design of machinery solution manual chapter 6

DESIGN OF MACHINERY SOLUTION MANUAL 6-1a-1

PROBLEM 6-1a

Statement: A ship is steaming due north at 20 knots (nautical miles per hour). A submarine is laying in wait 1/2mile due west of the ship. The sub fires a torpedo on a course of 85 degrees. The torpedo travelsat a constant speed of 30 knots. Will it strike the ship? If not, by how many nautical miles will itmiss? Hint: Use the relative velocity equation and solve graphically or analytically.

Units: naut_mile 1 knotsnaut_mile

hr

Given: Speed of ship Vs 20 knots s 90 deg

Speed of torpedo Vt 30 knots t 15 deg

Initial distance between ship and torpedo di 0.5 naut_mile

Note that, for compass headings, due north is 0 degrees, due east 90 degrees, and the angleincreases clockwise. However, in a right-handed Cartesian system, due north is 90 degrees (up)and due east is 0 degrees (to the right). The Cartesian system has been used above to define theship and torpedo headings.

Solution: See Mathcad file P0601a.

1. The key to this solution is to recognize that the only information of interest is the relative velocity of one vesselto the other. The ship captain wants to know the relative velocity of the torpedo versus the ship, Vts = Vt - Vs.In effect, we want to resolve the situation with respect to a moving coordinate system attached to the ship.

2. The figure below shows the initial positions of the torpedo and ship and their velocities.

shiptorpedo

Vs = 20 knots

0.5 n. mi.

Vt = 30 knots

3. The figure below shows the vector diagram that solves the relative velocity equation Vts = Vt - Vs. For thetorpedo to hit the moving ship, the relative velocity vector has to be perpendicular to the ship's velocity vector(if you were on the ship observing the torpedo, it would appear to be headed directly for you). As the velocitydiagram shows, the relative velocity vector is not perpendicular to the ship's velocity vector so it will miss andpass behind the ship.

22.89°V ts

V t

-Vs

4. Determine the distance by which the torpedo will miss the ship.

Time required for torpedo to travel 0.5 nautical miles due east

tt_eastdi

Vt cos t tt_east 62.117 s

Distance traveled by the torpedo due north in that time

dt_north Vt sin t tt_east dt_north 0.134naut_mile

Distance traveled by the ship due north in that time

ds_north Vs tt_east ds_north 0.345naut_mile

Distance by which the torpedo will miss the ship

dmiss ds_north dt_north dmiss 0.211naut_mile

DESIGN OF MACHINERY SOLUTION MANUAL 6-1b-1

PROBLEM 6-1b

Statement: A plane is flying due south at 500 mph at 35,000 feet altitude, straight and level. A second planeis initially 40 miles due east of the first plane, also at 35,000 feet altitude, flying straight and levelat 550 mph. Determine the compass angle at which the second plane would be on a collisioncourse with the first. How long will it take for the second plane to catch the first? Hint: Use therelative velocity equation and solve graphically or analytically.

Given:

Speed of first plane V1 500 mph 1 270 deg

Speed of second plane V2 550 mph

Initial distance between planes di 40 mi

Note that, for compass headings, due north is 0 degrees, due east 90 degrees, and the angleincreases clockwise. However, in a right-handed Cartesian system, due north is 90 degrees (up)and due east is 0 degrees (to the right). The Cartesian system has been used above to define theplane headings.

Solution: See Mathcad file P0601b.

1. The key to this solution is to recognize that the only information of interest is the relative velocity of one planeto the other. For a collision to occur, the relative velocity of the second plane with respect to the first must beperpendicular to the velocity vector of the first.

2. The figure below shows the initial positions of the two planes and their velocities.

V2 = 550 mph

40 mi.

V1 = 500 mph

plane 1

plane 2

3. The figure below shows the vector diagram that solves the relative velocity equation V2 = V1 + V21. Toconstruct this diagram, chose a convenient velocity scale and draw V1 to its correct length with the arrow headpointing straight down (indicating due south). From the tip of the vector, layoff a horizontal construction line tothe left (due west) an undetermined length. From the tail of the V1 vector, construct a circle whose radius isequal to the scaled length of vector V2. The intersection of the circle and the horizontal construction linedetermines the length of V21. Draw the arrowheads for V2 and V21 pointing toward the intersection of the circleand construction line. Label the horizontal vector V21 and the vector that joins the tail of V1 with the head of V21

as V2. The angle between V1 and V2 is the required direction for V2 in order that plane 2 collides with plane 1.

24.620°

2V

V21

V1

4. The angle can also be determined analytically from the velocity triangle asfollows.

acosV1

V2

24.620 deg

5. The time it will take for the second plane to catch the first is the time that it will take plane 2 to travel the 40 milesto the west.

tdi

V2 sin t 628.468 s t 10.474 min

DESIGN OF MACHINERY SOLUTION MANUAL 6-2-1

PROBLEM 6-2

Statement: A point is at a 6.5-in radius on a body in pure rotation with = 100 rad/sec. The rotation center isat the origin of a coordinate system. When the point is at position A, its position vector makes a 45deg angle with the X axis. At position B, its position vector makes a 75 deg angle with the X axis.Draw this system to some convenient scale and:a. Write an expression for the particle's velocity vector in position A using complex numbernotation, in both polar and Cartesian forms.b. Write an expression for the particle's velocity vector in position B using complex numbernotation, in both polar and Cartesian forms.c. Write a vector equation for the velocity difference between points B and A. Substitute thecomplex number notation for the vectors in this equation and solve for the velocity differencenumerically.d. Check the result of part c with a graphical method.

Given:

Rotation speed 100rad

sec

Vector angles A 45 deg B 75 deg

Vector magnitude R 6.5 in

Solution: See Mathcad file P0602.

1. Calculate the magnitude of the velocity at points A and B using equation 6.3.

V R V 650.000in

sec

2. Establish an X-Y coordinate frame and draw a circle with center at the origin and radius R.

3. Draw lines from the origin that make angles of 45 and 75 deg with respect to the X axis. Label the intersections ofthe lines with the circles as A and B, respectively. Make the line segment OA a vector by putting an arrowheadat A, pointing away from the origin. Label the vector RA. Repeat for the line segment OB, labeling it RB.

4. Choose a convenient velocity scale and draw the two velocity vectors VA and VB at the tips of RA and RB,respectively. The velocity vectors will be perpendicular to their respective position vectors.

4 82 6

500 in/sec

4

2

0

RA

BR

8

6

BVB VA

Y

A

X

Velocity scale:

Distance scale:

0

0 1 2 in

a. Write an expression for the particle's velocity vector in position A using complex number notation, in bothpolar and Cartesian forms.


Polar form: RA R ej A

RA 6.5 ej

4

VA R j ej A

VA 650 jej

4

Cartesian form: VA R j cos A j sin A

VA 459.619 459.619j( )in

sec

b. Write an expression for the particle's velocity vector in position B using complex number notation, in bothpolar and Cartesian forms.

Polar form: RB R ej B

RA 6.5 ej

4

VB R j ej B

VB 650 jej

75

180

Cartesian form: VB R j cos B j sin B

VB 627.852 168.232j( )in

sec

c. Write a vector equation for the velocity difference between points B and A. Substitute the complex numbernotation for the vectors in this equation and solve for the position difference numerically.

VBA VB VA VBA 168.232 291.387j( )in

sec

d. Check the result of part c with a graphical method. Solve the equation VB = VA + VBA using a velocity scaleof 250 in/sec per drawing unit.

Velocity scale:

0.673

V

1.166

VBA

B vO

VA

Y

500 in/sec

X

0

Velocity scale factor

kv 250in

sec

Horizontal component

VBAx 0.673 kv

VBAx 168.3in

sec

Vertical component

VBAy 1.166 kv VBAy 291.5in

sec

On the layout above the X and Y components of VBA are equal to the real and imaginary components calculated,confirming that the calculation is correct.


PROBLEM 6-3

Statement: A point A is at a 6.5-in radius on a body in pure rotation with = -50 rad/sec. The rotation center isat the origin of a coordinate system. At the instant considered its position vector makes a 45 degangle with the X axis. A point B is at a 6.5-in radius on another body in pure rotation with= +75rad/sec. Its position vector makes a 75 deg angle with the X axis. Draw this system to someconvenient scale and:a. Write an expression for the particle's velocity vector in position A using complex numbernotation, in both polar and Cartesian forms.b. Write an expression for the particle's velocity vector in position B using complex numbernotation, in both polar and Cartesian forms.c. Write a vector equation for the velocity difference between points B and A. Substitute thecomplex number notation for the vectors in this equation and solve for the position differencenumerically.d. Check the result of part c with a graphical method.

Given:

Rotation speeds 50rad

sec 75

rad

sec

Vector angles A 45 deg B 75 deg

Vector magnitude R 6.5 in

Solution: See Mathcad file P0603.

1. Calculate the magnitude of the velocity at points A and B using equation 6.3.

VA R VA 325.000in

sec

VB R VB 487.500in

sec

1. Establish an X-Y coordinate frame and draw a circle with center at the origin and radius R.

2. Draw lines from the origin that make angles of 45 and 75 deg with respect to the X axis. Label the intersections ofthe lines with the circles as A and B, respectively. Make the line segment OA a vector by putting an arrowheadat A, pointing away from the origin. Label the vector RA. Repeat for the line segment OB, labeling it RB.

3. Choose a convenient velocity scale and draw the two velocity vectors VA and VB at the tips of RA and RB,respectively. The velocity vectors will be perpendicular to their respective position vectors.

Velocity scale:

Distance scale:

R

2

20

4 B

4 6

RA

VA

BB

6

8

Y

V

A

8X

0

0 1

500 in/sec

2 in

a. Write an expression for the particle's velocity vector on body A using complex number notation, in both polarand Cartesian forms.


Polar form: RA R ej A

RA 6.5 ej

4

VA R j ej A

VA 325 jej

4

Cartesian form: VA R j cos A j sin A

VA 229.810 229.810j( )in

sec

b. Write an expression for the particle's velocity vector on body B using complex number notation, in both polarand Cartesian forms.

Polar form: RB R ej B

RA 6.5 ej

4

VB R j ej B

VB 487.5 jej

75

180

Cartesian form: VB R j cos B j sin B

VB 470.889 126.174j( )in

sec

c. Write a vector equation for the velocity difference between the points on bodies B and A. Substitute thecomplex number notation for the vectors in this equation and solve for the position difference numerically.

VBA VB VA VBA 700.699 355.984j( )in

sec

d. Check the result of part c with a graphical method. Solve the equation VB = VA + VBA using a velocity scaleof 250 in/sec per drawing unit.

Velocity scale:

2.803

V

1.424 V VA

vOB

BA

Y

500 in/sec

X

0

Velocity scale factor

kv 250in

sec

Horizontal component

VBAx 2.803 kv

VBAx 700.7in

sec

Vertical component

VBAy 1.424 kv VBAy 356.0in

sec

On the layout above the X and Y components of VBA are equal to the real and imaginary components calculated,confirming that the calculation is correct.


PROBLEM 6-4a

Statement: For the fourbar defined in Table P6-1, line a , find the velocities of the pin joints A and B, and of theinstant centers I1,3 and I2,4. Then calculate 3 and 4 and find the velocity of point P. Use agraphical method.

Given: Link lengths:

Link 1 d 6 in Link 2 a 2 in

Link 3 b 7 in Link 4 c 9 in

Crank angle: 30 deg Crank velocity: 10 rad sec1

Coupler point data:

Rpa 6 in 30 deg

Solution: See Figure P6-1 and Mathcad file P0604a.

1. Draw the linkage to scale in the position given, find the instant centers, distances from the pin joints to theinstant centers and the angles that links 3 and 4 make with the x axis.

6.8067

P

O2

B

A

I

OI

1,3

2,4 4

88.8372°

117.2861°

148.2007°

1.7118 4.2882

5.9966

SCALE0 1 IN

3.3384

From the layout above:

O2I24 1.7118 in O4I24 4.2882 in

AI13 3.3384 in BI13 5.9966 in PI13 6.8067 in

117.2861 deg 88.8372 deg

2. Use equation 6.7 and inspection of the layout to determine the magnitude and direction of the velocity at pointA.


VA a VA 20.0in

sec

VA 90 deg VA 120.0 deg

3. Determine the angular velocity of link 3 using equation 6.9a.

VA

AI13 5.991

rad

sec CW

4. Determine the magnitude of the velocity at point B using equation 6.9b. Determine its direction by inspection.VB BI13

VB 35.925in

sec

VB 90 deg VB 27.286 deg

5. Use equation 6.9c to determine the angular velocity of link 4.

VB

c 3.99

rad

sec CW

6. Use equation 6.9d and inspection of the layout to determine the magnitude and direction of the velocity at pointP.

VP PI13 VP 40.778in

sec

VP 148.2007 deg 90 deg VP 58.201 deg


PROBLEM 6-5a

Statement: A general fourbar linkage configuration and its notation are shown in Figure P6-1. The linklengths, coupler point location, and the values of2 and2 for the same fourbar linkages as usedfor position analysis in Chapter 4 are redefined in Table P6-1, which is the same as Table P4-1. Forrow a, find the velocities of the pin joints A and B, and coupler point P. Calculate 3 and4. Draw

the linkage to scale and label it before setting up the equations.Given: Link lengths:

Link 1 d 6 Link 2 a 2 Link 3 b 7 Link 4 c 9

Coupler point: Rpa 6 30 deg

Link 2 position and velocity: 30 deg 10

Two argument inverse tangent atan2 x y( ) 0.5 return x 0= y 0if

1.5 return x 0= y 0if

atany

x

return x 0if

atany

x

otherwise


1. Draw the linkage to scale and label it.

4

B'

2

143.660°

115.211°

CROSSED

O

88.837°

2A

3

y B

117.286°

O

4

x

OPEN

2. Determine the values of the constants needed for finding 4from equations 4.8a and 4.10a.

K1d

a K2

d

c

K1 3.0000 K2 0.6667

K3a

2b

2 c

2 d

2

2 ac K3 2.0000

A cos K1 K2 cos K3

B 2 sin

C K1 K2 1 cos K3

A 0.7113 B 1.0000 C 3.5566

3. Use equation 4.10b to find values of4 for the open and crossed circuits.

Open: 2 atan2 2 A B B2

4 A C 477.286 deg

360 deg 117.286 deg

Crossed: 2 atan2 2 A B B2

4 A C 216.340 deg

4. Determine the values of the constants needed for finding 3 from equations 4.11b and 4.12.


K4d

b K5

c2

d2

a2

b2

2 a b K4 0.8571 K5 0.2857

D cos K1 K4 cos K5 D 1.6774

E 2 sin E 1.0000

F K1 K4 1 cos K5 F 2.5906

5. Use equation 4.13 to find values of3 for the open and crossed circuits.

Open: 2 atan2 2 D E E2

4 D F 448.837 deg

360 deg 88.837deg

Crossed: 2 atan2 2 D E E2

4 D F 244.789 deg

6. Determine the angular velocity of links 3 and 4 for the open circuit using equations 6.18.

a

b

sin sin 5.991

a

c

sin sin 3.992

7. Determine the velocity of points A and B for the open circuit using equations 6.19.

VA a sin j cos

VA 10.000 17.321j VA 20 arg VA 120 deg

VB c sin j cos

VB 31.928 16.470j VB 35.926 arg VB 27.286 deg

8. Determine the velocity of the coupler point P for the open circuit using equations 6.36.

VPA Rpa sin j cos

VPA 31.488 17.337j

VP VA VPA

VP 21.488 34.658j VP 40.779 arg VP 58.201deg

9. Determine the angular velocity of links 3 and 4 for the crossed circuit using equations 6.18.

a

b

sin sin 0.662

a

c

sin sin 2.662

10. Determine the velocity of point B for the crossed circuit using equations 6.19.


VB c sin j cos

VB 14.195 19.295j VB 23.954 arg VB 126.340 deg

11. Determine the velocity of the coupler point P for the crossed circuit using equations 6.36.

VPA Rpa sin j cos

VPA 3.960 0.332j

VP VA VPA

VP 13.960 16.989j VP 21.989 arg VP 129.411deg


PROBLEM 6-6a

Statement: The general linkage configuration and terminology for an offset fourbar slider-crank linkage areshown in Fig P6-2. The link lengths and the values of 2 and2 are defined in Table P6-2. For rowa, find the velocities of the pin joints A and B and the velocity of slip at the sliding joint using agraphical method.


