 − ( jπ /10 ) ( t 2 + 4 ) e − jπ t /10
R
( 20 ) = 40e− jπ 20 /10 − ( jπ /10 ) ( 202 + 4 ) e − jπ 20 /10
R
= 40e− j 2π − jπ 40.4e − j 2π
( 20 ) = 40.000 − j126.920 = 133.074∠ − 72.5° m/s
R
3.3
Ans.
If automobile A is traveling south at 88 km/h and automobile B north 60° east at
64 km/h, what is the velocity difference between B and A? What is the apparent velocity
of B to the driver of A?
VA = 88∠ − 90° = −88ˆj km/h
V = 64∠30° = 55.42ˆi + 32ˆj km/h
B
VBA = VB − VA = 55.42ˆi + 120ˆj km/h
VBA = 132.16∠65.2° =132.16 N 24.8° E km/h
Ans
Naming B as car 3 and A as car 2, we have VB2 = VA since 2 is
translating. Then VB3 / 2 = VB3 − VB2 = VBA
VB3 / 2 = 132.16∠65.2° =132.179 N 24.8° E km /h
Ans.
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( 0.40 ) = jπ 2.5e jπ 0.40
R
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20
3.4
In the figure, wheel 2 rotates at 600 rev/min and drives wheel 3 without slipping. Find
the velocity difference between points B and A.
ω2 =
( 600 rev/min )( 2π
rad/rev )
60 s/min
= 20π rad/s cw
VAO2 = ω2 RAO2 = 2000π = 6275 mm /s
3.5
Two points A and B, located along the radius of a wheel (see figure), have speeds of 80
and 140 m/s, respectively. The distance between the points is RBA = 300 mm .
(a) What is the diameter of the wheel?
(b) Find VAB , VBA, and the angular velocity of the wheel.
( ) ( )
= ( −80ˆj) − ( −140ˆj) = 60ˆj m/s
VBA = VB − VA = −140ˆj − −80ˆj = −60ˆj m/s Ans.
VAB = VA − VB
ω2 =
RBO2
VBA
60 m/s
=
= 200 rad/s cw
RBA 0.300 m
VBO2
140 m/s
=
=
= 0.700 m
ω 2 200 rad/s
Ans.
Ans.
Dia = 2 RBO2 = 2 ( 0.7 m ) = 1.4 m = 1 400 mm Ans.
www.elsolucionario.net
VB = VA + VBA
VB = 5575∠90° mm/s Ans.
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21
3.6
A plane leaves point B and flies east at 560 km/h. Simultaneously, at point A, 320 km
southeast (see figure), a plane leaves and flies northeast at 625 km/h.
(a) How close will the planes come to each other if they fly at the same altitude?
(b) If they both leave at 6:00 p.m., at what time will this occur?
VA = 625∠45° km/h = 442ˆi + 442ˆj ; VB = 560ˆi
V = V − V = 118ˆi − 442ˆj
BA
B
A
At initial time R BA ( 0 ) = 320∠120° km = −160ˆi + 277ˆj
At later time
R BA (t ) = R BA ( 0 ) + VBAt = ( −160 + 118t ) ˆi + ( 277 − 442t ) ˆj
To find the minimum of this:
2
2
2
RBA
= ( −160 + 118t ) + ( 277 − 442t )
418 576t − 282 628 = 0 t = 0.677 h = 41 min or 6:41p.m. Ans.
R BA ( 0.677 ) = −80.17ˆi − 22.23ˆj = 83∠ − 165° km
Ans.
3.7
To the data of Problem 3.6, add a wind of 48 km/h from the west.
(a) If A flies the same heading, what is its new path?
(b) What change does the wind make in the results of Problem 3.6?
With the added wind VA = 490ˆi + 442ˆj = 659.89∠42° km/h
Since the velocity is constant, the new path is a straight line at N 48º E.
Ans.
Since the velocities of both planes change by the same amount, the velocity difference
VBA does not change. Therefore the results of Problem 3.6 do not change.
Ans.
3.8
The velocity of point B on the linkage shown in the figure is 40 m/s. Find the velocity of
point A and the angular velocity of link 3.
VA = VB + VAB
VA = 48.99∠ − 165° m/s
Ans.
VAB = 14.64∠ − 120° m/s
V
14.64
ω3 = AB =
= 36.60 rad/s ccw
RAB 0.400
Ans.
www.elsolucionario.net
2
dRBA
dt = 2 ( −160 + 118t )(118 ) + 2 ( 277 − 442t )( −442 ) = 0
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22
3.9
The mechanism shown in the figure is driven by link 2 at ω 2 = 45 rad/s ccw. Find the
angular velocities of links 3 and 4.
VB = VA + VBA = VO4 + VBO4
VBA = 358.5 mm /s ; VBO4 = 4619 mm/s .
VBA 358.5
=
= 1.43 rad/s ccw
RBA
250
V
4619
ω4 = BO4 =
= 15.40 rad/s ccw
RBO4
300
ω3 =
3.10
Ans.
Ans.
Crank 2 of the push-link mechanism shown in the figure is driven at ω2 = 60 rad / s cw.
Find the velocities of points B and C and the angular velocities of links 3 and 4.
VAO2 = ω 2 RAO2 = ( 60 rad/s )( 0.150 m ) = 9.0 m/s
VB = VA + VBA = VO4 + VBO4
VBA = 13.020 m/s ; VB = 11.360∠41° m/s
Ans.
www.elsolucionario.net
VAO2 = ω2 RAO2 = ( 45 rad/s )(100 mm ) = 4500 mm
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23
VC = VA + VCA = VB + VCB
VC = 3.830∠60° m/s
ω3 =
VBO4 11.360
VBA 13.020
=
= 43.40 rad/s cw ; ω 4 =
=
= 37.87 rad/s cw
0.300
RBO4
RBA 0.300
Ans.
Find the velocity of point C on link 4 of the mechanism shown in the figure if crank 2 is
driven at ω2 = 48 rad / s ccw. What is the angular velocity of link 3?
VAO2 = ω2 RAO2 = ( 48 rad/s )( 200 mm ) = 9600 mm/s
VB = VA + VBA = VO4 + VBO4
VBA 267.5
=
= 0.335 rad/s ccw
RBA
800
= VB + VCB ; VC = 7117.5∠ − 75.8° mm/s = 7.11 < 284°m/s
VBA = 267.5 mm/s ;
VC = VO4 + VCO4
3.12
ω3 =
Ans.
Ans.
The figure illustrates a parallel-bar linkage, in which opposite links have equal lengths.
For this linkage, show that ω3 is always zero and that ω 4 = ω 2 . How would you
describe the motion of link 4 with respect to link 2?
Refering to Fig. 2.19 and using r1 = r3 and r2 = r4 , we compare Eqs. (2.26) and (2.27) to
see that ψ = β . Then Eq. (2.29) gives θ 3 = 0 and its derivative is ω3 = 0 .
Ans.
Next we substitute Eq. (2.25) into Eq. (2.33) to see that γ = θ 2 . Then Fig. 2.19 shows
that, since link 3 is parallel to link 1 (θ 3 = 0 ) , then θ 4 = γ = θ 2 . Finally, the derivative of
this gives ω 4 = ω 2 .
Since ω 4 / 2 = ω 4 − ω 2 = 0 , link 4 is in curvilinear translation with respect to link 2.
Ans.
Ans.
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3.11
Ans.
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24
3.13
The figure illustrates the antiparallel or crossed-bar linkage. If link 2 is driven at ω 2 = 1
rad/s ccw, find the velocities of points C and D.
VB = VA + VBA = VO4 + VBO4
Construct the velocity image of link 3.
VC = 0.402∠151° m/s
VD = 0.290∠249° m/s
3.14
Ans.
Ans.
Find the velocity of point C of the linkage shown in the figure assuming that link 2 has an
angular velocity of 60 rad/s ccw. Also find the angular velocities of links 3 and 4.
VAO2 = ω2 RAO2 = ( 60 rad/s )(150 mm ) = 9000 mm/s = 9 m/s
VB = VA + VBA = VO4 + VBO4
VBA = 4235 mm/s ; VBO4 = 7940 mm/s
VBA 4235
=
= 28.24 rad/s cw
RBA 150
V
7940
ω4 = BO4 =
= 31.76 rad/s cw
250
RBO4
ω3 =
Construct the velocity image of link 3;
VC = 12677.5∠156.9° mm/s
= 1.267∠156.9° m/s
Ans.
Ans.
Ans.
www.elsolucionario.net
VAO2 = ω 2 RAO2 = (1 rad/s )( 0.300 m ) = 0.300 m/s
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25
3.15
The inversion of the slider-crank mechanism shown in the figure is driven by link 2 at
ω 2 = 60 rad/s ccw. Find the velocity of point B and the angular velocities of links
3 and 4.
VAO2 = ω 2 RAO2 = ( 60 rad/s )( 0.075 m ) = 4.500 m/s
ω3 =
VP3 A
RP3 A
=
4.259
= 22.0 rad/s ccw
0.194
Ans.
Construct the velocity image of link 3;
VB = 4.789∠96.5° m/s
3.16
Ans.
Find the velocity of the coupler point C and the angular velocities of links 3 and 4 of the
mechanism shown if crank 2 has an angular velocity of 30 rad/s cw.
VAO2 = ω2 RAO2 = ( 30 rad/s )( 75 mm ) = 2250 mm/s
VB = VA + VBA = VO4 + VBO4
Construct the velocity image of link 3;
VC = 2250∠126.9° mm/s
ω3 =
VBA 2250
=
= 18.00 rad/s ccw ;
RBA 125
Ans.
ω4 =
VBO4
RBO4
=
0.00
= 0.00
150
Ans.
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VP3 = VA + VP3 A = VP4 + VP3 / 4
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26
3.17
Link 2 of the linkage shown in the figure has an angular velocity of 10 rad/s ccw. Find
the angular velocity of link 6 and the velocities of points B, C, and D.
VB = VA + VBA
VB = 289.25∠180° mm/s
Construct velocity image of link 3.
VC = 605.5∠207.6° mm/s
VC = VA + VCA = VB + VCB
VD = VC + VDC = V/ O6 + VDO6 VD = 604.5∠206.2° mm/s
ω6 =
3.18
VDO6
RDO6
=
604.5
= 4.03 rad/s ccw
150
Ans.
Ans.
Ans.
Ans.
The angular velocity of link 2 of the drag-link mechanism shown in the figure is 16 rad/s
cw. Plot a polar velocity diagram for the velocity of point B for all crank positions.
Check the positions of maximum and minimum velocities by using Freudenstein’s
theorem.
The graphical construction is shown in the
position where θ 2 = 135° where the result is
VB = 5.76∠ − 7.2° m/s.
It is repeated at
increments of Δθ 2 = 15° . The maximum and
minimum velocities are VB ,max = 9.13∠ − 146.6°
m/s at θ 2 = 15° and VB ,min = 4.59∠63.7° m/s at
θ 2 = 225° , respectively.
Within graphical
accuracy these two positions approximatly verify
Freudenstein’s theorem.
A numeric solution for the same problem can be
found from Eq. (3.22) using Eqs. (2.25) through
(2.33) for position values. The accuracy of the values reported above have been verified
this way.
www.elsolucionario.net
VAO2 = ω2 RAO2 = (10 rad/s )( 63 mm ) = 630 mm/s
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27
3.19
Link 2 of the mechanism shown in the figure is driven at ω 2 = 36 rad/s cw. Find the
angular velocity of link 3 and the velocity of point B.
VAO2 = ω2 RAO2 = ( 36 rad/s )(125 mm ) = 4500 mm/s
VB = VA + VBA = VO4 + VBO4
3.20
Ans.
Ans.
Find the velocity of point C and the angular velocity of link 3 of the push-link
mechanism shown in the figure. Link 2 is the driver and rotates at 8 rad/s ccw.
VAO2 = ω 2 RAO2 = ( 8 rad/s )( 0.150 m ) = 1.200 m/s
VB = VA + VBA = V/ O4 + VBO4
Construct velocity image of link 3.
VC = VA + VCA = VB + VCB
VC = 3.848∠223.2° m/s
V
3.784
ω3 = BA =
= 15.14 rad/s ccw
RBA 0.250
Ans.
Ans.
www.elsolucionario.net
VB = 5070∠ − 56.3° mm/s
V
647.5
ω3 = BA =
= 3.23 rad/s ccw
RBA
200
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28
3.21
Link 2 of the mechanism shown in the figure has an angular velocity of 56 rad/s ccw.
Find the velocity of point C.
VB = VA + VBA = VO4 + VBO4
Construct velocity image of link3.
VC = 9.028∠137.8° m/s
VC = VA + VCA = VB + VCB
3.22
Ans.
Find the velocities of points B, C, and D of the double-slider mechanism shown in the
figure if crank 2 rotates at 42 rad/s cw.
VAO2 = ω2 RAO2
= ( 42 rad/s )( 50 mm ) = 2100 mm/s
VB = VA + VBA
VB = 1634∠180° mm/s
Construct velocity image of link 3.
VC = 1696.5∠154.2° mm/s
VC = VA + VCA = VB + VCB
VD = VC + VDC = 530.75∠90° mm/s
Ans.
Ans.
Ans.
www.elsolucionario.net
VAO2 = ω 2 RAO2 = ( 56 rad/s )( 0.150 m ) = 8.400 m/s
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29
The figure shows the mechanism used in a two-cylinder 60° V engine consisting, in part,
of an articulated connecting rod. Crank 2 rotates at 2000 rev/min cw. Find the velocities
of points B, C, and D.
ω2 =
VAO2
2000 rev/min ( 2π rad/rev )
= 209.4 rad/s
60 min/s
= ω2 RAO2
= ( 209.4 rad/s )( 50 mm ) = 10470 mm/s
VB = VA + VBA = 10647.5∠ − 120° mm/s
Construct velocity image of link 3.
VC = VA + VCA = VB + VCB = 12172.5∠ − 93.3° mm/s
VD = VC + VDC = 9467.5∠ − 60° mm/s
3.24
Ans.
Ans.
Ans.
Make a complete velocity analysis of the linkage shown in the figure given that
ω 2 = 24 rad/s cw. What is the absolute velocity of point B? What is its apparent velocity
to an observer moving with link 4?
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3.23
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30
VAO2 = ω2 RAO2 = ( 24 rad/s )( 200 mm ) = 4800 mm/s
Using the path of P3 on link 4, we write
VP3 = VA + VP3 A = VP4 + VP3 / 4
ω3 =
VP3 A
RP3 A
=
4045
= 6.16 rad/s cw
656.75
From this, or graphically, we complete the velocity image of link 3, from which
VB3 = 3945∠ − 39.7° mm/s
Ans.
Then, since link 4 remains perpendicular to link 3, we have ω 4 = ω3 and we find the
velocity image of link 4.
VB3 / 4 = 2582.5∠ − 12.4° mm/s
Ans.
Find VB for the linkage shown in the figure if V A = 300 mm/s.
www.elsolucionario.net
3.25
VA = 300 mm/s
Using the path of P3 on link 4, we write
VP3 = VA + VP3 A = V/ P4 + VP3 / 4
Next construct the velocity image of link 3 or VB = VP3 + VBP3 = VA + VBA
VB = 312.5∠ − 23.0° mm/s
Ans.
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31
The figure shows a variation of the Scotch-yoke mechanism. It is driven by crank 2 at
ω 2 = 36 rad/s ccw. Find the velocity of the crosshead, link 4.
VAO2 = ω 2 RAO2 = ( 36 rad/s )( 0.250 m ) = 9.0 m/s
Using the path of A2 on link 4, we write
VA2 = VA4 + VA2 / 4
(Note that the path is unknown for VA4 / 2 !)
VA4 = 4.658∠180° m/s
Ans.
All other points of link 4 have this same velocity; it is in translation.
3.27
Make a complete velocity analysis of the linkage shown in the figure for ω 2 = 72 rad/s
ccw.
VAO2 = ω2 RAO2 = ( 72 rad/s )( 38 mm ) = 2736 mm/s
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3.26
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32
VB = VA + VBA = VO4 + VBO4
Construct velocity image of link3. VC3 = VA + VCA = VB + VCB
VC3 = 1908∠203.2° mm/s
Ans.
Using the path of C3 on link 6, we next write
VC3 = VC6 + VC3 / 6 and VC6 = VC6O6 , from which VC6 = 1076.75∠241.4° mm/s
Ans.
From this, graphically, we can complete the velocity image of link 6, from which
VE = 1935∠ − 98.9° mm/s
VC O
Since link 5 remains perpendicular to link 6, ω5 = ω 6 = 6 6 = 9.67 rad/s cw
RC6O6
From these we can get VD5 = VC5 + VD5C5 = 1613∠210.1°
Ans.
Ans.
Slotted links 2 and 3 are driven independently at ω 2 = 30 rad/s cw and ω3 = 20 rad/s cw,
respectively. Find the absolute velocity of the center of the pin P4 carried in the two
slots.
VP2 = ω 2 RP2 A = ( 30 rad/s )( 0.054 9 m ) = 1.647 m/s
VP3 = ω3 RP3 B = ( 20 rad/s )( 0.102 6 m ) = 2.052 m/s
Identifying the pin as separate body 4 and, noticing the two paths it travels on bodies 2
and 3, we write
Ans.
VP4 = VP2 + VP4 / 2 = VP3 + VP4 / 3 = 2.355∠15.6° m/s
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3.28
Ans.
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33
The mechanism shown is driven such that VC = 250 mm/s to the right. Rolling contact is
assumed between links 1 and 2, but slip is possible between links 2 and 3. Determine the
angular velocity of link 3.
Using the
C2 on link 3, we write
VC2 = VC3 + VC2 / 3 and VC3 = VD3 + VC3 D3
ω3 =
3.30
VC3 D3
RC3 D3
=
105.65
= 1.569 rad/s cw
67.25
path of
Ans.
The circular cam shown is driven at an angular velocity of ω 2 = 15 rad/s ccw. There is
rolling contact between the cam and the roller, link 3. Find the angular velocity of the
oscillating follower, link 4.
VAO2 = ω2 RAO2 = (15 rad/s )( 31.25 mm ) = 468.75 mm/s
VD4 = VD2 + VD4 / 2 and VD4 = VE + VD4 E
ω4 =
VD4 E
RD4 E
=
381
= 4.355 rad/s ccw
87.5
Ans.
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3.29
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34
The mechanism shown is driven by link 2 at 10 rad/s ccw. There is rolling contact at
point F. Determine the velocity of points E and G and the angular velocities of links 3, 4,
5, and 6.
VBA = ω2 RBA = (10 rad/s )( 25 mm ) = 250 mm/s
VC = VB + VCB = VD + VCD
V
333.25
ω3 = CB =
= 3.333 rad/s ccw
Ans.
100
RCB
166.675
V
ω4 = CD =
= 3.333 rad/s ccw
Ans.
50
RCD
Construct velocity image of link 3. VE3 = VB + VE3 B = VC + VE3C = 251.5∠220.9° mm/s Ans.
Using the path of E3 on link 6, we write VE3 = VE6 + VE3 / 6 and VE6 = VH + VE6 H
ω6 =
VE6 H
RE6 H
=
121.45
= 3.774 rad/s cw
32.2
Ans.
Construct velocity image of link 6. VG = VE6 + VGE6 = VH + VGH = 298.25∠ − 57.1° mm/s Ans.
VF5 = VF6 ;
VE5 = VE3 ; ω5 =
VF5 E
RF5 E
=
319.5
= 25.56 rad/s cw
12.5
Ans.
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3.31
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35
The figure shows the schematic diagram for a two-piston pump. It is driven by a circular
eccentric, link 2, at ω 2 = 25 rad/s ccw. Find the velocities of the two pistons, links 6 and
7.
www.elsolucionario.net
3.32
VF2 = VF2 E = ω2 RFE = ( 25 rad/s )( 25 mm ) = 625 mm/s
Using the path of F2 on link 3, we write VF2 = VF3 + VF2 / 3 and VF3 = VG + VF3G
Construct velocity image of link 3.
VC = VF3 + VCF3 = VG + VCG and
VD = VF3 + VDF3 = VG + VDG . Then
VA = VC + VAC = 32.55∠180° mm/s
VB = VD + VBD = 144.5∠180° mm/s
Ans.
Ans.
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36
The epicyclic gear train shown is driven by the arm, link 2, at ω 2 = 10 rad/s cw.
Determine the angular velocity of the output shaft, attached to gear 3.
VB = VBA = ω2 RBA = (10 rad/s )( 75 mm ) = 750 mm/s
Using VD = 0 construct the velocity image of link 4 from which VC = 1500∠0° mm/s .
1500
V
ω3 = CA =
= 30.00 rad/s cw Ans.
50
RCA
3.34
The diagram shows a planar schematic approximation of an automotive front suspension.
The roll center is the term used by the industry to describe the point about which the auto
body seems to rotate with respect to the ground. The assumption is made that there is
pivoting but no slip between the tires and the road. After making a sketch, use the
concepts of instant centers to find a technique to locate the roll center.
By definition, the “roll center” (of the vehicle body ,link 2, with respect to the road, link
1,) is the instant center P12. It can be found by the repeated application of Kennedy’s
theorem as shown.
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3.33
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37
In the automotive industry it has become common practice to use only half of this
construction, assuming by symmetry that P12 must lie on the centerline of the vehicle.
Notice that this is true only when the right and left suspension arms are symmetrically
positioned. It is not true once the vehicle begins to roll as in a turn.
Having lost sight of the relationship to instant centers and Kennedy’s theorem, and
remembering only the shortened graphical construction on one side of the vehicle, many
in the industry are now confused about “the movement of the roll center along the
centerline of the vehicle”. They should be thinking about the fixed and moving centrodes
(Section 3.21) which are more horizontal than vertical!
Locate all instant centers for the linkage of Problem 3.22.
Instant centers P12 , P23 ,
P34 , P14 (at infinity), P35 ,
P56 , and P16 (at infinity) are
found by inspection. All
others are found by
repeated applications of
Kennedy’s theorem until
P46 .
One line can be found for
P46 ; however no second
line can be found since no
line can be drawn between
P14 and P16 (in finite
space). Now it must be
seen that P46 must be
infinitely remote because
the relative motion between
links 4 and 6 is translation;
the angle between links 4
and 6 remains constant.
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3.35
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38
3.36
Locate all instant centers for the mechanism of Problem 3.25.
3.37
Locate all instant centers for the mechanism of Problem 3.26.
Instant centers P12 , P23 , P34 (at
infinity), and P14 (at infinity) are
found by inspection. All others
are found by repeated applications
of Kennedy’s theorem except P13 .
One line ( P12 P23 ) can be found
for P13 ; however no second line
can be found since no line can be
drawn (in finite space) between
P14 and P34 . Now it must be seen
that P13 must be infinitely remote
because the relative motion
between links 1 and 3 is
translation; the angle between
links 1 and 3 remains constant.
www.elsolucionario.net
Instant centers P12 (at infinity), P23 ,
P34 (at infinity), and P14 are found
by inspection. All others are found
by
repeated
applications
of
Kennedy’s theorem.
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39
3.38
Locate all instant centers for the mechanism of Problem 3.27.
3.39
Locate all instant centers for the mechanism of Problem 3.29.
Instant centers P12 and P13 are found by
inspection. One line for P23 is found by
Kennedy’s theorem. The other is found by
drawing perpendicular to the relative velocity
of slipping at the point of contact between
links 2 and 3.
3.40
Locate all instant centers for the mechanism of Problem 3.30.
Instant centers P12 , P23 , P34 and P14 are found by
inspection. The other two are found by Kennedy’s
theorem
www.elsolucionario.net
Instant centers P12 , P23 , P34 , P14 , P35 , P56 (at
infinity), and P16 are found by inspection.
All others are found by repeated
applications of Kennedy’s theorem.
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41
Chapter 4
Acceleration
4.1
(
)
The position vector of a point is defined by the equation R = 4t − t 3 3 ˆi + 10ˆj where R is
in mm and t is in seconds. Find the acceleration of the point at t = 2 s.
(
)
R ( t ) = ( 4 − t 2 ) ˆi
R ( t ) = −2tˆi
4.2
R ( 2 s ) = −2 ( 2 ) ˆi = −4ˆi mm/s 2
Ans.
Find the acceleration at t = 3 s of a point which moves according to the equation
(
) ( )
R = t 2 − t 3 6 ˆi + t 3 3 ˆj . The units are meters and seconds.
(
) ( )
R ( t ) = ( 2t − t 2 2 ) ˆi + t 2 ˆj
R ( t ) = t 2 − t 3 6 ˆi + t 3 3 ˆj
R ( t ) = ( 2 − t ) ˆi + 2tˆj
4.3
R ( 3 s ) = ( 2 − 3) ˆi + 2 ( 3) ˆj = −ˆi + 6ˆj m/s 2
Ans.
The path of a point is described by the equation R = (t 2 + 4)e− jπ t / 10 where R is in
millimeters and t is in seconds. For t = 20 s, find the unit tangent vector for the path, the
normal and tangential components of the point’s absolute acceleration, and the radius of
curvature of the path.
R ( t ) = (t 2 + 4)e − jπ t /10
R ( t ) = 2te − jπ t /10 −
jπ 2
(t + 4)e− jπ t /10
10
jπ t ⎞ − jπ t /10 jπ t − jπ t /10 π 2 2
⎛
R (t ) = ⎜ 2 −
e
−
−
(t + 4)e− jπ t /10
⎟e
5 ⎠
5
100
⎝
Noticing, at t = 20 s, that e− jπ t /10 = e− j 2π = 1.0 , we find that
R ( 20 s ) = (202 + 4) = 404 mm
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R ( t ) = 4t − t 3 3 ˆi + 10ˆj
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42
R ( 20 s ) = 2 ( 20 ) −
jπ
(202 + 4) = 40.00 − j126.92 = 133.07∠ − 72.5° mm/s
10
jπ 20 ⎞ jπ 20 π 2
⎛
(202 + 4) = −37.873 − j 25.133 mm/s 2
−
R ( 20 s ) = ⎜ 2 −
⎟−
5
5
100
⎝
⎠
From the direction of the velocity we find the unit tangent and unit normal vectors
τˆ = 1∠ − 72.5° = cos − 72.5°ˆi + sin − 72.5°ˆj = 0.30058ˆi − 0.95376ˆj
ρˆ = τˆ × kˆ = sin − 72.5°ˆi − cos− 72.5°ˆj = −0.95376ˆi − 0.30058ˆj
From these, the components of the point’s absolute acceleration are
An = ρˆ iR = −0.95376ˆi − 0.30058ˆj i −37.873ˆi − 25.133ˆj mm/s 2 = 43.676 mm/s 2
(
)(
)
At = τˆ iR = ( 0.30058ˆi − 0.95376ˆj)i( −37.873ˆi − 25.133ˆj mm/s 2 ) = 12.586 mm/s 2
Ans.
Ans.
Ans.
Then, from Eq. (4.2) or Eq. (4.14), the radius of curvature is
2
(133.07 mm/s )
=−
2
= −405.4 mm
Ans.
43.676 mm/s 2
An
Where the negative sign indicates that the point is in the negative ρ̂ direction from the
center of curvature of the point’s path.
4.4
( )
The motion of a point is described by the equations x = 4t cos π t 3 and y = t 3 6 sin 2π t
where x and y are in meters and t is in seconds. Find the acceleration of the point at
t = 1.40 s .
R ( t ) = ⎡⎣ 4t cos π t 3 ⎤⎦ ˆi + ⎡⎣( t 3 6 ) sin 2π t ⎤⎦ ˆj
R ( t ) = ⎡⎣ 4 cos π t 3 − 12π t 3 sin π t 3 ⎤⎦ ˆi + ⎡⎣( t 2 2 ) sin 2π t + (π t 3 3) cos 2π t ⎤⎦ ˆj
R ( t ) = ⎡⎣ −48π t 2 sin π t 3 − 36π 2t 5 cos π t 3 ⎤⎦ ˆi + ⎡⎣t (1 − 2π 2t 2 3) sin 2π t + 2π t 2 cos 2π t ⎤⎦ ˆj
R (1.40 s ) = 1 112.620ˆi − 19.753ˆj = 1 112.796∠ − 1.0° ft/s 2
4.5
Ans.
Link 2 in the figure has an angular velocity of ω 2 = 120 rad/s ccw and an angular
acceleration of 4800 rad/ s2 ccw at the instant shown. Determine the absolute
acceleration of point A.
A A = A O2 + A nAO2 + A tAO2
= −ω 22 R AO2 + α 2kˆ × R AO2
2
= − (120 ) 0.500ˆi + 4 800kˆ × 0.500ˆi
A A = −7 200ˆi + 2 400ˆj m/s 2 = 7 589∠161.6° m/s 2
Ans.
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ρ =−
R
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43
4.6
Link 2 is rotating clockwise as shown in the figure. Find its angular velocity and
acceleration and the acceleration of its midpoint C.
A B = A A + A nBA + A tBA
Construct the acceleration polygon.
n
178.38 m/s 2
ABA
=
= 356.8 rad/s cw
RBA
0.5 m
Ans.
Note the ambiguous sign of this square root. The sense of ω cannot be determined from
the accelerations, but here is found from the problem statement.
51.03 m/s 2
At
α = BA =
= 102.1 rad/s 2 cw
Ans.
RBA
0.5 m
A C = 92.76∠44.0° m/s 2
4.7
Ans.
For the data given in the figure, find the velocity and acceleration of points B and C.
VB = VA + VBA
VBA = ω RBA = ( 24 rad/s )( 0.4 m ) = 9.6 m/s
Construct the velocity polygon.
VB = 3.6∠270° m/s
VC = 2.51∠12.1° m/s
Ans.
Ans.
A B = A A + A nBA + A tBA
n
ABA
= ω 2 RBA = ( 24 rad/s ) ( 0.4 m ) = 230.4 m/s 2
2
t
ABA
= α RBA = (160 rad/s 2 ) ( 0.4 m ) = 64 m/s 2
Construct the acceleration polygon.
A B = 118.53∠165° m/s 2
A C = 63∠240.3° m/s 2
Ans.
Ans.
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ω=
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44
4.8
For the straight-line mechanism shown in the figure, ω 2 = 20 rad/s cw
and
α 2 = 140 rad/s 2 cw . Determine the velocity and acceleration of point B and the angular
acceleration of link 3.
VA = VO2 + VAO2
VC = VA + VCA Construct the velocity polygon of link 3.
VB = 3.864∠270° m/s
A A = A O2 + A
n
AO2
+A
Ans.
t
AO2
n
= ω 22 RAO2 = ( 20 rad/s ) ( 0.100 m ) = 40.000 m/s 2
AAO
2
2
t
AAO
= α 2 RAO2 = (140 rad/s 2 ) ( 0.100 m ) = 14.000 m/s 2
2
n
A C = A A + A CA
+ A tCA
n
2
ACA
= VCA
RCA = ( 2.000 m/s )
2
( 0.100 m ) = 40.000 m/s 2
Construct the acceleration polygon of link 3.
A B = 6.340∠90° m/s 2
α3 =
Ans.
2
t
BA
A
14.000 m/s
=
= 140 rad/s 2 cw
0.100 m
RBA
Ans.
4.9 In the figure, the slider 4 is moving to the left with a constant velocity of 2.0 m/s. Find the
angular velocity and acceleration of link 2.
VA = VC + VAC = VO2 + VAO2 . Construct the velocity polygon of link 3.
ω2 =
VAO2
RAO2
=
3.864 m/s
= 38.64 rad/s cw
0.100 m
Ans.
www.elsolucionario.net
VAO2 = ω 2 RAO2 = ( 20 rad/s )( 0.100 m ) = 2.000 m/s
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45
A A = A C + A nAC + A tAC = A O2 + A nAO + A tAO
2
=V
n
AC
A
2
AC
2
/ RAC = ( 3.864 m/s ) / ( 0.100 m ) = 149.28 m/s 2
2
2
n
/ RAO2 = ( 3.864 m/s ) / ( 0.100 m ) = 149.28 m/s 2
AAO
= VAO
2
2
2
α2 =
RAO2
=
557.13 m/s 2
= 5 571.3 rad/s 2 cw
0.100 m
Ans.
Solve Problem 3.8, for the acceleration of point A and the angular acceleration of link 3.
Assume VB = constant. Therefore A B = 0 .
A A = A B + A nAB + A tAB
(14.64 m/s ) = 535.9 m/s2
V2
= AB =
RAB
0.400 m
2
n
AB
A
A A = 757.9∠15° m/s 2
α3 =
4.11
Ans.
t
AAB
535.9 m/s 2
=
= 1 339.7 rad/s 2 cw
0.400 m
RAB
For Problem 3.9, find the angular accelerations of links 3 and 4.
Assume ω 2 = constant. Therefore α 2 = 0 .
t
A A = A O2 + A nAO + A AO
2
n
AO2
A
2
= ω RAO2 = ( 45 rad/s ) (100 mm ) = 2025 m/s 2
2
2
2
Ans.
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4.10
t
AAO
2
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46
A B = A A + A nBA + A tBA = A O4 + A nBO + A tBO
4
4
A = V / RBA = ( 358.5 mm/s ) / ( 250 mm ) = 514.09 mm/s 2
n
BA
2
2
BA
(Ignore compared to other
parts.)
2
n
2
/ RBO4 = ( 4620 mm/s ) / ( 300 mm ) = 71148 mm/s 2
ABO
= VBO
4
4
α3 =
α4 =
t
ABO
4
RBO4
=
37102.5 in/s 2
= 123.7 rad/s 2 ccw
300 mm
Ans.
Ans.
For Problem 3.10, find the acceleration of point C and the angular accelerations of links 3
and 4.
www.elsolucionario.net
4.12
t
140817.5 m/s 2
ABA
=
= 563.3 rad/s 2 ccw
RBA
250 mm
Assume ω 2 = constant. Therefore α 2 = 0 .
t
A A = A O2 + A nAO + A AO
2
n
AO2
A
2
= ω RAO2 = ( 60 rad/s ) ( 0.150 m ) = 540.0 m/s 2
2
2
2
A B = A A + A nBA + A tBA = A O4 + A nBO + A tBO
4
4
A = V / RBA = (13.02 m/s ) / ( 0.300 m ) = 565.1 m/s 2
n
BA
2
2
BA
n
2
ABO
= VBO
/ RBO4 = (11.36 m/s ) / ( 0.300 m ) = 430.1 m/s 2
4
4
2
Construct the acceleration polygon of link 3.
A C = 209.6∠10.6° m/s 2
α3 =
α4 =
A
136.5 m/s
=
= 455.0 rad/s 2 ccw
0.300 m
RBA
t
ABO
4
RBO4
Ans.
2
t
BA
=
46.04 m/s 2
= 153.5 rad/s 2 cw
0.300 m
Ans.
Ans.
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47
4.13
For Problem 3.11, find the acceleration of point C and the angular accelerations of links 3
and 4.
t
A A = A O2 + A nAO + A AO
2
n
AO2
A
2
= ω RAO2
2
2
= ( 48 rad/s ) ( 200 mm ) = 18 432.0 in/s 2
2
A B = A A + A nBA + A tBA = A O4 + A nBO + A tBO
4
4
A = V / RBA = ( 267.5 mm/s ) / ( 800 mm ) = 89.45 mm/s 2
n
BA
2
2
BA
(Ignore compared to other
parts.)
2
n
2
/ RBO4 = ( 9365 mm/s ) / ( 400 mm ) = 219.26 m/s 2
ABO
= VBO
4
4
Construct the acceleration polygon of link 4.
A C = 931.35∠114.4° m/s 2
α3 =
α4 =
4.14
A
1,393, 250 mm/s
=
= 1 741.6 rad/s 2 ccw
800 mm
RBA
t
ABO
4
RBO4
Ans.
2
t
BA
=
1, 222,300 mm/s 2
= 3 055.8 rad/s 2 ccw
400 mm
Ans.
Ans.
Using the data of Problem 3.13, solve for the accelerations of points C and D and the
angular acceleration of link 4.
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Assume ω 2 = constant. Therefore α 2 = 0 .
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48
Assume ω 2 = constant. Therefore α 2 = 0 .
t
A A = A O2 + A nAO + A AO
2
n
AO2
A
2
= ω RAO2 = (1.0 rad/s ) ( 0.300 m ) = 0.300 m/s 2
2
2
2
A B = A A + A nBA + A tBA = A O4 + A nBO + A tBO
4
4
A = V / RBA = ( 0.263 5 m/s ) / ( 0.150 m ) = 0.462 9 m/s 2
2
2
BA
n
BA
n
2
/ RBO4 = ( 0.227 0 m/s ) / ( 0.300 m ) = 0.171 8 m/s 2
ABO
= VBO
4
4
2
Construct the acceleration polygon of link 3.
A C = 0.515 4∠ − 119.8° m/s 2
Ans.
A D = 0.492 5∠21.8° m/s 2
Ans.
4.15
t
ABO
4
RBO4
=
0.221 2 m/s 2
= 0.737 4 rad/s 2 cw
0.300 m
Ans.
For Problem 3.14, find the acceleration of point C and the angular acceleration of link 4.
Assume ω 2 = constant. Therefore α 2 = 0 .
t
A A = A O2 + A nAO + A AO
2
n
AO2
A
2
= ω RAO2 = ( 60.0 rad/s ) (150 mm ) = 540 m/s 2
2
2
2
A B = A A + A nBA + A tBA = A O4 + A nBO + A tBO
4
4
A = V / RBA = ( 4235 mm/s ) / (150 mm ) = 119.57 m/s 2
2
2
BA
n
BA
n
2
ABO
= VBO
/ RBO4 = ( 7940 mm/s ) / ( 250 mm ) = 252.174 m/s 2
4
4
2
Construct the acceleration polygon of link 3.
A C = 781.25∠ − 68.9° m/s 2
α4 =
t
BO4
A
RBO4
=
373.75 m/s 2
= 1 495 rad/s 2 ccw
0.25 m
Ans.
Ans.
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α4 =
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49
4.16
Solve Problem 3.16 for the acceleration of point C and the angular acceleration of link 4.
Assume ω 2 = constant. Therefore α 2 = 0 .
t
A A = A O2 + A nAO + A AO
2
A
2
= ω RAO2 = ( 30.0 rad/s ) ( 75 mm ) = 67.5 m/s 2
2
2
2
A B = A A + A nBA + A tBA = A O4 + A nBO + A tBO
4
4
A = V / RBA = ( 2250 mm/s ) / (125 mm ) = 40.5 m/s 2
n
BA
2
2
BA
n
2
ABO
= VBO
/ RBO4 = ( 0.0 mm/s ) / (125 mm ) = 0.0 m/s 2
4
4
2
Construct the acceleration polygon of link 3.
A C = 148.5∠216.9° m/s 2
α4 =
4.17
t
ABO
4
RBO4
=
108 m/s 2
= 720 rad/s 2 ccw
0.150 m
Ans.
Ans.
For Problem 3.17, find the acceleration of point B and the angular accelerations of links 3
and 6.
Assume ω 2 = constant. Therefore α 2 = 0 .
t
A A = A O2 + A nAO + A AO
2
n
AO2
A
2
= ω RAO2 = (10.0 rad/s ) ( 63 mm ) = 6300 mm/s 2
2
2
2
A B = A A + A nBA + A tBA
n
2
/ RBA = ( 468.75 mm/s ) / ( 250 mm ) = 878.9 mm/s 2
ABA
= VBA
2
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n
AO2
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50
A B = 5022.5∠0° mm/s 2
α3 =
Ans.
2
t
BA
4372.5 mm/s
A
=
= 17.49 rad/s 2 ccw
RBA
250 mm
Ans.
n
+ A tCA = A B + A nCB + A tCB
Construct the acceleration polygon of link 3, or A C = A A + A CA
A D = A C + A nDC + A tDC = A O6 + A nDO + A tDO
6
=V
2
DC
n
DC
A
6
/ RDC = (15 mm/s ) / (100 mm ) = 2.25 mm/s 2
2
(Ignore compared to other
parts.)
2
n
2
ADO
= VDO
/ RDO6 = ( 604.5 mm/s ) / (150 mm ) = 2436.13 mm/s 2
6
6
4.18
RDO6
1621.954 mm/s 2
=
= 10.81 rad/s 2 cw
150 mm
Ans.
For the data of Problem 3.18, what angular acceleration must be given to link 2 for the
position shown to make the angular acceleration of link 4 zero?
t
A B = A O4 + A nBO + A BO
4
=V
2
BO4
n
BO4
A
4
/ RBO4 = ( 5.764 m/s ) / ( 0.400 m ) = 83.056 m/s 2
2
A A = A B + A nAB + A tAB = A O2 + A nAO + A tAO
2
n
AB
A
2
= V / RAB = ( 5.004 m/s ) / ( 0.425 m ) = 58.93 m/s 2
2
AB
2
n
2
/ RAO2 = ( 5.600 m/s ) / ( 0.35 m ) = 89.60 m/s 2
AAO
= VAO
2
2
2
α2 =
t
AAO
2
RAO2
18.54 m/s 2
=
= 52.98 rad/s 2 ccw
0.350 m
Ans.
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α6 =
t
ADO
6
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51
4.19
For the data of Problem 3.19, what angular acceleration must be given to link 2 for the
angular acceleration of link 4 to be 100 rad/ s2 cw at the instant shown?
A B = A O4 + A nBO + A tBO
4
=V
2
BO4
n
BO4
A
4
/ RBO4 = ( 5070 mm/s ) / ( 200 mm ) = 128.52 m/s 2
2
t
= α 4 RBO4 = (100 rad/s 2 ) ( 200 mm ) = 20 m/s 2
ABO
4
2
n
AB
A
2
= V / RAB = ( 646.75 mm/s ) / ( 200 mm ) = 2091.43 mm/s 2 (Ignorable compared to other
2
2
AB
parts.)
2
n
2
AAO
= VAO
/ RAO2 = ( 4500 mm/s ) / (125 mm ) = 162 m/s 2
2
2
α2 =
4.20
t
AAO
2
RAO2
=
522575 mm/s 2
= 4 180.7 rad/s 2 ccw
125 mm
Ans.
Solve Problem 3.20 for the acceleration of point C and the angular acceleration of link 3.
Assume ω 2 = constant. Therefore α 2 = 0 .
t
A A = A O2 + A nAO + A AO
2
n
AO2
A
2
= ω RAO2 = ( 8.0 rad/s ) ( 0.150 m ) = 9.600 m/s 2
2
2
2
A B = A A + A nBA + A tBA = A O4 + A nBO + A tBO
4
4
A = V / RBA = ( 3.783 9 m/s ) / ( 0.250 m ) = 57.270 m/s 2
n
BA
2
2
BA
n
2
/ RBO4 = ( 3.483 7 m/s ) / ( 0.250 m ) = 48.544 m/s 2
ABO
= VBO
4
4
2
Construct the acceleration polygon of link 3.
A C = 63.41∠ − 22.9° m/s 2
α3 =
t
BA
Ans.
2
A
15.647 m/s
=
= 62.59 rad/s 2 cw
0.250 m
RBA
Ans.
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A A = A B + A nAB + A tAB = A O2 + A nAO + A tAO
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52
4.21
For Problem 3.21, find the acceleration of point C and the angular acceleration of link 3.
ω 2 = constant.
Assume
Therefore α 2 = 0 .
t
A A = A O2 + A nAO + A AO
2
n
AO2
A
2
= ω RAO2 = ( 56.0 rad/s ) ( 0.150 m ) = 470.4 m/s 2
2
2
2
A B = A A + A nBA + A tBA = A O4 + A nBO + A tBO
4
4
A = V / RBA = ( 6.966 m/s ) / ( 0.250 m ) = 194.09 m/s 2
2
2
BA
n
2
ABO
= VBO
/ RBO4 = (11.380 m/s ) / ( 0.250 m ) = 518.06 m/s 2
4
4
2
Construct the acceleration polygon of link 3.
A C = 450.6∠255.6° m/s 2
α3 =
4.22
Ans.
t
ABA
18.52 m/s 2
=
= 74.08 rad/s 2 cw
0.250 m
RBA
Ans.
Find the accelerations of points B and D of Problem 3.22.
Assume ω 2 = constant. Therefore α 2 = 0 .
t
A A = A O2 + A nAO + A AO
2
n
AO2
A
2
= ω RAO2 = ( 42.0 rad/s ) ( 50 mm ) = 88.2 m/s 2
2
2
A B = A A + A nBA + A tBA
2
n
2
ABA
= VBA
/ RBA = (1066 mm/s ) / ( 250 mm ) = 4.54 m/s 2
2
A B = 84.68∠0° in/s 2
Ans.
n
t
n
Construct the acceleration polygon of link 3, or A C = A A + A CA + A CA = A B + A CB + A tCB
A D = A C + A nDC + A tDC
n
2
ADC
= VDC
/ RDC = (1541.5 mm/s ) / ( 200 mm ) = 11.88 m/s 2
2
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n
BA
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53
A D = 40.4∠90° m/s 2
4.23
Ans.
Find the accelerations of points B and D of Problem 3.23.
Assume ω 2 = constant. Therefore α 2 = 0 .
t
A A = A O2 + A nAO + A AO
2
A
2
= ω RAO2 = ( 209.4 rad/s ) ( 50 mm ) = 2.19 m/s 2
2
2
2
A B = A A + A nBA + A tBA
n
2
ABA
= VBA
/ RBA = ( 5468.75 mm/s ) / (150 mm ) = 199.38 m/s 2
2
A B = 732.17∠240° m/s 2
Ans.
n
t
n
Construct the acceleration polygon of link 3, or A C = A A + A CA + A CA = A B + A CB + A tCB
A D = A C + A nDC + A tDC
n
2
ADC
= VDC
/ RDC = ( 6725 mm/s ) / (125 mm ) = 361.8 m/s 2
2
A D = 1.209∠120° m/s 2
4.24
Ans.
to 4.30 The nomenclature for this group of problems is shown in the figure, and the
dimensions and data are given in the accompanying table. For each problem,
determine θ3 , θ 4 , ω3 , ω 4 , α 3 , and α 4 . The angular velocity ω 2 is constant for each
problem, and a negative sign is used to indicate the clockwise direction. The dimensions
are given in millimeters.
This group of problems was solved on a programmable calculator. The position solution
values were found from Eqs. (2.25) through (2.32). The velocity values were found from
Eqs. (3.22). The acceleration values were found from Eqs. (4.31) and (4.32).
Prob.
4.24
4.25
4.26
4.27
4.28
4.29
4.30
θ 3 , deg
θ 4 , deg
ω3 , rad/s
ω 4 , rad/s
α 3 , rad/s 2
α 4 , rad/s 2
105.29
171.01
45.57
28.32
24.17
38.42
73.16
159.60
195.54
91.15
55.88
63.73
155.60
138.51
0.809 3
70.452 7
-4.000 0
-0.632 6
-1.516 7
-6.855 2
-0.505 1
0.525 0
47.566 8
-4.000 0
-2.155 7
1.712 9
-1.234 5
7.275 3
0.230 28
3 196.657 49
-1.120 22
7.822 40
41.414 93
62.500 44
-206.384 28
0.008 09
3 330.841 37
54.890 98
6.704 18
74.975 93
-96.514 21
-94.122 01
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n
AO2
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54
4.31
Crank 2 of the system shown in the figure has a speed of 60 rev/min ccw. Find the
velocity and acceleration of point B and the angular velocity and acceleration of link 4.
rad ⎞ ⎛ 1 min ⎞
⎛ rev ⎞ ⎛
⎟ ⎜ 2π
⎟⎜
⎟
rev ⎠ ⎝ 60 s ⎠
⎝ min ⎠ ⎝
= 6.283 rad/s
VA2 = VO2 + VA2O2 = VA4 + VA2 / 4
ω 2 = ⎜ 60
VA2O2 = ω2 RA2O2 = ( 6.283 rad/s )(175 mm )
= 1.099 m/s
Construct the velocity polygon.
V
910.75 mm/p
ω4 = A4O4 =
RA4O4
138.57 mm
Ans.
VB = VO4 + VBO4
VBO4 = ω4 RBO4 = ( 6.572 rad/s )( 700 mm )
VB = 4.6∠ − 19.1° m/s
Ans.
Since we know the path of A2 on link 4, we write
A A2 = A O2 + Α nA2O2 + Α At 2O2 = A A4 + A cA2 A4 + A nA2 / 4 + A tA2 / 4 ; A A4 = A O4 + Α nA4O4 + Α tA4O4
AAn2O2 = ω22 RA2O2 = ( 6.283 rad/s ) (175 mm ) = 6.9 m/s 2
2
A cA2 A4 = 2ω4 × VA2 / 4 = 2 ( 6.572 rad/s )( 616 mm/s ) = 8.096∠19.1° m/s 2
n
A2 / 4
A
=
VA22 / 4
ρA /4
2
( 616 mm/s )
=
∞
2
=0
Construct the acceleration polygon.
AAt 4O4 11969.5 mm/s 2
α4 =
=
= 86.38 rad/s 2 ccw
RA4O4
138.57 mm
Construct the acceleration image of link 4.
A B = 67.57∠187.5° m/s 2
4.32
Ans.
Ans.
The mechanism shown in the figure is a marine steering gear called Rapson’s slide. O2 B
is the tiller, and AC is the actuating rod. If the velocity of AC is 3 m/min to the left, find
the angular acceleration of the tiller.
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ω 4 = 6.572 rad/s cw
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55
VA4 =
( 3 m/min ) = 50 mm/s
( 60 s/min )
VA2 = VO2 + VA2O2 = VA4 + VA2 / 4
Construct the velocity polygon.
Assume VA4 = constant. Therefore AA4 = 0 .
Since we know the path of A4 on link 2, we write
A A2 = A O2 + Α nA2O2 + Α tA2O2 ; and A A4 = A A2 + A cA4 A2 + A nA4 / 2 + A tA4 / 2
( 43.3 mm/s )
=
2
=
c
A4 A2
= 2ω2 × VA4 / 2 = 2 ( 0.020 83 rad/s )( 25 mm/s ) = 1.042∠30.0° mm/s 2
A
A
AAn4 / 2 =
RA2O2
VA24 / 2
ρA /2
4
2078.5 mm
=
( 25 mm/s )
∞
= 0.9 mm/s 2
2
=0
Construct the acceleration polygon.
AAt 2O2 1.042 mm/s 2
α2 =
=
= 0.000 50 rad/s 2 cw
2078.5 mm
RA2O2
4.33
Ans.
Determine the acceleration of link 4 of Problem 3.26.
Assume ω 2 = constant. Therefore α 2 = 0 .
Since we know the path of A2 on link 4, we write
A A2 = A O2 + Α nA2O2 + Α At 2O2 = A A4 + A cA2 A4 + A nA2 / 4 + A tA2 / 4
AAn2O2 = ω 22 RA2O2 = ( 36.0 rad/s ) ( 0.250 m ) = 324.0 m/s 2
2
A cA4 A2 = 2ω 4 × VA2 / 4 = 2 ( 0.0 rad/s )( 6.588 m/s ) = 0 ; AAn4 / 2 =
VA22 / 4
ρA /4
2
Construct the acceleration polygon.
A A4 = 290.5∠180.0° m/s 2
=
( 6.588 in/s )
∞
2
=0
Ans.
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VA22O2
n
A2O2
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56
4.34
For Problem 3.27, find the acceleration of point E.
Assume ω 2 = constant. Therefore α 2 = 0 .
t
A A = A O2 + A nAO + A AO
2
n
AO2
A
2
= ω RAO2 = ( 72.0 rad/s ) ( 38 mm ) = 196.99 m/s 2
2
2
2
A B = A A + A nBA + A tBA = A O4 + A nBO + A tBO
4
4
A = V / RBA = ( 3780 mm/s ) / ( 263 mm ) = 54.33 m/s 2
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2
2
BA
n
BA
n
2
/ RBO4 = (1800 mm/s ) / (125 mm ) = 25.92 m/s 2
ABO
= VBO
4
4
2
Construct the acceleration polygon of link 3.
Since we know the path of C3 on link 6, we write
A C3 = A C6 + A Cc 3C6 + A Cn 3 / 6 + A Ct 3 / 6 and A C6 = A O6 + Α Cn 6O6 + Α Ct 6O6
A Cc 3C6 = 2ω6 × VC3 / 6 = 2 ( 9.673 rad/s )(1253.75 mm/s ) = 24.255 m/s 2 ;
n
C3 / 6
A
=
VC23 / 6
ρC / 6
(1253.75 mm/s )
=
n
C6O6
A
=
V
RC6O6
=0
∞
3
2
C6O6
2
(1076.75 mm/s )
=
111.325 mm
2
= 10.41 m/s 2
Construct the acceleration polygon of link 6.
A E = 180.82∠252.8° m/s 2
4.35
Ans.
Find the acceleration of point B and the angular acceleration of link 4 of Problem 3.24.
Assume ω 2 = constant. Therefore α 2 = 0 .
t
A A = A O2 + A nAO + A AO
2
n
AO2
A
2
= ω RAO2 = ( 24.0 rad/s ) ( 200 mm ) = 115.2 m/s 2
2
2
2
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Since we know the path of P3 on link 4, we write
A P3 = A A3 + Α nP3 A3 + Α tP3 A3 = A P4 + A cP3 P4 + A nP3 / 4 + A tP3 / 4
APn3 A3 = VP23 A3 / RP3 A3 = ( 4045 mm/s ) / ( 656.75 mm ) = 24.91 m/s 2
2
A cP3 P4 = 2ω4 × VP3 / 4 = 2 ( 6.159 rad/s )( 2592.5 mm/s ) = 31.934∠ − 102.4° m/s 2
n
P3 / 4
A
=
VP23 / 4
ρP / 4
3
( 2592.5 mm/s )
=
∞
2
=0
Construct the acceleration polygon of link 3.
A B = 123.765∠ − 25.7° m/s 2
Since links 3 and 4 remain perpendicular,
APt 3 A3 30070 mm/s 2
α4 = α3 =
=
= 45.78 rad/s 2 ccw
RP3 A3
656.75 mm
4.36
Ans.
Ans.
For Problem 3.25, find the acceleration of point B and the angular acceleration of link 3.
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57
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58
Assume VA = constant. Therefore A A = 0 .
Since we know the path of P3 on link 4, we write
A P3 = A A3 + Α nP3 A3 + Α tP3 A3 = A P4 + A cP3 P4 + A nP3 / 4 + A tP3 / 4
APn3 A3 = VP23 A3 / RP3 A3 = ( 68.375 mm/s ) / ( 225 mm ) = 20.78 mm/s 2
2
A cP3 P4 = 2ω4 × VP3 / 4 = 2 ( 0.3039 rad/s )( 307.75 mm/s ) = 187∠ − 102.8° mm/s 2
A
=
VP23 / 4
ρP / 4
( 307.75 mm/s )
=
∞
3
2
=0
Construct the acceleration polygon of link 3.
A B = 505.75∠ − 102.1° mm/s 2
APt 3 A3 280.5 mm/s 2
α3 =
=
= 1.247 rad/s 2 cw
RP3 A3
225 mm
4.37
Ans.
Ans.
Assuming that both links 2 and 3 of Problem 3.28 are rotating at constant speed, find the
acceleration of point P4 .
Since ω 2 = constant and ω3 = constant, α 2 = α 3 = 0 .
Since we know the paths of P4 on link 2 and on link 3, we write
A P2 = A O2 + A nP O + A Pt O ;
A P3 = A O3 + A nP O + A Pt O
2 2
n
P2O2
A
2 2
3 3
3 3
= ω RP2O2 = ( 30.0 rad/s ) ( 0.054 9 m ) = 49.410 m/s .
2
2
2
2
APn3O3 = ω32 RP3O3 = ( 20.0 rad/s ) ( 0.102 6 m ) = 41.039 m/s 2
2
A P4 = A P2 + A cP4 P2 + A nP4 / 2 + A tP4 / 2 = A P3 + A cP4 P3 + A nP4 / 3 + A tP4 / 3
A cP4 P2 = 2ω 2 × VP4 / 2 = 2 ( 30.0 rad/s )(1.683 m/s ) = 100.98∠ − 30.0° m/s 2
A cP4 P3 = 2ω3 × VP4 / 3 = 2 ( 20.0 rad/s )(1.155 m/s ) = 46.22∠225.0° m/s 2
n
P4 / 2
A
=
VP24 / 2
ρP / 2
4
(1.683 m/s )
=
∞
2
= 0;
Construct the acceleration polygon.
A P4 = 125.73∠ − 66.6° m/s 2
n
P4 / 2
A
=
VP24 / 2
ρP / 2
4
(1.155 m/s )
=
∞
2
=0
Ans.
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n
P3 / 4
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59
4.38
Solve Problem 3.32, for the accelerations of points A and B.
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Assume
ω 2 = constant. Therefore α 2 = 0 .
Since we know the path of F2 on link 3, we write
A F2 = A E2 + Α nF2 E2 + Α Ft 2 E2 = A F3 + A cF2 F3 + A nF2 / 3 + A tF2 / 3 and A F3 = A G3 + A nF3G3 + A tF3G3
AFn2 E2 = ω22 RF2 E2 = ( 25.0 rad/s ) ( 25 mm ) = 15.625 m/s 2
2
AFn3G3 =
VF23G3
RF3G3
=
(102.75 mm/s )
2
152 mm
= 69.46 m/s 2
A cF2 F3 = 2ω3 × VF2 / 3 = 2 ( 0.676 rad/s )( 616.5 mm/s ) = 833.5∠ − 80.5° mm/s 2
n
F2 / 3
A
=
VF22 / 3
ρF / 3
( 616.5 mm/s )
=
2
2
=0
∞
Construct the acceleration polygon of link 3.
A A = A C + A nAC + A tAC
and
A B = A D + A nBD + A tBD
2
n
/ RAC = (195.175 mm/s ) / (150 mm ) = 253.95 mm/s 2
AAC
= VAC
2
A A = 4230∠0° mm/s 2
Ans.
n
2
ABD
= VBD
/ RBD = (190.55 mm/s ) / (150 mm ) = 242.1 mm/s 2
2
A B = 20.152∠0° m/s 2
4.39
Ans.
For Problem 3.33, determine the acceleration of point C4 and the angular acceleration of
link 3 if crank 2 is given an angular acceleration of 2 rad/ s2 ccw.
A B = A A + A nBA + A tBA
n
ABA
= ω22 RBA = (10.0 rad/s ) ( 75 mm ) = 7500 mm/s 2
2
t
ABA
= α 2 RBA = ( 2.0 rad/s 2 ) ( 75 mm ) = 150 mm/s 2
A D4 = A rD4 /1 = A B + A nDB + A tDB
2
n
/ RDB = ( 750 mm/s ) / ( 25 mm ) = 225 m/s 2
ADB
= VDB
2
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60
Draw the acceleration image of link 4.
A C4 = 15∠91.1° m/s 2
Ans.
n
A C3 = A C4 + A Cr 3 / 4 = A A + A CA
+ A tCA
n
2
ACA
= VCA
/ RCA = (1500 mm/s ) / ( 50 mm ) = 45 m/s 2
Draw the acceleration image of link 3.
ACt 3 A3 300 mm/s 2
α3 =
=
= 6.0 rad/s 2 ccw
50 mm
RC3 A3
4.40
Determine the angular accelerations of links 3 and 4 of Problem 3.30.
Assume ω 2 = constant. Therefore α 2 = 0 .
Since we know the path of D4 on link 2, we write
A D2 = A A2 + A nD2 A2 + A Dt 2 A2
Ans.
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2
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61
ADn 2 A2 = ω22 RD2 A2 = (15.0 rad/s ) ( 31 mm ) = 6975 mm/s 2 .
2
A D4 = A D2 + A cD4 D2 + A nD4 / 2 + A tD4 / 2 = A E4 + A nD4 E4 + A tD4 E4
A cD4 D2 = 2ω2 × VD4 / 2 = 2 (15.0 rad/s )(1235.25 mm/s ) = 37.05∠80.8° m/s 2
ADn 4 / 2 = VD24 / 2 ρ D4 / 2 = (1235.25 mm/s )
ADn 4 E4 = VD24 E4 RD4 E4
( 62.5 mm ) = 24.413 m/s 2
2
= ( 381 mm/s ) ( 88 mm ) = 1.65 m/s 2
2
Draw the acceleration images of links 2 and 4.
ADt 4 E4 8810 mm/s 2
α4 =
=
= 100.1 rad/s 2 ccw
RD4 E4
88 mm
Ans.
A C3 = A C2 + A Cr 3 / 2 = A D3 + A Cn 3 D3 + A Ct 3 D3
α3 =
4.41
ACt 3 D3
RC3 D3
=
2
(13 mm ) = 84.45 m/s2
5792.5 mm/s 2
= 445.6 rad/s 2 ccw
13 mm
Ans.
For Problem 3.31, determine the acceleration of point G and the angular accelerations of
links 5 and 6.
Assume ω 2 = constant. Therefore α 2 = 0 .
A B = A A + A nBA + A BAt
n
ABA
= ω22 RBA = (10 rad/s ) ( 25 mm ) = 2.5 m/s 2
2
n
n
A C = A B + A CB
+ A tCB = A D + A CD
+ A tCD
2
n
/ RCB = ( 333.25 mm/s ) / (100 mm ) = 1.11 m/s 2
ACB
= VCB
2
n
2
ACD
= VCD
/ RCD = (166.5 mm/s ) / ( 50 mm ) = 0.56 m/s 2
2
Construct the acceleration image of link 3.
Since we know the path of E3 on link 6, we write
A E3 = A E6 + A cE3 E6 + A nE3 / 6 + A tE3 / 6 and A E6 = A H 6 + Α nE6 H 6 + Α tE6 H 6
A cE3 E6 = 2ω6 × VE3 / 6 = 2 ( 3.774 rad/s )( 272.25 mm/s ) = 2.05∠104.5° m/s 2 ;
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ACn3 D3 = VC23 D3 RC3 D3 = (1047.7 mm/s )
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62
AEn3 / 6 = VE23 / 6 / ρ E3 / 6 = ( 272.25 mm/s ) / ∞ = 0
2
AEn6 H 6 = VE26 H 6 RE6 H 6 = (121.5 mm/s ) 32.25 mm = 457.74 mm/s 2
2
Construct the acceleration image of link 6.
A G = 8785∠ − 64.5° mm/s 2
α6 =
AEt 6 H 6
RE6 H 6
=
3547.5 mm/s 2
= 110 rad/s 2 cw
32.25 mm
Ans.
Ans.
A F5 = A F6 + A rF5 / 6 = A E5 + A nF5 E5 + A tF5 E5
α5 =
4.42
AFt 5 E5
RF5 E5
0 in/s 2
=
=0
13 mm
2
(13 mm ) = 7.85 m/s2
Ans.
Find the inflection circle for motion of the coupler of the double-slider mechanism shown
in the figure. Select several points on the centrode normal and find their conjugate
points. Plot portions of the paths of these points to demonstrate for yourself that the
conjugates are indeed the centers of curvature.
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AFn5 E5 = VF25 E5 RF5 E5 = ( 319.5 mm/s )
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63
Find the inflection circle for motion of the coupler relative to the frame of the linkage
shown in the figure. Find the center of curvature of the coupler curve of point C and
generate a portion of the path of C to verify your findings.
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4.43
Since point C is on the inflection circle; its center of curvature is at infinity; and its point
path is a straight line in the vicinity of the position shown.
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64
For the motion of the coupler relative to the frame, find the inflection circle, the centrode
normal, the centrode tangent, and the centers of curvature of points C and D of the
linkage of Problem 3.13. Choose points on the coupler coincident with the instant center
and inflection pole and plot their paths.
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4.44
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65
On 450- by 600-mm paper, draw the linkage shown in the figure in full size, placing A′
at 150 mm from the lower edge and 175 mm from the right edge. Better utilization of the
paper is obtained by tilting the frame through about 15° as shown.
(a)
Find the inflection circle.
(b)
Draw the cubic of stationary curvature.
(c)
Choose a coupler point C coincident with the cubic and plot a portion of its
coupler curve in the vicinity of the cubic.
(d)
Find the conjugate point C ′ . Draw a circle through C with center at C ′ and
compare this circle with the actual path of C.
(e) Find Ball’s point. Locate a point D on the coupler at Ball’s point and plot a portion
of its path. Compare the result with a straight line.
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4.45
Drawn with a precise CAD system above, the circle around center C’ matches the coupler
curve near C within better than visual comparison can detect for the ±30° of crank
rotation shown. Similarly, Ball’s point D follows an almost perfect straight line over the
same range as shown.
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PART II
DESIGN OF MECHANISMS
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67
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69
Chapter 5
5.1
The reciprocating radial roller follower of a plate cam is to rise 50 mm with simple
harmonic motion in 180D of cam rotation and return with simple harmonic motion in the
remaining 180° . If the roller radius is 10 mm and the prime-circle radius is 50 mm,
construct the displacement diagram, the pitch curve, and the cam profile for clockwise
cam rotation.
5.2
A plate cam with a reciprocating flat-face follower has the same motion as in Problem
5.1. The prime-circle radius is 50 mm, and the cam rotates counterclockwise. Construct
the displacement diagram and the cam profile, offsetting the follower stem by 20 mm in
the direction that reduces the bending stress in the follower during rise.
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Cam Design
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70
Construct the displacement diagram and the cam profile for a plate cam with an
oscillating radial flat-face follower that rises through 30° with cycloidal motion in 150°
of counterclockwise cam rotation, then dwells for 30D , returns with cycloidal motion in
120° , and dwells for 60° . Determine the necessary length for the follower face,
allowing 5 mm clearance at each end. The prime-circle radius is 30 mm, and the
follower pivot is 120 mm to the right.
Notice that, with the prime circle radius given, the cam is undercut and the follower will
not reach positions 7 and 8. The follower face length shown is 200 mm but can be made
as short as 195 mm (position 9)from the follower pivot.
Ans.
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5.3
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71
A plate cam with an oscillating roller follower is to produce the same motion as in
Problem 5.3. The prime-circle radius is 60 mm, the roller radius is 10 mm, the length of
the follower is 100 mm, and it is pivoted at 125 mm to the left of the cam rotation axis.
The cam rotation is clockwise. Determine the maximum pressure angle.
From a graphical analysis, ϕ max = 39°
5.5
Ans.
For a full-rise simple harmonic motion, write the equations for the velocity and the jerk at
the midpoint of the motion. Also, determine the acceleration at the beginning and the end
of the motion.
Using Eqs. (5.18) we find
⎛θ 1⎞ πL
π πL
sin =
y′ ⎜ = ⎟ =
2 2β
⎝ β 2 ⎠ 2β
Ans.
⎛θ 1⎞
π 3L π
π 3L
sin = −
y′′′ ⎜ = ⎟ = −
2
2β 3
2β 3
⎝ β 2⎠
Ans.
⎛θ
⎞ π 2L
π 2L
y′′ ⎜ = 0 ⎟ =
cos 0 =
2β 2
⎝β
⎠ 2β 2
Ans.
⎛θ
⎞ π 2L
π 2L
cos π = −
y′′ ⎜ = 1⎟ =
2β 2
⎝β
⎠ 2β 2
Ans.
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5.4
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72
5.6
For a full-rise cycloidal motion, determine the values of θ for which the acceleration is
maximum and minimum. What is the formula for the acceleration at these points? Find
the equations for the velocity and the jerk at the midpoint of the motion.
5.7
⎛ θ 1 ⎞ 2π L
π 2π L
′′
y′′ ⎜ = ⎟ =
sin =
= ymax
2 β2
⎝ β 4⎠ β2
Ans.
⎛ θ 3 ⎞ 2π L
3π
2π L
′′
sin
=−
= ymin
y′′ ⎜ = ⎟ =
2
2
4
2
β
β
β
⎝
⎠
Ans.
⎛θ 1⎞ L
2L
y′ ⎜ = ⎟ = (1 − cos π ) =
β
⎝ β 2⎠ β
Ans.
⎛ θ 1 ⎞ 4π 2 L
4π 2 L
′′′
y ⎜ = ⎟=
cos π = −
β3
β3
⎝ β 2⎠
Ans.
A plate cam with a reciprocating follower is to rotate clockwise at 400 rev/min. The
follower is to dwell for 60° of cam rotation, after which it is to rise to a lift of 62.5 mm.
During 25 mm of its return stroke, it must have a constant velocity of 1000 mm/s.
Recommend standard cam motions from Sec. 5.7 to be used for high speed operation and
determine the corresponding lifts and cam rotation angles for each segment of the cam.
The curves shown are initially only
sketches and not drawn to scale.
They suggest the standard curve
types which might be chosen. The
actual choices are shown in the
table below.
ω=
( 400 rev/min )( 2π rad/rev )
( 60 s/min )
= 41.888 rad/s cw
To match the given velocity
condition in Seg. DE we must have
y 4 = y4′ω
1000 mm/s = y4′ ( 41.888 rad/s )
y4′ = − L4 β 4 = −23.87 mm/rad
− ( 25 mm ) β 4 = −23.87 mm/rad
−2 L5 β 5 = y4′ = −0.954 930 in/rad
β 4 = 1.047339 rad = 60°
Matching the first derivatives at D
and E we find
L5 = 11.9366β 5
(1)
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Using Eqs. (5.19), we know that acceleration is an extremum when jerk is zero. This
happens when cos 2πθ β = 0 or when θ β = 1 4 or θ β = 3 4 .
