Theory of Machines and Mechanisms, 4e
Uicker et al.
Solutions Manual to accompany
THEORY OF MACHINES
AND MECHANISMS
Fourth Edition
International Version
John J. Uicker, Jr.
Professor Emeritus of Mechanical Engineering
University of Wisconsin – Madison
Gordon R. Pennock
Associate Professor of Mechanical Engineering
Purdue University
Joseph E. Shigley
Late Professor Emeritus of Mechanical Engineering
The University of Michigan
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
PART 1
KINEMATICS AND MECHANISMS
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Theory of Machines and Mechanisms, 4e
Uicker et al.
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.
Since the variety is unbounded no standard solutions are shown here.
1.2
The link lengths of a planar four-bar linkage are 0.2, 0.4, 0.6 and 0.6 m. 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  0.2, l  0.6, p  0.4, q  0.6 ;
since 0.2  0.6  0.4  0.6 .
1.3
these
linkages
all
satisfy
Grashof’s
law
Ans.
Drag-link mechanism
Drag-link mechanism
Ans.
Crank-rocker mechanism
Crank-rocker mechanism
Ans.
Crank-rocker mechanism
Ans.
Double-rocker mechanism
A crank-rocker linkage has a 250 mm frame, a 62.5 mm crank, a 225 mm coupler, and a
187.5 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.
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Theory of Machines and Mechanisms, 4e
Extremum transmission angles:  min   1  53.1 max   3  98.1
Toggle positions: 2  40.1 2  59.14  228.6 4  90.9
1.4
In Fig. P1.4, point C is attached to the coupler; plot its complete path.
1.5
Find the mobility of each mechanism illustrated in Fig. P1.5.
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Uicker et al.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
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.
Notice that each coaxial pair of sliding ground joints is counted as only a single
prismatic pair.
1.6
Use the Kutzbach criterion to determine the mobility of the mechanism illustrated in Fig.
P1.6.
n  5, j1  5, j2  1; m  3  5  1  2  5  11  1
Ans.
Notice that the double pin is counted as two single j1 pins.
1.7
Find a planar mechanism with a mobility of one that contains a moving quaternary link.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
How many distinct variations of this mechanism can you find?
To have at least one quaternary 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
quaternary link. These seven are shown below:
Ans.
1.8
Use the Kutzbach criterion to detemine the mobility of the planar mechanism illustrated
in Fig. P1.8. Clearly number each link and label the lower pairs (j1) and higher pairs (j2)
on Fig. P1.8.
n  5, j1  5, j2  1; m  3  5  1  2  5  11  1
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Ans.
Theory of Machines and Mechanisms, 4e
1.9
Uicker et al.
For the mechanism illustrated in Fig. P1.9, determine the number of links, the number of
lower pairs, and the number of higher pairs. Using the Kutzbach criterion determine the
mobility. Is the answer correct? Briefly explain.
n  4, j1  3, j2  2; m  3  4  1  2  3  1 2   1
Ans.
If it is not evident that the input shown will increment this device in the direction shown,
then consider incrementing link 3 downward. Since it seems intuitive that this
determines the position of all other links, this verifies that mobility of one is correct.
1.10
Use the Kutzbach criterion to detemine the mobility of the planar mechanism illustrated
in Fig. P1.10. Clearly number each link and label the lower pairs (j1) and higher pairs (j2)
on Fig. P1.10. Treat rolling contact to mean rolling with no slipping.
n  5, j1  5, j2  1; m  3  5  1  2  5  11  1
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Ans.
Theory of Machines and Mechanisms, 4e
1.11
Uicker et al.
For the mechanism illustrated in Fig. P1.11 treat rolling contact to mean rolling with no
slipping. Determine the number of links, the number of lower pairs, and the number of
higher pairs. Using the Kutzbach criterion determine the mobility. Is the answer correct?
Briefly explain.
n  7, j1  8, j2  1; m  3  7  1  2 8  11  1
Ans.
This result appears to be correct. If all parts remain assembled and within the limits of
travel of the joints shown, then it appears that when any one member is locked the total
system becomes a structure.
1.12
Does the Kutzbach criterion provide the correct result for the planar mechanism
illustrated in Fig. P1.12? Briefly explain why or why not.
n  4, j1  2, j2  3; m  3  4  1  2  2   1 3  2
Ans.
This result appears to be correct. If any part except the wheel is moved, other parts are
required to follow. However, after all other parts are in a set position, the wheel is still
able to rotate because of slipping against the frame at A.
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Theory of Machines and Mechanisms, 4e
1.13
Uicker et al.
The mobility of the mechanism illustrated in Fig. P1.13 is m = 1. Use the Kutzbach
criterion to determine the number of lower pairs and the number of higher pairs. Is the
wheel rolling without slipping, or rolling and slipping, at point A on the wall?
Suppose that we identify the number of constraints at A by the symbol k. Then if we
account for all links and all other joints as follows, the Kutzbach criterion gives
n  5; j1  4; j2  1; jk  1; m  3  5  1  2  4   11  k 1  3  k;
Therefore, to have mobility of m  1 , we must have k  2 constraints at A. The wheel
must be rolling without slipping.
Ans.
1.14
Devise a practical working model of the drag-link mechanism.
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Theory of Machines and Mechanisms, 4e
1.15
Uicker et al.
Find the time ratio of the linkage of Problem 1.3.
From the values of  2 and  4 we find   188.5 and   171.5 .
Then, from Eq. (1.5), Q     1.099 .
Ans.
1.16
Plot the complete coupler curve of the Roberts' mechanism illustrated in Fig. 1.24b. Use
AB = CD = AD = 62.5 mm and BC = 31.25 mm.
1.17
If the crank of Fig. 1.11 is turned 25 revolutions counterclockwise, how far and in what
direction does the carriage move?
Screw and carriage move by (25 rev)/(6 rev/mm) = 4.17 mm to the right.
Carriage moves (7 rev)/(18 rev/mm) = 3.57 mm to the left with respect to the screw.
Net motion of carriage = 25/6 – 25/7 = 25/42 = 0.59 mm to the right.
Ans.
More in-depth study of such devices is covered in Chapter 9.
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Theory of Machines and Mechanisms, 4e
1.18
Uicker et al.
Show how the mechanism of Fig. 1.15b can be used to generate a sine wave.
With the length and angle of crank 2 designated as R and 2, respectively, the horizontal
motion of link 4 is x4  R cos 2  R sin  2  90  .
1.19
Devise a crank-and-rocker linkage, as in Fig. 1.14c, having a rocker angle of 60. The
rocker length is to be 0.50 m.
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Theory of Machines and Mechanisms, 4e
1.20
Uicker et al.
A crank-rocker four-bar linkage is required to have a time ratio Q = 1.2. The rocker is to
have a length of 62.5 mm and oscillate through a total angle of 60. Determine a
suitable set of link lengths for the remaining three links of the four-bar linkage.
Following the procedure of Example 1.4, the required time ratio gives
180  
Q
 1.2 and, therefore, we must have   16.36 .
180  
Then, with the X-line chosen at 30°, the drawing shown below (dimensioned 10 times
size) gives measured distances of RO4O2  r1  108.5 mm , RB2O2  r3  r2  160.5 mm , and
RB1O2  r3  r2  111 mm . From these we get one possible solution, which has link lengths
r1  111 mm, r2  24.8 mm, r3  135.8 mm, r4  62.5 mm
111 mm
160.5 mm
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Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
Chapter 2
Position and Displacement
2.1
Describe and sketch the locus of a point A that moves according to the equations
RAx  atcos  2t  , RAy  atsin  2 t  , and RAz  0 .
The locus is the spiral shown.
2.2
Ans.
Find the position difference to point P from point Q on the curve
y  x2  x  16 , where RPx  2 and RQx  4 .
RPy   2   2  16  10 ; R P  2ˆi  10ˆj
2
2
RQy   4   4  16  4 ;
RQ  4ˆi  4ˆj
R  R  R  2ˆi  14ˆj  14.142  98.1
PQ
P
Q
Ans.
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Theory of Machines and Mechanisms, 4e
2.3
Uicker et al.
The path of a moving point is defined by the equation y  2 x2  28 .
Find the position difference to point P from point Q if RPx  4 and
RQx  3 .
RPy  2  4   28  4 ;
2
R P  4ˆi  4ˆj
2
RQy  2  3  28  10 ; RQ  3ˆi  10ˆj
R  R  R  7ˆi  14ˆj  15.65263.4
PQ
2.4
P
Q
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
RPy  3  60   3 / 3  51 ;
R  3  3ˆi  51ˆj
3
P
R P  R P (3)  R P (0)  3ˆi  9ˆj  9.487  71.57
2.5
Ans.
If point A moves on the locus of Problem 2.1, find its displacement from t = 1.5 to t = 2.
R A 1.5  1.5a cos3 ˆi  1.5a sin 3 ˆj  1.5aˆi
R  2.0   2.0a cos 4 ˆi  2.0a sin 4 ˆj  2.0aˆi
A
ΔR A  R A  2.0   R A 1.5  3.5aˆi
2.6
Ans.
The position of a point is given by the equation R  100e j 2t . What is the path of the
point? Determine the displacement of the point from t = 0.10 to t = 0.60.
The point moves in a circle of radius 100 with center at the origin.
R  0.10  100e j 0.628  80.902ˆi  58.779ˆj
R  0.60   100e j 3.770  80.902ˆi  58.779ˆj
ΔR  R  0.60  R  0.10  161.804ˆi  117.557ˆj  200.0216
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Ans.
Ans.
Theory of Machines and Mechanisms, 4e
2.7

Uicker et al.

The equation R  t 2  4 e jt /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 orientation 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
   t /10 , then d / dt is negative and therefore
the position vector is rotating clockwise.
Ans.


R  0   02  4 e j 0  40
The position vector will next have the same
direction when  t /10  2 , that is, when t=20. Ans.
R  20    202  4  e j 2  4040
R  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t / 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9 / 30  1454  8.229ˆi  11.326ˆj
ΔR  R  3  R  0   6.229ˆi  11.326ˆj  12.92661.19
Ans.
  54  0  54 ccw
Ans.
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Theory of Machines and Mechanisms, 4e
2.9
Uicker et al.
Link 2 in Fig. P2.9 rotates according to the equation    t / 4 . Block 3 slides outward
on link 2 according to the equation r  t 2  2 . What is the absolute displacement
R P from t = 1 to t = 3? What is the apparent displacement R P ?
3
3/ 2
R P3  re j   t 2  2  e j t / 4
R P3 1  345  2.121ˆi  2.121ˆj
R  3  11135  7.778ˆi  7.778ˆj
P3
ΔR P3  R P3  3  R P3 1  9.899ˆi  5.657ˆj  11.402150.26 Ans.
R P3 / 2  re j 0   t 2  2  ˆi 2
R P3 / 2 1  3ˆi 2
R  3  11ˆi
P3 /2
2
ΔR P3 /2  R P3 /2  3  R P3 /2 1  8ˆi2
2.10
Ans.
A wheel with center at O rolls without slipping so that its center is displaced 250 mm to
the right. What is the displacement of point P on the periphery during this interval?
Since the wheel rolls without slipping,
RO   RPO .
  RO / RPO
 250 mm /150 mm  1.667 rad  95.51
For R PO ,
       270  95.51  174.49
RPO  150 mm174.49  149.3ˆi  14.4ˆj mm
ΔR P  ΔR O   RPO  R PO 
 250ˆi  149.3ˆi  14.4ˆj  150ˆj mm
RPO  6 in
2.11
R P  100.7ˆi  164.4ˆj mm  192.8 mm58.51 Ans.
A point Q moves from A to B along link 3 whereas link 2 rotates from  2  30 to
 2  120 . Find the absolute displacement of Q.
RQ3  0.3 m30  259.8ˆi  150.0ˆj mm
R  0.3 m120  150.0ˆi  259.8ˆj mm
Q3
ΔRQ3  RQ3  RQ3  409.8ˆi  109.8ˆj mm
ΔR
 R  600.0ˆi mm
Q5 /3
BA
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Theory of Machines and Mechanisms, 4e
Uicker et al.
RAO  RBO  0.3 m ; RBA  RO O  0.6 m ΔRQ5  ΔRQ3  ΔRQ5 / 3
2
4
4 2
ΔRQ5  190.2ˆi  109.8ˆj mm  219.6 mm30
2.12
Ans.
The linkage 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   45 .
The loop-closure equation is
R A  R B  R AB
Ans.
RAe j /12  RB  RAB e j   
 RB  RAB e j
RAB  500 mm,   15
Taking the imaginary components of this, we get
RA sin15   RAB sin 
sin 
sin  45
RA   RAB
 500 mm
 1365 mm
sin15
sin15
2.13
Ans.
The offset slider-crank mechanism is driven by the rotating crank 2. Write the loopclosure equation. Solve for the position of the slider 4 as a function of  2 .
RAO  20 mm , RBA  50 mm , and RCB  140 mm
RC  R A  R BA  RCB
RC  RAe j / 2  RBAe j2  RCBe j3
Taking real and imaginary parts,
RC  RBA cos 2  RCB cos3 and 0  RA  RBA sin  2  RCB sin 3
and, solving simultaneously, we get
  R  RBA sin  2 
 3  sin 1  A
 with 90  3  90
RCB


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Theory of Machines and Mechanisms, 4e
2
RC  RBA cos  2  RCB
  RA  RBA sin  2 
Uicker et al.
2
Ans.
 50 cos  2  19 200  1 000sin  2  2 500sin  2
2
2.14
For the mechanism illustrated in Fig. P2.14, define a set of vectors that is suitable for a
complete kinematic analysis of the mechanism. Label and show the sense and orientation
of each vector on Fig. P2.14. Write the vector loop equation(s) for the mechanism.
Identify suitable input(s) for the mechanism. Identify the known quantities, the unknown
variables, and any constraints. If you have identified constraints then write the constraint
equation(s).
One suitable set of two vector loop equations is

?
?
?

I
?
 C1
C 2
C 3
R 2  R3  R 4  R5  R1  0 and R 2  R3  R 44  R 24  R 22  0
The angle  2 is a reasonable input. Three constraint equations are required.
Ans.
24  4   (C2)
There are four unknowns 3 , 4 , 5 , and R24 .
Ans.
44  4 (C1)
22  2   (C3)
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Theory of Machines and Mechanisms, 4e
2.15
Uicker et al.
Assume rolling with no slip between pinion 5 and rack 4 in the mechanism illustrated in
Fig. P2.15. Define a set of vectors that is suitable for a complete kinematic analysis of
the mechanism. Label and show the sense and orientation of each vector on Fig. P2.15.
Write the vector loop equation(s) for the mechanism. Identify suitable input(s) for the
mechanism. Identify the known quantities, the unknown variables, and any constraints.
If you have identified constraints then write the constraint equation(s).
One suitable set of vectors is as shown. The vector loop equation is
ÖI
Ö?
?Ö
ÖÖ
ÖÖ
ÖÖ
R 2  R 3  R 6  R 4  R 15  R 1  0 with 5 5  R6
The angle  2 is a suitable input.
There are two unknown variables,  3 and R6.
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Ans.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
2.16
Uicker et al.
For the geared five-bar mechanism illustrated in Fig. P2.16, there is rolling with no
slipping between gears 2 and 5. Define a set of vectors that is suitable for a complete
kinematic analysis of the mechanism. Label and show the sense and orientation of each
vector on Fig. P2.16. Write the vector loop equation(s) for the mechanism. Identify
suitable input(s) for the mechanism. Identify the known quantities, the unknown
variables, and any constraints. If you have identified constraints then write the constraint
equation(s).
One suitable set of vectors is as shown. The vector loop equation is
ÖI
Ö?
Ö?
ÖC
ÖÖ
R2  R3  R4  R 5  R 1  0
with
2 2  55  0
The angle  2 is a suitable input.
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
2.17
Uicker et al.
For the mechanism illustrated in Fig. P2.17, gear 3 is pinned to link 4 at point B, and is
rolling without slipping on the semi-circular ground link 1. The radius of the semicircular ground link is 1 and the radius of gear 3 is 3 . Define a set of vectors that is
suitable for a complete kinematic analysis of the mechanism shown. Label and show the
sense and orientation of each vector in Fig. P2.17. Write the vector loop equation(s) for
the mechanism. Identify suitable input(s) for the mechanism. Identify the known
quantities, the unknown variables, and any constraints. If you have identified constraints
then write the constraint equation(s).
One suitable set of vectors is as shown. The vector loop equation is
ÖI
Ö?
Ö?
ÖÖ
R 2  R 4  R5  R 1  0
with
R2 2  33
The angle  2 is a suitable input.
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
2.18
Uicker et al.
For the mechanism illustrated in Fig. P1.6, define a set of vectors that is suitable for a
complete kinematic analysis of the mechanism. Label and show the sense and orientation
of each vector in Fig. P1.6. Write the vector loop equation(s) for the mechanism.
Identify suitable input(s) for the mechanism. Identify the known quantities, the unknown
variables, and any constraints. If you have identified constraints then write the constraint
equation(s).
One set of vectors suitable for a kinematic analysis of the mechanism is shown below.
The corresponding vector loop equations are

I
?
?
?
?
?
C
R1  R 2  R3  R13a  0 and R3  R 4  R15  R13b  0
with the constraint equation R13a  R13b  constant.
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
2.19
Uicker et al.
For the mechanism illustrated in Fig. P1.8, define a set of vectors that is suitable for a
complete kinematic analysis of the mechanism. Label and show the sense and orientation
of each vector in Fig. P1.8. Write the vector loop equation(s) for the mechanism.
Identify suitable input(s) for the mechanism. Identify the known quantities, the unknown
variables, and any constraints. If you have identified constraints then write the constraint
equation(s).
One set of vectors suitable for a kinematic analysis of the mechanism is shown here.
The corresponding set of vector loop equations is

I
?
?
R1  R 2  R 4  R 5  0 and
with the constraint equation 22  2   .

C
?
?
R1  R 22  R 3  R 35  0
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
2.20
Uicker et al.
For the mechanism illustrated in Fig. P1.9, define a set of vectors that is suitable for a
complete kinematic analysis of the mechanism. Label and show the sense and orientation
of each vector in Fig. P1.9. Write the vector loop equation(s) for the mechanism.
Identify suitable input(s) for the mechanism. Identify the known quantities, the unknown
variables, and any constraints. If you have identified constraints then write the constraint
equation(s).
One set of vectors suitable for a complete kinematic analysis of this mechanism is as
shown.
The corresponding set of vector loop equations is


?
I
R1  R3  R32  R 2  0 and
with the two constraint equations
34  4  90°
C1

?
? C1
C 2
?
I
R11  R 4  R34  R33  R32  R 2  0
and
33  4  180°
© Oxford University Press 2015. All rights reserved.
C2.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
2.21
Uicker et al.
For the mechanism illustrated in Fig. P1.10, define a set of vectors that is suitable for a
complete kinematic analysis of the mechanism. Label and show the sense and orientation
of each vector in Fig. P1.10. Write the vector loop equation(s) for the mechanism.
Identify suitable input(s) for the mechanism. Identify the known quantities, the unknown
variables, and any constraints. If you have identified constraints then write the constraint
equation(s).
One set of vectors suitable for a kinematic analysis of the mechanism is shown.
The corresponding set of vector loop equations is


I

R1  R11  R 2  R 3  0
with the constraint equation

and
I
 C1
?
?
R11  R 2  R34  R 4  R9  0
Ans.
34  3  
C1
Ans.
However, these equations do not analyze the angular displacement of the small wheel,
body 5. In order to do this, we might consider the apparent angular displacement as seen
by an observer fixed on vector 9 and viewing the point of contact between bodies 5 and
1. The non-slip condition would provide the constraint
11/9  5 5/9
1  1  9   5  5  9 
5 5   1  5  9  0
5 5  R9 9  0
Ans.
where 5 is the radius of wheel 5 and 5 is the angular displacement of body 5.
2.22
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 solution is shown here.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
2.23
Uicker et al.
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 ANSI/ISO FORTRAN 77, is supplied here as
an example. There are also no universally accepted standards for programming graphics.
Therefore the Tektronix PLOT10 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 77 PROGRAM TO PLOT THE COUPLER CURVE OF ANY CRANK-ROCKER
OR DOUBLE-CRANK FOUR-BAR LINKAGE, GIVEN ITS DIMESNIONS.
ORIGINALLY WRITTEN USING SUBROUTINES FROM TEKTRONIX PLOT10 FOR
DISPLAY ON 4010 SERIES DISPLAYS.
REF:J.J.UICKER,JR, G.R.PENNOCK, & J.E.SHIGLEY, ‘THEORY OF MACHINES
AND MECHANISMS,’ FOURTH EDITION, OXFORD UNIVERSITY PRESS, 2009.
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
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
CTH2=COS(TH2)
STH2=SIN(TH2)
C
C
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
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
2.24
Uicker et al.
For each linkage illustrated in Fig. P2.24, find the path of point P: (a) inverted slidercrank mechanism; (b) second inversion of the slider-crank mechanism; (c) Scott-Russell
straight-line mechanism; and (d) drag-link mechanism.
(a)
(c)
(b)
(d)
(a) RCA  40 mm , RBA  70 mm , RPC  80 mm ; (b) RCA  100 mm , RBA  50 mm ,
RPB  162.5 mm ; (c) RBA  RCB  RPB  125 mm ; (d) RDA  10 mm , RBA  20 mm ,
RCC  RDC  30 mm , RPB  40 mm .
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
2.25
Uicker et al.
Using the offset slider-crank mechanism in Fig. P2.13, find the crank angles
corresponding to the extreme values of the transmission angle.
As shown,   90  3 .
Also from the figure
e  r2 sin  2  r3 cos  .
Differentiating with
respect to  2 ,
d
r2 cos 2  r3 sin 
;
d 2
r cos 2
d
 2
.
d 2
r3 sin 
Now, setting d / d 2  0 , we get cos 2  0 .
Therefore, we conclude that  2    2k  1  / 2  90, 270,
2.26
Ans.
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  reaches a maximum or minimum when crank 2 is colinear
with the line O2 O4 . Show, analytically, that this statement is true.
From O4O2 A :
s 2  r12  r22  2r1r2 cos 2 .
Also, from ABO4 :
s 2  r32  r42  2r3r4 cos  .
Equating these we
differentiate with respect to  2
to obtain
d
2r1r2 sin  2  2r3r4 sin 
or
d 2
r r sin  2
d
 12
.
d 2 r3r4 sin 
Now, for
d
 0 , we have sin  2  0 . Thus,  2  0,  180,  360,
d 2
© Oxford University Press 2015. All rights reserved.
Q.E.D.
Theory of Machines and Mechanisms, 4e
2.27
Uicker et al.
Figure P2.27 illustrates a crank-and-rocker four-bar 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  100 mm ,
r2  50 mm , r3  125 mm , and r4  100 mm .
(a)
(b)
(c)
Find  2 and  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  2  29 ,
 4  58 and  2  248 ,  4  136 .
Ans.
(b) Then  4   4   4  78
Ans.
(c) Finally, from isosceles triangle
O4O2 B ,   29 and    68 .
Ans.
2.28
A double-rocker four-bar linkage has a dead-center position and may also have a limit
position (see Prob. 2.27). These positions occur when links 3 and 4 in Fig. P2.28 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
illustrated in Fig. P2.28, that r1  140 mm , r2  55 mm , r3  50 mm , and r4  120 mm .
Find  2 and  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
2  114.0 ,  4  162.8 and, symmetrically,
 2  114.0 ,  4  162.8 . There are
also two limit positions; these occur at
 2  56.5 , 4  133.1 and, symmetrically,
at  2  56.5 ,  4  133.1 .
Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
2.29
Uicker et al.
Figure P2.29 illustrates a slider-crank linkage 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 a time ratio of the drive
to return strokes. Also determine which direction the crank should rotate.
From the figure we can see that e   r3  r2  sin  2   r3  r2  sin  2  180 or

e 
e 
1 
 ,  2  180  sin 

 r3  r2 
 r3  r2 
 2  sin 1 
 e 
e 
1 
 drive   2   2  180  sin 1 
  sin 

 r3  r2 
 r3  r2 
Ans.
 e 
e 
1 
 return   2  360   2  180  sin 1 
Ans.
  sin 

 r3  r2 
 r3  r2 
Assuming driving is when B is sliding to the right, the crank should rotate clockwise. Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
Chapter 3
Velocity
3.1
The position vector of a point is given by the equation R  100e jt , where R is in meters.
Find the velocity of the point at t  0.40 s.
R  t   100e j t m
R  t   j 100e j t m/s
R  0.40s   j 100e j 0.40 m/s
 j 100  cos 0.40  j sin 0.40  m/s
 100 sin 72  j100 cos 72 m/s
R  0.40s   298.783  j97.080 m/s  314.159 m/s162
3.3
Ans.
If automobile A is traveling south at 70.4 km/h and automobile B north 60 east at 51.2
km/h, what is the velocity difference between B and A? What is the apparent velocity of
B to the driver of A?
VA  70.4 km/h  90  70.4ˆj km/h
V  51.2 km/h30  44.34ˆi  25.6ˆj km/h
B
VBA  VB  VA  44.34ˆi  96ˆj km/h
VBA  105.74 km/h65.2 =105.74 km/h N 24.8 E
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  105.74 km/h65.2 =105.74 km/h N 24.8 E
3.4
Ans.
In Fig. P3.4, wheel 2 rotates at 450 rev/min and drives wheel 3 without slipping. Find
the velocity difference between points B and A.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
2 
 450 rev/min  2
rad/rev 
60 s/min
 15 rad/s cw
VAO2  2 RAO2  1500  4710 mm/s
VB  4200 mm/s90
VBA  VB  VA
 4710ˆi  4200ˆj
Ans.
 6300 mm/s138.4
3.5
Two points A and B, located along the radius of a wheel (see Fig. P3.5), have speeds of
80 and 140 in/s, respectively. The distance between the points is RBA  .75 mm .
(a) What is the diameter of the wheel?
(b) Find VAB , VBA, and the angular velocity of the wheel.
   
  2ˆj   3.5ˆj  1.5ˆj m/s
VBA  VB  VA  3.5ˆj  2ˆj  1.5ˆj m/s Ans.
VAB  VA  VB
2 
RBO2
VBA 1.5 m/s

 20 rad/s cw
RBA 0.075 m
VBO2 3.5 m/s


 175 mm
2 20 rad/s
D  2RBO2  2 17.5 mm   350 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
3.6
Uicker et al.
A plane leaves point B and flies east at 448 km/h. Simultaneously, at point A, 320 km
southeast (see Fig. P3.6), a plane leaves and flies northeast at 499.2 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  499.245 km/h  352.96ˆi  352.96ˆj km/h ;
VB  448ˆi km/h
VBA  VB  VA  95ˆi  352.96ˆj km/h
At initial time R BA  0   320 120  160ˆi  276.8ˆj km
Later R (t )  R  0  V t   160  95t  ˆi   276.8  352.96t  ˆj km
BA
BA
BA
To find the minimum of this:
2
2
2
RBA
  160  95t    276.8  352.96t 
2
dRBA
dt  2  160  95t  95  2  276.8  352.96t  352.96  0
269211t  225 798  0 t  0.838 h  51 min or 6:51p.m.
RAB  320 km
R BA  0.845 h   79.68ˆi  21.44ˆj  82.56 km  165
3.7
Ans.
Ans.
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  400.96ˆi  352.96ˆj km/h  534.24 km/h41.4
Since the velocity is constant, the new path is a straight line at N 48.6º E.
Ans.
Since the velocities of both planes change by the same amount, the velocity difference
Ans.
VBA does not change. Therefore the results of Problem 3.6 do not change.
3.8
The velocity of point B on the linkage illustrated in Fig. P3.8 is 1 m/s. Find the velocity
of point A and the angular velocity of link 3.
RAB  0.1 m
VA  VB  VAB
VA  1.24 m/s  165
VAB  0.37 m/s  120
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
3 
VAB 0.37 m/s

 3.7 rad/s ccw
0.1 m
RAB
© Oxford University Press 2015. All rights reserved.
Uicker et al.
Ans.
Theory of Machines and Mechanisms, 4e
3.9
Uicker et al.
The mechanism illustrated in Fig. P3.9 is driven by link 2 at 2  45 rad/s ccw. Find the
angular velocities of links 3 and 4.
RAO  100 mm, RBA  250 mm, RO O  250 mm, and RBO  300 mm.
4 2
2
VAO2  2 RAO2   45 rad/s 100 mm   4 500 mm/s
4
VB  VA  VBA  VO4  VBO4
VBA  358.5 mm/s ; VBO4  4 619 mm/s .
VBA 358.5 in/s

 1.43 rad/s ccw
250 in
RBA
V
4 619 mm/s
4  BO4 
 15.40 rad/s ccw
300 mm
RBO4
3 
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Ans.
Ans.
Theory of Machines and Mechanisms, 4e
3.10
Uicker et al.
Crank 2 of the push-link mechanism illustrated in Fig. P3.10 is driven at
2  60 rad / s cw. Find the velocities of points B and C and the angular velocities of
links 3 and 4.
RAO  6 in, RBA  12 in, RO O  3 in, RBO  12 in, RDA  6 in, and RCD  4 in.
2
4 2
4
VAO2  2 RAO2   60 rad/s  6 in   360 in/s
VB  VA  VBA  VO4  VBO4
VBA  520.8 in/s ; VB  454.4 in/s41
VC  VA  VCA  VB  VCB
VC  153.2 in/s60
3 
3.11
Ans.
Ans.
VBO4 454.4 in/s
VBA 520.8 in/s

 37.87 rad/s cw

 43.40 rad/s cw ; 4 
12 in
RBA
12 in
RBO4
Ans.
Find the velocity of point C on link 4 of the mechanism illustrated in Fig. P3.11 if crank
2 is driven at 2  48 rad / s ccw. What is the angular velocity of link 3?
VAO2  2 RAO2   48 rad/s  200 mm   9 600.0 mm/s
VB  VA  VBA  VO4  VBO4
RAO  200 mm, RBA  800 mm, RO O  400 mm, RBO  400 mm, and RCO  300 mm.
2
4 2
VBA 268 mm/s

 0.335 rad/s ccw
800 mm
RBA
 VB  VCB ; VC  7 118 mm/s  75.8
VBA  268 mm/s ;
VC  VO4  VCO4
4
3 
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4
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
3.12
Uicker et al.
Figure P3.12 illustrates a parallel-bar linkage, in which opposite links have equal lengths.
For this linkage, demonstrate that 3 is always zero and that 4  2 . How would you
describe the motion of link 4 with respect to link 2?
Referring to Fig. 2.19 and using r1  r3 and r2  r4 , we compare Eqs. (2.26) and (2.27) to
see that    . Then Eq. (2.29) gives 3  0 and its derivative is 3  0 .
Ans.
Next we substitute Eq. (2.25) into Eq. (2.33) to see that    2 . Then Fig. 2.19 shows
that, since link 3 is parallel to link 1 3  0  , then  4     2 . Finally, the derivative of
this gives 4  2 .
Since 4/ 2  4  2  0 , link 4 is in curvilinear translation with respect to link 2.
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
3.13
Uicker et al.
Figure P3.13 illustrates the antiparallel or crossed-bar linkage. If link 2 is driven at
2  1 rad/s ccw, find the velocities of points C and D.
RAO  RBO  300 mm , RBA  RO O  150 mm , and RCA  RDB  75 mm
2
4
4 2
VAO2  2 RAO2  1 rad/s  300 mm   300 m/s
VB  VA  VBA  VO4  VBO4
Construct the velocity image of link 3.
VC  402.5 mm/s151
VD  290 mm/s249
3.14
Ans.
Ans.
Find the velocity of point C of the linkage illustrated in Fig. P3.14 assuming that link 2
has an angular velocity of 60 rad/s ccw. Also find the angular velocities of links 3 and 4.
RAO  RBA  150 mm, RO O  RBO  250 mm, and RCA  200 mm.
2
4 2
4
VAO2  2 RAO2   60 rad/s 150 mm   9 000 mm/s
VB  VA  VBA  VO4  VBO4
VBA  4 235 mm/s ; VBO4  7 940 mm/s
VBA 4 235 mm/s

 28.23 rad/s cw
150 in
RBA
V
7 940 mm/s
4  BO4 
 31.76 rad/s cw
250 mm
RBO4
3 
Construct the velocity image of link 3:
VC  12 680 mm/s156.9
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Ans.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
3.15
Uicker et al.
The inversion of the slider-crank mechanism illustrated in Fig. P3.15 is driven by link 2
at 2  60 rad/s ccw. Find the velocity of point B and the angular velocities of links 3 and
4.
RAO  75 mm, RBA  400 mm, and RO O  125 mm.
2
4 2
VAO2  2 RAO2   60 rad/s  75 mm   4500 mm/s
VP3  VA  VP3 A  VP4  VP3 / 4
4  3 
VP3 A
RP3 A

4260 mm/s
 22.0 rad/s ccw
194 mm
Ans.
Construct the velocity image of link 3:
VB  4790 mm/s96.5
3.16
Ans.
Find the velocity of the coupler point C and the angular velocities of links 3 and 4 of the
mechanism illustrated if crank 2 has an angular velocity of 30 rad/s cw.
RAO  75 mm, RBA  RCB  125 mm, RO O  250 mm, and RBO  150 mm.
2
4
2
4
VAO2  2 RAO2   30 rad/s  75 mm   2 250.0 mm/s
VB  VA  VBA  VO4  VBO4
Construct the velocity image of link 3:
VC  2 250.0 mm/s126.9
3 
VBO4
VBA 2 250.0 mm/s
0

0

 18.00 rad/s ccw ; 4 
RBO4 200.0 mm
125 mm
RBA
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Ans.
Ans.
Theory of Machines and Mechanisms, 4e
3.17
Uicker et al.
Link 2 of the linkage illustrated in Fig. P3.17 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.
RAO  62.5 mm,
2
RBA  250 mm,
RCB  200 mm,
RCA  RDC  100 mm,
RO O  200 mm, and RDO  150 mm.
6
2
6
VAO2  2 RAO2  10 rad/s  62.5 mm   625.0 mm/s
VB  VA  VBA
VB  289.25 mm/s180
Construct velocity image of link 3:
VC  VA  VCA  VB  VCB
VC  605.5 mm/s207.6
VD  VC  VDC  V
 O6  VDO6 VD  604.5 mm/s206.2
6 
3.18
VDO6
RDO6

Ans.
Ans.
Ans.
604.5 mm/s
 4.03 rad/s ccw
150 mm
Ans.
The angular velocity of link 2 of the drag-link mechanism illustrated in Fig. P3.18 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. RAO  350 mm, RBA  425 mm, RO O  100 mm, and RBO  400 mm.
2
4 2
4
The graphical construction is shown in the position where
 2  135 , where the result is VB  5760 mm/s  7.2 .
It is repeated at increments of  2  15 . The maximum
and
minimum
velocities
are
at
and
VB,max  9130 mm/s  146.6
 2  15
VB,min  4590 mm/s63.7 at  2  225 , respectively.
Within graphical accuracy these two positions
approximately 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
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Theory of Machines and Mechanisms, 4e
Uicker et al.
verified in this way.
3.19
Link 2 of the mechanism illustrated in Fig. P3.19 is driven at 2  36 rad/s cw. Find the
angular velocity of link 3 and the velocity of point B.
RAO  125 mm, RBA  RBO  200 mm, and RO O  175 mm.
2
4 2
4
VAO2  2 RAO2   36 rad/s 125 mm   4 500 mm/s
VB  VA  VBA  VO4  VBO4
VB  5 070 mm/s  56.3
V
647 mm/s
3  BA 
 3.23 rad/s ccw
200 mm
RBA
3.20
Ans.
Ans.
Find the velocity of point C and the angular velocity of link 3 of the push-link mechanism
illustrated in Fig. P3.20. Link 2 is the driver and rotates at 8 rad/s ccw.
RAO  150 mm, RBA  RBO  250 mm, RO O  75 mm, RCA  300 mm, and RCB  100 mm.
2
4
4 2
VAO2  2 RAO2  8 rad/s 150 mm   1200 mm/s
VB  VA  VBA  V
 O4  VBO4
Construct velocity image of link 3:
VC  VA  VCA  VB  VCB
VC  3847.5 mm/s  136.8
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Ans.
Theory of Machines and Mechanisms, 4e
3 
3.21
Uicker et al.
VBA 3785 mm/s

 15.14 rad/s ccw
250 mm
RBA
Ans.
Link 2 of the mechanism illustrated in Fig. P3.21 has an angular velocity of 56 rad/s ccw.
Find the velocity of point C.
RAO2  150 in, RBA  RBO  250 mm, RO O  100 mm, and RCA  300 mm.
4 2
4
VAO2  2 RAO2   56 rad/s 150 mm   8400 mm/s
VB  VA  VBA  VO4  VBO4
Construct the velocity image of link3:
VC  VA  VCA  VB  VCB
VC  927.5 mm/s137.8
3.22
Ans.
Find the velocities of points B, C, and D of the double-slider mechanism illustrated in
Fig. P3.22 if crank 2 rotates at 42 rad/s cw.
RAO  50 mm, RBA  250 mm, RCA  100 mm, RCB  175 mm, RDC  200 mm.
2
VAO2  2 RAO2
  42 rad/s  50 mm   2100 mm/s
VB  VA  VBA
VB  1635 mm/s180
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Construct velocity image of link 3:
VC  VA  VCA  VB  VCB
VD  VC  VDC  530 mm/s90
3.23
Uicker et al.
VC  1695 mm/s154.2
Ans.
Ans.
Figure P3.23 illustrates 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.
RAO  50 mm, RBA  RCB  150 mm, RCA  50 mm, RDC  125 mm.
2
2000 rev/min  2 rad/rev 
 209.4 rad/s
2 
60 s/min
VAO2  2 RAO2
  209.4 rad/s  50 mm   10 470 mm/s
VB  VA  VBA  10 648 mm/s  120
Construct velocity image of link 3:
VC  12 180 mm/s   93
VD  VC  VDC  9 468 mm/s  60
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Ans.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
3.24
Uicker et al.
Make a complete velocity analysis of the linkage illustrated in Fig. P3.24 given that
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?
RAO  200 mm, RO O  500 mm.
2
4 2
VAO2  2 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
3 
VP3 A
RP3 A

4045 mm/s
 6.16 rad/s cw
656.75 mm
From this, or graphically, we complete the velocity image of link 3, from which
VB3  3945 mm/s  39.7
Ans.
Then, since link 4 remains perpendicular to link 3, we have 4  3 and we find the
velocity image of link 4:
Ans.
VB3 /4  2582.5 mm/s  12.4
3.25
Find VB for the linkage illustrated in Fig. P3.25 if VA  300 mm/s.
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 mm/s  23.0
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Ans.
Theory of Machines and Mechanisms, 4e
3.26
Uicker et al.
Figure P3.26 illustrates a variation of the Scotch-yoke mechanism. The mechanism is
driven by crank 2 at 2  36 rad/s ccw. Find the velocity of the crosshead, link 4.
RAO  250 mm.
2
VAO2  2 RAO2   36 rad/s  250 mm   9000 mm/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  4657.5 m/s180
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 illustrated in Fig. P3.27 for 2  72
rad/s ccw.
RAO  RDC  37.5 mm, RBA  262.5 mm, RO O  150 mm, RBO  125 mm, RO O  175 mm,
2
4
2
4
6
2
and REO  200 mm.
6
VAO2  2 RAO2   72 rad/s  37.5 mm   2700 mm/s
VB  VA  VBA  VO4  VBO4
Construct velocity image of link 3: VC3  VA  VCA  VB  VCB
VC3  1908 mm/s203.2
Ans.
Using the path of C3 on link 6, we next write
VC3  VC6  VC3 / 6 and VC6  VC6O6 , from which VC6  1076.75 mm/s241.4
Ans.
From this, graphically, we can complete the velocity image of link 6, from which
VE  1934.75 mm/s  98.9
VC O
Since link 5 remains perpendicular to link 6, 5  6  6 6  9.67 rad/s cw
RC6O6
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Ans.
Ans.
Theory of Machines and Mechanisms, 4e
From these we can get VD5  VC5  VD5C5  1612.25 mm/s210.1
3.28
Uicker et al.
Ans.
The mechanism illustrated in Fig. P3.28 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 path of C2 on link 3, we write
VC2  VC3  VC2 / 3 and VC3  VD3  VC3D3
3 
3.29
VC3D3
RC3D3

105.65 mm/s
 1.569 rad/s cw
67.25 mm
Ans.
The circular cam illustrated in Fig. P3.29 is driven at an angular velocity of 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.
VBA  2 RBA  15 rad/s  31.25 mm   468.75 mm/s
VD4  VD2  VD4 / 2 and VD4  VE  VD4 E
4 
VD4 E
RD4 E

381 mm/s
 4.355 rad/s ccw
87.5 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
3.30
Uicker et al.
The mechanism illustrated in Fig. P3.30 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  2 RBA  10 rad/s  25 mm   250 mm/s
VC  VB  VCB  VD  VCD
V
333.25 mm/s
Ans.
 3.333 rad/s ccw
3  CB 
RCB
100 mm
V
166.75 mm/s
Ans.
 3.333 rad/s ccw
4  CD 
RCD
50 mm
Construct velocity image of link 3: VE3  VB  VE3B  VC  VE3C  251.5 mm/s220.9 Ans.
Using the path of E3 on link 6, we write VE3  VE6  VE3 / 6 and VE6  VH  VE6 H
6 
VE6 H
RE6 H

121.45 mm/s
 3.774 rad/s cw
32.2 mm
Ans.
Construct velocity image of link 6: VG  VE6  VGE6  VH  VGH  298.25 mm/s  57.1 Ans.
VF5  VF6 ;
VE5  VE3 ; 5 
VF5 E
RF5 E

319.5 mm/s
 25.56 rad/s cw
12.5 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
3.31
Uicker et al.
Figure P3.31 is a schematic diagram for a two-piston pump. The pump is driven by a
circular eccentric, link 2, at 2  25 rad/s ccw. Find the velocities of the two pistons,
links 6 and 7.
VF2  VF2 E  2 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 mm/s180
VB  VD  VBD  144.42 mm/s180
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
3.32
Uicker et al.
The epicyclic gear train illustrated in Fig. P3.32 is driven by the arm, link 2, at 2  10
rad/s cw. Determine the angular velocity of the output shaft, attached to gear 3.
VB  VBA  2 RBA  10 rad/s  75 mm   750 mm/s
Using VD  0 construct the velocity image of link 4 from which VC  1500 mm/s0 .
V
1500 mm/s
Ans.
 30.00 rad/s cw
3  CA 
RCA
50 mm
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
3.33
Uicker et al.
The diagram in Fig. P3.33 illustrates 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 I12. It can be found by the repeated application of Kennedy’s
theorem as shown.
In the automotive industry it has become common practice to use only half of this
construction, assuming by symmetry that I12 must lie on the vertical 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 (called the “jacking coefficient”!). They should be thinking
about the fixed and moving centrodes (Section 3.21), which are more horizontal than
vertical!
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
3.34
Uicker et al.
Locate all instant centers for the linkage of Problem 3.22.
Instant centers I12 , I 23 ,
I 34 , I14 (at infinity), I 35 ,
I 56 , and I16 (at infinity)
are found by inspection.
All others are found by
repeated applications of
Kennedy’s theorem except
I 46 .
One line can be found for
I 46 ; however, no second
line can be found by
Kennedy’s theorem since
no line can be drawn (in
finite space) between I14
and I16 . Now it must be
seen that I 46 must be
infinitely remote because
the
relative
motion
between links 4 and 6 is
translation; that is, the
angle between lines on
links 4 and 6 remains
constant.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
3.35
Uicker et al.
Locate all instant centers for the mechanism of Problem 3.25.
Instant centers I12 (at infinity), I 23 ,
I 34 (at infinity), and I14 are found
by inspection. All others are found
by repeated applications of
Kennedy’s theorem.
3.36
Locate all instant centers for the mechanism of Problem 3.26.
Instant centers I12 , I 23 , I 34 (at
infinity), and I14 (at infinity) are
found by inspection. All others
are found by repeated applications
of Kennedy’s theorem except I13 .
One line ( I12 I 23 ) can be found for
I13 ; however, no second line can
be found by Kennedy’s theorem
since no line can be drawn (in
finite space) between I14 and I 34 .
Now it must be seen that I13 must
be infinitely remote because the
relative motion between links 1
and 3 is translation; the angle
between links 1 and 3 remains
constant.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
3.37
Uicker et al.
Locate all instant centers for the mechanism of Problem 3.27.
Instant centers I12 , I 23 ,
I 34 , I14 , I 35 , I 56 (at
infinity), and I16 are
found by inspection. All
others are found by
repeated applications of
Kennedy’s theorem.
3.38
Locate all instant centers for the mechanism of Problem 3.28.
Instant centers I12 and I13 are
found by inspection.
One line for I 23 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.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
3.39
Uicker et al.
Locate all instant centers for the mechanism of Problem 3.29.
Instant centers I12 , I 23 , I 34 , and I14 are found by inspection. The other two are found by
use of Kennedy’s theorem.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
3.40
Uicker et al.
For the mechanism illustrated in Fig. P3.40, the input link 2 is in the position
RAO4  150 mm and is moving to the right at a velocity of VA  18.75 mm/s . Determine
the first-order kinematic coefficients for the mechanism in the given position, and
determine the angular velocities of links 3 and 4.
RBO4  RBA  150 mm
Let the following vectors be defined: r2  RAO4 e j , r3  RBAe j3 , and r4  RBO4 e j4 . Then
the loop-closure equation is r2  r3  r4  0 . The two scalar position equations are
r2  r3 cos 3  r4 cos  4  0
r3 sin 3  r4 sin  4  0
With the given data, at the position r2  150 mm , the solution is 3  60 and 4  120 .
Taking the derivative of the position equations with respect to input r2 gives
1  r3 sin 33  r4 sin  4 4  0
r3 cos 33  r4 cos  4 4  0
 r sin 3 r4 sin  4  3   1
or, in matrix format,  3
    
 r3 cos 3 r4 cos  4   4   0 
The determinant of the Jacobian is   r3r4 sin 4  3  and goes to zero when 3  4 or
when 3  4  180 .
The solutions for the first-order kinematic coefficients are
3  r4 cos4   3.85 103 rad/mm and 4  r3 cos3   3.85 103 rad/mm Ans.
The input velocity is given as r2  15.0 m/s .
3  3r2  72.17 rad/s (ccw) and 4  4r2  72.17 rad/s (cw)
Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
3.41
Uicker et al.
For the mechanism illustrated in Fig. P3.41 pinion 3 is rolling without slipping on rack 4
at point D. Input link 2 is in the position RGO4  250 mm , and the input velocity is
VG  75ˆi mm/s . Determine the first-order kinematic coefficients of the mechanism.
Find the angular velocities of both the rack 4 and the pinion 3.
Rack and pinion mechanism. RDG  3  125 mm .
Let the following vectors be defined as r2  RGO4 e j 0 , r4  RDO4 e j4 , and ρ3   j 3e j4 .
Then the loop-closure equation is r2  ρ3  r4  0 . The two scalar position equations are
r2  3 sin  4  cos  4 r4  0
 3 cos  4  sin  4 r4  0
At the position r2  250 mm , the solution is 4  150 and r4   3 tan 4  216.5 mm .
Taking the derivative of the position equations with respect to input r2 gives
1  3 cos  4 4  sin  4 r4 4  cos  4 r4  0
3 sin  4 4  cos  4 r4 4  sin  4 r4  0
or, simplifying by use of the position equations and putting into matrix format,
 0 cos  4   4  1 
 r sin    r    0
 
2
4 4 
The determinant of the Jacobian is   r2 cos 4 and goes to zero when 4  90 or r2  0 .
The solutions for the first-order kinematic coefficients are
4  sin 4   0.002 309 rad/mm and r4  r2   1.154 7 mm/mm
Ans.
The input velocity is given as r2  1875 mm/s . From this we can get
4  4r2  0.173 2 rad/s (cw)
Ans.
However, we must notice that vector ρ3 is not attached to link 3. To find 3 we start
with the constraint for rolling with no slip. If we designate rotation of link 3 by the angle
 3 then 3  3  4   r4 . Dividing this by t and taking the limit, we get the
angular velocity of the pinion, link 3
3  3  4  r4 3  0.519 6 rad/s (ccw)
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
3.42
Uicker et al.
For the mechanism of Example 2.9, see Fig. 2.34, the dimensions are R1  800 mm ,
R9  550 mm , and 3  500 mm . In the position where R2  750 mm , the input link 2 has
a velocity of VA  150ˆj mm/s . Determine the first-order kinematic coefficients for this
mechanism. Find the velocity of rack 4, and the angular velocity of pinion 3.
Using the vectors defined in Example 2.9, the complex algebra loop-closure equation is
jR2  jR9e j34  R34e j34  jR1  R4  0
and the two scalar position equations are
 R9 sin 34  R34 cos 34  R4  0
R2  R9 cos 34  R34 sin 34  R1  0
At R2  750 mm with the given dimensions these give R34  259.8 mm and R4  606.2 mm
with the no-slip condition that R34  33 .
Taking the derivative of the position equations with respect to input R2 gives
  R4  0
 cos 34 R34
 0
1  sin 34 R34
  33 .
with the condition that R34
From these, the first-order kinematic coefficients are
  1.154 7 mm/mm , 3  2.3 103 rad/mm , and R4  0.577 35 mm/mm
R34
The velocity of the rack is V   R R ˆi  86.6 ˆi mm/s
Ans.
The angular velocity of the pinion is 3  3R2  0.3464 rad/s ccw.
Ans.
4
4
2
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Ans.
Theory of Machines and Mechanisms, 4e
3.43
Uicker et al.
For the mechanism illustrated in Fig. P3.43, in the current position RAO4  250 mm , and
the input velocity is V  125ˆi mm/s . Determine the first-order kinematic coefficients
A
of the mechanism. Find the angular velocity of link 3 and the slipping velocity between
links 3 and 4.
RPA  125 mm and APO4  90.
Let the following vectors be defined as R AO4  r2e j 0 , R PA  r3e j3 , R PO4  jr4e j3 . Then
the loop-closure equation is
r2  r3e j3  jr4e j3  0
The two scalar equations are
r2  r3 cos 3  r4 sin 3  0
r3 sin 3  r4 cos 3  0
which, at the input position r2  250 mm , has the solution 3  cos1   r3 r2   240 and
r4  r3 tan 3  216.5 mm .
Taking the derivative of the position equations with respect to input r2 gives
1  r3 sin 33  r4 cos 33  sin 3r4  0
r3 cos 33  r4 sin 33  cos 3r4  0
or, simplifying by use of the position equations and putting into matrix format,
sin 3  3   1
 0

    
 r2  cos 3   r4   0 
The determinant of the Jacobian is   r2 sin 3 and goes to zero when 3  0 at r2  125 mm .
From the solution of these equations, the first-order kinematic coefficients are
Ans.
3  cos3   0.002 309 rad/mm and r4  r2   1.1547 mm/mm
For the given input velocity of r2  125 mm/s ,
the angular velocity of link 3 is 3  3  3r2  0.289 rad/s (cw)
and the slipping velocity is V3/4  r4  r4r2  144.34 mm/s .
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
3.44
Uicker et al.
For the mechanism illustrated in Fig. 3.30 there is rolling contact at point F. The input
has an angular velocity of 2  10 rad/s ccw and there is rolling contact between links 5
and 6 at point F. Determine the first-order kinematic coefficients for links 3, 4, 5, and 6.
Find the angular velocities for links 3, 4, 5, and 6 and the velocities of points E and G.
Let the following vectors be defined:
RCD  r4e j4 ,
R BA  r2e j2 ,
R EB  12.5r3e j3  j37.5e j3 mm,
RCB  r3e j3 ,
R DA  r1e j 0 ,
R HA  25  j37.5 mm,
and
R EH  j12.5e j6  r6e j6 .
Then there are two loop-closure equations
R BA  R CB  R CD  R DA  0
R BA  R EB  R EH  R HA  0
and four corresponding scalar equations
r2 cos  2  r3 cos 3  r4 cos  4  r1  0
r2 sin  2  r3 sin 3  r4 sin  4  0
r2 cos  2  ½ r3 cos 3  1.5sin 3  0.5sin  6  r6 cos  6  25  0
r2 sin  2  ½ r3 sin 3  1.5cos 3  0.5cos  6  r6 sin  6  37.5  0
Numerical solution of these with the dimensions specified at the position 2  180 gives
the current position as 3  28.955, 4  75.522, 6  14.478, r6  29.65 mm.
Taking derivatives of these four equations with respect to input  2 gives
r2 sin  2  r3 sin 33  r4 sin  4 4  0
r2 cos  2  r3 cos 33  r4 cos  4 4  0
r2 sin  2  ½ r3 sin 33  1.5cos 33  0.5cos  6 6  r6 sin  6 6  cos  6 r6  0
r2 cos  2  ½ r3 cos 33  1.5sin 33  0.5sin  6 6  r6 cos  6 6  sin  6 r6  0
Numerical solution gives the solution as 3  0.33334 rad/rad, 4  0.33334 rad/rad,
6  0.37751 rad/rad, and r6  27.24 mm/rad.
The no-slip condition gives the displacement constraint r6  5  5  6  from which
we find r6  5 5  6  , which gives 5  6  r6 5  2.55663 rad/rad.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
Therefore the first-order kinematic coefficients are
Ans.
3  0.3333 rad/rad, 4  0.3333 rad/rad, 5  2.5566 rad/rad, 6  0.3775 rad/rad.
The angular velocities are
3  32  3.333 rad/s ccw, 4  42  3.333 rad/s ccw, 5  52  25.566 rad/s (cw),
and 6  62  3.775 rad/s (cw).
Ans.
The positions of point E and G are
xE  r2 cos  2  ½ r3 cos 3  37.5sin 3
yE  r2 sin  2  ½ r3 sin 3  37.5cos 3
xG  25  25sin  6  75cos  6
yG  37.5  25cos  6  75sin  6
The derivatives of these give the first-order kinematic coefficients
xE  r2 sin  2  ½ r3 sin 33  37.5cos 33  19 mm/rad
yE   r2 cos  2  ½ r3 cos 33  37.5sin 33  16.468 mm/rad
xG  25cos  6 6  75sin  6 6  16.22 mm/rad
yG  25sin  6 6  75cos  6 6  25.05 mm/rad
And the velocities are
VE  xE &2 î + yE 2 ˆj  190 î  164.67ˆj  251.47mm / s   139.09
V  x &2 î  y &ˆj  6.487 î  10.022ˆj  298.45mm / s  57.09
G
G
G
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Ans.
Theory of Machines and Mechanisms, 4e
3.45
Uicker et al.
For the mechanism illustrated in Fig. P3.45, input link 2 is moving vertically upwards
with a velocity of VA  187.5 mm/s . Pinion 4 has a radius of 25 mm and is rolling without
slipping on rack 3 at point B. The distance from point E to point B is equal to the
distance from point B to pin A. The distance from O4 to A is 50 mm. Determine the firstorder kinematic coefficients for the rack 3 and the pinion 4, and find the angular velocity
of rack 3 and pinion 4 and the velocity of point E. Also find the velocity along rack 3 of
the point of contact between links 3 and 4 (that is, point B).
Let
the
following
j3
vectors
R BA  r3e , and R BO4  r4e
j4
be
defined:
R AO4  jr2 ,
j3
 jr4e .
Then the loop-closure equation is R AO4  R BA  R BO4  0
and the scalar equations are
r3 cos 3  r4 sin 3  0
r2  r3 sin 3  r4 cos 3  0
The position solution for the given data at r2  50 mm is
3  120, r3  43.3 mm .
Taking derivatives of these equations with respect to input r2 gives
cos 3r3  r3 sin 33  r4 cos 33  0
25  sin 3r3  r3 cos 33  r4 sin 33  0
or, simplifying by use of the position equations and putting into matrix format,
cos 3 r2   r3   0 

   

 sin 3 0  3   25
The determinant of the Jacobian is   r2 sin 3 and goes to zero when 3  0 or 180 .
From these, the first-order kinematic coefficients are r3  1 sin 3 and 3  1  r2 tan 3 
The no-slip condition gives the displacement constraint r3  4  4  3  from
which we find r3  4 4  3  , which gives 4  3  r3 4 .
Therefore the first-order kinematic coefficients are
r3  15470 mm/mm, 3  0.0115 rad/mm, 4  0.0346 rad/mm.
Ans.
For r2  187.5 mm/s , the angular velocities of links 3 and 4 are
3  3r2  2.165 rad/s ccw and 4  4r2  6.495 rad/s (cw) .
Ans.
Given that REA = 2r3 = 86.6 mm, the position of point E is R E  xE  jyE  jr2  86.6e j3
xE  86.6cos 3  43.3 mm and yE  r2  86.6sin 3  25 mm
The derivative with respect to input r2 gives
xE  86.6sin 33  0.866 03 mm/mm and
yE  1  3.464cos33  0.500 mm/mm
xE  xE r2  162.38 mm/s and yE  yE r2  93.75 mm/s
The velocity of point E is VE  187.5 mm/s150
Ans.
The velocity along rack 3 of the point of contact between links 3 and 4 is
VB4 /3  r3  r3r2   1.154 70 mm/mm 187.5 mm/s   216.5 mm/s
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
VB4 /3  216.5 mm/s  60
3.46
Ans.
For the mechanism illustrated in Fig. P3.46, the dimensions are RAO2  250 mm and
RPO4  500 mm . At the position illustrated, where O4O2 A  30 , RPA  RAO4 , and
RPB  RBA , the angular velocity of the input link 2 is ω2 = 5 rad/s cw. Determine the
first-order kinematic coefficients for links 3, 4, and 5. Then find: (i) the angular
velocities of links 3 and 4; (ii) the velocity of link 5; and (iii) the velocity of point P fixed
in link 4.
Using instant centers, the first-order kinematic coefficients for link 3 and link 4 are
RI I
250.00 mm
 0.500 rad/rad
3  23 12 
Ans.
500.00 mm
RI23I13
 4 
For link 5
RI24 I12
RI24 I14
xB  0,
144.35 mm
 0.500 rad/rad
288.70 mm
Ans.
rB  yB  RI25 I12  216.50 mm/rad
Ans.

From these, with 2  5 rad/s (cw),
(i)
3  32  2.50 rad/s ccw, 4  42  2.50 rad/s ccw
(ii)
V  r   1 082.5ˆj mm/s
B
(iii)
B
2
Ans.
Ans.
rP  4 RPI14   0.500 rad/rad  500.00 mm   250.00 mm/rad
VP  rP2  1 250.0 in/s120
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
3.47
Uicker et al.
For the mechanism illustrated in Fig. P3.47, the input link 2 is moving parallel to the Xaxis with a constant velocity VB  375 mm/s to the right. At the instant indicated, the
angle θ4 = 60º. (i) Determine the first-order kinematic coefficients for links 3 and 4, and
find the angular velocities of links 3 and 4. (ii) Determine the conditions for the
determinant of the coefficient matrix of part (i) to be zero; then sketch the mechanism in
the position where the determinant is zero.
RBA  RAO4  100 mm.
The two scalar loop-closure equations are
r2  RBA cos 3  RAO4 cos  4  0
y2  RBA sin 3  RAO4 sin  4  0
The solution at the current geometry, with r2  50 mm, y2  86.6 mm, is 4  60 and 3  90.
(i) Taking the derivative with respect to input r2 gives
1  RBA sin 33  RAO4 sin  4 4  0
 RBA cos 33  RAO4 cos  4 4  0
which in matrix format becomes
RAO4 sin  4  3   1
 RBA sin 3

    
  RBA cos 3  RAO4 cos  4   4   0 
The determinant of the Jacobian matrix is   RBA RAO4 sin 4  3   5 000  m2 .
The first-order kinematic coefficients for links 3 and 4 are
3  RAO4 cos4   0.010 rad/mm and 4   RBA cos 3   0
The angular velocities are
3  3r2  3.75 rad/s (cw)
Ans.
Ans.
4  4r2  0
(ii)
The conditions for which   0 are that 3  4 or 3  4  180 ; e.g., this will
happen when 3  4  68.907 and r2  71.975 mm as shown below.
and
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
3.48
Uicker et al.
For the mechanism illustrated in Fig. P2.15, the dimensions are RAO2  50 mm ,
RBA  150 mm , and RCO5  62.5 mm . In the position indicated, the angle BAO2 is 150°
and the distance RBC  80 mm . The input link 2 is vertical and the angular velocity is
2  10 rad/s cw . (i) Show the locations of all instant centers. (ii) Using instant centers,
determine the first-order kinematic coefficients of link 3, rack 4, and pinion 5. (iii)
Determine the angular velocity of link 3, the velocity of rack 4, and the angular velocity
of pinion 5.
(ii) The first-order kinematic coefficients are
RI I
50 mm
3  23 12 
 0.385 rad/rad
RI23 I13 130 mm
r4  RI24 I12  29 mm/rad
5 
RI25 I12
RI25 I15

146 mm
 0.462 rad/rad
316 mm
Ans.
Ans.
Ans.
(iii) The requested velocities are
3  32   0.385 rad/rad  10 rad/s   3.85 rad/s (ccw)
Ans.
V4  r42   29 mm/rad  10rad/s   290 mm/s    90 
Ans.
4  42   0.462 rad/rad  10 rad/s   4.62 rad/s (cw)
Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
3.49
Uicker et al.
For the mechanism illustrated in Fig. P2.16, the dimensions are RBA  177 mm and
RBC  150 mm . The radius of gear 2 is 2  25 mm and the radius of gear 5 is
5  50 mm . In the position indicated, the angular velocity 2  5 rad/s ccw .
Determine the first-order kinematic coefficients of links 3, 4, and 5. Find the angular
velocities of links 3, 4, and 5.
The two scalar loop closure equations are
RO5O2  5 cos 5  RBC cos  4  RBA cos 3  2 cos  2  0
5 sin 5  RBC sin  4  RBA sin 3  2 sin  2  0
with the rolling contact constraint equation 55   2 2 .
At the position 2  90 with the given dimensions the solution is 3  45, 4  90, 5  0.
The derivatives of these equations with respect to input  2 are
 5 sin 55  RBC sin  4 4  RBA sin 33   2 sin  2  0
5 cos 55  RBC cos  4 4  RBA cos 33  2 cos  2  0
with the constraint 55   2 .
In matrix form, these appear as
 RBA sin 3  RBC sin  4  3   5 sin 55  2 sin  2    2  sin 5  sin  2  



  R cos 
RBC cos  4   4    5 cos 55  2 cos  2    2  cos 5  cos  2  
3
 BA
The determinant is   RBA RBC sin 3  4   18 750  m2 .
At 2  90 with the given dimensions the first-order kinematic coefficients are
3  2 RBC sin  (4  5   sin 4  2    0.200 rad/rad
Ans.
4  2 RBC sin  (3  5   sin 3  2    0
Ans.
5   2 5  0.500 rad/rad
Ans.
The angular velocities are
3  32  0.200 rad/rad  5 rad/s   1.00 rad/s (cw)
4  42  0
5  52  0.500 rad/rad  5 rad/s   2.50 rad/s (cw)
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
3.50
Uicker et al.
For the mechanism illustrated in Fig. P2.17, the radius of wheel 3 is 3  15 mm and the
other dimensions are RO2O5  140 mm , RBA  110 mm and RAO5  52 mm . For the given
position, link 4 is parallel to the X-axis and link 5 is coincident with the Y-axis. Also, the
input link 2 has an angular velocity of 2  15 rad/s cw . Determine the first-order
kinematic coefficients for links 3, 4, and 5. Find the angular velocities of links 3, 4, and
5.
The two scalar loop closure equations are
RO2O5  RBO2 cos  2  RBA cos  4  RAO5 cos 5  0
RBO2 sin  2  RBA sin  4  RAO5 sin 5  0
with the rolling contact constraint equation 3  3  2    1  1  2  , which,
since 1  0 , reduces to 33   1  3  2 .
At the position 2  120 with RBO2  60 mm, 1  45 mm, and the given dimensions,
the solution is 4  0, 5  90.
The derivatives of the loop-closure equations with respect to input  2 are
 RBO2 sin  2  RBA sin  4 4  RAO5 sin 55  0
RBO2 cos  2  RBA cos  4 4  RAO5 cos 55  0
and the constraint equation derivative gives 33   1  3  .
In matrix form, these appear as
RAO5 sin 5   4   RBO2 sin  2 
 RBA sin  4

   

  RBA cos  4  RAO5 cos 5  5    RBO2 cos  2 
The determinant is   RBA RAO5 sin 5  4   5 720  m2 .
At 2  120 with the given dimensions the first-order kinematic coefficients are
3   1  3  3  4.000 rad/rad
Ans.
4  RBO RAO sin 5  2    0.273 rad/rad
Ans.
5  RBA RBO sin 2  4    1.000 rad/rad
Ans.
2
5
2
The angular velocities are
3  32  4.000 rad/rad  15 rad/s   60.00 rad/s (cw)
Ans.
4  42  0.273 rad/rad  15 rad/s   4.09 rad/s (ccw)
Ans.
5  52  1.000 rad/rad  15 rad/s   15.00 rad/s (cw)
Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
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 meters and t is in seconds. Find the acceleration of the point at t  3 s.


R  t    4  t 2  ˆi
R  t   4t  t 3 3 ˆi  10ˆj
R  t   2tˆi
4.2
R  3 s   2  3 ˆi  6ˆi m/s2
Ans.
Find the acceleration at t  2 s of a point that moves according to the equation

  
R = t 2  t 3 6 ˆi  t 3 3 ˆj . The units are mm and seconds.

  
R  t  =  2t  t 2 2  ˆi  t 2ˆj
R  t  = t 2  t 3 6 ˆi  t 3 3 ˆj
R  t  =  2  t  ˆi  2tˆj
R  2 s  =  2  2  ˆi  2  2  ˆj  4ˆj mm/s2
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
4.3
Uicker et al.
The path of a point is described by the equation R = (t 2  4)e j t /10 where R is in
metre and t is in seconds. For t = 15 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 j t /10
j 2
(t  4)e j t /10
10
j t   j t /10 j t  j t /10  2 2

e


R t  =  2 
(t  4)e j t /10
e
5 
5
100

 j t /10
 j1.5
e
 j , we find that
Noticing, at t = 15 s, that e
2
R 15 s  = (15  4)  229 m
R  t  = 2te j t /10 
j
(152  4) j  71.94  30.00 j  77.95 m/s22.6
10
2
j 15 
j 15

(152  4) j  18.850  j 20.601 m/s 2
R  20 s  =  2 
j

j


5 
5
100

From the direction of the velocity we find the unit tangent and unit normal vectors
uˆ t  1.022.6  cos  22.6 ˆi  sin  22.6 ˆj  0.92296ˆi  0.38489ˆj
uˆ n  kˆ  uˆ t   sin  22.6 ˆi  cos  22.6 ˆj  0.38489ˆi  0.92296ˆj
R  20 s  = 2 15 j 
Ans.
From these, the components of the point’s absolute acceleration are
An  uˆ n R  0.38489ˆi  0.92296ˆj 18.850ˆi  20.601ˆj ft/s 2  26.269 m/s2
Ans.
At  uˆ t
Ans.



R   0.92296ˆi  0.38489ˆj 18.850ˆi  20.601ˆj ft/s   9.468 m/s
2
2
Then, from Eq. (4.2) or Eq. (4.14), the radius of curvature is

R
2
 77.95 m/s 
2

 231.3 m
Ans.
An
26.269 m/s2
where the negative sign indicates that the center of curvature is in the negative uˆ n
direction from the point.
4.4
The motion of a point is described by the equations x  4t cos  t 3 and y   t 3 6  sin 2 t
where x and y are in meters and t is in seconds. Find the acceleration of the point at
t  1.26 s .
R  t    4t cos  t 3  ˆi   t 3 6  sin 2 t  ˆj
R  t   4 cos  t 3  12 t 3 sin  t 3  ˆi   t 2 2  sin 2 t +  t 3 3 cos 2 t  ˆj
R  t   48 t 2 sin  t 3  36 2t 5 cos  t 3  ˆi  t 1  2 2t 2 3 sin 2 t +2 t 2 cos 2 t  ˆj
R 1.26 s   1 128.379ˆi  10.626ˆj  1 128.429 m/s2180.54
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
4.5
Uicker et al.
Link 2 in Fig. P4.5 has an angular velocity of 2  120 rad/s ccw and an angular
acceleration of 4 800 rad/s2 ccw at the instant indicated. Determine the absolute
acceleration of point A.
RAO2  500 mm
A A  AO2  A nAO2  AtAO2
 22 R AO2   2kˆ  R AO2

 
 
  120 rad/s  500 mmˆi mm  4 800 rad/s 2kˆ  500ˆi mm
2
A A  7200 mˆi  2400 mˆj m/s2  7589.5 m/s2161.6
4.6

Ans.
Link 2 is rotating clockwise as illustrated in Fig. P4.6. Find its angular velocity and
acceleration and the acceleration of its midpoint C.
RBA  500 mm
A B  A A  AnBA  AtBA
Construct the acceleration polygon.
2  
n
ABA
178.4 m/s 2

 18.9 rad/s cw
RBA
0.5 m
Ans.
Note the ambiguous sign of this square root. The sense of  cannot be determined from
the accelerations, but here is found from the problem statement.
At
51 m/s 2
 2  BA 
 102 rad/s2 cw
Ans.
0.5 m
RBA
AC  92.76m/s244.0
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
4.7
Uicker et al.
For the data given in Fig. P4.7, find the velocity and acceleration of points B and C.
RBA  400 mm, RCA  250 mm, RCB  200 mm
VB  VA  VBA
VBA  2 RBA   24 rad/s  400 mm   9600 mm/s
Construct the velocity polygon.
VB  3600 mm/s270
VC  2510 mm/s12.1
Ans.
Ans.
A B  A A  AnBA  AtBA
n
ABA
 22 RBA   24 rad/s   400 mm   230.4m/s2
2
t
ABA
  2 RBA  160 rad/s2   400 mm   63.99 m/s2
Construct the acceleration polygon.
A B  118.53 m/s2165
Ans.
AC  63.06 m/s2240.3
Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
4.8
Uicker et al.
For the straight-line mechanism illustrated in Fig. P4.8, 2  20 rad/s cw and
 2  140 rad/s2 cw . Determine the velocity and acceleration of point B and the angular
acceleration of link 3.
RAO2  RCA  RBA  100 mm
VA  VO2  VAO2
VAO2  2 RAO2   20 rad/s 100 mm   2000 mm/s
VC  VA  VCA
Construct the velocity image of link 3.
VB  3864mm/s270
A A  AO2  A
n
AO2
A
Ans.
t
AO2
n
AAO
 22 RAO2   20 rad/s  100 mm   40,000 mm/s2
2
2
t
AAO
  2 RAO2  140 rad/s2  100 m   14,000 mm/s2
2
n
t
AC  A A  ACA
 ACA
n
2
ACA
 VCA
RCA   2000 mm/s 
2
100 mm   40,000.0 mm/s2
Construct the acceleration image of link 3.
A B  6340 mm/s290
3 
t
ABA
14000 mm/s 2

 140 rad/s 2 cw
RBA
100 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
4.9
Uicker et al.
In Fig. P4.8, the slider 4 is moving to the left with a constant velocity of 200 mm/s. Find
the angular velocity and angular acceleration of link 2.
VA  VC  VAC  VO2  VAO2 .
Construct the velocity image of link 3.
V
386.4 mm/s
 3.864 rad/s ccw
2  AO2 
100 mm
RAO2
Ans.
A A  AC  AnAC  AtAC  AO2  AnAO  AtAO
2
n
AC
A
V
2
AC
2
/ RAC   386.4 mm/s  / 100 mm   1493 mm/s2
2
2
n
AAO
 VAO
/ RAO2   386.4 mm/s  / 100 mm   1493 mm/s 2
2
2
2
2 
4.10
t
AAO
2
RAO2

5572 mm/s 2
 55.72 rad/s 2 cw
100 mm
Ans.
Solve Problem 3.8 using constant input velocity, for the acceleration of point A and the
angular acceleration of link 3.
A A  A B  AnAB  AtAB
 366 mm/s   1339.5 mm/s2
V2
 AB 
RAB
100 mm
2
n
AB
A
A A  75.78 in/s215
3 
t
AB
Ans.
2
A
1339.5mm/s

 13.39 rad/s 2 cw
100 mm
RAB
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
4.11
Uicker et al.
For Problem 3.9, using constant input velocity, find the angular accelerations of links 3
and 4.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2   45 rad/s  100 mm   202 500 mm/s 2
2
2
2
A B  A A  AnBA  AtBA  AO4  AnBO  AtBO
4
4
A  V / RBA   358.5 mm/s  /  250.0 mm   514.09 mm/s2 (Ignore compared to other
n
BA
2
BA
2
parts.)
2
n
2
ABO
 VBO
/ RBO4   4 619 mm/s  /  300 mm   71 117 mm/s2
4
4
3 
4 
t
ABA
140 818 mm/s 2

 563.3 rad/s 2 ccw
RBA
250 mm
t
ABO
4
RBO4
37 103 mm/s 2

 123.7 rad/s 2 ccw
300 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
4.12
Uicker et al.
For Problem 3.10, using constant input velocity, find the acceleration of point C and the
angular accelerations of links 3 and 4.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2   60 rad/s  150 mm   21 600 in/s2
2
2
2
A B  A A  AnBA  AtBA  AO4  AnBO  AtBO
4
4
A  V / RBA  13,020 mm/s  /  300 mm   540,000 mm/s 2
n
BA
2
2
BA
n
2
ABO
 VBO
/ RBO4  11,360 mm/s  /  300 mm   430,75 mm/s2
4
4
2
Construct the acceleration image of link 3.
AC  209,615 mm/s210.6
3 
4 
t
BA
A
136,500 mm/s

 455.0 rad/s 2 ccw
300 mm
RBA
t
ABO
4
RBO4
Ans.
2

46,050 mm/s 2
 153.5 rad/s 2 cw
300 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
4.13
Uicker et al.
For Problem 3.11, using constant input velocity, find the acceleration of point C and the
angular accelerations of links 3 and 4.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2   48 rad/s   200 mm   460 800 mm/s 2
2
2
2
A B  A A  AnBA  AtBA  AO4  AnBO  AtBO
4
4
A  V / RBA   268 mm/s  / 800 mm   89.78 mm/s 2 (Ignore compared to other parts.)
n
BA
2
2
BA
2
n
ABO
 VBO
/ RBO4   9 365 mm/s  /  400 mm   219 258 mm/s 2
4
4
2
Construct the acceleration image of link 4.
AC  931 350 mm/s2114.4
3 
4 
t
BA
A
1 393 250 mm/s

 1 741.6 rad/s 2 ccw
RBA
800 mm
t
ABO
4
RBO4
Ans.
2

1 222 300 mm/s 2
 3 055.8 rad/s 2 ccw
400 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
4.14
Uicker et al.
Using the data of Problem 3.13 and assuming constant input velocity, solve for the
accelerations of points C and D and the angular acceleration of link 4.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2  1 rad/s   300 mm   300 mm/s 2
2
2
2
A B  A A  AnBA  AtBA  AO4  AnBO  AtBO
4
4
A  V / RBA   263.5 mm/s  / 150 mm   462.87 mm/s2
n
BA
2
2
BA
2
n
ABO
 VBO
/ RBO4   227 mm/s  /  300 mm   171.77 mm/s 2
4
4
2
Construct the acceleration image of link 3.
AC  515.4 mm/s2  119.8
Ans.
A D  492.5 mm/s 21.8
Ans.
2
4 
t
ABO
4
RBO4
221.2 mm/s 2

 0.737 3 rad/s 2 cw
300 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
4.15
Uicker et al.
For Problem 3.14, using constant input velocity, find the acceleration of point C and the
angular acceleration of link 4.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2   60.0 rad/s  150 mm   540 000 mm/s2
2
2
2
A B  A A  AnBA  AtBA  AO4  AnBO  AtBO
4
4
A  V / RBA   4 235 mm/s  / 150 mm   119 570 mm/s 2
n
BA
2
2
BA
2
n
ABO
 VBO
/ RBO4   7 940 mm/s  /  250.0 mm   252 170 mm/s 2
4
4
2
Construct the acceleration image of link 3.
AC  781 250 mm/s2  68.9
4 
t
ABO
4
RBO4

373 750 mm/s 2
 1 495 rad/s 2 ccw
250.0 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
4.16
Uicker et al.
Solve Problem 3.16, using constant input velocity, for the acceleration of point C and the
angular acceleration of link 4.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2   30.0 rad/s   75 mm   67 500 mm/s 2
2
2
2
A B  A A  AnBA  AtBA  AO4  AnBO  AtBO
4
4
A  V / RBA   2 250.0 mm/s  / 125 mm   40 500 mm/s 2
2
2
BA
n
BA
n
2
ABO
 VBO
/ RBO4   0.0 mm/s  / 125.0 mm   0.0 mm/s2
4
4
2
Construct the acceleration image of link 3.
AC  148 500 mm/s2216.9
4 
t
BO4
A
RBO4

108 000 mm/s 2
 720 rad/s 2 ccw
150 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
4.17
Uicker et al.
For Problem 3.17 using constant input velocity, find the acceleration of point B and the
angular accelerations of links 3 and 6.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2  10.0 rad/s   62.5 mm   6250 mm/s2
2
2
2
A B  A A  AnBA  AtBA
n
2
ABA
 VBA
/ RBA   467.5 mm/s  /  250 mm   874.75 mm/s2
2
A B  5022.5 mm/s20
3 
t
BA
Ans.
2
A
4372.5 mm/s

 17.49 rad/s 2 ccw
250 mm
RBA
Ans.
n
t
Construct the acceleration image of link 3, or AC  A A  ACA
 ACA
 A B  AnCB  AtCB
A D  AC  AnDC  AtDC  AO6  AnDO  AtDO
6
n
DC
A
V
2
DC
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   2436.14 mm/s2
6
6
6 
t
ADO
6
RDO6

1621.9 mm/s 2
 10.81 rad/s 2 cw
150 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
4.18
Uicker et al.
For the data of Problem 3.18, what angular acceleration must be given to link 2 for the
position indicated to make the angular acceleration of link 4 zero?
t
A B  AO4  AnBO  A BO
4
n
BO4
A
V
2
BO4
4
/ RBO4   5760 mm/s  /  400 mm   3 317.8 in/s2
2
A A  A B  AnAB  AtAB  AO2  AnAO  AtAO
2
2
A  V / RAB   5005 mm/s  /  425 mm   58940 mm/s2
n
AB
2
2
AB
n
2
AAO
 VAO
/ RAO2   5600 mm/s  /  350 mm   89600 mm/s2
2
2
2
2 
4.19
t
AAO
2
RAO2

18540 mm/s 2
 52.97 rad/s 2 ccw
350 mm
Ans.
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 indicated?
A B  AO4  AnBO  AtBO
4
n
BO4
A
V
2
BO4
4
/ RBO4   5 070 mm/s  /  200 mm   128 525 mm/s 2
2
t
ABO
  4 RBO4  100 rad/s2   200 in   20 000 mm/s2
4
A A  A B  AnAB  AtAB  AO2  AnAO  AtAO
2
2
A  V / RAB   647 mm/s  /  200 mm   2 093 mm/s 2 (Ignore compared to other parts.)
n
AB
2
AB
2
n
2
AAO
 VAO
/ RAO2   4 500 mm/s  / 125 mm   162 000 mm/s 2
2
2
2
2 
t
AAO
2
RAO2
522 575 mm/s 2

 4 180.6 rad/s 2 ccw
125 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
4.20
Uicker et al.
Solve Problem 3.20 using constant input velocity for the acceleration of point C and the
angular acceleration of link 3.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2  8.0 rad/s  150 mm   9600 mm/s2
2
2
2
A B  A A  AnBA  AtBA  AO4  AnBO  AtBO
4
4
A  V / RBA   3785 mm/s  /  250 mm   57305 mm/s2
n
BA
2
2
BA
2
n
ABO
 VBO
/ RBO4   3483.75 mm/s  /  250 mm   48545 mm/s2
4
4
2
Construct the acceleration image of link 3.
AC  63410 mm/s2  22.9
3 
4.21
t
ABA
15,647.5 mm/s 2