Link 2 a 1.4 in Link 3 b 4 in

Offset c 1 in 45 deg 10 rad sec1


1. Draw the linkage to scale and indicate the axes of slip and transmission as well as the directions of velocities ofinterest.

Direction of VBA

Axis of slip andDirection of VB

45.000°2

O2

Direction of VA

A

Y

179.856° 1.000

X

Axis of transmission

3 B

2. Use equation 6.7 to calculate the magnitude of the velocity at point A.

VA a VA 14.000in

sec VA 90 deg VA 135 deg

3. Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B. The equation to be solvedgraphically is

VB = VA + VBA

a. Choose a convenient velocity scale and layout theknown vector VA.b. From the tip of VA, draw a construction line with thedirection of VBA, magnitude unknown.c. From the tail of VA, draw a construction line withthe direction of VB, magnitude unknown.d. Complete the vector triangle by drawing VBA fromthe tip of VA to the intersection of the VB constructionline and drawing VB from the tail of VA to theintersection of the VBA construction line.

1.975

VB

BAV

VA

135.000°

X

Y

0 5 units/sec

4. From the velocity triangle we have:


Velocity scale factor: kv5 in sec

1

in

VB 1.975 in kv VB 9.875in

sec VB 180 deg

5. Since the slip axis and the direction of the velocity of point B are parallel, Vslip = VB.


PROBLEM 6-7a

Statement: The general linkage configuration and terminology for an offset fourbar slider-crank linkage areshown in Fig P6-2. The link lengths and the values of 2 and2 are defined in Table P6-2. For rowa, find the velocities of the pin joints A and B and the velocity of slip at the sliding joint using theanalytic method. Draw the linkage to scale and label it before setting up the equations.


Link 2 (O2 A) a 1.4 Link 3 (AB) b 4 Offset (yB) c 1

Crank angle 45 deg Crank angular velocity 10



4.990d =

0.144° 45.000°2

O2

3.010

3(CROSSED)B'

2d =

A

Y

179.856° 1.000

X

3 (OPEN)

1

B

2. Determine3 and d using equations 4.16 and 4.17.

Crossed: asina sin c

b

0.144 deg

d2 a cos b cos d2 3.010

Open: asina sin c

b

180.144 deg

d1 a cos b cos d1 4.990

3. Determine the angular velocity of link 3 using equation 6.22a:

Open a

b

cos cos 2.475

Crossed a

b

cos cos 2.475

4. Determine the velocity of pin A using equation 6.23a:

VA a sin j cos

VA 9.899 9.899i VA 14.000 arg VA 135.000 deg


5. Determine the velocity of pin B using equation 6.22b:

Open VB1 a sin b sin VB1 9.875

Crossed VB2 a sin b sin VB2 9.924

The angle of VB is 0 deg if VB is positive and 180 deg if VB negative.

6. The velocity of slip is the same as the velocity of pin B.


PROBLEM 6-8a

Statement: The general linkage configuration and terminology for an inverted fourbar slider-crank linkage areshown in Fig P6-3. The link lengths and the values of 2 and2 and are defined in Table P6-3.For row a , using a graphical method, find the velocities of the pin joints A and B and the velocity ofslip at the sliding joint. Draw the linkage to scale.

Given: Link lengths:Link 1 d 6 in Link 2 a 2 in

Link 4 c 4 in 90 deg 30 deg 10 rad sec1



127.333°

Axis of slip andDirection of VB

Axis of transmission and direction of VBA

b

1.793

0

30.000°

2

a

d

A

Direction of VA

90.0°y

B

142.666°

c

04

x


VA a VA 20.000in

sec VA 90 deg VA 120 deg

3. Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B on link 3. The equation to besolved graphically is

VB3 = VA3 + VBA3

0.771

1.846

VB

VAY

BAV

52.667°

X

10 in/sec0

a. Choose a convenient velocity scale and layout theknown vector VA.b. From the tip of VA, draw a construction line with thedirection of VBA, magnitude unknown.c. From the tail of VA, draw a construction line with thedirection of VB, magnitude unknown.d. Complete the vector triangle by drawing VBA from thetip of VA to the intersection of the VB construction lineand drawing VB from the tail of VA to the intersection ofthe VBA construction line.



1

in


VB3 0.771 in kv VB3 7.710in

sec VB3 52.667 deg

VBA3 1.846 in kv VBA3 18.460in

sec

5. Determine the angular velocity of link 3 using equation 6.7.

From the linkage layout above: b 1.793 in and 142.666 deg

VBA3

b 10.296

rad

sec CW

The way in which link 3 slides in link 4 requires that

6. Determine the magnitude and sense of the vector VB4 using equation 6.7.

VB4 c VB4 41.182in

sec

VB4 90 deg VB4 52.666 deg

7. Note that VB3 and VB4 are in the same direction in this case. The velocity of slip is

Vslip VB3 VB4 Vslip 33.472in

sec

slip VB4 180 deg slip 232.666deg


PROBLEM 6-9a

Statement: The general linkage configuration and terminology for an inverted fourbar slider-crank linkage areshown in Fig P6-3. The link lengths and the values of 2 and2 and are defined in Table P6-3.For row a , using an analytic method, find the velocities of the pin joints A and B and the velocity ofslip at the sliding joint. Draw the linkage to scale and label it before setting up the equations.

Given: Link lengths:Link 1 d 6 Link 2 a 2

Link 4 c 4 90 deg 30 deg 10



atany

x

return x 0if

atany

x

otherwise



127.333°b

0

79.041°

B'

30.000°

2

A

a

d

90.0°y

B

169.040°

142.666°

c

04

x

2. Determine the values of the constants needed for finding 4 from equations 4.8a and 4.10a.

P a sin sin a cos d cos P 1.000

Q a sin cos a cos d sin Q 4.268

R c sin R 4.000 T 2 P T 2.000

S R Q S 0.268 U Q R U 8.268

3. Use equation 4.22c to find values of 4 for the open and crossed circuits.

OPEN 2 atan2 2 S T T2

4 SU 142.667 deg

CROSSED 2 atan2 2 S T T2

4 SU 169.041 deg

4. Use equation 4.18 to find values of3 for the open and crossed circuits.


OPEN 232.667 deg

CROSSED 259.041 deg

5. Determine the magnitude of the instantaneous "length" of link 3 from equation 4.20a.

OPEN b1a sin c sin

sin b1 1.793

CROSSED b2a sin c sin

sin b2 1.793

6. Determine the angular velocity of link 4 using equation 6.30c:

OPEN a cos

b 1 c cos 10.292

CROSSED a cos

b 2 c cos 3.639


VA a sin j cos

VA 10.000 17.321i VA 20.000 arg VA 120.000 deg

8. Determine the velocity of point B on link 4 using equation 6.31:

OPEN VB4x1 c sin VB4x1 24.966

VB4y1 c cos VB4y1 32.734

VB41 VB4x12

VB4y12

VB41 41.168

VB1 atan2 VB4x1 VB4y1 VB1 52.667 deg

CROSSED VB4x2 c sin VB4x2 2.767

VB4y2 c cos VB4y2 14.289

VB42 VB4x22

VB4y22

VB42 14.555

VB2 atan2 VB4x2 VB4y2 VB2 79.041 deg

9. Determine the slip velocity using equation 6.30a:

OPEN Vslip1a sin b 1 sin c sin

cos Vslip1 33.461

CROSSED Vslip2a sin b 2 sin c sin

cos Vslip2 33.461


PROBLEM 6-10a

Statement: The link lengths, gear ratio (), phase angle (), and the values of 2 and2 for a geared fivebar

from row a of Table P6-4 are given below. Draw the linkage to scale and graphically find 3 and 4,using a graphical method.


Link 1 f 6 in Link 3 b 7 in Link 5 d 4 in

Link 2 a 1 in Link 4 c 9 in

Gear ratio, phase angle, and crank angle:

2 30 deg 60 deg 10 rad sec1


1. Determine the angle of link 5 using the equation in Figure P6-4.

150 deg

2. Choose the pitch radii of the gears. Since the gear ratio is positive, an idler must be used between gear 2 andgear 5. Let the idler be the same diameter as gear 5, and let all three gears be in line.

f r2 3 r5 r2

r5

r5f

3 r5 1.200 in

r2 r5 r2 2.400 in

3. Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest.

B

O2 O 52

Direction of VBA

Direction of VBC

Direction of VC

Direction of VA

4

3

C

A 5

4. Use equation 6.7 to calculate the magnitude of the velocity at points A and C.

VA a VA 10.00in

sec 60 deg 90 deg

20.000rad

sec

VC d VC 80.00in

sec 150 deg 90 deg


5. Use equation 6.5 to (graphically) determine the magnitudes of the relative velocity vectors VBA and VBC. Theequation to be solved graphically is the last of the following three.

VB = VA + VBA VB = VC + VBC VA + VBA = VC + VBC

a. Choose a convenient velocity scale and layout the known vector VA.b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown.c. From the tail of VA, layout the known vector VC.d. From the tip of VC, draw a construction line with the direction of VBC, magnitude unknown.e. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VBC constructionline and drawing VBC from the tip of VC to the intersection of the VBA construction line.

0 50 in/sec

VBAV

BC

3.051

4.562

CV

VA

Y

X



1

in

VBA 4.562 in kv VBA 228.100in

sec

VBC 3.051 in kv VBC 152.550in

sec

7. Determine the angular velocity of links 3 and 4using equation 6.7.

VBA

b 32.586

rad

sec

VBC

c 16.950

rad

sec

Both links are rotating CCW.


PROBLEM 6-11a

Statement: The general linkage configuration and terminology for a geared fivebar linkage are shown in FigureP6-4. The link lengths, gear ratio (), phase angle (), and the values of 2 and 2 are defined in

Table P6-4. For row a, find 3 and 4, using an analytic method. Draw the linkage to scale andlabel it before setting up the equations.


Link 1 d 4 Link 2 a 1 Link 3 b 7

Link 4 c 9 Link 5 f 6 Input angle 60 deg

Gear ratio 2.0 Phase angle 30 deg 10



atany

x

return x 0if

atany

x

otherwise



5

B`

34

173.6421°

115.4074°

B 4

3

O2

2

y

5

177.7152°

124.0501°

C 150.0000°

xO

2. Determine the values of the constants needed for finding 3 and 4 from equations 4.24h and 4.24i.

A 2 c d cos a cos f A 36.646

B 2 c d sin a sin B 20.412

C a2

b2

c2

d2

f2

2 a fcos 2 d a cos f cos

2 ad sin sin

C 37.431

D C A D 0.785

E 2 B E 40.823

F A C F 74.077


G 2 b d cos a cos f G 28.503

H 2 b d sin a sin H 15.876

K a2

b2

c2

d2

f2

2 a fcos 2 d a cos f cos

2 ad sin sin

K 26.569

L K G L 1.933

M 2 H M 31.751

N G K N 55.072

3. Use equations 4.24h and 4.24i to find values of 3 and 4 for the open and crossed circuits.

OPEN 2 atan2 2 L M M2

4 LN 173.642 deg

2 atan2 2 D E E2

4 D F 177.715 deg

CROSSED 2 atan2 2 L M M2

4 LN 115.407 deg

2 atan2 2 D E E2

4 D F 124.050 deg

4. Determine the position and angular velocity of gear 5 from equations 4.23c and 6.32c

150.000 deg

20.000 CCW

Angular velocity of links 3 and 4 from equations 6.33

OPEN 2 sin a sin d sin

b cos 2 cos

32.585 CCW

a sin b sin d sin

c sin

16.948 CCW

CROSSED 2 sin a sin d sin

b cos 2 cos

75.191 CW

a sin b sin d sin

c sin

59.554 CW


PROBLEM 6-12

Statement: Find all of the instant centers of the linkages shown in Figure P6-5.

Solution: See Figure P6-5 and Mathcad file P0612.

a. This is a fourbar slider-crank with n 4 .

1. Determine the number of instant centers for this mechanism using equation 6.8a.

Cn n 1( )

2 C 6

2. Draw the linkage to scale and identify those ICs that can be found by inspection.

AY

2,3

1,2

XO 2

23,4

1,4 at infinity

3

4B

C

3. Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 and I2,4

I1,3: I1,2-I2,3 and I1,4-I3,4 I2,4: I1,2-I1,4 and I2,3-I3,4

A

1,2

1,4 at infinity

O

2

22,3

2,4

1,4 at infinity

3,4

B

3

4

C

1,3

1

2

3

4

b. This is a fourbar with planetary motion (roller 3), n 4 .


1,2 and 2,4 and 1,4

2,3 at contact point (behind link 4)

2

O2

4

1

1,3

A

3

Cn n 1( )

2 C 6

2. Draw the linkage to scale and identify those ICs thatcan be found by inspection, which in this case, is all ofthem.

c. This is a fourbar with n 4 .

1. Determine the number of instant centers for thismechanism using equation 6.8a.


Cn n 1( )

2 C 6


4

1,2 1,4

2,3

O2

2

A

3

C

3,4

O4

B



4

1,2 1,4

1,3 at infinity1,3 at infinity

1

2

32,4 at infinity

2,4 at infinity

2,3

O2

2 A

1,3 at infinity

3

C

3,4

O4

B

2,4 at infinity

2,4 at infinity

1,3 at infinity

4

d. This is a fourbar with n 4 .


Cn n 1( )

2 C 6


2

1,22,3

2O

1,4 at infinity

3,4

A

3

4




1,4 at infinity

1

2

3

42,4

1,4 at infinity

2

1,22,3

2O

1,3

3,4

A

3

4

e. This is a threebar with n 3 .


Cn n 1( )

2 C 3


O

1,2

2

2

O31,3

3

3. Use Kennedy's Rule and a linear graph to find the remaining IC, I2,3

I2,3: I1,2-I1,3

2,3

Common normal

O

1,2

2

2

O31,3

3

1

3 2

f. This is a fourbar with n 4 .


Cn n 1( )

2 C 6



1,2 at infinity

3

VA

1,4 at infinity

B

3,4

4

2,3

2A

C3. Use Kennedy's Rule and a linear graph to find the remaining 2

ICs, I1,3 and I2,4


1,2 at infinity

VA

1,4 at infinity

2,4 at infinity3,4

B

43

1,3 2

2,3

2A

3

C

4

1


PROBLEM 6-13



a. This is a fourbar inverted slider-crank with n 4 .


Cn n 1( )

2 C 6


11,4

1,2

2,3

1

3

4

2

3,4 at infinity



11,4

1,3

1,2

2,42,3

1

3

4

2

3,4 at infinity

3,4 at infinity3

4

1

2

b. This is a sixbar with slider, n 6 .


Cn n 1( )

2 C 15


1

1

2 3

1,2

2,3 1,6 at infinity

4

5 1

3,4; 3,5; 4,5

1,4

6

5,63. Use Kennedy's Rule and a linear graph to find the remaining7 ICs.

I1,3: I1,2-I2,3 and I1,4-I3,4

I1,5: I1,6-I5,6 and I1,4-I4,5

I2,5: I1,2-I1,5 and I2,4-I4,5




1

1,6 at infinity4,6

2,6

1,6 at infinity

to 2,4

to 2,4 1,5

2,5

1

2

1,2

3

2,3

1,3

1,6 at infinity

4

5 1

1,4

3,4; 3,5; and 4,5

65,6

1,6 at infinity

3,6

6 2

3

4

5

1

c. This is a sixbar with slider and roller with n 6 .


Cn n 1( )

2 C 15


3

4

1

1,4 at infinity

1,6

3,4

1

2

56

1

5,62,5

1,2

2,3

3. Use Kennedy's Rule and a linear graph to find theremaining 8 ICs.

I2,6: I1,2-I1,6 and I2,5-I5,6

I1,5: I1,6-I5,6 and I1,2-I2,5

I4,5: I1,4-I1,5 and I2,4-I2,5

I3,6: I3,6-I5,6 and I2,3-I2,6



3

4

1

1,4 at infinity

1,6

3,4

4

1,3

4,6

3,6

3,5

5,6

2,6

1,4 at infinity

1

1

2

5

2,51,5

6

4,5

1,2

2,32,4

1,4 at infinity

6 2

35

1

d. This is a sixbar with slider and roller with n 6 .


Cn n 1( )

2 C 15



1,6 at infinity

1,2

1,4

1

2

3

5

2,3; 2,5; and 3,5

14

3,4

6

1

5,6

3. Use Kennedy's Rule and a linear graph to find the remaining 7 ICs.




I4,6: I1,6-I1,4 and I4,5-I5,6

1,6 at infinity

1,3

1,6 at infinity

1,2

1,4

1,6 at infinity

2,6

1

2

3

5

2,3; 2,5; and 3,5

14

3,4

4,5

2,4

4,6

6

1

5,6

1,5

6 2

35

4

3,6

1


PROBLEM 6-14



a. This is a pin-jointed fourbar with n 4 .