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73
−π L3 ( 2 β 3 ) = y4′ = −0.954 930 in/rad
L3 = 15.1981β 3
(2)
Matching the second derivatives at C we find
−131.707 ( 62.5 ) β 22 = −π 2 L3 4β 32
β 22 = 133.4476β32 L3
(3)
5.8
Repeat Problem 5.7 except with a dwell that is to be for 20° of cam rotation.
The procedure is the same as for Problem 5.7. The results are:
Seg. Type
Eq.
L, mm
β, rad
AB dwell
--0
0.349 066
th
BC 8 order poly.
(5.20)
62.5
1.852 072
CD half harmonic
(5.26)
5.93
0.390 658
DE constant velocity
--25
1.047 198
EA half cycloidal
(5.31)
31.56
2.644 191
5.9
β, deg
20.000
106.116
22.383
60.000
151.501
If the cam of Problem 5.7 is driven at constant speed, determine the time of the dwell and
the maximum and minimum velocity and acceleration of the follower for the cam cycle.
The time duration of the dwell is Δt = β1 ω = 1.047 198 rad 41.888 rad/s = 0.025 s Ans.
Working from the equations listed, the maximum and minimum values of the derivatives
in each segment of the cam are as follows:
Seg.
Eq.
′
′
′′
′′
ymax
ymin
ymax
ymin
AB
--0
0
0
BC
(5.20)
102.48
0
280.346
CD
(5.26)
0
-23.86
0
DE
---23.87
-23.87
0
EA
(5.31)
0
-23.873
12.62
′ ω = (102.48 mm/rad )( 41.888 rad/s ) = 4.292 m/s
y max = ymax
0
-280.346
-280.346
0
0
Ans.
′ ω = ( −23.87 mm/rad )( 41.888 rad/s ) = 1 m/s
y min = ymin
Ans.
′′ ω 2 = ( 280.346 mm/rad ) ( 41.888 rad/s ) = 0.4919 m/s 2
ymax = ymax
Ans.
′′ ω 2 = ( −280.346 mm/rad ) ( 41.888 rad/s ) = −0.4919 m/s 2
ymin = ymin
Ans.
2
2
2
2
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For geometric continuity, we have
(4)
L1 + L2 = L3 + L4 + L5 or
L3 + L5 = 37.5 mm
β1 + β 2 + β 3 + β 4 + β 5 = 2π or
β 2 + β 3 + β 5 = 4.188 790 rad
(5)
Equations (1) to (5) are now solved simultaneously for β 2 , L3 , β 3 , L5 , and β 5 . The
results are summarized in the following table:
Seg. Type
Eq.
L, mm
β, rad
β, deg
AB dwell
--0
1.047 198
60.000
BC 8th order poly.
(5.20)
62.5
1.083 747
62.094
CD half harmonic
(5.26)
2.03
0.133 763
7.664
DE constant velocity
--25
1.047 198
60.000
EA half cycloidal
(5.31)
35.47
2.971 280
170.242
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74
5.10
A plate cam with an oscillating follower is to rise through 20° in 60D of cam rotation,
dwell for 45° , then rise through an additional 20° , return, and dwell for 60° of cam
rotation. Assuming high-speed operation, recommend standard cam motions from
Section 5.7 to be used, and determine the lifts and cam-rotation angles for each segment
of the cam.
From the sketches shown (not drawn to scale), the
curves types identified in the table below were
chosen.
Next, equating the second derivatives at D, the
remaining entries in the table were found.
L
L
−5.26830 32 = −5.26830 42
β3
β4
β
L
= 4 = 2.000
β
L3
β 4 = 2β3
(
)
β 3 + β 4 = 1 + 2 β 3 = 195°
β 3 = 80.772°
β 4 = 114.228°
Seg.
AB
BC
CD
DE
EA
5.11
Type
cycloidal
dwell
8th order poly.
8th order poly.
dwell
Eq.
(5.19)
--(5.20)
(5.23)
---
L, deg
20.000
0
20.000
40.000
0
β, rad
1.047 198
0.785 399
1.409 731
1.993 661
1.047 198
β, deg
60.000
45.000
80.772
114.228
60.000
Determine the maximum velocity and acceleration of the follower for Problem 5.10,
assuming that the cam is driven at a constant speed of 600 rev/min.
ω = ( 600 rev/min )( 2π rad/rev ) ( 60 s/min ) = 62.832 rad/s
Working from the equations listed, the maximum and minimum values of the derivatives
in each segment of the cam are as follows:
Seg.
Eq.
′
′
′′
′′
ymax
ymin
ymax
ymin
AB
(5.19)
0.666 667
0
2.000 000
BC
--0
0
0
CD
(5.20)
0.440 004
0
0.925 344
DE
(5.23)
0
-0.622 264
0.925 402
EA
--0
0
0
′ ω = ( 0.666 667 rad/rad )( 62.832 rad/s ) = 41.89 rad/s
y max = ymax
′′ ω 2 = ( 2.000 rad/rad )( 62.832 rad/s ) = 7 896 rad/s 2
ymax = ymax
2
-2.000 000
0
-0.924 344
-0.925 402
0
Ans.
Ans.
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2
4
2
3
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75
5.12
The boundary conditions for a polynomial cam motion are as follows: for θ = 0 , y = 0 ,
and y ′ = 0 ; for θ = β , y = L, and y′ = 0 . Determine the appropriate displacement
equation and the first three derivatives of this equation with respect to the cam rotation
angle. Sketch the corresponding diagrams.
Since there are four boundary conditions,
we choose a cubic polynomial
( β) ( β) ( β)
3C
C
2C
θ
θ
y′ =
β+
β ( β )+
β( β)
y = C0 + C1 θ
+ C2 θ
2
+ C3 θ
3
2
1
2
3
Then from the boundary conditions:
0
1
2
3
3
2
C0 = 0
C1 = 0
C2 = 3 L
C3 = −2 L
Therefore the equation and its three derivatives are:
2
3
2
3
⎡
⎤
= 3L θ
− 2L θ
= L ⎢3 θ
−2 θ
y θ
⎥
β
β
β
β
β
⎣
⎦
2
2
⎡
⎤
= 6L θ
− 6L θ
= 6L ⎢ θ
− θ
y′ θ
β
β β
β β
β⎣ β
β ⎥⎦
θ ⎤
6L
6L ⎡
12 L 2 θ
y′′ θ
β = β2 −
β = β 2 ⎢⎣1 − 2 β ⎥⎦
β
( ) ( ) ( ) ( ) ( )
( ) ( ) ( )
( )( )
( )
( )
( )
y′′′ ( θ ) = − 12L
β
β
3
5.13
Ans.
Ans.
Ans.
Ans.
Determine the minimum face width using 2.5 mm allowances at each end, and determine
the minimum radius of curvature for the cam of Problem 5.2.
Referring to Problem 5.2 for the data and figure,
β = 180° = π rad
R0 = 50 mm
L = 50 mm
From Eqs. (5.18) and (5.21) for simple harmonic motion,
′ = π L ( 2 β ) = 25 mm
′ = −π L ( 2 β ) = −25 mm
ymax
ymax
From Eq. (5.35)
′ + allowances
Face width=y′max − ymin
Face width= ( 25 mm ) − ( −25 mm ) + 2 ( 2.5 m ) = 55 mm
From Eq. (5.33)
Ans.
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( β = 0) = C = 0
C
y′ ( θ = 0 ) =
β
β =0
y ( θ = 1.0 ) = C + C = L
β
3C
2C
y′ ( θ = 1.0 ) =
β
β+
β =0
y θ
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76
ρ = R0 + y + y′′
L⎛
L
πθ ⎞ ⎛ L
πθ ⎞
⎜ 1 − cos
⎟ + ⎜ cos
⎟ = R0 + (constant)
β ⎠ ⎝2
β ⎠
2⎝
2
ρ = ( 50 mm ) + ( 50 mm ) 2 = 75 mm
= R0 +
Determine the maximum pressure angle and the minimum radius of curvature for the cam
of Problem 5.1.
Referring to Problem 5.1 for the data and figure,
β = 180° = π rad
R0 = 50 mm
Rr = 10 mm
L = 50 mm
For simple harmonic motion, Eq. (5.18) can be substituted into Eq. (5.43) to give
sin θ
. This can be differentiated and dφ dθ set to zero to find the angle
tan φ =
3 − cos θ
θ = 70.53° at which φmax = 19.47° . However, it is much simpler to use the nomogram of
Fig. 5.28 we find φmax = 20° directly. For the accuracy needed, the nomogram is
considered sufficient.
Ans.
From Fig. 5.30a, using R0 L = 1.0 , we get ( ρ min + Rr ) R0 = 1.43 .
This gives
ρ min = 1.43R0 − Rr = 1.43 ( 50 mm ) − (10 mm ) = 61.5 mm
5.15
Ans.
A radial reciprocating flat-face follower is to have the motion described in Problem 5.7.
Determine the minimum prime-circle radius if the radius of curvature of the cam is not to
be less than 12.5 mm. Using this prime-circle radius, what is the minimum length of the
follower face using allowances of 3.75 mm on each side?
From
Problem
P5.9,
′ = 102.48 mm/rad ,
ymax
′ = −23.87 mm/rad ,
ymin
′′ = −280.346 mm/rad 2
ymin
Therefore, from Eq. (5.34),
′′ − y = (12.5 mm ) − ( −280.346 mm ) − ( 62.5 mm ) = 230.34 mm
R0 > ρ min − ymin
Ans.
Also, from Eq. (5.35),
′ − ymin
′ + allowances = (102.48 mm ) − ( −23.87 mm ) + 2 ( 3.75 mm ) = 133.85 mm
Face width = ymax
Ans.
5.16
Graphically construct the cam profile of Problem 5.15 for clockwise cam rotation.
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5.14
Ans.
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77
5.17
A radial reciprocating roller follower is to have the motion described in Problem 5.7.
Using a prime-circle radius of 500 mm, determine the maximum pressure angle and the
maximum roller radius that can be used without producing undercutting.
We will use the nomogram of Fig. 5.28 to find the maximum pressure angle in each
segment of the cam. Calculations are shown in the following table. Asterisks are used to
signify values used with the nomogram to adjust half-return curves to equivalent full
return curves, and to adjust baseline.
Seg.
φmax , deg
R0* , mm
L* , mm
R0* L*
β * , deg
8.0
137.4
7.0
62.1
15.3
340.5
12
1
3
Ans.
Also we use Figs. 5.32 and 5.33 to check for undercutting. Again, asterisks are used to
denote values that are adjusted for use with the charts. Note that doubling as was done
for use of the nomogram is not necessary since we do have figures for half-harmonic and
half-cycloidal cam segments. Note also that segment EA need not be checked since
undercutting only occurs in segments with negative acceleration.
β , deg
L, mm
Seg.
R0* , mm
R0* L
Rrmax ,
( ρmin + Rr ) R0*
mm
BC
500
62.5
8.0
62.1
0.725
362.5
CD
560.4675
2.0325
275.8
7.7
0.680
380
To avoid undercutting for the entire cam, Rr < 362.5 mm .
5.18
Ans.
Graphically construct the cam profile of Problem 5.17 using a roller radius of 20 mm.
The cam rotation is to be clockwise.
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BC
500
62.5
CD
558.435
4.065
EA
500
70.935
For the total cam, φmax = 12 deg
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78
A plate cam rotates at 300 rev/min and drives a reciprocating radial roller follower
through a full rise of 75 mm in of cam rotation. Find the minimum radius of the primecircle if simple harmonic motion is used and the pressure angle is not to exceed 25D .
Find the maximum acceleration of the follower.
Using
φmax = 25°
β = 180° , Fig. 5.28 gives
and
R0 = 0.75 L = 0.75 ( 75 mm ) = 56 mm
ω=
( 300 rev/min )( 2π
′′ =
ymax
rad/rev )
60 s/min
π 2 ( 0.075 m )
R0 L = 0.75 .
Ans.
= 31.416 rad/s
π 2L
=
= 0.037 5 m/rad 2
2
2
2β
2 (π rad )
′′ ω 2 = ( 0.037 5 m/rad 2 ) ( 31.416 rad/s ) = 37.0 m/s 2
ymax = ymax
Ans.
Repeat Problem 5.19 except that the motion is cycloidal.
Fig. 5.28 gives R0 L = 0.95 . Therefore R0 = 0.95L = 0.95 ( 75 mm ) = 71 mm
Ans.
2
5.20
′′ =
ymax
2π L
β
2
=
2π ( 0.075 m )
(π
rad )
2
= 0.047 7 m/rad 2
′′ ω 2 = ( 0.047 7 m/rad 2 ) ( 31.416 rad/s ) = 47.1 m/s 2
ymax = ymax
Ans.
Repeat Problem 5.19 except that the motion is eighth-order polynomial.
Fig. 5.28 gives R0 L = 0.95 . Therefore R0 = 0.95L = 0.95 ( 75 mm ) = 71 mm
Ans.
2
5.21
′′ =
ymax
5.2683L
β
2
=
5.2683 ( 0.075 m )
(π
rad )
2
= 0.040 0 m/rad 2
′′ ω 2 = ( 0.040 0 m/rad 2 ) ( 31.416 rad/s ) = 39.5 m/s 2
ymax = ymax
2
5.22
Therefore
Ans.
Using a roller diameter of 20 mm, 180° determine whether the cam of Problem 5.19 will
be undercut.
Using R0 L = 0.75 and β = 180° , Fig. 5.30a gives ( ρ min + Rr ) R0 = 1.55 .
ρ min = 1.55 ( 56 mm ) − ( 20 mm ) = 66.8 mm > 0 ; this cam is not undercut.
Ans.
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5.19
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79
Equations (5.36) and (5.37) describe the profile of a plate cam with a reciprocating flatface follower. If such a cam is to be cut on a milling machine with cutter radius Rc ,
determine similar equations for the center of the cutter.
In complex polar notation, using
Eq. (5.32) and using u and v to
denote the local rectangular part
coordinates of the cam shape, the
loop closure equation is
ue jθ + jve jθ = jR0 + jy + y′ + jRc
Dividing this by e jθ
u + jv = j ( R0 + Rc + y ) e − jθ + y′e − jθ
Now separating this into real and
imaginary parts we find
u = ( R0 + Rc + y ) sin θ + y′ cos θ
v = ( R0 + Rc + y ) cos θ − y′ sin θ Ans.
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5.23
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81
Chapter 6
Spur Gears
6.1
Find the module of a pair of gears having 32 and 84 teeth, respectively, whose center
distance is 92 mm.
mN 2 mN 3 (32 + 84)m
+
=
= 92 mm
2
2
2
92 × 2
∴m =
= 1.586 mm/tooth
116
6.2
Find the number of teeth and the circular pitch of a 150 mm pitch diameter gear whose
module is 3 mm/tooth.
(2 R)
150 mm
=
= 50 teeth
m
3mm / tooth
p = π m = π 3 = 9.42 mm/tooth
N=
6.3
Ans.
Ans.
Ans.
Determine the module of a pair of gears having 18 and 40 teeth, respectively, whose
center distance is 58 mm.
mN 2 mN 3 m (18 + 40 )
+
=
= 58.0 mm
2
2
2
2 ( 58.0 mm )
m=
= 2.0 mm/tooth
58 teeth
R2 + R3 =
6.4
6.5
Ans.
Find the number of teeth and the circular pitch of a gear whose pitch diameter is 200 mm
if the module is 8 mm/tooth.
N = (2 R) m = ( 200.0 mm ) ( 8 mm/tooth ) = 25 teeth
Ans.
p = π m = π ( 8 mm/tooth ) = 25.13 mm/tooth
Ans.
Find the module and the pitch diameter of a 40-tooth gear whose circular pitch is
88 mm/tooth.
m = p /π =
88 mm/ tooth
π
= 28 mm/ tooth
D = 2 R = m N p = 28 mm/ tooth × 40 tooth = 1120 mm
Ans.
Ans.
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R2 + R3 =
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82
The pitch diameters of a pair of mating gears are 90 mm and 210 mm, respectively. If the
module is 1.5 mm/tooth how many teeth are there on each gear?
( 90 mm ) = 60 teeth
2 R2 D2
=
=
m
m 1.5 mm/ tooth
( 210 mm ) = 140 teeth
2R
D
N3 = 3 = 3 =
m
m (1.5 mm/tooth)
N2 =
6.7
6.8
6.9
Ans.
Find the module and the pitch diameter of a gear whose circular pitch is 40 mm/tooth if
the gear has 36 teeth.
m = p π = ( 40 mm/tooth ) π = 12.732 mm/tooth
Ans.
D = 2 R = mN = (12.732 mm/tooth )( 36 teeth ) = 458.4 mm
Ans.
The pitch diameters of a pair of gears are 60 mm and 100 mm, respectively. If their
module is 2.5 mm/tooth, how many teeth are there on each gear?
N 2 = 2 R2 m = D2 m = ( 60 mm ) ( 2.5 mm/tooth ) = 24 teeth
Ans.
N 3 = 2 R3 m = D3 m = (100 mm ) ( 2.5 mm/tooth ) = 40 teeth
Ans.
What is the diameter of a 33-tooth gear if its circular pitch is 22 mm/tooth?
D = 2 R = pN π = ( 22 mm/tooth )( 33 teeth ) π = 231.21 mm
6.10
Ans.
Ans.
A shaft carries a 30-tooth, 8 mm/tooth module gear that drives another gear at a speed of
480 rev/min. How fast does the 30-tooth gear rotate if the shaft center distance
is 225 mm?
R2 = mN 2 2 = (8)(30) 2 = 120 mm
R3 = ( R2 + R3 ) − R2 = 225 mm − 120 mm = 105 mm
ω2 =
6.11
R3
105 mm
ω3 =
( 480 rev/min ) = 420 rev/min
R2
120 mm
Ans.
Two gears having an angular velocity ratio of 3:1 are mounted on shafts whose centers
are 136 mm apart. If the module of the gears is 4 mm/tooth, how many teeth are there on
each gear?
⎛ ω ⎞
3
( R2 + R3 ) = (136 mm ) = ⎜1 + 2 ⎟ R2 = ⎜⎛1 + ⎟⎞ R2 = 4 R2
⎝ 1⎠
⎝ ω3 ⎠
R2 = 34.0 mm
R3 = ( R3 + R2 ) − R2 = 102.0 mm
2 R2 2 ( 34.0 mm )
=
= 17 teeth
m
4 mm/tooth
2 R 2 (102.0 mm )
N3 = 3 =
= 51 teeth
4 mm/tooth
m
N2 =
Ans.
Ans.
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6.6
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83
6.12
A gear having a module of 4 mm/tooth and 21 teeth drives another gear at a speed of 240
rev/min. How fast is the 21-tooth gear rotating if the shaft center distance is 156 mm?
R2 = mN 2 2 = ( 4 mm/tooth )( 21 teeth ) 2 = 42 mm
R3 = ( R2 + R3 ) − R2 = (156 mm ) − 42 mm = 114 mm
ω2 =
Ans.
A 6 mm/tooth module, 24-tooth pinion is to drive a 36-tooth gear. The gears are cut on
the 20° full-depth involute system. Find and tabulate the addendum, dedendum,
clearance, circular pitch, base pitch, tooth thickness, base circle radii, length of paths of
approach and recess, and contact ratio.
a = 1m = 1× 6 mm/ tooth = 6 mm
d = 1.25 m = 1.25 × 6 mm / tooth = 7.5 mm
c = d − a = ( 7.5 mm ) − ( 6 mm ) = 1.5 mm
Ans.
Ans.
Ans.
p = π m = π (6 mm/ tooth) = 18.84 mm/tooth
pb = p cos φ = (18.84 mm/tooth ) cos 20° = 17.7 mm/tooth
Ans.
Ans.
t = p 2 = (18.84 mm/tooth ) 2 = 9.42 mm
Ans.
mN 2 ( 6 mm/ tooth )( 24 teeth )
=
= 72 mm
2p
2
mN 3 ( 6 mm/ tooth )( 36 teeth )
R3 =
=
= 108 mm
2
2
r2 = R2 cos φ = ( 72 mm ) cos 20° = 67.657 mm ;
R2 =
r3 = R3 cos φ = (108 mm ) cos 20° = 101.48 mm
CP = 14.98 mm [measured or by Eq. (6.10)]
PD = 14.17 mm [measured or by Eq. (6.11)]
CD (14.98 mm ) + (14.17 mm )
mc =
=
= 1.648 teeth avg.
pb
17.7 mm/tooth
Ans.
Ans.
Ans.
Ans.
Ans.
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6.13
R3
114 mm
ω3 =
( 240 rev/min ) = 651.4 rev/min
R2
42 mm
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84
6.14
A 5 mm/tooth module, 15-tooth pinion is to mate with a 30-tooth internal gear. The
gears are 20° full-depth involute. Make a drawing of the gears showing several teeth on
each gear. Can these gears be assembled in a radial direction? If not, what remedy
should be used?
With this condition the involute
curves of the internal gear are
extended radially to meet the
addendum circle and this results
in converging radii; therefore the
gears can not be assembled in the
radial direction.
Ans.
One remedy is to reduce the internal gear addendum to match the base circle radius.
However, the internal gear is then non-standard. A better remedy is to increase the
module to 6 mm/tooth so that the standard addendum circle of the internal gear is 69 mm.
6.15
A 10 mm/tooth module 17-tooth pinion and a 50-tooth gear are paired. The gears are cut
on the 20° full-depth involute system. Find the angles of approach and recess of each
gear and the contact ratio.
a = 1m = 1× 10 mm/ tooth = 10 mm
p = π m = π × 10 mm/tooth = 31.4 mm/tooth
pb = p cos φ = ( 31.4 mm/tooth ) cos 20° = 29.506 mm/tooth
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Since the addendum circle of
internal gear 3 is of lesser radius
(70
mm)
than
its
base
circle(70.475 mm), contact is
initiated to the left of point A
before proper involute contact is
possible.
This is similar to
undercutting but on the internal
gear it is called fouling.
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85
mN 3 10 mm/ tooth× 50 teeth
mN 2 10 mm/tooth ×17 teeth
=
= 250 mm
=
= 85 mm R3 =
2
2
2
2
r2 = R2 cos φ = ( 85 mm ) cos 20° = 79.87 mm
R2 =
r3 = R3 cos φ = ( 250 mm ) cos 20° = 234.92 mm
CP = 25.9 mm [Eq. (6.10)]
PD = 22.35 mm [Eq. (6.11)]
CP 25.9 mm
CP
25.9 mm
α2 =
=
= 0.324 rad = 18.58° α 3 =
=
= 0.110 rad = 6.32° Ans.
r2
79.87 mm
r3
234.92 mm
PD 22.35 mm
PD 22.35 mm
β2 =
=
= 0.280 rad = 16.04° β 3 =
=
= 0.095 rad = 5.45° Ans.
r2
79.87 in
r3
234.92 mm
6.16
CD ( 25.9 mm ) + ( 22.35 mm )
=
= 1.63 teeth avg.
pb
29.506 mm/tooth
Ans.
A gearset with a module of 5 mm/tooth has involute teeth with 22½˚ pressure angle, and
has 19 and 31 teeth, respectively. (Such gears came from an older standard and are now
obsolete.) They have 1.0m for the addendum and 1.35m for the dedendum. (In SI, tooth
sizes are given in modules, m, and a = 1.0m means 1 module, not 1 meter.) Tabulate the
addendum, dedendum, clearance, circular pitch, base pitch, tooth thickness, base circle
radius, and contact ratio.
a = 1.0m = 5.0 mm
d = 1.35m = 1.35 ( 5 mm/tooth ) = 6.75 mm
Ans.
Ans.
c = d − a = 1.75 mm
p = π m = π ( 5 mm/tooth ) = 15.708 mm/tooth
Ans.
Ans.
pb = p cos φ = (15.708 mm/tooth ) cos 22.5° = 14.512 mm/tooth
Ans.
t = p 2 = 7.854 mm
R2 = N 2 m 2 = (19 teeth )( 5 mm/tooth ) 2 = 47.500 mm
Ans.
R3 = N 3 m 2 = ( 31 teeth )( 5 mm/tooth ) 2 = 77.500 mm
r2 = R2 cos φ = ( 47.500 mm ) cos 22.5° = 43.884 mm
Ans.
r3 = R3 cos φ = ( 77.500 mm ) cos 22.5° = 71.601 mm
Ans.
CP = 11.325 mm [Eq. (6.10)]
PD = 10.640 mm [Eq. (6.11)]
+
11.325
mm
10.640
mm
) (
) = 1.51 teeth avg.
CD (
=
mc =
pb
14.512 mm/tooth
Ans.
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mc =
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86
6.17
A gear with a module of 8 mm/tooth and 22 teeth is in mesh with a rack; the pressure
angle is 25˚. The addendum and dedendum are 1.0m and 1.25m, respectively. (In SI,
tooth sizes are given in modules, m, and a = 1.0m means 1 module, not 1 meter.) Find
the lengths of the paths of approach and recess and determine the contact ratio.
a = 1.0m = 8.0 mm
p = π m = π ( 8 mm/tooth ) = 25.133 mm/tooth
pb = p cos φ = ( 25.133 mm/tooth ) cos 25° = 22.778 mm/tooth
R2 = N 2 m 2 = ( 22 teeth )( 8 mm/tooth ) 2 = 88.0 mm
6.18
Ans.
Ans.
Ans.
Repeat Problem 6.15 using the 25˚ full depth system.
a = 1m = 1×10 mm/ tooth = 10 mm
p = π m = π ×10 mm/ tooth = 31.4 mm/tooth
pb = p cos φ = ( 31.4 mm/tooth ) cos 25° = 28.458 mm/tooth
N m 50 teeth × 10 mm/tooth
N 2 m 17 teeth × 10 mm/tooth
=
= 85 mm R3 = 3 =
= 250 mm
2
2
2
2
r2 = R2 cos φ = ( 85 mm ) cos 25° = 77.04 mm
R2 =
r3 = R3 cos φ = ( 250 mm ) cos 25° = 226.57 mm
CP = 21.875 mm [Eq. (6.10)]
PD = 19.675 mm [Eq. (6.11)]
CP 21.875 mm
CP 21.875 mm
α2 =
=
= 0.284 rad = 16.27° α 3 =
=
= 0.097 rad = 5.53° Ans.
r2
77.04 mm
r3
226.57 mm
PD 0.787 mm
PD 19.675 mm
β2 =
=
= 0.255 rad = 14.63° β 3 =
=
= 0.087 rad = 4.97° Ans.
r2
77.04 mm
r3
226.57 mm
mc =
CD ( 21.875 mm ) + (19.675 mm )
=
= 1.46 teeth avg.
pb
28.458 mm/tooth
Ans.
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CP = a sin φ = 18.930 mm [Fig. 6.10]
PD = 16.243 mm [Eq. (6.11)]
CD (18.930 mm ) + (16.243 mm )
mc =
=
= 1.54 teeth avg.
pb
22.778 mm/tooth
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87
Draw a 12 mm/tooth module, 26-tooth, 20˚ full-depth involute gear in mesh with a rack.
(a)
Find the lengths of the paths of approach and recess and the contact ratio.
(b)
Draw a second rack in mesh with the same gear but offset 3 mm away from the
gear center. Determine the new contact ratio. Has the pressure angle changed?
(a)
(b)
6.20
CP = a sin φ = 12 mm sin 20° = 35.085 mm [see figure]
PD = 28.714 mm [Eq. (6.11)]
CD ( 35.085 mm ) + ( 28.714 mm )
mc =
=
= 1.80 teeth avg.
35.407 mm/tooth
pb
Ans.
Ans.
Ans.
Since the pressure angle is a property which has determined the shapes of the
teeth on both the rack and the pinion, moving the rack by 3 mm does not change
the tooth shapes or the pressure angle. The modified contact ratio is:
C ′P′ = a′ sin φ ′ = 9 mm sin 20° = 26.314 mm [see figure]
Ans.
P′D′ = 28.714 mm [Eq. (6.11)]
Ans.
C ′D′ ( 26.314 mm ) + ( 28.714 mm )
=
= 1.55 teeth avg.
Ans.
mc′ =
35.407 mm/tooth
pb′
through 6.24 Shaper gear cutters have the advantage that they can be used for either
external or internal gears and also that only a small amount of runout is necessary at the
end of the stroke. The generating action of a pinion shaper cutter can easily be simulated
by employing a sheet of clear plastic. The figure illustrates one tooth of a 16-tooth
pinion cutter with 20° pressure angle as it can be cut from a plastic sheet. To construct
the cutter, lay out the tooth on a sheet of drawing paper. Be sure to include the clearance
at the top of the tooth. Draw radial lines through the pitch circle spaced at distances
equal to one fourth of the tooth thickness as shown in the figure. Now fasten the sheet of
plastic to the drawing and scribe the cutout, the pitch circle, and the radial lines onto the
sheet. The sheet is then removed and the tooth outline trimmed with a razor blade. A
small piece of fine sandpaper should then be used to remove any burrs.
www.elsolucionario.net
6.19
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88
Problem No.
6.20
6.21
6.22
6.23
6.24
No. of Teeth
10
12
14
20
36
The diagram used to
make the plastic template
for
Problems
6.20
through 6.24 is shown at
the left.
The drawing generated
for Problem 6.20 is also
shown at left. Note how
the tip of the shaper
cutter do slightly cut
away the material at the
flank of the tooth so that
the tooth is a small
amount narrower here
than at its thickest radius.
This is the meaning of
the term undercut.
Problems 6.21 to 6.24
are similar.
6.25
A 10-mm/tooth module gear has 17 teeth, a 20° pressure angle, an addendum of 1.0m,
and a dedendum of 1.25m. (In SI, tooth sizes are given in modules, m, and a = 1.0m
means 1 module, not 1 meter.) Find the thickness of the teeth at the base circle and at the
addendum circle. What is the pressure angle corresponding to the addendum circle?
At the pitch circle:
rp = R = mN 2 = (10 mm/tooth )(17 teeth ) 2 = 85.0 mm
t p = π R N = π ( 85.0 mm ) 17 = 15.708 mm
inv φ = inv 20° = 0.014 904
At the base circle:
r = rb = R cos φ = ( 85.0 mm ) cos 20° = 79.874 mm
inv ϕ = inv 0° = 0.0
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To generate a gear with the cutter, only the pitch circle and the addendum circle need be
drawn. Divide the pitch circle into spaces equal to those used on the template and
construct radial lines through them. The tooth outlines are then obtained by rolling the
template pitch circle upon that of the gear and drawing the cutter tooth lightly for each
position. The resulting generated tooth upon the gear will be evident. The following
problems all employ a standard 25 mm/tooth module 20° full depth template constructed
as described above. In each case you should generate a few teeth and estimate the
amount of undercutting
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89
From Eq. (6.16)
⎡ tp
⎤
+ inv φ − inv ϕ ⎥
t = 2r ⎢
⎣ 2R
⎦
6.26
Ans.
ϕ a = cos −1 ( rb ra ) = cos −1 ( 79.874 mm 95.0 mm ) = 32.78°
Ans.
⎡ 15.708 mm
⎤
ta = 2 ( 95.0 mm ) ⎢
+ 0.014 904 − 0.071 844 ⎥ = 6.737 mm
⎣ 2 ( 85.0 mm )
⎦
Ans.
A 15-tooth pinion has 20 mm/tooth module 20° full-depth teeth. Calculate the thickness
of the teeth at the base circle. What are the tooth thickness and the pressure angle at the
addendum circle?
At the pitch circle:
rp = R = m N 2 = ( 20 mm/ tooth ) × (15 teeth ) 2 = 150 mm
tp =
πR
=
π (150 )
= 31.4 mm
15
N
inv φ = inv 20° = 0.014 904
At the base circle:
r = rb = R cos φ = (150 mm ) cos 20° = 140.95 mm
inv ϕ = inv 0° = 0.0
From Eq. (6.16)
⎡ tp
⎤
+ inv φ − inv ϕ ⎥
t = 2r ⎢
⎣ 2R
⎦
⎡ 31.4 mm
⎤
t = 2 (140.95 mm ) ⎢
+ 0.014 904 − 0.0 ⎥ = 33.706 mm
⎣ 2 (150 mm )
⎦
At the addendum circle:
ra = R + a = 150 mm + 20 mm = 170 mm
Ans.
ϕ = cos −1 ( rb ra ) = cos −1 (140.95 mm 170 mm ) = 33.99°
Ans.
⎡ 31.4 mm
⎤
t = 2 (170 mm ) ⎢
+ 0.014 904 − 0.081 018⎥ = 13.11 mm
⎣ 2 (150 mm )
⎦
Ans.
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⎡ 15.708 mm
⎤
t = 2 ( 79.874 mm ) ⎢
+ 0.014 904 − 0.0 ⎥ = 17.142 mm
⎣ 2 ( 85.0 mm )
⎦
At the addendum circle:
ra = R + a = 85.0 mm + 10.0 mm = 95.0 mm
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90
6.27
A tooth is 20 mm thick at a pitch circle radius of 200 mm and a pressure angle of 25°.
What is the thickness at the base circle?
At the base circle:
r = rb = R cos φ = ( 200 mm ) cos 25° = 181.26 mm
6.28
Ans.
A tooth is 40 mm thick at the pitch radius of 400 mm and a pressure angle of 20˚. At
what radius does the tooth become pointed?
⎡ tp
⎤
+ inv φ − inv ϕ ⎥ = 0
t = 2r ⎢
⎣ 2R
⎦
inv ϕ = t p 2 R + inv φ = 40 mm/ 2 ( 400 mm ) + 0.014 904 = 0.064904
r = rb cos ϕ = ( 400 mm ) cos 20° cos 31.784° = 442.19 mm
6.29
Ans.
A 25˚ involute, 2.5 mm/tooth module pinion has 18 teeth. Calculate the tooth thickness
at the base circle. What are the tooth thickness and pressure angle at the addendum
circle?
At the pitch circle:
rp = R = mN 2 = ( 2.5 mm/ tooth ) × (18 teeth ) 2 = 22.5 mm
t p = π R N = π ( 22.5 mm ) (18 teeth ) = 3.925 mm
inv φ = inv 25° = 0.029 975
At the base circle:
r = rb = R cos φ = ( 22.5 mm ) cos 25° = 20.39 mm
inv ϕ = inv 0° = 0.0
⎡ tp
⎤
+ inv φ − inv ϕ ⎥
t = 2r ⎢
⎣ 2R
⎦
⎡ 3.925 mm
⎤
t = 2 ( 20.39 mm ) ⎢
+ 0.029 975 − 0.0 ⎥ = 4.779 mm
⎣ 2 ( 22.5 mm )
⎦
Ans.
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inv ϕ = inv 0° = 0.0
From Eq. (6.16)
⎡ tp
⎤
+ inv φ − inv ϕ ⎥
t = 2r ⎢
⎣ 2R
⎦
⎡ 20 mm
⎤
t = 2 (181.26 mm ) ⎢
+ 0.029 975 − 0.0 ⎥ = 28.99 mm
⎣ 2 ( 200 mm )
⎦
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91
At the addendum circle:
ra = R + a = 22.5 mm + 2.5 mm = 25 mm
6.30
ϕ = cos −1 ( rb ra ) = cos −1 ( 20.39 mm 25 mm ) = 35.35°
Ans.
⎡ 3.925 mm
⎤
t = 2 ( 25 mm ) ⎢
+ 0.029 975 − 0.092 339 ⎥ = 1.243 mm
⎣ 2 ( 22.5 mm )
⎦
Ans.
A nonstandard 10-tooth 3 mm/tooth module pinion is to be cut with a 22½˚ pressure
angle. (Such gears came from an older standard and are now obsolete.) What maximum
addendum can be used before the teeth become pointed?
At the pitch circle:
rp = R = mN 2 = ( 3mm/ tooth ) × (10 teeth ) 2 = 15 mm
At the addendum circle:
⎡ tp
⎤
+ inv φ − inv ϕ ⎥ = 0
t = 2r ⎢
⎣ 2R
⎦
inv ϕ = t p 2 R + inv φ = 4.71 mm 2 (15 mm ) + 0.021 514 = 0.178 514 ; ϕ = 42.766°
r = rb cos ϕ = (15 mm ) cos 22.5° cos 42.766° = 18.877 mm
a = r − R = 18.877 mm − 15 mm = 3.877 mm
6.31
Ans.
The accuracy of cutting gear teeth can be measured by fitting hardened and ground pins
in diametrically opposite tooth spaces and measuring the distance over the pins. For a
2.5 mm/tooth module 20˚ full-depth involute system 96 tooth gear:
(a)
Calculate the pin diameter that will contact the teeth at the pitch lines if there is to
be no backlash.
(b)
What should be the distance measured over the pins if the gears are cut
accurately?
(a)
R = mN 2 = ( 2.5 mm/tooth ) × ( 96 teeth ) 2 = 120 mm
rb = R cos φ = (120 mm ) cos 20° = 112.76 mm
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t p = π R N = π (15 mm ) (10 teeth ) = 4.71 mm
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92
ρ = rb tan φ = (112.76 ) tan 20° = 41.04 mm
α = π 2 N = π 2 ( 96 teeth ) = 0.016362 rad = 0.937 5°
s = rb tan (φ + α ) − ρ = (112.76 mm ) tan ( 20.937 5° ) − ( 41.04 mm ) = 2.103 mm
d = 2 s = 2 ( 2.103 mm ) = 4.206 mm
(b)
Ans.
rs = rb cos (φ + α ) = (112.76 mm ) cos20.9375°=120.73 mm
distance over pins = 2 ( rs + s ) = 2 (120.73 mm + 2.103 mm ) = 245.666 mm
A set of interchangeable gears with a 6 mm/tooth module is cut on the 20˚ full-depth
involute system. The gears have tooth numbers of 24, 32, 48, and 96. For each gear,
calculate the radius of curvature of the tooth profile at the pitch circle and at the
addendum circle.
N , teeth
R = mN 2, mm
rb = R cos φ , mm
ρ p = rb tan φ , mm
24
32
48
96
72
96
144
288
67.66
90.21
135.31
270.63
24.62
32.83
49.25
98.50
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6.32
Ans.
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93
6.33
N , teeth
ra , mm
rb ra
ϕ a = cos −1 ( rb ra ) , deg
ρ a = rb tan ϕa , mm
24
32
48
96
78
102
150
294
0.86743
0.88441
0.90206
0.92051
29.84
27.82
25.56
23.00
38.812
47.602
64.713
114.87
Calculate the contact ratio of a 17-tooth pinion that drives a 73-tooth gear. The gears are
4 mm/tooth module and cut on the 20˚ fine pitch system.
a = m = 4 mm/tooth = 4 mm
pb = π m cos φ = (π × 4 mm/tooth ) cos 20° = 11.8 mm/tooth
mN 3 4 mm/tooth× 73 teeth
mN 2 4 mm/ tooth ×17 teeth
=
= 34 mm R3 =
=
= 146 mm
2
2
2
2
CP = 10.71 mm [measured or by Eq. (6.10)]
PD = 8.942 mm [measured or by Eq. (6.11)]
CD (10.71 mm ) + ( 8.942 mm )
=
= 1.665 teeth avg.
Ans.
mc =
11.8 mm/tooth
pb
6.34
A 25˚ pressure angle 11-tooth pinion is to drive a 23-tooth gear. The gears have a
module of 3 mm/tooth and have stub teeth. What is the contact ratio?
a = 0.8 m = 0.8 × 3mm/ tooth = 2.4 mm (Notice the stub teeth).
pb = π m cos φ = (π × 3mm/ tooth ) cos 25° = 8.54 mm/tooth
mN 3 3mm/ tooth× 23 teeth
mN 2 3mm/ tooth×11 teeth
=
= 16.5 mm R3 =
=
= 34.5 mm
2
2
2
2
CP = 5.0142 mm [measured or by Eq. (6.10)]
PD = 4.5847 mm [measured or by Eq. (6.11)]
CD ( 5.0142 mm ) + ( 4.5847 )
=
= 1.124 teeth avg.
Ans.
mc =
8.54 mm/tooth
pb
R2 =
6.35
A 22-tooth pinion mates with a 42-tooth gear. The gears are full depth, have a module of
1.5 mm/tooth, and are cut with a 17½˚ pressure angle. (Such gears came from an older
standard and are now obsolete.) Find the contact ratio.
a = m = 1.5 mm/tooth = 1.5 mm
pb = (π m ) cos φ = (π × 1.5 mm/tooth ) cos17.5° = 4.49 mm/tooth
mN 3 1.5 mm/tooth× 42 teeth
mN 2 1.5 mm/tooth× 22 teeth
=
= 31.5 mm
=
= 16.5 mm R3 =
2
2
2
2
CP = 4.1833 mm [measured or by Eq. (6.10)]
PD = 3.7772 mm [measured or by Eq. (6.11)]
CD ( 4.1833 mm ) + ( 3.7772 mm )
=
= 1.772 teeth avg.
Ans.
mc =
4.49 mm/tooth
pb
R2 =
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R2 =
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94
6.36
The center distance of two 24-tooth, 20° pressure angle, full-depth involute spur gears
with module of 12 mm/tooth is increased by 3 mm in over the standard distance. At what
pressure angle do the gears operate?
The original two gears are
identical.
mN
2
12 mm/tooth× 24 teeth
=
2
= 144 mm
When the gear centers are
separated to a non-standard
distance, the base circles do
not change. The line of
contact adjusts to remain
tangent to the new locations
of the two base circles. The
pressure angle changes.
Since the two base circles do
not change,
r2 = r3 = (144 mm ) cos 20°
= 135.315 mm
From the figure we can see that the shaft center distance is related to the new pressure
angle as follows
r
r +r
r2
+ 3 = 2 3
cos φ ′ cos φ ′ cos φ ′
r +r
135.315 mm + 135.315 mm
φ ′ = cos −1 2 3 = cos−1
= 21.56°
R2′ + R3′
291 mm
R2′ + R3′ =
6.37
Ans.
The center distance of two 18-tooth, 25° pressure angle, full-depth involute spur gears
with module of 8 mm/tooth is increased by 1.5 mm in over the standard distance. At
what pressure angle do the gears operate?
Consult the figure and the discussion with the solution of Prob. 6.36.
mN 8 mm/ tooth× 18 teeth
R2 = R3 =
=
= 72 mm
2
2
r2 = r3 = ( 72 mm ) cos 25° = 65.254 mm
φ ′ = cos −1
r2 + r3
65.254 mm + 65.254 mm
= cos −1
= 26.24°
R2′ + R3′
145.5 mm
Ans.
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R2 = R3 =
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95
6.38
A pair of mating gears have 1.25 mm/tooth module and are generated on the 20˚ fulldepth involute system. If the tooth numbers are 15 and 50, what maximum addendums
may they have if interference is not to occur?
mN 3 1.25 mm/tooth× 50 teeth
mN 2 1.25 mm/tooth× 15 teeth
R2 =
=
= 9.375 mm R3 =
=
= 31.25 mm
2
2
2
2
r2 = R2 cos φ = ( 9.375 mm ) cos 20° = 8.8096 mm
r3 = R3 cos φ = ( 31.25 mm ) cos 20° = 29.365 mm
a2 ≤ r22 + ( R2 + R3 ) sin 2 φ − R2 = 7.0769 mm
Ans.
a3 ≤ r32 + ( R2 + R3 ) sin 2 φ − R3 = 1.2363 mm
Ans.
2
2
6.39
A set of gears is cut with a 112.5 mm/tooth circular pitch and a 17½˚ pressure angle.
(Such gears came from an older standard and are now obsolete.) The pinion has 20 fulldepth teeth. If the gear has 240 teeth, what maximum addendum may it have in order to
avoid interference?
P = π p = π (112.5 mm/tooth ) = 0.0279 teeth/mm
R2 =
N
20 teeth
N2
240 teeth
=
= 358.423 mm R3 = 3 =
= 4301.075 mm
2 P 2 ( 0.0279 teeth/mm )
2 P 2 ( 0.0279 teeth/mm )
r3 = R3 cos φ = ( 4301.075 mm ) cos17.5° = 4102 mm
a3 ≤ r32 + ( R2 + R3 ) sin 2 φ − R3 = 33.62 mm
2
6.40
Ans.
Using the method described for Problems P6.20 through P6.24, cut a 25 mm/tooth
module 20˚ pressure angle full-depth involute rack tooth from a sheet of clear plastic.
Use a nonstandard clearance of 0.35 m in order to obtain a stronger fillet. This template
can be used to simulate the generating action of a hob. Now, using the variable-centerdistance system, generate an 11-tooth pinion to mesh with a 25-tooth gear without
interference. Record the values found for center distance, pitch radii, pressure angle,
gear blank diameters, cutter offset, and contact ratio. Note that more than one
satisfactory solution exists.
One solution may be found by the procedure shown in the numeric example in Section
6.11 under the title of Center Distance Modifications. It proceeds as follows:
φ = 20° , m = 25 mm/tooth , p = π m = 78.5 mm/tooth ,
a = m = 25 mm , d = 1.35m = 33.75 mm , c = 0.35m = 8.75 mm ,
N 2 = 11 teeth ,
N 3 = 25 teeth ,
mN 3 25 mm/tooth× 25 teeth
mN 2 25 mm/tooth×11 teeth
=
= 312.5 mm
R2 =
=
= 137.5 mm R3 =
2
2
2
2
r2 = R2 cos φ = 129.2 mm
r3 = R3 cos φ = 293.65 mm
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From Eq. 6.12, using Eqs. 6.10 and 6.11,
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96
t2 = 2e tan φ + p 2 = 45.76 mm
t3 = p 2 = 39.25 mm
N 2 ( t2 + t3 ) − 2π R2
φ ′ = 22.70°
+ inv φ = 0.022 115 rad ,
2 R2 ( N 2 + N 3 )
R cos φ
R cos φ
R2′ = 2
R3′ = 3
= 140 mm
= 318.3 mm
cos φ ′
cos φ ′
R2′ + R3′ = 458.3 mm
Working depth = 49.45 mm
R2′ + a2′ = 170.87 mm
R3′ + a3′ = 336.92 mm
CP′ = 42.40 mm
PD′ = 57.76 mm
mc′ = CD′ pb = 1.36 teeth avg.
inv φ ′ =
6.41
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Using the template cut in Problem 6.40 generate an 11-tooth pinion to mesh with a 44tooth gear with the long-and-short-addendum system. Determine and record suitable
values for gear and pinion addendum and dedendum and for the cutter offset and contact
ratio. Compare the contact ratio with that of standard gears.
φ = 20° , m = 25 mm/ tooth , p = π m = 78.5 mm/tooth ,
a = m = 25 mm , d = 1.35m = 33.75 mm , c = 0.35m = 8.75 mm ,
N 2 = 11 teeth ,
N 3 = 44 teeth ,
mN 3 25 mm/ tooth× 44 teeth
mN 2 25 mm/tooth×11 teeth
R2 =
=
= 137.5 mm R3 =
=
= 550 mm
2
2
2
2
r3 = R3 cos φ = 516.83 mm
r2 = R2 cos φ = 129.2 mm
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e = a − R2 sin 2 φ = 0.356 6 in
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97
a3′ ≤ r32 + ( R2 + R3 ) sin 2 φ − R3 = 17.8 mm
2
Ans.
Since the working depth of standard gears is retained, the new dedendum of the gear is
d3′ = m + 1.35m − a3′ = 40.95 mm
Ans.
Then, retaining the same clearance and pitch point P,
d 2′ = m + 1.35m − a2′ = 26.55 mm
a2′ = d3′ − 0.35m = 32.2 mm ,
Ans.
Assuming a standard rack cutter with a = m = 25 mm , Fig. 6.26 shows the offset is
e2 = a − d 2′ = −1.55 mm ,
e3 = a − d3′ = −15.95 mm
Ans.
From Eqs. (6.9), (6.10), and (6.11) the contact ratio is
CP′ = 47.03 mm
PD′ = 62.975 mm
′
′
mc = CD pb = 1.49 teeth avg.
Ans.
It is not easy to find a “contact ratio” for standard gears since these would have
undercutting over the range CC’. The distance C’D’ is slightly less than the distance CD,
but has eliminated the interference.
6.42
A pair of spur gears with 9 and 36 teeth are to be cut with a 20° full-depth cutter with
module of 8 mm/tooth.
(a)
Determine the amount that the addendum of the gear must be decreased in order
to avoid interference.
(b)
If the addendum of the pinion is increased by the same amount, determine the
contact ratio.
φ = 20° , m = 8 mm/tooth , p = π m = 25.12 mm/tooth ,
N 2 = 9 teeth ,
N 3 = 36 teeth ,
mN 3 8 mm/tooth× 36 teeth
mN 2 8 mm/tooth× 9 teeth
=
= 144 mm
R2 =
=
= 36mm R3 =
2
2
2
2
r2 = R2 cos φ = 33.83 mm
r3 = R3 cos φ = 135.315 mm
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Since,
with
standard
gears,
point C is to the
left of point A,
there
is
interference
and
undercutting. This
problem can be
eliminated
using
the long and short
addendum system
as shown in the
figure at the left. Since the interference is near point C, we reduce the addendum of the
gear until point C’ is coincident with point A. Combining Eqs. (6.10) and (6.12) we can
show that
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98
(a)
From Eqs. (6.10) and (6.12)
a3′ ≤ r32 + ( R2 + R3 ) sin 2 φ − R3 = 4.8575 mm
(b)
6-43
Δ = m − a3′ = 3.1425 mm
Ans.
a2′ = m + Δ = 11.1425 mm
a3′ = m − Δ = 4.8575 mm
From Eqs. (6.9), (6.10), and (6.11) the contact ratio is
CP′ = 12.825 mm
PD′ = 21.667 mm
mc′ = CD′ p cos φ = 1.46 teeth avg.
Ans.
A standard 20˚ pressure angle full-depth 25 mm/tooth module 20-tooth pinion drives a
48-tooth gear. The speed of the pinion is 500 rev/min. Using the position of the point of
contact along the line of action as the abscissa, plot a curve showing the sliding velocity
at all points of contact. Notice that the sliding velocity changes sign when the point of
contact passes through the pitch point.
φ = 20° , m = 25 mm / tooth , a = m = 25 mm , ω 2 = 500 rev/min = 52.360 rad/s ,
N 2 = 20 teeth ,
N 3 = 48 teeth ,
R2 = m N 2 2 = 250 mm
R3 = mN 3 2 = 600 mm
r2 = R2 cos φ = 234.92 mm
r3 = R3 cos φ = 563.82 mm
Defining X to be the distance from the point of contact to the pitch point along the line of
action, then, since this is the distance to the instant center, the sliding velocity at the point
of contact is
VX 3 / 2 = X (ω3 − ω2 ) = X ( R2 R3 + 1) ω2 = 1854.4 X mm/s
and, using Eqs. (6.9) and (6.10), X varies between X init = CP = 64.475 mm and
X final = PD = 57.45 mm .
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2
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99
Helical Gears
A pair of parallel-axis helical gears has 14½° normal pressure angle, module of
4 mm/tooth, and 45° helix angle. The pinion has 15 teeth, and the gear has 24 teeth.
Calculate the transverse and normal circular pitch, the normal diametral pitch, the pitch
diameters, and the equivalent tooth numbers.
N 2 = 15 teeth ,
mt = 4 mm/ tooth ,
pt = π mt = 12.56 mm/tooth ,
N 3 = 24 teeth ,
mn = mt cosψ = 2.828 mm/tooth ,
pn = pt cosψ = 8.8812 mm/tooth, ,
R3 = mt N 3 ( 2 ) = 48 mm,
Ans.
N e 2 = N 2 cos ψ = 42.43 teeth ,
N e3 = N 3 cos ψ = 67.88 teeth
Ans.
R2 = mt N 2 ( 2 ) = 30 mm ,
3
6-45
3
A set of parallel-axis helical gears are cut with a 20° normal pressure angle and a 30°
helix angle. They have module of 1.5 mm/tooth and have 16 and 40 teeth, respectively.
Find the transverse pressure angle, the normal circular pitch, the axial pitch, and the pitch
radii of the equivalent spur gears.
N 2 = 16 teeth ,
N 3 = 40 teeth ,
φt = tan −1 [ tan φn cosψ ] = 22.796° ,
Ans.
R2 = mt N 2 ( 2 ) = 12 mm,
R3 = mt N 3 ( 2 ) = 30 mm ,
Re 2 = R2 cos 2 ψ = 16 mm ,
Re3 = R3 cos 2 ψ = 40 mm
Ans.
A parallel-axis helical gearset is made with a 20° transverse pressure angle and a 35°
helix angle. The gears have module of 2.5 mm/tooth and have 15 and 25 teeth,
respectively. If the face width is 19 mm, calculate the base helix angle and the axial
contact ratio.
Ans.
ψ b = tan −1 [ tanψ cos φt ] = 33.34° ,
mt = 2.5 mm/tooth,
mx = F tanψ pt = 1.69 teeth avg
6-47
Ans.
pt = π mt = 4.71 mm/tooth ,
px = pt tanψ = 8.157 mm/tooth ,
mt = 1.5 mm/tooth ,
pn = pt cosψ = 4.0789 mm/tooth,
6-46
Ans.
Ans.
pt = π mt = 7.85 mm/tooth,
Ans.
A set of helical gears is to be cut for parallel shafts whose center distance is to be about
88 mm to give a velocity ratio of approximately 1.80. The gears are to be cut with a
standard 20° pressure angle hob whose module is 3 mm/tooth. Using a helix angle of
30°, determine the transverse values of the module and circular pitch and the tooth
numbers, pitch radii, and center distance.
ω 2 ω3 = R3 R2 = 1.8 ,
R2 + R3 = R2 + 1.8 R2 = 2.8 R2 ≈ 88 mm ,
R3 = 1.8 R2 ,
R2 ≈ 31.43 mm, R3 ≈ 56.57 mm,
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6-44
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100
mn = 3mm / tooth,
pt = π mt = 10.876 mm/tooth
Ans.
mt = mn cosψ = 3.464 mm/tooth ,
N 2 = 2 R2 / mt ≈ 18.1 teeth,
N 3 = 2 R3 / mt ≈ 32.66 teeth,
Therefore we will use N 2 = 18 teeth and N 3 = 33 teeth
Ans.
R2 = mt N 2 ( 2 ) = 31.176 mm ,
R3 = mt N 3 ( 2 ) = 57.156 mm ,
R2 + R3 = 88.33 mm
Ans.
A 16-tooth helical pinion is to run at 1800 rev/min and drive a helical gear on a parallel
shaft at 400 rev/min. The centers of the shafts are to be spaced 275 mm apart. Using a
helix angle of 23° and a pressure angle of 20°, determine the values for the tooth
numbers, pitch diameters, normal circular and module, and face width.
ω 2 ω3 = R3 R2 = 1800 400 = 4.5
R2 + R3 = R2 + 4.5 R2 = 5.5 R2 = 275 mm , R2 = 50 mm, R3 = 225 mm ,
N 2 = 16 teeth ,
N 3 = ( R3 R2 ) N 2 = 4.5 N 2 = 72 teeth
Ans.
mt = 2 R2 N 2 = 6.25 mm/tooth ,
pt = π mt = 19.625 mm/tooth,
mn = mt cosψ = 5.7531mm/tooth ,
pn = π mn = 18.06 mm/tooth,
Ans.
Ans.
px = pt tanψ = 46.233 mm/tooth
F ≈ ( 2 teeth ) px = 92.466 mm
Therefore, we may choose F = 93 mm .
6-49
R3 = 4.5R2
Ans.
Ans.
The catalog description of a pair of helical gears is as follows: 14½° normal pressure
angle, 45° helix angle, module of 3 mm/tooth, 25 mm face width, and normal module of
2.12 mm/tooth. The pinion has 12 teeth and a 37.5 mm pitch diameter, and the gear has
32 teeth and a 100 mm pitch diameter. Both gears have full-depth teeth, and they may be
purchased either right- or left-handed. If a right-hand pinion and left-hand gear are
placed in mesh, find the transverse contact ratio, the normal contact ratio, the axial
contact ratio, and the total contact ratio.
R2 = ( 37.5 mm ) 2 = 18.75 mm ,
a = 1xmt = 3 mm,
tan φt = tan φn cosψ = 0.365 7 ,
pb = pt cos φt = 8.852 mm/tooth,
r2 = R2 cos φt = 17.62 mm,
CP = 7.69 mm [Eq. (6.10)],
mt = CD pb = 1.60 teeth avg,
R3 = (100 mm ) 2 = 50 mm ,
pt = π mt = 9.42 mm/tooth ,
φt = tan −1 ( 0.365 7 ) = 20.09° ≈ 20.00° ,
r3 = R3 cos φt = 46.98 mm
PD = 6.55 mm [Eq. (6.11)],
Ans.
ψ b = tan −1 [ tanψ cos φt ] = 43.22° ,
mn = mt cos 2 ψ b = 3.01 teeth avg
Ans.
mx ≥ F tanψ pt = 2.65 teeth avg ,
m = mx + mt = 4.25 teeth avg
Ans.
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6-48
Ans.
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101
6-51
In a medium-sized truck transmission a 22-tooth clutch-stem gear meshes continuously
with a 41-tooth countershaft gear. The data show normal module of 3.5 mm/tooth, 18½°
normal pressure angle, 23½° helix angle and a 28 mm face width. The clutch-stem gear
is cut with a left-hand helix, and the countershaft gear is cut with a right-hand helix.
Determine the normal and total contact ratio if the teeth are cut full-depth with respect to
the normal diametral pitch.
tan φt = tan φn cosψ = 0.364 9 ,
φt = tan −1 ( 0.364 9 ) = 20.04° ≈ 20.00°
mn = 3.5 mm/tooth ,
mt = mn cosψ = 3.21 mm/tooth ,
R2 = mt N 2 2 = 35.31 mm ,
r2 = R2 cos φt = 33.18 mm,
CP = 8.78 mm [Eq. (6.10)],
pb = pt cos φt = 9.47 mm/tooth,
a = 1× m = 3.5 mm ,
pt = π mt = 10.08 mm/tooth ,
R3 = mt N 3 2 = 65.8 mm ,
r3 = R3 cos φt = 61.83 mm
PD = 8.05 mm [Eq. (6.11)],
mt = CD pb = 1.77 teeth avg , ,
ψ b = tan −1 [ tanψ cos φt ] = 22.22° ,
mn = mt cos 2 ψ b = 2.06 teeth avg
Ans.
mx = F tanψ pt = 1.21 teeth avg,
m = mx + mt = 2.98 teeth avg
Ans.
A helical pinion is right-hand, has 12 teeth, has a 60° helix angle, and is to drive another
gear at a velocity ratio of 3.0. The shafts are at a 90° angle, and the normal module of the
gears is 3 mm/tooth. Find the helix angle and the number of teeth on the mating gear.
What is the shaft center distance?
ψ 3 = Σ − ψ 2 = 30° RH ,
N 3 = (ω 2 ω3 ) N 2 = 36 teeth
R2 = mn N 2 ( 2 cosψ 2 ) = 36 mm,
R3 = mn N 3 ( 2 cosψ 3 ) = 62.35 mm
R2 + R3 = 98.35 mm
6-52
Ans.
Ans.
A right-hand helical pinion is to drive a gear at a shaft angle of 90°. The pinion has 6
teeth and a 75° helix angle and is to drive the gear at a velocity ratio of 6.5. The normal
module of the gear is 2 mm/tooth. Calculate the helix angle and the number of teeth on
the mating gear. Also determine the pitch diameter of each gear.
ψ 3 = Σ −ψ 2 = 15° RH ,
N 3 = (ω 2 ω3 ) N 2 = 39 teeth
Ans.
R2 = mn N 2 ( 2 cosψ 2 ) = 23.18 mm, ,
R3 = mn N 3 ( 2 cosψ 3 ) = 40.37 mm
Ans.
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6-50
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102
Gear 2 in the figure is to rotate clockwise and drive gear 3 counterclockwise at a velocity
ratio of 2. Use a normal module of 5 mm/tooth, a shaft center distance of about 250 mm,
and the same helix angle for both gears. Find the tooth numbers, the helix angles, and the
exact shaft center distance.
ψ 2 = ψ 3 = Σ 2 = 25° ,
ω 2 ω3 = R3 R2 = 2.0 , R3 = 2.0 R2 ,
Ans.
R2 + R3 = R2 + 2.0 R2 = 3.0 R2 ≈ 250 mm ,
R2 ≈ 83.33 mm, R3 ≈ 166.66 mm ,
cosψ 2 R2
≈ 30.209 ,
N2 = 2
mn
cosψ 3 R3
≈ 60.418
N3 = 2
mn
Therefore we choose
N 2 = 30 teeth ,
Ans.
N 3 = 60 teeth ,
Ans.
R2 = mn N 2 ( 2 cosψ 2 ) = 82.75 mm ,
R3 = mn N 3 ( 2 cosψ 3 ) = 165.5 mm ,
R2 + R3 = 248.25 mm
Ans.
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6-53
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103
Bevel Gears
A pair of straight-tooth bevel gears is to be manufactured for a shaft angle of 90°. If the
driver is to have 18 teeth and the velocity ratio is to be 3:1, what are the pitch angles?
N 3 = (ω 2 ω3 ) N 2 = 3 N 2 = 54 teeth ,
N 2 = 18 teeth ,
γ 2 = tan
−1
( N2
N 3 ) = 18.43° ,
γ 3 = 90° − γ 2 = 71.57°
Ans.
6-55
A pair of straight-tooth bevel gears has a velocity ratio of 1.5 and a shaft angle of 75°.
What are the pitch angles?
⎡
⎤
sin Σ
γ 2 = tan −1 ⎢
Ans.
⎥ = 28.78° , γ 3 = 75° − γ 2 = 46.22°
⎣ (ω 2 ω3 ) + cos Σ ⎦
6-56
A pair of straight-tooth bevel gears is to be mounted at a shaft angle of 120°. The pinion
and gear are to have 15 teeth and 33 teeth, respectively. What are the pitch angles?
⎡
⎤
sin Σ
γ 2 = tan −1 ⎢
Ans.
⎥ = 27.00° , γ 3 = 120° − γ 2 = 93.00°
+
Σ
N
N
cos
(
)
3
2
⎣
⎦
6-57
A pair of straight-tooth bevel gears with module of 12 mm/tooth have 19 teeth and 28
teeth, respectively. The shaft angle is 90°. Determine the pitch diameters, pitch angles,
addendum, dedendum, face width, and pitch diameters of the equivalent spur gears.
R2 = mN 2 2 = 114 mm ,
R3 = mN 3 2 = 168 mm ,
Ans.
γ 2 = tan
γ 3 = 90° − γ 2 = 55.84° ,
Ans.
−1
( N2
N 3 ) = 34.16° ,
Using Table 8.1: m90 = mG = N 3 N 2 = 1.474 , a3 = 9 mm ,
Whole depth = 2.188 m = 26.26 mm ,
d3 = Whole depth − a3 = 17.26 mm
Working depth = 2 m = 24 mm ,
c = 0.188m + 0.05 mm = 2.3 mm
a2 = Working depth − a3 = 15 mm
d 2 = Whole depth − a2 = 11.26 mm
Cone distance, A = R2 sin γ 2 = 203 mm Let F ≈ 0.3A = 60.9 mm , say F = 60 mm
Re 2 = R2 cos γ 2 = 137.77 mm
Re3 = R3 cos γ 3 = 299.19 mm
Ans.
Ans.
Ans.
Ans.
Ans.
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6-54
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104
A pair of straight-tooth bevel gears with module of 3 mm/tooth have 17 teeth and 28
teeth, respectively, and a shaft angle of 105°. For each gear, calculate the pitch diameter,
pitch angle, addendum, dedendum, face width, and the equivalent tooth numbers. Make
a sketch of the two gears in mesh. Use the standard tooth proportions as for a 90° shaft
angle.
R2 = mN 2 2 = 25.5 mm,
⎡
R3 = mN 3 2 = 42 mm,
⎤
sin Σ
⎥ = 34.83° , γ 3 = 105° − γ 2 = 70.17°
+
Σ
N
N
cos
(
)
⎣ 3 2
⎦
Using Table 8.1: mG = N 3 N 2 = 1.647, m90 = 1.996 , a3 = 1.966 mm ,
Whole depth = 2.188 m = 6.564 mm ,
d3 = Whole depth − a3 = 4.598 mm
Working depth = 2 m = 6 mm ,
c = 0.188 m + 0.05 mm = 0.614 mm
a2 = Working depth − a3 = 4.034 mm
d 2 = Whole depth − a2 = 2.53 mm
Cone distance, A = R2 sin γ 2 = 44.65 mm Let F ≈ 0.3A = 13.39 mm , say F = 14 mm
Re 2 = R2 cos γ 2 = 31 mm
Re3 = R3 cos γ 3 = 123.8 mm
N e 2 = 2 Re 2 / m = 20.66 teeth ,
N e3 = 2 Re3 / m = 82.53 teeth
γ 2 = tan −1 ⎢
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
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6-58
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105
Worms and Worm Gears
6-59
A worm having 4 teeth and a lead of 25 mm in drives a worm gear at a velocity ratio of
7.5. Determine the pitch diameters of the worm and worm gear for a center distance of
45 mm.
px = A N 2 = 25 mm 4 teeth = 6.25 mm/tooth
N 3 = (ω 2 ω3 ) N 2 = 7.5 ( 4 teeth ) = 30 teeth
R3 = N 3 px 2π = 29.86 mm
Specify a suitable worm and worm gear combination for a velocity ratio of 60 and a
center distance of 160 mm in using an axial pitch of 12.5 mm/tooth.
N 3 = (ω2 ω3 ) N 2 = 60 (1 tooth ) = 60 teeth
Use N 2 = 1 tooth ,
R3 = N 3 px 2π = 119.43 mm
6-61
R2 = 160 − R3 = 40.57 mm
Ans.
A triple-threaded worm drives a worm gear having 40 teeth. The axial pitch is 30 mm
and the pitch diameter of the worm is 45 mm. Calculate the lead and lead angle of the
worm. Find the helix angle and pitch diameter of the worm gear.
A = N 2 px = 3 teeth ( 30 mm/tooth ) = 90 mm
Ans.
λ = tan −1 ( A 2π R2 ) = tan −1 ( 90 mm π 1.45 mm ) = 32.49°
ψ = λ = 32.49°
R3 = N 3 px 2π = 40 teeth ( 30 mm/tooth ) 2π = 191 mm
6-62
Ans.
Ans.
Ans.
Ans.
A triple threaded worm with a lead angle of 20˚ and an axial pitch of 10 mm/tooth drives
a worm gear with a velocity reduction of 15 to 1. Determine for the worm gear: (a) the
number of teeth, (b) the pitch diameter, and (c) the helix angle. (d) Determine the pitch
diameter of the worm. (e) Compute the center distance.
A = N 2 px = 3 teeth (10 mm/tooth ) = 30 mm
R2 = A ( 2π tan λ ) = 30 mm 2π tan 20° = 13.12 mm
Ans.
R3 = (ω2 ω3 ) R2 = 15 (13.12 mm ) = 196.87 mm
Ans.
N 3 = 2π R3 px = 123.6 teeth
ψ = λ = 20.0°
R2 + R3 = 13.12 mm + 196.87 mm = 209.99 mm
Ans.
Ans.
Ans.
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6-60
R2 = 45 − R3 = 15.14 mm
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107
Chapter 7
Mechanism Trains
Find the speed and direction of gear 8 in the figure. What is the kinematic coefficient of
the train?
N 2 N 4 N 5 N 7 18 15 33 16
5
=
=+
88
N 3 N 5 N 6 N8 44 33 36 48
ω8 = θ 8/′ 2ω 2 = ( +5 88 )(1 200 rev/min ccw ) = 68.18 rev/min ccw
θ 8′/ 2 =
7.2
Ans.
Ans.
Part (a) of the figure gives the pitch diameters of a set of spur gears forming a train.
Compute the kinematic coefficient of the train. Determine the speed and direction of
rotation of gears 5 and 7.
R2 R4 175 mm 225 mm
7
R R
7 750 mm 225 mm
21
θ 7′ / 2 = θ5/′ 2 5 6 =
= + Ans.
=
=+
R3 R5 375 mm 750 mm
50
R6 R7 50 225 mm 400 mm
80
Ans.
ω5 = θ 5/′ 2ω 2 = ( +7 50 )(120 rev/min cw ) = 16.80 rev/min cw
θ5/′ 2 =
ω 7 = θ 7′ / 2ω 2 = ( +21 80 )(120 rev/min cw ) = 31.50 rev/min cw
7.3
Ans.
Part (b) of the figure shows a gear train consisting of bevel gears, spur gears, and a worm
and worm gear. The bevel pinion is mounted on a shaft which is driven by a V-belt on
pulleys. If pulley 2 rotates at 1200 rev/min in the direction shown, find the speed and
direction of rotation of gear 9.
R2 N 4 N 6 N8 150 mm 18 20 3
3
=
=
R3 N 5 N 7 N 9 250 mm 38 48 36 304
ω9 = θ 9′/ 2ω 2 = ( 3 304 )(1 200 rev/min ) = 11.84 rev/min cw
θ9′/ 2 =
Ans.