 62.59 rad/s 2 cw
RBA
250 mm
Ans.
Ans.
For Problem 3.21, using constant input velocity, find the acceleration of point C and the
angular acceleration of link 3.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2   56.0 rad/s  150 mm   470400 mm/s 2
2
2
2
A B  A A  AnBA  AtBA  AO4  AnBO  AtBO
4
4
A  V / RBA   6965 mm/s  /  250 mm   194045 mm/s2
n
BA
2
2
BA
2
n
ABO
 VBO
/ RBO4  11380 mm/s  /  250 mm   518017.5 mm/s 2
4
4
2
Construct the acceleration image of link 3.
AC  450600 mm/s2255.6
3 
t
BA
Ans.
2
A
18520 mm/s

 74.08 rad/s 2 cw
250 mm
RBA
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
4.22
Uicker et al.
Find the accelerations of points B and D of Problem 3.22 using constant input velocity.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2   42.0 rad/s   50 mm   88200 mm/s2
2
2
2
A B  A A  AnBA  AtBA
n
2
ABA
 VBA
/ RBA  1066 mm/s  /  250 mm   4547.5 mm/s2
2
A B  2117 in/s20
Ans.
n
t
Construct the acceleration image of link 3, or AC  A A  ACA
 ACA
 A B  AnCB  AtCB
A D  AC  AnDC  AtDC
2
n
ADC
 VDC
/ RDC  1541.5 mm/s  /  200 mm   11880 mm/s2
2
A D  49400 mm/s290
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
4.23
Uicker et al.
Find the accelerations of points B and D of Problem 3.23 using constant input velocity.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2   209.4 rad/s   50 mm   2 192 420 mm/s2
2
2
2
2
n
ABA
 VBA
/ RBA   5469 mm/s  / 150 mm   199 400 mm/s2
A B  A A  AnBA  AtBA
2
A B  732 180 mm/s2240
Ans.
n
t
Construct the acceleration image of link 3, or AC  A A  ACA
 ACA
 A B  AnCB  AtCB
A D  AC  AnDC  AtDC
2
n
ADC
 VDC
/ RDC   6 725 mm/s  / 125 mm   361 805 mm/s2
2
A D  1 209 300 mm/s2120
4.24
Ans.
to 4.30 The nomenclature for this group of problems is illustrated in Fig. P4.24, and the
dimensions and data are given in Table P4.24 to P4.30. For each problem,
determine 3 , 4 , 3 , 4 , 3 , and  4 . The angular velocity  2 is constant for each
problem, and a negative sign is used to indicate the clockwise direction. The dimensions
of even-numbered problems are given in inches and odd-numbered problems 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.
3 , deg  4 , deg
4.24
4.25
4.26
4.27
4.28
4.29
4.30
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
3 , rad/s
4 , rad/s
3 , rad/s2
 4 , rad/s2
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
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
4.31
Uicker et al.
Crank 2 of the system illustrated in Fig. P4.31 has a constant speed of 60 rev/min ccw.
Find the velocity and acceleration of point B and the angular velocity and acceleration of
link 4.
RO4O2  300 mm, RAO2  175 mm, RBO4  700 mm

rev 
rad  1 min 
2   60
 2

  6.283 rad/s
 min  rev  60 s 
VA2  VO2  VA2O2  VA4  VA2 / 4
VA2O2  2 RA2O2   6.283 rad/s 175 mm   1 099.5 mm/s
Construct the velocity polygon.
V
910.75 mm/s
4  A4O4 
 6.571 rad/s cw
138.6 mm
RA4O4
Ans.
VB  VO4  VBO4
VBO4  4 RBO4   6.571 rad/s  700 mm   4 600 mm/s
VB  4 600 mm/s  19.1
Since we know the path of A2 on link 4, we write
A A2  AO2  ΑnA2O2  Α At 2O2  A A4  AcA2 A4  AnA2 / 4  AtA2 / 4 ; A A4  AO4  ΑnA4O4  ΑtA4O4
Ans.
AAn2O2  22 RA2O2   6.283 rad/s  175 mm   6 908.3 mm/s2
2
AcA2 A4  24 × VA2 /4  2  6.571 rad/s  616 mm/s   8 095.5 mm/s219.1
n
A2 /4
A

VA22 /4
 A /4
2
 616 mm/s 


2
0
Construct the acceleration polygon.
AAt 4O4 11 970 mm/s 2
4 

 86.36 rad/s 2 ccw
138.6 mm
RA4O4
Construct the acceleration image of link 4.
A B  67 568 mm/s2187.5
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
4.32
Uicker et al.
Determine the acceleration of link 4 of Problem 3.26 assuming constant input velocity.
Since we know the path of A2 on link 4, we write
A A2  AO2  ΑnA2O2  Α At 2O2  A A4  AcA2 A4  AnA2 / 4  AtA2 / 4
AAn2O2  22 RA2O2   36.0 rad/s   250 mm   324000 mm/s2
2
AcA4 A2  24 × VA2 /4  2  0.0 rad/s  6587.5 mm/s   0 ;
n
A4 /2
A

VA22 /4
 A /4
2
 6587.5 mm/s 


2
0
Next we construct the acceleration polygon.
A A4  2905000 mm/s2180.0
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
4.33
Uicker et al.
For Problem 3.27 using constant input velocity, find the acceleration of point E.
t
A A  AO2  AnAO  A AO
2
n
AO2
A
2
  RAO2   72.0 rad/s   37.5 mm   194400 mm/s2
2
2
2
A B  A A  AnBA  AtBA  AO4  AnBO  AtBO
4
4
A  V / RBA   3780 mm/s  /  262.5 mm   54432.5 mm/s2
2
2
BA
n
BA
2
n
ABO
 VBO
/ RBO4  1800 mm/s  / 125 mm   25920 mm/s 2
4
4
2
Construct the acceleration image of link 3.
Since we know the path of C3 on link 6, we write
AC3  AC6  ACc 3C6  ACn 3 / 6  ACt 3 / 6 and AC6  AO6  ΑCn 6O6  ΑCt 6O6
ACc 3C6  26 × VC3 /6  2  9.673 rad/s 1253.75 mm/s   24254 mm/s 2
n
C3 /6
A

VC23 /6
C /6
3
n
C6O6
A

2
C6O6
V
RC6O6
1253.75 mm/s 

2
0

1046.75 mm/s 

111.325 mm
2
 10415 mm/s 2
Construct the acceleration image of link 6.
A E  180822.5 mm/s2252.8
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
4.34
Uicker et al.
Find the acceleration of point B and the angular acceleration of link 4 of Problem 3.24
using constant input velocity.
n
AAO
 22 RAO2   24.0 rad/s   200 mm   115200 mm/s2
2
2
Since we know the path of P3 on link 4, we write
A P3  A A3  ΑnP3 A3  ΑtP3 A3  A P4  AcP3P4  AnP3 / 4  AtP3 / 4
APn3 A3  VP23 A3 / RP3 A3   4045 mm/s  /  656.75 mm   24915 mm/s2
2
AcP3P4  24 × VP3 /4  2  6.159 rad/s  2592.5 mm/s   31937.5 in/s2  102.4
n
P3 /4
A

VP23 /4
 P /4
3
 2592.5 mm/s 


2
0
Construct the acceleration image of link 3.
A B  123765 mm/s2  25.7
Since links 3 and 4 remain perpendicular,
APt 3 A3 30070 mm/s 2

 45.78 rad/s 2 ccw
 4  3 
RP3 A3
656.75 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
4.35
Uicker et al.
For Problem 3.25, using constant input velocity, find the acceleration of point B and the
angular acceleration of link 3.
Since we know the path of P3 on link 4, we write
A P3  A A3  ΑnP3 A3  ΑtP3 A3  A P4  AcP3P4  AnP3 / 4  AtP3 / 4
APn3 A3  VP23 A3 / RP3 A3   68.375 mm/s  /  225 mm   20.775 mm/s2
2
AcP3P4  24 × VP3 /4  2  0.3039 rad/s  307.75 mm/s   187 mm/s 2 102.8
n
P3 /4
A

VP23 /4
 P /4
 307.75 mm/s 


3
2
0
Construct the acceleration image of link 3.
A B  505.75 mm/s2  102.1
3 
APt 3 A3
RP3 A3

280.5 mm/s 2
 1.247 rad/s 2 cw
225 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
4.36
Uicker et al.
Solve Problem 3.31 for the accelerations of points A and B assuming constant input
velocity.
Since we know the path of F2 on link 3, we write
A F2  A E2  ΑnF2 E2  Α Ft 2 E2  A F3  AcF2 F3  AnF2 / 3  AtF2 / 3 and A F3  AG3  AnF3G3  AtF3G3
AFn2 E2  22 RF2 E2   25.0 rad/s   25 mm   15625 mm/s 2
2
n
F3G3
A

VF23G3
RF3G3
102.75 mm/s 

152 mm
2
 69.45 mm/s 2
AcF2 F3  23 × VF2 /3  2  0.676 rad/s  616.5 mm/s   833.5 mm/s2  80.5
n
F2 /3
A

VF22 /3
 F /3
 616.5 mm/s 


2
2
0
Construct the acceleration image of link 3.
A A  AC  AnAC  AtAC
A B  A D  AnBD  AtBD
and
n
2
AAC
 VAC
/ RAC  195.175 mm/s  / 150 mm   254 mm/s2
2
A A  169.2 in/s20
n
BD
A
V
2
BD
Ans.
/ RBD  190.55 mm/s  / 150 mm   242 mm/s
2
2
A B  20152.5 mm/s20
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
4.37
Uicker et al.
For Problem 3.32, 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  AnBA  AtBA
n
ABA
 22 RBA  10.0 rad/s   75 mm   7500 mm/s2
2
t
ABA
  2 RBA   2.0 rad/s2   75 mm   150 mm/s 2
A D4  ArD4 /1  A B  AnDB  AtDB
2
n
ADB
 VDB
/ RDB   750 mm/s  /  25 mm   22500 mm/s 2
2
Draw the acceleration image of link 4.
AC4  15002.5 mm/s291.1
AC3  AC4  A
r
C3 / 4
 AA  A  A
n
CA
Ans.
t
CA
2
n
ACA
 VCA
/ RCA  1500 mm/s  /  50 mm   45000 mm/s 2
2
Draw the acceleration image of link 3.
ACt A 300 mm/s 2
3  3 3 
 6.0 rad/s 2 ccw
RC3 A3
50 m
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
4.38
Uicker et al.
Determine the angular accelerations of links 3 and 4 of Problem 3.29 assuming constant
input velocity.
Since we know the path of D4 on link 2, we write
A D2  A A2  AnD2 A2  A Dt 2 A2
ADn2 A2  22 RD2 A2  15.0 rad/s   31.25 mm   7031.25 mm/s2
2
A D4  A D2  AcD4 D2  AnD4 / 2  AtD4 / 2  A E4  AnD4 E4  AtD4 E4
AcD4 D2  22 × VD4 /2  2 15.0 rad/s 1235.25 mm/s   37057.5 mm/s280.8
ADn4 /2  VD24 /2 D4 /2  1235.25 mm/s 
ADn4 E4  VD24 E4 RD4 E4
 62.5 mm   24414.5 mm/s2
2
  381 mm/s  87.5 mm   1659.75 mm/s2
2
Draw the acceleration images of links 2 and 4.
ADt 4 E4 8810 mm/s 2
4 

 100.7 rad/s 2 ccw
87.5 mm
RD4 E4
Ans.
AC3  AC2  ACr 3 / 2  A D3  ACn 3D3  ACt 3D3
ACn3D3  VC23D3 RC3D3  1047.75 mm/s 
3 
ACt 3 D3
RC3 D3

2
12.5 mm  87825 mm/s2
5792.5 mm/s 2
 463.4 rad/s 2 ccw
12.5 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
4.39
Uicker et al.
For Problem 3.30 using constant input velocity, determine the acceleration of point G and
the angular accelerations of links 5 and 6.
A B  A A  AnBA  A BAt
n
ABA
 22 RBA  10 rad/s   25 mm   2500 mm/s2
2
n
t
n
t
AC  A B  ACB
 ACB
 A D  ACD
 ACD
n
2
ACB
 VCB
/ RCB   333.25 mm/s  / 100 mm   1111 mm/s 2
2
2
n
ACD
 VCD
/ RCD  166.5 mm/s  /  50 mm   555.5 mm/s2
2
Construct the acceleration image of link 3.
Since we know the path of E3 on link 6, we write
A E3  A E6  AcE3E6  AnE3 /6  AtE3 /6 and A E6  A H6  ΑnE6 H6  ΑtE6 H6
AcE3E6  26 × VE3 /6  2  3.774 rad/s  272.25 mm/s   2055.75 mm/s2104.5
AEn3 /6  VE23 /6 /  E3 /6   292.75 mm/s  /   0
2
AEn6 H6  VE26 H6 RE6 H6  121.5 mm/s  32.25 mm  458.25 mm/s2
2
Construct the acceleration image of link 6.
AG  8785 mm/s2  64.5
6 
t
E6 H 6
A
RE6 H6

3547.5 mm/s 2
 110.2 rad/s 2 cw
32.25 in
Ans.
Ans.
A F5  A F6  ArF5 / 6  A E5  AnF5 E5  AtF5 E5
AFn5 E5  VF25 E5 RF5 E5   319.5 mm/s 
5 
AFt 5 E5
RF5 E5
2
12.5 mm  8167.5 mm/s2
0 mm/s 2

0
12.5 mm
© Oxford University Press 2015. All rights reserved.
Ans
Theory of Machines and Mechanisms, 4e
4.40
Uicker et al.
Continue with Problem 3.40 and find the second-order kinematic coefficients of links 3
and 4. Assuming an input acceleration of AA2  18750 mm/s2 find the angular
accelerations of links 3 and 4.
RBO4  RBA  150 mm
From the solution of Problem 3.40 we have the derivative of the loop-closure equations
with respect to the input r2. In matrix form this is
 r3 sin 3 r4 sin  4  3  1 
 r cos  r cos       0
 
3
4
4 4
3
From these we found the determinant of the Jacobian and the first-order kinematic
derivatives. At the position shown these are   r3r4 sin 4  3   19485.625 mm2 ,
3  r4 cos4   3.849 103 rad/mm , and 4  r3 cos3   3.849 103 rad/mm .
The next derivative of the above equations with respect to input r2, in matrix form, gives
 r3 sin 3 r4 sin  4  3  r3 cos 332  r4 cos  4 42 

   
2
2 
 r3 cos 3 r4 cos  4   4  r3 sin 33  r4 sin  4 4 
The solution to this set of equations is
3 1  r4 cos  4
     r cos 
3
 4
 3

r4 sin  4   r3 cos 3  r4 cos  4  32 
 
r3 sin 3   r3 sin 3  r4 sin  4   42 
 32 
r42
1  r3r4 cos 3   4 

 
r32
r3r4 cos 3   4    42 

For the specified position these give values of 3  8.5536 106 rad/mm2
Ans.
and 4  8.5536 10 rad/mm .
From these we find angular accelerations of
3  3r2  3r22  2 934.9 rad/s2 (cw)
Ans.
6
and
2
4  4r2   r  2 934.9 rad/s ccw
2
4 2
2
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
4.41
Uicker et al.
Continue with Problem 3.49 and find the second-order kinematic coefficients of links 3,
4, and 5. Assuming constant angular velocity for link 2 find the angular accelerations of
links 3, 4, and 5.
From the solution of Problem 3.49 we have the
derivative of the loop-closure equations with respect to
the input  2 . In matrix form this is
 RBA sin 3  RBC sin  4  3    2  sin 5  sin  2  


  R cos 
RBC cos  4   4   2  cos 5  cos  2  
3
 BA
with the constraint 55   2 and the determinant
  RBA RBC sin 3  4  .
At the position shown these give values of
  18 750  m2
3  2 RBC sin  (4  5   sin 4  2    0.200 rad/rad
4  2 RBC sin  (3  5   sin 3  2    0
5   2 5  0.500 rad/rad
The next derivative of these equations gives
 RBA sin 3
  R cos 
3
 BA
 RBC sin  4  3   RBA cos 3

RBC cos  4   4   RBA sin 3
RBC cos  4  32 
 
RBC sin  4   42 
   cos 5   2 cos  2  5 
 2
 
   2 sin 5   2 sin  2   1 
with the derivative of the constraint giving 5  0 . The solution of the above equations
gives
2
 32 
RBC
3 1   RBA RBC cos 3   4 
 
    
2
 RBA
RBA RBC cos 3   4    42 
 4

  R cos  4  5  RBC cos  2   4   5 
 2  BC

  RBA cos 3  5  RBA cos  2  3    1 
At the position shown the values of the second-order kinematic derivatives are
3  0.240 rad/rad 2 , 4  0.150 rad/rad 2 , and 5  0 .
Ans.
With 2  5 rad/s  const , the requested angular accelerations are
3  322  6.0 rad/s2 ccw ,
 4  422  3.75 rad/s2 ccw ,
© Oxford University Press 2015. All rights reserved.
5  522  0 .
Ans.
Theory of Machines and Mechanisms, 4e
4.42
Uicker et al.
Continue with Problem 3.50 and find the second-order kinematic coefficients of links 3,
4, and 5. Assuming constant angular velocity for link 2 find the angular accelerations of
links 3, 4, and 5.
From the solution of Problem 3.50 we have the derivative of the loop-closure equations
with respect to the input  2 . In matrix form this is
RAO5 sin 5   4   RBO2 sin  2 
 RBA sin  4

   

  RBA cos  4  RAO5 cos 5  5    RBO2 cos  2 
with the constraint 33   1  3  and the determinant   RBA RAO5 sin 5  4  .
At the position shown these give values of   5 720  m2
3   1  3  3  4.000 rad/rad
4  RBO RAO sin 5  2    0.273 rad/rad
2
5
5  RBA RBO sin  (2  4    1.000 rad/rad
2
The next derivative of these equations gives
RAO5 sin 5   4   RBA cos  4  RAO5 cos 5   42   RBO2 cos  2 
 RBA sin  4

   
 2  

  RBA cos  4  RAO5 cos 5  5   RBA sin  4  RAO5 sin 5  5   RBO2 sin  2 
with the derivative of the constraint giving 3  0 . The solution of the above equations
gives
2
  42  1  RAO5 RBO2 cos  2  5  
RAO
 4 1  RBA RAO5 cos  4  5 
5
  

    
2
 RBA RAO5 cos  4  5   52    RBA RBO2 cos  2   4  
RBA
 5

At the position shown the values of the second-order kinematic derivatives are
3  0 , 4  0.000 35 rad/rad 2 , and 5  0.420 rad/rad2 .
Ans.
With 2  15 rad/s  const , the requested angular accelerations are
3  322  0 ,
 4  422  0.078 8 rad/s2 ccw ,
5  522  94.41 rad/s2 (cw) . Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
4.43
Uicker et al.
Find the inflection circle for motion of the coupler of the double-slider mechanism
illustrated in Fig. P4.43. 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.
RBA  125 mm
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
4.44
Uicker et al.
Find the inflection circle for motion of the coupler relative to the frame of the linkage
illustrated in Fig. P4.44. 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.
RcA  62.5 mm, RAO2  22.5 mm, RBO4  87.5 mm, RPO4  29.25 mm
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.
Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
4.45
Uicker et al.
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 nearby portions of their paths.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
4.46
Uicker et al.
The planar four-bar linkage illustrated in Fig. P4.46 has link dimensions RO4O2  50 mm ,
RAO2  20 mm , RBA  63 mm , and RBO4  30 mm . For the position indicated, link 2 is
30 counterclockwise from the ground link O2O4 and the angular velocity and angular
acceleration of the coupler link AB are 3  5 rad/s ccw and 3  20 rad/s2 cw ,
respectively. For the instantaneous motion of the coupler link AB show: (a) the velocity
pole I, the pole tangent T, and the pole normal N; (b) the inflection circle and the Bresse
circle; (c) the acceleration center of the coupler link AB. Then determine; (d) the radius
of curvature of the path of coupler point C where RCB  25 mm , (e) the magnitude and
direction of the velocity of coupler point C, (f) the magnitude and direction of the angular
velocity of link 2, (g) the magnitude and direction of the velocity of the pole, (h) the
magnitude and direction of the acceleration of coupler point C, and (i) the magnitude and
direction of the acceleration of the velocity pole.
(a) The pole I is coincident with the instant center I13 shown in the figure below. The
instant center I 24 and the collineation axis are as shown in the figure. From Bobillier's
theorem, the angle from the collineation axis to the first ray (say link 2) is measured as
84° cw. This is equal to the angle from the second ray (link 4) to the pole tangent T; that
is, 84° cw. Therefore, the pole tangent T is as shown in the figure and the pole normal N,
which is perpendicular to the pole tangent T, is also shown.
(b) The inflection point J A for point A on link 3 can be obtained from the Euler-Savary
equation; that is,
13.8 mm   9.5 mm
R2
 AI 
20 mm
RAOA
2
RAJ A
The location of the inflection point J A is shown in the figure.
Similarly, the inflection point J B for point B on link 3 can be obtained from the EulerSavary equation; that is,
 56.2 mm   105.3 mm
R2
 BI 
30 mm
RBOB
2
RBJ B
The location of the inflection point J B is shown on the figure.
Knowing the pole normal and the two inflection points, the inflection circle can be
drawn. The inflection circle for the motion of link 3 with respect to 1, the inflection pole
J, and the center of the inflection circle (denoted as point O) are shown on the figure.
Note that the pole normal N points from the pole I toward the inflection pole J and the
pole tangent T is 90 clockwise from the pole normal. The diameter of the inflection
circle for the motion 3/1 is measured as
RJI  49.2 mm
Ans.
The diameter of the Bresse circle is
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
 5 rad/s  49.2 mm  61.5 mm
32
b
RJI 
20 rad/s 2
3
2
Ans.
Since the angular acceleration of the coupler link is clockwise (that is, negative), the
Bresse circle must lie on the positive side of the pole tangent, as shown on the figure.
(c) The point of intersection of the inflection circle and the Bresse circle (other than pole I) is the
acceleration center  of the coupler link; see the figure.
(d) From the Euler-Savary equation, the radius of curvature of the coupler point C is
 37.4 mm   16.4 mm
RCI2


85.2 mm
RCJC
2
C  RCO
C
Ans.
The location of the center of curvature of the path of point C (that is, OC ) is as shown on
the figure.
(e) The velocity of coupler point C is
VC  3 RCI   5 rad/s  37.4 mm   187 mm/s
Ans.
The direction of the velocity vector of point C is as shown on the figure.
(f) The angular velocity of link 2 can be written as
RI I
13.7 mm
5 rad/s  3.43 rad/s (cw)
2  23 13 3 
RI23I12
20 mm
(g) The velocity of the pole I is
v  3 RJI   5 rad/s  49.2 mm  246 mm/s
Ans.
Ans.
Since the angular velocity of the coupler link is positive (counterclockwise) the velocity
of the pole must be negative; that is, in the direction opposite to the pole tangent T (as
shown on the figure).
(h) The acceleration of coupler point C can be written as
AC  RC 34   32
  73.6 mm 
 5 rad/s 
4
  20 rad/s 2 
2
 2 356 mm/s2
Ans.

I
The angle from the line
to the pole normal N is measured as 38.25 ccw, as shown
in the figure. Therefore, the direction of the acceleration vector of point C is 38.25 ccw
from the line connecting  to C, as shown in the figure.
(i) The acceleration of the pole I can be written as
2
AI  v3  32 RJI   5 rad/s   49.2 mm   1 230 mm/s 2
The acceleration of the pole is directed along the pole normal N as shown on the figure.
The angle from the horizontal axis to the pole normal N is measured as 33.0.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
As a check, the acceleration of the pole can be written as
AI  RI  34   32
  38 mm 
 5 rad/s 
4
  20 rad/s 2 
2
=1 217 mm/s2
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Ans.
Theory of Machines and Mechanisms, 4e
4.47
Uicker et al.
Consider the double-slider mechanism in the position given in Problem 3.8. Point B
moves with a constant velocity VB  1000 mm/s to the left as illustrated in the figure.
The angular velocity and angular acceleration of coupler link AB are 3  3.66 rad/s ccw
and 3  13.40 rad/s2 cw , respectively. For the absolute motion of coupler link AB in
the specified position, draw the inflection circle and the Bresse circle. Then determine:
(a) the radius of curvature of the path of point C, which is a point in link 3 midway
between points A and B; and
(b) the magnitude and direction of the velocity of the velocity pole I. Using the
acceleration pole determine:
(c) the magnitude and direction of the acceleration of the pole I;
(d) the magnitude and direction of the velocity of points A and C; and
(e) the magnitude and direction of the acceleration of points A and C.
The pole I is the point coincident with the instant center I13 at the intersection of the
vertical line through point B and the line perpendicular to the direction of motion of
slider 2 through point A. Since point A moves on a straight line, the center of curvature
for point A is at infinity. Hence, the inflection point J A is coincident with point A.
Similarly, the inflection point J B is coincident with point B since point B also moves on
a straight line and the center of curvature of point B is at infinity. Knowing the two
inflection points J A and J B and the pole I, the center O of the inflection circle is
obtained as the intersection of the perpendicular bisectors of IJ A and IJ B as shown in the
figure below. The centrode normal passes through I and O and intersects the inflection
circle at inflection point J. The diameter of the inflection circle is measured as
RJI  386.25 mm .
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Theory of Machines and Mechanisms, 4e
Uicker et al.
We keep in mind that the centrode normal N points from I toward J and the centrode
tangent is 90 clockwise from the centrode normal. The diameter of the Bresse circle is
 3.66 rad/s  (386.25 mm)  1386.25 mm
32
b
RJI 
13.40 rad/s2
3
Since the angular acceleration of link 3 (that is,  3 ) is clockwise (negative), the Bresse
circle must lie on the positive side of the centrode tangent. The Bresse circle is
positioned as shown in the figure.
2
As a check, note that point B, fixed in link 3, moves on a straight line with a constant
velocity. Hence point B must be the acceleration center for the absolute motion of link 3.
With the above construction of the inflection circle and the Bresse circle, the acceleration
center  (the point of intersection of the inflection circle and the Bresse circle)
coincides with point B.
The angle from the line I  to the centrode normal N is measured as 45 ccw . To
check, in Eq. (4.48),
2
1  3
1 13.40 rad/s
 tan
 45 ccw
  tan
2
32
 3.66 rad/s 
(a) The radius of curvature of point C, from the Euler-Savary equation, is
R2
(301.325 mm) 2
C  RCC  CI 
 10939.25 mm
RCJC
8.3 mm
Ans.
Since the center of curvature C  for point C does not lie on the paper, the direction is
indicated by an arrow on the figure.
(b) The magnitude of the velocity of the pole I is
  RJI 3  (386.25 mm)  3.66 rad/s   1413.75 mm/s
Ans.
Since the angular velocity of link 3 is positive (counterclockwise) the pole velocity  is
in the negative pole tangent direction as shown in the figure.
(c) The magnitude of the acceleration of the pole I is
AI  3  (1413.75 mm/s)  3.66 rad/s   51.75 mm/s2
Ans.
The acceleration of the pole points along the positive pole normal as shown in the figure.
(d) The magnitude and direction of the velocity of points A and C are found as follows.
Since point A is fixed in link 3 the velocity of point A is
Ans.
VA  3 RAI   3.66 rad/s  (334.5 mm)  1224.275 mm/s
The direction of the velocity of A is perpendicular to the line RAI as shown in the figure.
Since point C is fixed in link 3 the velocity of point C is
Ans.
VC  3 RCI   3.66 rad/s  (301.25 mm)  1102.575 mm/s
The direction of the velocity of C is perpendicular to the line RCI as shown in the figure.
(e) The magnitude and direction of the acceleration of points A and C are found as
follows:
From Eq. (4.49), the acceleration of point A is given by
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Theory of Machines and Mechanisms, 4e
Uicker et al.
AA  RA 34   32
 100 mm  (3.66 rad/s) 4  (13.40 rad/s 2 ) 2
Ans.
 1894.75 mm/s 2
The angle between the line  I and the centrode normal is 45 ccw. Therefore, the
acceleration of point A is directed at an angle of 45 ccw from the line  A as shown in
the figure.
The acceleration of point C can be written as
AC  RC 34   32
  50 mm  (36.6 rad/s) 4  (1 340 rad/s 2 ) 2
Ans.
 947.5 mm/s 2
The acceleration of point C is at an angle of 45 ccw from line  C as shown in the
figure.
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Theory of Machines and Mechanisms, 4e
4.48
Uicker et al.
For the mechanism of Problem 3.17, the input link 2 is rotating with an angular velocity
For the
2  15 rad/s ccw and an angular acceleration  2  320.93 rad/s2 cw .
instantaneous motion of the connecting rod 3, find:
(a) the inflection circle and the Bresse circle;
(b) the location of the acceleration pole;
(c) the center of curvature of the path traced by the coupler point C fixed in link 3;
(d) the magnitude and direction of the velocity and acceleration of points A, B, and C;
(e) the magnitude and direction of the velocity and acceleration of the inflection pole J.
(a) The velocity pole I for the connecting rod 3 is coincident with the instant center I13 .
Since point B on link 3 travels on a straight line, it is an inflection point; that is, JB
coincides with point B (the inflection circle for the motion 3/1 has to pass through point
B). The inflection point for point A on link 3 can be obtained from the Euler-Savary
equation; that is,
 334.25 mm   1787.5 mm
R2
 AI 
RAOA
62.5 mm
2
RAJ A
The inflection circle is drawn through points I, JA, and JB as shown in the figure below.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
The diameter of the inflection circle is measured as
Ans.
RJI  2213.75 mm
The centrode tangent T and the centrode normal N are also shown in the figure. Note that
the centrode normal passes through the pole I and the center of the inflection circle O and
intersects the inflection circle at the inflection pole J. The centrode normal points from I
to J and the centrode tangent is 90° clockwise from the centrode normal.
In order to draw the Bresse circle, the angular velocity and angular acceleration of link 3
must be known. The method of kinematic coefficients (see Sections 3.12 and 4.12) is
used here to determine 3 and  3 .
The vectors for the slider-crank portion of the mechanism are shown in the figure. The
vector loop equation can be written as
R 2  R3  R 4  R1  0
The X and Y components can be written as
R2 cos 2  R3 cos 3  R4  0
R2 sin 2  R3 sin 3  R1  0
Differentiating these with respect to the input  2 gives
 R2 sin 2  R3 sin 33  R4  0
R2 cos 2  R3 sin 33  0
where 3  d3 d2 and R4  dR4 d2 are the first-order kinematic coefficients of links
3 and 4, respectively. Writing these equations in matrix form gives
  R3 sin 3

 R3 cos 3
1  3   R2 sin  2 
   

0   R4    R2 cos  2 
The determinant of the coefficient matrix is   R3 cos 3 . The length of the input link is
R2  62.5 mm and the length of the coupler link is R3  250 mm. The slider offset is
R1  37.5 mm . For the given input position  2  135o , the coupler angle (found from
trigonometry) is 3  19.07 . Substituting the known data gives
3.267 1  3   1.768 

   

 9.451 0   R4   1.768
Therefore, the first-order kinematic coefficients of link 3 and link 4 are
3  0.187 1 rad/rad and R4  28.925 mm/rad
The angular velocity of link 3 is
3  32  (0.187 1 rad/rad)(15 rad/s)  2.81 rad/s (ccw)
Differentiating the above matrix equation with respect to the input variable  2 gives
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Theory of Machines and Mechanisms, 4e
Uicker et al.
  R3 sin 3 1  3   R2 cos  2  R3 cos 332 


   
2

 R3 cos 3 0   R4   R2 sin  2  R3 sin 33 
Then substituting the numerical data gives
3.267 1  3   1.437 
 9.451 0   R    1.653 

  4  

Using Cramer's rule, the second-order kinematic coefficients of the mechanism are
3  0.175 rad/rad 2 and R4  50.25 mm/rad 2
The angular acceleration of link 3 can be written (see Table 4.2) as
       2
3
3
2
3
2
 (0.187 1 rad/rad)(320.93 rad/s 2 )  (0.175 rad/rad 2 )(15 rad/s)2
 20.67 rad/s 2 (cw)
The negative sign indicates that the angular acceleration of link 3 is clockwise.
The diameter of the Bresse circle for the motion 3/1 can now be written as
2
2.81 rad/s 

32
Ans.
b  RJI
  2213.75 mm 
 845.75 mm
3
20.67 rad/s 2
Since the angular acceleration of link 3 is clockwise (negative) the Bresse circle must lie
on the positive side of the centrode tangent as shown in the figure.
(b) The acceleration center  for the absolute motion of link 3 is the point of
intersection of the inflection circle and the Bresse circle. The angle from the line  I to
the centrode normal N is measured as 69.09 ccw . As a check, from Eq. (4.48), the
angle  is given by
  tan 1
2
3
1 20.67 rad/s

tan
 69.09  ccw 
2
32
 2.81 rad/s 
(c) From the Euler-Savary equation, the radius of curvature of point C can be written as
RCI2
(323.75 mm) 2
C  RCOC 

 86.5 mm
1210.25 mm
RCJC
The center of curvature OC of point C is as shown in the figure.
(d) The velocity of point A is
VA  3 RAI  (2.81 rad/s)(334.25 mm)  937.75 mm/s
As a cross-check, the velocity of point A can also be found as
VA  2 RAO2  (15 rad/s)(62.5 mm)  937.55 mm/s.
The direction of velocity of point A is 135° as shown in the figure.
The velocity of point B is
VB  3 RBI  (2.81 rad/s)(154.5 mm)  433.75 mm/s
The direction of velocity of point B is 180° as shown in the figure.
The velocity of point C is
VC  3 RCI  (2.81 rad/s)(323.75 mm)  908.5 mm/s
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Ans.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
The direction of velocity of point C is 152.39° as shown in the figure.
From Eq. (4.49), the acceleration of point A can be written as
AA  R A 34  32  1109.75 mm
 2.81 rad/s 
4
 (20.67 rad/s 2 )2  24550 mm/s2 Ans.
With   69.09o ccw, the direction of the acceleration of point A is 9.94° as shown in the
figure. The acceleration of point B can be written as
 2.81 rad/s 
AB  R B 34  32  932.5 mm
 (20.67 rad/s 2 )2  20628.5 mm/s2
4
Ans.
The direction of the acceleration of point B is as shown in the figure. The acceleration of
point C can be written as
AC  R C 34  32  1113.5 mm
 2.81 rad/s 
4
 (20.67 rad/s 2 ) 2  24632 mm/s2
Ans.
The direction of the acceleration of point C is 4.79° as shown in the figure.
(e) The velocity of the inflection pole J is the same as the velocity of the pole I.
Therefore, the velocity of the inflection pole J is
Ans.
v  3 RJI   2.81 rad/s  2213.95 in   6212.75 mm/s
The direction of the velocity of inflection pole J is perpendicular to line IJ as shown in
the figure.
The acceleration of the inflection pole J is
AJ  R J 34  32  2068 mm
 2.81 rad/s 
4
 (20.67 rad/s 2 )2  45747.5 mm/s2
Ans.
The acceleration of the inflection pole J makes an angle of   69.09 with the line  J .
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
4.49
Uicker et al.
Figure P3.32 illustrates an epicyclic gear train driven by the arm, link 2, with an angular
velocity 2  3.33 rad/s cw and an angular acceleration  2  15 rad/s2 ccw . Define
point E as a point on the circumference of the planet gear 4 horizontal to the right of
point B such that the angle DBE  90. For the absolute motion of the planet gear 4,
draw the inflection circle and the Bresse circle on a scaled drawing of the epicyclic gear
train. Then determine:
(a) The location of the acceleration center of the planet gear.
(b) The radii of curvature of the paths of points B and E.
(c) The locations of the centers of curvature of the paths of points B and E.
(d) The magnitudes and the directions of the velocities of points B and E and the pole I.
(e) The magnitudes and the directions of the accelerations of points B and E and the pole I.
Since the problem is for the motion 4/1 where 4 is the planet gear which is in internal
rolling contact with the fixed ring gear 1 then the pole I is coincident with the instant
center I14 which is the point of contact between the two gears. Note that the fixed
centrode is gear 1 and the moving centrode is gear 4. Recall that the centrode normal N
points from the fixed centrode toward the moving centrode (in the neighborhood of the
pole I). Therefore, the centrode normal N points vertically downward. Also, recall that
the Euler-Savary equation can be written as
1
1
1


RJI RIOF RIOM
The center of curvature of the fixed centrode OF is coincident with the center of the
fixed gear, that is, point A, and the center of curvature of the moving centrode OM is
coincident with the center of the planet gear; that is, point B. Recall that I with respect to
OF and I with respect to OM are both vertically upward; therefore, the radii of curvature
of the fixed and the moving centrodes, respectively, are
1  RIOF  100 mm and 4  RIOM  25 mm
Substituting into the Euler-Savary equation gives
1
1
1
3



RJI 100 mm 25 mm 100 mm
Therefore, the diameter of the inflection circle is
RJI  100 3  33.33 mm
The inflection circle is shown in the figure below. Recall that the centrode tangent T is
90 clockwise from the centrode normal N and is, therefore, horizontal and is positive to
the left.
As a check: The inflection point for point C is coincident with the inflection pole J;
therefore, the radius of curvature of the path of point C, from the Euler-Savary equation,
is
R2
(50 mm) 2
C  RCOC  CI 
 150 mm
RCJC 16.66 mm
To determine the angular velocity and the angular acceleration of the planet gear 4, the
rolling contact equation between the planet gear 4 and the fixed ring gear 1 (see Chapter
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Theory of Machines and Mechanisms, 4e
Uicker et al.
3, Example 3.9) can be written as

 4   2
100 mm
  1 
 4
1   2
25
4
We use the positive sign here since there is internal rolling contact between the planet
gear 4 and the ring gear 1. Differentiating this equation with respect to the input position,
the rolling contact equation, in terms of first-order kinematic coefficients, is written as

 4   2
  1 4
 

1
2
4
Since the input is the arm (link 2) then  2  1 , and since the ring gear 1 is fixed then
1  0 . Therefore, the first- and second-order kinematic coefficients of the planet gear 4
from the above equation are
4   3 rad/rad
and
 4  0
The angular velocity of the planet gear 4 is
4  4 2  (3 rad/rad)(3.33 rad/s)  10 rad/s (ccw)
The angular acceleration of the planet gear 4 is
 4  4 2  422  (3 rad/rad) 15 rad/s   0  45 rad/s2 (cw)
Therefore, the diameter of the Bresse circle for the motion 4/1 is
2
10 rad/s 

42
b  RJI
  33.33 mm 
 74 mm
4
45 rad/s 2
Since the angular acceleration of planet gear 4 is clockwise (negative), the Bresse circle
lies on the positive side of the centrode tangent T. The Bresse circle is as shown in the
figure.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
(a) The acceleration center  for the absolute motion of planet gear is the intersection
of the inflection circle and the Bresse circle. The acceleration center  is shown in the
figure. The angle from the line I  to the centrode normal N is measured as
  24.23 ccw
Check: The angle  is given by the relation
  tan 1
2
4
1 45 rad/s

 24.23o ccw
tan
2
2
4
10 rad/s 
(b) and (c) The radius of curvature of the path of point B, from the Euler-Savary
equation, is
2
25 mm 

RBI2
Ans.

 75.75 mm
 B  RBOB 
RBJ B
8.25 mm
Note that the center of curvature OB of point B is coincident with point A and the
inflection point J B is coincident with the inflection pole J. The radius of curvature of the
path of point E, from the Euler-Savary equation, is
2
35.25 mm 

REI2
Ans.
 E  REOE 

 105.75 mm
11.75 mm
REJ E
The center of curvature OE of point E is shown on the figure.
(d) The magnitude and direction of the velocity of points B, E and the pole I.
The velocity of point B is
VB  4 RBI  10 rad/s  25 mm   25 mm/s
Ans.
The direction of the velocity of point B is horizontal to the right as shown in the figure.
The velocity of point E is
Ans.
VE  4 REI  10 rad/s  35.25 mm   352.5 mm/s
The direction of the velocity of point E is perpendicular to line IE and is inclined at an
angle of 45° from the horizontal.
The velocity of the pole I can be written as
Ans.
  4 RJI  10 rad/s  33.25 mm   332.5 mm/s
The direction of the velocity of the pole I is along the negative centrode tangent T since
the angular velocity of link 4 is counterclockwise; that is, positive.
(e) The acceleration of point B can be written as
AB  R B 44   42  12.75 mm 
10 rad/s 
4
 (45 rad/s2 )2  1400 mm/s 2
Ans.
With   24.23o ccw the direction of the acceleration of point B is as shown in the figure.
The acceleration of point E can be written as
AE  R E 44   42   37.75 mm 
10 rad/s 
4
 (45 rad/s2 )2  4139.5 mm/s2
Ans.
With   24.23 ccw the direction of the acceleration of point E is as shown in the
figure. The acceleration of the pole I is
Ans.
AI  4  (333.25 mm/s) 10 rad/s   3332.5 mm/s2
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Theory of Machines and Mechanisms, 4e
Uicker et al.
The acceleration of the pole is directed along the positive pole normal as shown in the
figure.
Check: The acceleration of the pole I can also be obtained from the equation
AI  R I 44   42   30.4 mm 
10 rad/s 
4
 (45 rad/s 2 )2  3332.5 mm/s 2
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
4.50
Uicker et al.
On 450 by 600 mm paper, draw the linkage illustrated in Fig. P4.50 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 indicated.
(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.
RAA  25 mm, RBA  125 mm, RBA  43.75 mm, RBB  81.25 mm
Drawn with a precise CAD system above, the circle around center C  matches the
coupler curve near C to 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.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
Chapter 5
Multi-Degree-of-Freedom Planar Linkages
5.1
The slotted links 2 and 3 are driven independently at constant speeds of 2  30 rad/s cw
and 3  20 rad/s cw, respectively. Find the absolute velocity and acceleration of the
center of the pin P4 carried in the two slots.
XB = 100 mm; YB = 25 mm.
Identifying the pin as separate body 4 and, noticing the two paths it travels on bodies 2
and 3, we write
VP2  2 RP2 A   30 rad/s  54.9 mm   1647 mm/s
VP3  3 RP3B   20 rad/s 102.6 mm   2052 mm/s
VP4  VP2  VP4 /2  VP3  VP4 /3
Construct the velocity polygon
VP4  2355 mm/s15.6
A P3  AO3  AnP O  A Pt O
A P2  AO2  AnP O  A Pt O ;
2 2
Ans.
2 2
3 3
3 3
APn2O2  22 RP2O2   30.0 rad/s   54.9 mm   49410 mm/s 2 .
2
APn3O3  32 RP3O3   20.0 rad/s  102.6 mm   41040 mm/s2
2
A P4  A P2  AcP4 P2  AnP4 / 2  AtP4 / 2  A P3  AcP4 P3  AnP4 / 3  AtP4 / 3
AcP4 P2  22 × VP4 /2  2  30.0 rad/s 1683 mm/s   100980 mm/s 2  30.0
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
AcP4 P3  23 × VP4 /3  2  20.0 rad/s 1155 mm/s   46200 mm/s2225.0
n
P4 /2
A

VP24 /2
 P /2
4
1683 mm/s 


2
0;
n
P4 /2
A

VP24 /2
 P /2
1155 mm/s 


4
2
0
Construct the acceleration polygon.
A P4  125730 mm/s2  66.6
Ans.
For comparison, let us now solve the same problem by use of kinematic coefficients. The
loop-closure constraint equations can be written as
r2 cos  2  r3 cos 3  xB  0
r2 sin  2  r3 sin 3  yB  0
Recognizing that  2 and  3 are the two independent degrees of freedom, and that r2 and
r3 are dependent position unknowns. Since these appear linearly (which is not true in
other problems), the loop-closure equations can be written in matrix form as follows:
cos  2  cos 3   r2   X B 

 sin 
 sin 3   r3   YB 
2

For the given position 2  60 and 3  135 , and the dimensions are X B  100 mm and
YB  25 mm. The determinant of this set is    sin 3  2   0.966 , and the solutions
for the two unknown position values are r2  54.9 mm and r3  102.6 mm . The position
coordinates of the center of the pin are
xP  r2 cos  2  xB  r3 cos 3  27.45 mm
yP  r2 sin  2   yB  r3 sin 3  47.55 mm
Taking the derivatives of the loop-closure equations with respect to both  2 and  3 , in
turn, we find the following two sets of equations for the first-order kinematic coefficients.
r3 sin 3 
cos  2  cos 3   r22 r23   r2 sin  2

 sin 



 sin 3   r32 r33   r2 cos  2 r3 cos 3 
2

and the solutions for these are
 r22 r23  1  r2 cos 3   2 
r r    
r2
 32 33 



r3 cos 3   2  
At the current position, the numeric values of the first-order kinematic coefficients are
 r22 r23   14.71 mm/rad 106.2 mm/rad 



 r32 r33   56.84 mm/rad 27.49 mm/rad 
r3
Using these, the first-order kinematic coefficients for the center of the pin are
xP 2  r22 cos  2  r2 sin  2  r32 cos 3  40.19 mm/rad
yP 2  r22 sin  2  r2 cos  2  r32 sin 3  40.19 mm/rad
xP 3  r23 cos  2  r33 cos 3  r3 sin 3  53.11 mm/rad
yP 3  r23 sin  2  r33 sin 3  r3 cos 3  91.98 mm/rad
With the given independent input velocities of 2  30 rad/s cw and 3  20 rad/s cw,
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Theory of Machines and Mechanisms, 4e
Uicker et al.
the velocity of the pin P4 is
xP  xP 22  xP 33  2268 mm/s
yP  yP 22  yP 33  634 mm/s
V  x ˆi  y ˆj  2268ˆi  634ˆj mm/s  2355 mm/s15.62
P
P
P
Ans.
Taking the second derivatives of the loop-closure equations with respect to both  2 and
 3 , in turn, we find the following three sets of equations for the second-order kinematic
coefficients:
 r223
 r233
 
cos  2  cos 3   r222
 sin 


 r323
 r333
 
 sin 3   r322
2

r23 sin  2  r32 sin 3 2r33 sin 3  r3 cos 3 
 2r  sin  2  r2 cos  2
  22

 2r22 cos  2  r2 sin  2 r23 cos  2  r32 cos 3 2r33 cos 3  r3 sin 3 
and the solutions for these are
 r223
 r233
  1 2r22 cos 3  2   r2 sin 3  2  r23 cos 3  2   r32
2r33

r222

r r r    
2r22
r23  r32 cos 3  2  2r33 cos 3  2   r3 sin 3  2 
 322 323 333 

At the current position, the numeric values of the second-order kinematic coefficients are
 r223
 r233
  62.787 mm/rad 2 87.307 mm/rad 2 56.92 mm/rad 2 
 r222

 r  r  r    
2
125.195 mm/rad 2 117.31 mm/rad 2 
 322 323 333   30.46 mm/rad
The second-order kinematic coefficients for the center of the pin are
 cos  2  2r22 sin  2  r2 cos  2  r322
 cos 3  21.538 mm/rad 2
xP 22  r222
 sin  2  2r22 cos  2  r2 sin  2  r322
 sin 3  21.538 mm/rad 2
yP 22  r222
 cos  2  r23 sin  2  r323
 cos 3  r32 sin 3  48.334 mm/rad 2
xP 23  r223
 sin  2  r23 cos  2  r323
 sin 3  r32 cos 3  128.719 mm/rad 2
yP 23  r223
 cos  2  r333
 cos 3  2r33 sin 3  r3 cos 3  28.461 mm/rad 2
xP33  r233
 sin  2  r333
 sin 3  2r33 cos 3  r3 sin 3  49.296 mm/rad 2
yP33  r233
With the given independent input velocities and accelerations of 2  30 rad/s cw,
3  20 rad/s cw, and  2  3  0 , the acceleration of the pin P4 is
xP  xP 2 2  xP 3 3  xP 2222  2 xP2323  xP3332  50 mm/s 2
yP  yP 2 2  yP 3 3  yP 2222  2 yP2323  yP3332  115360 mm/s
A  x ˆi  y ˆj  50ˆi  115360ˆj mm/s2  125730 mm/s2  66.6
Ans.
It should be noted that the exact match between the graphic and the analytic solutions
achieved for this problem is not at all typical, nor can such matches be expected. Graphic
results are usually far less accurate than analytic ones. The reason for the agreement
achieved here is that a very precise CAD system was used for the graphic constructions.
P
P
P
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
5.2
Uicker et al.
For the five-bar linkage in the position illustrated in Fig. P5.2, the angular velocity of
link 2 is 15 rad/s cw and the angular velocity of link 5 is 15 rad/s cw. Determine the
angular velocity of link 3 and the apparent velocity VB4 /5 .
RO2O5  200 mm23.1 , RAO2  300 mm , and RBA  200 mm
Let us define r1  RO2O5  200 mm, 1  23.1, r2  RAO2  300 mm, r3  RBA  200 mm,
and r4  RBO5 . Then the loop-closure equation can be written as
r11  r22  r33  r45  0
with horizontal and vertical components of
r1 cos 1  r2 cos  2  r3 cos 3  r4 cos 5  0
r1 sin 1  r2 sin  2  r3 sin 3  r4 sin 5  0
Solution of these position equations give two unknowns, r4  300 mm and 3  203.1 .
Derivatives of the loop-closure equations with respect to each of the independent degrees
of freedom,  2 and  5 , give the first-order kinematic coefficients
 r3 sin 3  cos 5  32 35   r2 sin  2 r4 sin 5 




 r3 cos 3  sin 5   r42 r45   r2 cos  2 r4 cos 5 
The determinant of the Jacobian is   r3 cos 3  5  and this will go to zero whenever
3  5   2k  1  2 ; that is, whenever link 3 is perpendicular to link 5. At the current
position,   159.94 mm. The solutions for the first-order kinematic coefficients are
r4

32 35  1  r2 cos 5   2 





 r42 r45    r2 r3 sin 3   2  r3r4 sin 3  5  
At the current position, with the given data, the values for the first-order kinematic
coefficients are
32 35  1.875 74 rad/rad 1.875 74 rad/rad 
 r  r     225.25 mm/rad
225.25 mm/rad 
 42 45  
Therefore, with 2  5  15 rad/s, we have
3  32 2  35 5  0 and VB4 /5  r42 2  r45 5  0
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
5.3
Uicker et al.
For the five-bar linkage in the position illustrated in Fig. P5.2, the angular velocity of
link 2 is 2  25 rad/s ccw and the apparent velocity VB4 /5 is 5000 mm/s upward along
link 5. Determine the angular velocities of links 3 and 5.
Here we can continue the solution of Problem 5.2. However, the problem is now
expressed in terms of two different input variables,  2 and r4 , as independent degrees of
freedom. Therefore, we can use the same vectors and the same loop-closure equations.
However, we must now take derivatives with respect to  2 and r4 to find the first-order
kinematic coefficients. The result, in matrix form, is
 r3 sin 3 r4 sin 5  32 34   r2 sin  2 cos 5 




 r3 cos 3 r4 cos 5  52 54   r2 cos  2 sin 5 
The determinant of the Jacobian is now   r3r4 sin 3  5  , which goes to zero whenever
link 3 is aligned with link 5. The solutions for the first-order kinematic coefficients are
r4
 
0
32 34  1  r2 r4 sin 5   2 
8.3276 103 rad/mm 


        r r sin   
3
 52 54 
 2 3  3 2  r3 cos 3  5   1.000 00 rad/rad 4.4396 10 rad/mm 
With the given input velocities, 2  25 rad/s and r4  5000 mm/s , the requested
velocities are
3  32 2  34 r4  41.64 rad/s (cw) and 5  52 2  54 r4  47.20 rad/s ccw
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
5.4
Uicker et al.
For Problem 5.2, assuming that the two given input velocities are constant, determine the
angular acceleration of link3 at the instant indicated.
Starting with the equations of Problem 5.2 for the first-order kinematic coefficients,
 r3 sin 3  cos 5  32 35   r2 sin  2 r4 sin 5 




 r3 cos 3  sin 5   r42 r45   r2 cos  2 r4 cos 5 
we can take derivatives with respect to each independent variable to find equations for
the second-order kinematic derivatives
 325
 355
 
 r3 sin 3  cos 5  322

  

 
r425
r455
 r3 cos 3  sin 5   r422
 r3 cos 3322  r2 cos  2 r3 cos 332 35  sin  5 r42 r3 cos  3 352  2sin  5r45  r4 cos  5 


2
2
 r3 sin 332  r2 sin  2 r3 sin 332 35  cos 5 r42 r3 sin 335  2 cos 5 r45  r4 sin 5 
With satisfaction, we notice that the Jacobian is identical with that of Problem 5.2. The
solution to these equations gives
 325
 355
 
322
 r r r  
 422 425 455 
2

r3 sin 3  5 32 35  r42
r3 sin 3  5 352  2r45
1  r3 sin 3  5  32  r2 sin 5   2 


2 2
2
2 2

r3 32  r2 r3 cos 3   2 
r3 cos 3  5 32 35  r3 sin 3  5  r42 r3 35  2r3 sin 3  5  r45  r3 r4 cos 3  5 
Substituting the numeric data, including results of Problem 5.2, we get
 325
 355
   2.641 69 rad/rad 2 4.050 06 rad/rad 2 5.458 43 rad/rad 2 
322


 r 

   579.95 mm/rad 2
r455
534.56 mm/rad 2
1518.19 mm/rad 2 
 422 r425
Therefore, given that 2  5  15 rad/s (cw) and  2  5  0, the angular acceleration
of link 3 is
 22  2325
 25  355
 52  0
3  32  2  35 5  322
Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
5.5
Uicker et al.
Figure P5.5 illustrates link 2 rotating at a constant angular velocity of 10 rad/s ccw while
the sliding block 3 slides toward point A at the constant rate of 125 mm/s. At the instant
indicated RPA = 100 mm. Find the absolute velocity and absolute acceleration of point P
of block 3.
RAO2  75 mm and RBA  150 mm
With the dimensions given, O2AP is a 3-4-5 right triangle and R PO2  125ˆj mm .
For velocity, we write
VP3  VP2  VP3 /2
VP2  VO2  ω 2  R P2O2

 

 10kˆ rad/s  125ˆj mm  1 250ˆi mm/s
Also, we are given VP3 /2  125 mm/s . From these we construct the velocity polygon
shown in the figure, and from this we measure
VP3  1 179.25 mm/s  175.13
Ans.
For acceleration, we write
A P3  A P2  A cP3 P2  A nP3 /2  AtP3 /2
A P2  AO2  A nP2O2  A Pt2O2


2
APn2O2   10 rad/s  125ˆj mm  12 500ˆj mm/s 2
APc3 P2  22VP3 /2  2 10 rad/s 125 mm/s   2 500 mm/s 2
APn3 /2  VP23 /2   VP23 /2   0
APt 3 /2  0
From these we construct the acceleration polygon shown in the figure, and from this we
measure
A P3  11 180 in/s2  79.70
Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
5.6
Uicker et al.
For Problem 5.5, determine the value of the sliding velocity VP3 /2 that minimizes the
absolute velocity of point P of block 3. Find the value of VP3 /2 that minimizes the
absolute acceleration of point P of block 3.
By careful inspection of the velocity polygon of Problem 5.5 we can see that the absolute
velocity is minimized when it becomes perpendicular to VP3 /2 . Reconstructing the
velocity polygon in this condition, as shown in the figure, we find
VP3  1 000.0 mm/s  143.13
and
VP3 /2  750.0 mm/s  53.13
Ans.
Similarly, the absolute acceleration of P3 is minimized when it becomes perpendicular to
A cP3 P2 . Reconstructing the acceleration polygon in this condition, as shown in the figure,
we find A P3  10 000 mm/s2  53.13 and AcP3P2  7 500 mm/s2  143.13 . Then, from
this, we can calculate
APc3 P2
7 500 mm/s 2
 375.0 mm/s
Ans.
22 2 10.00 rad/s 
Notice how the visualization of the inherent geometry has dramatically simplified this
problem, compared to a totally mathematical approach. Notice also that VP3 /2 must
VP3 /2 

increase in both cases.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
5.7
Uicker et al.
The two-link planar robots illustrated in Fig. P5.7 have the link lengths
RAO2  RBO4  300 mm and RPA  RPB  400 mm . The two robots are carrying a small
At the instant indicated the angular positions are 2  45 and
3/2  15 . (Notice that the angle 3/2  3  2 is given because that is the angle
controlled by the motor in joint A.) A second robot having identical dimensions is
stationed at position O4, 1000 mm to the right. What angular positions must the two
joints  4 and 5/4 of the second robot have at this moment to allow it to take over
possession of the object P?
object labeled P.
Dimensions are:
RAO2  300 mm , RPA  400 mm .
The loop-closure constraint equations at this instant allow us to write
1000  300cos  4  400cos 5  400cos 30  300cos 45  0
300sin  4  400sin 5  400sin 30  300sin 45  0
These can be rearranged to read
cos 5  0.750cos  4  1.10364
sin 5  0.750sin  4  1.03033
Now, by squaring and adding, we eliminate the variable 5.
1.0  0.562 5  1.655 47cos 4  1.545 50sin 4  2.279 60
1.655 47cos 4 1.545 50sin 4  1.842 10  0
or
Next, by defining x  tan 4 2  , and by use of the standard identities, this becomes
1.65547 1  x 2   1.545 50  2 x   1.842 10 1  x 2   0
0.18663x2  3.09100x  3.49757  0
or
The roots of this equation give
x  15.340 54 and x  1.221 64
and from the definition of x these give two values of 4
4  172.54 and 4  101.39
Now, returning these to the above equations, we can solve for values of 5
5  111.10 and 5  162.84
Of these, the second value of each pair fits our figure. Therefore,
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
4  101.39
5.8
and
5/4  61.45
Ans.
For the transfer of the object described in Problem 5.7 it is necessary that the velocities of
point P of the two robots match. If the two input velocities of the first robot are
2  10 rad/s cw and 3/2  15 rad/s ccw, what angular velocities must be used for 4
and 5/4 ?
First we find
3  2  3/2  10 rad/s (cw)  15 rad/s (ccw)  5 rad/s (ccw)
Then, the velocity of point P is given by
VP  VO2  VAO2  VPA  VO4  VBO4  VPB
VAO2  2 RAO2  10 rad/s  300 mm   3000 mm/s
VPA  3 RPA   5 rad/s  400 mm   2000 mm/s
From these data and equations, we can construct the velocity polygon shown in the
figure. This allows us to find data for the following calculations:
4 
VBO4
5 
VPB 686.5 mm/s

 1.716 rad/s ccw
400 mm
RPB
RBO4

1350.5 mm/s
 4.502 rad/s cw
300 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
5/4  5  4  1.716 rad/s    4.502 rad/s   6.218 rad/s (ccw)
Ans.
If an analytical solution is preferred, we start with the robot on the left, where we find
3  2  3/2  10 rad/s (cw)  15 rad/s (ccw)  5 rad/s (ccw)

 

  5.0kˆ rad/s  ×  400cos 30ˆi  400sin 30ˆj mm   1000ˆi  1732.05ˆj mm/s
VA  ω 2 × R AO2  10.0kˆ rad/s × 300cos 45ˆi  300sin 45ˆj mm  2121.32ˆi  2121.32ˆj mm/s
VPA  ω3 × R PA
VP  VA  VPA  1121.32ˆi  389.27ˆj mm/s
Similarly, for the robot on the right, we have

 

  kˆ rad/s  ×  400 cos162.84ˆi  400 sin162.84ˆj mm 
VB  ω 4 × R BO4  4kˆ rad/s × 300 cos101.39ˆi  300 sin101.39ˆj mm
VPB  ω5 × R PB
5
VP  VB  VPB   294.09i  59.25 j mm  4   118.02i  382.18 j mm  5
Next, by setting the two equations for VP equal to each other, and then separating the î
and ĵ components, we obtain a set of two equations for 4 and 5
 294.09 mm 118.02 mm  4  1121.32 mm/s 
 59.25 mm 382.18 mm      389.27 mm/s 

 5 

The solutions to these equations give
4   4.502 rad/s (cw) 
    1.716 rad/s (ccw) 

 5 
5/4  5  4  1.716 rad/s    4.502 rad/s   6.218 rad/s (ccw)
Ans.
Ans
We see that these results agree precisely with those obtained above by the graphical
approach. It must be pointed out that this is not usual for graphic solutions, but is the
result of the high-precision CAD system used here.
As yet a third approach to the solution of this problem, we can find the instant centers of
velocity. In doing this we follow exactly the approach shown in Example 5.5 in the text.
Since we are given velocities for 2 and 3 , the location of instant center I13 is defined
by
2 RI23I13 10.0 rad/s


 2.0 or RI23 I13  2.0RI23 I12
3 RI23 I12 5.0 rad/s
Therefore I13 takes the position shown in the figure below. When the other instant
centers are found through the Aronhold-Kennedy theorem, this results in the instant
centers shown for I 24 , I 34 and for I 25 , I 35 .
Once the remaining instant centers are found, we may find the information requested in
the problem. We find
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Theory of Machines and Mechanisms, 4e
4 
RI24 I12
5 
RI25 I12
RI24 I14
RI25 I15
Uicker et al.
2 
818.84 mm
 10 rad/s   4.502 rad/s (cw)
1818.84 mm
2 
193.87 mm
 10 rad/s   1.716 rad/s (ccw)
1129.51 mm
5/4  5  4  1.716 rad/s    4.502 rad/s   6.218 rad/s (ccw)
Ans.
Ans.
Again, the very high degree of agreement is a consequence of the precision of the CAD
system used in finding the distances between instant centers.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
5.9
Uicker et al.
For the transfer of the object described in Problem 5.7 it is necessary that the velocities of
point P of the two robots match. If the two input velocities of the first robot are
2  10 rad/s cw and 3/2  10 rad/s ccw, what angular velocities must be used for 4
and 5/4 ?
First we find
3  2  3/2  10 rad/s (cw)  10 rad/s (ccw)  0 rad/s
Notice that this implies that link 3 is in translation relative to the ground.
Then, the velocity of point P is given by
VP  VO2  VAO2  VPA  VO4  VBO4  VPB
VAO2  2 RAO2  10 rad/s  300 mm   3000 mm/s
VPA  3 RPA   0 rad/s  400 mm   0.0 in/s
From these data and equations, we can construct the velocity polygon shown in the
figure. This allows us to find data for the following calculations:
4 
VBO4
5 
VPB 2844.5 mm/s

 7.111 rad/s ccw
400 mm
RPB
RBO4

3020 mm/s
 10.067 rad/s cw
300 mm
5/4  5  4   7.111 rad/s    10.067 rad/s   17.178 rad/s (ccw)
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
If an analytical solution is preferred, we start with the robot on the left, where we find
3  2  3/2  10 rad/s (cw)  10 rad/s (ccw)  0 rad/s

 

  0kˆ rad/s  ×  400cos 30ˆi  400sin 30ˆj in   0.000 00ˆi  0.000 00ˆj mm/s
VA  ω 2 × R AO2  10.0kˆ rad/s × 300cos 45ˆi  300sin 45ˆj in  2121.32ˆi  2121.32ˆj mm/s
VPA  ω3 × R PA
VP  VA  VPA  2.121 32ˆi  2.121 32ˆj mm/s
Similarly, for the robot on the right, we have

 

  kˆ rad/s  ×  400 cos162.84ˆi  400 sin162.84ˆj mm 
VB  ω 4 × R BO4  4kˆ rad/s × 300 cos101.39ˆi  300 sin101.39ˆj mm
VPB  ω5 × R PB
5
VP  VB  VPB   294.09i  59.25 j mm  4   118.02i  382.18 j mm  5
Next, by setting the two equations for VP equal to each other, and then separating the î
and ĵ components, we obtain a set of two equations for 4 and 5
 294.09 mm 118.02 mm  4   2121.32 mm/s 
 59.25 mm 382.18 mm      2121.32 mm/s 

 5 

The solutions to these equations give
4   10.067 rad/s (cw) 
    7.111 rad/s (ccw) 

 5 
5/4  5  4   7.111 rad/s    10.067 rad/s   17.178 rad/s (ccw)
Ans.
Ans
We see that these results agree precisely with those obtained above by the graphical
approach. It must be pointed out that this is not usual for graphic solutions, but is the
result of the high-precision CAD system used here.
As yet a third approach to the solution of this problem, we can find the instant centers of
velocity. In doing this we follow exactly the approach shown in Example 5.5 in the text.
Since we are given velocities for 2 and 3 , the location of instant center I13 is defined
by
2 RI23 I13 10.0 rad/s


  or RI23I13  
0.0 rad/s
3 RI23I12
Therefore I13 goes to infinity in the direction shown in the figure below. This indicates
that link 3 is in translation with respect to the ground, which we could see when we found
that 3  0 . When the other instant centers are found through the Aronhold-Kennedy
theorem, this results in the instant centers shown for I 24 , I 34 and for I 25 , I 35 . However,
we find that the lines toward instant center I 24 are essentially parallel, and that instant
center I 24 appears to also be at infinity. This implies that link 4 is in translation with
respect to link 2, which means that 4  2  10 rad/s (cw) .
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
Once the remaining instant centers are found, we may find the information requested in
the problem. We find
R
4  I24 I12 2  10.000 rad/s (cw)
Ans.
RI24 I14
5 
RI25 I12
RI25 I15
2 
18.503 60 in
 10 rad/s   7.112 rad/s (ccw)
26.017 20 in
5/4  5  4   7.112 rad/s    10.000 rad/s   17.112 rad/s (ccw)
Ans.
Notice that the precision is not perfect this time, in spite of the use of a high-precision
CAD system. However, this probably stems from the possibility that the apparent
parallelism and intersections at infinity were likely not perfect. Still, the precision is
amazing for a graphical solution.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
5.10
Uicker et al.
For the transfer of the object described in Problem 5.7 it is necessary that the velocities of
point P of the two robots match. If the two input velocities of the first robot are
2  10 rad/s cw and 3/2  0, what angular velocities must be used for 4 and 5/4 ?
First we find
3  2  3/2  10 rad/s (cw)  0 rad/s  10 rad/s (cw)
Notice that this implies that link 3 and link 2 rotate together as a single unit.
Then, the velocity of point P is given by
VP  VO2  VAO2  VPA  VO4  VBO4  VPB
VAO2  2 RAO2  10 rad/s  300 mm   3000 mm/s
VPA  3 RPA  10 rad/s  400 mm   4000 mm/s
From these data and equations, we can construct the velocity polygon shown in the
figure. This allows us to find data for the following calculations:
4 
VBO4
RBO4

6359.5 m/s
 21.198 rad/s cw
300 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
5 
Uicker et al.
VPB 7160.5 mm/s

 17.901 rad/s ccw
400 mm
RPB
5/4  5  4  17.901 rad/s    21.198 rad/s   39.099 rad/s ccw
Ans.
If an analytical solution is preferred, we start with the robot on the left, where we find

 

  10.0kˆ rad/s  ×  400cos 30ˆi  400sin 30ˆj in   80.000 00ˆi  138 56ˆj mm/s
VA  ω 2 × R AO2  10.0kˆ rad/s × 300cos 45ˆi  300sin 45ˆj in  2121.32ˆi  2121.32ˆj mm/s
VPA  ω3 × R PA
VP  VA  VPA  164.852 80ˆi  223.416 80ˆj in/s
Similarly, for the robot on the right, we have

 

  kˆ rad/s  × 16 cos162.84ˆi  16 sin162.84ˆj in 
VB  ω 4 × R BO4  4kˆ rad/s × 12 cos101.39ˆi  12 sin101.39ˆj in
VPB  ω5 × R PB
5
VP  VB  VPB   11.763 60i  2.370 00 j in  4   4.720 80i  15.287 20 j in  5
Next, by setting the two equations for VP equal to each other, and then separating the î
and ĵ components, we obtain a set of two equations for 4 and 5
 294.09 mm 118.02 mm  4   4121.3 mm/s 
 59.25 mm 382.18 mm      5585.42 mm/s 

 5 

The solutions to these equations give
4   21.198 rad/s (cw) 
    17.901 rad/s ccw 

 5 
5/4  5  4  17.901 rad/s    21.198 rad/s   39.099 rad/s ccw
Ans.
Ans
We see that these results agree precisely with those obtained above by the graphical
approach. It must be pointed out that this is not usual for graphic solutions, but is the
result of the high-precision CAD system used here.
As yet a third approach to the solution of this problem, we can find the instant centers of
velocity. In doing this we follow exactly the approach shown in Example 5.5. However,
since we have equal velocities for 2 and 3 , the location of instant center I13 is defined
by
2 RI23 I13 10.0 rad/s


 1.0
3 RI23 I12 10.0 rad/s
and therefore I13 becomes coincident with I12 as shown in the figure below. When the
other instant centers are found through the Kennedy-Aronhold theorem, this also results
in coincident instant centers for I 24 , I 34 and for I 25 , I 35 as shown in the figure. Under
these input velocity conditions, links 2 and 3 act as a single solid unit.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
Once the remaining instant centers are found, however, we may still find the information
requested in the problem. We find
R
1892.675 mm
4  I24 I12 2 
Ans.
 10 rad/s   21.198 rad/s (cw)
892.85 mm
RI24 I14
5 
RI25 I12
RI25 I15
2 
694.075 mm
 10 rad/s   17.901 rad/s (ccw)
387.725 mm
Ans.
Again, the perfect agreement results from the precision of the CAD system used in
finding the distances between instant centers.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
5.11
Uicker et al.
To successfully transfer an object between two robots, as described in Problems 5.7 and
5.8, it is helpful if the accelerations are also matched at point P. Assuming that the two
input accelerations are  2  3  0 at this instant for the robot on the left, what angular
accelerations must be given to the two input joints of the robot on the right to achieve
this?
Starting after the solutions of Problem 5.8 (the velocity analysis) is completed, the
condition for the acceleration of point P is written as
t
t
t
t
A P  AO2  A nAO2  A AO
 A nPA  A PA
 A O4  A nBO4  A BO
 A nPB  A PB
2
4
n
AO2
A

2
VAO
2
RAO2
 3000 mm/s 

300 mm
2
 30000 mm/s
2
 2000 mm/s   1000 mm/s2
VPA

400 mm
RPA
2
n
BO4
A
2
n
APA


2
VBO
4
RBO4
1350.6 mm/s 

300 mm
2
 6080.5 mm/s 2
2
 686.5 mm/s   1178.25 mm/s 2
VPB

400 mm
RPB
2
n
APB

Notice that a difficulty arises in the graphic solution of the above acceleration equation in
that the two unknowns do not arise consecutively in the equation. Nevertheless, recalling
that vector addition is commutative (independent of order), we can proceed with the
vectors appearing out of order, as is shown in the dotted lines in the figure above. Once
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
the solution with the dotted lines is completed, we have obtained the correct magnitudes
and directions of the two unknown tangential components. However, we do not have a
valid acceleration polygon unless we now arrange the components in their correct order,
according to the original acceleration equation, as is shown in the dashed lines in the
figure. If this is not done, the acceleration image point B cannot be labeled, and the
absolute acceleration of B and acceleration images of links 4 and 5 cannot be correctly
shown.
Whether or not the vectors are arranged in their correct order, however, we can proceed
with the solution for the two unknown angular accelerations as follows:
4 
t
ABO
4
RBO4

27583 mm/s 2
 91.943 rad/s 2 ccw
300 mm
t
APB
15127 mm/s 2
4 

 37.818 rad/s 2 ccw
400 mm
RPB
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Ans.
Ans.
Theory of Machines and Mechanisms, 4e
PART 2
DESIGN OF MECHANISMS
© Oxford University Press 2015. All rights reserved.
Uicker et al.
Theory of Machines and Mechanisms, 4e
Uicker et al.
Chapter 6
Cam Design
6.1
The reciprocating radial roller follower of a plate cam is to rise 40 mm with simple
harmonic motion in 180 of cam rotation and return with simple harmonic motion in the
remaining 180 . If the roller radius is 7.5 mm and the prime-circle radius is 40 mm,
construct the displacement diagram, the pitch curve, and the cam profile for clockwise
cam rotation.
6.2
A plate cam with a reciprocating flat-face follower has the same motion as in Problem
6.1. The prime-circle radius is 40 mm, and the cam rotates counterclockwise. Construct
the displacement diagram and the cam profile, offsetting the follower stem by 15 mm in
the direction that reduces the bending stress in the follower during rise.
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Theory of Machines and Mechanisms, 4e
6.3
Uicker et al.
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 30 , returns with cycloidal motion in
120 , and dwells for 60 . Determine the necessary length for the follower face, allowing
6.25 mm clearance at the free end. The prime-circle radius is 37.5 mm, and the follower
pivot is 150 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 250 mm but can be made
as short as 243.75 mm (position 9) from the follower pivot.
Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
6.4
Uicker et al.
A plate cam with an oscillating roller follower is to produce the same motion as in
Problem 6.3. The prime-circle radius is 75 mm, the roller radius is 12.5 mm, the length
of the follower is 125 mm, and it is pivoted at 156.25 mm to the right of the cam rotation
axis. The cam rotation is clockwise. Determine the maximum pressure angle.
From a graphical analysis, max  39
6.5
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. (6.12) and (6.11) we find
 1 L
 L
sin 
y    
2 2
  2  2
 1
 3L 
 3L
y      3 sin   3
2
2
2
  2
2
2

  L
 L
cos 0 
y   0  
2
2 2

 2

  2L
 2L
cos
y   1 



2
2 2

 2

y


y


y


y

1 L
 

2  2
1
 3L
    3 3
2
2
  2L 2
 0 

2
 2

 2L
 1   2  2
2

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Ans.
Ans.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
6.6
Uicker et al.
For a full-rise cycloidal motion, determine the values of  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.
Using Eqs. (6.13), we know that acceleration is an extremum when jerk is zero. This
occurs when cos 2   0 ; that is, when    1 4 or when    3 4 ,
  1  2 L
 2 L

y     2 sin  2  ymax
2

  4 
  3  2 L
3
2 L

y     2 sin
  2  ymin
2

  4 
 1 L
2L
y     1  cos   

  2 
  1  4 2 L
4 2 L


cos

y    
3
3
  2
6.7
Ans.
Ans.
Ans.
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 50 mm.
During 20 mm of its return stroke, it must have a constant velocity of 800 mm/s.
Recommend standard cam motions from Section 6.7 to be used for high-speed operation
and determine the corresponding lifts and cam rotation angles for each segment of the
cam.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
The curves shown are initially only sketches and not drawn to scale. They suggest the
standard curve types that might be chosen. The actual choices are shown in the table
below.
 400 rev/min  2 rad/rev   41.888 rad/s cw

 60 s/min 
To match the required velocity condition in Seg. DE we must have
y4  y4
800.000 mm/s  y4  41.888 rad/s 
y4  19.098 600 mm/rad   L4  4    20.000 mm   4
 4  1.047 198 rad  60.000
Matching the first derivatives at D and E we find
2L5 5  y4  19.098 593 mm/rad
 L3  23   y4  19.098 593 mm/rad
Matching the second derivatives at C we find
5.26830  2.5000  22   2 L3 432
L5  9.549 2975 mm/rad
(1)
L3  12.158 5423 mm/rad
(2)
22  106.758 07832 L3  8.780 5003 (3)
For geometric continuity, we have
(4)
L1  L2  L3  L4  L5 or
L3  L5  30.000 000 mm
(5)
1   2  3   4  5  2 or
2  3  5  4.188 790 rad
Equations (1) through (5) are now solved simultaneously for  2 , L3 ,  3 , L5 , and  5 .
The results are summarized in the following table:
Seg. Type
Eq.
L, mm
rad
, deg
AB dwell
--0
1.047 198
60.000
th
BC 8 order poly.
(6.14)
50.000
1.089 824
62.442
CD half harmonic
(6.20)
1.648
0.135 268
7.750
DE constant velocity
--20.000
1.047 198
60.000
EA half cycloidal
(6.25)
28.352
2.963 698
169.808
6.8
Repeat Problem 6.7 except with a dwell for 20 of cam rotation.
The procedure is the same as for Problem 6.7. The results are:
Seg. Type
Eq.
L, mm
rad
AB dwell
--0
0.349 066
th
BC 8 order poly.
(6.14)
50.000
1.870 958
CD half harmonic
(6.20)
4.872
0.398 667
DE constant velocity
--20.000
1.047 198
EA half cycloidal
(6.25)
25.128
2.617 296
© Oxford University Press 2015. All rights reserved.
, deg
20.000
107.198
22.842
60.000
149.960
Theory of Machines and Mechanisms, 4e
6.9
Uicker et al.
If the cam of Problem 6.7 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 t  1   1.047 198 rad 41.888 rad/s  0.025 s Ans.
Working from the equations listed, the maximum and minimum values of the kinematic
coefficients in each segment of the cam are as follows:
Seg.
Eq.
 mm/rad
 mm/rad2
 mm/rad
 mm/rad2
ymin
ymin
ymax
ymax
AB
BC
CD
DE
EA
ymax
--0
0
0
(6.14)
241.704
0
221.783
(6.20)
0
0
19.137
--0
19.099
19.099
(6.25)
0
10.141
19.133
    241.704 mm/rad  41.888 rad/s   10 124 mm/s
 ymax
0
221.783
221.783
0
0
Ans.
    19.137 mm/rad  41.888 rad/s   800.0 mm/s
ymin  ymin
Ans.
  2   221.783 mm/rad  41.888 rad/s   389 141 mm/s 2
ymax  ymax
Ans.
  2   221.783 mm/rad  41.888 rad/s   389 141 mm/s 2
ymin  ymin
Ans.
2
2
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Theory of Machines and Mechanisms, 4e
6.10
Uicker et al.
A plate cam with an oscillating follower is to rise through 20 in 60 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 6.7 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 curve 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
3
4
 42 L4

 2.000
 32 L3
 4  23


3   4  1  2 3  195
3  80.772
 4  114.228
Seg.
AB
BC
CD
DE
EA
Type
cycloidal
dwell
8th order poly.
8th order poly.
dwell
Eq.
(6.13)
--(6.14)
(6.17)
---
L, deg
20.000
0
20.000
40.000
0
rad
, deg
1.047 198
0.785 399
1.409 731
1.993 661
1.047 198
60.000
45.000
80.772
114.228
60.000
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Theory of Machines and Mechanisms, 4e
6.11
Uicker et al.
Determine the maximum velocity and acceleration of the follower for Problem 6.10,
assuming that the cam is driven at a constant speed of 600 rev/min.
Working from the equations listed, the maximum and minimum values of the kinematic
coefficients in each segment of the cam are as follows:
Seg.
Eq.




ymin
ymin
ymax
ymax
AB
(6.13)
0.666 667
0
2.000 000
BC
--0
0
0
CD
(6.14)
0.440 004
0
0.925 344
DE
(6.17)
0
0.925 402
0.622 264
EA
--0
0
0
   600 rev/min  2 rad/rev   60 s/min   62.832 rad/s
2.000 000
0
0.924 344
0.925 402
0
    0.666 667 rad/rad  62.832 rad/s   41.888 rad/s
ymax  ymax
Ans.
  2   2.000 rad/rad  62.832 rad/s   7 896 rad/s2
ymax  ymax
Ans.
2
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Theory of Machines and Mechanisms, 4e
6.12
Uicker et al.
The boundary conditions for a polynomial cam motion are as follows: for   0 , y  0 ,
and y  0 ; for    , 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
     
3C
C
2C


y 

   
 
y  C0  C1 
2
 C2 
3
 C3 
2
1
3
2
Then from the boundary conditions:
   0  C  0
C
y    0  

 0
y    1.0   C  C  L

3C
2C
y    1.0  


 0
y 
0
1
2
3
3
2
C0  0
C1  0
C2  3L
C3  2 L
Therefore the equation and its three derivatives are:
2
3
3
  2





y
  3L   2 L   L 3   2  
2
2
 

6
L
6
L
6
L





y
             
6L
6L 
 
12 L 2 
y 
  2 
   2 1  2  

         
     
  
 
 
 
y     12L


3
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Ans.
Ans.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
6.13
Uicker et al.
Determine the minimum face width using 2-mm allowances at each end and determine
the minimum radius of curvature for the cam of Problem 6.2.
Referring to Problem 6.2 for the data and figure,
  180   rad
R0  40.0 mm
L  40.000 mm
From Eqs. (6.12) and (6.15) for simple harmonic motion,
   L  2   20.000 mm/rad
   L  2   20.000 mm/rad
ymax
ymin
From Eq. (6.30)
  allowances
Face width  ymax  ymin
Face width   20.000 mm    20.000 mm   2  2 mm   44.0 mm
Ans.
From Eq. (6.28):
  R0  y  y
 L
L
 
   cos
  R0  (constant)
2
 
 2
   40.0 mm    40.000 mm  2  60.0 mm
 R0 
6.14
L

1  cos
2

Ans.
Determine the maximum pressure angle and the minimum radius of curvature for the cam
of Problem 6.1.
Referring to Problem 6.1 for the figure and data,
  180   rad
R0  40.000 mm
Rr  7.500 mm
L  40.000 mm
For simple harmonic motion, Eq. (6.12) can be substituted into Eq. (6.33) to give
sin 
tan  
. This can be differentiated and d d set to zero to find the angle
3  cos
  70.53 at which max  19.47 . However, it is much simpler to use the nomogram of
Fig. 6.28 to find max  20 directly. For the accuracy needed, the nomogram is
considered sufficient.
Ans.
From Fig. 6.30a, using R0 L  1.0 , we get   min  Rr  R0  1.43 . This gives
 min  1.43R0  Rr  1.43  40 mm    7.50 mm   49.7 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
6.15
Uicker et al.
A radial reciprocating flat-face follower is to have the motion described in Problem 6.7.
Determine the minimum prime-circle radius if the radius of curvature of the cam is not to
be less than 10 mm. Using this prime-circle radius, what is the minimum length of the
follower face using allowances of 3 mm on each side?
From
Problem
6.9,
  241.704 mm/rad ,
ymax
  19.137 mm/rad ,
ymin
  221.783 in/rad
ymin
Therefore, from Eq. (6.29),
  y  10.000 mm    221.783 mm    50.000 mm   181.783 mm Ans.
R0  min  ymin
2
Also, from Eq. (6.30),
  ymin
  allowances   241.704 mm   19.137 mm  2  3.0 mm  266.84 mm Ans.
Face width  ymax
6.16
Graphically construct the cam profile of Problem 6.15 for clockwise cam rotation.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
6.17
Uicker et al.
A radial reciprocating roller follower is to have the motion described in Problem 6.7.
Using a prime-circle radius of 400 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. 6.28 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 fullreturn curves, and to adjust the prime-circle baseline.
Seg.
max , deg
 * , deg
L* , mm
R0* , mm
R0* L*
BC
400.000
50.000
CD
446.704
3.296
EA
400.000
56.704
For the total cam, max  12.
8.0
135.5
7.0
62.4
15.5
339.6
12
1
3
Ans.
Also we use Figs. 6.32 and 6.33 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 have figures for half-harmonic and
half-cycloidal cam segments. Note also that segment EA need not be checked since
undercutting occurs only in segments with negative acceleration.
 , deg    Rr  R0* Rrmax mm
Seg.
L, in
R0* L
R0* , mm
min
BC
CD
400.000
448.352
50.000
1.648
8.0
272.1
62.1
7.7
To avoid undercutting for the entire cam, Rr  290 mm .
6.18
0.725
0.680
290
304
Ans.
Graphically construct the cam profile of Problem 6.17 using a roller radius of 15 mm.
The cam rotation is to be clockwise.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
6.19
Uicker et al.
A plate cam rotates at 300 rev/min and drives a reciprocating radial roller follower
through a full rise of 75 mm in 180° of cam rotation. Find the minimum radius of the
prime-circle if simple harmonic motion is used and the pressure angle is not to exceed
25 . Find the maximum acceleration of the follower.
Using max  25 and   180 , Fig. 6.28 gives R0 L  0.75 . Therefore
R0  0.75L  0.75  75 mm   56.25 mm

 300 rev/min  2
 
ymax
rad/rev 
60 s/min
 2  75 mm 
Ans.
 31.416 rad/s
 2L

 37.5 mm/rad 2
2
2
2
2  rad 
  2   37.5 mm/rad 2   31.416 rad/s   37000 mm/s2
ymax  ymax
2
6.20
Ans.
Repeat Problem 6.19 except that the motion is cycloidal.
Figure 6.28 gives R0 L  0.95 . Therefore R0  0.95L  0.95  75 mm   71.25 mm Ans.
 
ymax
2 L

2

2  75 mm 

rad 
2
 47.7475 mm/rad 2
  2   47.7475 mm/rad 2   31.416 rad/s   47125 mm/s2
ymax  ymax
2
6.21
Ans.
Repeat Problem 6.19 except that the motion is eighth-order polynomial.
Figure 6.28 gives R0 L  0.95 . Therefore R0  0.95L  0.95  75 mm   71.25 mm Ans.
 
ymax
5.2683L

2

5.2683  75 mm 

rad 
2
 40.025 mm/rad 2
  2   40.025 mm/rad 2   31.416 rad/s   39500 mm/s 2
ymax  ymax
2
6.22
Ans.
Using a roller diameter of 0.80 in, determine whether the cam of Problem 6.19 will be
undercut.
Using R0 L  0.75 and   180 , Fig. 6.30a gives  min  Rr  R0  1.55 .
min  1.55  2.25 in    0.80 in   2.688 in  0 ; thus, this cam is not undercut.
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Ans.
Theory of Machines and Mechanisms, 4e
6.23
Uicker et al.
Equations (6.30) and (6.31) 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. (6.27) and using u and v to denote the local
rectangular part coordinates of the cam shape, the loop-closure equation is
ue j  jve j  jR0  jy  y  jRc
Dividing this by e j
u  jv  j  R0  Rc  y  e j  ye j
Now separating this into real and imaginary parts we find
u   R0  Rc  y  sin   y cos
v   R0  Rc  y  cos  y sin  Ans.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
6.24
& 6.25 Since programming languages vary so much, particularly with the use of
graphics, no attempt is made to show a “standard” solution for these problems.
6.26
A plate cam with an offset reciprocating roller follower is to have a dwell of 60° and then
rise in 90° to another dwell of 120°, after which it is to return in 90° of cam rotation. The
radius of the base circle is 40 mm, the radius of the roller follower is 15 mm, and the
follower offset is 20 mm. For the rise motion 60    150, the equation of the
displacement (the lift) is to be


y  40(  sin  )
where y is in millimeters and  is the cam rotation angle in radians. (i) Find equations
for the first- and second-order kinematic coefficients of the lift y for this rise motion. (ii)
Sketch the displacement diagram and the first- and second-order kinematic coefficients
for the follower motion described. Comment on the suitability of this rise motion in the
context of the other displacements specified. At the cam angle θ = 120°, determine the
following: (iii) the location of the point of contact between the cam and follower,
expressed in the moving Cartesian coordinate system attached to the cam; (iv) the radius
of the curvature of the pitch curve and the radius of curvature of the cam surface; and (v)
the pressure angle of the cam. Is this pressure angle acceptable?
(i)
From the equation given for the lift, the first- and second-order kinematic
coefficients are
1

y  40   cos   mm/rad and y  40sin  mm/rad 2


(ii)
Sketches of the first-order and the second-order kinematic coefficients of the
displacement diagram are shown here.
0
y'' (cm/rads)
y' (cm/rad)
6
5
4
3
2
1
0
60
80
100
120
140
-0.5 60
80
100
120
140
-1
-1.5
-2
-2.5
-3
-3.5
-4
-4.5
θ (degree)
θ (degree)
The rise motion specified is not suitable between dwells on either side since the first- and
second-order kinematic coefficients are not zero at the beginning and end of the rise. A
cycloidal rise curve would be preferable, but would have higher values of acceleration
and would lead to higher forces.
At the cam angle θ = 120º, we have     60  60   3 rad .