Cn n 1( )

2 C 6


1,2

3,4

2

3

4

1

1,4

2,3

3. Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3 andI2,4


1,2

3,4

2

3

4

1

2,3

1,41,3

2

3

4

2,4

1

b. This is a fourbar inverted slider-crank, n 4 .


Cn n 1( )

2 C 6

2. Draw the linkage to scale and identify those ICs that can be found by inspection, which in this case, is all ofthem.

1,4

3,4 at infinity

1

2

3

4

2,3

1,2

3. Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3and I2,4


1,2

2,3

1,4

1,3

3,4 at infinity

1

24

3 1

3,4 at infinity

2,44

3

2

1,2 at infinity

3,4

1

2

3

2,3

4

1,4 at infinity

c. This is a fourbar double slider with n 4 .


Cn n 1( )

2 C 6




1,2 at infinity

3,4

2,4 at infinity

1

2

3

2,3

4

1,4 at infinity

1,3

3

4 2

1

I1,3: I1,2-I2,3 and I1,4-I3,4

I2,4: I1,2-I1,4 and I2,3-I3,4

d. This is a pin-jointed fourbar with n 4 .


Cn n 1( )

2 C 6


3,42,3

1

23

41,2

1,4

3. Use Kennedy's Rule and a linear graph to find the remaining 2 ICs,I1,3 and I2,4


3,42,3

2,4

1

23

4

1,2

1,3

4 2

1,4 3

1

e. This is a fourbar effective slider-crank with n 4 .


Cn n 1( )

2 C 6


2,3

1,4 at infinity

3

4

1

3,4

1,2

2

3. Use Kennedy's Rule and a linear graph to find the remaining 2 ICs,I1,3 and I2,4


2,3


3

4

1

3,4

1,3

1,2

2

3

4 2

2,4 1


f. This is a fourbar cam-follower with n 4 .


Cn n 1( )

2 C 6


1,2 at infinity

34

1

1,4

3,4

2,3

2

3. Use Kennedy's Rule and a linear graph to find the remaining 2 ICs, I1,3and I2,4



34

1

1,4

3,42,4

2,3

2

4

3

2

1,31

g. This is a fourbar slider-crank with n 4 .


Cn n 1( )

2 C 6


3

1,2

1

2

2,3

4

3,4

1,4 at infinity

3. Use Kennedy's Rule and a linear graph to find the remaining 2ICs, I1,3 and I2,4

I1,3: I1,2-I2,3 and I1,4-I3,4

I2,4: I1,2-I1,4 and I2,3-I3,4

3

1,2

1

2,4 2

2,31,4 at infinity

4

3,4

1,4 at infinity

3

1,3

4 2

1

1,4

3,4 at infinity

1

3

4

2,3 1,2

1

2

h. This is a fourbar inverted slider-crank with n 4 .


Cn n 1( )

2 C 6




1,3

1,4

3,4 at infinity

1

4

3

2,3 1,2

1

2

3,4 at infinity2,4

3,4 at infinity

4

3

2

1

I1,3: I1,2-I2,3 and I1,4-I3,4

I2,4: I1,2-I1,4 and I2,3-I3,4

i. This is a fourbar slider-crank (hand pump) with n 4 .


Cn n 1( )

2 C 6


1

1,4

43,4

1,2 at infinity 2

3

2,3



1

2,4 1,2 at infinity

1,4

43,4

1,2 at infinity

1,2 at infinity1,3

2

2,3

3

3

1

4 2


PROBLEM 6-15



a. This is a pin-jointed fourbar with n 4 .


Cn n 1( )

2 C 6


4

1,2 O2

3,4

B

2,3

23

B

1,4

4O

3. Use Kennedy's Rule and a linear graph to findthe remaining 2 ICs, I1,3 and I2,4

I1,3: I1,2-I2,3 and I1,4-I3,4

I2,4: I1,2-I1,4 and I2,3-I3,4

4

1,4

1,3

3,4 and 2,4

1,2

2,3

2

324

3

1

b. This is a pin-jointed fourbar with n 4 .


4

1,21,4

O4

2,3

2A

O2

3

3,4

BCn n 1( )

2 C 6

2. Draw the linkage to scale and identify those ICs thatcan be found by inspection, which in this case, is allof them.

3. Use Kennedy's Rule and a linear graph to find theremaining 2 ICs, I1,3 and I2,4

I1,3: I1,2-I2,3 and I1,4-I3,4


I2,4: I1,2-I1,4 and I2,3-I3,4

2,3

3,4

1,2

2,4

O2

2

A

1,4

O4

4

3B

3

24

1

1,3

c. This is an eightbar (three slider-cranks with a common crank) with n 8 .


1,2

5

2,3; 2,4; 2,5;3,4; 3,5; and 4,5

1,7 at infinity

1,6 at infinity

3,6

6

13

4,7

47 2

5,8

8

1,8 at infinity

Cn n 1( )

2 C 28

2. Draw the linkage to scale and identify those ICs that canbe found by inspection.


I1,5: I1,2-I2,5 and I1,8-I5,8

I2,6: I1,2-I1,6 and I2,3-I3,6

I2,7: I1,2-I1,7 and I2,4-I4,7

I3,7: I1,3-I1,7 and I3,4-I4,7 I6,8: I2,7-I2,8 and I3,7-I3,8 I3,8: I1,3-I3,8 and I3,5-I5,8




1,2

5

2,63,8

2,3; 2,4; 2,5; 3,4; 3,5; and 4,5

1,7 at infinity

2,7

1,6 at infinity

3,6 1

6

32,8

4,7

1,44,6

47 2

To 1,51,3

To 1,5

3,7

1,8 at infinity

5,8

8

Note that, for clarity, not all ICs are shown.


d. This is a pin-jointed fourbar with n 4 .


1,2 1,4

2

2,3

3

1

E

3,4

4

Cn n 1( )

2 C 6

2. Draw the linkage to scale and identify those ICsthat can be found by inspection.

3. Use Kennedy's Rule and a linear graph to findthe remaining 2 ICs, I1,3 and I2,4

I1,3: I1,2-I2,3 and I1,4-I3,4

I2,4: I1,2-I1,4 and I2,3-I3,4

1,2 1,4

2

2,3

2,4 at infinity

3

1

E

3

11,3 at infinity

3,4

4

4 2

e. This is an eightbar with n 8 .


1,2

2

2,5

2,3

4,6

5

3

1,886,8 7,8

6 7

4,74

1,4

4,5

3,4C 6Cn n 1( )

2

2. Draw the linkage to scale and identify those ICs thatcan be found by inspection.

3. The remaining 17 ICs are at infinity.

1,2

3,4

1,4 at infinity

1

4

2,3

2

3

f. This is an offset crank-slider withn 4 .

1. Determine the number of instantcenters for this mechanism usingequation 6.8a.

Cn n 1( )

2

C 6

2. Draw the linkage to scale andidentify those ICs that can befound by inspection.




1,2

3,4

1,4 at infinity

1,4 at infinity

1

4

2,3

22,4

3

1,3

3

24

1

1,4

2

2,5

1,4

4

5

3,4

3

5,6

6

1,6

2,3

g. This is a sixbar with n 6 .

1. Determine the number of instant centers for thismechanism using equation 6.8a.

Cn n 1( )

2 C 15

2. Draw the linkage to scale and identify those ICs thatcan be found by inspection.


I1,3: I1,4-I4,3 and I1,2-I3,2

2,4

2,6

2

1,2

3,6

1,4

4,5 and 1,3

4,6 at infinity

5

4

3,4

3

3,5 at infinity

2,3 and 1,5

5,6

6

1,6

3,5 at infinity

4,5I1,5: I1,6-I5,6 and I1,2-I2,5

I3,5: I1,3-I1,5 and I3,2-I2,5

I2,4: I1,2-I1,4 and I3,4-I2,3

I2,6: I1,6-I1,2 and I2,5-I5,6

I4,6: I1,4-I1,6 and I2,4-I2,6

I4,5: I4,6-I5,6 and I3,4-I3,5

I3,6: I3,4-I4,6 and I3,5-I5,6


PROBLEM 6-16a

Statement: The linkage in Figure P6-5a has the dimensions and crank angle given below. Find 3, VA, VB, and

VC for the position shown for 2 = 15 rad/sec in the direction shown. Use the velocity differencegraphical method.

Given: Link lengths: Coupler point data:Link 2 (O2 to A) a 0.80 in Distance from A to C p 1.33 in

Link 3 (A to B) b 1.93 in Angle BAC 38.6 deg

Offset c 0.38 in Crank angle: 34.3 deg

Input crank angular velocity 15 rad sec1

CCW

Solution: See Figure P6-5a and Mathcad file P0616a.


Direction of VBA

Direction of VA

A

OX

Y

2

34.300°

38.600°

C

BDirection of VB

154.502°

Direction of VCA


VA a VA 12.00in

sec 34.3 deg 90 deg


A

2.298

2.197

VB

VBA

V

124.300°

X

Y

0 5 in/secVB = VA + VBA

a. Choose a convenient velocity scale andlayout the known vector VA.b. From the tip of VA, draw a construction linewith the direction of VBA, magnitude unknown.c. From the tail of VA, draw a construction linewith the direction of VB, magnitude unknown.d. Complete the vector triangle by drawing VBAfrom the tip of VA to the intersection of the VBconstruction line and drawing VB from the tail ofVA to the intersection of the VBA constructionline.




1

in

VB 2.298 in kv VB 11.490in

sec B 180 deg


sec


VBA

b 5.692

rad

sec

6. Determine the magnitude and sense of the vector VCA using equation 6.7.

VCA p VCA 7.570in

sec

CA 154.502 180 38.6 90( ) deg CA 76.898 deg

7. Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C. The equation to be solvedgraphically is

A

1.130

V

VCA

V

153.280°

C

X

0Y

5 in/sec

VC = VA + VCA

a. Choose a convenient velocity scale and layoutthe known vector VA.b. From the tip of VA, layout the (now) knownvector VCA.c. Complete the vector triangle by drawing VC fromthe tail of VA to the tip of the VCA vector.



1

in

VC 1.130 in kv VC 5.650in

sec C 153.28 deg


PROBLEM 6-16b


VC for the position shown for 2 = 15 rad/sec in the direction shown. Use the instant centergraphical method.

Given: Link lengths: Coupler point data:

Link 2 (O2 to A) a 0.80 in Distance from A to C p 1.33 in




CCW

Solution: See Figure P6-5a and Mathcad file P0616b.


2.109

1,2

1,4 at infinity

O

2

22,3

2,4

A 2.019

1,4 at infinity

3,4

B

3

4

0.993

116.732°

C

1,3

From the layout:

AI13 2.109 in

BI13 2.019 in

CI13 0.993 in

C 360 116.732( ) deg

2. Use equation 6.7 and inspection of thelayout to determine the magnitude anddirection of the velocity at point A.

VA a

VA 12.000in

sec



VA

AI13 5.690

rad

sec CW

4. Determine the magnitude of the velocity at point B using equation 6.9b. Determine its direction by inspection.

VB BI13 VB 11.488in

sec

VC 180 deg

5. Determine the magnitude of the velocity at point C using equation 6.9b. Determine its direction by inspection.

VC CI13 VC 5.650in

sec

VC C 90 deg VC 153.268 deg

DESIGN OF MACHINERY SOLUTION MANUAL 6-16c-1

PROBLEM 6-16c


VC for the position shown for 2 = 15 rad/sec in the direction shown. Use an analytical method.

Given: Link lengths: Coupler point data:

Link 2 (O2 to A) a 0.80 in Distance from A to C Rca 1.33 in




CCW

0.380"

A

OX

Y

2

34.300°

38.600°

B

154.502°

CSolution: See Figure P6-5a and Mathcad file P0616c.


2. Determine3 and d using equation 4.17.

asina sin c

b

154.502 deg

d1 a cos b cos d1 2.403 in


a

b

cos cos 5.691

rad

sec


VA a sin j cos

VA 6.762 9.913j( )in

sec VA 12.000

in

sec arg VA 124.300 deg


VB a sin b sin

VB 11.490in

sec VB 11.490

in

sec arg VB 180.000 deg

6. Determine the velocity of the coupler point C for the open circuit using equations 6.36.

VCA Rca sin j cos

VCA 1.716 7.371j( )in

sec

VC VA VCA

VC 5.047 2.542j( )in

sec VC 5.651

in

sec arg VC 153.268 deg

Note that3 is defined at point B for the slider-crank and at point A for the pin-jointed fourbar. Thus, to use

equation 6.36a for the slider-crank, 180 deg must be added to the calculated value of 3.


PROBLEM 6-17a

Statement: The linkage in Figure P6-5c has the dimensions and effective crank angle given below. Find 3,4,

VA, VB, and VC for the position shown for 2 = 15 rad/sec in the direction shown. Use the velocitydifference graphical method.

Given: Link lengths: Coupler point:

Link 2 (point of contact to A) a 0.75 in Distance A to C p 1.2 in

Link 3 (A to B) b 1.5 in Angle BAC 30 deg

Link 4 (point of contact to B) c 0.75 in

Link 1 (between contact points) d 1.5 in Crank angle: 77 deg


Solution: See Figure P6-5c and Mathcad file P0617a.

1. Although the mechanism shown in Figure P6-5c is not entirely pin-jointed, it can be analyzed for the positionshown by its effective pin-jointed fourbar, which is shown below. Draw the linkage to a convenient scale.Indicate the directions of the velocity vectors of interest.

77.000°

Effective link 4

Direction of VCA

Direction of VBA

30.000°

Direction of VA

A

O2 O4

b

77.000°ca

d

Effective link 2

2 4

CY

3 Direction of VB

B

X

0 0.5 1 in


VA a VA 11.250in

sec 77 deg 90 deg

3. Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relativevelocity VBA, and the angular velocity of link 3. The equation to be solved graphically is

VB = VA + VBA

a. Choose a convenient velocity scale and layout the known vector VA.b. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown.c. From the tail of VA, draw a construction line with the direction of VB, magnitude unknown.d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB constructionline and drawing VB from the tail of VA to the intersection of the VBA construction line.


0 5 in/sec

167.000°

Direction of VB

VAVB

Direction of

X

YVBA


VB VA VB 11.250in

sec 167 deg

VBA 0in

sec

5. Determine the angular velocity of links 3 and 4 using equation 6.7.

VBA

b 0.000

rad

sec

VB

c 15.000

rad

sec

6. Determine the magnitude of the vector VCA using equation 6.7.

VCA p VCA 0.000in

sec


VC = VA + VCA

Normally, we would draw the velocity triangle represented by this equation. However, since VCA is zero,

VC VA VC 11.250in

sec

C C 167.000deg


PROBLEM 6-17b


VA, VB, and VC for the position shown for 2 = 15 rad/sec in the direction shown. Use the instantcenter graphical method.


Link 2 (point of contact to A) a 0.75 in Distance A to C p 1.2 inLink 3 (A to B) b 1.5 in Angle BAC 30 degLink 4 (point of contact to B) c 0.75 in



CCW

Solution: See Figure P6-5c and Mathcad file P0617b.

1. Although the mechanism shown in Figure P6-5c is not entirely pin-jointed, it can be analyzed for the positionshown by its effective pin-jointed fourbar, which is shown below. Draw the linkage to scale in the positiongiven, find the instant centers, distances from the pin joints to the instant centers and the angles that links 3 and4 make with the x axis.

4

1,3 at infinity

2,3

30.000°2,4 at infinity

2,4 at infinity

1,2

1,3 at infinity

O2

2 A

77.000°

1,3 at infinity

3

C

3,4

1,4

O4

B

2,4 at infinity

2,4 at infinity

77.000°

1,3 at infinity

Since I1,3

is at infinity, link 3 is not rotating ( 0 rad sec1

). Thus, the velocity of every point on link 3 is

the same.


77.000 deg 0.000 deg


VA a VA 11.250in

sec


3. Determine the magnitude of the velocity at point B knowing that it is the same as that of point A.

VB VA VB 11.250in

sec




VB

c 15

rad

sec CCW

5. Determine the magnitude of the velocity at point C knowing that it is the same as that of point A.

VC VA VC 11.250in

sec

VC VA VC 167.000 deg


PROBLEM 6-17c


VA, VB, and VC for the position shown for 2 = 15 rad/sec in the direction shown. Use ananalytical method.


