DESIGN OF MACHINERY - 6th Ed.
SOLUTION MANUAL 4-6a-1
PROBLEM 4-6a
Statement:
The link lengths and value of 2 for some fourbar linkages are defined in Table P4-1.
The linkage configuration and terminology are shown in Figure P4-1. For row a, draw
the linkage to scale and graphically find all possible solutions (both open and crossed)
for angles 3 and 4. Determine the Grashof condition.
Given:
Solution:
Link 1
d  6  in
Link 2
a  2  in
Link 3
b  7  in
Link 4
c  9  in
θ2  30 deg
See figure below for one possible solution. Also see Mathcad file P0406a.
1.
Lay out an xy-axis system. Its origin will be the link 2 pivot, O2.
2.
Draw link 2 to some convenient scale at its given angle.
3.
Draw a circle with center at the free end of link 2 and a radius equal to the given length of link 3.
4.
Locate pivot O4 on the x-axis at a distance from the origin equal to the given length of link 1.
5.
Draw a circle with center at O4 and a radius equal to the given length of link 4.
6.
The two intersections of the circles (if any) are t he two solutions to the position analysis problem,
crossed and open. If the circles don't intersect, there is no solution.
7.
Draw links 3 and 4 in their two possible positions (shown as solid for open and dashed for crossed in the
figure) and measure their angles 3 and 4 with respect to the x-axis. From the solution below,
OPEN
θ31  88.84  deg
θ41  117.29 deg
CROSSED
θ32  360  deg  115.21 deg
θ32  244.790  deg
θ42  360  deg  143.66 deg
θ42  216.340  deg
DESIGN OF MACHINERY - 6th Ed.
8.
SOLUTION MANUAL 4-6a-2
y
Check the Grashof condition.
B
OPEN
3
4
88.837°
2
O2
117.286°
A
115.211°
O4
143.660°
CROSSED
B'
Condition( a b c d ) 
S  min ( a b c d )
L  max( a b c d )
SL  S  L
PQ  a  b  c  d  SL
return "Grashof" if SL  PQ
return "Special Grashof" if SL = PQ
return "non-Grashof" otherwise
Condition( a b c d )  "Grashof"
x
DESIGN OF MACHINERY - 6th Ed.
SOLUTION MANUAL 4-7a-1
PROBLEM 4-7a
Statement:
The link lengths and value of 2 for some fourbar linkages are defined in Table P4-1.
The linkage configuration and terminology are shown in Figure P4-1. For row a, find all
possible solutions (both open and crossed) for angles 3 and 4 using the vector loop
method. Determine the Grashof condition.
Given:
Solution:
1.
Link 1
d  6  in
Link 2
a  2  in
Link 3
b  7  in
Link 4
c  9  in
See Mathcad file P0407a.
Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a.
K1 
d
K2 
a
K1  3.0000
2
d
K3 
c
K2  0.6667
 
 
B  2  sin θ 
C  K   K  1   cos θ   K
2
2

B  1.0000
C  3.5566
3

θ41  2   atan2 2  A B 

B  4  A  C 
2
θ41  242.714  deg
θ41  θ41  360  deg

θ41  602.714  deg

Crossed: θ42  2   atan2 2  A B 

B  4  A  C 
2
θ42  216.340  deg
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12.
K4 
2
d
K5 
b
2
2
c d a b
2 a b
 
 
E  2  sin θ 
F  K   K  1   cos θ   K
4
2
K4  0.8571
D  1.6774
E  1.0000
2
1
2
K5  0.2857
D  cos θ2  K1  K4 cos θ2  K5
4.
2 a c
Use equation 4.10b to find values of 4 for the open and crossed circuits.
Open:
3.
2
A  0.7113
2
1
2
a b c d
K3  2.0000
A  cos θ2  K1  K2 cos θ2  K3
2.
θ2  30 deg
F  2.5906
5
Use equation 4.13 to find values of 3 for the open and crossed circuits.

Open:

θ31  2   atan2 2  D E 

E  4  D F 
2
θ31  271.163  deg
2
DESIGN OF MACHINERY - 6th Ed.
SOLUTION MANUAL 4-7a-2
θ31  θ31  360  deg
Crossed:
5.