 3

 sin 60   48.0 mm
y  40(  sin  )  40 

 

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Theory of Machines and Mechanisms, 4e
Uicker et al.
1

1

y  40   cos    40   cos 60   32.7 mm/rad




y  40sin   40sin 60  34.6 mm/rad 2
The absolute coordinates of the trace point are
X 0    20 mm,
Y0 
2
 R  Rr    2

 40 mm  15 mm    20 mm 
2
2
 51.2 mm
X  X 0  20 mm, Y  Y0  y  51.2 mm  48.0 mm  99.2 mm
The cam coordinates of the trace point (the pitch curve) and derivatives are
u   X cos  Y sin     20 mm  cos120   99.2 mm  sin120  75.9 mm
v   X sin   Y cos     20 mm  sin120   99.2 mm  cos120  66.9 mm
u   X sin   Y cos  y sin   38.6 mm/rad
v   X cos  Y sin   y cos   92.3 mm/rad
w  u2  v2  100.0 mm/rad
u   X cos  Y sin   2 y cos  y sin   13.87 mm/rad 2
v   X sin   Y cos   2 y sin   y cos   2.76 mm/rad 2
(iii)
From Eq. (6.38), the cam coordinates of the point of contact (the cam surface) are
v
u
 61.1 mm
 62.1 mm
ucam  u  Rr
vcam  v  Rr
Ans.
w
w
(iv)
From Eq. (6.39), the radius of curvature of the pitch curve is
w3

 72.2 mm
Ans.
uv  vu
(v)
From Eq. (6.39) the pressure angle is
 v 
 u 
  7.3
cos      sin     cos   0.9919
Ans.
 w 
 w 
This pressure angle is less than 30° and is acceptable.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
6.27
A plate cam with an offset reciprocating roller follower is to be designed using the input,
the rise and fall, and the output motion shown in Table P6.27. The radius of the base circle
is to be 30 mm, the radius of the roller follower is 12.5 mm, and the follower offset (or
eccentricity) is to be 15 mm.
Table P6.27 Displacement Information for Plate Cam With Reciprocating Roller Follower
Cam angle range
0° - 20°
20° - 110°
110° - 120°
120° - 200°
200° - 270°
270° - 360°
Rise or Fall (mm)
0
+25
0
+5
0
-30
Follower motion
Dwell
Full-rise simple harmonic motion
Dwell
Full-rise cycloidal motion
Dwell
Full-return cycloidal motion
Comment on the suitability of the motions specified. At the cam angle   50,
determine the following: (i) the first-, second-, and third-order kinematic coefficients of
the lift curve, (ii) the coordinates of the point of contact between the roller follower and
the cam surface, expressed in the Cartesian coordinate system rotating with the cam, (iii)
the radius of curvature of the pitch curve, (iv) the unit tangent and the unit normal vectors
to the pitch curve, and (v) the pressure angle of the cam.
The portion of the motions which specifies simple harmonic motion is not suitable for
high-speed operation since there will be discontinuities in the second derivatives at both
ends of that segment where it interfaces with dwells. Cycloidal motion would correct
this problem but would give a higher peak acceleration. Still, simple harmonic motion is
specified.
(i)
Eqs. (6.12), with   50  20  30, L  25 mm,   90   2 rad, gives
  25 mm 

L
1  cos   6.25 mm
1  cos

 
2
2 
3
 L 

  25 mm  sin  21.65 mm/rad
y 
sin

2
3
y
Ans.
 2L


cos
 2  25 mm  cos  25.0 mm/rad 2
2

2
3
3
 L 

 4  25 mm  sin  86.60 mm/rad 3
y   3 sin

2
3
y 
Ans.
Ans.
Therefore the fixed coordinates of the tracepoint are
R0  R  Rr  30 mm  12.5 mm  42.5 mm, Y0  R02   2 
 42.5 mm  15 mm
2
X    15 mm, Y  Y0  y  39.76 mm  6.25 mm  46.01 mm
The cam coordinates of the trace point (the pitch curve) and derivatives are
u   X cos  Y sin    15 mm  cos50   46.01 mm  sin 50  44.89 mm
v   X sin   Y cos    15 mm  sin 50   46.01 mm  cos50  18.08 mm
u   X sin   Y cos  y sin   34.67 mm/rad
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2
 39.76 mm
Theory of Machines and Mechanisms, 4e
Uicker et al.
v   X cos  Y sin   y cos   30.97 mm/rad
w  u2  v2  46.49 mm/rad
u   X cos  Y sin   2 y cos   y sin   2.10 mm/rad 2
v   X sin   Y cos   2 y sin   y cos   35.18 mm/rad 2
(ii)
From Eq. (6.38), the cam coordinates of the point of contact (the cam surface) are
v
u
Ans.
 8.76 mm
 36.56 mm
ucam  u  Rr
vcam  v  Rr
w
w
(iii) From Eq. (6.39), the radius of curvature of the pitch curve is
Ans.
  w3  uv  vu   87.02 mm
(iv)
The unit tangent and the unit normal vectors to the pitch curve are
t
uˆ   u w ˆi   v w  ˆj  0.746ˆi  0.666ˆj
uˆ n   v w  ˆi   u w  ˆj  0.666ˆi  0.746ˆj
(v)
From Eq. (6.39) the pressure angle is
cos     v w  sin    u w  cos   0.9897
  8.2
Ans.
Ans.
Ans.
This pressure angle is less than 30° and is acceptable for the specified cam angle   50.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
6.28
Uicker et al.
A plate cam with a radial reciprocating roller follower is to be designed using the input,
the rise and fall, and the output motion shown in Table P6.28. The base circle diameter is
3 in and the diameter of the roller is 1 in. Displacements are specified as follows.
Table P6.28 Displacement Information for Plate Cam With Reciprocating Roller Follower
Input  (deg)
Lift L (in)
Output y
0  90
Cycloidal Rise
3.0
90  105
Dwell
0
105  195
Cycloidal Fall
-3.0
195  210
Dwell
0
210  270
Simple Harmonic Rise
2.0
270  285
Dwell
0
285  345
Simple Harmonic Fall
-2.0
345  360
Dwell
0
Plot the lift curve (the displacement diagram), and the profile of the cam. (i) Comment
on the lift curves at appropriate positions of the cam, (for example, when the cam angle is
  0,   45,   180,   210,   225, and   300 ). (ii) Identify on your
cam profile the location(s) and the value of the largest pressure angle. Would this
pressure angle cause difficulties for a practical cam-follower system? (iii) Identify on
your cam profile the location(s) of discontinuities in position, velocity, acceleration,
and/or jerk. Are these discontinuities acceptable (why or why not)? (iv) Identify on your
cam profile any regions of positive radius of curvature of the cam profile. Are these
regions acceptable (why or why not)? (v) For the values given in Table P6.28, what
design changes would you suggest to improve the cam design?
The lift curve (displacement diagram is shown in Fig. 1.
Figure 1. The lift curve or displacement diagram.
The cam profile is plotted in Fig. 2.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
Figure 2. The cam profile.
(i)
Because of the choice of harmonic motion rise and return curves, there are
discontinuities in acceleration at   210,   270,   285, and   345. Because
of adjacent dwells, cycloidal motion would be preferable, although it would lead to
slightly higher peak acceleration in these segments.
(ii)
The pressure angle, see Sec. 6.10, should be less than 30. In this design, the
pressure angle is more than the accepted value at the cam angles
  16  64,
and
131  180,
216  256,
299  341
Therefore, this cam profile is not a good design. The high values of the pressure angle
may be due to the selection of the displacement curves and the diameter of the base circle
and the diameter of the follower.
(iii) Position discontinuities never occur. Discontinuities in the derivatives will occur
only at transitions between dwell segments and lifting/returning segments of motion.
Discontinuities in the derivatives are undesirable. There is an acceleration discontinuity
at the beginning and end of the simple harmonic motions, both rise and return. There is a
jerk discontinuity at the beginning and end of the cycloidal motions, both rise and return.
Whether these discontinuities are acceptable depends on the intended speed of operation,
and on the masses and stiffnesses involved.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
(iv)
The radius of curvature of a cam profile should always be negative for a good
cam design. Positive curvature means that the cam has a concave surface and there is the
possibility that the follower may lose contact with the cam. If the radius of curvature of
the cam profile is positive then the radius of curvature of the cam must be greater than the
radius of the follower. In the proposed design, the positive values of the radius of
curvature of the cam are always greater than 0.5 in (i.e., the radius of the follower). The
radius of curvature of the cam is positive for the cam angles
  3  25 , 170  192 , 215  220 , and 261  270
The radius of curvature is positive, and smaller than the radius of the roller follower, for
the cam angles
215  216
  9  13 , 181  187 , and
Note that the radius of curvature of the cam is zero between the cam angles 214 and 215
degrees, meaning that pointing has occurred. Also, it could imply that undercutting has
occurred. Also, with the exception of where the radius of curvature of the cam goes to
zero, there is an inflection point at the boundary of each range of angles for which the
radius of curvature is positive.
(v)
Possible design changes to the cam-follower system. (a) Increasing the radius of
the prime circle (with the same lift curve) in general would reduce the pressure angle. (b)
Change the profiles to match acceleration at the transistion (blend) points to eliminate
acceleration discontinuities. (c) Changing both SHM profiles to cycloidal would make
accelerations continuous but would also increase accelerations (and the pressure angle) in
the middle parts of the rise and the return profiles. (d) We may want to increase the
diameter of the roller if the contact stresses are high. (e) We could explore numerically
the effects on forces of changing the offset (or the eccentricity)  . These are not obvious
from observation.
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Theory of Machines and Mechanisms, 4e
Continue using the same displacement information and the same design parameters as in
Problem 6.28. Use a spreadsheet to determine and plot the following for a complete
rotation of the cam: (i) the first-order kinematic coefficients of the follower center; (ii)
the second-order kinematic coefficients of the follower center; (iii) the third-order
kinematic coefficients of the follower center; (iv) the lift curve (displacement diagram);
(v) the radius of curvature of the cam surface; and (vi) the pressure angle of the camfollower system. Is the pressure angle suitable for a practical cam-follower system?
(i)
The first-order kinematic coefficients for the cam design are shown in Fig. 3.
First order Kinematics Coeeficients
6
xf c'
yf c'
4
First order KC - in
2
0
-2
-4
-6
0
50
100
150
200
250
300
350
400
 (Cam angle) - degrees
Figure 3. The first-order kinematic coefficients.
(ii)
The second-order kinematic coefficients for the cam design are shown in Fig. 4.
Second order Kinematics Coeeficients
15
x f c pp"
yf c"
10
5
Second order KC - in
6.29
Uicker et al.
0
-5
-10
-15
0
50
100
150
200
250
300
 (Cam angle) - degrees
Figure 4. The second-order kinematic coefficients.
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350
400
Theory of Machines and Mechanisms, 4e
(iii)
Uicker et al.
The third-order kinematic coefficients for the cam design are shown in Fig. 5.
Third order Kinematics Coeeficients
40
x f c p"'
y f c "'
30
20
Third order KC - in
10
0
-10
-20
-30
-40
0
50
100
150
200
250
300
350
400
 (Cam angle) - degrees
Figure 5. The third-order kinematic coefficients.
(iv)
The lift curve (displacement diagram) for the cam design is shown in Fig. 6.
Position Profile
3
2.5
Lift - in
2
1.5
1
0.5
0
0
50
100
150
200
250
 (Cam angle) - degrees
Figure 6. The lift curve (displacement diagram).
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300
350
400
Theory of Machines and Mechanisms, 4e
(v)
Uicker et al.
The radius of curvature of the cam surface is shown in Fig. 7.
Figure 7. The radius of curvature of the cam surface.
The radius of curvature of a cam profile should always be negative for a good cam
design. The positive curvature means that the cam has a concave surface and there is the
possibility that the follower may lose contact with the cam. If the radius of curvature of
the cam profile is positive then the radius of curvature of the cam must be greater than the
radius of the follower. In the proposed design, the positive values of the radius of
curvature of the cam are always greater than 0.5 in (i.e., the radius of the follower).
Note that the radius of curvature of the cam surface goes to infinity when the cam angle
is 4°, 26°, 168°, 191°, 225°, and 330°; i.e., there are six inflection points on the cam
surface. Also, note the discontinuity at the start and the end of the simple harmonic
profile at 210°, 270°, 285°, and 345°. This discontinuity is due to the simple harmonic
profiles and for this reason simple harmonic profiles are not generally recommended for
high-speed cam-follower systems.
(vi)
The pressure angle for the cam-follower system is shown in Fig. 8.
Pressure Angle
50
45
40
 Pressure angle - degrees
35
30
25
20
15
10
5
0
0
50
100
150
200
250
300
350
400
 (Cam angle) - degrees
Figure 8. The pressure angle of the cam-follower system.
The recommended value of the pressure angle is less than 30 degrees. In this design, the
pressure angle is more than the accepted value at the cam angles  2  18  62,
134  178, 220  256, and 302  336 . Therefore, this cam profile is not a good
design. The high values of the pressure angle may be due to the selection of the
displacement curves and the dimensions of the cam and the diameter of the roller.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
Chapter 7
Spur Gears
7.1
Find the module of a pair of gears having 32 and 84 teeth, respectively, whose center
distance is 87 mm.
mN 2 mN 3 m  32  84 


 87 mm
2
2
2
2  87 mm 
m
 1.5 mm/tooth
116 teeth
R2  R3 
7.2
Find the number of teeth and the circular pitch of a 150-mm pitch diameter gear whose
module is 2.5 mm/tooth.
N  2 R m  150 mm   2.5 mm/tooth   60 teeth
p   m    2.5 mm/tooth   7.854 mm/tooth
7.3
Ans.
Ans.
Ans.
Determine the module pitch of a pair of gears having 18 and 40 teeth, respectively, whose
center distance is 90.625 mm.
 N  N3 
R2  R3   2
m
2


 58 
90.625    m
 2 
2  90.625
 m=
58
3.125 mm/tooth
Ans.
7.4
Find the number of teeth and the circular pitch of a gear whose pitch diameter is 10.5 mm
in if module is 3.125 mm/tooth.
2R
 125 mm / 3.125 mm per tooth
m
p =  m  3.14  3.125 = 9.8125 mm/tooth
N
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
7.5
Find the module and the pitch diameter of a 40-tooth gear whose circular pitch is 37.7
mm/tooth.
m  p    37.7 mm/tooth    12 mm/tooth
D  2 R  mN  12 mm/tooth  40 teeth   480 mm
7.6
p

9.815
3.14
= 3.125 mm/tooth
36  3.125
NM
=
 35.83mm
D

3.14
Ans.
Ans.
The pitch diameters of a pair of gears are 62 mm and 100 mm, respectively. If their
module is 2 mm per/tooth, how many teeth are there on each gear?
N 2  2 R2 / m  D2 / m   62 mm / 2 mm per tooth   31 teeth
Ans.
N 3  2 R3 / m  D3 / m  100 mm / 2 mm per tooth   50 teeth
Ans.
What is the diameter of a 33-tooth gear if its module is 2 mm/tooth?
D  2 R  mN   2 mm/tooth  33 teeth   66 mm
7.10
Ans.
where P is circular pitch
=
7.9
Ans.
Find the module and the pitch diameter of a gear whose circular pitch is 9.815 mm/tooth
if the gear has 36 teeth.
m
7.8
Ans.
Ans.
The pitch diameters of a pair of mating gears are 42 mm and 102 mm, respectively. If the
module is 1.5 mm/tooth how many teeth are there on each gear?
N 2  2 R2 m  D2 m   42 mm  1.5 mm/tooth   28 teeth
N 3  2 R3 m  D3 m  102 mm  1.5 mm/tooth   68 teeth
7.7
Uicker et al.
Ans.
A shaft carries a 30-tooth, 3-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 105
mm?
R2  mN 2 2   3 mm/tooth   30 teeth  2  45 mm
R3   R2  R3   R2  105 mm  45 mm  60 mm
2 
R3
60 mm
3 
 480 rev/min   640 rev/min
R2
45 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
7.11
Uicker et al.
Two gears having an angular velocity ratio of 3:1 are mounted on shafts whose centers
are 150 mm apart. If the module of the gears is 3 mm/tooth, how many teeth are there on
each gear?
  
3
 R2  R3   150 mm   1  2  R2  1   R2  4 R2
 1
 3 
R2  37.5 mm in
N2  2R2 / m  2  37.5 mm / 3 mm / tooth   25 teeth
Ans.
N3  2R3 / m  2 112.5 mm/ 3 mm/ tooth   75 teeth
Ans.
R3   R3  R2   R2  112.5 mm
7.12
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 170 mm?
R2  mN 2 2   4  21/ 2 
 42 mm
R3   R2  R3   R2  170 mm   42 mm  128 mm
2 
7.13
R3
128 mm
3 
 240 rev/min   731.4 rev/min
R2
42 mm
Ans.
A 6.35 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, pitch circle radii, base circle radii,
length of paths of approach and recess, and contact ratio.
a  m  6.35 mm / tooth  6.35 mm
d  1.25 m  1.25  6.35 mm / tooth  7.9375 mm
c  d  a   7.9375 mm    6.35 mm   1.5875 mm
Ans.
Ans.
Ans.
p  πm =(π  6.35 mm / tooth)  19.94 mm/tooth
pb  p cos   19.94 mm/tooth  cos 20  18.737mm/tooth
Ans.
Ans.
t  p 2  19.94 mm/tooth  2  9.97 mm
Ans.
6.35 mm/tooth  24 teeth
 76.2 mm
2
6.35 mm/tooth  36 teeth
 114.3 mm
2
2
R2  mN 2 2 
R3  mN 3
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
r2  R2 cos    76.2 mm  cos 20  71.6 mm ; r3  R3 cos   114.3 mm  cos 20  107.4 mm Ans.
CP  15.875 mm [measured or by Eq. (7.10)]
PD  15 mm [measured or by Eq. (7.11)]
CP  PD 15.875 mm   15 mm 
mc 

 1.647 teeth avg.
18.737 mm/tooth
pb
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
7.14 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?
Since the addendum circle of internal gear 3 is of lesser radius (71.12 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 an internal gear it is called
fouling.
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 cannot 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 6.35 mm/tooth so that the addendum circle of the internal gear is 63.5 mm.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
7.15
Uicker et al.
A 10 mm/tooth module, pitch 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  m  10 mm/tooth = 10 mm
p  πm    10 mm/tooth = 31.4 mm
pb  p cos    31.4 mm/tooth  cos 20  29.5 mm/tooth
10 mm/tooth 17 teeth
 85 mm R3  mN
2
r2  R2 cos    85 mm  cos 20  79.87 mm
R2  mN
2
2
3
2
10 mm/tooth  50 teeth
 250 mm
2
r3  R3 cos    250 mm  cos 20  234.923 mm
CP  26.314 mm [Eq. (7.10)]
PD  22.707 mm [Eq. (7.11)]
CP 26.314 mm
CP 26.314 mm
2 

 0.110 rad  6.32 Ans.

 0.324 rad  18.58  3 
r2
79.87 mm
r3
234.923 mm
PD 22.707 mm
PD 22.707 mm
2 

 0.095 rad  5.45 Ans.

 0.280 rad  16.04  3 
r2
79.87 mm
r3
234.973 mm
mc 
7.16
CP  PD  26.314 mm    22.707 mm 

 1.66 teeth avg.
29.5 mm/tooth
pb
A gearset with a module of 5 mm/tooth has involute teeth with 22½° pressure angle, and
has 19 and 31 teeth, respectively. They have 1.0m for the addendum and 1.25m for the
dedendum.* 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   6.75 mm
c  d  a  1.75 mm
p   m    5 mm/tooth   15.708 mm/tooth
pb  p cos   15.708 mm/tooth  cos 22.5  14.512 mm/tooth
t  p 2  7.854 mm
R2  N 2 m 2  19 teeth  5 mm/tooth  2  47.500 mm
R3  N 3m 2   31 teeth  5 mm/tooth  2  77.500 mm
r2  R2 cos    47.500 mm  cos 22.5  43.884 mm
r3  R3 cos    77.500 mm  cos 22.5  71.601 mm
CP  11.325 mm [Eq. (7.10)]
PD  10.640 mm [Eq. (7.11)]
CP  PD 11.325 mm   10.640 mm 
mc 

 1.51 teeth avg.
pb
14.512 mm/tooth
*
Ans.
In SI, tooth sizes are given in modules, m, and a = 1.0m means 1 module, not 1 meter.
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
7.17
Uicker et al.
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.* Find the
lengths of the paths of approach and recess and determine the contact ratio.
a  1.0m  8.0 mm
p   m    8 mm/tooth   25.133 mm/tooth
pb  p cos    25.133 mm/tooth  cos 25  22.778 mm/tooth
R2  N 2 m 2   22 teeth  8 mm/tooth  2  88.0 mm
CP  a sin   18.930 mm [Fig. 7.10]
PD  16.243 mm [Eq. (7.11)]
CD 18.930 mm   16.243 mm 
mc 

 1.54 teeth avg.
22.778 mm/tooth
pb
7.18
Ans.
Ans.
Ans.
Repeat Problem 7.15 using the 25° full-depth system.
a  m  10 mm/tooth = 10 mm
p   m   10 mm/tooth = 31.4 mm/tooth
pb  p cos    31.4 mm/tooth  cos 25  28.458 mm/tooth
10 mm/tooth  17 teeth
10 mm/tooth  50 teeth
 85 mm R3  m N 3 / 2 
 250 mm
2
2
r2  R2 cos    85 mm  cos 25  77 mm r3  R3 cos    250 mm  cos 25  226.576 mm
R2  m N 2 / 2 
CP  22.225 mm [Eq. (7.10)]
PD  19.989 mm [Eq. (7.11)]
CP 22.225 mm
CP 22.225 mm

 0.288 rad  16.50  3 
2 

 0.098 rad  5.61 Ans.
r2
77 mm
r3
226.576 mm
PD 19.989 mm
PD 19.989 mm
2 

 0.088 rad  5.04 Ans.

 0.259 rad  14.85 3 
r2
77 mm
r3
226.576 mm
CD  22.225 mm   19.989 mm 
Ans.
mc 

 1.48 teeth avg.
pb
28.458 mm/tooth
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
7.19
Uicker et al.
Draw a 12.7 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 3.175 mm in further
away from the gear center. Determine the new contact ratio. Has the pressure
angle changed?
12.7 mm
3.175 mm
(a)
(b)
CP  a sin   12.7 mm sin 20  37.132 mm [see figure]
PD  30.3784 mm [Eq. (7.11)]
CD  37.132 mm    30.3784 mm 
mc 

 1.80 teeth avg.
37.49 mm/tooth
pb
Ans.
Ans.
Ans.
Since the pressure angle is a property that has determined the shapes of the teeth
on both the rack and the pinion, moving the rack by 3.175 mm does not change
the tooth shapes or the pressure angle. The modified contact ratio is:
C P  a sin    9.525 mm sin 20  27.849 mm [see figure]
Ans.
Ans.
PD  30.378 mm [Eq. (7.11)]
C D  27.849 mm    30.378 mm 
Ans.
mc 

 1.55 teeth avg.
37.49 mm/tooth
pb
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
7.20
Uicker et al.
through 7.24 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. Next, fasten the sheet of
plastic to the drawing and scribe the cutout, the pitch circle, and the radial lines onto the
sheet. Then remove the sheet and trim the tooth outline with a razor blade. Then use a
small piece of fine sandpaper to remove any burrs.
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 1-tooth/in diametral pitch 20 full-depth template
constructed as described above. In each case you should generate a few teeth and
estimate the amount of undercutting
Problem No.
7.20
7.21
7.22
7.23
7.24
No. of Teeth
10
12
14
20
36
The diagram used to make
the plastic template for
Problems 7.20 through 7.24
is shown at the left.
The drawing generated for
Problem 7.20 is also shown
at left. Note how the tip(s)
of the shaper cutter 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 7.21 to 7.24 are
similar.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
7.25
Uicker et al.
A 10-mm/tooth module gear has 17 teeth, a 20° pressure angle, an addendum of 1.0m,
and a dedendum of 1.25m.* 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   R N    85.0 mm  17  15.708 mm
inv   inv 20  0.014 904
At the base circle:
r  rb  R cos    85.0 mm  cos 20  79.874 mm
inv   inv 0  0.0
From Eq. (7.16)
t

t  2r  p  inv   inv  
 2R

 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
*
Ans.
 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.
In SI, tooth sizes are given in modules, m, and a = 1.0m means 1 module, not 1 meter.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
7.26
Uicker et al.
A 15-tooth pinion has 16.9 mm/tooth module, 20° full-depth involute 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  MN 2  16.9 mm/tooth  15 teeth  2  126.75 mm
t p   R N   126.75 mm  15 teeth   26.533 mm
inv   inv 20  0.014 904
At the base circle:
r  rb  R cos   126.75 mm  cos 20  119.1 mm
inv   inv 0  0.0
From Eq. (7.16)
t

t  2r  p  inv   inv  
 2R

 26.533 mm

t  2 119.1 mm  
 0.014 904  0.0   28.48 mm
 2 126.75 mm 

At the addendum circle:
ra  R  a  126.75 mm  16.9 mm  143.65 mm
7.27
Ans.
  cos 1  rb ra   cos 1 119.1 mm 143.65 mm   33.99
Ans.
 26.533 mm

t  2 143.65 mm  
 0.014 904  0.081 018  11.076 mm
 2 126.75 mm 

Ans.
A tooth is 19.9 mm thick at a pitch circle radius of 200 mm in and a pressure angle of
25°. What is the thickness at the base circle?
At the base circle:
r  rb  R cos    200 mm  cos 25  181.26 mm
inv   inv 0  0.0
From Eq. (7.16)
 tp

t  2r 
 inv   inv  
 2R

 19.9 mm

t  2 181.26 mm  
 0.029 975  0.0   28.9 mm
 2  200 mm 

© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
7.28
Uicker et al.
A tooth is 39.9 mm thick at the pitch radius of 400 mm and has a pressure angle of 20°.
At what radius does the tooth become pointed?
t

t  2r  p  inv   inv    0
 2R

inv   t p 2 R  inv   39.9 mm 2  400 mm   0.014 904  0.064779
r  rb cos    400 mm  cos 20 cos 31.647  441.53 mm
7.29
Ans.
A 25˚ full-depth involute, 2.1 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  2.1 mm/tooth  18 teeth  2  18.9 mm
t p   R N   18.9 mm  18 teeth   3.297 mm
inv   inv 25  0.029 975
At the base circle:
r  rb  R cos   18.9 mm  cos 25  17.129 mm
inv   inv 0  0.0
t

t  2r  p  inv   inv  
 2R

 3.297 mm

t  2 17.129 mm  
 0.029 975  0.0  4 mm
 2 18.9 mm 

At the addendum circle:
ra  R  a  18.9 mm  2.1 mm  21 mm
Ans.
  cos 1  rb ra   cos 1 17.129 mm 21 mm   35.35
Ans.
 3.297 mm

t  2  21 mm  
 0.029 975  0.092 339   1.04 mm
 2 18.9 mm 

Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
7.30 A nonstandard 10-tooth 3.1 mm/tooth module involute pinion is to be cut with a 22½°
pressure angle. What maximum addendum can be used before the teeth become pointed?
At the pitch circle:
rp  R  MN 2   3.1 mm/tooth 10 teeth  2  15.5 mm
t p   R N   15.5 mm  10 teeth   4.867 mm
At the addendum circle:
 tp

t  2r 
 inv   inv    0
 2R

inv   t p 2 R  inv   4.867 mm 2 15.5 mm   0.021 514  0.18214 ;   42.772
r  rb cos   15.5 mm  cos 22.5 cos 42.772  19.5 mm
a  r  R  19.5 mm  15.5 mm  4 mm
7.31
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   120 mm  cos 20  112.76 mm
  rb tan   112.76 mm  tan 20  41 mm
   2 N   2  96 teeth   0.016362 rad  0.937 5
s  rb tan        112.76 mm  tan  20.937 5    41 mm   2.14 mm
d  2 s  2  2.14 mm   4.28 mm
(b)
Ans.
rs  rb cos      112.76 mm  cos20.9375=120.73 mm
distance over pins  2  rs  s   2 120.73 mm  2.14 mm   245.74 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
7.32
7.33
Uicker et al.
A set of interchangeable gears with 6.35 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  , mm
 p  rb tan  , mm
24
32
48
96
76.2
101.6
152.4
304.8
71.6
95.47
143.2
286.4
26.06
34.74
52.12
104.24
N , teeth
ra , mm
rb ra
 a  cos 1  rb ra  , deg
 a  rb tan a , mm
24
32
48
96
82.55
107.95
158.75
311.15
0.86741
0.88442
0.90211
0.92052
29.84
27.82
25.56
23.00
41.07
50.36
68.50
121.56
Calculate the contact ratio of a 17-tooth pinion that drives a 73-tooth gear. The gears are
0.26 mm/tooth module and cut on the 20° fine pitch system.
a  m  0.26 mm/tooth  0.26 mm
pb   m  cos    0.26 mm/tooth  cos 20  0.767 mm/tooth
MN 3 0.26  73 teeth
MN 2 0.26  17 teeth

 9.49 mm

 2.21 mm R3 
2
2
2
2
CP  0.708 mm [measured or by Eq. (7.10)]
PD  0.591 mm [measured or by Eq. (7.11)]
CD  0.708 mm    0.591 mm 

 1.693 teeth avg.
mc 
0.767 mm/tooth
pb
R2 
7.34
Ans.
A 25° pressure angle 11-tooth pinion is to drive a 23-tooth gear. The gears have a
module of 3.175 mm/tooth and have stub teeth. What is the contact ratio?
a  0.8 m  0.8  3.175 mm/tooth = 2.54 mm (Notice the stub teeth.)
pb  πm cos     3.175 mm/tooth  cos 25  9 mm/tooth
MN 3 3.175  23 teeth
MN 2 3.175  11

 36.5125 mm

 17.4625 mm R3 
2
2
2
2
CP  5.3 mm [measured or by Eq. (7.10)]
PD  4.85 mm [measured or by Eq. (7.11)]
CD  5.3 mm    4.85 mm 
Ans.

 1.127 teeth avg.
mc 
pb
9 mm/tooth
R2 
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
7.35
Uicker et al.
A 22-tooth pinion mates with a 42-tooth gear. The gears have full-depth involute teeth,
have a module of 1.5 mm/teeth, and are cut with a 17½° pressure angle.* Find the contact
ratio.
a  M  1.5 mm/tooth  1.5 mm
pb   m  cos     1.5 mm/tooth  cos17.5  4.49 mm/tooth
MN3 1.5  42 teeth
MN 2 1.5  22 teeth
R2 

 31.5 mm

 16.5 mm R3 
2
2
2
2
CP  4.43 mm [measured or by Eq. (7.10)]
PD  3.997 mm [measured or by Eq. (7.11)]
CD  4.43 mm    3.997 mm 

 1.877 teeth avg.
mc 
4.49 mm/tooth
pb
7.36
Ans.
The center distance of two 24-tooth, 20 pressure angle, full-depth involute spur gears
with module of 12.7 mm/tooth is increased by 3.17 mm over the standard distance. At
what pressure angle do the gears operate?
The original
identical.
two
gears
are
mN
2
12.7  24 teeth

2
 152.4 mm
R2  R3 
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  152.4 mm  cos 20
 143.2 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   cos   cos  
r r
143.2 mm  143.2 mm
 21.56
   cos 1 2 3  cos 1
R2  R3
307.96 mm
R2  R3 
*
Such gears came from an older standard and are now obsolete.
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
7.37
Uicker et al.
The center distance of two 18-tooth, 25 pressure angle, full-depth involute spur gears
with module of 8.5 mm/tooth is increased by 1.5875 mm in over the standard distance.
At what pressure angle do the gears operate?
Consult the figure and the discussion with the solution of Problem 7.36.
MN 8.5 mm/tooth 18 teeth
R2  R3 

 76.5 mm
2
2
r2  r3   76.5 mm  cos 25  69.3 mm
   cos 1
7.38
r2  r3
69.3 mm  69.3 mm
 cos 1
 25.82
153.98 mm
R2  R3
Ans.
A pair of mating gears have 1 mm/tooth module and are generated on the 20° full-depth
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 mm/tooth  50 teeth
MN 2 1 mm/tooth  15 teeth

 7.5 mm R3 

 25 mm
2
2
2
2
r2  R2 cos    7.5 mm  cos 20  7 mm
R2 
r3  R3 cos    25 mm  cos 20  23.5 mm
From Eq. (7.12), using Eqs. (7.10) and (7.11),
a2  r22   R2  R3  sin 2   R2  5.99 mm
Ans.
a3  r32   R2  R3  sin 2   R3  1.1 mm
Ans.
2
2
7.39
A set of gears is cut with a 114 mm/tooth circular pitch and a 17½° pressure angle.* The
pinion has 20 full-depth teeth. If the gear has 240 teeth, what maximum addendum may
it have in order to avoid interference?
M  P    114 mm/tooth /   36.3 mm/tooth
MN3 36.3 mm/tooth  240 teeth
MN 2 36.3 mm/tooth  20 teeth

 4356 mm

 363 mm R3 
2
2
2
2
r3  R3 cos    4356 mm  cos17.5  4154.39 mm
R2 
a3  r32   R2  R3  sin 2   R3  34.272 mm
2
*
Such gears came from an older standard and are now obsolete.
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
7.40
Uicker et al.
Using the method described for Problems 7.20 through 7.24, 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-center-distance
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
7.11 under the title of Center-Distance Modification. It proceeds as follows:
  20 , M  25 mm/tooth , p  πm  78.5 mm/tooth ,
a  m  25 mm , d  1.35 m  33.75 mm , c  0.35 m  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
R2 

 137.5 mm R3 

 312.5 mm
2
2
2
2
r2  R2 cos   129.2 mm
r3  R3 cos   293.65 mm
e  a  R2 sin 2   9 mm
t2  2e tan   p 2  46.67 mm
t3  p 2  39.25 mm
inv   
N 2  t2  t3   2 R2
2 R2  N 2  N 3 
 inv   0.022 115 rad ,
   22.70
R cos 
R2 cos 
 142.85 mm
R3  3
 324.67 mm
cos  
cos  
R2  R3  467.52 mm
Working depth  50.2 mm
R2  a2  174.28 mm
R3  a3  343.69 mm
CP  43.1 mm
PD  58.68 mm
mc  CD pb  1.36 teeth avg.
R2 
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
7.41
Uicker et al.
Using the template cut in Problem 7.40 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.
  20 , m  25 mm/tooth , p   m  78.5 mm/tooth ,
a  m  25 mm , d  1.35 m  33.75 mm , c  0.35 m  8.75 mm ,
N 2  11 teeth ,
N 3  44 teeth ,
MN3 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   516.83 mm
r2  R2 cos   129.2 mm
41.59
4159
18.1
32 .7
26.98
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. (7.10) and (7.12) we can show that
a3  r32   R2  R3  sin 2   R3  18.1 mm
2
Ans.
Since the working depth of standard gears is retained, the new dedendum of the gear is
Ans.
d3  m  1.35 m  a3  41.59 mm in
Then, retaining the same clearance and pitch point P,
Ans.
a2  d3  0.85 P  32.7 mm ,
d 2  m  1.35 m  a2  26.98 mm
Assuming a standard rack cutter with a  m  25 mm , Fig. 7.26 shows the offset is
Ans.
e2  a  d 2  1.582 mm ,
e3  a  d3  16.197 mm
From Eqs. (7.9), (7.10), and (7.11) the contact ratio is
CP  47.779 mm
PD  63.982 mm
Ans.
mc  CD pb  1.49 teeth avg.
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.
7.42
A pair of involute spur gears with 9 and 36 teeth are to be cut with a 20 full-depth cutter
with module of 8.5 mm/tooth.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
(a)
(b)
Uicker et al.
Determine the amount that the addendum of the gear must be decreased in order
to avoid interference.
If the addendum of the pinion is increased by the same amount, determine the
contact ratio.
  20 , M  8.5 mm/tooth , p   m  26.69 mm/tooth ,
N 2  9 teeth ,
N 3  36 teeth ,
mN 3 8.5 mm/tooth  36 teeth
mN 2 8.5 mm/tooth  9 teeth
R2 

 38.25 mm R3 

 153 mm
2
2
2
2
r2  R2 cos   35.94 mm
r3  R3 cos   143.77 mm
(a)
From Eqs. (7.10) and (7.12)
a3  r32   R2  R3  sin 2   R3  4.935 mm
2
(b)
7.43
  m - a3  3.565 mm
Ans.
a2  m    12.065 mm
a3  m    4.935 mm
From Eqs. (7.9), (7.10), and (7.11) the contact ratio is
CP  13 mm
PD  22 mm
mc  CD p cos   1.40 teeth avg.
Ans.
A standard 20° pressure angle full-depth involute 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 indicating 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.
  20 , m  25 mm/tooth , a  m  25 mm ,  2  500 rev/min  52.360 rad/s ,
N 2  20 teeth ,
R2  mN 2 2  250 mm
N 3  48 teeth ,
R3  mN 3 2  600 mm
r3  R3 cos   563.82 mm
r2  R2 cos   234.92 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 3  2   X  R2 R3  1 2  1854.4 X mm/s
and, using Eqs. (7.9) and (7.10), X varies between
X init  CP  65.5 mm and
X final  PD  58.37 mm .
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
mm/s
5000
2500
-75
-50
-25
25
50
75
-mm
-2500
-5000
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
Chapter 8
Helical Gears, Bevel Gears, Worms, and Worm Gears
8.1
A pair of parallel-axis helical gears has 14½° normal pressure angle, module of
3mm/teeth, and 45° helix angle. The pinion has 15 teeth, and the gear has 24 teeth.
Calculate the transverse and normal circular pitch, the normal module of 3 mm/teeth, the
pitch radii, and the equivalent tooth numbers.
N2  15 teeth ,
m  3 mm/teeth ,
pt   m  9.43 mm/tooth ,
8.2
pn  pt cos  6.668 mm/tooth ,
Ans.
Ans.
R2  mN2  2   22.5 mm ,
R3  mN3  2   36 mm ,
Ans.
Ne 2  N2 cos3   42.43 teeth ,
Ne3  N3 cos3   67.88 teeth
Ans.
A set of parallel-axis helical gears are cut with a 20° normal pressure angle and a 30°
helix angle. They have module of 1.8 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.
N2  16 teeth ,
t  tan
1
 tan n
cos   22.796 ,
N3  40 teeth ,
m = 1.8 mm/tooth
pt   m  5.657 mm/tooth ,
Ans.
Ans.
R2  mN2  2   14.4 mm ,
px  pt tan  9.798 mm/tooth ,
R3  mN3  2   36 mm ,
Re 2  R2 cos2   19 mm ,
Re3  R3 cos2   47.62 mm
Ans.
pn  pt cos  4.899 mm/tooth ,
8.3
N3  24 teeth ,
A parallel-axis helical gearset is made with a 20° transverse pressure angle and a 35°
helix angle.The gears have module of 2mm/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.
 b  tan 1  tan cos t   33.34 ,
Ans.
m  2 mm/tooth ,
mx  F tan pt  2.006 teeth avg
pt   m  6.2857 mm/tooth ,
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
8.4
Uicker et al.
A set of helical gears is to be cut for parallel shafts whose center distance is to be about
88.9 mm in 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 2.1 mm/teeth. Using a helix angle of
30°, determine the transverse values of the diametral and circular pitches and the tooth
numbers, pitch radii, and center distance.
2 3  R3 R2  1.8 ,
R3  1.8R2 ,
R2  R3  R2  1.8R2  2.8R2  88.9 mm , R2  31.75 mm, R3  57.15 mm ,
m = 2.1 mm/tooth ,
m
2.1
Ans.
 mt =

 2.425 mm/tooth ,
pt   m  7.85 mm/tooth
cos cos 20
2R
2R
2  31.75 mm
2  31.75 mm
N2  2 
 26.18 say 26 , Now, N3  3 
 47.15 say 47 ,
mt 2.425 mm/tooth
mt 2.425 mm/tooth
Therefore we will use N2  26 teeth and N3  47 teeth
Ans.
mtN3 2.425  47
mtN 2 2.425  26
Ans.

 56.987 mm ,
R2 

 31.525 mm , R3 
2
2
2
2
Ans.
R2  R3  88.512 mm
8.5
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 25° and a pressure angle of 20°, determine the values for the tooth
numbers, pitch radii, normal circular and module as well as and face width.
2 3  R3 R2  1800 400  4.5
R3  4.5R2
R2  R3  R2  4.5R2  5.5R2  275 mm , R2  50 mm, R3  225 mm ,
Ans.
N2  16 teeth ,
Ans.
N3   R3 R2  N2  4.5N2  72 teeth
2  50
 6.25 mm/tooth , mt = mn / cos  6.25 / cos 25  6.896 mm/tooth ,
16
pt  π mn = 3.14  6.25  19.625 mm/tooth ,
pn   Pn  0.711 8 in/tooth ,
m2  2R 2 / N 2 
px  pt tan  19.625 / tan 25  42.08 mm/tooth F   2 teeth  px  2  42.08  84.16 mm
Therefore, we may choose F  84 mm .
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
8.6
The catalog description of a pair of helical gears is as follows: 14½° normal pressure
angle, 45° helix angle, module of 3.175 mm/tooth, 25 mm in face width, and normal
module of 2.24 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 ,
R3  100 mm  2  50 mm ,
a  m  3.175 mm ,
pt  πm = 9.9695 mm/tooth ,
tan t  tan n cos  0.365 7 ,
t  tan 1  0.365 7   20.09  20.00 ,
pb  pt cos t  9.9695 cos14.5  9.36819 mm/tooth ,
r2  R2 cos t  7.69 mm ,
r3  R3 cos t  46.98 mm
CP  7.69 mm [Eq. (7.10)],
PD  6.55 mm [Eq. (7.11)],
mt  CD pb  1.54 teeth avg ,
Ans.
 b  tan
mn  mt cos  b  2.91 teeth avg
Ans.
m  mx  mt  4.09 teeth avg
Ans.
1
 tan cos t   43.22 ,
mx  tan pt  2.55 teeth avg ,
8.7
Uicker et al.
2
In a medium-sized truck transmission a 22-tooth clutch-stem gear meshes continuously
with a 41-tooth countershaft gear. The data indicate normal module of 3.34 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 t  tan n cos  0.364 9 ,
t  tan 1  0.364 9  20.04  20.00
Pn  7.600 teeth/in ,
a  mn  3.34 mm ,
mt  πmt  11.429 mm/tooth ,
mt  mn cos  3.64 mm/tooth ,
R2  mN2  2   40 mm ,
r2  R2 cos t  37.58 mm ,
CP  8.55 mm [Eq. (7.10)],
pb  pt cos t  9.85 mm/tooth ,
 b  tan 1  tan cos t   22.22 ,
mx  F tan pt  1.08 teeth avg ,
R3  mN3  2   74.62 mm ,
r3  R3 cos t  70.12 mm
PD  6.657 mm [Eq. (7.11)],
mt  CD pb  1.53 teeth avg ,
mn  mt cos2  b  1.79 teeth avg
Ans.
m  mx  mt  2.87 teeth avg
Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
8.8
Uicker et al.
A helical pinion is right-handed, 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 2.5 mm/tooth. Find the helix angle and the number of teeth on the
mating gear. What is the shaft center distance?
 3    2  30 RH ,
N3  2 3  N2  36 teeth
R2  mN2 2cos 2  30 mm ,
R3  mN3  2cos 3   51.96 mm
R2  R3  81.96 mm
8.9
8.10
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 2.5 mm/tooth. Calculate the helix angle and the number of teeth on
the mating gear. Also determine the pitch radius of each gear.
 3    2  15 RH ,
N3  2 3  N2  39 teeth
Ans.
R2  mN2  2cos 2   28.97 mm ,
R3  mN3  2cos 3   50.47 mm
Ans.
Gear 2 in Fig P8.10 is to rotate clockwise and drive gear 3 counterclockwise at a velocity
ratio of 2. Use a normal module of 4.25mm/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.
 2   3   2  25 ,
2 3  R3 R2  2.0 , R3  2.0R2 ,
Ans.
R2  R3  R2  2.0R2  3.0R2  250 mm ,
R2  83.33 mm, R3  166.66 mm ,
N2  2cos 2 R2 / mn  35.5 teeth ,
N3  2cos 3 R3 / mn  71.1 teeth
Therefore we choose
N2  36 teeth ,
Ans.
N3  72 teeth ,
Ans.
R2  mN2  2cos 2   84.4 mm ,
R3  mN3  2cos 3   168.8 mm ,
R2  R3  253.2 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
8.11
8.12
A pair of straight-tooth bevel gears are to be manufactured for a shaft angle of 90°. If the
driver is to have 16 teeth and the velocity ratio is to be 3:1, what are the pitch angles?
N2  16 teeth ,
N3  2 3  N2  3N2  48 teeth ,
 2  tan 1  N2 N3   18.43 ,
 3  90   2  71.57


sin 
  23.41 ,
 2 3   cos  
 3  60   2  36.59
Ans.
A pair of straight-tooth bevel gears is to be mounted at a shaft angle of 120°. The pinion
and gear are to have 16 and 36 teeth, respectively. What are the pitch angles?


sin 
  26.33 ,  3  120   2  93.67
  N3 N 2   cos  
 2  tan 1 
8.14
Ans.
A pair of straight-tooth bevel gears has a velocity ratio of 1.5 and a shaft angle of 60°.
What are the pitch angles?
 2  tan 1 
8.13
Uicker et al.
Ans.
A pair of straight-tooth bevel gears with module of 12.7mm/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 radii of the equivalent spur gears.
R2  mN2  2   120.65 mm ,
R3  mN3  2   177.8 mm ,
Ans.
 2  tan 1  N2 N3   34.16 ,
 3  90   2  55.84 ,
Ans.
Using Table 8.2: m90  mG  N3 N2  1.474 , a3  9.55 mm ,
d3  Whole depth  a3  18.24 mm
Whole depth  55.58 2  27.79 mm ,
Working depth  2 m  25.4 mm ,
c  0.122 m  0.05 mm  2.43 mm
a2  Working depth  a3  15.85 mm
d2  Whole depth  a2  64.5 mm
Ans.
Ans.
Ans.
Cone distance,  R2 sin  2  214.8 mm Let F  0.3  4.465 mm , say F = 64.5mm Ans.
Re 2  R2 cos  2  145.8 mm
Re3  R3 cos  3  316.7 mm
Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
8.15
Uicker et al.
A pair of straight-tooth bevel gears with module of 3.2 mm/tooth have 18 teeth and 30
teeth, respectively, and a shaft angle of 105°. For each gear, calculate the pitch radius,
pitch angle, addendum, dedendum, face width, and equivalent number of teeth. Make a
sketch of the two gears in mesh. Use standard tooth proportions as for a 90° shaft angle.
R2  mN2  2   28.8 mm ,
R3  mN3  2   48 mm ,


sin 
  34.45 ,  3  105   2  70.55


N
N
cos


 3 2

Using Table 8.2: mG  N3 N2  1.667, m90  2.032 , a3  2.0675 mm ,
d3  Whole depth  a3  4.8793 mm
Whole depth  6.9469 mm ,
Working depth  2.188  3.2  6.4 mm , c  0.188 m  0.05 mm  0.6516 mm
a2  Working depth  a3  4.3325 mm d2  Whole depth  a2  2.6219 mm
Cone distance,  R2 sin  2  50.52 mm Let F  0.3  15.16 mm , say F = 15 mm
Re 2  R2 cos  2  34.64 mm
Re3  R3 cos  3  143 mm
Ne 2  2mRe 2  21.65 teeth ,
Ne3  2mRe3  89.375 teeth
 2  tan 1 
8.16
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
A worm having 4 teeth and a lead of 25 mm 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 44.5
mm.
px 
N2  25 mm 4 teeth  6.25 mm/tooth
N3  2 3  N2  7.5  4 teeth   30 teeth
R3  N3 px 2  29.86 mm
R2  44.5  R3  14.64 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
8.17
8.18
Uicker et al.
Specify a suitable worm and worm gear combination for a velocity ratio of 50 and a
center distance of 150 mm. Use an axial pitch of 12.5 mm/tooth.
Use N2  1 tooth ,
N3  2 3  N2  50 1 teeth   50 teeth
R3  N3 px 2  99.52 mm
R2  150 mm  R3  50.48 mm
A double-threaded worm drives a worm gear having 40 teeth. The axial pitch is 31.75
mm and the pitch diameter of the worm is 44.45mm. Calculate the lead and lead angle of
the worm. Find the helix angle and pitch diameter of the worm gear.
 N2 px  2 teeth  31.75 mm/tooth   63.5 mm
  tan 1  2 R2   tan 1  63.5 mm  44.45 mm   24.453
    24.453
R3  N3 px 2  40 teeth  31.75 mm/tooth  2  202.23 mm
8.19
Ans.
Ans.
Ans.
Ans.
Ans.
A double-threaded worm with a lead angle of 20° and an axial pitch of 12.7 mm/tooth
drives a worm gear with a velocity reduction of 16 to 1. Determine the following for the
worm gear: (a) the number of teeth, (b) the pitch radius, and (c) the helix angle. (d)
Determine the pitch radius of the worm. (e) Compute the center distance.
 N2 px  2 teeth 12.7 mm/tooth   25.4 mm
 2 tan    25.4 mm 2 tan 20  11.11 mm
R3  2 3  R2  16 11.11 mm   177.76 mm
Ans.
N3  2 R3 px  87.9 teeth
    20.0
R2  R3  11.11 mm  177.76 mm  188.87 mm
Ans.
Ans.
Ans.
R2 
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
Chapter 9
Mechanism Trains
9.1
Find the speed and direction of gear 8 in Fig. P9.1. What is the first-order kinematic
coefficient of the train?
N 2 N 4 N5 N7 18 15 33 16 5


N3 N5 N 6 N8 44 33 36 48 88
8  82 2   5 881 200 rev/min ccw   68.18 rev/min ccw
82 
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
9.2
Uicker et al.
Figure P9.2 gives the pitch diameters of a set of spur gears forming a train. Compute the
first-order kinematic coefficient of the train. Determine the speed and direction of
rotation of gears 5 and 7.
750mm
375mm
225mm
175mm
225mm
400mm
R R
R2 R4 175 mm 225 mm 7
7 750 mm 225 mm 21
Ans.

72  52 5 6 


R6 R7 50 225 mm 400 mm 80
R3 R5 375 mm 750 mm 50
Ans.
5  52 2   7 50120 rev/min cw   16.80 rev/min cw
52 
7  72 2   21 80120 rev/min cw   31.50 rev/min cw
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
9.3
Uicker et al.
Figure P9.3 illustrates a gear train consisting of bevel gears, spur gears, and a worm and
worm gear. The bevel pinion is mounted on a shaft that is driven by a V-belt on pulleys.
If pulley 2 rotates at 1200 rev/min in the direction indicated, find the speed and direction
of rotation of gear 9.
R2 N 4 N6 N8 6 in 18 20 3
3


R3 N5 N 7 N9 10 in 38 48 36 304
9  92 2   3 3041 200 rev/min   11.84 rev/min cw
92 
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
9.4
Uicker et al.
Use the truck transmission of Fig. 9.2 and an input speed of 4 000 rev/min to find the
drive shaft speed for each forward gear and for the reverse gear.
First gear:
92 
N 2 N 6 17 17 289


N3 N9 43 43 1 849
9  92 2   289 1 849  4 000 rev/min   625.2 rev/min
Second gear:
82 
N 2 N5 17 27 153


N3 N8 43 33 473
8  82 2  153 473 4 000 rev/min   1 293.9 rev/min
Third gear:
72 
Ans.
N 2 N 4 17 36 51


N3 N7 43 24 86
7  72 2   51 86 4 000 rev/min   2 372.1 rev/min
22  1.0
Fourth gear:
2  22 2  1.0 4 000 rev/min   4 000 rev/min
Reverse gear:
Ans.
92  
Ans.
Ans.
N 2 N6 N11
17 17 22
3 179


43 18 43
16 641
N3 N10 N9
9  92 2   3 179 16 641 4 000 rev/min   764.1 rev/min
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
9.5
Uicker et al.
Figure P9.5 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 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? Note that the problem has many solutions.
We also set N6  20 teeth (minimum).
Ans.
Since the largest speed reduction will be obtained with gears 2-5-6-9,
N N
20 20
450 rev/min  137 rev/min
C ,min  2 6  A 
N5 N9
N5 45
From this we get N5  29.197 , and we choose N5  30 teeth.
Ans.
Next, using distance units of circular pitch, the distance between shafts B and C is
Ans.
BC  N5  N8  N6  N9  N7  N10  65 teeth . N8  BC  N5  35 teeth .
Similarly the distance between shafts A and B is
AB  N2  N5  N3  N6  N4  N7  50 teeth . N3  AB  N6  30 teeth.
Ans.
Since the minimum is 20 teeth and since AB  N4  N7  50 teeth we see that
20  N4 , N7  30 and we choose
N4  N7  25 teeth
Ans.
N10  BC  N7  40 teeth .
and, finally,
Ans.
With all tooth numbers known, we can now find the output shaft speed for each gear
arrangement. These are:

Arrangement First-order kinematic coefficient, CA
Output shaft speed, C , rev/min
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
0.571
0.296
0.416
1.285
0.667
0.938
0.857
0.444
0.625
© Oxford University Press 2015. All rights reserved.
257.1
133.3
187.5
578.6
300.0
421.9
385.7
200.0
281.3
Theory of Machines and Mechanisms, 4e
9.6
Uicker et al.
The internal gear (gear 7) in Fig. P9.6 turns at 60 rev/min ccw. What are the speed and
direction of rotation of arm 3?
N 2 N 4 N6 20 teeth 40 teeth 36 teeth 20


N 4 N5 N7 40 teeth 18 teeth 154 teeth 77
  3 60 rev/min  3 20
 7

 ; 3  81.1 rev/min ccw
2  3
0  3
77
72 

72/3
9.7
Ans.
If the arm in Fig. P9.6 rotates at 400 rev/min ccw, find the speed and direction of rotation
of internal gear 7.
N 2 N 4 N6 20 teeth 40 teeth 36 teeth 20


N 4 N5 N7 40 teeth 18 teeth 154 teeth 77
  3 7  400 rev/min 20
 7


;   296.1 rev/min ccw
2  3 0  400 rev/min 77 7
72 

72/3
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
9.8
Uicker et al.
In Fig. P9.8a, shaft C is stationary. If gear 2 rotates at 600 rev/min ccw, what are the
speed and direction of rotation of shaft B?
N2
3
18 teeth
3
3  32 2    600 rev/min ccw   450 rev/min cw


N3
24 teeth
4
4
N N
18 teeth 20 teeth 3
85  5 7 

N6 N8 42 teeth 40 teeth 14
  3 0  450 rev/min
3
  8
Ans.

 ; 5  1 650 rev/min ccw
85/3
5  3 5  450 rev/min 14
32  
9.9
In Fig. P9.8a, consider shaft B as stationary. If shaft C is driven at 450 rev/min ccw,
what are the speed and direction of rotation of shaft A?
N5 N7 18 teeth 20 teeth 3


N6 N8 42 teeth 40 teeth 14
  3 450 rev/min  3 3
  8

 ; 3  572.7 rev/min ccw
85/3
5  3
0  3
14
N
24 teeth
572.7 rev/min ccw  763.6 rev/min cw
 A  2   3 3  
N2
18 teeth
85 
9.10
Ans.
In Fig. P9.8a, determine the speed and direction of rotation of shaft C under the
following conditions:
(a)
Shafts A and B both rotate at 450 rev/min ccw; and
(b)
Shaft A rotates at 450 rev/min cw and shaft B rotates at 450 rev/min ccw.
N5 N7 18 teeth 20 teeth 3


N6 N8 42 teeth 40 teeth 14
N
18 teeth
450 rev/min ccw  337.5 rev/min cw
3   2 2  
N3
24 teeth
85 
(a)
8  3
8  337.5 rev/min
3 ;   168.8 rev/min cw Ans.


8
5  3 450 rev/min  337.5 rev/min 14
N
18 teeth
450 rev/min cw  337.5 rev/min ccw
3   2 2  
N3
24 teeth
  3
8  337.5 rev/min
3 ;   361.6 rev/min ccw Ans.
  8


85/3
8
5  3 450 rev/min  337.5 rev/min 14
 
85/3
(b)
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
9.11
Uicker et al.
In Fig. P9.8b, 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?
N 2 N 4 N5 20 teeth 28 teeth 16 teeth
5


N 4 N5 N6 28 teeth 16 teeth 108 teeth 27
  3 0  3
5
5
  6
3    2
62/3


2  3 2  3 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.
62 
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
9.12
Uicker et al.
The Lévai type-L train illustrated in Fig. 9.10 has N2 = 16T, N4 = 19T, N5 = 17T, N6 =
24T, and N7 = 95T. Internal gear 7 is fixed. Find the speed and direction of rotation of
the arm if gear 2 is driven at 150 rev/min cw.
N 2 N 4 N6 16 teeth 19 teeth 24 teeth 384


N 4 N5 N7 19 teeth 17 teeth 95 teeth 1 615
  3
0  3
384
3  46.79 rev/min ccw
 7


;
2  3 150 rev/min  3 1 615
72 

72/3
9.13
Ans.
The Lévai type-A train of Fig. 9.10 has N2 = 20T and N4 = 32T.
(a)
If the module is m = 8 mm/tooth, 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 15 rev/min ccw, find the speed and
direction of rotation of the arm.
(a)
(b)
N5  N2  2 N4  20 teeth  2  32 teeth   84 teeth
Ans.
R3  m  N2  N4  2  8 mm/tooth  20 teeth  32 teeth  2  208 mm
Ans.
N 2 N 4 20 teeth 32 teeth
5


N 4 N5 32 teeth 84 teeth
21
  3 15 rev/min  3
5
 5


0  3
21
2  3
52 

52/3
3  12.12 rev/min ccw
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
9.14
Uicker et al.
The tooth numbers for the automotive differential illustrated in Fig. 9.22 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
2 
1 200 rev/min   377.8 rev/min
N3
54 teeth
N N
16 teeth 11 teeth
 1
65   5 4  
N4 N6
11 teeth 16 teeth
  3 6  377.8 rev/min
  6
6  755.6 rev/min
65/3

 1
5  3 0  377.8 rev/min
3 
9.15
Ans.
A vehicle using the differential illustrated in Fig. 9.22 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 9.14. 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)
car 
vcar


 48 km/h 1000 m/km   60 min/h   33.33 rad/min
 24 m 
For the right and left wheels, respectively:
 33.33 rad/min  24 m 1000 mm/m  750 mm   658.108 rev/min Ans.
v
6  R 
r
 2 rad/rev  375 / 2 mm 
vL  33.33 rad/min  24 m 1000 mm/m  750 mm 

 700.566 rev/min Ans.
r
 2 rad/rev  375/2 mm 
N N
16 teeth 11 teeth
65   5 4  
 1
11 teeth 16 teeth
N4 N6
  3 658.108 rev/min  3
  6

 1 3  679.34 rev/min
65/3
Ans.
5  3 700.566 rev/min  3
5 
(b)
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
9.16
Uicker et al.
Figure P9.16 illustrates 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 mandrel shaft.
(a)
Make a table demonstrating all possible gear arrangements, beginning with the
slowest 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/tooth, what must be the shaft center
distances?
N2 = 16T, N3 = 36T, N4 = 25T, N5 = 64T, N6 = 66T, N7 = 17T, N8 = 55T, N9 =
79T, N10 = 41T
(a)
(b)
Gears
B , rev/min
C , 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  N2  N5  2   5 mm/tooth 16 teeth  64 teeth  2  200 mm
Ans.
BC  m  N7  N9  2   5 mm/tooth 17 teeth  79 teeth  2  240 mm
Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
9.17
Uicker et al.
Shaft A in Fig. P9.17 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?
N2 = 16T, N3 = 18T, N4 = 16T, N5 = 18T, N6 = 50T
N 2 N3 N5
16 teeth 18 teeth 18 teeth
9


N3 N 4 N 6
18 teeth 16 teeth 50 teeth
25
  A 0  A
9
 A 6
62/

 ;
 A   9 34  2
2   A 2   A
25
This train is Lévai type F.
62  
9.18
Ans.
Ans.
In Problem 9.17, shaft B rotates at 150 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
A    9 25 2   9 25150 rev/min cw   54.0 rev/min cw
5
+1
+16/9
+25/9
6
+1
-1
0
Ans.
3   47 81 9 25 2    47 225150 rev/min cw   31.3 rev/min ccw
Ans.
4   25 9 9 25 2  1.0150 rev/min cw   150.0 rev/min cw
Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
9.19
Uicker et al.
Bevel gear 2 is driven by the engine in the reduction unit illustrated in Fig. P9.19. Bevel
planets 3 mesh with crown gear 4 and are pivoted on the spider (arm), which is connected
to propeller shaft B. Find the percentage speed reduction.
N2 = 36T, N3 = 21T, N4 = 52T; crown gear 4 is fixed
N 2 N3
36 teeth 21 teeth
9


N3 N 4
21 teeth 52 teeth
13
  B 0  B
9
 B 4

 ;
 42/
 B   9 22  2
2  B 2  B
13
Speed reduction to 9/22 = 40.9% is speed reduction of 59.1%.
 42  
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
9.20
Uicker et al.
In the clock mechanism illustrated in the Fig. P9.20, a pendulum on shaft A drives an
anchor (see Fig. 1.12c). The pendulum period is such that one tooth of the 30T
escapement 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.
(a)
Show that the train values are such that the minute hand rotates once every hour
and that the hour hand rotates once every 12 hours.
(b)
How many turns does the drum on shaft F make every day?
(a)
 B  1.0 rev/min
N B NC
8 teeth 8 teeth
1


NC N D 60 teeth 64 teeth 60
 B  1 60 rev/min
D   DB
tD  60 min/rev
 
 DB
N D N E  1  28 teeth 8 teeth
1
 

N E N H  60  42 teeth 64 teeth 720
720 min/rev
 B  1 720 rev/min
tH 
 12 hr/rev
H   HB
60 min/hr
Ans.
   DB

 HB
(b)
N D  1  8 teeth
1
 

N F  60  96 teeth 720
 B  1 720 rev/min  60 min/hr  24 hr/day   2 rev/day
F   FB
Ans.
   DB

 FB
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
Chapter 10
Synthesis of Linkages
10.1
A function varies from 0 to 1. Find the Chebychev spacing for two, three, four, five, and
six precision positions.
With x0  0.0 and xN 1  10.0 , Eq. (10.5) becomes:
x j  5.0  5.0cos
j\N
1
2
3
4
5
6
2
0.146447
0.853553
3
0.066987
0.500000
0.933013
(2 j  1)
2N
4
0.038060
0.308658
0.691342
0.961940
j  1, 2,..., N
5
0.024472
0.206107
0.500000
0.793893
0.975528
© Oxford University Press 2015. All rights reserved.
6
0.017037
0.146447
0.370591
0.629409
0.853553
0.982963
Theory of Machines and Mechanisms, 4e
10.2
Uicker et al.
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    180     1.20
and, from this, our design must have   16.40. Therefore we construct the point C
such that the central angle B1CB2  2  32.80. Using this point C as the center of a
circle ensures that any point O2 on this circle will have the angle
B1O2 B2    16.40 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  1 324.956 mm and O2 B2  r3  r2  801.946 mm and from these we find
Ans.
r2  261.50 mm
Ans.
r3  1 063.45 mm
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
10.3
Uicker et al.
Determine a set of link lengths for a slider-crank linkage such that the stroke is 400 mm
and the time ratio is 1.25.
After laying out the distance B1B2 = 400 mm, we see that the time ratio is
Q  180    180     1.25 and, from this, our design must have   20.00.
Therefore we construct the point C such that the central angle B1CB2  2  40.00.
Using this point C as the center of a circle ensures that any point O2 on this circle will
have the angle B1O2 B2    20.00 and thus will be a possible solution point. One
typical solution uses the point O2 shown.
400
392.25
740.5
Choosing the point O2 shown we measure the two distances O2 B1  r3  r2  740.5 mm
and O2 B2  r3  r2  392.25 mm and from these we find
r2  176.5 mm
r3  568.75 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
10.4
Uicker et al.
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 .
B1O4 B2    45 with BO4 = 500 mm, we see that the time
ratio is Q  180    180     1.25 and, from this, we find that   20.00.
After laying out the angle
Therefore we construct the point C such that the central angle B1CB2  2  40.00.
Then, using this point C as the center of a circle ensures that any point O2 on this circle
will have the angle B1O2 B2    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
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
10.5
Uicker et al.
A crank-and-rocker mechanism is to have a rocker of 1800 mm 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    75 with BO4 = 1800 mm, we see that the time
ratio is Q  180    180     1.32 and, from this, we find that   24.83.
Therefore we construct the point C such that the central angle B1CB2  2  49.66.
Then, using this point C as the center of a circle ensures that any point O2 on this circle
will have the angle B1O2 B2    24.83 and thus will be a possible solution point. One
typical solution uses the point O2 shown.
1907.25
3807.85
2434.25
Choosing the point O2 shown we measure the three distances O2O4  2434.25 mm ,
O2 B1  r3  r2  3807.85 mm , and O2 B2  r3  r2  1907.25 mm and from these we find
r1  2434.25 mm
r2  950.25 mm
r3  2857.5 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
10.6
Uicker et al.
Design a crank and coupler to drive rocker 4 in Fig. P10.6 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    60 with BO4 = 400 mm, we see that the time
ratio is Q  180    180     1.20 and, from this, we find that   16.36.
Therefore we construct the point D such that the central angle B1DB2  2  32.72.
Then, using this point D as the center of a circle ensures that any point O2 on this circle
will have the angle B1O2 B2    16.36 and thus will be a possible solution point. One
typical solution uses the point O2 shown.
549
895.75
895.75
626
Choosing the point O2 shown we measure the three distances O2O4  626 mm ,
O2 B1  r3  r2  895.75 mm , and O2 B2  r3  r2  549 mm and from these we find
r1  626 mm
r2  173.25 mm
r3  722.5 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
10.7
Uicker et al.
Design a crank and rocker for a six-link mechanism such that the slider in Fig. P10.6 for
Problem 10.6 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 .
After laying out the angle
B1O4 B2    2sin 1  400 1200   38.94 with BO4 = 1 200
mm, we see that the time ratio is Q  180    180     1.12 and, from this, we find
that   10.19.
Therefore we construct the point D such that the central angle B1DB2  2  20.38.
Then, using this point D as the center of a circle ensures that any point O2 on this circle
will have the angle B1O2 B2    10.19 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  1 979 mm ,
O2 B1  r3  r2  2 634 mm , and O2 B2  r3  r2  1 942 mm and from these we find
r1  1 979 mm
r2  346 mm
r3  2 288 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
10.8
Uicker et al.
Figure P10.8 illustrates 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.
10.9
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
O2 A  AB  O4 B  O2O4 . A 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.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
10.10 For Fig. P10.10, synthesize a linkage to move AB from position 1 to position 2 and
return.
10.11 Synthesize a mechanism to move AB successively through positions 1, 2, and 3 of Fig.
P10.11.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
10.12 through 10.21* Figure P10.12 illustrates 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 illustrated in Table P10.12 to P10.31. Plot a curve of the desired function
and a curve of the actual function that the linkage generates. Compute the maximum
error between them in percent.
Problem
number
Function,
y  f  x
Range
x0  0 ,
deg
y0 0 ,
deg
r2 r1
r3 r1
r4 r1
Max.
Error, deg
10.12, 10.22
log10 x
1 x  2
52.628
259.077
-3.352
0.845
3.485
0.0037
10.13, 10.23
sin x
0  x  2
-62.263
75.606
1.834
2.238
-0.693
0.1900
10.14, 10.24
tan x
0  x  4
269.709
124.189
-2.660
7.430
8.685
0.0380
10.15, 10.25
ex
0  x 1
241.644
40.422
-3.499
0.878
3.399
0.0258
10.16, 10.26
1x
1 x  2
33.804
120.213
-0.385
1.030
0.384
0.0161
10.17, 10.27
x1.5
0  x 1
-5.171
211.689
0.625
1.309
-0.401
0.1460
10.18, 10.28
x2
0  x 1
-29.321
233.836
2.523
3.329
-0.556
0.0673
10.19, 10.29
x 2.5
0  x 1
-88.313
44.492
-1.801
0.908
1.274
0.4120
10.20, 10.30
x3
0  x 1
-85.921
37.637
-1.606
0.925
1.107
0.5095
10.21, 10.31
x2
1  x  1
-21.180
-53.670
-0.610
0.565
0.380
2.3400
10.22 through 10.31 Repeat Problems 10.12 through 10.21 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.
*
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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Theory of Machines and Mechanisms, 4e
Uicker et al.
10.32 Figure P10.32 illustrates a coupler curve that can be generated by a four-bar linkage (not
illustrated). 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 precision positions, 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 traverses 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.
This coupler curve for point C was found from the Hrones and Nelson atlas, page 150.
The hesitation is shown by the following plot of the first-order kinematic coefficient 6 .
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Theory of Machines and Mechanisms, 4e
Uicker et al.
10.33 Synthesize a four-bar linkage to obtain a coupler curve having an approximate straightline segment. Then, using the suggestion included in Fig. 10.42b or Fig. 10.44b,
synthesize a dwell motion. Using an input crank angular velocity of unity, plot the
velocity of rocker 6 versus the input crank displacement.
The Hrones and Nelson atlas contains a wide variety of coupler curves similar to the one
shown; 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 6 .
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Theory of Machines and Mechanisms, 4e
Uicker et al.
10.34 Synthesize a dwell mechanism using the idea suggested in Fig. 10.42a 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 here; 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 6 .
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Theory of Machines and Mechanisms, 4e
Uicker et al.
Chapter 11
Spatial Mechanisms
11.1
Use the Kutzbach criterion to determine the mobility of the SSC linkage illustrated in Fig.
P11.1. Identify any idle freedoms and state how they can be removed. What is the
nature of the path described by point B?
RBA  RO3O  75 mm, RBO3  150 mm, and 2 = 30
n  3 , j2  3 , j3  1 , j1  j4  j5  0
m  6  3  1  4 1  3  2   2
Ans.
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.
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Theory of Machines and Mechanisms, 4e
11.2
Uicker et al.
For the SSC linkage illustrated in Fig. P11.1 express the position of each link in vector
form.
RO3O  75ˆi mm , R BA  75cos 2ˆi  75sin 2kˆ  64.952 08ˆi  37.5kˆ mm ,
R AO  RAO ˆj mm , RBO3  150 mm .
Ans.
Substituting these into RO3O  R BO3  R AO  R BA gives
R AO  450  cos 2  1ˆj  144.899ˆj mm ,
Ans.
R BO3  75  cos  2  1 ˆi  450  cos  2  1ˆj  75sin  2kˆ
Ans.
 10.048ˆi  144.889ˆj  37.5kˆ mm
11.3
For the linkage of Fig. P11.1 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  2ˆj  ω × R
B
3
2
BO3
BA
Assuming that the idle freedom is not active we can set ω3 R BO3  0 , where
ω3  3x ˆi  3y ˆj  3z kˆ and ω2  2 ˆj . Expanding these and using the position data from
Problem 11.2 gives four simultaneous equations:
37.5   2   0 
10.048 144.889
 0
 37.5
0
37.5
144.899  3x   0 


 0
0
37.5
10.048  3y   50 

 z 

0
64.952 144.889 10.048
 3   0 
Solving these gives ω  2.570ˆj rad/s
Ans.
2
ω3  1.158ˆi  0.086ˆj  0.643kˆ rad/s ,
VB  96.35ˆi  50ˆj  166.9kˆ mm/s ,
3  1.327 rad/s
Ans.
VB  199 mm/s
Ans.
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Theory of Machines and Mechanisms, 4e
11.4
Uicker et al.
Solve Problem 11.3 using graphic techniques.
The position and velocity solutions are shown in the figure below. After the top and
front views are drawn to scale, first and second auxiliary views are drawn to view rod
O3B in true length and end views, respectively.
Next a velocity polygon is drawn with origin at point B. The velocity VA is drawn true
length, downward in the front view. The direction of VBA is added horizontal in the front
view and perpendicular to link 2 in the top view. This direction is projected to the first
auxiliary view where it intersects the line of VB, which is perpendicular to rod 3 in this
view. This completes the velocity polygon for the equation VB  VA  VBA . Projecting
this to all other views, we can measure the true lengths of VB from the second auxiliary
view, and VBA from the top view.
The angular velocities are then found from
2 
VBA 305 mm/s

 4.067 rad/s
75 mm
RBA
3 
VBO3
RBO3

222 mm/s
 1.480 rad/s
150 mm
VB  222 mm/s
Ans.
Ans.
Ans.
The results show typical graphical error when compared with the analytical solution in
Problem 11.3.
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Theory of Machines and Mechanisms, 4e
11.5
Uicker et al.
For the spherical RRRR illustrated in Fig. P11.5, use vector algebra to make complete
velocity and acceleration analyses at the position given.
RO2O  175kˆ mm, RO4O  50ˆi mm, R AO2  75ˆi mm, R BO4  225ˆj mm, R BA  125ˆi  225ˆj 175kˆ mm,
ω  60kˆ rad/s.
2
RO4O2  50ˆi  175kˆ mm ,
R AO2  75ˆi mm ,
R BO4  225ˆj mm .
Substituting these into R AO2  R BA  RO4O2  R BO4 gives
R BA  125ˆi  225ˆj  175kˆ mm , RBA  311.25 mm
The velocity analysis proceeds as follows:
VA  VAO2  ω2 × R AO2  60kˆ rad/s × 75ˆi mm  4 500ˆj mm/s
VB  VBO4  ω4 × R BO4

 

  ˆi  ×  225ˆj mm   225 kˆ
4
Ans.
4
Since the two revolutes at A and B have axes that 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  3ˆi and
V  ω × R   ˆi × 125ˆi  225ˆj  175kˆ  175 ˆj  225 kˆ
BA
3
BA
3


3
3
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Theory of Machines and Mechanisms, 4e
Uicker et al.
Substituting these into VB  VA  VBA or 2254kˆ  4 500ˆj  1753ˆj  2253kˆ , and
equating components gives
Ans.
ω3  ω4   4 500 175  25.714ˆi rad/s ,
Ans.
VBA  4 500ˆj  5 785kˆ mm/s , and VB  5 785kˆ mm/s .
To find accelerations we first calculate
Ans.
AtAO2  α 2 × R AO2  0
AnAO2  22 R AO2  270 000ˆi mm/s2 ,
An   2 R  148 775ˆj mm/s2 ,
At  α × R  225 kˆ
4
BO4
4
BO4
BO4
4
BO4
Remembering that link 3 rotates about point O we also find
AnAO  3 ×  3 × R AO   115 725kˆ mm/s 2 ,
At  α × R  175 y ˆi  175 x  75 z  ˆj  75 ykˆ
AO
A
n
BO
AtBO
3
3
3
AO
3
3
 3 ×  3 × R BO   148 775ˆj mm/s ,
 α × R  225 z ˆi  50 z ˆj   225 x  50 y  kˆ
2
3
3
BO
3
3
Substituting these into A A  A
n
AO2
A
n
AO
3
A
equating components gives
270 000 
1753y
1753x
0
0  115 725
753y
,
t
AO
and A B  AnBO4  AnBO  AtBO , and
2253z ,
0
503z ,
753z , 148 775  148 775
, and
225 4 
2253x
From these we solve for α 3  1 543ˆj rad/s2 and α 4  343ˆi rad/s2 .
A  148 775ˆj  77 175kˆ mm/s2
B
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503y
.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
11.6
Uicker et al.
Solve Problem 11.5 using graphic techniques.
To avoid confusion, the position and velocity solutions are shown in a separate figure
above. The acceleration solution is shown below. The results agree with those of the
previous analytic solution of Problem 11.5.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
11.7
Uicker et al.
Solve Problem 11.5 using transformation matrix techniques.
Following the conventions of Sec. 11.6 and Fig. 11.11, the Denavit-Hartenberg parameter
values are:
i,j
ai , j
i, j
i, j
si , j
1,2
2,3
3,4
4,1
0
0
0
0
23.20°
94.90°
77.47°
90.00°
1  0
2  101.54º
3  67.30º
4  90.00º
Using Eqs. (11.12) and (11.15) we find
0
0
0
1
0 0.91914 0.39394 0

T12  
0 0.39394 0.91914 0


0
0
1
0
 0.20005 0.08369 0.97620
 0.97979 0.01709 0.19932
T23  

0.99635 0.08542
0

0
0
0

0
 0.20005

 0.90056
0
, T13  
 0.38598
0


1
0

0 1 0
 0.38591 0.20015 0.90057 0 
0 0 1
 0.92254 0.08372 0.37671 0 


T34 
, T14  
1 0 0

0.97618 0.21695 0 
0



0
0
0
1

0 0 0
Next, from Eqs. (11.22) and (11.25),
0.91914 0.39394
0
0 1 0 0 

 0.91914
1 0 0 0 
0
0
 , D2  
D1  
0 0 0 0 
 0.39394
0
0



0
0
0

0 0 0 0 
0
0
0
0
0.08369 0.97620
0.37679 0.21685
0.92252
0
0
0
0
0 
,
0

1
0
0 
.
0

1
0
0 
,
0

0
0.21685 0 
0
 0
0 0 1 0 
 0

0 0 0 0 
0.97620 0 
0
.
D3  
, D4  
0.21685 0.97620
1 0 0 0 
0
0




0
0
0
 0
0 0 0 0 
Now, from Eq. (11.26), D11  D22  D33  D44  0 , we get the following set of
equations:
0.97620 0  2   0 
 0
0


0.39394 0.21685 1    0     0  rad/s

 3   1  
0.91914
60 
0
0  4   1
From these we find 2  65.275 rad/s , 3  0 , and 4  25.714 rad/s . From these and
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Theory of Machines and Mechanisms, 4e
Uicker et al.
Eq. (11.27) we find the velocity matrices
0 25.714 0 
0 25.714
 0 60 0 0 
 0
 0
 60 0 0 0 
 0


0
0
0
0
0 0
 rad/s, 3  
rad/s,  4  
2  
 25.714 0
 25.714 0 0
 0 0 0 0
0
0





0
0
0
0 0
 0
 0
 0 0 0 0
These can be used with Eq. (11.28) to find the velocities of all moving points.
0
0 
rad/s.
0

0
Acceleration analysis follows similar steps. From Eq. (11.29) we get the following set of
equations:
0.97620
0  2   1 543
 0
0.39394 0.21685 1     0  rad/s 2

 3 

0
0  4   0 
0.91914
From these we find 2  0 , 3  1 580 rad/s2 , and 4  343 rad/s2 . From these and Eq.
(11.30) we find the acceleration matrices
0
0
0 0 0 0 
0 0
 0 0 343 0 
0 0 0 0 
0 0 1 543 0 
 0 0 0 0
2


 rad/s2.