Link 2 (point of contact to A) a 0.75 in Distance A to C Rca 1.2 in

Link 3 (A to B) b 1.5 in Angle BAC 30 deg

Link 4 (point of contact to B) c 0.75 in





atany

x

return x 0if

atany

x

otherwise

Solution: See Figure P6-5c and Mathcad file P0617c.


30.000°

77.000°

Effective link 2

O2

2

a

A

d

b

Y

3

C

77.000°

Effective link 4

O4

X

4

c

B

0 0.5 1 in


K1d

a K2

d

c K3

a2

b2

c2

d2

2 ac

K1 2.0000 K2 2.0000 K3 1.0000

A cos K1 K2 cos K3

B 2 sin

C K1 K2 1 cos K3

A 1.2250 B 1.9487 C 2.3251

3. Use equation 4.10b to find values of4 for the open circuit.

2 atan2 2 A B B2

4 A C 437.000 deg


360 deg 77.000deg


K4d

b K5

c2

d2

a2

b2

2 a b K4 1.0000

K5 2.0000


E 2 sin E 1.9487

F K1 K4 1 cos K5 F 0.0000

5. Use equation 4.13 to find values of3 for the open circuit.

2 atan2 2 D E E2

4 D F 360.000 deg

360 deg 0.000deg


a

b

sin sin 0.000

rad

sec

a

c

sin sin 15.000

rad

sec


VA a sin j cos

VA 10.962 2.531j( )in

sec VA 11.250

in


VB c sin j cos

VB 10.962 2.531j( )in

sec VB 11.250

in


8. Determine the velocity of the coupler point C for the open circuit using equations 6.36.

VCA Rca sin j cos

VCA 0.000in

sec

VC VA VCA

VC 10.962 2.531j( )in

sec VC 11.250

in



PROBLEM 6-18a

Statement: The linkage in Figure P6-5f has the dimensions and coupler angle given below. Find 3, VA, VB,and VC for the position shown for VA = 10 in/sec in the direction shown. Use the velocitydifference graphical method.

Given: Link lengths and angles: Coupler point:

Link 3 (A to B) b 1.8 in Distance A to C p 1.44 in

Coupler angle 128 deg Angle BAC 49 deg

Slider 4 angle 59 deg Input slider velocity VA 10 in sec1

Solution: See Figure P6-5f and Mathcad file P0618a.


128.000°

3

59.000°

A 2

b 49.000°

VA

B

Y

4

Direction of VB

Direction of VBA

Direction of VCA

X

0 0.5 1 in

C

2. The magnitude and sense of the velocity at point A.

VA 10.000in

sec 180 deg


3.436

59.000°

4.784

AV

38.000°

0 5 in/sec Y

X

VB

VBA

VB = VA + VBA

a. Choose a convenient velocity scale andlayout the known vector VA.b. From the tip of VA, draw a construction linewith the direction of VBA, magnitude unknown.c. From the tail of VA, draw a construction linewith the direction of VB, magnitude unknown.d. Complete the vector triangle by drawingVBA from the tip of VA to the intersection of theVB construction line and drawing VB from thetail of VA to the intersection of the VBAconstruction line.




1

in

59 degVB 3.438 in kv VB 17.190in

sec

38 degVBA 4.784 in kv VBA 23.920in

sec


VBA

b 13.289

rad

sec


VCA p VCA 19.136in

sec

CA 128 49 90( ) deg CA 11.000 deg


VC = VA + VCA

a. Choose a convenient velocity scale and layout the known vector VA.b. From the tip of VA, layout the (now) known vector VCA.c. Complete the vector triangle by drawing VC from the tail of VA to the tip of the VCA vector.

3.827

1.902

Y

V

11.000°

A

0 5 in/sec

22.572°V

CAV

C

X



1

in

C 22.572 degVC 1.902 in kv VC 9.510in

sec


PROBLEM 6-18b

Statement: The linkage in Figure P6-5f has the dimensions and coupler angle given below. Find 3, VA, VB,and VC for the position shown for VA = 10 in/sec in the direction shown. Use the instant centergraphical method.


Link 3 (A to B) b 1.8 in Distance A to C p 1.44 in



Solution: See Figure P6-5f and Mathcad file P0618b.


1,2 at infinity

A

1.293

0.716

1,3

VA

3

1,4 at infinity

2,4 at infinity3,4

B

4

67.428°

2,3

2

0.753

C

From the layout:

AI13 0.753 in BI13 1.293 in

CI13 0.716 in C 67.428 deg

2. Determine the angular velocity of link 3using equation 6.9a.

VA

AI13 13.280

rad

sec CW

3. Determine the magnitude of the velocity at point B usingequation 6.9b. Determine its direction by inspection.

VB BI13 VB 17.17in

sec

VB VB 59.000 deg


VC CI13 VC 9.509in

sec

VC C 90 deg VC 22.572 deg


PROBLEM 6-18c

Statement: The linkage in Figure P6-5f has the dimensions and coupler angle given below. Find 3, VA, VB,and VC for the position shown for VA = 10 in/sec in the direction shown. Use an analytical method.


Link 3 (A to B) b 1.8 in Distance A to C Rca 1.44 in



Solution: See Figure P6-5f and Mathcad file P0618c.

1. Draw the mechanism to scale and define a vector loop using the fourbar slider-crank derivation in Section 6.7 asa model.

49 .000°

59.000°

VA

R2

b

R 4

Y

3

4R3

B

128.000°

A 2X

0 0.5 1 in

C

2. Write the vector loop equation, differentiate it, expand the result and separate into real and imaginary parts tosolve for3 and VB.

R2 R3 R4

a ej

b ej

c ej

where a is the distance from the origin to point A, a variable; b is the distance from A to B, a constant; and c isthe distance from the origin to point B, a variable. Angle 2 is zero,3 is the angle that AB makes with the x axis,

and4 is the constant angle that slider 4 makes with the x axis. Differentiating,

tad

dj b e

j

tcd

d

ej

Substituting the Euler equivalents,

tad

db sin j cos

tcd

d

cos j sin

Separating into real and imaginary components and solving for3 and VB. Note that dc/dt = VB and da/dt = VA

VA tan

b sin tan cos 13.288

rad

sec

VBVA b sin

cos VB 17.180

in

sec


VB VB cos j sin arg VB 59.000 deg

3. Determine the velocity of the coupler point C using equations 6.36.

VCA Rca sin j cos

VCA 18.783 3.651j( )in

sec

VA VA

VC VA VCA

VC 8.783 3.651j( )in

sec VC 9.512

in



PROBLEM 6-19

Statement: The cam-follower in Figure P6-5d has O2A = 0.853 in. Find V4, Vtrans, and Vslip for the position

shown with 2 = 20 rad/sec in the direction (CCW) shown.

Given: 20 rad sec1

O2A a 0.853 in

Assumptions: Rolling contact (no sliding)

Solution: See Figure P6-5d and Mathcad file P0619.


Direction of VA

0.853


2

2O

A

Axis of slip

3

4B

Direction of V4


VA a VA 17.060in

sec

3. Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A. The equationto be solved graphically is

VA = Vtrans + Vslip

0.870

1.523

0.768

trans

Vslip

V

AV

Y

XV4

0 10 in/sec

a. Choose a convenient velocity scale and layout theknown vector VA.b. From the tip of VA, draw a construction line with thedirection of Vslip , magnitude unknown.c. From the tail of VA, draw a construction line with thedirection of Vtrans, magnitude unknown.d. Complete the vector triangle by drawing Vslip fromthe tip of Vtrans to the tip of VA and drawing Vtrans fromthe tail of VA to the intersection of the Vslip constructionline.



1

in Vtrans 0.768 in kv Vtrans 7.680

in

sec

Vslip 1.523 in kv Vslip 15.230in

sec V4 0.870 in kv V4 8.700

in

sec


PROBLEM 6-20

Statement: The cam-follower in Figure P6-5e has O2A = 0.980 in and O3A = 1.344 in. Find 3, Vtrans, and Vslip

for the position shown with 2 = 10 rad/sec in the direction (CW) shown.

Given: 10 rad sec1

Distance from O2

to A: a 0.980 in

Distance from O3 to A: b 1.344 in

Assumptions: Roll-slide contact

Solution: See Figure P6-5e and Mathcad file P0620.



1.344

Direction of VA3

Direction of VA2

O

0.980

2 A

2

O33

Axis of slip

2. Use equation 6.7 to calculate the magnitude of thevelocity at point A.

VA a VA 9.800in

sec

3. Use equation 6.5 to (graphically) determine themagnitude of the velocity components at point A.The equation to be solved graphically is

VA2 = Vtrans + VA2slip

a. Choose a convenient velocity scale and layout the known vector VA.b. From the tip of VA, draw a construction line with the direction of Vslip, magnitude unknown.c. From the tail of VA, draw a construction line with the direction of Vtrans, magnitude unknown.d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA and drawing Vtrans fromthe tail of VA to the intersection of the Vslip construction line.

1.599

1.134

2.048

A2slip

2.598

VV

V

A3

A3slip

VA2

Y

transV

X

0 5 in/sec



1

in

VA2slip 1.134 in kv VA2slip 5.670in

sec

Vtrans 1.599 in kv Vtrans 7.995in

sec

VA3slip 2.048 in kv VA3slip 10.240in

sec

VA3 2.598 in kv VA3 12.990in

sec


VA3

b 9.665

rad

sec CCW

6. The relative slip velocity is Vslip VA3slip VA2slip Vslip 4.570in

sec


PROBLEM 6-21a

Statement: The linkage in Figure P6-6b has L1 = 61.9, L2 = 15, L3 = 45.8, L4 = 18.1, L5 = 23.1 mm. 2 is 68.3 degin the xy coordinate system, which is at -23.3 deg in the XY coordinate system. The X componentof O2C is 59.2 mm. Find, for the position shown, the velocity ratio VI5,6/VI2,3 and the mechanicaladvantage from link 2 to link 6. Use the velocity difference graphical method.


Link 1 d 61.9 mm Link 2 a 15.0 mm

Link 3 b 45.8 mm Link 4 c 18.1 mm

Link 5 e 23.1 mm Offset f 59.2 mm from O2

Crank angle: 45 deg Global XY system

Coordinate rotation angle 23.3 deg Global XY system to local xy system

Solution: See Figure P6-6b and Mathcad file P0621a.


Direction of VB

23 .3 °

1

Direction of VA

3

O2

245.0°A

2,3

yY

BDirection of VBA

4

1

O4

x

Direction of VCB

Direction of VC

5,6

5 1

6

C

X

2. Choose an arbitrary value for the magnitude ofthe velocity at I2,3 (point A). Let link 2 rotateCCW.

VA 10mm

sec

VC 45 deg 90 deg

VC 135.000 deg

3. Use equation 6.5 to (graphically) determine themagnitude of the velocity at point B, the magnitudeof the relative velocity VBA, and the angular velocityof link 3. The equation to be solved graphically is

VB = VA + VBA


2.248

2.053

VB

VBA

VA

167.553°

X

Y0 5 mm/sec


Velocity scale factor:

kv5 mm sec

1

in

VB 2.248 in kv

VB 11.2mm

sec

VB 360 167.558( ) deg



VC = VB + VCB

a. Choose a convenient velocity scale and layout the (now) known vector VB.b. From the tip of VB, draw a construction line with the direction of VCB, magnitude unknown.c. From the tail of VB, draw a construction line with the direction of VC, magnitude unknown.d. Complete the vector triangle by drawing VCB from the tip of VB to the intersection of the VC constructionline and drawing VC from the tail of VB to the intersection of the VCB construction line.

VB

VBA

VA

1.128

V

V

CB

C

X

Y0 5 mm/sec


Velocity scale factor: kv5 mm sec

1

in

VC 1.128 in kv VC 5.6mm

sec VC 270 deg

7. The ratio VI5,6

/VI2,3

isVC

VA0.564

8. Use equations 6.12 and 6.13 to derive an expression for the mechanical advantage for this linkage where theinput is a rotating crank and the output is a slider.

FinTin

rin=

Pin

rin in=

Pin

VA=

FoutPout

Vout=

Pout

VC=

mAFout

Fin=

Pout

VC

VA

Pin= mA

VA

VC mA 1.773


PROBLEM 6-21b

Statement: The linkage in Figure P6-6b has L1 = 61.9, L2 = 15, L3 = 45.8, L4 = 18.1, L5 = 23.1 mm. 2 is 68.3 degin the xy coordinate system, which is at -23.3 deg in the XY coordinate system. The X componentof O2C is 59.2 mm. Find, for the position shown, the velocity ratio VI5,6/VI2,3 and the mechanicaladvantage from link 2 to link 6. Use the instant center graphical method.


Link 1 d 61.9 mm Link 2 a 15.0 mm

Link 3 b 45.8 mm Link 4 c 18.1 mm

Link 5 e 23.1 mm Offset f 59.2 mm from O2


Coordinate rotation angle 23.3 deg Global XY system to local xy system

Solution: See Figure P6-6b and Mathcad file P0621b.


50.121

44.594

1

3

O2

2

A2,3

Y

B

22.683

4

1

O4

5,6

11 .377

5 1

6

C

X

1,5

1,3

From the layout:

AI13 44.594 mm BI13 50.121 mm

BI15 22.683 mm CI15 11.377 mm

2. Choose an arbitrary value for the magnitude of thevelocity at I2,3 (point A). Let link 2 rotate CCW.

VA 10mm

sec

VC 90 deg

VC 135.000 deg

3. Determine the angular velocity of link 3 using equation6.9a.

VA

AI13 0.224

rad

sec CW


VB BI13 VB 11.239mm

sec


VB

BI15 0.495

rad

sec CW


VC CI15 VC 5.637mm

sec downward

7. The ratio VI5,6

/VI2,3

isVC

VA0.56



FinTin

rin=

Pin

rin in=

Pin

VA=

FoutPout

Vout=

Pout

VC=

mAFout

Fin=

Pout

VC

VA

Pin= mA

VA

VC mA 1.77


PROBLEM 6-22a

Statement: The linkage in Figure P6-6d has L2 = 15, L3 = 40.9, L5 = 44.7 mm. 2 is 24.2 deg in the XY coordinatesystem. Find, for the position shown, the velocity ratio VI5,6/VI2,3 and the mechanical advantagefrom link 2 to link 6. Use the velocity difference graphical method.


Link 2 a 15.0 mm Link 3 b 40.9 mm

Link 5 c 44.7 mm Offset f 0 mm from O2

Crank angle: 24.2 deg

Solution: See Figure P6-6d and Mathcad file P0622a.


5

Direction of VB

O

Direction of VA

2,3

2

1

A

2

3

14

CDirection of VBA

5,6

1

6B

2. Choose an arbitrary value for the magnitude of the velocity at I2,3 (point A). Let link 2 rotate CCW.

VA 10mm

sec VC 90 deg VC 114.200 deg


VB = VA + VBA


5 mm/sec

1.073

VBA

VBX

VA

Y0




1

in

VB 1.073 in kv VB 5.4mm

sec VB 180 deg

5. The ratio VI5,6

/VI2,3

isVB

VA0.54


FinTin

rin=

Pin

rin in=

Pin

VA=

FoutPout

Vout=

Pout

VB=

mAFout

Fin=

Pout

VB

VA

Pin= mA

VA

VB mA 1.86


PROBLEM 6-22b

Statement: The linkage in Figure P6-6d has L2 = 15, L3 = 40.9, L5 = 44.7 mm. 2 is 24.2 deg in the XY coordinatesystem. Find, for the position shown, the velocity ratio VI5,6/VI2,3 and the mechanical advantagefrom link 2 to link 6. Use the instant center graphical method.

Given: Link lengths:Link 2 a 15.0 mm Link 3 b 40.9 mm

Link 5 c 44.7 mm Offset f 0 mm from O2


Solution: See Figure P6-6d and Mathcad file P0622b.


5

48.561

O

2,3

2

1

A

2

3

14

C

26.0755,6

1

6B

1,5

From the layout:

AI15 48.561 mm BI15 26.075 mm

2. Choose an arbitrary value for the magnitude of the velocity at I2,3 (point A). Let link 2 rotate CCW.

VA 10mm

sec VC 90 deg VC 114.200 deg


VA

AI15 0.206

rad

sec CW


VB BI15 VB 5.370mm

sec to the left

5. The ratio VI5,6

/VI2,3

isVB

VA0.54


FinTin

rin=

Pin

rin in=

Pin

VA= Fout

Pout

Vout=

Pout

VC=

mAFout

Fin=

Pout

VC

VA

Pin= mA

VA

VB mA 1.86


PROBLEM 6-23

Statement: Generate and draw the fixed and moving centrodes of links 1 and 3 for the linkage in Figure P6-7a.