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7.1
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108
Use the truck transmission of Fig. 7.2 and an input speed of 3000 rev/min to find the
drive shaft speed for each forward gear and for reverse gear.
First gear:
θ 9′/ 2 =
N 2 N 6 17 17
289
=
=+
1 849
N 3 N 9 43 43
ω9 = θ 9′/ 2ω 2 = ( 289 1 849 )( 3 000 rev/min ) = 468.9 rev/min
N N 17 27
153
Second gear:
θ 8′/ 2 = 2 5 =
=+
473
N 3 N8 43 33
ω8 = θ 8/′ 2ω 2 = (153 473)( 3 000 rev/min ) = 970.4 rev/min
N N
17 36
51
Third gear:
θ 7′ / 2 = 2 4 =
=+
N 3 N 7 43 24
86
ω 7 = θ 7′ / 2ω 2 = ( 51 86 )( 3 000 rev/min ) = 1 779.1 rev/min
Fourth gear:
Ans.
θ 2′/ 2 = 1.0
ω 2 = θ 2′ / 2ω 2 = (1.0 )( 3 000 rev/min ) = 3 000 rev/min
17 17 22
3 179
N N N
θ 9′/ 2 = − 2 6 11 = −
=−
Reverse gear:
43 18 43
16 641
N 3 N10 N 9
ω9 = θ 9′/ 2ω 2 = ( −3 179 16 641)( 3 000 rev/min ) = −573.1 rev/min
7.5
The figure illustrates the gears in a speed-change gearbox used in machine tool
applications. By sliding the cluster gears on shafts B and C, nine speed changes can be
obtained. The problem of the machine tool designer is to select tooth numbers for the
various gears so as to produce a reasonable distribution of speeds for the output shaft.
The smallest and largest gears are gears 2 and 9, respectively. Using 20 teeth and 45
teeth for these gears, determine a set of suitable tooth numbers for the remaining gears.
What are the corresponding speeds of the output shaft? Notice that the problem has
many solutions.
Ans.
We also set N 6 = 20 teeth (minimum).
Since the largest speed reduction will be obtained with gears 2-5-6-9,
20 20
N N
450 rev/min = 137 rev/min
ωC ,min = 2 6 ω A =
N5 N9
N 5 45
Ans.
From this we get N 5 = 29.197 , and we choose N 5 = 30 teeth .
Next, using distance units of circular pitch, the distance between shafts B and C is
Ans.
BC = N 5 + N8 = N 6 + N 9 = N 7 + N10 = 65 teeth . N8 = BC − N 5 = 35 teeth
Similarly the distance between shafts A and B is
Ans.
AB = N 2 + N 5 = N 3 + N 6 = N 4 + N 7 = 50 teeth . N 3 = AB − N 6 = 30 teeth .
Since the minimum is 20 teeth and since AB = N 4 + N 7 = 50 teeth we see that
20 ≤ N 4 , N 7 ≤ 30 and we choose
N 4 = N 7 = 25 teeth
Ans.
and, finally,
N10 = BC − N 7 = 40 teeth
Ans.
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7.4
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109
With all tooth numbers known, we can now find the output shaft speed for each gear
arrangement. These are
Arrangement
Output shaft speed, ωC , rev/min
Kinematic coefficient, θ C′ / A
2-5-5-8
2-5-6-9
2-5-7-10
3-6-5-8
3-6-6-9
3-6-7-10
4-7-5-8
4-7-6-9
4-7-7-10
257.1
133.3
187.5
578.6
300.0
421.9
385.7
200.0
281.3
The internal gear (gear 7) in the figure turns at 60 rev/min ccw. What are the speed and
direction of rotation of arm 3?
N 2 N 4 N 6 20 teeth 40 teeth 36 teeth 20
=
=
N 4 N 5 N 7 40 teeth 18 teeth 154 teeth 77
ω − ω3 60 rev/min − ω3 20
θ 7′ / 2 = 7
=
=
; ω3 = 81.1 rev/min ccw
ω 2 − ω3
0 − ω3
77
θ 7′ / 2 =
7.7
Ans.
If the arm in the figure rotates at 300 rev/min ccw, find the speed and direction of
rotation of internal gear 7.
N 2 N 4 N 6 20 teeth 40 teeth 36 teeth 20
=
=
N 4 N 5 N 7 40 teeth 18 teeth 154 teeth 77
ω − ω3 ω 7 − 300 rev/min 20
=
=
θ 7′ / 2 = 7
; ω = 222.1 rev/min ccw
ω 2 − ω3 0 − 300 rev/min 77 7
θ 7′ / 2 =
7.8
Ans.
In part (a) of the figure, shaft C is stationary. If gear 2 rotates at 800 rev/min ccw, what
are the speed and direction of rotation of shaft B?
N2
18 teeth
3
3
=−
=−
ω3 = θ 3/′ 2ω 2 = − ( 800 rev/min ccw ) = 600 rev/min cw
N3
24 teeth
4
4
N N
18 teeth 20 teeth 3
θ 8′/ 5 = 5 7 =
=
N 6 N8 42 teeth 40 teeth 14
ω − ω3 0 + 600 rev/min
3
θ 8′/ 5 = 8
=
= ; ω3 = 2 200 rev/min cw
Ans.
ω5 − ω3 ω5 + 600 rev/min 14
θ 3/′ 2 = −
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7.6
0.571
0.296
0.416
1.285
0.667
0.938
0.857
0.444
0.625
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110
7.9
In part (a) of the figure, consider shaft B as stationary. If shaft C is driven at 380 rev/min
ccw, what are the speed and direction of rotation of shaft A?
N 5 N 7 18 teeth 20 teeth 3
=
=
N 6 N8 42 teeth 40 teeth 14
ω − ω3 380 rev/min − ω3 3
=
= ; ω3 = 483.6 rev/min ccw
θ 8′/ 5 = 8
ω5 − ω3
0 − ω3
14
N
24 teeth
483.6 rev/min ccw = 644.8 rev/min cw
ω 2 = − 3 ω3 = −
N2
18 teeth
θ 8′/ 5 =
In part (a) of the figure, determine the speed and direction of rotation of shaft C if (a)
shafts A and B both rotate at 360 rev/min ccw and (b) shaft A rotates at 360 rev/min cw
and shaft B rotates at 360 rev/min ccw.
θ 8′/ 5 =
(a)
(b)
7.11
N 5 N 7 18 teeth 20 teeth 3
=
=
N 6 N8 42 teeth 40 teeth 14
N
18 teeth
ω3 = − 2 ω 2 = −
360 rev/min ccw = 270 rev/min cw
N3
24 teeth
ω − ω3
ω8 + 270 rev/min
3
ω8 = 135 rev/min cw Ans.
=
= ;
θ 8′/ 5 = 8
ω5 − ω3 360 rev/min + 270 rev/min 14
N
18 teeth
ω3 = − 2 ω 2 = −
360 rev/min cw = 270 rev/min ccw
N3
24 teeth
ω − ω3
ω8 − 270 rev/min
3
θ 8′/ 5 = 8
=
= ; ω8 = 289.3 rev/min ccw Ans.
ω5 − ω3 360 rev/min − 270 rev/min 14
In part (b) of the figure, gear 2 is connected to the input shaft. If arm 3 is connected to
the output shaft, what speed reduction can be obtained? What is the sense of rotation of
the output shaft? What changes could be made in the train to produce the opposite sense
of rotation for the output shaft?
5
N 2 N 4 N 5 20 teeth 28 teeth 16 teeth
=
=
N 4 N 5 N 6 28 teeth 16 teeth 108 teeth 27
ω − ω3 0 − ω3
5
5
ω3 = − ω 2
=
=
θ 6′/ 2 = 6
ω 2 − ω3 ω 2 − ω3 27
22
The speed reduction is 17 22 = 77.3%
Ans.
The sense of the output rotation is opposite to the input sense.
Ans.
The opposite sense of rotation for the output shaft can be produced by replacing gears 4
and 5 by a single 44-tooth gear.
Ans.
θ 6′/ 2 =
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7.10
Ans.
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111
7.12
The Lévai type-L train shown in Fig. 7.10 has N2 = 16T, N4 = 19T, N5 = 17T, N6 = 24T,
and N3 = 95T. Internal gear 7 is fixed. Find the speed and direction of rotation of the
arm if gear 2 is driven at 100 rev/min cw.
N 2 N 4 N 6 16 teeth 19 teeth 24 teeth 384
=
=
N 4 N 5 N 7 19 teeth 17 teeth 95 teeth 1 615
ω − ω3
0 − ω3
384
ω3 = 31.19 rev/min ccw
=
=
;
θ 7′ / 2 = 7
ω 2 − ω3 −100 rev/min − ω3 1 615
θ 7′ / 2 =
The Lévai type-A train of Fig. 7.10 has N2 = 20T and N4 = 32T.
(a)
If the module is 6 mm, find the number of teeth on gear 5 and the crank arm
radius.
(b)
If gear 2 is fixed and internal gear 5 rotates at 10 rev/min ccw, find the speed and
direction of rotation of the arm.
(a)
(b)
7.14
N 5 = N 2 + 2 N 4 = 20 teeth + 2 ( 32 teeth ) = 84 teeth
Ans.
R3 = m ( N 2 + N 4 ) 2 = ( 6 mm/tooth )( 20 teeth + 32 teeth ) 2 = 156 mm
Ans.
N 2 N 4 20 teeth 32 teeth
5
=
=−
N 4 N 5 32 teeth 84 teeth
21
ω − ω3 10 rev/min − ω3
5
=
=−
θ 5/′ 2 = 5
ω 2 − ω3
0 − ω3
21
θ 5/′ 2 =
ω3 = 8.08 rev/min ccw
Ans.
The tooth numbers for the automotive differential shown in Fig. 7.19 are N2 = 17T, N3 =
54T, N4 = 11T, and N5 = N6 = 16T. The drive shaft turns at 1200 rev/min. What is the
speed of the right wheel if it is jacked up and the left wheel is resting on the road surface?
N2
17 teeth
ω2 =
(1 200 rev/min ) = 377.8 rev/min
N3
54 teeth
N N
16 teeth 11 teeth
θ 6′/ 5 = − 5 4 = −
= −1
11 teeth 16 teeth
N4 N6
ω − ω3 ω 6 − 377.8 rev/min
ω 6 = 755.6 rev/min
θ 6′/ 5 = 6
=
= −1
ω5 − ω3 0 − 377.8 rev/min
ω3 =
Ans.
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7.13
Ans.
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112
A vehicle using the differential shown in Fig. 7.19 turns to the right at a speed of 48 km/h
on a curve of 24 m radius. Use the same tooth numbers as in Problem 7.14. The tire
diameter is 375 mm. Use 1500 mm as the distance between treads.
(a)
Calculate the speed of each rear wheel.
(b)
Find the rotational speed of the ring gear.
(a)
ωcar =
vcar
ρ
=
( 48 Rm/h )(1000 m/km ) ( 60 min/h ) = 33.3 rad/min
( 24 m )
For the right and left wheels, respectively:
( 33.3 rad/min )( 24 m ⋅1000 mm/m − 750 mm ) = 657.51 rev/min
v
ω6 = R =
r
( 2π rad/rev )( 375 2 mm )
vR ( 33.3 rad/min )( 24 m ⋅1000 mm/m + 750 mm )
=
= 699.93 rev/min
r
( 2π rad/rev )( 375 2 mm )
N N
16 teeth 11 teeth
θ 6′/ 5 = − 5 4 = −
= −1
11 teeth 16 teeth
N4 N6
ω − ω3 657.51 rev/min − ω3
=
= −1 ω3 = 678.72 rev/min
θ 6′/ 5 = 6
ω5 − ω3 699.93 rev/min − ω3
ω5 =
(b)
7.16
Ans.
Ans.
Ans.
The figure shows a possible arrangement of gears in a lathe headstock. Shaft A is driven
by a motor at a speed of 720 rev/min. The three pinions can slide along shaft A so as to
yield the meshes 2 with 5, 3 with 6, or 4 with 8. The gears on shaft C can also slide so as
to mesh either 7 with 9 or 8 with 10. Shaft C is the mandril shaft.
Make a table showing all possible gear arrangements, beginning with the slowest
(a)
speed for shaft C and ending with the highest, and enter in this table the speeds of
shafts B and C.
(b)
If the gears all have a module of 5 mm, what must be the shaft center distances?
(a)
(b)
Gears
ω B , rev/min
ωC , rev/min
2-5-7-9
4-8-7-9
3-6-7-9
2-5-8-10
4-8-8-10
3-6-8-10
180.0
327.3
589.1
180.0
327.3
589.1
38.7
70.4
126.8
241.5
439.0
790.2
AB = m ( N 2 + N 5 ) 2 = ( 5 mm/tooth )(16 teeth + 64 teeth ) 2 = 200 mm
Ans.
BC = m ( N 7 + N 9 ) 2 = ( 5 mm/tooth )(17 teeth + 79 teeth ) 2 = 240 mm
Ans.
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7.15
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113
7.17
Shaft A in the figure is the output and is connected to the arm. If shaft B is the input and
drives gear 2, what is the speed ratio? Can you identify the Lévai type for this train?
N 2 N3 N5
16 teeth 18 teeth 18 teeth
9
=−
=−
18 teeth 16 teeth 50 teeth
25
N3 N 4 N 6
ω −ωA 0 −ωA
9
θ 6′/ 2 = 6
ω A = ( 9 34 ) ω 2
=
=− ;
ω2 − ω A ω2 − ω A
25
This train is Lévai type-F.
θ 6′/ 2 = −
In Problem 7.17, shaft B rotates at 100 rev/min cw. Find the speed of shaft A and of
gears 3 and 4 about their own axes.
Step
Arm
2
3
4
Locked
+1
+1
+1
+1
Arm fixed
0
+16/9
-128/81
+16/9
Total
+1
+25/9
-47/81
+25/9
ω A = + ( 9 25 ) ω 2 = ( 9 25 )(100 rev/min cw ) = 36.0 rev/min cw
7.19
Ans.
5
+1
+16/9
+25/9
6
+1
-1
0
Ans.
ω3 = ( −47 81)( 9 25 ) ω 2 = − ( 47 225 )(100 rev/min cw ) = 20.9 rev/min ccw
Ans.
ω 4 = ( 25 9 )( 9 25 ) ω 2 = (1.0 )(100 rev/min cw ) = 100.0 rev/min cw
Ans.
Bevel gear 2 is driven by the engine in the reduction unit shown in the figure. Bevel
planets 3 mesh with crown gear 4 and are pivoted on the spider (arm), which is connected
to propeller shaft B. Find the percent speed reduction.
N 2 N3
36 teeth 21 teeth
9
=−
=−
21 teeth 52 teeth
13
N3 N 4
9
ω − ωB 0 − ωB
ω B = ( 9 22 ) ω 2
=
=− ;
θ 4′/ 2 = 4
13
ω2 − ω B ω2 − ω B
Speed reduction to 9/22 = 40.9% is speed reduction of 59.1%.
θ 4′/ 2 = −
Ans.
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7.18
Ans.
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114
In the clock mechanism shown in the figure, a pendulum on shaft A drives an anchor (see
Fig. 1.9c). The pendulum period is such that one tooth of the 30T escape wheel on shaft
B is released every 2 s, causing shaft B to rotate once every minute. In the figure, note
that the second (to the right) 64T gear is pivoted loosely on shaft D and is connected by a
tubular shaft to the hour hand.
Show that the train values are such that the minute hand rotates once every hour
(a)
and that the hour hand rotates once every 12 hours.
(b)
How many turns does the drum on shaft F make every day?
(a)
ω B = 1.0 rev/min
8 teeth 8 teeth
1
N B NC
=
=
N C N D 60 teeth 64 teeth 60
ω D = θ D′ / Bω B = 1 60 rev/min
Δt D = 60 min/rev
θ D′ / B =
N D N E ⎛ 1 ⎞ 28 teeth 8 teeth
1
=⎜ ⎟
=
N E N H ⎝ 60 ⎠ 42 teeth 64 teeth 720
720 min/rev
Δt H =
= 12 hr/rev
ω H = θ H′ / Bω B = 1 720 rev/min
60 min/hr
Ans.
θ H′ / B = θ D′ / B
(b)
N D ⎛ 1 ⎞ 8 teeth
1
=⎜ ⎟
=
N F ⎝ 60 ⎠ 96 teeth 720
ω F = θ F′ / Bω B = (1 720 rev/min )( 60 min/hr )( 24 hr/day ) = 2 rev/day
Ans.
θ F′ / B = θ D′ / B
Ans.
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7.20
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115
Chapter 8
Synthesis of Linkages
8.1
A function varies from 0 to 10. Find the Chebychev spacing for two, three, four, five,
and six precision positions.
With x0 = 0.0 and xn +1 = 10.0 , Eq. (8.5) becomes:
j\n
1
2
3
4
5
6
8.2
2
1.46447
8.53553
(2 j − 1)π
2n
3
0.66987
5.00000
9.33013
4
0.38060
3.08658
6.91342
9.61940
j = 1, 2,..., n
5
0.24472
2.06107
5.00000
7.93893
9.75528
6
0.17037
1.46447
3.70591
6.29409
8.53553
9.82963
Determine the link lengths of a slider-crank linkage to have a stroke of 600 mm and a
time ratio of 1.20.
After laying out the distance B1B2 =
600 mm, we see that the time ratio
is Q = (180° + α ) (180° − α ) = 1.20
and, from this, our design must
have α = 16.40°.
Therefore we construct the point C
such that the central angle
B1CB2 = 2α = 32.80°. Using this
point C as the center of a circle
assures that any point O2 on this
circle will have the angle
B1O2 B2 ≡ α = 16.40° and thus will
be a possible solution point. One
typical solution uses the point O2 shown.
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x j = 5.0 − 5.0 cos
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116
Choosing
the
point
O2
shown
we
measure
the
two
distances
O2 B1 = r3 + r2 = 1 324.956 mm and O2 B2 = r3 − r2 = 801.946 mm and from these we find
r2 = 261.50 mm
Ans.
r3 = 1 063.45 mm
Ans.
Determine a set of link lengths for a slider-crank linkage such that the stoke is 400 mm
and the time ratio is 1.25.
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8.3
After laying out the distance B1B2 = 400 mm, we see that the time ratio is
Q = (180° + α ) (180° − α ) = 1.25 and, from this, our design must have α = 20.00°.
Therefore we construct the point C such that the central angle B1CB2 = 2α = 40.00°.
Using this point C as the center of a circle assures that any point O2 on this circle will
have the angle B1O2 B2 ≡ α = 20.00° and thus will be a possible solution point. One
typical solution uses the point O2 shown.
Choosing the point O2 shown we measure the two distances O2 B1 = r3 + r2 = 745 mm and
O2 B2 = r3 − r2 = 392 and from these we find
r2 = 175.5 mm
r3 = 569.5 mm
Ans.
Ans.
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117
The rocker of a crank-rocker linkage is to have a length of 500 mm and swing through a
total angle of 45 with a time radio of 1.25. Determine a suitable set of dimensions for
r1 , r2 , and r3 .
After laying out the angle B1O4 B2 ≡ φ = 45° with BO4 = 500 mm, we see that the time
ratio is Q = (180° + α ) (180° − α ) = 1.25 and, from this, we find that α = 20.00°.
Therefore we construct the point C such that the central angle B1CB2 = 2α = 40.00°.
Then, using this point C as the center of a circle assures that any point O2 on this circle
will have the angle B1O2 B2 ≡ α = 20.00° and thus will be a possible solution point. One
typical solution uses the point O2 shown.
Choosing the point O2 shown we measure the three distances O2O4 = 556 mm ,
O2 B1 = r3 + r2 = 878 mm , and O2 B2 = r3 − r2 = 588 mm and from these we find
r1 = 556 mm
r2 = 145 mm
r3 = 733 mm
Ans.
Ans.
Ans.
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8.4
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118
A crank-and-rocker mechanism is to have a rocker 1.8 m in length and a rocking angle of
75° . If the time ratio is to be 1.32, what are a suitable set of link lengths for the
remaining three links?
After laying out the angle B1O4 B2 ≡ φ = 75° with BO4 = 1.8 m = 1800 mm, we see that
the time ratio is Q = (180° + α ) (180° − α ) = 1.32 and, from this, we find that
α = 24.83°.
Therefore we construct the point C such that the central angle B1CB2 = 2α = 49.66°.
Then, using this point C as the center of a circle assures that any point O2 on this circle
will have the angle B1O2 B2 ≡ α = 24.83° and thus will be a possible solution point. One
typical solution uses the point O2 shown.
Choosing the point O2 shown we measure the three distances O2O4 = 2434 mm ,
O2 B1 = r3 + r2 = 3808 mm , and O2 B2 = r3 − r2 = 1907 mm and from these we find
r1 = 2434 mm
r2 = 950.5 mm
r3 = 2857.5 mm
8.6
Ans.
Ans.
Ans.
Design a crank and coupler to drive rocker 4 in the figure such that slider
6 will reciprocate through a distance of 400 mm with a time radio of 1.20. Use
a = r4 = 400 mm and r5 = 600 mm with r4 vertical at midstroke. Record the location of
O2 and dimensions r2 and r3 .
After laying out the angle B1O4 B2 ≡ φ = 60° with BO4 = 400 mm, we see that the time
ratio is Q = (180° + α ) (180° − α ) = 1.20 and, from this, we find that α = 16.36°.
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8.5
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Therefore we construct the point D such that the central angle B1 DB2 = 2α = 32.72°.
Then, using this point D as the center of a circle assures that any point O2 on this circle
will have the angle B1O2 B2 ≡ α = 16.36° and thus will be a possible solution point. One
typical solution uses the point O2 shown.
Choosing the point O2 shown we measure the three distances O2O4 = 626 mm ,
O2 B1 = r3 + r2 = 896 mm , and O2 B2 = r3 − r2 = 549 mm and from these we find
r1 = 626 mm
r2 = 173.5 mm
r3 = 722.5 mm
8.7
Ans.
Ans.
Ans.
Design a crank and rocker for a six-link mechanism such that the slider in the figure for
Problem 8.6 reciprocates through a distance of 800 mm with a time ratio of 1.12, use
a = r4 = 1 200 mm and r5 = 1 800 mm. Locate O4 such that rocker 4 is vertical when the
slider is at midstroke. Find suitable coordinates for O2 and lengths for r2 and r3 .
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119
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120
After laying out the angle B1O4 B2 ≡ φ = 2sin −1 ( 400 1200 ) = 38.94° with BO4 = 1 200
mm, we see that the time ratio is Q = (180° + α ) (180° − α ) = 1.12 and, from this, we find
that α = 10.19°.
Therefore we construct the point D such that the central angle B1 DB2 = 2α = 20.38°.
Then, using this point D as the center of a circle assures that any point O2 on this circle
will have the angle B1O2 B2 ≡ α = 10.19° and thus will be a possible solution point. One
typical solution uses the point O2 shown.
r1 = 1 979 mm
r2 = 346 mm
r3 = 2 288 mm
8.8
Ans.
Ans.
Ans.
Design a crank-rocker mechanism with optimum transmission angle, a unit time ratio,
and a rocker angle of 45° using a rocker 250 mm in length. Use the chart of Fig. 8.5 and
γ min = 50°.
From Fig. 8.5, using γ min = 50° , we get r2 r1 = 0.335 , r3 r1 = 0.610 , and r4 r1 = 0.856 .
With r4 = 250 mm , these give
r1 = 292 mm
r2 = 98 mm
r3 = 178 mm
Ans.
Ans.
Ans.
The small deviations of the figure below from the design specifications are the result of
inaccuracies in reading the design chart.
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Choosing the point O2 shown we measure the three distances O2O4 = 1 979 mm ,
O2 B1 = r3 + r2 = 2 634 mm , and O2 B2 = r3 − r2 = 1 942 mm and from these we find
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121
The figure shows two positions of a folding seat used in the aisles of buses to
accommodate extra passengers. Design a four-bar linkage to support the seat so that it
will lock in the open position and fold to a stable closing position along the side of the
aisle.
The open position is a toggle position with no force tending to open or close the 3-4-5 triangle. Thus a small catch only allowing joint A to rotate very slightly past the 180°
position at A2 will keep the seat open.
8.10
Design a spring-operated four-bar linkage to support a heavy lid like the trunk lid of an
automobile. The lid is to swing through an angle of 80° from the closed to the open
position. The springs are to be mounted so that the lid will be held closed against a stop,
and they should also hold the lid in a stable open position without the use of a stop.
One typical solution has
A
O2 A = AB = O4 B = O2O4 .
stop for the closed position may
be provided with point B
slightly below the line O4 A .
The open position is held stable
by the choice of the spring free
length.
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8.9
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8.11
For part (a) of the figure, synthesize a linkage to move AB from position 1 to position 2
and return.
8.12
For part (b) of the figure synthesize a mechanism to move AB successively through
positions 1, 2, and 3.
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122
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123
through 8.221 The figure shows a function-generator linkage in which the motion of
rocker 2 corresponds to x and the motion of rocker 4 to the function y = f (x). Use four
precision points and Chebychev spacing and synthesize a linkage to generate the
functions shown in the accompanying table. Plot a curve of the desired function and a
curve of the actual function which the linkage generates. Compute the maximum error
between them in percent.
Problem
number
Function,
y = f ( x)
Range
x0 ∼ ψ 0 ,
deg
y0 ∼ φ 0 ,
deg
r2 r1
r3 r1
r4 r1
Max.
Error, deg
8.13, 8.23
log10 x
1≤ x ≤ 2
52.628
259.077
-3.352
0.845
3.485
0.0037
8.14, 8.24
sin x
0 ≤ x ≤π 2
-62.263
75.606
1.834
2.238
-0.693
0.1900
8.15, 8.25
tan x
0≤ x ≤π 4
269.709
124.189
-2.660
7.430
8.685
0.0380
8.16, 8.26
ex
0 ≤ x ≤1
241.644
40.422
-3.499
0.878
3.399
0.0258
8.17, 8.27
1x
1≤ x ≤ 2
33.804
120.213
-0.385
1.030
0.384
0.0161
8.18, 8.28
x1.5
0 ≤ x ≤1
-5.171
211.689
0.625
1.309
-0.401
0.1460
8.19, 8.29
x2
0 ≤ x ≤1
-29.321
233.836
2.523
3.329
-0.556
0.0673
8.20, 8.30
x 2.5
0 ≤ x ≤1
-88.313
44.492
-1.801
0.908
1.274
0.4120
8.21, 8.31
x3
0 ≤ x ≤1
-85.921
37.637
-1.606
0.925
1.107
0.5095
8.22, 8.32
x2
−1 ≤ x ≤ 1
-21.180
-53.670
-0.610
0.565
0.380
2.3400
8.23
through 8.32 Repeat Problems 8.13 through 8.22 using the overlay method.
The overlay method can be used to confirm the above solutions. Other nearby solutions
are also possible, but are too numerous to display here.
8.33
1
The figure illustrates a coupler curve which can be generated by a four-bar linkage (not
shown). Link 5 is to be attached to the coupler point, and link 6 is to be a rotating
member with O6 as the frame connection. In this problem we wish to find a coupler
curve from the Hrones and Nelson atlas or by point-position reduction, such that, for an
appreciable distance, point C moves through an arc of a circle. Link 5 is then
proportioned so that D lies at the center of curvature of this arc. The result is then called
a hesitation motion because link 6 will hesitate in its rotation for the period during which
point C transverses the approximate circular arc. Make a drawing of the complete
linkage and plot the velocity-displacement diagram for 360° of displacement of the input
link.
Solutions for these problems were among the earliest computer work in kinematic synthesis and results are shown in
F. Freudenstein, “Four-bar Function Generators,” Machine Design, vol. 30, no. 24, pp. 119-123, 1958.
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8.13
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This coupler curve for point C was found using the Hrones and Nelson Atlas, page 150.
The hesitation is shown by the following plot of the first order kinematic coefficient φ6′ .
8.34
Synthesize a fourbar linkage to
obtain a coupler
curve having an
approximate
straight-line
segment.
Then,
using
the
suggestion
included in Fig.
8.29b or 8.31b,
synthesize a dwell
motion. Using an
input
crank
angular velocity of unity, plot the velocity of rocker 6 versus the input crank
displacement.
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124
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The Hrones and Nelson Atlas contains a wide variety of coupler curves similar to the one
shown above; this one is from page 93. The dwell in the rotation of link 6 is shown by
the following plot of the first order kinematic coefficient φ6′ .
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125
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126
Synthesize a dwell mechanism using the idea suggested in Fig. 8.29a and the Hrones and
Nelson atlas. Rocker 6 is to have a total angular displacement of 60° . Using this
displacement as the abscissa, plot a velocity diagram of the motion of the rocker to
illustrate the dwell motion.
The Hrones and Nelson Atlas contains a wide variety of coupler curves similar to the one
shown above; this one is from page 93. The dwell in the rotation of link 6 is shown by
the following plot of the first order kinematic coefficient φ6′ .
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8.35
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127
Chapter 9
Spatial Mechanisms
9.1
Use the Kutzbach criterion to determine the mobility of the GGC linkage shown in the
figure. Identify any idle freedoms and state how they can be removed. What is the
nature of the path described by point B?
n = 3 , j2 = 3 , j3 = 1 , j1 = j4 = j5 = 0
m = 6 ( 3 − 1) − 4 (1) − 3 ( 2 ) = 2
There is one idle freedom, the rotation of link 3 about its own axis. This idle freedom may
be eliminated by employing a two freedom pair, such as a universal joint, in place of one of
the two spheric pairs, either at B or at O3.
Ans.
The path described by point B is the curve of intersection of a cylinder of radius BA about
the y axis and a sphere of radius BO3 centered at O3.
Ans.
9.2
For the GGC linkage shown express the position of each link in vector form.
R O3O = 75ˆi mm , R BA = 75cos θ 2 ˆi + 75sin θ 2kˆ = 64.952ˆi + 37.500kˆ mm ,
R AO = RAO ˆj mm , RBO3 = 150 mm .
Ans.
Substituting these into R O3O + R BO3 = R AO + R BA gives
R AO = 11 250 ( cosθ 2 + 1)ˆj = 144.889ˆj mm , Ans.
R BO3 = 75 ( cos θ 2 − 1) ˆi + 11 250 ( cos θ 2 + 1)ˆj + 75sin θ 2kˆ
= −10.048ˆi + 144.889ˆj + 37.500kˆ mm
9.3
Ans.
With VA = −50ˆj mm/s , use vector analysis to find the angular velocities of links 2 and 3
and the velocity of point B at the position specified.
The velocity of point B is given by VB = VBO3 = VA + VBA , or
V = ω × R = −50ˆj + ω × R
B
3
2
BO3
BA
Assuming that the idle freedom is not active we have ω3 iR BO3 = 0 , where ω 2 = ω 2 ˆj and
ω = ω x ˆi + ω y ˆj + ω z kˆ .
3
3
3
3
Expanding these and using the position data from Problem P9.2 gives four simultaneous
equations:
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Ans.
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128
−10.048 144.889 37.500 ⎤ ⎡ω 2 ⎤ ⎡ 0 ⎤
⎡ 0
⎢ −37.500
0
37.500 −144.889 ⎥⎥ ⎢⎢ω3x ⎥⎥ ⎢⎢ 0 ⎥⎥
⎢
=
⎢ 0
−37.500
−10.048 ⎥ ⎢ω3y ⎥ ⎢ −50.000 ⎥
0
⎢
⎥⎢ z⎥ ⎢
⎥
0
⎣ 64.952 144.889 10.048
⎦ ⎢⎣ω3 ⎥⎦ ⎣ 0 ⎦
Ans.
Solving these gives ω = −2.576ˆj rad/s
2
ω3 = 1.330 rad/s
ω3 = 1.161ˆi − 0.086ˆj + 0.644kˆ rad/s ,
VB = −96.592ˆi − 50.000ˆj + 167.302kˆ mm/s , VB = 0.200 m/s
Ans.
Ans.
Solve Problem P9.3 using graphical techniques.
9.5
For the spherical RRRR shown in the figure, use vector algebra to make complete
velocity and acceleration analyses at the position given.
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9.4
R AO2 = −75ˆi mm ,
R O4O2 = 50ˆi − 175kˆ mm ,
R BO4 = 225ˆj mm .
Substituting these into R AO2 + R BA = R O4O2 + R BO4 gives
R BA = 125ˆi + 225ˆj − 175kˆ mm ,
RBA = 311.25 mm
The velocity analysis proceeds as follows:
VA = VAO2 = ω 2 × R AO2 = −60kˆ rad/s × −75ˆi mm = 4500ˆj mm/s
VB = VBO4 = ω 4 × R BO4
(
) (
)
= (ω ˆi ) × ( 225ˆj mm ) = 225ω kˆ
4
4
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129
Since the two revolutes at A and B have axes which intersect at O, this is a spherical
linkage; therefore triangle AOB rotates about O as a rigid link with point O stationary.
From this we see that the axis of rotation of link 3 passes through O and is perpendicular
to both VA and VB . Therefore ω3 = ω3ˆi and
V = ω × R = ω ˆi × 125ˆi + 225ˆj − 175kˆ = 175ω ˆj + 225ω kˆ
BA
3
BA
3
(
)
3
3
Substituting these into VB = VA + VBA or 225ω4kˆ = 4500ˆj + 175ω3ˆj + 225ω3kˆ , and
equating components gives
ω3 = ω 4 = − 4500 175 = −25.714ˆi rad/s ,
VBA = −4500ˆj − 5785kˆ mm/s ,
and
V = −5785kˆ mm/s .
B
BO4
4
BO4
A tAO2 = α 2 × R AO2 = 0
A tBO4 = α 4 × R BO4 = 225α 4kˆ
Remembering that link 3 rotates about point O we also find
A nAO = ω3 × ( ω3 × R AO ) = −115.725kˆ m/s 2 ,
A t = α × R = 175α y ˆi − (175α x + 75α z ) ˆj + 75α y kˆ
AO
3
AO
3
3
3
3
A nBO = ω3 × ( ω3 × R BO ) = −148.775ˆj m/s 2 ,
A tBO = α 3 × R BO = −225α 3z ˆi + 50α 3z ˆj + ( 225α 3x − 50α 3y ) kˆ
Substituting these into A A = A nAO2 = A nAO + A tAO
equating components gives
270 =
0.175α 3y
,
0=
− 0.175α 3x
− 0.075α 3z ,
and A B = A nBO4 = A nBO + A tBO , and
0=
− 0.225α 3z ,
−148.775 = −148.775
+ 0.050α 3z ,
+ 0.075α 3y , and
0.225α 4 =
0.225α 3x
From these we solve for α 3 = 1 543ˆj rad/s 2 and α 4 = −343ˆi rad/s 2 .
0 = −115.725
9.6
Solve Problem P9.5 using graphical techniques.
− 0.050α 3y .
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To find accelerations we first calculate
A nAO2 = −ω22 R AO2 = 270ˆi m/s 2 ,
A n = −ω 2 R = 148.775ˆj m/s 2 ,
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130
9.7
Solve Problem P9.5 using transformation matrix techniques.
Following the conventions of Sec. 9.6 and Fig. 9.11, the Denavit-Hartenberg parameter
values are:
i,j
ai , j
si , j
αi, j
θi, j
1,2
2,3
3,4
4,1
0
0
0
0
Using Eqs. (9.12) and (9.15) we find
0
0
0⎤
⎡1
⎢0 0.91914 0.39394 0 ⎥
⎥
T12 = ⎢
⎢0 −0.39394 0.91914 0 ⎥
⎢
⎥
0
0
1⎦
⎣0
-23.20°
-94.90°
-77.47°
-90.00°
φ1 = 0
φ2 = −101.54º
φ3 = −67.30º
φ4 = −90.00º
0
0
0
0
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To avoid confusion, the position and velocity solutions are shown in a separate figure on
the previous page. The acceleration solution is shown below. The results agree with those
of the previous analytic solution of Problem P9.5.
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⎡ −0.20005 −0.08369 0.97620 0 ⎤
⎢ −0.97979 0.01709 −0.19932 0 ⎥
⎥,
T23 = ⎢
⎢
−0.99635 −0.08542 0 ⎥
0
⎢
⎥
0
0
0
1⎦
⎣
⎡ −0.20005 −0.08369 0.97620 0 ⎤
⎢ −0.90056 −0.37679 −0.21685 0 ⎥
⎥,
T13 = ⎢
⎢ 0.38598 −0.92252
0
0⎥
⎢
⎥
0
0
0
1⎦
⎣
⎡ 0.38591 0.20015 0.90057 0 ⎤
⎡0 −1 0 0 ⎤
⎢ −0.92254 0.08372 0.37671 0 ⎥
⎢
⎥
⎥ , T14 = ⎢0 0 −1 0 ⎥ .
T34 = ⎢
⎢
⎢1 0 0 0 ⎥
−0.97618 0.21695 0 ⎥
0
⎢
⎥
⎢
⎥
0
0
0
1⎦
⎣
⎣0 0 0 1 ⎦
Next, from Eqs. (9.22) and (9.25),
−0.91914 0.39394 0 ⎤
0
⎡0 −1 0 0 ⎤
⎡
⎢1 0 0 0 ⎥
⎢ 0.91914
0
0
0 ⎥⎥
⎢
⎥
⎢
, D2 =
D1 =
,
⎢0 0 0 0 ⎥
⎢ −0.39394
0
0
0⎥
⎢
⎥
⎢
⎥
0
0
0
0⎦
⎣0 0 0 0 ⎦
⎣
−0.21685 0 ⎤
0
⎡ 0
⎡0 0 −1 0 ⎤
⎢ 0
⎥
⎢0 0 0 0 ⎥
−0.97620 0 ⎥
0
⎢
⎥.
, D4 = ⎢
D3 =
⎢0.21685 0.97620
⎢1 0 0 0 ⎥
0
0⎥
⎢
⎥
⎢
⎥
0
0
0⎦
⎣ 0
⎣0 0 0 0 ⎦
Now, from Eq. (9.26), D1φ1 + D2φ2 + D3φ3 + D4φ4 = 0 , we get the following set of equations:
0.97620 0 ⎤ ⎡φ2 ⎤ ⎡ 0 ⎤
⎡ 0
⎡0⎤
⎢ 0.39394 −0.21685 −1⎥ ⎢φ ⎥ = ⎢ 0 ⎥ φ = ⎢ 0 ⎥ rad/s
⎢
⎥⎢ 3⎥ ⎢ ⎥ 1 ⎢ ⎥
⎢⎣ 0.91914
⎢⎣60 ⎥⎦
0
0 ⎥⎦ ⎢⎣φ4 ⎥⎦ ⎢⎣ −1⎥⎦
From these we find φ2 = 65.275 rad/s , φ3 = 0 , and φ4 = 25.714 rad/s . From these
Eq. (9.27) we find the velocity matrices
0 25.714
0 25.714 0 ⎤
⎡ 0 60 0 0 ⎤
⎡ 0
⎡ 0
⎢ 0
⎥
⎢
⎢ −60 0 0 0 ⎥
0
0
0
0
0
0⎥
⎥ rad/s, ω3 = ⎢
ω2 = ⎢
rad/s, ω 4 = ⎢
⎢ 0 0 0 0⎥
⎢ −25.714 0
⎢ −25.714 0
0
0
0⎥
⎢
⎥
⎢
⎥
⎢
0
0
0
0
0⎦
⎣ 0 0 0 0⎦
⎣ 0
⎣ 0
These can be used with Eq. (9.28) to find the velocities of all moving points.
Acceleration analysis follows similar steps. From Eq. (9.29) we get the following
equations:
and
0⎤
0 ⎥⎥
rad/s.
0⎥
⎥
0⎦
set of
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131
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132
0.97620 0 ⎤ ⎡φ2 ⎤ ⎡ −1 543⎤
⎡ 0
⎢ 0.39394 −0.21685 −1⎥ ⎢φ ⎥ = ⎢ 0 ⎥ rad/s 2
⎢
⎥⎢ 3⎥ ⎢
⎥
⎢⎣ 0.91914
0
0 ⎦⎥ ⎣⎢φ4 ⎦⎥ ⎣⎢ 0 ⎦⎥
From these we find φ2 = 0 , φ3 = −1 580 rad/s 2 , and φ4 = 343 rad/s 2 . From these and Eq.
(9.30) we find the acceleration matrices
0
0⎤
⎡0 0 0 0⎤
⎡0 0
⎡ 0 0 343 0 ⎤
⎢ 0 0 0 0⎥
⎢0 0 0 0⎥
⎢0 0 −1 543 0 ⎥
2
⎢
⎥
⎥ rad/s2.
⎢
⎥
α2 =
rad/s , α 4 = ⎢
,α =
⎢ −343 0 0 0 ⎥
⎢ 0 0 0 0 ⎥ 3 ⎢0 1 543
0
0⎥
⎢
⎥
⎢
⎥
⎢
⎥
0
0⎦
⎣ 0 0 0 0⎦
⎣0 0
⎣0 0 0 0⎦
These can be used with Eq. (9.31) to find the accelerations of all moving points.
Solve Problem P9.5 except with θ 2 = 90° .
The position vectors for this new position are:
x ˆ
y ˆ
z ˆ
R BA = RBA
i + RBA
j + RBA
k,
R AO2 = −75ˆj mm ,
R = 225sin θ ˆj + 225cos θ kˆ .
4
BO4
R O4O2 = 50ˆi − 175kˆ mm ,
4
Substituting these into R AO2 + R BA = R O4O2 + R BO4 and separating components gives
y
z
x
RBA
= 50 , RBA
= 225sin θ 4 + 75 , RBA
= 225cos θ 4 − 175 , and squaring and adding these
gives
2
502 + 2252 − 3150 cos θ 4 + 1752 + 1350sin θ 4 + 752 = RBA
= 3875
If we now define χ = tan (θ 4 2 ) and use the identities cos θ 4 = (1 − χ 2 ) (1 + χ 2 ) and
sin θ 4 = 2 χ (1 + χ 2 ) , then the above equation can be reduced to
2850 χ 2 + 2700 χ − 3450 = 0
The root of interest here is χ = 0.72419 which corresponds to θ 4 = 71.823° . With this
the four position vectors are
R O4O2 = 50ˆi − 175kˆ mm ,
R BA = 50ˆi + 288.755ˆj − 104.8kˆ mm ,
R AO2 = −75ˆj mm ,
R = 213.7751ˆj + 70.2kˆ mm
BO4
The velocity analysis proceeds as in problem P9.5:
VA = VAO2 = ω 2 × R AO2 = −60kˆ rad/s × −75ˆj mm = −4500ˆi mm/s
VB = VBO4 = ω 4 × R BO4
(
) (
)
= (ω ˆi ) × ( 213.775ˆj + 70.2kˆ mm ) = −70.2ω ˆj + 213.775ω kˆ
4
4
4
Since the two revolutes at A and B have axes which intersect at O, this is a spherical
linkage; therefore triangle AOB rotates about O as a rigid link with point O stationary.
From this we see that the axis of rotation of link 3 passes through O and is perpendicular
to both VA and VB . Therefore ω3 = 0.95010ω3ˆj + 0.31195ω3kˆ and
V = ω × R = −189.65ω ˆi + 15.6ω ˆj − 47.5ω kˆ
BA
3
BA
3
3
3
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9.8
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133
Substituting these into VB = VA + VBA and equating components gives ω3 = −23.726 rad/s
and ω = 5.273 rad/s , and from these we get ω = −22.542ˆj − 7.401kˆ rad/s ,
3
4
ω 4 = 5.273ˆi rad/s , VBA = 4500ˆi − 370ˆj + 1127kˆ mm/s , and VB = −370ˆj + 1127kˆ mm/s .
To find accelerations we first calculate
A nAO2 = −ω22 R AO2 = 270ˆj m/s 2 ,
A n = −ω 2 R = −5943ˆj − 1951kˆ mm/s 2 ,
BO4
4
BO4
A tAO2 = α 2 × R AO2 = 0 ,
A tBO4 = α 4 × R BO4 = −70.2α 4 ˆj + 213.775α 4kˆ
Remembering that link 3 rotates about point O we also find
A nAO = ω3 × ( ω3 R AO ) = 33.3ˆj − 101.45kˆ m/s 2 ,
A t = α × R = (175α y + 75α z ) ˆi − 175α x ˆj − 75α x kˆ ,
3
AO
A
n
BO
3
AO
3
3
3
= ω3 × ( ω3 × R BO ) = −28.15ˆi m/s ,
2
Substituting these into
A A = A nAO2 = A nAO + A tAO
equating components gives
0=
175α 3y +75α 3z , 0
270, 000 = 33,300 −175α 3x
0 = −101, 450 −75α 3x
,
and A B = A nBO4 = A nBO + A tBO , and
+70.2α 3y
= −28150
−5943 − 70.2α 4 =
−70.2α 3x
− 1951 + 213.775α 4 =
+213.775α 3x −50α 3y
From these we solve for α 3 = −1 302ˆi + 49ˆj − 115kˆ rad/s 2 and α 4 = −1 304ˆi rad/s 2 .
9.9
,
−212.775α 3z ,
+50α 3z ,
.
Determine the advance-to-return time ratio for Problem P9.5. What is the total angle of
oscillation of link 4?
Time ratio = 181.1/178.9 = 1.012
Δθ 4 = 46.5°
Ans.
Ans.
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A tBO = α 3 × R BO = ( 70.2α 3y − 213.775α 3z ) ˆi + ( −70.2α 3x + 50α 3z ) ˆj + ( 213.775α 3x − 50α 3y ) kˆ
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134
For the spherical RRRR linkage shown, determine whether the crank is free to turn
through a complete revolution. If so, find the angle of oscillation of link 4 and the
advance-to-return time ratio.
Time ratio = 187/173 = 1.081
Δθ 4 = 38°
9.11
Ans.
Ans.
Use vector algebra to make complete velocity and acceleration analyses of the linkage at
the position specified.
The position vectors were found from a graphical analysis done on a CAD software
system; (see problem P9.12). For the position θ 2 = 120° they are as follows:
R = 236.975ˆj − 111.744kˆ mm ,
R
= 225ˆi − 150kˆ mm ,
BO4
O4O2
R AO2 = −18.750ˆi + 32.476ˆj mm ,
R BA = 243.750ˆi + 204.499ˆj − 261.744kˆ mm .
The velocity analysis for this position, with ω 2 = 36kˆ rad/s , proceeds as follows:
VA = ω 2 × R AO2 = −1.169ˆi − 0.675ˆj m/s , VB = ω 4 ˆi × R BO4 = 0.111744ω 4 ˆj + 0.236975ω 4kˆ ,
Since all revolute axes intersect at O, this is a spherical mechanism and triangle AOB
(link 3) rotates about O. Thus the axis of rotation of link 3 passes through O and is
perpendicular to VA and VB . Calling this axis uˆ , ( ω3 = ω3uˆ ) ,
uˆ = ( V × V ) V × V = −0.462932ˆi + 0.801731ˆj − 0.378051kˆ
B
A
B
A
VBA = ω3 × R BA = −0.132537ω3ˆi − 0.213320ω3ˆj − 0.290091ω3kˆ
Substituting these into VB = VA + VBA , equating components, and solving gives
ω3 = −8.820 rad/s and ω 4 = 10.797 rad/s . From these
ω = 4.083ˆi − 7.071ˆj + 3.334kˆ rad/s , and ω = 10.797ˆi rad/s .
3
4
For acceleration analysis we first calculate
A nAO2 = ω2 × ω2 × R AO2 = 24.300ˆi − 42.089ˆj rad/s 2 , A tAO2 = α 2 × R AO2 = 0 ,
(
)
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9.10
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135
(
)
A nBO4 = ω4 × ω4 × R BO4 = −27.626ˆj + 13.027kˆ rad/s 2 ,
A tBO4 = α 4 ˆi × R BO4 = 0.111744α 4 ˆj + 0.236975α 4kˆ
Remembering that link 3 rotates about point O we also find
A nAO = ω3 × ( ω3 × R AO ) = 2.251ˆi − 3.898ˆj − 11.023kˆ rad/s 2 ,
A n = ω × ( ω × R ) = −22.117ˆi − 10.447ˆj + 4.926kˆ rad/s 2 ,
BO
A
t
AO
3
3
BO
= α 3 × R AO = ( 0.150α 3y − 0.032α 3z ) ˆi − ( 0.019α 3z + 0.150α 3x ) ˆj + ( 0.032α 3x + 0.019α 3y ) kˆ ,
A tBO = α 3 × R BO = − ( 0.112α 3y + 0.237α 3z ) ˆi + ( 0.225α 3z + 0.112α 3x ) ˆj + ( 0.237α 3x − 0.225α 3y ) kˆ
Next, from A A = A nAO2 + A tAO2 = A nAO + A tAO and A B = A nBO4 + A tBO4 = A nBO + A tBO , we
separate components and obtain
24.300 = 2.251 + 0.150α 3y − 0.032α 3z ;
0 = −22.117 − 0.112α 3y − 0.237α 3z ;
0 = −11.023 + 0.032α 3x + 0.019α 3y ; 13.027 + 0.237α 4 = 4.926 + 0.237α 3x − 0.225α 3y ;
From these α = 273ˆi + 115ˆj − 148kˆ rad/s 2 and α = 130ˆi rad/s 2 .
3
9.12
4
Solve Problem P9.11 using graphical techniques.
The position and velocity solutions are shown here with the acceleration solution on the
next page. The results verify those of the analytical solution in problem P9.11.
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−42.089 = −3.898 − 0.150α 3x − 0.019α 3z ; −27.626 + 0.112α 4 = −10.447 + 0.112α 3x + 0.225α 3z ;
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9.13
Solve Problem P9.11 using transformation matrix techniques.
The Denavit-Hartenberg parameters are:
a12 = 0, α12 = −14.04°, θ12 = φ1 = −60.00°, s12 = 0,
a23 = 0, α 23 = −104.41°, θ 23 = φ2 = −69.52°, s23 = 0,
a34 = 0, α 34 = −49.34°, θ 34 = φ3 = −93.18°, s34 = 0,
a41 = 0, α 41 = −90.00° θ 41 = φ4 = −64.75° s41 = 0.
From Eqs. (9.12) and (9.15) the transformation matrices are:
⎡ 0.50000 0.84015 0.21005 0 ⎤
⎢ −0.86603 0.48506 0.12127 0 ⎥
⎥
T12 = ⎢
⎢
0
−0.24260 0.97014 0 ⎥
⎢
⎥
0
0
0
1⎦
⎣
⎡ −0.61206 −0.39312 0.68618 0 ⎤
⎢ −0.75747 0.04214 −0.65151 0 ⎥
⎥
T13 = ⎢
⎢ 0.22720 −0.91852 −0.32356 0 ⎥
⎢
⎥
0
0
0
1⎦
⎣
⎡ 0.42650 −0.90449 0 0 ⎤
⎢ 0
0
−1 0 ⎥⎥
⎢
T14 =
⎢ 0.90449 0.42650 0 0 ⎥
⎢
⎥
0
0 1⎦
⎣ 0
Next, from Eqs. (9.22) and (9.25),
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136
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137
−1 0 0 ⎤
0
−0.97015 0.12127 0 ⎤
⎡
⎥
⎢
0 0 0⎥
0.97015
0
−0.21005 0 ⎥⎥
D2 = ⎢
⎢ −0.12127 0.21005
0 0 0⎥
0
0⎥
⎥
⎢
⎥
0 0 0⎦
0
0
0
0⎦
⎣
0
0.32357 −0.65151 0 ⎤
⎡ 0 0 −1 0 ⎤
⎡
⎢0 0 0 0⎥
⎢ −0.32357
⎥
0
−0.68618 0 ⎥
⎥
D3 = ⎢
D4 = ⎢
⎢1 0 0 0 ⎥
⎢ 0.65151 0.68618
0
0⎥
⎢
⎥
⎢
⎥
0
0
0
0⎦
⎣0 0 0 0⎦
⎣
and from Eq. (9.26) we get the following set of equations
⎡ 0.21005 0.68618 0 ⎤ ⎡φ2 ⎤
⎡0⎤
⎡ 0 ⎤
⎢
⎥
⎢
⎥
⎢ ⎥
⎢
⎥
⎢ 0.12127 −0.65151 −1⎥ ⎢φ3 ⎥ = − ⎢ 0⎥ φ1 = ⎢ 0 ⎥ rad/s
⎢⎣ 0.97015 −0.32357 0 ⎥⎦ ⎣⎢φ4 ⎦⎥
⎢⎣1 ⎥⎦
⎢⎣ −36 ⎥⎦
from which we find φ2 = −33.670 rad/s , φ3 = 10.307 rad/s , and φ4 = −10.798 rad/s .
With these values and Eqs. (9.27) we find the velocity matrices
0 −10.798 0 ⎤
⎡ 0 −36 0 0 ⎤
⎡ 0 −3.335 −4.083 0 ⎤
⎡ 0
⎢36 0 0 0 ⎥
⎢3.335
⎥
⎢
0
0
0
0 ⎥⎥
0
7.072 0 ⎥
⎢
⎥
⎢
⎢
,
=
=
,
ω2
ω
ω4 =
.
⎢ 0 0 0 0 ⎥ 3 ⎢ 4.083 −7.072
⎢10.798 0
0
0⎥
0
0⎥
⎢
⎥
⎢
⎥
⎢
⎥
0
0
0⎦
0
0
0⎦
⎣ 0
⎣ 0 0 0 0⎦
⎣ 0
These can be used with Eq. (9.28) to find the velocities of all moving points.
The acceleration analysis follows parallel steps using Eqs. (9.29) and (9.30)
φ2 = −152 rad/s 2
⎡ 0.21005 0.68618 0 ⎤ ⎡φ2 ⎤ ⎡ −183.0 ⎤
⎢ 0.12127 −0.65151 −1⎥ ⎢φ ⎥ = ⎢ 254.6 ⎥ rad/s 2 ;
φ3 = −220 rad/s 2
⎢
⎥⎢ 3⎥ ⎢
⎥
φ4 = −130 rad/s 2
⎣⎢ 0.97015 −0.32357 0 ⎦⎥ ⎣⎢φ4 ⎦⎥ ⎣⎢ −76.4 ⎦⎥
1
⎡0 0 0 0⎤
⎡ 0 0 −130 0 ⎤
⎡ 0 148 −273 0 ⎤
⎢0 0 0 0⎥
⎢ −148 0 −115 0 ⎥
⎢ 0 0 0 0⎥
⎢
⎥
⎢
⎥
⎥.
,
α2 =
α3 =
, α4 = ⎢
⎢130 0 0 0 ⎥
⎢0 0 0 0⎥
⎢ 273 115 0 0 ⎥
⎢
⎥
⎢
⎥
⎢
⎥
0
0 0⎦
⎣0 0 0 0⎦
⎣ 0 0 0 0⎦
⎣ 0
These results correlate with and verify those of Problems P9.11 and P9.12.
9.14
The figure shows the top, front, and auxiliary views of a spatial slider-crank RGGP
linkage. In the construction of many such mechanisms provision is made to vary the
angle β; thus the stroke of slider 4 becomes adjustable from zero, when β = 0, to twice
the crank length, when β = 90°. With β = 30°, use vector algebra to make a complete
velocity analysis of the linkage at the given position.
The position vectors were found from a graphical analysis done on a CAD software
system; (see problem P9.15). For the position θ 2 = 240° they are as follows:
R = 164.72ˆi mm ,
R = 21.65ˆi + 37.5ˆj − 25 000kˆ mm ,
AO
BO
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⎡0
⎢1
D1 = ⎢
⎢0
⎢
⎣0
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138
R BA = 143.07ˆi − 37.5ˆj + 25kˆ mm .
The velocity analysis for this position, with ω 2 = 24 rad/s , proceeds as follows:
ω2 = 20.784610ˆi − 12.0ˆj rad/s , VA = ω2 × R AO = 300ˆi + 519.61ˆj + 1039.23kˆ mm/s ,
VBA = ω3 × R BA = (ω3y + 37.5ω3z ) ˆi + (143.07ω3z − ω3x ) ˆj + ( −37.5ω3x − 143.07ω3y ) kˆ , VB = VB ˆi .
Substituting these into VB = VA + VBA and separating into components,
VB = 300
0.0 = 519.6 −
+
ω3y + 37.5
ω3x
+ 143.07ω3z
ω3z
0.0 = 143.07ω3x − 37.5ω3y + 25ω3z
The four equations can now be solved to give
ω3 = 2.309 401ˆi + 6.658 519ˆj − 3.228 373kˆ rad/s and VB = 345.4ˆi mm/s
9.15
Solve Problem P9.14 using graphical techniques.
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0.0 = 1039.2 − 37.5ω3x − 143.07ω3y
However, this is a set of only three equations and there are four unknown variables. This
results from the fact that the linkage has two degrees of freedom and the connecting rod
is free to rotate about the axis AB. If we assume that this second “idle freedom” is
inactive, then ω3 iR BA = 0 .
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139
Solve Problem P9.14 using transformation matrix techniques.
One choice for the Denavit-Hartenberg parameters gives:
α12 = 90° ,
θ12 = φ1 = −30° ,
s12 = 0 ,
a12 = 0 ,
a14 = 0 ,
α14 = β = 30° ,
θ14 = 0 ,
s14 = φ4 .
From Eqs. (9.12) and (9.11) we obtain
⎡ cos φ1 0 sin φ1 0 ⎤
⎡ 0 ⎤ ⎡ 2sin φ1 ⎤
⎢ 0 ⎥ ⎢ −2 cos φ ⎥
⎢ sin φ 0 − cos φ 0 ⎥
1
1
1⎥
⎢
⎥
RA = T12 ⎢ ⎥ = ⎢
;
,
T12 =
⎢2⎥ ⎢ 0 ⎥
⎢ 0 1
0
0⎥
⎢ ⎥ ⎢
⎥
⎢
⎥
0
1⎦
⎣1 ⎦ ⎣ 1 ⎦
⎣ 0 0
0
0
0
⎡1 0
⎤
⎡0 ⎤ ⎡
⎤
⎢0 ⎥ ⎢ −φ sin β ⎥
⎢ 0 cos β − sin β −φ sin β ⎥
4
⎥.
⎥,
RB = T14 ⎢ ⎥ = ⎢ 4
T14 = ⎢
⎢0 ⎥ ⎢ φ4 cos β ⎥
⎢ 0 sin β cos β φ4 cos β ⎥
⎢ ⎥ ⎢
⎥
⎢
⎥
0
1
1
⎣1 ⎦ ⎣
⎦
⎣0 0
⎦
From these and the length of the connecting rod we write
2
2
2
t
2
RBA
= ( RB − RA ) ( RB − RA ) = ( −2sin φ1 ) + ( 2 cos φ1 − φ4 sin β ) + (φ4 cos β ) = 1502
This reduces to φ42 − 4φ4 sin β cos φ1 − 32 = 0 which has a solution
φ4 = 2sin β cos φ1 + 4sin 2 β cos 2 φ1 + 32
By differentiating the above equation with respect to time we obtain
2φ4φ4 − 4φ4 sin β cos φ1 + 4φ1φ4 sin β sin φ1 = 0
which has for a solution
2φ4 sin β sin φ1
φ4 =
φ
( 2sin β cos φ1 − φ4 ) 1
and, from Eqs. (9.22), (9.23), (9.25), and (9.27),
0 ⎤
⎡0 −1 0 0 ⎤
⎡0 0 0
⎢1 0 0 0 ⎥
⎢ 0 0 0 sin β ⎥
⎥,
⎥;
D4 = ⎢
D1 = ⎢
⎢0 0 0 0 ⎥
⎢ 0 0 0 − cos β ⎥
⎢
⎥
⎢
⎥
0 ⎦
⎣0 0 0
⎣0 0 0 0 ⎦
⎡ 0 −φ1
⎢
φ 0
ω2 = ⎢ 1
⎢0 0
⎢
⎣⎢ 0 0
0 0⎤
⎥
0 0⎥
,
0 0⎥
⎥
0 0 ⎦⎥
⎡0
⎢
0
ω4 = ⎢
⎢0
⎢
⎣⎢0
0 0 − sin βφ4 ⎤
⎥
0 0 cos βφ4 ⎥
.
⎥
0 0
⎥
0 0
0
⎦⎥
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9.16
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140
β = 30° , φ1 = −30° , and φ1 = 24 rad/s , the above formulae give
φ4 = 164.7 mm , φ4 = 345.4 mm/s , and
Now, with
0
⎡
⎤
⎡1039.2 mm/s ⎤
⎢ −82.35 mm ⎥
⎢ −600 mm/s ⎥
⎢
⎥
⎥,
RB =
, RA = ⎢
⎢142.65 mm ⎥
⎢
⎥
0
⎢
⎥
⎢
⎥
1
0
⎣
⎦
⎣
⎦
0
⎡
⎤
⎢ −172.7 mm/s ⎥
⎥.
RB = ⎢
⎢ 299.125 mm/s ⎥
⎢
⎥
0
⎣
⎦
These results agree with those of Probs. P9.14 and P9.15 once the Denavit-Hartenberg
coordinate directions are considered. Note, however, that the loop-closure equation was
never used. No coordinate system was fixed to link 3 and no velocity of link 3 was
found. This is because of our lack of information about the degree of freedom
representing the spin of link 3 about the line AB.
9.17
Solve Problem P9.14 with β = 60° using vector algebra.
The position vectors were found from a graphical analysis done on a CAD software
system; (see problem P9.18). For the position θ 2 = 240° they are as follows:
R = 37.5ˆi + 21.65ˆj − 25kˆ mm ,
R = 183.8ˆi mm ,
AO
BO
R BA = 146.3ˆi − 21.65ˆj + 25kˆ mm .
The velocity analysis for this position, with ω 2 = 24 rad/s , proceeds as follows:
ω = 12.0ˆi − 20.784 610ˆj rad/s , V = ω × R = 519.6ˆi + 300ˆj + 1039.23kˆ mm/s ,
2
VBA = ω3 × R BA
A
2
AO
= (ω + 21.65ω ) ˆi + (146.3ω3z − ω3x ) ˆj + ( −21.65ω3x − 146.3ω3y ) kˆ , VB = VB ˆi .
y
3
z
3
Substituting these into VB = VA + VBA and separating into components we get
ω3y + 21.65ω3z
VB = 519.6 +
0.0 = 300
−
ω3x
+ 146.3ω3z
0.0 = 1039.23 − 21.65ω3x − 146.3ω3y
However, this is a set of only three equations and there are four unknown variables. This
results from the fact that the linkage has two degrees of freedom and the connecting rod
is free to rotate about the axis AB. If we assume that this second “idle freedom” is
inactive, then ω3 iR BA = 0 .
0.0 = 146.3ω3x − 21.65ω3y + ω3z
The four equations can now be solved to give
ω3 = 1.333 333ˆi + 6.905 692ˆj − 1.822 629kˆ rad/s and VB = 652.82ˆi mm/s
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⎡ −25 mm ⎤
⎢ 43.3 mm ⎥
⎥,
RA = ⎢
⎢
⎥
0
⎢
⎥
1
⎣
⎦
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9.18
Solve Problem P9.14 with β = 60° using graphical techniques.
9.19
Solve Problem P9.14 with β = 60° using transformation matrix techniques.
One choice for the Denavit-Hartenberg parameters gives:
α12 = 90° ,
θ12 = φ1 = −30° ,
s12 = 0 ,
a12 = 0 ,
a14 = 0 ,
α14 = β = 60° ,
θ14 = 0 ,
s14 = φ4 .
From Eqs. (9.12) and (9.11) we obtain
⎡ cos φ1 0 sin φ1 0 ⎤
⎡ 0 ⎤ ⎡ 2sin φ1 ⎤
⎢ sin φ 0 − cos φ 0 ⎥
⎢ 0 ⎥ ⎢ −2 cos φ ⎥
1
1
1⎥
⎢
⎥
T12 =
RA = T12 ⎢ ⎥ = ⎢
;
,
⎢2⎥ ⎢ 0 ⎥
⎢ 0 1
0
0⎥
⎢ ⎥ ⎢
⎥
⎢
⎥
0
1⎦
⎣1 ⎦ ⎣ 1 ⎦
⎣ 0 0
0
0
0
⎡1 0
⎤
⎡0 ⎤ ⎡
⎤
⎢ 0 cos β − sin β −φ sin β ⎥
⎢0 ⎥ ⎢ −φ sin β ⎥
4
⎥.
⎥,
RB = T14 ⎢ ⎥ = ⎢ 4
T14 = ⎢
⎢0 ⎥ ⎢ φ4 cos β ⎥
⎢ 0 sin β cos β φ4 cos β ⎥
⎢
⎥
⎢ ⎥ ⎢
⎥
1
0
1
⎣1 ⎦ ⎣
⎦
⎣0 0
⎦
From these and the length of the connecting rod we write
2
2
2
t
2
= ( RB − RA ) ( RB − RA ) = ( −2sin φ1 ) + ( 2 cos φ1 − φ4 sin β ) + (φ4 cos β ) = 1502
RBA
This reduces to φ42 − 4φ4 sin β cos φ1 − 32 = 0 which has a solution
φ4 = 2sin β cos φ1 + 4sin 2 β cos 2 φ1 + 32
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141
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142
⎡ 0 −φ1
⎢
φ 0
ω2 = ⎢ 1
⎢0 0
⎢
⎣⎢ 0 0
0 0⎤
⎥
0 0⎥
,
0 0⎥
⎥
0 0 ⎦⎥
⎡0
⎢
0
ω4 = ⎢
⎢0
⎢
⎣⎢0
0 0 − sin βφ4 ⎤
⎥
0 0 cos βφ4 ⎥
.
⎥
0 0
⎥
0 0
0
⎦⎥
β = 60° , φ1 = −30° , and φ1 = 24 rad/s , the above formulae give
φ4 = 183.8 mm , φ4 = 652.8 mm/s , and
Now, with
⎡ −25 mm ⎤
⎢ 43.3 mm ⎥
⎥,
RA = ⎢
⎢
⎥
0
⎢
⎥
1
⎣
⎦
0
⎡
⎤
⎢ −159.175 mm ⎥
⎥,
RB = ⎢
⎢ 91.9 mm ⎥
⎢
⎥
1
⎣
⎦
⎡1039.225 mm/s ⎤
⎢ −600 mm/s ⎥
⎥,
RA = ⎢
⎢
⎥
0
⎢
⎥
0
⎣
⎦
0
⎡
⎤
⎢ −565.35 mm/s ⎥
⎥.
RB = ⎢
⎢ 326.4 mm/s ⎥
⎢
⎥
0
⎣
⎦
These results agree with those of Probs. P9.17 and P9.18 once the Denavit-Hartenberg
coordinate directions are considered.
9.20
The figure shows the top, front, and profile views of an RGRC crank and oscillatingslider linkage. Link 4, the oscillating slider, is rigidly attached to a round rod that rotates
and slides in the two bearings. (a) Use the Kutzbach criterion to find the mobility of this
linkage. (b) With crank 2 as the driver, find the total angular and linear travel of link 4.
(c) Write the loop-closure equation for this mechanism and use vector algebra to solve it
for all unknown position data.
a)
The RGRC linkage has n = 4, j1 = 2, j2 = 1, j3 = 1. The Kutzbach criterion gives
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By differentiating the above equation with respect to time we obtain
2φ4φ4 − 4φ4 sin β cos φ1 + 4φ1φ4 sin β sin φ1 = 0
which has for a solution
2φ4 sin β sin φ1
φ4 =
φ
( 2sin β cos φ1 − φ4 ) 1
and, from Eqs. (9.22), (9.23), (9.25), and (9.27),
0 ⎤
⎡0 −1 0 0 ⎤
⎡0 0 0
⎢1 0 0 0 ⎥
⎢ 0 0 0 sin β ⎥
⎥,
⎥;
D4 = ⎢
D1 = ⎢
⎢0 0 0 0 ⎥
⎢ 0 0 0 − cos β ⎥
⎢
⎥
⎢
⎥
0 ⎦
⎣0 0 0
⎣0 0 0 0 ⎦
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143
m = 6 ( n − 1) − 5 j1 − 4 j2 − 3 j3 = 6(4 − 1) − 5(2) − 4(1) − 3(1) = 1
Ans.
Since vectors do not show the rotation θ4, matrix methods were necessary and are
shown in problem P9.23. See c) for the vector solution. Together, they show:
−135° < θ 4 < −45° ;
Δθ 4 = 90° .
Ans.
182.85 mm < yB < 382.85 mm ;
ΔyB = 200 mm .
Ans.
R B + R AB = R Q + R AQ
y ˆj + x ˆi + y ˆj + z kˆ = −100ˆi − 100sin θ ˆj + 100 cos θ kˆ
b)
c)
B
AB
AB
2
AB
2
Separating components,
x AB = −100 ,
yB + y AB = −100sin θ 2 ,
and, from the length of link 3,
2
2
2
x AB
+ y 2AB + z AB
= RAB
( −100 ) + ( −100sin θ 2 − yB ) + (100 cos θ 2 )
2
2
2
z AB = 100 cos θ 2
= 3002
9.21
yB = −100sin θ 2 + 100 175 + sin 2 θ 2
Ans.