, 3 
rad/s ,  4  
2 
0 0 0 0 
 343 0 0 0 
0 1 543
0
0






0
0
0 0
 0 0 0 0
0 0 0 0 
These can be used with Eq. (11.31) to find the accelerations of all moving points.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
11.8
Uicker et al.
Solve Problem 11.5 except with   90 .
The position vectors for this new position are:
x ˆ
y ˆ
z ˆ
i  RBA
j  RBA
k , RO4O2  50ˆi  175kˆ mm , R BO4  225sin 4ˆj  225cos4kˆ .
R AO2  75ˆj mm , R BA  RBA
Substituting these into R AO2  R BA  RO4O2  R BO4 and separating components gives
x
y
z
 225sin 4  75 , RBA
RBA
 50 , RBA
 225cos4  175 , and squaring and adding these gives
2
502  2252  78 750 cos 4  1752  33 750sin 4  752  RBA
 96 875
If we now define   tan  4 2  and use the identities cos 4  1   2  1   2  and
sin  4  2  1   2  , then the above equation can be reduced to 19 2  18  23  0 .
The root of interest here is   0.72419 , which corresponds to  4  71.823 . With this
the four position vectors are
R AO2  75ˆj mm , R BA  50ˆi  288.772ˆj  104.810kˆ mm , RO4O2  50ˆi 175kˆ mm , R BO4  213.772ˆj  70.189kˆ mm
The velocity analysis proceeds as in Problem 11.5:

  
 

VA  VAO2  ω2 × R AO2  60kˆ rad/s × 75ˆj mm  4 500ˆi mm/s
VB  VBO4  ω4 × R BO4  4ˆi × 213.772ˆj  70.189kˆ mm  70.1894 ˆj  213.7724kˆ

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  0.950103ˆj  0.311953kˆ and
VBA  ω3 × R BA  189.6503ˆi  15.6003ˆj  47.5003kˆ
Substituting these into VB  VA  VBA and equating components gives
3  23.726 rad/s and 4  5.273 rad/s , and from these we get
Ans.
ω3  22.542ˆj  7.401kˆ rad/s , ω4  5.273ˆi rad/s ,
ˆ
ˆ
ˆ
ˆ
ˆ
VBA  4 500i  370 j  1 127k mm/s , and VB  370 j  1 127k mm/s .
Ans.
To find accelerations we first calculate
AnAO2  22 R AO2  270 000ˆj mm/s2 ,
AtAO2  α 2 × R AO2  0 ,
An   2 R  5 943ˆj  1 951kˆ mm/s 2 , At  α × R  70.20 ˆj  213.78 kˆ
4
BO4
4
BO4
BO4
4
BO4
4
Remembering that link 3 rotates about point O we also find
AnAO  3 ×  3R AO   33 300ˆj  101 450kˆ mm/s 2 ,
At  α × R  175 y  75 z  ˆi  175 x ˆj  75 xkˆ ,
AO
A
n
BO
AtBO
3
AO
3
3
3
3
 3 ×  3 × R BO   28 150ˆi mm/s2 ,
 α × R   70.20 y  213.78 z  ˆi   70.20 x  50 z  ˆj   213.78 x  50 y  kˆ
3
BO
3
Substituting these into A A  A
3
3
n
AO2
A
n
AO
A
t
AO
and A B  A
3
3
n
BO4
A
n
BO
A
t
BO
3
, and equating
components gives
0  1753y  753z , 0  28 150  70.203y  213.783z ,
270 000  33 300 1753x
, 5 943  70.20 4  70.203x  503z , 0  101 450  753x , 1 951  213.784  213.783x  503y .
From these we solve for α  1 302ˆi  49ˆj  115kˆ rad/s2 and α  1 304ˆi rad/s2 . Ans.
3
ˆ mm/s 2
A A  270 000ˆj mm/s2 , AtBO  85 750ˆj  280 750k
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4
Ans.
Theory of Machines and Mechanisms, 4e
11.9
Uicker et al.
Determine the advance-to-return time ratio for Problem 11.5. What is the total angle of
oscillation of link 4?
Time ratio = 181.1/178.9 = 1.012,
 4  46.5
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Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
11.10 For the spherical RRRR linkage illustrated in Fig. P11.10, 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.
RO2O  150 mm, RO4O  225 mm, RAO2  37.5 mm, RBO4  262.5 mm, RBA  412.5 mm,
  120 , and ω  30kˆ rad/s.
2
Time ratio = 187/173 = 1.081,

 4  38
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Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
11.11 Use vector algebra to make complete velocity and acceleration analyses of the linkage of
Fig. P11.10 at the position specified.
The position vectors were found from a graphical analysis done on a CAD software
system (see Problem 11.12). For the position  2  120 they are as follows:
R
 225ˆi  150kˆ mm ,
R  236.975ˆj  111.75kˆ mm ,
O4O2
R AO2
 18.75ˆi  32.475ˆj mm ,
BO4
R BA  243.75ˆi  204.5ˆj  261.75kˆ mm .
The velocity analysis for this position, with ω2  30kˆ rad/s , proceeds as follows:
V  ω × R  974.275ˆi  562.5ˆj mm/s , V   ˆi × R  111.75 ˆj  236.975 kˆ .
A
2
4
B
AO2
BO4
4
4
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 ,  3  3uˆ  ,
uˆ   V × V  V × V  11.5733ˆi  20.0432ˆj  9.45121kˆ
B
A
B
A
VBA  3 × R BA  132.5253ˆi  213.3253ˆj  290.13kˆ
Substituting these into VB  VA  VBA , equating components, and solving gives
3  8.820 rad/s and 4  10.797 rad/s . From these
Ans.
3  4.083ˆi  7.071ˆj  3.334kˆ rad/s , and 4  10.797ˆi rad/s .
For acceleration analysis we first calculate
AnAO2  2 × 2 × R AO2  25300ˆi  42089ˆj mm/s2 , AtAO2  2 × R AO2  0 ,
AnBO4  4

× 
4
× R BO4

  25126ˆj  13027kˆ mm/s
2
,
AtBO4   4ˆi × R BO4  111.75 4 ˆj  236.975 4kˆ
Remembering that link 3 rotates about point O we also find
AnAO  3 ×  3 × R AO   2251ˆi  3898ˆj  11023kˆ mm/s2 ,
An   ×   × R   22117ˆi  10447ˆj  4926kˆ mm/s2 ,
BO
A
t
AO
3
3
BO
 3 × R AO  1503y  323z  ˆi  193z  1503x  ˆj   323x  193y  kˆ ,
AtBO  3 × R BO   1123y  2373z  ˆi   2253z  1123x  ˆj   2373x  2253y  kˆ
Next, from A A  AnAO2  AtAO2  AnAO  AtAO and A B  AnBO4  AtBO4  AnBO  AtBO , we
separate components and obtain
24300  2251  1503y  323z ;
0  22117 1123y  2373z ;
42089  3898  1503x  193z ; 27626  112 4  10447  1123x  2253z ;
13027  237 4  4926  2373x  2253y ;
0  11023  323x  193y ;
From these α3  273ˆi  115ˆj  148kˆ rad/s2 and α 4  130ˆi rad/s2 .
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
11.12
Uicker et al.
Solve Problem 11.11 using graphic techniques.
The position and velocity solutions are shown first with the acceleration solution on the
next diagram. The results verify those of the analytical solution in Problem 11.11.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
11.13
Uicker et al.
Solve Problem 11.11 using transformation matrix techniques.
The Denavit-Hartenberg parameters are:
a12  0, 12  14.04, 12  1  60.00, s12  0,
a23  0,  23  104.41,  23  2  69.52, s23  0,
a34  0,  34  49.34,  34  3  93.18, s34  0,
a41  0,  41  90.00  41  4  64.75 s41  0.
From Eqs. (11.12) and (11.15) the transformation matrices are:
 0.50000 0.84015 0.21005 0 
 0.86603 0.48506 0.12127 0 

T12  

0.24260 0.97014 0 
0


0
0
0
1

 0.61206 0.39312 0.68618
 0.75747 0.04214 0.65151
T13  
 0.22720 0.91852 0.32356

0
0
0

0
0 
0

1
0.42650 0.90449 0 0 
 0
1 0 
0
T14  
0.90449 0.42650 0 0


0
0 1
 0
Next, from Eqs. (11.22) and (11.25),
0.97015 0.12127 0 
0
0 1 0 0 

1 0 0 0 
 0.97015
0.21005 0 
0



D1 
D2 
 0.12127 0.21005
0 0 0 0 
0
0




0
0
0
0

0 0 0 0 
0
0.32357 0.65151 0
0 0 1 0 

 0.32357

0 0 0 0 
0.68618 0
0


D3 
D4  
 0.65151 0.68618
1 0 0 0 
0
0




0
0
0
0

0 0 0 0 
and from Eq. (11.26) we get the following set of equations
 0.21005 0.68618 0  2 
0 
 0 
0.12127 0.65151 1     0     0  rad/s

 3
  1 



 0.97015 0.32357 0  4 
1 
 36
from which we find 2  33.670 rad/s , 3  10.307 rad/s , and 4  10.798 rad/s .
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
With these values and Eqs. (11.27) we find the velocity matrices
0 10.798
 0 36 0 0 
 0 3.335 4.083 0 
 0
36 0 0 0 
3.335


0
0
0
0
7.072 0 
 , 3  
, 4  
2  
 0 0 0 0
 4.083 7.072
10.798 0
0
0
0





0
0
0
0
0
 0 0 0 0
 0
 0
These can be used with Eq. (11.28) to find the velocities of all moving points.
0
0 
.Ans.
0

0
Ans.
The acceleration analysis follows parallel steps using Eqs. (11.29) and (11.30)
2  152 rad/s 2
 0.21005 0.68618 0  2   183.0 
0.12127 0.65151 1     254.6  m/s 2 ;
3  220 rad/s 2

 3 

 0.97015 0.32357 0  4   76.4 
4  130 rad/s 2
1
0 0 0 0 
 0 148 273 0 
 0 0 130 0 
0 0 0 0 
 148 0 115 0 


,
 ,  4   0 0 0 0 .
2  
3  
Ans.
0 0 0 0 
130 0 0 0 
 273 115 0 0 






0
0 0
 0
0 0 0 0 
 0 0 0 0
Although the global axes have changed because of the Denavit-Hartenberg conventions,
these results correlate with and verify those of Problems 11.11 and 11.12.
Ans.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
11.14 Figure P11.14 illustrates the top, front, and auxiliary views of a spatial slider-crank RSSP
linkage. In the construction of many such mechanisms provision is made to vary the
angle ; thus the stroke of slider 4 becomes adjustable from zero, when  = 0, to twice
the crank length, when  = 90. With  = 30, use vector algebra to make a complete
velocity analysis of the linkage at the given position.
RAO  50 mm, RBA  150 mm,    240 , ω  24ˆi rad/s
The position vectors were found from a graphical analysis done on a CAD software
system (see Problem 11.15). For the position  2  240 they are as follows:
R  21.651ˆi  37.500ˆj  25.000kˆ mm , R  164.720ˆi mm ,
AO
R BA
BO
 143.069ˆi  37.500ˆj  25.000kˆ mm .
The velocity analysis for this position, with 2  24 rad/s , proceeds as follows:
ω2  20.785ˆi  12.000ˆj rad/s , VA  ω2 × R AO  300.000ˆi  519.615ˆj  1 039.230kˆ mm/s ,
V  ω ×R  25.000 y  37.500 z ˆi  143.069 z  25.000 x ˆj  37,500 x  143.069 y kˆ , V  V ˆi .
BA
3
BA

3
3
 
3
3
 
3
3

B
B
Substituting these into VB  VA  VBA and separating into components,
 25.0003y  37.5003z
VB  300.000
0.0  519.615  25.0003x
 143.0693z
0.0  1 039.230  37.5003x  143.0693y
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 we can set 3 R BA  0 to get a fourth equation:
0.0  143.0693x  37.5003y  25.0003z
The four equations can now be solved to give
ω3  2.309ˆi  6.659ˆj  3.228kˆ rad/s and VB  345.400ˆi mm/s
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
11.15 Solve Problem 11.14 using graphical techniques.
The graphic solution is shown in the figure above. The results verify those found in
Problem 11.14. They are
3 
11.16
VBA 1 159.6 mm/s

 7.731 rad/s , and VB  345.5 mm/s
150 mm
RBA
Solve Problem 11.14 using transformation matrix techniques.
One choice for the Denavit-Hartenberg parameters gives:
12  90 ,
12  1  30 ,
s12  0 ,
a12  0 ,
s14  4 .
14    30 ,
14  0 ,
a14  0 ,
From Eqs. (11.12) and (11.11) we obtain
cos 1 0 sin 1 0 
 0   2sin 1 
 sin  0  cos  0 
 0   2 cos  
1
1
1
,
T12  
RA  T12    
;
 0 1
2  0 
0
0


  

0
1
 0 0
1   1 
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
0
0
0
1 0

0 

0 cos   sin   sin  
0    sin  
4
,
.
RB  T14     4
T14  
0 sin  cos  4 cos  
0   4 cos  


  

0
1
1
0 0

1  

From these and the length of the connecting rod we write
t
2
2
2
2
  RB  RA   RB  RA    2sin 1    2cos 1  4 sin    4 cos    62
RBA
This reduces to 42  44 sin  cos 1  32  0 , which has a solution
4  2sin  cos 1  4sin 2  cos 2 1  32
By differentiating the above equation with respect to time we obtain
244  44 sin  cos 1  414 sin  sin 1  0
which has for a solution
24 sin  sin 1
4 

 2sin  cos 1  4  1
and, from Eqs. (11.22), (11.23), (11.25), and (11.27),
0 
0 1 0 0 
0 0 0
1 0 0 0 
0 0 0 sin  
,
;
D1  
D4  
0 0 0 0 
0 0 0  cos  




0 
0 0 0 0 
0 0 0
 0 1 0 0 
0 0 0  sin 4 




0 0 0 cos 4 
1 0 0 0 


2 
4 
,
.
 0 0 0 0
0 0 0





0
 0 0 0 0 
0 0 0

Now, with   30 , 1  30 , and 1  24 rad/s , the above formulae give
4  164.7 mm , 4  345.4 mm/s , and
Ans.
0
0
1 039.225 mm/s 
 25.000 mm 




 600.000 mm/s 
 43.300 mm 
 82.350 mm 
 172.700 mm/s 
 , RB  
,R  
 . Ans.
 , RB  
RA  

142.650 mm  A 
 299.125 mm/s 


0
0








1
1
0
0








These results agree with those of Problems 11.14 and 11.15 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.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
11.17
Uicker et al.
Solve Problem 11.14 with  = 60 using vector algebra.
The position vectors were found from a graphical analysis done on a CAD software
system (see Problem 11.18). For the position  2  240 they are as follows:
R  37.500ˆi  21.650ˆj  25.000kˆ mm , R  183.809ˆi mm ,
BO
AO
R BA  146.309ˆi  21.651ˆj  25.000kˆ mm .
The velocity analysis for this position, with 2  24 rad/s , proceeds as follows:
ω2  12.000ˆi  20.784ˆj rad/s , VA  ω2 × R AO  519.615ˆi  300.000ˆj  1 039.230kˆ mm/s , Ans.
V  ω ×R  25.000 y  21.651 z ˆi  146.309 z  25.000 x ˆj  21.651 x  146.309 y kˆ , V  V ˆi .
BA
3
BA

3
3
 
3
3
 
3
3

B
B
Substituting these into VB  VA  VBA and separating into components we get
 25.0003y  21.6503z
VB  519.615
0.0  300.000  25.0003x
 146.3093z
0.0  1 039.230  21.6513x  146.3093y
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 R BA  0 .
0.0  146.3093x  21.6513y  25.0003z
The four equations can now be solved to give
ω3  1.333ˆi  6.905ˆj  1.823kˆ rad/s and VB  652.821ˆi mm/s
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
11.18
Uicker et al.
Solve Problem 11.14 with  = 60 using graphical techniques.
The graphic solution is shown in the figure above. The results verify those found in
Problem 11.17. They are
V
1 089.75 mm/s
3  BA 
 7.265 rad/s , and VB  652.851 mm/s
Ans.
150 mm
RBA
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Theory of Machines and Mechanisms, 4e
Uicker et al.
11.19 Solve Problem 11.14 with  = 60 using transformation matrix techniques.
One choice for the Denavit-Hartenberg parameters gives:
12  90 ,
12  1  30 ,
s12  0 ,
a12  0 ,
s14  4 .
14    60 ,
14  0 ,
a14  0 ,
From Eqs. (11.13) and (11.12) we obtain
cos 1 0 sin 1 0 
 0   2sin 1 
 sin  0  cos  0 
 0   2 cos  
1
1
1


T12 
RA  T12    
,
;
 0 1
2  0 
0
0


  

0
1
 0 0
1   1 
0
0
0
1 0

0 

0 cos   sin   sin  
0    sin  
4
,
.
RB  T14     4
T14  
0 sin  cos  4 cos  
0   4 cos  


  

0
1
1
0 0

1  

From these and the length of the connecting rod we write
t
2
2
2
2
RBA
  RB  RA   RB  RA    2sin 1    2cos 1  4 sin    4 cos    62
This reduces to 42  44 sin  cos 1  32  0 , which has a solution
4  2sin  cos 1  4sin 2  cos 2 1  32
By differentiating the above equation with respect to time we obtain
244  44 sin  cos 1  414 sin  sin 1  0
which has for a solution
24 sin  sin 1
4 

 2sin  cos 1  4  1
and, from Eqs. (11.22), (11.23), (11.25), and (11.27),
0 
0 0 0
0 1 0 0 
0 0 0 sin  
1 0 0 0 
;
,
D1  
D4  
0 0 0  cos  
0 0 0 0 




0 
0 0 0
0 0 0 0 
 0 1 0 0 
0 0 0  sin 4 




1 0 0 0 
0 0 0 cos 4 


2 
4 
,
.
 0 0 0 0
0 0 0





0
 0 0 0 0 
0 0 0

  60 , 1  30 , and 1  24 rad/s , the above formulae give
4  183.800 mm , 4  652.800 mm/s , and
Now, with
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Theory of Machines and Mechanisms, 4e
Uicker et al.
0
0


1 039.225 mm/s 
 25.0 mm 


 159.175 mm 
 600.0 mm/s 
 43.3 mm 
 565.35 mm/s 
 , RB  
, R  
 , RB  
.
RA  


 91.900 mm  A 

 326.40 mm/s 
0
0








1
1
0
0








These results agree with those of Problems 11.17 and 11.18 once the Denavit-Hartenberg
coordinate directions are considered.
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Theory of Machines and Mechanisms, 4e
11.20
Uicker et al.
Figure P11.20 illustrates the top, front, and profile views of an RSRC crank and
oscillating-slider 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.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
100
RAO  100 mm in, RBA  300 mm in, 2  40 , and ω  48ˆi rad/s.
(a)
The RSRC linkage has n = 4, j1 = 2, j2 = 1, j3 = 1. The Kutzbach criterion gives
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Theory of Machines and Mechanisms, 4e
Uicker et al.
m  6  n  1  5 j1  4 j2  3 j3  6(4  1)  5(2)  4(1)  3(1)  1
Ans.
(b)
Since vectors do not show the rotation , matrix methods were necessary and are
shown in Problem 11.23. See (c) for the vector solution. Together, they show:
Ans.
135   4  45 ;
 4  90 .
182.85 mm  yB  382.85 mm ;
Ans.
yB  400 mm .
(c)
R B  R AB  RQ  R AQ
yB ˆj  xAB ˆi  yAB ˆj  z ABkˆ  100ˆi  100sin 2ˆj  100cos 2kˆ
Separating components,
yB  yAB  100sin 2 ,
xAB  100 ,
and, from the length of link 3,
2
2
2
xAB
 y 2AB  z AB
 RAB
 100   100sin 2  yB   100cos2 
2
2
2
z AB  100cos 2
 (300)2
yB2  200sin 2 yB  2800  0
yB  100sin 2  100 175  sin 2 2
R AB  100ˆi  100 175  sin 2 2 ˆj  100cos 2kˆ
For  2  40 , R B  yB ˆj  208ˆj mm , R AB  100ˆi  272.275ˆj  76.6kˆ mm .
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Ans.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
11.21 Use vector algebra to find VB, 3, and 4 for Problem 11.20.
First we identify, for  2  40 , that RQ  100ˆi mm , R AQ  64.275ˆj  76.6kˆ mm ,
R B  208ˆj mm , and R AB  100ˆi  272.275ˆj  76.6kˆ mm .
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 as
ˆ 3/ 4  0.608ˆi  0.794kˆ . Then the angular velocity of link 3 can be written as
ω
With this done, we can find
ω3  ω4  ω3/ 4  0.6083/ 4ˆi  4 ˆj  0.7943/ 4kˆ .
and
V  V ˆj ,
V  ω × R  3677ˆj  3085.375kˆ mm/s ,
A
2
B
AQ
VAB  ω3 × R AB
B
 (76.64  216.153/4 )ˆi  125.9753/4 ˆj  1004  165.5753/4  kˆ . Now,
setting VB  VAB  VA and equating components, we get the following equations:
76.64  216.153/4  0
VB
 125.9753/4  147.081
1004  165.5753/4  123.415
which can be solved to give 4  19.444 rad/s , 3/ 4  6.891 rad/s , VB  2808.95 mm/s .
Ans.
ω3  4.190ˆi  19.444ˆj  5.471kˆ rad/s , ω4  19.444ˆj rad/s , VB  2808.95ˆj mm/s .
Note that if ω were written as ω   x ˆi   y ˆj   z kˆ , then the above set of simultaneous
3
3
3
3
3
equations would have four unknowns and could not be solved.
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Theory of Machines and Mechanisms, 4e
11.22
Uicker et al.
Solve Problem 11.21 using graphical techniques.
The graphic solution is shown in the figure above. The results verify those found in
Problem 11.21. 3 and 4 are not apparent in the graphic method, but
VB  2807.5 mm/s .
Ans.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
11.23 Solve Problem 11.21 using transformation matrix techniques.
The Denavit-Hartenberg parameters from the global coordinate system to joint A are:
a12  0 ,
12  0 ,
12   2  40 ,
s12  100 mm ,
and, proceeding in the other direction around the loop, to joint A, they are:
16  90 ,
16  0 ,
s16  0 ,
a16  0 ,
 65  0 ,
a65  0 ,
s65  yB ,
 65  0 ,
 54  0 ,
a54  0 ,
54   4 ,
s54  0 ,
 43  90 ,
a43  0 ,
 43  0 ,
s43  0 ,
a3 B  0 ,
3B  3 ,
s3 B  0 ,
3B  0 ,
aBA  300 mm ,
 BA  0 ,
 BA  90 ,
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sBA  0 .
Theory of Machines and Mechanisms, 4e
Uicker et al.
From Eqs. (11.12) we obtain for a first path to A:
0 
cos  2  sin  2 0
 sin 
cos  2 0
0 
2

,
T12 
 0
0
1 100 


0
0
1 
 0
and along the other path to joint A, from Eqs. (11.12) and (11.15), we get:
1 0 0 0 
0 0 1 0 
,
T16  
0 1 0 0 


0 0 0 1 
1
0
T65  
0

0
0
1
0
0
cos 4
 sin 
4
T54  
 0

 0
1
0
T43  
0

0
0
0 
,
yB 

1
0
0
1
0
 sin  4
cos 4
0
0
0 0
0 1
1 0
0 0
1 0
0 0
T15  
0 1

0 0
0
0
1
0
 cos 4
 0
T14  
  sin  4

 0
0
0 
,
0

1
 cos 4
 0
T13  
  sin  4

 0
0
0 
,
0

1
 cos  3 cos  4

sin  3
T1B  
  cos  3 sin  4

0

cos 3
 sin 
3
T3 B  
 0

 0
 sin  3
cos 3
0
0
0
 1
TBA  
0

0
0
0 
 sin 3 cos  4

  cos 
0 300 
3
, T1 A  
  sin 3 sin  4
1
0 


0
1 
0

1
0
0
0
0
0
1
0
0
0 
,
0

1
0
1
0
0
0
yB 
,
0

1
0
1
0
0
0
yB 
,
0

1
0 sin  4
1
0
0 cos  4
0
0
0
yB 
,
0

1
 sin  4
0
 cos 4
0
 sin  3 cos  4
cos  3
sin  3 sin  4
0
cos 3 cos  4
sin 3
 cos 3 sin  4
0
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sin  4
0
cos  4
0
sin  4
0
cos  4
0
0
yB 
,
0

1
300sin 3 cos  4 
yB  300 cos 3 
.
300sin 3 sin  4 

1

Theory of Machines and Mechanisms, 4e
Uicker et al.
At the terminations of these two paths, the position of point A must agree. Therefore,
0   100   sin 3 cos  4
cos 3 cos  4 sin  4 300sin 3 cos  4  0 
cos  2  sin  2 0
 sin 




yB  300 cos 3  0
cos  2 0
0   0    cos 3
sin 3
0
2


 0
0
1 100   0    sin 3 sin  4  cos 3 sin  4 cos  4 300sin 3 sin  4  0 


 
 
0
0
1  1  
0
0
0
1
 0
 1 
 100 cos  2   300sin 3 cos  4 
 100sin    y  300 cos  
2
3 

 B
 100   300sin 3 sin  4 

 

1
1

 

Noting from the figure that 90  3  0 and 180   4  0 , we can solve these three
equations for the position results
3   sin 1

1  cos 2  2 3

 4   tan 1 1 cos 2 
yB  100 7  sin 2 2  100sin 2
and, at  2  40 , these give 3  24.83 ,  4  52.55 , and yB  208 mm .
For velocity analysis we begin by using Eqs. (11.22) and (11.25) 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 
yB cos  4 
0
0 1 0 0 
 0  cos  4
1 0 0 0 
cos 

0
0
 sin  4
4
,
,
Q3  
D3  T13Q3T131  
0 0 0 0 
 0
sin  4
0
 yB sin  4 




0
0
0
 0

0 0 0 0 
0
1
Q4  
0

0
1
0
0
0
0
0
Q6  
0

0
150
0
0
0
0
0
0
0
0
0
0
0
0
0 
,
0

0
0
0 
,
1

0
0
0
1
D4  T14Q4T14  
 1

0
0
0
0
0
1
0
0
0
0
0 
,
0

0
0
0
D6  T16Q6T161  
0

0
0
0
0
0
0
0
0
0
0
1 
.
0

0
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Theory of Machines and Mechanisms, 4e
Uicker et al.
Next we write from Eq. (11.27)
 0  2 0 0 


 2 0 0 0

 2  D2 2 
 0 0 0 0


 0 0 0 0 
and, along the other path,
 0
yB cos 4 3 
 cos 4 3
4


cos  4 3
0
yB
 sin  4 3


3  D6 yB  D4 4  D3 3 
  4
sin  4 3
0
 yB sin  4 3 


0
0
0
 0

 0

0
 4  D6 yB  D4 4  
  4

 0
0
0
0
0
4 0 

0 yB 
0 0

0 0 
Since the velocity of point A must agree along the two paths
 100 cos  2 
 100 cos  2 
 100sin  
 100sin  
2
2

 3 
2
 100 
 100 




1
1




 100sin  2 2   100sin  2 cos  43  yB cos  43  4 4


 

 100 cos  2 2    100 cos  2 cos  43  100sin  43  yB 

 100 cos  2 4  100sin  2 sin  43  yB sin  43 
0

 

0
0

 

At the position where  2  40 , 3  24.83 ,  4  52.55 , yB  208 mm , and
 2  48.0 rad/s , these equations can be solved for 3  6.891 rad/s ,  4  19.444 rad/s ,
and yB  112.357 in/s . With these values we can evaluate
4.191 19.444 34.865 
0 19.444
0 
 0
 0
 4.191


0
5.471 112.357 
0
0
0
112.357 
3  
and  4  
 19.444 5.471
 19.444 0
0
45.513 
0
0 




0
0
0 
0
0
0 
 0
 0
from which we write the vector forms of the results:
VB  2808ˆj mm/s , ω3  5.471ˆi  19.444ˆj  4.191kˆ rad/s , and ω4  19.444ˆj rad/s .
Ans.
Note that the global x1 and z1 axis orientations in this solution differ from those of
Problem 11.21 because of the conventions of the Denavit-Hartenberg parameters. This is
also the reason that the components of 3 seem switched.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
Chapter 12
Robotics
12.1
For the SCARA robot shown in Fig. P12.1, 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 1  30 , 2  60 , 3  50 mm , and 4  0 . Also find
the absolute position of the tool point that has coordinates x5 = y5 = 0, z5 = 35 mm.
a12  a23  250 mm , a34  a45  0 , 12  34  45  0 ,  23  180 , 12  1 ,  23  2 , 34  0 ,  45  4 ,
s12  300 mm , s23  0 , s34  3 , and s45  50 mm .
See the solution to Problem 12.2 for the formulae before numerical evaluation.
 0.866 0.500 0 433  0   433 mm 
 0.500 0.866 0 0   0   0 



R1  T15 R5  
 0
1 200  35 mm  165 mm 
0


 

0
0 1  1   1 
 0
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
12.2
Uicker et al.
Repeat Problem 12.1 using arbitrary (symbolic) values for the joint variables.
a12  a23  250 mm , a34  a45  0 , 12  34   45  0 ,  23  180 , 12  1 ,  23  2 ,
34  0 ,  45  4 , s12  300 mm , s23  0 , s34  3 , s45  50 mm
cos 1  sin 1
 sin  cos 
1
1
T12  
 0
0

0
 0
1
0
T34  
0

0
0 250 cos 1 mm 
0 250sin 1 mm 
1
300 mm 

0
1

0
0 
3 

1
cos 1  2  sin 1  2 

sin 1  2   cos 1  2 
T13  T12T23  

0
0

0
0

0
1
0
0
0
0
1
0
cos 1  2  sin 1  2 

sin 1  2   cos 1  2 
T14  T13T34  

0
0

0
0

cos 2 sin 2
 sin   cos 
2
2
T23  
 0
0

0
 0
cos 4  sin 4
 sin  cos 
4
4
T45  
 0
0

0
 0
0
0 
0
0 
1 50 mm 

0
1 
0 250 cos 1  250 cos 1  2  mm 

0 250sin 1  250sin 1  2  mm 

1
300 mm

0
1

0 250 cos 1  250 cos 1  2  mm 

0 250sin 1  250sin 1  2  mm 

300  3 mm
1

0
1

cos 1  2  4  sin 1  2  4 

sin 1  2  4   cos 1  2  4 
T15  T14T45  

0
0

0
0

 0 
 0 

R5  
35 mm 


 1 
0 250 cos 1 mm 
0 250sin 1 mm 

1
0

0
1

0 250cos 1  250cos 1  2  mm 

0 250sin 1  250sin 1  2  mm 
Ans.

250  3 mm
1

0
1

 250 cos 1  250 cos 1  2  mm 


250sin 1  250sin 1  2  mm 
R1  T15 R5  
Ans.


215  3 mm


1


© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
12.3
Uicker et al.
For the gantry robot shown in Fig. P12.3, 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 1  450 mm , 2  181.25 mm , 3  50 mm , and 4  0 .
Also find the absolute position of the tool point that has coordinates x5 = y5 = 0, z5 =
43.75 mm.
a12  a23  a34  a45  0 , 12  90 ,  23  90 , 34   45  0 , 12   23  90 ,  34  0 ,  45  4 ,
s12  1 , s23  2 , s34  3 , and
s45  50 mm
.
See the solution to Problem 12.4 for the formulae before numerical evaluation.
0 1 0 181.25 mm 
 181.25 mm 
0 0 1 100 mm 
 143.75 mm 


T15  
R1  
1 0 0 450 mm 
 450 mm 




1
1
0 0 0



© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
12.4
Uicker et al.
Repeat Problem 12.3 using arbitrary (symbolic) values for the joint variables.
a12  a23  a34  a45  0 12  90  23  90 34   45  0 12   23  90  34  0
,
,
,
,
,
,
 45  4 s12  1 s23  2 s34  3
,
,
,
, s45  50 mm
0
1
T12  
0

0
0
0
1
0
1
0
T34  
0

0
0
1
0
0
0
0 
1 

1
0 0
0 0 
1 3 

0 1
0 1
0 0

1 0

0 0
0
1
T23  
0

0
1
0
0
0
0
1
0
0
2 
0 
T13  T12T23
1 

1
  sin 4  cos 4
 0
0
T15  T14T45  
 cos 4  sin 4

0
 0
0




0


R5 
 43.75 mm 


1


1
0
0
0
0
0 
2 

1
cos 4  sin 4 0
 sin  cos  0
4
4
T45  
 0
0
1

0
0
 0
0 1 0
0 0 1
T14  T13T34  
1 0 0

0 0 0
0
0
1
0
0
2

1 3  50 mm 

0
1

0
1

0 
0 
50 mm 

1 
2 
3 
1 

1 
2


   93.75 mm 

R1  T15 R5   3


1


1


© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
12.5
Uicker et al.
For the SCARA robot of Problem 12.1 in the position described, find the instantaneous
velocity and acceleration of the same tool point, x5 = y5 = 0, z5 = 35 mm, if the actuators
have (constant) velocities of 1  0.20 rad/s , 2  0.35 rad/s , and 3  4  0 .
0
1
D1  Q1  
0

0
1
0
0
0
0
0
0
0
0
0
1
D3  T13Q3T13  
0

0
0
0 
0

0
0 0
0 0
0 0
0 0
0
1
1
D2  T12Q2T12  
0

0
0
 1
1
D4  T14Q4T14  
0

0
0
0 
1

0
 0 0.200
0.200
0
 2  1  D11  
 0
0

0
 0
 0 0.150 0
 0.150 0 0
4  3  D33  
 0
0 0

0 0
 0
0
0
0
0
0
0
0
0
0
0 
0

0
 4   3  D33  3 D3  D33  3
0

0

0

0
0
0
0
0
0 15.16 mm/s 2 

0 8.75 mm/s 2 

0
0

0
0

 43.75 mm/s 
 10.83 mm/s 

R5  5 R1  


0


0


1
0
0
0
0 125 mm/rad 
0 216.5 mm/rad 

0
0

0
0

0
0

0 433 mm/rad 

0
0

0
0

0
0.150 0 43.75 mm/s 
 0


0
0.150 0 0 75.78 mm/s 

3  2  D22 
 0

0
0 0
0



0
0 0
0
 0

45.75 mm/s 

0
0.150
0
45.75
mm/s




75.78 mm/s 
0.150 0 0 75.78 mm/s 

  4  D44 
 5
 0

0
0 0
0



0
0 0
0

 0

 2  1  D11  1 D1  D11  1
0
0

0

0
1
0
0
0
0
0
0
0
 3   2  D22  2 D2  D22  2
0

0

0

0
0
0
0
0
0

0

0

0
0
0
0
0
0 15.16 mm/s 2 

0 8.75 mm/s 2 

0
0

0
0

 5   4  D44  4 D4  D44  4
0 15.16 mm/s 2 

0 8.75 mm/s 2 

0
0

0
0

 13.5 mm/s 2 


2.2 mm/s 2 
R5   5  55  R1  


0


0


© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
12.6
Uicker et al.
For the gantry robot of Problem 12.3 in the position described, find the instantaneous
velocity and acceleration of the same tool point, x5 = y5 = 0, z5 = 43.75 mm, if the
actuators have (constant) velocities of 1  2  0 , 3  40 mm/s , and 4  20 rad/s .
0
0
D1  Q1  
0

0
0
0 
1

0
0 0
0 0
1
D3  T13Q3T13  
0 0

0 0
0
0
0
0
0
0
0
0
0
0
0
0
0
1
0

0
0
0
1
D2  T12Q2T12  
0

0
0
0
0
0
0
0
1
D4  T14Q4T14  
1

0
0
0
0
0
1
0 
0

0
1 450 
0
0 
0 180 

0
0 
0
0
0
0
3  2  D22  0
2  1  D11  0
0
0
4  3  D33  
0

0
0
0
0
0
0 0 
0 40 
0 0 

0 0 
 0
 0
5  4  D44  
500

 0
0 500 9000 
40 
0 0
0 0 3600 

0 0
0 
 2  1  D11  1D1  D11 1  0
3   2  D22  2 D2  D22 2  0
 4  3  D33  3 D3  D33 3  0
5   4  D44  4 D4  D44 4  0
0


 40 mm/s 

R5  5 R1  


0


0


0 
0 
R5   5  55  R1   
0 
 
0 
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
12.7
Uicker et al.
The SCARA robot of Problem 12.1 is to be guided along a path for which the origin of
the end effector O5 follows the straight line given by
RO (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 effector is to remain constant
with kˆ 5  kˆ 1 (vertically downward) and î 5 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 the problem statement we construct the figure shown for the path described.
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 Problem 12.2,
cos 1  2  4  sin 1  2  4  0 250 cos 1  250 cos 1  2 


sin 1  2  4   cos 1  2  4  0 250sin 1  250sin 1  2  
T15  T14T45  
.


0
0
250  3
1


0
0
0
1


Equating these gives
(a )
250 cos 1  250 cos 1  2   100  40t  100 1.000  0.400t 
(b )
250sin 1  250sin 1  2   75  30t  75 1.000  0.400t 
(c )
250  3  50
1  2  4    36.87
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(d )
Theory of Machines and Mechanisms, 4e
Uicker et al.
Squaring and adding Eqs. (a) and (b) gives
125 000  62 500cos 2  15 625 1.000  0.800t  0.160t 2 
2  cos1  0.04t 2  0.20t  1.75
180  2  0
Ans.
where the quadrant was found from the figure of the robot.
Expanding the trigonometric functions and recognizing that 2 is now known, Eqs. (a)
and (b) become
250 1  cos 2  cos 1  250  sin 2  sin 1  100 1.000  0.400t 
250  sin 2  cos 1  250 1  cos 2  sin 1  75 1.000  0.400t 
which can be solved for sin 1 and cos 1
sin 1  3 1  cos 2   4sin 2  
cos 1  4 1  cos 2   3sin 2  
where   20 1  cos 2  1.000  0.400t 
From the ratio of these we get
 3 1  cos 2   4sin 2 
1  tan 1 

 4 1  cos 2   3sin 2 
Ans.
where the quadrant of 1 is found by considering the signs of the numerator and
denominator separately.
With both 1 and 2 known, Eqs. (c) and (d) give
Ans.
3  200 mm
Ans.
4  1  2  36.87
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Theory of Machines and Mechanisms, 4e
12.8
Uicker et al.
The gantry robot of Problem 12.3 is to travel a path for which the origin of the end
effector O5 follows the straight line given by
RO (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 effector 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 Problem 12.4 and the problem statement we can write
2
  sin 4  cos 4 0
 1
 0
0
1 3  50  0
T15  

 cos 4  sin 4 0
 0
1

 
0
0
1
 0
 0
0
0
1
0
0 120t  300 
1
150 
0 90t  225 

0
1

Equating individual elements and solving, we get
1  90t  225 mm
2  120t  300 mm
3  100 mm
4  90
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Ans.
Ans.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
12.9
Uicker et al.
The end effector of the SCARA robot of Problem 12.1 is working against a force loading
of 10ˆi1  5tkˆ 1 N and a constant torque loading of 25kˆ 1 mm  N as it follows the trajectory
described in Problem 12.7. Find the torques required at the actuators, as functions of
time, to achieve the motion described.
Using the 1 , 2 , 3 , and 4 values from Problem 12.7 and formulae from Problems 12.2
and 12.5,
0 1 0 0 
0 1 0 250sin 1 
1 0 0 0 
1 0 0 250 cos  
1

D2  T12Q2T121  
D1  Q1  
0 0 0 0 
0 0 0

0




0
0 0 0

0 0 0 0 
0
0


0
1
1
D4  T14Q4T14  
0
1


0
0
From these the Jacobian and loads are
0
0
0
0

0

0
0
0


1

1
1
0
J 

0 250sin 1 0 250sin 1  250sin 1  2  
0 250 cos 1 0 250 cos 1  250 cos 1  2  


1
0
0
0

Now, from Eq. (12.19), we get
1  25 mm  N
0
0
1
D3  T13Q3T13  
0

0
0
0
0
0
0
0
0
0
1
0
0
0
0 250sin 1  250sin 1  2  

0 250 cos 1  250 cos 1  2  

0
0

0
0

0




0


 25 mm  N 
F 

 10 N 


0


 5N 
 2  25  2500sin 1 mm  N
 3  5 N
 4  25  2500sin 1  2500sin 1  2  mm  N
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Ans.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
12.10 The end effector of the gantry robot of Problem 12.3 is working against a force loading of
20ˆi1  10tˆj1 lb and a constant torque loading of 22.5ˆj1 NM as it follows the trajectory
described in Problem 12.8. Find the torques required at the actuators, as functions of time,
to achieve the motion described.
Using the 1 , 2 , 3 , and 4 values from Problem 12.8 and formulae from Problems 12.4
and 12.6,
0 0 0 0 
0 0 0 1 
0 0 0 0 
0 0 0 0


D1  Q1  
D2  T12Q2T121  
0 0 0 1 
0 0 0 0




0 0 0 0 
0 0 0 0
0 0 0 0 
0 0 1 1.107  0.406 
0 0 0 1
0 0 0 1.356  0.542 


D4  T14Q4T141  
D3  T13Q3T131  
0 0 0 0 
0
1 0 0





0
0 0 0 0 
0 0 0

From these, the Jacobian and loads are
0
0
0 0 0



0 0 0



1
 22.6 N M 




0 0 0

0
0
J 
F 


 89 N 
0 1 0 1.107  0.406t 
 44.5t N 
0 0 1 1.356  0.542t 




0
0


1 0 0

Now, from Eq. (12.19), we get
Ans.
1  0
Ans.
 2  89 N
Ans.
 3  44.5t N
 4  2.26  5.424t  54.24t 2 NM
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Ans.
Theory of Machines and Mechanisms, 4e
PART 3
DYNAMICS OF MACHINES
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Theory of Machines and Mechanisms, 4e
Uicker et al.
Chapter 13
Static Force Analysis
13.1
Figure P13.1 illustrates 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 orientations.
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Theory of Machines and Mechanisms, 4e
© Oxford University Press 2015. All rights reserved.
Uicker et al.
Theory of Machines and Mechanisms, 4e
13.2
Uicker et al.
What moment M12 must be applied to the crank of the mechanism illustrated in Fig.
P13.2 if P  4005 N ?
RAO2  75 mm, RBA  350 mm.
Kinematic analysis:
  sin 1  r sin     sin 1  75 mmsin105 350 mm   11.95
Force analysis:
P  4005ˆi N
F  P  F ˆj  F

14
34
 cosˆi  sin ˆj  4005 Nˆi  F ˆj  F  0.978ˆi  0.207ˆj  0
14
F34  4005 N 0.978  4094 N
4005 N  0.978F34  0
F14  0.207 F34  0

34

F14  0.207  4094 N   845.5 N
F32  F34  4094 N 0.978ˆi  0.207ˆj  4005ˆi  845.5ˆj N
M  M
12

 

 r2  F32  M12  75 mm cos105ˆi  sin105ˆj × 4005ˆi  845.5ˆj N  0
M12  277.98 N.M  0
M12  277.98kˆ N.M
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Ans.
Theory of Machines and Mechanisms, 4e
13.3
Uicker et al.
If M12  452 N  m for the mechanism illustrated in Fig. P13.2, what force P is required to
maintain static equilibrium?
RAO2  75 mm, RBA  350 mm
Kinematic analysis:
Recall   11.95 from Problem 13.2.
x  r cos  cos   75 mmcos105  350 mmcos11.95  323 mm
Force analysis:
 MOz  xF14  M12  0
F14  M12 x  452 N.M 323 mm  139938 N
Recall the force polygon on link 4 from Problem 13.2.
P  F14 tan   1399.38 N tan11.95  6510.35 N
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Ans.
Theory of Machines and Mechanisms, 4e
13.4
Uicker et al.
Find the frame reactions and torque M12 necessary to maintain equilibrium of the fourbar linkage shown in Fig. P13.4a.
Kinematic analysis:
R AO2  87.5 mm210  75.775ˆi  43.75ˆj mm R BO4  150 mm135.53  107.05ˆi  105.075ˆj mm
R BA  150 mm82.83  18.725ˆi  148.825ˆj mm RCO4  100 mm135.53  71.375ˆi  70.05ˆj mm
445 N
445 N
Force analysis:
 MO4  RCO4 × P  R BO4 × F34  0
 71.375ˆi  70.05ˆj mm  × 311.732ˆi  317.574ˆj N 
  107.05ˆi  105.075ˆj in  ×  cos82.83ˆi  sin 82.83ˆj F
34
0
 45.203 N  M  119.325 inF34  kˆ  0
F34  9.4702 N  m
F
i4
 P  F34  F14  0
Ans.
 Fi 2  F32  F12  0
F34  372.941 N82.83  46.569ˆi  370.022ˆj N
F  265.153ˆi  52.447ˆj N  270.288 N168.81
14
F12  46.569ˆi  370.022ˆj N  372.941 N82.83
Ans.
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Theory of Machines and Mechanisms, 4e
M
O2
Uicker et al.
 M12  R AO2 × F32  0

 

M12kˆ  75.75ˆi  43.75ˆj mm × 46.569ˆi  370.021ˆj N  0
M12  26.408 N  M
13.5
M12  26.408kˆ N  M
Ans.
What torque must be applied to link 2 of the linkage illustrated in Fig. P13.4b to maintain
static equilibrium?
22.5 N
RAO2  87.5 mm; RBA  RBO4  150 mm; RCO4  100 mm; RO2O4  50 mm; RDO4  175 mm
Kinematic analysis:
R AO2  87.5 mm240  43.75ˆi  75.75ˆj 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
222.5 N
222.5 N
Force analysis:
 MO4  R DO4  P  R BO4  F34  0
 155.425ˆi  80.425ˆj in  ×  222.5ˆi N    133.225ˆi  68.925ˆj mm  ×  cos105.26ˆi  sin105.26ˆj F
34
 18.176 N  M  110.375 inF34  kˆ  0
F34  4.116 N  M
F34  162.109 N105.26  42.66ˆi  156.391ˆj N
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Theory of Machines and Mechanisms, 4e
M
O2
 M12  R AO2 × F32  0

Uicker et al.
 

M12kˆ  43.75ˆi  75.751ˆj mm × 42.66ˆi  156.39ˆj N  0
M12  10.23 N  M
M12  10.23kˆ N  M
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Ans.
Theory of Machines and Mechanisms, 4e
13.6
Uicker et al.
Sketch a complete free-body diagram of each link of the linkage illustrated in Fig. P13.6.
What force P is necessary for equilibrium?
RAO2  100 mm, RBA  150 mm, RBO4  125 mm, RCO4  200 mm, RCD  400 mm, and RO2O4  60 mm.
Kinematic analysis:
R AO2  100 mm90  100ˆj mm
R  150 mm  4.86  149ˆi  13ˆj mm
BA
R DC  400 mm  20.44  375ˆi  140ˆj mm
Force analysis:
 MO2  M12  R AO2 × F32  0

 
R BO4  125 mm44.3  89ˆi  87ˆj mm
R  200 mm44.3  143ˆi  140ˆj mm
CO4

90kˆ N  m  100ˆj mm × cos  4.86ˆi  sin  4.86ˆj F32  0
 90 N  m  99.640 mmF32  kˆ  0
M
O4
F32  903 N  4.86
 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
F54  472 N  20.44=44ˆi  165ˆj N
F  Pˆi  443 Nˆi  165 Nˆj  F ˆj=0
P  443ˆi N
86 N  m  181 mmF54  kˆ  0

16
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Ans.
Theory of Machines and Mechanisms, 4e
13.7
Uicker et al.
Determine the torque M12 required to drive slider 6 of Fig. P13.7 against a load of
P  112.5 N at a crank angle of   30 , or as specified by your instructor.
250 mm
150 mm
RAO2  62.5 mm; RBO4  400 mm; RBC  200 mm.
Kinematic analysis:
R AO2  62.5 mm30  54.125ˆi  31.25ˆj mm
R AO4  7.466 in73.37  2.136ˆi  7.154ˆj in
 400 mm73.37  114.45ˆi  383.275ˆj mm RCB  200 mm175.20  199.3ˆi  16.725ˆj mm
R BO4
Force analysis:
 F  Pˆi  F16ˆj  cos175.20ˆi  sin175.20ˆj F56  0


F56   P cos175.20  1112.5 N 0.996 1116.416 N
F  1116.416 N175.20  112.5ˆi  93.419ˆj N
56
M
O4
 R BO4 ×F54  R AO4 ×F34  0
114.45ˆi  383.275ˆj mm  × 12.5ˆi  93.419ˆj N  + 53.4ˆi  178.85ˆj mm  ×  cos163.37ˆi  sin163.37ˆj F
34
 4439.954 N  M  186.65 mmF34  kˆ  0
M
O2
F34  2341.679 N163.37=  2243.735ˆi  670.165ˆj N
 M12  R AO2  F32  0

 

M12kˆ  54.125ˆi  31.25ˆj mm  2243.73ˆi  670.165ˆj N  0
M12  106.38 N  M
13.8
M12  106.38kˆ N  M
Ans.
Sketch complete free-body diagrams for the four-bar linkage illustrated in Fig. P13.8.
What torque M12 must be applied to link 2 to maintain static equilibrium at the position
shown?
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Theory of Machines and Mechanisms, 4e
Uicker et al.
Kinematic analysis:
RCO4  350 mm  109.05  114ˆi  331ˆj mm
R AO2  200 mm60  100ˆi  173ˆj mm
R BA  400 mm  46.06  278ˆi  288ˆj mm , RCA  700 mm  46.06  486ˆi  504ˆj mm
Force analysis:
Since the lines of action of all constraint forces cannot be found from two- and threeforce members, the force F34 is resolved into radial and transverse components,

. Then
F34r and F34
M


F34  129 N  19.05=122ˆi  42ˆj N
45kˆ N  m + 350kˆ mmF34  0


M

 M14 + RCO4 ×F34  45kˆ N  m + 114ˆi  331ˆj mm × cos 19.05ˆi  sin  19.05ˆj F34  0
O4

+ R CA × F43r  0
 R BA  P  R CA × F43
A
 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
F  Fr  F  48ˆi  257ˆj N  261 N100.58
43
43
43
Now the lines of action for other forces may be found as shown.
 F  F43r  F43  P  F23  0
74ˆi  215ˆj N  122ˆi  42ˆj N  350ˆi N  F  0 , F =398ˆi  257ˆj N  474 N  32.85
M
O2

M12  94.55 N  m
13.9
23
23
 

 M12  R AO2 ×F32  M12kˆ  100ˆi  173ˆj mm × 398ˆi  257ˆj N  0
M12  94.55kˆ N  m
Ans.
Sketch free-body diagrams of each link and show in Fig. P13.9 all the forces acting. Find
the magnitude and direction of the moment that must be applied to link 2 to drive the
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Theory of Machines and Mechanisms, 4e
Uicker et al.
linkage against the forces shown.
Kinematic analysis:
R AO2  200 mm30  173.2ˆi  100ˆj mm RCO4  500 mm84.34  1.973ˆi  19.902ˆj in
R  700 mm67.81  264.375ˆi  648.15ˆj mm R  700 mm34.61  23.045ˆi  15.903ˆj in
BA
 350 mm84.34  34.525ˆi  348.3ˆj mm
R DO4
CA
445 N
445 N
890 N
890 N
Force analysis:
Since the lines of action of all constraint forces cannot be found from two- and threeforce members, the force F34 is resolved into radial and transverse components,

F34r and F34
. Then
M
O4
 R DO4 × PD  R CO4 ×F34  0
34.5251ˆi  348.3ˆj mm ×  858.85ˆi  231.4ˆj N  +  49.325ˆi  497.55ˆj mm ×  cos 5.66ˆi  sin 5.66ˆj F

34
307kˆ N  M  499.95kˆ mmF34  0
M
A

F34
 614 lN  5.66=609.65ˆi  62.3ˆj N

 R BA × PB  R CA × F43
 R CA × F43r  0
 264.375ˆi  648.15ˆj mm  ×  445ˆi N  + 576.125ˆi  397.5755ˆj mm  ×  609.65ˆi  62.3ˆj N 
+  576.125ˆi  397.575ˆj mm  ×  cos  95.66ˆi  sin  95.66ˆj F  0
r
43
288.36kˆ N  M  278.347kˆ N  M  534.15kˆ mmF43r  0 ,
F43r  1059.1N  95.66=  102.35ˆi  1054.65ˆj N

F43  F43r  F43
 712ˆi  992.35ˆj N  1219.3 N  125.66
Now the lines of action for other forces may be found as shown.
 F  F43r  F43  PB  F23  0
102.35ˆi  1054.65ˆj N  609.65ˆi  62.3ˆj N  445ˆi N  F  0 ,
23
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0
Theory of Machines and Mechanisms, 4e
Uicker et al.
F23 =1157ˆi  992.35ˆj N  1526.35 N40.62
 MO2  M12  R AO2 ×F32  0; M12kˆ  173.2ˆi  100ˆj mm × 1157ˆi  992.35ˆj N  0

 

M12  56kˆ N  M
M12  56 N  M
Ans.
13.10 Figure P13.10 illustrates 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.
RAO2  150 mm, RCA  600 mm, RO4O2  600 mm, RCO4  RBA  400 mm, and
RBC  300 mm.
Kinematic analysis:
R AO2  150 mm  30  130ˆi  75ˆj mm
R  400 mm16.00  385ˆi  110ˆj mm
BA
RCO4  400 mm124.56  227ˆi  329ˆj mm
R  600 mm42.38  443ˆi  404ˆj mm
CA
Force analysis:
 M A  R BA ×PB  RCA ×PC  RCA ×F43  0
 385ˆi  110ˆj mm  ×  354ˆi  354ˆj N  +  443ˆi  404ˆj mm  × 1 800ˆi N  +  443ˆi  404ˆj mm  ×  cos124.56ˆi  sin124.56ˆj F
43
174.876kˆ N  m  726.000kˆ N  m  594kˆ mmF43r  0 ,
F  927124.56 N=  526ˆi  763ˆj N
43
F  F
43
 PB  PC  F23  0
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Theory of Machines and Mechanisms, 4e
Uicker et al.
526ˆi  763ˆj N  354ˆi  354ˆj N+1 800ˆi N  F23  0 ,
F =  920ˆi  1 117ˆj N  1 447 N 129.48
23
M
O2
 M12  R AO2 ×F32  0;

 

M12kˆ  130ˆi  75ˆj mm × 920ˆi  1 117ˆj N  0
M12  214.21kˆ N  m
M12  214.21 N  m
Ans.
13.11 Draw a free-body diagram of each of the members of the mechanism illustrated in Fig.
P13.11 and find the magnitudes and the directions 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.
534 N
801 N
RAO2  50 mm; RCA  125 mm; RO4O2  RCO4  100 mm; RDO4  75 mm; RDC  50 mm; RBA  175 mm;
RBC  62.5 mm.
Kinematic analysis:
R AO2  50 mm180  50ˆi mm
R  175 mm55.98  97.9ˆi  145ˆj mm
BA
RCO4  100 mm124.23  56.25ˆi  82.675ˆj mm
R  125 mm41.41  93.75ˆi  82.675ˆj mm
CA
R DO4  75 mm95.27  6.9ˆi  74.675ˆj mm
534 N
534 N
801 N
Force analysis:
Since the lines of action of all constraint forces cannot be found from two- and threeforce members, the force F34 is resolved into radial and transverse components,
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Theory of Machines and Mechanisms, 4e
Uicker et al.

. Then
F34r and F34
M
O4
 R DO4 × PD  R CO4 ×F34  0
 6.9ˆi  74.675ˆj mm ×  694.2ˆi  400.5ˆj N  +  56.25ˆi  82.675ˆj mm ×  cos34.23ˆi  sin 34.23ˆj F

34

F34
 489.5 N34.23=404.95ˆi  275.9ˆj N
49kˆ N  M  100kˆ mmF34  0
M
A
=0

 R BA × PB + R CA × F43
 R CA × F43r  0
97.9ˆi  145ˆj mm  ×  534ˆi N  + 93.75ˆi  82.675ˆj mm  ×  405ˆi  276ˆj N 
+  93.75ˆi  82.675ˆj mm  ×  cos  55.77ˆi  sin  55.77ˆj F  0
r
43
77.43kˆ N  M  7.565kˆ N  M  124kˆ mmF43r  0 , F43r  685.3 N  55.77=387.15ˆi  565.15ˆj N
F34  F43  17.8ˆi  841ˆj N  841 N88.79
F43  F43r  F43  17.8ˆi  841ˆj N  841 N  91.21 ,
Ans.
Now the lines of action for other forces may be found as shown.
17.8ˆi  841ˆj N  534ˆi N  F23  0 ,
 F  F43  PB  F23  0 ,
F =53.4ˆi  841ˆj N  1005.7 N56.73 ,
F  F  124ˆi  189ˆj lb  226 lb 123.27 Ans.
23
32
23
4ˆi  189ˆj lb  156ˆi  90ˆj lb  F14  0 ,
 F  F34  PD  F14  0 ,
F14 =676.4ˆi  1241.55ˆj N  1415.1 N  61.42 , F12  F32  534ˆi  841ˆj N  1005.7 N56.73
M  M  R ×F  0;
M kˆ  50ˆi mm × 534ˆi  841ˆj N = 0

O2
12
AO2
M12  42 N  M
32
12

 
M12  42kˆ N  M

Ans.
13.12 Determine the magnitude and direction of the torque that must be applied to link 2 to
maintain static equilibrium.
445 N
222.5 N
RAO2  75 mm; RCA  350 mm; RBA  175 mm; RBC  200 mm.
Kinematic analysis:
R AO2  75 mm90  75ˆj mm RCA  350 mm  12.37  341.275ˆi  75ˆj mm
R  175 mm  34.93  143.475ˆi  100.225ˆj mm
BA
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Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
445 N
222.5 N
Force analysis:
 M A  R BA × PB  RCA × PC  RCA ×F14  0
143.475ˆi 100.225ˆj mm ×  222.5ˆj N  + 341.875ˆi  75ˆj mm ×  445ˆi N  + 341.875ˆi  75ˆj in  × ˆj F
14
F14  4.45 N90=4.45ˆj N
31.927kˆ N  M  33.375kˆ N  M  350kˆ mmF14  0
F  P
 PC  F14  F23  0
222.5ˆj N  445ˆi N  4.45ˆj N  F23  0 ,
B
M
O2
 M12  R AO2  F32  0
M12  33.375 N  M
F23 =445ˆi  226.95ˆj N  498.4 N  27.02
M kˆ + 75ˆj mm × 445ˆi  226.95ˆj N  0
12

 
M12  33.375kˆ N  M
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
Ans.
=0
Theory of Machines and Mechanisms, 4e
Uicker et al.
13.13 Figure P13.13 shows a 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 the position  2  49 . Minimum outreach is 10.5 m at
 2  132. Other dimensions are given in the legend to Fig. P13.13. For the maximum
outreach position and a grab load of 10 T (under standard gravity), find the bearing
reactions at A, B, O2 , and O4 , as well as the moment M12 required. Notice that the
photograph shows a counterweight on link 2; neglect this weight and also the weights of
the members.
RAO2  14.7 m; RBA  6.5 m; RBO4  19.3 m; RCA  22.3 m; RCB  16 m. (Photograph and
dimensions by permission from B.V. Machinefabriek Figee, Haarlem, Holland)
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Theory of Machines and Mechanisms, 4e
Uicker et al.
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 , RCA  22.300 m14.39  21.600ˆi  5.543ˆj m
Force analysis:
Note that a metric ton is a unit of mass whereas a more appropriate unit for rating a crane
would be force capacity. Nevertheless, the weight of a metric ton in standard gravity is
W  mg  1 000 kg  9.81 m/s2   9.810 kN . Therefore, the stated load on the crane is F  98.100 kN .
M
A
 R BA ×F43  R CA ×F  0
 6.494ˆi  0.269ˆj m  ×  cos59.70ˆi  sin 59.70ˆj F +  21.600ˆi  5.543ˆj m ×  98.100ˆj kN   0
43
5.471kˆ mF43  2 119kˆ kN  m  0 ,
F  38759.70 kN=195ˆi  334ˆj kN ,
43
F  F
 F  F23  0
195ˆi  334ˆj kN  98.1ˆj kN  F23  0 ,
F =  195ˆi  236ˆj kN  307 kN  129.59 ,
F14  F34  387 kN59.70=195ˆi  334ˆj kN , Ans.
43
23
M
O2
 M12  R AO2 ×F32  0 ;
M12  113 kN  m
F12 =  F32 = 195ˆi  236ˆj kN  307 kN129.59 , Ans.
M12kˆ  9.644ˆi  11.094ˆj m × 195ˆi  236ˆj kN  0

M12  113kˆ kN  m
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

Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
13.14 Repeat Problem 13.13 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 , RCA  22.300 m15.83  21.454ˆi  6.083ˆj m
Force analysis:
Note that a metric ton is a unit of mass whereas a more appropriate unit for rating a crane
would be force capacity. Nevertheless, the weight of a metric ton in standard gravity is
W  mg  1 000 kg  9.81 m/s2   9.810 kN . Therefore, the stated load on 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 ,
F  362 kN120.35=  183ˆi  312ˆj kN ,
43
F  F
 F  F23  0
F23 =183ˆi  214ˆj kN  282 kN  49.51 ,
43
M
O2
 M12  R AO2 ×F32  0,
M12  109 kN  m
F14  F34  362 kN120.35=  183ˆi  312ˆj kN Ans.
183ˆi  312ˆj kN  98.1ˆj kN  F  0
23
F12 =  F32 =183ˆi  214ˆj kN  282 kN 49.51 , Ans.
M kˆ  9.836ˆi  10.924ˆj m × 183ˆi  214ˆj kN  0
12

M12  109kˆ kN  m
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

Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
13.15 Repeat Problem 13.7 assuming coefficients of Coulomb friction c  0.20 between links
1 and 6 and c  0.10 between links 3 and 4. Determine the torque M12 necessary to
drive the system, including friction, against the load P.
250 mm
150 mm
RAO2  62.5 mm; RBO4  400 mm; RBC  200 mm.
See the figure and solution for Problem 13.7 for the kinematic and frictionless solutions.
For friction between links 1 and 6, the friction angle is   tan 1  0.20   11.31 . Since
the impending motion VC6 /1 is to the left the friction force f16  c F16n is toward the right.
Also, since the non-friction normal force F16n is downward (from the solution for Problem
13.7), the total force F16 acts at the angle 90  11.31  78.69 . Therefore,
F  Pˆi  cos 78.69ˆi  sin  78.69ˆj F ˆj  cos175.20ˆi  sin175.20ˆj F  0



16


56
1112.5 N  cos 78.69F16  cos175.20F56  0 , sin  78.69F16  sin175.20F56  0
F  96.89 N , F  1135.484 N , F  1135.484 N175.20  1131.5ˆi  95ˆj N . For friction
16
56
56
between links 3 and 4, the friction angle is   tan 1  0.10  5.71 . Since the impending
motion VA3 / 4 is upward the friction force f34  c F34n is upward. Also, since the nonfriction normal force F34n is toward the left (from the solution for Problem 13.7), the total
force F34 acts at the angle 163.37  5.71  157.66 . Therefore,
M
O4
 R BO4 ×F54  R AO4 ×F34  0
114.45ˆi  383.275ˆj mm × 1131.5ˆi  95ˆj N  53.4ˆi  178.85ˆj mm × cos157.66ˆi  sin157.66ˆj F
34
0
 444.55 N  M  185.725 mmF34  kˆ  0 , F34  2393.6 N157.66=  2213.95ˆi  909.815ˆj N
M
O2



 M12  R AO2 ×F32  0 , M12kˆ  54.125 ˆi  31.25 ˆj mm × 2213.95 ˆi  909.815 ˆj N  0
M12  118.22 N  M
M12  118.22kˆ N  M
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Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
13.16 Repeat Problem 13.12 assuming a coefficient of static friction   0.20 between links 1
and 4. Determine the torque M12 necessary to overcome friction.
445 N
222.5 N
RAO2  75 mm; RCA  350 mm; RBA  175 mm; RBC  200 mm.
See the figure and solution for Problem 13.12 for the kinematic and frictionless solution.
For friction between links 1 and 4, the friction angle is   tan 1  0.20   11.31 . Since
the impending motion VC4 /1 is to the right the friction force f14  c F14n is toward the left.
Also, since the non-friction normal force F14n is upward (from the solution for Problem
13.12), the total force F14 acts at the angle 90  11.31  101.31 . 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  ×  cos101.31ˆi  sin101.31ˆj F  0
14
31.923kˆ N  M  33.375kˆ N  M  349.95kˆ mmF14  0 F14  4.147 N101.31=  0.814ˆi  3.964ˆj N
F  P
 PC  F14  F23  0
222.5ˆj N  445ˆi N  0.814ˆi  3.964ˆj N  F23  0 , F23 =445.8ˆi  226.54ˆj N  500 N  26.94
M  M  R  F  0 , M kˆ  75ˆj mm  445.8ˆi  226.54ˆj N  0
B

O2
12
M12  33.435 N
AO2
32
12

 
M12  33.435kˆ N  M
© Oxford University Press 2015. All rights reserved.

Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
13.17 In each case shown, pinion 2 is the driver, gear 3 is an idler, and the gears have module of
4.20 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
18 hp to the gearset. For (b) and (c), pinion 2 rotates at 900 rev/min and transmits 25 hp
to the gearset.
mN3 4.20  34 teeth
mN 2 4.20 18 teeth
R3 

 71.4 mm

 37.8 mm
2
2
2
2
 600 rev/min  2   62.832 rad/s cw ,
R
37.8 mm
2 
3  2 2 
62.832 rad/s  33.264 rad/s ccw
60 s/min
R3
71.4 mm
(a)
R2 
F23t 
18 hp  734.25 N  M/s/hp   5565 N
P

R33  71.4 mm 103 m   33.264 rad/s 
F23  F23t cos   5565 N cos20  5922.15 N , F43  F23  5922.15 N
F  F
23
 F43  F13  0
F13  F  F  5565 N  5565 N  11130 N
t
23
t
43
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
mN3 4.20 mm/tooth36 teeth
mN 2 4.20 mm/tooth 18 teeth

 75.6 mm

 37.8 mm R3 
2
2
2
2
 900 rev/min  2   94.248 rad/s ccw ,   R2   37.8 mm 94.248 rad/s  47.124 rad/s cw
2 
3
2
60 s/min
R3
75.6 mm
(b)
R2 
F23t 
 25 hp  734.25 N  M/s/hp   5152.5 N
P

R33  75.6 mm 10.3 m  47.124 rad/s 
F23  F23t cos   5152.5 N cos20  5483 N , F43  F23  5483 N
 F  F23  F43  F13  0
F13    F23  F43 
   5483 N  20  5483 110 
 4634.7 N225
mN 2 4.20 18

 37.8 mm
2
2
 900 rev/min  2   94.248 rad/s ccw ,
2 
60 s/min
mN3 4.20  36

 75.6 mm
2
2
R
37.8 mm
94.248 rad/s  47.124 rad/s cw
3  2 2 
R3
75.6 mm
R2 
F23t 
 25 hp  734.25 N  M/s/hp   5152.5 N
P

R33
 75.6 mm  47.124 rad/s 
F  F
23
 F43  F13  0
Ans.
R3 
(c)
F23  F23t cos   5152.5 N cos20  5483 N ,
Ans.
F43  F23  5483 N
Ans.
F13    F23  F43 
   5483 N  20  5483 N  70 
 9931.5 N135
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
13.18 A 15-tooth spur pinion with module of 5 mm/tooth and 20 pressure angle, rotates at 600
rev/min, and drives a 60-tooth gear. The drive transmits 18 kW. Construct a free-body
diagram of each gear showing upon it the tangential and radial components of the forces
and their proper directions.
mN3  5 mm/tooth  60 teeth
mN 2  5 mm/tooth 15 teeth

 150 mm

 37.500 mm R3 
2
2
2
2
 600 rev/min  2   62.832 rad/s
R
1.500 in
2 
3  2  2 
62.832 rad/s  15.708 rad/s
60 s/min
R3
6.000 in
R2 
F32t 
18 kW 1000 N  m/s/kW 1000 mm/m   1 910 N , r
P

F32  F32t tan   695 N
R33
150 mm  62.832 rad/s 
F23t  F32t  1 910 N
F23r  F32r  695 N
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
13.19 A 16-tooth pinion on shaft 2 rotates at 1 720 rev/min and transmits 5 hp to the doublereduction gear train. All gears have 20 pressure angle. The distances between centers
of the bearings and gears for shaft 3 are shown in Fig. P13.19. Find the magnitude and
direction of the radial force that each bearing exerts against the shaft.
4.20 m,
50 mm
200 mm
50 mm
3.175 mm,
mN 2
3.175 mm/teeth 16 teeth
mN A
3.175 mm/teeth  64 teeth

 25.4 mm RA 

 101.6 mm
2
2
2
2
mN A
mN B
4.20 mm/teeth  36 teeth
4.20mm/teeth  24 teeth

 75.6 mm
RB 

 50.4 mm R4 
2
2
2
2
R2 
2 
1 720 rev/min  2   180.118 rad/s
3 
R2
25.4 mm
180.118 rad/s  45.029 rad/s
2 
RA
101.6 mm
60 s/min
R
50.4 mm
45.029 rad/s  30.020 rad/s
4  B 3 
R4
75.6 mm
 5 hp  734.25 NM/s/hp   802.47 N , F  F t cos   855 N
P
F23t 

23
23
RA3 101.6 mm  45.029 rad/s 
F43t 
RA t 101.6 mm
802.47 N  1617.67 N
F23 
50.4 mm
RB
F43  F43t cos   1721.49 N
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
Choosing a coordinate system with origin at C as shown we have
FA  F23  855 N20  802.47ˆi  293.79ˆj N
R A  101.6ˆj  50.4kˆ mm
R  50.4ˆj  254kˆ mm
F  F  1721.49 N  20  1617.67ˆi  527.64ˆj N
B
43
B
FC  F ˆi  F ˆj
F  F x ˆi  F y ˆj
D
M
C
x
C
y
C
RC  0
D
D
R D  304.8kˆ mm
 R A ×FA  R B ×FB  R D ×FD  0
 101.6ˆj  50.4kˆ mm × 803.47ˆi  293.79ˆj N  50.4ˆj  254kˆ mm × 1617.67ˆi  587.84ˆj N  304.8kˆ mm ×  F ˆi  F ˆj  0
 14.79ˆi  40.72ˆj  81.43kˆ N  M   148.4ˆi  407.6ˆj  81.43kˆ N  M    304.8F ˆi  304.8F ˆj mm  0
x
D
y
D
FD  350 lb163.42  1495.2ˆi  445ˆj N Ans.
FDx  1495.2 N, FDy  445 N
F  F
A
 FB  FC  FD  0
803.47ˆi  293.79ˆj N  1617.67ˆi  587.84ˆj N  F  1495.2ˆi  445ˆj N   0 ,
C
FC  961.2 N188.86  952.3ˆi  146.85ˆj N
Ans.
© Oxford University Press 2015. All rights reserved.
y
D
x
D
Theory of Machines and Mechanisms, 4e
Uicker et al.
13.20 Solve Problem 13.17 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 module is
4.20 mm/teeth for each case.
Since the pressure angles and the helix angle are related by cos   tan n tan t ,
t  tan 1  tan n cos   tan 1  tan 20 cos30  22.80
mN3 4.20 mm/teeth  34 teeth
mN 2 4.20 mm/teeth 18 teeth

 71.4 mm

 37.8 mm R3 
2
2
2
2
 600 rev/min  2   62.832 rad/s cw ,   R2   37.8 mm 62.832 rad/s  33.264 rad/s ccw
2 
3
2
60 s/min
R3
71.4 mm
(a)
R2 
F23t 
18 hp  734.25 N  M/s/hp   5607 N
P

R33
 71.4 mm  33.264 rad/s 
F23r  F23t tan t   5607 N  tan 22.80  2358.5 N ,
F23a  F23t tan   5607 N  tan 30  3235.15 N
F23  5607ˆi  2358.5ˆj  3235.15kˆ N
F43  5607ˆi  2358.5ˆj  3235.15kˆ N
F  F F F  0
F  11218.45ˆi N

 M R ˆj×F
23
43
3
23
13
13
Ans.
 R3ˆj  F43  M13  0
 71.4ˆj mm × 5607ˆi  2358.5ˆj  3235.15kˆ N   71.4ˆj mm  × 5607ˆi  2358.5ˆj  3235.15kˆ lb  M
 71.4ˆj mm × 5607ˆi  2358.5ˆj  3235.15kˆ N   71.4ˆj mm  × 5607ˆi  2358.5ˆj  3235.15kˆ N  M
M13  458.238ˆi N  M This moment must be supplied by the shaft bearings.
© Oxford University Press 2015. All rights reserved.
Ans.
13
0
13
0
Theory of Machines and Mechanisms, 4e
Uicker et al.
mN3 4.20 mm/teeth  36 teeth
mN 2 4.20 mm/teeth 18 teeth

 75.6 mm

 37.8 mm R3 
2
2
2
2
 900 rev/min  2   94.248 rad/s ccw ,   R2   37.8 mm 94.248 rad/s  47.124 rad/s cw
2 
3
2
60 s/min
R3
75.6 mm
(b)
R2 
F23t 
 25 hp  734.25 N  M/s/hp   5152.5 N
P

R33
 75.6 mm  47.124 rad/s 
F23r  F23t tan t   5152.5 N  tan 22.80  2184.95 N ,
F23a  F23t tan   5152.5 N  tan 30  2999.3 N
F23  5152.5ˆi  2184.95ˆj  2999.3kˆ N
F43  2184.95ˆi  5152.5ˆj  2999.3kˆ N
F13  3ˆi  3ˆj N
Ans.
 F  F23  F43  F13  0
M R ˆj×F  R ˆi ×F  M  0

3
23
3
43
13
75.6ˆj mm × 5152.5ˆi  2184.95ˆj  2999.3kˆ N   756ˆi mm ×  2184.95ˆi  5152.5ˆj  2999.3kˆ N  M
13
M13  224.95ˆi  224.95ˆj N  M This moment must be supplied by the shaft bearings. Ans.
mN3 4.20 mm/teeth  36 teeth
mN 2 4.20 mm/teeth 18 teeth

 75.6 mm

 37.8 mm R3 
2
2
2
2
 900 rev/min  2   94.248 rad/s ccw ,   R2   37.8 mm 94.248 rad/s  47.124 rad/s cw
2 
3
2
60 s/min
R3
75.6 mm
(c)
R2 
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
F23t 
Uicker et al.
 25 hp  734.25 N  M/s/hp   5152.5 N
P

R33
 75.6 mm  47.124 rad/s 
F23r  F23t tan t   5152.5 N  tan 22.80  2184.95 N ,
F23a  F23t tan   5607 N  tan 30  2999.3 N
F  2184.95ˆi  5152.5ˆj  2999.3kˆ N
F23  5152.5ˆi  2184.95ˆj  2999.3kˆ N
43
F13  7378.1ˆi  7378.1ˆj N
F  F  F
 M R ˆj×F
23
3
 F13  0
ˆ
23  R3i ×F43  M13  0
43
Ans.
75.6ˆj mm × 5152.5ˆi  2184.95ˆj  2999.3kˆ N   75.6ˆi mm ×  2184.95ˆi  5152.5ˆj  2999.3kˆ N  M
13
M13  224.95ˆi  224.95ˆj N  M This moment must be supplied by the shaft bearings. Ans.
13.21 Analyze the gear shaft of Example 13.8 and find the bearing reactions FC and FD .
The solution is shown in Fig. 13.20c.
623 N
525 N
690 N
58.3 N
94.82 mm
1117 N
316 N
98 .5 mm
1223.75 N
761 N
1806.7 N
207 N
623 N
FC  525ˆi  623ˆj  1171kˆ N
F  316ˆi  6901kˆ N
D
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
13.22 In each of the bevel gear drives illustrated in Fig. P13.22, bearing A takes both thrust load
and radial load, whereas 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.
32 mm
50 mm
26.25 mm
17.25 mm
42.5 mm
62.5 mm
2.5 mm
(a)
34.5 mm
4.2 mm
50 mm
  tan 1  32 teeth 16 teeth   63.43
F32t  T2 R2  20 N  M 17.25 mm  1161.45 N
F32r  F32t tan  cos   189.12 N
F23  189.1ˆi  1161.45ˆj  377.8kˆ N
M
A
F32a  F32t tan  sin   377.8 N
 R BA ×FB  R PA ×F23  T3  0
 50kˆ mm ×  F ˆi  F ˆj  34.5ˆi  59kˆ mm ×  189.12.5ˆi 1161.45ˆj  377.8kˆ N   T kˆ  0
50 mmF ˆi  50 mmF ˆj  68.49ˆi 1.889ˆj  40kˆ N  M   T kˆ  0
x
B
y
B
y
B
3
x
B
3
FB  37.825ˆi  1370.6ˆj N
FA  226.95ˆi  209.15ˆj  378.25kˆ N
T3  40kˆ N  M
F  F
A
(b)
 FB  F23  0
Ans.
Ans.
  tan 1 18 teeth 24 teeth   36.87
F32t  T2 R2  26.7 N  M 32 mm  834.375 N
F32r  F32t tan  cos   242.97 N
F32  242.97ˆi  834.375ˆj  182kˆ N
M
A
F32a  F32t tan  sin   182 N
 R BA ×FB  R PA ×F32  T2  0
50kˆ mm ×  F ˆi  F ˆj   32ˆi  20kˆ mm ×  242.97ˆi  834.375ˆj 182kˆ N    26.7kˆ N  M   0
 50 mmF ˆi  50 mmF ˆj   16.687ˆi  0.89ˆj  26.7kˆ N  M    26.7kˆ N  M   0
x
B
y
B
y
B
x
B
FB  17.8ˆi  333.75ˆj lb
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
F  F
A
 FB  F23  0
Uicker et al.
FA  222.5ˆi  1170.35ˆj  182.45kˆ N
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
18.75 mm
43.75 mm
13.23 Figure P13.23 illustrates a gear train composed of a pair of helical gears and a pair of
straight-tooth bevel gears. Shaft 4 is the output of the train and delivers 6 hp 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, whereas bearing F is to take only radial load, determine
the force that each bearing exerts against shaft 4.
12.5 mm,
3m,
21.875 mm,
21.875 mm,
2m,
The diameters of the bevel gears at their large faces are
R3  mN3 2  3 mm/tooth  2 teeth   30 mm
R4  mN4 2  3 mm/tooth×2 teeth   60 mm
  tan 1  R4 R3   63.43
  tan 1  R3 R4   26.57
The average pitch radii are
R4,avg  R4  12.5sin   51.3 mm
R3,avg  R3  12.5sin   25.65 mm
4 
 370 rev/min  2   38.746 rad/s
60 s/min
 6 hp  734.25 N  M/s/hp   2216 N
P
F34t 

R4,avg4  51.3 mm  38.746 rad/s 
F34r  F34t tan  cos   360.45 N
F34  720.9ˆi  360.45ˆj  2216kˆ N
M
E
F34a  F34t tan  sin   720.9 N
 R FE ×FF  R PE ×F34  T4  0
60ˆi mm ×  F ˆj  F kˆ   18ˆi  51.3ˆj mm ×  720.9ˆi  360.45ˆj  2216kˆ N    113.69ˆi N  M   0
 60 mmF ˆj  60 mmF kˆ   113.69ˆi  40ˆj  30.48kˆ in  lb    113.69ˆi N  M   0
y
F
z
F
F  F
E
z
F
y
F
 FF  F34  0
FF  489.5ˆj  640.8kˆ N
FE  720.9ˆi  849.95ˆj  1575.3kˆ N
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Ans.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
13.24 Using the data of Problem 13.23, 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/teeth  35 teeth  2   35 mm
F23t  F43t R3,avg RS  2216 N  25.65 mm 36.45 mm  1559.4 N
F23r  F23t tan   1559.4 N  tan 20  567.575 N
F23a  F23t tan  1559.4 N  tan 30  900.32 N
F  567.575ˆi  900.32ˆj  1559.4kˆ N
23
M
C
 R DC ×FD  R PC ×F43  R RC ×F23  0
 43.75ˆj mm  ×  F ˆi  F kˆ    25.65ˆi  67.425ˆj mm  × 720.9ˆi  360.45ˆj  2216kˆ N 
  36.45ˆi  21.875ˆj in  ×  567.575ˆi  900.32ˆj  1559.4kˆ N   0
 43.75 mmF ˆi  43.75 mmF kˆ   149.4ˆi  56.85ˆj  39.38kˆ N  m  34ˆi  56.85ˆj  45.278kˆ N  m  0
x
D
z
D
F  F
C
z
D
x
D
 FD  F23  F43  0
FD  133.5ˆi  4191.9kˆ N
FC  284.8ˆi  538.45ˆj  418.3kˆ N
Since the thrust force is in the jˆ direction, C should be a thrust bearing.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
13.25 Use the method of virtual work to solve the slider-crank mechanism of Problem 13.2.
  sin 1  r sin     sin 1  75 mm  sin105 350 mm  11.95
x  r cos  cos   75 mmcos105  350 mmcos11.95  342.4 mm
The first-order kinematic coefficient is
x  dx d  RI24 I12  x tan   342.4 mm tan11.95  72.467 mm
M12  Px  4005 N  72.467 mm   290.23 N  M cw
Ans.
13.26 Use the method of virtual work to solve the four-bar linkage of Problem 13.5.
222.5 N
RAO2  87.5 mm; RBA  RBO4  150 mm; RCO4  100 mm; RO2O4  50 mm; RDO4  175 mm.
The first-order kinematic coefficient is
4  d4 d2  RI24 I12 RI24 I14  64.420 mm 114.425 mm  0.563
M14  RDO4 P sin152.64  175 mm  222.5 N  sin152.64  17.89 N  M cw
M12  M14 d4 d2  M144   17.89 N  M cw  0.563  10.072 N  M ccw
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Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
13.27 Use the method of virtual work to analyze the crank-shaper linkage of Problem 13.7.
Given that the load remains constant at P  100ˆi lb, find and plot a graph of the crank
torque M12 for all positions in the cycle using increments of 30 for the input crank.
xAO4  RAO4 cos 4  2.5cos 2
 6  2.5sin  2 

 2.5cos 2 
yAO4  RAO4 sin  4  6  2.5sin  2
 4  tan 1 
RAO4  42.25  30sin  2
yI24 I14  RAO4 sin 4
 4 
yBC  8sin 5  16  16sin  4
5  sin 1  2  2sin  4 
d 4 yI24 I14  6

 1  6sin  4 RAO4
d 2
yI24 I14
yI46C  yBC  xC xCB 
 8sin 5  8cos 5  16 cos  4  8cos 5  
 8  sin 5  2 cos  4 tan 5 
dxC d4  yI46 I14  16  8sin 5  16cos 4 tan 5
M12   dxC d 4  d 4 d 2  P
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Theory of Machines and Mechanisms, 4e
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Values for one cycle are shown in the following table.
 2 (deg.)
 4 (deg.)
RAO4 (mm)
d 4 d 2
5 (deg.)
dxC d4 (mm)
M12 (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
0.147 93
0.240 15
0.281 97
0.294 12
0.281 97
0.240 15
0.147 93
0.04593
0.41416
0.71429
0.41416
0.04593
0.147 93
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
393.175
392.875
396.775
400
394
373.65
345.275
333.65
368.05
400
392.575
394.35
393.75
25.88
41.98
49.78
52.35
49.43
39.93
22.73

67.83
127.143
72.35
8.06
25.88
The values of M 12 from this table are graphed as follows:
100
50
-50
-100
-150
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Theory of Machines and Mechanisms, 4e
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13.28 Use the method of virtual work to solve the four-bar linkage of Problem 13.10.
3  d3 d2  RI
RB  dR B
RC  dRC
RI23I13  75 mm 688 mm  0.1089
d2  3kˆ  R BI13  0.1089kˆ   568 mm135.33   61.881 mm45.33
d   kˆ  R  0.1089kˆ   662 mm124.56   72.153 mm34.56
2
23 I12
3
M12  PB RB  PC RC
CI13
  500 N135   61.881 mm45.33   1 800 N0   72.153 mm34.56 
 30.940 cos89.67 N  m+129.876 cos 34.56 N  m
M12  107 N  m cw
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Theory of Machines and Mechanisms, 4e
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13.29 A car (link 2) that weighs 8900 N is slowly backing a 4450 N trailer (link 3) up a 30
inclined ramp as illustrated in Fig. P13.29. The car wheels are of 325 mm radius, and the
trailer wheels have 250 mm radius; the center of the hitch ball is also 13 in 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 Fig. 13.29. 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.
125
0m
m
600 mm
800 mm
1200 mm
900 mm
2000 mm
4450 N
8900 N
For the trailer:
RG3B  945ˆi  863.15ˆj mm  1279.9 mm42.41
 M  1250 mmF kˆ  R  W  0
13
B
G3 B
3
F13  3364.2 N120  1682ˆi  2914.75ˆj N
F  F
13
Ans.
 F23  W3  0
F23  2278.49 N42.41  1682ˆi  1535.25ˆj N
For the car:
 MP  2000 mmF12Rkˆ  800 mm 8900 N  kˆ  2900ˆi  325ˆj mm  F32  0

F  F
F
12
 f 12ˆi  F12R  F32  0
F12R  5513.55ˆj N
F F  4921.7ˆj N
12
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
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
  f12 F  1682 N 4921.7 N
F
12
Uicker et al.
f12  1682ˆi N
Ans.
  0.34
Ans.
13.30 Repeat Problem 13.29 assuming that the car has rear-wheel drive rather than front-wheel
drive.
The entire solution is identical with that of Problem 13.29 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
Ans.
  f12 F12R  1682 N 5513.5 N
  0.31
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13.31 The low-speed disk cam with oscillating flat-faced follower illustrated in Fig. P13.31 is
driven at a constant shaft speed. The displacement curve for the cam has a full-rise
cycloidal motion, defined by Eq. (6.13), with parameters L  30 ,   30 , and a primecircle radius Ro  30 mm ; the instant pictured is at 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 demonstrated. Use
the virtual work approach to determine the moment M12 required on the crankshaft at
the instant shown to produce this motion.
RO2O3  50 mm; RB  42 mm; RC  150 mm.
The moment on link 3 caused by the output load is
M13  RCO3 ×FC  150 mm 8 N  sin  135 kˆ  0.849kˆ N  m
From Eq. (6.13b),
2  30 
112.5 
L
y  1  cos

1  cos 2
  0.200

  150 
150 
From virtual work
M12   d3 d 2 M13   yM13  0.200 0.849kˆ N  m  0.170kˆ N  m


Ans.
13.32 Repeat Problem 13.31 for the entire lift portion of the cycle, finding M12 as a function of
2.
From Problem 13.31
M13  RCO3  FC  150 mm 8 N  sin  135  kˆ  0.849kˆ N  m
360 2 
2  30 
L
1  cos

1  cos
  0.200 1  cos 2.4 2 

  150 
150 
M12   yM13  0.200 1  cos 2.4 2  0.849kˆ N  m  0.170 1  cos 2.4 2  kˆ N  m Ans.
y 


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Theory of Machines and Mechanisms, 4e
Uicker et al.
13.33 A disk 3 of radius R is being slowly rolled under a pivoted bar 2 driven by an applied
torque T as illustrated in Fig. P13.33. Assume that a coefficient of static friction of 
exists 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.
d cos 
d
 FB
x cos 
x
Rd
T  F23 R sin   FB
sin 
 MC  0
X
But, for geometric compatibility, R  x sin  . Therefore,
2
T  FB d sin 2   FB d  R x 
M
O2
0
F32  FB
Ans.
Also,
R
sec 
1  tan 2 
1
R
R
R
1
sin 
tan 
tan 
tan 2 
Motion is still possible as long as tan    , or as long as
x
x  R 1  2 1
Ans.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
13.34 Links 2 and 3 are pinned together at B and a constant vertical load P  800 kN is applied
at B as illustrated in Fig. P13.34. Link 2 is fixed in the ground at A and link 3 is pinned to
the ground at C. The length of link 2 is 8 m and it has a 150 mm solid square crosssection. The length of link 3 is 0.5 m and it has a solid circular cross-section with
diameter D. Both links are made from a steel with a modulus of elasticity E  207 GPa
and a compressive yield strength Syc  200 MPa. Using the theoretical values for the end
condition constants of each link, determine: (i) the slenderness ratio, the critical load, and
the factor of safety guarding against buckling of link 2; and (ii) the minimum diameter
Dmin of link 3 if the static factor of safety guarding against buckling is to be N  2 .
(i) The area moment of inertia and the radius of gyration of link 2, respectively, are
4
bh3 150 mm  150 mm 
I2 

 0.421 88 104 m
12
12
4
I
0.150 m
b /12
b
k2  2 


 0.043 30 m
2
A2
b
12
12
Therefore, the slenderness ratio of link 2 is
L
8m
Sr2  2 
 184.75
Ans.
k2 0.043 30 m
The end-condition constant for link 2 (with fixed-pinned ends) is C2  2. Therefore, the
slenderness ratio at the point of tangency is
3
 Sr  D  
2
2  2   207 109 Pa 
2C2 E

 202.14
S yc
200 106 Pa
Comparing these gives Sr2   Sr  D . Therefore, the Johnson parabolic equation must be
2
used to determine the critical load of link 2.
The critical unit load of link 2 is
2
Pcr
 S yc Sr2  1
2
 S yc  

A2
 2  C2 E
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Theory of Machines and Mechanisms, 4e
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Substituting the given information into this equation, the critical load of link 2 is
Pcr2  2.62 106 N
Ans.
The axial compressive load F2 is the component of the vertical load P acting along the
central axis of the link 2 as shown in this figure:
30°
F2
P
60°
F3
Therefore, the axial compressive load in link 2 is
F2  P cos30o  692.8 kN
Similarly, the axial compressive load in link 3 is
F3  P sin 30o  400 kN
The factor of safety guarding against buckling for link 2 is
Pcr
2.62 106 N
N2  2 
 3.782
F2 692.8 103 N
Therefore, link 2 is safe against this axial compressive load.
Ans.
(ii) The end condition constant for link 3 (with pinned-pinned ends) is C3  1 . The
slenderness ratio at the point of tangency for link 3 is
 Sr  D
3
2C3 E
2 1 207 109 Pa


 142.93
S yc
200 106 Pa
(1)
The cross-sectional area and the area moment of inertia of link 3 are
 D2
 D4
A3 
and I 3 
4
64
Therefore, the radius of gyration for the link 3, in terms of the diameter D, is
I
D
k3  3 
A3 4
Therefore, the slenderness ratio for link 3, in terms of diameter D, is
L 0.5 m 2 m
Sr  3 

k3 D 4
D
To determine the minimum diameter of link 3 to prevent buckling, the critical load must
be derived in terms of the diameter D for both the Euler column formula and the Johnson
parabolic equation. The critical load can be written as
Pcr3  N3 F3
3
Substituting N3  2 and F3 from above, the critical load is
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Theory of Machines and Mechanisms, 4e
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Pcr3  2  400 kN  800 kN
CASE (1). The critical load in terms of the diameter D from the Euler column formula is
 C  2E  
N 
Pcr3  A3  3 2    4.011 44 1011 4  D 4
m 
 Sr 3  
Equating with the load F3 gives
Pcr3  4.011 44 1011 N m4 D4  800 kN


Therefore, the minimum diameter from the Euler column formula is
DEuler  0.038 m
(2)
CASE (2). The critical load can be written in terms of the diameter D from the Johnson
parabolic equation as
2

 S yc Sr  1 


Pcr  A3 S yc  


 2   C3 E 
3
3


or as
Pcr3  1.570 8 108 Pa  D2  15 377.3 N
Equating with the load F3 gives
Pcr3  1.570 8 108 Pa  D2  15 377.3 N  800 kN
Therefore, the minimum diameter of the link 3 from the Johnson parabolic equation is
DJohnson  0.072 m
(3)
Note that the diameter given by the Euler column formula, Eq.(2), is smaller than the
diameter from the Johnson parabolic equation, Eq. (3), i.e., ( DEuler  DJohnson ).
In certain cases, the claim that the bigger of the two diameters is the correct answer
may not be true. Therefore, to determine the minimum diameter of the column we need
to decide which of the two criteria is valid. The slenderness ratio must be compared with
the slenderness ratio at the point of tangency for both diameters.
Tthe slenderness ratio of link 3 is
2m
Sr3E 
 52.63
DEuler
Comparing with Eq. (1), the conclusion is
Sr3E   Sr D3
Therefore, the Euler column formula is not appropriate. So Dmin  0.038 m. From Eq.
(3), the slenderness ratio of link 3 can be written as
2m
Sr3J 
 27.78
DJohnson
The conclusion is
Sr3J   Sr D
3
Therefore, the Johnson parabolic equation is the valid equation. The minimum diameter
of the column 3 from the Johnson parabolic equation is
Dmin  DJohnson  0.072 m  72 mm
Ans.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
13.35 The horizontal link 2 is subjected to the load F  150 kN at C as illustrated in Fig.
P13.35. The link is supported by the solid circular aluminum link 3. The lengths of the
links are L2  RCA  5 m, RBA  3 m, and L3  RBD  3 m. The end D of link 3 is fixed in
the ground and the opposite end B is pinned to link 2 (that is, the effective length of the
link is LEFF  0.5 L3 ). For aluminum, the yield strength is Syc = 370 MPa and the
modulus of elasticity is E = 207 GPa. Determine the diameter d of the solid circular
cross-section of link 3 to ensure that the static factor of safety is N = 2.5.
The cross-sectional area and the area moment of inertia of link 3, respectively, are
A = π d2/ 4
and
I = π d 4 / 64
Therefore, the radius of gyration of the link is
I
 d 4 64 d


A
d2 4 4
Using the effective length LEFF  0.5 L3 , the slenderness ratio of the link is
k
Sr  LEFF k  0.5  3 m  d / 4   6 m d
(1)
The slenderness ratio at the point of tangency is
 Sr  D  
2E Syc   2  207 109 Pa 370 106 Pa  105.09
(2)
Taking moments about A gives
3 m P  5 m F cos60
where P is the compressive load acting at B on link 3. Solving this equation, the
compressive load is
 5 m 150 000 N 0.5  125 000 N
P
3m
The factor of safety guarding against buckling of link 3 is defined as
N  Pcr P
Substituting N = 2.5, the critical unit load is
Pcr  2.5 125 000 N   312 500 N
Using the Euler column formula, the critical unit load can be written as
Pcr   2 E
A Sr 2
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Theory of Machines and Mechanisms, 4e
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Which, with the available data and Eq. (1), can be written as
2
9
312 500 N   207 10 Pa 

2
d2 4
6 m d 
Rearranging this equation gives
d 4  701.12108 m4
Therefore, the diameter of link 3 is
d  0.0515 m  51.5 mm
Using the Johnson parabolic equation, the critical unit load can be written as
2
Pcr  S  1  Sy Sr 
y
A
E  2  
(3)
which can be written as
2
 370 106 10 m d 
312 500 N
1
6
 370 10 Pa 


d2 4
207 109 Pa 
2

Rearranging this equation gives
d 2  5.60 103 m2
Therefore, the diameter of link 3 is
d  0.074 8 m  74.8 mm
(4)
To check which answer is valid, that is, Eq. (3) or Eq. (4), recall that the slenderness
ratio, from Eq. (1), is
Sr = 6 m/ d
To check the Euler column formula, the slenderness ratio is
(Sr)EULER = 6 m / (0.0515 m) = 116.5
Comparing this answer with Eq. (2) indicates that
(Sr)EULER > (Sr)D that is 116.5 > 105.09
Therefore, the Euler column formula is a valid equation. The correct diameter of link 3 is
d = 51.5 mm
Ans.
Next we check the validity of the Johnson parabolic equation. The slenderness ratio is
(Sr)JOHNSON = 6 m / (0.0748 m) = 80.21
Therefore
(Sr)JOHNSON < (Sr)D; that is, whether 80.21 < 105.09, which is not possible. Therefore, the
Johnson parabolic equation is not valid.
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13.36 The horizontal link 2 is subjected to the inclined load F  8 000 N at C as illustrated in
Fig. P13.36. The link is supported by a solid circular cross-section link 3 whose length
L3  BD  5 m . The end D of the vertical link 3 is fixed in the ground link and the end B
supports link 2 (that is, the effective length of the link is LEFF  0.5 L3 ). Link 3 is a steel
with a compressive yield strength Syc  370 MPa and a modulus of elasticity
E  207 GPa . Determine the diameter d of link 3 to ensure that the factor of safety
guarding against buckling is N  2.5. Also, answer the following statements true or false
and briefly give your reasons. (i) The slenderness ratio at the point of tangency between
the Euler column formula and the Johnson parabolic equation does not depend on the
geometry of the column. (ii) Under the same loading conditions, a link with pinnedpinned ends will give a higher factor of safety against buckling than an identical link with
fixed-fixed ends. (iii) If the slenderness ratio Sr  ( Sr ) D at the point of tangency then the
critical unit load does not depend on the yield strength of the column material.
The cross-sectional area and the area moment of inertia of link 3, respectively, are
A = π d2/ 4
and I = π d 4 / 64
Therefore, the radius of gyration of the link is
k
I

A
 d 4 64 d

 d2 4 4
Taking moments about A gives
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 4 m P  5 m  F cos 60
where P is the compressive load acting at B on link 3. Rearranging this equation, the
compressive load P is
P   5 m 8 000 N  0.5  4 m   5 000 N
Using the effective length LEFF  0.5L3 , the slenderness ratio of the link is
Sr  LEFF / k  0.5  5 m   d 4  10 m d
(1)
The slenderness ratio at the point of tangency is
 Sr  D    2 E
S yc 
12
   2  207 109 Pa 370 106 Pa 
12
 105.087
The factor of safety guarding against buckling can be written as
N = Pcr / P
Therefore, the critical unit load is
Pcr  NP  2.5  5 000 N  12 500 N
and the critical unit load for this factor of safety is
Pcr 12 500 N

A
 d2 4
From the Euler column formula, the critical unit load can be written as
Pcr  2 E
 2
A
Sr
Equating Eqs. (2) and (3) gives
12 500 N  2  207 109 Pa

2
 d2 4
10 m d 
Rearranging and solving, the diameter of the link using the Euler formula is
d  0.0297 m  29.7 mm
(2)
(3)
(4)
From the Johnson parabolic equation, the critical unit load can be written as
2
Pcr
1S S 
 Sy   y r 
A
E  2 
Substituting the known data gives
2
 370 106 Pa 10 m d 
12 500 N
1
6
370
10
Pa





 d2 4
207 109 Pa 
2

Rearranging and solving, the diameter of the link using the Johnson parabolic equation is
d  0.0676 m  67.6 mm
(5)
To determine the correct diameter, that is, Eq. (4) or Eq. (5), the slenderness ratio from
Eq. (1) is
Sr  10 m d
Using Eq. (4), the slenderness ratio, from the Euler column formula, is
(Sr)EULER = 10 m / 0.0297 m = 336.7
Since 336.7 > 105.087, (Sr)EULER > (Sr)D is valid. Therefore, the diameter of link 3 is
d = 29.7 mm
Ans.
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To check the Johnson parabolic equation: Using Eq. (5), the slenderness ratio, from the
Johnson parabolic equation, is
(Sr)JOHNSON = 10 m / 0.0676 m = 147.93
Since 147.93 > 105.087, (Sr)JOHNSON > (Sr)D, which is not possible. Therefore, the
Johnson parabolic equation is not valid.
Statement (i) is true.
The reason is that the slenderness ratio at the point of tangency is defined as
Ans.
( Sr ) D    2 E / S yc 
12
which is a function of only the material properties of the link (and does not depend on the
geometry of the link).
Statement (ii) is false.
Ans.
The factor of safety is defined as N = Pcr / P. Therefore, a higher value for the critical
load Pcr will give a higher value for the factor of safety. The critical load Pcr is greater for
fixed-fixed ends (C = 4) than for pinned-pinned ends (C = 1) from both the Euler column
formula and the Johnson parabolic equation.
Statement (iii) is true.
Ans.
2
P  E
When Sr > (Sr)D, then we must use the Euler column formula; that is, cr  2 , which
A
Sr
does not depend on the yield strength of the material.
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13.37 A load PA is acting at A and a load PB is acting at B of the horizontal link 3, as illustrated
in Fig. P13.37. Link 3 is pinned to the vertical link 2 at O and link 2 is fixed in the
ground link 1 at D.
The lengths are AO  1200 mm, OB  600 mm, and
DO  1800 mm. Both links have solid circular cross-sections with diameter D  50 mm
and are made from a steel alloy with a compressive yield strength Syc  585.65 mPa, a
tensile yield strength Syt  516.75 mPa, and modulus of elasticity E  206.7 103 mPa.
Assuming that links 2 and 3 are in static equilibrium and using the theoretical value for
the end-condition constant for link 2, determine: (i) the magnitude of the force PB that is
acting as shown at B if PA  133.5 RN, (ii) the critical load, critical unit load, and the
factor of safety to guard against buckling for link 2, and (iii) the diameter of a solid
circular cross-section for link 2 that will ensure the factor of safety guarding against
buckling of the link is N = 4.
(i) The free body diagram of link 3 is as shown in the figure below.
Taking moments about O gives
M
O
 1200 mm  PAy   600 mm  PBy  0 which can be
written as
1200 mm PA cos30   600 mm  PB cos30
Therefore, the reaction force at B is
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PB  2 PA
Substituting the given data into this equation, the reaction force at B is
PB  133  5 RN  267 RN
Taking the sum of the forces in the y-direction on link 3 gives
 F y  PAy  PBy  F23y  0
which can be written as
F23y  PA cos30  PB cos30
Substituting Eq. (1) gives
F23y  3PA cos30  3 133.5 RN  cos30  346.84 RN
(1)
Ans.
(2)
Taking the sum of the forces in the x-direction gives
 F x  PAx  PBx  F23x  0
which can be written as
F12x  PA sin 30  PB sin 30
Substituting Eq. (1), the x-component of the reaction force at O is
F23x  200.25 RN
(ii)
A
I

4

The cross-sectional area and the area moment of inertia of link 2, respectively, are
D2 
D4 

4
(50 mm)2  1963.5 mm2
and

(50 mm)4  306796.875 mm4
64
64
The radius of gyration of link 2 is
I
306796.875 mm 4

 12.5 mm
A
1963.5 mm 2
The slenderness ratio of link 2 can be definded as
L 1800 mm
Sr  
 150
k 12.5 mm
and the slenderness ratio of link 2 at the point of tangency can be written as
2CE
 Sr  D  
S yc
k
(3)
where the end condition constant for fixed-pinned end conditions (using the theoretical
value) is C = 2; that is, the effective length for fixed-pinned ends is LEFF  0.5L .
Therefore, the slenderness ratio of link 2 at the point of tangency is
2  2  206.7 103 mPa
 118.04
(4)
585.65 mPa
In order to determine the critical load on link 2, we must first determine if this link is an
Euler column or a Johnson column. The criterion for using the Johnson parabolic
equation is
Sr   Sr  D
 Sr  D  
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From Eqs. (3) and (4) this implies that 150  118.04 , which is a contradiction. Therefore,
link 2 is an Euler column. The critical load on link 2 from the Euler column equation can
be written as
 C 2 E 
(5)
Pcr  A  2 
 Sr 
Substituting the known values into this equation, the critical load on link 2 is
3
2  206.7 10 mPa 
3
Ans.
Pcr  1963.5 mm  
  335.695 10 RN or 355.695 RN
2
150


The critical unit load on link 2 is
Pcr 355.695 103 N

 181.15 mPa
Ans.
A
1963.5 mm2
Link 2 is in compression as shown in the figure.
The definition of the factor of safety for link 2 is
P
N  cry
F32
where, from Eq. (2), F32y  77 942 lb is the compressive load at point O on link 2.
Therefore, the factor of safety for link 2 is
355.695 N
N
 1.025
Ans.
346840 N
(iii) For the circular cross-section of link 2 and a factor of safety N = 4, the critical
load for buckling can be written as
Pcr
N
4
346840 N
Therefore, the required critical load for the buckling is
Pcr  346840 N  1387.36 RN
(6)
The cross-section area and the area moment of inertia of the solid circular link 2,
respectively, are
A   D2 4 and I   D4 64
Therefore, the radius of gyration of the link is
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I
 D 4 64 D


4
A
 D2 4
The slenderness ratio of link 2 is
L 1800 mm 7200 mm
(7)
Sr  

k
D4
D
First consider the Euler column formula. Substituting Eqs. (6) and (7) into Eq. (5), the
critical load is
 C 2 E   D 2  2 2  206.7 103 mPa 
Pcr  A  2  

  1387.36 RN
4   7200 mm D 2 
 Sr 


k
Rearranging this equation gives
4 1387360 N(7200 mm) 2
D4 
 9556506.65 mm 2
2 3 206700 mPa
Therefore, the diameter of the cross-section of the link 2 is D  68.25 mm ,
Substituting this into Eq. (7), the slenderness ratio of link 2 is
7200 mm 7200 mm
Sr 

 105.5
(8)
D
68.25 mm
Compaing Eq. (8) with Eq. (4) gives that Sr   Sr  D ; that is, 105.5  118.04 . Therefore,
the condition for link 2 to be an Euler column is not satisfied; that is, the assumption that
the link is an Euler column is not correct.
Now we assume that link 2 is a Johnson column; the Johnson parabolic equation is
2

1  S y Sr  
Pcr  A  S y 

 
CE  2  


Substituting Eq. (7) into this equation gives
2
 S yc  7200 mm D  
1
 2
Pcr  D  S yc 

 
4
2  206760 mPa 
2

 
Rearranging this equation gives
2
 4P
 S yc  7200 mm   1
1
2
cr

D 

 
2  206760 mPa 
2
 
  S yc
Substituting the known values into this equation gives
2
 4.1387.36 RN
1
1
 585.65 mPa  7200 mm  

 4781.25 mm2
D2  

 
2  206760 mPa 
2

  585.65 mPa

Therefore, the diameter of link 2 from the Johnson parabolic equation is D  69.146 mm .
Substituting this into Eq. (7), the slenderness ratio of link 2 is
7200 mm 7200 mm
Sr 

 104.1
D
69.146
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Comparing with Eq. (4) now shows that Sr   Sr D ; that is, that 104.1  118.04 .
Therefore, the assertion that the column is a Johnson column is valid. The diameter of
Ans.
link 2 to guard against buckling is D  2.77 in .
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Theory of Machines and Mechanisms, 4e
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13.38 The horizontal link 2 is subjected to the load P = 5 000 N as illustrated in Fig. P13.38.
This link is supported by the vertical link 3, which has a constant circular cross-section.
The lengths are AC  5 m, AB  4 m, and DB  L3  5 m. For the vertical link 3, the end
D is fixed in the ground link and the end B supports link 2 (that is, the effective length of
link 3 is LEFF  0.5 L3 ). The yield strength and the modulus of elasticity for the aluminum
link 3 are Sy = 370 MPa and E = 207 GPa, respectively. Determine the diameter d of link
3 to ensure that the static factor of safety guarding against buckling is N = 2.5.
The cross-sectional area and the second moment of area of link 3 can be written as
A = π d 2 / 4 and I = π d 4 / 64
Therefore, the radius of gyration of the link is
k
I

A
 d 4 64 d

d2 4 4
Using the effective length LEFF  0.5 L3 , the slenderness ratio of the link is
Sr  LEFF k  0.5  5 m  d 4   10 m d
The slenderness ratio at the point of tangency is
 Sr  D    2 E
Sy 
12
   2  207 109 Pa 370 106 Pa 
12
 105.09
Taking moments about A of all forces on link 2 we find
M A  4 m  F32y  5 m  P sin 53.13  0
And therefore the vertical load on link 3 is
F32y  P sin 53.13 5 m 4 m  5 000 N
The factor of safety guarding against buckling of link 3 can be written as
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Theory of Machines and Mechanisms, 4e
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N  Pcr P  2.5
Therefore, the critical unit load is
Pcr  NP  2.5  5 000 N=12 500 N
(i) Using the Euler column formula, the critical unit load can be written as
Pcr  2 E
 2
A
Sr
which can be written as
2
9
12 500 N   207 10 Pa 

2
d2 4
10 m d 
Rearranging this equation gives
d 4  7.79 107 m4
Therefore, the diameter of link 3 is
d  0.0297 m  29.7 mm
(ii) Using the Johnson parabolic equation, the critical unit load can be written as
2
Pcr
1  S y Sr 
 Sy  

A
E  2 
which can be written as
2
(1)
 370 106 Pa 10 m d 
12 500 N
1
6
370
10
Pa





 d2 4
207 109 Pa 
2

Rearranging this equation gives
d 2  4.57 103 m2
Therefore, the diameter of link 3 is
d  0.0676 m  67.6 mm
(2)
To check which answer is valid, that is, Eq. (1) or Eq. (2), recall that the slenderness ratio
is defined as
Sr = 10 m/ d
To check the Euler column formula: The slenderness ratio is
(Sr)EULER = 10 m/ (0.0297 m) = 336.7
or (Sr)EULER > (Sr)D since the values are 336.7 > 105.09. Therefore, the Euler column
formula is the valid equation. The correct diameter of the link is
d = 29.7 mm
Ans.
Using the Johnson parabolic equation, the slenderness ratio is (Sr)JOHNSON = 10 / 0.0676 =
147.93 or (Sr)JOHNSON > (Sr)D. Since 147.93 < 105.09 is not possible, therefore the
Johnson parabolic equation is not valid.
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Theory of Machines and Mechanisms, 4e
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13.39 The horizontal link 2 is pinned to the vertical wall at A and pinned to link 3 at B as
illustrated in Fig. P13.39. The opposite end of link 3 is pinned to the wall at C. A
vertical force P  25 kN is acting on link 2 at B. Link 2 has a 20 × 30 mm solid
rectangular cross-section and link 3 has a 40 × 40 mm solid square cross-section. The
length of link 3 is BC  1.2 m and the angle ABC  30 . The two links are made from
a steel alloy with a tensile yield strength Syt  190 MPa , a compressive yield strength
Syc  205 MPa , and a modulus of elasticity E  207 GPa . Using the theoretical value for
the end-condition constant for link 3, determine: (i) the value of the slenderness ratio at
the point of tangency between the Euler column formula and the Johnson parabolic
formula; (ii) the critical load and the factor of safety guarding against buckling of link 3;
and (iii) the minimum width of the square cross-section of link 3 for the factor of safety to
guard against buckling to be N = 1.
The free-body diagram of link 2 is as shown below.
F12Y
F32Y
B
2
A
F12X
F32X
P
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Theory of Machines and Mechanisms, 4e
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Taking moments about A gives
 M A  RBA F32y  RBA P  0
Therefore, the y-component of the reaction force at B is
F32y  P  25 kN
The free-body diagram of link 3 is as shown in the next figure.
F23Y
A
B
F23X
3
F13Y
F13X
C
Taking moments about C gives
 M C  RBA F23y  RAC F23x  0
or
RBA y
F23  tan 60F23y
RAC
Therefore, the x- and y-components of the reaction force at B are
F23x   tan 60 25 kN  43.3 kN
and
F23y   F32y  25 kN
The magnitude of the tensile load T on link 2 at B is
T  F32x   F23x  43.3 kN
F23x 
(1)
The magnitude of the compressive load Papp on link 3 at B is
Papp  F23 
F   F 
x 2
23
y 2
23
 50 kN
(2)
Link 2 is subjected to the tensile load T, which creates a tensile stress  in the link. The
factor of safety guarding against yielding for the link is defined as
N  S yt 
(3)
where the tensile stress can be written as
 T A
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(4)
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and the cross-sectional area of the link is
(5)
A  (0.02 m)(0.03 m)  6 104 m2
Substituting Eqs. (1) and (5) into Eq. (4), the tensile stress is
43.3 103 N
(6)
 72.17 MPa

6 104 m2
Substituting Eq. (6) and the tensile yield strength into Eq. (3), the factor of safety
guarding against yielding for link 2 is
190 MPa
N
 2.63
72.17 MPa
(i)
 Sr  D
The slenderness ratio of link 3, at the point of tangency, can be written as
2CE

S yc
Using the theoretical value (for pinned-pinned ends), the end-condition constant for the
link is
(7)
C 1
Substituting E  207 GPa , Syc  205 MPa , and Eq. (7) into Eq. (6), the slenderness
ratio, at the point of tangency, is
 Sr  D  
2 1 207 109 Pa
 141.18
205 106 Pa
(ii)
The cross-sectional area of link 3 is
2
A  b  (0.040 m)2  1.6 103 m2
The second moment of area of the link is
b4 (0.040 m)4
I

 2.13 107 m4
12
12
and the radius of gyration of the link is
Ans.
(8)
(9)
(10)
I
2.13 107 m 4
k

 1.155 102 m
3
2
A
1.6 10 m
Therefore, the slenderness ratio is
1.2 m
L
Sr  
 103.92
(11)
k 1.155 102 m
In order to determine the critical load for link 3, we must first determine if this column is
an Euler column or a Johnson column. The criterion for using the Johnson parabolic
equation is
Sr   Sr  D
From Eqs. (8) and (11) we have Sr   Sr  D that is, 103.92  141.18 . Therefore, link 3 is
a Johnson column. The critical load for link 3 (that is, the Johnson parabolic equation)
can be written as
2

1  S yc Sr  
Pcr  A  S yc 
(12)

 
CE  2  


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Theory of Machines and Mechanisms, 4e
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Substituting the known values into this equation, the critical load is
2

 205 106 Pa 103.92  
1
3
2
6
Pcr  1.6 10 m  205 10 Pa 

 
1 207 109 Pa 
2

 
Therefore, the critical load is
Ans.
Pcr  239.14 kN
The definition of the factor of safety guarding against buckling is
P
N  cr
Papp
(13)
(14)
where Papp is the compressive load on column 3 at B and is given by Eq. (2). Substituting
Eqs. (2) and (13) into Eq. (14), the factor of safety guarding against buckling is
239.14 kN
N
 4.78
Ans.
50 kN
(iii)
The critical load for a factor of safety N  1 is Pcr  NPapp  1 50 kN  50 kN .
If we assume that link 3 is an Euler column, then the critical load on link 3 (that is, the
Euler column equation) can be written as
C 2 EA
(15)
Pcr 
Sr2
From Eqs. (9) and (10), the radius of gyration of link 3 is
I
b 4 /12
b


2
A
b
12
and from Eq. (11), the slenderness ratio of link 3 is
L
Sr   12 L / b
(16)
k
Substituting Eqs. (9) and (16) into Eq. (15), the critical load on link 3 can be written as
C 2 Eb 2
Pcr 
12 L2 / b 2
Rearranging this equation, the minimum width of the link can be written as
k
1/4
 12L2 Pcr 
b

2
 C E 
Substituting the known values into this equation gives
1/4
 12  1.2 m 2  50 103 N 
b
  0.025 5 m  25.5 mm
 1  2  207 109 Pa 


To check whether the assumption of an Euler column is correct, from Eq. (16), the new
slenderness ratio of link 3 is
Sr  12L / b  12 1.2 m  / (25.5 103 m)  163.02
Comparing this result with Eq. (8) we have Sr   Sr  D ; that is, 163.01  141.18 . So this
verifies that link 3 is indeed an Euler column.
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Uicker et al.
If we assume that link 3 is a Johnson column, then substituting Eqs. (9) and (16) into
Eq. (12), the critical load can be written as
2


1  S yc ( 12 L / b)  
2 
Pcr  b S yc 


 

2
CE 
 

or as
2
12 L2  S yc 
2
Pcr  S ycb 


CE  2 
Rearranging this equation, the new width of the link can be written as
2

12 L2  S yc  
b   Pcr 

  S yc
CE  2  


Substituting the known values into this equation, the width is

b  50 103 N 


12 1.2 m   205 106 Pa 2 
6
  (205 10 Pa )  26 mm
2
1 207 109 Pa 
 
2

Now we must check if the assertion that the link is a Johnson column is correct. From
Eq. (16), the new slenderness ratio of link 3 is
Sr  12L / b  12 1.2 m  / (2.60 102 m)  159.88
Comparing this result with Eq. (8) we have Sr   Sr  D ; that is, 159.88  141.18 . This
means that the assumption that link 3 is a Johnson column is invalid. Link 3 is an Euler
column and the minimum width of the square cross-section (in order for the factor of
safety to guard against buckling to be N = 1) is b = 25.5 mm.
Ans.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
13.40 The link BC = 1.2 m and 25 mm square cross-section is fixed in the vertical wall at C and
pinned at B to a circular steel cable AB with diameter d = 20 mm as illustrated in Fig.
P13.40. The distance AC = 0.7 m. The mass m of a container, suspended from pin B,
produces a gravitational load at B that results in the moment at point C in the wall
M C  8 000 Nm ccw. The yield strength and modulus of elasticity of the steel cable AB
and the steel link BC are Sy  370 MPa and E  207 GPa, respectively. Given that m =
2 000 kg, determine: (i) the tension in the cable AB and the factor of safety guarding
against tensile failure; (ii) the compressive load acting in link BC; (iii) the factor of safety
guarding against buckling of link BC. (Use the theoretical value of the end-condition
constant assuming that the link has fixed-pinned ends.) If M C  10 000 Nm ccw, then
determine the maximum mass of a container that can be suspended from pin B before
buckling of link BC will begin (that is, the factor of safety guarding against buckling
failure is N  1).
(i)
The angle between the cable and the link is
 0.7 
  tan 1 
  30.256
 1.2 
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Theory of Machines and Mechanisms, 4e
Uicker et al.
The cable AB and the link BC are in static equilibrium. The free body diagram of link BC
is as shown in the following figure.
Summing moments about C at the wall gives
 M C M C  L sin 30.256P  Lmg  0
(1)
where P is the tension in the cable AB. Rearranging Eq. (1), the tension can be written as
mg  M C L
P
sin 30.256
Substituting the given data into this equation, the tension is
2 000 kg  9.81 m/s 2  8 000 Nm 1.2 m
P
 25 708 N
Ans.
sin 30.256
The axial stress in the cable can be written as
P
P
A  
A  d2 4
Substituting the given data into this equation, the axial stress in the cable is
25 708 N
A 
 81.83 MPa
2
  0.020 m  4
The factor of safety guarding against tensile failure in the cable can be written as
S
370 MPa
 4.52
N y 
Ans.
 A 81.83 MPa
(2)
(ii)
The compressive load in link BC can be obtained by summing forces in the xdirection; that is,
x
 F  Pc  P cos30.256  0
where Pc is the compressive load in the link. Therefore, the compressive load is
Pc  P cos30.256  22 206 N
Ans.
(iii) To determine whether the link is an Euler column or a Johnson column, we first
find the slenderness ratio at the point of tangency between the Euler column formula and
the Johnson parabolic equation
2 EC
 Sr  D  
Sy
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Theory of Machines and Mechanisms, 4e
Uicker et al.
where the theoretical value of the end-condition constant corresponding to fixed-pinned
end conditions is C  2 . Therefore, the slenderness ratio at the point of tangency D is
 Sr  D  
2  207 109 Pa  2
 148.62
370 106 Pa
The radius of gyration of the link can be written as
I
bh3 12
h


A
bh
12
Which, for the given data, is
0.025 m
k
 0.007 22 m
12
The slenderness ratio of the link can be written as
L
1.2 m
Sr  
 166.2
k 0.007 22 m
Since Sr   Sr  D , therefore the link is an Euler column. The critical load using the Euler
k
column formula can be written as
C 2 AE 2 2 (0.025 m)2 (207 109 Pa)
Pcr 

 92 452 N
Sr 2
(166.2)2
Therefore, the factor of safety guarding against buckling of the link is
P
92 367 N
Ans.
N  cr 
 4.16
Pc 22 206 N
Combining Eqs. (1) and (2), the compressive load exerted on the link, as a function of
the mass of the container, is
mg  M C L
mg  M C L
Pc  P cos30.256 
cos30.256 
(3)
sin 30.256
tan 30.256
For a factor of safety guarding against buckling N = 1, the critical load must be equal to
the compressive load; that is,
mg  M C L
Pcr  Pc 
tan 30.256
Rearranging this equation, the mass of the container can be written as
tan 30.256Pcr  M C L
m
g
Substituting the given data into this equation, the mass of the container is
tan 30.256(92 367 N)  10 000 Nm 1.2 m
m
 6 342 kg
Ans.
(4)
9.81 m/s2
Substituting Eq. (4) into Eq. (3), the compressive load is
mg  M C L (6 342 kg)(9.81 m/s 2 )  10 000 Nm 1.2 m

 92 370 N
Pc 
tan 30.256
tan 30.256
Note that the compressive load in the link is equal to the critical load. Also, note that it is
important to show that this answer cannot be obtained using the factor of safety found in
part (iii) because the moment has changed; that is, mnew  2 000 kg  4.16  8 320 kg.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
13.41 A vertically upward force F is applied at C of the horizontal link 4 as illustrated in Fig.
P13.41. The link is pinned to the ground at B and pinned to the vertical link 2 at A. The
lengths AB  600 mm and AC 1500 mm and links 2 and 4 are made from a steel alloy
with a compressive yield strength Syc  413.4 mPa and a modulus of elasticity
E  206.7 103 mPa. Link 2 has a hollow circular cross-section with an outside diameter
D  50 mm, wall thickness t  6.25 mm, and length L  1500 mm. Using the theoretical
value for the end-condition constant for link 2, determine: (i) the value of the slenderness
ratio at the point of tangency between Euler’s column formula and Johnson’s parabolic
formula; (ii) the critical load acting on link 2; and (iii) the force F for the factor of safety
of link 2 to guard against buckling to be N  1 . (iv) If link 2 has a solid circular crosssection with diameter D  75 mm and F  89 RN then determine the maximum length
of link 2 for the factor of safety to guard against buckling to be N  2 .
(i)
The cross-sectional area and the second moment of area for the hollow circular
link 2, respectively, are

A   D2  d 2


4    50 mm    37.5 mm 
2
2
 4  859 mm
2
and
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Theory of Machines and Mechanisms, 4e


Uicker et al.
I   D4  d 4 64    50 mm    37.5 mm  64  209.73 mm4
4
4
Therefore, the radius of gyration for the link is
k
I
209.73 mm4

 15.62 mm
A
859 mm2
and the slenderness ratio of the link is
L 1500 mm
Sr  
 96
k 15.62 mm
The slenderness ratio at the point of tangency can be written as
2CE
 Sr  D  
Sy
where the theoretical value for the end-condition constant for link 2 is C = 2 . Therefore,
the slenderness ratio at the point of tangency is
 Sr  D  
2  2  206.7 103 mPa
 140.50
413.4 mPa
Ans.
(ii)
The critical load is determined by first finding if the link is to be considered an
Euler column or a Johnson column. The criterion for using the Johnson parabolic
equation is Sr   Sr  D . In this example, we have 96 < 140.50. Therefore, the link is
regarded as a Johnson column. The critical load from the Johnson parabolic equation is
2

1  S y Sr  
Pcr  A  S y 

 
CE  2  


Substituting the known values into this equation gives the critical load as
2

1
 413.4 mPa  96  
2
Ans.
Pcr  859 mm  413.4 mPa 

   281.297 RN
2  206.7 103 mPa 
2
 

(iii) The definition of the factor of safety of the link is
N  Pcr PAPP
From the given factor of safety for the link, this equation can be written as
N  Pcr PAPP  1
Therefore, the applied force at point A can be written as
PAPP  Pcr 1  281297 N
To determine the force F acting at C in link 4, we take moments about B, which gives
 M B  900 mm  F  600 mm  PAPP  0
Therefore, the force F acting at point C is
600 mm  PAPP 600 mm  281297 N

 187531.9 N or 187.53 RN
F
900 mm
900 mm
Ans.
(iv)
Link 2 is a solid circular column with factor of safety of N = 2. To determine the
applied force we take moments about B; that is,
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Theory of Machines and Mechanisms, 4e
M
B
Uicker et al.
0
Therefore, the applied force is
900 mm  F 900 mm  89 RN
PAPP 

 133.5 RN
600 mm
600 mm
From the definition of the factor of safety
N  Pcr PAPP  2
Therefore, the critical load is
Pcr  2  PAPP  2 133.5 RN  267 RN
The cross-sectional area is
2
A   D2 4    75 mm  4  4418.75 mm2
The second moment of area is
4
I   D4 64    75 mm  64  1554687.5 mm4
The radius of gyration is
15546875 mm 4
I

 18.75 mm
4418.75 mm 2
A
First, the critical unit load from the Johnson parabolic equation is
2
Pcr 
1  S y Sr  
  Sy 

 
A 
CE  2  


With known values, this equation can be written as
2

Sr 2
267 RN
 413.4 mpa  
  413.4 mPa 

 
4418.75 mm2 
2.206.7 103 mPa 
2
 
Solving this equation gives the slenderness ratio from the Johnson parabolic formula as
Sr  184.10
Second, the critical unit load from the Euler column formula is
Pcr C 2 E

A
Sr 2
Substituting the known values into this equation gives
267 RN
2   2 206.7 103 mPa

4418.75 mm2
Sr 2
Therefore, the slenderness ratio from the Euler formula is
Sr  264.14
A comparison of the two slenderness ratios shows that
264.14  184.10
In other words
(Sr)EULER > (Sr)JOHNSON
The length of link 2 can be written as
L  k  Sr
where the slenderness ratio that is used in this equation is for the Euler column formula.
Therefore, the length of the link is
k
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Theory of Machines and Mechanisms, 4e
Uicker et al.
Ans.
L  18.75 mm  264.14  4952.625 mm  4.952 m
As an alternative method to determine the critical load, consider the effective length of
the link; that is, LEFF. From the end-condition constant
C = (1/α)2
This defines α = 0.707. Therefore, the effective length of the link can be written as
LEFF   L  0.707 1500 mm  1060.5 mm
The slenderness ratio of the link is
L
1060.5 mm
Sr  EFF 
 67.88
15.625 mm
k
The slenderness ratio at the point of tangency is
 Sr  D  
2E
206.7 103 mPa

 99.32
Sy
413.4 mPa
The critical load is determined by first finding if the link is an Euler column or a Johnson
column. The criterion for the Johnson column is
Sr   Sr  D
In this example, we have
67.88 < 99.32
Therefore, the Johnson parabolic equation must be applied. The equation can be written
as
2

1  S y Sr  
Pcr  A  S y  
 
E  2  


Substituting the known values into this equation, the critical load is
2

1
 413.4 mPa  67.88  
Pcr  859 mm2  413.4 mPa 

   281.3 N
206.7 103 mPa 
2
 

Note that this answer is in good agreement with Eq. (1).
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Theory of Machines and Mechanisms, 4e
Uicker et al.
13.42 A force F is acting at C perpendicular to link 2 as illustrated in Fig. P13.42. The end A is
pinned to the ground and the supporting link 3 is pinned to link 2 at B and pinned to the
ground at D. The lengths are AC  200 mm and AD  150 mm. Link 3 has a circular
cross-section with diameter D  5 mm. Both links are a steel alloy with a compressive
yield strength Syc  420 MPa and modulus of elasticity E  206 GPa. Using the
theoretical value for the end-condition constant for link 3, determine: (i) the slenderness
ratio at the point of tangency between Euler’s column formula and Johnson’s parabolic
formula; (ii) the critical load and the critical unit load acting on the link; and (iii) the
force F for the factor of safety to guard against buckling to be N  1 . If the force
F  3 000 N then for link 3 determine: (a) the critical load for the factor of safety to
guard against buckling to be N  1; and (b) the slenderness ratio.
(i)
From the pinned-pinned end conditions, the end-condition constant for link 3 is
C = 1.
(1)
Therefore, the slenderness ratio of link 3 at the point of tangency is
 Sr  D  
2CE
2 1 206 109 Pa

 98.4
S yc
420 106 Pa
Ans.
(2)
(ii)
In order to determine the critical load on link 3, we first determine if this link is an
Euler column or a Johnson column. The Euler and Johnson criterion, respectively, are
Sr   Sr  D and Sr   Sr  D
(3)
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Theory of Machines and Mechanisms, 4e
The cross-sectional area of link 3 is
A   D2 4   (0.005 m)2 4  1.963 105 m2
and the second moment of area of link 3 is
I   D4 64   (0.005 m)4 64  3.068 1011 m4
The radius of gyration of link 3 is
I
3.068 1011 m4

 1.25 103 m
A
1.963 105 m2
The slenderness ratio of link 3 is
0.15 m
L
Sr  
 120
k 1.25 103 m
Substituting Eqs. (2) and (5) into Eq. (3) gives
Sr   Sr  D ; that is, 120  98.4
Uicker et al.
(4)
k
(5)
Therefore, link 3 is an Euler column. The critical load on link 3 from the Euler column
equation can be written as
 C 2 E 
(6)
Pcr  A  2 
 Sr 
Substituting the known values into this equation, the critical load on link 3 is
2
9
2 1    206  10 Pa 
5
Ans.
Pcr  1.963 10 m 
  2 772 N
1202


Therefore, the critical unit load on link 3 is
Pcr
2 772 N

 141.2 106 Pa
Ans.
A 1.963 105 m2
(iii) The free body diagram of link 2 is shown in the figure. Taking moments about pin
A can be written as
 M A  (RCAx F y  RCAy F x )  (RBAx F32y  RBAy F32x )  0
or as
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Theory of Machines and Mechanisms, 4e
Uicker et al.
( RCA cos2 )( F sin F )  ( RCA sin 2 )( F cos F )  ( RBA cos2 )( F32 sin 3 )  ( RBA sin 2 )( F32 cos 3 )  0
or as
RCA F sin( F  2 )  RBA F32 sin(3  2 )  0
Rearranging this equation, the force is
R F sin( 2  3 )
F  BA 32
(7)
RCA sin( F   2 )
The free-body diagram of link 3 is shown in the following figure. Note that link 3 is in
compression.
The reaction force FB  F23   F32 ; therefore, the magnitude of the reaction force FB is
equal to the magnitude of the internal force F32 ; that is, Eq. (7) can be written as
R F sin ( 2  3 )
F  BA B
(8)
RCA sin( F   2 )
The factor of safety guarding against buckling for link 3 can be written as
P
N  cr
(9)
FB
Rearranging Eq. (9), the compressive load on link 3 can be written as
P
2 772 N
FB  cr 
 2 772 N
(10)
1
N
Substituting 2  120, 3  60,  F  210, RCA  0.2 m, RBA  0.15 m, and putting Eq.
(10) into Eq. (8), the force is
0.15 m  2 772 N  sin(120  60)
F
 1 800.6 N
Ans.
 0.2 m  sin(210  120)
(a)
Rearranging Eq. (7) gives
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Theory of Machines and Mechanisms, 4e
Uicker et al.
RCA F sin( F   2 )
RBA sin( 2  3 )
Substituting F  3 000 N and the given data into this equation gives
0.2 m  3 000 N  sin(210  120)
FB  F32 
 4 618.8 N
(11)
 0.15 m  sin(120  60)
Rearranging Eq. (9) and substituting Eq. (11) and the factor of safety N = 1 into the
resulting equation, the critical load applied on link 3 is
Ans.
(12)
Pcr  NFB  1 4 618.8 N  4 618.8 N
F32 
(b)
If we assume that the link is an Euler column:
slenderness ratio of link 3 can be written as
Rearranging Eq. (6), the
C 2 EA
Pcr
Substituting Eqs. (1), (4), and (12) into Eq. (13), the slenderness ratio of link 3 is
 Sr Euler 
 Sr Euler 

(13)

1  2 206 109 Pa 1.963 105 m2
 92.97
4 618.8 N
Next, if we assume that the link is a Johnson column, the Johnson parabolic equation is
2
Pcr 
1  S y Sr  
  Sy 

 
A 
CE  2  


Rearranging this equation, the slenderness ratio can be written as
 Sr Johnson 
P
2 
S y  cr

Sy 
A

 CE

or as
2
4 618.8 N 

6
(14)
1 206 109 Pa  92.3
 420 10 Pa 
6
2 
5
420 10 Pa 
1.963 10 m 
From Eq. (2), the slenderness ratio of link 3 at the point of tangency is  Sr D  98.4 .
 Sr Johnson 
Note that
 Sr Johnson   Sr D ; that is, 92.3  98.4
which is the correct answer. Therefore, the link must be a Johnson column. The correct
value for the slenderness ratio is given by Eq. (14); that is,
Sr   Sr Johnson  92.3
Ans.
As a check, we note that
 Sr Euler   Sr D ; that is, 92.97  98.4 , which is not possible. Therefore, the link must be
a Johnson column. So again the correct value for the slenderness ratio is
Sr   Sr Johnson  92.3 .
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Theory of Machines and Mechanisms, 4e
Uicker et al.
Chapter 14
Dynamic Force Analysis*
14.1
The steel bell crank illustrated in Fig. P14.1 is used as an oscillating cam follower. Using
76.42 RN/ m3 for the density of steel, find the mass moment of inertia of the lever about
an axis through O.
75 mm
25mm
mm
12.5
9.375 mm
18.75 mm
150 mm
For the vertical arm, using Appendix A, Table 5,
m  wht   18.75 mm  93.75 mm  9.375 mm   76.42 RN/m3   1.32 N
2
2
IG  m  a 2  c 2  12  1.32 N 9804 mm/s 2  18.75 mm    93.75 mm   12  0.10446 N  mm  s 2
IO  IG  md 2  0.10446 N  mm  s 2  1.32 N 9804 mm/s 2   37.5 mm   0.29726 N  mm  s 2
2
For the horizontal arm, using Appendix A, Table 5,
m  wht   18.75 mm 150 mm  9.375 mm   76.42 KN/m3   2.12 N
2
2
IG  m  a 2  c 2  12   2.12 N / 9804 mm/s 2  18.75 mm   150 mm   12  0.41774 N  mm  s 2
2
IO  IG  md 2  0.41774 N  mm  s 2   2.12 N / 9804 mm / s 2   93.75 mm   1.98002 N  mm  s 2
For the roller, using Appendix A, Table 5,
*
Unless instructed otherwise, solve all problems without friction and without gravitational loads.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
m   r 2t    12.5 mm  12.5 mm   76.42 RN/m3   0.493 N
2
IG  mr 2 2   0.493 N 9804 mm/s 2  12.5 mm  2  4 103 N  mm  s 2
2
IO  IG  md 2  4 103 N  mm  s 2   0.493 N 9804 mm/s 2  150 mm   1.1529 N  mm  s 2
2
For the composite lever
m  1.32 N    2.12 N    0.493 N   3.933 N
IO   0.29726 N  mm  s 2   1.98002 N  mm  s 2   1.1529 N  mm  s 2   3.4302 N  mm  s 2
Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
14.2
Uicker et al.
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.
Dimensions are in millimeters.
For the rectangular bar, using Appendix A, Table 5,
m  wht    0.050 m  0.300 m  0.005 m   7.80 Mg/m3   0.585 kg
2
2
IG  m  a 2  c 2  12   0.585 kg   0.050 m    0.300 m   12  0.004 509 kg  m2


IO  IG  md 2  0.004 509 kg  m2   0.585 kg  0.125 m   0.013 650 kg  m2
2
For the two circular disks, using Appendix A, Table 5,
2
m  2 r 2t   2  0.025 m   0.020 m   7.80 Mg/m3   0.613 kg
IG  mr 2 2   0.613 kg  0.025 m  2  0.000 191 kg  m2
2
IO  IG  md 2  0.000 191 kg  m2   0.613 kg  0.250 m   0.038 480 kg  m2
2
For the composite lever
m   0.585 kg    0.613 kg   1.198 kg
IO   0.013 650 kg  m2    0.038 480 kg  m2   0.052 130 kg  m2
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
14.3
Uicker et al.
Find the external torque that must be applied to link 2 of the four-bar linkage shown to
drive it at the given velocity.
RAO2  75 mm, RA4O2  175 mm, RBA  200 mm, RBO  150 mm, RG3 A  100 mm,
 3.47 N, IG2  2.8702 N mm s2 , IG3  1.7132 N mm s2 ,
 1.246 N mm s2 , ω2  180kˆ rad/s,  2  0, 2  4 950kˆ rad/s2 , 4  8 900kˆ rad/s2 ,
G 3  1896ˆi + 225ˆj m/s2 , G 4  684ˆi + 225ˆj m/s 2 , .
RG3o4  75 mm,
IG4
w3o4
 3.15 N,
w4
The D’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
f 4  m4 AG4
 618.5ˆi  71.2ˆj N  623 N186.77
t 3   I G3 α 3
 244.75ˆi  80.1ˆj N  258.1 N198.21
t 4   I G4 α 4
   3.15 N 9.66 m/s 2  1896ˆi +225ˆj m/s 2 

  1.7132 N  mm  s 2  4 950kˆ rad/s 2

   3.47 N 9.66 m/s 2   684ˆi +225ˆj m/s 2 

  1.242 N  mm  s 2  8 900kˆ rad/s 2

 11125kˆ N  mm
 8455kˆ N  mm
h3  t3 f3  14240 N  mm   623 N   13.625 mm ,
h4  t4 f 4  1125 N  mm   258.1 N   42.85 mm
Next, the free-body diagrams are drawn with the inertia forces applied. Since the lines of
action for the forces on the free-body diagrams cannot be discovered from two- and threeforce 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
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
with the radial and transverse axes of link 3. This results from the perpendicularity of
links 3 and 4.)
M
O4
 RG4O4 ×f4  t 4  R BO4 ×F34t  0
 45ˆi  60ˆj mm  ×  244.75ˆi  80.1ˆj N   1125kˆ N  mm    90ˆi  120ˆj mm  ×  0.800ˆi  0.600ˆj F
182895kˆ N  mm  11125kˆ N  mm   150kˆ mm F
t
34
t
34
0
0
F34t  44 lb
 M  R ×f  t  R ×F  0
80ˆi  60ˆj in  ×  618.55ˆi  71.2ˆj lb    8455kˆ N  mm   160ˆi  120ˆj mm  × 0.600ˆi  0.800ˆj F
A
G3 A
3
3
BA
r
43
31.417kˆ N  mm   8455kˆ N  mm   200kˆ mm F
r
43
r
43
0
0
F34  89ˆi  209.15ˆj N  226.95 N67.26
F43r  115.7 N
F  F
 f3  F23  0
F23  707.55ˆi  284.8ˆj N  760.95 N21.82
43
M
O2
 R AO2 ×F32  M12  0
M12  854.4kˆ N  mm
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
14.4
Uicker et al.
Crank 2 of the four-bar linkage illustrated in Fig. P14.4 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.
RAO2  50 mm, RO4O2  325 mm,
RBA  425 mm, RBO4  200 mm,
RG3 A  212.5 mm,
 11.79 N, w4  29.9 N, IG2  2.66 N mm s2 , IG3  6.74 N mm s2 ,
IG4  59.07 N mm s2 , ω2  200kˆ rad/s,  2  0, 3  6 500kˆ rad/s2 , 4  2 4 0kˆ r a d 2/ s
G 3  948ˆi + 78.3ˆj m/s 2 , G 4  240ˆi + 633ˆj m/s2 ,
RG4o4  100 mm,
w3o4
The D’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
f 4  m4 AG4
 1157ˆi  97.9ˆj N  1161.45 N  4.74
t 3   I G3 α 3
 743.15ˆi  1958ˆj N  2096 N69.24
t 4   I G4 α 4
  11.79 N 9.66 m/s 2   948ˆi +78.6ˆj m/s 2 

   6.7417 N  mm  s 2  6 500kˆ rad/s 2

   29.9 N 9.66 m/s 2   240ˆi  633ˆj m/s 2 

   59.1 N  mm  s 2  240kˆ rad/s 2

 43832.5kˆ N  mm
 14128.75kˆ N  mm
h3  t3 f3   43832.5 N  mm  1161.45 N   37.725 mm ,
h4  t4 f 4  141228.75 N  mm   2096 N   6.775 mm
Next, the free-body diagrams with inertia forces are drawn. Since the lines of action for
the forces on the free-body diagrams cannot be discovered from two- and three-force
member concepts, the force F34 is divided into radial and transverse components.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
M
O4
Uicker et al.
 RG4O4 ×f4  t 4  R BO4 ×F34t  0
 36.6ˆi  93.1ˆj mm  ×  747.15ˆi  1958ˆj N   14128.75kˆ N  mm   73ˆi  186.175ˆj mm  ×  0.931ˆi  0.365ˆj F
 2407.451kˆ N  mm  14128.75kˆ N  mm   200kˆ mm F
t
34
t
34
0
0
F34t  84.55 N
M
A
 RG3 A ×f3  t 3  R BA ×F43t  R BA ×F43r  0
181.35ˆi  110.77ˆj mm  × 1157ˆi  97.9ˆj N    43832.5kˆ N  mm  362.7ˆi  221.55ˆj mm  ×  75.65ˆi  31.15ˆj N
  362.7ˆi  221.55ˆj mm  ×  0.365ˆi  0.931ˆj F  0
r
43
 145.913kˆ N  m   43.83kˆ N  m  18.37kˆ N  m   0.29717kˆ m F
r
43
0
F34  26.7ˆi  293.7ˆj N  293.7 N  95.06 Ans.
F43r  280.35 N
F  F
 f4  F14  0
F14  720.9ˆi  1668.7ˆj N  1815.6 N  113.27
34
F  F
43
 f3  F23  0
F23  1183.7ˆi  195.8ˆj N  1201.5 N  170.57
F  F
32
Ans.
 F12  0
Ans.
F12  1183.7ˆi  195.8ˆj N  1201.5 N9.43 Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
M
O2
 R AO2  F32  M12  0
Uicker et al.
M12  54.846kˆ N  m
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
14.5
Uicker et al.
For the angular velocity of crank 2 given in Fig. P14.5, find the reactions at each joint and
the external torque applied to the crank.
m3  1.54 kg, m4  1.30 kg,
IG2  0.039 8 kg  m2 ,
IG3  0.012 2 kg  m2 ,
ω2  2 kˆ 1
0
α 2  0,
α3  7.670kˆ rad/s2 , AG3  2 384ˆi  1 486ˆj m/s2 , AG4  2 393ˆi m/s2 .
RBA  300 mm,
RAO2  75 mm,
RG3 A  112 mm,
The D’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f 4  m4 AG4
f3  m3 AG3
  1.54 kg   2 384ˆi  1 486ˆj m/s 2 
  1.30 kg   2 393ˆi m/s 2 
 3 671ˆi  2 288ˆj N  4 326 N31.94
 3 111ˆi N  3 111 N0.00
t 3   I G3 α 3

ˆ rad/s 2
   0.012 2 kg  m 2  7 670k
t 4   IG4 α 4  0

ˆ mm  N
 85 904k
h3  t3 f3  85 904 mm  N   4 326 lb   19.86 mm , h4  t4 f 4  0
Next, the free-body diagrams with inertia forces are drawn and the solution proceeds.
M
A
 RG3 A ×f3  t 3  R BA ×f4  R BA ×F14  0
110.2ˆi 19.8ˆj mm  × 3 671ˆi  2 288ˆj N   85 904kˆ mm  N    295.3ˆi  53.0ˆj mm  × 3 111ˆi N 
  295.3ˆi  53.0ˆj mm  × F ˆj  0
324 902kˆ mm  N   85 904kˆ mm  N   164 985kˆ mm  N    295.3kˆ mm  F  0
14
14
F14  1 273ˆj N  1 273 N  90
Ans.
F  3 111ˆi  1 273ˆj N  3 361 N157.75 Ans.
F14  1 273 N
F  f  F  F  0
F  F  f  F  0
F  F  F  0
 M  R ×F  M
4
O2
14
34
43
3
23
32
12
AO2
32
12
34
F23  6 782ˆi  1 015ˆj N  6 858 N 171.49 Ans.
F  6 782ˆi  1 015ˆj N  6 858 N  171.49 Ans.
12
0
M12  305.84kˆ N  m
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
14.6
Uicker et al.
Figure P14.6 illustrates a slider-crank mechanism with an external force FB applied to the
piston. For the given crank velocity, find the reaction forces in the joints and the crank
torque.
RAO2  75 mm, RBA  300 mm, RG2o2  31.25 mm, RG3 A  87.5 mm, w2  4.22 N,
w3  15.575 N, w4  11.125 N, IG2  0.4105 N  mm s2 , IG3  12.2375 N  mm s2 ,
ω  160kˆ rad/s, α  0, α  3 090kˆ rad/s2 , A  792 m/s21500 ,
2
3
2
G2
AG3  1839 m/s 158.3 , AG4  1.884 m/s 180 , FB  3560 N 1800 ,
2
0
2
0
The D’Alembert inertia forces and offsets are:
f 2  m2 AG2
   4.2 N 9804 mm/s 2  685.8ˆi  396ˆj m/s 2  t 2   IG2 α 2  0
 298.15ˆi  173.55ˆj N  347 N  30.00
h2  t2 f 2  0
f3  m3 AG3
   9804 mm/s 2   1708ˆi  680ˆj m/s 2 
 2754.5ˆi  1094.7ˆj N  2963.7 N  21.70
t 3   I G3 α 3

  12.2375 N  mm  s 2  3 090kˆ rad/s 2
f 4  m4 AG4
  11.125 N 9804 mm/s 2   1.884ˆi mm/s 2 
 2171.6ˆi N  2171.6 N0.00

t 4   IG4 α 4  0
 37.83kˆ N  m
h3  t3 f3   37825 N  mm   2963.7 N   12.9 mm ,
h4  t4 f 4  0
Next, the free-body diagrams are drawn with inertia forces and the solution proceeds.
M
A
 RG3 A ×f3  t 3  R BA ×f4  R BA ×FB  R BA ×F14  0
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
86.82ˆi  11.12ˆj mm  ×  2754.5ˆi 1094.7ˆj N   37825kˆ N  mm    297.65ˆi  375ˆj mm  ×  2171.6ˆi N 
  297.65ˆi  37.5ˆj mm  ×  3560ˆi N    297.65ˆi  375ˆj mm  × F ˆj  0
14
 125.16kˆ N  m  37.82kˆ N  m   81.43kˆ N  m  133.5kˆ N  m  297.65kˆ mm F
14
F14  120.15ˆj N  120.15 N90
F14  120.15 N
F  f
4
Ans.
 FB  F14  F34  0
F34  1388.4ˆi  120.15ˆj N  1392.85 N  4.95
F  F
43
F  F  F
M  R
32
Ans.
 f3  F23  0
F23  1366.15ˆi  974.5ˆj N  1677.65 N144.50
O2
0
12
AO2
0
 F32  M12  0
Ans.
F12  1664.3ˆi  1148.1ˆj N  2020.3 n145.40 Ans.
M  12015kˆ N  mm
Ans.
12
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
14.7
Uicker et al.
The following data apply to the four-bar linkage illustrated in Fig. P14.7: RAO2  0.3 m ,
RO4O2  0.9 m , RBA  1.5 m , RBO4  0.8 m , RCA  0.85 m , C  33 , RDO4  0.4 m ,
 D  53, RG2O2  0, RG3 A  0.65 m,   16, RG4O4  0.45 m,   17, m2  5.2 kg,
m3  65.8 kg, m4  21.8 kg, IG2  2.3 kg  m2 , IG3  4.2 kg  m2 , IG4  0.51 kg  m2 . A
kinematic analysis at   60 , ω  12kˆ rad/s ccw , and α  0 , gives   0.7 ,
2
2
2
3
 4  20.4 , 3  85.6 rad/s cw ,  4  172 rad/s cw , AG3  96.4259 m/s2 , and
2
2
AG4  97.8270 m/s2 . Find all pin reactions and the torque applied to link 2.
The D’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
f 4  m4 AG4
 1 210ˆi  6 227ˆj N  6 343 N79
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
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
Next, the free-body diagrams are drawn with inertia forces. Since the lines of action for
the forces on the free-body diagrams cannot be discovered from two- and three-force
member concepts, the force F34 is divided into radial and transverse components.
M
O4
 RG4O4 ×f4  t 4  R BO4 ×F34t  0
 0.357ˆi  0.273ˆj m  ×  2 132ˆj N   88kˆ N  m    0.750ˆi  0.279ˆj m  ×  0.348ˆi  0.937ˆj F
761kˆ N  m  88kˆ N  m   0.800kˆ m F
t
34
t
34
0
0
F34t  1 061 N
M
A
 RG3 A ×f3  t 3  R BA ×F43t  R BA ×F43r  0
 0.622ˆi  0.187ˆj m  × 1 210ˆi  6 227ˆj N   360kˆ N  m   1.499ˆi  0.019ˆj m  ×  370ˆi  996ˆj N 
 1.499ˆi  0.019ˆj m  ×  0.937ˆi  0.348ˆj F  0
r
43
3 650kˆ N  m  360kˆ N  m  1 501kˆ N  m   0.505kˆ m F
r
43
34
4
14
F34  10 600ˆi  2 800ˆj N  10 964 N14.8
F  10 600ˆi  4 932ˆj N  11 691 N 155.0
43
3
23
F23  9 390ˆi  3 427ˆj N  9 996 N  20.0
32
12
F43r  10 913 N
F  F  f  F  0
F  F  f  F  0
F  F  F  0
 M  R ×F  M
O2
0
AO2
32
12
14
0
F12  9 390ˆi  3 427ˆj N  9 996 N  20.0
M  2 954kˆ N  m
12
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Ans.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
14.8
Uicker et al.
Solve Problem 14.7 with an additional external force FD  12 kN0 acting at point D.
Here, the method of superposition is used for the solution. The force components of
Problem 14.7 are denoted with primes and the additional force increments with double
primes. The figures here show the incremental forces only.
 M
O4
 R DO4 ×FD  R BO4 ×F34  0
 0.114ˆi  0.383ˆj m  × 12 000ˆi N    0.750ˆi  0.279ˆj m  ×  0.999ˆi  0.013ˆj F34  0
 4 600kˆ N  m  0.269kˆ m F   0
F34  17 081 N
F34  17 079ˆi  217ˆj N  17 081 N  179.3
F  5 079ˆi  217ˆj N  5 084 N2.4
Ans.
F34  6 479ˆi  2 584ˆj N  6 975 N158.3
F  5 521ˆi  4 715ˆj N  7 260 N  139.5 Ans.
F23  17 079ˆi  217ˆj N  17 081 N  179.3
F23  7 689ˆi  3 644ˆj N  8 509 N  154.6 Ans.
F12  17 079ˆi  217ˆj N  17 087 N  179.3
F12  7 689ˆi  3 644ˆj N  8 509 N  154.6 Ans.
M12  1 451kˆ N  m
Ans.
34
14
  4 405kˆ N  m
M12
14
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
14.9
Uicker et al.
Make a complete kinematic and dynamic analysis of the four-bar linkage of Problem 14.7
using the same data, but with 2  170, 2  12 rad/s ccw,  2  0, and an external
force FD  8.9464.3 kN acting at point D.
Kinematic Analysis:

 
VA  ω2 × 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   3kˆ rad/s  × 1.304ˆi  0.740ˆj m   4kˆ rad/s  ×  0.109ˆi  0.793ˆj m 

 
 
 0.625ˆi  3.545ˆj m/s  0.7403ˆi  1.3043 ˆj m  0.7934 ˆi  0.1094 ˆj m
ω3  3.020kˆ rad/s
ω4  3.607kˆ rad/s
V  2.859ˆi  0.393ˆj m/s  2.886 m/s172.16
B

2
A A   22 R AO2  α 2 × R AO2   12 rad/s  × 0.295ˆi  0.052ˆj m
 42.543ˆi  7.502ˆj m/s 2  43.200 m/s 2  10
© Oxford University Press 2015. All rights reserved.

Ans.

Theory of Machines and Mechanisms, 4e
Uicker et al.
A B  A A  32 R BA  α3 × R BA  42 R BO4  α 4 × R BO4


 
 
  1.419ˆi  10.311ˆj m/s    kˆ rad/s  ×  0.109ˆi  0.793ˆj m 
  32.069ˆi  3.940ˆj m/s    0.740 ˆi  1.304 ˆj m    0.793 ˆi  0.109 ˆj m 
 42.543ˆi  7.502ˆj m/s 2  11.893ˆi  6.749ˆj m/s 2   3kˆ rad/s 2 × 1.304ˆi  0.740ˆj m
2

2
4
2
3
α3  0.390kˆ rad/s
3
AG  A A   R G A  α 3 × R G A
3

2
3
4
4
α 4  40.816kˆ rad/s
2
3
3
 
 
2
Ans.

 42.543ˆi  7.502ˆj m/s  4.149ˆi  4.234ˆj m/s 2  0.390kˆ rad/s 2 × 0.455ˆi  0.464ˆj m
2
AG3  38.575ˆi  11.913ˆj m/s2  40.373 m/s2  17.16
AG  42 R G O  α 4 × RG O
4

4 4
4 4
 

 0.932ˆi  5.780ˆj m/s 2  40.816kˆ rad/s 2 × 0.072ˆi  0.444ˆj m

Ans.