Solution: See Figure P6-7a and Mathcad file P0623.

1. Draw the linkage to scale, find the instant center I1,3, and repeat for several positions of the linkage. The locusof points I1,3 is the fixed centrode.

FIXED CENTRODE

2. Invert the linkage, grounding link 3. Draw the linkage to scale, find the instant center I1,3, and repeat for severalpositions of the linkage. The locus of points I1,3 is the moving centrode. (See next page.)


MOVING CENTRODE


PROBLEM 6-24

Statement: The linkage in Figure P6-8a has the dimensions and crank angle given below. Find f, VA, and VB

for the position shown for = 15 rad/sec clockwise (CW). Use the velocity difference graphicalmethod.

Given: Link lengths: Crank angle:

Link 2 (O2 to A) a 116 mm 37 deg Global XY system

Link 3 (A to B) b 108 mm Input crank angular velocity

Link 4 (B to O4) c 110 mm 15 rad sec

1

Link 1 (O2 to O4) d 174 mm

Coordinate rotation angle 25 deg Global XY system to local xy system



22.319°

Direction of VA

70.133°3

Direction of VB

d

O2

B

4

2

Y

A

37.000°

Direction of VBA

O4

X


VA a VA 1740.0mm

sec A 37 deg 90 deg


VB = VA + VBA



Y

0.952

1.498112.319°

VBA

1000 mm/sec0

VB

VA

53.000°

X



1

in


sec B 112.319 deg

VBA 1.498 in kv VBA 1498.0mm

sec


VBA

b 13.87

rad

sec


VB

c 8.655

rad

sec


PROBLEM 6-25

Statement: The linkage in Figure P6-8a has the dimensions and crank angle given below. Find f, VA, and VB

for the position shown for = 15 rad/sec clockwise (CW). Use the instant center graphicalmethod.





1





2,3

157.681°

125.463

3O2

1,3

1,2

68.638

3,4

B

4

2,4

2

Y

A

1,4

O4

X

From the layout:

AI13 125.463 mm

BI13 68.638 mm

157.681 deg

2. Use equation 6.7 and inspection of thelayout to determine the magnitude anddirection of the velocity at point A.

VA a VA 1740.0mm

sec



VA

AI13 13.869

rad

sec CW


VB BI13 VB 951.915mm

sec



VB

c 8.654

rad

sec CCW


PROBLEM 6-26

Statement: The linkage in Figure P6-8a has the dimensions and crank angle given below. Find4, VA, and VB

for the position shown for = 15 rad/sec clockwise (CW). Use an analytical method.




Link 4 (B to O4) c 110 mm 15 rad sec1

CW




atany

x

return x 0if

atany

x

otherwise


362.000°

2

O2

B

4

2

yA

x

O4


2. Determine the values of the constants needed forfinding4 from equations 4.8a and 4.10a.

K1d

a K1 1.5000

K2d

c K2 1.5818

K3a

2b

2 c

2 d

2

2 ac K3 1.7307

A cos K1 K2 cos K3

B 2 sin

C K1 K2 1 cos K3

A 0.0424 B 1.7659 C 2.0186

3. Use equation 4.10b to find values of4 for the crossed circuit.

2 atan2 2 A B B2

4 A C 182.681 deg


K4d

b K5

c2

d2

a2

b2

2 a b K4 1.6111 K5 1.7280


E 2 sin E 1.7659

F K1 K4 1 cos K5 F 0.0589

5. Use equation 4.13 to find values of3 for the crossed circuit.


2 atan2 2 D E E2

4 D F 275.133 deg


a

b

sin sin 13.869

rad

sec

a

c

sin sin 8.654

rad

sec

7. Determine the velocity of points A and B for the crossed circuit using equations 6.19.

VA a sin j cos

VA 1536.329 816.881j( )mm

sec VA 1740.000

mm


VB c sin j cos

VB 44.524 950.875j( )mm

sec VB 951.917

mm



PROBLEM 6-27

Statement: The linkage in Figure P6-8a has the dimensions given below. Find and plot4, VA, and VB in the

local coordinate system for the maximum range of motion that this linkage allows if2 = 15 rad/secclockwise (CW).


Link 2 (O2 to A) a 116 mm Link 3 (A to B) b 108 mm

Link 4 (B to O4) c 110 mm Link 1 (O2 to O4) d 174 mm


CW



atany

x

return x 0if

atany

x

otherwise



362.000°

2

O2

B

4

2

yA

x

O4

2. Determine the range of motion for this non-Grashoftriple rocker using equations 4.33.

arg1a

2d

2 b

2 c

2

2 ad

b c

a d

arg2a

2d

2 b

2 c

2

2 ad

b c

a d

arg1 1.083 arg2 0.094

2toggle acos arg2 2toggle 95.4deg

The other toggle angle is the negative of this. Thus,

2toggle 2toggle 1 deg 2toggle


K1d

a K2

d

c

K1 1.5000 K2 1.5818

K3a

2b

2 c

2 d

2

2 ac K3 1.7307

A cos K1 K2 cos K3

B 2 sin

C K1 K2 1 cos K3


2 atan2 2 A B B 2 4 A C



K4d

b K5

c2

d2

a2

b2

2 a b K4 1.6111 K5 1.7280

D cos K1 K4 cos K5

E 2 sin

F K1 K4 1 cos K5


2 atan2 2 D E E 2 4 D F


a

b

sin sin

a

c

sin sin


VA a sin j cos

VAx Re VA VAy Im VA

VB c sin j cos

VBx Re VB VBy Im VB

9. Plot the angular velocity of the output link, 4, and the x and y components of the velocities at points A and B.

100 75 50 25 0 25 50 75 100100

80

60

40

20

0

20

40

60ANGULAR VELOCITY OF LINK 4

Ang

ular

Vel

ocity

,rad

/sec

sec

rad

deg


100 75 50 25 0 25 50 75 1002000

1000

0

1000

2000VELOCITY COMPONENTS, POINT A

Join

tVel

ocity

,mm

/sec

VAx sec

mm

VAy sec

mm

deg

100 75 50 25 0 25 50 75 1003000

2000

1000

0

1000

2000

3000

4000VELOCITY COMPONENTS, POINT B

Join

tVel

ocity

,mm

/sec

VBx sec

mm

VBy sec

mm

deg


PROBLEM 6-28

Statement: The linkage in Figure P6-8b has the dimensions and crank angle given below. Find 4, VA, and VB

for the position shown for 2 = 20 rad/sec counterclockwise (CCW). Use the velocity differencegraphical method.





1



Solution: See Figure P6-8b and Mathcad file P0628.1. Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest.

O4

O2

Direction of VA

2X

2

Y

A

y

3

x

4

Direction of VBA

Direction of VB

B

2. Use equation 6.7 to calculate the magnitudeof the velocity at point A.

VA a

VA 800.000mm

sec

A 90 deg

3. Use equation 6.5 to (graphically) determinethe magnitude of the velocity at point B,the magnitude of the relative velocity VBA,and the angular velocity of link 3. Theequation to be solved graphically is VB = VA + VBA


147.000°

1.790"

1.293"

VB

85.486°

0

AV

VBA

400 mm/sec

X

6.406°

Y



kv400 mm sec

1

in


sec

B 186.406 deg

VBA 1.293 in kv VBA 517.2mm

sec


VBA

b 5.388

rad

sec

6. Determine the angular velocity of link 4 using equation 6.7. VB

c 5.869

rad

sec


PROBLEM 6-29


for the position shown for 2 = 20 rad/sec counterclockwise (CCW). Use the instant centergraphical method.

Given: Link lengths: Crank angle:a 40 mm 57 deg Global XY system

Link 2 (O2 to A)b 96 mm Input crank angular velocity

Link 3 (A to B)c 122 mm 20 rad sec

1Link 4 (B to O4)

d 162 mmLink 1 (O2 to O4)


Solution: See Figure P6-8b and Mathcad file P0629.


O2

2X

2

Y

A

y

3

B

1,3

148.700

133.192

96.406°

O4x

4

From the layout:

AI13 148.700 mm BI13 133.192 mm

96.406 deg

2. Use equation 6.7 and inspection of the layout to determinethe magnitude and direction of the velocity at point A.

VA a VA 800.0mm

sec



VA

AI13 5.380

rad

sec CW


VB BI13 VB 716.6mm

sec



VB

c 5.874

rad

sec CCW


PROBLEM 6-30


for the position shown for 2 = 20 rad/sec counterclockwise (CCW). Use an analytical method.





1





atany

x

return x 0if

atany

x

otherwise


O2

2X

2

Y

A

y

3

93.000

xO4

4

B



K1d

a K2

d

c

K1 4.0500 K2 1.3279

K3a

2b

2 c

2 d

2

2 ac K3 3.4336

A cos K1 K2 cos K3 B 2 sin

C K1 K2 1 cos K3

A 0.5992 B 1.9973 C 7.6054


2 atan2 2 A B B2

4 A C 2 132.386 deg


K4d

b K5

c2

d2

a2

b2

2 a b K4 1.6875 K5 2.8875


E 2 sin E 1.9973

F K1 K4 1 cos K5 F 1.1265


2 atan2 2 D E E2

4 D F 2 31.504 deg



a

b

sin sin 5.385

rad

sec

a

c

sin sin 5.868

rad

sec


VA a sin j cos

VA 798.904 41.869j( )mm

sec VA 800.000

mm


VB c sin j cos

VB 528.774 482.608j( )mm

sec VB 715.900

mm



PROBLEM 6-31

Statement: The linkage in Figure P6-8b has the dimensions and crank angle given below. Find and plot4, VA,

and VB in the local coordinate system for the maximum range of motion that this linkage allows if 2= 20 rad/sec counterclockwise (CCW).


Link 2 (O2 to A) a 40 mm Link 3 (A to B) b 96 mm

Link 4 (B to O4) c 122 mm Link 1 (O2 to O4) d 162 mm


CCW



atany

x

return x 0if

atany

x

otherwise


X93.000

O2

xO4

4

2

2

Y

A

y

3B


2. Determine Grashof condition.

Condition S L P Q( ) SL S L

PQ P Q

"Grashof"return SL PQif

"Special Grashof"return SL PQ=if

"non-Grashof"return otherwise

Condition a d b c( ) "Grashof" Crank-rocker

0 deg 1 deg 360 deg


K1d

a K2

d

c K3

a2

b2

c2

d2

2 ac

K1 4.0500 K2 1.3279 K3 3.4336


C K1 K2 1 cos K3


2 atan2 2 A B B 2 4 A C


K4d

b K5

c2

d2

a2

b2

2 a b K4 1.6875 K5 2.8875

D cos K1 K4 cos K5


E 2 sin

F K1 K4 1 cos K5


2 atan2 2 D E E 2 4 D F


a

b

sin sin

a

c

sin sin

8. Determine the velocity of points B and C for the open circuit using equations 6.19.

VA a sin j cos

VAx Re VA VAy Im VA

VB c sin j cos

VBx Re VB VBy Im VB

9. Plot the angular velocity of the output link, 4, and the magnitudes of the velocities at points B and C.

0 45 90 135 180 225 270 315 36010

5

0

5


Ang

ular

Vel

ocity

,rad

/sec

sec

rad

deg


0 45 90 135 180 225 270 315 3601000

500

0

500


Join

tVel

ocity

,mm

/sec

VAx sec

mm

VAy sec

mm

deg

0 45 90 135 180 225 270 315 3601000

600

200

200

600


Join

tVel

ocity

,mm

/sec

VBx sec

mm

VBy sec

mm

deg


PROBLEM 6-32

Statement: The offset slider-crank linkage in Figure P6-8f has the dimensions and crank angle given below.Find VA, and VB for the position shown if 2 = 25 rad/sec CW. Use the velocity differencegraphical method.


Link 2 a 63 mm Crank angle: 51 deg

Link 3 b 130 mm Input crank angular velocity

Offset c 52 mm 25 rad sec1

CW

Solution: See Figure P6-8f and Mathcad file P0632.


51.000°

Direction of VB

Direction of VBA

O2

2

B

1

3

4

ADirection of VA

2. Use equation 6.7 to calculate the magnitude of thevelocity at point A.

VA a VA 1575.0mm

sec

A 51 deg 90 deg

3. Use equation 6.5 to (graphically) determine themagnitude of the velocity at point B. Theequation to be solved graphically is

VB = VA + VBA


BA

2.208

V

VB

0 1000 mm/secY

X

VA



1

in

VB 2.208 in kv VB 2208mm

sec

VB 270 deg


PROBLEM 6-33

Statement: The offset slider-crank linkage in Figure P6-8f has the dimensions and crank angle given below.Find VA, and VB for the position shown for 2 = 25 rad/sec CW. Use the instant center graphicalmethod.



Link 3 b 130 mm Offset c 52 mm


CW



166.309

118.639

2

O2

1

3

4B

A

1,3


AI13 118.639 mm BI13 166.309 mm


VA a VA 1575.0mm

sec



VA

AI13 13.276

rad

sec CCW


VB BI13 VB 2207.8mm

sec

VB 270 deg


PROBLEM 6-34

Statement: The offset slider-crank linkage in Figure P6-8f has the dimensions and crank angle given below.Find VA, and VB for the position shown for 2 = 25 rad/sec CW. Use an analytical method.



Link 3 b 130 mm Local xy coordinate system

Offset c 52 mm Input crank angular velocity

25 rad sec1


141.000°

52.000

x

2

O2

1

3

4

Y

B

A

X, y

1. Draw the linkage to a convenient scale.

2. Determine3 and d using equations 4.16 for the crossed circuit.

asina sin c

b

44.828 deg

d a cos b cos d 141.160 mm


a

b

cos cos 13.276

rad

sec


VA a sin j cos

VA 991.180 1224.005i( )mm

sec VA 1575.000

mm


In the global coordinate system, VA arg VA 90 deg VA 39.000 deg


VB a sin b sin VB 2207.849mm

sec

VB VB

In the global coordinate system, VB arg VB 90 deg VB 90.000 deg


PROBLEM 6-35

Statement: The offset slider-crank linkage in Figure P6-8f has the dimensions and crank angle given below.Find and plot VA, and VB in the global coordinate system for the maximum range of motion

that this linkage allows if 2 = 25 rad/sec CW.


Link 2 a 63 mm Offset c 52 mm

Link 3 b 130 mm



1. Draw the linkage to a convenient scale. The coordinate rotation angle is 90 deg

52.000

x

2

O2

1

3

4

Y

B

A

X, y2

2. Determine the range of motion for this slider-crank linkage.

0 deg 2 deg 360 deg

3. Determine3 using equations 4.16 for the crossed circuit.

asina sin c

b


a

b

cos cos

5. Determine the x and y components of the velocity of pin A using equation 6.23a:

VA a sin j cos

VAx Re VA VAy Im VA

In the global coordinate system,

VAX VAy VAY VAx


VBx a sin b sin

In the global coordinate system,

VBY VBx

7. Plot the x and y components of the velocity of A. (See next page.)


0 60 120 180 240 300 3602000

1000

0

1000

2000VELOCITY OF POINT A

Velo

city

,mm

/sec

VAX sec

mm

VAY sec

mm

deg

7. Plot the velocity of point B.

0 60 120 180 240 300 3603000

2000

1000

0

1000

2000VELOCITY OF POINT B

Velo

city

,mm

/sec

VBY sec

mm

deg


PROBLEM 6-36

Statement: The linkage in Figure P6-8d has the dimensions and crank angle given below. Find VA, VB, and

Vbox for the position shown for 2 = 30 rad/sec clockwise (CW). Use the velocity differencegraphical method.

Given: Link lengths:Link 2 a 30 mm Link 3 b 150 mm

Link 4 c 30 mm Link 1 d 150 mm



CW



2. Use equation 6.7 to calculate the magnitude ofthe velocity at point A.

Direction of VA3A

O2

2

1

BDirection of VB

O4 4

Direction of VBA

VboxVA a VA 900.000mm

sec

A 58 deg 90 deg

3. Use equation 6.5 to (graphically) determine themagnitude of the velocity at point B, themagnitude of the relative velocity VBA, and theangular velocity of link 3. The equation to besolved graphically is

VB = VA + VBA


BV

Y

3 2.000°

Direction of VBA

VA

X0 500 mm/sec


VB VA

VB 900.000mm

sec

B 32 deg VBA 0in

sec


VBA

b 0.000

rad

sec

VB

c 30.000

rad

sec

6. Determine the magnitude of the vector Vbox . This is a special case Grashof mechanism in the parallelogramconfiguration. Link 3 does not rotate, therefore all points on link 3 have the same velocity. The velocity Vbox isthe horizontal component of VA.