R AB = −100ˆi − 100 175 + sin 2 θ 2 ˆj + 100 cos θ 2kˆ
For θ 2 = 40° , R B = yB ˆj = 208ˆj mm , R AB = −100ˆi − 272.275ˆj + 76.6kˆ mm .
Ans.
Ans.
Use vector algebra to find VB, ω3, and ω4 for Problem P9.20.
First we identify for θ 2 = 40° , that R Q = −100ˆi in , R AQ = −64.275ˆj + 76.6kˆ mm ,
R B = 208ˆj mm , and R AB = −100ˆi − 272.275ˆj + 76.6kˆ m .
Next we note that the rotation axis of the revolute at B is j × R = 76.6ˆi + 100kˆ and,
AB
normalizing this, we can express the apparent angular velocity ω3/ 4 axis by
ˆ 3/ 4 = 0.608ˆi + 0.794kˆ . Then the angular velocity of link 3 can be written as:
ω
ω3 = ω 4 + ω3/ 4 = 0.608ω3/ 4 ˆi + ω 4 ˆj + 0.794ω3/ 4kˆ .
With this done, we can find
V = ω × R = 3677ˆj + 3085.375kˆ mm/s ,
V = V ˆj ,
and
A
2
AQ
B
B
VAB = ω3 × R AB = (76.6ω4 + 216.15ω3/ 4 )ˆi − 125.975ω3/ 4 ˆj + (100ω4 − 165.575ω3/ 4 ) kˆ . Now,
setting VB + VAB = VA and separating components, we get the following equations
76.6ω4 + 216.15ω3/ 4 = 0
VB
− 125.975ω3/ 4 = 3677
100ω4 − 165.575ω3/ 4 = 3085.375
which can be solved to give ω 4 = 19.444 rad/s , ω3/ 4 = −6.891 rad/s , VB = 2808.95 mm/s .
ω = −4.190ˆi + 19.444ˆj − 5.471kˆ rad/s , ω = 19.444ˆj rad/s , V = 2808.95ˆj mm/s . Ans.
4
3
B
Note that, if ω3 were written as ω3 = ω ˆi + ω3y ˆj + ω3z kˆ , then the above set of simultaneous
equations would have would have four unknowns and could not be solved.
x
3
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yB2 + 200sin θ 2 yB − 2800 = 0
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9.22
Solve Problem P9.21 using graphical techniques.
9.23
Solve Problem P9.21 using transformation matrix techniques.
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144
The Denavit-Hartenberg parameters from the global coordinate system to joint A are:
a12 = 0 ,
α12 = 0 ,
θ12 = θ 2 = 40° ,
s12 = −100 mm ,
and, proceeding in the other direction around the loop, to joint A, they are:
α16 = −90° ,
θ16 = 0 ,
s16 = 0 ,
a16 = 0 ,
a65 = 0 ,
α 65 = 0 ,
θ 65 = 0 ,
s65 = yB ,
a54 = 0 ,
α 54 = 0 ,
θ 54 = θ 4 ,
s54 = 0 ,
a43 = 0 ,
α 43 = 90° ,
θ 43 = 0 ,
s43 = 0 ,
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145
⎡1
⎢0
T65 = ⎢
⎢0
⎢
⎣0
0
1
0
0
⎡ cos θ 4
⎢ sin θ
4
T54 = ⎢
⎢ 0
⎢
⎣ 0
⎡1
⎢0
T43 = ⎢
⎢0
⎢
⎣0
⎡1 0
⎢0 0
T15 = ⎢
⎢0 −1
⎢
⎣0 0
0⎤
0 ⎥⎥
,
yB ⎥
⎥
1⎦
0
0
1
0
− sin θ 4
cosθ 4
0
0
0 0
0 −1
1 0
0 0
0
0
1
0
⎡ cos θ 4
⎢ 0
T14 = ⎢
⎢ − sin θ 4
⎢
⎣ 0
0⎤
0 ⎥⎥
,
0⎥
⎥
1⎦
⎡ cos θ 4
⎢ 0
T13 = ⎢
⎢ − sin θ 4
⎢
⎣ 0
0⎤
0 ⎥⎥
,
0⎥
⎥
1⎦
⎡ cos θ 3
⎢ sin θ
3
T3 B = ⎢
⎢ 0
⎢
⎣ 0
− sin θ 3
cos θ 3
0
0
⎡0
⎢ −1
TBA = ⎢
⎢0
⎢
⎣0
0 0 ⎤
⎡ sin θ 3 cos θ 4
⎥
⎢ − cosθ
0 −12 ⎥
3
, T1 A = ⎢
⎢ − sin θ 3 sin θ 4
1 0 ⎥
⎥
⎢
0
0 1 ⎦
⎣
1
0
0
0
0
0
1
0
0⎤
0 ⎥⎥
,
0⎥
⎥
1⎦
⎡ cosθ 3 cos θ 4
⎢
sin θ 3
T1B = ⎢
⎢ − cos θ 3 sin θ 4
⎢
0
⎣
0
1
0
0
0⎤
yB ⎥⎥
,
0⎥
⎥
1⎦
0
1
0
0
0⎤
yB ⎥⎥
,
0⎥
⎥
1⎦
0 sin θ 4
1
0
0 cos θ 4
0
0
0⎤
yB ⎥⎥
,
0⎥
⎥
1⎦
− sin θ 4
0
− cos θ 4
0
− sin θ 3 cos θ 4
cos θ 3
sin θ 3 sin θ 4
0
cos θ 3 cos θ 4
sin θ 3
− cos θ 3 sin θ 4
0
sin θ 4
0
cosθ 4
0
sin θ 4
0
cosθ 4
0
0⎤
yB ⎥⎥
,
0⎥
⎥
1⎦
12sin θ 3 cosθ 4 ⎤
yB − 12 cosθ 3 ⎥⎥
.
−12sin θ 3 sin θ 4 ⎥
⎥
1
⎦
At the terminations of these two paths, the position of point A must agree. Therefore
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α3B = 0 ,
θ3B = θ3 ,
s3 B = 0 ,
a3 B = 0 ,
aBA = 300 mm ,
α BA = 0 ,
θ BA = −90 ,
sBA = 0 .
From Eqs. (9.12) we obtain for the first path to A:
⎡ cos θ 2 − sin θ 2 0 0 ⎤
⎢ sin θ
cosθ 2 0 0 ⎥⎥
2
⎢
T12 =
,
⎢ 0
0
1 −4 ⎥
⎢
⎥
0
0 1⎦
⎣ 0
and along the other path to joint A, from Eqs. (9.12) and (9.15), we get:
⎡1 0 0 0 ⎤
⎢0 0 1 0⎥
⎥,
T16 = ⎢
⎢ 0 −1 0 0 ⎥
⎢
⎥
⎣0 0 0 1 ⎦
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146
⎡ cos θ 2 − sin θ 2 0 0 ⎤ ⎡ −4 ⎤ ⎡ sin θ 3 cosθ 4
⎢ sin θ
cos θ 2 0 0 ⎥⎥ ⎢⎢ 0 ⎥⎥ ⎢⎢ − cos θ 3
2
⎢
=
⎢ 0
0
1 −4 ⎥ ⎢ 0 ⎥ ⎢ − sin θ 3 sin θ 4
⎢
⎥⎢ ⎥ ⎢
0
0 1 ⎦⎣ 1 ⎦ ⎣
0
⎣ 0
⎡ −4 cosθ 2 ⎤ ⎡ 12sin θ 3 cos θ 4 ⎤
⎢ −4sin θ ⎥ ⎢ y − 12 cos θ ⎥
2⎥
3 ⎥
⎢
=⎢ B
⎢ −4 ⎥ ⎢ −12sin θ 3 sin θ 4 ⎥
⎢
⎥ ⎢
⎥
1
1
⎣
⎦ ⎣
⎦
cosθ 3 cosθ 4
sin θ 3
− cos θ 3 sin θ 4
0
sin θ 4
0
cos θ 4
0
12sin θ 3 cosθ 4 ⎤ ⎡ 0⎤
yB − 12 cosθ 3 ⎥⎥ ⎢⎢ 0⎥⎥
−12sin θ 3 sin θ 4 ⎥ ⎢ 0⎥
⎥⎢ ⎥
1
⎦ ⎣1 ⎦
Noting from the figure that −90° < θ 3 < 0 and −180° < θ 4 < 0 , we can solve these three
equations for the position results
(
θ 4 = − tan
(1 cosθ 2 )
−1
1 + cos 2 θ 2 3
)
yB = 4 7 + sin 2 θ 2 − 4sin θ 2
and, at θ 2 = 40° , these give θ 3 = −24.83° , θ 4 = −52.55° , and yB = 208 mm .
For velocity analysis we begin by using Eqs. (9.22) and (9.25) to find
⎡ 0 −1 0 0 ⎤
⎡ 0 −1 0 0 ⎤
⎢1 0 0 0 ⎥
⎢1 0 0 0 ⎥
⎢
⎥
⎥,
,
Q1 =
D1 = Q1 = ⎢
⎢0 0 0 0⎥
⎢0 0 0 0 ⎥
⎢
⎥
⎢
⎥
⎣0 0 0 0⎦
⎣0 0 0 0 ⎦
0
yB cos θ 4 ⎤
⎡ 0 − cos θ 4
⎡ 0 −1 0 0 ⎤
⎢cosθ
⎥
⎢1 0 0 0 ⎥
− sin θ 4
0
0
4
−1
⎢
⎥,
⎢
⎥
,
Q3 =
D3 = T13Q3T13 =
⎢ 0
⎢0 0 0 0 ⎥
− yB sin θ 4 ⎥
sin θ 4
0
⎢
⎥
⎢
⎥
0
0
0
⎣ 0
⎦
⎣0 0 0 0 ⎦
⎡0
⎢1
Q4 = ⎢
⎢0
⎢
⎣0
−1
0
0
0
⎡0
⎢0
Q6 = ⎢
⎢0
⎢
⎣0
6
0
0
0
0
0
0
0
0
0
0
0
0⎤
0 ⎥⎥
,
0⎥
⎥
0⎦
0⎤
0 ⎥⎥
,
1⎥
⎥
0⎦
⎡0
⎢0
D4 = T14Q4T14−1 = ⎢
⎢ −1
⎢
⎣0
0
0
0
0
1
0
0
0
0⎤
0 ⎥⎥
,
0⎥
⎥
0⎦
⎡0
⎢0
−1
D6 = T16Q6T16 = ⎢
⎢0
⎢
⎣0
0
0
0
0
0
0
0
0
0⎤
1 ⎥⎥
.
0⎥
⎥
0⎦
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θ 3 = − sin −1
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147
⎡ 0
⎢
0
ω 4 = D6 yB + D4θ 4 = ⎢
⎢ −θ 4
⎢
⎣⎢ 0
0 θ4 0 ⎤
⎥
0 0 yB ⎥
0 0 0⎥
⎥
0 0 0 ⎦⎥
Since the velocity of point A must agree along the two paths
⎡ −4 cos θ 2 ⎤
⎡ −4 cos θ 2 ⎤
⎢ −4sin θ ⎥
⎢ −4sin θ ⎥
2⎥
2⎥
⎢
= ω3 ⎢
ω2
⎢ −4 ⎥
⎢ −4 ⎥
⎢
⎥
⎢
⎥
1
1
⎣
⎦
⎣
⎦
⎡ −4sin θ 2θ 2 ⎤ ⎡ 4sin θ 2 cos θ 4θ 3 + yB cosθ 4θ 3 − 4θ 4 ⎤
⎢
⎥ ⎢
⎥
⎢ −4 cos θ 2θ 2 ⎥ = ⎢ −4 cos θ 2 cos θ 4θ 3 + 4sin θ 4θ 3 + yB ⎥
⎢
⎥ ⎢ 4 cosθ 2θ 4 − 4sin θ 2 sin θ 4θ 3 − yB sin θ 4θ 3 ⎥
0
⎢
⎥ ⎢
⎥
0
0
⎢⎣
⎥⎦ ⎢⎣
⎥⎦
At the position where θ 2 = 40° , θ 3 = −24.83° , θ 4 = −52.55° , yB = 208 mm , and
θ 2 = −48.0 rad/s , these equations can be solved for θ 3 = 6.891 rad/s , θ 4 = 19.444 rad/s ,
and yB = 2808.95 mm/s . With these values we can evaluate
0 19.444
0 ⎤
−4.191 19.444 34.865 ⎤
⎡ 0
⎡ 0
⎢ 4.191
⎥
⎢
0
0
0
112.357 ⎥⎥
0
5.471 112.357 ⎥
and ω 4 = ⎢
ω3 = ⎢
from
⎢ −19.444 −5.471
⎢ −19.444 0
0
0 ⎥
0
45.513 ⎥
⎢
⎥
⎢
⎥
0
0
0 ⎦
0
0
0 ⎦
⎣ 0
⎣ 0
which we write the vector forms of the results:
VB = 2808.95ˆj mm/s , ω3 = −5.471ˆi + 19.444ˆj + 4.191kˆ rad/s , and ω 4 = 19.444ˆj rad/s .Ans.
Note that the global x1 and z1 axis orientations in this solution differ from those of
problem P9.21 because of the conventions of the Denavit-Hartenberg parameters. This is
the reason that the components of ω3 seem switched.
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Next we write from Eq. (9.27)
⎡ 0 −θ 2 0 0 ⎤
⎢
⎥
θ 2 0 0 0⎥
⎢
ω 2 = D2θ 2 =
⎢ 0 0 0 0⎥
⎢
⎥
⎢⎣ 0 0 0 0 ⎥⎦
and, along the other path,
⎡ 0
θ4
yB cos θ 4θ 3 ⎤
− cos θ 4θ 3
⎢
⎥
− sin θ 4θ 3
yB
cos θ 4θ 3
0
⎢
⎥
ω3 = D6 yB + D4θ 4 + D3θ 3 =
⎢ −θ 4
sin θ 4θ 3
0
− yB sin θ 4θ 3 ⎥
⎢
⎥
0
0
0
⎢⎣ 0
⎥⎦
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149
Chapter 10
Robotics
10.1 For the SCARA robot shown in the figure, find the transformation matrix T15 relating the
position of the tool coordinate system to the ground coordinate system when the joint
actuators are set to the values φ1 = 30° , φ2 = −60° , φ3 = 50 mm , φ4 = 0 . Also find the
See the solution to problem P10.2 for the formulae before numerical evaluation.
⎡ 433 ⎤
⎡ 0.866 −0.500 0 433⎤
⎢ 0 ⎥
⎢ −0.500 −0.866 0 0 ⎥
⎥
⎥
T15 = ⎢
R1 = T15 R5 = ⎢
⎢162.5⎥
⎢ 0
0
−1 200 ⎥
⎢
⎥
⎢
⎥
0
0 1 ⎦
⎣ 1 ⎦
⎣ 0
10.2 Repeat Problem 10.1 using arbitrary (symbolic) values for the joint variables.
Ans.
a12 = a23 = 250 mm , a34 = a45 = 0 , α12 = α 34 = α 45 = 0 , α 23 = 180° , θ12 = φ1 , θ 23 = φ2 ,
θ 34 = 0 , θ 45 = φ4 , s12 = 300 mm , s23 = 0 , s34 = φ3 , s45 = 50 mm
⎡ cos φ1 − sin φ1
⎢ sin φ cos φ
1
1
T12 = ⎢
⎢ 0
0
⎢
0
⎣ 0
⎡1
⎢0
T34 = ⎢
⎢0
⎢
⎣0
0 250 cos φ1 ⎤
0 250sin φ1 ⎥⎥
1
300 ⎥
⎥
0
1
⎦
0⎤
0 ⎥⎥
φ3 ⎥
⎥
1⎦
⎡cos (φ1 + φ2 ) sin (φ1 + φ2 )
⎢
sin (φ1 + φ2 ) − cos (φ1 + φ2 )
T13 = T12T23 = ⎢
⎢
0
0
⎢
0
0
⎢⎣
0
1
0
0
⎡ cos φ2 sin φ2
⎢ sin φ − cos φ
2
2
T23 = ⎢
⎢ 0
0
⎢
0
⎣ 0
⎡ cos φ4 − sin φ4
⎢ sin φ cos φ
4
4
T45 = ⎢
⎢ 0
0
⎢
0
⎣ 0
0
0
1
0
0
0
−1
0
0 250 cos φ1 ⎤
0 250sin φ1 ⎥⎥
⎥
0
−1
⎥
0
1
⎦
0⎤
0 ⎥⎥
50 ⎥
⎥
1⎦
250 cos φ1 + 250 cos (φ1 + φ2 ) ⎤
⎥
250sin φ1 + 250sin (φ1 + φ2 ) ⎥
⎥
300
⎥
1
⎥⎦
0
0
1
0
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absolute position of the tool point which has coordinates x5 = y5 = 0, z5 = 37.5 mm.
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150
⎡cos (φ1 + φ2 ) sin (φ1 + φ2 )
⎢
sin (φ1 + φ2 ) − cos (φ1 + φ2 )
T14 = T13T34 = ⎢
⎢
0
0
⎢
0
0
⎣⎢
0 250 cos φ1 + 250 cos (φ1 + φ2 ) ⎤
⎥
0 250sin φ1 + 250sin (φ1 + φ2 ) ⎥
⎥
−1
300 − φ3
⎥
0
1
⎦⎥
⎡ 0 ⎤
⎢ 0 ⎥
⎥
R5 = ⎢
⎢37.5⎥
⎢
⎥
⎣ 1 ⎦
0 250 cos φ1 + 10 cos (φ1 + φ2 ) ⎤
⎥
0 250sin φ1 + 10sin (φ1 + φ2 ) ⎥
⎥
10 − φ3
−1
⎥
0
1
⎥⎦
⎡ 250 cos φ1 + 250 cos (φ1 + φ2 ) ⎤
⎢
⎥
250sin φ1 + 250sin (φ1 + φ2 ) ⎥
R1 = T15 R5 = ⎢
⎢
⎥
212.5 − φ3
⎢
⎥
1
⎣⎢
⎦⎥
10.3 For the gantry robot shown in the figure, find the transformation matrix T15 relating the
position of the tool coordinate system to the ground coordinate system when the joint
actuators are set to the values φ1 = 450 mm , φ2 = 180 mm , φ3 = 50 mm , φ4 = 0 . Also
find the absolute position of the tool point which has coordinates x5 = y5 = 0, z5 = 45 mm.
See the solution to problem P10.4 for the formulae before numerical evaluation.
⎡ 180 mm ⎤
⎡0 −1 0 180 ⎤
⎢0 0 −1 −100 ⎥
⎢ −145 mm ⎥
⎢
⎥
⎥
T15 =
R1 = ⎢
⎢ 450 mm ⎥
⎢1 0 0 450 ⎥
⎢
⎥
⎢
⎥
1 ⎦
1
⎣0 0 0
⎣
⎦
10.4 Repeat Problem 10.3 using arbitrary (symbolic) values for the joint variables.
Ans.
a12 = a23 = a34 = a45 = 0 α12 = 90° α 23 = −90° α 34 = α 45 = 0 θ12 = θ 23 = 90° θ 34 = 0
,
,
,
,
,
,
θ 45 = φ4 s12 = φ1 s23 = φ2 s34 = φ3 s45 = 50 mm
,
,
,
,
⎡0 0 1 0 ⎤
⎡ 0 0 −1 0 ⎤
⎢1 0 0 0 ⎥
⎢1 0 0 0 ⎥
⎢
⎥
⎥
T23 = ⎢
T12 =
⎢ 0 −1 0 φ2 ⎥
⎢ 0 1 0 φ1 ⎥
⎢
⎥
⎢
⎥
⎣0 0 0 1 ⎦
⎣0 0 0 1 ⎦
⎡1 0 0 0 ⎤
⎡ cos φ4 − sin φ4 0 0 ⎤
⎢0 1 0 0 ⎥
⎢ sin φ cos φ 0 0 ⎥
4
4
⎢
⎥
⎥
T45 = ⎢
T34 =
⎢ 0
⎢0 0 1 φ3 ⎥
0
1 50 ⎥
⎢
⎥
⎢
⎥
0
0 1⎦
⎣ 0
⎣0 0 0 1 ⎦
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⎡ cos (φ1 + φ2 − φ4 ) sin (φ1 + φ2 − φ4 )
⎢
sin (φ1 + φ2 − φ4 ) − cos (φ1 + φ2 − φ4 )
T15 = T14T45 = ⎢
⎢
0
0
⎢
0
0
⎢⎣
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151
φ2 ⎤
0
−1 −φ3 − 50 ⎥⎥
φ1 ⎥
0
⎥
0
1 ⎦
⎡0
⎢0
T14 = T13T34 = ⎢
⎢1
⎢
⎣0
−1
0
0
0
0
−1
0
0
φ2 ⎤
−φ3 ⎥⎥
φ1 ⎥
⎥
1 ⎦
Ans.
⎡ φ2 ⎤
⎢ −φ − 95⎥
⎥
R1 = T15 R5 = ⎢ 3
⎢ φ1 ⎥
⎢
⎥
⎣ 1 ⎦
Ans.
10.5 For the SCARA robot of Problem 10.1 and in the position described, find the
instantaneous velocity and acceleration of the same tool point, x5 = y5 = 0, z5 = 1.5 in, if
the actuators have (constant) velocities of φ1 = 0.20 rad/s , φ2 = −0.35 rad/s , and
φ = φ = 0 .
3
4
⎡0
⎢1
D1 = Q1 = ⎢
⎢0
⎢
⎣0
−1
0
0
0
0
0
0
0
⎡0
⎢0
D3 = T13Q3T13−1 = ⎢
⎢0
⎢
⎣0
0⎤
0 ⎥⎥
0⎥
⎥
0⎦
0 0
0 0
0 0
0 0
0⎤
0 ⎥⎥
−1⎥
⎥
0⎦
⎡ 0 −0.200
⎢0.200
0
ω 2 = ω1 + D1φ1 = ⎢
⎢ 0
0
⎢
0
⎣ 0
0.150
⎡ 0
⎢ −0.150 0
ω4 = ω3 + D3φ3 = ⎢
⎢ 0
0
⎢
0
⎣ 0
0
0
0
0
0
0
0
0
⎡ 0 −1 0
⎢1 0 0
D2 = T12Q2T12−1 = ⎢
⎢0 0 0
⎢
⎣0 0 0
⎡0 1 0
⎢ −1 0 0
D4 = T14Q4T14−1 = ⎢
⎢0 0 0
⎢
⎣0 0 0
⎡ 0
⎢ −0.150
ω3 = ω2 + D2φ2 = ⎢
⎢ 0
⎢
⎣ 0
125 ⎤
−216.5⎥⎥
0 ⎥
⎥
0 ⎦
0 ⎤
433⎥⎥
0 ⎥
⎥
0 ⎦
0.150 0 −43.75⎤
0 0 75.775 ⎥⎥
0 0
0 ⎥
⎥
0 0
0 ⎦
0⎤
0 ⎥⎥
0⎥
⎥
0⎦
0.150 0 −43.75⎤
−43.75⎤
⎡ 0
⎥
⎢
75.775 ⎥
−0.150 0 0 75.775 ⎥⎥
ω5 = ω4 + D4φ4 = ⎢
⎢ 0
0 0
0 ⎥
0 ⎥
⎥
⎢
⎥
0 0
0 ⎦
0 ⎦
⎣ 0
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⎡0 −1 0 φ2 ⎤
⎢0 0 −1 0 ⎥
⎥
T13 = T12T23 = ⎢
⎢1 0 0 φ1 ⎥
⎢
⎥
⎣0 0 0 1 ⎦
⎡ − sin φ4 − cos φ4
⎢ 0
0
T15 = T14T45 = ⎢
⎢ cos φ4 − sin φ4
⎢
0
⎣ 0
⎡ 0 ⎤
⎢ 0 ⎥
⎥
R5 = ⎢
⎢ 45 mm ⎥
⎢
⎥
⎣ 1 ⎦
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152
α 2 = α1 + D1φ1 + (ω1 D1 − D1ω1 ) φ1
⎡0
⎢0
=⎢
⎢0
⎢
⎣0
α 3 = α 2 + D2φ2 + (ω 2 D2 − D2ω 2 ) φ2
0⎤
0 ⎥⎥
0⎥
⎥
0⎦
α 4 = α 3 + D3φ3 + (ω3 D3 − D3ω3 ) φ3
0
0
0
0
0
0
0
0
0 −0.606 ⎤
0 −0.350 ⎥⎥
0
0 ⎥
⎥
0
0 ⎦
α 5 = α 4 + D4φ4 + (ω 4 D4 − D4ω 4 ) φ4
⎡0
⎢0
=⎢
⎢0
⎢
⎣0
0
0
0
0
0 0 −0.606 ⎤
⎡0 0 0 −0.606 ⎤
⎢0 0 0 −0.350 ⎥
0 0 −0.350 ⎥⎥
⎥
=⎢
⎢0 0 0
0 0
0 ⎥
0 ⎥
⎥
⎢
⎥
0 0
0 ⎦
0 ⎦
⎣0 0 0
⎡ −13.525 mm/s 2 ⎤
⎡ −43.75 mm/s ⎤
⎢
⎥
⎢10.825 mm/s ⎥
−2.2 mm/s 2 ⎥
⎢
⎥
R5 = ω5 R1 = ⎢
R5 = (α 5 + ω5ω5 ) R1 =
⎢
⎥
⎢
⎥
0
0
⎢
⎥
⎢
⎥
0
0
⎣
⎦
⎣⎢
⎦⎥
10.6 For the gantry robot of Problem 10.3 and in the position described, find the instantaneous
velocity and acceleration of the same tool point, x5 = y5 = 0, z5 = 45 mm, if the actuators
have (constant) velocities of φ1 = φ2 = 0 , φ3 = 40 mm/s , and φ4 = 20 rad/s .
⎡0
⎢0
D1 = Q1 = ⎢
⎢0
⎢
⎣0
0 0⎤
0 0 ⎥⎥
0 1⎥
⎥
0 0⎦
⎡0 0
⎢0 0
−1
D3 = T13Q3T13 = ⎢
⎢0 0
⎢
⎣0 0
ω = ω + D φ = 0
2
1
⎡0
⎢0
−1
D2 = T12Q2T12 = ⎢
⎢0
⎢
⎣0
⎡0
⎢0
−1
D4 = T14Q4T14 = ⎢
⎢1
⎢
⎣0
ω = ω + D φ = 0
0
0
0
0
0
0
0
0
0⎤
−1⎥⎥
0⎥
⎥
0⎦
3
1 1
⎡0 0 0 0 ⎤
⎢0 0 0 −40 ⎥
⎥
ω 4 = ω3 + D3φ3 = ⎢
⎢0 0 0 0 ⎥
⎢
⎥
⎣0 0 0 0 ⎦
α 2 = α1 + D1φ1 + (ω1 D1 − D1ω1 ) φ1 = 0
α = α + D φ + (ω D − D ω ) φ = 0
4
3
3 3
3
3
3
3
3
2
0
0
0
0
0
0
0
0
0 1⎤
0 0 ⎥⎥
0 0⎥
⎥
0 0⎦
−1 450 ⎤
0
0 ⎥⎥
0 −180 ⎥
⎥
0
0 ⎦
2 2
⎡ 0 0 −20 9000 ⎤
⎢0 0 0
−40 ⎥⎥
⎢
ω5 = ω 4 + D4φ4 =
⎢ 20 0 0 −3600 ⎥
⎢
⎥
0 ⎦
⎣0 0 0
α = α + D φ + (ω D − D ω ) φ = 0
3
2
2 2
2
2
2
2
2
α 5 = α 4 + D4φ4 + (ω 4 D4 − D4ω 4 ) φ4 = 0
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⎡0
⎢0
=⎢
⎢0
⎢
⎣0
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153
0
⎡
⎤
⎢ −40 mm/s ⎥
⎥
R5 = ω5 R1 = ⎢
⎢
⎥
0
⎢
⎥
0
⎣
⎦
⎡0⎤
⎢ ⎥
= (α + ω ω ) R = ⎢ 0 ⎥
R
5
5
5 5
1
⎢0⎥
⎢ ⎥
⎣0⎦
10.7 The SCARA robot of Problem 10.1 is to be guided along a path for which the origin of
the end effecter O5 follows the straight line given by
R O (t ) = ( 40t + 100 ) ˆi1 + ( 30t + 75 ) ˆj1 + 50kˆ 1 mm
5
with t varying from 0.0 to 5.0 s; the orientation of the end effecter is to remain constant
with kˆ 5 = −kˆ 1 (vertically downward) and î5 radially outward from the base of the robot.
Find expressions for how each of the actuators must be driven, as functions of time, to
achieve this motion.
From this we write the transformation T15.
⎡0.800 0.600 0 100 + 40t ⎤
⎢0.600 −0.800 0 75 + 30t ⎥
⎥
T15 = ⎢
⎢ 0
0
50 ⎥
−1
⎢
⎥
0
0
1
⎣ 0
⎦
However, as shown in the solution for Prob. P10.2,
⎡ cos (φ1 + φ2 − φ4 ) sin (φ1 + φ2 − φ4 ) 0 250 cos φ1 + 250 cos (φ1 + φ2 ) ⎤
⎢
⎥
sin (φ1 + φ2 − φ4 ) − cos (φ1 + φ2 − φ4 ) 0 250sin φ1 + 250sin (φ1 + φ2 ) ⎥
⎢
.
T15 = T14T45 =
⎢
⎥
−1
0
0
250 − φ3
⎢
⎥
0
0
0
1
⎣⎢
⎦⎥
Equating these gives
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From the problem statement we construct the figure shown for the path described.
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154
250 cos φ1 + 250 cos (φ1 + φ2 ) = 100 + 40t = 100 (1.000 + 0.400t )
250sin φ1 + 250sin (φ1 + φ2 ) = 75 + 30t = 75 (1.000 + 0.400t )
250 − φ3 = 50
φ1 + φ2 − φ4 = θ = 36.87°
Squaring and adding Eqs. (a) and (b) gives
5000 + 5000 cos φ2 = 625 (1.000 + 0.800t + 0.040t 2 )
φ2 = cos −1 ( −0.875 + 0.050t + 0.020t 2 )
−180° ≤ φ2 ≤ 0
(a)
(b)
(c )
(d )
Ans.
Where the quadrant was found from the figure of the robot.
Expanding the trigonometric functions and recognizing that φ2 is now known, Eqs. (a)
and (b) become
250 (1 + cos φ2 ) cos φ1 − 250 ( sin φ2 ) sin φ1 = 100 (1.000 + 0.400t )
250 ( sin φ2 ) cos φ1 + 250 (1 + cos φ2 ) sin φ1 = 75 (1.000 + 0.400t )
sin φ1 = ⎡⎣3 (1 + cos φ2 ) − 4sin φ2 ⎤⎦ Δ
cos φ1 = ⎡⎣ 4 (1 + cos φ2 ) + 3sin φ2 ⎤⎦ Δ
where Δ = 500 (1 + cos φ2 ) (1.000 + 0.400t )
From the ratio of these we get
⎡ 3 (1 + cos φ2 ) − 4sin φ2 ⎤
φ1 = tan −1 ⎢
⎥
⎣ 4 (1 + cos φ2 ) + 3sin φ2 ⎦
Ans.
where the quadrant of φ1 is found by considering the signs of the numerator and
denominator separately.
With both φ1 and φ2 known, Eqs. (c ) and (d) give
φ3 = 200 mm
Ans.
φ4 = φ1 + φ2 − 36.87°
Ans.
10.8 The gantry robot of Problem 10.3 is to travel a path for which the origin of the end
effecter O5 follows the straight line given by
R O (t ) = (120t + 300 ) ˆi1 − 150ˆj1 + ( 90t + 225 ) kˆ 1 mm
5
with t varying from 0.0 to 4.0 s; the orientation of the end effecter is to remain constant
with kˆ 5 = −ˆj1 (vertically downward) and ˆi5 = ˆi1 . Find expressions for the positions of
each of the actuators, as functions of time, for this motion.
From Prob. 10.4 and the problem statement we can write
φ2 ⎤ ⎡1 0 0 300 + 120t ⎤
⎡ − sin φ4 − cos φ4 0
⎢ 0
0
−1 −φ3 − 50 ⎥⎥ ⎢⎢0 0 −1 −150 ⎥⎥
⎢
=
T15 =
⎢ cos φ4 − sin φ4 0
φ1 ⎥ ⎢0 1 0 225 + 90t ⎥
⎢
⎥ ⎢
⎥
0
0
1 ⎦ ⎣0 0 0
1
⎣ 0
⎦
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which can be solved for sin φ1 and cos φ1
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155
Equating individual elements and solving, we get
φ1 = 225 + 90t mm
φ2 = 300 + 120t mm
φ3 = 100 mm
φ4 = −90°
Ans.
Ans.
Ans.
Ans.
10.9 The end effecter of the SCARA robot of Problem 10.1 is working against a force loading
of 44.5ˆi1 + 22.25tkˆ 1 N and a constant torque loading of 2.825kˆ 1 N − m as it follows the
trajectory described in Problem 10.7. Find the torques required at the actuators, as
functions of time, to achieve the motion described.
0⎤
⎡0 1
⎢
⎥
0⎥
−1 0
−1
D4 = T14Q4T14 = ⎢
⎢0 0
−1⎥
⎢
⎥
0⎦
⎣⎢ 0 0
From these the Jacobian and loads are
0
0
0
⎡0
⎤
⎢0
⎥
0
0
0
⎢
⎥
⎢1
⎥
−1
1
0
J =⎢
⎥
⎢ 0 44.5sin φ1 0 −44.5sin φ1 − 44.5sin (φ1 + φ2 ) ⎥
⎢ 0 −44.5cos φ1 0 44.5cos φ1 + 44.5cos (φ1 + φ2 ) ⎥
⎢
⎥
0
0
−1
⎢⎣ 0
⎥⎦
Now, from Eq. (10.19), we get
τ 1 = 2.825 N − m
τ 2 = 2.825 + 11.3sin φ1 N − m
τ 3 = −22.25 N
⎡0
⎢0
D3 = T13Q3T13−1 = ⎢
⎢0
⎢
⎣0
0
0
0
0
0
0
0
0
τ 4 = −2.825 − 11.3sin φ1 − 11.3sin (φ1 + φ2 ) N − m
0
0
0
0
from Probs. 10.2 and
0 44.5sin φ1 ⎤
0 −44.5cos φ1 ⎥⎥
⎥
0
0
⎥
0
0
⎦
−44.5sin φ1 − 44.5sin (φ1 + φ2 ) ⎤
⎥
44.5cos φ1 + 44.5cos (φ1 + φ2 ) ⎥
⎥
0
⎥
0
⎦⎥
0
⎡
⎤
⎢
⎥
0
⎢
⎥
⎢ 2.825 N − m ⎥
F =⎢
⎥
⎢ 44.5 N ⎥
⎢
⎥
0
⎢
⎥
⎢⎣ 22.25 N ⎥⎦
Ans.
Ans.
Ans.
Ans.
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Using the φ1 , φ2 , φ3 , and φ4 values from Prob. 10.7 and formulae
10.5,
⎡0 −1
⎡0 −1 0 0 ⎤
⎢1 0
⎢1 0 0 0 ⎥
⎥
D1 = Q1 = ⎢
D2 = T12Q2T12−1 = ⎢
⎢0 0
⎢0 0 0 0 ⎥
⎢
⎢
⎥
⎣0 0
⎣0 0 0 0 ⎦
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156
10.10 The end effecter of the gantry robot of Problem 10.3 is working against a force loading of
20ˆi1 + 10tˆj1 N and a constant torque loading of 5ˆj1 N ⋅ m as it follows the trajectory
Using the φ1 , φ2 , φ3 , and φ4 values from Prob. 10.8 and formulae from Probs. 10.4 and
10.6,
⎡0 0 0 1 ⎤
⎡0 0 0 0 ⎤
⎢0 0 0 0 ⎥
⎢0 0 0 0 ⎥
−1
⎥
⎢
⎥
D1 = Q1 =
D2 = T12Q2T12 = ⎢
⎢0 0 0 0 ⎥
⎢0 0 0 1 ⎥
⎢
⎥
⎢
⎥
⎣0 0 0 0 ⎦
⎣0 0 0 0 ⎦
⎡0 0 −1 0.225 + 0.090t ⎤
⎡0 0 0 0 ⎤
⎢0 0 0 −0.300 − 0.120t ⎥
⎢0 0 0 −1⎥
−1
−1
⎥
⎢
⎥
D4 = T14Q4T14 = ⎢
D3 = T13Q3T13 =
⎢1 0 0
⎥
⎢0 0 0 0 ⎥
0
⎢
⎥
⎢
⎥
0
⎣0 0 0
⎦
⎣0 0 0 0 ⎦
From these, the Jacobian and loads are
0
⎡ 0 ⎤
⎡0 0 0
⎤
⎢5 N ⋅ m ⎥
⎢0 0 0
⎥
−1
⎢
⎥
⎢
⎥
⎢ 0 ⎥
⎢0 0 0
⎥
0
J =⎢
F =⎢
⎥
⎥
⎢ 20 N ⎥
⎢ 0 1 0 0.225 + 0.090t ⎥
⎢ 10t N ⎥
⎢ 0 0 −1 −0.300 − 0.120t ⎥
⎢
⎥
⎢
⎥
0
⎢⎣ 0 ⎥⎦
⎢⎣1 0 0
⎥⎦
Now, from Eq. (10.19), we get
τ1 = 0
Ans.
τ 2 = 20 N
Ans.
τ 3 = −10t N
Ans.
τ 4 = −0.500 − 1.200t − 1.200t 2 N ⋅ m
Ans.
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described in Problem 10.8 Find the torques required at the actuators, as functions of
time, to achieve the motion described.
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PART III
DYNAMICS OF MACHINES
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149
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151
Chapter 11
Static Force Analysis
The figure shows four mechanisms and the external forces and torques exerted on or by
the mechanisms. Sketch the free-body diagram of each part of each mechanism. Do not
attempt to show the magnitudes of the forces, except roughly, but do sketch them in their
proper locations and directions.
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11.1
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152
11.2
What moment M12 must be applied to the crank of the mechanism shown if
P = 0.9 kN ?
Kinematic analysis:
φ = sin −1 ⎡⎣( r sin θ ) ⎤⎦ = sin −1 ⎡⎣( 75 mm sin105° ) 350 mm ⎤⎦ = 11.95°
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153
Force analysis:
P = −900ˆi N
∑ F = P + F14 ˆj + F34 cos φ ˆi − sin φ ˆj = −900 Nˆi + F14ˆj + F34 0.978ˆi − 0.207ˆj = 0
(
−900 N + 0.978F34 = 0
F14 − 0.207 F34 = 0
(
)
(
)
F34 = 900 N 0.978 = 920 N
F14 = 0.207 ( 920 N ) = 190 N
)
+ 75 mm ( cos105ˆi + sin105ˆj) × ( −900ˆi + 190ˆj N ) = 0
F32 = −F34 = 920 N −0.978ˆi + 0.207ˆj = −900ˆi + 190ˆj N
∑M = M
12
+ r2 × F32 = M12
M12 + 61.5kˆ N ⋅ m = 0
Ans.
If M12 = 100 N ⋅ m for the mechanism shown, what force P is required to maintain static
equilibrium?
Kinematic analysis:
Recall φ = 11.95° from Prob. P11.2.
x = r cos θ + cos φ = 75 mm cos105° + 350 mm cos11.95° = 323 mm
Force analysis:
∑ M Oz = xF14 − M 12 = 0
F14 = M 12 x = 100N ⋅ m 0.323 m = 310 N
Recall the force polygon on link 4 from Prob. P11.2.
P = F14 tan φ = 310 N tan11.95° = 1 463 N
11.4
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11.3
M12 = −61.5kˆ N ⋅ m
Ans.
Find the frame reactions and torque M12 necessary to maintain equilibrium of the fourbar linkage shown in the figure.
Kinematic analysis:
R AO2 = 88 mm∠210° = −75.775ˆi − 43.75ˆj mm
R = 150 mm∠82.83° = 18.725ˆi + 148.825ˆj mm
BA
R BO4 = 150 mm∠135.53° = −107.04ˆi + 105.07ˆj mm
R = 100 mm∠135.53° = −71.375ˆi + 70.05ˆj mm
CO4
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154
( −71.375ˆi + 70.05ˆj mm ) × (311.72ˆi + 317.57ˆj N )
+ ( −107.04ˆi + 105.07ˆj mm ) × ( cos82.83°ˆi + sin 82.83°ˆj) F
34
=0
( −45.2 N − m − 119.325 mmF34 ) kˆ = 0
F34 = −372.94 N∠82.83° = −46.55ˆi − 370.02ˆj N
F34 = −9.47 N − m
∑M
O2
= M12 + R AO2 × F32 = 0
(
) (
)
M 12kˆ + −75.775ˆi − 43.75ˆj mm × −46.55ˆi − 370.02ˆj N = 0
M 12 = 26.408 N − m
11.5
M12 = 26.408kˆ N − m
Ans.
What torque must be applied to link 2 of the linkage shown to maintain static
equilibrium?
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Force analysis:
∑ M O4 = R CO4 × P + R BO4 × F34 = 0
Kinematic analysis:
R AO2 = 87.5 mm∠240° = −43.75ˆi − 75.775ˆj mm R BO4 = 150 mm∠152.64° = −133.225ˆi + 68.925ˆj mm
R BA = 150 mm∠105.26° = −39.475ˆi + 144.7ˆj mm R DO4 = 175 mm∠152.64° = −155.425ˆi + 80.425ˆj mm
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155
Force analysis:
∑ M O4 = R DO4 × P + R BO4 × F34 = 0
( −155.425ˆi + 80.425ˆj mm ) × ( 222.5ˆi N )
+ ( −133.225ˆi + 68.925ˆj mm ) × ( cos105.26°ˆi + sin105.26°ˆj) F
34
=0
( −18.176 N − m − 110.375 mmF34 ) kˆ = 0
F34 = −4.116 N − m
∑M
O2
F34 = −162.109 N∠105.26° = 42.666ˆi − 156.39ˆj N
= M12 + R AO2 × F32 = 0
(
) (
)
M 12kˆ + −43.75ˆi − 75.775ˆj mm × −42.666ˆi + 156.39ˆj N = 0
M 12 = 10.23 N − m
Ans.
Sketch a complete free-body diagram of each link of the linkage shown. What force P is
necessary for equilibrium?
Kinematic analysis:
R AO2 = 100 mm∠90° = 100ˆj mm
R = 150 mm∠ − 4.86° = 149ˆi − 13ˆj mm
BA
R BO4 = 125 mm∠44.3° = 89ˆi + 87ˆj mm
R = 200 mm∠44.3° = 143ˆi + 140ˆj mm
CO4
R DC = 400 mm∠ − 20.44° = 375ˆi − 140ˆj mm
Force analysis:
∑ M O2 = M12 + R AO2 × F32 = 0
(
) (
)
90kˆ N ⋅ m + 100ˆj mm × cos− 4.86°ˆi + sin − 4.86°ˆj F32 = 0
( 90 N ⋅ m − 99.640 mmF32 ) kˆ = 0
F32 = 903 N∠ − 4.86°
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11.6
M12 = 10.23kˆ N − m
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156
∑M
O4
= R BO4 × F34 + R CO4 × F54 = 0
(89ˆi + 87ˆj mm ) × ( cos175.14°ˆi + sin175.14°ˆj) 903 N
+ (143ˆi + 140ˆj mm ) × ( cos− 20.44°ˆi + sin − 20.44°ˆj) F
54
=0
(86 N ⋅ m − 181 mmF54 ) kˆ = 0
∑
16
Ans.
Determine the torque M12 required to drive slider 6 of the figure against a load of
P = 445 N at a crank angle of θ = 30° , or as specified by your instructor.
Kinematic analysis:
R AO2 = 63 mm∠30° = 54.125ˆi + 31.25ˆj mm
R AO4 = 186.65 mm∠73.37° = 53.4ˆi + 178.85ˆj mm
R BO4 = 400 mm∠73.37° = 114.45ˆi + 383.275ˆj mm R CB = 200 mm∠175.20° = −199.3ˆi + 16.725ˆj mm
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11.7
F54 = 472 N∠ − 20.44°=44ˆi − 165ˆj N
P = 443ˆi N
F = Pˆi − 443 Nˆi + 165 Nˆj + F ˆj= 0
Force analysis:
∑ F = Pˆi + F16ˆj + cos175.20°ˆi + sin175.20ˆj F56 = 0
(
)
F56 = − P cos175.20 = −445 N −0.996 = 446.56 N ; F56 = 446.56 N∠175.20° = −445ˆi + 37.34ˆj N
∑M
O4
= R BO4 × F54 + R AO4 × F34 = 0
(114.45ˆi + 383.275ˆj mm ) × ( 445ˆi − 37.34ˆj N )
+ ( 53.4ˆi + 178.85ˆj mm ) × ( cos163.37°ˆi + sin163.37°ˆj) F
34
( −177.58 N − m + 186.65 mmF34 ) kˆ = 0
=0
F34 = 936.67 N∠163.37°= − 897.49ˆi + 268.07ˆj N
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157
∑M
O2
= M12 + R AO2 × F32 = 0
(
) (
)
M 12kˆ + 54.125ˆi + 31.25ˆj mm × 897.49ˆi − 268.07ˆj N = 0
M 12 = 43.22 N − m
Ans.
Sketch complete free-body diagrams for the illustrated four-bar linkage. What torque
M12 must be applied to 2 to maintain static equilibrium at the position shown?
Kinematic analysis:
R AO2 = 200 mm∠60° = 100ˆi + 173ˆj mm
R CO4 = 350 mm∠ − 109.05° = −114ˆi − 331ˆj mm
R BA = 400 mm∠ − 46.06° = 278ˆi − 288ˆj mm , R CA = 700 mm∠ − 46.06° = 486ˆi − 504ˆj mm
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11.8
M12 = 43.22kˆ N − m
Force analysis:
Since the lines of action of all constraint forces can not be found from two- and threeforce members, the force F34 is resolved into radial and transverse components,
F34r and F34θ . Then
∑M
O4
(
−45kˆ N ⋅ m+350kˆ mmF34θ = 0
∑M
A
) (
)
= M14 + R CO4 × F34θ = −45kˆ N ⋅ m+ −114ˆi − 331ˆj mm × cos− 19.05°ˆi + sin − 19.05°ˆj F34θ = 0
F34θ = 129 N∠ − 19.05°=122ˆi − 42ˆj N
θ
= R BA × P + R CA × F43
+ R CA × F43r = 0
( 278ˆi − 288ˆj mm ) × ( −350ˆi N ) + ( 486ˆi − 504ˆj mm ) × ( −122ˆi + 42ˆj N )
+ ( 486ˆi − 504ˆj mm ) × ( cos 70.95°ˆi + sin 70.95°ˆj) F = 0
r
43
−101kˆ N ⋅ m − 41kˆ N ⋅ m+624kˆ mmF43r = 0 , F43r = 228 N∠70.95°=74ˆi + 215ˆj N
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158
θ
F43 = F43r + F43
= −48ˆi + 257ˆj N = 261 N∠100.58°
Now the lines of action for other forces may be found as shown.
∑ F = F43r + F43θ + P + F23 = 0
74ˆi + 215ˆj N − 122ˆi + 42ˆj N − 350ˆi N + F = 0 , F =398ˆi − 257ˆj N = 474 N∠ − 32.85°
∑M
23
O2
= M12 + R AO2 × F32 = 0
(
) (
23
)
M 12kˆ + 100ˆi + 173ˆj mm × −398ˆi + 257ˆj N = 0
M12 = −94.55kˆ N ⋅ m
M 12 = −94.55 N ⋅ m
Sketch free-body diagrams of each link and show all the forces acting. Find the
magnitude and direction of the moment that must be applied to link 2 to drive the linkage
against the forces shown.
Kinematic analysis:
R AO2 = 100 mm∠30° = 86.6ˆi + 50ˆj mm
R CO4 = 250 mm∠84.34° = 24.675ˆi + 248.775ˆj mm
R BA = 350 mm∠67.81° = 132.2ˆi + 324.075ˆj mm R CA = 350 mm∠34.61° = 288.075ˆi + 198.775ˆj mm
R
= 175 mm∠84.34° = 17.275ˆi + 174.15ˆj mm
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11.9
Ans.
DO4
Force analysis:
Since the lines of action of all constraint forces can not be found from two- and threeforce members, the force F34 is resolved into radial and transverse components,
F34r and F34θ . Then
∑M
O4
= R DO4 × PD + R CO4 × F34θ = 0
(17.275ˆi + 174.15ˆj mm ) × ( −858.85ˆi + 231.4ˆj N ) + ( 24.675ˆi + 248.775ˆj mm )
× ( cos − 5.66°ˆi + sin − 5.66°ˆj) F = 0
θ
34
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159
155.94kˆ N − m − 249.975kˆ mmF34θ = 0
∑M
A
F34θ = 614.1 N∠ − 5.66°=609.65ˆi − 62.3ˆj N
θ
= R BA × PB + R CA × F43
+ R CA × F43r = 0
(132.2ˆi + 324.075ˆj mm ) × ( −445ˆi N ) + ( 288.075ˆi + 198.775ˆj mm ) × ( −609.65ˆi + 62.3ˆj N )
+ ( 288.075ˆi + 198.775ˆj mm ) × ( cos − 95.66°ˆi + sin − 95.66°ˆj) F = 0
r
43
146.448kˆ N − m + 141.363kˆ N − m − 267.075kˆ mmF43r = 0 ,
F r = 1059.1 N∠ − 95.66°= − 102.35ˆi − 1054.65ˆj N
43
θ
F43 = F43r + F43
= −712ˆi − 992.35ˆj N = 1219.3∠ − 125.66°
Now the lines of action for other forces may be found as shown.
∑ F = F43r + F43θ + PB + F23 = 0
−102.35ˆi − 1054.65ˆj N − 609.65ˆi + 62.3ˆj N − 445ˆi N + F = 0 ,
23
∑M
O2
= M12 + R AO2 × F32 = 0
(
) (
)
M 12kˆ + 86.6ˆi + 50ˆj mm × −1157ˆi − 992.35ˆj N = 0
M 12 = 28.476 N − m
M12 = 28.476kˆ N − m
Ans.
11.10 The figure shows a four-bar linkage with external forces applied at points B and C. Draw
a free-body diagram of each link and show all the forces acting on each. Find the torque
that must be applied to link 2 to maintain equilibrium.
Kinematic analysis:
R AO2 = 75 mm∠ − 30° = 65ˆi − 38ˆj mm
R = 200 mm∠16.00° = 192ˆi + 55ˆj mm
BA
R CO4 = 200 mm∠124.56° = −113ˆi + 165ˆj mm
R = 300 mm∠42.38° = 222ˆi + 202ˆj mm
CA
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F23 =1157ˆi + 992.35ˆj N = 1526.35 N∠40.62°
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160
Force analysis:
∑ M A = R BA × PB + R CA × PC + R CA × F43 = 0
(192ˆi + 55ˆj mm ) × ( −354ˆi + 354ˆj N ) + ( 222ˆi + 202ˆj mm ) × (1 800ˆi N )
+ ( 222ˆi + 202ˆj mm ) × ( cos124.56°ˆi + sin124.56°ˆj) F = 0
43
87.438kˆ N ⋅ m − 363.000kˆ N ⋅ m + 297kˆ mmF43r = 0 ,
F = 927∠124.56° N= − 526ˆi + 763ˆj N
43
∑F = F
+ PB + PC + F23 = 0
−526ˆi + 763ˆj N − 354ˆi + 354ˆj N+1 800ˆi N + F23 = 0 ,
F = − 920ˆi − 1 117ˆj N = 1 447 N∠ − 129.48°
43
23
∑M
O2
= M12 + R AO2 × F32 = 0
(
) (
)
M 12 = −107.57 N ⋅ m
M12 = −107.57kˆ N ⋅ m
Ans.
11.11 Draw a free-body diagram of each of the members of the mechanism shown in the figure,
and find the magnitude and the direction of all the forces and moments. Compute the
magnitude and direction of the torque that must be applied to link 2 to maintain static
equilibrium.
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M 12kˆ + 65ˆi − 38ˆj mm × 920ˆi + 1 117ˆj N = 0
Kinematic analysis:
R CO4 = 200 mm∠124.23° = −112.5ˆi + 165.35ˆj mm
R AO2 = 100 mm∠180° = −100ˆi mm
R BA = 350 mm∠55.98° = 195.85ˆi + 290.075ˆj mm R CA = 250 mm∠41.41° = 187.5ˆi + 165.35ˆj mm
R
= 150 mm∠95.27° = −13.775ˆi + 149.375ˆj mm
DO4
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161
Force analysis:
Since the lines of action of all constraint forces can not be found from two- and threeforce members, the force F34 is resolved into radial and transverse components,
F34r and F34θ . Then
∑M
O4
= R DO4 × PD + R CO4 × F34θ = 0
( −13.775ˆi + 149.375ˆj mm ) × ( −694.2ˆi + 400.5ˆj N ) + ( −112.5ˆi + 165.35ˆj mm )
× ( cos 34.23°ˆi + sin 34.23°ˆj) F = 0
θ
34
99.779kˆ N − m − 200kˆ mmF34θ = 0
∑M
F34θ = 489.5 N∠34.23°=404.95ˆi + 275.9ˆj N
θ
= R BA × PB + R CA × F43
+ R CA × F43r = 0
A
(195.85ˆi + 290.075ˆj mm ) × ( −534ˆi N ) + (187.5ˆi + 165.35ˆj mm ) × ( −404.95ˆi − 275.9ˆj N )
+ (187.5ˆi + 165.35ˆj mm ) × ( cos − 55.77°ˆi + sin − 55.77°ˆj) F = 0
157.296kˆ N − m + 15.481kˆ N − m − 248.025kˆ mmF43r = 0
F r = 685.3 N∠ − 55.77°=387.15ˆi − 565.15ˆj N
43
θ
F43 = F + F43
= −17.8ˆi − 841.05ˆj N = 841.05 N∠ − 91.21°
Ans.
Now the lines of action for other forces may be found as shown.
∑ F = F43 + PB + F23 = 0
−17.8ˆi − 841.05ˆj N − 534ˆi N + F23 = 0 ,
F23 =551.8ˆi + 841.05ˆj N = 1005.7 N∠56.73° Ans.
F = −F = 17.8ˆi + 841.05ˆj N = 841.05 N∠88.79°
Ans.
r
43
34
43
∑F = F
+ PD + F14 = 0
17.8ˆi + 841.05ˆj N − 694.2ˆi + 400.5ˆj N + F14 = 0 ,
F =676.4ˆi − 1241.55ˆj N = 1415.1 N∠ − 61.42°
34
14
F32 = −F23 = −551.8ˆi − 841.05ˆj N = 1005.7 N∠ − 123.27°
F = −F = 551.8ˆi + 841.05ˆj N = 1005.7 N∠56.73°
12
∑M
32
O2
= M12 + R AO2 × F32 = 0
(
) (
Ans.
Ans.
Ans.
)
M 12kˆ + −100ˆi mm × −551.8ˆi − 841.05ˆj N = 0
M 12 = −85.428 N − m
M12 = −85.428kˆ N − m
Ans.
11.12 Determine the magnitude and direction of the forces that must be applied to link 2 to
maintain static equilibrium.
Kinematic analysis:
R AO2 = 75 mm∠90° = 75ˆj mm
R BA
R CA = 350 mm∠ − 12.37° = 341.875ˆi − 75ˆj mm
= 175 mm∠ − 34.93° = 143.475ˆi − 100.225ˆj mm
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r
43
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162
Force analysis:
∑ M A = R BA × PB + R CA × PC + R CA × F14 = 0
(143.475ˆi − 100.225ˆj mm ) × ( 222.5ˆj N ) + ( 341.875ˆi − 75ˆj mm ) × ( −445ˆi N )
+ ( 341.875ˆi − 75ˆj mm ) × ( ˆj) F = 0
32.431kˆ N − m − 33.9kˆ N − m + 350kˆ mmF14 = 0
F14 = 4.45 N∠90°=4.45ˆj N
∑F = P
+ PC + F14 + F23 = 0
222.5ˆj N − 445ˆi N + 4.45ˆj N + F23 = 0 ,
B
∑M
O2
= M12 + R AO2 × F32 = 0
M 12 = −33.9 N − m
F23 =445ˆi − 226.95ˆj N = 498.5 N∠ − 27.02°
M kˆ + 75ˆj mm × −445ˆi + 226.95ˆj N = 0
12
(
) (
M12 = −33.9kˆ N − m
)
Ans.
11.13 The photograph shows the Figee floating crane with leminscate boom configuration. Also
shown is a schematic diagram of the crane. The lifting capacity is 16 T (with 1 T = 1
metric ton =1 000 kg) including the grab which is about 10 T. The maximum outreach is
30 m, which corresponds to θ 2 = 49° . Minimum outreach is 10.5 m at θ 2 = 49°. Other
dimensions are given in the figure caption. For the maximum outreach position and a
grab load of 10 T, find the bearing reactions at A, B, O2 , and O4 , and the moment M12
required. Notice that the photograph shows a counterweight on link 2; neglect this
weight and also the weights of the members.
Kinematic analysis:
R AO2 = 14.700 m∠49.00° = 9.644ˆi + 11.094ˆj m , R BO4 = 19.300 m∠59.70° = 9.739ˆi + 16.663ˆj m
R BA = 6.500 m∠2.37° = 6.494ˆi + 0.269ˆj m
R CA = 22.300 m∠14.39° = 21.600ˆi + 5.543ˆj m
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14
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Force analysis:
Note that a metric ton is a unit of mass; the weight of a metric ton in standard gravity is
W = mg = (1 000 kg ) ( 9.81 m/s 2 ) = 9.810 kN . Therefore the rated load of the crane is
F = 98.100 kN .
∑ M A = R BA × F43 + R CA × F = 0
( 6.494ˆi + 0.269ˆj m ) × ( cos 59.70°ˆi + sin 59.70°ˆj) F + ( 21.600ˆi + 5.543ˆj m ) × ( −98.100ˆj kN ) = 0
43
5.471kˆ mF43 − 2 119kˆ kN ⋅ m = 0 ,
F43 = 387∠59.70° kN=195ˆi + 334ˆj kN
F14 = −F34 = 387 kN∠59.70°=195ˆi + 334ˆj kN
Ans.
∑F = F
+ F + F23 = 0
195ˆi + 334ˆj kN − 98.1ˆj kN + F23 = 0
F = − 195ˆi − 236ˆj kN = 307 kN∠ − 129.59°
43
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163
23
F12 = − F32 = − 195ˆi − 236ˆj kN = 307 kN∠ − 129.59°
∑M
O2
= M12 + R AO2 × F32 = 0
(
) (
Ans.
)
M 12kˆ + 9.644ˆi + 11.094ˆj m × −195ˆi − 236ˆj kN = 0
M 12 = −47 kN ⋅ m
M12 = −47kˆ kN ⋅ m
Ans.
11.14 Repeat Problem 11.13 for the minimum outreach position.
Kinematic analysis:
R AO2 = 14.700 m∠132.00° = −9.836ˆi + 10.924ˆj m , R BO4 = 19.300 m∠120.35° = −9.751ˆi + 16.656ˆj m
R BA = 6.500 m∠3.81° = 6.486ˆi + 0.432ˆj m
R CA = 22.300 m∠15.83° = 21.454ˆi + 6.083ˆj m
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Force analysis:
Note that a metric ton is a unit of mass; the weight of a metric ton in standard gravity is
W = mg = (1 000 kg ) ( 9.81 m/s 2 ) = 9.810 kN . Therefore the rated load of the crane is
F = 98.100 kN .
∑ M A = R BA × F43 + R CA × F = 0
( 6.486ˆi + 0.432ˆj m ) × ( cos120.35°ˆi + sin120.35°ˆj) F + ( 21.454ˆi + 6.083ˆj m ) × ( −98.1ˆj kN ) = 0
43
5.815kˆ mF43 − 2 105kˆ kN ⋅ m = 0
F43 = 362 kN∠120.35°= − 183ˆi + 312ˆj kN
F14 = −F34 = 362 kN∠120.35°= − 183ˆi + 312ˆj kN
Ans.
∑F = F
−183ˆi + 312ˆj kN − 98.1ˆj kN + F23 = 0
+ F + F23 = 0
F23 =183ˆi − 214ˆj kN = 282 kN∠ − 49.51°
F = − F =183ˆi − 214ˆj kN = 282 kN∠ − 49.51°
43
12
∑M
32
O2
= M12 + R AO2 × F32 = 0
(
) (
Ans.
)
M 12kˆ + −9.836ˆi + 10.924ˆj m × −183ˆi + 214ˆj kN = 0
M 12 = 109 kN ⋅ m
M12 = 109kˆ kN ⋅ m
Ans.
11.15 Repeat Problem 11.7 assuming coefficients of Coulomb friction μc = 0.20 between links
1 and 6 and μc = 0.10 between links 3 and 4. Determine the torque M12 necessary to
drive the system, including friction, against the load P.
See the figure and solution for Problem P11.7 for the kinematic and frictionless solution.
For friction between links 1 and 6, the friction angle is φ = tan −1 ( 0.20 ) = 11.31° . Since
the impending motion VC6 /1 is to the left the friction force f16 = μ c F16n is toward the right.
Also, since the non-friction normal force F16n is downward (from the solution for Prob.
P11.7), the total force F16 acts at the angle −90° + 11.31° = −78.69° . Therefore,
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164
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165
∑ F = Pˆi + ( cos− 78.69°ˆi + sin − 78.69°ˆj) F
16
(
)
ˆj + cos175.20°ˆi + sin175.20°ˆj F = 0
56
445 N + cos− 78.69° F16 + cos175.20° F56 = 0 , and sin − 78.69° F16 + sin175.20° F56 = 0
F16 = 38.76 N
F56 = 454.19 N
F = 454.19 N∠175.20° = −452.6ˆi + 38ˆj N
56
For friction between links 3 and 4, the friction angle is φ = tan −1 ( 0.10 ) = 5.71° . Since
the impending motion VA3 / 4 is upward the friction force f34 = μ c F34n is upward. Also,
since the non-friction normal force F34n is toward the left (from the solution for Prob.
P11.7), the total force F34 acts at the angle 163.37° − 5.71° = 157.66° . Therefore,
O4
= R BO4 × F54 + R AO4 × F34 = 0
(114.45ˆi + 383.275ˆj mm ) × ( 452.6ˆi − 38ˆj N )
+ ( 53.4ˆi + 178.85ˆj mm ) × ( cos157.66°ˆi + sin157.66°ˆj) F
34
( −180.176 N − m + 185.725 mmF34 ) kˆ = 0
∑M
O2
F34 = 957.45 N∠157.66°= − 885.58ˆi + 363.94ˆj N
= M12 + R AO2 × F32 = 0
(
=0
) (
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∑M
)
M 12kˆ + 54.125ˆi + 31.25ˆj mm × 885.58ˆi − 363.94ˆj N = 0
M 12 = 48.117 N − m
M12 = 48.117kˆ N − m
Ans.
11.16 Repeat Problem 11.12 assuming a coefficient of static friction μ = 0.15 between links 1
and 4. Determine the torque M12 necessary to overcome friction.
See the figure and solution for Problem P11.12 for the kinematic and frictionless
solution. For friction between links 1 and 4, the friction angle is φ = tan −1 ( 0.15 ) = 8.53° .
Since the impending motion VC4 /1 is to the right the friction force f14 = μ c F14n is toward
the left. Also, since the non-friction normal force F14n is upward (from the solution for
Prob. P11.12), the total force F14 acts at the angle 90° + 8.53° = 98.53° . Therefore,
∑M
A
= R BA × PB + R CA × PC + R CA × F14 = 0
(143.475ˆi − 100.225ˆj mm ) × ( 222.5ˆj N ) + ( 341.875ˆi − 75ˆj mm ) × ( −445ˆi N )
+ ( 341.875ˆi − 75ˆj mm ) × ( cos 98.53°ˆi + sin 98.53°ˆj) F = 0
14
32.431kˆ N − m − 33.9kˆ N − m + 326.95kˆ mmF14 = 0 , F14 = 4.42 N∠98.53°= − 0.654ˆi + 4.374ˆj N
∑F = P
+ PC + F14 + F23 = 0
222.5ˆj N − 445ˆi N − 0.654ˆi + 4.374ˆj N + F23 = 0 ,
B
∑M
O2
= M12 + R AO2 × F32 = 0
M 12 = −33.94 N − m
F23 =445.445ˆi − 226.95ˆj N = 500.18 N∠ − 26.98°
M kˆ + 75ˆj mm × −445.445ˆi + 226.95ˆj N = 0
12
(
) (
M12 = −33.94kˆ N − m
)
Ans.
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166
11.17 In each case shown, pinion 2 is the driver, gear 3 is an idler, the gears have module of
4 mm/tooth and 20° pressure angle. For each case, sketch the free-body diagram of
gear 3 and show all forces acting. For (a) pinion 2 rotates at 600 rev/min and transmits
13.43 kw to the gearset. For (b) and (c), pinion 2 rotates at 900 rev/min and transmits
18.65 kw to the gearset.
N m 34 teeth × 4 mm/tooth
N 2 m 18 teeth × 4 mm/tooth
=
= 36 mm R3 = 3 =
= 68 mm
2
2
2
2
( 600 rev/min )( 2π ) = 62.832 rad/s cw , ω = R2 ω = 36 mm 62.832 rad/s = 33.264 rad/s ccw
ω2 =
3
2
60 s/min
68 mm
R3
13.43 × 1000 w × (1 N − m) × (1000 mm/m)
P
=
= 5.937 kN
F23t =
R3ω3
( 68 mm )( 33.264 rad/s )
F23 = F23t cos φ = 5937 N cos20° = 6318 N , F43 = F23 = 6318 N
∑F = F
23
+ F43 + F13 = 0
F13 = F23t + F43t = 5937 + 5937 = 11.874 kN
Ans.
N m 36 teeth × 4 mm/tooth
N 2 m 18 teeth × 4 mm/tooth
= 72 mm
=
= 36 mm R3 = 3 =
2
2
2
2
( 900 rev/min )( 2π ) = 94.248 rad/s ccw , ω = R2 ω = 36 mm 94.248 rad/s = 47.124 rad/s cw
ω2 =
3
2
72 mm
60 s/min
R3
P
(18.65 kw) × (1000 w/kw) × (1 N − m/w) × (1000 mm/m)
F23t =
=
= 5497 N
R3ω3
( 72 mm )( 47.124 rad/s )
b) R2 =
F23 = F23t cos φ = 5497 N cos20° = 5849 N , F43 = F23 = 5849 N
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a) R2 =
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167
F13 = − ( F23 + F43 )
∑F = F
23
+ F43 + F13 = 0
= − ( 5849 N∠ − 20° + 5849 N∠110° )
= 4945 N∠225°
Ans.
N m 36 teeth × 4 mm/tooth
N 2 m 18 teeth × 4 mm/tooth
= 72 mm
=
= 36 mm R3 = 3 =
2
2
2
2
( 900 rev/min )( 2π ) = 94.248 rad/s ccw , ω = R2 ω = 36 mm 94.248 rad/s = 47.124 rad/s cw
ω2 =
3
2
60 s/min
R3
72 mm
(18.65 kw )(1000 w/kw )(1 N − m/w ×1000 mm/m ) = 5497 N
P
F23t =
=
R3ω3
( 75 mm )( 47.124 rad/s )
F23 = F23t cos φ = 5497 N cos20° = 5849 N , F43 = F23 = 5849 N
F13 = − ( F23 + F43 )
∑F = F
23
+ F43 + F13 = 0
= − ( 5849 N∠ − 20° + 5849 N∠ − 70° )
= 10.6 N∠ − 45°
Ans.
11.18 A 15-tooth spur pinion has a module of 5 mm/tooth and 20° pressure angle, rotates at
600 rev/min, and drives a 60-tooth gear. The drive transmits 18.65 kw. Construct a freebody diagram of each gear showing upon it the tangential and radial components of the
forces and their proper directions.
N m 60 teeth × 5 mm/tooth
N 2 m 15 teeth × 5 mm/tooth
=
= 37.5 mm R3 = 3 =
= 150 mm
2
2
2
2
( 600 rev/min )( 2π ) = 62.832 rad/s
37.5 mm
R
62.832 rad/s = 15.708 rad/s
ω3 = 2 ω2 =
ω2 =
60 s/min
150 mm
R3
R2 =
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c) R2 =
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168
P
R3ω3
=
(18.65 kw )(1000 w/kw )(1 N − m/w ) × (1000 mm/m ) = 7915 N, F r = F t tan φ = 2880.8 N
32
32
( 37.5 mm )( 62.832 rad/s )
F23t = F32t = 7915 N
F23r = F32r = 2880.8 N
11.19 A 16-tooth pinion on shaft 2 rotates at 1 720 rev/min and transmits 3.73 kw to the
double-reduction gear train. All gears have 20° pressure angle. The distances between
centers of the bearings and gears for shaft 3 are shown in the figure. Find the magnitude
and direction of the radial force that each bearing exerts against the shaft.