AG4  19.054ˆi  2.841ˆj m/s2  19.265 m/s2  8.48
Ans.
Dynamic Analysis:
The D’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
f 4  m4 AG4
 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/s2

   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
Next, the free-body diagrams with inertia forces are drawn. Since the lines of action for
the forces on the free-body diagrams cannot be discovered from two- and three-force
member concepts, the force F34 is divided into radial and transverse components.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
M
O4
Uicker et al.
 RG4O4 ×f4  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
 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
BA
t
43
r
43
BA
r
43
1 534kˆ N  m   2kˆ N  m   3 618kˆ N  m   1.192kˆ in  F43r  0
F34  4 173ˆi  1 189ˆj N  4 339 N  164.10
F43r  1 747 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
14
32
12
Ans.
F14  711ˆi  6 929ˆj N  6 966 N  84.14 Ans.
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.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.10 Repeat Problem 14.9 using 2  200, 2  12 rad/s ccw,  2  0, and an external force
FC  8.4945 kN acting at point C.
Kinematic Analysis:
 

VA  ω2 × 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   3kˆ rad/s  × 1.198ˆi  0.902ˆj m   4kˆ rad/s  ×  0.016ˆi  0.799ˆj m 

 
 
 1.231ˆi  3.383ˆj m/s  0.9023ˆi  1.1983 ˆj m  0.7994 ˆi  0.0164 ˆj m
ω3  2.846kˆ rad/s
ω4  1.671kˆ rad/s
V  1.336ˆi  0.027ˆj m/s  1.337 m/s178.84
B


Ans.
2
A A   22 R AO2  α 2 × R AO2   12 rad/s  × 0.282ˆi  0.103ˆj m

 40.595ˆi  14.775ˆj m/s 2  43.200 m/s 220
A B  A A  32 R BA  α 3 × R BA  42 R BO4  α 4 × R BO4
 

 
 
ˆ rad/s  ×  0.016ˆi  0.799ˆj m 
  0.045ˆi  2.233ˆj m/s    k
  30.937ˆi  9.702ˆj m/s    0.902 ˆi  1.198 ˆj m    0.799 ˆi  0.016 ˆj m 
ˆ rad/s 2 × 1.198ˆi  0.902ˆj m
 40.595ˆi  14.775ˆj m/s 2  9.703ˆi  7.306ˆj m/s 2   3k
2

2
4
2
3
α3  8.760kˆ rad/s
2
3
4
α 4  48.558kˆ rad/s
© Oxford University Press 2015. All rights reserved.
4
2
Ans.
Theory of Machines and Mechanisms, 4e
AG3  A A  32 R G3 A  α 3 × R G3 A

Uicker et al.
 
 
 
 40.595ˆi  14.775ˆj m/s 2  3.169ˆi  4.204ˆj m/s 2  8.760kˆ rad/s 2 × 0.391ˆi  0.519ˆj m

AG3  41.972ˆi  7.146ˆj m/s  42.576 m/s 9.66
2
AG4  42 R G4O4  α 4 × R G4O4

2
 
 
 0.343ˆi  1.209ˆj m/s 2  48.558kˆ rad/s 2 × 0.123ˆi  0.433ˆj m
Ans.

AG4  21.365ˆi  4.754ˆj m/s2  21.887 m/s212.54
Ans.
Dynamic Analysis:
The D’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
f 4  m4 AG4
   65.8 kg   41.972ˆi  7.146ˆj m/s 2 
   21.8 kg   21.365ˆi  4.754ˆj m/s 2 
 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

   4.200 kg  m 2  8.760kˆ rad/s2


   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 with inertia forces are drawn. Since the lines of action for
the forces on the free-body diagrams cannot be discovered from two- and three-force
member concepts, the force F34 is divided into radial and transverse components.
M
O4
 RG4O4 ×f4  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
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0
Theory of Machines and Mechanisms, 4e
Uicker et al.
 214kˆ N  m   25kˆ N  m   0.800kˆ m F34t  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
F43r  1 260 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  274ˆi  1 266ˆj N  1 295 N  77.80 Ans.
F14  192ˆi  1 370ˆj N  1 383 N82.01 Ans.
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.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.11 At  2  270 , 2  18 rad/s ccw ,  2  0 , a kinematic analysis of the linkage whose
geometry is given in Problem 14.7 gives 3  46.6 ,  4  80.5 , 3  178 rad/s2 cw ,
 4  256 rad/s2 cw , AG  112 m/s222.7 , AG  119 m/s2352.5 . An external force
3
4
FD  8.94 kN64.3 acts at point D. Make a complete dynamic analysis of the linkage.
The D’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
   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 AG4
   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
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Theory of Machines and Mechanisms, 4e
Uicker et al.
Next, the free-body diagrams with inertia forces are drawn. Since the lines of action for
the forces on the free-body diagrams cannot be discovered from two- and three-force
member concepts, the force F34 is divided into radial and transverse components.
M
O4
 RG4O4 ×f4  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  0.789ˆj 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
14
32
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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Theory of Machines and Mechanisms, 4e
Uicker et al.
14.12 The following data apply to a four-bar linkage similar to the one illustrated in Fig. P14.7:
RAO2  120 mm , RO4O2  300 mm , RBA  320 mm , RBO4  250 mm , RCA  360 mm ,
C  15 , RDO4  0 ,  D  0 , RG2O2  0 , RG3 A  200 mm ,   8 , RG4O4  125 mm ,
  0, m2  0.5 kg, m3  4 kg, m4  1.5 kg, IG  0.005 N  m  s2 , IG  0.011 N  m  s2 ,
3
2
IG4  0.002 3 N  m  s2 . A kinematic analysis at 2  90 and 2  32 rad/s ccw with  2  0
gave 3  23.9. 4  91.7, 3  221 rad/s2 ccw,  4  122 rad/s2 ccw, AG3  88.6 m/s2255,
and AG4  32.6 m/s2244. Using an external force FC  632 N342 acting at point C,
make a complete dynamic analysis of the linkage.
The D’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
   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
 2.431kˆ N  m

f 4  m4 AG4
  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
 0.281kˆ N  m
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
Theory of Machines and Mechanisms, 4e
Uicker et al.
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 with inertia forces are drawn. Since the lines of action for
the forces on the free-body diagrams cannot be discovered from two- and three-force
member concepts, the force F34 is divided into radial and transverse components.
M
O4
 RG4O4 ×f4  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
BA
t
43
BA
r
43
r
43
 48kˆ N  m   2.431kˆ N  m   191kˆ N  m   1.623kˆ N  m   0.296kˆ in  F
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
0
F34  2ˆi  496ˆj N  496 N  89.71
Ans.
F14  24ˆi  452ˆj N  453 N93.02
Ans.
F23  690ˆi  643 lb  944 N  137
Ans.
F12  690ˆi  643 lb  944 N  137
Ans.
M12  82.83kˆ N  m
Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.13 Repeat Problem 14.12 at  2  260 . Analyze both the kinematics and the dynamics of
the system at this position.
Kinematic Analysis:
 

VA  ω2 × 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  0.667ˆj m/s   3kˆ rad/s  ×  0.138ˆi  0.289ˆj m   4kˆ rad/s  ×  0.183ˆi  0.171ˆj m 

 
 
 3.782ˆi  0.667ˆj m/s  0.2893ˆi  0.1383 ˆj m  0.1714 ˆi  0.1834 ˆj m

ω3  10.554kˆ rad/s
ω4  4.314kˆ rad/s
V  0.736ˆi  0.789ˆj m/s  1.079 m/s46.99
B

2
A A   22 R AO2  α 2 × R AO2    32 rad/s  × 0.021ˆi  0.118ˆj m
Ans.

 21.338ˆi  121.013ˆj m/s 2  122.880 m/s 280
A B  A A  32 R BA  α 3 × R BA  42 R BO4  α 4 × R BO4

 
 
  3.402ˆi  3.174ˆj m/s    kˆ rad/s  ×  0.183ˆi  0.171ˆj m 
 
 21.338ˆi  121.013ˆj m/s 2  15.374ˆi  32.158ˆj m/s 2   3kˆ rad/s 2 × 0.138ˆi  0.289ˆj m
2

2
4
 2.562ˆi  92.029ˆj m/s    0.289 ˆi  0.138 ˆj m   0.171 ˆi  0.183 ˆj m
2
3
3
α3  199.622kˆ rad/s2
4
α 4  352.356kˆ rad/s2
AG3  A A   R G3 A  α 3 × R G3 A

4
2
3
 
 
 
 21.338ˆi  121.013ˆj m/s 2  6.718ˆi  21.240ˆj m/s2  199.622kˆ rad/s 2 × 0.060ˆi  0.191ˆj m
AG3  52.748ˆi  87.796ˆj m/s2  102.423 m/s259.00
AG4  42 R G4O4  α 4 × R G4O4

 

 1.701ˆi  1.587ˆj m/s 2  352.356kˆ rad/s 2 × 0.091ˆi  0.085ˆj m
Ans.

Ans.

AG4  31.651ˆi  30.477ˆj m/s2  43.939 m/s243.92
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
Dynamic Analysis:
The D’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f 4  m4 A G4
f3  m3 A G3
  1.5 kg   31.651ˆi  30.477 ˆj m/s 2 
   4.0 kg   52.748ˆi  87.796ˆj m/s 2 
 47ˆi  46ˆj N  66 N  136.08
 211ˆi  351ˆj N  410 N  121.00
t 3   I G3 α 3

ˆ rad/s 2
   0.011 N  m  s 2  199.622k

ˆ Nm
 2.196k
t 4   I G4 α 4

ˆ rad/s 2
   0.002 3 N  m  s 2  352.356k

ˆ Nm
 0.810k
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 with inertia forces are drawn. Since the lines of action for
the forces on the free-body diagrams cannot be discovered from two- and three-force
member concepts, the force F34 is divided into radial and transverse components.
M
O4
 RG4O4 ×f4  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
8.212kˆ N  m  0.810kˆ N  m   0.250kˆ m F
t
34
t
34
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
BA
r
43
r
43
19kˆ N  m   2.196kˆ N  m   226kˆ N  m  3.629kˆ N  m    0.305kˆ in  F
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
0
F34  505ˆi  422ˆj N  659 N  39.88
Ans.
F14  458ˆi  468ˆj N  655 N134.39
Ans.
F23  115ˆi  124 N  169 N46.97
Ans.
F12  115ˆi  124 N  169 N46.97
Ans.
M12  11.03kˆ N  m
Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.14 Repeat Problem 14.13 at  2  300 .
Kinematic Analysis:
 

ˆ rad/s × 0.060ˆi  0.104ˆj m
VA  ω 2 × R AO2  32k

 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   3kˆ rad/s  ×  0.093ˆi  0.306ˆj m   4kˆ rad/s  ×  0.147ˆi  0.202ˆj m 

 
 
 3.326ˆi  1.920ˆj m/s  0.3063ˆi  0.0933 ˆj m  0.2024 ˆi  0.1474 ˆj m

ω3  1.597kˆ rad/s
ω4  14.071kˆ rad/s
V  2.844ˆi  2.066ˆj m/s  3.515 m/s36.04
B

2
A A   22 R AO2  α 2 × R AO2    32 rad/s  × 0.060ˆi  0.104ˆj m
Ans.

 61.440ˆi  106.417ˆj m/s 2  122.880 m/s 2120
A B  A A  32 R BA  α3 × R BA  42 R BO4  α 4 × R BO4

 
 
  29.105ˆi  39.995ˆj m/s    kˆ rad/s  ×  0.147ˆi  0.202ˆj m 
 
 61.440ˆi  106.417ˆj m/s 2  0.237ˆi  0.780ˆj m/s 2   3kˆ rad/s 2 × 0.093ˆi  0.306ˆj m
2

2
4
 90.782ˆi 145.632ˆj m/s    0.306 ˆi  0.093 ˆj m    0.202 ˆi  0.147 ˆj m
2
3
α3  671.302kˆ rad/s
AG3  A A   R G3 A  α 3 × R G3 A
2
3

3
4
4
α 4  567.507kˆ rad/s
2
 
 
2
 
 61.440ˆi  106.417ˆj m/s 2  0.079ˆi  0.504ˆj m/s2  671.302kˆ rad/s 2 × 0.031ˆi  0.198ˆj m
AG3  71.399ˆi  85.103ˆj m/s2  111.087 m/s250.00
AG4   R G4O4  α 4 × R G4O4

2
4
 
 
 14.547ˆi  20.022ˆj m/s 2  567.507kˆ rad/s 2 × 0.073ˆi  0.101ˆj m

Ans.

AG4  71.865ˆi  21.406ˆj m/s  74.986 m/s 16.59
2
Ans.
2
Dynamic Analysis:
The D’Alembert inertia forces and offsets are:
t 2   IG2 α 2  0
f2  m2 AG2  0
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
h2  t2 f 2  0
f3  m3 AG3
f 4  m4 AG4
 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
  1.5 kg   71.865ˆi  21.406ˆj m/s 2 

 7.384kˆ N  m

   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 with inertia forces are drawn. Since the lines of action for
the forces on the free-body diagrams cannot be discovered from two- and three-force
member concepts, the force F34 is divided into radial and transverse components.
M
O4
 RG4O4 ×f4  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
13.273kˆ N  m  1.305kˆ N  m   0.250kˆ m F
t
34
0
t
34
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
 46kˆ N  m  7.384kˆ N  m   219kˆ N  m  11.170kˆ N  m    0.256kˆ in  F
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
23
32
12
0
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.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.15 Analyze the dynamics of the offset slider-crank mechanism illustrated in Fig. P14.15
using the following data: a  0.06 m , RAO2  0.1 m, RAB  0.38 m, RCA  0.4 m,
C  32,
  22,
RG3 A  0.26 m,
m2  2.5 kg,
m3  7.4 kg,
m4  2.5 kg,
IG2  0.005 N  m  s , IG3  0.013 6 N  m  s , 2  120, and 2  18 rad/s cw, with  2  0,
FB  2 000ˆi N, FC  1 000ˆi N. . Assume a balanced crank and no friction forces.
2
2
Kinematic Analysis:
 

VA  ω2 × 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    kˆ rad/s  ×  0.351ˆi  0.147ˆj m 
B
3

 
 1.559ˆi  0.900ˆj m/s  0.1473ˆi  0.3513 ˆj m
ω3  2.567kˆ rad/s


VB  1.182ˆi m/s
A A   22 R AO2  α 2 × R AO2   18 rad/s  × 0.050ˆi  0.087ˆj m
2
Ans.

 16.200ˆi  28.059ˆj m/s 2  32.400 m/s 2  60
A B  A A  32 R BA  α3 × R BA

 
 
A ˆi  13.890ˆi  27.093ˆj m/s    0.147 ˆi  0.351 ˆj m

AB ˆi  16.200ˆi  28.059ˆj m/s2  2.310ˆi  0.966ˆj m/s 2  3kˆ rad/s 2 × 0.351ˆi  0.147ˆj m

2
3
B
A B  25.156ˆi m/s2
α3  77.188kˆ rad/s2
AG3  A A   R G3 A  α 3 × R G3 A

2
3
3
 
 
 
 16.200ˆi  28.059ˆj m/s 2  1.713ˆi  0.021ˆj m/s 2  77.188kˆ rad/s2  0.260ˆi  0.003ˆj m
AG3  14.466ˆi  7.969ˆj m/s2  16.516 m/s2  28.85
© Oxford University Press 2015. All rights reserved.
Ans.

Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
Dynamic Analysis:
The D’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
f 4  m4 A B
   7.4 kg  14.466ˆi  7.969ˆj m/s 2 
 107ˆi  59ˆj N  122 N151.15
t 3   I G3 α 3

   0.013 6 N  m  s 2  77.188kˆ rad/s 2
   2.5 kg   25.156ˆi m/s 2 
 63ˆi N  63 N180

t 4   IG4 α 4  0
 1.050kˆ N  m
h3  t3 f3  1.050 N  m  122 N   0.009 m , h4  t4 f 4  0
Next, the free-body diagrams with inertia forces are drawn. Here the lines of action for the
forces on the free-body diagrams can all be discovered from two- and three-force member
concepts.
M
A
 RG3 A ×f3  t 3  RCA ×FC  R BA ×f4  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  0.147ˆj m  ×  63ˆi N    0.351ˆi  0.147ˆj m  ×  2 000ˆi N    0.351ˆi  0.147ˆj m  × 1.000ˆj  F
0
14
0
15kˆ N  m    1.050kˆ N  m    65kˆ N  m    9.261kˆ N  m    294kˆ N  m   0.351kˆ in  F
F14  639 N
F  F  f  F  F  0
F  F  f  F  F  0
F  F  F  0
 M  R ×F  M  0
O2
F14  639ˆj N  639 N90.00
14
Ans.
14
4
B
34
F34  2 063ˆi  639ˆj N  2 160 N  17.21 Ans.
43
3
C
23
F23  3 170ˆi  698ˆj N  3 246 N  12.42
Ans.
32
12
F12  3 170ˆi  698ˆj N  3 246 N  12.42
Ans.
M12  241kˆ N  m
Ans.
AO2
32
12
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.16 Analyze the system of Problem 14.15 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 M 12 versus  2 .
Kinematic Analysis:
jR1  R2e j2  R3e j3  R4
j 0.060  0.100e j2  0.380e j3  RB
0.060  0.100sin  2  0.380sin 3  0
RG3  jR1  R2e j2  RG3 Ae
j 3  
3   sin 1  0.158  0.263sin  2 
RB  0.100cos 2  0.380cos3
RG3  j 0.060  0.100e j2  0.260e
j 3  22
RG3  0.100cos 2  0.260cos 3  22   j 0.060  0.100sin  2  0.260sin 3  22 
The first-order kinematic coefficients are found as follows:
j 0.100e j2  j 0.380e j33  R4
0.100cos 2  0.380cos33  0
0.100sin  2  0.380sin 33  R4
3  0.263cos 2 cos3
RB  0.100  sin  2  cos 2 tan 3 
j   22
RG3  j 0.100e j2  j 0.260e  3 3
RG3  0.100sin  2  0.260sin 3  22 3   j 0.100cos 2  0.260cos 3  22 3 
Similarly, the second-order kinematic coefficients are as follows:
0.100e j2  j 0.380e j33  0.380e j332  RB
0.100sin  2  0.380cos33  0.380sin 332  0
0.100cos 2  0.380sin 33  0.380cos332  RB
3  0.263sin  2 cos3  tan 332 ,
RB   0.100cos 3   2   0.38032  cos3
j   22
j   22
RG3  0.100e j2  j 0.260e  3 3  0.260e  3 32
RG3   0.100cos 2  0.260sin  3  22  3  0.260cos  3  22  32 
 j  0.100sin  2  0.260cos  3  22  3  0.260sin  3  22  32 
Dynamic Analysis:
By virtual work we can formulate the dynamic input torque requirement as:
M12  f3 RG3  t 3 3kˆ  f4 RB  FB RB
The individual elements of this equation are:
f3  m3 AG3  m3RG322    2 397.6 kg/s2  RG3
f3   240cos 2  623sin  3  22  3  623cos  3  22  32 
 j  240sin  2  623cos  3  22  3  623sin  3  22  32 
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Theory of Machines and Mechanisms, 4e
Uicker et al.
2
f3 R G   240 cos  2  623sin  3  22   3  623 cos  3  22   3   0.100 sin  2  0.260 sin  3  22   3 
3
  0.100 cos  2  0.260 cos  3  22  3 
f3 RG3  62sin 3  22   2 3 1  3   62cos 3  22   2 3  16233
t 3   IG3 α3   IG3322kˆ    4.406 N  m 3kˆ
t  kˆ    4.406 N  m   
  240 sin  2  623 cos  3  22   3  623sin  3  22   3
2
3
3
3 3
f4  m4 A B  m4 RB   810 kg/s2  RB
2
2
f4  81cos 3   2   30832  ˆi N cos3
f4 RB  sin 3   2  8.1cos 3   2   30.832  N  m cos2 3


FB  500 1  sgn  cos 2 tan 3  sin 2  ˆi N=500 1+sgn sin 3  2  cos 3  ˆi N


FB RB  50 1+sgn sin 3   2  cos3  sin 3   2  cos3 N  m
Finally, putting these pieces together, we obtain:
M 12  62sin  3  22   2  3 1   3   62 cos  3  22   2  3  158 3 3
 sin  3   2  8.1cos  3   2   30.8 32  cos 2  3


 50 1+sgn sin  3   2  cos  3  sin  3   2  cos  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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Theory of Machines and Mechanisms, 4e
Uicker et al.
14.17 A slider-crank mechanism similar to that of Problem 14.15 has a  0, RAO2  100 mm,
RBA  450 mm, RCB  0, C  0, RG3 A  200 mm,   0, w2  14.68 N, w3  34.26 N,
w4  11.57 N,
IG3  58.96 N  mm  s 2 ,
IG2  9.79 N  mm  s2 ,
and
600.75 N  mm.
Corresponding to 2  120 and 2  24 rad/s cw with  2  0, a kinematic analysis
gives   10.9, R  392 mm,   89.3 rad/s2 ccw, A  40600ˆi mm/s2 , and
3
3
B
B
AG3  40600ˆi  22600ˆj mm/s2 . Assume link 2 is balanced and find F14 and F23 .
The D’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
   34.26 N 9650 mm/s 2   40600ˆi  22600ˆj mm / s 2 
 144.18ˆi  80.23ˆj N  165 N150.90
f 4  m4 A B
  11.57 N 9650 mm/s 2   40600ˆi mm / s 2 
 48.68ˆi N  48.68 N180
t 3   I G3 α 3

   58.96 N  mm  s 2  89.3kˆ rad/s 2
 5265.46kˆ N  mm

h3  t3 f3  5265.5kˆ in  lb
M
A

 165 N   31.9 mm ,
t 4   IG4 α 4  0
h4  t4 f 4  0
 RG3 A ×f3  t 3  R BA ×f4  R BA ×F14  0
 220.9ˆi  42.5ˆj mm  ×  144.2ˆi  802ˆj N    52655kˆ N  mm    441.9ˆi  85ˆj mm  ×  48.68ˆi N 
  441.9ˆi  85ˆj mm  × 1.000ˆj  F  0
14
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Theory of Machines and Mechanisms, 4e
Uicker et al.
11596.7kˆ N  mm    5265.5kˆ N  mm    4139.6kˆ N  mm    441.9kˆ mm  F
14
F14  4.94 N
F  F
14
 f4  f3  F23  0
0
F14  4.94ˆj N  4.94 N90.00
Ans.
F23  192.86ˆi  85.2 N  210.8 N  23.83
Ans.
14.18 Repeat Problem 14.17 for 2  240. The results of a kinematic analysis are 3  10.9,
R  392 mm,   112 rad/s2 ccw, A  35200ˆi mm/s2 , and
3
B
AG3  31600ˆi  27700ˆj mm/s .
B
2
The D’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
   34.26 N 9650 mm/s 2   3160ˆi  27700ˆj mm/s 2 
 112.21ˆi  98.34ˆj N  158.1 N  138.76
f 4  m4 A B
  11.57 N / 9650 mm / s 2   35200ˆi mm / s 2 
 42.2ˆi N  42.2 N180
t 3   I G3 α 3

   58.96 N  mm  s 2  112kˆ rad/s 2
 6603.8kˆ N  mm

h3  t3 f3  6603.8kˆ N  mm

t 4   IG4 α 4  0
 149.21 N   44.25 mm , h
4
 t4 f 4  0
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Theory of Machines and Mechanisms, 4e
M
A
Uicker et al.
 RG3 A ×f3  t 3  R BA ×f4  R BA ×F14  0
 220.95ˆi  42.52ˆj mm  ×  112.2ˆi  142.84ˆj N    6603.8kˆ N  mm    441.9ˆi  85.1ˆj mm  ×  42.2 N 
  441.9ˆi  85.1ˆj mm  × 1.000ˆj  F  0

14
 
 
 

16958.95kˆ N  mm  6603.8kˆ N  mm  3587.8kˆ N  mm  441.9kˆ mm F14  0
F14  15.3 N
F  F
14
 f4  f3  F23  0
F14  15.3ˆj N  15.3 N  90.00
Ans.
F23  154.37ˆi  113.65 N  191.7 N36.36
Ans.
14.19 A slider-crank mechanism, as in Problem 14.15, has a  0.008 m, RAO2  0.25 m,
RBA  1.25 m,
C  38,
RCA  1.0 m, ,
RG3 A  0.75 m,
  18,
m2  10 kg,
m3  140 kg, m4  50 kg, IG2  2.0 N  m  s , and IG3  8.42 N  m  s , and has a balanced
2
2
crank. Make a complete kinematic and dynamic analysis of this system at 2  120 with
2  6 rad/s ccw and  2  0 , using FB  50 kN180 and FC  80 kN  60.
Kinematic Analysis:

 
VA  ω2 × R AO2  6kˆ rad/s × 0.125ˆi  0.217ˆj m

 1.299ˆi  0.750ˆj m/s  1.500 m/s  150
VB  VA  ω3 × R BA
V ˆi   1.299ˆi  0.750ˆj m/s    kˆ rad/s  × 1.227ˆi  0.225ˆj m 
B
3

 
 1.299ˆi  0.750ˆj m/s  0.2253ˆi  1.2273 ˆj m
ω3  0.611kˆ rad/s

VB  1.162ˆi m/s
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.

A A   22 R AO2  α 2 × R AO2    6 rad/s  × 0.125ˆi  0.217ˆj m
2

 4.500ˆi  7.794ˆj m/s 2  9.000 m/s 2  60
A B  A A  32 R BA  α3 × R BA

 
 
A ˆi   4.041ˆi  7.710ˆj m/s    0.225 ˆi  1.227 ˆj m 

AB ˆi  4.500ˆi  7.794ˆj m/s 2  0.459ˆi  0.084ˆj m/s 2  3kˆ rad/s 2 × 1.227ˆi  0.225ˆj m

2
3
B
3
A B  5.455ˆi m/s2
α3  6.284kˆ rad/s2
AG3  A A   R G3 A  α3 × R G3 A

2
3
 
 
 
 4.500ˆi  7.794ˆj m/s 2  0.247ˆi  0.133ˆj m/s 2  6.284kˆ rad/s 2 × 0.660ˆi  0.357ˆj m
AG3  6.496ˆi  3.514ˆj m/s  16.516 m/s   28.41
2
2
Dynamic Analysis:
The D’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3
f 4  m4 A B
  140 kg   6.496ˆi  3.514ˆj m/s 2 
   50 kg   5.455ˆi m/s 2 
 909ˆi  492ˆj N  1 034 N151.59
t 3   I G3 α 3

   8.42 N  m  s 2  6.284kˆ rad/s 2

 273ˆi N  273 N180
t 4   IG4 α 4  0
 52.911kˆ N  m
h3  t3 f3   52.911 N  m  1 034 N   0.051 m , h4  t4 f 4  0
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Ans.

Ans.
Theory of Machines and Mechanisms, 4e
M
A
Uicker et al.
 RG3 A ×f3  t 3  RCA ×FC  R BA ×f4  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 
 1.227ˆi  0.225ˆj m  ×  273ˆi N   1.227ˆi  0.225ˆj m  ×  50 000ˆi N   1.227ˆi  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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Theory of Machines and Mechanisms, 4e
Uicker et al.
14.20 Cranks 2 and 4 of the cross-linkage illustrated in Fig. P14.20 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, w3  17.8 N,
IG2  IG4  7 N  mm  s2 , and
IG3  55.29 N  mm  s2. Corresponding to the position shown, and with 2  10 rad/s ccw and
 2  0, a kinematic analysis gives results of 3  1.43 rad/s cw, 4  11.43 rad/s cw,
3   4  84.8 rad/s2 ccw, and AG3  7.776ˆi  7.374ˆj m/s2 .
Find the driving torque
and the pin reactions with FC  133.5ˆj N.
The D’Alembert inertia forces and offsets are:
f2  m2 AG2  0
t 2   IG2 α 2  0
h2  t2 f 2  0
f3  m3 AG3

  17.8 N  7.776ˆi  7.374ˆj mm/s 2
  9.65 m/s 
f4  m4 AG4  0
2
 14.33ˆi  13.58ˆj N  19.74 N  136.52
t 3   I G3 α 3

   55.29 N  mm  s 2  84.8kˆ rad/s 2

t 4   I G4 α 4

   7 N  mm  s 2  84.8kˆ rad/s 2
 4688.74kˆ N  mm
 594.3kˆ N  mm
h3  t3 f3   4688.74 N  mm  19.74 N   237.48 mm , h4  0
© Oxford University Press 2015. All rights reserved.

Theory of Machines and Mechanisms, 4e
Uicker et al.
Next, the free-body diagrams with inertia forces are drawn. Since the lines of action for
the forces on the free-body diagrams cannot 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
 594.3kˆ N  mm    21.42ˆi  148.45ˆj mm  ×  0.990ˆi  0.143ˆj F
t
34
0
F34t  4.99 N
 M  R ×f  t  R ×F  R ×F  R ×F  0
 235.72ˆi  185.57ˆj m  ×  14.32ˆi  13.59ˆj N    4690.6kˆ N  mm    471.4ˆi  371.15ˆj in  ×  133.5ˆj lb 
  353.57ˆi  278.375ˆj mm  ×  4.49ˆi  0.716ˆj N    353.57 ˆi  278.375ˆj mm  ×  0.143ˆi  0.990ˆj F  0
A
G3 A
3
3
CA
C
BA
t
43
BA
r
43
r
43
 5861.7kˆ N  mm   4690.3kˆ N  mm  62935.2kˆ N  mm  999kˆ N  mm  389.75kˆ mm F
r
43
F43r  191.13 N
F  F
F  F
F  F
34
 F14  0
43
 f3  FC  F23  0
32
 F12  0
F34  22.38ˆi  189.92ˆj N  191.26 N  96.7 Ans.
F  22.38ˆi  189.92ˆj N  191.26 N83.3 Ans.
14
F23  8.05ˆi  42.8ˆj N  43.56 N  100.67 Ans.
F  8.05ˆi  42.8ˆj N  43.56 N  100.67 Ans.
12
© Oxford University Press 2015. All rights reserved.
0
Theory of Machines and Mechanisms, 4e
M
O2
 R AO2 ×F32  M12  0
Uicker et al.
M12  2163.8kˆ N  mm
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.21 Find the driving torque and the pin reactions for the mechanism of Problem 14.20 under
the same dynamic conditions, but with crank 4 as the driver.
Given the same dynamic conditions, the D’Alembert forces and torques are the same as in
Problem 14.20. 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:
 MB  RG3B ×f3  t3  RCB ×FC  R AB ×F23  0
117.85ˆi + 92.8ˆj mm   14.32ˆi 13.59ˆj N  4690.6 N.mm  117.85ˆi  92.8ˆj mm  133.5ˆj N
  353.575ˆi  278.37ˆj mm    0.500ˆi  0.866ˆj F  0
 2931.2kˆ N  mm   4690.6kˆ N  mm   15733.97kˆ N  mm  445.37kˆ mm F  0
23
23
F23  39.275 N
F  F
F  F
F  F
32
 F12  0
23
 f3  FC  F43  0
34
 F14  0
F23  19.64ˆi  34ˆj N  39.27 N  120
F  19.64ˆi  34ˆj N  39.27 N  120
12
Ans.
Ans.
F43  33.95ˆi  181.1ˆj N  184.27 N79.38 Ans.
F43  33.951ˆi  181.1ˆj N  184.27 N79.38 Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
M
O4
Uicker et al.
M14  1754.4kˆ N  mm
 R BO4 ×F34  t 4  M14  0
Ans.
14.22 A kinematic analysis of the mechanism of Problem 14.20 at  2  210 with
2  10 rad/s ccw and  2  0 gave 3  14.7, 4  164.7, 3  4.73 rad/s ccw,
4  5.27 rad/s cw, 3  4  10.39 rad/s cw, and AG  7.820.85 m / s2 . Compute the crank
torque and the pin reactions for this posture using the same force FC as in Problem 14.20.
3
The D’Alembert inertia forces and offsets are:
t 2   IG2 α 2  0
f2  m2 AG2  0
h2  t2 f 2  0
f3  m3 AG3

  17.8 N  7.29ˆi  2.77ˆj m/s 2
  9.65 m/s 
f4  m4 AG4  0
2
 13.43ˆi  5.12ˆj N  14.37 N  159.15
t 3   I G3 α 3

   7 N  mm  s 2  10.39kˆ rad/s 2
t 4   I G4 α 4


   7 N  mm  s 2  10.39kˆ rad/s 2

 72.8kˆ N  mm
h3  t3 f3   72.86 N  mm  14.37 N   5.1 mm , h4  0
Next, the free-body diagrams are drawn. Since the lines of action for the forces on the
free-body diagrams cannot be discovered from two- and three-force member concepts, the
force F34 is divided into radial and transverse components.
 72.8kˆ N  mm
M
O4
 t 4  R BO4 ×F34t  0
 72.86kˆ N  mm    144.7ˆi  39.47ˆj mm  ×  0.263ˆi  0.965ˆj F
t
34
0
t
F34
 0.485 N
M

t
r
 RG3 A ×f3  t 3  RCA × FC  R BA × F43
 R BA × F43
0
290.12ˆi  76.32ˆj mm × 13.83ˆi  5.12ˆj N   72.86kˆ N  mm   580.25ˆi  152.65ˆj mm × 133.5ˆj
A
 

 

 

 
 


 435.2ˆi  114.48ˆj mm × 0.129ˆi  0.467 ˆj N  435.2ˆi  114.48ˆj mm × 0.965ˆi  0.263ˆj F  0
r
43
 459.685kˆ N  mm   72.86kˆ N  mm    77430kˆ N  mm    188.57kˆ N  mm    224.92kˆ mm F
r
43
F43r  347 N
F  F
43
 f3  FC  F23  0
F34  333.75ˆi  89ˆj N  347 N  15.20 Ans.
F  347ˆi  48.95ˆj N  351.5 N8.02 Ans.
23
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0
Theory of Machines and Mechanisms, 4e
M
O2
 R AO2 ×F32  M12  0
Uicker et al.
M12  19691.25kˆ N  mm
Ans.
14.23 Figure P14.23 illustrates 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  400 mm,
RO4O2  RBA  1000 mm,
RG3 A  800 mm, and
RBO4  1400 mm,
RG4O4  500 mm.
Also w3  987.9 N, w4  925.6 N,
IG4  29370 N  mm  s2 , and the crank is balanced.
IG3  25142.5 N  mm  s2 , and
Make a kinematic and dynamic
for a complete rotation of the crank using 2  10 rad/s ccw,
FC  222.5ˆi  3942.7ˆj N for 90  2  300 and FC  0 otherwise.
100 m
m
analysis
400 m
m
Kinematic Analysis
R2e j2  R3e j3  R1  R4e j4
400e j2  1000e j3  1000  1400e j4
400cos 2  1000cos 3  1000  1400cos 4 400sin 2  1000sin 3  1400sin 4
Eliminating  3 we find  4 from the roots of the quadratic
 425  200cos2  tan 2 4 2  1400sin 2  tan 4 2  1200cos2  3075  0
Then 3  tan 1 175sin 4  50sin 2  175cos 4  125  50cos 2 
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Ans.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
RG3  400e j2  800e j3
RG4  1000  500e j4

RC  400e j2  1403.5e  3
The first-order kinematic coefficients are:
j 400e j2  j1000e j33  j1400e j44
400sin 2 1000sin 33  1400sin 44
j   4.086
3  350sin 4  2  875sin 4  3 
RG3  j 400e
j 2
Ans.
Ans.
400cos2  1000cos33  1400cos44
4  250sin 2  3  875sin 4  3 
Ans.
 j800e 3 3  400  sin 2  50sin 33   j 400  cos 2  50cos 33  Ans.
j
RG4  j500e j44  500sin 44  j500cos 44
Ans.
j   4.086
3
RC  j 400e j2  j1403.57e  3
Ans.
  400sin  2  1403.57 sin 3  4.086  3   j  400 cos  2  1403.57 cos 3  4.086  3 
The second-order kinematic coefficients are:
400e j2  j1000e j33  1000e j332  j1400e j44  1400e j442
400cos 2  1000sin 33  1000cos332  1400sin 44  1400cos442
400cos 2  1000sin 33  1000cos332  1400sin 44  1400cos442
3  350cos 4  2   875cos 4  3 32  122542  875sin 4  3 
4  250cos 4  2   62532  875cos 4  3 42  875sin 4  3 
RG3  400e
j 2
j
Ans.
Ans.
j
 j800e 3 3  800e 3 32
  400 cos  2  800sin 33  800 cos 3332  j  400sin  2  800 cos 33  800sin 332  Ans.
RG4  j500e j 4  500e j 42   500sin  4 4  500 cos  4 42   j  500 cos  4 4  500sin  4 42  Ans.
4
4
Dynamic Analysis
By virtual work we can formulate the dynamic input torque requirement as:
M12  f3 RG3  t 3 3kˆ  f4 RG4  t 4  4kˆ  FC RC
The individual elements of this equation are:
2
f3  m3 AG3  m3RG3 22    987.9 9804 mm/s 2  10 rad/s  RG3  10 N/mmRG3
  4094 cos  2  8188 sin 33  8188 cos 3332  j  4094 sin  2  8188 cos 33  8188 sin 332  N
f3 RG3   4094cos  2  8188sin 33  8188cos 3332  400sin  2  800sin 33  N  mm
  4094sin  2  8188cos 33  8188sin 332   400cos  2  800 cos 33  N  mm
f3 RG3   32575.97 N  m   sin 3  2 3 1  3   cos 3  2 3  233
t 3   IG3 α3   IG3322kˆ    2514.25 N  m 3kˆ
t  kˆ    2514.25 N  m   
3
3
3 3
f4  m4 AG4  m4 RG4     925.6 N 9804 mm/s 2  10 rad/s  RG4  9.44 N/inRG4
2
2
2
  4797sin  4 4  4797 cos  4 42   j  4797 cos  4 4  4797sin  4 42  N
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
f4 RG4   4797sin  4 4  4797 cos  4 42   500sin  4 4  N  mm
  4797 cos  4 4  4797sin  4 42   500cos 4 4  N  mm
f4 RG4  2298.5544 N  m
t   I α   I   2kˆ    2937 N  m  kˆ
4
G4
4
G4 4
2
4
t 4 4kˆ  293744 N  m
FC RC   4450 N  cos120  400sin  2  1403.57sin 3  4.086 3 mm 
  4450 N  sin120 400cos  2  1403.57 cos 3  4.086 3 mm 
FC RC  1780sin 2  120  6245.9sin 3  124.086 3 N  m
Reassembling the elements we must remember that force FC is nonzero for only a portion
of the cycle. Therefore,
M12   3275.9    sin       1      cos        2     2514.2      2298.55    2937   N  m
3
2
3
3
3
2
3
3
3
3
3
4
4
4
4
=3275.9 cos 3   2  3  3275.9sin 3   2 3 1  3   906633  5335 4 4 N  m
for the entire cycle and, for 90   2  300 , an additional increment is added:
Ans.
M12  1780sin 2  120  6245.9 n 3  124.086 3 N  m
This input torque requirement is shown in the following plot. Notice the small
discontinuities in the curve when force FC begins and ends its effect.
11,000
5,500
5,500
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.24 Figure P14.24 illustrates 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  303.7 N  mm  s2 ;
flywheel
shaft,
gear,
pinion,
I  1.724 N  mm  s2 ;
I  19.135 N  mm  s2 ;
2
2
I  0.388 N  mm  s ; motor, I  9.612 N  mm  s . If the motor has a starting torque of
8343.75 N  mm, what is the angular acceleration of the flywheel shaft at the instant the
motor is turned on ?
225 mm
50 mm
If we identify the motor shaft as 2 and the flywheel shaft as 3 then
I 2  9.612 N  mm  s2  0.388 N  mm  s2  10 N  mm  s 2
I3  303.7 N  mm  s2  1.724 N  mm  s 2  19.135 N  mm  s 2  324.559 N  mm  s2
3    R2 R3  2
3    R2 R3  2
M
M
3
 R3 F23  I 33
F23    R2 R32  I3 2
2
 M12  R2 F32  I 2 2
2
M12  I 2 2  R2 F32   I 2   R2 R3  I 3   2


Now, substituting the numeric values,
2
8343.75 N  mm  10 N  mm  s 2   25 mm 112.5 mm  324.559 N  mm  s 2   2   26.028 N  mm  s 2   2
2  8343.75 N  mm 26.028 N  mm  s2  320.56 rad/s2
3    R2 R3 2    25 mm 112.5 mm  320.56 rad/s2  71.24 rad/s2
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.25 The disk cam of Problem 13.31 is driven at a constant input shaft speed of
2  25 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.
IG3  m3k32   0.030 kg  0.035 m   0.000 036 75 kg  m2
2
For full-rise cycloidal cam motion, Eq. (5.19),
2  30 
112.5 
L
y  1  cos

1  cos 2
  0.200

  150 
150 
 360 30 sin 2 112.5  0.480
150

2
1502
2
3  y22  0.480  25 rad/s   300 rad/s 2
y 
2 L
sin
2

t3   IG33    0.000 036 75 kg  m2  300 rad/s 2   0.011 025 N  m
By virtual work,
M12   y t3  R CO3 × FC kˆ 


 0.200 0.011 025 N  m   0.150 m 8 N  sin  45    0.168 N  m ccw


© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.26 Repeat Problem 14.25 with a shaft speed of 2  50 rad/s ccw.
IG3  m3k32   0.030 kg  0.035 m   0.000 036 75 kg  m2
2
For full-rise cycloidal cam motion, Eq. (5.19),
2  30 
112.5 
L
y  1  cos

1  cos 2
  0.200

  150 
150 
 360 30 sin 2 112.5  0.480

150

1502
2
3  y22  0.480  50 rad/s   1 200 rad/s 2
y 
2 L
2
sin
2

t3   IG33    0.000 036 75 kg  m2  1 200 rad/s 2   0.044 1 N  m
By virtual work,
M12   y t3  R CO3 × FC kˆ 


 0.200 0.044 1 N  m   0.150 m 8 N  sin  45    0.161 N  m ccw


© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.27 A rotating drum is pivoted at O2 and is decelerated by the double-shoe brake mechanism
illustrated in Fig. P14.27. The mass of the drum is 1023.5 N and its radius of gyration is
1414.5 mm. The brake is actuated by force P  445ˆj N , and it is assumed that the
contacts between the two shoes and the drum act at points C and D, where the
coefficients of Coulomb friction are   0.350. Determine the angular acceleration of
the drum and the reaction force at the fixed pivot F12 .
F  445P  1780 N
 M   300ˆi mm ×  Pˆj   75ˆj mm ×  F ˆi   0
F  1780ˆi  445ˆj N  1833.4 N14.04
F  P  F  F  0
5
65
65
5
65
35
35
The friction angle is   tan
1
 0.350  19.29 .
 M  550ˆj mm ×F   125ˆi  225ˆj mm  ×   cos ˆi  sin ˆj F
3
53
23
0
F32  3644.5ˆi  1277ˆj N  3858 N19.29
 M4  625ˆj mm ×F64  125ˆi  225ˆj mm × cos ˆi  sin ˆj F24  0
F23  3858 N



 

F42  6136.5ˆi  2149.35ˆj N  6501.45b  160.71
F24  6501.45 N
IG2  m2 k22  1023.5 N 9804.4 mm/s 2  141.5 mm   2123.1 N  mm  s 2
2
M
F
2
O2

 
 
 

 200ˆi mm × 3644.5ˆi  1277.2ˆj N  200ˆi mm × 6136.5ˆi  2149.4ˆj N  IG2 α 2
α 2  322kˆ rad/s2
Ans.
 F12  F32  F42  0
F12  2496.45ˆi  872.2ˆj N  2643.3 N19.29
Ans.
Note that gravitational effects are not yet included. If gravity acts in the jˆ 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
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Theory of Machines and Mechanisms, 4e
Uicker et al.
were known, however, these weights 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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Theory of Machines and Mechanisms, 4e
Uicker et al.
14.28 For the mechanism illustrated in Fig. P14.28, the dimensions are RG2O4  0.15 m,
REG2  0.20 m, and the length of link 4 is 0.20 m, symmetric about O4. The ground
bearing is midway between E and G2. There is a torque T2 acting on the input link 2, and
a torque T4 acting on link 4. Link 2 is in translation with a velocity of
V  0.114 8ˆj m/s and an acceleration of A  0.35ˆj m/s2 and the line connecting
2
2
mass centers G2 and G3 is horizontal. The kinematic coefficients are 3  11.5 rad/m ,
  2 m/m , and R43
  40 m/m2 (where R 43 is the vector from G3
3  380 rad/m2 , R43
to G ). The acceleration of the mass center of link 3 is A  1.053ˆi  0.432ˆj m/s2 . The
4
G3
masses and second moments of mass of the moving links are m2  m4  0.5 kg,
m3  1 kg, IG 2  IG 4  2 kg  m2 , and IG 3  5 kg  m2 . Assume that gravity acts in the
negative Z direction, and that the effects of friction can be neglected. Determine the
unknown internal reaction forces, and the unknown torques T2 and T4 .
The free-body diagram for link 2 is shown in Figure 1. Recall that gravity acts in the
negative Z-direction and the effects of friction can be neglected.
Since the center of mass G2 is translating in the Y-direction, the sum of the external
forces in the X-direction acting on link 2 shows
F12X  F32X  P X  m2 AGX2  0
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Theory of Machines and Mechanisms, 4e
Uicker et al.
Figure 1. Free-body diagram of link 2.
Since friction can be neglected, the sum of the external forces in the Y-direction acting on
link 2 gives
F32Y  PY  m2 AGY2
Substituting the given information, the Y-component of the reaction force between links 2
and 3 is
Ans. (2)
F32Y   0.5 kg  (0.35 m/s2 )   40 N  sin120  34.816 N
Since link 2 is not rotating then the angular acceleration  2  0 . Therefore, the sum of the
external moments on link 2 acting about G2 can be written as
 R7 F12X  R9 P X  T2  0
(3)
Therefore, there are still 4 unknowns for link 2, namely the forces F12X , F32X , and F32Y and
the torque T2 .
The free-body diagram for link 3 is shown in Figure 2.
Figure 2. Free-body diagram of link 3.
The sum of the external forces in the X-direction acting on link 3 can be written as
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Theory of Machines and Mechanisms, 4e
Uicker et al.
F23X  F43X  m3 AGX3
(4)
The sum of the external forces in the Y-direction acting on link 3 can be written as
F23Y  F43Y  m3 AGY3
(5)
The sum of the external moments acting about the center of mass G3 can be written as
R43 F43  RG2G3 F23Y  IG33
(6)
The vector R 43 points from the center of mass of link 3 to the location of the reaction
force F43, and the vector R G2G3 points from the center of mass of link 3 to the center of
mass of link 2.
Equations (4), (5), and (6) contain two new unknown variables, namely the internal
reaction force F43 and the location of this force (i.e., R43). Note that the force F43 is
perpendicular to the slot since friction is neglected. Therefore, F43X and F43Y are not
independent unknowns (the angle is known). Therefore, there are 6 equations and 6
unknown variables, namely the forces F12X , F32X , F32Y , F43 , the torque T2, and the distance
R43.
These six unknowns can now be solved for by inspection. Substituting Eq. (2) and
the given acceleration of the center of mass G2 into Eq. (5), the Y-component of the
internal reaction force between links 3 and 4 is
F43Y  F43 sin 60  1 kg  (0.432 m/s2 )  (34.816 N)  34.384 N
Ans.
Therefore, the force between links 3 and 4 is
34.384 N
Ans.
F43 
 39.70 N
sin 60
Substituting known values into Eq. (4), the internal reaction force between links 2 and
3 is
Ans.
F23X  ( 39.70 N) cos 60  1 kg  1.053 m/s2   20.91 N
Substituting known values into Eq. (6) gives
R34 (39.70 N)  (0.259 81 m)(34.816 N)  5 kg  m2 (0.983 rad/s2 )


Rearranging this equation, the unknown distance is
13.961 N  m
R43 
 0.351 66 m
39.70 N
Therefore, the distance from the ground pin O4 to the line of action of the internal
reaction force F43 is
Z  R43  0.300 m  0.351 66  0.300 m  51.7 mm
Since the distance Z is less than the length of link 4 then link 3 is sliding along link 4 (i.e.,
there is sliding contact and not tipping). The internal reaction force between links 3 and 4
acts within the physical limits of link 4.
Substituting known values into Eq. (1) gives
F12X  20.91 N   40 N  cos 60  0
Ans.
Rearranging this equation, the unknown force is
F12X  40.91 N
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Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
Substituting known values into Eq. (3) gives
 0.1 m (40.91 N)   0.2 m (40 N) cos120  T2  0
Rearranging this equation, the unknown torque acting on link 2 is
T2  0.091 Nm
The free-body diagram for link 4 is shown in Figure 3.
Ans.
Figure 3. Free-body diagram of link 4.
Since G4 is coincident with the ground pivot O4 , the sum of the external forces in the Xdirection acting on link 4 can be written as
F34 cos 60  F14X  m4 AGX4  0
(7)
The sum of the external forces in the Y-direction acting on link 4 can be written as
(8)
F34 sin 60  F14Y  0
The sum of the external moments acting about the center of mass of link 4 can be written
as
(9)
ZF34  T4  I G4  4
Equations (7), (8), and (9) contain three new unknown variables, namely the internal
reaction forces F14X , F14Y , and the torque T4. These three unknown variables can now be
solved for as follows. Substituting known values into Eq. (7), the X-component of the
force between links 1 and 4 is
F14X  19.85 N
Ans.
Substituting known values into Eq. (8), the Y-component of the internal reaction force
between links 1 and 4 is
F14Y  34.38 N
Ans.
Substituting known values into Eq. (9), the torque acting on link 4 is
T4   2 kg  m2  (0.983 rad/s 2 )  (0.051 7 m)(39.70 N)  4.018 Nm
Ans.
The negative sign indicates that the torque acting on link 4 is clockwise.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
14.29 For the mechanism illustrated in Fig. P14.29, the kinematic coefficients are
The
3  2 rad/m, 3  6.928 rad/m2 , R4  1.732 m/m, and R4  8 m/m2 .
velocity and the acceleration of the input link 2 are V  5ˆj m/s and A  20ˆj m/s2
2
2
and the force acting on link 2 is F  200ˆj N. The length of link 3 is RBA  1 m and the
distance RG3G2  0.5 m. A linear spring is attached between points O and A with a free
length L  0.5 m and a spring constant K  2 500 N/m. A viscous damper with a
damping coefficient C  45 N  s/m is connected between the ground and link 4. The
masses and mass moments of inertia of the links are m2  0.75 kg, m3  2.0 kg,
m4  1.5 kg, IG2  0.25 N  m  s2 , IG3  1.0 N  m  s 2 and IG4  0.35 N  m  s2 .
Assume
that gravity acts in the negative Y-direction (as illustrated in Fig. P14.29) and the effects
of friction in the mechanism can be neglected.
(i) Determine the first-order kinematic coefficients of the linear spring and the viscous
damper.
(ii) Determine the equivalent mass of the mechanism.
(iii) Determine the magnitude and direction of the horizontal force P that is acting on link
4.
A double-slider mechanism.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
(i) The vectors for the linear spring are shown in Fig. 2a.
Figure 2a. Vectors for the linear spring.
The vector loop for the linear spring can be written as
?
I
R S  R2  0
From which, the magnitudes can be written as the scalar equation
RS  R2
Differentiating this with respect to the input position R2 , the first-order kinematic
coefficient of the spring is
RS  1 m/m
Ans. (1)
Note that the sign is positive because, for a negative input, the length of the linear spring
is decreasing. Also, note that the first-order kinematic coefficient of the mass center of
input link 2 is
YG2  RS  1 m/m
The vectors for the viscous damper are shown in Fig. 2b.
Fig. 2b. Vectors for the viscous damper.
The vector loop for the damper can be written as

?

R9  R C  R 4  0
Since all components of this equation are horizontal, this gives
R9  RC  R4  0
Differentiating with respect to the input position R2 gives
 RC  R4  0
Rearranging and substituting the given data, the first-order kinematic coefficient of the
viscous damper is
RC   R4  1.732 m/m
Ans. (2)
The positive sign agrees with our intuition since, for a positive input, the length of the
viscous damper is increasing.
(ii) The equivalent mass of the mechanism can be written as
4
mEQ   Aj
(3)
j 2
For link 2:
A2  m2  X G2  YG2   IG 22
2
2
2
The vector loop for the center of mass of the input link can be written as
I
??
RG  R 2  0
2
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Theory of Machines and Mechanisms, 4e
Uicker et al.
The X and Y components of this equation give
X G2  0 and YG 2  R2
Differentiating these with respect to the input position R2 give
X G 2  0 and YG2  R2  1 m/m
Substituting these values into Eq. (4) gives
A2   0.75 kg   02  12    0.25 kg  m2   0   0.75 kg
2

(5)

2
2
A3  m3 X G 3  YG 3  I G 3 3 2
For link 3:
The vector loop for the center of mass of link 3 can be written as
I
??
(6)

R G3  R 2  R G3G2
The X and Y components are
X G3   RG3G2 cos 3  0.25 m
YG3  R2  RG3G2 sin 3  0.433 m
and
Differentiating these with respect to the input position R2 give
X G 3  RG3G2 sin 33  0.866 m/m
YG3  1  RG3G2 cos 33  0.5 m/m
and
Substituting these and other known data into Eq. (6) gives
2
A3   2.0 kg  (0.866 m/m) 2   0.5 m/m    1.0 kg  m2  (2 rad/m) 2  6 kg


2
2
A4  m4 X G 4  YG 4  I G 4 4 2
For link 4:

(7)

(8)
Note from given data that X G 4  R4  1.732 m/m; therefore, Eq. (8) can be written as
A4  1.5 kg  [ 1.732 m/m   02 ]   0.35 kg  m2   0   4.5 kg
2
2
(9)
Therefore, substituting Eqs. (5), (7), and (9) into Eq. (3), the equivalent mass of the
mechanism is
Ans. (10)
mEQ  0.75 kg  6 kg  4.5 kg  11.25 kg
(iii)
The power equation for the mechanism can be written as
dT dU dW f
F  V2  P  V4 


dt
dt
dt
Substituting the time rate of change of energy terms into the right-hand side gives
4
4
4
j 2
j 2
j 2
F  V2  P  V4   Aj R2 R2   B j R23   m j gYG j R2  K s  RS  RS 0  RS R2  CRC2 R 2
2
The linear velocity of link 2 and the force acting on link 2 are both in the same direction
(that is, both downward). Assuming that the force P is in the same direction as the
velocity of link 4 (that is, to the right), then the above equation can be written as
4
4
4
j 2
j 2
j 2
FV2  PV4   Aj R2 R2   B j R23   m j gYG j R2  K s  RS  RS 0  RS R2  CRC2 R 2
2
The velocity of the input link 2 is V2  R2 and the velocity of link 4 is V4  R4 ; therefore,
this can be written as
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Theory of Machines and Mechanisms, 4e
4
Uicker et al.
4
4
FR2  PR4   Aj R2 R2   B j R   m j gYG j R2  K s  RS  RS 0  RS R2  CRC2 R 2
j 2
3
2
j 2
2
j 2
Dividing by the input velocity R2 throughout gives the equation of motion for the
mechanism, that is
4
4
4
(11)
F  PR4   Aj R2   B j R22   m j gYG  K s  RS  RS 0  RS  CRC2 R2
j 2
j 2
j
j 2
where the first-order kinematic coefficient of link 4 is given as R4  1.732 m/m . The
sum of the Bj terms can be written as
4
1 4 dA
(12)
 Bj   j
2
j 2
j 2
d 2
B2  m2 ( X G 2 X G2  YG2 YG2 )  IG222
For link 2:
and this has a value of
B2   0.75 kg  0  0   0.25 kg  m2  0   0




(13)
B3  m3 X G 3 X G3  YG3YG3  I G333
For link 3:
which has a value of
B3   2 kg [(0.866 m/m)(4 m/m2 )  (0.5 m/m)(0)]  1 kg  m2  (2 rad/m)(0.928 rad/m2 )  20.78 kg/m (14)


B4  m4 X G 4 X G4  YG4 YG4  IG444
For link 4:
which has a value of
B4  1.5 kg  [(1.732 m/m)(8 m/m2 )  (0)(0)]  0.35 kg  m2  0 0   20.78kg/m (15)


Substituting Eqs. (13), (14), and (15) into Eq. (12) gives
4
B
j 2
j
 B2  B3  B4  0  20.78 kg/m  20.78 kg/m  41.56 kg/m
(16)
The change in potential energy due to gravity is
4
 m gY 
j 2
j
Gj
For link 2:
m2 gYG2  (0.75 kg)(9.81 m/s2 ) 1 m/m   7.36 N
For link 3:
m3 gYG3  (2 kg)(9.81 m/s2 )(2 m/m)  39.24 N
For link 4:
m4 gYG4  (1.5 kg)(9.81 m/s2 )(0)  0
Summing these three values gives
4
 m gY 
j 2
j
Gj
 7.36 N  39.24 N  0  31.80 N
Then substituting Eqs. (1), (2), (10), (16), and (17) into Eq. (11) gives
F  PR4  11.25 kg  R2   41.56 kg/m R22  31.80 N  KS  RS  RS 0  (1 m/m)  C(1.732 m/m)2 R2
Rearranging this equation, the force acting on link 4 can be written as
1
  F  11.25 kg  R2   41.56 kg/m  R22  31.80 N  K S  RS  RS 0   3CR2 
P

R4
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Theory of Machines and Mechanisms, 4e
Uicker et al.
The input velocity is R2  5 m/s , the input acceleration is R2  20 m/s2 , and the force
is F  200 N. Substituting these values and the known data into this equation, the force
acting on link 4 can be written as
P
200 N  11.25 kg  (20 m/s2 )   41.56 kg/m  (5 m/s)2  31.80 N

1


1.732 m/m 
  2500 N/m  0.866 m  0.5 m   3  45 Ns/m  (5 m/s) 
or as
P
1
 200 N  225 N  1 039 N  31.80 N  915.0 N  675.0 N 
1.732
Therefore, the force acting on link 4 is
Ans.
P  705.66 N
The negative sign indicates that the force P (acting on link 4) is acting to the left; that is,
in the opposite direction to the velocity of the output link 4. Recall that the force P was
originally assumed to be acting to the right.
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Theory of Machines and Mechanisms, 4e
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14.30 Consider the four-bar linkage of Problem P14.20 modified as illustrated in Fig. P14.30.
The linkage includes a spring and a viscous damper as shown. The spring has a stiffness
k  2.14 N/mm and a free length R0  112.5 mm. The viscous damper has a damping
coefficient C  0.04.45 N  /mm. The external force acting at point C of coupler link 3 is
FC  556.25 N vertically downward (that is, in the negative Y-direction). The input
crank is rotating with a constant angular velocity ω2 = 10 rad/s ccw and the angular
acceleration of link 3 is 3  84.8 rad/s2 ccw; the acceleration of the mass center of
coupler link 3 is A  7750ˆi  7375ˆj mm/s2 . Use the masses and the second moments of
G3
mass as specified in Problem 14.20 with the exception that the weight of link 3 is w3 =
445 N. Assume that the locations of the centers of mass of links 2 and 4 are coincident
with the ground pivots O2 and O4 , respectively, and the center of mass of link 3 is as
indicated by G3 in Fig. P14.30. Also, assume that gravity acts vertically downward (that
is, in the negative Y-direction) and the effects of friction in the mechanism can be
neglected.
1. Write the equation of motion for the mechanism.
2. Determine the equivalent mass moment of inertia of the mechanism.
3. Determine the driving torque T2 on the input crank 2 from the equation of motion.
The mechanism modified from Problem 14.20.
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Theory of Machines and Mechanisms, 4e
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From Problem 14.20, the given data are
RAO2  150 mm
RO4O2  450 mm
RAB  450 mm
RBO4  150 mm
RAC  600 mm
2  10 rad/s ccw
3  1.43 rad/s cw
4  11.43 rad/s cw
RG3 A  300 mm
w3  445 N
I G3  55.29 N  mm  s 2
I G2  I G4  7 N  mm  s 2
2  0
 3  84.8 rad/s 2 ccw
 4  84.8 rad/s 2 ccw
FC  556.25ˆj N
0 shows vectors that are used throughout the solution.
g  9660 mm/s 2
The first-order kinematic coefficient of link 3 can be written as
 1.43 rad/s
3  33  3 
 0.143 rad/rad
2
10 rad/s
The first-order kinematic coefficient of link 4 can be written as
 11.43 rad/s
 4  4 
 1.143 rad/rad
2
10 rad/s
The angular acceleration of link 3 can be written as
3  32 2  3 2
Rearranging this, the second-order kinematic coefficient for link 3 can be written as
 3  3 2 84.8 rad/s2  (0.143 rad/rad)(0)
3  33 

 0.84 rad/rad 2
2
2
2
10 rad/s 
Similarly, the second-order kinematic coefficient of link 4 is
    84.8 rad/s2  (0.143 rad/rad)(0)
 4  4 2 4 2 
 0.84 rad/rad 2
2
2
10 rad/s 
Since the mass centers G2 and G4 are located at the fixed pivots O2 and O4, respectively,
the first- and second-order kinematic coefficients of these mass centers are
xG2  0
xG 2  0
xG2  0
yG2  0
yG 2  0
yG2  0
xG4  18 in
xG 4  0
xG4  0
yG4  0
yG 4  0
yG4  0
To find the first- and second-order kinematic coefficients for the center of mass G3, the
vector loop for the center of mass of link 3 can be written as
RG3  R 2  R33
where R2 = 150 mm, θ2 = 60º, R33 = 300 mm, and θ33 = θ3 = 321.8º. The X and Y
components of the vector equation for the center of mass of link 3 are
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Theory of Machines and Mechanisms, 4e
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xG 3  R2 cos2  R33 cos3  150 mm  cos 60   300 mm  cos321.8  310.75 mm
yG3  R2 sin 2  R33 sin 3  150 mm  sin 60   300 mm  sin 321.8  55.62 mm
The first-order kinematic coefficients for the center of mass of link 3 are
xG 3   R2 sin 2  R33 sin 33   150 mm  sin 60  300 mm  sin321.8(0.143 rad/rad)  156.43 mm/rad
yG 3  R2 cos 2  R33 cos33  150 mm cos60   300 mm  cos321.8(0.143 rad/rad)  41.29 mm/rad
The second-order kinematic coefficients for the center of mass of link 3 can be written as
xG3   R2 cos 2  R33 cos 332  R33 sin 33
yG3   R2 sin 2  R33 sin 332  R33 cos 33
Therefore,
xG3   150 mm cos60  300 in  cos321.8(0.143 rad/rad)2   300 mm sin321.8(0.848 rad/rad2 )  77.5 mm/rad 2
yG3   150 mm sin 60   300 mm sin321.8(0.143 rad/rad)2   300 mm cos321.8(0.848 rad/rad)  73.8 mm/rad 2
4
To determine
 Aj , note that
j 2
4
A
j 2
j
 I EQ (that is, the equivalent mass moment of
inertia). Therefore, the units must be in  lb  s2 .
For link 2:
A2  m2 ( xG22  yG22 )  IG222  m2 (0  0)  7 N  mm  s2 (1 rad/rad)2  7 N  mm  s 2
For link 3:
A3  m3 ( xG23  yG23 )  I G 332

445 N
2
[(156.43 mm/rad)2   41.29 mm/rad  ]   55.29 N  mm  s 2  (0.143 rad/rad)2  121.7 N  mm  s2
2
9653 mm/s
For link 4:
A4  m4 ( xG24  yG24 )  IG442  m4 (0  0)   7 N  mm  s2  (1.143 rad/rad)2  9.15 N  mm  s 2
Therefore, the equivalent mass moment of inertia of the mechanism is
4
I EQ   Aj  A2  A3  A4  7 N  mm  s2  121.1 N  mm  s2  9.15 N  mm  s 2  137.85 N  mm  s 2
Ans.
j 2
4
To determine
 B j , note that
j 2
4
 Bj 
j 2
1 4 dAj
; therefore, the units must be in  lb  s2 .

2 j 2 d 2
For link 2:
B2  m2 ( xG 2 xG 2  yG 2 yG 2 )  I G 222  m2 (0  0)   7 N  mm  s 2  (1 rad/rad)(0)  0
For link 3:
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B3  m3 ( xG 3 xG 3  yG 3 yG 3 )  I G 333

445 N
[(156.43 mm/rad)(77.5 mm/rad 2 )  (41.29 mm/rad)(73.8mm/rad 2 )]
9653 mm/s 2
  55.29 N  mm  s 2  ( 0.143 rad/rad)(0.848 rad/rad 2 )
 48.505 N  mm  s 2
For link 4:
B4  m4 ( xG 4 xG 4  yG 4 yG 4 )  I G 444
 m4 (0  0)   7 N  mm  s 2  (1.143 rad/rad)(0.848 rad/rad 2 )  6.786 N  mm  s 2
Therefore, the sum of these coefficients is
4
B
j 2
j
 B2  B3  B4  0  48.505 N  mm  s 2  6.786 N  mm  s 2  55.291 N  mm  s 2
The power equation can be written as
dT dU dW f
(1)


P
dt
dt
dt
The left-hand side of the power equation can be written
(2)
P  T22  FC  VC
The unknown torque T2 is taken to be positive in the same direction as the input angular
velocity (that is, counterclockwise). The velocity of point C can be written as
VC  ( xC ˆi  yC ˆj)2
(3)
The first-order kinematic coefficients for the path of point C can be obtained from the
vector equation
RC  R 2  R3
The X and Y components of this vector equation are
xC  R2 cos 2  R3 cos 3
yC  R2 cos 2  R3 sin 3
Therefore, the first-order kinematic coefficients for point C are
xC   R2 sin 2  R3 sin 33   150 mm sin 60   600 mm  sin321.8(0.143 rad/rad)  182.963 mm/rad (4a)
yC  R2 cos 2  R3 cos33  150 mm cos60   600 mm  cos321.8(0.143 rad/rad)  7.5735 mm/rad (4b)
Substituting Eqs. (4) into Eq. (3), the velocity of point C can be written as
VC   182.963 mm/rad  ˆi   7.5735 mm/rad  ˆj 2
Therefore, the power due to the vertically downward force at point C is
FC  VC    556.25 N  ˆj  ( xC ˆi  yC ˆj)2    556.25 N  yC 2
   556.25 N  (7.5735 mm/rad)2    4213 N  mm/rad  2
(5)
The negative sign indicates that the vertical force and the vertical component of the
velocity of point C are in opposite directions (that is, that the vertical component of the
velocity of point C is upwards).
Substituting Eq. (5) into Eq. (2), the net power is
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Theory of Machines and Mechanisms, 4e
P  T22   4213 N  mm/rad  2
Uicker et al.
(6)
Now consider the right-hand side of the power equation, see Eq. (1). In general, the time
rate of change of kinetic energy can be written as
4
4
dT
  Aj   B j 3
dt j 2
j 2
However, the generalized inputs for this problem are    2 ,   2  2 , and
  2   2 . Therefore, this equation can be written as
4
4
dT
  Aj2 2   B j 23
dt j 2
j 2
The constant angular velocity of the input link is 2  10 rad/s ccw. Therefore, the time
rate of change of the kinetic energy is
dT
 [137.838 N  mm  s 2  (0)   55.291 N  mm  s 2  (10 rad/s) 2 ]2    5529.1 N  mm  2
dt
(7)
The time rate of change of the potential energy due to gravity is
4
dU g
  m j gyG j  m3 gyG 3 2   445 N  41.2875 mm/rad  2  1837.293 N  mm  2 (8)
dt
j 2
The vector loop equation for the spring can be written as
R 2  R S  R10  0
The X and Y components are
(9a)
R2 cos  2  RS cos  S  0
(9b)
R2 sin  2  RS sin S  R10  0
From Eq. (9a) the stretched length of the spring (for this position of the mechanism) is
Rs  xA  R2 cos 2  150 mm  cos 60  75 mm
Differentiating Eqs. (9) with respect to the input position gives
 R2 sin  2  RS cos S  RS sin SS  0
R2 cos  2  RS sin S  RS cos S S  0
Substituting the known information gives
 150 mm  sin 60o  RS  0
Therefore, the first-order kinematic coefficient for the spring is
Rs   150 mm  sin 60  129.75 mm/rad
The time rate of change of the potential energy in the spring is
dU s
 k ( Rs  Rso ) Rs2   2.136 N/mm  (75 mm  112.5 mm)(129.75 mm/rad)2  10.392 N  mm/rad  2
dt
(10)
The vector loop equation for the viscous damper can be written as
R 4  RC  R11  0
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The X and Y components of this equation are
R4 cos 4  RC cosC  R11 cos11  0
R4 sin 4  RC sin C  R11 sin 11  0
Differentiating these with respect to the input position gives
 R4 sin 44  RC cos C  RC sin CC  0
R4 cos 44  RC sin C  RC cos CC  0
Substituting the known information (with the angle 4  261.8, and the first-order
kinematic coefficient 4  1.143 rad/rad ) gives
 150 mm  sin 261.8(1.143 rad/rad)  RC cos180  0
Therefore, the first-order kinematic coefficient for the damper is
RC  150 mm  sin 261.8(1.143 rad/rad)  169.75 mm/rad
The positive sign indicates that the length of the vector R C is increasing for positive
input. The velocity of point B at the end of the damper is
VB  RC2  169.75 mm/rad  (10 rad/s)  1697.5 mm/s
The time rate of change of the dissipative effect of the damper is
dW f
dt
 CRc22 2   0.0445 N  s/mm  (169.75 mm/rad)2 10 rad/s  2  12.784 N  mm/rad  2
(11)
Therefore, from Eqs. (7), (8), (10), and (11), the right hand side of the power equation,
Eq. (1), can be written as
dT dU dW f
 
   5.5291 N  m/rad  2  1.8373 N  m/rad  2  10.392 N  m/rad  2  12.784 N  m/rad  2
(12)
dt dt dt
  30.5424 N  m/rad  2
Substituting Eqs. (6) and (12) into Eq. (1) gives
T22   4.213 N  m/rad  2   30.5424 N  m/rad  2
(13)
The equation of motion is obtained by dividing both sides of Eq. (13) by the input ngular
velocity ω2. Therefore, the equation of motion for this problem can be written as
T2  4.213 N  m  30.5424 N  m
Ans.
The driving torque acting on the input crank is
T2  34.755 N  m
The positive sense indicates that the driving torque acting on the input crank is in the
same direction as the input angular velocity. Therefore, the driving torque acting on the
input crank is
T2  34.7554 N  m ccw
Ans.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
14.31 For the Scotch-yoke mechanism in the position illustrated in Fig. P14.31, an external
force P  125ˆj N is acting on link 4, and an unknown torque T2 is acting on the input
link 2. The length R2  1 m , the angle   30 , and the angular velocity and acceleration
of link 2 are ω  15 kˆ rad/s and α  2 kˆ rad/s2 , respectively. The accelerations of the
2
2
centers of mass of the links are AG2  5.4ˆi  11.3ˆj m/s2 , AG3  10.8ˆi  22.6ˆj m/s2 , and
AG4  22.6jˆ m/s2 . The centers of mass of links 2 and 3 are at the geometric centers of
links 2 and 3, respectively. The masses and mass moments of inertia of the links are
m2  5 kg , m3  5 kg , m4  15 kg , IG2  0.02 N  m  s2 , IG3  0.12 N  m  s2 , and
IG4  0.08 N  m  s2 . Gravity is acting vertically downwards (that is, g  9.81 m/s2 in the
negative Y-direction). Assume that friction in the mechanism can be neglected. (i) Draw
free-body diagrams of all moving links of the mechanism. (ii) Write the governing
equations for all moving links. List all unknown variables. (iii) Determine the
magnitudes and directions of all internal reaction forces. (iv) Determine the magnitude
and the direction of the torque T2 . (v) Indicate the point(s) of contact of link 4 with
ground link 1.
A Scotch-yoke mechanism.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
(i) The free-body diagram for link 2 is shown in Fig. 1.
F12Y
F12X
O2
T2
R5
G2
F32Y
W2
R2
G3
F32X
Figure 1. The free-body diagram for link 2.
Ans.
The sum of the forces acting on link 2 in the X-direction can be written as
 F X  F12X  F32X  m2 AGX
(1)
The sum of the forces acting on link 2 in the Y-direction can be written as
 F Y  F12Y  F32Y  W2  m2 AGY2 .
(2)
2
The sum of the moments acting on link 2 about point O2 can be written as
M
O2
R2X F32Y  R2Y F32X  R5X W2  T2  IG2 2  m2 ( R5X AGY2  R5Y AGX2 )
(3)
Therefore, there are three equations and five unknowns for the free-body diagram of link
2. The unknowns are the four reaction forces F12X , F12Y , F32X , F32Y and the crank torque T2 .
The free-body diagram for link 3 is shown is Fig. 2.
F23Y
F23X
G3
W3
F43
R7
Figure 2. The free-body diagram for link 3.
Ans.
The sum of the forces acting on link 3 in the X-direction can be written as
 F X F23X  m3 AGX3
(4)
The sum of the forces acting on link 3 in the Y-direction can be written as
 F Y F23Y  m3 AGY3
(5)
The sum of the moments acting on link 3 about the center of mass G3 can be written as
M
G3
R7 F43  IG33
Since link 3 cannot rotate, the angular acceleration is  3  0 . Therefore, this equation
can be written as
R7 F43  0
(6)
Equations (4), (5), and (6) contain two new unknowns, F43 and R7 . Therefore, there are
now a total of six equations and seven unknowns.
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Theory of Machines and Mechanisms, 4e
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If we assume that link 4 is only sliding on ground link 1, then the free-body diagram for
link 4 is as shown in Fig. 3.
G4
R8
R9
F34
W4
F14
P
Figure 3. The free-body diagram for link 4 when sliding only.
The sum of the forces acting on link 4 in the X-direction can be written as
 F X F14  m4 AGX4
Since link 4 can only accelerate in the Y-direction, that is, since AGX4  0 , this becomes
F14  0
The sum of the forces acting on link 4 in the Y-direction can be written as
 F Y F34  P  W4  m4 AGY4
Ans. (7)
(8)
The sum of the moments acting on link 4 about the center of mass G 4 can be written as
M
G4
R8 F34  R9 F14  I G 4 4
Since link 4 can not rotate, that is, since  4  0 , this becomes
R8 F34  R9 F14  0
(9)
From Equations (7) and (9), the distance R9   , which means that link 4 attempts to tip.
For tipping, the free body diagram of link 4 is modified as shown in Figure 4.
G4
R8 F34
R10
F14T
W4 R11
F14B
P
Figure 4. The free-body diagram for link 4 with tipping.
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Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
The sum of the forces acting on link 4 in the X-direction can be written as
 F X F14T  F14 B  m4 AGX4  0
(10)
The sum of the forces acting on link 4 in the Y-direction can be written as
 F Y F34  P  W4  m4 AGY4
(11)
Note that Eq. (11) is the same as Eq. (8).
The sum of the moments acting on link 4 about the center of mas G 4 can be written as
R8 F34  R10 F14T  R11F14B  0
(12)
Equations (10), (11) and (12) contain two new unknowns F14T and F14B . Therefore, there
are a total of nine equations and nine unknowns.
(iii) We will solve this problem by the method of inspection.
Substituting m3  5 kg and AGX3  10.8 m/s2 into Eq.(4), we have
F23X  (5 kg)(10.8 m/s2 )  54 N
Ans. (13)
Substituting m2  5 kg , AGX2  5.4 m/s2 , and F23X  54 N into Eq.(1), we have
F12X  (5kg)(5.4 m/s2 )  (54 N)  81 N
Ans. (14)
Equation (6) implies that either R7  0 or F43  0 . Since there is contact between links 3
and 4, the internal reaction force F43 cannot be zero. Therefore
R7  0
(15)
Substituting m4  15 kg , AGY4  22.6 m/s2 , W4  m4 g  (15 kg)(9.81 m/s2 )  147 N , and
P  125 N into Eq. (11) we have
F34  (15 kg)(22.6 m/s2 )  147 N 125 N  361 N
Ans. (16)
Substituting m3  5 kg , AG 3Y  22.6 m/s , W3  m3 g  (5)(9.8) kg  m/s  49 N , and
F34  361 N into Eq. (5), we have
2
2
F23Y  (5 kg)(22.6 m/s2 )  49 N  361 N  523 N
Substituting m2  5 kg ,
Ans. (17)
A  11.3 m/s , W2  m2 g  (5 kg)(9.81 m/s )  49 N , and
Y
G2
2
2
F23Y  523 N into Eq. (2), we have
F12Y  (5 kg)(11.3 m/s2 )  49 N  523 N  628.5 N
From Eq. (12), we have
R10 F14T  R11F14B  R8 F34
Using Eqs. (10) and (19), we have
RF
F14T   8 34 ,
R11  R10
and
F14B 
R8 F34
R11  R10
Substituting R8  R2X  R7  R2X  R2 sin 30  0.5 m , R10  0.4 m ,
F34  361 N into Eq. (20), we have
(0.5 m)(361 N)
F14T  
 361 N
0.9 m  0.4 m
and
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Ans. (18)
(19)
(20)
R11  0.9 m and
Ans. (21a)
Theory of Machines and Mechanisms, 4e
F14 B 
Uicker et al.
(0.5 m)(361 N)
 361 N
0.9 m  0.4 m
(iv) From Eq. (3), we have
T2  IG2  2  m2 ( R5X AGY2  R5Y AGX2 )  R5X W2  R2X F23Y  R2Y F23X
Ans. (21b)
(22a)
1
Substituting W2  49 N, IG2  0.02 N  m  s2 ,  2  2 rad/s 2 , m2  5 kg , R5X  R2 sin30  0.25 m ,
2
1
R5Y   R2 cos30  0.433 m , AGX2  5.4 m/s2 , AGY2  11.3 m/s2 , R2X  R2 sin 30  0.5 m ,
2
Y
R2   R2 cos30  0.866 m , F23X  54 N , and F23Y  523 N into Eq. (22a), we have
T2  (0.02 N  m  s 2 )(2 rad/s 2 )  (5 kg) (0.25 m)(11.3 m/s 2 )  (0.433 m)(5.4 m/s 2 ) 
(0.25 m)(49 N)  (0.5 m)(523 N)  ( 0.866 m)( 54 N) N  m
 229.46 N.
Ans. (22b)
(v) Since F14T  361 N , therefore F41T  361 N ; this means that link 4 is pushing to the
right on ground link 1 at the upper-right corner of the slot. Also, since F14B  361 N ,
therefore F41B  361 N ; this means that link 4 is pushing to the left on ground link 1 at
the lower-left corner of the slot.
Ans.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
14.32 For the parallelogram four-bar linkage in the position illustrated in Fig. P14.32, the
angular velocity and acceleration of the input link 2 are 2  2 rad / s ccw and
 2  1 rad / s2 ccw, respectively. The distances RBO2  RAO4  0.2 m, RBA  RO2O4  0.3 m,
and RCB  0.1 m. The force FC  100 N acts at point C on link 3 in the X-direction and a
counterclockwise torque T4  10 N  m acts on link 4. The masses and the second
moments of mass are
m2  m4  0.5 kg,
IG2  IG4  2 kg  m2 ,
m3  1 kg,
and
IG3  5 kg  m . The mass centers of links 2 and 4 are coincident with pins O2 and O4 and
2
the mass center of link 3 is coincident with pin A. Gravity acts into the page (in the
negative Z-direction) and friction can be neglected. The first and second–order kinematic
coefficients of the mass center of link 3 are X G 3  0.141 m/rad, YG3  0.141 m/rad,
X G3  0.141 m/rad 2 , and YG3  0.141 m/rad 2 . (i) Determine the acceleration of the
mass center of link 3. (ii) Draw the free-body diagrams for links 2, 3, and 4. List all
unknown variables. (iii) Determine the magnitudes and the directions of the internal reaction
forces F23 and F43 . (iv) Determine the magnitude and the direction of the input torque T2 .
Figure P14.32 A parallelogram four-bar linkage.
(i)
The acceleration of the mass center G3 can be written as
2
A  X G3 22  X G 3 2  0.141 m   2 rad/s     0.141 m 1 rad/s2   0.423 m/s 2