Vbox VA cos A Vbox 763.243mm

sec


PROBLEM 6-37

Statement: The linkage in Figure P6-8d has the dimensions and effective crank angle given below. Find VA,

VB, and Vbox in the global coordinate system for the position shown for 2 = 30 rad/sec CW. Usean analytical method.


Link 2 a 30 mm Link 3 b 150 mm


Crank angle: 58 deg




atany

x

return x 0if

atany

x

otherwise

3A

O2

2

1

B

O4 4

Vbox




K1d

a K2

d

c

K1 5.0000 K2 5.0000

K3a

2b

2 c

2 d

2

2 ac K3 1.0000


C K1 K2 1 cos K3

A 6.1197 B 1.6961 C 2.8205


2 atan2 2 A B B2

4 A C 418.000 deg

360 deg 58.000deg


K4d

b K5

c2

d2

a2

b2

2 a b K4 1.0000 K5 5.0000


E 2 sin E 1.6961

F K1 K4 1 cos K5 F 0.0000



2 atan2 2 D E E2

4 D F 360.000 deg

360 deg 0.000deg


a

b

sin sin 0.000

rad

sec

a

c

sin sin 30.000

rad

sec


VA a sin j cos

VA 763.243 476.927j( )mm

sec VA 900.000

mm


VB c sin j cos

VB 763.243 476.927j( )mm

sec VB 900.000

mm


8. Determine the velocity Vbox. Since link 3 does not rotate (this is a special case Grashof linkage in theparallelogram mode), all points on it have the same velocity. Therefore,

Vbox VA Vbox 900.000mm

sec


PROBLEM 6-38

Statement: The linkage in Figure P6-8d has the dimensions and effective crank angle given below. Find andplot VA, VB, and Vbox in the global coordinate system for the maximum range of motion that this

linkage allows if 2 = 30 rad/sec CW.


Link 2 a 30 mm Link 3 b 150 mm





atany

x

return x 0if

atany

x

otherwise


3A

O2

2

1

B

O4 4

Vbox


2. Determine the range of motion for this special-case Grashofdouble crank.

0 deg 2 deg 360 deg

3. Determine the values of the constants needed for finding 4 fromequations 4.8a and 4.10a.

K1d

a K2

d

c

K1 5.0000 K2 5.0000

K3a

2b

2 c

2 d

2

2 ac K3 1.0000


C K1 K2 1 cos K3


2 atan2 2 A B B 2 4 A C


K4d

b K5

c2

d2

a2

b2

2 a b K4 1.0000 K5 5.0000

D cos K1 K4 cos K5

E 2 sin

F K1 K4 1 cos K5


2 atan2 2 D E E 2 4 D F



a

b

sin sin

a

c

sin sin


VA a sin j cos

VA VA

VB c sin j cos

VB VB

9. Plot the angular velocity of the output link, 4, and the magnitudes of the velocities at points A and B.

Since this is a special-case Grashof linkage in the parallelogram configuration,3 = 0 and 4 = 2 for all values of

2. Similarly, VA, VB, and Vbox all have the same constant magnitude through all values of 2.

5 deg( ) 30.000rad

sec 135 deg( ) 30.000

rad

sec

VA 5 deg( ) 900.000mm

sec VA 135 deg( ) 900.000

mm

sec

VB 5 deg( ) 900.000mm

sec VB 135 deg( ) 900.000

mm

sec


PROBLEM 6-39

Statement: The linkage in Figure P6-8g has the dimensions and crank angle given below. Find 4, VA, and VB

for the position shown for 2 = 15 rad/sec clockwise (CW). Use the velocity difference graphicalmethod.


Link 2 (O2 to A) a 49 mm Link 2 (O2 to C) a' 49 mm

Link 3 (A to B) b 100 mm Link 5 (C to D) b' 100 mm

Link 4 (B to O4) c 153 mm Link 6 (D to O6) c' 153 mm

Link 1 (O2 to O4) d 87 mm Link 1 (O2 to O6) d' 87 mm



CW


Solution: See Figure P6-8g and Mathcad file P0639.


2

O

Direction of VBA

O

4C

5

O4

Direction of VB

B3

Y

29.000°

X

Direction of VA

D

62

A

6

2. Use equation 6.7 to calculate the magnitude of the velocity at point B.

VB a VB 735.0mm

sec B 29 deg 90 deg

3. Use equation 6.5 to (graphically) determine the magnitude of the velocity at point A, the magnitude of the relativevelocity VAB, and the angular velocity of link 3. The equation to be solved graphically is

VA = VB + VAB

a. Choose a convenient velocity scale and layout the known vector VB.b. From the tip of VB, draw a construction line with the direction of VAB, magnitude unknown.c. From the tail of VB, draw a construction line with the direction of VA, magnitude unknown.d. Complete the vector triangle by drawing VAB from the tip of VB to the intersection of the VA constructionline and drawing VA from the tail of VB to the intersection of the VAB construction line.


1.517

BA

AV

V

0

Y

V B

500 mm/sec

0.124°

X



1

in

VA 1.517 in kv VA 758.5mm

sec A 0.124 deg


VA

c 4.958

rad

sec


PROBLEM 6-40


for the position shown for 2 = 15 rad/sec clockwise (CW). Use the instant center graphicalmethod.








CW


2

97.094

O

O

4

100.224 1,3

89.876°

O4

5

C

B3

Y

X6

D

2

A

6


1. Draw the linkage to scale in the position given, find theinstant centers, distances from the pin joints to the instantcenters and the angles that links 3 and 4 make with the xaxis.

From the layout:

AI13 97.094 mm BI13 100.224 mm

89.876 deg

2. Use equation 6.7 and inspection of the layout todetermine the magnitude and direction of the velocityat point A.

VA a VA 735.0mm

sec



VA

AI13 7.570

rad

sec CW


VB BI13 VB 758.694mm

sec



VB

c 4.959

rad

sec CW


PROBLEM 6-41


for the position shown for 2 = 15 rad/sec clockwise (CW). Use an analytical method.






Crank angle: 148 deg Local xy system




atany

x

return x 0if

atany

x

otherwise


2

148.000°

O

2

4

x

O4

5

C O

B3

Y

X6

yD

A

6



K1d

a K2

d

c

K1 1.7755 K2 0.5686

K3a

2b

2 c

2 d

2

2 ac K3 1.5592

A cos K1 K2 cos K3

B 2 sin

C K1 K2 1 cos K3

A 0.5821 B 1.0598 C 4.6650


2 atan2 2 A B B2

4 A C 208.876 deg


K4d

b K5

c2

d2

a2

b2

2 a b K4 0.8700 K5 0.3509


E 2 sin E 1.0598

F K1 K4 1 cos K5 F 2.2367



2 atan2 2 D E E2

4 D F 266.892 deg


a

b

sin sin 7.570

rad

sec

a

c

sin sin 4.959

rad

sec

7. Determine the velocity of points B and A for the crossed circuit using equations 6.19.

VA a sin j cos

VA 389.491 623.315j( )mm

sec VA 735.000

mm


VB c sin j cos

VB 366.389 664.362j( )mm

sec VB 758.694

mm



PROBLEM 6-42

Statement: The linkage in Figure P6-8g has the dimensions and crank angle given below. Find and plot 4, VA,

and VB in the local coordinate system for the maximum range of motion that this linkage allows if 2= 15 rad/sec clockwise (CW).









atany

x

return x 0if

atany

x

otherwise


2

O

2

4

x

O4

5

C O

B3

Y

X6

yD

2

A

6


2. Determine the range of motion for this non-Grashof triplerocker using equations 4.33.

arg1a

2d

2 b

2 c

2

2 ad

b c

a d arg1 0.840

arg2a

2d

2 b

2 c

2

2 ad

b c

a d arg2 6.338



2toggle 2toggle 1 deg 360 deg 2toggle


K1d

a K2

d

c K3

a2

b2

c2

d2

2 ac

K1 1.7755 K2 0.5686 K3 1.5592


C K1 K2 1 cos K3


2 atan2 2 A B B 2 4 A C



K4d

b K5

c2

d2

a2

b2

2 a b K4 0.8700 K5 0.3509

D cos K1 K4 cos K5 E 2 sin

F K1 K4 1 cos K5


2 atan2 2 D E E 2 4 D F


a

b

sin sin

a

c

sin sin


VB a sin j cos

VBx Re VB VBy Im VB

VA c sin j cos

VAx Re VA VAy Im VA

9. Plot the angular velocity of the output link, 4, and the x and y components of the velocities at points B and A.

0 45 90 135 180 225 270 315 360100

80

60

40

20

0

20

40


Ang

ular

Vel

ocity

,rad

/sec

sec

rad

deg


0 45 90 135 180 225 270 315 3601000

500

0

500


Join

tVel

ocity

,mm

/sec

VBx sec

mm

VBy sec

mm

deg

0 45 90 135 180 225 270 315 3603000

2000

1000

0

1000

2000

3000


Join

tVel

ocity

,mm

/sec

VAx sec

mm

VAy sec

mm

deg


PROBLEM 6-43

Statement: The 3-cylinder radial compressor in Figure P6-8c has the dimensions and crank angle given below.Find piston velocities V6, V7, and V8 for the position shown for 2 = 15 rad/sec clockwise (CW).Use the velocity difference graphical method.



Links 3, 4, and 5 b 70 mm



CW

Cylinder angular spacing 120 deg

Direction of V7

Direction of V62

Direction of V6

7

14

Y

623

Direction of V72

Direction of V8

Direction of V82

8

X5

Solution: See Figure P6-8c and Mathcad file P0643.

1. Draw the linkage to a convenient scale. Indicate thedirections of the velocity vectors of interest.

2. Use equation 6.7 to calculate the magnitude of thevelocity at rod pin on link 2.

V2 a V2 285.0mm

sec

2 53 deg 90 deg

3. Use equation 6.5 to (graphically) determine themagnitude of the velocity at pistons 6, 7, and 8.. Theequations to be solved graphically are

V7 = V2 + V72 V6 = V2 + V62 V8 = V2 + V82

a. Choose a convenient velocity scale and layout the known vector V2.b. From the tip of V2, draw a construction line with the direction of V72, magnitude unknown.c. From the tail of V2, draw a construction line with the direction of V7, magnitude unknown.d. Complete the vector triangle by drawing V72 from the tip of V2 to the intersection of the V7 construction lineand drawing V7 from the tail of V2 to the intersection of the V72 construction line.e. Repeat for V6 and V8.

Y 0.4171.463

V

V V

V

8

2

82

72

0

V V662

200 mm/sec

1.046

V7

X



1

in

V7 1.046 in kv V7 209.2mm

sec

V7 270 deg

V6 0.417 in kv V6 83.4mm

sec

V6 150 deg

V8 1.463 in kv V8 292.6mm

sec V8 210 deg


PROBLEM 6-44

Statement: The 3-cylinder radial compressor in Figure P6-8c has the dimensions and crank angle given below.Find V6, V7, and V8 for the position shown for 2 = 15 rad/sec clockwise (CW). Use an analyticalmethod.

Given: Link lengths:Link 2 a 19 mm Offset c 0 mm

Links 3, 4, and 5 b 70 mm Crank angle: 37 deg


CW Local x'y' system


37.000°

x'

1

y''4

7

x''

6

y'''

3

Y

2

x'''

X, y'

8

5



2. Determine4 and d' using equation 4.17.

asina sin c

b

170.599 deg (x'y' system)

d' a cos b cos d' 3.316 in


a

b

cos cos 3.296

rad

sec

4. Determine the velocity of the rod pin on link 2 using equation 6.23a:

V2 a sin j cos

V2 171.517 227.611i( )mm

sec V2 285.000

mm

sec arg V2 53.000 deg

In the global coordinate system, V2 arg V2 90 deg V2 143.000 deg

5. Determine the velocity of piston 7 using equation 6.22b:

V7 a sin b sin

V7 209.204mm

sec V7 209.204

mm

sec arg V7 0.000deg


6. Determine3 and d'' using equation 4.17.

120 deg 157.000 deg (x''y'' system)

asina sin c

b

173.912 deg

d'' a cos b cos d'' 2.052 in


a

b

cos cos 3.769

rad

sec



V2 a sin j cos

V2 111.358 262.344i( )mm

sec V2 285.000

mm

sec arg V2 67.000deg

In the global coordinate system, V2 arg V2 150 deg V2 217.000deg


V6 a sin b sin

V6 83.378mm

sec V6 83.378

mm

sec arg V6 0.000deg


10. Determine5 and d''' using equation 4.17.

120 deg 277.000 deg (x'''y''' system)

asina sin c

b

195.629 deg

d''' a cos b cos d''' 2.745 in


a

b

cos cos 0.515

rad

sec


V2 a sin j cos

V2 282.876 34.733i( )mm

sec V2 285.000

mm




V8 a sin b sin

V8 292.592mm

sec V8 292.592

mm




PROBLEM 6-45

Statement: The 3-cylinder radial compressor in Figure P6-8c has the dimensions and crank angle given below.Find and plot V6, V7, and V8 for one revolution of the crank if 2 = 15 rad/sec clockwise (CW).



Links 3, 4, and 5 b 70 mm


CW


21

y''

7

x'

4

y'''

6

x''

35

Y

8

X, y'

x'''


1. Draw the linkage to scale and label it. Note thatthere are three local coordinate systems.

2. Determine the range of motion for this slider-cranklinkage. This will be the same in each coordinateframe.

0 deg 2 deg 360 deg

3. Determine4 using equation 4.17.

asina sin c

b

(x'y' system)


a

b

cos cos


V7 a sin b sin


asina sin c

b

(x''y'' system)


a

b

cos cos


V6 a sin b sin

9. Determine5 and d''' using equation 4.17.

asina sin 2 c

b

(x'''y''' system)


a

b

cos 2 cos



V8 a sin 2 b sin

12. Plot the velocities of pistons 6, 7, and 8.

0 60 120 180 240 300 360400

200

0

200

400VELOCITY OF PISTONS 6, 7, AND 8

Crank Angle, deg

Velo

city

,mm

/sec

V6 sec

mm

V7 sec

mm

V8 sec

mm

deg


PROBLEM 6-46

Statement: Figure P6-9 shows a linkage in one position. Find the instantaneous velocities of points A, B, andP if link O2A is rotating CW at 40 rad/sec.


Link 2 a 5.00 in Link 3 b 4.40 in

Link 4 c 5.00 in Link 1 d 9.50 in

Coupler point: Rpa 8.90 in 56 deg

Crank angle and speed: 50 deg 40 rad sec1



atany

x

return x 0if

atany

x

otherwise


B

3

2 50.000°

O2

A

14.000°

O41

4

y

P

Y

x

X

1. Draw the linkage to scale and label it. All calculatedangles are in the local xy system.

2. Determine the values of the constants needed forfinding4 from equations 4.8a and 4.10a.

K1d

a K2

d

c

K1 1.9000 K2 1.9000

K3a

2b

2 c

2 d

2

2 ac K3 2.4178

A cos K1 K2 cos K3

B 2 sin

C K1 K2 1 cos K3

A 0.0607 B 1.5321 C 2.4537


2 atan2 2 A B B2

4 A C 473.008 deg

360 deg 113.008 deg


K4d

b K5

c2

d2

a2

b2

2 a b K4 2.1591 K5 2.4911


E 2 sin E 1.5321

F K1 K4 1 cos K5 F 0.1539


2 atan2 2 D E E2

4 D F 370.105 deg


360 deg 10.105deg


a

b

sin sin 41.552

rad

sec

a

c

sin sin 26.320

rad

sec


VA a sin j cos

VA 153.209 128.558j( )in

sec VA 200.000

in


VB c sin j cos

VB 121.130 51.437j( )in

sec VB 131.598

in


8. Determine the velocity of the coupler point P for the open circuit using equations 6.36.

VPA Rpa sin j cos

VPA 338.121 149.797j( )in

sec

VP VA VPA

VP 184.912 21.239j( )in

sec VP 186.128

in

sec arg VP 173.448deg


PROBLEM 6-47

Statement: Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver tocalculate and plot the magnitude and direction of the velocity of point P in Figure 3-37a as afunction of2 for a constant2 = 1 rad/sec CCW. Also calculate and plot the velocity of point P

versus point A.