R2 =
N m 64 teeth × 3 mm/tooth
N 2 m 16 teeth × 3 mm/tooth
= 96 mm
=
= 24 mm RA = A =
2
2
2
2
N m 36 teeth × 4 mm/tooth
N B m 24 teeth × 4 mm/tooth
= 72 mm
=
= 48 mm R4 = A =
2
2
2
2
(1 720 rev/min )( 2π ) = 180.118 rad/s ω = R2 ω = 24 mm 180.118 rad/s = 45.029 rad/s
ω2 =
3
2
60 s/min
96 mm
RA
R
48 mm
ω4 = B ω3 =
45.029 rad/s = 30.020 rad/s
R4
72 mm
RB =
F23t =
P
RAω3
=
( 3.73 kw )(1000 w/kw )(1 N − m/w ) × (1000 mm/m ) = 862.8 N ,
( 96 mm )( 45.029 rad/s )
F23 = F23t cos φ = 918.17 N
R
96 mm
862.8 N = 1725.6 N
F43t = A F23t =
48 mm
RB
F43 = F43t cos φ = 1836.3 N
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F32t =
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Choosing a coordinate system with origin at C as shown we have
R A = −96ˆj + 48kˆ mm
FA = F23 = 918.17 N∠20° = 862.8ˆi + 314.03ˆj lb
R B = 48ˆj + 240kˆ mm
FB = F43 = 1836.3 N∠ − 20° = 1725.56ˆi − 628ˆj N
F = F x ˆi + F y ˆj
R =0
C
C
C
C
FD = F ˆi + FDy ˆj
∑M
R D = 288kˆ mm
x
D
C
= R A × FA + R B × FB + R D × FD = 0
( −96ˆj + 48kˆ mm ) × (862.8ˆi + 314.03ˆj N ) + ( 48ˆj + 240kˆ mm ) × (1725.56ˆi − 628ˆj N )
+ ( 288kˆ mm ) × ( F ˆi + F ˆj) = 0
( −15ˆi + 41.35ˆj + 82.72kˆ N − m ) + (150.7ˆi + 414ˆj − 82.72kˆ N − m ) + ( −288F ˆi + 288F ˆj mm ) = 0
x
D
y
D
y
D
FDx = −1495 N, FDy = 445 N
∑F = F
A
x
D
FD = 1557.5 N∠163.42° = −1495ˆi + 445ˆj N Ans.
+ FB + FC + FD = 0
(862.8ˆi + 314.03ˆj N ) + (1495ˆi − 445ˆj N ) + F + ( −1495ˆi + 445ˆj N ) = 0
C
FC = 961.2 N∠188.86° = −952.3ˆi − 146.85ˆj N Ans.
11.20 Solve Problem 11.17 if each pinion has right-hand helical teeth with a 30° helix angle
and a 20° pressure angle. All gears in the train are helical, and, of course, the normal
module is 4 mm/tooth for each case.
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170
N m 34 teeth × 4 mm/tooth
N 2 m 18 teeth × 4 mm/tooth
= 68 mm
=
= 36 mm R3 = 3 =
2
2
2
2
( 600 rev/min )( 2π ) = 62.832 rad/s cw , ω = R2 ω = 36 mm 62.832 rad/s = 33.264 rad/s ccw
ω2 =
3
2
60 s/min
R3
68 mm
(13.42 kw ×1000 w/kw ×1 N − m/w ×1000 mm/m )( 550 ft ⋅ lb/s/hp )(12 in/ft ) = 5932 N
P
=
F23t =
R3ω3
( 68 mm )( 33.268 rad/s )
a) R2 =
Since the pressure angles and the helix angle are related by cosψ = tan φn tan φt ,
φt = tan −1 ( tan φn cosψ ) = tan −1 ( tan 20° cos 30° ) = 22.80°
F23r = F23t tan φt = ( 5932 N ) tan 22.80° = 2493.5 N , F23a = F23t tanψ = ( 5932 N ) tan 30° = 3424.8 N
F = 5932ˆi − 2493.5ˆj + 3424.8kˆ N
F = 5923ˆi + 2493.5ˆj − 3424.8kˆ N
23
43
+ F13 = 0
ˆ
23 − R3 j × F43 + M13 = 0
23
43
3
F13 = −11.864ˆi N
Ans.
( 68ˆj mm ) × (5932ˆi − 2493.5ˆj + 3424.8kˆ N ) − ( 68ˆj mm ) × (5932ˆi + 2493.5ˆj − 3424.8kˆ N ) + M
( 68ˆj mm ) × (5932ˆi − 2493.5ˆj + 3424.8kˆ N ) − ( 68ˆj mm ) × (5932ˆi + 2493.5ˆj − 3424.8kˆ N ) + M
M13 = −465.45ˆi N − m This moment must be supplied by the shaft bearings.
13
=0
13
=0
Ans.
N m 36 teeth × 4 mm/tooth
N 2 m 18 teeth × 4 mm/tooth
=
= 36 mm R3 = 3 =
= 72 mm
2
2
2
2
( 900 rev/min )( 2π ) = 94.248 rad/s ccw , ω = R2 ω = 36 mm 94.248 rad/s = 47.124 rad/s cw
ω2 =
3
2
60 s/min
R3
72 mm
b) R2 =
F23t =
P
R3ω3
=
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∑F = F + F
∑ M =R ˆj × F
(18.65 kw )(1000 w/kw ) × (1000 w/kw ×1000 mm/m ) = 5496.7 N
( 72 mm )( 47.124 rad/s )
Since the pressure angles and the helix angle are related by cosψ = tan φn tan φt ,
φt = tan −1 ( tan φn cosψ ) = tan −1 ( tan 20° cos 30° ) = 22.80°
F23r = F23t tan φt = ( 5496.7 N ) tan 22.80° = 2310.59 N , F23a = F23t tanψ = ( 5496.7 N ) tan 30° = 3173.5 N
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171
F43 = −2310.59ˆi + 5496.7ˆj + 3173.5kˆ N
F13 = −3186.11ˆi − 3186.11ˆj N
Ans.
F23 = 5496.7ˆi − 2310.59ˆj − 3173.5kˆ N
∑F = F + F
∑ M =R ˆj × F
23
3
+ F13 = 0
ˆ
23 + R3 i × F43 + M13 = 0
43
( 72ˆj mm ) × (5496.7ˆi − 2310.59ˆj − 3173.5kˆ N ) + ( 72ˆi mm )
× ( −2310.59ˆi + 5496.7ˆj + 3173.5kˆ N ) + M = 0
13
N m 36 teeth × 4 mm/tooth
N 2 m 18 teeth × 4 mm/tooth
= 72 mm
=
= 36 mm R3 = 3 =
2
2
2
2
( 900 rev/min )( 2π ) = 94.248 rad/s ccw , ω = R2 ω = 36 mm 94.248 rad/s = 47.124 rad/s cw
ω2 =
3
2
60 s/min
R3
72 mm
P
(18.65 kw) × (1000 w/kw) × (1 N − m/w) × (1000 mm/m)
=
= 5496.7 N
F23t =
R3ω3
( 72 mm )( 47.124 rad/s )
c) R2 =
Since the pressure angles and the helix angle are related by cosψ = tan φn tan φt ,
φt = tan −1 ( tan φn cosψ ) = tan −1 ( tan 20° cos 30° ) = 22.80°
F23r = F23t tan φt = ( 5496.7 N ) tan 22.80° = 2310.59 N ,
F23a = F23t tanψ = ( 5496.7 N ) tan 30° = 3173.5 N
F23 = 5496.7ˆi − 2310.59ˆj − 3173.5kˆ N
F43 = 2310.59ˆi − 5496.7ˆj + 3173.5kˆ N
F = −7807.29ˆi + 7807.29ˆj N
F =F +F +F = 0
∑
∑ M =R ˆj × F
23
43
3
23
13
13
Ans.
+ R3ˆi × F43 + M13 = 0
( 72ˆj mm ) × (5496.7ˆi − 2310.59ˆj − 3173.5kˆ N ) + ( −72ˆi mm )
× ( 2310.59ˆi − 5496.7ˆj + 3173.5kˆ N ) + M = 0
13
M13 = 228.49ˆi − 228.49ˆj N − m This moment must be supplied by the shaft bearings. Ans.
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M13 = 228.49ˆi + 228.49ˆj N − m This moment must be supplied by the shaft bearings. Ans.
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172
11.21 Analyze the gear shaft of Example 11.8 and find the bearing reactions FC and FD .
The solution is shown in Fig. 11.20c.
11.22 In each of the bevel gear drives shown in the figure, bearing A takes both thrust load and
radial load, while bearing B takes only radial load. The teeth are cut with a 20° pressure
angle. For (a) T2 = −20ˆi N − m and for (b) T2 = −26.7kˆ N − m . Compute the bearing
loads for each case.
a) Γ = tan −1 ( 32 teeth 16 teeth ) = 63.43°
F32t = T2 R2 = 20 N − m/0.018 m = 1111 N
F32r = F32t tan φ cos Γ = 180.87 N
F23 = −180.87ˆi + 1111ˆj + 361.66kˆ N
F32a = F32t tan φ sin Γ = 361.66 N
A
= R BA × FB + R PA × F23 + T3 = 0
( −50kˆ mm ) × ( F ˆi + F ˆj) + (35ˆi − 59kˆ mm ) × ( −180.87ˆi + 1111ˆj + 361.66kˆ N ) + T kˆ = 0
(50 mmF ˆi − 50 mmF ˆj) + ( 70ˆi − 1.9ˆj + 40.68kˆ N − m ) + T kˆ = 0
x
B
y
B
y
B
x
B
T3 = −40.68kˆ N − m
∑F = F
A
+ FB + F23 = 0
b) γ = tan −1 (18 teeth 24 teeth ) = 36.87°
F32r = F32t tan φ cos γ = 242.95 N
F32 = 242.95ˆi − 834.375ˆj + 182.21kˆ N
∑M
A
3
3
FB = −37.825ˆi − 1370.6ˆj N
FA = 218.695ˆi + 259.0ˆj − 361.66kˆ N
Ans.
Ans.
F32t = T2 R2 = 26.7 N − m 32 mm = 834.375 N
F32a = F32t tan φ sin γ = 182.21 N
= R BA × FB + R PA × F32 + T2 = 0
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∑M
(50kˆ mm ) × ( F ˆi + F ˆj) + ( −32ˆi − 20kˆ mm ) × ( 242.95ˆi − 834.375ˆj + 182.21kˆ N ) + ( −26.7kˆ N − m ) = 0
( −50 mmF ˆi + 50 mmF ˆj) + ( −16.95ˆi + 0.9ˆj + 26.7kˆ N − m ) + ( −26.7kˆ N − m ) = 0
x
B
y
B
∑F = F
A
y
B
x
B
+ FB + F23 = 0
FB = −17.8ˆi − 333.75ˆj N
FA = −222.5ˆi + 1170.35ˆj − 182.45kˆ N
Ans.
Ans.
11.23 The figure shows a gear train composed of a pair of helical gears and a pair of straight
bevel gears. Shaft 4 is the output of the train and delivers 4.5 kw to the load at a speed of
370 rev/min. All gears have pressure angles of 20° . If bearing E is to take both thrust
load and radial load, while bearing F is to take only radial load, determine the forces that
each bearing exert against shaft 4.
The diameters of the bevel gears at their large ends are
R4 = mN 4 2 = 3 mm/tooth × ( 40 teeth ) 2 = 60 mm
R3 = mN 3 2 = 3 mm/tooth× ( 20 teeth ) 2 = 30 mm
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173
Γ = tan −1 ( R4 R3 ) = 63.43°
γ = tan −1 ( R3 R4 ) = 26.57°
The average pitch radii are
R4,avg = R4 − 0.500sin Γ = 59.55 mm
R3,avg = R3 − 0.500sin γ = 29.78 mm
ω4 =
( 370 rev/min )( 2π ) = 38.746 rad/s
60 s/min
4.5 kw ×1000 w/kw ×1 N − m/w ×1000 mm/m
P
=
= 1950 N
F34t =
R4,avgω4
( 59.55 mm )( 38.746 rad/s )
F34r = F34t tan φ cos Γ = 317.46 N
F34 = −634.78ˆi + 317.46ˆj − 1950kˆ N
∑M
E
F34a = F34t tan φ sin Γ = 634.78 N
= R FE × FF + R PE × F34 + T4 = 0
( 60ˆi mm ) × ( F ˆj + F kˆ ) + (18ˆi − 59.56ˆj mm ) × ( −634.78ˆi + 317.46ˆj − 1950kˆ N ) + ( −115.5ˆi N − m ) = 0
( −60 mmF ˆj + 60 mmF kˆ ) + (115.5ˆi + 40.7ˆj − 31kˆ N − m ) + ( −115.5ˆi N − m ) = 0
y
F
z
F
∑F = F
E
y
F
+ FF + F34 = 0
FF = 430.98ˆj + 564.192kˆ N
FE = 634.78ˆi − 748.34ˆj + 1386.97kˆ N
Ans.
Ans.
11.24 Using the data of Problem 11.23, find the forces exerted by bearings C and D onto shaft
3. Which of these bearings should take the thrust load if the shaft is to be loaded in
compression?
The pitch radius of the helical gear S is
RS = mN S 2 = 2 mm/tooth × 35 teeth 2 = 35 mm
F23t = F43t R3,avg RS = 1950 N ( 29.78 mm/35 mm ) = 1659.17 N
F23r = F23t tan φ = (1659.17 N ) tan 20° = 603.89 N
F = −603.89ˆi + 957.92ˆj + 1659.17kˆ N
F23a = F23t tanψ = (1659.17 N ) tan 30° = 957.92 N
23
∑M
C
= R DC × FD + R PC × F43 + R RC × F23 = 0
( 44ˆj mm ) × ( F ˆi + F kˆ ) + ( −29.78ˆi + 78.21ˆj mm ) × ( 634.78ˆi − 317.96ˆj + 1950kˆ N )
+ ( 35ˆi + 22ˆj mm ) × ( −603.89ˆi + 957.92ˆj + 1659.17kˆ N ) = 0
( 44 mmF ˆi − 44 mmF kˆ ) + (152.5ˆi + 58ˆj − 40kˆ N − m ) + (36.5ˆi − 58ˆj + 46.81kˆ N − m ) = 0
x
D
z
D
∑F = F
C
z
D
x
D
+ FD + F23 + F43 = 0
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z
F
FD = 133.5ˆi − 4192kˆ N
FC = −284.8ˆi − 538.45ˆj + 418.3kˆ N
Since the thrust force is in the −ˆj direction, C should be a thrust bearing.
Ans.
Ans.
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174
11.25 Use the method of virtual work to solve the slider-crank mechanism of Problem 11.2.
φ = sin −1 ⎡⎣( r sin θ ) ⎤⎦ = sin −1 ⎡⎣( 75 mm sin105° ) 350 mm ⎤⎦ = 11.95°
x = r cos θ + cos φ = 75 mm cos105° + 350 mm cos11.95° = 323 mm
The first-order kinematic coefficient is
x′ = dx dθ = yP24 = x tan φ = 323 mm tan11.95° = 68.36 mm
Ans.
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M 12 = Px′ = 900 N ( 68.36 mm ) = 61.5 N ⋅ m cw
11.26 Use the method of virtual-work to solve the four-bar linkage of Problem 11.5.
RP24O2 = 64.425 mm , RP24O4 = 114.425 mm , RDO4 = 175 mm∠152.64°
The first-order kinematic coefficient is
θ 4′ = dθ 4 dθ 2 = RP24O2 RP24O4 = 64.425 mm/114.425 mm = 0.563
M 14 = RDO4 P sin152.64° = 175 mm ( 222.5 N ) sin152.64° = 17.89 N − m cw
M 12 = − M 14 dθ 4 dθ 2 = − M 14θ 4′ = − (17.89 N − m cw )( 0.563) = 10.07 N − m ccw
Ans.
11.27 Use the method of virtual work to analyze the crank-shaper linkage of Problem 11.7.
Given that the load remains constant at P = 445ˆi N, find and plot a graph of the crank
torque M12 for all positions in the cycle using increments of 30° for the input crank.
x AO4 = RAO4 cos θ 4 = 63cos θ 2
⎛ 150 + 63sin θ 2 ⎞
⎟
⎝ 63cos θ 2 ⎠
θ 4 = tan −1 ⎜
y AO4 = RAO4 sin θ 4 = 150 + 63sin θ 2
RAO4 = 1056.25 + 750sin θ 2
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dθ 4 yP24O4 − 150
=
= 1 − 150sin θ 4 RAO4
dθ 2
yP24O4
yP24O4 = RAO4 sin θ 4
θ 4′ =
yBC = 200sin θ5 = 400 − 400sin θ 4
θ 5 = sin −1 ( 2 − 2sin θ 4 )
yP46C = yBC ( xC xCB )
= 200sin θ5 ⎡⎣( 200 cos θ5 − 400 cos θ 4 ) ( 200 cos θ 5 ) ⎤⎦
= 200 ( sin θ5 − 2 cos θ 4 tan θ5 )
dxC dθ 4 = yP46O4 = 400 + 200sin θ5 − 400 cos θ 4 tan θ5
M 12 = ( dxC dθ 4 )( dθ 4 dθ 2 ) P
Values for one cycle are shown in the following table.
θ 2 (deg.)
θ 4 (deg.)
RAO4 (mm)
dθ 4 dθ 2
θ 5 (deg.)
dxC dθ 4 (mm)
M 12 (N − m)
0
30
60
90
120
150
180
210
240
270
300
330
360
67.38
73.37
81.30
90.00
98.70
106.63
112.62
114.50
108.05
90.00
71.95
65.50
67.38
162.5
189.15
206.5
212.5
206.5
189.15
162.5
130.5
100.85
87.5
100.85
130.5
162.5
0.14793
0.24015
0.28197
0.29412
0.28197
0.24015
0.14793
-0.04593
-0.41416
-0.71429
-0.41416
-0.04593
0.14793
-8.85
-4.80
-1.32
0.00
-1.32
-4.80
-8.85
-10.37
-5.65
0.00
-5.65
-10.37
-8.85
393.18
392.88
396.79
400
393.99
373.65
395.48
333.64
368.04
400
392.57
394.35
393.18
26.29
42.65
50.57
53.18
50.21
40.56
23.09
-6.93
-68.89
-129.14
-73.49
-8.19
26.29
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175
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176
The values of M 12 from this table are graphed as follows:
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11.28 Use the method of virtual work to solve the four-bar linkage of Problem 11.10.
θ 3′ = dθ 3 dθ 2 = RP
RP23P13 = 75 mm 688 mm = −0.1089
dθ 2 = θ 3′kˆ × R BP13 = −0.1089kˆ × ( 568 mm∠135.33° ) = 61.881 mm∠45.33°
dθ = θ ′kˆ × R = −0.1089kˆ × ( 662 mm∠124.56° ) = 72.153 mm∠34.56°
23 P12
R′B = dR B
R′C = dR C
2
3
CP13
M 12 = PB iR′B + PC iR′C
= ( 500 N∠135° )i( 61.881 mm∠45.33° ) + (1 800 N∠0° )i( 72.153 mm∠34.56° )
= 30.940 cos89.67° N ⋅ m+129.876 cos 34.56° N ⋅ m
M 12 = 107 N ⋅ m cw
Ans.
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177
For the trailer:
R G3 B = 945ˆi + 863.15ˆj mm = 1280 mm∠42.41°
∑ M = 1250 mmF kˆ + R × W = 0
13
B
G3 B
3
F13 = 3364.2 N∠120° = −1682.1ˆi + 2914.75ˆj N
∑F = F
13
+ F23 + W3 = 0
Ans.
F23 = 2278.4 N∠42.41° = 1682.1ˆi + 1535.25ˆj N
For the car:
∑ M P = 2000 mmF12Rkˆ − 800 mm (8900 N ) kˆ + 2900ˆi + 325ˆj mm × F32 = 0
(
∑F = F
F
12
+ f 12 ˆi + F12R + F32 = 0
μ ≥ f12 F = 1682.1 N 4921.7 N
F
12
)
F12R = 5513.5ˆj N
F F = 4921.7ˆj N
Ans.
f12 = 1682.1ˆi N
Ans.
μ ≥ 0.342
Ans.
12
Ans.
11.30 Repeat Problem 11.29 assuming that the car has rear-wheel drive rather than front-wheel
drive.
The entire solution is identical with that of Problem P11.29 except that friction force f12
acts on the rear wheel of the car instead of on the front wheel. The solution process and
all values are the same until the final step. Then
μ ≥ 0.305
μ ≥ f12 F12R = 1682.1 N 5513.5 N
Ans.
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11.29 A car (link 2) which weighs 8900 N is slowly backing a 4450 N trailer (link 3) up a 30°
inclined ramp as shown in the figure. The car wheels are of 325 mm radius, and the
trailer wheels have 250 mm radius; the center of the hitch ball is also 325 mm above the
roadway. The centers of mass of the car and trailer are located at G2 and G3 ,
respectively, and gravity acts vertically downward in the figure. The weights of the
wheels and friction in the bearings are considered negligible. Assume that there are no
brakes applied on the car or on the trailer, and that the car has front-wheel drive.
Determine the loads on each of the wheels and the minimum coefficient of static friction
between the driving wheels and the road to avoid slipping.
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178
11.31 The low-speed disk cam with oscillating flat-faced follower shown in the figure is driven
at a constant shaft speed. The displacement curve for the cam has a full-rise cycloidal
motion, defined by Eq. (5.19) with parameters L = 30° , β = 30° , and a prime circle
radius Ro = 30 mm ; the instant pictured is at θ 2 = 112.5° . A force of FC = 8 N is
applied at point C and remains at 45° from the face of the follower as shown. Use the
virtual work approach to determine the moment M12 required on the crankshaft at the
instant shown to produce this motion.
From Eq. (5.19b),
2πθ ⎞ 30° ⎛
112.5° ⎞
L⎛
y′ = ⎜ 1 − cos
⎟=
⎜1 − cos 2π
⎟ = 0.200
β⎝
β ⎠ 150° ⎝
150° ⎠
From virtual work
M12 = − dθ 3 dθ 2 M13 = − y′M13 = −0.200 −0.849kˆ N ⋅ m = 0.170kˆ N ⋅ m
(
)
Ans.
11.32 Repeat Problem 11.31 for the entire lift portion of the cycle, finding M12 as a function of
θ2.
From Problem P11.31
M13 = R CO3 × FC = (150 mm )( 8 N ) sin − 135°kˆ = −0.849kˆ N ⋅ m
360°θ 2 ⎞
2πθ ⎞ 30° ⎛
L⎛
⎜ 1 − cos
⎟=
⎜ 1 − cos
⎟ = 0.200 (1 − cos 2.4θ 2 )
β⎝
β ⎠ 150° ⎝
150° ⎠
M12 = − y′M13 = −0.200 (1 − cos 2.4θ 2 ) −0.849kˆ N ⋅ m = 0.170 (1 − cos 2.4θ 2 ) kˆ N ⋅ m Ans.
y′ =
(
)
11.33 A disk 3 of radius R is being slowly rolled under a pivoted bar 2 driven by an applied
torque T as shown in the figure. Assume a coefficient of static friction of μ between the
disk and ground and that all other joints are frictionless. A force F is acting vertically
downward on the bar at a distance d from the pivot O2 . Assume that the weights of the
links are negligible in comparison to F. Find an equation for the torque T required as a
function of the distance X = RCO2 , and an equation for the final distance X that is
reached when friction no longer allows further movement.
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The moment on link 3 caused by the output load is
M13 = R CO3 × FC = (150 mm )( 8 N ) sin − 135°kˆ = −0.849kˆ N ⋅ m
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179
d cos θ
d
= FB
X cos θ
X
Rd
T = F23 R sin θ = FB
sin θ
∑ MC = 0
X
But, for geometric compatibility, R = X sin θ . Therefore,
O2
F32 = FB
=0
T = FB d sin 2 θ = FB d ( R X )
2
Ans.
Also,
R
secθ
1 + tan 2 θ
1
=R
=R
=R
+1
sin θ
tan θ
tan θ
tan 2 θ
Motion is still possible as long as tan θ ≤ μ , or as long as
X=
X ≥ R 1 μ 2 +1
Ans.
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∑M
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181
Chapter 12
Dynamic Force Analysis (Planar)1
12.1
The steel bell crank shown in the figure is used as an oscillating cam follower. Using
For the vertical arm, using Appendix Table 5,
m = wht ρ = (18.75 mm )( 93.75 mm )( 9.375 mm ) ( 7.83 Mg/m3 ) = 129 g
2
2
I G = m ( a 2 + c 2 ) 12 = (129 g/9810 mm/s 2 ) ⎡⎣(18.75 mm ) + ( 93.75 mm ) ⎤⎦ 12 = 10.02 g ⋅ mm ⋅ s 2
I O = I G + md 2 = 10.02 g ⋅ mm ⋅ s 2 + (129 g/98.10 mm/s 2 ) ( 37.5 mm ) = 28.51 g ⋅ mm ⋅ s 2
2
For the horizontal arm, using Appendix Table 5,
m = wht ρ = ( 9.375 mm )(18.75 mm )(150 mm ) ( 7.83 Mg/m3 ) = 206 g
2
2
I G = m ( a 2 + c 2 ) 12 = ( 206 g/9810 mm/s 2 ) ⎡⎣(18.75 mm ) + (150 mm ) ⎤⎦ 12 = 39.99 g ⋅ mm ⋅ s 2
I O = I G + md 2 = 39.99 g ⋅ mm ⋅ s 2 + ( 206 g/9810 mm/s 2 ) ( 84.375 mm ) = 189.48 g ⋅ mm ⋅ s 2
2
For the roller, using Appendix Table 5,
2
m = π r 2t ρ = π (12.5 mm ) (12.5 mm ) ( 7.83 Mg/m3 ) = 48 g
I G = mr 2 2 = ( 48 g 9810 mm/s 2 ) (12.5 mm ) 2 = 0.382 g ⋅ mm ⋅ s 2
2
I O = I G + md 2 = 0.382 g ⋅ mm ⋅ s 2 + ( 48 g 9810 mm/s 2 ) (150 mm ) = 110.47 g ⋅ mm ⋅ s 2
2
For the compositie lever
m = (129 g ) + ( 206 g ) + ( 48 g ) = 383 g
I O = ( 28.51 g ⋅ mm ⋅ s 2 ) + (189.48 g ⋅ mm ⋅ s 2 ) + (110.47 g ⋅ mm ⋅ s 2 ) = 328.46 g ⋅ mm ⋅ s 2
= 0.33 kg ⋅ mm ⋅ s 2
Ans.
1
Unless stated otherwise, solve all problems without friction and without gravitational loads.
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7.83 Mg/ m3 for the density of steel, find the mass moment of inertia of the lever about
an axis through O.
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182
12.2
A 5- by 50- by 300-mm steel bar has two round steel disks, each 50 mm in diameter and
20 mm long, welded to one end as shown. A small hole is drilled 25 mm from the other
end. The density of steel is 7.80 Mg/ m3 . Find the mass moment of inertia of this
weldment about an axis through the hole.
For the rectangular bar, using Appendix Table 5,
m = wht ρ = ( 0.050 m )( 0.300 m )( 0.005 m ) ( 7.80 Mg/m3 ) = 0.585 kg
2
2
I G = m ( a 2 + c 2 ) 12 = ( 0.585 kg ) ⎡( 0.050 m ) + ( 0.300 m ) ⎤ 12 = 0.004 509 kg ⋅ m 2
⎣
⎦
I O = I G + md 2 = 0.004 509 kg ⋅ m 2 + ( 0.585 kg )( 0.125 m ) = 0.013 650 kg ⋅ m 2
2
For the two circular disks, using Appendix Table 5,
2
m = 2π r 2t ρ = 2π ( 0.025 m ) ( 0.020 m ) ( 7.80 Mg/m3 ) = 0.613 kg
I G = mr 2 2 = ( 0.613 kg )( 0.025 m ) 2 = 0.000 191 kg ⋅ m 2
2
I O = I G + md 2 = 0.000 191 kg ⋅ m 2 + ( 0.613 kg )( 0.250 m ) = 0.038 480 kg ⋅ m 2
For the compositie lever
m = ( 0.585 kg ) + ( 0.613 kg ) = 1.198 kg
I O = ( 0.013 650 kg ⋅ m 2 ) + ( 0.038 480 kg ⋅ m 2 ) = 0.052 130 kg ⋅ m 2
12.3
Ans.
Find the external torque which must be applied to link 2 of the four-bar linkage shown to
drive it at the given velocity.
The D’Alembert inertia forces and offsets are:
f 2 = −m2 AG2 = 0
t 2 = − I G2 α 2 = 0
h2 = t2 f 2 = 0
f3 = − m3 A G3
f 4 = −m4 A G4
= −62.1ˆi − 7.37ˆj kg = 63.56 kg∠186.77°
t 3 = − I G3 α 3
= −22.4ˆi − 7.37ˆj kg = 26.33 kg∠198.21°
t 4 = − I G4 α 4
= − ( 0.3214 kg 9.81 m/s 2 ) (1896ˆi +225ˆj m/s 2 )
(
= − (17.47 g ⋅ cm ⋅ s 2 ) 4 950kˆ rad/s 2
)
= − ( 0.3214 kg 9.81 m/s 2 ) ( 684ˆi +225ˆj m/s 2 )
(
= − (12.91 g ⋅ cm ⋅ s 2 ) −8 900kˆ rad/s 2
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2
)
= 114.94kˆ kg ⋅ cm
= −86.48kˆ kg ⋅ cm = −0.8648kˆ kg ⋅ m
h3 = t3 f3 = ( 0.8648 kg − m ) ( 63.56 kg ) = 13.6 mm, h4 = t4 f 4 = (1.149 kg − m ) ( 26.33 kg ) = 43 mm
Next, the free body diagrams are drawn. Since the lines of action for the forces on the
free body diagrams can not be discovered from two- and three-force member concepts,
the force F34 is divided into radial and transverse components. (Notice that it is totally
coincidental that the reaction components are also exactly aligned with the radial and
transverse axes of link 3.)
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183
O4
= R G4O4 × f 4 + t 4 + R BO4 × F34t = 0
( −45ˆi + 60ˆj mm ) × ( −22.4ˆi − 7.37ˆj kg ) + (1.149kˆ kg − m ) + ( −90ˆi + 120ˆj mm )( 0.800ˆi + 0.600ˆj) F
(1.676kˆ kg − m ) + (1.149kˆ kg − m ) + ( 0.15kˆ m ) F34t = 0
t
34
=0
F34t = 19.98 kg
∑M = R ×f + t + R ×F = 0
(80ˆi + 60ˆj mm ) × ( −62.1ˆi − 7.37ˆj kg ) + ( −0.8648kˆ kg − m ) + (160ˆi + 120ˆj mm ) × ( 0.600ˆi − 0.800ˆj) F
A
3
G3 A
3
BA
r
43
(3.1364kˆ kg − m ) + ( −0.8648kˆ kg − m ) + ( 0.2kˆ m ) F43r = 0
F43r = −11.36 kg
∑F = F + f
∑M = R
43
O2
12.4
3
AO2
r
43
=0
+ F23 = 0
F34 = 9ˆi + 21.34ˆj kg = 23.15 kg∠67.26°
F = 72.18ˆi + 29ˆj kg = 77.63 kg∠21.82°
× F32 + M12 = 0
M12 = −21.47kˆ N − m or 2.167 kg − m Ans.
23
Crank 2 of the four-bar linkage shown in the figure is balanced. For the given angular
velocity of link 2, find the forces acting at each joint and the external torque that must be
applied to link 2.
The D’Alembert inertia forces and offsets are:
f 2 = −m2 A G2 = 0
t 2 = − I G2 α 2 = 0
h2 = t2 f 2 = 0
f3 = − m3 A G3
= − (1.2 kg 9.81 m/s 2 ) ( −948ˆi +78.6ˆj m/s 2 )
= 115.96ˆi − 9.61ˆj kg = 116.36 kg∠ − 4.74°
f 4 = −m4 A G4
= − ( 3 kg 9.81 m/s 2 ) ( −240ˆi − 633ˆj m/s 2 )
= 73.39ˆi + 193.58ˆj kg = 207 kg∠69.24°
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∑M
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184
t 3 = − I G3 α 3
(
= − ( 0.0066 N − m − s 2 ) −6 500kˆ rad/s 2
)
t 4 = − I G4 α 4
(
= − ( 0.059 N − m − s 2 ) −240kˆ rad/s 2
)
= 42.9kˆ N − m
= 14.16kˆ N − m
h3 = t3 f3 = ( 42.9 N − m ) (116.36 × 9.81 N ) = 37.58 mm ,
h4 = t4 f 4 = (14.16 N − m ) ( 207 × 9.81 N ) = 6.97 mm
∑M
O4
= R G4O4 × f 4 + t 4 + R BO4 × F34t = 0
( 36.52ˆi + 93.1ˆj mm ) × ( 720ˆi + 1899ˆj N ) + (14.16kˆ N − m ) + ( 73ˆi + 186ˆj mm )( 0.931ˆi − 0.365ˆj) F
( 2.32kˆ N − m ) + (14.16kˆ N − m ) + ( −0.2kˆ m ) F34t = 0
t
34
=0
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Next, the free body diagrams are drawn. Since the lines of action for the forces on the
free body diagrams can not be discovered from two- and three-force member concepts,
the force F34 is divided into radial and transverse components.
F34t = 82.4 N
∑M
A
= R G3 A × f3 + t 3 + R BA × F43t + R BA × F43r = 0
(181.35ˆi + 110.77ˆj mm ) × (1157ˆi − 97.9ˆj N ) + ( 44.52kˆ N − m ) + (362.7ˆi + 221.55ˆj mm ) × ( −75.65ˆi + 31.15ˆj N )
+ ( 362.7ˆi + 221.55ˆj mm ) × ( −0.365ˆi − 0.931ˆj) F = 0
r
43
( −148.2kˆ N − m ) + ( 44.52kˆ N − m ) + (18.66kˆ N − m ) + ( −297.175kˆ mm ) F43r = 0
F34 = −26.7ˆi − 293.7ˆj N = 293.7 N∠ − 95.06°
Ans.
F = −720.9ˆi − 1668.75ˆj N = 1815.6 N∠ − 113.27° Ans.
F43r = −280.35 N
∑F = F + f + F = 0
∑F = F + f + F = 0
∑F = F + F = 0
∑M = R ×F + M
O2
34
4
14
43
3
23
32
12
AO2
32
14
12
=0
F23 = −1183.7ˆi − 195.8ˆj N = 1201.5 N∠ − 170.57° Ans.
F12 = 1183.7ˆi + 195.8ˆj N = 1201.5 N∠9.43°
Ans.
M = −49.6kˆ N − m
Ans.
12
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185
For the angular velocity of crank 2 given in the figure, find the reactions at each joint and
the external torque to be applied to the crank.
The D’Alembert inertia forces and offsets are:
f 2 = −m2 A G2 = 0
t 2 = − I G2 α 2 = 0
h2 = t2 f 2 = 0
f 4 = −m4 A G4
f3 = − m3 A G3
= − (1.298 kg 9.81 m/s 2 ) ( −2394ˆi m/s 2 )
= − (1.54 kg 9.81 m/s 2 )( −2385ˆi − 1487 ˆj m/s 2 )
= 374.4ˆi + 233.4ˆj kg = 441.7 kg∠31.95°
= 43,330.7 N∠31.95°
t 3 = − I G3 α 3
(
= − ( 0.012 N − m ⋅ s 2 ) −7 670kˆ rad/s 2
)
= 316.75ˆi kg = 316.75 kg∠0.00°
= 3107.3 N∠0.00°
t 4 = − I G4 α 4 = 0
= 92kˆ N − m
h3 = t3 f3 = ( 92 N − m ) ( 43, 330.7 N ) = 21.23 mm , h4 = t4 f 4 = 0
Next, the free body diagrams are drawn and the solution proceeds.
∑M
A
= R G3 A × f3 + t 3 + R BA × f 4 + R BA × F14 = 0
(110.725ˆi − 19.875ˆj mm ) × (3675.7ˆi + 2291.75ˆj N ) + (92kˆ N − m )
+ ( 295.275ˆi − 53.025ˆj mm ) × ( 316.75 × 9.81ˆi N ) + ( 295.275ˆi − 53.025ˆj mm ) × ( ˆj) F
(331.99kˆ N − m ) + (92kˆ N − m ) + (167kˆ N − m ) + ( 0.295kˆ m ) F
14
F14 = −1975.8 N
∑F = f + F + F = 0
∑F = F + f + F = 0
∑F = F + F = 0
∑M = R ×F + M
4
O2
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12.5
14
34
43
3
23
32
12
AO2
32
14
=0
=0
F14 = −1975.8ˆj N = 1975.8 N∠ − 90°
F = −3107.3ˆi + 1975.8ˆj lb = 3675.7 N∠147.50°
34
Ans.
Ans.
F23 = −6777.3ˆi − 315.95ˆj N = 6788.25 N∠ − 177.33° Ans.
F = −6777.3ˆi − 315.95ˆj N = 6788.25 N∠ − 177.33° Ans.
12
12
=0
M12 = 348kˆ N − m
Ans.
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186
The figure shows a slider-crank mechanism with an external force FB applied to the
piston. For the given crank velocity, find all the reaction forces in the joints and the
crank torque.
The D’Alembert inertia forces and offsets are:
f 2 = −m2 A G2
= − ( 0.43 kg 9.81 m/s 2 )( −685.8ˆi + 396ˆj m/s 2 )
t 2 = − I G2 α 2 = 0
= 30.4ˆi − 17.7ˆj kg = 35.4 kg∠ − 30.00°
h2 = t2 f 2 = 0
f 4 = − m4 A G4
f3 = − m3 A G3
= − (1.59 kg 9.81 m/s 2 )( −1708.8ˆi + 680.1ˆj m/s 2 )
= 276.9ˆi − 110.2ˆj kg = 298 kg∠ − 21.70°
t 3 = − I G3 α 3
(
= − ( 0.012 Nms 2 ) −3 090kˆ rad/s 2
= − (1.13 kg 9.81 m/s 2 ) ( −1884ˆi m/s 2 )
= 217ˆi kg = 217 kg∠0.00°
)
www.elsolucionario.net
12.6
t 4 = − I G4 α 4 = 0
= 37kˆ N − m
h3 = t3 f3 = ( 37 N − m ) ( 298 × 9.81 N ) = 12.65 mm , h4 = t4 f 4 = 0
Next, the free body diagrams are drawn and the solution proceeds.
∑M
A
= R G3 A × f3 + t 3 + R BA × f 4 + R BA × FB + R BA × F14 = 0
( 0.087ˆi + 0.011ˆj m ) × ( 2716.4ˆi − 1081ˆj N ) + ( 37kˆ N − m ) + ( 0.298ˆi + 0.0375ˆj m ) × ( 2128.7ˆi N )
+ ( 0.298ˆi + 0.0375ˆj m ) × ( −3560ˆi N ) + ( 0.298ˆi + 0.0375ˆj m ) × ( ˆj) F = 0
14
( −123.93kˆ N − m ) + (37kˆ N − m ) + ( −79.83kˆ N − m ) + (133.5kˆ N − m ) + ( 0.298kˆ m ) F
14
F14 = 111.6 N
∑F = f + F + F + F = 0
∑F = F +f + F = 0
∑F = F + F = 0
∑M = R ×F + M = 0
4
O2
14
B
43
3
32
12
AO2
34
23
32
12
=0
F14 = 111.6ˆj N = 111.6 N∠90°
Ans.
F34 = 1388ˆi − 111.6ˆj N = 1393 N∠ − 4.95° Ans.
F23 = −1366ˆi + 975ˆj N = 1678 N∠144.50°
Ans.
F12 = −1664ˆi + 1148ˆj N = 2020 N∠145.40°
M = 12.2kˆ N − m
Ans.
12
Ans.
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187
The following data apply to the four-bar linkage shown in the figure: RAO2 = 0.3 m ,
RO4O2 = 0.9 m , RBA = 1.5 m , RBO4 = 0.8 m , RCA = 0.85 m , θC = 33° , RDO4 = 0.4 m ,
θ D = 53° , RG2O2 = 0 , RG3 A = 0.65 m , α = 16° , RG4O4 = 0.45 m , β = 17° ,
m2 = 5.2 kg , m3 = 65.8 kg , m4 = 21.8 kg , IG2 = 2.3 kg ⋅ m 2 , IG3 = 4.2 kg ⋅ m 2 ,
IG4 = 0.51 kg ⋅ m 2 . A kinematic analysis at θ 2 = 53° , ω2 = 12kˆ rad/s ccw , and α2 = 0 ,
θ3 = 0.7° ,
θ 4 = 20.4° ,
α 3 = 85.6 rad/s 2 cw ,
α 4 = 172 rad/s 2 cw ,
gave
AG3 = 96.4∠259° m/s 2 , and AG4 = 97.8∠270° m/s 2 . Find all the pin reactions and the
torque to be applied to link 2.
The D’Alembert inertia forces and offsets are:
f 2 = −m2 A G2 = 0
t 2 = − I G2 α 2 = 0
h2 = t2 f 2 = 0
f3 = − m3 A G3
f 4 = −m4 A G4
= −1 210ˆi − 6 227ˆj N = 6 343 N∠259°
t 3 = − I G3 α 3
= 2 132ˆj N = 2 132 N∠90°
t 4 = − I G4 α 4
= − ( 65.8 kg ) ( −18.394ˆi − 94.629ˆj m/s 2 )
(
= − ( 4.200 kg ⋅ m 2 ) −85.6kˆ rad/s 2
)
= − ( 21.8 kg ) ( −97.800ˆj m/s 2 )
(
= − ( 0.51 kg ⋅ m 2 ) −172kˆ rad/s 2
)
= 88kˆ N ⋅ m
= 360kˆ N ⋅ m
h3 = t3 f3 = ( 360 N ⋅ m ) ( 6 343 N ) = 0.057 m , h4 = t4 f 4 = ( 88 N ⋅ m ) ( 2 132 N ) = 0.041 m
Next, the free body diagrams are drawn. Since the lines of action for the forces on the
free body diagrams can not be discovered from two- and three-force member concepts,
the force F34 is divided into radial and transverse components.
∑M
O4
= R G4O4 × f 4 + t 4 + R BO4 × F34t = 0
( 0.390ˆi + 0.225ˆj m ) × ( 2 132ˆj N ) + ( 88kˆ N ⋅ m ) + ( 0.779ˆi + 0.180ˆj m ) × ( 0.226ˆi − 0.974ˆj) F
t
34
=0
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12.7
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188
(831kˆ N ⋅ m ) + (88kˆ N ⋅ m ) + ( −0.800kˆ m ) F34t = 0
F34t = 1 149 N
∑M
A
= R G3 A × f3 + t 3 + R BA × F43t + R BA × F43r = 0
( 0.631ˆi + 0.154ˆj m ) × ( −1 210ˆi − 6 227ˆj N ) + (360kˆ N ⋅ m ) + (1.498ˆi − 0.059ˆj m ) × ( −259ˆi + 1 119ˆj N )
+ (1.498ˆi − 0.059ˆj m ) × ( −0.974ˆi − 0.226ˆj) F = 0
r
43
F43r = −4 365 N
∑F = F +f + F = 0
∑F = F +f + F = 0
∑F = F + F = 0
∑M = R ×F + M
O2
12.8
F34 = −3 993ˆi − 2 104ˆj N = 4 513 N∠ − 152.21°
Ans.
Ans.
34
4
14
F14 = 3 993ˆi − 28ˆj N = 3 993 N∠ − 0.40°
43
3
23
F23 = −2 783ˆi + 4 123ˆj N = 4 974 N∠124.02° Ans.
32
12
AO2
32
12
F12 = −2 783ˆi + 4 123ˆj N = 4 974 N∠124.02° Ans.
M12 = 1 411kˆ N ⋅ m
Ans.
=0
Solve Problem 12.7 with an additional external force FD = 12∠0° kN acting at point D.
The method of superposition is used for the solution. The force components of Problem
P12.7 are denoted with primes and the additional force increments with double primes.
The figure here shows the incremental forces only.
∑ M′′
O4
= R DO4 × FD′′ + R BO4 × F34′′ = 0
( 0.162ˆi + 0.366ˆj m ) × (12 000ˆi N ) + ( 0.779ˆi + 0.180ˆj m )( −0.999ˆi + 0.040ˆj) F34′′ = 0
( −4 386kˆ N ⋅ m ) + ( 0.211kˆ m ) F ′′ = 0
F34′′ = 20 772 N
F34′′ = −20 756ˆi + 822ˆj N = 20 772 N∠177.73°
F′′ = 8 756ˆi − 822ˆj N = 8 794 N∠ − 5.36°
F34 = −24 749ˆi − 1 282ˆj N = 24 782 N∠182.97° Ans.
F = 12 749ˆi − 850ˆj N = 12 777 N∠ − 3.81° Ans.
F23′′ = −20 756ˆi + 822ˆj N = 20 772 N∠177.73°
F23 = −23 539ˆi + 4 945ˆj N = 24 053 N∠168.14° Ans.
F12′′ = −20 756ˆi + 822ˆj N = 20 772 N∠177.73°
F12 = −23 539ˆi + 4 945ˆj N = 24 053 N∠168.14° Ans.
M12 = 6 532kˆ N ⋅ m
Ans.
34
14
′′ = 5 121kˆ N ⋅ m
M12
14
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( −3 745kˆ N ⋅ m ) + (360kˆ N ⋅ m ) + (1 661kˆ N ⋅ m ) + ( −0.395kˆ in ) F43r = 0
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189
Make a complete kinematic and dynamic analysis of the four-bar linkage of Problem 12.7
using the same data, but with θ 2 = 170°, ω 2 = 12 rad/s ccw, α 2 = 0, and an external
force FD = 8.94∠64.3° kN acting at point D.
Kinematic Analysis:
) (
(
VA = ω 2 × R AO2 = 12kˆ rad/s × −0.295ˆi + 0.052ˆj m
)
= −0.625ˆi − 3.545ˆj m/s = 3.600 m/s∠ − 100°
VB = VA + ω3 × R BA = ω 4 × R BO4
= ( −0.625ˆi − 3.545ˆj m/s ) + (ω 3kˆ rad/s ) × (1.304ˆi + 0.740ˆj m ) = (ω 4kˆ rad/s ) × ( 0.109ˆi + 0.793ˆj m )
(
) (
) (
= −0.625ˆi − 3.545ˆj m/s + −0.740ω 3ˆi + 1.304ω 3 ˆj m = −0.793ω 4 ˆi + 0.109ω 4 ˆj m
ω3 = 3.020kˆ rad/s
ω 4 = 3.607kˆ rad/s
V = −2.859ˆi + 0.393ˆj m/s = 2.886 m/s∠172.16°
B
(
2
A A = −ω 22 R AO2 + α 2 × R AO2 = − (12 rad/s ) × −0.295ˆi + 0.052ˆj m
)
Ans.
)
= 42.543ˆi − 7.502ˆj m/s 2 = 43.200 m/s 2 ∠ − 10°
A B = A A − ω32 R BA + α 3 × R BA = −ω 42 R BO4 + α 4 × R BO4
)(
(
) (
) (
= ( −1.419ˆi − 10.311ˆj m/s ) + (α kˆ rad/s ) × ( 0.109ˆi + 0.793ˆj m )
= ( 32.069ˆi − 3.940ˆj m/s ) + ( −0.740α ˆi + 1.304α ˆj m ) = ( −0.793α ˆi + 0.109α ˆj m )
= 42.543ˆi − 7.502ˆj m/s 2 + −11.893ˆi − 6.749ˆj m/s 2 + α 3kˆ rad/s 2 × 1.304ˆi + 0.740ˆj m
2
)
2
4
2
3
α 3 = −0.390kˆ rad/s
3
AG = A A − ω R G A + α3 × R G A
3
2
3
(
4
4
α 4 = −40.816kˆ rad/s
2
3
3
) (
) (
2
)(
= 42.543ˆi − 7.502ˆj m/s 2 + −4.149ˆi − 4.234ˆj m/s 2 + −0.390kˆ rad/s 2 × 0.455ˆi + 0.464ˆj m
A G3 = 38.575ˆi − 11.913ˆj m/s = 40.373 m/s ∠ − 17.16°
2
AG = −ω R G O + α 4 × R G O
4
(
2
4
4 4
4 4
) (
2
)(
= 0.932ˆi − 5.780ˆj m/s2 + −40.816kˆ rad/s2 × −0.072ˆi + 0.444ˆj m
A G4 = 19.054ˆi − 2.841ˆj m/s 2 = 19.265 m/s 2 ∠ − 8.48°
Dynamic Analysis:
The D’Alembert inertia forces and offsets are:
t 2 = − I G2 α 2 = 0
f 2 = −m2 A G2 = 0
h2 = t2 f 2 = 0
Ans.
)
Ans.
)
Ans.
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12.9
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190
f3 = − m3 A G3
f 4 = −m4 A G4
= −2 538ˆi + 784ˆj N = 2 657 N∠162.84°
t 3 = − I G3 α 3
= −415ˆi + 62ˆj N = 420 N∠171.52°
t 4 = − I G4 α 4
= − ( 65.8 kg ) ( 38.575ˆi − 11.913ˆj m/s 2 )
(
= − ( 4.200 kg ⋅ m 2 ) −0.390kˆ rad/s 2
= − ( 21.8 kg ) (19.054ˆi − 2.841ˆj m/s 2 )
)
(
= − ( 0.51 kg ⋅ m 2 ) −40.816kˆ rad/s 2
)
= 21kˆ N ⋅ m
= 1.638kˆ N ⋅ m
h3 = t3 f3 = (1.638 N ⋅ m ) ( 2 657 N ) = 0.001 m , h4 = t4 f 4 = ( 21 N ⋅ m ) ( 420 N ) = 0.050 m
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Next, the free body diagrams are drawn. Since the lines of action for the forces on the
free body diagrams can not be discovered from two- and three-force member concepts,
the force F34 is divided into radial and transverse components.
∑M
O4
= R G4O4 × f 4 + t 4 + R DO4 × FD + R BO4 × F34t = 0
( −0.072ˆi + 0.444ˆj m ) × ( −415ˆi + 62ˆj N ) + ( 21kˆ N ⋅ m ) + ( −0.284ˆi + 0.282ˆj m ) × ( 3 877ˆi + 8 056ˆj N )
+ ( 0.109ˆi + 0.793ˆj m ) × ( 0.991ˆi − 0.136ˆj ) F = 0
t
34
(180kˆ N ⋅ m ) + ( 21kˆ N ⋅ m ) + ( −3 379kˆ N ⋅ m ) + ( −0.800kˆ m ) F34t = 0
F34t = −3 972 N
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191
∑M = R ×f + t + R ×F + R ×F = 0
( 0.455ˆi + 0.464ˆj m ) × ( −2 538ˆi + 784ˆj N ) + (1.638kˆ N ⋅ m ) + (1.304ˆi + 0.740ˆj m ) × ( 3 935ˆi − 542ˆj N )
+ (1.304ˆi + 0.740ˆj m ) × ( −0.136ˆi − 0.991ˆj) F = 0
A
G3 A
3
3
t
43
BA
r
43
BA
r
43
(1 534kˆ N ⋅ m ) + ( 2kˆ N ⋅ m ) + ( −3 618kˆ N ⋅ m ) + ( −1.192kˆ in ) F43r = 0
F43r = −1 747 N
F34 = −4 173ˆi − 1 189ˆj N = 4 339 N∠ − 164.10°
∑F = F + f + F + F = 0
∑F = F + f + F = 0
∑F = F + F = 0
∑M = R ×F + M = 0
4
D
43
3
23
32
12
O2
AO2
32
F14 = 711ˆi − 6 929ˆj N = 6 966 N∠ − 84.14° Ans.
14
12
F23 = −1 635ˆi − 1 972 N = 2 562 N∠ − 129.65°
Ans.
F12 = −1 635ˆi − 1 972 N = 2 562 N∠ − 129.65°
Ans.
M12 = 668kˆ N ⋅ m
Ans.
12.10 Repeat Problem 12.9 using θ 2 = 200°, ω 2 = 12 rad/s ccw, α 2 = 0, and an external force
FC = 8.49∠45° kN acting at point C.
Kinematic Analysis:
) (
(
VA = ω 2 × R AO2 = 12kˆ rad/s × −0.282ˆi − 0.103ˆj m
)
= 1.231ˆi − 3.383ˆj m/s = 3.600 m/s∠ − 70°
VB = VA + ω3 × R BA = ω 4 × R BO4
= (1.231ˆi − 3.383ˆj m/s ) + (ω 3kˆ rad/s ) × (1.198ˆi + 0.902ˆj m ) = (ω 4kˆ rad/s ) × ( 0.016ˆi + 0.799ˆj m )
(
) (
) (
= 1.231ˆi − 3.383ˆj m/s + −0.902ω 3ˆi + 1.198ω 3 ˆj m = −0.799ω 4 ˆi + 0.016ω 4 ˆj m
ω3 = 2.846kˆ rad/s
ω 4 = 1.671kˆ rad/s
V = −1.336ˆi + 0.027ˆj m/s = 1.337 m/s∠178.84°
B
(
A A = −ω 22 R AO2 + α 2 × R AO2 = − (12 rad/s ) × −0.282ˆi − 0.103ˆj m
2
)
Ans.
)
= 40.595ˆi + 14.775ˆj m/s 2 = 43.200 m/s 2 ∠20°
A B = A A − ω32 R BA + α 3 × R BA = −ω 42 R BO4 + α 4 × R BO4
)(
(
) (
) (
= ( −0.045ˆi − 2.233ˆj m/s ) + (α kˆ rad/s ) × ( 0.016ˆi + 0.799ˆj m )
= ( 30.937ˆi + 9.702ˆj m/s ) + ( −0.902α ˆi + 1.198α ˆj m ) = ( −0.799α ˆi + 0.016α ˆj m )
= 40.595ˆi + 14.775ˆj m/s 2 + −9.703ˆi − 7.306ˆj m/s2 + α 3kˆ rad/s2 × 1.198ˆi + 0.902ˆj m
2
)
2
4
2
3
3
α 3 = −8.760kˆ rad/s 2
A G3 = A A − ω R G3 A + α 3 × R G3 A
(
2
3
4
4
α 4 = −48.558kˆ rad/s 2
) (
) (
) (
= 40.595ˆi + 14.775ˆj m/s 2 + −3.169ˆi − 4.204ˆj m/s 2 + −8.760kˆ rad/s 2 × 0.391ˆi + 0.519ˆj m
A G3 = 41.972ˆi + 7.146ˆj m/s = 42.576 m/s ∠9.66°
2
2
Ans.
)
Ans.
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34
Ans.
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192
A G4 = −ω 42 R G4O4 + α 4 × R G4O4
(
) (
) (
= 0.343ˆi − 1.209ˆj m/s 2 + −48.558kˆ rad/s 2 × −0.123ˆi + 0.433ˆj m
)
A G4 = 21.365ˆi + 4.754ˆj m/s = 21.887 m/s ∠12.54°
2
2
Ans.
Dynamic Analysis:
The D’Alembert inertia forces and offsets are:
f 2 = −m2 A G2 = 0
t 2 = − I G2 α 2 = 0
h2 = t2 f 2 = 0
f3 = − m3 A G3
f 4 = −m4 A G4
= − ( 65.8 kg ) ( 41.972ˆi + 7.146ˆj m/s 2 )
= − ( 21.8 kg ) ( 21.365ˆi + 4.754ˆj m/s 2 )
(
= − ( 4.200 kg ⋅ m 2 ) −8.760kˆ rad/s 2
)
(
= − ( 0.51 kg ⋅ m 2 ) −48.558kˆ rad/s 2
)
= 37kˆ N ⋅ m
= 25kˆ N ⋅ m
h3 = t3 f3 = ( 37 N ⋅ m ) ( 2 802 N ) = 0.013 m , h4 = t4 f 4 = ( 25 N ⋅ m ) ( 477 N ) = 0.052 m
Next, the free body diagrams are drawn. Since the lines of action for the forces on the
free body diagrams can not be discovered from two- and three-force member concepts,
the force F34 is divided into radial and transverse components.
www.elsolucionario.net
= −2 762ˆi − 470ˆj N = 2 802 N∠ − 170.34° = −466ˆi − 104ˆj N = 477 N∠ − 167.46°
t 3 = − I G3 α 3
t 4 = − I G4 α 4
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193
∑M
O4
= R G4O4 × f 4 + t 4 + R BO4 × F34t = 0
( −0.123ˆi + 0.433ˆj m ) × ( −466ˆi − 104ˆj N ) + ( 25kˆ N ⋅ m ) + ( 0.016ˆi + 0.799ˆj m ) × ( 0.999ˆi − 0.020ˆj) F
t
34
( 214kˆ N ⋅ m ) + ( 25kˆ N ⋅ m ) + ( −0.800kˆ m ) F34t = 0
=0
F34t = 299 N
∑M = R ×f + t + R ×F + R ×F + R ×F = 0
( 0.391ˆi + 0.519ˆj m ) × ( −2 762ˆi − 470ˆj N ) + ( 37kˆ N ⋅ m ) + ( 0.291ˆi + 0.799ˆj m ) × ( 6 003ˆi + 6 003ˆj N )
+ (1.198ˆi + 0.902ˆj m ) × ( −299ˆi + 6ˆj N ) + (1.198ˆi + 0.902ˆj m ) × ( −0.020ˆi − 0.999ˆj) F = 0
A
G3 A
3
3
CA
C
BA
t
43
BA
r
43
r
43
(1 250kˆ N ⋅ m ) + (37kˆ N ⋅ m ) + ( −3 050kˆ N ⋅ m ) + ( 277kˆ N ⋅ m ) + ( −1.179kˆ in ) F43r = 0
∑F = F + f + F = 0
∑F = F + f + F + F = 0
∑F = F + F = 0
∑M = R ×F + M = 0
34
4
14
43
3
C
32
12
O2
AO2
23
32
12
F23 = −2 968ˆi − 6 799 N = 7 419 N∠ − 113.58°
Ans.
F12 = −2 968ˆi − 6 799 N = 7 419 N∠ − 113.58°
Ans.
M12 = 1 612kˆ N ⋅ m
Ans.
12.11 At θ 2 = 270° , ω 2 = 18 rad/s ccw , α 2 = 0 , a kinematic analysis of the linkage whose
geometry is given in Problem 12.7 gives θ3 = 46.6° , θ 4 = 80.5° , α 3 = 178 rad/s 2 cw ,
α 4 = 256 rad/s 2 cw , AG3 = 112∠22.7° m/s 2 , AG4 = 119∠352.5° m/s 2 .
An external
force FD = 8.94∠64.3° kN acts at point D. Make a complete dynamic analysis of the
linkage.
The D’Alembert inertia forces and offsets are:
f 2 = −m2 A G2 = 0
t 2 = − I G2 α 2 = 0
h2 = t2 f 2 = 0
f3 = − m3 A G3
= − ( 65.8 kg ) (103.324ˆi + 43.221ˆj m/s 2 )
= −6 799ˆi − 2 844ˆj N = 7 370 N∠ − 157.30°
t 3 = − I G3 α 3
(
= − ( 4.200 kg ⋅ m 2 ) −178kˆ rad/s 2
)
f 4 = −m4 A G4
= − ( 21.8 kg ) (117.982ˆi − 15.533ˆj m/s 2 )
= −2 572ˆi + 339ˆj N = 2 594 N∠172.50°
t 4 = − I G4 α 4
(
= − ( 0.51 kg ⋅ m 2 ) −256kˆ rad/s 2
)
= 748kˆ N ⋅ m
= 131kˆ N ⋅ m
h3 = t3 f3 = ( 748 N ⋅ m ) ( 7 370 N ) = 0.101 m , h4 = t4 f 4 = (131 N ⋅ m ) ( 2 594 N ) = 0.050 m
Next, the free body diagrams are drawn. Since the lines of action for the forces on the
www.elsolucionario.net
F34 = 274ˆi − 1 266ˆj N = 1 295 N∠ − 77.80° Ans.
F14 = 192ˆi + 1 370ˆj N = 1 383 N∠82.01° Ans.
F43r = −1 260 N
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194
∑M
O4
= R G4O4 × f 4 + t 4 + R DO4 × FD + R BO4 × F34t = 0
( −0.059ˆi + 0.446ˆj m ) × ( −2 572ˆi + 339ˆj N ) + (131kˆ N ⋅ m ) + ( −0.276ˆi + 0.290ˆj m ) × ( 3 877ˆi + 8 056ˆj N )
+ ( 0.131ˆi + 0.789 ˆj m ) × ( 0.986ˆi − 0.164ˆj ) F = 0
t
34
(1 127kˆ N ⋅ m ) + (131kˆ N ⋅ m ) + ( −3 348kˆ N ⋅ m ) + ( −0.800kˆ m ) F34t = 0
F34t = −2 613 N
∑M = R ×f + t + R ×F + R ×F = 0
( 0.300ˆi + 0.577ˆj m ) × ( −6 799ˆi − 2 844ˆj N ) + ( 748kˆ N ⋅ m ) + (1.031ˆi + 1.089ˆj m ) × ( 2 578ˆi − 429ˆj N )
+ (1.031ˆi + 1.089ˆj m ) × ( −0.164ˆi − 0.986ˆj) F = 0
A
G3 A
3
3
BA
t
43
r
43
BA
r
43
(3 070kˆ N ⋅ m ) + ( 748kˆ N ⋅ m ) + ( −3 250kˆ N ⋅ m ) + ( −0.839kˆ in ) F43r = 0
F43r = 677 N
∑F = F + f + F + F = 0
∑F = F +f + F = 0
∑F = F + F = 0
∑M = R ×F + M = 0
O2
34
4
D
43
3
23
32
12
AO2
32
14
12
F34 = −2 466ˆi + 1 097ˆj N = 2 699 N∠156.01° Ans.
F14 = −1 411ˆi − 9 153ˆj N = 9 261 N∠ − 98.76° Ans.
F23 = 9 265ˆi + 1 747 lb = 9 428 N∠10.68°
Ans.
F12 = 9 265ˆi + 1 747 lb = 9 428 N∠10.68°
Ans.
M12 = 2 780kˆ N ⋅ m
Ans.
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free body diagrams can not be discovered from two- and three-force member concepts,
the force F34 is divided into radial and transverse components.
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195
12.12 The following data apply to the four-bar linkage shown in the figure: RAO2 = 120 mm ,
RO4O2 = 300 mm , RBA = 320 mm , RBO4 = 250 mm , RCA = 360 mm , θC = 15° ,
RDO4 = 0 , θ D = 0° , RG2O2 = 0 , RG3 A = 200 mm , α = 8° , RG4O4 = 125 mm , β = 0° ,
m2 = 0.5 kg , m3 = 4 kg , m4 = 1.5 kg , IG2 = 0.005 N ⋅ m ⋅ s 2 , IG3 = 0.011 N ⋅ m ⋅ s 2 ,
IG4 = 0.002 3 N ⋅ m ⋅ s 2 . A kinematic analysis at θ 2 = 90° and ω 2 = 32 rad/s ccw with
α 2 = 0 gave θ3 = 23.9° , θ 4 = 91.7° , α 3 = 221 rad/s 2 ccw , α 4 = 122 rad/s 2 ccw ,
Using an external force
AG3 = 88.6∠255° m/s 2 , and AG4 = 32.6∠244° m/s 2 .
FC = 632∠342° N acting at point C, make a complete dynamic analysis of the linkage.
The D’Alembert inertia forces and offsets are:
t 2 = − I G2 α 2 = 0
f 2 = −m2 A G2 = 0
= − ( 4.0 kg ) ( −22.931ˆi − 85.581ˆj m/s 2 )
= 92ˆi + 342ˆj N = 354 N∠75°
t 3 = − I G3 α 3
(
= − ( 0.011 kg ⋅ m 2 ) 221kˆ rad/s 2
)
f 4 = −m4 A G4
= − (1.5 kg ) ( −14.291ˆi − 29.301ˆj m/s 2 )
= 21ˆi + 44ˆj N = 49 N∠64°
t 4 = − I G4 α 4
(
= − ( 0.002 3 kg ⋅ m 2 ) 122kˆ rad/s 2
)
= −2.431kˆ N ⋅ m
= −0.281kˆ N ⋅ m
h3 = t3 f3 = ( 2.431 N ⋅ m ) ( 354 N ) = 0.007 m , h4 = t4 f 4 = ( 0.281 N ⋅ m ) ( 49 N ) = 0.006 m
Next, the free body diagrams are drawn. Since the lines of action for the forces on the
free body diagrams can not be discovered from two- and three-force member concepts,
the force F34 is divided into radial and transverse components.
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h2 = t2 f 2 = 0
f3 = − m3 A G3
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196
∑M
O4
= R G4O4 × f 4 + t 4 + R BO4 × F34t = 0
( −0.004ˆi + 0.125ˆj m ) × ( 21ˆi + 44ˆj N ) + ( −0.281kˆ N ⋅ m ) + ( −0.008ˆi + 0.249ˆj m ) × ( 0.999ˆi + 0.030ˆj) F
t
34
( −2.844kˆ N ⋅ m ) + ( −0.281kˆ N ⋅ m ) + ( −0.250kˆ m ) F34t = 0
=0
F34t = −13 N
∑M = R ×f + t + R ×F + R ×F + R ×F = 0
( 0.170ˆi + 0.106ˆj m ) × ( 92ˆi + 342ˆj N ) + ( −2.431kˆ N ⋅ m ) + ( 0.280ˆi + 0.226ˆj m ) × ( 601ˆi − 195ˆj N )
+ ( 0.292ˆi + 0.130ˆj m ) × (12.5ˆi N ) + ( 0.292ˆi + 0.130ˆj m ) × ( 0.030ˆi − 0.999ˆj) F = 0
A
G3 A
3
3
CA
C
t
43
BA
r
43
BA
r
43
F43r = −496 N
∑F = F +f + F = 0
∑F = F + f + F + F = 0
∑F = F + F = 0
∑M = R ×F + M = 0
O2
34
4
14
43
3
C
32
12
AO2
23
32
12
F34 = 2ˆi − 496ˆj N = 496 N∠ − 89.71°
Ans.
F14 = −24ˆi + 452ˆj N = 453 N∠93.02°
Ans.
F23 = −690ˆi − 643 N = 944 N∠ − 137°
Ans.
F12 = −690ˆi − 643 N = 944 N∠ − 137°
Ans.
M12 = 82.83kˆ N ⋅ m
Ans.
12.13 Repeat Problem 12.12 at θ 2 = 260° . Analyze both the kinematics and the dynamics of
the system at this position.
Kinematic Analysis:
) (
(
VA = ω 2 × R AO2 = 32kˆ rad/s × −0.021ˆi − 0.118ˆj m
)
= 3.782ˆi − 0.667ˆj m/s = 3.840 m/s∠ − 10°
VB = VA + ω3 × R BA = ω 4 × R BO4
= ( 3.782ˆi − 0.667 ˆj m/s ) + (ω 3kˆ rad/s ) × ( 0.138ˆi + 0.289ˆj m ) = (ω 4 kˆ rad/s ) × ( −0.183ˆi + 0.171ˆj m )
(
) (
) (
= 3.782ˆi − 0.667 ˆj m/s + −0.289ω 3ˆi + 0.138ω 3 ˆj m = −0.171ω 4 ˆi − 0.183ω 4 ˆj m
)
ω3 = 10.554kˆ rad/s
ω 4 = −4.314kˆ rad/s
V = 0.736ˆi + 0.789ˆj m/s = 1.079 m/s∠46.99°
Ans.
B
(
2
A A = −ω 22 R AO2 + α 2 × R AO2 = − ( 32 rad/s ) × −0.021ˆi − 0.118ˆj m
)
= 21.338ˆi + 121.013ˆj m/s 2 = 122.880 m/s 2 ∠80°
A B = A A − ω32 R BA + α 3 × R BA = −ω 42 R BO4 + α 4 × R BO4
(
) (
) (
)(
= ( 3.402ˆi − 3.174ˆj m/s ) + (α kˆ rad/s ) × ( −0.183ˆi + 0.171ˆj m )
( 2.562ˆi + 92.029ˆj m/s ) + ( −0.289α ˆi + 0.138α ˆj m ) = ( −0.171α ˆi − 0.183α ˆj m )
= 21.338ˆi + 121.013ˆj m/s2 + −15.374ˆi − 32.158ˆj m/s 2 + α 3kˆ rad/s 2 × 0.138ˆi + 0.289ˆj m
2
2
4
2
3
3
4
4
)
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( 48kˆ N ⋅ m ) + ( −2.431kˆ N ⋅ m ) + ( −191kˆ N ⋅ m ) + ( −1.623kˆ N ⋅ m ) + ( −0.296kˆ in ) F43r = 0
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197
α 4 = −352.356kˆ rad/s 2
α 3 = −199.622kˆ rad/s 2
A G3 = A A − ω R G3 A + α 3 × R G3 A
(
2
3
) (
) (
Ans.