2
AGY3  YG3 22  YG3 2  0.141 m   2 rad/s     0.141 m 1 rad/s 2   0.705 m/s 2


(ii)
The free-body diagram of link 2 is shown in Fig. 1.
X
G3
Figure 1. Free-body diagram of link 2.
The sum of the forces in the X-direction acting on link 2 can be written as
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Ans.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
F
X
F12X  F32X  0
(1)
The sum of the forces in the Y-direction can be written as
 F Y F12Y  F32Y  0
(2)
The sum of the moments acting about the mass center G2 can be written as
 M G2 RB cos B  F32Y  RB sin B  F32X  T2  IG22
(3)
The unknown variables in Eqs. (1), (2), and (3) are F12X , F12Y , F32X , F32Y , and T2 . Therefore,
the total number of unknown variables is five and there are only three equations.
The free body diagram of Link 3 is shown in Fig. 2.
Figure 2. Free-body diagram of link 3.
The sum of the forces in the X-direction acting on link 3 can be written as
 F X F23X  F43X  FC  m3 AGX3
The sum of the forces in the Y-direction acting on link 3 can be written as
 F Y F23Y  F43Y  m3 AGY3
Ans.
(4)
(5)
The sum of the external moments acting about the mass center G3 can be written as
M
G3
RBB cos  BB  F23Y   RBB sin  BB  F23X   0
(6a)
The new unknown variables in Eqs. (4), (5), and (6a) are F43X and F43Y . Therefore, the
total number of equations is six and the total number of unknown variables is seven; that
is, F12X , F12Y , F32X , F32Y , T2 , F43X and F43Y . Note that  BB  0; therefore, Eq. (6a) can be
written as
RBB F23Y  0
(6b)
 
Therefore, either
RBB  0
or
F23Y  0
(6c)
The free body diagram of link 4 is shown in Fig. 3.
Figure 3. Free-body diagram of link 4.
The sum of the forces in the X-direction acting on link 4 can be written as
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Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
F
 F14X  F34X  0
X
(7)
The sum of the forces in the Y-direction acting on link 4 can be written as
 F Y F34Y  F14Y  0
(8)
The sum of the moments acting on link 4 about the mass center G4 can be written as
M
G4
RA cos  A  F34Y   RA sin  A  F34X   T4  I G4 4
(9)
The new unknown variables in Eqs. (7), (8), and (9) are F14X and F14Y . Therefore, there are
a total of nine equations and nine unknown variables; that is, F12X , F12Y , F32X , F32Y , T2 ,F43X ,F43Y ,
Ans.
F14X and F14Y .
(iii) The solution procedure will be the method of inspection. From Eq. (6c), the
reaction force
(10)
F23Y  0
Substituting Eq. (10) into Eq. (5), the reaction force
F43Y  m3 AGY3  F23Y  1 kg(0.705 m/s2 )  0  0.705 N
(11)
Rearranging Eq. (9), the reaction force F34X can be written as
F 
X
34

I G4  4  T4  RA cos  4  F34Y 
 RA sin  4
2kg  m 2 (1 rad/s 2 )  10 N  m  0.2 m  0.707  0.705 N 
0.2 m  0.707 
(12)
 55.87 N
Therefore, the total reaction force is
F43  55.87 N 180.720
X
23
Solving Eq. (7), the reaction force F
Ans. (13)
can be written as
F  m3 A  F  FC
X
23
X
G3
X
43
(14)
 1 kg  0.423 m/s 2    55.87 N   100 N  43.71 N
Therefore, the total reaction force is
F23  43.71 N1800
(iv)
Rearranging Eq. (3), the input torque T2 can be written as
Ans. (15)
T2  I G2  2  RB sin  2  F32X   RB cos  2  F32Y 
 2 kg  m
2
1 rad/s   0.2 m  0.707  43.71 N  0.2 m  0.707  0  8.18 N  m
2
The positive sign indicates that the input torque is counterclockwise.
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Ans.
Ans.
(16)
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.33 For the mechanism in the position shown, the velocity and acceleration of link 2 are
V2  10ˆi m/s and A2  10ˆi m/s2 , respectively. The length of link 3 is RCG3  3.5 m and
the distance ROG3  2.5 m. The first- and second-order kinematic coefficients of link 3
are 3  0.200 rad/m and 3  0.1386 rad/m2 . The force FC  10 N acts in the
negative Y direction at point C on link 3 and the line of action of an unknown force P
acting on link 2 is parallel to the X-axis as illustrated in Fig. P14.33. The masses and the
second moments of mass of links 2 and 3 are m2  3 kg, m3  5 kg, I G2  1.5 kg  m2 ,
and I G3  7.5 kg  m2 . Gravity acts in the negative Y-direction and there is no friction in
the mechanism. (i) Draw the free-body diagrams of links 2 and 3. (ii) Write the
governing equations for links 2 and 3. List all unknown variables. (iii) Determine the
magnitudes and directions of all internal reaction forces. (iv) Determine the magnitude
and direction of the force P acting on link 2. (v) Indicate the point(s) of contact of link 2
with ground link 1.
Figure P14.33 A planar mechanism.
(i)
The free-body diagram of link 2 is shown in Fig. 1.
Figure 1. Free-body diagram of link 2.
(ii)
Ans.
The sum of the forces acting on link 2 in the X-direction can be written as
Ans. (1)
 F X F32X  P  m2 AGX2
Note that the direction of the external force P is assumed to be in the positive X-direction.
The sum of the forces acting on link 2 in the Y-direction can be written as
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Theory of Machines and Mechanisms, 4e
F
Uicker et al.
Y
F12Y
 F32Y W2  m2 AGY
Ans. (2)
2
Note that AGX2  10 m/s2 and AGY2  0 .
The sum of the moments acting on link 2 about the mass center G2 can be written as
M
R12X F12Y  RG3G2W2  IG3 2  m2 ( RGXG AGY2  RGY G AGX )
G2
3 2
3 2
Ans. (3)
2
Therefore, there are three equations and five unknowns for the free-body diagram of link
2. The unknowns are the three reaction forces F32X , F32Y , F12Y , the distance R12X (that is, the
location of F12Y ), and the external force P.
The free-body diagram of link3 is shown is Fig. 2.
Figure 2. Free-body diagram of link 3.
Ans.
The sum of the forces acting on link 3 in the X-direction can be written as
 F X F23X  F13X  m3 AGX3
Ans. (4)
The sum of the forces acting on link 3 in the Y-direction can be written as
 F Y F23Y  F13Y  FC  W3  m3 AGY3
Ans. (5)
Note that the acceleration of the center of gravity of link 3 is the same as the acceleration
of link 2 (that is, AGX3  A2  10 m/s2 and AGY3  0 ).
The sum of the moments acting on link 3 about the center of mass G3 can be written as
M
G3
R32 cos 3 F13Y  R32 sin 3 F13X  RC cos 3 FC  I G33
X
13
Ans. (6)
Y
13
Equations (4), (5) and (6) contain two new unknowns, F and F . Therefore, there are
a total of six equations and seven unknowns.
Since links 1 and 3 have contact at a pin in a slot, the direction of the reaction force must
be perpendicular to the slot. This means only the magnitude of the reaction force F13 is
unknown.
The X and Y components of this reaction force can be written as
F13X  F13 cos(3  90) and F13Y  F13 sin(3  90)
(7a)
This provides a seventh equation allowing the seven unknowns to be solved. Also, since
the internal reaction force F13 is perpendicular to the slot then Eq. (6) can be written as
R32 F13  RC cos 3 FC  I G33
(7b)
The angular acceleration of link 3 can be written as
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Theory of Machines and Mechanisms, 4e
Uicker et al.
3  3 R2  3 R22
(8a)
where R2  V2  10 m/s and R2   A2  10 m/s2 . This agrees with the observation that
the input velocity R2 must be positive; that is, the vector R 2 is increasing in length for
this position. Substituting this information and the known kinematic coefficients for link
3 (that is, 3  0.200 rad/m and 3  0.1386 rad/m2 ) into Eq. (8a), the angular
acceleration of link 3 is
(8b)
3  (0.200 rad/m)(10 m/s2 )  (0.1386 rad/m2 )(10 m/s)2  15.86 rad/s2
Note that the angular acceleration of link 3 has a negative value; that is, the angular
acceleration of link 3 is clockwise. Also, note that the angular velocity of link 3 is
counterclockwise for this position.
(iii) Substituting the known values into Eq. (7b) gives
(9a)
(2.5 m) F13  (3.5 m) cos(30)10 N  (7.5 kg  m2 )(15.86 rad/s2 )
Therefore, the reaction force is
Ans. (9b)
F13   35.46 N
The negative sign indicates that the internal reaction force F13 is acting downward; that
is, in the opposite direction to the assumed direction shown in Fig. 2. Therefore, the
point of contact between link 3 and link 1 is on the top side of the ground pin O.
Substituting the known values into Eq. (5) gives
(10a)
F23Y   35.46 N  sin(60)  10 N  (5 kg)(9.81 m/s2 )  0
Therefore, the reaction force is
F23Y  89.76 N
Substituting the known values into Eq. (4) gives
F23X   35.46 N  cos(60)  (5 kg)(10 m/s 2 )
Ans. (10b)
(11a)
Therefore, the reaction force is
F23X  67.73 N
(iv)
Substituting the known values into Eq. (1) gives
67.73 N  P  (3 kg)(10 m/s2 )
Therefore, the applied force is
P  97.73 N
Substituting the known values into Eq. (2) gives
F12Y  89.76 N  (3 kg)(9.81 m/s2 )  0
Therefore, the reaction force is
F12Y  119.19 N
Substituting the known values into Eq. (3) gives
R12X (119.19 N)  RG3G2 (3 kg)(9.81 m/s 2 )  0
(v)
Ans. (11b)
(12a)
Ans. (12b)
(13a)
Ans. (13b)
(14a)
Therefore, the distance is
R12X  0.25RG3G2
(14b)
The negative sign indicates that the location of the internal reaction force F12Y is to the left
of the mass center of link 3. Given that the distance
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Theory of Machines and Mechanisms, 4e
Uicker et al.
RG3G2  0.5 m
(15)
Then the distance from the mass center of link 3 to the point of application of the normal
force is
R12X  0.25  0.5 m   0.125 m
Ans. (16)
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Theory of Machines and Mechanisms, 4e
Uicker et al.
14.34 Consider the slider-crank mechanism of Problem 14.5. The designer proposes to modify
this mechanism by including a linear spring and a viscous damper as illustrated in Fig.
P14.34. The spring, with a stiffness kS  3.5 kN/m and an unstretched length r0  75 mm,
is attached from the ground to point E on the input link 2. In the given position, the
spring is parallel to the X-axis. The damper with coefficient C  1.25 kN  s/m is attached
between the ground pivot O2 and pin B on link 4. At the input position 2  45 , the
motor driving input link 2 is applying a torque T2  6.7 N  m ccw, causing the link to
rotate with an angular velocity ω2  100 rad/s ccw and an angular acceleration
α 2  10 rad/s2 ccw. An external horizontal force FB is acting at pin B on link 4, as
illustrated in Fig. P14.34. The variable positions, velocities, and accelerations of the
mechanism have been determined and are provided in Table P14.34.
Table P14.34
2
3
RAO2
deg
deg
45  10.18
mm
75
RBA
mm
300
REO2
mm
125
3
rad/s
 17.96
V4
m/s
-6.25
3
A4
2
rad/s
 1736.20
m/s 2
-534
Assume: (i) Gravity acts vertically downward (that is, in the negative Y-direction). (ii)
The location of the center of mass of link 2 is coincident with the ground pivot O2 and
the center of mass of link 4 is coincident with pin B. The location of the center of mass
of link 3 is as indicated in Fig. P14.5. (iii) The effects of friction in the mechanism can
be neglected. (iv) The weight and mass moment of inertia of each link are as given in
Fig. P14.5. Determine: (i) the first- and second-order kinematic coefficients of the
mechanism that are necessary for the power equation. (ii) the equivalent mass moment of
inertia of the mechanism. (iii) the equation of motion for the mechanism. (iv) the
magnitude and direction of the external force FB acting on link 4 when the mechanism is
in the given position.
Figure P14.34. The mechanism modified from Problem 14.5.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
The vectors for a kinematic analysis of the mechanism are shown in Fig. 1.
Figure 1. The vectors for analysis of the mechanism.
(i)
The first-order kinematic coefficient of link 3 can be written as
 17.96 rad/s
 0.179 6 rad/rad
3  3 
100 rad/s
2
The first-order kinematic coefficient of link 4 can be written as
V
6.25 m/s
(1)
 62.5 mm/rad
R4  4 
2 100 rad/s
The angular acceleration of link 3 can be written as
3  322  3 2
Rearranging this equation, the second-order kinematic coefficient of link 3 can be written
as
3  3 2 (1 736.20 rad/s2 )  (0.1796 rad/rad)(10 rad/s2 )


 0.173 8 rad/rad 2
3 
2
2
2
(100 rad/s)
Similarly, the linear acceleration of link 4 can be written as
A4  R422  R4 2
Therefore, the second-order kinematic coefficient of link 4 is
A  R
534 000 mm/s 2  (625 mm/rad)(10 rad/s 2 )
R4  4 2 4 2 
 54 mm/rad 2
2
(100 rad/s)
2
Since the mass center of link 2, G2, is located at the fixed pivot O2, their first- and
second-order kinematic coefficients are
xG 2  0 , xG2  0
yG 2  0 , and yG2  0 .
To determine the first- and second-order kinematic coefficients for the center of mass of
link 3: The vector loop for point G3, see Fig. 1, can be written as
??
I
C
R G3  R 2  R 33
where the magnitude of the vector R 33 is given as 112.5 mm and the angle 33  3 . This
implies that the first-order kinematic coefficient 33  3 and the second-order kinematic
coefficient 33  3 . The X and Y components of the above equation are
X G3  R2 cos2  R33 cos3 =163.76 mm
YG3  R2 sin 2  R33 sin 3  33.145 mm
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Theory of Machines and Mechanisms, 4e
Uicker et al.
The first-order kinematic coefficients of the center of mass of link 3 are
X G 3   R2 sin 2  R33 sin 33  56.605 mm/rad
YG3  R2 cos 2  R33 cos 33  33.145 mm/rad
The second-order kinematic coefficients of the center of mass of link 3 are
X G3   R2 cos2  R33 cos332  R33 sin 33  53.148 mm/rad 2
YG3   R2 sin 2  R33 sin 332  R33 cos33  33.145 mm/rad 2
The first- and second-order kinematic coefficients of the center of mass of link 4 are
xG4  R4  53.340 mm/rad2
xG 4  R4  62.558 mm/rad
yG 4  0
yG4  0
The first-order kinematic coefficient for the damper is
RC  R4  62.558 mm/rad
The first-order kinematic coefficient for the spring can be obtained from the vector loop
for point E, see Fig. 1; that is,
C
??
R E  R 22
where the magnitude of the vector R 22 is given as 5 in and the angle 22  2  180  225 .
This implies that the first-order kinematic coefficient 22  1 rad/rad . The X component
of point E is
X E  R22 cos22 =  88.388 mm
Therefore, the first-order kinematic coefficient of point E is
X E   R22 sin 222  88.388 mm/rad
Note that the first-order kinematic coefficient for the spring is
RS   X E  88.388 mm/rad
4
(ii) To determine
 Aj : Note that
j 2
4
A
j 2
j
 I EQ ; (that is, the equivalent mass moment of
inertia) therefore, the units must be mm  N  s2 .
For link 2:
A2  m2 ( xG22  yG22 )  I G2 22
 m2 (0  0)   0.039 kg  m 2  (1 rad/rad) 2  0.039 kg  m 2  39.0 mm  N  s 2
For link 3:
A3  m3 ( xG23  yG23 )  I G332
 1.54 kg  [(0.056 605 m/rad)2  (0.033 145 m/rad) 2 ]   0.012 kg  m2  (0.179 6 rad/rad) 2
 0.007 013 kg  m2  7.013 mm  N  s 2
For link 4:
A4  m4 ( xG24  yG24 )  I G4 42
 1.30 kg  [(0.062 558 mm/rad)2  (0) 2 ]  0  0.005 088 kg  m 2  5.088 mm  N  s2
Therefore, the sum of the coefficients is
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Theory of Machines and Mechanisms, 4e
4
A
j 2
j
Uicker et al.
 A2  A3  A4  39.0 mm  N  s2  7.013 mm  N  s 2  5.088 mm  N  s 2  51.101 mm  N  s 2 Ans.
1 4 dAj
To determine  B j : Note that  B j  
; therefore, the units must be mm  N  s2 .
2 j 2 d 2
j 2
j 2
For link 2:
B2  m2 ( xG 2 xG2  yG 2 yG2 )  I G222  m2 (0  0)   0.039 kg  m2  (1 rad/rad)(0)  0
4
4
For link 3:
B3  m3 ( xG 3 xG3  yG 3 yG3 )  I G333
 1.54 kg  [(0.056 605 m/rad)(0.053 148 m/rad)  (0.033 145 m/rad)(0.033 145 m/rad)]
  0.012 2 kg  m 2  (0.179 6 rad/rad)(0.173 8 rad/rad 2 )
 0.002 560 kg  m 2  2.560 mm  N  s 2
For link 4:
B4  m4 ( xG 4 xG4  yG 4 yG4 )  I G4 4 4
 1.30 kg  [(0.062 558 m/rad)(0.053 340 in/rad 2 )  0]  0
 0.004 338 kg  m 2  4.338 mm  N  s 2
Therefore, the sum of the coefficients is
4
B
j 2
j
 B2  B3  B4  0  2.560 mm  N  s 2  4.338 mm  N  s 2  6.898 mm  N  s 2
The power equation can be written as
dT dU dW f
P


dt
dt
dt
The time rate of change of the kinetic energy can be written as
4
4
dT
  Aj   B j 3
dt j 2
j 2
Ans.
(iii)
(2)
where the generalized inputs for this problem are   2 ,   2  2 , and   2   2 .
Therefore, the time rate of change of the kinetic energy is
4
dT 4
  Aj2 2   B j 23  [51.101 mm  N  s2 (10 rad/s2 )  6.898 mm  N  s 2 (100 rad/s)2 ]2
dt j 2
j 2
That is,
dT
  69.491 N  m  2
dt
The time rate of change of the potential energy can be written as
4
dU
  m j gyG j  K S ( RS  RO ) RS  m3 gyG 2  K S ( RS  RO ) RS 2
3
dt
j 2
The time rate of change of the potential energy due to gravity is
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Theory of Machines and Mechanisms, 4e
dU g
dt
Uicker et al.
 m3 gyG 2  w3 yG 2
3
3
 1.54 kg  9.81 m/s 2  (0.033 145 m/rad)2   0.500 7 N  m  2
The time rate of change of the potential energy due to the linear spring can be written as
dU S
 K S ( RS  RS 0 ) RS 2
dt
 3.500 N/mm[125 mm  cos 45  75 mm](0.088 388 m/rad)2   4.141 8 N  m  2
Therefore, the time rate of change of the total potential energy is
dU dU g dU S


dt
dt
dt
  0.500 7 N  m  2   4.141 8 N  m  2   3.641 1 N  m  2
The time rate of change of the dissipative effects due to the viscous damper is
dW f
 CRC 2 2  1.25 N  s/mm(62.558 mm/rad)2 (100 rad/s)2   489.188 N  m  2
dt
Note that the time rate of change of the dissipative effects due to the viscous damper is a
positive value. This must always be true for this term on the right-hand side of the power
equation.
Therefore, the right-hand side of the power equation, see Eq. (2), can be written as
dT dU dW f


  69.491 N  m  2   3.641 1 N  m  2   489.188 N  m  2
(3)
dt dt
dt
  555.038 N  m  2
Note that the most influential term is the time rate of change of the dissipative effects due
to the viscous damper. This implies that the damping coefficient C  7 lb  s/in is a very
large value.
The left-hand side of the power equation, see Eq. (2), can be written as
(4)
P  T2  ω2  FB  VB   6.7 N  m  2  FBX ( xB 2 )
Note that the torque T2 is acting in the same direction as the angular velocity of link 2
(that is, counterclockwise) and the external force acting on the piston FB is assumed
positive when acting in the same direction as the velocity of the piston (link 4) (that is, in
the negative X-direction). The first-order kinematic coefficient of point B can be
obtained from the point path vector equation, or by noting that xB  R4 . From Eq. (1),
the first-order kinematic coefficient of link 4 is R4  62.5 mm/rad . Therefore, the firstorder kinematic coefficient of point B is
xB  62.5 mm/rad
(5)
Substituting Eq. (5) into Eq. (4) gives
P   6.7 N  m  2  FBX (0.062 5 m)2
(6)
Finally, equating the two equations, Eqs. (3) and (6), the power equation can be written as
(7)
 6.7 N  m 2  FBX (0.062 5 m)2  555.038 N  m 2
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Theory of Machines and Mechanisms, 4e
Uicker et al.
The equation of motion is obtained by dividing both sides of the power equation, Eq. (7),
by the input angular velocity 2 . Therefore, the equation of motion can be written as
Ans. (8)
6.7 N  m  (0.062 5 m) FBX  555.038 N  m
(iv)
Rearranging Eq. (8), the external force acting on the piston is
Ans.
FBX  8 773.4 N
The negative sign indicates that the external force acting on the piston (link 4) is acting to
the left, that is, in the negative X-direction. Therefore, the assumption made in Eq. (4)
was correct; that is, the force is acting in the same direction as the velocity of the piston
(link 4).
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Theory of Machines and Mechanisms, 4e
Uicker et al.
14.35 For the mechanism in the position illustrated in Fig. P14.35, the velocity and acceleration
of the input link 2 are V2  7ˆi m/s and A2  2ˆi m/s2 , respectively. The first- and
second-order kinematic coefficients of links 3 and 4 are 3  0, 4  1.0 rad/m,
3  1.0 rad/m2 , and 4  3.0 rad/m2 . The radius of massless link 4, which is rolling on
the ground link, is R4  1 m, the length of link 3 is RG2 A  6 m, and the radius of the
ground link is R1  2 m. The free length and spring rate of the spring, the damping
constant of the viscous damper, the masses, and the mass moments of inertia of links 2
and 3 (about their mass centers) are as shown in Table P14.35. Assume that gravity acts
in the negative Y direction and the effects of friction can be neglected. Determine: (i) the
kinematic coefficients rS , rC , xG 3 , yG 3 , xG3 , and yG3 . ; (ii) the equivalent mass of the
mechanism; (iii) the equation of motion for the mechanism in symbolic form; (iv) the
magnitude and direction of the horizontal external force P that is acting on link 2.
Table P14.35
Ro
K
C
m2
m3
I G2
I G3
m
N/m
Ns/m
kg
kg
3
25
15
1.20
0.80
kg-m2
0.25
kg-m2
0.10
Figure P14.35 A planar mechanism.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
The vectors for kinematic analysis of the mechanism are shown in Fig. 1.
Figure 1. Vectors for a kinematic analysis of the mechanism.
The vector loop equation for the mechanism can be written as
R 2  R3  R 7  0
where link 7 is an arm connecting the ground link to the center of the wheel, link 4. The
X and Y components of this equation are
R2 cos2  R3 cos3  R7 cos7  0
R2 sin 2  R3 sin 3  R7 sin 7  0
(i)
Differentiating these equations with respect to the input position gives
cos2  R3 sin 33  R7 sin 77  0
sin 2  R3 cos 33  R7 cos77  0
Differentiating again with respect to the input position gives
 R3 sin 33  R3 cos3 32 R7 sin 77  R7 cos7 72  0
R3 cos 33  R3 sin 3 32 R7 cos77  R7 sin 7 72  0
Solving for the first-order kinematic coefficients we get
3  0 and 7  0.333 rad/m
Solving for the second-order kinematic coefficients, they are
3  1 rad/m2 and 7  1 rad/m2
The rolling contact equation between the wheel, link 4, and the ground link can be written
as
  7
R1
 4
R4
1  7
The correct sign is negative because there is external contact between link 4 and the
ground link. Differentiating this equation with respect to the input position gives
    7
R1
 4
1   7
R4
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Theory of Machines and Mechanisms, 4e
Uicker et al.
Substituting the known values into this equation, the first-order kinematic coefficient for
link 4 is
4  1 rad/m
Differentiating the above equation with respect to the input position gives
    
R1
 4 7
R4
1 7
Substituting the known values into this equation, the second-order kinematic coefficient
for link 4 is
4  3 rad/m2
The first-order kinematic coefficient of the spring is
Ans.
RS  R2  1 m/m
Note that the answer is positive because the length of the spring increases for a positive
change in the input position. The first-order kinematic coefficient of the damper can be
written as
Ans. (1)
RC   R2  1 m/m
Note that the answer is negative because the change in the length of the vector RC  RAO1
decreases for positive change in the input position.
Check: Since link 4 is rolling on the ground link at point E then point E is the instant
center I14 . Therefore, the velocity of point A (which is directed in the positive Xdirection) can be written as
(2)
VA  4 RI14 A  (4 R2 ) RI14 A  (1 rad/m  7 m/s)(1 m)  7 m/s
The first-order kinematic coefficient of the damper is defined as
V
 7 m/s 
RC   A   
  1 m/m
R2
 7 m/s 
The correct sign is negative because the change in length of the vector RC  RAO1
decreases as the change in length of the input vector R2 increases.
The vector equation for the center of mass of link 3 can be written as
RG3  R 2  R33
The X and Y components of this equation are
xG3  R2 cos  2  R33 cos 3 and yG3  R2 sin 2  R33 sin 3
Differentiating these equations with respect to the input position, the first-order kinematic
coefficients for the center of mass of link 3 are
xG 3  cos 2  R33 sin 33 and yG 3  sin 2  R33 cos 33
(3)
Substituting the known data into this equation, the first-order kinematic coefficients for
the center of mass of link 3 are
xG 3  1 m/m and yG 3  0
Ans.
Differentiating Eq. (3) with respect to the input position, the second-order kinematic
coefficients of the center of mass of link 3 can be written as
xG3   R33 sin 33  R33 cos3 332 and
yG3  R33 cos 33  R33 sin 3 332
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Theory of Machines and Mechanisms, 4e
Uicker et al.
Substituting the known data into this equation, the second-order kinematic coefficients
for the center of mass of link 3 are
Ans.
xG3  1.5 m/m2 and
yG3  2.598 m/m2
The X and Y components of the acceleration of the mass center of link 3 are
AGX3  xG3 R22  xG 3 R2  (1.5 m/m2 )(7m/s)2  (1 m/m)(2 m/s2 )  75.5 m/s2
AYG3  yG3 R22  yG 3 R2  (2.598 m/m2 )(7 m/s)2  (0)(2 m/s2 )  127.302 m/s2
Note that the mass center of link 3 has X and Y components, therefore, the path of the
mass center of link 3 is not a horizontal straight line. The magnitude and direction of the
acceleration of the mass center of link 3 are
AG3  148.0 m/s2239.33
(ii)
The equivalent mass of the mechanism can be written as
n
n
j 2
j 2
mEQ   Aj   m j ( xG j 2  yG j 2 )  I G j  j2
Substituting known data into Eq. (4) gives
A2  1.2 kg[(1 m/m)2  02 ]  0.25 kg  m2 (0)2  1.2 kg
Substituting known data into Eq. (4) gives
A3  0.8 kg[(1 m/m)2  02 ]  0.10 kg  m2 (0)2  0.8 kg
Since link 4 is massless then
A4  0
Therefore, the equivalent mass of the mechanism is
mEQ  1.20 kg  0.80 kg  0  2.00 kg
(iii)
(4)
Ans.
The power equation for the mechanism can be written as
4
4
4
j=2
j 2
j 2
P  V2  [ Aj R2   B j R22 ]R2   m j gyG j R2  K ( RS  R0 ) RS R2  CRC2 R22
Assume that the force P acting on link 2 is positive in the same direction as the positive
input velocity. Canceling the input velocity V2  R2 , the equation of motion for the
mechanism can be written as
n
4
4
j 2
j 2
j 2
P   Aj R2   B j R22   m j gyG j  K (R S  R0 ) Rs  CRC2 R2
Ans. (5)
The coefficients B j can be written as
B j  m j ( xG j xG j  yG j yG j )  I G  j j
j
Substituting known data into Eq. (6) gives
B2  1.20 kg[(1 m/m)(0)  (0)(0)]  (0.25 kg  m2 )(0)(0)  0
Substituting known data into Eq. (6) gives
B3  0.80 kg[(1 m/m)(1.5 m/m)  (0)(2.598 m/m2 )]  0.10 kg  m2 (0)(1 rad/m2 )  1.20 kg/m
Since link 4 is massless,
B4  0
Therefore, the sum of the coefficients B j is
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(6)
Theory of Machines and Mechanisms, 4e
4
B
j 2
j
Uicker et al.
0  1.20 kg/m  0  1.20 kg/m
(7)
The effects of gravity can be written as
4
 m gy
j 2
j
Gj
 g[m2 yG 2  m3 yG 3  m4 yG 4 ]  9.81 m/s 2 [(1.20 kg)(0)  (0)(0)  (4 kg)(0)]  0
The terms for the spring and the damper in Eq. (5) are
K ( RS  R0 )Rs  25 N/m(5.196 m  3 m)(1 m/m)  54.9 N  m/m
(iv)
CRC2 R2  (15 N  s/m)(1 m/m)2 (7 m/s)  105 N  m/m
The external force P acting on link 2 can be written from Eq. (5) as
4
4
j 2
j 2
(8a)
(8b)
P  mEQ R2   B j R22   m j gyG j  K (R S  R0 ) Rs  CRc2 R2
Substituting the given data and Eqs. (1), (2), (7), and (8) into this equation, the magnitude
of the external force P acting on link 2 can be written as
Ans.
P  (2.0 kg)(2 m/s2 )  (1.2 kg/m)(7 m/s) 2  0  54.9 N  105 N  97.1 N
The positive sign indicates that the assumption that the external force P is acting to the
right (that is, in the same direction as the input velocity) is correct. Therefore, the external
force P is acting to the right.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
14.36 For the mechanism in the position illustrated in Fig. P14.36, link 3 is horizontal. The
constant angular velocity of the input link 2, which rolls without slipping on the inclined
plane, is 2  20 rad/s cw. The first- and second-order kinematic coefficients of links 3
and 4 are 3  0.125 rad/rad, R4  1.299 m/rad, 3  0, and R4  0.094 m/rad 2 . The
radius of link 2 is R  1.5 m, the length of link 3 is RBA  RG4G2  6 m, and RAOS  2.5 m.
The free length of the spring is 3 m, the spring rate is k  25 N/m, and the damping
constant of the viscous damper is C  15 N·s/m. The masses and mass moments of
inertia of links 2 and 4 are m2  7 kg, m4  4 kg, IG  18 kg·m2 , and IG4  22 kg·m2 .
2
Assume that the mass of link 3 is negligible compared with the masses of links 2 and 4,
the effects of friction can be neglected, and gravity acts vertically downward as illustrated
in Fig. P14.36. Determine: (i) the first- and second-order kinematic coefficients of the
mass centers of links 2 and 4, (ii) the equivalent mass moment of inertia of the
mechanism, and (iii) the magnitude and direction of the input torque acting on link 2.
Figure P14.36 A planar mechanism.
Vectors for the mass centers of links 2 and 4 are shown in Fig. 1.
Figure 1. Vectors for the mass centers of links 2 and 4.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
The vector equation for the mass center of link 2, see Fig. 1, can be written as
?


R G2  R 9  R 7
The X and Y components of this equation are
X G2  R9 cos 9  R7 cos 7 and YG2  R9 sin 9  R7 sin 7
Differentiating with respect to the input position  2 gives
X G 2  R9 cos 9 and YG2  R9 sin 9
(1)
Substituting 9  150 and R9  1.5 m/rad gives
X G 2  1.5cos150o  1.299 m
YG2  R9 sin 9  1.5sin150o  0.75 m
and
Ans.
The rolling contact equation between link 2 and the inclined plane (link 1) can be written
in terms of the first-order kinematic coefficients as
R9   22   2  1.5 m
Therefore, the second-order kinematic coefficient is
R9  0
Differentiating Eqs. (1) with respect to the input position  2 gives
Ans. (2)
X G2  R9 cos 9  0 and YG2  R9 sin 9  0
The vectors for the mass center of link 4 are shown in Fig. 1. The first-order kinematic
coefficients of the mass center of link 4 can be written as
Ans.
X G 4   R4  X G 2  1.299 m and YG4  0
The second-order kinematic coefficients of the mass center of link 4 can be written as
X G4   R4  0.094 m and YG4  0
Ans.
(ii)
The power equation for the mechanism can be written as
4
4
j 2
j 2
T2  ω2  [ I EQ 2   B j22 ]2   m j gyG j 2  K (rs  r0 )rs2  Crc222
The input torque is taken positive in the same direction as the given input angular velocity
(that is, clockwise). Then canceling the input angular velocity, the equation of motion for
the mechanism can be written as
4
4
j 2
j 2
T2  I EQ 2   B j22   m j gyG j  K (rs  r0 )rs  Crc22
The equivalent mass moment of inertia of the mechanism can be written as
I EQ   A   m j ( xG 2  yG 2 )  I G  2
j
j
j
j
j
Substituting known data for link 2 into this equation gives
A2  (7 kg)((1.299 m/rad)2  (0.75 m/rad)2 )  18 kg  m2 (1 rad/rad)2  33.75 kg  m2 /rad 2
Substituting known data for link 3 into Eq. (4) gives
A3  0
Substituting known data for link 4 into Equation (4) gives
A4  (4 kg)((1.299 m/rad)2  02 )  22 kg  m2 (0) 2  6.75 kg  m2 /rad2
Therefore, the equivalent mass moment of inertia of the mechanism is
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(3)
(4)
Theory of Machines and Mechanisms, 4e
Uicker et al.
4
I EQ   Aj 33.75 kg  m 2  0  6.75 kg  m 2  40.5 kg  m 2
Ans.
j 2
(iii)
The coefficient B j can be written as
B j  m j ( xG j xG j  yG j yG j )  I G j  j j
(5)
Substituting known data for link 2 into Eq. (5) gives
B2  7 kg((1.299 m/rad)(0)  (0.75 m/rad)(0))  (18 kg  m2 )(1 rad/rad)(0)  0
Substituting known values for link 3 into Eq. (5) gives
B3  0
Substituting known data for link 4 into Eq. (5) gives
B4  4 kg((1.299 m/rad)(0.094 m/rad 2 )  (0)(0))  (22 kg  m2 )(0)(0)  0.488 kg  m2
Therefore, the coefficient is
4
B
j 2
j
0  0  0.488 kg  m 2  0.488 kg  m 2
The effects of gravity can be written as
4
 m gy
j 2
j
Gj
 [m2 gyG 2  m3 gyG 3  m4 gyG 4 ]
 9.81 m/s 2 [(7 kg)(0.75 m/rad)  (0)( yG 3 )  (4 kg)(0)]  51.5 N  m/rad
The velocity of the mass center of link 2 down the inclined plane is
VG2  2 R   20 rad/s 1.5 m   30 m/s
which agrees with the first-order kinematic coefficients in Eq. (2); that is,
 1.299 m/rad    0.75 m/rad  2  1.5 m/rad  2  30 m/s
VG2 
2
2
Therefore, the first-order kinematic coefficient of the spring is
rS  R  1.5 m / rad
and is negative because the length of the spring is decreasing for positive input motion.
Therefore
K (rS  r0 )rS  25 N/m(2.5 m  3 m)(1.5 m/rad)  18.75 N  m/rad
The first-order kinematic coefficient of the damper can be written as
rC  r4  1.299 m/rad
Therefore
Crc22  (15 N  s/m)(1.299 m/rad)2 (20 rad/s)  506.22 N  m/rad
From Eq. (3). the input torque can be written as
4
4
j 2
j 2
T2  I EQ 2   B j22   m j gyG j  K (rS  r0 )rS  CrC22
 (40.5 kg  m 2 )(0)  (0.488 kg  m 2 )(20 rad/s) 2  51.5 N  m  18.75 N  m  506.22 N  m Ans.
 240.77 N  m
The negative sign indicates that the torque is in the opposite direction to the input angular
velocity (which is specified as clockwise). Therefore, the input torque must be acting
counterclockwise.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
14.37 Figure P14.37 illustrates 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 to reduce the bearing
reactions, caused by the rotating eccentric cranks, to zero. If these weights are to be
mounted 75 mm from the axis of rotation, how much must they weigh?
50
15
15
15
50
m
m
15
0
0m
m mm
m
0m
5 m
745
m
m
50
m
mm
m
50
75
M
M
0m
mm
m
m
mm
y
A
  50 mm  mB rB2 2   200 mm  mC rC2 2   500 mm  mE rE2 2   650 mm  mF rF2 2   700 mm  FG  0
y
G
  700 mm  FA   650 mm  mB rB2 2   500 mm  mC rC2 2   200 mm  mE rE2 2   50 mm  mF rF2 2  0
Dividing by  50 mm   2 and substituting numeric values gives
5625 mm  m   400500 mm  N    73125 mm  m
  73125 mm  m   400500 mm  N    5625 mm  m
2
2
2
2
B
F
0
F
0
2
B
2
Solving simultaneously gives
wB  wF  mB  5.932 N
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Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.38 Figure P14.38 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
50
15 0
0m
30
50
mm
mm
m
m
50 m
m
0m
15
50
M
y
A
y
F
mm

 


  
  500ˆi mm  ×  kˆ  m r    650ˆi mm  ×  sin  ˆj  cos  kˆ  m r   0
  650ˆi mm  ×  sin  ˆj  cos  kˆ  m r    500ˆi mm  ×  ˆj m r 
  200ˆi mm  ×  kˆ  m r    50ˆi mm  ×  sin  ˆj  cos  kˆ  m r   0
 50ˆi mm × sin  B ˆj  cos  Bkˆ mB rB2 2  200ˆi mm × ˆj mC rC2 2
2
D D
M
m
2
E
B
2
D D
B
2
B B
2
E E
E
2
2
2
C C
2
E
E
2
E E
2
2
Dividing by  50 mm   2 , substituting numeric values, and equating vector components
gives

5625 mm2  mB cos  B   667500 N  mm2    73125 mm2  mE cos  E  0
 5625 mm  m sin    267000 N  mm   73125 mm  m sin   0
 73125 mm  m cos   267000 N  mm    5625 mm  m cos  0
  73125 mm  m sin    667500 N  mm    5625 mm  m sin   0
2
2
B
2
B
2
2
B
2
B
B
E
E
E
2
B
2
E
2
E
E
Solving simultaneously gives
mB cos  B  2.968 N , mB sin  B  8.9 N , mE cos  E  8.9 N , mE sin  E  2.968 N
mB  9.38 N ,  B  108.43 , mE  9.38 N ,  E  161.57
Ans.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
14.39 Solve Problem 14.38 with the angle between the two throws reduced from 90 to 0.
M
y
A

 


  
  500ˆi mm  ×  kˆ  m r    650ˆi mm  ×  sin  ˆj  cos  kˆ  m r   0
  650ˆi mm  ×  sin  ˆj  cos  kˆ  m r    500ˆi mm  ×  kˆ  m r 
  200ˆi mm  ×  kˆ  m r    50ˆi mm  ×  sin  ˆj  cos  kˆ  m r   0
 50ˆi mm × sin  B ˆj  cos  Bkˆ mB rB2 2  200ˆi mm × kˆ mC rC2 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
2
E E
E
2
2
Dividing by  50 mm   2 , substituting numeric values, and equating vector components
gives

5625 mm2  mB cos  B   934500 N  mm2    73125 mm2  mE cos  E  0
 73125 mm  m sin   0
 5625 mm  m sin 
 73125 mm  m cos  934500 N  in    5625 mm  m cos  0
  73125 mm  m sin 
  5625 mm  m sin   0
2
2
B
B
2
E
2
B
E
E
2
B
Solving simultaneously gives
mB cos  B  11.868 N ,
mE sin  E  0.000 lb
2
B
2
E
B
mB sin  B  0.000 N ,
mB  11.868 N ,  B  180.00 , mE  11.868 N ,  E  180.00
© Oxford University Press 2015. All rights reserved.
E
E
mE cos  E  11.868 N ,
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.40 The connecting rod illustrated in Fig. P14.40 weighs 35.155 N and is pivoted on a knife
edge and caused to oscillate as a pendulum. The rod is observed to complete 54.5
oscillations in 1 min. Determine the mass moment of inertia of the rod about its own
center of mass.
43.75 mm
56.25 mm
350 mm
rG  100 mm , w  35.155 N ,   60 s 54.5 cycles  1.101 s/cycle
From Eq. (14.101),
2
2
IO  mgrG  2    35.155 N 100 mm 1.101 s/cycle 2 rad/cycle   107.91 N  mm  s2
IG  IO  mrG2  107.91 N  mm  s2   35.155 N 9650 mm/s 2  100 mm   71.53 N  mm  s2
2
Ans.
14.41 A gear is suspended on a knife edge at the rim as illustrated in Fig. P14.41 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 axis of rotation are coincident. If the weight of the gear is 178
N, find the mass moment of inertia and the radius of gyration of the gear.
200mm
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
From Eq. (14.101),
2
2
IO  mgrG  2   178 N  200 mm 1.08 s/cycle 2 rad/cycle   10515.75 N  mm  s2
IG  IO  mrG2  1051.75 N  mm  s2  178 N 9650 mm/s2   200 mm   313.95 N  mm  s2
2
Ans.
kG  IG m  313.95 N  mm  s
2
178 N 9650 mm/s   130.45 mm
2
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.42 Figure P14.42 illustrates 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
 2
l
I  Wbl 
 
 4 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 Wbl  W  Wb  rG and the mass moment of inertia is
IO  I  Wbl 2 g . Now, using Eq. (14.101)
2
Wbl 2 W  Wb 
   Wbl

I
grG 
 
4 2
g
g
 2 
Rearranging this we get
 2
l
I  Wbl 
 
 4 2 g 


2
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Q.E.D.
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.43 If the weight of the pendulum arm is not neglected in Problem 14.42, 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
 2 
W  l
W 
I l
Wb  a    Wb  a  

2
2  g
3  
 4 
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
The total mass moment of inertia is
2
Wbl 2 Wal 2 Wa  l  
Wbl 2 Wal 2
IO  I 



  I
3g
g
g  2  
g
 12 g
Using Eq. (14.101)
W l2 W l2 
W l   
I  b  a   Wbl  a  

g
3g 
2   2 
which can now be rearranged to read
 2 
W  l
W 
I  l  2  Wb  a    Wb  a  
2  g
3 
 4 
2
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Q.E.D.
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.44 Wheel 2 in Fig. P14.44 is a round disk that 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 2 with link 3 fully
extended. Develop an expression for the angular velocity 3 that link 3 would acquire if
the wheel were suddenly stopped.
Consider link 3 alone. The momentum before and after are
L  m3  R  r  2ˆi
L  m3r3ˆi
The angular momentum before and after about point G are
HG   m3 3 r 23kˆ
HG   m3 3 r 22kˆ
The angular momentum before and after about point B are
H B  H G  rˆj× L
H  H  rˆj× L
B
G
  m3 3 r 22kˆ  m3  Rr  r 2  2kˆ
 m3  Rr  4 r 2 3 2kˆ
  m3 3 r 23kˆ  m3r 23kˆ
  4 m3 3 r 23kˆ
Since there is no angular impulse on link 3 about point B, H B  H B
3  1  3R 4r  2
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.45 Repeat Problem 14.44, 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 3 is zero?
Consider link 3 alone. The momentum before and after are
L  m3  R  r  2ˆi
L  m3r3ˆi
The angular momentum before and after about point G are
HG   m3 3 r 23kˆ
HG   m3 3 r 22kˆ
The angular momentum before and after about point B are
H B  H G 2  rˆj× L
H B  H G1  rˆj× L1
  m3 3 r 22kˆ  m3  Rr  r 2  2kˆ
  m3 3 r 23kˆ  m3r 23kˆ
 m3   Rr  4 r 2 3 2kˆ
  4 m3 3 r 23kˆ
Since there is no angular impulse on link 3 about point B, H B  H B
3  1  3R 4r  2
3  0 for r  3R 4
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.46 Figure P14.46 illustrates a planetary gear-reduction unit that utilizes 3.63mm/tooth spur
gears cut on the 20 full-depth system. All parts are steel with density 8 10 N/mm3 .
The arm is rectangular and is 100 mm wide by 350 mm long with a 100 mm diameter
central hub and two 75 mm diameter planetary hubs. The segment separating the planet
gears is a 0.5 × 100 mm diameter cylinder. The inertia of the gears can be obtained by
treating them as cylinders equal in diameter to their respective pitch circles. The input to
the reducer is driven with 25 with a torque of 297 N-m at 600 rev/min. The mass
moment of inertia of the resisting load is 648.5875 N  mm  s2 . 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?
37.5 mm
25 mm
37.5 mm
34.375
mm
37.5 mm
34.375mm
34.375 mm
31.25
337.5 mm
40.625 mm
12. 5 mm
5 0 mm
25 mm
50 mm
40.625 mm
268.75 mm
200 mm
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200 mm
Theory of Machines and Mechanisms, 4e
Uicker et al.
400 mm
mN1 3.63 mm/tooth 104 teeth
mN 2 3.63mm/tooth  35 teeth

 188.76 mm R2 

 63.525 mm
2
2
2
2
mN3 3.63 mm/tooth  52 teeth
mN 4 3.63 mm/tooth 17 teeth
R3 

 94.38 mm R4 

 30.85 mm
2
2
2
2
Tout  T1A  297 N  m
Assume a symmetric arrangement of (typically = 3) planets, symmetrically arranged on
an pronged planet carrier. Then the tangential component of the force F2 A for each planet
is
T
297 N  m
 2371.8 N
F2tA  out 
RA 125.23 mm 
For equilibrium of each planet
 M 2  R3 F43 cos 20  R2 F12 cos 20  0 R3 F43  R2 F12
R1 
F
 F12 cos 20  F43 cos 20  FAt 2  0
F12 
R3
FAt 2
R

R

cos
20
 2 3
t
2
F12  F43  1  R2 R3  F12  FAt 2 cos 20
F43   R2 R3  F12
F12  1508.5 / N
F43  1014.6 N
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Theory of Machines and Mechanisms, 4e
F
r
2
  F12 sin 20  F43 sin 20  FAr2  0
FAr2   F12  F43  sin 20
FA2 
Uicker et al.
F   F 
2
t
A2
r 2
A2
FAr2  169.1
FA2  2376.3
N
N
Ans.
Assuming that the m-pronged planet carrier is arranged symmetrically, there is no net
Ans.
force on the input or output shafts;
F14  F1A  0
The input torque is
Ans.
Tin  T14  28.92 N  m
 M 4  R4 F34 cos 20  T14  0
Balancing the moments on the casing
 M1  R2 F21 cos 20   400 mm F11  0
F11  222.5 N
This force F11 must be absorbed by the mounting bolts.
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Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.47 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
14.46 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  8 105 N/mm3  35037.5   2212.5  2 37.5212.5  215.75287.5  133 N
m2  8 105 N/mm3  62.5234.5   92.75234.5   50212.5   18.75281.25  109.47 N
m4  8 105 N/mm3  30.25234.5  8 N
The centroidal mass moments of inertia are:
I A  8 105 N/mm3[4  350  375  3502  1002  12   50412.5 2  2 37.5412.5 2
 2 37.5212.5 123.252  2 15.75487.5 2  2 15.75287.5 123.52 ]  1.479 N  m 2
I 2  8 105 N/mm3  62.5434.5 2   92.75434.5 2   50412.5 2   18.75481.25 2   0.3949 N  m2
I 4  8 105 N/mm3  30.25434.5 2   3.6434 103 N  m2
The angular accelerations are found by the tabular method (see Section 9.7):
Step Number
Frame 1 Arm A Planets 2, 3
Sun 4
1. Gears fixed to arm




2. Arm fixed

0
 104 35
104 35 52 17 
0

  69 35
 6 003 595
3. Total
The output torque is Tout  292.14 N  m .
The input torque is Tin  2.5  28.92 N  m   72.31 N  m .
Balancing the sun gear and input shaft:
 M 4  2R4 F34t  Tin  I44
2  30.35 mm  F34t  72.31 N  m   3.6434 103 N  m2 9650 mm/s 2   6003 595 
F34t  1190.8  0.0627
Balancing the arm and output shaft:
 M A  Tout  2RA F2tA  I A A
292.14 N  m  2 123.2 mm  F2tA = 1.479 N  m2 9650 mm/s 2  
F2tA  1185.48  0.6225
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Theory of Machines and Mechanisms, 4e
Uicker et al.
Balancing a typical planet:
 M 2  R3 F43t  R2 F12t  I 22
 92.85 mm 1190.8  0.0627    62.5 in  F12t   0.3949 N  m2
9650 mm / s2   69 35 
F12t  1769.3  1.3839
F
t
2
F12t  F43t  FAt 2  m2 AG2
1769.3  1.3839   1190.8  0.0627   1185.48  0.6225   109.47 N
9650 mm/s2   21.93 
  512 rad/s2
Now, reassembling the above results,
F34t  1190.8  0.6274  1158.78 N
F34r  F34t tan 20  421.86 N
F12t  1769.3  1.3839  1060.88 N
F12r  F12t tan 20  386.26 N
F2tA  1158.78  0.6225  1504 N
F2rA  386.26  421.86  35.6 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
 1504.5 N
The forces in the mounting bolts to restrain the unbalanced frame moment are:
F11  2R1F21t 400 mm  985.23 N
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Ans.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.48 The gear-reduction unit of Problem 14.46 is running at 600 rev/min when the motor is
suddenly turned off, without changing the resisting-load torque. Solve Problem 14.46 for
this condition.
Here we can use the free-body diagrams from Problem 14.46 and the mass data and
angular motion relationships from Problem 14.47. Then, proceeding as in Problem 14.47,
but with Tin = 0, we balance the sun gear and input shaft:
 M 4  2R4 F34t  Tin  I44
 60.7 mm  F34t  3.6434 103 N  m2
9650 mm/s2   6003 595 
F34t  0.6274
Balancing the arm and output shaft:
 M A  Tout  2RA F2tA  I A A
0.2921 N  m  2 123.2 mm  F2tA = 1.4796 N  m2 9650 mm/s2  
F2tA  1185.48  0.6225
Balancing a typical planet:
 M 2  R3 F43t  R2 F12t  I 22
 92.84 mm  0.6274    62.5 mm  F12t   0.3949 N  m2
9650 mm/s 2   69 35 
F12t  1.3839
F
t
2
F12t  F43t  FAt 2  m2 AGt 2
1.3839    0.6274   1185.48  0.6225   109.47 N
  342 rad/s2
A  600 rev/min  62.8 rad/s
F
r
2
9650 mm/s2   123.2 
AGr 2   RAA2  486425 mm/s2
  F12r  F43r  FAr2  m2 AGr 2  109.47 N 9650 mm/s2  486425 mm/s2   5518 N
Reassembling the above results,
F34t  0.6279  21.36 N
F34r  F34t tan 20  8 N
F12t  1.3839  473.48 N
F12r  F12t tan 20  172.215 N
F2tA  1185.48  0.6225  972.77 N
F2rA  172.21  8  5682.65 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
 5762.75 N
The forces in the mounting bolts to restrain the unbalanced frame moment are:
F11  2R1F21t 400 mm  439.66 N
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Ans.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
14.49 The differential gear train illustrated in Fig. P14.49 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 module of gears 1 and 5 are both 3.175 mm/tooth, while the
module of gears 3 and 4 are both 4.23 mm/tooth. All gears have the 20 pressure angles
and are each 18.75 mm thick, and all are made of steel with density 7.9 103 g / mm3.
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
t
t
 F14
each of the bearings. (Hint: It is reasonable to assume through symmetry that F13
.
It is also necessary to recognize that only compressive loads, not tension, can be
transmitted between gear teeth.)
m2  7.98 103 g/mm3  1002 18.75  47.97 N
m3  7.92 103 g/mm3  752 18.75  26.96 N
IGyy2   47.97 N  42 2  0.2399 N  m2
IGxx3   26.96 N  32 2  0.0759 N  m2
IGyy3   26.96 N  752 4  0.0379 N  m2
IGzz3  IGyy3  0.0379 N  m2
m4  m3  26.96 N
IGxx4  IGxx3  0.0759 N  m2
IGyy4  IGyy3  0.0379 N  m2
IGzz4  IGyy4  0.0379 N  m2
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Theory of Machines and Mechanisms, 4e
Uicker et al.
m5  0
ω5  500ˆj rev/min  52.36ˆj rad/s
ω2  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 ,
α3  69.81 52.36kˆ  3 655kˆ rad/s2 , α 4  69.81 52.36kˆ  3 655kˆ rad/s2 , α5  0 .




Link 5: Note that we assume the mass of link 5 is negligible. Also, assuming no thrust
bearing at the fixed pivot, F15y  F35y  F45y  0 .
F
F
M
M
M
x
5
 F15x  F35x  F45x  0
z
5
 F15z  F35z  F45z  0
x
5
 M15x  d5 F15z  0
y
5
 M15y  4F35z  4F45z  0
z
5
 M15z  M 35z  M 45z  d5 F15x  0
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Theory of Machines and Mechanisms, 4e
Uicker et al.
7.275
7 .2 75
5.45
5. 45
5. 45
5. 45
7. 27 5
7.275
Link 4: Note that the forces on bevel gear teeth are related by Eq. (13.20) where, in this
case, cos  4  0.8 , sin  4  0.6 , and   20 . Noting that
A  ω × ω ×100ˆi mm  100 2ˆi mm  274150ˆi mm/s 2
5
G4


5
5
x
4
 5.45F14z  5.45F24z  F54x  m4 AGx4   26.96 N 9650 mm/s2  274150 mm/s2   765.4 N
y
4
 7.275F14z  7.275F24z  0
z
4
 F14z  F24z  F54z  0
F
F
F
F24z  F14z
Using Eqs. (14.110) for the moment equations,
 M 4x  75F14z  75F24z  0
M
M
y
4
0
z
4
 5.45F14z  5.45F24z  M 54z  IGzz4  4z  ( IGxx4  IGyy4 )4x4y  6460 mm/s2
M 54z  6460 mm/s2
Link 3: Again cos  3  0.8 , sin  3  0.6 , and   20 and using Eqs. (14.110) we get
similar results
AG3  ω5 × ω5 × 100ˆi in  10052ˆi mm  274150ˆi mm/s 2


x
3
 5.45F13z  5.45F23z  F53x  m3 AGx3   26.96 N 9650 mm/s2  274150 mm/s2   765.845 N
y
3
 7.275F13z  7.275F23z  0
z
3
  F13z  F23z  F53z  0
F
F
F
M
M
M
F23z  F13z
x
3
 75F13z  75F23z  0
y
3
0
z
3
 5.45F13z  5.45F23z  M 53z  IGzz33z  ( IGxx3  IGyy3 )3x3y  6460 mm/s2
M 53z  6460 mm /s 2
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Theory of Machines and Mechanisms, 4e
Uicker et al.
7.275
5.45
5.45
7.275
Link 2: This time cos  2  0.6 , sin  2  0.8 , and   20 .
F
F
F
M
M
M
x
2
 F12x  5.45F32z  5.45F42z  0
y
2
  F12y  7.275F32z  7.275F42z  0
z
2
 F12z  F32z  F42z  0
x
2
 d2 F12z  0
F12z  0
y
2
 M12y  100F32z  100F42z  0
F32z  F42z  133.5 N  m 100 m m  1335 N
z
2
 100  7.275F32z   100  7.275F42z   d2 F12x  0
Reviewing these again shows
F32z  F42z  667.5 N
Finally, collecting all results, we have:
F12  388.7ˆj N
F15  0
F35  1057.3ˆi  1335kˆ N
F  1057.3ˆi  1335kˆ N
45
F12x  0 , F12y  388.7 N
M12  133.5ˆj N  m
M  267ˆj N  m
15
M35  28.74kˆ N  m
M  28.749kˆ N  m
45
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Ans.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
14.50 Figure P14.50 illustrates 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 15.575 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 0.178 N/mm 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  ˆj . Make a graph of the height R versus the rotational speed
D
 rev/min, assuming that changes in speed are slow.
0
15
m
15
400 mm
m
0
40
0
m
m
m
m
The free-body diagram below shows only one of the arms, body 2, containing 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  400  2 150cos    400  300cos  mm
The total force in the spring is
k  RD  RD 0   1 lb/in  400 mm  300cos  mm  275 mm   22.25  53.4cos  N
Since this total force must balance two flyball arms, force FA of the free-body diagram is
FA  11.17  26.7 cos  N
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Theory of Machines and Mechanisms, 4e
Uicker et al.
Taking moments about point B,
150cos mm m2 300sin    2  150sin  mm m2 g  150sin  mm11.125  26.7cos N   0
Dividing by 150sin  mm
 53.4cos  m2 2  m2 g 11.125 N  26.7 cos N  0
Substituting values and rearranging we get
 0.484 N  s2  cos 2  26.7  26.7 cos  6 1  cos  N , cos 2  55.155 1  cos  rad2 /s2
  7.427 rad/s
1  cos 
1  cos 
 70.919 rev/min
,
cos 
cos 
RD  400  300cos  mm Ans.
mm
400
300
200
100
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Uicker et al.
Chapter 15
Vibration Analysis
15.1
Derive the differential equation of motion for each of the systems illustrated in Fig. P15.1
and write the formula for the natural frequency n for each system.
(a)
 F  k  x  y   mx
(b)
 F  F  cx  kx  mx
(c)
 F  cx  k1x  k2 x  mx
mx  kx  ky
Ans.
n  k m
mx  cx  kx  F
Ans.
n  k m
mx  cx   k1  k2  x  0
n 
 k1  k2 
m
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Theory of Machines and Mechanisms, 4e
(d)
(e)
(f)
 F  k x  k  x  y   mx
1
2
 F  c  x  y   kx  mx
Uicker et al.
mx   k1  k2  x  k2 y
n 
 k1  k2 
Ans.
m
mx  cx  kx  cy
n  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   k3k4 x   k3  k4 
k3k4
x  mx

k
3
4
 F  k x  k x  k
1
2

kk 
mx   k1  k2  3 4  x  0
k3  k 4 

kk
k1  k2  3 4
k3  k 4
n 
m
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Ans.
Theory of Machines and Mechanisms, 4e
15.2
Uicker et al.
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 we use a trial solution of:
x  A sin nt  B cos nt
with initial value of:
x(0)  B
(1)
x  n A cos nt  n B sin nt
x  0   n A
(2)
x  n2 A sin nt  n2 B cos nt
x  0   n2 B
(3)
x  n3 A cos nt  n3 B sin nt
x  0   n3 A
(4)
(a)
(b)
x  x0 , x  0 . Use Eqs. (1) and (2); A  0, B  x0
x  x0 cos nt
x  0, x  v0 . Use Eqs. (1) and (2); A  v0 n , B  0
Ans.
x   v0 n  sin nt
(c)
Ans.
x  x0 , x  a0 . Use Eqs. (1) and (3); B  x0 , B   a0 
2
n
These are inconsistent unless a0  n2 x0 . Second given condition is not useful.
One more initial condition required, such as x  0   v0 from which A  v0 n .
x   v0 n  sin nt  x0 cos nt
x  x02   v0 n  sin nt   where   tan 1  x0n v0 
2
(d)
x  x0 , x  b0 . Use Eqs. (1) and (4); B  x0 , A   b0 
x   b0 n3  sin nt  x0 cos nt
x  x02   b0 n3  sin nt   where   tan 1  x0n3 b0 
2
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Ans.
3
n
Ans.
Theory of Machines and Mechanisms, 4e
15.3
Uicker et al.
A system like Fig. 15.5 has m = 1 kg and
x  20cos 8 t   / 4  mm. Determine the following:
an
equation
of
motion
(a)
The spring constant k
(b)
The static deflection  st
(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)
n  k m  8 rad/s
k  n2 m  8 rad/s  1 kg   631.65 N/m
2
Ans.
(c)
F mg 1 kg   9.81 m/s 


 0.015 53 m  15.53 mm
631.65 N/m
k
k
  2 n   2 rad/rev  8 rad/s   0.250 s/rev
Ans.
(d)
f  1   4 rev/s  4 Hz
Ans.
2
(b)
(e)
 st 
Ans.
  8 t   4  8t  0.25  8  0.20   0.25   1.35 rad  243
x   8 rad/s  0.020 m  sin 8 t   4   0.503sin  243 m/s  0.448 m/s Ans.
x   8 rad/s   0.020 m  cos 8 t   4   12.633cos  243 m/s 2  5.735 m/s2 Ans.
2
(f)
F  kx   631.65 N/m  0.020 m  cos 243  12.633cos 243 N  5.735 N
x  0.020cos 243 m , x  0.503sin 243 m/s , x  12.633cos 243 m/s2
F  12.633cos 243 N
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Ans.
Theory of Machines and Mechanisms, 4e
15.4
Uicker et al.
The weight W1 in Fig. P15.4 drops through the distance h and collides with W2 with
plastic 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 W1 g  2 gh  W1  W2  v2 g
or v2  2 ghW1 W1  W2  . Therefore, at t   0 , x  0 , x  v2 .
Note that, at x  0 , the spring force 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 n  kg W . Then
x  A cos nt   B sin nt   W1 k
x   An sin nt   Bn cos nt 
and with the initial conditions stated above A  W1 k and B  v2 n . Therefore
x   W1 k  cos nt    v2 n  sin nt   W1 k 
Transforming to a single transient term we get
x  X sin nt     W1 k where
X
2
2
W1 k    v2 n 
W k 
and   tan 1  1 
 v2  n 
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Ans.
Theory of Machines and Mechanisms, 4e
15.5
Uicker et al.
The vibrating system illustrated in Fig. P15.5 has k1  k3  850 N/m , k2  1 850 N/m ,
and W  40 N . What is the natural frequency in hertz?
Springs 1 and 2 both experience 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
 k3 x  mx
 kk

mx   1 2  k3  x  0
 k1  k2

The natural frequency is
 k1k2

 k3 

k k

n   1 2
W g
850 N/m 1850 N/m   850 N/m
850 N/m  1850 N/m
40 N 9.81 m/s 2
 18.74 rad/s  2.983 Hz Ans.
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Theory of Machines and Mechanisms, 4e
15.6
Uicker et al.
Figure P15.6 illustrates a weight W = 80.1 N connected to a pivoted rod which is
assumed to be weightless and very rigid. A spring having a rate of k =13.35 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.
150 mm
300 mm
 M  a   ka  
W  W g  2
ka 2
a kg 150 mm


n 
2
W g
W 300 mm

2
n

W
2
g    ka 2   W
13.35 N/M   9650 mm/s2 
80.1 N
2 rad/rev
 0.313 s/rev
20.05 rad/s
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 20.05 rad/s
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
15.7 Figure P15.7 illustrates 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.
(a)
M W 
  a  k1a  a  k2a  W g  2
W
2
g    k1  k2  a 2  W   0

2

g   k1  k2  a
 1   0

W


2

g   k1  k2  a
Ans.
n 
 1

W


(b) 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
Ans.
a   k1  k2  a W
15.8
(a) Write the differential equation for the system illustrated in Fig. P15.8 and find
the natural frequency.
(b)
Find the response x if y is a step input of height y0 .
(c)
Find the relative response z  x  y to the step input of part (b).
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Theory of Machines and Mechanisms, 4e
(a)
 F  k x  k  x  y   mx
1
2
mx   k1  k2  x  k2 y
(b)
Uicker et al.
n 
 k1  k2 
m
Ans.
The complementary solution is
x  A cos nt  B sin nt
For a particular solution, arrange the equation to the form
x  n2 x   k2 m  y
Since, for a step input the right-hand side is constant, try x  C , x  0 .
n2C   k2 m  y
C  k2 y  k1  k2 
The complete solution is x  x  x
x  A cos nt  B sin nt  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  n A sin nt  n B cos nt
At t  0 , x  0
0  n A  0   n B 1
B0
Therefore, the complete response is
x   k2 y0  k1  k2  cos nt  k2 y0  k1  k2 
x
(c)
k2 y0
1  cos nt 
k1  k2
k  k  y
k2 y0
1  cos nt   1 2 0
k1  k2
k1  k2
k y  k y cos nt
z 1 0 2 0
k1  k2
Ans.
z  x y 
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Ans.
Theory of Machines and Mechanisms, 4e
15.9
Uicker et al.
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 that follows.
(b)
What are the amplitudes in each era?
(c)
Sketch a time plot of the displacement.
(a)
n 
 35 000 N/m  1.2 kg   171 rad/s
1.2 kg  x   35 000 N/m x  50 N
First era: 0  t  0.040 s :
From Eq. (15.21)
x   F k 1  cos nt    50 N 35 000 N/m 1  cos171t 
x   0.001 429 m 1  cos171t 
x  1.429 mm 1  cos171t 
Ans.
Also x   0.244 m/s  sin171t   244 mm/s  sin171t
At the end of the first era t  0.040 s , nt  6.831 rad  391.4
x  0.000 209 m  0.209 mm ,
x  0.127 m/s  127 mm/s
Second era: t  0.040 s :
From Eqs. (15.16) and (15.17)
X 0  x02   v0 n  
2
1.2x  35 000x  0
 0.209 mm 2  127 mm/s 171 rad/s 2  0.773 mm
  tan 1  v0 n x0   tan 1 127 mm/s  171 rad/s  0.209 mm   74.3
(b)
x  X 0 cos nt      0.773 mm  cos 171t  74.3
Ans.
First era; 0  t  0.040 s :
Second era; t  0.040 s :
Ans.
Ans.
X  1.429 mm
X  0.773 mm
(c)
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
15.10 Figure P15.10 illustrates a round shaft whose torsional spring constant is kt in  lb/rad
connecting two wheels having mass moments of inertia I1 and I 2 . Show that the system
is likely to vibrate torsionally with a frequency of
n 
kt  I1  I 2 
I1 I 2
Designating the angular positions of the two wheels by 1 and  2 , respectively, and
summing moments on each, we get
I11  kt1  kt 2  0
 M1  kt 1 2   I11
M
2
 kt 1   2   I 2 2
I 2 2  kt1  kt 2  0
Next, we assume a solution of the form
 j  C j cos nt   
for each inertia with j = 1, 2.
Substituting these gives
n2 I1C1 cos nt     kt C1 cos nt     kt C2 cos nt     0
n2 I 2C2 cos nt     kt C1 cos nt     kt C2 cos nt     0
Dividing each by cos nt    and writing these in matrix form they become
 n2 I1  kt
  C1 
kt

  0
2
 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,
 n2 I1  kt  n2 I2  kt    kt  kt   0
This expands to a quadratic equation in  n2
I1I 2 n2   kt  I1  I 2  n2  0
2
Solving this, we get four roots:
n  0
n  
kt  I1  I 2 
I1 I 2
Q.E.D.
Note that the other frequency of n  0 shows the capability for rigid body rotation
since the entire shaft with wheels is free to rotate.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
15.11 A motor is connected to a flywheel by a 15.625 mm diameter steel shaft 900 mm long, as
shown. Using the methods of Chapter 15, it can be demonstrated that the torsional spring
constant of the shaft is 531.1 N  M/rad. The mass moments of inertia of the motor and
flywheel are 2.712 and 6.328 N  M  s2 , respectively. The motor is turned on for 2 s,
and during this period it exerts a constant torque of 22.6 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 o f 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   , T  22.6 NM , and t  2 s . Substituting,
 I1  I 2    H 0  0  22.6 NM  dt
9.04 NM  s2     22.6 NM  t
t
   2.5 rad/s2  t for 0  t  2 s
(b)
From Problem 15.10
n 
(c)
k t  I1  I 2 
I1 I 2

At t = 2 s   5.0 rad/s  47.75 rev/min
 531.1 NM/rad   2.712 NMs 2  6.328 NMs 2 
 2.712 NMs  6.328 NMs 
2
2
 16.726 rad/s Ans.
First era; 0  t  2 s :
The differential equations are:
I11  kt1  kt 2  T
I 2 2  kt1  kt 2  0
After using the conditions that, at t = 0, 1   2  0 the solutions become
I2
T  I2 t 2 
B cos nt 
  
I1
I1  I 2  kt 2 
Tt 2
 2  A  B cos nt 
2  I1  I 2 
1  A 
Then using the conditions that, at t = 0, 1   2  0 , we find
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
B  A 
Uicker et al.
I1 I 2
T
kt  I1  I 2 2
and the solutions become
I2
T
T  I2 t 2 
1  
cos



I
I
t


  
1
2
n
kt  I1  I 2 2
I1  I 2  kt 2 
I1 I 2
T
Tt 2
2  
1  cos nt  
kt  I1  I 2 2
2  I1  I 2 
But we are interested in the relative motion, the twist in the shaft, which is
T I 2 1  cos  nt 
1   2 
kt
 I1  I 2 
So the amplitude during the first era is
2
22.6 NM   6.328 NMs 

I2
T


 0.029 rad  1.707
kt  I1  I 2   531.1 NM/rad   9.04 NMs 2 
Ans.
For the completion of the first era we can compute that, at t = 2.0 s,
Thus,
nt  16.7 rad/s  2.0 s   33.45 rad  1 916.7 and cos nt  0.449 .
1  5.0302 rad , and 1  5.0302 rad . These are the initial displacements for the
second era.
Second era; t  2 s :
The differential equations now become:
I11  kt1  kt 2  0
I 2 2  kt1  kt 2  0
Following a similar procedure to that above, we eventually obtain
1   2  0.0555cos nt   19.6
Therefore the amplitude of the second era is
  0.0555 rad  3.180
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Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
15.12 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 illustrated in Fig. P15.12. What is the final amplitude of the motion?
k  mn2   44.5 n 9650 mm/s 2   2 rad/s   4.552 n/25 mm  0.182n/mm
2
F1 k  53.4 n 0.182 n/mm  293.4 mm
F2 k  26.7 n 0.182 n/mm  146.7 mm
n t1   2 rad/s  0.25 s   1.571 rad  90.0 , n t2   2 rad/s  0.25 s   1.571 rad  90.0
The final amplitude is X = 464 mm.
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
15.13 An undamped vibrating system has a spring scale of 35.6 N/mm 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 illustrated in Fig. P15.13. Use the phase-plane method.
n  k m  35.6 n/mm  50 lb 9650 mm/s2   39.298 rad/s
n t   39.298 rad/s  0.040 s   1.572 rad  90 ,
F k  55.625 n 35.6 n/mm  1.5625 mm
The final amplitude is 4.42 mm.
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Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
15.14 A vibrating system has a spring k = 3 936 N/m and a weight of W = 20 N. Plot the
response of this system to the forcing function illustrated in Fig. P15.14:
(a)
Using three steps
(b)
Using six steps
Fmax  200 N
n  k m  3 936 N/m  20 N 9.81 m/s2   43.937 rad/s
(a)
Three-step solution: n t   43.937 rad/s  0.1 s 3  1.465 rad  83.91
F1 k  0.083 m , F2 k  0.250 m , F3 k  0.417 m , F k  0.500 m
(b)
Six-step solution: n t   43.937 rad/s  0.1 s 6   0.732 rad  41.96
F1 k  0.042 m , F2 k  0.125 m , F3 k  0.208 m , F4 k  0.292 m ,
F5 k  0.375 m , F6 k  0.458 m , F k  0.500 m
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Theory of Machines and Mechanisms, 4e
15.15 (a)
(b)
(c)
(d)
Uicker et al.
What is the value of the coefficient of critical damping for a spring-mass-damper
system in which k = 64 kN/m and m = 36 kg?
If the actual damping is 20% of critical, what is the natural frequency of the
system?
What is the period of the damped system?
What is the value of the logarithmic decrement?
n  k m  64 000 N/m 36 kg  42.164 rad/s
cc  2mn  2  36 kg  42.164 rad/s   3 035.8 N  s/m
Ans.
(b)
(c)
d  n 1   2  42.164 rad/s 1  0.202  41.312 rad/s
  2 d  2 41.312 rad/s  0.152 s/cycle
Ans.
Ans.
(d)
  2
(a)
1   2  2  0.2 
1  0.202  1.283
Ans.
15.16 A vibrating system has a spring k = 3.6 kN/m and a mass m = 16 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.
n  k m  3 600 N/m 16 kg  15.0 rad/s
Using Eq. (15.32) with  N  ln 1.0 0.25  1.386 and N  4.80 s
  N
 Nn   1.386  4.80 s 15.0 rad/s   0.019 25
c   2mn  0.01925  2 15 kg 15.0 rad/s  8.663 N  s/m
Ans.
Ans.
15.17 A vibrating system has k = 53.44 N/mm, W = 445 N, and damping equal to 20% of
critical.
(a)
What is the damped natural frequency d of the system?
(b)
What are the period and the logarithmic decrement?
n  k m  53.4 N/mm  445 N 9650 mm/s2   34.03 rad/s
(a)
(b)
d  n 1   2  34.03 rad/s 1  0.202  33.34 rad/s
  2 d  2 33.34 rad/s  0.188 s/cycle
Ans.
Ans.
  2
Ans.
1   2  2  0.20
1  0.202  1.283
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Theory of Machines and Mechanisms, 4e
Uicker et al.
15.18 Solve Problem 15.14 using damping equal to 15% of critical.
n  k m  3 936 N/m  20 N 9.81 m/s2   43.937 rad/s
d  n 1   2  43.937 rad/s 1  0.152  43.440 rad/s
Six-step solution: d t   43.440 rad/s  0.10 s 6  0.724 rad  41.48
F1 k  0.042 m , F2 k  0.125 m , F3 k  0.208 m , F4 k  0.292 m ,
F5 k  0.375 m , F6 k  0.458 m , F k  0.500 m
In each step of d t  0.724 rad the reduction in amplitude is
X n 41.48 X n  e  0.724 rad 
1 2
 0.896 . Therefore,
x0  0.0417 m , 0  90
x1  0.896  0.0417 m   0.0376 m
x1  0.1138 m , 1  77.34
x2  0.896  0.1138 m  0.1020 m
x2  0.1653 m , 2  59.98
x3  0.896  0.1653 m  0.1481 m
x3  0.1916 m , 3  42.86
x4  0.896  0.1916 m   0.1716 m
x4  0.1926 m , 4  27.01
x5  0.896  0.1926 m   0.1726 m
x5  0.1719 m , 5  13.53
x6  0.896  0.1719 m   0.1539 m
x6  0.1394 m , 6  12.64
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Theory of Machines and Mechanisms, 4e
Uicker et al.
15.19 A damped vibrating system has an undamped natural frequency of 10 Hz and a weight of
3560 N. The damping ratio is 0.15. Using the phase-plane method, determine the response
of the system to the forcing function illustrated in Fig. P15.19.
n  10 revs/s  2 rad/rev   62.832 rad/s
d  n 1   2  62.832 rad/s 1  0.152  62.121 rad/s
d t   62.121 rad/s  0.01 s   0.621 rad  35.59
k  mn2   3560 N 9650 mm/s 2   62.832 rad/s   339.1 N/mm
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 d t  0.621 rad the reduction in amplitude is
X n35.59 X n  e  0.621 rad  1  0.910 . Therefore,
x1  23.875 mm
x0  26.25 mm , 0  90
x2  27.025 mm
x1  29.7 mm , 1  43.53
x2  43.5 mm , 2  59.95
x3  39.6 mm
x3  58.5 mm , 3  113.94
x4  53.15 mm
x4  51.25 mm , 4  199.90
2
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x1  22.75 mm
x2  24.6 mm
x3  36.025 mm
x4  48.375 mm
Theory of Machines and Mechanisms, 4e
Uicker et al.
15.20 A vibrating system has a spring rate of 3 000 N/m, a damping coefficient of 100 N  s/m,
and a weight of 800 N. It is excited by a harmonically varying force F0  50 N at a
frequency of 60 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  W g  800 N   9.81 m/s2   81.549 kg
k
3 000 N/m

 6.065 rad/s
m
81.549 kg
c
100 N  s/m

 0.101

2mn 2  81.549 kg   6.065 rad/s 
n 
  60 cycles/min  2 rad/cycle   60 s/min 

 1.036
n
6.065 rad/s
F0 k
X
1  
 n2    2   n 
2
2
2
50 N 3 000 N/m
X
1  1.036    2  0.1011.036
2 2
  tan 1
 0.075 m  75 mm
Ans.
2
2  n
2  0.1011.036
 tan 1
 109.25
2
2
1   n
1  1.0362
Ans.
15.21 A spring-mounted mass has k = 44.5 N/mm, c = 1.424 / NS/mm, and weighs 1557.5 N.
This system is excited by a force having an amplitude of 890 N at a frequency of 2 Hz.
Find the amplitude and phase angle of the resulting vibration and plot several cycles of
the force-time and displacement-time diagrams.
m  w g  1557.5 n.  9650 mm/s2   0.161 Ns2 /mm
 44.5 N/mm   0.161 ns2 /mm   16.605 rad/s
  c  2mn   1.424 Ns/mm   2  0.16 Ns 2 /mm 16.605 rad/s   0.266
 n   2 Hz  2 rad/cycle  16.605 rad/s  0.757
n  k m 
X 
X 
F0 k
1  
2

2
n

2
  2   n 
2
890 N 44.5 N/mm
1  0.757 
2
  tan 1
2
  2  0.266  0.757 
 34.075 mm
Ans.
2
2  n
2  0.266  0.757
 tan 1
 43.26
1   2 n2
1  0.7572
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Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
15.22 When a 26700-N press is mounted upon structural-steel floor beams, it causes them to
deflect18.75 mm. If the press has a reciprocating unbalance of 2002.5. N and it operates
at a speed of 72 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  26700 N 18.75 mm  1.424 N/mm
m  26700 N 9650 mm/s2  2.767 Ns2 /mm
n  k m  1424 N/mm 2.76 Ns2 /mm  22.714 rad/s
 n   72 rev/min  2 rad/rev 60 s/min  22.689 rad/s  0.332
Assuming no damping, Eq. (15.62) gives
1
1
1


 1.124
T
2
2
2
2




1
1
0.332
2
2
n
1  

n

Ftr  TF0  1.124  2002.5 N  2250.8 N
Fig. 15.37 shows that, with  n  0.332 , small changes in either damping or  n will do
little to reduce transmissibility. Therefore the mounting cannot be improved. The
primary opportunity for improvement would be to reduce the unbalance.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
15.23 Four vibration mounts are used to support a 450-kg machine that has a rotating unbalance
of 0.35 kg  m and runs at 250 rev/min. The vibration mounts have damping equal to 36
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. (15.63)
T  0.20 

2
/ n2  1   2  0.30 /  n 
2
1   /      2  0.30 / n 


2
2
2
n

2
  / n 
2
1  0.36  / n 
2
2
1   / n 2   0.36  / n 2
 

2
2
2
2
2
0.20 1   / n    0.36  / n    / n  1  0.36  / n 




2
2
2
4
2
0.04 1   / n    0.36  / n    / n  1  0.36  / n  




0.36  / n   0.96  / n   0.0656  / n   0.04  0
6
4
2
Numerically searching for the root we find
2
 / n  0.410 39
 / n   0.168 423
  250 rev/min  26.180 rad/s
n  63.792 83 rad/s
k  m  450 kg  63.792 83 rad/s   1 831 286 N/m
2
2
n
Ans.
Now, from Eq. (15.56)
 / n 
 0.168 423
mX