Link 2 (O2 to A) a 1.00 Link 3 (A to B) b 2.06

Link 4 (B to O4) c 2.33 Link 1 (O2 to O4) d 2.22

Coupler point: Rpa 3.06 31 deg

Crank speed: 1 rad sec1



atany

x

return x 0if

atany

x

otherwise



2c

O21

d

O4

x

2 a

A

y

3

b

4

4

p

B

P

2. Determine the range of motion for this Grashof crank rocker.

0 deg 2 deg 360 deg


K1d

a K2

d

c

K1 2.2200 K2 0.9528

K3a

2b

2 c

2 d

2

2 ac K3 1.5265


C K1 K2 1 cos K3


2 atan2 2 A B B 2 4 A C


K4d

b K5

c2

d2

a2

b2

2 a b K4 1.0777 K5 1.1512


F K1 K4 1 cos K5


2 atan2 2 D E E 2 4 D F



a

b

sin sin

a

c

sin sin

8. Determine the velocity of point A using equations 6.19.

VA a sin j cos VA VA

9. Determine the velocity of the coupler point P using equations 6.36.

VPA Rpa sin j cos

VP VA VPA

10. Plot the magnitude and direction of the velocity at coupler point P.

Magnitude: VP VP Direction: VP1 arg VP

VP if VP1 0 VP1 2 VP1

0 45 90 135 180 225 270 315 3600

1

2

3MAGNITUDE

Velo

city

,mm

/sec

VP

deg

0 45 90 135 180 225 270 315 3600

100

200

300

DIRECTION

Vect

orAn

gle,

deg

VP deg

deg


PROBLEM 6-48

Statement: Figure P6-11 shows a linkage that operates at 500 crank rpm. Write a computer program or use anequation solver to calculate and plot the magnitude and direction of the velocity of point B at 2-degincrements of crank angle. Check your result with program FOURBAR.

Units: rpm 2 rad min1


Link 2 (O2 to A) a 2.000 in Link 3 (A to B) b 8.375 in

Link 4 (B to O4) c 7.187 in Link 1 (O2 to O4) d 9.625 in

Input crank angular velocity 500 rpm 52.360rad sec1



atany

x

return x 0if

atany

x

otherwise


1

O4

4

3A

2O

22

B



0 deg 1 deg 360 deg

3. Determine the values of the constants needed for finding 4 fromequations 4.8a and 4.10a.

K1d

a K2

d

c

K1 4.8125 K2 1.3392

K3a

2b

2 c

2 d

2

2 ac K3 2.7186


C K1 K2 1 cos K3


2 atan2 2 A B B 2 4 A C


K4d

b K5

c2

d2

a2

b2

2 a b K4 1.1493 K5 3.4367


F K1 K4 1 cos K5


2 atan2 2 D E E 2 4 D F



a

b

sin sin

a

c

sin sin

8. Determine the velocity of point B using equations 6.19.

VB c sin j cos

VB VB VB1 arg VB

9. Plot the magnitude and angle of the velocityat point B.

0 60 120 180 240 300 3600

50

100

150MAGNITUDE OF VELOCITY AT B

Velo

city

,in/

sec

VB sec

in

deg

VB if VB1 0 VB1 VB1

0 60 120 180 240 300 36010

15

20

25

30

35

40

45

50DIRECTION OF VELOCITY AT B

Angl

e,de

g

VB deg

deg


PROBLEM 6-49

Statement: Figure P6-12 shows a linkage and its coupler curve. Write a computer program or use an equationsolver to calculate and plot the magnitude and direction of the velocity of the coupler point P at2-deg increments of crank angle over the maximum range of motion possible. Check your resultswith program FOURBAR.






Crank speed: 20 rpm



atany

x

return x 0if

atany

x

otherwise


3

2

4

A

O2 O4

x

2

3

B

Py1. Draw the linkage to scale and label it.

2. Using the geometry defined in Figure 3-1a in the text, determinethe input crank angles (relative to the line O2O4) at which links 2and 3, and 3 and 4 are in toggle.

acosa

2d

2 b c( )

2

2 a d

158.286 deg

2 deg


K1d

a K2

d

c

K1 0.6930 K2 0.5726 K3a

2b

2 c

2 d

2

2 ac K3 1.1317


C K1 K2 1 cos K3


2 atan2 2 A B B 2 4 A C


K4d

b K5

c2

d2

a2

b2

2 a b K4 1.5281 K5 0.2440



F K1 K4 1 cos K5


2 atan2 2 D E E 2 4 D F


a

b

sin sin

a

c

sin sin


VA a sin j cos


VPA Rpa sin j cos

VP VA VPA

10. Plot the magnitude and direction of the coupler point P.

Magnitude: VP VP Direction: VP arg VP

200 150 100 50 0 50 100 150 2000

50

MAGNITUDE

Velo

city

,mm

/sec

VP sec

in

deg

200 150 100 50 0 50 100 150 200200

0

200DIRECTION

Vect

orAn

gle,

deg

VP deg

deg


PROBLEM 6-50







Crank speed: 80 rpm



atany

x

return x 0if

atany

x

otherwise



A

23

P

O2

y

x

34

B

O4

2. Determine the range of motion for this non-Grashof triple rocker using equations 4.33.

arg1a

2d

2 b

2 c

2

2 ad

b c

a d arg1 1.228

arg2a

2d

2 b

2 c

2

2 ad

b c

a d arg2 0.439






K1d

a K2

d

c K3

a2

b2

c2

d2

2 ac

K1 2.5814 K2 2.5814 K3 2.0181


C K1 K2 1 cos K3


2 atan2 2 A B B 2 4 A C


K4d

b K5

c2

d2

a2

b2

2 a b K4 1.2000 K5 2.6244


F K1 K4 1 cos K5


2 atan2 2 D E E 2 4 D F


a

b

sin sin

a

c

sin sin


VA a sin j cos


VPA Rpa sin j cos

VP VA VPA


Magnitude: VP VP Direction: VP arg VP

(See next page)


150 100 50 0 50 100 1500

5

10

15

20MAGNITUDE

Velo

city

,mm

/sec

VP sec

in

deg

150 100 50 0 50 100 150200

100

0

100

200DIRECTION

Vect

orAn

gle,

deg

VP deg

deg


PROBLEM 6-51







Crank speed: 80 rpm



atany

x

return x 0if

atany

x

otherwise



A

B

O2

3

2

4

y

P

xO 4

2. Determine the range of motion for this non-Grashof triple rocker using equations 4.33.

arg1a

2d

2 b

2 c

2

2 ad

b c

a d arg1 1.451

arg2a

2d

2 b

2 c

2

2 ad

b c

a d arg2 0.568






K1d

a K2

d

c

K1 2.5278 K2 2.1412

K3a

2b

2 c

2 d

2

2 ac K3 3.3422

A cos K1 K2 cos K3

B 2 sin

C K1 K2 1 cos K3


2 atan2 2 A B B 2 4 A C


K4d

b K5

c2

d2

a2

b2

2 a b K4 2.6765

K5 3.6465

D cos K1 K4 cos K5

E 2 sin

F K1 K4 1 cos K5


2 atan2 2 D E E 2 4 D F


a

b

sin sin

a

c

sin sin


VA a sin j cos


VPA Rpa sin j cos

VP VA VPA



Magnitude: VP VP

Direction: VP arg VP

60 40 20 0 20 40 600

5

10

15

20MAGNITUDE

Velo

city

,in/

sec

VP sec

in

deg

60 40 20 0 20 40 60100

50

0

50

100

150

200DIRECTION

Crank Angle, deg

Vect

orAn

gle,

deg

VP deg

deg


PROBLEM 6-52

Statement: Figure P6-15 shows a power hacksaw that is an offset slider-crank mechanism that has thedimensions given below. Draw an equivalent linkage diagram; then calculate and plot the velocityof the saw blade with respect to the piece being cut over one revolution of the crank, which rotatesat 50 rpm.




Link 3 b 170 mm

Input crank angular velocity 50 rpm


1. Draw the equivalent linkage to a convenient scale and label it.

c

4B 3

O2

2

b

a

x

2

A

y


0 deg 2 deg 360 deg

3. Determine3 using equations 4.16 for the crossed circuit.

asina sin c

b


a

b

cos cos


VB a sin b sin

7. Plot the magnitude of the velocity of B. (See next page.)


0 60 120 180 240 300 360400

200

0

200

400

600VELOCITY OF POINT B

Velo

city

,mm

/sec

VB sec

mm

deg


PROBLEM 6-53

Statement: Figure P6-16 shows a walking-beam indexing and pick-and-place mechanism that can be analyzedas two fourbar linkages driven by a common crank. Calculate and plot the absolute velocities ofpoints E and P and the relative velocity between points E and P for one revolution of gear 2.


Given: Link lengths ( walking-beam linkage):

Link 2 (O2 to A) a' 40 mm Link 3 (A to D) b' 108 mm

Link 4 (O4 to D) c' 40 mm Link 1 (O2 to O4) d' 108 mm

Link lengths (pick and place linkage):

Link 5 (O5 to B) a 13 mm Link 7 (B to C) b 193 mm

Link 6 (C to O6) c 92 mm Link 1 (O5 to O6) d 128 mm

Rocker point E: u 164 mm Crank speed: 10 rpm

Gears 4 & 5 phase angle 143 deg



atany

x

return x 0if

atany

x

otherwise


1. Draw the walking-beam linkage to scale and label it.

D

80°58 mm

x'

c'd'

O4

30 mm

P

b'

A

y'

a'

O2X

Y

2. Determine the range of motion for this mechanism.

0 deg 2 deg 360 deg (local x'y' coordinate system)

3. This part of the mechanism is a special-case Grashof in the parallelogram configuration. As such, the couplerdoes not rotate, but has curvilinear motion with ever point on it having the same velocity. Therefore, it is onlynecessary to calculate the X-component of the velocity at point A in order to determine the velocity of thecylinder center, P.


VA a' sin j cos

VAx Re VA

In the global X-Y coordinate frame,

VP VAx

4. Draw the pick and place linkage to scale and label it.

6

C

O6

5 B

O5

a

7

1

bc

d

E

5. Establish the relationship between5 and 2. Note that gear 5 is driven by gear 4 and that their ratio is -1 (i.e.,they rotate in opposite directions with the same speed). Also, because the walking beam fourbar is specialGrashof, 4 = 2. Thus,

and


K1d

a K1 9.8462 K2

d

c K2 1.3913

K3a

2b

2 c

2 d

2

2 ac K3 5.1137

A cos K1 K2 cos K3

B 2 sin

C K1 K2 1 cos K3



2 atan2 2 A B B 2 4 A C


K4d

b K5

c2

d2

a2

b2

2 a b K4 0.6632 K5 9.0351

D cos K1 K4 cos K5

E 2 sin

F K1 K4 1 cos K5


2 atan2 2 D E E 2 4 D F

10. Determine the angular velocity of links 6 and 7 using equations 6.18.

a

b

sin sin

a

c

sin sin

10. Determine the velocity of the rocker point E using equations 6.34.

VE u sin j cos

11. Transform this into the global XY system.

VEx Re VE VEy Im VE

VEX VEx VEY VEx

VEXY VEX j VEY

12. Calculate and plot the velocity of E relative to P.

VEP VEXY VP


0 30 60 90 120 150 180 210 240 270 300 330 3600

20

40

60

80

100RELATIVE VELOCITY

Crank Angle, deg

Velo

city

,mm

/sec

VEP sec

mm

deg


PROBLEM 6-54

Statement: Figure P6-17 shows a paper roll off-loading mechanism driven by an air cylinder. In the positionshown, it has the dimensions given below. The V-links are rigidly attached to O4A. The aircylinder is retracted at a constant velocity of 0.2 m/sec. Draw a kinematic diagram of themechanism, write the necessary equations, and calculate and plot the angular velocity of the paperroll and the linear velocity of its center as it rotates through 90 deg CCW from the position shown.

Given: Link lengths and angles: Paper roll location from O4:

Link 4 (O4 to A) c 300 mm u 707.1 mm

Link 1 (O2 to O4) d 930 mm 181 deg

Link 4 initial angle 62.8 deg with respect to local x axis

Input cylinder velocity adot 200 mm sec1


1. Draw the mechanism to scale and define a vector loop using the fourbar derivation in Section 6.7 as a model.

3 2

1

ab

d

A O

f

y

2 O2

O4

R 1

RR

R

23

4

A

24

V-Link

Roll Center

x

45.000°

707.107

4

46.000°

4c O

2. Write the vector loop equation, differentiate it, expand the result and separate into real and imaginary parts tosolve for f, 2, and 4.

R1 R4 R2 R3 (a)

d ej

c ej

a ej

b ej

(b)

where a is the distance from the origin to the cylinder piston, a variable; b is the distance from the cylinderpiston to A, a constant; and c is the distance from 4 to point A, a constant. Angle 1 is zero, 3 = 2, and 4 isthe variable angle that the rocker arm makes with the x axis. Solving the position equations:

Let f a b then, making this substitution and substituting the Euler equivalents,

d c cos j sin f cos j sin (c)

Separating into real and imaginary components and solving for2 and f,

atan2 d c cos c sin (d)


f c sin

sin (e)

Differentiate equation b.

j c ej

t

fdd

ej

j f ej

(f)

Substituting the Euler equivalents,

c sin j cos t

fdd

cos j sin f sin j cos (g)

Separating into real and imaginary components and solving for4 . Note that df/dt = adot.

adot

c sin (h)

3. Plot 4 over a range of4 of

1 deg 90 deg

60 80 100 120 140 1600

0.2

0.4

0.6

0.8

1

1.2ANGULAR VELOCITY, LINK 4

Link 4 Angle, deg

Ang

ular

Vel

ocity

,rad

/sec

sec

rad

deg

4. Determine the velocity of the center of the paper roll using equation 6.35. The direction is in the local xycoordinate system.

VU u sin j cos

VU VU VU arg VU

5. Plot the magnitude and direction of the velocity of the paper roll center. (See next page.)


60 80 100 120 140 1600

200

400

600

800MAGNITUDE OF PAPER CENTER VELOCITY

Rocker Arm Position, deg

Velo

city

,mm

/sec

VU sec

mm

deg

60 80 100 120 140 16040

20

0

20

40

60

80DIRECTION OF PAPER CENTER VELOCITY

Rocker Arm Position, deg

Vect

orAn

gle,

deg

VU deg

deg


PROBLEM 6-55a

Statement: Figure P6-18 shows a powder compaction mechanism. Calculate its mechanical advantage for theposition shown.


Link 2 (A to B) a 105 mm Offset c 27 mm

Link 3 (B to D) b 172 mm

Distance to force application:

Link 2 (AC) rin 301 mm

Position of link 2: 44 deg Let 1 rad sec1



A

2

3

4

B

C

D

44.000°

X

Y


asina sin c

b

164.509 deg


a

b

cos cos

4. Determine the velocity of pin D using equation 6.22b:

VD a sin b sin Positive upward

5. Calculate the velocity of point C using equation 6.23a:

VC rin sin j cos

VC VC

6. Calculate the mechanical advantage using equation 6.13.

mAVC

VD mA 3.206


PROBLEM 6-55b

Statement: Figure P6-18 shows a powder compaction mechanism. Calculate and plot its mechanical advantageas a function of the angle of link AC as it rotates from 15 to 60 deg.


Link 2 (A to B) a 105 mm Offset c 27 mm

Link 3 (B to D) b 172 mm


Link 2 (AC) rin 301 mm

Initial and final positions of link 2: 15 deg 60 deg Let 1 rad sec1

Solution: See Figure P6-18 and Mathcad file P0655b.


A

2

3

4

B

C

D

X

Y

2


1 deg


asina sin c

b


a

b

cos cos


VD a sin b sin Positive upward

6. Calculate the velocity of point C using equation 6.23a:


VC rin sin j cos

VC VC

7. Calculate the mechanical advantage using equation 6.13.

mA VC VD

15 20 25 30 35 40 45 50 55 600

1

2

3

4

5

6

7

8

9

10

11

12MECHANICAL ADVANTAGE

Crank Angle, deg

Mec

hani

calA

dvan

tage

mA

deg


PROBLEM 6-56

Statement: Figure P6-19 shows a walking beam mechanism. Calculate and plot the velocity Vout for onerevolution of the input crank 2 rotating at 100 rpm.