) (
= 21.338ˆi + 121.013ˆj m/s 2 + −6.718ˆi − 21.240ˆj m/s 2 + −199.622kˆ rad/s 2 × 0.060ˆi + 0.191ˆj m
A G3 = 52.748ˆi + 87.796ˆj m/s = 102.423 m/s ∠59.00°
2
A G4 = −ω 42 R G4O4 + α 4 × R G4O4
(
) (
2
) (
= 1.701ˆi − 1.587ˆj m/s 2 + −352.356kˆ rad/s 2 × −0.091ˆi + 0.085ˆj m
)
Ans.
)
A G4 = 31.651ˆi + 30.477ˆj m/s 2 = 43.939 m/s 2 ∠43.92°
Ans.
Dynamic Analysis:
h2 = t2 f 2 = 0
f3 = − m3 A G3
f 4 = −m4 A G4
= −211ˆi − 351ˆj N = 410 N∠ − 121.00°
t 3 = − I G3 α 3
= −47ˆi − 46ˆj N = 66 N∠ − 136.08°
t 4 = − I G4 α 4
= − ( 4.0 kg ) ( 52.748ˆi + 87.796ˆj m/s 2 )
(
= − ( 0.011 N ⋅ m ⋅ s 2 ) −199.622kˆ rad/s 2
= 2.196kˆ N ⋅ m
= − (1.5 kg ) ( 31.651ˆi + 30.477ˆj m/s 2 )
)
(
= − ( 0.002 3 N ⋅ m ⋅ s 2 ) −352.356kˆ rad/s 2
)
= 0.810kˆ N ⋅ m
h3 = t3 f3 = ( 2.196 N ⋅ m ) ( 410 N ) = 0.005 m , h4 = t4 f 4 = ( 0.810 N ⋅ m ) ( 66 N ) = 0.012 m
Next, the free body diagrams are drawn. Since the lines of action for the forces on the
free body diagrams can not be discovered from two- and three-force member concepts,
the force F34 is divided into radial and transverse components.
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The D’Alembert inertia forces and offsets are:
t 2 = − I G2 α 2 = 0
f 2 = −m2 A G2 = 0
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198
∑M
O4
= R G4O4 × f 4 + t 4 + R BO4 × F34t = 0
( −0.091ˆi + 0.085ˆj m ) × ( −47ˆi − 46ˆj N ) + ( 0.810kˆ N ⋅ m ) + ( −0.183ˆi + 0.170ˆj m ) × ( 0.682ˆi + 0.731ˆj) F
t
34
(8.212kˆ N ⋅ m ) + ( 0.810kˆ N ⋅ m ) + ( −0.250kˆ m ) F34t = 0
=0
F34t = 36 N
∑M = R ×f + t + R ×F + R ×F + R ×F = 0
( 0.060ˆi + 0.191ˆj m ) × ( −211ˆi − 351ˆj N ) + ( 2.196kˆ N ⋅ m ) + ( 0.066ˆi + 0.354ˆj m ) × ( 601ˆi − 195ˆj N )
+ ( 0.138ˆi + 0.289ˆj m ) × ( −25ˆi − 26ˆj N ) + ( 0.138ˆi + 0.289ˆj m ) × ( 0.731ˆi − 0.682ˆj) F = 0
A
G3 A
3
3
CA
C
BA
t
43
r
43
BA
r
43
F43r = −658 N
∑F = F +f + F = 0
∑F = F + f + F + F = 0
∑F = F + F = 0
∑M = R ×F + M = 0
O2
34
4
14
43
3
C
32
12
AO2
23
32
12
F34 = 505ˆi − 422ˆj N = 659 N∠ − 39.88°
Ans.
F14 = −458ˆi + 468ˆj N = 655 N∠134.39°
Ans.
F23 = 115ˆi + 124 lb = 169 N∠46.97°
Ans.
F12 = 115ˆi + 124 lb = 169 N∠46.97°
Ans.
M12 = 11.03kˆ N ⋅ m
Ans.
12.14 Repeat Problem 12.13 at θ 2 = 300° .
Kinematic Analysis:
) (
(
VA = ω 2 × R AO2 = 32kˆ rad/s × 0.060ˆi − 0.104ˆj m
)
= 3.326ˆi + 1.920ˆj m/s = 3.840 m/s∠30°
VB = VA + ω3 × R BA = ω 4 × R BO4
= ( 3.326ˆi + 1.920ˆj m/s ) + (ω 3kˆ rad/s ) × ( 0.093ˆi + 0.306ˆj m ) = (ω 4kˆ rad/s ) × ( −0.147ˆi + 0.202ˆj m )
(
) (
) (
= 3.326ˆi + 1.920ˆj m/s + −0.306ω 3ˆi + 0.093ω 3 ˆj m = −0.202ω 4 ˆi − 0.147ω 4 ˆj m
)
ω3 = 1.597kˆ rad/s
ω 4 = −14.071kˆ rad/s
V = 2.844ˆi + 2.066ˆj m/s = 3.515 m/s∠36.04°
Ans.
B
(
2
A A = −ω 22 R AO2 + α 2 × R AO2 = − ( 32 rad/s ) × 0.060ˆi − 0.104ˆj m
)
= −61.440ˆi + 106.417ˆj m/s 2 = 122.880 m/s 2 ∠120°
A B = A A − ω32 R BA + α 3 × R BA = −ω 42 R BO4 + α 4 × R BO4
(
) (
) (
)(
= ( 29.105ˆi − 39.995ˆj m/s ) + (α kˆ rad/s ) × ( −0.147ˆi + 0.202ˆj m )
( −90.782ˆi + 145.632ˆj m/s ) + ( −0.306α ˆi + 0.093α ˆj m ) = ( −0.202α ˆi − 0.147α ˆj m )
= −61.440ˆi + 106.417ˆj m/s2 + −0.237ˆi − 0.780ˆj m/s 2 + α 3kˆ rad/s 2 × 0.093ˆi + 0.306ˆj m
2
)
2
4
2
3
α 3 = −671.302kˆ rad/s
2
3
4
α 4 = −567.507kˆ rad/s
4
2
Ans.
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(19kˆ N ⋅ m ) + ( 2.196kˆ N ⋅ m) + ( −226kˆ N ⋅ m ) + (3.629kˆ N ⋅ m ) + ( −0.305kˆ in ) F43r = 0
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199
A G3 = A A − ω32 R G3 A + α 3 × R G3 A
(
) (
) (
) (
= −61.440ˆi + 106.417ˆj m/s 2 + −0.079ˆi − 0.504ˆj m/s 2 + −671.302kˆ rad/s 2 × 0.031ˆi + 0.198ˆj m
A G3 = 71.399ˆi + 85.103ˆj m/s = 111.087 m/s ∠50.00°
2
A G4 = −ω 42 R G4O4 + α 4 × R G4O4
(
2
) (
) (
= 14.547ˆi − 20.022ˆj m/s 2 + −567.507kˆ rad/s 2 × −0.073ˆi + 0.101ˆj m
)
Ans.
)
A G4 = 71.865ˆi + 21.406ˆj m/s 2 = 74.986 m/s 2 ∠16.59°
Ans.
Dynamic Analysis:
h2 = t2 f 2 = 0
f3 = − m3 A G3
f 4 = −m4 A G4
= −286ˆi − 340ˆj N = 444 N∠ − 130.00°
t 3 = − I G3 α 3
= −108ˆi − 32ˆj N = 112 N∠ − 163.41°
t 4 = − I G4 α 4
= − ( 4.0 kg ) ( 71.399ˆi + 85.103ˆj m/s 2 )
(
= − ( 0.011 N ⋅ m ⋅ s 2 ) −671.302kˆ rad/s 2
= 7.384kˆ N ⋅ m
= − (1.5 kg ) ( 71.865ˆi + 21.406ˆj m/s 2 )
)
(
= − ( 0.002 3 N ⋅ m ⋅ s 2 ) −567.507kˆ rad/s 2
)
= 1.305kˆ N ⋅ m
h3 = t3 f3 = ( 7.384 N ⋅ m ) ( 444 N ) = 0.017 m , h4 = t4 f 4 = (1.305 N ⋅ m ) (112 N ) = 0.012 m
Next, the free body diagrams are drawn. Since the lines of action for the forces on the
free body diagrams can not be discovered from two- and three-force member concepts,
the force F34 is divided into radial and transverse components.
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The D’Alembert inertia forces and offsets are:
f 2 = −m2 A G2 = 0
t 2 = − I G2 α 2 = 0
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200
∑M
O4
= R G4O4 × f 4 + t 4 + R BO4 × F34t = 0
( −0.073ˆi + 0.101ˆj m ) × ( −108ˆi − 32ˆj N ) + (1.305kˆ N ⋅ m ) + ( −0.147ˆi + 0.202ˆj m ) × ( 0.809ˆi + 0.588ˆj) F
t
34
(13.273kˆ N ⋅ m ) + (1.305kˆ N ⋅ m ) + ( −0.250kˆ m ) F34t = 0
=0
F34t = 58 N
∑M = R ×f + t + R ×F + R ×F + R ×F = 0
( 0.032ˆi + 0.197ˆj m ) × ( −286ˆi − 340ˆj N ) + ( 7.384kˆ N ⋅ m ) + ( 0.013ˆi + 0.360ˆj m ) × ( 601ˆi − 195ˆj N )
+ ( 0.094ˆi + 0.306ˆj m ) × ( −47ˆi − 34ˆj N ) + ( 0.094ˆi + 0.306ˆj m ) × ( 0.588ˆi − 0.809ˆj) F = 0
A
G3 A
3
3
CA
C
BA
t
43
BA
r
43
r
43
F43r = −603 N
∑F = F +f + F = 0
∑F = F + f + F + F = 0
∑F = F + F = 0
∑M = R ×F + M = 0
O2
34
4
14
43
3
C
32
12
AO2
32
23
12
F34 = 401ˆi − 453ˆj N = 606 N∠ − 48.50°
Ans.
F14 = −293ˆi + 486ˆj N = 567 N∠121.13°
Ans.
F23 = 86ˆi + 81 N = 119 N∠43.26°
Ans.
F12 = 86ˆi + 81 N = 119 N∠43.26°
Ans.
M12 = 13.85kˆ N ⋅ m
Ans.
12.15 Analyze the dynamics of the offset slider-crank mechanism shown in the figure using the
following data: a = 0.06 m , RAO2 = 0.1 m , RAB = 0.38 m , RCA = 0.4 m , θ 2 = 32° ,
RG3 A = 0.26 m , α = 22° , m2 = 2.5 kg , m3 = 7.4 kg , m4 = 2.5 kg , IG2 = 0.005 N ⋅ m ⋅ s 2 ,
I = 0.013 6 N ⋅ m ⋅ s 2 , θ = 120° , and ω = 18 rad/s cw with α = 0 , F = −2 000ˆi N ,
2
2
G3
2
B
FC = −1 000ˆi N . Assume a balanced crank and no friction forces.
Kinematic Analysis:
) (
(
VA = ω 2 × R AO2 = −18kˆ rad/s × −0.050ˆi + 0.087ˆj m
)
= 1.559ˆi + 0.900ˆj m/s = 1.800 m/s∠30°
VB = VA + ω3 × R BA
V ˆi = (1.559ˆi + 0.900ˆj m/s ) + (ω kˆ rad/s ) × ( 0.351ˆi − 0.147 ˆj m )
B
3
(
) (
= 1.559ˆi + 0.900ˆj m/s + 0.147ω 3 ˆi + 0.351ω 3 ˆj m
ω3 = −2.567kˆ rad/s
)
VB = 1.182ˆi m/s
(
Ans.
2
A A = −ω 22 R AO2 + α 2 × R AO2 = − (18 rad/s ) × −0.050ˆi + 0.087ˆj m
)
= 16.200ˆi − 28.059ˆj m/s 2 = 32.400 m/s 2 ∠ − 60°
A B = A A − ω32 R BA + α 3 × R BA
(
) (
) (
)(
AB ˆi = 16.200ˆi − 28.059ˆj m/s 2 + −2.310ˆi + 0.966ˆj m/s 2 + α 3kˆ rad/s 2 × 0.351ˆi − 0.147ˆj m
)
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( 46kˆ N ⋅ m ) + ( 7.384kˆ N ⋅ m ) + ( −219kˆ N ⋅ m ) + (11.170kˆ N ⋅ m ) + ( −0.256kˆ in ) F43r = 0
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201
(
) (
AB ˆi = 13.890ˆi − 27.093ˆj m/s 2 + 0.147α 3ˆi + 0.351α 3ˆj m
α 3 = 77.188kˆ rad/s
A G3 = A A − ω R G3 A + α 3 × R G3 A
(
2
3
)
A B = 25.156ˆi m/s 2
2
) (
) (
) (
= 16.200ˆi − 28.059ˆj m/s 2 + −1.713ˆi + 0.021ˆj m/s 2 + 77.188kˆ rad/s 2 × 0.260ˆi − 0.003ˆj m
A G3 = 14.466ˆi − 7.969ˆj m/s 2 = 16.516 m/s 2 ∠ − 28.85°
Ans.
)
Ans.
Dynamic Analysis:
The D’Alembert inertia forces and offsets are:
t 2 = − I G2 α 2 = 0
f 2 = −m2 A G2 = 0
h2 = t2 f 2 = 0
f3 = − m3 A G3
f 4 = −m4 A B
= − ( 7.4 kg ) (14.466ˆi − 7.969ˆj m/s 2 )
(
= − ( 0.013 6 N ⋅ m ⋅ s 2 ) 77.188kˆ rad/s 2
= −63ˆi N = 63 N∠180°
)
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= −107ˆi + 59ˆj N = 122 N∠151.15°
t 3 = − I G3 α 3
= − ( 2.5 kg ) ( 25.156ˆi m/s 2 )
t 4 = − I G4 α 4 = 0
= −1.050kˆ N ⋅ m
h3 = t3 f3 = (1.050 N ⋅ m ) (122 N ) = 0.009 m , h4 = t4 f 4 = 0
∑M
A
= R G3 A × f3 + t 3 + R CA × FC + R BA × f 4 + R BA × FB + R BA × F14 = 0
( 0.260ˆi − 0.003ˆj m ) × ( −107ˆi + 59ˆj N ) + ( −1.050kˆ N ⋅ m ) + ( 0.395ˆi + 0.065ˆj m ) × ( −1 000ˆi N )
+ ( 0.351ˆi − 0.147 ˆj m ) × ( −63ˆi N ) + ( 0.351ˆi − 0.147 ˆj m ) × ( −2 000ˆi N ) + ( 0.351ˆi − 0.147 ˆj m ) × (1.000ˆj ) F
14
=0
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202
(15kˆ N ⋅ m ) + ( −1.050kˆ N ⋅ m ) + ( 65kˆ N ⋅ m ) + ( −9.261kˆ N ⋅ m ) + ( −294kˆ N ⋅ m ) + ( 0.351kˆ in ) F
=0
F14 = 639 N
Ans.
14
F14 = 639ˆj N = 639 N∠90.00°
∑F = F + f + F + F = 0
∑F = F + f + F + F = 0
∑F = F + F = 0
∑M = R ×F + M = 0
O2
14
4
B
34
F34 = 2 063ˆi − 639ˆj N = 2 160 N∠ − 17.21° Ans.
43
3
C
23
F23 = 3 170ˆi − 698 N = 3 246 N∠ − 12.42°
Ans.
32
12
F12 = 3 170ˆi − 698 N = 3 246 N∠ − 12.42°
Ans.
M12 = −241kˆ N ⋅ m
Ans.
AO2
32
12
12.16 Analyze the system of Problem 12.15 for a complete rotation of the crank. Use FC = 0
and FB = −1 000ˆi N when VB is toward the right, but use FB = 0 when VB is toward
the left. Plot a graph of M12 versus θ 2 .
j 0.060 + 0.100e jθ 2 + 0.380e jθ3 = RB
0.060 + 0.100sin θ 2 + 0.380sin θ 3 = 0
θ 3 = − sin −1 ( 0.158 + 0.263sin θ 2 )
RB = 0.100 cosθ 2 + 0.380 cos θ 3
R G3 = jR1 + R2 e jθ 2 + RG3 Ae
j (θ 3 +α )
R G3 = j 0.060 + 0.100e jθ 2 + 0.260e
j (θ 3 + 22° )
R G3 = ⎡⎣0.100 cosθ 2 + 0.260 cos (θ 3 + 22° ) ⎤⎦ + j ⎡⎣ 0.060 + 0.100sin θ 2 + 0.260sin (θ 3 + 22° ) ⎤⎦
The first-order kinematic coefficients are found as follows:
j 0.100e jθ2 + j 0.380e jθ3θ 3′ = R4′
−0.100sin θ 2 − 0.380sin θ 3θ 3′ = R4′
0.100 cosθ 2 + 0.380 cosθ 3θ 3′ = 0
RB′ = −0.100 ( sin θ 2 − cos θ 2 tan θ 3 )
θ 3′ = −0.263cosθ 2 cosθ 3
R′G3 = j 0.100e jθ 2 + j 0.260e
j (θ 3 + 22° )
θ 3′
R′G3 = ⎡⎣ −0.100sin θ 2 − 0.260sin (θ 3 + 22° )θ 3′ ⎤⎦ + j ⎡⎣0.100 cos θ 2 + 0.260 cos (θ 3 + 22° )θ 3′ ⎤⎦
Similarly, the second-order kinematic coefficients are as follows:
−0.100e jθ2 + j 0.380e jθ3θ 3′′ − 0.380e jθ3θ 3′2 = RB′′
−0.100sin θ 2 + 0.380 cosθ 3θ 3′′ − 0.380sin θ 3θ 3′2 = 0
−0.100 cosθ 2 − 0.380sin θ 3θ 3′′ − 0.380 cosθ 3θ 3′2 = RB′′
θ 3′′ = 0.263sin θ 2 cosθ 3 + tan θ 3θ 3′2 ,
R′′G3 = −0.100e jθ2 + j 0.260e
j (θ 3 + 22° )
RB′′ = − ⎡⎣ 0.100 cos (θ 3 − θ 2 ) + 0.380θ 3′2 ⎤⎦ cosθ 3
θ 3′′ − 0.260e j (θ + 22°)θ 3′2
3
R′′G3 = ⎡⎣ −0.100 cos θ 2 − 0.260sin (θ 3 + 22° )θ 3′′ − 0.260 cos (θ 3 + 22° )θ 3′2 ⎤⎦
+ j ⎡⎣ −0.100sin θ 2 + 0.260 cos (θ 3 + 22° )θ 3′′ − 0.260sin (θ 3 + 22° )θ 3′2 ⎤⎦
By virtual work we can formulate the dynamic input torque requirement as:
M12 = f3 iR′G3 + t 3 iθ 3′kˆ + f 4 iR′B + FB iR′B
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jR1 + R2 e jθ 2 + R3e jθ3 = R4
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203
The individual elements of this equation are:
f3 = − m3 A G3 = − m3 R′′G3ω 22 = − ( 2 397.6 kg/s 2 ) R′′G3
f3 = ⎡⎣ 240 cos θ 2 + 623sin (θ 3 + 22° )θ 3′′ + 623cos (θ 3 + 22° )θ 3′2 ⎤⎦
+ j ⎡⎣ 240sin θ 2 − 623cos (θ 3 + 22° )θ 3′′ + 623sin (θ 3 + 22° )θ 3′2 ⎤⎦
f3 i R ′G = ⎡⎣ 240 cos θ 2 + 623 sin (θ 3 + 22° ) θ 3′′ + 623 cos (θ 3 + 22° ) θ 3′
2
3
⎤⎦ [ −0.100 sin θ 2 − 0.260 sin (θ 3 + 22° ) θ 3′ ]
2
+ ⎡⎣ 240 sin θ 2 − 623 cos (θ 3 + 22° ) θ 3′′ + 623 sin (θ 3 + 22° ) θ 3′ ⎤⎦ [ 0.100 cos θ 2 + 0.260 cos (θ 3 + 22° ) θ 3′ ]
f3 iR ′G3 = −62sin (θ 3 + 22° − θ 2 )θ 3′ (1 − θ 3′ ) − 62 cos (θ 3 + 22° − θ 2 )θ 3′′ + 162θ 3′θ 3′′
t = − I α = − I θ ′′ω 2kˆ = − ( 4.406 N ⋅ m )θ ′′kˆ
3
3
G3
G3 3
2
3
t 3 iθ 3′kˆ = − ( 4.406 N ⋅ m )θ 3′θ 3′′
f 4 = ⎡⎣81cos (θ 3 − θ 2 ) + 308θ 3′2 ⎤⎦ ˆi N cosθ 3
f 4 iR′B = sin (θ 3 − θ 2 ) ⎡⎣8.1cos (θ 3 − θ 2 ) + 30.8θ 3′2 ⎤⎦ N ⋅ m cos 2 θ 3
{
}
FB = 500 ⎣⎡1 + sgn ( cos θ 2 tan θ3 − sin θ 2 ) ⎦⎤ ˆi N=500 1+sgn ⎣⎡sin (θ3 − θ 2 ) cosθ3 ⎦⎤ ˆi N
{
}
FB iR′B = 50 1+sgn ⎡⎣sin (θ 3 − θ 2 ) cosθ 3 ⎤⎦ sin (θ 3 − θ 2 ) cosθ 3 N ⋅ m
Finally, putting these pieces together, we obtain:
M 12 = −62sin (θ 3 + 22° − θ 2 )θ 3′ (1 − θ 3′ ) − 62 cos (θ 3 + 22° − θ 2 )θ 3′′ + 158θ 3′θ 3′′
+ sin (θ 3 − θ 2 ) ⎡⎣8.1cos (θ 3 − θ 2 ) + 30.8θ 3′2 ⎤⎦ cos 2 θ 3
{
}
+ 50 1+sgn ⎡⎣sin (θ 3 − θ 2 ) cosθ 3 ⎤⎦ sin (θ 3 − θ 2 ) cosθ 3 N ⋅ m
The plot of this torque requirement is shown below. The sinusoidal curve in the first half
of the cycle is caused primarily by the mass of the connecting rod; the mass of the piston
is included also. The applied force FB causes the rise in the second half of the cycle.
Note that the mass of link 3 causes dynamic torque which helps to overcome up to one
third of the applied force effect.
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f 4 = −m4 A B = − m4 R′′Bω 22 = − ( 810 kg/s 2 ) R′′B
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204
12.17 A slider-crank mechanism similar to that of Problem 12.15 has a = 0 , RAO2 = 0.1 m ,
RBA = 0.45 m , RCA = 0 , θC = 0 , RG3 A = 0.2 m , α = 0 , m2 = 1.5 kg , m3 = 3.5 kg ,
m4 = 1.2 kg ,
IG2 = 0.01 N ⋅ m ⋅ s 2 ,
IG3 = 0.060 N ⋅ m ⋅ s 2 ,
and
M12 = 60 N ⋅ m .
Corresponding to θ 2 = 120° and ω 2 = 24 rad/s cw with α 2 = 0 , a kinematic analysis
gave θ = −9° , R = 0.374 m , α = 89.3 rad/s 2 ccw , A = 40.6ˆi m/s 2 , and
3
3
B
B
2
AG3 = 40.6ˆi − 22.6ˆj m / s . Assume link 2 is balanced and find F14 and F23 .
The D’Alembert inertia forces and offsets are:
t 2 = − I G2 α 2 = 0
f 2 = −m2 A G2 = 0
= − ( 3.5 kg ) ( 40.6ˆi − 22.6ˆj m / s
2
f 4 = −m4 A B
= − (1.2 kg ) ( 40.6ˆi m/s 2 )
)
= −49ˆi N = 49 N∠180°
= −142ˆi + 79ˆj N = 163 N∠150.90°
t 3 = − I G3 α 3
(
= − ( 0.060 N ⋅ m ⋅ s 2 ) 89.3kˆ rad/s 2
)
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h2 = t2 f 2 = 0
f3 = − m3 A G3
t 4 = − I G4 α 4 = 0
= −5.358kˆ N ⋅ m
h3 = t3 f3 = ( 5.358 N ⋅ m ) (163 N ) = 0.033 m , h4 = t4 f 4 = 0
∑M
A
= R G3 A × f3 + t 3 + R BA × f 4 + R BA × F14 = 0
( 0.196ˆi − 0.038ˆj m ) × ( −142ˆi + 79ˆj N ) + ( −5.358kˆ N ⋅ m ) + ( 0.442ˆi − 0.087ˆj m ) × ( −49ˆi N )
+ ( 0.442ˆi − 0.087 ˆj m ) × (1.000ˆj ) F = 0
14
(10.039kˆ N ⋅ m ) + ( −5.358kˆ N ⋅ m ) + ( −4.243kˆ N ⋅ m ) + ( 0.442kˆ m ) F
14
F14 = −1 N
∑F = F
14
+ f 4 + f3 + F23 = 0
=0
F14 = −1ˆj N = 1 N∠ − 90.00°
Ans.
F23 = 191ˆi − 78 N = 206 N∠ − 22.21°
Ans.
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205
12.18 Repeat Problem 12.17 for θ 2 = 240° . The results of a kinematic analysis are θ3 = 11.1° ,
RB = 0.392 m , α 3 = 112 rad/s 2 cw , A B = 35.2ˆi m / s 2 , and AG3 = 31.6ˆi + 27.7ˆj m / s 2 .
The D’Alembert inertia forces and offsets are:
t 2 = − I G2 α 2 = 0
f 2 = −m2 A G2 = 0
h2 = t2 f 2 = 0
f3 = − m3 A G3
f 4 = −m4 A B
= − (1.2 kg ) ( 35.2ˆi m/s 2 )
= − ( 3.5 kg ) ( 31.6ˆi + 27.7ˆj m / s 2 )
= −42ˆi N = 42 N∠180°
= −111ˆi − 97ˆj N = 147 N∠ − 138.76°
t 3 = − I G3 α 3
(
= − ( 0.060 N ⋅ m ⋅ s 2 ) −112kˆ rad/s 2
)
t 4 = − I G4 α 4 = 0
= 6.720kˆ N ⋅ m
∑M
A
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h3 = t3 f3 = ( 6.720 N ⋅ m ) (147 N ) = 0.046 m , h4 = t4 f 4 = 0
= R G3 A × f3 + t 3 + R BA × f 4 + R BA × F14 = 0
( 0.196ˆi + 0.038ˆj m ) × ( −111ˆi − 97ˆj N ) + ( 6.720kˆ N ⋅ m ) + ( 0.442ˆi + 0.087ˆj m ) × ( −42ˆi N )
+ ( 0.442ˆi + 0.087 ˆj m ) × (1.000ˆj ) F = 0
(
) (
14
) (
) (
)
−22.961kˆ N ⋅ m + 6.720kˆ N ⋅ m + 3.637kˆ N ⋅ m + 0.442kˆ in F14 = 0
F14 = 29 N
∑F = F
14
+ f 4 + f3 + F23 = 0
F14 = 29ˆj N = 29 N∠90.00°
Ans.
F23 = 153ˆi + 68 N = 168 N∠24.11°
Ans.
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206
12.19 A slider-crank mechanism, as in Problem 12.15, has a = 0.008 m , R AO2 = 0.25 m ,
RBA = 1.25 m ,
RCA = 1.0 m , θC = −38° ,
RG3 A = 0.75 m , α = −18° , m2 = 10 kg ,
m3 = 140 kg , m4 = 50 kg , IG2 = 2.0 N ⋅ m ⋅ s 2 , and IG3 = 8.42 N ⋅ m ⋅ s 2 , and has a
balanced crank. Make a complete kinematic and dynamic analysis of this system at
with ω 2 = 6 rad/s 2 ccw
FC =80∠ − 60° kN .
θ 2 = 120°
Kinematic Analysis:
α 2 = 0 , using
and
) (
(
VA = ω 2 × R AO2 = 6kˆ rad/s × −0.125ˆi + 0.217ˆj m
FB =50∠180° kN
and
)
= −1.299ˆi − 0.750ˆj m/s = 1.500 m/s∠ − 150°
VB = VA + ω3 × R BA
V ˆi = ( −1.299ˆi − 0.750ˆj m/s ) + (ω kˆ rad/s ) × (1.227ˆi − 0.225ˆj m )
B
(
) (
= −1.299ˆi − 0.750ˆj m/s + 0.225ω 3ˆi + 1.227ω 3 ˆj m
ω3 = 0.611kˆ rad/s
)
VB = −1.162ˆi m/s
(
2
A A = −ω 22 R AO2 + α 2 × R AO2 = − ( 6 rad/s ) × −0.125ˆi + 0.217ˆj m
Ans.
)
= 4.500ˆi − 7.794ˆj m/s 2 = 9.000 m/s 2 ∠ − 60°
A B = A A − ω32 R BA + α 3 × R BA
(
) (
) (
A ˆi = ( 4.041ˆi − 7.710ˆj m/s ) + ( 0.225α ˆi + 1.227α ˆj m )
)(
AB ˆi = 4.500ˆi − 7.794ˆj m/s 2 + −0.459ˆi + 0.084ˆj m/s 2 + α 3kˆ rad/s 2 × 1.227ˆi − 0.225ˆj m
)
2
3
B
α 3 = 6.284kˆ rad/s
A G3 = A A − ω R G3 A + α 3 × R G3 A
(
2
3
3
A B = 5.455ˆi m/s 2
2
) (
) (
) (
= 4.500ˆi − 7.794ˆj m/s 2 + −0.247ˆi + 0.133ˆj m/s 2 + 6.284kˆ rad/s 2 × 0.660ˆi − 0.357ˆj m
A G3 = 6.496ˆi − 3.514ˆj m/s 2 = 16.516 m/s 2 ∠ − 28.41°
Dynamic Analysis:
The D’Alembert inertia forces and offsets are:
t 2 = − I G2 α 2 = 0
f 2 = −m2 A G2 = 0
h2 = t2 f 2 = 0
f3 = − m3 A G3
= − (140 kg ) ( 6.496ˆi − 3.514ˆj m/s 2 )
= −909ˆi + 492ˆj N = 1 034 N∠151.59°
f 4 = −m4 A B
= − ( 50 kg ) ( 5.455ˆi m/s 2 )
= −273ˆi N = 273 N∠180°
Ans.
)
Ans.
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3
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207
t 3 = − I G3 α 3
(
= − ( 8.42 N ⋅ m ⋅ s 2 ) 6.284kˆ rad/s 2
)
t 4 = − I G4 α 4 = 0
= −52.911kˆ N ⋅ m
∑M
A
= R G3 A × f3 + t 3 + R CA × FC + R BA × f 4 + R BA × FB + R BA × F14 = 0
( 0.660ˆi − 0.357ˆj m ) × ( −909ˆi + 492ˆj N ) + ( −52.911kˆ N ⋅ m ) + ( 0.263ˆi − 0.301ˆj m ) × ( 40 000ˆi − 69 282ˆj N )
+ (1.227 ˆi − 0.225ˆj m ) × ( −273ˆi N ) + (1.227 ˆi − 0.225ˆj m ) × ( −50 000ˆi N ) + (1.227 ˆi − 0.225ˆj m ) × (1.000ˆj ) F
( 0.207kˆ N ⋅ m ) + ( −52.911kˆ N ⋅ m ) + ( −6 157kˆ N ⋅ m ) + ( −61.425kˆ N ⋅ m ) + ( −11 250kˆ N ⋅ m ) + (1.227kˆ in ) F
F14 = 14 280 N
∑F = F +f + F + F = 0
∑F = F + f + F + F = 0
∑F = F + F = 0
∑M = R ×F + M = 0
O2
14
=0
14
=0
F14 = 14 280ˆj N = 14 280 N∠90.00°
Ans.
14
4
B
34
F34 = 50 273ˆi − 14280ˆj N = 52 262 N∠ − 15.86°
Ans.
43
3
C
23
F23 = 11 182ˆi + 54 510ˆj N = 55 645 N∠78.41°
Ans.
32
12
F12 = 11 182ˆi + 54 510ˆj N = 55 645 N∠78.41°
Ans.
M12 = −9 240kˆ N ⋅ m
Ans.
AO2
32
12
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h3 = t3 f3 = ( 52.911 N ⋅ m ) (1 034 N ) = 0.051 m , h4 = t4 f 4 = 0
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208
12.20 Cranks 2 and 4 of the cross-linkage shown in the figure are balanced. The dimensions of
the linkage are RAO2 = 150 mm , RO4O2 = 450 mm , RBA = 450 mm , RBO4 = 150 mm ,
RCA = 600 mm , and RG3 A = 300 mm . Also, m3 = 1.8 kg , IG2 = IG4 = 0.007 N ⋅ m ⋅ s 2 ,
IG3 = 0.055 N ⋅ m ⋅ s 2 .
and
Corresponding
to
the
position
shown,
and
with
ω 2 = 10 rad/s ccw and α 2 = 0 , a kinematic analysis gave as results ω3 = 1.43 rad/s cw ,
ω4 = 11.43 rad/s cw , α 3 = α 4 = 84.7 rad/s 2 cw , and AG3 = 14.28ˆi + 21.1ˆj m/s 2 . Find
the driving torque and the pin reactions if FC = −133.5ˆj N .
The D’Alembert inertia forces and offsets are:
f 2 = −m2 A G2 = 0
t 2 = − I G2 α 2 = 0
f3 = − m3 A G3
(
= − (1.8 kg ) 14.28ˆi + 21.1ˆj m/s 2
)
f 4 = −m4 A G4 = 0
= −25.7ˆi − 37.98ˆj N = 45.84 N∠ − 124.10°
t 3 = − I G3 α 3
(
= − ( 0.007 N ⋅ m ⋅ s 2 ) −84.7kˆ rad/s 2
)
t 4 = − I G4 α 4
(
= − ( 0.007 N ⋅ m ⋅ s 2 ) −84.7kˆ rad/s 2
)
= 0.593kˆ N ⋅ m
f3 = ( 0.593 N ⋅ m ) ( 45.84 N ) = 12.93 mm , h4 = 0
= 0.593kˆ N ⋅ m
h3 = t3
Next, the free body diagrams are drawn. Since the lines of action for the forces on the
free body diagrams can not be discovered from two- and three-force member concepts,
the force F34 is divided into radial and transverse components.
∑M
O4
= t 4 + R BO4 × F34t = 0
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h2 = t2 f 2 = 0
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209
( 0.593kˆ N ⋅ m ) + ( −21.42ˆi − 148.45ˆj mm )( 0.990ˆi − 0.143ˆj) F
t
34
=0
F = −3.96 N
t
34
∑M = R ×f + t + R ×F + R ×F + R ×F = 0
( 0.23ˆi − 0.18ˆj m ) × ( −25.7ˆi − 37.98ˆj N ) + ( 0.593kˆ N − m ) + ( 0.47ˆi − 0.37ˆj m ) × ( −133.5ˆj N )
+ ( 0.35ˆi − 0.28ˆj m ) × ( 3.92ˆi − 0.56ˆj N ) + ( 0.35ˆi − 0.28ˆj m ) × ( 0.143ˆi + 0.990ˆj) F = 0
A
G3 A
3
3
CA
C
t
43
BA
BA
r
43
r
43
( −14.24kˆ N ⋅ m ) + ( 0.6kˆ N − m ) + ( −63.96kˆ N − m ) + ( 0.9kˆ N − m ) + (0.39kˆ m ) F43r = 0
F43r = 195.8 N
∑F = F + f
∑M = R
43
O2
3
AO2
+ FC + F23 = 0
F34 = −31.18ˆi − 188.8ˆj N = 191.35 N∠ − 99.38°
F = −5.28ˆi − 18.82ˆj N = 19.55 N∠ − 105.67°
× F32 + M12 = 0
M12 = −0.74kˆ N − m
23
Ans.
Given the same dynamic conditions, the D’Alembert forces and torques are the same as
in Problem P12.20. However, crank 2 is now a two-force member with no applied
moment. Therefore the free body diagrams appear as:
The solution now proceeds as follows:
∑ M B = R G3B × f3 + t 3 + R CB × FC + R AB × F23 = 0
( −0.117ˆi + 0.093ˆj m ) × ( −25.7ˆi − 37.98ˆj N ) + ( 0.593kˆ N ⋅ m ) + ( 0.117ˆi − 0.093ˆj m ) × ( −133.5ˆj N )
+ ( −0.35ˆi + 0.28ˆj m ) × ( −0.500ˆi − 0.866ˆj) F = 0
23
( 7.119kˆ N ⋅ m) + ( 0.603kˆ N ⋅ m ) + ( −15.93kˆ N ⋅ m ) + ( 0.445kˆ m ) F
23
F23 = 18.15 N
=0
F23 = −9.075ˆi − 15.72ˆj N = 18.15 N∠ − 120°
Ans.
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12.21 Find the driving torque and the pin reactions for the mechanism of Problem 12.20 under
the same dynamic conditions, but with crank 4 as the driver.
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210
∑F = F + f
∑M = R
23
O4
+ FC + F43 = 0
3
BO4
× F34 + t 4 + M14 = 0
F43 = 35.4ˆi + 188.08ˆj N = 191.38 N∠79.34°
M = 0.639kˆ N ⋅ m
Ans.
Ans.
14
12.22 A kinematic analysis of the mechanism of Problem 12.20 at θ 2 = 210° with
ω 2 = 10 rad/s ccw and α 2 = 0 gave θ3 = 14.7° , θ 4 = 164.7° , ω3 = 4.73 rad/s ccw ,
ω4 = 5.27 rad/s cw , α 3 = α 4 = 10.39 rad/s cw , and AG3 = 7.8∠20.85° m / s 2 . Compute
the crank torque and the pin reactions for this phase of the motion using the same force
FC as in Problem 12.20.
The D’Alembert inertia forces and offsets are:
f 2 = −m2 A G2 = 0
t 2 = − I G2 α 2 = 0
f3 = − m3 A G3
(
= − (1.8 kg ) 7.29ˆi + 2.78ˆj m/s 2
)
f 4 = −m4 A G4 = 0
= −13.42ˆi − 5.11ˆj N = 14.37 N∠ − 159.15°
t 3 = − I G3 α 3
(
= − ( 0.055 N ⋅ m ⋅ s 2 ) −10.39kˆ rad/s 2
)
t 4 = − I G4 α 4
(
= − ( 0.007 N ⋅ m ⋅ s 2 ) −10.39kˆ rad/s 2
)
= 0.0727kˆ N ⋅ m
f3 = ( 0.571 N ⋅ m ) (14.37 N ) = 39.73 mm , h4 = 0
= 0.571kˆ N ⋅ m
h3 = t3
Next, the free body diagrams are drawn. Since the lines of action for the forces on the
free body diagrams can not be discovered from two- and three-force member concepts,
the force F34 is divided into radial and transverse components.
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h2 = t2 f 2 = 0
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211
∑M
O4
= t 4 + R BO4 × F34t = 0
( 0.0727kˆ N ⋅ m ) + ( −0.144ˆi + 0.039ˆj m )( 0.263ˆi + 0.965ˆj) F
t
34
=0
F = 0.485 N
t
34
∑M = R ×f + t + R ×F + R ×F + R ×F = 0
( 0.29ˆi + 0.076ˆj m ) × ( −13.42ˆi − 5.11ˆj N ) + ( 0.0727kˆ N ⋅ m ) + ( 0.58ˆi + 0.153ˆj m ) × ( −133.5ˆj N )
+ ( 0.435ˆi + 0.114ˆj m ) × ( −0.129ˆi − 0.467ˆj N ) + ( 0.435ˆi + 0.114ˆj m ) × ( 0.965ˆi − 0.263ˆj) F = 0
( −0.128kˆ N ⋅ m ) + ( 0.074kˆ N ⋅ m ) + ( −78.65kˆ N ⋅ m ) + ( −0.191kˆ N ⋅ m ) + ( −0.225kˆ m ) F43r = 0
A
G3 A
3
3
CA
C
t
43
BA
BA
r
43
r
43
∑F = F + f
∑M = R
43
O2
3
AO2
+ FC + F23 = 0
× F32 + M12 = 0
F34 = −334.86ˆi + 89ˆj N = 347 N∠164.80°
Ans.
F23 = −322.27ˆi + 230.13ˆj N = 396 N∠144.47° Ans.
M = 54.8kˆ N − m
Ans.
12
12.23 The figure shows a linkage with an extended coupler having an external force of FC
acting during a portion of the cycle. The dimensions of the linkage are RAO2 = 0.14 m ,
RO4O2 = RBA = 1 m ,
RBO4 = 1.4 m ,
RG3 A = 0.8 m , and
RG4O4 = 0.5 m .
Also
m3 = 100.78 kg , m4 = 94.4 kg , IG3 = 25.14 N ⋅ m ⋅ s 2 , and IG4 = 29.37 N ⋅ m ⋅ s 2 , and the
crank is balanced. Make a kinematic and dynamic analysis for a complete rotation of the
crank using ω 2 = 10 rad/s ccw , FC = −2.225ˆi + 3.94ˆj kN for 90° ≤ θ 2 ≤ 300° and
FC = 0 for other angles.
For position analysis, the loop closure equation is:
R2 e jθ2 + R3e jθ3 = R1 + R4 e jθ4
0.4e jθ2 + 1e jθ3 = 1 + 1.4e jθ4
0.4 cos θ 2 + 1cos θ3 = 1 + 1.4 cos θ 4
0.4sin θ 2 + 1sin θ3 = 1.4sin θ 4
Eliminating θ 3 we find θ 4 from the roots of the quadratic
( 0.425 − 0.2 cos θ 2 ) tan 2 θ 4 2 + (1.4sin θ 2 ) tan θ 4 2 + (1.2 cos θ 2 − 3.075) = 0
Then θ3 = tan −1 ⎡⎣( 0.175sin θ 4 − 0.05sin θ 2 ) ( 0.425cos θ 4 + 0.125 − 0.05cos θ 2 ) ⎤⎦
R G3 = 0.4e
jθ 2
+ 0.8e
jθ3
R C = 0.4e jθ2 + 1.403e
R G4 = 1 + 0.5e
jθ 4
j (θ3 − 4.086° )
The first-order kinematic coefficients are found by differentiating the above:
j 0.4e jθ2 + j1e jθ3θ3′ = j1.4e jθ4θ 4′
0.4 cos θ 2 + 1cos θ3θ3′ = 1.4 cos θ 4θ 4′
−0.4sin θ 2 − 1sin θ3θ3′ = −1.4sin θ 4θ 4′
Ans.
Ans.
Ans.
Ans.
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F43r = −78 lb
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212
θ3′ = −0.35sin (θ 4 − θ 2 ) 0.875sin (θ 4 − θ 3 )
θ 4′ = 0.25sin (θ 2 − θ3 ) 0.875sin (θ 4 − θ3 ) Ans.
jθ
jθ
R′G3 = j 0.4e 2 + j 0.8e 3θ ′ = −0.4 ( sin θ 2 + 2sin θ3θ3′ ) + j 0.4 ( cos θ 2 + 2 cos θ3θ 3′ )
3
jθ 4
R′G4 = j 0.5e θ 4′ = −0.5sin θ 4θ 4′ + j 0.5cos θ 4θ 4′
R′C = j 0.4e jθ2 + j1.403e
Ans.
Ans.
j (θ3 − 4.086° )
θ3′
Ans.
= [ −0.4 sin θ 2 − 1.403sin (θ 3 − 4.086° ) θ 3′ ] + j [ 0.4 cos θ 2 + 1.403cos (θ 3 − 4.086° ) θ 3′ ]
The second-order kinematic coefficients are found by differentiating again:
−0.4e jθ2 + j1e jθ3θ3′′ − 1e jθ3θ3′2 = j1.4e jθ4θ 4′′ − 1.4e jθ4θ 4′2
−0.4 cos θ 2 − 1sin θ3θ3′′ − 1cos θ3θ3′2 = −1.4sin θ 4θ 4′′ − 1.4 cos θ 4θ 4′2
−0.4sin θ 2 + 1cos θ3θ3′′ − 1sin θ3θ3′2 = 1.4 cos θ 4θ 4′′ − 1.4sin θ 4θ 4′2
θ3′′ = ⎡⎣0.35cos (θ 4 − θ 2 ) + 0.875cos (θ 4 − θ3 )θ3′2 − 1.225θ 4′2 ⎤⎦ 0.875sin (θ 4 − θ3 )
Ans.
θ 4′′ = ⎡⎣0.25cos (θ 4 − θ 2 ) + 0.625θ3′2 − 0.875cos (θ 4 − θ3 )θ 4′2 ⎤⎦ 0.875sin (θ 4 − θ3 )
jθ
jθ
jθ
R′′G3 = −0.4e 2 + j 0.8e 3θ ′′ − 0.8e 3θ ′ 2
3
3
(
) (
= −0.4 cos θ − 0.8sin θ θ ′′ − 0.8 cos θ θ ′3 2 + j −0.4 sin θ + 0.8 cos θ θ ′′ − 0.8sin θ θ ′ 2
2
3 3
2
3 3
3 3
jθ
jθ
2
2
R′′G4 = j 0.5e θ 4′′ − 0.5e θ 4′ = −0.5sin θ 4θ 4′′ − 0.5 cos θ 4θ 4′ + j 0.5 cos θ 4θ 4′′ − 0.5sin θ 4θ 4′2
4
3 3
(
4
) (
) Ans.
)
Ans.
By virtual work we can formulate the dynamic input torque requirement as:
M 12 = f3 iR′G3 + t 3 iθ 3′kˆ + f 4 iR′G4 + t 4 iθ 4′kˆ + FC iR′C
The individual elements of this equation are:
2
f3 = − m3 A G3 = − m3 R′′G3 ω22 = − (100.78 kg )(10 rad/s ) R′′G3 = −1007.8 × 9.81 NR′′G3
(
) (
)
= 4094 cos θ + 8188 sin θ θ ′′ + 8188 cos θ θ ′3 2 + j 4094 sin θ − 8188 cos θ θ ′′ + 8188 sin θ θ ′ 2 N
2
3 3
3 3
2
3 3
3 3
(
)
+ ( 4094sin θ − 8188cos θ θ ′′ + 8188sin θ θ ′ ) ( 0.4 cos θ + 0.8cos θ θ ′ ) N ⋅ m
2
3 3
3 3
f3 iR′G3 = 4094 cos θ + 8188sin θ θ ′′ + 8188cos θ θ ′3 2 ( −0.4sin θ 2 − 0.8sin θ 3θ 3′ ) N ⋅ m
2
3 3
3 3
2
2
f3 iR′G3 = ( 3327.5 N ⋅ m ) ⎡⎣ − sin (θ3 − θ 2 ) θ3′ (1 − θ3′ ) + cos (θ 3 − θ 2 )θ 3′′ − 2θ3′θ 3′′⎤⎦
t 3 = − I G3 α 3 = − I G3θ3′′ω22kˆ = − ( 2553.8 N ⋅ m ) θ3′′kˆ
t iθ ′kˆ = − ( 2553.8 N ⋅ m ) θ ′θ ′′
3
3
3 3
f 4 = −m4 A G4 = −m4 R′′G4 ω = − ( 94.4 kg )(10 rad/s ) R′′G4 = −944 × 9.81 NR′′G4
2
2
2
= ( 4797 sin θ 4θ 4′′ + 4797 cos θ 4θ 4′2 ) + j ( −4797 cos θ 4θ 4′′ + 4797 sin θ 4θ 4′2 ) N
f 4 iR′G4 = ( 4797 sin θ 4θ 4′′ + 4797 cos θ 4θ 4′2 ) ( −0.5sin θ 4θ 4′ ) N ⋅ m
+ ( −4797 cos θ 4θ 4′′ + 4797 sin θ 4θ 4′2 ) ( 0.5cos θ 4θ 4′ ) N ⋅ m
3 3
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Ans.
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213
f 4 iR′G4 = −2436.3θ 4′θ 4′′ N ⋅ m
t = − I α = − I θ ′′ω 2kˆ = − ( 2983 N ⋅ m ) θ ′′kˆ
4
G4
4
G4
4
2
4
t 4 iθ 4′kˆ = −2983θ 4′θ 4′′ N ⋅ m
FC iR′C = ( 4450 N ) cos120° ⎡⎣ −0.4sin θ 2 − 1.403sin (θ 3 − 4.086° ) θ 3′ m ⎤⎦
+ ( 4450 N ) sin120° ⎡⎣0.4 cos θ 2 + 1.403cos (θ 3 − 4.086° ) θ3′ m ⎤⎦
FC iR′C = −1808sin (θ 2 − 120° ) − 6344sin (θ3 − 124.086° ) θ3′ N ⋅ m
Reassembling the elements we must remember that force is only nonzero for a portion of
the cycle. Therefore,
M 12 = ( 3327 ) [ − sin (θ − θ ) θ ′ (1 − θ ′ ) + cos (θ − θ ) θ ′′ − 2θ ′θ ′′] − ( 2554 ) θ ′θ ′′ − 21 560θ ′θ ′′ − 26 400θ ′θ ′′ N ⋅ m
2
3
3
3
2
3
3
3
3
3
4
4
4
4
=3327 cos (θ 3 − θ 2 ) θ 3′′ − 3327 sin (θ 3 − θ 2 ) θ 3′ (1 − θ 3′ ) − 9209θ 3′θ 3′′ − 5419θ 4′θ 4′′ N ⋅ m
for the entire cycle and, for 90° ≤ θ 2 ≤ 300° , an additional increment is added:
Ans.
ΔM 12 = −1808sin (θ 2 − 120° ) − 6344 sin (θ 3 − 124.086° ) θ 3′ N ⋅ m
This input torque requirement is shown in the following plot. Notice the very small
discontinuities in the curve when force FC begins and ends its effect.
12.24 The figure shows a motor geared to a shaft on which a flywheel is mounted. The mass
moments of inertia of the parts are as follows: flywheel, I = 0.308 N ⋅ m ⋅ s 2 ; flywheel
shaft, I = 0.1724 N ⋅ cm ⋅ s 2 ; gear, I = 1.9 N ⋅ cm ⋅ s 2 ; pinion, I = 0.0388 N ⋅ cm ⋅ s2 ;
motor, I = 0.9612 N ⋅ cm ⋅ s 2 . If the motor has a starting torque of 8.34 N ⋅ m , what is the
angular acceleration of the flywheel shaft at the instant the motor is turned on ?
If we identify the motor shaft as 2 and the flywheel shaft as 3 then
I 2 = 0.009612 N ⋅ m ⋅ s 2 + 0.000388 N ⋅ m ⋅ s 2 = 0.01 N ⋅ m ⋅ s 2
I 3 = 0.308 N ⋅ m ⋅ s 2 + 0.001724 N ⋅ m ⋅ s 2 + 0.019 N ⋅ m ⋅ s 2 = 0.3287 N ⋅ m ⋅ s 2
θ 3 = − ( R2 R3 )θ 2
∑M
3
= R3 F23 = I 3θ 3
θ 3 = − ( R2 R3 )θ 2
F23 = − ( R2 R32 ) I 3θ 2
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3
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214
∑M
2
2
M 12 = I 2θ 2 + R2 F32 = ⎡ I 2 + ( R2 R3 ) I 3 ⎤ θ 2
⎣
⎦
= M 12 − R2 F32 = I 2θ 2
Now, substituting the numeric values,
2
8.475 N − m = ⎡ 0.01 N ⋅ m ⋅ s 2 + ( 25 mm 112.5 mm ) 0.3287 N ⋅ m ⋅ s 2 ⎤ θ 2 = ( 0.02623 N ⋅ m ⋅ s 2 ) θ 2
⎣
⎦
2
2
θ 2 = 8.475 N ⋅ m 0.02623 N ⋅ m ⋅ s = 323.103 rad/s
θ3 = − ( R2 R3 )θ 2 = − ( 25 mm 112.5 mm ) 323.103 rad/s 2 = 71.80 rad/s 2
Ans.
12.25 The disk cam of Problem 14.31 is driven at a constant input shaft speed of
ω 2 = 20 rad/s ccw. Both the cam and the follower have been balanced so that the
centers of mass of each are located at their respective fixed pivots. The mass of the cam
is 0.075 kg with radius of gyration of 30 mm, and for the follower the mass is 0.030 kg
with radius of gyration of 35 mm. Determine the moment M12 required on the camshaft
at the instant shown in the figure to produce this motion.
I G3 = m3k32 = ( 0.030 kg )( 0.035 m ) = 0.000 036 75 kg ⋅ m 2
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2
For full-rise cycloidal cam motion, Eq. (5.19),
2πθ ⎞ 30° ⎛
112.5° ⎞
L⎛
y′ = ⎜ 1 − cos
⎟=
⎜1 − cos 2π
⎟ = 0.200
150° ⎠
β⎝
β ⎠ 150° ⎝
( 360° )( 30° ) sin 2π 112.5° = −0.480
150°
β
β2
(150°)2
2
θ 3 = y′′θ 22 = −0.480 ( 20 rad/s ) = −192 rad/s 2
y′′ =
2π L
sin
2πθ
=
t3 = − I G3θ 3 = − ( 0.000 036 75 kg ⋅ m 2 )( −192 rad/s 2 ) = 0.007 056 N ⋅ m
By virtual work,
M 12 = − y′ ⎡t3 + R CO3 × FC ikˆ ⎤
⎣
⎦
= −0.200 ⎡⎣0.007 056 N ⋅ m + ( 0.150 m )( 8 N ) sin − 45° ⎤⎦ = 0.168 N ⋅ m ccw
(
)
12.26 Repeat Problem 12.25 with a shaftspeed of ω 2 = 40 rad/s ccw.
I G3 = m3 k32 = ( 0.030 kg )( 0.035 m ) = 0.000 036 75 kg ⋅ m 2
2
For full-rise cycloidal cam motion, Eq. (5.19),
2πθ ⎞ 30° ⎛
112.5° ⎞
L⎛
y′ = ⎜ 1 − cos
⎟=
⎜1 − cos 2π
⎟ = 0.200
150° ⎠
β⎝
β ⎠ 150° ⎝
( 360° )( 30° ) sin 2π 112.5° = −0.480
150°
β
β2
(150°)2
2
θ 3 = y′′θ 22 = −0.480 ( 40 rad/s ) = −768 rad/s 2
y′′ =
2π L
sin
2πθ
=
t3 = − I G3θ 3 = − ( 0.000 036 75 kg ⋅ m 2 )( −768 rad/s 2 ) = 0.028 224 N ⋅ m
Ans.
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215
By virtual work,
M 12 = − y′ ⎣⎡t3 + R CO3 × FC ikˆ ⎦⎤
= −0.200 ⎡⎣0.028 224 N ⋅ m + ( 0.150 m )( 8 N ) sin − 45°⎤⎦ = 0.164 N ⋅ m ccw
(
)
Ans.
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12.27 A rotating drum is pivoted at O2 and is decelerated by the double-shoe brake mechanism
shown in the figure. The mass of the drum is 105 kg and its radius of gyration is
142 mm. The brake is actuated by force P = −445ˆj N , and it is assumed that the contact
between the two shoes and the drum act at points C and D, where the coefficients of
coulomb friction are μ = 0.300. Determine the angular acceleration of the drum and the
reaction force at the fixed pivot F12 .
F = 4 P = 1780 N
∑ M = ( −300ˆi mm ) × ( − Pˆj) + ( 75ˆj mm ) × ( F ˆi ) = 0
F = −1778.63ˆi + 445ˆj N = 1833.4 N ∠14.04°
∑F = P + F + F = 0
5
5
65
65
35
65
35
The friction angle is φ = tan
−1
( 0.300 ) = 16.70° .
∑ M = ( 550ˆj mm ) × F + ( −125ˆi + 225ˆj mm ) × ( − cos φ ˆi − sin φ ˆj) F
3
F23 = 3408.7 N
53
23
=0
F32 = 3264.9ˆi + 979.5ˆj N = 3408.7 N ∠16.70°
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216
∑ M = ( 625ˆj mm ) × F + (125ˆi + 225ˆj mm ) × ( cos φ ˆi + sin φ ˆj) F
4
64
24
F24 = 6194.4 N
=0
F42 = −5933ˆi − 1780ˆj N = 6194.4 N ∠ − 163.30°
I G2 = m2 k22 = (105 kg )( 0.1415 m ) = 2.156 N ⋅ m ⋅ s 2
2
∑M
∑F
2
O2
(
) (
) (
) (
)
= −0.2ˆi m × 3261.85ˆi + 979ˆj N + 0.2ˆi m × −5931.85ˆi − 1780ˆj N = I G2 α 2
= F12 + F32 + F42 = 0
α 2 = −261kˆ rad/s 2
F12 = 2668.2ˆi + 800.5ˆj lb = 2785.7 N ∠16.70°
Ans.
Ans.
Note that gravitational effects are not yet included. If standard gravity acts in the −jˆ
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direction then the ĵ component is 410 lb. Since the main bearing at O2 supports this
weight, it does not affect the friction forces and can be added by superposition. If
weights of the other parts were known, however, these weight might have some small
effect on the friction forces and the braking forces, and would have to be included
simultaneously. Superposition could not be applied.
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217
Chapter 13
Dynamic Force Analysis (Spatial)1
The figure shows a two-throw opposed-crank crankshaft mounted in bearings at A and G.
Each crank has an eccentric weight of 26.7 N, which may be considered as located at a
radius of 50 mm from the axis of rotation, and at the center of each throw (points C and
E). It is proposed to locate weights at B and F in order to reduce the bearing reactions,
caused by the rotating eccentric cranks, to zero. If these weights are to be mounted at
75 mm from the axis of rotation, how much must they weigh?
∑M
y
A
∑M
y
G
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13.1
= ( 50 mm ) mB rB2ω 2 − ( 200 mm ) mC rC2ω 2 + ( 500 mm ) mE rE2ω 2
− ( 650 mm ) mF rF2ω 2 + ( 700 mm ) FG = 0
= ( 700 mm ) FA − ( 650 mm ) mB rB2ω 2 + ( 500 mm ) mC rC2ω 2
− ( 200 mm ) mE rE2ω 2 + ( 50 mm ) mF rF2ω 2 = 0
Dividing by ( 50 mm ) ω 2 and substituting numeric values gives
( 5625 mm ) m + ( 96.12 Nm ) − ( 73125 mm ) m = 0
− ( 73125 mm ) m + ( 96.12 Nm ) + ( 5625 mm ) m = 0
2
2
2
B
F
2
2
B
2
F
Solving simultaneously gives
mB = mF = 5931.85 N
1
Unless stated otherwise, all problems are solved without friction and without gravitational loads.
Ans.
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218
The figure illustrates a two-throw crankshaft, mounted in bearings at A and F, with the
cranks spaced 90° apart. Each crank may be considered to have an eccentric weight of
26.7 N at the center of the throw and 50 mm from the axis of rotation. It is proposed to
eliminate the rotating bearing reactions, which the crank would cause, by mounting
additional correction weights on 75 mm arms at points B and E. Calculate the magnitudes
and angular locations of these weights.
∑M
y
A
(
) (
)
(
) ( )
+ ( 500ˆi mm ) × ( kˆ ) m r ω + ( 650ˆi mm ) × ( sin θ ˆj + cos θ kˆ ) m r ω = 0
= ( −650ˆi mm ) × ( sin θ ˆj + cos θ kˆ ) m r ω + ( −500ˆi mm ) × ( − ˆj) m r ω
+ ( −200ˆi mm ) × ( kˆ ) m r ω + ( −50ˆi mm ) × ( sin θ ˆj + cos θ kˆ ) m r ω = 0
= 50ˆi mm × sin θ B ˆj + cos θ B kˆ mB rB2ω 2 + 200ˆi mm × −ˆj mC rC2ω 2
2
D D
∑M
y
F
2
E
B
B
2
D D
2
B B
2
E E
E
2
2
2
C C
2
E
E
2
E E
2
2
Dividing by ( 50 mm ) ω 2 , substituting numeric values, and equating vector components
gives
− ( 5625 mm 2 ) mB cos θ B − (160.2 Nm 2 ) − ( 73125 mm 2 ) mE cos θ E = 0
( 5625 mm ) m sin θ − ( 64 Nm ) + ( 73125 mm ) m sin θ = 0
( 73125 mm ) m cos θ + ( 64 Nm ) + ( 5625 mm ) m cosθ = 0
− ( 73125 mm ) m sin θ + (160.2 Nm ) − ( 5625 mm ) m sin θ = 0
2
2
B
2
B
E
2
2
B
2
B
2
E
2
B
B
E
E
2
E
E
Solving simultaneously gives
mB cos θ B = −2.968 N , mB sin θ B = 8.9 N , mE cos θ E = −8.9 N , mE sin θ E = 2.968 N
mB = 9.381 N , θ B = 108.43° , mE = 9.381 N , θ B = 161.57°
Ans.
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13.2
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219
Solve Problem 13.2 with the angle between the two throws reduced from 90° to 0°.
∑M
y
A
(
) (
)
(
) ( )
+ ( 500ˆi mm ) × ( kˆ ) m r ω + ( 650ˆi mm ) × ( sin θ ˆj + cos θ kˆ ) m r ω = 0
= ( −650ˆi mm ) × ( sin θ ˆj + cos θ kˆ ) m r ω + ( −500ˆi mm ) × ( kˆ ) m r ω
+ ( −200ˆi mm ) × ( kˆ ) m r ω + ( −50ˆi mm ) × ( sin θ ˆj + cos θ kˆ ) m r ω = 0
= 50ˆi mm × sin θ B ˆj + cos θ B kˆ mB rB2ω 2 + 200ˆi mm × kˆ mC rC2ω 2
2
D D
∑M
y
F
2
E
B
2
D D
2
B B
B
2
E E
E
2
2
2
C C
2
E
E
2
E E
2
2
Dividing by ( 50 mm ) ω 2 , substituting numeric values, and equating vector components
gives
− ( 5625 mm 2 ) mB cos θ B − ( 224.28 Nm 2 ) − ( 73125 mm 2 ) mE cos θ E = 0
+ ( 73125 mm ) m sin θ = 0
( 5625 mm ) m sin θ
( 73125 mm ) m cos θ + ( 224.28 Nm ) + ( 5625 mm ) m cosθ = 0
− ( 73125 mm ) m sin θ
− ( 5625 mm ) m sin θ = 0
2
2
B
B
E
2
2
B
2
B
2
E
E
E
2
B
B
E
E
Solving simultaneously gives
mB cos θ B = −11.87 N , mB sin θ B = 0 N , mE cos θ E = −11.87 N , mE sin θ E = 0 N
mB = 11.87 N , θ B = 180.00° , mE = 11.87 N , θ B = 180.00°
13.4
Ans.
The connecting rod shown in the figure weighs 35 N and is pivoted on a knife edge and
caused to oscillate as a pendulum. The rod is observed to complete 64.5 oscillations in 1
min. Determine the mass moment of inertia of the rod about its own center of mass.
rG = 100 mm , mg = 35 N , τ = 60 s 64.5 cycles = 0.930 s/cycle
From Eq. (13.2),
2
2
I O = mgrG (τ 2π ) = ( 35 N )(100 mm )( 0.930 s/cycle 2π rad/cycle ) = 0.07676 Nms 2
I G = I O − mrG2 = 0.07676 Nms 2 − ( 35 N 9.81 m/s 2 ) ( 0.1 m ) = 0.041 Nms 2
2
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13.3
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220
13.5
A gear is suspended on a knife edge at the rim as shown in the figure and caused to
oscillate as a pendulum. Its period of oscillation is observed to be 1.08 s. Assume that
the center of mass and the center of the axis of rotation are coincident. If the weight of
the gear is 182.5 N, find the mass moment of inertia and the radius of gyration of the
gear.
I G = I O − mrG2 = 1.0794 Nms 2 − (182.5 N 9.81 m/s 2 ) ( 0.2 m ) = 0.3353 Nms 2
Ans.
kG = I G m = 0.3353 N − m ⋅ s 2 (182.5 N 9.81 m/s 2 ) = 0.134 m
Ans.
2
13.6
The figure shows a wheel whose mass moment of inertia I is to be determined. The
wheel is mounted on a shaft in bearings with very low frictional resistance to rotation. At
one end of the shaft and on the outboard side of the bearings is connected a rod with a
weight Wb secured to its end. It is possible to measure the mass moment of inertia of the
wheel by displacing the weight Wb from its equilibrium and permitting the assembly to
oscillate. If the weight of the pendulum arm is neglected, show that the mass moment of
inertia of the wheel can be obtained from the equation
⎛ τ2
l⎞
− ⎟
I = Wbl ⎜
⎜ 4π 2 g ⎟
⎝
⎠
Using W for the weight of the wheel, the location of the center of mass of the assembly is
rG where WrG = Wb ( l − rG ) or Wb l = (W + Wb ) rG and the mass moment of inertia is
I O = I + Wb l 2 g . Now, using Eq. (13.2)
2
Wbl 2 (W + Wb )
⎛ τ ⎞ Wb lτ
=
I+
grG ⎜
⎟ =
4π 2
g
g
⎝ 2π ⎠
Rearranging this we get
⎛ τ2
l⎞
− ⎟
I = Wbl ⎜
⎜ 4π 2 g ⎟
⎝
⎠
2
13.7
Q.E.D.
If the weight of the pendulum arm is not neglected in Problem 13.6, but is assumed to be
uniformly distributed over the length l, show that the mass moment of inertia of the
wheel can be obtained from the equation
⎡ τ2 ⎛
W ⎞ l⎛
W ⎞⎤
I =l⎢
Wb + a ⎟ − ⎜ Wb + a ⎟ ⎥
⎜
2
2 ⎠ g⎝
3 ⎠ ⎦⎥
⎣⎢ 4π ⎝
where Wa is the weight of the arm.
Using W for the weight of the wheel and rG for the location of the center of mass of the
assembly,
WrG = Wb ( l − rG ) + Wa ( l 2 − rG ) or
(W + Wb + Wa ) rG = Wbl + Wa l 2
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From Eq. (13.2),
2
2
I O = mgrG (τ 2π ) = (182.5 N )( 200 mm )(1.08 s/cycle 2π rad/cycle ) = 1.0794 Nms 2
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221
The total mass moment of inertia is
2
Wb l 2 ⎡Wa l 2 Wa ⎛ l ⎞ ⎤
Wb l 2 Wa l 2
IO = I +
+⎢
+
+
⎜ ⎟ ⎥=I+
2 ⎝ 2 ⎠ ⎦⎥
3g
g
g
⎣⎢ 12 g
Using Eq. (13.2)
2
W l2 W l2 ⎛
W l ⎞⎛ τ ⎞
I + b + a = ⎜ Wb l + a ⎟ ⎜
⎟
3g ⎝
2 ⎠ ⎝ 2π ⎠
g
which can now be rearranged to read
⎡ τ2 ⎛
W ⎞ l⎛
W ⎞⎤
I = l ⎢ 2 ⎜ Wb + a ⎟ − ⎜ Wb + a ⎟ ⎥
2 ⎠ g⎝
3 ⎠⎦
⎣ 4π ⎝
Wheel 2 in the figure is a round disk which rotates about a vertical axis z through its
center. The wheel carries a pin B at a distance R from the axis of rotation of the wheel,
about which link 3 is free to rotate. Link 3 has its center of mass G located at a distance r
from the vertical axis through B, and it has a weight W3 and a mass moment of inertia
I G about its own mass center. The wheel rotates at an angular velocity ω2 with link 3
fully extended. Develop an expression for the angular velocity ω3 that link 3 would
acquire if the wheel were suddenly stopped.
Consider link 3 alone. The momentum before and after are
L1 = − m3 ( R + r ) ω 2 ˆi
L 2 = − m3 rω3ˆi
The angular momentum before and after about point G are
H G1 = ( m3 3) r 2ω 2kˆ
H G 2 = ( m3 3) r 2ω3kˆ
The angular momentum before and after about point B are
H B1 = H G1 + rˆj × L1
H B 2 = H G 2 + rˆj × L 2
= ( m3 3) r 2ω 2kˆ + m3 ( Rr + r 2 ) ω 2kˆ
= ( m3 3) r 2ω3kˆ + m3 r 2ω3kˆ
= m3 ( Rr + 4 r 2 3) ω 2kˆ
= ( 4 m3 3) r 2ω3kˆ
Since there is no angular impulse on link 3 about point B, H B 2 = H B1
⎛
⎝
ω3 = ⎜ 1 +
13.9
3R ⎞
⎟ ω2
4r ⎠
Ans.
Repeat Problem 13.8, except assume that the wheel rotates with link 3 radially inward.
Under these conditions, is there a value for the distance r for which the resulting angular
velocity ω3 is zero?
Consider link 3 alone. The momentum before and after are
L1 = − m3 ( R − r ) ω 2 ˆi
L 2 = m3 rω3ˆi
The angular momentum before and after about point G are
H G1 = ( m3 3) r 2ω 2kˆ
H G 2 = ( m3 3) r 2ω3kˆ
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13.8
Q.E.D.
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222
The angular momentum before and after about point B are
H B1 = H G1 − rˆj × L1
H B 2 = H G 2 − rˆj × L 2
= ( m3 3) r 2ω 2kˆ − m3 ( Rr − r 2 ) ω 2kˆ
= ( m3 3) r 2ω3kˆ + m3 r 2ω3kˆ
= m3 ( − Rr + 4 r 2 3) ω 2kˆ
= ( 4 m3 3) r 2ω3kˆ
Since there is no angular impulse on link 3 about point B, H B 2 = H B1
⎛ 3R ⎞
⎟ω2
⎝ 4r ⎠
ω3 = 0 for r = 3R 4
ω3 = ⎜ 1 −
Ans.