θ32  2   atan2 2  D E 
θ31  631.163  deg

E  4  D F 
2
Check the Grashof condition.
Condition( a b c d ) 
S  min ( a b c d )
L  max( a b c d )
SL  S  L
PQ  a  b  c  d  SL
return "Grashof" if SL  PQ
return "Special Grashof" if SL = PQ
return "non-Grashof" otherwise
Condition( a b c d )  "Grashof"
θ32  244.789  deg
DESIGN OF MACHINERY - 6th Ed.
SOLUTION MANUAL 4-9a-1
PROBLEM 4-9a
Statement:
The link lengths, value of 2, and offset for some fourbar slider-crank linkages are defined
in Table P4-2. The linkage configuration and terminology are shown in Figure P4-2. For
row a, draw the linkage to scale and graphically find all possible solutions (both open and
crossed) for angles 3 and slider position d.
Given:
Solution:
Link 2
a  1.4 in
Link 3
b  4  in
Offset
c  1  in
θ2  45 deg
See figure below for one possible solution. Also see Mathcad file P0409a.
1.
Lay out an xy-axis system. Its origin will be the link 2 pivot, O2.
2.
Draw link 2 to some convenient scale at its given angle.
3.
Draw a circle with center at the free end of link 2 and a radius equal to the given length of link 3.
4.
Draw a horizontal line through y = c (the offset).
5.
The two intersections of the circle with the horizontal line (if any) are the two solutions to the
position analysis problem, crossed and open. If t he circle and line don't intersect, there is no
solution.
6.
Draw link 3 and the slider block in their two possible positions (shown as solid for open and dashed for
crossed in the figure) and measure the angle 3 and length d for each circuit. From the solution below,
θ31  360  deg  179.856  deg
θ31  180.144  deg
θ32  0.144  deg
d 2  3.010  in
d 1  4.990  in
DESIGN OF MACHINERY - 6th Ed.
SOLUTION MANUAL 4-9a-2
Y
d2 =
3.010
d1 =
4.990
3(CROSSED)
B'
0.144°
A
2
O2
45.000°
3 (OPEN)
B
179.856° 1.000
X
DESIGN OF MACHINERY - 6th Ed.
SOLUTION MANUAL 4-10a-1
PROBLEM 4-10a
Statement:
The link lengths, value of 2, and offset for some fourbar slider-crank linkages are defined
in Table P4-2. The linkage configuration and terminology are shown in Figure P4-2. For
row a, using the vector loop method, find all possible solutions (both open and crossed) for
angles 3 and slider position d.
Given:
Solution:
Link 2
a  1.4 in
Link 3
b  4  in
Offset
c  1  in
θ2  45 deg
See Figure P4-2 and Mathcad file P0410a.
Y
d2 =
3.010
d1 =
4.990
3(CROSSED)
B'
A
2
0.144°
45.000°
3 (OPEN)
B
179.856° 1.000
X
O2
1.
Determine 3 and d using equations 4.16 and 4.17.
Crossed:
 
 a sin θ2  c 
θ32  asin

b


θ32  0.144  deg
d 2  a  cos θ2  b  cos θ32
d 2  3.010  in
 
Open:
 
 
 a  sin θ2  c 
θ31  asin 
π
b


θ31  180.144  deg
d 1  a  cos θ2  b  cos θ31
d 1  4.990  in
 
 
DESIGN OF MACHINERY - 6th Ed.
SOLUTION MANUAL 4-23-1
PROBLEM 4-23
Statement:
For the linkage in Figure P4-10, calculate and plot the angular displacement of links 3 and 4
and the path coordinates of point P with respect to the angle of the input crank O2A for one
revolution.
B
y
3
b
P
p
4
A
2
a
2
4
c
d
Given:
x
1
O2
O4
Link lengths:
Ground link
d  2.22
Crank
a  1.0
Coupler
b  2.06
Rocker
c  2.33
Coupler point data:
p  3.06
Solution:
δ  31.00  deg
See Figure P4-10 and Mathcad file P0423.
1.
Define one revolution of t he input crank: θ2  0  deg 1  deg  360  deg
2.
Use equation 4.8a and 4.10 to calculate 4 as a function of 2 (for the open circuit).
K1 
d
K2 
a
2
d
K3 
c
K1  2.2200
K2  0.9528
   
B θ   2  sin θ 
 
C θ   K   K
2
2 a c
K3  1.5265
A θ2  cos θ2  K1  K2 cos θ2  K3
2
 
2


2


   
θ41 θ2  2   atan2 2  A θ2 B θ2 
3.
1
2
  
 1  cos θ2  K3

 2  4A θ Cθ  
B θ2
2
2
If the calculated value of 4 is greater than 2, subtract 2 from it.
 
  
 
 
θ4 θ2  if θ41 θ2  2  π θ41 θ2  2  π θ41 θ2
4.
2
a b c d
Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12.
2
DESIGN OF MACHINERY - 6th Ed.
K4 
SOLUTION MANUAL 4-23-2
2
d
K5 
b
   
E θ   2  sin θ 
2
2
c d a b
2
K4  1.0777
2 a b
 
F θ   K   K
K5  1.1512
D θ2  cos θ2  K1  K4 cos θ2  K5
2
5.
2
2
1
  
 1  cos θ2  K5
Use equation 4.13 to find values of 3 for the open circuit.


 


   
θ31 θ2  2   atan2 2  D θ2 E θ2 
6.
4

 2  4Dθ Fθ  
E θ2
2
2
If the calculated value of 3 is greater than 2, subtract 2 from it.
 
  
 
 
θ3 θ2  if θ31 θ2  2  π θ31 θ2  2  π θ31 θ2
Plot 3 and 4 as functions of the crank angle 2 (measured from the ground link).
Angular Displacement of Coupler
 240
 260
Coupler angle, deg
7.
 
θ3 θ2  280
deg
 300
 320
 340
0
60
120
180
θ2
deg
Crank angle, deg
240
300
360
DESIGN OF MACHINERY - 6th Ed.
SOLUTION MANUAL 4-23-3
Angular Displacement of Rocker
 200
Rocker angle, deg
 220
   240
θ4 θ 2
deg
 260
 280
0
60
120
180
240
300
360
θ2
deg
Crank angle, deg
8.
Use equation 4.31 to define the x- and y-components of the vector RP.
RP  RA  RPA
  
 
 p   cos θ  δ  j  sin θ  δ 
θ   acosθ   p cosθ θ   δ
RA  a  cos θ2  j  sin θ2
RPA
RPx
9.
3
2
3
2
3
2
 
 
   
RPy θ2  a  sin θ2  p  sin θ3 θ2  δ
Plot the coordinates of the coupler point in the local xy coordinate system.
DESIGN OF MACHINERY - 6th Ed.
SOLUTION MANUAL 4-23-4
COUPLER POINT CURVE
3
2.5
Y
2
1.5
1
0.5
1.5
2
2.5
3
X
3.5
4