 0.186 81
2
2
mu e
2 2
2
1   / n    0.36  / n 
1   0.168 423  0.72  0.168 423


2
X  0.186 81 0.35 kg  m  450 kg  0.145 mm
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
15.24 A 600-mm long steel shaft is simply supported by two bearings at A and C as illustrated
in Fig. P15.24. Flywheels 1 and 2 are attached to the shaft at locations B and D,
respectively. Flywheel 1 at location B weighs 50 N, flywheel 2 at location D weighs 20
N, and the weight of the shaft can be neglected. The known stiffness coefficients are
k11  20 000 N/m, k12  50 000 N/m, and k22  40 000 N/m. Determine: (i) the first
and second critical speeds of the shaft using the exact solution and the first critical speed
using (ii) the Dunkerley and (iii) Rayleigh-Ritz approximations. (iv) If flywheel 2 is then
placed at location B and flywheel 1 is placed at location D, determine the first critical
speed of the new system using the Dunkerley approximation.
(i)
The exact solutions for the first and second critical speeds of the shaft are
(a11m1  a22 m2 )  (a11m1  a22 m2 ) 2  4m1m2 (a11a22  a12 a21 )
1 22
2
The influence coefficients are the reciprocals of the stiffness coefficients; that is,
and a jk  akj  1 k jk
aii  1 kii
1
,
2
1

Therefore, the influence coefficients are
a11  1  2 104 N/m   5 105 m/N ,
and
a22  1  4 104 N/m   2.5 105 m/N
a12  1  5 104 N/m   2 105 m/N
The masses of the two flywheels are
50 N
20 N
m1 
 5.10 kg and m2 
 2.04 kg
2
9.81 m/s
9.81 m/s 2
Substituting Eqs. (3) and (4) into Eq. (1) gives
(1)
(2)
(3a)
(3b)
(4)
30.6 105 s2  (30.62 s 4  208.080 s 4 ) 1010
 (15.3  13.49) 105 s 2
2
2
1 2
Using the positive sign for the first critical speed and the negative sign for the second
critical speed gives
Ans.
1  86.10 rad/s and 2  235.05 rad/s
1
,
2
1

© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
(ii)
Using the Dunkerley approximation, the first critical speed of the shaft with the
two flywheels can be written as
1
(5)
 a11m1  a 22 m2
2
1
Substituting Eqs. (3) and the masses into Eq. (5) gives
1
 5 105 m/N  5.10 kg   2.5 105 m/N  2.04 kg   3.06 10 4 s 2
2
1
Therefore, the first critical speed of the shaft is
1  57.17 rad / s
Ans.
(iii) Using the Rayleigh-Ritz approximation, the first critical speed of the shaft
the two flywheels can be written as
g (W1 x1  W2 x2 )
12 
(W1 x12  W2 x22 )
The total deflections of the shaft at the mass particles can be written as
x1  a11W1  a12W2 and x2  a12W1  a22W2
Substituting the known values and Eq. (2) into Eqs. (7) the total deflections are
x1  2.9 103 m and x2  1.5 103 m
Substituting Eqs. (8) and the known values into Eq. (6) gives
9.81 m/s 2 [(50 N)(2.9 103 m)  (20 N)(1.5 103 m)]
12 
[(50 N)(2.9 103 m)2  (20 N)(1.5 103 m) 2 ]
1.72 m/s 2
 3 688 rad 2 /s 2
466 106 m
Therefore, the first critical speed of the shaft is
1  60.72 rad/s
or
with
(6)
(7)
(8)
12 
Ans.
(iv)
When the two flywheels are interchanged then the Dunkerley approximation, see
Eq. (5), can be written as
1
(9)
 a11m1new  a 22 m2new
2
1
Note that the influence coefficients of the shaft do not change (even though the two
flywheels were interchanged). Therefore, substituting these values into Eq. (9) gives
1
 5 105 m/N  2.04 kg   2.5 105 m/N  5.10 kg   2.30 10 4 s 2
2
1
Therefore, the first critical speed of the shaft is
1  66.01 rad/s
Ans.
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Theory of Machines and Mechanisms, 4e
Uicker et al.
15.25 The first critical speeds of a rotating shaft with two mass disks, obtained from three
different mathematical techniques, are 110 rad/s, 112 rad/s, and 100 rad/s, respectively.
(i) Which values correspond to the first critical speed of the shaft from the exact solution,
the Dunkerley approximation, and the Rayleigh-Ritz approximation? (ii) If the influence
coefficients are a11  a22  104 m/N and the masses of the two disks are the same, that is,
m1 = m2 = m, then use the Dunkerley approximation to calculate the mass m. (iii) If the
influence coefficients are a11  a22  104 m/N and the masses of the two disks are
specified as m1 = m2 = m = 0.5 kg, use the Rayleigh-Ritz approximation to calculate the
influence coefficient a12.
(i) The first critical speed from the Rayleigh-Ritz approximation is an upper bound;
therefore, the value 1  112 rad/s corresponds to the answer from the Rayleigh-Ritz
approximation. The Dunkerley approximation gives a lower limit to the first critical
speed; therefore, the value 1  100 rad/s corresponds to the answer from the Dunkerley
approximation. The value of the first critical speed from the exact method is
Ans.
  110 rad/s .
1
(ii) Since the first critical speed and the influence coefficients are given then the
Dunkerley approximation can be used to calculate the two masses; that is,
(1)
1 12  a11m1  a22 m2
Substituting the given information and the first critical speed from the table into Eq. (1)
gives
2
1 100 rad/s   104 m/N  m  104 m/N  m
Therefore, the mass is
1
Ans.
 0.5 kg
m
 2 10
4
m/N  100 rad/s 
2
(iii) The Rayleigh-Ritz equation can be written as
g (W1 x1  W2 x2 )
12 
(W1 x12  W2 x22 )
Since the two masses are the same then this equation can be written as
12  g ( x1  x2 )  x12  x22 
(2)
The total deflections of the shaft at locations 1 and 2 can be written as
x1  a11W1  a12W2 and x2  a12W1  a22W2
(3)
Since the influence coefficients a11  a22 and a12  a21 and the masses m1 = m2 = m then
the deflections x1  x2  x . Therefore, Eq. (2) can be written as
12  g x
(4)
Substituting the known data and Eqs. (3) into Eq. (4) gives
9.81 m/s2
2
112 rad/s  
(0.5 kg)(9.81 m/s2 )(104 m/N  a12 )
Solving for the influence coefficient gives
a12  5.94 105 m/N
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
15.26 A steel shaft is simply supported by two rolling element bearings at A and B as illustrated
in Fig. P15.26. The length of the shaft is 1.45 m and two flywheels with weight 300 N
are attached to the shaft at the locations shown. One flywheel is 0.35 m to the right of
the left bearing at A and the other flywheel is 0.35 m to the left of the right bearing at B.
The weight of the shaft can be neglected. The influence coefficients are specified as
a11  126 105 mm/N and a21  92.5 105 mm/N. (i) Determine the first and second
critical speeds of the shaft using the exact solution. Determine the first critical speed of
the shaft using: (ii) the Dunkerley approximation and (iii) the Rayleigh-Ritz equation.
(i) The exact solutions for the first and second critical speeds of the shaft can be written
(a11m1  a22 m2 )  (a11m1  a22 m2 )2  4(a11a22  a12 a21 )m1m2
,

2
12 22
From the symmetry of the loading, we find the influence coefficients
a11  a22  1.26 106 m/N
From Maxwell's reciprocity theorem, we get the influence coefficients
a21  a12  0.925 106 m/N
The mass of the flywheels are
300 N
m  m1  m2 
 30.581 N  s 2 /m
2
9.81 m/s
Substituting Eqs. (2), (3), and (4) into Eq. (1), the exact solutions can be written as
1 1
  a11  a21  m
,
12 22
Equation (5) can be written as
1
12 , 22 
 a11  a21  m
Substituting the numerical values into Eq. (6), the exact solutions can be written as
106
12 , 22 
(1.26 m/N  0.925 m/N)  30.581 N  s 2 /m 
1
1
(1)
(2)
(3)
(4)
(5)
(6)
(7)
Using the positive sign in the denominator of Eq. (7), the first critical speed of the shaft is
obtained from the relation
106
106
12 

 1.4966 104 rad 2 / s 2
2
2
 2.185 m/N  30.581 N  s /m 66.819 s


Therefore, the first critical speed of the shaft is
ω1  122.3 rad / s
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Ans.
(8)
Theory of Machines and Mechanisms, 4e
Uicker et al.
Similarly, using the negative sign in the denominator of Eq. (7), the second critical speed
of the shaft can be obtained from the relation
106
106
12 

 9.761104 rad 2 / s 2
2
2
 0.335 m/N  30.581 N  s /m 10.245 s


Therefore, the second critical speed of the shaft is
Ans.
2  312.4 rad/s
Note that the second critical speed is about three times the first critical speed.
(ii) The Dunkerley approximation to the first critical speed of the shaft can be written as
1
  a11  a22  m  2a11m
(9)
2
1
Substituting the numerical values into Eq. (9), the Dunkerley approximation to the first
critical speed of the shaft is
1
 2 1.26 106 m/N  30.581 N  s 2 /m   77.064 106 s 2
2
1
Therefore, the Dunkerley approximation to the first critical speed of the shaft is
Ans.
1  113.9 rad/s
Note that the the Dunkerley approximation to the first critical speed of the shaft is less
than the exact answer, see Eq. (8); that is, the Dunkerley approximation always gives a
lower bound.
(iii) The Rayleigh-Ritz equation can be written as
 W x  W2 x2 
12  g  1 12
(10)
2
W1 x1  W2 x2 
where the deflections are
x1  a11W1  a12W2  300 N(1.260  0.925) 106 m/N  655.5 106 m
and
x2  a21W1  a22W2  300 N(0.925  1.260) 106 m/N  655.5 106 m
Substituting these values into Eq. (10), the Rayleigh-Ritz equation can be written as
9.81 m/s 2
 2Wx 
1
2
ω1  g 
g  
 1.4966 104 rad 2 /s 2
2
6
 2Wx 
 x  655.5 10 m
Therefore, the Rayleigh-Ritz equation to the first critical speed of the shaft is
Ans.
1  122.3 rad/s
Note that the Rayleigh-Ritz approximation to the first critical speed of the shaft gives the
same as the exact answer, see Eq. (8). In general, the Rayleigh-Ritz equation will give a
slightly greater value than the exact answer; that is, the Rayleigh-Ritz equation will give
an upper bound.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
15.27 A steel shaft is simply supported by two rolling element bearings at A and C as illustrated
in Fig. P15.27. The length of the shaft is 0.6 m and two flywheels are attached to the
shaft at the locations B and D as shown. The flywheel at location B weighs 200 N and the
flywheel at location D weighs 90 N. The weight of the shaft can be neglected. It was
found that with flywheel 1 alone, the first critical speed of the shaft is 800 rad/s and with
flywheel 2 alone, the first critical speed of the shaft is 1 200 rad/s. (i) Determine the first
critical speed for the two mass system. (ii) If the two flywheels are interchanged (that is,
flywheel 2 is placed at location B and flywheel 1 is placed at location D), determine the
first critical speed of the new system using the Dunkerley approximation.
(i) Using the Dunkerley approximation, the first critical speed of the shaft with the two
flywheels can be written as
1 ω12  a11m1  a22 m2
(1)
or as
1
1
1
 2  2
(2)
2
1
11 22
From the given data, the critical speeds are 11  800 rad/s and 22  1200 rad/s.
Therefore, Eq. (2) can be written as
1
1
1
1 
 1
4 2


 
(3a)
 10 s
2
2
2
1 800 rad/s  1 200 rad/s   64 144 
or as
12  44.308 104 rad 2 /s2
Therefore, the first critical speed of the shaft is
1  665.6 rad/s
(3b)
Ans.
(4)
(ii) When the two flywheels are interchanged then Eq. (1) can be written as
1 12  a11m1new  a22 m2new
(5)
Note that the influence coefficients of the shaft (by definition) do not change (even
though the two flywheels were interchanged). Therefore, the influence coefficient
a11  1 112 m1
(6a)


which can be written as
1
a11 
 7.6641108 m/N
2
2
800 rad/s  (200 N / 9.81 m/s )
© Oxford University Press 2015. All rights reserved.
(6b)
Theory of Machines and Mechanisms, 4e
Uicker et al.
Similarly, the influence coefficient
2
a22  1 22
m2


which can be written as
1
a22 
 7.5694 108 m/N
2
2
1 200 rad/s  (90 N / 9.81 m/s )
Substituting Eqs. (6b) and (7b) into Eq. (5) gives
1
 90 N 
 200 N 
 (7.6641108 m/N) 
 (7.5694 108 m/N) 
2
2 
2 
ω1
 9.81 m/s 
 9.81 m/s 
From this equation, the first critical speed of the shaft is 1  667.2 rad/s. Ans.
© Oxford University Press 2015. All rights reserved.
(7a)
(7b)
(8)
Theory of Machines and Mechanisms, 4e
Uicker et al.
15.28 A steel shaft, which is 1250 mm in length, is simply supported by two bearings at B and
D as illustrated in Fig. P15.28. Flywheels 1 and 2 are attached to the shaft at A and C,
respectively. The flywheel at location A weighs 66.75 N the flywheel at location C
weighs 133.5 N, and the weight of the shaft can be neglected. The stiffness coefficients
are specified as k11  3560 N/mm and k22  720 m/mm . (i) Determine the first critical
speed for the two-mass system using the Dunkerley approximation. (ii) If the two
flywheels are interchanged (that is, flywheel 2 is placed at location A and flywheel 1 is
placed at location C), determine the first critical speed of the new system.
(i) Using the Dunkerley approximation, the first critical speed of the shaft with the two
flywheels can be written as
(1)
1 12  a11m1  a22 m2
The influence coefficients are the inverse of the spring stiffness coefficients.
aii  1 kii
Therefore, the influence coefficients are
a11  2.8 104 mm/N and a22  1.4 104 mm/N
Substituting these values and the masses into Eq. (1) gives
1
 66.75 N 
 135.5 N 
  2.8 104 mm/N  
 1.4 104 mm/N  
 3.9 106 s 2
2
2  
2 
1
 9650 mm/s 
 9650 mm/s 
which gives
1  507.3 rad/s
Ans.
(ii)
When the two flywheels are interchanged then Eq. (1) becomes
2
(2)
1 1  a11m1new  a22 m2new
Note that the influence coefficients of the shaft do not change (even though the two
flywheels are interchanged). Therefore, substituting values into Eq. (2) gives
1
 133.5 N 
 66.75 N 
  2.8 104 mm/N  
 1.4 104 mm/N  
 4.08 106 s 2
2
2  
2 
1
 9650 mm/s 
 9650 mm/s 
which gives
1  453.7 rad/s
Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
Chapter 16
Dynamics of Reciprocating Engines
16.1
A one-cylinder, four-stroke engine has a compression ratio of 7.6 and develops brake
power of 2.25 kW at 3 000 rev/min. The crank length is 22 mm with a 60-mm bore.
Develop and plot a rounded indicator diagram using a card factor of 0.90, a mechanical
efficiency of 72%, a suction pressure of 100 kPa and a polytropic exponent of 1.30.
A   D2 4    0.060 m  4  0.002 827 m2
2
v  2rA  2  0.022 m  0.002 827 m2 1 000 L/m3  0.124 4 L  124.4 mL
v1  v R  R  1  124.4 mL   7.6  7.6  1  143.2 mL
v2  v1  v  143.2 mL  124.4 mL  18.8 mL
C  v2 v  18.8 mL 124.4 mL  0.1511  15.11%
 2.25 kW  60 s/min  1 000 N  m/  kW  s    0.001 kPa  m 2 / N 
pb 
 724 kPa
 0.044 m   0.002 827 m2  3 000 rev/min 2 rev/work stroke 
pi  pb em  724 kPa 0.72  1 005 kPa
p1  100 kPa
R  1 pi
7.6  1 1005 kPa
 p1  1.30  1 1.3
 100 kPa  447 kPa
p4   k  1 k
R  R fc
7.6  7.6 0.90
As in Example 16.1, we calculate the values:
X(%) v(mL) pc(kPa) pe(kPa)
0
18.8
1401
6266
5
25.0
966
4321
10
31.2
724
3238
15
37.5
572
2557
20
43.7
468
2094
25
49.9
394
1761
30
56.1
338
1512
35
62.3
295
1319
40
68.6
261
1165
45
74.8
233
1041
50
81.0
210
938
55
87.2
191
852
60
93.4
174
779
65
99.7
160
717
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
70
75
80
85
90
95
100
Uicker et al.
105.9
112.1
118.3
124.5
130.8
137.0
143.2
148
138
128
120
113
106
100
662
615
573
536
503
474
447
Then we sketch and round the following diagram:
4500
4000
3500
Pressure p , kPa
3000
2500
2000
1500
1000
500
0
0
20
40
60
Displacement X , %
© Oxford University Press 2015. All rights reserved.
80
100
Theory of Machines and Mechanisms, 4e
16.2
Uicker et al.
Construct a rounded indicator diagram for a four-cylinder, four-stroke gasoline engine
having a 85-mm bore, a 90-mm stroke, and a compression ratio of 6.25. The operating
conditions to be used are 22.4 kW at 1 900 rev/min. Use a mechanical efficiency of 72%,
a card factor of 0.90, a suction pressure of 100 kPa, and a polytropic exponent of 1.30.
A   D2 4    0.085 m  4  0.001 806 m2
2
v  A  0.090 m  0.001 806 m2 1 000 L/m3  0.162 54 L  162.54 mL
v1  v R  R  1  162.54 mL  6.25  6.25  1  193.5 mL
v2  v1  v  193.5 mL  162.54 mL  30.96 mL
C  v2 v  30.96 mL 162.54 mL  0.1905  19.05%
 22.4 kW  60 s/min  1 000 N  m/  kW  s    0.001 kPa  m 2 / N 
pb 
 8 704 kPa
 0.090 m   0.001 806 m2  1 900 rev/min 2 rev/work stroke 
pi  pb em  8 704 kPa 0.72  12 090 kPa
p1  100 kPa
R  1 pi
6.25  1 12 090 kPa
 p1  1.30  1
 100 kPa  4 719 kPa
p4   k  1 k
R  R fc
6.251.3  6.25
0.90
As in Example 16.1, we calculate the values:
X(%) v(mL) pc(kPa) pe(kPa)
0
31.0
1 083 51 102
5
39.1
800 37 744
10
47.2
626 29 526
15
55.3
509 24 019
20
63.5
426 20 100
25
71.6
364 17 186
30
79.7
317 14 944
35
87.8
279 13 172
40
96.0
249 11 741
45 104.1
224 10 564
50 112.2
203
9 580
55 120.4
185
8 748
60 128.5
170
8 036
65 136.6
157
7 420
70 144.7
146
6 883
75 152.9
136
6 411
80 161.0
127
5 994
85 169.1
119
5 622
90 177.2
112
5 289
95 185.4
106
4 990
100 193.5
100
4 719
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
Then we sketch and round the following diagram:
40000
35000
Pressure p , kPa
30000
25000
20000
15000
10000
5000
0
0
10
20
30
40
50
60
Displacement X , %
© Oxford University Press 2015. All rights reserved.
70
80
90
100
Theory of Machines and Mechanisms, 4e
16.3
Uicker et al.
Construct an indicator diagram for a V6 four-stroke 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%, a card factor of 0.88, a suction
pressure of 100 kPa, and a polytropic exponent of 1.30.
A   D2 4    0.100 m  4  0.007 854 m2
2
v  A  0.090 m  0.007 854 m2 1 000 L/m3  0.707 L  707 mL
v1  v R  R  1  707 mL  8.40 8.40  1  803 mL
v2  v1  v  803 mL  707 mL  96 mL
C  v2 v  96 mL 707 mL  0.1358  13.58%
150 000 W 6 cyl  60 s/min   0.001 kPa  m2 / N 
pb 
 965 kPa
 0.090 m   0.007 854 m2   4 400 rev/min 2 rev/work stroke 
pi  pb em  965 kPa 0.72  1 340 kPa
p1  100 kPa
R  1 pi
8.40  1 1 340 kPa
 p1  1.30  1
 100 kPa  550 kPa
p4   k  1 k
R  R fc
8.401.3  8.40 0.88
As in Example 16.1, we calculate the values:
X(%) v(mL) pc(kPa) pe(kPa)
0
96
1 590
8 751
5
131
1 056
5 812
10
167
774
4 259
15
202
603
3 315
20
237
488
2 687
25
272
408
2 243
30
308
348
1 914
35
343
302
1 661
40
378
266
1 462
45
414
237
1 302
50
449
213
1 170
55
484
193
1 061
60
520
176
968
65
555
161
888
70
590
149
820
75
626
138
760
80
661
129
708
85
696
120
661
90
732
113
620
95
767
106
583
100
802
100
550
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Then we sketch and round the following diagram:
© Oxford University Press 2015. All rights reserved.
Uicker et al.
Theory of Machines and Mechanisms, 4e
16.4
Uicker et al.
A single-cylinder, two-stroke 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%, a suction pressure of 100 kPa, and a polytropic exponent of 1.30.
A   D2 4    0.080 m  4  0.005 027 m2
2
v  A  0.070 m  0.005 027 m2 1 000 L/m3  0.352 L  352 mL
v1  v R  R  1  352 mL  7.0  7.0  1  411 mL
v2  v1  v  411 mL  352 mL  59 mL
C  v2 v  59 mL 352 mL  0.1662  16.62%
pb 
30 000 W  60 s/min   0.001 kPa  m 2 / N 
 0.070 m   0.005 027 m2   4 500 rev/min 1 rev/work stroke 
 1 137 kPa
pi  pb em  1 137 kPa 0.65  1 749 kPa
p1  100 kPa
R  1 pi
7.0  1 1 749 kPa
 p1  1.30  1 1.3
 100 kPa  673 kPa
p4   k  1 k
R  R fc
7.0  7.0 0.990
As in Example 16.1, we calculate the values:
x(%) v(mL) pc(kPa) pe(kPa)
0
59
1 259
8 473
5
76
894
6019
10
94
682
4 592
15
111
546
3 672
20
129
451
3 034
25
147
382
2 569
30
164
329
2 217
35
182
288
1 942
40
199
256
1 722
45
217
229
1 542
50
235
207
1 394
55
252
188
1 268
60
270
173
1 162
65
287
159
1 070
70
305
147
991
75
323
137
921
80
340
128
860
85
358
120
805
90
375
112
756
95
393
106
712
100
411
100
673
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Then we sketch and round the following diagram:
© Oxford University Press 2015. All rights reserved.
Uicker et al.
Theory of Machines and Mechanisms, 4e
16.5
Uicker et al.
The engine of Problem 16.1 has a connecting rod 80 mm long and a mass of 0.100 kg,
with the mass center 15 mm from the crankpin end. Piston mass is 0.175 kg. Find the
bearing reactions and the crankshaft torque during the expansion stroke corresponding to
a piston displacement of X = 30% (  t  60 ). To find pe , see the answer to Problem
16.1 in Appendix B.
 0.080 m , m3  0.100 kg , m4  0.175 kg ,
A
 0.015 m ,
m3 A  m3
B
  0.100 kg  0.065 m  0.080 m  0.081 25 kg ,
m3B  m3
A
  0.100 kg  0.015 m  0.080 m  0.018 75 kg ,
B
 
A
 0.065 m ,
   3 000 rev/min  2 rad/rev   60s/min   314.16 rad/s , r  0.022 m ,
r
  0.022 m   0.080 m   0.275 , r 2   0.022 m  314.16 rad/s   2 171 m/s 2 ,
2
  t  60 ,
 sin     0.022 m  cos60   0.080 m  1   0.275sin 60  0.088 7 m ,
2
X  30% , pe  1 512 kPa (from Prob. 16.1), A    0.060 m  4  0.002 827 m2 ,
P  pe A  1 512 kPa   0.002 827 m2   4 274 N
x  r cos 
1 r
2
2
2
r
x  r 2 (cos t  cos 2t )    2 171 m/s 2   cos 60  0.275cos120   787 m/s 2


 0.2752

r
r2
tan   sin t 1  2 sin 2 t   0.275sin 60 1 
sin 2 60   0.244 9
2
 2



F41    m3B  m4  x  P  tan  ˆj


   0.018 75 kg  0.175 kg  787 m/s2  4 274 N  0.244 9ˆj


 1 013ˆj N
F34   m4 x  P  ˆi   m3B  m4  x  P  tan  ˆj
  0.175 kg  787 m/s 2  4 274 N  ˆi  1 013ˆj N


 4 136ˆi  1 013ˆj N  4 258 N 13.8

Ans.


Ans.

F32  [ m3 Ar 2 cos t   m3 B  m4  x  P ]ˆi  m3 Ar 2 sin t   m3 B  m4  x  P  tan  ˆj


  0.081 25 kg  2 171 m/s 2 cos 60  4 274 N  ˆi  153  1 013 ˆj N


 4 186ˆi  1 166ˆj N  4 345 N164.4
T21  x  m3B  m4  x  P  tan kˆ   0.088 7 m 1 013 N  kˆ  89.82kˆ N  m
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
16.6
Uicker et al.
Repeat Problem 16.5, but do the computations for the compression cycle ( ωt  660 ).
 0.080 m , m3  0.100 kg , m4  0.180 kg ,
A
 0.010 m ,
m3 A  m3
B
  0.100 kg  0.065 m  0.080 m  0.081 25 kg ,
m3B  m3
A
  0.100 kg  0.015 m  0.080 m  0.018 75 kg ,
B
 
A
 0.070 m ,
   3 000 rev/min  2 rad/rev   60s/min   314.16 rad/s , r  0.022 m ,
r
  0.022 m   0.080 m   0.275 , r 2   0.022 m  314.16 rad/s   2 171 m/s 2 ,
2
  t  660 ,
 sin     0.022 m cos660   0.080 m  1   0.275sin 660  0.088 7 m ,
2
X  30% , pc  338 kPa (from Prob. 16.1), A    0.060 m  4  0.002 827 m2 ,
P  pc A   338 kPa   0.002 827 m2   956 N
x  r cos 
1 r
2
2
2
r
x  r 2 (cos t  cos 2t )    2 171 m/s2   cos 660  0.275cos1320   787 m/s2


 0.2752

r
r2
tan   sin t 1  2 sin 2 t   0.275sin 660 1 
sin 2 660   0.230 36
2
 2



ˆ
F41    m3B  m4  x  P  tan  j


   0.012 5 kg  0.175 kg  787 m/s 2  956 N   0.230 36  ˆj


 186ˆj N
F34   m4 x  P  ˆi   m3B  m4  x  P  tan  ˆj
  0.175 kg  787 m/s 2  956 N  ˆi  186ˆj N


 818ˆi  186ˆj N  839 N12.8

Ans.


Ans.

F32  [ m3 Ar 2 cos t   m3 B  m4  x  P ]ˆi  m3 Ar 2 sin t   m3 B  m4  x  P  tan  ˆj


  0.081 25 kg  2 171 m/s 2 cos 660  1 105 N  ˆi   153  186  ˆj N


 1 017ˆi  339ˆj N  1 071 N 161.5
T21  x  m3B  m4  x  P  tan kˆ   0.088 7 m   34 N  kˆ  3.02kˆ N  m
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
16.7
Uicker et al.
Make a complete force analysis of the engine of Problem 16.5. Plot a graph of the
crankshaft torque versus crank angle for 720 of crank rotation.
 0.080 m , m3  0.100 kg , m4  0.175 kg ,
A
 0.010 m ,
m3 A  m3
B
  0.100 kg  0.065 m  0.080 m  0.081 25 kg ,
m3 B  m3
A
  0.100 kg  0.015 m  0.080 m  0.018 75 kg ,
B
 
A
 0.070 m ,
   3 000 rev/min  2 rad/rev   60s/min   314.16 rad/s , r  0.022 m ,
r
  0.022 m   0.080 m   0.275 , r 2   0.022 m  314.16 rad/s   2 171 m/s 2 ,
2
A    0.060 m  4  0.002 827 m2 , p  pe and/or pc as taken from Prob. 16.1,
2
P  Ap   0.002 827 m2  p ,
x  r cos t 
2
2
r

1   sin t    0.022 m  cos    0.080 m  1   0.275sin  


r
x  r 2 (cos t  cos 2t )    2 171 m/s 2   cos   0.275cos 2 


r
r2
tan   sin t 1  2 sin 2 t   0.275sin  1  0.037 8sin 2  
 2

F14   m3 B  m4  x  P  tan 
  0.193 75 kg  x  P  tan 
T21   m3B  m4  x  P  x tan   F14 x
t, 
x, m
X,%
P, N
x, m/s 2
tan 
F14 , N
T21 , N·m
0
15
30
45
60
75
90
105
120
135
150
165
180
195
210
225
240
255
0.102
0.101
0.098
0.094
0.089
0.083
0.077
0.071
0.067
0.063
0.060
0.059
0.058
0.059
0.060
0.063
0.067
0.071
0
2.27
8.43
18.12
30.23
43.59
57.01
69.47
80.23
88.83
95.03
98.76
100.00
98.76
95.03
88.83
80.23
69.47
14 276
15 218
10 115
6 412
4 249
3 042
2 326
1 888
1 615
1 444
1 340
1 283
1 264
1 264
1 264
1 264
1 264
1 264
-2 768
-2 614
-2 179
-1 535
-787
-45
597
1 078
1 384
1 535
1 582
1 580
1 574
1 580
1 582
1 535
1 384
1 078
0
0.071 36
0.138 80
0.198 13
0.244 91
0.275 00
0.285 40
0.275 00
0.244 91
0.198 13
0.138 80
0.071 36
0
-0.071 36
-0.138 80
-0.198 13
-0.244 91
-0.275 00
0
1 048
1 345
1 211
1 003
834
697
577
461
345
229
113
0
-112
-218
-309
-375
-405
0
106
132
114
89
69
54
41
31
22
14
7
0
-7
-13
-19
-25
-29
© Oxford University Press 2015. All rights reserved.
,
Theory of Machines and Mechanisms, 4e
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
0.077
0.083
0.089
0.094
0.098
0.101
0.102
0.101
0.098
0.094
0.089
0.083
0.077
0.071
0.067
0.063
0.060
0.059
0.058
0.059
0.060
0.063
0.067
0.071
0.077
0.083
0.089
0.094
0.098
0.101
0.102
57.01
43.59
30.23
18.12
8.43
2.27
0
2.27
8.43
18.12
30.23
43.59
57.01
69.47
80.23
88.83
95.03
98.76
100.00
98.76
95.03
88.83
80.23
69.47
57.01
43.59
30.23
18.12
8.43
2.27
0
Uicker et al.
1 264
1 264
1 264
1 264
1 264
1 264
774
283
283
283
283
283
283
283
283
283
283
283
283
287
300
324
361
422
521
681
950
1 434
2 262
3 402
3 961
597
-45
-787
-1 535
-2 179
-2 614
-2 768
-2 614
-2 179
-1 535
-787
-45
597
1 078
1 384
1 535
1 582
1 580
1 574
1 580
1 582
1 535
1 384
1 078
597
-45
-787
-1 535
-2 179
-2 614
-2 768
-0.285 40
-0.275 00
-0.244 91
-0.198 13
-0.138 80
-0.071 36
0
0.071 36
0.138 80
0.198 13
0.244 91
0.275 00
0.285 40
0.275 00
0.244 91
0.198 13
0.138 80
0.071 36
0
-0.071 36
-0.138 80
-0.198 13
-0.244 91
-0.275 00
-0.285 40
-0.275 00
-0.244 91
-0.198 13
-0.138 80
-0.071 36
0
-394
-345
-272
-192
-117
-54
0
-16
-19
-2
32
75
114
135
135
115
82
42
0
-42
-84
-123
-154
-173
-181
-185
-196
-226
-256
-207
0
T21 (N∙m) vs. ωt (deg.)
© Oxford University Press 2015. All rights reserved.
-30
-29
-24
-18
-11
-5
0
-2
-2
0
3
6
9
10
9
7
5
2
0
-2
-5
-8
-10
-12
-14
-15
-17
-21
-25
-21
0
Theory of Machines and Mechanisms, 4e
16.8
Uicker et al.
The engine of Problem 16.3 uses a connecting rod 300 mm long. The masses are
m3 A  0.90 kg, m3B  0.30 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% ( t  63.2 ). The pressure should be obtained from the
indicator diagram, Fig. AP16.3 in Appendix B.
 0.300 m , m3 A  0.80 kg , m3B  0.30 kg , m4  1.64 kg ,
   4 400 rev/min  2 rad/rev   60s/min   460.8 rad/s , r  0.045 m ,
r
  0.045 m   0.300 m   0.150 , r 2   0.045 m  460.8 rad/s   9 554 m/s 2 ,
2
  t  63.2 ,
 sin     0.045 m cos63.2   0.300 m  1   0.15sin 63.2  0.317 6 m ,
2
X  30.0% , pe  1 914 kPa (from Prob. 16.3), A    0.100 m  4  0.007 854 m2 ,
P  pe A  1 914 kPa   0.007 854 m2   15 033 N
x  r cos 
1 r
2
2
2
r
x  r 2 (cos t  cos 2t )    9 554 m/s 2   cos 63.2  0.150cos126.4   3 457 m/s2


 0.152

r
r2
tan   sin t 1  2 sin 2 t   0.15sin 63.2 1 
sin 2 63.2   0.135
2
 2



ˆ
F41    m3B  m4  x  P  tan  j


   0.30 kg  1.64 kg  3 457 m/s 2  15 033 N  0.135ˆj


 1 124ˆj N
F34   m4 x  P  ˆi   m3B  m4  x  P  tan  ˆj
   1.64 kg  3 457 m/s 2  15 033 N  ˆi  1 124ˆj N


ˆ
ˆ
 9 364i 1 124 j N  9 431 N  6.8

Ans.

Ans.
F32  [m3 Ar 2 cos t   m3B  m4  x  P ]ˆi  m3 Ar 2 sin t   m3B  m4  x  P  tan  ˆj


  0.80 kg  9 554 m/s 2 cos 63.2  8 326 N  ˆi 
 0.80 kg 9 554 m/s  sin 63.2  1 124 N ˆj




 4 880ˆi  7 946ˆj N  9 325 N121.6
T21  x  m3B  m4  x  P  tan kˆ   0.317 6 m 1 124 N  kˆ  357kˆ N  m
© Oxford University Press 2015. All rights reserved.
2
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
16.9
Uicker et al.
Repeat Problem 16.8, but do the computations for the same position in the compression
cycle ( t  656.8 ).
 0.300 m , m3 A  0.90 kg , m3B  0.30 kg , m4  1.64 kg ,
   4 400 rev/min  2 rad/rev   60s/min   460.8 rad/s , r  0.045 m ,
r
  0.045 m   0.300 m   0.150 , r 2   0.045 m  460.8 rad/s   9 554 m/s 2 ,
2
  t  656.8 ,
2
2
x  r cos  1   r   sin     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. 16.3), A    0.100 m  4  0.007 854 m2 ,
2
P  pc A   348 kPa   0.007 854 m2   2 733 N
x  r 2 (cos t  cos 2t )    9 554 m/s 2   cos 656.8  0.150 cos1313.6   3 457 m/s 2
r


 0.152

r
r2
tan   sin t 1  2 sin 2 t   0.15sin 656.8 1 
sin 2 656.8   0.135
2
 2



F41    m3B  m4  x  P  tan  ˆj
     0.30 kg  1.64 kg  3 457 m/s 2  2 733 N   0.135  ˆj


ˆ
 536 j N
F34   m4 x  P  ˆi   m3B  m4  x  P  tan  ˆj
   1.64 kg  3 457 m/s 2  2 733 N  ˆi  536ˆj N


ˆ
ˆ
 2 936i  536 j N  2 985 N 169.6

Ans.

Ans.
F32  [m3 Ar 2 cos t   m3B  m4  x  P]ˆi  m3 Ar 2 sin t   m3B  m4  x  P  tan  ˆj






  0.90 kg  9 554 m/s 2 cos 656.8  4 250 N  ˆi   0.90 kg  9 554 m/s 2 sin 656.8  536 N ˆj


 7 850ˆi  7 139ˆj N  10 611 N  42.3
T21  x  m3B  m4  x  P  tan kˆ   0.317 6 m  536 N  kˆ  170kˆ N  m
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
16.10 Additional data for the engine of Problem 16.4 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.
 0.110 m , m3  0.13 kg , m4  0.24 kg ,
A
 0.015 m ,
m3 A  m3
B
  0.13 kg  0.095 m  0.110 m  0.112 kg ,
m3B  m3
A
  0.13 kg  0.015 m  0.110 m  0.018 kg ,
B
 
A
 0.095 m ,
   4 500 rev/min  2 rad/rev   60 s/min   471.24 rad/s , r  0.035 m ,
r
  0.035 m   0.110 m   0.318 , r 2   0.035 m  471.24 rad/s   7 772 m/s 2 ,
2
A   D2 4    0.080 m  4  0.005 027 m2 , p  pe or pc as taken from Prob. 16.4,
2
P  Ap   0.005 027 m2  p ,
x  r cos t 
1 r

2
sin 2 t   0.035 m  cos    0.110 m  1   0.318sin   ,
r
x  r 2 (cos t  cos 2t )    7 772 m/s 2   cos   0.318cos 2 


r
r2
tan   sin t 1  2 sin 2 t   0.318sin  1  0.050 62sin 2  
 2

F14   m3B  m4  x  P  tan 
T21   m3B  m4  x  P  x tan 


  0.005 027 m2 p   2005 N  cos   0.318cos 2   x tan 


© Oxford University Press 2015. All rights reserved.
2
Theory of Machines and Mechanisms, 4e
t, 
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
31 945
35 849
24 949
16 102
10 845
7 812
6 022
4 943
4 263
3 831
3 567
3 429
1 943
510
531
568
635
733
897
1 160
1 609
2 394
3 706
5 508
31 945
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
Uicker et al.
x, m
tan 
F14 , N
T21 , N·m
0.145 00
0.143 43
0.138 91
0.131 93
0.123 24
0.113 74
0.104 28
0.095 62
0.088 24
0.082 43
0.078 29
0.075 82
0.075 00
0.075 82
0.078 29
0.082 43
0.088 24
0.095 62
0.104 28
0.113 74
0.123 24
0.131 93
0.138 91
0.143 43
0.145 00
0
0.082 63
0.161 10
0.230 68
0.286 02
0.321 86
0.334 29
0.321 86
0.286 02
0.230 68
0.161 10
0.082 63
0
-0.082 63
-0.161 10
-0.230 68
-0.286 02
-0.321 86
-0.334 29
-0.321 86
-0.286 02
-0.230 68
-0.161 10
-0.082 63
0
0
2 757
3 688
3 387
2 906
2 525
2 226
1 936
1 597
1 211
803
398
0
-157
-314
-458
-560
-581
-513
-384
-265
-225
-266
-249
0
0
395.4
512.3
446.9
358.2
287.2
232.2
185.1
140.9
99.8
62.9
30.2
0
-11.9
-24.6
-37.8
-49.4
-55.5
-53.5
-43.7
-32.6
-29.7
-36.9
-33.8
0
600
500
400
300
200
100
0
0
-100
90
180
T21 (N∙m) vs. ωt (deg)
© Oxford University Press 2015. All rights reserved.
270
360
Theory of Machines and Mechanisms, 4e
Uicker et al.
16.11 The four-stroke engine of Problem 16.1 has a stroke of 66 mm and a connecting rod
length of 183 mm. The mass of the rod is 0.386 kg, and the center of mass center is 42
mm from the crankpin. The piston assembly has mass of 0.576 kg. Make a complete
force analysis for one cylinder of this engine for 720 of crank rotation. Use 110 kPa for
the exhaust pressure and 70 kPa for the suction pressure. Plot a graph to show the
variation of the crankshaft torque with the crank angle. Use Fig. 16.23 for the pressures.
 0.183 m , m3  0.386 kg , m4  0.576 kg ,
A
 0.042 m ,
B
m3 A  m3
B
  0.386 kg  0.141 m  0.183 m  0.297 4 kg ,
m3B  m3
A
  0.386 kg  0.042 m  0.183 m  0.088 6 kg ,
 
A
 0.141 m ,
   3 000 rev/min  2 rad/rev   60s/min   314.16 rad/s , r  0.033 m ,
r
  0.033 m   0.183 m  0.180 , r 2   0.033 m  314.16 rad/s   3 257 m/s 2 ,
2
A    0.060 m  4  0.002 83 m2 , p  pe or pc as taken from Example 16.1,
2
P  Ap   0.002 83 m2  p ,
2
2
r

x  r cos t  1   sin t    0.033 m  cos    0.183 m  1   0.180sin   ,


r
x  r 2 (cos t  cos 2t )    3 257 m/s 2   cos   0.180cos 2 


r
r2
tan   sin t 1  2 sin 2 t   0.180sin  1  0.016 2sin 2  
 2

F14   m3B  m4  x  P  tan 
T21   m3B  m4  x  P  x tan 
t, 
P, N
0
15
30
45
60
75
90
105
120
135
150
165
180
195
210
14 279
16 066
10 688
6 792
4 445
3 230
2 432
1 976
1 649
1 461
1 357
1 288
1 263
1 263
1 263
x, m/s 2
-3 843
-3 654
-3 114
-2 303
-1 335
-335
586
1 351
1 922
2 303
2 528
2 638
2 671
2 638
2 528
x, m
tan 
F14 , N
T21 ,N·m
0.216
0.215
0.211
0.205
0.197
0.189
0.180
0.172
0.164
0.158
0.154
0.151
0.150
0.151
0.154
0
0.046 64
0.090 36
0.128 31
0.157 78
0.176 49
0.182 92
0.176 49
0.157 78
0.128 31
0.090 36
0.046 64
0
-0.046 64
-0.090 36
0
636
778
675
561
531
516
507
462
384
274
142
0
-137
-266
0
136.8
164.3
138.4
110.6
100.3
92.9
87.2
75.7
60.6
42.3
21.4
0
-20.7
-40.9
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
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
1 263
1 263
1 263
1 263
1 263
1 263
1 263
1 263
1 263
773
283
283
283
283
283
283
283
283
283
283
283
283
289
305
327
371
440
543
723
993
1 521
2 378
3 588
3 962
Uicker et al.
2 303
1 922
1 351
586
-335
-1 335
-2 303
-3 114
-3 654
-3 843
-3 654
-3 114
-2 303
-1 335
-335
586
1 351
1 922
2 303
2 528
2 638
2 671
2 638
2 528
2 303
1 922
1 351
586
-335
-1 335
-2 303
-3 114
-3 654
-3 843
0.158
0.164
0.172
0.180
0.189
0.197
0.205
0.211
0.215
0.216
0.215
0.211
0.205
0.197
0.189
0.180
0.172
0.164
0.158
0.154
0.151
0.150
0.151
0.154
0.158
0.164
0.172
0.180
0.189
0.197
0.205
0.211
0.215
0.216
-0.128 31
-0.157 78
-0.176 49
-0.182 92
-0.176 49
-0.157 78
-0.128 31
-0.090 36
-0.046 64
0
0.046 64
0.090 36
0.128 31
0.157 78
0.176 49
0.182 92
0.176 49
0.157 78
0.128 31
0.090 36
0.046 64
0
-0.046 64
-0.090 36
-0.128 31
-0.157 78
-0.176 49
-0.182 92
-0.176 49
-0.157 78
-0.128 31
-0.090 36
-0.046 64
0
-358
-401
-381
-302
-184
-59
34
73
54
0
-100
-161
-160
-95
11
123
208
246
233
177
95
0
-95
-179
-238
-260
-236
-171
-88
-17
1
-28
-54
0
-56.6
-65.7
-65.6
-54.4
-34. 7
-11.7
7.0
15.4
11.7
0
-21.5
-34.1
-32.8
-18.8
2.0
22.1
35.8
40.4
36.8
27.3
14.3
0
-14.4
-27.6
-37.6
-42.6
-40.6
-30.7
-16.7
-3.3
0.3
-5.9
-11.6
0
150
100
50
0
0
90
180
270
360
450
-50
100
T21 (N∙m) vs. ωt (deg)
© Oxford University Press 2015. All rights reserved.
540
630
720
Theory of Machines and Mechanisms, 4e
Uicker et al.
Chapter 17
Balancing
17.1
Determine the bearing reactions at A and B for the system illustrated in Fig. P17.1 if the
speed is 350 rev/min. Determine the magnitude and the angular orientation of the
balancing mass if it is located at a radius of 50 mm.
R1  25 mm, R2  35 mm, R3  40 mm, m1  2 kg, m2  1.5 kg, m3  3 kg.
   350 rev/min  2 rad/rev   60 s/min   36.652 rad/s
F1  m1R1 2   2 kg  0.025 m  36.652 rad/s   67.168 N
2
F2  m2 R2 2  1.5 kg  0.035 m  31.416 rad/s   70.527 N
2
F3  m3 R3 2   3 kg  0.040 m  31.416 rad/s   161.204 N
F  67.168 N90  67.168ˆj N
2
1
F2  70.527 N  165  68.123ˆi  18.254ˆj N
F  161.204 N  75  41.723ˆi  155.711ˆj N
3
F  F  F
1
2
 F3  26.401ˆi  106.796ˆj N  110.011 N  103.9
Since all rotating masses are in a single plane, the correction mass must be in that plane.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
M
A
Uicker et al.
 0.200kˆ m × 110.011 N  103.9  1.000kˆ m ×FB  0
FB  22.00276.1 N
M
A
Ans.
 0.800kˆ m × 110.011 N  103.9  1.000kˆ m ×FA  0
FA  88.00876.1 N
Ans.
FC   F  110.011 N76.1
FC  mC RC 2  mC  0.050 m  31.416 rad/s   110.011 N
2
mC  FC
 R    110.011 N   0.050 m 31.416 rad/s    1.638 kg
2
2
C
C  76.1
Ans.
Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
17.2
Uicker et al.
Figure P17.2 illustrates 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 350 rev/min.
A counterweight is to be located at a radius of 250 mm. Find the value of the weight and
its angular orientation.
150 mm
300 mm
R1  200 mm, R2  300 mm, R3  150 mm, w1  0.556 N, w2  0.417 N, w3  0.834 N.
   350 rev/min  2 rad/rev   60 s/min   36.652 rad/s
 0.556 N  200 mm  36.652 rad/s
F1  m1R1 
2
2
 15.486 N
9650 mm / s
2
0.417 N  300 mm  36.652 rad/s 

2
F2  m2 R2 
 17.42 N
9650 mm/s2
2
0.834 N 150 mm  36.652 rad/s 

2
F3  m3 R3 
 17.42 N
9650 mm2
F1  15.486 N90  15.486ˆj N
F  17.42 N  135  12.317ˆi  12.317ˆj N
2
2
F3  17.42 N  30  15.0891ˆi  8.713ˆj N
 F  F  F  F  2.767ˆi  5.544ˆj N  6.198 N  63.5
1
2
3
Since all rotating masses are in a single plane, the correction mass must be in that plane.
FC   F  6.198 lb116.5
FC  mC RC 2  mC  250 mm 36.652 rad/s   6.198 N
2
6.198 N  9650 mm/s2 
FC

 0.178 N
mC 
RC 2 250 N  31.416 rad/s 2
Ans.
C  116.5
Ans.
Without the correction mass
 M A 450 mm kˆ ×FB  300 mm kˆ ×  F  0
FB  4.134 N116.5
 MB  450 mm kˆ  FA 150 mm kˆ   F  0
Ans.
FA  2.064 N116.5
Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
17.3
Uicker et al.
Figure P17.3 illustrates two weights connected to a rotating shaft and mounted outboard
of bearings A and B. If the shaft rotates at 150 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.
100 mm
50 mm
R1  100 mm, R2  150 mm, w1  17.8 N, w2  13.35 N.
  150 rev/min  2 rad/rev   60 s/min   15.708 rad/s
17.8 mm 100 mm 15.709 rad/s 
F1  m1R1 
2
2
9650 mm/s
17.8 N 150 mm15.708 rad/s
F2  m2 R2 
2
2
 45.514 N
2
2
9650 mm/s
 51.201 N
F1  45.514 N90  45.514ˆj N
F2  51.201 N  135  36.205ˆi  36.205ˆj N
 F  F  F  36.205ˆi  9.309ˆj N  37.384 N165.6
1
M
B
2
  150 mm  kˆ ×  F   100 mm  kˆ × FA


 1396.187 mm  N  ˆi   5430.78 mm  N  ˆj   100 mm  kˆ × FAx ˆi  FAy ˆj  0
FA  54.307ˆi  13.959ˆj lb  56.074 N  14.4
F    F   F  18.102ˆi  4.65ˆj N  18.690 N165.6
B
A
Ans.
Ans.
The correction should be
FC   F  37.384 N165.6
This can be done by mass removal at a radius of 125 mm of
37.384 N  9650 mm/s2 
FC

 11.699 N
mC 
2
RC 2
125 mm 15.708 rad/s 
C  14.4
Ans.
Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
17.4
Uicker et al.
For a speed of 250 rev/min, calculate the magnitudes and relative angular orientations of
the bearing reactions at A and B for the two-mass system illustrated in Fig. P17.4.
R1  60 mm, R2  40 mm, m1  2 kg, m2  1.5 kg.
   250 rev/min  2 rad/rev   60 s/min   26.180 rad/s
F1  m1R1 2   2 kg  0.060 m  26.180 rad/s   82.247 N
2
F2  m2 R2 2  1.5 kg  0.040 m  26.180 rad/s   41.123 N
F  82.247 N90  82.247ˆj N
2
1
F2  41.123 N  90  41.123ˆj N
 MB  0.250kˆ m × 82.246ˆj N  0.550kˆ m × 41.123ˆj N  0.500kˆ m ×FA  0

 
 
 
© Oxford University Press 2015. All rights reserved.
 

Theory of Machines and Mechanisms, 4e
17.5
Uicker et al.
The rotating system illustrated in Fig. P17.5 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 150
rev/min.
  150 rev/min  2 rad/rev   60 s/min   15.708 rad/s
F1  m1R1 2  1 kg  0.060 m 15.708 rad/s   14.804 N
2
F2  m2 R2 2   3 kg  0.060 m 15.708  44.413 N
F  14.804 N90  14.804ˆj N
2
1
F2  44.413 N  90  44.413ˆj N
 MB  0.300kˆ m × 14.804ˆj N  0.900kˆ m × 44.413ˆj N  1.200kˆ m ×FA  0

 
 
 
 

FA  29.609ˆj N  29.609 N90.0
FB    F1  F2  FA   0
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Ans.
Theory of Machines and Mechanisms, 4e
17.6
Uicker et al.
The rotating shaft illustrated in Fig. P15.5 supports two masses m1 and m2 whose weights
are 17.8 N and 22.25 N, 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 350 rev/min.
   350 rev/min  2 rad/rev   60 s/min   36.652 rad/s
17.8 N 100 mm  36.683 rad/s
F1  m1R1 
2
2
 247.789 lb
9650 mm/s
2
22.25 N  75 mm  36.683 rad/s 

2
F2  m2 R2 
 232.303 N
9650 in/s2
F1  247.789 N90  247.789ˆj N
F  232.303 N  90  232.303ˆj N
2
2
M
B
  50 mm  kˆ × F1   250 mm  kˆ × F2   325 mm  kˆ × FA


 12389.57 mm  N  ˆi   58076.06 mm  N  ˆi   325 in  kˆ × FAx ˆi  FAy ˆj  0
FA  140.575ˆj N  140.575 N90
F    F  F  F   156.061ˆj N  156.061 lb  90
B
1
2
A
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Ans.
Ans.
Theory of Machines and Mechanisms, 4e
17.7
Uicker et al.
The shaft illustrated in Fig. P17.7 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 1.1125 N, 0.834 N,
and 1.39 N, 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 N mm and their angular orientations.


m1R1  1.1125 N  125 inˆj  139.062ˆj N  mm
m2R 2   0.834 N 100 mm  150  83.437 N  mm  150  72.256ˆi  41.718ˆj N  mm
m3R3  1.39 N 125 mm  60  173.828 N  mm  60  86.914ˆi  150.542ˆj N  mm
Using Eqs. (17.6) and (17.7),
m1R1 450 mm 875 mm  71.519ˆj N  mm
m R 675 mm 875 mm  55.743ˆi  32.186ˆj N  mm
2
2
m3R3 200 mm 875 mm  19.865ˆi  34.411ˆj N  mm
m R  35.878ˆi  4.922ˆj N  mm  36.211 N  mm  7.8
Ans.
mRR R  50.535ˆi  58.121ˆj N  mm  77.019 N  mm131.0
Ans.
L
L
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Theory of Machines and Mechanisms, 4e
17.8
Uicker et al.
The shaft of Problem 17.7 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.
mLR L  36.211 N  mm172.2
mR R R  77.019 N  mm  49.0
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Ans.
Ans.
Theory of Machines and Mechanisms, 4e
17.9
Uicker et al.
The shaft illustrated in Fig. P17.7 is to be balanced by subtracting masses in the two
correction planes L and R. The three masses are m1  6 g, m2  8 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.


m1R1   6 g  125 mmˆj  750.000ˆj g  mm
m2 R 2  8 g 150 mm  150  1 200 g  mm  150  1 039.230ˆi  600.000ˆj g  mm
m R   5 g 100 mm  60  500 g  mm  60  250.000ˆi  433.013ˆj g  mm
3
3
Using Eqs. (17.6) and (17.7),
m1R1 900 mm 1 125 mm  600.000ˆj g  mm
m R 1 050 mm 1 125 mm  969.948ˆi  560.000ˆj g  mm
2
2
m3R3 300 mm 1 125 mm  66.667ˆi  115.470ˆj g  mm
The masses to be removed are:
mL R L  903.281ˆi  75.470ˆj g  mm  906.428 g  mm  175.2
m R  114.051ˆi  207.543ˆj g  mm  236.816 g  mm  61.2
R
R
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Ans.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
17.10 Repeat Problem 17.9 if masses are to be added in the two correction planes.
mL R L  903.281ˆi  75.470ˆj g  mm  906.428 g  mm4.8
m R  114.051ˆi  207.543ˆj g  mm  236.816 g  mm118.8
R
R
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Ans.
Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
17.11 Solve the two-plane balancing problem as stated in Section 17.8.
This is an experimental procedure and is explained in Section 17.8; no further solution
process is shown here.
17.12 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
because of 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.
X A  5142 , XB  3  22 ,
m L  12210 , X AL  8160 , XBL  4260 ,
m R  6  70 , X AR  274 , XBR  4.5  80 .
These gave the following results from a programmable calculator run:
M L  6.05234 , M R  5.9865.2
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Ans.
Theory of Machines and Mechanisms, 4e
Uicker et al.
Chapter 18
Cam Dynamics
18.1
In Fig. P18.1a, the mass m is constrained to move only in the vertical direction. The
circular cam has an eccentricity of 50 mm, a speed of 25 rad/s, and the weight of the
mass is 26.7 N. Neglecting friction, find the angle    t at the instant the cam jumps.
 F  W  F  my
my  mg  F
y   2 y0 cos t
y  y0 1  cos t 
F  m 2 y0 cos t  mg
Jump begins when F = 0; that is, when
m 2 y0 cos t  mg  0
9650 mm/s 2
mg
cos t  


 0.308 8
2
m 2 y0
 25 rad/s   50 mm 
  t  cos1  0.308 8  107.99
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Ans.
Theory of Machines and Mechanisms, 4e
18.2
Uicker et al.
In Fig. P18.1a, the mass m is driven up and down by the eccentric cam and it has a
weight of 8 lb. The cam eccentricity is 0.75 in. Assume no friction.
(a)
Derive the equation for the contact force.
(b)
Find the cam velocity  corresponding to the beginning of the cam jump.
(a)
(b)
From the solution to Problem 18.1, we have for the contact force
F  m 2 y0 cos t  mg
Jump begins when cos t  1 and F = 0: that is, when
m 2 y0  mg  0

g
386 in/s 2

 22.69 rad/s
y0
 0.75 in 
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Ans.
Ans.
Theory of Machines and Mechanisms, 4e
18.3
Uicker et al.
In Fig. P18.1a, the slider has a mass of 2.5 kg. The cam is a simple eccentric and causes
the slider to rise 30 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 18.1, we have for the contact force
F  m 2 y0 cos t  mg
Jump begins when cos t  1 and F = 0; that is, when m 2 y0  mg  0

g

y0
9.81 m/s 2
 18.08 rad/s  172.7 rev/min
 0.030 m 
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Ans.
Theory of Machines and Mechanisms, 4e
18.4
(a)
(b)
(a)
Uicker et al.
The cam-and-follower system illustrated in Fig. P18.1b has k  0.9 kN / m ,
m  0.80 kg, y  15  15 cos  t mm, and   60 rad / s. The retaining
spring is assembled with a preload of 2.4 N.
Compute the maximum and minimum values of the contact force.
If the follower is found to jump off the cam, compute the angle  t corresponding
to the very beginning of jump.
Let Fc = contact force, and P = preload.
my  ky  P  Fc
 F  Fc  ky  P  my
y  0.015  0.015cos 60t m ,
y  54cos 60t m/s2
Fc   0.80 kg   54cos 60t m/s 2    900 N/m  0.015  0.015cos 60t m   2.4 N
(b)
 15.9  29.7 cos 60t N
Fc,max  15.9  29.7 N  45.6 N ,
Fc,min  0
Jump begins when Fc = 0; that is, when
  60t  cos1  15.9 N 29.7 N   122.37
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Ans.
Ans.
Theory of Machines and Mechanisms, 4e
18.5
(a)
(b)
(a)
Uicker et al.
Figure P18.1b illustrates 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 N/mm, and the spring preload is 26.7 N corresponding to the y  0 position.
The weight of the mass is 160.2 N.
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.
At what speed in revolutions per minute would jump begin?
Just as in Problem 18.4, if we let Fc = contact force and P = preload:
my  ky  P  Fc
 F  Fc  ky  P  my
Using first-order kinematic coefficients and assuming that the input shaft speed is
constant, then y  y 2 and
Fc  160.2 N 9650 mm/s2  y 2   7.12 N/mm  y  26.7 N
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   t  75 or point B
  t  180 or point C   t  330 , the three points where the contact
force drops discontinuously, depending on whether  is large enough for the
contact force to indicate a negative value.
(b)
For point A   t  75 , L  50 mm ,   150  2.618 rad . From Eq. (6.6a),
y  25 mm and, from Eq. (6.6c), y  1.167 in/rad 2 . Therefore,
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
Fc , A  160.2 N 9650 mm/s 2  y 2   7.12 N/mm  y  26.7 N
  0.485 N  s 2   2  204.7 N
Thus, Fc , A  0 for   20.559 rad/s
For point B
  t  180 ,
L  50 mm ,   150  2.618 rad .
From Eq.
(6.12a), y  50 mm and, from Eq. (6.21c), y  36 mm/rad . Therefore,
2
Fc , B  160.2 N 9650 mm/s 2  y 2   7.12 N/mm  y  26.7 N
  0.485 N  s 2   2  382.7 N
Thus, Fc , B  0 for   25.308 rad/s
For point C   t  330  , y  y  0 , and Fc,C  26.7 N for all values of  .
Of these cases, jump begins at A when   20.559 rad/s  196.3 rev/min .
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Ans.
Theory of Machines and Mechanisms, 4e
18.6
Uicker et al.
A cam-and-follower mechanism is illustrated in abstract form in Fig. P18.1b. 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 , and then return to the starting position in
the remaining 180 of cam rotation, also with harmonic motion. The spring is assembled
with a 20-N preload and it has a rate of 4.25 kN/m. The follower mass is 18 kg.
Compute the cam speed in revolutions per minute at which jump would begin.
Just as in Problem 18.4, if we let Fc = contact force and P = preload:
my  ky  P  Fc
 F  Fc  ky  P  my
Using first-order kinematic coefficients and assuming that the input shaft speed is
constant, then y  y 2 and
Fc  18 kg  y 2   4 250 N/m  y  20 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   t  180  , L  0.025 m ,
  180  3.1416 rad .
From Eq. (6.15c), y  0.025 m and y  0.0125 m/rad 2 .
Therefore,
Fc  18 kg  y 2   4 250 N/m  y  20 N
  0.630 N  s 2   2  132 N
Thus, Fc  0 for   23.688 rad/s  226.2 rev/min.
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Ans.
Theory of Machines and Mechanisms, 4e
18.7
Uicker et al.
Figure P18.7 illustrates a lever OAB driven by a cam cut to give the roller a rise of 37.5
mm with parabolic motion and a parabolic return with no dwells. The lever and roller are
to be assumed weightless, and there is no friction. Calculate the jump speed if
l  125 mm
Taking moments about the fixed pivot
2
 M O  FA  2 mg  m  2  
FA   500 mm  m  mg
FA  4m   2mg
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   t  90 , L  37.5 mm ,
  180  3.1416 rad .
From
y  15.2 mm/rad . Therefore,
Eqs.
(6.6a)
and
(6.6c),
y  18.75 mm
and
2
  y 2
  15.2 mm/rad 2 125 mm   2  0.121 6 2
FA   500 mm  m  0.1216   2  m  9650 mm/s 2 
  60.8 mm  m 2  m  9650 mm/s 2 
Thus, FA  0 for   12.6 rad/s  120.3 rev/min.
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Ans.
Theory of Machines and Mechanisms, 4e
18.8
Uicker et al.
A cam-and-follower system similar to the one in Fig. 18.6 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 equation
F  0.325  10.75 y, where y is in meters and F is in kilonewtons. Dimensions
corresponding to Fig. 18.6 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.
  600 rev/min  62.832 rad/s
For simple harmonic rise motion, we use Eqs. (6.12) with L  0.020 m and   150 .
For the first part of the parabolic return motion, following Example 6.1,
2
y  0.020 1  2     m , y    0.080    m , y   0.080  2  0.008 106 m


For the second part of the parabolic return motion,
2
y  0.040 1     m , y    0.080  1     m , y  0.080  2  0.008 106 m
Then we can use Eq. (18.11)
F23y  325  10 750 y  14 000  y  0.0125   1.6  y 2  N
 500  3 250 y  6 317 y N
and Eqs. (18.9) and (18.13)
a tan   y   y
T12  a tan  F23y   yF23y
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Theory of Machines and Mechanisms, 4e
Uicker et al.
  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
© Oxford University Press 2015. All rights reserved.
-2.181
-4.089
-5.503
-6.284
-6.390
-5.871
-4.836
-3.422
-1.768
0
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
Theory of Machines and Mechanisms, 4e
18.9
Uicker et al.
Repeat Problem 18.8 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   0.025.
  900 rev/min  94.248 rad/s
For simple harmonic rise motion, we use Eqs. (6.12) with L  0.020 m and   150 .
For the first part of the parabolic return motion, following Example 6.1,
2
y  0.020 1  2     m , y    0.080    m , y   0.080  2  0.008 106 m


For the second part of the parabolic return motion,
2
y  0.040 1     m , y    0.080  1     m , y  0.080  2  0.008 106 m
Then we can use Eq. (18.11)
110  10 750 y  14 000  y  0.0125   1.6  y 2  N
y
F23 
1  1.666 667 y  0.033 333 tan  sgn y
285  24 750 y  14 212 y N
1  1.666 667 y  0.033 333 tan  sgn y
and Eqs. (18.9) and (18.13)
a tan   y   y
T12  a tan  F23y   yF23y

© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
  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
© Oxford University Press 2015. All rights reserved.
-1.85
-3.59
-5.05
-6.08
-6.54
-6.35
-5.49
-4.04
-2.13
0
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
Theory of Machines and Mechanisms, 4e
Uicker et al.
18.10 A plate cam drives a reciprocating roller follower through the distance L = 31.25 mm
with parabolic motion in 120 of cam rotation, dwells for 30 , and returns with cycloidal
motion in 120 , followed by a dwell for the remaining cam angle. The external load on
the follower is F14  160.2 N during the rise and zero during the dwells and the return.
In the notation of Fig. 18.6, R = 75 mm, r = 25 mm, lB  150 mm,
lC  200 mm, and k  26.7 N/mm. The spring is assembled with a preload of 166.875 N
when the follower is at the bottom of its stroke. The weight of the follower is 8 N, 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    60 , we use Eqs. (6.5a) (6.5c) with L  31.25 mm and   120 .
y  62.5    mm , y  59.675    mm , y  28.5 mm
2
For 60    120 , we use Eqs. (6.6a) – (6.6c) with L  31.25 mm and   120 .
2
y  31.25 1  2 1      mm , y  59.675 1     mm , y  28.5 mm


For 150    270 , we use Eqs. (6.13) with L  31.25 mm and   120 .
Then we can use Eq. (18.11)
F23y  F14  166.875  687.5 y  406.63 y N
and Eqs. (18.9) and (18.13)
a tan   y   y
T12  a tan  F23y   yF23y
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
  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
© Oxford University Press 2015. All rights reserved.
54.8
120.0
206.3
324.0
75.3
93.2
79.6
45.1
0
0
0
-18.7
-26.5
-61.2
-156.7
-206.2
-130.2
-27.2
0
0
0
0
0
0
0
Theory of Machines and Mechanisms, 4e
Uicker et al.
18.11 Repeat Problem 18.10 if friction exists with   0.04 and the cycloidal return takes place
in 180.
For 0    60 , we use Eqs. (6.6a) – (6.6c) with L  31.25 mm and   120 .
y  62.5    mm , y  59.675    mm , y  28.5 mm
2
For 60    120 , we use Eqs. (6.6a) – (6.6c) with L  28.5 mm and   120 .
2
y  31.25 1  2 1      mm , y  59.675 1     mm , y  28.5 mm


For 150    330 , we use Eqs. (6.13) with L  31.25 mm and   180 .
Then we can use Eq. (18.11)
F  166.875  667.5 y  406.63 y N
F23y  14
1   5.6 y  16.8 tan  sgn y
and Eqs. (18.9) and (18.13)
a tan   y   y
T12  a tan  F23y   yF23y
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
  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
© Oxford University Press 2015. All rights reserved.
55
122
211
332
77
95
80
45
0
0
0
-10
-31
-54
-76
-94
-106
-104
-83
-51
-21
-4
0
0
0
Theory of Machines and Mechanisms, 4e
Uicker et al.
Chapter 19
Flywheels, Governors, and Gyroscopes
19.1
Table P19.1 lists the output torque for a one-cylinder engine running at 4 500 rev/min.
(a)
Find the mean output torque.
(b)
Determine the mass moment of inertia of an appropriate flywheel using
Cs  0.018 .
Table P19.1 Torque data for Problem 19.1
i
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
i
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
i
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
i
deg
540
550
560
570
580
590
600
610
620
630
640
650
660
670
680
690
700
710
Using n = 72 and h = 4π/72, we enter the data from Table P19.1 into Simpson’s
rule to find U 2  U1  890.7 N  m .
Tm  U 2  U1   4   890.7 N  m   4 rad   70.88 N  m
(b)
Ti
Nm
0
-344
-540
-577
-572
-643
-793
-893
-836
-605
-379
-264
-300
-368
-334
-198
-56
-2
  4 500 rev/min  471.24 rad/s
Ans.
2
I  U 2  U1   Cs 2    890.7 N  m   0.018  471.24 rad/s    0.223 N  m  s2 Ans.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
19.2
Uicker et al.
Using the data of Table 19.2, determine the moment of inertia for a flywheel for a twocylinder 90 V engine having a single crank. Use Cs = 0.010 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 250 mm? Use  = 7.8
Mg/m3 as the density of steel.
Table 19.2 Torque data for a four-cylinder, four-cycle internal combustion engine
i
Ttotal
T
T 180
T +360
T 540
deg
Nm
m
m
Nm
Nm
0
0
0
0
0
0
15
311.5
-11.9
-9.5
-11.9
278.2
30
232.52
-22.9
-13.9
-22.9
172.7
45
270.3
-31.1
-9.9
-32.5
196.8
60
240.3
-35.9
0.9
-39.5
165.8
75
204.7
-34.5
14
-41.3
142.9
90
176.8
-26.9
26.9
-40.3
136.6
105
134.6
-14
34.5
-34.7
120.4
120
118.6
-0.9
35.9
-30.3
123.4
135
89.3
9.9
31.1
-30.5
99.9
150
59.2
13.9
22.9
-60.9
35
165
20.5
9.5
11.9
-84.5
-42.7
Using n = 48 and h = 4π/48, we integrate the data from columns 2-5 of Table 19.2 by
Simpson’s rule to find U 2  U1  394.32 N  m .
  4 600 rev/min  481.71 rad/s
2
I  U 2  U1   Cs 2    394.32 N  m   0.0100  481.71 rad/s    0.1699 kg  m2
m  2I R2  2  0.1699 kg  m2   0.350 m   2.7744 kg
V  m /   2.7744 kg 7 800 kg/m3  0.000 356 m3
2
t  V A  0.000 356 m3    0.350 m    0.000924 m  0.924 mm


2
19.3
i
deg
0
15
30
45
60
75
90
Ans.
Using the data of Table 19.1, 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.
. Table 19.1 Example 19.1: Torque data for Fig. 19.3
i
i
i
i
Ti
Ti
Ti
Ti
Ti
deg
deg
deg
deg
N.m
N.m
N .m
N .m
N .m
0
150
59.2
300
-0.9
450
26.9
600
-39.5
311.5
165
20.5
315
9.9
465
34.5
615
-41.3
232.5
180
0
330
13.9
480
35.9
630
-40.3
270.3
195
-11.9
345
9.5
495
31.1
645
-34.7
240.32 210
-22.9
360
0
510
22.9
660
-30.3
204.7
225
-31.1
375
-9.5
525
11.9
675
- 30.5
176.8
240
-35.9
390
-13.9
540
0
690
-60.9
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
105
120
135
134.6
118.6
89.3
255
270
285
-34.5
-26.9
-14
Uicker et al.
405
420
435
-9.9
0.9
14
555
570
585
-11.9
-22.9
-32.5
705
-84.5
Using n = 48 and h = 4π/48, we integrate the data from Table 19.1 by Simpson’s rule to
find U 2  U1  388.27 Nm .
Tm  U 2  U1   4    388.27 N  m   4 rad   30.89 N  m
  2 400 rev/min  251.3 rad/s
2
I  U 2  U1   Cs 2    388.27 N.m   0.03 251.3 rad/s    0.2 m  s 2
© Oxford University Press 2015. All rights reserved.
Ans.
Ans.
Theory of Machines and Mechanisms, 4e
19.4
Uicker et al.
The load torque required by a 200-tonne punch press is displayed in Table P19.4 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.
(a)
Determine the mean motor torque required at the flywheel shaft and the motor
horsepower needed, assuming a constant torque-speed characteristic for the
motor.
(b)
Find the moment of inertia needed for the flywheel.
i
deg
0
10
20
30
40
50
60
70
80
(a)
Ti
m
95.3
95.3
95.3
95.3
95.3
143.2
286.1
572.3
763
Table P19.4 Torque data for Problem 19.4
i
i
Ti
Ti
deg
deg
m
Nm
90
877.5
180
200.3
100
925.3
190
181.2
110
944.3
200
162.2
120
953.8
210
152.6
130
934.8
220
124
140
858.5
230
114.5
150
391
240
104.9
160
238.5
250
95.3
170
219.4
260
95.3
270
280
290
300
310
320
330
340
350
Ti
m
95.3
95.3
95.3
95.3
95.3
95.3
95.3
95.3
95.3
Using n = 36 and h = 2π/36, we integrate the data from Table P19.4 by Simpson’s
rule to find U  1857.875 N  m .
Ans.
Tm  U  2   1857.875 N  m   2 rad   295.7 Nm
  2 400 rev/min  251.3 rad/s
P  T 
(b)
i
deg
 295.7 N  m  2 400 rev/min  2
rad/rev 
 6 073 HP
Ans.
734.25 N  m/min/HP
The torque data show a constant requirement of 95.34 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  1658.18 N  m
2
Ans.
I  U  Cs 2   1658.18 N  m   0.075 251.3 rad/s    0.349 N  s 2


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Theory of Machines and Mechanisms, 4e
Find Tm for the four-cylinder engine whose torque displacement is that of Fig. 19.4.
333.75
NM
19.5
Uicker et al.
222.5
111.25
Table 19.2 Torque data for a four-cylinder, four-cycle internal combustion engine
i
Ttotal
T
T 180
T +360
T 540
deg
m
m
Nm
Nm
Nm
0
0
0
0
0
15
311.5
-11.9
-9.5
-11.9
278.2
30
232.5
-22.9
-13.9
-22.9
172.7
45
270.3
-31.1
-9.9
-32.5
196.8
60
240.3
-35.9
0.9
-39.5
165.7
75
204.7
-34.5
14
-41.3
142.9
90
176.8
-26.9
26.9
-40.3
136.6
105
134.6
-14
34.5
-34.7
120.4
120
118.6
-0.9
35.9
-30.3
123.4
135
89.3
9.9
31.1
-30.5
99.9
150
59.2
13.9
22.9
-60.9
35
165
20.5
9.5
11.9
-84.5
-42.7
Using n = 12 and h = π/12, we integrate the data from column 6 of Table 19.2 by
Simpson’s rule to find U 2  U1  388.3 N  m .
Tm  U 2  U1      388.3 N  m   rad   123.6 N  m
© Oxford University Press 2015. All rights reserved.
Ans.
Theory of Machines and Mechanisms, 4e
19.6
Uicker et al.
A pendulum mill is illustrated schematically in Fig. P19.6. 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 = 436 N; its principal mass moments of
inertia are I s  13.46 N  m  s2 and I  9.79 N  m  s2 . The length of the muller axle is
n
l  RGA  1000 mm  X iYi and the radius of the muller at its center of mass is
i 1
RGB  250 mm . Assuming that the vertical shaft is to be inclined at   30 and will be
driven at a constant angular velocity of  p  240 rev/min , find the crushing force
between the muller and the pan. Also determine the minimum angular velocity p
required to ensure contact between the muller and the pan.


 p  240 rev/min  25.133 rad/s
ωs  ω4/ 3  s sin  ˆi  cos ˆj
ω3   p ˆj  25.133 rad/sˆj
ω4  ω3  ω4/ 3  s sin  ˆi   p  s cos  ˆj
VG  ω3  R GA
VG  ω 4  R GB

  p ˆj  1000sin  ˆi  1000 cos  ˆj
 1000sin  p kˆ

 s sin  ˆi   p  s cos   ˆj   10 cos  ˆi  10sin  ˆj
  250s  250cos  p  kˆ
Equating these with   30 we find s    4sin   cos  p  2.866 p . Then
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
ω
p
Uicker et al.
ˆ
×ωs    p ˆj× s  sin  ˆi  cos  ˆj   ps sin  kˆ   4 sin 2   sin  cos    p2kˆ  1.433 p2k
I S  mr 2 2   4361 N 9650 mm/s2   250 mm  2  564.7 N  m  s2
2
2
2
I  mr 2 4  ml 2   4361N 9650 mm/s2   250 mm  4  1000 mm    444.755 Nms 2


Now Eq. (19.36) shows
p
 s

s
cos   p   s
 M  I  I  I
s




 p cos 

 4sin 2   sin  cos  N  m  s 2  2 kˆ
14.12
430.64
M



p
  4sin   cos    p 









2 2
 M  14.12sin   4sin   cos    430.64sin  cos   N  m  s  pkˆ
2 2
 M  173.29 N  m  s  pkˆ  10946.555 N  m kˆ
Now formulating the externally applied moments,
 M  Wˆj× RGA  Fcˆi × R BA
 M  4361 Nˆj× sin  ˆi  cos ˆj1000 mm  F ˆi × sin  ˆi  cos ˆj1000 mm   cos ˆi  sin  ˆj 250 mm
 M   4361 200sin  N  m kˆ  F  4cos   sin   250 nn kˆ 
c
c
 M   2180.5 N  m kˆ  741F
c
mm kˆ
1
n
and setting the two expressions equal
 M   2180.5 N  m kˆ  741Fc mm kˆ  10946.555 N  m kˆ
we can now solve for the crushing force
Fc  14769.55 kN
If we start before setting the angular velocity, then we have
 M  173.29 N  m  s2 p2kˆ   2180.5 N  m kˆ  741.025Fc mm kˆ
Ans.
Now by setting Fc to zero we can determine the minimum angular velocity p required to
ensure contact between the muller and the pan:
 p  2180.5 N  m 173.29 N  m  s2  3.547 rad/s  33.87 rev/min
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Ans.
Theory of Machines and Mechanisms, 4e
19.7
Uicker et al.
Use the gyroscopic formulae of this chapter to solve again the problem presented in
Example 14.9 of Chapter 14.
From the given data we can identify
ω p  ω2  5kˆ rad/s
ωs  ω3  350ˆi  5kˆ rad/s
I s  mk 2   4.5 kg  0.050 m   0.011 3 kg  m2
2
From Eq. (19.37)
 M  I ω


×ωs   0.011 3 kg  m2 5kˆ rad/s  350ˆi  5kˆ rad/s   19.8ˆj N  m


This brings us precisely to the formulation of Eq. (2) of Example 14.9. 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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Theory of Machines and Mechanisms, 4e
19.8
Uicker et al.
The oscillating fan illustrated in Fig. P19.8 precesses sinusoidally according to the
equation  p   sin1.5t , where   30 ; the fan blade spins at  s  1800ˆi rev/min . The
weight of the fan and motor armature is 23.36 N, and other masses can be assumed
negligible; gravity acts in the jˆ direction. The principal mass moments of inertia are
I s  7.23 N  mm  s2 and I  2.78 N  mm  s2 ; the center of mass is located at
RGC  100 mm to the front of the precession axis. Determine the maximum moment
M z that must be accounted for in the clamped tilting pivot at C.


 s  1800ˆi rev/min  188.5ˆi rad/s
ω p  1.5 sin ˆi  cos  ˆj rad/s
ω
p
 ωs   1.5  sin ˆi  cos  ˆj rad/s   188.5ˆi rad/s   282.7 cos  kˆ rad/s 2  244.9kˆ rad/s2
p
 s

s
cos    p   s
 M  I  I  I
s










1.5 rad/s


2
2
cos   244.9kˆ rad/s 2  1.77 N  m
 M  7.23 N  mm  s  4.45 N  mm  s
188.5 rad/s


On the other side of the equation, the external moments are
z
z
z
 M  M kˆ  RGC W  sin  ˆi  cos  ˆj  M kˆ  100 mm W cos  kˆ  M kˆ  2.02 N  m kˆ


Now, equating the two we can solve for the moment M z .
M z  0.24kˆ N  m
M z kˆ  2.02 N  m kˆ  1.77kˆ N  m
Ans.
Here we see that the gyroscopic moment is almost large enough to support the weight of
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Theory of Machines and Mechanisms, 4e
Uicker et al.
the fan motor.
19.9
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. (19.36) or (19.37) 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 direction of rotation of the engine and the direction of the turn. Eq. (19.37)
shows that the effect is likely to be very small since, for any reasonable rate of turn,
ω p  ωs  will be at least an order of magnitude smaller than  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.
© Oxford University Press 2015. All rights reserved.
Theory of Machines and Mechanisms, 4e
Uicker et al.
19.10 A large and very high-speed turbine is to operate at an angular velocity of
  18 000 rev/min and will have a rotor with a principal mass moment of inertia of
I s  25 N  m  s2 . 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 gyroscopic
loads on its bearings. Estimate the size of these additional loads.
ωs  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    25 N  m  s   0.022kˆ rad/s   0.5461kˆ N  m
2
p
s
2
s
p
2
s
Ans.
Thus, if the bearings were separated by only 125es, they would experience less than one
additional pound of loading. This is totally negligible.
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