13.10 The figure illustrates a planetary gear-reduction unit which utilizes 3.5 mm/tooth module
spur gears cut on the 20° full-depth system. The input to the reducer is driven with 18.65
kw at 600 rev/min. The mass moment of inertia of the resisting load is 0.6588 N ⋅ m ⋅ s 2 .
Calculate the bearing reactions on the input, output, and planetary shafts. As a designer,
what forces would you use in designing the mounting bolts? Why?
N1m 104 teeth × 3.5
=
= 182 mm
2
2
N m 52 teeth × 3.5
R3 = 3 =
= 91 mm
2
2
R1 =
Tout = T1 A =
N 2 m 35 teeth × 3.5
=
= 61.25 mm
2
2
N m 17 teeth × 3.5
R4 = 4 =
= 29.75 mm
2
2
R2 =
(18.65 kw )(1000 Nm/s/kw )( 600 /min ) = 296.97 Nm
( 600 rev/min )( 2π rad/rev )
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Ans.
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223
Assume a symmetric arrangement of m (typically m = 3) planets, symmetrically arranged
on an m-pronged planet carrier. Then the tangential component of the force F2 A for each
planet is
T
296.97 Nm
= 2410.47 m N
F2tA = out =
mRA m ( 0.1232 m )
For the equilibrium of each planet
∑ M 2 = R3 F43 cos 20° − R2 F12 cos 20° = 0 R3 F43 = R2 F12
∑F
= F12 cos 20° + F43 cos 20° − FAt 2 = 0
F12 =
R3
FAt 2
cos
20
R
+
R
°
( 2 3)
t
2
F12 + F43 = (1 + R2 R3 ) F12 = FAt 2 cos 20°
F43 = ( R2 R3 ) F12
r
2
F43 = 1014.6 m N
= − F12 sin 20° + F43 sin 20° + FAr2 = 0
FAr2 = ( F12 − F43 ) sin 20°
FA 2 =
(F ) + (F )
2
t
A2
r 2
A2
FAr2 = 169.1 m N
FA 2 = 2376.3 m N
Ans.
Assuming that the m-pronged planet carrier is arranged symmetrically then there is no
Ans.
F14 = F1 A = 0
net force on the input or output shafts;
The input torque is
Tin = T14 = 29.38 N ⋅ m
Ans.
∑ M 4 = mR4 F34 cos 20° − T14 = 0
Balancing the moments on the casing
∑ M1 = mR2 F21 cos 20° − (16 in ) F11 = 0
F11 = 222.5 N
Ans.
This force F11 must be absorbed by the mounting bolts.
13.11 It frequently happens in motor-driven machinery that the greatest torque is exerted when
the motor is first turned on, because of the fact that some motors are capable of delivering
more starting torque than running torque. Analyze the bearing reactions of Problem 13.10
again, but this time use a starting torque equal to 250% of the full-load torque. Assume a
normal-load torque and a speed of zero. How does this starting condition affect the forces
on the mounting bolts?
Note that data are given for m = 2 planets. The masses of the moving elements are:
mA = 7.81 Mg/m3 [0.1× 0.35 × 0.0375 + π (0.05) 2 × (0.0125) + 2π (0.0375) 2 × (0.0125)
+ 2π (0.01575) 2 × (0.0875)] = 133 N
m2 = 7.81 Mg/m3 [π (0.0625) 2 × (0.0345) + π (0.0927) 2 (0.0345) + π (0.05) 2 (0.0125)
− π (0.01875) 2 × (0.08125)] = 109.47 N
m4 = 7.81 Mg/m3 ⎡⎣(π × (0.03) 2 × 0.0345 ) ⎤⎦ = 8 N
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∑F
F12 = 1508.55 m N
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224
The centroidal mass moments of inertia are:
I A = 7.81 Mg/m 3 [100 ⋅ 350 ⋅ 37.5 ( 3502 + 1002 ) 12 + π 50412.5 2 + 2π 37.5412.5 2
+ 2π 37.5212.5 ⋅123.252 + 2π 15.75487.5 2 + 2π 15.75287.5 ⋅123.252 ] = 1.48 N ⋅ m 2
I 2 = 7.81 Mg/m3 (π 62.5434.5 2 + π 92.75434.5 2 + π 50412.5 2 − π 18.75481.25 2 ) = 0.395 N ⋅ m 2
I 4 = 7.81 Mg/m3 (π 30.25434.5 2 ) = 0.0036 N ⋅ m 2
Step Number
Frame 1 Arm A
Planets 2,3
Sun 4
1. Gears fixed to arm
α
α
α
α
2. Arm fixed
−α
0
− (104 35 ) α
(104 35 )( 52 17 )α
0
α
− ( 69 35 ) α
( 6 003 595 )α
3. Total
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The angular accelerations are found by the tabluar method (See Section 10.7):
The output torque is Tout = 296.74 N ⋅ m .
The input torque is Tin = 2.5 ( 29.38 Nm ) = 73.45 N − m .
Balancing the sun gear and input shaft
∑ M 4 = 2 R4 F34t − Tin = I 4α 4
2 ( 30.25 mm ) F34t − 73.45 N ⋅ m = ( 0.0036 N − m 2 9.81 m/s 2 ) ( 6003 595 ) α
F34t = 1190.82 + 0.01410α
Balancing the arm and output shaft
∑ M A = Tout − 2 RA F2tA = I Aα A
296.74 Nm − 2 ( 0.1232 m ) F2tA= (1.48 N ⋅ m2 9.81 m/s 2 ) α
F2tA = 1185.48 − 0.1399α
Balancing a typical planet
∑ M 2 = R3 F43t − R2 F12t = I 2α 2
( 0.093 m )(1190.82 + 0.01410α ) − ( 62.5 mm ) F12t = ( 0.394 Nm2
9.81 m/s 2 ) ( −69 35 ) α
F12t = 1769.32 + 0.3110α
∑F
t
2
=F12t + F43t − FAt 2 = m2 AG2
(1769.32 + 0.3110α ) + (1190.82 + 0.01410α ) − (1185.48 − 0.1399α ) = (109.47 N
α = −512 rad/s 2
Now, reassembling the above results,
F34t = 1190.82 + 0.01410α = 1158.78 N
F34r = F34t tan 20° = 421.86 N
F12t = 1769.32 + 0.3110α = 1060.88 N
F12r = F12t tan 20° = 386.26 N
F2tA = 1185.48 − 0.1399α = 1504.1 N
F2rA = 386.26 − 421.86 = −35.6 N
9.81 m/s 2 ) ( −4.928α )
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225
The input and output bearing reactions are zero.
F2 A =
The load on the planet shaft is
Ans.
(F ) +(F )
2
t
2A
r 2
2A
= 1504.1 N
The forces in the mounting bolts to restrain the unbalanced frame moment are:
F11 = 2 R1 F21t 0.4 m = 985.23 N
Ans.
Ans.
13.12 The gear-reduction unit of Problem 13.10 is running at 600 rev/min when the motor is
suddenly turned off, without changing the resisting-load torque. Solve Problem 13.10 for
this condition.
Here we can use the free body diagrams from Problem 13.10 and the mass data and
angular motion relationships from Problem 13.11. Then, proceeding as in Problem 13.11,
but with Tin = 0, we balance the sun gear and input shaft
∑ M 4 = 2 R4 F34t − Tin = I 4α 4
9.81 m/s 2 ) ( 6003 595 ) α
F34t = 0.0606α
Balancing the arm and output shaft
∑ M A = Tout − 2 RA F2tA = I Aα A
296.7 N ⋅ m − 2 ( 0.123 m ) F2tA= (1.48 Nm 2 9.81 m/s 2 ) α
F2tA = 1206 − 0.613α
Balancing a typical planet
∑ M 2 = R3 F43t − R2 F12t = I 2α 2
( 0.092.85 m )( 0.0606α ) − ( 0.0625 m ) F12t = ( 0.395 Nm2
9.81 m/s 2 ) ( −69 35 ) α
F12t = 1.36α
∑F
t
2
=F12t + F43t − FAt 2 = m2 AGt 2
(1.36α ) + ( 0.0606α ) − (1206 − 0.613α ) = (109.5 N
ω A = 600 rev/min = 62.8 rad/s
∑F
r
2
9.81 m/s 2 ) ( −4.928α ) α = 342 rad/s 2
AGr 2 = − RAω A2 = −486.4 m/s 2
= − F12r + F43r + FAr2 = m2 AGr 2 = (109.5 N 9.81 m/s 2 )( −486.4 m/s 2 ) = −5429 N
Reassembling the above results,
F34t = 0.0606α = 20.72 N
F34r = F34t tan 20° = 7.54 N
F12t = 1.36α = 465.12 N
F12r = F12t tan 20° = 169.29 N
F2tA = 1206 − 0.613α = 996.35 N
F2rA = 169.29 − 7.54 = 161.75 N
The input and output bearing reactions are zero.
The load on the planet shaft is
F2 A =
(F ) + (F )
t
2A
2
r
2A
2
= 1009.4 N
The forces in the mounting bolts to restrain the unbalanced frame moment are:
F11 = 2 R1 F21t 0.4 m = 439.66 N
Ans.
Ans.
Ans.
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( 0.061 m ) F34t = ( 0.0036 Nm2
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226
13.13 The differential gear train shown in the figure has gear 1 fixed and is driven by rotating
shaft 5 at 500 rev/min in the direction shown. Gear 2 has fixed bearings constraining it to
rotate about the positive y axis, which remains vertical; this is the output shaft. Gears 3
and 4 have bearings connecting them to the ends of the carrier arm which is integral with
shaft 5. The pitch diameters of gears 1 and 5 are both 200 mm, while the pitch diameters
of gears 3 and 4 are both 150 mm. All gears have the 20° pressure angles and are each
18.75 mm thick, and all are made of steel with density 7.81 Mg/m3. The mass of shaft
5 and all gravitational loads are negligible. The output shaft torque loading is
T = −133.5ˆj N − m as shown. Note that the coordinate axes shown rotate with the input
shaft 5. Determine the driving torque required, and the forces and moments in each of the
m2 = 7.81 Mg/m 3 (π 0.12 ⋅ 0.01875 ) = 48 N
I Gyy2 = ( 48 N ) 0.12 2 = 0.24 Nm 2
m3 = 7.81 Mg/m3 (π 0.0752 ⋅ 0.01875 ) = 27 N
I Gxx3 = ( 27 N ) 0.0752 2 = 0.0759 Nm 2
I Gyy3 = ( 27 N ) 0.0752 4 = 0.0379 Nm 2
I Gzz3 = I Gyy3 = 0.0379 Nm 2
m4 = m3 = 27 N
I Gxx4 = I Gxx3 = 0.0759 Nm 2
I Gyy4 = I Gyy3 = 0.0379 N ⋅ m 2
I Gzz4 = I Gyy4 = 0.0379 Nm 2
m5 ≈ 0
ω5 = 500ˆj rev/min = 52.36ˆj rad/s
ω 2 = 2ω5 = 104.72ˆj rad/s
Noting that the primary xyz axes rotate at an angular velocity of ω5
ω = 69.81ˆi + 52.36ˆj rad/s
ω = −69.81ˆi + 52.36ˆj rad/s
3
4
Recognizing that dˆi dt = ω5 × ˆi = −52.36kˆ rad/s , the absolute accelerations are α 2 = 0 ,
α 3 = 69.81 −52.36kˆ = −3 655kˆ rad/s 2 , α 4 = −69.81 −52.36kˆ = 3 655kˆ rad/s 2 , α 5 = 0 .
(
)
(
)
Link 5: Note that we assume the mass of link 5 negligible. Also, assuming no thrust
bearing at the fixed pivot, F15y = F35y = F45y = 0 .
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t
t
bearings. (Hint: It is reasonable to assume through symmetry that F13
= F14
. It is also
necessary to recognize that only compressive loads, not tension, can be transmitted
between gear teeth.)
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∑F
∑F
∑M
∑M
∑M
x
5
= F15x − F35x + F45x = 0
z
5
= F15z − F35z + F45z = 0
x
5
= M 15x − d5 F15z = 0
y
5
= M 15y − 4 F35z − 4 F45z = 0
z
5
= M 15z + M 35z − M 45z + d5 F15x = 0
Link 4: Note that the forces on bevel gear teeth are related by Eq. (14.20) where, in this
case, cos γ 4 = 0.8 , sin γ 4 = 0.6 , and φ = 20° . Noting that
A = ω × ω × 0.1ˆi m = −0.1ω 2 ˆi m = −274.15ˆi m/s 2
5
G4
(
5
)
5
x
4
= 0.218 F14z + 0.218 F24z − F54x = m4 AGx4 = ( 27 N 9.81 m/s 2 )( −274.15 m/s 2 ) = −754.5 N
y
4
= 0.291F14z − 0.291F24z = 0
z
4
= F14z + F24z − F54z = 0
∑F
∑F
∑F
F24z = F14z
Using Eqs. (13.11) for the moment equations,
x
z
z
∑ M 4 = −3F14 + 3F24 = 0
y
∑ M4 = 0
yy
z
z
z
z
zz z
xx
x y
2
∑ M 4 = 0.218 F14 − 0.218 F24 + M 54 = IG α 4 − ( IG − IG )ω4 ω4 = 6.46 m/s
4
4
4
z
M 54
= 6.46 m/s 2
Link 3: Again cos γ 3 = 0.8 , sin γ 3 = 0.6 , and φ = 20° and, using Eqs. (13.11) we get
similar results
A G3 = ω5 × ω5 × −0.1ˆi in = 0.1ω52 ˆi in = 274.15ˆi m/s 2
(
)
x
3
= −0.218 F13z − 0.218 F23z + F53x = m3 AGx3 = ( 27 N 9.81 m/s 2 )( 274.15 m/s 2 ) = 754.5 N
y
3
= 0.291F13z − 0.291F23z = 0
z
3
= − F13z − F23z + F53z = 0
∑F
∑F
∑F
x
z
z
∑ M 3 = 0.075F13 − 0.075 F23 = 0
y
∑ M3 = 0
F23z = F13z
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227
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228
yy
z
z
z
z
zz z
xx
x y
2
∑ M 3 = −0.218 F13 + 0.218 F23 − M 53 = IG α 3 − ( IG − IG )ω3 ω3 = −6.46 m/s
3
3
3
z
M 53
= 6.46 m/s 2
∑F
∑F
∑F
∑M
∑M
∑M
x
2
= F12x + 0.218 F32z − 0.218 F42z =0
y
2
= − F12y + 0.291F32z + 0.291F42z =0
z
2
= F12z + F32z − F42z =0
x
2
= d 2 F12z = 0
F12z = 0
y
2
= − M 12y + 0.1F32z + 0.1F42z = 0
F32z + F42z = 135.6 Nm 0.1 m = 1356 N
z
2
= −0.1( 0.291F32z ) + 0.1( 0.291F42z ) + d 2 F12x = 0
Reviewing these again shows
F32z = F42z = 667.5 N
Finally, collecting all results, we have:
F12 = −388.7ˆj N
F12x = 0 , F12y = 388.7 N
F15 = 0
M12 = −135.6ˆj N ⋅ m
M = 271.2ˆj N ⋅ m
F35 = −1057.3ˆi − 1356kˆ N
F = 1057.3ˆi + 1356kˆ N
M 35 = 29.19kˆ N ⋅ m
M = −29.19kˆ N ⋅ m
45
15
45
Ans.
Ans.
Ans.
Ans.
13.14 The figure shows a flyball governor. Arms 2 and 3 are pivoted to block 6, which remains
at the height shown but is free to rotate around the y axis. Block 7 also rotates about and
is free to slide along the y axis. Links 4 and 5 are pivoted at both ends between the two
arms and block 7. The two balls at the ends of links 2 and 3 weigh 15.57 N each, and all
other masses are negligible in comparison; gravity acts in the -jˆ direction. The spring
between links 6 and 7 has a stiffness of 178 N/m and would be unloaded if block 7 were
at a height of R D = 275ˆj mm. All moving links rotate about the y axis with angular
velocities of ω ˆj . Make a graph of the height R versus the rotational speed ω in
D
rev/min, assuming that changes in speed are slow.
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Link 2: This time cos γ 2 = 0.6 , sin γ 2 = 0.8 , and φ = 20° .
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229
The free body diagram below shows only one of the arms, body 2, containing only one of
the two flyballs. Force F42 comes from body 4 which is a two-force member, thus
defining its line of action. The force FA comes from the spring. The position of the
bottom of the spring is
RD = 0.4 − 2 ( 0.15cos θ ) = 0.4 − 0.3cos θ
The total force in the spring is
k ( RD − RD 0 ) = 178 N/m ( 0.4 − 0.3cos θ m − 0.275 m ) = 0.125 − 0.3cos θ N
Taking moments about point B,
( 0.15cos θ ) m2 ( 0.3sin θ ) ω 2 − ( 0.15sin θ ) m2 g − ( 0.15sin θ )( 0.0625 − 0.15cos θ ) = 0
Dividing by 0.15sin θ
( 0.3cos θ ) m2ω 2 − m2 g − 0.0625 + 0.15cos θ = 0
Substituting values and rearranging we get
0.00272 cos θω 2 = 0.15 − 0.15cos θ = 0.15 (1 − cos θ ) ,
ω = 7.427
1 − cos θ
1 − cos θ
rad/s = 70.919
rev/min ,
cos θ
cos θ
cos θω 2 = 55.147 (1 − cos θ )
RD = 0.4 − 0.3cos θ
Ans.
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Since this total force must balance two flyball arms, force FA of the free body diagram is
FA = 0.0625 − 0.15cos θ
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231
Chapter 14
Vibration Analysis
Derive the differential equation of motion for each of the systems shown in the figure and
write the formula for the natural frequency ω n for each system.
(a)
∑ F = −k ( x − y ) = mx
mx + kx = ky
Ans.
ωn = k m
(b)
∑ F = F − cx − kx = mx
mx + cx + kx = F
Ans.
ωn = k m
(c)
∑ F = −cx − k1 x − k2 x = mx
mx + cx + ( k1 + k2 ) x = 0
ωn =
( k1 + k2 )
m
Ans.
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14.1
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232
(d)
∑ F = −k1 x − k2 ( x − y ) = mx
mx + ( k1 + k2 ) x = k2 y
ωn =
( k1 + k2 )
Ans.
m
∑ F = −c ( x − y ) − kx = mx
(f)
mx + cx + kx = cy
ωn = k m
Ans.
Both springs 3 and 4 experience the same spring force F34, and each is deflected
by an amount consistent with its own rate, F34/k3 or F34/k4, respectively. The total
deflection is x = ( F34 k3 ) + ( F34 k4 ) or F34 = ( k3 k4 x ) ( k3 + k4 )
∑ F = −k x − k x − k
1
2
k3 k 4
x = mx
3 + k4
⎡
kk ⎤
mx + ⎢ k1 + k2 + 3 4 ⎥ x = 0
k3 + k 4 ⎦
⎣
ωn =
k1 + k2 +
m
k3 k 4
k3 + k 4
Ans.
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(e)
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233
Evaluate the constants of integration of the solution to the differential equation for an
undamped free system, using the following sets of starting conditions:
(a)
x = x0 , x = 0
(b)
x = 0, x = v0
(c)
x = x0 , x = a0
(d)
x = x0 , x = b0
For each case, transform the solution to a form containing a single trigonometric term.
For each case use a trial solution of:
x = A sin ω n t + B cos ω nt
with initial value of:
x(0) = B
(1)
x = ω n A cos ω n t − ω n B sin ω nt
x ( 0 ) = ω n A
(2)
x = −ω n2 A sin ω nt − ω n2 B cos ω n t
x ( 0 ) = −ω n2 B
(3)
x = −ω n3 A cos ω nt + ω n3 B sin ω nt
x ( 0 ) = −ω n3 A
(4)
(a)
(b)
x = x0 , x = 0 . Use Eqs. (1) and (2); A = 0, B = x0
x = x0 cos ω n t
x = 0, x = v0 . Use Eqs. (1) and (2); A = v0 ω n , B = 0
Ans.
x = ( v0 ω n ) sin ω nt
(c)
Ans.
x = x0 , x = a0 . Use Eqs. (1) and (3); B = x0 , B = − a0 ω
2
n
These ae inconsistent unless a0 = −ω n2 x0 .
(d)
x = x0 , x = b0 . Use Eqs. (1) and (4); B = x0 , A = − b0 ω
x = ( −b0 ω n3 ) sin ω n t + x0 cos ω n t
Ans.
3
n
x = x02 + ( b0 ω n3 ) sin (ω nt + ψ ) where ψ = tan −1 ( x0ω n3 b0 )
2
Ans.
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14.2
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234
A system like Fig. 14.5 has m = 1 kg and an equation of motion
x = 20 cos ( 8π t − π / 4 ) mm. Determine:
(a)
The spring constant k
(b)
The static deflection δst
(c)
The period
(d)
The frequency in hertz
(e)
The velocity and acceleration at the instant t = 0.20 s
(f)
The spring force at t = 0.20 s
Plot a phase diagram to scale showing the displacement, velocity, acceleration, and
spring-force phasors at the instant t = 0.20 s.
(a)
ω n = k m = 8π rad/s
k = ω n2 m = ( 8π rad/s ) (1 kg ) = 631.65 N/m
2
(b)
(c)
(d)
(e)
F mg (1 kg ) ( 9.81 m/s
=
=
k
k
631.65 N/m
τ = 2π ω n = 2π ( 8π ) = 0.250 s
δ st =
2
) = 0.015 53 m = 15.53 mm
f = 1 τ = 4 Hz
Ans.
Ans.
Ans.
Ans.
θ = 8π t − π 4 = ( 8t − 0.25 ) π = ⎡⎣8 ( 0.20 ) − 0.25⎤⎦ π = 1.35π rad = 243°
x = −8π ( 20 ) sin ( 8π t − π 4 ) = −503sin ( 243° ) = 448 mm/s
Ans.
x = − ( 8π ) ( 20 ) cos ( 8π t − π 4 ) = −12 633cos ( 243° ) = 5 735 mm/s 2
Ans.
2
(f)
F = kx = ( 0.631 65 N/mm )( 20 mm ) cos 243° = 12.633cos 243° N = −5.735 N Ans.
(g)
x = −12.633cos 243° m/s 2
x = 0.020 cos 243° m , x = −0.503sin 243° m/s , F = 12.633cos 243° N
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14.3
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235
The weight W1 in the figure drops through the distance h and collides with W2 with
inelastic impact (a coefficient of restitution of zero). Derive the differential equation of
motion of the system, and determine the amplitude of the resulting motion of W2 .
Define t ′ = 0 at the instant of impact. At the beginning of impact we have v1 = 2 gh .
By conservation of momentum, m1v1 = ( m1 + m2 ) v2 . Thus
v2 =
(W + W2 ) v or
W1
2 gh = 1
2
g
g
W1 2 gh
. Therefore, at t ′ = 0 , x = 0 , x = v2 .
W1 + W2
Note that, at x = 0 , the spring force now includes a reaction to W2 .
And so
∑ F = −kx + W = (W g ) x where, for convenience, we have defined W = W + W .
1
1
2
the force balance we get the differential equation of motion
(W g ) x + kx = W1
From
Ans.
with natural frequency of ω n = kg W . Then
x = A cos ω nt ′ + B sin ω nt ′ + W1 k
x = − Aω n sin ω nt ′ + Bω n cos ω n t ′
and with the initial conditions stated above A = − W1 k and B = v2 ω n . Therefore
x = − (W1 k ) cos ω nt ′ + ( v2 ω n ) sin ω nt ′ + (W1 k )
Transforming to a single transient term we get
x = X sin (ω n t ′ −ψ ) + W1 k where
X=
2
2
(W1 k ) + ( v2 ω n )
⎛W k ⎞
and ψ = tan −1 ⎜ 1 ⎟
⎝ v2 ω n ⎠
Ans.
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14.4
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236
14.5
The vibrating system shown in the figure has k1 = k3 = 875 N/m , k2 = 1 750 N/m , and
W = 40 N . What is the natural frequency in hertz?
Springs 1 and 2 bothexperience the same spring force F12, and each is deflected by an
amount consistent with its own rate, F12/k1 or F12/k2, respectively. The total deflection is
x = ( F12 k1 ) + ( F12 k2 ) or F12 = ( k1k2 x ) ( k1 + k2 )
∑ F = −F
12
⎛ kk
⎞
mx + ⎜ 1 2 + k3 ⎟ x = 0
⎝ k1 + k2
⎠
− k3 x = mx
⎛ k1k2
⎞
+ k3 ⎟
⎜
k +k
⎠=
ωn = ⎝ 1 2
W g
14.6
( 875)(1750 ) + 875
875 + 1750
40 9.81
= 18.91 rad/s = 3.010 Hz
Ans.
The figure shows a weight W = 66.75 N connected to a pivoted rod which is assumed to
be weightless and very rigid. A spring having a rate of k = 10,680 N/m is connected to
the center of the rod and holds the system in static equilibrium at the position shown.
Assuming that the rod can vibrate with a small amplitude, determine the period of the
motion.
(W A g )θ + ( ka )θ = −W A
(10, 680 N/m ) ( 9.81 m/s )
= 19.81 rad/s
∑ M = −a ⋅ ( kaθ ) − A ⋅W = (W g ) A θ
2
ka 2
a kg 150 mm
ωn =
=
=
2
W A g A W 300 mm
τ=
2π
ωn
=
2π rad/rev
= 0.317 s/rev
19.81 rad/s
2
2
2
66.75 N
Ans.
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The natural frequency is
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237
The figure shows an upside-down pendulum of length l retained by two springs
connected a distance a from the pivot. The springs have been positioned such that the
pendulum is in static equilibrium when it is in the vertical position.
(a)
For small amplitudes, find the natural frequency of this system.
(b)
Find the ratio l/a at which the system becomes unstable.
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14.7
∑ M = W ⋅ Aθ − a ⋅ k aθ − a ⋅ k aθ = (W g ) A θ
2
1
2
(W A g )θ + ( k + k ) a θ − W Aθ = 0
2
2
1
θ +
2
2
⎤
g ⎡ ( k1 + k2 ) a
− 1⎥ θ = 0
⎢
A⎣
WA
⎦
2
⎤
g ⎡ ( k1 + k2 ) a
− 1⎥
Ans.
ωn =
⎢
A⎣
WA
⎦
The system becomes unstable whenever the natural frequency becomes the square root of
a negative number (imaginary). At such values the system is not oscillatory. This
happens whenever
A a ≥ ( k1 + k2 ) a W
Ans.
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238
14.8
(b)
(c)
Write the differential equation for the system shown in the figure and find the
natural frequency.
Find the response x if y is a step input of height y0 .
Find the relative response z = x - y to the step input of part (b).
(a)
∑ F = −k x − k ( x − y ) = mx
1
2
mx + ( k1 + k2 ) x = k2 y
(b)
ωn =
( k1 + k2 )
m
Ans.
The complimentary solution is
x′ = A cos ω n t + B sin ω nt
For a particular solution, arrange the equation to the form
x + ω n2 x = ( k2 m ) y
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(a)
x′′ = 0 .
Since, for a step input the right-hand side is constant, try x′′ = C , 2
ω n C = ( k2 m ) y
C = k2 y ( k1 + k2 )
The complete solution is x = x′ + x′′
x = A cos ω nt + B sin ω n t + k2 y ( k1 + k2 )
At t = 0 , x = 0 and y = y0 .
0 = A (1) + B ( 0 ) + k2 y0 ( k1 + k2 )
A = − k2 y0 ( k1 + k2 )
x = −ω n A sin ω n t + ω n B cos ω nt
At t = 0 , x = 0
0 = −ω n A ( 0 ) + ω n B (1)
B=0
Therefore, the complete response is
x = − k2 y0 ( k1 + k2 ) cos ω n t + k2 y0 ( k1 + k2 )
x=
(c)
k 2 y0
(1 − cos ωnt )
k1 + k2
(k + k ) y
k2 y0
(1 − cos ωnt ) − 1 2 0
k1 + k2
k1 + k2
k y + k y cos ω n t
z=− 1 0 2 0
k1 + k2
Ans.
z = x− y =
Ans.
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239
14.9
An undamped vibrating system consists of a spring whose scale is 35 kN/m and a mass of
1.2 kg. A step force F = 50 N is exerted on the mass for 0.040 s.
(a)
Write the equations of motion of the system for the era in which the force acts and
for the era after.
(b)
What are the amplitudes in each era?
(c)
Sketch a displacement time plot of the motion.
(a)
ωn =
( 35 000 N/m ) (1.2 kg ) = 171 rad/s
1.2 x + 35000 x = 50
1st era; 0 ≤ t ≤ 0.040 s :
From Eq. (14.21)
x = ( F k )(1 − cos ω n t ) = ( 50 N 35 000 N/m )(1 − cos171t )
X 0 = x02 + ( v0 ω n ) =
2
( 0.209 mm )2 + (127 mm/s 171 rad/s )2
Ans.
= 0.773 mm
φ = tan −1 ( v0 ω n x0 ) = tan −1 ⎡⎣(127 mm/s ) (171 rad/s )( 0.209 mm ) ⎤⎦ = 74.3°
(b)
x = X 0 cos (ω nt − φ ) = ( 0.773 mm ) cos (171t − 74.3° )
1st era; 0 ≤ t ≤ 0.040 s :
X = 1.429 mm
X = 0.773 mm
2nd era; t ≥ 0.040 s :
Ans.
Ans.
Ans.
(c)
14.10 The figure shows a round shaft whose torsional spring constant is kt lb ⋅ in/rad
connecting two wheels having the mass moments of inertia I1 and I 2 . Show that the
system is capable of vibrating torsionally with a frequency of
⎡k (I + I )⎤
ωn = ⎢ t 1 2 ⎥
I1I 2
⎣
⎦
1/ 2
Designating the angular positions of the two wheels by θ1 and θ 2 , respectively, and
summing moments on each, we get
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x = ( 0.001 429 m )(1 − cos171t )
x = (1.429 mm )(1 − cos171t )
Also x = 0.244sin171t m/s = 244sin171t mm/s
At the end of the 1st era t = 0.040 s , ω n t = 6.831 rad = 391.4°
x = 0.000 209 m = 0.209 mm ,
x = 0.127 m/s = 127 mm/s
2nd era; t ≥ 0.040 s :
1.2 x + 35 000 x = 0
From Eqs. (14.16) and (14.17)
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240
∑M
∑M
1
2
I1θ1 + ktθ1 − ktθ 2 = 0
I 2θ2 − ktθ1 + ktθ 2 = 0
= − kt (θ1 − θ 2 ) = I1θ1
= k (θ − θ ) = I θ
t
1
2
2 2
Next, assume a solution of the form
θ j = C j cos (ω nt − φ )
for each inertia with j = 1, 2. Substituting these gives
−ω n2 I1C1 cos (ω nt − φ ) + kt C1 cos (ω nt − φ ) − kt C2 cos (ω n t − φ ) = 0
−ω n2 I 2C2 cos (ω nt − φ ) − kt C1 cos (ω nt − φ ) + kt C2 cos (ω nt − φ ) = 0
⎡ −ω n2 I1 + kt
⎤ ⎡ C1 ⎤
− kt
⎢
⎥⎢ ⎥ =0
2
−ω n I 2 + kt ⎦ ⎣C2 ⎦
⎣ − kt
For this set of equations to have a non-trivial solution for C1 and C2, the determinant of
the coefficient matrix must vanish. Therefore,
( −ωn2 I1 + kt )( −ωn2 I 2 + kt ) − ( −kt )( −kt ) = 0
This expands to a quadratic equation in ω n2
I1 I 2 (ω n2 ) − kt ( I1 + I 2 ) ω n2 = 0
2
Solving this, we get
ωn = ±
ω n = ±0
kt ( I1 + I 2 )
I1 I 2
Q.E.D.
14.11 A motor is connected to a flywheel by a 15.63 mm-diameter steel shaft 900 mm long, as
shown. Using the methods of this chapter, it can be demonstrated that the torsional
spring constant of the shaft is 522.875 N–m/rad. The mass moments of inertia of the
motor and flywheel are 2.67 and 6.23 Nms2, respectively. The motor is turned on for 2 s,
and during this period it exerts a constant torque of 22.25 N–m on the shaft.
(a)
What speed in revolutions per minute does the shaft attain?
(b)
What is the natural circular frequency of vibration of the system?
(c)
Assuming no damping, what is the amplitude of the vibration of the system in
degrees during the first era? During the second era?
This is a difficult problem, but too interesting and challenging not to include.
(a)
The angular impulse equation is
t
H = H 0 + ∫ Tdt
0
Since the motor starts from rest, its initial angular momentum is H 0 = 0 . We also
see that H = ( I1 + I 2 ) ω , T = 22.25 N − m , and t = 2 s . Substituting,
( I1 + I 2 ) ω = H 0 + ∫0 ( 22.25 N − m ) dt
(8.9 Nms2 ) ω = ( 22.25 N − m ) t
t
ω = ( 2.5 rad/s 2 ) t for 0 ≤ t ≤ 2 s
At t = 2 s ω = 5.0 rad/s = 47.75 rev/min Ans.
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Dividing each by cos (ω nt − φ ) and writing these in matrix form they become
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241
From Problem P14.10
ωn =
(c)
k t ( I1 + I 2 )
I1 I 2
=
( 522.875 N − m/rad ) ( 2.67 Nms 2 + 6.23 Nms 2 )
( 2.67 Nms )( 6.23 Nms )
2
2
= 16.726 rad/s
Ans.
1st era; 0 ≤ t ≤ 2 s :
The differential equations are:
I1θ1 + ktθ1 − ktθ 2 = T
I θ − k θ + k θ = 0
2 2
t 1
t 2
After using the conditions that, at t = 0, θ1 = θ2 = 0 the solutions become
I2
T ⎛ I2 t 2 ⎞
B cos ω n t +
⎜ + ⎟
I1
I1 + I 2 ⎝ kt 2 ⎠
Tt 2
θ 2 = A + B cos ω nt +
2 ( I1 + I 2 )
θ1 = A −
Then using the conditions that, at t = 0, θ1 = θ 2 = 0 , we find
I1 I 2
T
B = −A =
kt ( I1 + I 2 )2
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(b)
and the solutions become
I2
T
T ⎛ I2 t 2 ⎞
θ1 = −
ω
+
+
cos
I
I
t
(
)
⎜ + ⎟
1
2
n
kt ( I1 + I 2 )2
I1 + I 2 ⎝ kt 2 ⎠
I1 I 2
T
Tt 2
θ2 = −
(1 − cos ω nt ) +
2 ( I1 + I 2 )
kt ( I1 + I 2 )2
But we are interested in the relative motion, the twist in the shaft, which is
T I 2 (1 − cos ω n t )
θ1 − θ 2 =
kt
( I1 + I 2 )
So the amplitude during the first era is
6.23 Nms 2 )
(
22.25 Nm )
(
I2
T
γ=
=
= 0.029 rad = 1.707°
kt ( I1 + I 2 ) ( 522.875 Nm/rad ) ( 8.9 Nms 2 )
Ans.
For the completion of the first era we can compute that, at t = 2.0 s,
ω nt = (16.7 rad/s )( 2.0 s ) = 33.45 rad = 1 916.7° and cos ω nt = −0.449 .
Thus,
θ1 = 5.0302 rad , and θ1 = 5.0302 rad . These are the initial displacements for the
second era.
2nd era; t ≥ 2 s :
The differential equations now become:
I1θ1 + ktθ1 − ktθ 2 = 0
I θ − k θ + k θ = 0
2 2
t 1
t 2
Following a similar procedure to that above, we eventually obtain
θ1 − θ 2 = 0.0555cos (ω nt ′ − 19.6° )
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242
Therefore the amplitude of the second era is
γ = 0.0555 rad = 3.180°
Ans.
14.12 The weight of the mass of a vibrating system is 44.5 N, and it has a natural frequency of
1 Hz. Using the phase-plane method, plot the response of the system to the force
function shown in the figure. What is the final amplitude of the motion?
k = mωn2 = ( 44.5 N 9.81 m/s 2 ) ( 2π rad/s ) = 178.89 N/m
2
The final amplitude is X = 464 mm.
Ans.
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F1 k = 53.4 N 178.89 N/m = 0.298 m
F2 k = −26.7 N 178.89 N/m = −0.149 m
ω n Δt1 = ( 2π rad/s )( 0.25 s ) = 1.571 rad = 90.0° , ω n Δt2 = ( 2π rad/s )( 0.25 s ) = 1.571 rad = 90.0°
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243
14.13 An undamped vibrating system has a spring scale of 35.6 kN/m and a weight of 222.5 N.
Find the response and the final amplitude of vibration of the system if it is acted upon by
the forcing function shown in the figure. Use the phase-plane method.
ωn = k m = 35.6 kN/m ( 222.5 N 9.81 m/s 2 ) = 39.618 rad/s
The final amplitude is 4.42 mm.
Ans.
14.14 A vibrating system has a spring k = 71.2 kN/m and a weight of W = 356 N. Plot the
response of this system to the forcing function shown in the figure:
(a)
Using three steps
(b)
Using six steps
ωn = k m = 71.2 kN/m ( 356 N 9.81 m/s 2 ) = 44.294 rad/s
(a)
Three step solution: ωn Δt = ( 44.294 rad/s )( 0.1 s 3) = 1.476 rad = 84.63°
F1 k = 2 mm , F2 k = 6.25 mm , F3 k = 10.425 mm , F k = 12.5 mm
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ωn Δt = ( 39.618 rad/s )( 0.040 s ) = 1.585 rad = 90° , F k = 55.63 N 35.6 kN/m = 1.5625 mm
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244
(b)
Six step solution: ωn Δt = ( 44.294 rad/s )( 0.1 s 6 ) = 0.738 rad = 42.3°
F1 k = 1 mm , F2 k = 3.125 mm , F3 k = 5.2 mm , F4 k = 7.3 mm ,
F5 k = 9.375 mm , F6 k = 11.45 mm , F k = 12.5 mm
(c)
(d)
What is the value of the coefficient of critical damping for a spring-mass-damper
system in which k = 56 kN/m and m = 40 kg?
If the actual damping is 20 percent of critical, what is the natural circular
frequency of the system?
What is the period of the damped system?
What is the value of the logarithmic decrement?
(a)
ω n = k m = 56 N/m 40 kg = 1.183 rad/s
(a)
(b)
cc = 2mω n = 2 ( 40 kg )(1.183 rad/s ) = 94.657 N ⋅ s/m
Ans.
(b)
(c)
ω d = ω n 1 − ς 2 = 1.183 rad/s 1 − 0.202 = 1.159 rad/s
τ = 2π ω d = 2π 1.159 rad/s = 5.420 s/cycle
Ans.
Ans.
(d)
δ = 2πς
1 − ς 2 = 2π ( 0.2 )
1 − 0.202 = 1.283
Ans.
14.16 A vibrating system has a spring k = 3.5 kN/m and a mass m = 15 kg. When disturbed, it
was observed that the amplitude decayed to one-fourth of its original value in 4.80 s.
Find the damping coefficient and the damping factor.
ω n = k m = 3 500 N/m 15 kg = 15.275 rad/s
Using Eq. (14.32) with δ N = ln (1.0 0.25 ) = 1.386 and Nτ = 4.80 s
( Nτω n ) = 1.386 ⎡⎣( 4.80 s )(15.275 rad/s )⎤⎦ = 0.0189
Ans.
c = ς 2mω n = 0.0189 ⋅ 2 ⋅15 kg ⋅15.275 rad/s = 8.664 N ⋅ s/m
Ans.
ς = δN
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14.15
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245
14.17 A vibrating system has k = 53.4 kN/m, W = 400.5 N, and damping equal to 20 percent of
critical.
(a)
What is the damped natural frequency ωd of the system?
(b)
What are the period and the logarithmic decrement?
ωn = k m = 53.4 kN/m ( 400.5 N 9.81 m/s 2 ) = 36.16 rad/s
(a)
(b)
ωd = ωn 1 − ς 2 = 36.16 rad/s 1 − 0.202 = 35.43 rad/s
τ = 2π ωd = 2π 35.43 rad/s = 0.177 s/cycle
Ans.
Ans.
δ = 2πς
Ans.
1 − ς 2 = 2π ⋅ 0.20
1 − 0.202 = 1.283
14.18 Solve Problem 14.14 using damping equal to 15 percent of critical.
ωd = ωn 1 − ς 2 = 44.29 rad/s 1 − 0.152 = 43.788 rad/s
Six step solution: ωd Δt = ( 43.788 rad/s )( 0.10 s 6 ) = 0.729 rad = 41.78°
F1 k = 1 mm , F2 k = 3.125 mm , F3 k = 5.2 mm , F4 k = 7.3 mm ,
F5 k = 9.375 mm , F6 k = 11.45 mm , F k = 12.5 mm
In each step of ω d Δt = 0.724 rad the reduction in amplitude is
X n + 41.48° X n = e
−
ς ( 0.724 rad )
1−ς 2
= 0.896 . Therefore,
x0 = 1 mm , φ0 = −90°
x1′ = 0.896 ( 0.1 mm ) = 0.0896 mm
x1 = 2.845 mm , φ1 = −77.34°
x2′ = 0.896 ( 2.845 mm ) = 2.549 mm
x2 = 4.132 mm , φ2 = −59.98°
x3′ = 0.896 ( 4.132 mm ) = 3.703 mm
x3 = 4.79 mm , φ3 = −42.86°
x4′ = 0.896 ( 4.79 mm ) = 4.292 mm
x4 = 4.815 mm , φ4 = −27.01°
x5′ = 0.896 ( 4.815 mm ) = 4.314 mm
x5 = 4.297 mm , φ5 = −13.53°
x6′ = 0.896 ( 0.1719 in ) = 3.85 mm
x6 = 3.485 mm , φ6 = 12.64°
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ωn = k m = 71.2 kN/m ( 356 N 9.81 m/s 2 ) = 44.29 rad/s
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14.19 A damped vibrating system has an undamped natural frequency of 10 Hz and a weight of
3.56 kN. The damping ratio is 0.15. Using the phase-plane method, determine the
response of the system to the forcing function shown in the figure.
ω n = 10 revs/s ( 2π rad/rev ) = 62.832 rad/s
ω d = ω n 1 − ς 2 = 62.832 rad/s 1 − 0.152 = 62.121 rad/s
ω d Δt = ( 62.121 rad/s )( 0.01 s ) = 0.621 rad = 35.59°
k = mωn2 = ( 3.56 kN 9.81 m/s 2 ) ( 62.832 rad/s ) = 1432.66 kN/m
2
F1 k = 26.25 mm , F2 k = 39.375 mm , F3 k = 13.125 mm , F4 k = −13.125 mm ,
In each step of ω d Δt = 0.621 rad the reduction in amplitude is
ς ( 0.621 rad )
X n +35.59° X n = e 1−ς = 0.910 . Therefore,
x1′ = 23.875 mm
x0 = 26.25 mm , φ0 = −90°
x1 = 29.7 mm , φ1 = −43.53°
x2′ = 27.025 mm
x2 = 43.5 mm , φ2 = 59.95°
x3′ = 39.6 mm
x3 = 58.4 mm , φ3 = 113.94°
x4′ = 53.15 mm
x4 = 51.25 mm , φ4 = 199.90°
2
x1′′ = 22.75 mm
x2′′ = 24.6 mm
x3′′ = 36.025 mm
x4′′ = 48.375 mm
14.20 A vibrating system has a spring rate of 534 kN/m, a damping factor of 9790 Ns/m, and a
weight of 3.56 kN. It is excited by a harmonically varying force F0 = 445 N at a
frequency of 435 cycles per minute.
(a)
Calculate the amplitude of the forced vibration and the phase angle between the
vibration and the force.
(b)
Plot several cycles of the displacement-time and force-time diagrams.
(a)
m = 3.56 kN 9.81 m/s 2 = 362.89 N ⋅ s 2 /m
ωn = k m = 534 kN/m 362.89 N ⋅ s 2 /m = 38.36 rad/s
ς = c ( 2mωn ) = 9790 Ns/m
(( 2) (362.89 Ns /m ) ⋅ 38.36 rad/s ) = 0.351
2
ω ωn = ( 435 cycles/min ⋅ 2π rad/cycle 60 s/min ) 38.36 rad/s = 1.187
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−
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247
F0 k
X=
(1 − ω
ω n2 ) + ( 2ς ω ω n )
2
2
2
445 N 534 kN/m
X=
= 0.875 mm
Ans.
2ς ω ωn
2 ⋅ 0.351 ⋅1.187
= tan −1
= 116.14°
2
2
1 − ω ωn
1 − 1.187 2
Ans.
(1 − 1.187 ) + ( 362.89 ×1.187 )
2 2
φ = tan −1
2
ω n = k m = 525 000 N/m 360 kg = 38.188 rad/s
ς = c ( 2mω n ) = 9 640 N ⋅ s/m ( 2 ⋅ 360 kg ⋅ 38.188 rad/s ) = 0.351
ω ω n = ( 4.80 Hz ⋅ 2π rad/cycle ) 38.188 rad/s = 0.790
X=
X=
F0 k
(1 − ω
2
ω n2 ) + ( 2ς ω ω n )
2
2
450 N 525 000 N/m
(1 − 0.790 ) + ( 2 ⋅ 0.351⋅ 0.790 )
2 2
φ = tan −1
= 0.001 280 m = 1.280 mm
Ans.
2
2ς ω ω n
2 ⋅ 0.351 ⋅ 0.790
= tan −1
= 55.80°
2
2
1 − ω ωn
1 − 0.7902
Ans.
14.22 When a 2724 kg press is mounted upon structural-steel floor beams, it causes them to
deflect 18.75 mm. If the press has a reciprocating unbalance of 190.7 kg and it operates
at a speed of 80 rev/min, how much of the force will be transmitted from the floor beams
to other parts of the building? Assume no damping. Can this mounting be improved?
k = 2724 kg 18.75 mm = 145.28 kg/mm
m = 2724 kg 9.81 m/s 2 = 277.67 kgs 2 /m
ωn = k m = 145.28 kg/mm 277.67 kgs 2 /m = 22.87 rad/s
ω ωn = ( 80 rev/min ⋅ 2π rad/rev 60 s/min ) 22.87 rad/s = 0.366
Assuming no damping, Eq. (14.62) gives
1
1
1
T=
=
=
= 1.155
2
2
2
1
1
0.366
−
−
ω
ω
2
2 2
n
1− ω ω
(
n
)
Ftr = TF0 = 1.155 × 190.7 kg = 220.26 kg
Fig. 14.37 shows that, with ω ωn = 0.366 , small changes in either damping or ω n will do
little to reduce transmissibility. Therefore the mounting cannot be improved. The
primary improvement would be to reduce the unbalance.
www.elsolucionario.net
14.21 A spring-mounted mass has k = 525 kN/m, c = 9640 N ⋅ s / m, and m = 360 kg. This
system is excited by a force having an amplitude of 450 N at a frequency of 4.80 Hz.
Find the amplitude and phase angle of the resulting vibration and plot several cycles of
the force-time and displacement-time diagrams.
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248
14.23 Four vibration mounts are used to support a 450-kg machine that has a rotating unbalance
of 0.35 kg ⋅ m and runs at 300 rev/min. The vibration mounts have damping equal to 30
percent of critical. What must the spring constant of the mounting be if 20 percent of the
exciting force is transmitted to the foundation? What is the resulting amplitude of motion
of the machine?
From Eq. (14.63)
T = 0.20 =
(ω 2 / ωn2 ) 1 + ( 2 ⋅ 0.30ω / ω n )2 =
2
⎡1 − ω 2 / ω 2 ⎤ + ( 2 ⋅ 0.30ω / ω )2
n ) ⎦⎥
n
⎣⎢ (
(ω / ω n ) 2
1 + 0.36 (ω / ω n )
2
2
⎡1 − (ω / ω )2 ⎤ + 0.36 (ω / ω )2
n ⎥
n
⎣⎢
⎦
2
⎧⎪
⎪
2 2
2⎫
4
2
0.04 ⎨ ⎡1 − (ω / ω n ) ⎤ + 0.36 (ω / ω n ) ⎬ = (ω / ω n ) ⎡1 + 0.36 (ω / ω n ) ⎤
⎢
⎥
⎢
⎣
⎦
⎣
⎦⎥
⎪⎩
⎪⎭
0.36 (ω / ω n ) + 0.96 (ω / ω n ) + 0.0656 (ω / ω n ) − 0.04 = 0
6
4
2
Numerically searching for the root we find
(ω / ωn )2 = 0.168 423 rad 2 /s2
ω / ω n = 0.410 rad/s
k = mω n2 = 450 kg ( 0.168 423 rad 2 /s 2 ) = 75.790 N/m
Ans.
Now, from Eq. (14.57)
mX
=
mu e
(ω / ω n )
2
2
=
⎡1 − ω / ω n )2 ⎤ + 0.36 (ω / ω n )2
⎣ (
⎦
X = 0.489 ( 0.35 kg ⋅ m ) 450 kg = 0.380 mm
( 0.410 )
= 0.489
2
⎡⎣1 − ( 0.410 ) ⎤⎦ + 0.36 ( 0.410 )
Ans.
www.elsolucionario.net
2
2
2
2
0.20 ⎡1 − (ω / ω n ) ⎤ + 0.36 (ω / ω n ) = (ω / ω n ) 1 + 0.36 (ω / ω n )
⎣⎢
⎦⎥
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249
Chapter 15
Dynamics of Reciprocating Engines
A one-cylinder, four-cycle engine has a compression of 7.6 and develops 2244 w at 3 000
rev/min. The crank length is 22 mm with a 59.4 mm bore. Develop and plot a rounded
indicator diagram using a card factor of 0.90, a mechanical efficiency of 72 percent, a
suction pressure of 0.1 mPa, and a polytropic exponent of 1.30.
A = π D 2 4 = π ( 59.4 mm ) 4 = 2769.76 mm 2
2
Δv = AA = 2 ⋅ 22 mm ⋅ 2769.76 mm 2 = 121,869 mm3
v1 = Δv r ( r − 1) = 121,869 mm3 ⋅ 7.6 ( 7.6 − 1) = 140,334 mm3
v2 = v1 − Δv = 140,334 mm3 − 121,869 mm3 = 18, 465 mm3
C = v2 Δv = 18, 465 mm3 121,869 mm3 = 0.1515 = 15.15%
pb =
( 44 mm ) ( 2769.76 mm
2
( 2244 w )
) ( 3 000 rev/min
2 rev/work stroke )
= 702.78 kPa
pi = pb em = 702.78 kPa 0.72 = 976 kPa
p1 = 0.1 mPa
7.6 − 1 976 kPa
r − 1 pi
+ p1 = (1.30 − 1) 1.3
+ 0.1 mPa (100 kPa) = 439.58 kPa
p4 = ( k − 1) k
7.6 − 7.6 0.90
r − r fc
As in Example 15.1, we calculate the values
x(%)
v(kmm3)
pc(kPa)
pe(kPa)
0
18.34 1410.42
6121.44
5
24.41
974.31
4228.65
10
30.45
730.76
3171.63
15
36.51
577.49
2506.44
20
42.58
473.18
2053.69
25
48.63
398.12
1727.93
30
54.68
341.84
1483.62
35
60.75
298.25
1294.42
40
66.79
263.60
1144.09
45
72.86
235.50
1022.11
50
78.92
212.29
921.39
55
84.97
192.85
836.99
www.elsolucionario.net
15.1
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250
60
65
70
75
80
85
90
95
100
91.03
97.08
103.14
109.20
115.25
121.31
127.37
133.42
139.48
176.35
162.18
149.92
139.21
129.78
121.42
113.97
107.29
101.28
765.36
703.91
650.69
604.19
563.26
526.99
494.66
465.68
439.58
15.2
Construct a rounded indicator diagram for a four-cylinder, four-cycle gasoline engine
having a 84.4 mm bore, a 88 mm stroke, and a compression ratio of 6.25. The operating
conditions to be used are 22.44 kw at 1 900 rev/min. Use a mechanical efficiency of 72
percent, a card factor of 0.90, and a polytropic exponent of 1.30.
A = π D 2 4 = π ( 84.4 mm ) 4 = 5591.84 mm 2
2
Δv = AA = 88 mm ⋅ 5591.84 mm 2 = 492.081 kmm3
v1 = Δv r ( r − 1) = 492.081 kmm3 ⋅ 6.25 ( 6.25 − 1) = 585.81 kmm 3
v2 = v1 − Δv = 585.81 kmm3 − 492.081 kmm3 = 93.73 kmm3
C = v2 Δv = 93.73 kmm3 492.081 kmm3 = 0.1905 = 19.05%
pb =
( 22.44 kN 4 cyl )
( 3.50 in ) ( 5591.84 mm 2 ) (1 900 rev/min
pi = pb em = 687.97 kPa 0.72 = 955.5 kPa
p1 = 101.28 kPa
2 rev/work stroke )
= 687.97 kPa
www.elsolucionario.net
Then we sketch and round the following diagram
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251
6.25 − 1 955.5 kPa
r − 1 pi
+ p1 = (1.30 − 1)
+ 101.28 kPa = 466.31 kPa
k
6.251.3 − 6.25 0.90
r − r fc
As in Example 15.1, we calculate the values
x(%
v(kmm3)
pc(kPa)
pe(kPa)
)
0
93.18
1096.78
5051.18
5
117.65
810.09
3730.84
10
142.11
633.71
2918.54
15
165.78
515.5
2374.12
20
191.03
431.39
1986.77
25
215.5
368.85
1698.74
30
239.97
320.74
1477.15
35
264.42
282.71
1302.03
40
288.89
252.00
1160.56
45
313.34
226.73
1044.19
50
337.81
205.62
946.98
55
362.28
187.75
864.71
60
386.73
172.47
794.29
65
411.20
159.25
733.43
70
435.65
147.73
680.35
75
460.125
137.60
633.70
80
484.578
128.64
592.44
85
509.05
120.66
555.69
90
533.51
113.52
522.81
95
557.96
107.09
493.21
100
582.43
101.28
466.31
Then we sketch and round the following diagram
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p4 = ( k − 1)
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252
Construct an indicator diagram for a V6 four-cycle gasoline engine having a 100-mm
bore, a 90-mm stroke, and a compression ratio of 8.40. The engine develops 150 kW at
4 400 rev/min. Use a mechanical efficiency of 72 percent, a card factor of 0.88, and a
polytropic exponent of 1.30.
A = π D 2 4 = π ( 0.100 m ) 4 = 0.007 854 m 2
2
Δv = AA = 0.090 m ⋅ 0.007 854 m 2 = 0.000 707 m3
v1 = Δv r ( r − 1) = 0.000707 m3 ⋅ 8.40 ( 8.40 − 1) = 0.000 802 m 3
v2 = v1 − Δv = 0.000 802 m3 − 0.000 707 m3 = 0.000 096 m3
C = v2 Δv = 0.000 096 m3 0.000 707 m3 = 0.1358 = 13.58%
pb =
(150 000 W 6 cyl )( 60 s/min )
= 964 kPa
( 0.090 m ) ( 0.007 854 m 2 ) ( 4 400 rev/min 2 rev/work stroke )
pi = pb em = 964 kPa 0.72 = 1 339 kPa
p1 = 100 kPa
8.40 − 1 1 339 kPa
r − 1 pi
+ p1 = (1.30 − 1)
+ 100 kPa = 550 kPa
p4 = ( k − 1) k
8.401.3 − 8.40 0.88
r − r fc
As in Example 15.1, we calculate the values
pc(kPa)
pe(kPa)
x(%
v(m3)
)
0
0.000096
1 590
8 751
5
0.000131
1 056
5 812
10
0.000167
774
4 259
15
0.000202
603
3 315
20
0.000237
488
2 687
25
0.000272
408
2 243
30
0.000308
348
1 914
35
0.000343
302
1 661
40
0.000378
266
1 462
45
0.000414
237
1 302
50
0.000449
213
1 170
55
0.000484
193
1 061
60
0.000520
176
968
65
0.000555
161
888
70
0.000590
149
820
75
0.000626
138
760
80
0.000661
129
708
85
0.000696
120
661
90
0.000732
113
620
95
0.000767
106
583
100
0.000802
100
550
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15.3
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253
15.4
A single-cylinder, two-cycle gasoline engine develops 30 kW at 4 500 rev/min. The
engine has an 80-mm bore, a stroke of 70 mm, and a compression ratio of 7.0. Develop a
rounded indicator diagram for this engine using a card factor of 0.990, a mechanical
efficiency of 65 percent, a suction pressure of 100 kPa, and a polytropic exponent of
1.30.
A = π D 2 4 = π ( 0.080 m ) 4 = 0.005 027 m 2
2
Δv = AA = 0.070 m ⋅ 0.005 027 m 2 = 0.000 352 m3
v1 = Δv r ( r − 1) = 0.000 352 m 3 ⋅ 7.0 ( 7.0 − 1) = 0.000 411 m3
v2 = v1 − Δv = 0.000 411 m3 − 0.000 352 m3 = 0.000 059 m3
C = v2 Δv = 0.000 059 m3 0.000 352 m3 = 0.1662 = 16.62%
pb =
30 000 W ( 60 s/min )
( 0.070 m ) ( 0.005 027 m 2 ) ( 4 500 rev/min 1 rev/work stroke )
= 1 137 kPa
pi = pb em = 1137 kPa 0.65 = 1 749 kPa
p1 = 100 kPa
r − 1 pi
7.0 − 1 1 749 kPa
p4 = ( k − 1) k
+ p1 = (1.30 − 1) 1.3
+ 100 kPa = 673 kPa
r − r fc
7.0 − 7.0 0.990
As in Example 15.1, we calculate the values
x(%)
v(m3)
pc(kPa)
pe(kPa)
0 0.000059
1 259
8 473
5 0.000076
894
6019
10 0.000094
682
4 592
15 0.000111
546
3 672
20 0.000129
451
3 034
25 0.000147
382
2 569
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Then we sketch and round the following diagram
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254
0.000164
0.000182
0.000199
0.000217
0.000235
0.000252
0.000270
0.000287
0.000305
0.000323
0.000340
0.000358
0.000375
0.000393
0.000411
329
288
256
229
207
188
173
159
147
137
128
120
112
106
100
2 217
1 942
1 722
1 542
1 394
1 268
1 162
1 070
991
921
860
805
756
712
673
Then we sketch and round the following diagram
15.5
The engine of Problem 15.1 has a connecting rod 78.125 mm long and a weight of 0.056
kg, with the mass center 10 mm from the crankpin end. Piston weight is 0.98 kg. Find
the bearing reactions and the crankshaft torque during the expansion stroke
corresponding to a piston displacement of X = 30 percent ( ω t = 60° ). To find pe , see
the answer to Problem 15.3 in the answers to selected problems.
A = 0.078125 m , m3 = 0.056 kg 9.81 m/s 2 = 0.0057 kg ⋅ s 2 /m ,
m4 = 0.18 kg 9.81 m/s 2 = 0.01834 kg ⋅ s 2 /m , A A = 0.01 m , A B = A − A A = 0.068125 m ,
m3 A = m3A B A = ( 0.0057 kg ⋅ s 2 /m ) ( 0.068125 m ) 0.078125 m = 0.00497 kg ⋅ s 2 /m ,
m3 B = m3A A A = ( 0.0057 kg ⋅ s 2 /m ) ( 0.01 m ) 0.078125 m = 0.0064 kg ⋅ s 2 /m ,
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30
35
40
45
50
55
60
65
70
75
80
85
90
95
100
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255
ω = ( 3 000 rev/min )( 2π rad/rev ) ( 60s/min ) = 314.16 rad/s , r = 0.022 m ,
r A = ( 0.022 m ) ( 0.078125 m ) = 0.280 , rω 2 = ( 0.022 m )( 314.16 rad/s ) = 2171.32 m/s 2 ,
2
θ = ω t = 60° ,
x = r cos θ + l 1 − ( r l )( sin θ )
2
= ( 0.022 m ) cos 60° + ( 0.078125 m ) 1 − ( 0.28 )( sin 60° ) = 0.086725 m
,
2
X = 30% , pe = 1483.62 kPa (from Prob. 15.1), A = π ( 59.4 mm ) 4 = 2769.76 mm 2 ,
2
r
⎛
⎞
x = − rω 2 ⎜ cos ωt + cos 2ωt ⎟ = − ( 2171.32 m/s 2 ) ( cos 60° + 0.280 cos120° ) = −783.75 m/s 2
l
⎝
⎠
2
⎛
⎞
r
r
tan φ = sin ω t ⎜1 + 2 sin 2 ω t ⎟ = 0.28sin 60° (1 + 0.282 sin 2 60° 2 ) = 0.250
l
⎝ 2l
⎠
x + P ⎤⎦ tan φ ˆj
F41 = − ⎡⎣( m3B + m4 ) = − ⎡ − 0.0064 kg ⋅ s 2 /m + 0.01834 kg ⋅ s 2 /m 783.75 m/s 2 + 418 kg ⎤ 0.250ˆj
⎢⎣
⎥⎦
= −104.4ˆj kg
Ans.
F34 = ( m4 x + P ) ˆi − ⎡⎣( m3B + m4 ) x + P ⎤⎦ tan φ ˆj
= ⎡ − 0.01834 kg ⋅ s 2 /m 783.75 m/s 2 + 418 kg ⎤ ˆi − 104.4ˆj kg
⎣⎢
⎦⎥
= 418.59ˆi − 104.4ˆj kg = 431.75 kg∠ − 14.0°
Ans.
(
(
)
)
{
}
= ⎡( 0.00497 kg ⋅ s 2 /m )( 2171.32 m/s 2 ) cos 60° − 416.77 kg ⎤ ˆi + 116.22ˆj kg
⎣⎢
⎦⎥
F32 = [ m3 A rω 2 cos ωt − ( m3 B + m4 ) x − P ]ˆi + m3 A rω 2 sin ωt + [( m3 B + m4 ) x + P ] tan φ ˆj
= −408.15ˆi + 116.22ˆj kg = 424 kg∠164.1°
x + P ⎤⎦ tan φ kˆ = ( 0.086725 m )(104.4 kg ) kˆ = 88.8kˆ N − m
T21 = x ⎡⎣( m3B + m4 ) 15.6
Ans.
Ans.
Repeat Problem 15.5, but do the computations for the compression cycle ( ωt = 660° ).
A = 0.078125 m , m3 = 0.056 kg 9.81 m/s 2 = 0.0057 kg ⋅ s 2 /m ,
m4 = 0.18 kg 9.81 m/s 2 = 0.01834 kg ⋅ s 2 /m , A A = 0.01 m , A B = A − A A = 0.068125 m ,
m3 A = m3A B A = ( 0.0057 kg ⋅ s 2 /m ) ( 0.068125 m ) 0.078125 m = 0.00497 kg ⋅ s 2 /m ,
m3 B = m3A A A = ( 0.0057 kg ⋅ s 2 /m ) ( 0.01 m ) 0.078125 m = 0.0064 kg ⋅ s 2 /m ,
ω = ( 3 000 rev/min )( 2π rad/rev ) ( 60s/min ) = 314.16 rad/s , r = 0.022 m ,
r A = ( 0.022 m ) ( 0.078125 m ) = 0.280 , rω 2 = ( 0.022 m )( 314.16 rad/s ) = 2171.32 m/s 2 ,
2
www.elsolucionario.net
P = pe A = (1483.62 kPa ) ( 0.002769 m 2 ) = 418 kg
= 0.002769 m2
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256
θ = ω t = 660° ,
x = r cos θ + l 1 − ( r l )( sin θ )
2
= ( 0.022 m ) cos 660° + ( 0.078125 m ) 1 − ( 0.28 )( sin 660° ) = 0.086725 m
,
2
X = 30% , pc = 49.614 lb/in 2 (from Prob. 15.1), A = π ( 0.0594 m ) 4 = 0.002769 m 2 ,
2
P = pc A = ( 341.84 kPa ) ( 0.002769 m 2 ) = 99.88 kg
r
⎛
⎞
x = − rω 2 ⎜ cos ωt + cos 2ωt ⎟ = − ( 2171.32 m/s 2 ) ( cos 660° + 0.280 cos1320° ) = −783.75 m/s 2
l
⎝
⎠
2
⎛
⎞
r
r
tan φ = sin ω t ⎜1 + 2 sin 2 ω t ⎟ = 0.28sin 660° (1 + 0.282 sin 2 660° 2 ) = −0.250
l
⎝ 2l
⎠
x + P ⎤⎦ tan φ ˆj
F41 = − ⎡⎣( m3B + m4 ) = − ⎡ − 0.0064 kg ⋅ s 2 /m + 0.01834 kg ⋅ s 2 /m 783.75 m/s2 + 99.88 kg ⎤ ( −0.250 ) ˆj
⎢⎣
⎥⎦
Ans.
= 20.88ˆj kg
F34 = ( m4 x + P ) ˆi − ⎡⎣( m3B + m4 ) x + P ⎤⎦ tan φ ˆj
= ⎡ − 0.01834 kg ⋅ s 2 /m 782.75 m/s 2 + 99.88 kg ⎤ ˆi + 20.88ˆj kg
⎣⎢
⎦⎥
= 85.35ˆi + 20.88ˆj kg = 88 kg∠14°
Ans.
)
(
)
{
}
F32 = [ m3 A rω 2 cos ωt − ( m3 B + m4 ) x − P ]ˆi + m3 A rω 2 sin ωt + [( m3 B + m4 ) x + P ] tan φ ˆj
(
)(
)
= ⎡ 0.00497 kg ⋅ s 2 /m 217.32 m/s 2 cos 660° − 85.33 kg ⎤ ˆi − 37.68ˆj kg
⎣⎢
⎦⎥
= −75.82ˆi − 37.68ˆj kg = 84.89 kg∠206°
x + P ⎤⎦ tan φ kˆ = ( 0.086725 m )( −20.88 kg ) kˆ = −18.2kˆ N − m
T21 = x ⎡⎣( m3B + m4 ) 15.7
Ans.
Ans.
Make a complete force analysis of the engine of Problem 15.5. Plot a graph of the
crankshaft torque versus crank angle for 720° of crank rotation.
A = 0.078125 m , m3 = 0.056 kg 9.81 m/s 2 = 0.0057 kg ⋅ s 2 /m ,
m4 = 0.18 kg 9.81 m/s 2 = 0.01834 kg ⋅ s 2 /m , A A = 0.01 m , A B = A − A A = 0.068125 m ,
m3 A = m3A B A = ( 0.0057 kg ⋅ s 2 /m ) ( 0.068125 m ) 0.078125 m = 0.00497 kg ⋅ s 2 /m ,
m3 B = m3A A A = ( 0.0057 kg ⋅ s 2 /m ) ( 0.01 m ) 0.078125 m = 0.0064 kg ⋅ s 2 /m ,
ω = ( 3 000 rev/min )( 2π rad/rev ) ( 60s/min ) = 314.16 rad/s , r = 0.022 m ,
r A = ( 0.022 m ) ( 0.078125 m ) = 0.280 , rω 2 = ( 0.022 m )( 314.16 rad/s ) = 2171.32 m/s 2 ,
2
A = π ( 0.0594 m ) 4 = 0.002769 m 2 , p = pe or pc as taken from Prob. 15.1,
2
P = Ap = ( 0.002769 m 2 ) p ,
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(
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257
2
2
⎛r
⎞
x = r cos ωt + l 1 − ⎜ sin ωt ⎟ = ( 0.022 m ) cos θ + ( 0.078125 m ) 1 − ( 0.28sin θ ) ,
⎝l
⎠
r
x = − rω 2 (cos ωt + cos 2ωt ) = − ( 2171.32 m/s 2 ) ( cos θ + 0.280 cos 2θ )
l
⎛
⎞
r
r2
tan φ = sin ω t ⎜1 + 2 sin 2 ω t ⎟ = 0.28sin θ (1 + 0.0392sin 2 θ )
l
⎝ 2l
⎠
T21 = ⎡⎣( m3B + m4 ) x + P ⎤⎦ x tan φ
)
0
15
30
45
60
75
90
105
120
135
150
165
180
195
210
225
240
255
270
285
300
315
330
345
360
375
390
405
420
435
450
1103
1507
1007
643
429
307
235
191
164
146
136
130
128
29.5
29.5
29.5
29.5
29.5
29.5
29.5
29.5
29.5
29.5
29.5
29.5
0
0
0
0
0
0
-2763.5
-2609
-2172
-1527
-777
-35
605
1082
1382
1527
1567
1562
1554
1562
1567
1527
1382
1082
605
-35
-777
-1527
-2172
-2609
-2763.5
-2609
-2172
-1527
-777
-35
605
100
99
96.3
92
86.7
80.87
75
695
64.85
61.1
58.4
56.8
56.25
56.8
58.4
61.1
64.85
69.5
75
80.87
86.7
92
96.3
99
100
99
96.3
92
86.7
80.87
75
0
0.072 66
0.141 37
0.201 87
0.249 62
0.280 35
0.290 98
0.280 35
0.249 62
0.201 87
0.141 37
0.072 66
0
-0.072 66
-0.141 37
-0.201 87
-0.249 62
-0.280 35
-0.290 98
-0.280 35
-0.249 62
-0.201 87
-0.141 37
-0.072 66
0
0.072 66
0.141 37
0.201 87
0.249 62
0.280 35
0.29098
0
105.78
136.2
123.94
99.43
85.81
71.73
59.47
47.67
35.41
23.61
11.80
0
-4.54
-8.63
-12.26
-14.07
-14.07
-12.26
-8.17
-7.26
0
1.82
1.82
0
-3.63
-5.90
-5.90
-3.63
0
3.63
T21 , N–m
0
104.29
130.74
113.45
85.99
69.16
53.67
41.24
30.85
21.69
13.69
6.67
0
-2.49
-4.97
-7.34
-9.15
-9.83
-9.04
-6.55
-6.22
0
1.81
1.58
0
-3.73
-5.88
-5.54
-3.28
0
2.59
www.elsolucionario.net
(
= ⎡ 0.002769 m 2 p − ( 42.69 kg )( cos θ + 0.280 cos 2θ ) ⎤ x tan φ
⎢⎣
⎥⎦
tan φ
P, kg
x, mm
F14 , kg
ωt, °
x, m/s 2
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465
480
495
510
525
540
555
570
585
600
615
630
645
660
675
690
705
720
15.8
0
0
0
0
0
29.5
30
31
33.6
37.7
44
54
71
99
148
232
347
1103
1057
1382
1527
1567
1562
1554
1562
1567
1527
1382
1082
605
-35
-777
-1527
-2172
-2609
-2763.5
69.5
64.8
61.1
58.4
56.8
56.25
56.8
58.4
61.1
64.85
69.5
75
80.87
86.7
92
96.3
99
100
0.280 35
0.249 62
0.201 87
0.141 37
0.072 66
0
-0.072 66
-0.141 37
-0.201 87
-0.249 62
-0.280 35
-0.290 98
-0.280 35
-0.249 62
-0.201 87
-0.141 37
-0.072 66
0
5.9
6.8
5.9
4.54
2.27
0
-4.54
-8.63
-12.7
-16.3
-18.2
-19
-19.5
-20.8
-24
-26.8
-21.3
0
4.07
4.41
3.73
2.59
1.24
0
-2.48
-5.08
-7.79
-105
-12.65
-14.35
-15.8
-17.96
-21.9
-25.65
-21.24
0
The engine of Problem 15.3 uses a connecting rod 300 mm long. The masses are
m3 A = 0.80 kg, m3 B = 0.38 kg, and m4 = 1.64 kg. Find all the bearing reactions and the
crankshaft torque for one cylinder of the engine during the expansion stroke at a piston
displacement of X = 30 percent ( ω t = 63.2° ). The pressure should be obtained from the
indicator diagram, Fig. AP15.3 in the answers to selected problems.
A = 0.300 m , m3 A = 0.80 kg , m3 B = 0.38 kg , m4 = 1.64 kg ,
ω = ( 4 400 rev/min )( 2π rad/rev ) ( 60s/min ) = 460.8 rad/s , r = 0.045 m ,
r A = ( 0.045 m ) ( 0.300 m ) = 0.150 , rω 2 = ( 0.045 m )( 460.8 rad/s ) = 9 554 m/s 2 ,
2
www.elsolucionario.net
258
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259
x = r cosθ + l 1 − ( r l )( sin θ )
θ = ω t = 63.2° ,
2
= ( 0.045 m ) cos 63.2° + ( 0.300 m ) 1 − ( 0.15sin 63.2° ) = 0.317 6 m
2
X = 30.0% , pe = 1 914 kPa (from Prob. 15.3), A = π ( 0.100 m ) 4 = 0.007 854 m 2 ,
2
r
x = − rω 2 (cos ω t + cos 2ω t ) = − ( 9 554 m/s 2 ) ( cos 63.2° + 0.150 cos126.4° ) = −3 457 m/s 2
l
⎛
⎞
r
r2
tan φ = sin ω t ⎜1 + 2 sin 2 ω t ⎟ = 0.15sin 63.2° (1 + 0.152 sin 2 63.2° 2 ) = 0.135
l
⎝ 2l
⎠
F41 = − ⎡⎣( m3B + m4 ) x + P ⎤⎦ tan φ ˆj
= − ⎡ − ( 0.38 kg + 1.64 kg ) 3 457 m/s 2 + 15 033 N ⎤ 0.135ˆj
⎣
⎦
Ans.
= −1 087ˆj N
F34 = ( m4 x + P ) ˆi − ⎡⎣( m3B + m4 ) x + P ⎤⎦ tan φ ˆj
= ⎡ − (1.64 kg ) 3 457 m/s 2 + 15 033 N ⎤ ˆi − 1 087ˆj N
⎣
⎦
ˆ
ˆ
= 9 364i − 1 087 j N = 9 426 N∠ − 6.6°
Ans.
{
}
x − P ]ˆi + m3 A rω 2 sin ω t + [( m3 B + m4 ) x + P ] tan φ ˆj
F32 = [ m3 Arω 2 cos ω t − ( m3 B + m4 ) (
)
= ⎡( 0.80 kg ) 9 554 m/s 2 cos 63.2° − 8 050 N ⎤ ˆi +
{ 0.80 kg ( 9 554 m/s ) sin 63.2° + 1 087 N} ˆj
(
)
⎣
⎦
= −4 604ˆi + 7 909ˆj N = 9 152 N∠120.2°
x + P ⎤⎦ tan φ kˆ = ( 0.317 6 m )(1 087 N ) kˆ = 345kˆ N ⋅ m
T21 = x ⎡⎣( m3B + m4 ) 15.9
2
Ans.
Ans.
Repeat Problem 15.8, but do the computations for the same position in the compression
cycle ( ω t = 656.8° ).
A = 0.300 m , m3 A = 0.80 kg , m3 B = 0.38 kg , m4 = 1.64 kg ,
ω = ( 4 400 rev/min )( 2π rad/rev ) ( 60s/min ) = 460.8 rad/s , r = 0.045 m ,
r A = ( 0.045 m ) ( 0.300 m ) = 0.150 , rω 2 = ( 0.045 m )( 460.8 rad/s ) = 9 554 m/s 2 ,
2
θ = ω t = 656.8° ,
x = r cosθ + l 1 − ( r l )( sin θ )
2
= ( 0.045 m ) cos 656.8° + ( 0.300 m ) 1 − ( 0.15sin 656.8° ) = 0.317 6 m
2
X = 30.0% , pc = 348 kPa (from Prob. 15.3), A = π ( 0.100 m ) 4 = 0.007 854 m 2 ,
2
P = pc A = ( 348 kPa ) ( 0.007 854 m 2 ) = 2 733 N
x = − rω 2 (cos ω t + cos 2ω t ) = − ( 9 554 m/s 2 ) ( cos 656.8° + 0.150 cos1313.6° ) = −3 457 m/s 2
r
l
www.elsolucionario.net
P = pe A = (1 914 kPa ) ( 0.007 854 m 2 ) = 15 033 N
www.elsolucionario.net
260
⎛
⎞
r
r2
tan φ = sin ω t ⎜1 + 2 sin 2 ω t ⎟ = 0.15sin 656.8° (1 + 0.152 sin 2 656.8° 2 ) = −0.135
l
⎝ 2l
⎠
F41 = − ⎡⎣( m3B + m4 ) x + P ⎤⎦ tan φ ˆj
= − ⎡ − ( 0.38 kg + 1.64 kg ) 3 457 m/s 2 + 2 733 N ⎤ ( −0.135 ) ˆj
⎣
⎦
ˆ
= −574 j N
F34 = ( m4 x + P ) ˆi − ⎡⎣( m3B + m4 ) x + P ⎤⎦ tan φ ˆj
= ⎡ − (1.64 kg ) 3 457 m/s 2 + 2 733 N ⎤ ˆi − 574ˆj N
⎣
⎦
ˆ
ˆ
= −2 936i − 574 j N = 2 992 N∠191.0°
{
}
Ans.
Ans.
F32 = [ m3 A rω 2 cos ω t − ( m3B + m4 ) x − P ]ˆi + m3 A rω 2 sin ω t + [( m3B + m4 ) x + P ] tan φ ˆj
)
{ 0.80 kg ( 9 554 m/s ) sin 656.8° + 574 N} ˆj
(
)
⎣
⎦
= 7 696ˆi − 5 534ˆj N = 9 479 N∠ − 35.7°
x + P ⎤⎦ tan φ kˆ = ( 0.317 6 m )( 574 N ) kˆ = 182kˆ N ⋅ m
T21 = x ⎡⎣( m3B + m4 ) 2
Ans.
Ans.
15.10 Additional data for the engine of Problem 15.4 are l3 = 110 mm, RG 3 A = 15 mm,
m4 = 0.24 kg, and m3 = 0.13 kg. Make a complete force analysis of the engine and plot
a graph of the crankshaft torque versus crank angle for 360° of crank rotation.
A = 0.110 m , m3 = 0.13 kg , m4 = 0.24 kg , A A = 0.015 m , A B = A − A A = 0.095 m ,
m3 A = m3A B A = ( 0.13 kg )( 0.095 m ) 0.110 m = 0.112 kg ,
m3 B = m3A A A = ( 0.13 kg )( 0.015 m ) 0.110 m = 0.018 kg ,
ω = ( 4 500 rev/min )( 2π rad/rev ) ( 60 s/min ) = 471.24 rad/s , r = 0.035 m ,
r A = ( 0.035 m ) ( 0.110 m ) = 0.318 , rω 2 = ( 0.035 m )( 471.24 rad/s ) = 7 772 m/s 2 ,
2
A = π D 2 4 = π ( 0.080 m ) 4 = 0.005 027 m 2 , p = pe or pc as taken from Prob. 15.1,
2
P = Ap = ( 0.005 027 m 2 ) p ,
x = r cos ω t + l 1 − ( r l ) sin 2 ω t = ( 0.035 m ) cos θ + ( 0.110 m ) 1 − ( 0.318sin θ ) ,
2
r
x = − rω 2 (cos ω t + cos 2ω t ) = − ( 7 772 m/s 2 ) ( cosθ + 0.318cos 2θ )
l
⎛
⎞
r
r2
tan φ = sin ω t ⎜1 + 2 sin 2 ω t ⎟ = 0.318sin θ (1 + 0.050 62sin 2 θ )
l
⎝ 2l
⎠
T21 = ⎡⎣( m3B + m4 ) x + P ⎤⎦ x tan φ
(
)
= ⎡ 0.005 027 m 2 p − ( 2005 N )( cos θ + 0.318cos 2θ ) ⎤ x tan φ
⎣⎢
⎦⎥
2
www.elsolucionario.net
(
= ⎡( 0.80 kg ) 9 554 m/s 2 cos 656.8° + 4 250 N ⎤ ˆi +
www.elsolucionario.net
ωt, °
P, N
0
15
30
45
60
75
90
105
120
135
150
165
180
195
210
225
240
255
270
285
300
315
330
345
360
24 230
180 213
24 502
15 941
10 788
7 797
6 022
4 937
4 258
3 916
3 569
3 428
3 383
508
528
568
633
734
895
1 156
1 604
2 368
3 640
5 329
24 230
x, m/s 2
-10 245
-9 649
-7 968
-5 496
-2 650
130
2 473
4 153
5 123
5 496
5 495
5 366
5 299
5 366
5 495
5 496
5 123
4 153
2 473
130
-2 650
-5 496
-7 968
-9 649
-10 245
x, m
tan φ
F14 , N
T21 , N· m
0.145 000
0.143 434
0.138 910
0.131 928
0.123 241
0.113 735
0.104 283
0.095 617
0.088 241
0.082 431
0.078 288
0.075 819
0.075 000
0.075 819
0.078 288
0.082 431
0.088 241
0.095 617
0.104 283
0.113 735
0.123 241
0.131 928
0.138 910
0.143 434
0.145 000
0
0.082 631
0.161 104
0.230 683
0.286 015
0.321 855
0.334 288
0.321 855
0.286 015
0.230 683
0.161 104
0.082 631
0
-0.082 631
-0.161 104
-0.230 683
-0.286 015
-0.321 855
-0.334 288
-0.321 855
-0.286 015
-0.230 683
-0.161 104
-0.082 631
0
0
14 685
3 616
3 350
2 890
2 520
2 226
1 934
1 596
1 230
803
398
0
-156
-313
-458
-559
-581
-512
-383
-263
-219
-255
-234
0
0
2 106
502
442
356
287
232
185
141
101
63
30
0
-12
-25
-38
-49
-56
-53
-44
-32
-29
-35
-34
0
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261
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262
15.11 The four-cycle engine of Problem 15.1 has a stroke of 65 mm and a connecting rod
length of 180 mm. The weight of the rod is 0.386 kg, and the center of mass center is
41.5 mm from the crankpin. The piston assembly weighs 0.58 kg. Make a complete force
analysis for one cylinder of this engine for 720º of crank rotation. Use 0.11 mPa for the
exhaust pressure and 68.9 kPa for the suction pressure. Plot a graph to show the variation
of the crankshaft torque with the crank angle. Use Fig. 15.23 for the pressures.
A = 180 mm , m3 = 0.386 kg 9.81 m/s 2 = 0.039347 kg ⋅ s 2 /m ,
m4 = 0.58 kg 9.81 m/s 2 = 0.059123 kg ⋅ s 2 /m , A A = 41.5 mm , A B = A − A A = 138.5 mm ,
m3 A = m3A B A = ( 0.039347 kg ⋅ s 2 /m ) (138.5 mm ) 180 mm = 0.030275 kg ⋅ s 2 /m ,
m3 B = m3A A A = ( 0.039347 kg ⋅ s 2 /m ) ( 41.5 mm ) 180 mm = 0.009071 kg ⋅ s 2 /m ,
ω = ( 3 000 rev/min )( 2π rad/rev ) ( 60s/min ) = 314.16 rad/s , r = 32.5 mm ,
r A = ( 32.5 mm ) (180 mm ) = 0.181 , rω 2 = ( 32.5 mm )( 314.16 rad/s ) = 3.21 m/s 2 ,
2
A = π ( 59.375 mm ) 4 = 2767.4 mm 2 , p = pe or pc as taken from Example 15.1,
P = Ap = ( 2767.4 mm 2 ) p ,
2
2
⎛r
⎞
x = r cos ωt + l 1 − ⎜ sin ωt ⎟ = ( 32.5 mm ) cos θ + (180 mm ) 1 − ( 0.181sin θ ) ,
⎝l
⎠
r
x = − rω 2 (cos ωt + cos 2ωt ) = − ( 3.21 m/s 2 ) ( cos θ + 0.181cos 2θ )
l
⎛
⎞
r
r2
tan φ = sin ω t ⎜1 + 2 sin 2 ω t ⎟ = 0.181sin θ (1 + 0.0163sin 2 θ )
l
⎝ 2l
⎠
T21 = ⎡⎣( m3B + m4 ) x + P ⎤⎦ x tan φ
(
)
= ⎡ 2767.4 mm 2 p − ( 217.48 kg )( cos θ + 0.181cos 2θ ) ⎤ x tan φ
⎢⎣
⎥⎦
ωt, °
P, kg
0
15
30
45
60
75
90
105
120
135
150
165
1380.8
1958.5
1312
838.9
556.7
395
299.1
240.2
203.1
174.5
165.2
157.1
x, m/s 2
-3786.7
-3599.87
-3067.45
-2268.125
-1314.2
-328.6
579
1332
1893
2268
2488
2597
x, mm
tan φ
F14 , kg
T21 , kg· m
212.5
211.2
207.4
201.5
194
185.6
177
168.8
161.5
154
151
148.4
0
0.046 782
0.090 646
0.128 713
0.158 277
0.177 056
0.183 499
0.177 056
0.158 277
0.128 713
0.090 646
0.046 782
0
80.35
100
87.9
74.1
66
62
58.5
52.7
42.4
30.4
15.6
0
16.9
20.7
17.7
14.4
12.25
10.99
9.88
8.47
6.5
4.6
2.3
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2
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180
195
210
225
240
255
270
285
300
315
330
345
360
375
390
405
420
435
450
465
480
495
510
525
540
555
570
585
600
615
630
645
660
675
690
705
720
154.5
31.7
31.7
31.7
31.7
31.7
31.7
31.7
31.7
31.7
31.7
31.7
31.7
19.2
19.2
19.2
19.2
19.2
19.2
19.2
19.2
19.2
19.2
19.2
29
29.5
31.25
32.6
37.9
45.1
56.25
74.1
104.9
157.6
246.9
368.3
1380.8
2628
2597
2488
2268
1893
1332
579
-329
-1314
-2268
-3067
-3600
-3787
-3600
-3067
-2268
-1314
-329
579
1332
1893
2268
2488
2597
2628
2597
2488
2268
1893
1332
579
-329
-1314
-2268
-3067
-3600
-3786.7
147.5
148.4
151
154
161.5
168.8
177
185.6
194
201.5
207
211.2
212.5
211.2
207
201.5
194
185.6
177
168.8
161.5
154
151
148.4
147.5
148.4
151
154
161.5
168.8
177
185.6
194
201.5
207.4
211.2
212.5
0
-0.046 782
-0.090 646
-0.128 713
-0.158 277
-0.177 056
-0.183 499
-0.177 056
-0.158 277
-0.128 713
-0.090 646
-0.046 782
0
0.046 782
0.090 646
0.128 713
0.158 277
0.177 056
0.183 499
0.177 056
0.158 277
0.128 713
0.090 646
0.046 782
0
-0.046 782
-0.090 646
-0.128 713
-0.158 277
-0.177 056
-0.183 499
-0.177 056
-0.158 277
-0.128 713
-0.090 646
-0.046 782
0
0
-9.3
-18.3
-23.7
-25.45
-21.9
-12.9
-1.78
8.93
15.6
16.1
9.82
0
-10.7
-16.96
-17.4
-11.2
-0.45
10.7
19.2
23.2
22.3
16.96
8.93
0
-9.82
-18.3
-24.1
-26.3
-24.1
-17.4
-9.37
-2.68
-0.47
-3.57
-5.8
0
0
-1.44
-2.74
-3.68
-4.1
-3.67
-2.3
-0.3
1.76
3.17
3.3
2.1
0
-2.2
-3.55
-3.49
-2.14
-0.1
1.89
3.27
3.77
3.43
2.58
1.35
0
-1.43
-2.73
-3.69
-4.25
-4.05
-3.1
-1.7
-0.48
-0.1
-0.74
-1.23
0
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264
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265
Chapter 16
Balancing
16.1
Determine the bearing reactions at A and B for the system shown in the figure if the speed
is 300 rev/min. Determine the magnitude and the angular orientation of the balancing
mass if it is located at a radius of 50 mm.
F1 = m1 R1ω 2 = ( 2 kg )( 0.025 m )( 31.416 rad/s ) = 49.348 N
2
F2 = m2 R2ω 2 = (1.5 kg )( 0.035 m )( 31.416 rad/s ) = 51.815 N
2
F3 = m3 R3ω 2 = ( 3 kg )( 0.040 m )( 31.416 rad/s ) = 118.435 N
F = 49.348 N∠90° = 49.349ˆj N
2
1
F2 = 51.815 N∠ − 165° = −50.050ˆi − 13.411ˆj N
F = 118.435 N∠ − 75° = 30.653ˆi − 114.400ˆj N
3
∑F = F + F
1
2
+ F3 = −19.397ˆi − 78.462ˆj N = 80.824 N∠ − 103.9°
Since all rotating masses are in a single plane, the correction mass must be in that plane.
FC = −∑ F = 80.824 N∠76.1°
FC = mC RCω 2 = mC ( 0.050 m )( 31.416 rad/s ) = 80.824 N
2
mC = FC
( R ω ) = (80.824 N ) ⎡⎣( 0.050 m )( 31.416 rad/s ) ⎤⎦ = 1.638 kg
2
2
C
θ C = 76.1°
16.2
Ans.
Ans.
The figure shows three weights connected to a shaft that rotates in bearings at A and B.
Determine the magnitude of the bearing reactions if the shaft speed is 300 rev/min. A
counterweight is to be located at a radius of 250 mm. Find the value of the weight and its
angular orientation.
ω = ( 300 rev/min )( 2π rad/rev ) ( 60 s/min ) = 31.416 rad/s
( 56.75 g )( 200 mm )( 31.416 rad/s )
F1 = m1 R1ω =
2
2
( 9.81 m/s )
2
= 1.14 kg
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ω = ( 300 rev/min )( 2π rad/rev ) ( 60 s/min ) = 31.416 rad/s
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266
( 42.56 g )( 300 mm )( 31.416 rad/s )
F2 = m2 R2ω =
2
2
2
( 9.81 m/s )
( 85.13 g )(150 mm )( 31.416 rad/s )
F3 = m3 R3ω =
2
2
= 1.26 kg
2
= 1.26 kg
( 9.81 m/s )
F1 = 1.14 kg∠90° = 1.14ˆj kg
F2 = 1.26 kg∠ − 135° = −0.89ˆi − 0.89ˆj kg
F = 1.26 kg∠ − 30° = 1.09ˆi − 0.63ˆj kg
3
∑F = F + F
1
2
+ F3 = 0.2ˆi − 0.38ˆj kg = 0.44 kg∠ − 63.5°
Since all rotating masses are in a single plane, the correction mass must be in that plane.
FC = −∑ F = 0.44 kg∠116.5°
FC = mC RCω 2 = mC ( 250 mm )( 31.416 rad/s ) = 0.44 kg
0.44 kg ( 9.81 m/s 2 )
FC
mC =
=
= 17.49 g
RCω 2 0.25 m ( 31.416 rad/s )2
Ans.
θ C = 116.5°
Ans.
Without the correction mass
450 mm kˆ × FB + 300 mm kˆ × ∑ F = 0
450 mm kˆ × F + 300 mm kˆ × ∑ F = 0
B
16.3
The figure shows two weights connected to a rotating shaft and mounted outboard of
bearings A and B. If the shaft rotates at 120 rev/min, what are the magnitudes of the
bearing reactions at A and B? Suppose the system is to be balanced by reducing a weight
at a radius of 125 mm. Determine the amount and the angular orientation of the weight
to be removed.
ω = (120 rev/min )( 2π rad/rev ) ( 60 s/min ) = 12.566 rad/s
(1.82 kg )(100 mm )(12.566 rad/s )
F1 = m1 R1ω =
2
2
= 2.93 kg
2
9.81 m/s
(1.36 kg )(150 mm )(12.566 rad/s )
F2 = m2 R2ω =
2
2
9.81 m/s
ˆ
F1 = 2.93 kg∠90° = 2.93 j kg
F = 3.28 kg∠ − 135° = −2.319ˆi − 2.319ˆj kg
2
= 3.28 kg
2
∑ F = F + F = −2.319ˆi + 0.611ˆj kg = 2.44 kg∠165.6°
∑ M = ( −150 mm ) kˆ × ∑ F + ( −100 mm ) kˆ × F
1
B
2
A
(
)
= ( 0.907 N − m ) ˆi + ( 3.529 N − m ) ˆj + ( −100 mm ) kˆ × FAx ˆi + FAy ˆj = 0
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2
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267
FA = 3.545ˆi − 0.911ˆj kg = 3.66 kg∠ − 14.4°
F = − ( F + F ) = −1.182ˆi + 0.303ˆj kg = 1.22 kg∠165.6°
B
A
Ans.
∑
Ans.
The correction should be
FC = −∑ F = −2.44 kg∠165.6°
This can be done be mass removal at a radius of 5.0 in of
−2.44 kg ( 9.81 m/s 2 )
FC
=
= −1.193 kg
mC =
RCω 2 125 mm (12.566 rad/s )2
Ans.
θ C = 165.6°
16.4
Ans.
For a speed of 220 rev/min, calculate the magnitudes and relative angular orientations of
the bearing reactions at A and B for the two-mass system shown.
ω = ( 220 rev/min )( 2π rad/rev ) ( 60 s/min ) = 23.038 rad/s
F1 = m1 R1ω 2 = ( 2 kg )( 0.060 m )( 23.038 rad/s ) = 63.692 N
F2 = m2 R2ω 2 = (1.5 kg )( 0.040 m )( 23.038 rad/s ) = 31.846 N
F = 63.692 N∠90° = 63.692ˆj N
2
1
F2 = 31.846 N∠ − 90° = −31.846ˆj N
∑ M B = −0.250kˆ m × 63.692ˆj N + −0.550kˆ m × −31.846ˆj N + −0.500kˆ m × FA = 0
) (
(
) (
) (
) (
)
FA = 3.185ˆj N = 3.185 N∠90.0°
F = − ( F + F + F ) = −35.031ˆj N = 35.031 N∠ − 90.0°
B
16.5
1
2
Ans.
Ans.
A
The rotating system shown in the figure has R1 = R2 = 60 mm, a = c = 300 mm,
b = 600 mm, m1 = 1 kg, and m2 = 3 kg. Find the bearing reactions at A and B and their
angular orientations measured from a rotating reference mark if the shaft speed is 100
rev/min.
ω = (100 rev/min )( 2π rad/rev ) ( 60 s/min ) = 10.472 rad/s
F1 = m1 R1ω 2 = (1 kg )( 0.060 m )(10.472 rad/s ) = 6.580 N
2
F2 = m2 R2ω 2 = ( 3 kg )( 0.060 m )( 23.038 rad/s ) = 19.739 N
F = 6.580 N∠90° = 6.580ˆj N
2
1
F2 = 19.739 N∠ − 90° = −19.739ˆj N
∑ M B = −0.300kˆ m × 6.580ˆj N + −0.900kˆ m × −19.739ˆj N + −1.200kˆ m × FA = 0
(
) (
FA = 13.159ˆj N = 13.159 N∠90.0°
FB = − ( F1 + F2 + FA ) = 0
) (
) (
) (
)
Ans.
Ans.
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2
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268
16.6
The rotating shaft shown in the figure supports two masses m1 and m2 whose weights are
1.82 kg and 2.27 kg, respectively. The dimensions are R1 = 100 mm, R2 = 75 mm,
a = 50 mm, b = 200 mm, and c = 75 mm. Find the magnitudes of the rotating-bearing
reactions at A and B and their angular orientations measured from a rotating reference
mark if the shaft speed is 360 rev/min.
ω = ( 360 rev/min )( 2π rad/rev ) ( 60 s/min ) = 37.699 rad/s
(1.82 kg )(100 mm )( 37.699 rad/s )
F1 = m1 R1ω =
2
2
( 2.27 kg )( 75 mm )( 37.699 rad/s )
F2 = m2 R2ω =
2
2
= 26.37 kg
2
9.81 m/s
2
9.81 m/s
ˆ
F1 = 26.37 kg∠90° = 26.37 j kg
F = 24.66 kg∠ − 90° = −24.66ˆj kg
= 24.66 kg
∑M
B
= ( −50 mm ) kˆ × F1 + ( −250 mm ) kˆ × F2 + ( −325 mm ) kˆ × FA
(
)
= (13.328 N − m ) ˆi + ( −62.394 N − m ) ˆi + ( −325 mm ) kˆ × FAx ˆi + FAy ˆj = 0
FA = 15.17ˆj kg = 15.17 kg∠90°
F = − ( F + F + F ) = −16.84ˆj kg = 16.84 kg∠ − 90°
1
B
16.7
2
A
Ans.
Ans.
The shaft shown in the figure is to be balanced by placing masses in the correction planes
L and R. The weights of the three masses m1 , m2 , and m3 are 113.5 g, 85.125 g, and
113.5 g, respectively. The dimensions are R1 = 125 mm, R2 = 100 mm, R3 = 125 mm,
a = 25 mm, b = e = 200 mm, c = 250 mm, and d = 225 mm. Calculate the magnitudes
of the corrections in ounce-inches and their angular orientations.
(
)
m1R1 = (113.5 g ) 125 mmˆj = 14,187.5ˆj g ⋅ mm
m2 R 2 = ( 85.125 g )(100 mm∠ − 150° ) = 8,512.5 g ⋅ mm∠ − 150° = −7,372ˆi − 4, 256.25ˆj g ⋅ mm
m R = (113.5 g )(125 mm∠ − 60° ) = 14,187.5 g ⋅ mm∠ − 60° = 7, 093.75ˆi − 12, 286.7ˆj g ⋅ mm
3
3
Using Eqs. (16.6) and (16.7),
m1R1 450 mm 875 mm = 7, 296.6ˆj g ⋅ mm
m R 675 mm 875 mm = −5, 687ˆi − 3, 283.69ˆj g ⋅ mm
2
2
m3 R 3 200 mm 875 mm = 1, 621.63ˆi − 2,808.41ˆj g ⋅ mm
m R = 4, 065.4ˆi − 1, 204.5ˆj g ⋅ mm = 4, 240.6 g ⋅ mm∠ − 16.5°
Ans.
mR R R = −3, 789.39ˆi + 3,558.48ˆj g ⋅ mm = 5,198.3 g ⋅ mm∠136.8°
Ans.
L
L
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2
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269
16.8
The shaft of Problem 16.7 is to be balanced by removing weight from the two correction
planes. Determine the corrections to be subtracted in ounce-inches and their angular
orientations.
mL R L = −4, 240.64 g ⋅ mm∠163.5°
mR R R = −5,193.3 g ⋅ mm∠ − 43.2°
16.9
Ans.
Ans.
The shaft shown in the figure is to be balanced by subtracting masses in the two
correction planes L and R. The three masses are m1 = 6 g, m2 = 7 g, and m3 = 5 g . The
dimensions are R1 = 125 mm, R2 = 150 mm, R3 = 100 mm, a = 25 mm, b = 300 mm,
c = 600 mm, d = 150 mm, and e = 75 mm. Calculate the magnitudes and angular
locations of the corrections.
(
)
m2 R 2 = ( 7 g )(150 mm∠ − 150° ) = 1 050 g ⋅ mm∠ − 150° = −909.327ˆi − 525.000ˆj g ⋅ mm
m R = ( 5 g )(100 mm∠ − 60° ) = 500 g ⋅ mm∠ − 60° = 250.000ˆi − 433.013ˆj g ⋅ mm
3
3
Using Eqs. (16.6) and (16.7),
m1R1 900 mm 1 125 mm = 600.000ˆj g ⋅ mm
m R 1 050 mm 1 125 mm = −848.705ˆi − 490.000ˆj g ⋅ mm
2
2
m3 R 3 300 mm 1 125 mm = 66.667ˆi − 115.470ˆj g ⋅ mm
The masses to be removed are:
mL R L = −782.038ˆi − 5.470ˆj g ⋅ mm = 782.057 g ⋅ mm∠ − 179.6°
m R = 122.711ˆi − 202.543ˆj g ⋅ mm = 236.816 g ⋅ mm∠ − 58.8°
R
R
Ans.
Ans.
16.10 Repeat Problem 16.9 if masses are to be added in the two correction planes.
mL R L = 782.038ˆi + 5.470ˆj g ⋅ mm = 782.057 g ⋅ mm∠0.4°
m R = −122.711ˆi + 202.543ˆj g ⋅ mm = 236.816 g ⋅ mm∠121.2°
R
R
16.11 Solve the two-plane balancing problem as stated in Section 16.8.
Since this is an experimental procedure, no “standard” solution is shown here.
Ans.
Ans.
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m1R1 = ( 6 g ) 125 mmˆj = 750.000ˆj g ⋅ mm
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270
16.12 A rotor to be balanced in the field yielded an amplitude of 5 at an angle of 142° at the
left-hand bearing and an amplitude of 3 at an angle of −22° at the right-hand bearing due
to unbalance. To correct this, a trial mass of 12 was added to the left-hand correction
plane at an angle of 210° from the rotating reference. A second run then gave left-hand
and right-hand responses of 8∠160° and 4∠260° , respectively. The first trial mass was
then removed and a second mass of 6 added to the right-hand correction plane at an angle
of −70°. The responses to this were 2∠74° and 4.5∠ − 80° for the left- and right-hand
bearings, respectively. Determine the original unbalances.
Ans.
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X A = 5∠142° , X B = 3∠ − 22° ,
m L = 12∠210° , X AL = 8∠160° , X BL = 4∠260° ,
m R = 6∠ − 70° , X AR = 2∠74° , X BR = 4.5∠ − 80° .
These gave the following results from a TI-59 programmable calculator run:
M L = 6.05∠234° , M R = 5.98∠65.2°
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271
Chapter 17
Cam Dynamics
In part (a) of the figure, the mass m is constrained to move only in the vertical direction.
The eccentric cam has an eccentricity of 50 mm, a speed of 20 rad/s, and the weight of
the mass is 3.63 kg. Neglecting friction, find the angle θ = ω t at the instant the cam
jumps.
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17.1
∑ F = −W + F = my
my + mg = F
y = y0 (1 − cos ω t )
y = ω 2 y0 cos ω t
F = mω 2 y0 cos ω t + mg
Jump begins when F = 0, that is when
mω 2 y0 cos ω t + mg = 0
9.81 m/s 2
mg
cos ωt = −
=
−
= −0.4905
2
mω 2 y0
( 20 rad/s ) ( 50 mm )
θ = ωt = cos −1 ( −0.4905 ) = 119.37°
17.2
Ans.
In part (a) of the figure, the mass m is driven up and down by the eccentric cam and it has
a weight of 4.54 kg. The cam eccentricity is 25 mm. Assume no friction.
(a)
Derive the equation for the contact force.
(b)
Find the cam velocity ω corresponding to the beginning of the cam jump.
(a)
(b)
From the solution to problem P17.1, we have for the contact force
F = mω 2 y0 cos ω t + mg
Jump begins when cos ω t = −1 and F = 0, that is when
− mω 2 y0 + mg = 0
ω=
g
9.81 m/s 2
=
= 19.809 rad/s
y0
( 25 mm )
Ans.
Ans.
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272
17.3
In part (a) of the figure, the slider has a mass of 2.5 kg. The cam is a simple eccentric
and causes the slider to rise 25 mm with no friction. At what cam speed in revolutions
per minute will the slider first lose contact with the cam? Sketch a graph of the contact
force at this speed for 360° of cam rotation.
From problem P17.1, we have for the contact force
F = mω 2 y0 cos ω t + mg
Jump begins when cos ω t = −1 and F = 0, that is when − mω 2 y0 + mg = 0
17.4
9.81 m/s 2
= 19.81 rad/s = 189.2 rev/min
( 0.025 m )
Ans.
The cam-and-follower system shown in part (b) of the figure has k = 1 kN/m ,
m = 0.90 kg, y = 15 − 15 cos ω t mm, and ω = 60 rad / s. The retaining spring is
assembled with a preload 2.5 N.
(a)
Compute the maximum and minimum values of the contact force.
(b)
If the follower is found to jump off the cam, compute the angle ω t corresponding
to the very beginning of jump.
(a)
Let Fc = contact force, and P = preload.
my + ky + P = Fc
∑ F = Fc − ky − P = my
y = 0.015 − 0.015cos 60t m ,
y = 54 cos 60t m/s 2
Fc = ( 0.90 kg ) ( 54 cos 60t m/s 2 ) + (1 000 N/m )( 0.015 − 0.015cos 60t m ) + 2.5 N
(b)
= 17.5 + 33.6 cos 60t N
Fc ,max = 17.5 + 33.6 N = 51.1 N ,
Fc ,min = 0
Jump begins when Fc = 0; that is, when
θ = 60t = cos −1 ( −17.5 N 33.6 N ) = 121.39°
Ans.
Ans.
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ω=
g
=
y0
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273
Part (b) of the figure shows the mathematical model of a cam-and-follower system. The
motion machined into the cam is to move the mass to the right through a distance of 50
mm with parabolic motion in 150° of cam rotation, dwell for 30°, return to the starting
position with simple harmonic motion, and dwell for the remaining 30° of cam angle.
There is no friction or damping. The spring rate is 7.12 kN/m, and the spring preload is
2.72 kg, corresponding to the y = 0 position. The weight of the mass is 16.34 kg.
(a)
Sketch a displacement diagram showing the follower motion for the entire 360°
of cam rotation. Without computing numerical values, superimpose graphs of the
acceleration and cam contact force onto the same axes. Show where jump is most
likely to begin.
(b)
At what speed in revolutions per minute would jump begin?
(a)
Just as in problem P17.4, if we let Fc = contact force, and P = preload
my + ky + P = Fc
∑ F = Fc − ky − P = my
Using first order kinematic coefficient, and assuming that the input shaft speed is
constant, then y = y′′ω 2 and
Fc = (16.34 kg 9.81 m/s 2 ) y′′ω 2 + ( 7.12 kN/m ) y + 2.72 kg 9.81 m/s 2
Going through the different phases of the motion defined above, we can sketch
the approximate curve shown for the cam contact force
This sketch shows that jump is very possible at point A (θ = ω t = 75° ) or point B
(θ = ω t = 180° ) or point C (θ = ω t = 330° ) , the three points where the contact
force drops discontinuously, depending on whether ω is large enough for the
contact force to indicate a negative value.
(b)
For point A (θ = ω t = 75° ) , L = 50 mm , β = 150° = 2.618 rad . From Eq. (5.10),
y = 25 mm and, from Eq. (5.12), y′′ = −29.175 mm/rad 2 . Therefore,
Fc , A = (16.34 kg 9.81 m/s 2 ) y′′ω 2 + ( 7.12 kN/m ) y + 2.72 kg 9.81 m/s 2
= ( −0.48505 N ⋅ s 2 ) ω 2 + 19.06 kg 9.81 m/s 2
Thus, Fc , A = 0 for ω ≥ 20.559 rad/s
For point B
(θ = ω t = 180° ) ,
L = 50 mm , β = 150° = 2.618 rad .
From Eq.
(5.21a), y = 50 mm and, from Eq. (5.21c), y′′ = −36 mm/rad 2 . Therefore,
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17.5
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274
Fc , B = (16.34 kg 9.81 m/s 2 ) y′′ω 2 + ( 7.12 kN/m ) y + 2.72 kg 9.81 m/s 2
= ( −0.48505 N ⋅ s 2 ) ω 2 + 39 kg 9.81 m/s 2
Thus, Fc , B = 0 for ω ≥ 25.308 rad/s
For point C (θ = ω t = 330° ) , y = y′′ = 0 , and Fc ,C = 2.72 kg for all values of ω .
Of these cases, jump begins at A when ω = 20.559 rad/s = 196.3 rev/min .
A cam-and-follower mechanism is shown in abstract form in part (b) of the figure. The
cam is cut so that it causes the mass to move to the right a distance of 25 mm with
harmonic motion in 150° of cam rotation, dwell for 30° , then return to the starting
position in the remaining 180° of cam rotation, also with harmonic motion. The spring is
assembled with a 22-N preload and it has a rate of 4.4 kN/m. The follower mass is 17.5
kg. Compute the cam speed in revolutions per minute at which jump would begin.
Just as in problem P17.4, if we let Fc = contact force, and P = preload
my + ky + P = Fc
∑ F = Fc − ky − P = my
Using first order kinematic coefficient, and assuming that the input shaft speed is
constant, then y = y′′ω 2 and
Fc = (17.5 kg ) y′′ω 2 + ( 4 400 N/m ) y + 22 N
Going through the different phases of the motion defined above, shows that jump is most
likely at the transition from the dwell to the full return simple-harmonic motion since, at
that position, y′′ and Fc suddenly drop. For that position (θ = ω t = 180° ) , L = 0.025 m ,
β = 180° = 3.1416 rad .
From Eqs. (5.21), y = 0.025 m and y′′ = −0.036 m/rad 2 .
Therefore,
Fc = (17.5 kg ) y′′ω 2 + ( 4400 N/m ) y + 22 N
= ( −0.630 N ⋅ s 2 ) ω 2 + 132 N
Thus, Fc = 0 for ω ≥ 14.475 rad/s = 138.2 rev/min
17.7
Ans.
The figure shows a lever OAB driven by a cam cut to give the roller a rise of 1 in with
parabolic motion and a parabolic return with no dwells. The lever and roller are to be
assumed as weightless, and there is no friction. Calculate the jump speed if l = 125 mm
in and the mass weighs 2.27 kg.
Taking moments about the fixed pivot
2
∑ M O = AFA − 2Amg = m ( 2A ) θ
F = 4mAθ + 2mg
A
FA = (1.152 N ⋅ s 2 ) θ + 4.54 kg
Going through the different phases of the motion defined above, shows that jump is most
likely at the transition from the concave to the convex parabolic rise motion since, at that
position, y′′ and Fc suddenly drop. For that position (θ = ω t = 90° ) , L = 25 mm ,
β = 180° = 3.1416 rad .
From
Eqs.
(5.10)
and
(5.12),
y = 12.5 mm
and
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17.6
Ans.
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275
y′′ = −10.125 mm/rad 2 . Therefore,
θ = y′′ω 2 A = ( −10.125 mm/rad 2 125 mm ) ω 2 = −0.081ω 2
FA = (1.152 N ⋅ s 2 ) ( −0.081) ω 2 + 4.54 kg 9.81 m/s 2
= ( −0.09345 N ⋅ s 2 ) ω 2 + 4.54 kg 9.81 m/s 2
17.8
Ans.
A cam-and-follower system similar to the one of Fig. 17-6 uses a plate cam driven at a
speed of 600 rev/min and employs simple harmonic rise and parabolic return motions.
The events are rise in 150° , dwell for 30° , and return in 180° . The retaining spring has
a rate k = 14 kN/m with a precompression of 12.5 mm. The follower has a mass of 1.6
kg.
The external load is related to the follower motion y by the relation
F = 0.325 − 10.75 y, where y is in meters and F is in kilonewtons. Dimensions
corresponding to Fig. 17.6 are R = 20 mm, r = 5 mm, lB = 60 mm, and lC = 90 mm.
Using a rise of L = 20 mm and assuming no friction, plot the displacement, cam-shaft
torque, and radial component of the cam force for one complete revolution of the cam.
ω = 600 rev/min = 62.832 rad/s
For simple harmonic rise motion, we use Eqs. (5.18) with L = 0.020 m and β = 150° .
For the first part of the parabolic return motion, following Example 5.1,
2
y = 0.020 ⎡1 − 2 (θ π ) ⎤ m , y′ = − ( 0.080 π )θ π m , y′′ = − 0.080 π 2 = −0.008 106 m
⎣
⎦
For the second part of the parabolic return motion,
2
y = 0.040 (1 − θ π ) m , y′ = − ( 0.080 π )(1 − θ π ) m , y′′ = 0.080 π 2 = 0.008 106 m
Then we can use Eq. (17.11)
F23y = 325 − 10 750 y + 14 000 ( y + 0.0125 ) + 1.6 ( y′′ω 2 ) N
= 500 + 3 250 y + 6 317 y′′ N
and Eqs. (17.9) and (17.13)
a tan φ = y ω = y′
T12 = −a tan φ F23y = − y′F23y
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Thus, FA = 0 for ω ≥ 21.8 rad/s = 208.4 rev/min
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276
θ = ω t , deg
y, m
y′ , m/s
y′′ , m/s2
F23y , N
T12 , N·m
0
0
0
0.000 489
0.001 910
0.004 122
0.006 910
0.010 000
0.013 090
0.015 878
0.018 090
0.019 511
0.020 000
0.003 708
0.007 053
0.009 708
0.011 413
0.012 000
0.011 413
0.009 708
0.007 053
0.003 708
0
165
180
0.020 000
0.020 000
0
0
195
210
225
240
255
270
0.019 722
0.018 889
0.017 500
0.015 556
0.013 056
0.010 000
-0.002 122
-0.004 244
-0.006 366
-0.008 488
-0.010 610
-0.012 732
285
300
315
330
345
360
0.006 944
0.004 444
0.002 500
0.001 111
0.000 278
0
-0.010 610
-0.008 488
-0.006 366
-0.004 244
-0.002 122
0
551.2
591.0
588.1
579.8
566.9
550.6
532.5
514.4
498.1
485.2
476.9
474.0
565.0
565.0
565.0
513.8
512.9
510.2
505.7
499.4
491.2
481.3
583.7
573.8
565.6
559.3
554.8
552.1
551.2
591.0
0
15
30
45
60
75
90
105
120
135
150
0.008 106
0.014 400
0.013 695
0.011 650
0.008 464
0.004 450
0
-0.004 450
-0.008 464
-0.011 650
-0.013 695
-0.014 400
0
0
0
-0.008 106
-0.008 106
-0.008 106
-0.008 106
-0.008 106
-0.008 106
-0.008 106
0.008 106
0.008 106
0.008 106
0.008 106
0.008 106
0.008 106
0.008 106
0.014 400
0
0
1.088
2.165
3.219
4.239
5.212
6.128
7.430
6.088
4.801
3.561
2.355
1.172
0
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-2.181
-4.089
-5.503
-6.284
-6.390
-5.871
-4.836
-3.422
-1.768
0
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17.9
Repeat Problem 17.8 with the speed of 900 rev/min, F = 0.110 + 10.75 y kN, where y is
in meters, and the coefficient of sliding friction is μ = 0.025.
ω = 900 rev/min = 94.248 rad/s
For simple harmonic rise motion, we use Eqs. (5.18) with L = 0.020 m and β = 150° .
For the first part of the parabolic return motion, following Example 5.1,
2
y = 0.020 ⎡1 − 2 (θ π ) ⎤ m , y′ = − ( 0.080 π )θ π m , y′′ = − 0.080 π 2 = −0.008 106 m
⎣
⎦
For the second part of the parabolic return motion,
2
y = 0.040 (1 − θ π ) m , y′ = − ( 0.080 π )(1 − θ π ) m , y′′ = 0.080 π 2 = 0.008 106 m
Then we can use Eq. (17.11)
110 + 10 750 y + 14 000 ( y + 0.0125 ) + 1.6 ( y′′ω 2 ) N
y
F23 =
1 + (1.666 667 y − 0.033 333) tan φ sgn y′
285 + 24 750 y + 14 212 y′′ N
1 + (1.666 667 y − 0.033 333) tan φ sgn y′
and Eqs. (17.9) and (17.13)
a tan φ = y ω = y′
T12 = −a tan φ F23y = − y′F23y
=
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277
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278
θ = ω t , deg
y, m
y′ , m/s
y′′ , m/s2
F23y , N
T12 , N·m
0
0
0
0.000 489
0.001 910
0.004 122
0.006 910
0.010 000
0.013 090
0.015 878
0.018 090
0.019 511
0.020 000
0.003 708
0.007 053
0.009 708
0.011 413
0.012 000
0.011 413
0.009 708
0.007 053
0.003 708
0
165
180
0.020 000
0.020 000
0
0
195
210
225
240
255
270
0.019 722
0.018 889
0.017 500
0.015 556
0.013 056
0.010 000
-0.002 122
-0.004 244
-0.006 366
-0.008 488
-0.010 610
-0.012 732
285
300
315
330
345
360
0.006 944
0.004 444
0.002 500
0.001 111
0.000 278
0
-0.010 610
-0.008 488
-0.006 366
-0.004 244
-0.002 122
0
400
490
498
509
521
533
545
556
565
572
576
575
780
780
780
665
660
641
608
562
529
428
664
586
522
471
433
410
400
490
0
15
30
45
60
75
90
105
120
135
150
0.008 106
0.014 400
0.013 695
0.011 650
0.008 464
0.004 450
0
-0.004 450
-0.008 464
-0.011 650
-0.013 695
-0.014 400
0
0
0
-0.008 106
-0.008 106
-0.008 106
-0.008 106
-0.008 106
-0.008 106
-0.008 106
0.008 106
0.008 106
0.008 106
0.008 106
0.008 106
0.008 106
0.008 106
0.014 400
0
0
1.40
2.72
3.87
4.77
5.61
5.45
8.45
6.22
4.43
3.00
1.84
0.87
0
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-1.85
-3.59
-5.05
-6.08
-6.54
-6.35
-5.49
-4.04
-2.13
0
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17.10 A plate cam drives a reciprocating roller follower through the distance L = 31.25 mm
with parabolic motion in 120D of cam rotation, dwells for 30° , and returns with cycloidal
motion in 120D , then dwells for the remaining cam angle. The external load on the
follower is F14 = 160 N during rise and zero during the dwells and the return.
In the notation of Fig. 17-6, R = 75 mm, r = 25 mm, lB = 150 mm,
lC = 200 mm, and k = 26.7 kN/m. The spring is assembled with a preload of 17 kg when
the follower is at the bottom of its stroke. The weight of the follower is 0.82 kg, and the
cam velocity is 140 rad/s. Assuming no friction, plot the displacement, the torque
exerted on the cam by the shaft, and the radial component of the contact force exerted by
the roller against the cam surface for one complete cycle of motion.
For 0 ≤ θ ≤ 60° , we use Eqs. (5.5) – (5.7) with L = 31.25 mm and β = 120° .
y = 62.5 (θ β ) mm , y′ = 59.675 (θ β ) mm , y′′ = 28.5 mm
2
For 60° ≤ θ ≤ 120° , we use Eqs. (5.10) – (5.12) with L = 31.25 mm and β = 120° .
2
y = 62.5 ⎡1 − 2 (1 − θ β ) ⎤ mm , y′ = 59.675 (1 − θ β ) mm , y′′ = −28.5 mm
⎣
⎦
For 150° ≤ θ ≤ 270° , we use Eqs. (5.22) with L = 31.25 mm and β = 120° .
Then we can use Eq. (17.11)
F23y = F14 + 17 9.81 + ( 26.7 kN/m × 1000 ) y + 41.48 y′′ N
and Eqs. (17.9) and (17.13)
a tan φ = y ω = y′
T12 = −a tan φ F23y = − y′F23y
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279
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280
θ = ω t , deg
y, m
y′ , m/s
y′′ , m/s2
F23y , N
T12 , N·m
0
0
0
0.039 063
0.156 250
0.351 563
0.625 000
0.298 416
0.596 831
0.895 247
1.193 662
75
90
105
120
0.898 438
1.093 750
1.210 938
1.250 000
0.895 247
0.596 831
0.298 416
0
135
150
165
180
195
210
225
240
255
270
285
300
315
330
345
360
1.250 000
1.250 000
1.234 424
1.136 444
0.921 924
0.625 000
0.328 076
0.113 556
0.015 576
0
0
0
0
0
0
0
0
0
-0.174 808
-0.596 831
-1.018 854
-1.193 662
-1.018 854
-0.596 831
-0.174 808
0
0
0
0
0
0
0
37.5
177.7
183.5
201.1
230.4
271.4
63.1
104.1
133.4
151.0
156.8
225.0
225.0
225.0
107.0
44.4
60.1
131.3
202.4
218.1
155.5
37.5
37.5
37.5
37.5
37.5
37.5
37.5
177.7
0
15
30
45
60
0
1.139 863
1.139 863
1.139 863
1.139 863
1.139 863
-1.139 863
-1.139 863
-1.139 863
-1.139 863
-1.139 863
0
0
0
-1.266 070
-1.790 493
-1.266 070
0
1.266 070
1.790 493
1.266 070
0
0
0
0
0
0
0
1.139 863
0
0
-18.7
-26.5
-61.2
-156.7
-206.2
-130.2
-27.2
0
0
0
0
0
0
0
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54.8
120.0
206.3
324.0
75.3
93.2
79.6
45.1
0
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17.11 Repeat Problem 17-10 if friction exists with μ = 0.04 and the cycloidal return takes
place in 180°.
For 0 ≤ θ ≤ 60° , we use Eqs. (5.5) – (5.7) with L = 31.25 mm and β = 120° .
y = 62.5 (θ β ) mm , y′ = 59.675 (θ β ) mm , y′′ = 28.5 mm
2
For 60° ≤ θ ≤ 120° , we use Eqs. (5.10) – (5.12) with L = 31.25 mm and β = 120° .
2
y = 31.25 ⎡1 − 2 (1 − θ β ) ⎤ mm , y′ = 59.675 (1 − θ β ) mm , y′′ = −28.5 mm
⎣
⎦
For 150° ≤ θ ≤ 330° , we use Eqs. (5.22) with L = 31.25 mm and β = 180° .
Then we can use Eq. (17.11)
F + 17 kg 9.81 m/s 2 + 26700 N/my + 41.48 Ny′′
F23y = 14
1 + ( 5.6 y − 16.8 ) tan φ sgn y′
and Eqs. (17.9) and (17.13)
a tan φ = y ω = y′
T12 = −a tan φ F23y = − y′F23y
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281
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282
θ = ω t , deg
y, m
y′ , m/s
y′′ , m/s2
F23y , N
T12 , N·m
0
0
0
0.039 063
0.156 250
0.351 563
0.625 000
0.298 416
0.596 831
0.895 247
1.193 662
75
90
105
120
0.898 438
1.093 750
1.210 938
1.250 000
0.895 247
0.596 831
0.298 416
0
135
150
165
180
195
210
225
240
255
270
285
300
315
330
345
360
1.250 000
1.250 000
1.245 305
1.213 957
1.136 444
1.005 624
0.828 639
0.625 000
0.421 361
0.244 376
0.113 556
0.036 043
0.004 695
0
0
0
0
0
-0.053 307
-0.198 944
-0.397 887
-0.596 831
-0.742 468
-0.795 775
-0.742 468
-0.596 831
-0.397 887
-0.198 944
-0.053 307
0
0
0
38
178
185
204
235
278
65
135
152
152
157
225
225
188
188
157
136
127
127
133
139
139
129
106
75
38
38
38
178
0
15
30
45
60
0
1.139 863
1.139 863
1.139 863
1.139 863
1.139 863
-1.139 863
-1.139 863
-1.139 863
-1.139 863
-1.139 863
0
0
0
-0.397 887
-0.689 161
-0.795 775
-0.689 161
-0.397 887
0
0.397 887
0.689 161
0.795 775
0.689 161
0.397 887
0
0
0
1.139 863
0
0
-10
-31
-54
-76
-94
-106
-104
-83
-51
-21
-4
0
0
0
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55
122
211
332
77
95
80
45
0
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283
Chapter 18
Flywheels
Table P18.1 lists the output torque for a one cylinder engine running at 4 600 rev/min.
(a)
Find the mean output torque.
(b)
Determine the mass moment of inertia of an appropriate flywheel using
Cs = 0.025 .
Torque data for Problem 18.1.
θi
deg
0
10
20
30
40
50
60
70
80
90
100
110
120
130
140
150
160
170
(a)
Ti
N⋅m
0
17
812
963
1 016
937
774
641
697
849
1 031
1 027
902
712
607
594
544
345
θi
deg
180
190
200
210
220
230
240
250
260
270
280
290
300
310
320
330
340
350
Ti
N⋅m
0
-344
-540
-576
-570
-638
-785
-879
-814
-571
-324
-190
-203
-235
-164
-7
150
145
θi
deg
360
370
380
390
400
410
420
430
440
450
460
470
480
490
500
510
520
530
Ti
N⋅m
0
-145
-150
7
164
235
203
490
424
571
814
879
785
638
570
576
540
344
θi
deg
540
550
560
570
580
590
600
610
620
630
640
650
660
670
680
690
700
710
Ti
N⋅m
0
-344
-540
-577
-572
-643
-793
-893
-836
-605
-379
-264
-300
-368
-334
-198
-56
-2
Using n = 72 and h = 4π/72, we enter the data from Table P18.1 into Simpson’s
rule to find U 2 − U1 = 890.7 N ⋅ m .
Tm = (U 2 − U1 ) ( 4π ) = ( 890.7 N ⋅ m ) ( 4π rad ) = 70.88 N ⋅ m
(b)
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18.1
Ans.
ω = 4 600 rev/min = 481.7 rad/s
2
I = (U 2 − U1 ) ( Csω 2 ) = ( 890.7 N ⋅ m ) ⎡⎣( 0.025 )( 481.7 rad/s ) ⎤⎦ = 0.154 N ⋅ m ⋅ s 2 Ans.
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284
18.2
Using the data of Table 18.2, determine the moment of inertia for a flywheel for a twocylinder 90° V engine having a single crank. Use Cs = 0.012 5 and a nominal speed of
4 600 rev/min. If a cylindrical or disk-type flywheel is to be used, what should be the
thickness if it is made of steel and has an outside diameter of 400 mm? Use ρ = 7.8
Mg/m3 as the density of steel.
R = 200 mm
ρ = 7800 kg/m3
2
m = 2 I R 2 = 2 × 0.136 N − m ⋅ s 2 ( 200 mm ) = 6.8 kg
V = m / ρ = 6.8 kg 7800 kg/m3 = 872 mm3
2
t = V A = 872 mm3 ⎡π ⋅ (196.85 mm ) ⎤ = 6.85 mm
⎣
⎦
18.3
Ans.
Using the data of Table 18.1, find the mean output torque and the flywheel inertia
required for a three-cylinder in-line engine corresponding to a nominal speed of 2 400
rev/min. Use Cs = 0.03.
Using n = 48 and h = 4π/48, we integrate the data from Table 18.1 by Simpson’s rule to
find U 2 − U1 = 394.38 N − m .
Tm = (U 2 − U1 ) ( 4π ) = ( 394.38 N − m ) ( 4π rad ) = 31.38 N ⋅ m
Ans.
ω = 2 400 rev/min = 251.3 rad/s
2
I = (U 2 − U1 ) ( Csω 2 ) = ( 394.38 N − m ) ⎡⎣( 0.03)( 251.3 rad/s ) ⎤⎦ = 0.208 N − m ⋅ s 2
Ans.
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Using n = 48 and h = 4π/48, we integrate the data from columns 2-5 of Table 18.2 by
Simpson’s rule to find U 2 − U1 = 394.38 N − m .
ω = 4 600 rev/min = 481.7 rad/s
2
I = (U 2 − U1 ) ( Csω 2 ) = ( 394.38 N − m ) ⎡⎣( 0.0125 )( 481.7 rad/s ) ⎤⎦ = 0.136 N − m ⋅ s 2
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285
The load torque required by a 200-tonne punch press is displayed in Table P18.4 for one
revolution of the flywheel. The flywheel is to have a nominal angular velocity of 2 400
rev/min and to be designed for a coefficient of speed fluctuation of 0.075.
Determine the mean motor torque required at the flywheel shaft and the motor
(a)
horsepower needed, assuming a constant torque-speed characteristic for the
motor.
Find the moment of inertia needed for the flywheel.
(b)
θi
deg
0
10
20
30
40
50
60
70
80
(a)
(b)
18.5
Ti
N−m
96.84
96.84
96.84
96.84
96.84
145.4
290.6
581.3
775
Torque data for Problem 18-4.
θi
θi
Ti
Ti
deg
deg
N−m
N−m
90
891
180
203.5
100
940
190
184
110
959
200
165
120
969
210
155
130
949.5
220
126
140
872
230
116.3
150
397
240
106.5
160
242
250
96.84
170
222.8
260
96.84
θi
deg
270
280
290
300
310
320
330
340
350
Ti
N−m
96.84
96.84
96.84
96.84
96.84
96.84
96.84
96.84
96.84
Using n = 36 and h = 2π/36, we integrate the data from Table P18.4 by Simpson’s
rule to find U = 1887.1 N − m .
Ans.
Tm = U ( 2π ) = (1887.1 N − m ) ( 2π rad ) = 300.49 N − m
ω = 2 400 rev/min = 251.3 rad/s
( 300.49 N − m )( 2 400 rev/min )( 2π rad/rev ) = 6071 HP
P = Tω =
Ans.
746 watt/HP
The torque data shows a constant requirement of 96.84 N–m, probably friction, in
addition to the torque for the punching operation. If this constant torque is
subtracted from the data in the table (for speed fluctuation), and the integration
repeated, then we get, U = 1684.26 N − m
2
I = U ( Csω 2 ) = (1684.26 N − m ) ⎡( 0.075 )( 251.3 rad/s ) ⎤ = 0.355 N − m ⋅ s 2 Ans.
⎣
⎦
Find Tm for the four-cylinder engine whose torque displacement is that of Fig. 18.4.
Using n = 12 and h = π/12, we integrate the data from column 6 of Table 18.2 by
Simpson’s rule to find U 2 − U1 = 394.38 N − m .
Tm = (U 2 − U1 ) (π ) = ( 394.38 N − m ) (π rad ) = 125.598 N − m
Ans.
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18.4
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287
Chapter 20
20.1
A pendulum mill is shown schematically in the figure. In such a mill, grinding is done by
a conical muller that is free to spin about a pendulous axle that, in turn, is connected to a
powered vertical shaft by a Hooke universal joint. The muller presses against the inner
wall of a heavy steel pan, and it rolls around the inside of the pan without slipping. The
weight of the muller is W = 445 kg; its principal mass moments of inertia are
I s = 1.37 kg − m ⋅ s 2 and I = 0.99 kg − m ⋅ s 2 . The length of the muller axle is
l = RGA = 40 in and the radius of the muller at its center of mass is RGB = 0.25 m .
Assuming that the vertical shaft is to be inclined at θ = 30° and will be driven at a
constant angular velocity of ω p = 240 rev/min , find the crushing force between the
muller and the pan. Also determine the minimum angular velocity ωp required to ensure
contact between the muller and the pan.
(
)
ω p = 240 rev/min = 25.133 rad/s
ω s = ω 4 / 3 = ω s sin θ ˆi + cos θ ˆj
ω3 = ω p ˆj = 25.133 rad/sˆj
ω 4 = ω3 + ω 4 / 3 = ω s sin θ ˆi + (ω p + ω s cosθ ) ˆj
VG = ω3 × R GA
VG = ω 4 × R GB
(
= ω p ˆj × 40sin θ ˆi + 40 cosθ ˆj
)
= ⎡⎣ω s sin θ ˆi + (ω p + ω s cos θ ) ˆj⎤⎦ × ( −10 cos θ ˆi + 10 sin θ ˆj)
= (10ω s + 10 cos θω p ) kˆ
= −40sin θω p kˆ
Equating these with θ = 30° we find ω s = − ( 4sin θ + cosθ ) ω p = −2.866ω p . Then
(ω
p
× ω s ) = ω p ˆj × ω s ( sin θ ˆi + cos θ ˆj ) = − ω p ω s sin θ kˆ = ( 4 sin 2 θ + sin θ cos θ ) ω 2p kˆ = 1.433ω p2 kˆ
I S = mr 2 2 = ( 445 kg )( 0.25 m ) 2 = 13.91 kg − m ⋅ s 2
2
2
2
I = mr 2 4 + ml 2 = ( 445 kg ) ⎡( 0.25 m ) 4 + (1 m ) ⎤ = 451.95 kg − m ⋅ s 2
⎣
⎦
Now Eq. (20.6) shows
ωp
⎡ s
⎤
s
cos θ ⎥ ω p × ω s
∑ M = ⎢I + I − I
ωs
⎣⎢
⎦⎥
(
)
(
)
⎡
⎤
ω p cos θ
⎥ 4 sin 2 θ + sin θ cos θ kg − m ⋅ s 2 ω 2 kˆ
∑ M = ⎢13.91 + 438.04
p
− ( 4sin θ + cos θ ) ω p ⎥
⎢
⎣
⎦
(
)
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Gyroscopes
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288
2 2
∑ M = ⎡⎣13.91sin θ ( 4 sin θ + cos θ ) − 438.04sin θ cos θ ⎤⎦ kg − m ⋅ s ω p kˆ
2 2
∑ M = −17.68 kg − m ⋅ s ω p kˆ = −1116.79 kg − m kˆ
Now formulating the externally applied moments,
∑ M =Wˆj × R GA − Fc ˆi × R BA
∑ M =445 kgˆj × ( sin θ ˆi + cos θ ˆj)1 m − F ˆi × ⎡⎣( sin θ ˆi + cosθ ˆj)1 m + ( cos θ ˆi − sin θ ˆj) 0.25 m ⎤⎦
∑ M = − 444.92sin θ kg − m kˆ − F ⎡⎣( 4 cos θ − sin θ ) 0.25 m kˆ ⎤⎦
∑ M = − 222.46 kg − m kˆ − 0.74F m kˆ
c
c
c
and setting the two expressions equal
∑ M = − 222.46 kg − m kˆ − 0.74Fc m kˆ = −1116.79 kg − m kˆ
we can now solve for the crushing force
Fc = 1506.82 kg
If we start before setting the angular velocity, then we have
∑ M = −17.68 kg − m ⋅ s2ω 2pkˆ = − 222.46 kg − m kˆ − 0.74 Fc m kˆ
Now by setting Fc to zero we can determine the minimum angular velocity ωp required to
ensure contact between the muller and the pan.
ω p = 222.46 kg − m 17.68 kg − m ⋅ s 2 = 3.547 rad/s = 33.87 rev/min
20.2
Ans.
Use the gyroscopic formulae of this chapter to solve again the problem presented in
Example 16.3 of Chapter 16.
From the given data we can identify
ω p = ω 2 = 5kˆ rad/s
ω s = ω3 = 350ˆi + 5kˆ rad/s
I s = mk 2 = ( 4.5 kg )( 0.050 m ) = 0.011 3 kg ⋅ m 2
2
From Eq. (20.7)
∑ M = I (ω
(
)
× ω s ) = 0.011 3 kg ⋅ m 2 ⎡5kˆ rad/s × 350ˆi + 5kˆ rad/s ⎤ = 19.8ˆj N ⋅ m
⎣
⎦
This brings us precisely to the formulation of Eqs. (2) of Example 16.3. From there on
the solution procedure, and the results, are identical. Notice how much more simply this
approach can be accomplished.
Q.E.D.
s
p
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Ans.
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289
20.3
The oscillating fan shown in Fig. P20.3 precesses sinusoidally according to the equation
θ p = β sin1.5t , where β = 30° ; the fan blade spins at ω s = 1800ˆi rev/min . The weight of
the fan and motor armature is 2.38 kg, and other masses can be assumed negligible;
gravity acts in the −jˆ direction. The principal mass moments of inertia are
I s = 7.38 × 10−4 kg − m ⋅ s 2 and I = 2.84 × 10−4 kg − m ⋅ s 2 ; the center of mass is located
at RGC = 4 in to the front of the precession axis. Determine the maximum moment Mz
that must be accounted for in the clamped tilting pivot at C.
(
)
ω s = 1800ˆi rev/min = 188.5ˆi rad/s
ω p = 1.5 sinβ ˆi + cos β ˆj rad/s
(ω
× ω s ) = ⎡⎣1.5 ( sinβ ˆi + cos β ˆj) rad/s ⎤⎦ × (188.5ˆi rad/s ) = −282.7 cos β kˆ rad/s 2 = −244.9kˆ rad/s 2
ωp
⎡ s
⎤
s
cos θ ⎥ ω p × ω s
∑ M = ⎢I + I − I
ωs
⎣⎢
⎦⎥
p
(
(
)
)
(
)
(
)
On the other side of the equation, the external moments are
z
z
z
∑ M = M kˆ + R GC × W − sin β ˆi − cos β ˆj = M kˆ − 0.1 m W cos β kˆ = M kˆ − 0.21 kg − m kˆ
(
)
Now, equating the two we can solve for the moment Mz.
M z kˆ − 0.21 kg − m kˆ = −0.18kˆ kg − m
M z = 0.030kˆ kg − m
Ans.
Here we see that the gyroscopic moment is almost large enough to support the weight of
the fan motor.
20.4
The propeller of an outboard motorboat is spinning at high speed and is caused to precess
by steering to the right or left. Do the gyroscopic effects tend to raise or lower the rear of
the boat? What is the effect and is it of noticeable size?
In this case ω s refers to the angular velocity of the propeller and is directed fore or aft
depending on the direction of rotation. ω p is vertical and refers to the angular velocity
of the turn. The moment required to maintain the turn is proportional to ( ω p × ω s ) as
shown in Eq. (20.6) or (20.7) and this axis is lateral on the boat. Therefore the moment
(or its reaction) can tend to raise or lower the rear of the boat. The direction depends on
both the the direction of rotation of the engine and the direction of the turn. Eq. (20.7)
shows that the effect is likely to be very small since, for any reasonable rate of turn,
( ω p × ω s ) will be at least an order of magnitude smaller than ω s2 which is the order of
the usual accelerations of the engine. In a very extreme case, a knowledgeable person
might be able to detect this moment, but most would not. It would never be a danger.
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1.5 rad/s
⎡
⎤
−4
2
−4
2
cos β ⎥ −244.9kˆ rad/s 2 = −0.18kˆ kg − m
∑ M = ⎢ 7.38 × 10 kg − m ⋅ s + 4.54 × 10 kg − m ⋅ s
188.5 rad/s
⎣
⎦
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290
A large and very high-speed turbine is to operate at an angular velocity of
ω = 18 000 rev/min and will have a rotor with a principal mass moment of inertia of
I s = 2.55 kg − m ⋅ s 2 . It has been suggested that because this turbine will be installed at
the north pole with its axis horizontal, perhaps the rotation of the earth will cause
significant gyroscopic loads on it bearings. Estimate the size of these additional loads.
ω s = 18 000ˆi rev/min = 1885ˆi rad/s
ω = 1.0ˆj rev/day = 0.0000115ˆj rad/s
p
( ω × ω ) = ( 0.0000115ˆj rad/s ) × (1885ˆi rad/s ) = −0.022kˆ rad/s
I ( ω × ω ) = ( 2.55 kg − m ⋅ s ) ( −0.022kˆ rad/s ) = −0.056kˆ kg − m
2
p
s
2
s
p
s
2
Ans.
Thus, if the bearings were only separated by 0.125 m, they would experience less than
half a kg additional loading. This is totally negligible.
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20.5