Crank speed: 100 rpm



atany

x

return x 0if

atany

x

otherwise



2O

3A

2

14

26.00°

y

Y

O4

x

B

PX


0 deg 2 deg 360 deg


K1d

a K2

d

c

K1 2.2200 K2 0.9528

K3a

2b

2 c

2 d

2

2 ac K3 1.5265

A cos K1 K2 cos K3


B 2 sin

C K1 K2 1 cos K3


2 atan2 2 A B B 2 4 A C


K4d

b K5

c2

d2

a2

b2

2 a b K4 1.0777

K5 1.1512

D cos K1 K4 cos K5

E 2 sin

F K1 K4 1 cos K5


2 atan2 2 D E E 2 4 D F


a

b

sin sin

a

c

sin sin


VA a sin j cos


VPA Rpa sin j cos

VP VA VPA

10. Plot the X-component (global coordinate system) of the velocity of the coupler point P.

Coordinate rotation angle: 26 deg

Vout Re VP cos Im VP sin


0 45 90 135 180 225 270 315 36010

5

0

5

10

15Vout

Crank Angle, deg

Velo

city

,in/

sec

Vout sec

in

deg


PROBLEM 6-57a

Statement: Figure P6-20 shows a crimping tool. For the dimensions given below, calculate its mechanicaladvantage for the position shown.


Link 2 (AB) a 0.80 in Link 3 (BC) b 1.23 in

Link 4 (CD) c 1.55 in Link 1 (AD) d 2.40 in


Link 2 (AB) rin 4.26 in Link 4 (CD) rout 1.00 in

Initial position of link 2: 49 deg


1. Draw the mechanism to scale and label it.

1Fin

49.000°

Fout

4

D

2 2A B

3 C


K1d

a K2

d

c

K1 3.0000 K2 1.5484

K3a

2b

2 c

2 d

2

2 ac K3 2.9394

A cos K1 K2 cos K3

B 2 sin

C K1 K2 1 cos K3

A 0.4204 B 1.5094 C 4.2675

3. Use equation 4.10b to find value of 4 for the open circuit.

2 atan2 2 A B B2

4 A C 236.482 deg


K4d

b K5

c2

d2

a2

b2

2 a b K4 1.9512 K5 2.8000



E 2 sin E 1.5094

F K1 K4 1 cos K5 F 0.8241


2 atan2 2 D E E2

4 D F 325.961 deg

6. Referring to Figure 6-10, calculate the values of the anglesand.

374.961 deg

If > 360 deg, subtract 360 deg from it.

if 360 deg 360 deg 14.961 deg

89.479 deg

If > 90 deg, subtract it from 180 deg.

if 90 deg 180 deg 89.479 deg

7. Using equation 6.13e, calculate the mechanical advantage of the linkage in the position shown.

mAc sin

a sin

rin

rout mA 31.969


PROBLEM 6-57b

Statement: Figure P6-20 shows a crimping tool. For the dimensions given below, calculate and plot itsmechanical advantage as a function of the angle of link AB as it rotates from 60 to 45 deg.


Link 2 (AB) a 0.80 in Link 3 (BC) b 1.23 in

Link 4 (CD) c 1.55 in Link 1 (AD) d 2.40 in


Link 2 (AB) rin 4.26 in Link 4 (CD) rout 1.00 in

Range of positions of link 2: 60 deg 45 deg



1FinFout

4

D

2

2 2A B

3 C

2. Calculate the range of 2 in the local coordinate system (required to calculate 3 and 4).

1 deg


K1d

a K2

d

c

K1 3.0000 K2 1.5484

K3a

2b

2 c

2 d

2

2 ac K3 2.9394

A cos K1 K2 cos K3

B 2 sin

C K1 K2 1 cos K3


2 atan2 2 A B B 2 4 A C



K4d

b K5

c2

d2

a2

b2

2 a b K4 1.9512

K5 2.8000

D cos K1 K4 cos K5

E 2 sin

F K1 K4 1 cos K5


2 atan2 2 D E E 2 4 D F


8. Using equation 6.13e, calculate and plot the mechanical advantage of the linkage over the given range.

mA c sin

a sin

rin

rout

45 48 51 54 57 6010

20

30

40

50

60MECHANICAL ADVANTAGE vs HANDLE ANGLE

Handle Angle, deg

Mec

hani

calA

dvan

tage

mA

deg


PROBLEM 6-58

Statement: Figure P6-21 shows a locking pliers. Calculate its mechanical advantage for the position shown.Scale any dimensions needed from the diagram.


1. Draw the linkage to scale in the position given and find the instant centers.

F 1,2

F

O2

3

2,3

A

O

2

4

1,4

B4

P

3,4 and 1,3

1P

2. Note that the linkage is in a toggle position (links 2 and 3 are in line) and the angle between links 2 and 3 is 0 deg.From the discussion below equation 6.13e in the text, we see that the mechanical advantage for this linkage inthis position is theoretically infinite.


PROBLEM 6-59a

Statement: Figure P6-22 shows a fourbar toggle clamp used to hold a workpiece in place by clamping it at D.The linkage will toggle when link 2 reaches 90 deg. For the dimensions given below, calculate itsmechanical advantage for the position shown.

Given: Link lengths:Link 2 (O2A) a 70 mm Link 3 (AB) b 35 mm

Link 4 (O4B) c 34 mm Link 1 (O2O4) d 48 mm


Link 2 (O2C) rin 138 mm Link 4 (O4D) rout 82 mm

Initial position of link 2: 104 deg Global XY system


1. Draw the mechanism to scale and label it. To establish the position of O4 with respect to O2 (in the globalcoordinate frame), draw the linkage in the toggle position with2 = 90 deg. The fixed pivot O4 is then 48 mm from

O2 and 34 mm from B' (see layout).

y

135.069°

x 2

O4

A

3

O2

B 4

B'

X

D

D'

Linkage in toggle position

A'

C

Y

C'

2. Calculate the value of2 in the local coordinate system (required to calculate3 and4).

Rotation angle of local xy system to global XY system: 135.069 deg

31.069 deg


K1d

a K1 0.6857 K2

d

c K2 1.4118

K3a

2b

2 c

2 d

2

2 ac K3 1.4989


A cos K1 K2 cos K3

B 2 sin

C K1 K2 1 cos K3

A 0.4605 B 1.0321 C 0.1189


2 atan2 2 A B B2

4 A C 129.480 deg


K4d

b K5

c2

d2

a2

b2

2 a b K4 1.3714

K5 1.4843


E 2 sin E 1.0321

F K1 K4 1 cos K5 F 0.4804


2 atan2 2 D E E2

4 D F 196.400 deg


165.331 deg


if 90 deg 180 deg 14.669 deg

66.920 deg


if 90 deg 180 deg 66.920 deg

8. Using equation 6.13e, calculate the mechanical advantage of the linkage in the position shown.

mAc sin

a sin

rin

rout mA 2.970


PROBLEM 6-59b

Statement: Figure P6-22 shows a fourbar toggle clamp used to hold a workpiece in place by clamping it at D.The linkage will toggle when link 2 reaches 90 deg. For the dimensions given below, calculate andplot its mechanical advantage as a function of the angle of link AB as link 2 rotates from 120 to 90deg (in the global coordinate system).


Link 2 (O2A) a 70 mm Link 3 (AB) b 35 mm



Link 2 (O2C) rin 138 mm Link 4 (O4D) rout 82 mm

Range of positions of link 2: 120 deg 90.1 deg Global XY system


1. Draw the mechanism to scale and label it. To establish the position of O4 with respect to O2 (in the globalcoordinate frame), draw the linkage in the toggle position with2 = 90 deg. The fixed pivot O4 is then 48 mm from

O2 and 34 mm from B' (see layout).

y

Linkage in initial position

135.069°

x

O4

2

C

3A

B

O2

B'

X

D'

Linkage in final position

A'

4

D

Y

C'


Rotation angle of local xy system to global XY system: 135.069 deg

1 deg


K1d

a K1 0.6857 K2

d

c K2 1.4118


K3a

2b

2 c

2 d

2

2 ac K3 1.4989

A cos K1 K2 cos K3

B 2 sin

C K1 K2 1 cos K3


2 atan2 2 A B B 2 4 A C


K4d

b K5

c2

d2

a2

b2

2 a b K4 1.3714

K5 1.4843

D cos K1 K4 cos K5

E 2 sin

F K1 K4 1 cos K5


2 atan2 2 D E E 2 4 D F


8. Using equation 6.13e, calculate and plot the mechanical advantage of the linkage over the given range.

mA c sin

a sin

rin

rout


90 95 100 105 110 115 1200

10

20

30

40

50

mA

deg


PROBLEM 6-60

Statement: Figure P6-23 shows a surface grinder. The workpiece is oscillated under the spinning grindingwheel by the slider-crank linkage that has the dimensions given below. Calculate and plot thevelocity of the grinding wheel contact point relative to the workpiece over one revolution of thecrank.



Link 2 (O2 to A) a 22 mm Offset c 40 mm

Link 3 (A to B) b 157 mm

Grinding wheel diameter d 90 mm

Input crank angular velocity 120 rpm CCW

Grinding wheel angular velocity 3450 rpm CCW



2

c2

A

O

23

4

B

5


0 deg 2 deg 360 deg


asina sin c

b


a

b

cos cos


VB a sin b sin Positive to the right

6. Calculate the velocity of the grinding wheel contact point using equation 6.7:


VGd

2 VG 16.258

m

sec Directed to the right

7. The velocity of the grinding wheel contact point relative to the workpiece, which has velocity VB, is

VGB VG VB

0 60 120 180 240 300 36015.9

16

16.1

16.2

16.3

16.4

16.5

16.6RELATIVE VELOCITY AT CONTACT POINT

Crank Angle, deg

Velo

city

,m/s

ec

VGB sec

m

deg


PROBLEM 6-61

Statement: Figure P6-24 shows an inverted slider-crank mechanism. Given the dimensions below, find 2,3,

4, VA4, Vtrans, and Vslip for the position shown with VA2 = 20 in/sec in the direction shown.


Link 2 (O2A) a 2.5 in Link 4 (O4A) c 4.1 in Link 1 (O2O4) d 3.9 in

Measured angles:

75.5 deg trans 26.5 deg slip 116.5 deg

Velocity of point A on links 2 and 3: VA2 20 in sec1

VA3 VA2



3

O4

O

A

11.500°

Direction of VA2

4

1

2

Y

2


Direction of VA4

Axis of slip

X

2. Use equation 6.7 to calculate the angular velocity of link 2.

VA2

a 8.000

rad

sec CW

3. Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A on links 2 and3.. The equation to be solved graphically is

VA3 = Vtrans + VA3slip

a. Choose a convenient velocity scale and layout the known vector VA3.b. From the tip of VA3, draw a construction line with the direction of VA3slip, magnitude unknown.c. From the tail of VA3, draw a construction line with the direction of Vtrans, magnitude unknown.d. Complete the vector triangle by drawing VA3slip from the tip of Vtrans to the tip of VA3 and drawing Vtransfrom the tail of VA3 to the intersection of the VA3slip construction line.


1.509 VA3

VA4

1.686


2.064

A4slip

VA3slip

V trans

V

Y 0

X

Axis of slip

10 in/sec



1

in

VA4 1.686 in kv VA4 16.9in

sec


sec


sec


VA4

c 4.1

rad

sec CW

Because link 3 slides within link 4,3 = 4.


PROBLEM 6-62

Statement: Figure P6-25 shows a drag-link mechanism with dimensions. Write the necessary equations andsolve them to calculate and plot the angular velocity of link 4 for an input of2 = 1 rad/sec.Comment on the uses for this mechanism.


Link 2 (L2) a 1.38 in Link 3 (L3) b 1.22 in

Link 4 (L4) c 1.62 in Link 1 (L1) d 0.68 in


CW



atany

x

return x 0if

atany

x

otherwise



2

4

2

OO2 4

y3

A

x

B

2. Determine the range of motion for this Grashof double crank.

0 deg 2 deg 360 deg


K1d

a K2

d

c

K1 0.4928 K2 0.4198

K3a

2b

2 c

2 d

2

2 ac K3 0.7834

A cos K1 K2 cos K3

B 2 sin

C K1 K2 1 cos K3



2 atan2 2 A B B 2 4 A C


K4d

b K5

c2

d2

a2

b2

2 a b K4 0.5574

K5 0.3655

D cos K1 K4 cos K5

E 2 sin

F K1 K4 1 cos K5


2 atan2 2 D E E 2 4 D F


a

b

sin sin

a

c

sin sin

9. Plot the angular velocity of link 4.

0 60 120 180 240 300 3602.5

2

1.5

1

0.5ANGULAR VELOCITY, LINK 4

Crank Angle, deg

Ang

ular

Vel

ocity

,rad

/sec

sec

rad

deg


PROBLEM 6-63

Statement: Figure P6-25 shows a drag-link mechanism with dimensions. Write the necessary equations andsolve them to calculate and plot the centrodes of instant center I2,4.

Given: Measured link lengths:

Link 2 (L2) L2 1.38 in Link 3 (L3) L3 1.22 in

Link 4 (L4) L4 1.62 in Link 1 (L1) L1 0.68 in




atany

x

return x 0if

atany

x

otherwise


1. Draw the linkage to scale and label it. Instant center I2,4 is at the intersection of line AB with line O2O4. To getthe first centrode, ground link 2 and let link 3 be the input. Then we have

a L3 b L4 c L1 d L2

2

4

x

2

4

OO2 4

3A

y

B

2. Determine the range of motion for this Grashof double rocker. From Figure 3-1a on page 80, one toggle angle is

acosa

2d

2 b c( )

2

2 a d

124.294 deg

The other toggle angle is the negative of this. The range of motion is

1 deg


K1d

a K2

d

c

K1 1.1311 K2 2.0294


K3a

2b

2 c

2 d

2

2 ac K3 0.7418

A cos K1 K2 cos K3

B 2 sin

C K1 K2 1 cos K3


2 atan2 2 A B B 2 4 A C

5. Calculate the coordinates of the intersection of line BC with line AD. This will be the instant center I2,4.

Line BC: y x tan

Line AD: y 0 x d( ) tan

Eliminating y and solving for the x- and y-coordinates of the intersection,

x24 d tan

tan tan y24 x24 tan

6. Plot the fixed centrode.

10 5 0 5 1010

5

0

5

10FIXED CENTRODE

x-Coordinate, in

y-C

oord

inat

e,in

y24 in

x24 in

7. Invert the linkage, making C and D the fixed pivots. Then,

a L1 b L2 c L3 d L4


0 deg 0.5 deg 360 deg


4O2

2

O

4

2

y

A

3

B

x4


K1d

a K2

d

c

K1 2.3824 K2 1.3279

K3a

2b

2 c

2 d

2

2 ac K3 1.6097

A cos K1 K2 cos K3

B 2 sin

C K1 K2 1 cos K3


2 atan2 2 A B B 2 4 A C

11. Calculate the coordinates of the intersection of line BC with line AD. This will be the instant center I2,4.

Line AD: y x tan

Line BC: y 0 x d( ) tan

Eliminating y and solving for the x- and y-coordinates of the intersection,

x24 d tan

tan tan y24 x24 tan

6. Plot the moving centrode. (See next page.)


10 5 0 5 1010

5

0

5

10MOVING CENTRODE

x-Coordinate, in

y-C

oord

inat

e,in

y24 in

x24 in


PROBLEM 6-64

Statement: Figure P6-26 shows a mechanism with dimensions. Use a graphical method to calculate thevelocities of points A, B, and C and the velocity of slip for the position shown.

Given: Link lengths and angles:

Link 1 (O2O4) d 1.22 in Angle O2O4 makes with X axis 56.5 deg

Link 2 (O2A) a 1.35 in Angle 2 makes with X axis 14 deg

Link 4 (O4B) e 1.36 in

Link 5 (BC) f 2.69 in

Link 6 (O6C) g 1.80 in Angle O6C makes with X axis 88 deg

Angular velocity of link 2 20 rad sec1

CW



132.661°


Axis of slip

0.939

Direction of VA3

O 2

23

A

Y

O4

B

4

X

2. Use equation 6.7 to calculate the magnitude of the velocity at point A on links 2 and 3.

VA3 a VA3 27.000in

sec VA3 90 deg VA3 76.0 deg

3. Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A on link 3. Theequation to be solved graphically is

VA3 = Vtrans + Vslip

a. Choose a convenient velocity scale and layout the known vector VA3.b. From the tip of VA3, draw a construction line with the direction of Vslip, magnitude unknown.c. From the tail of VA3, draw a construction line with the direction of Vtrans, magnitude unknown.d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA3 and drawing Vtrans fromthe tail of VA3 to the intersection of the Vslip construction line.


V

1.295

2.369 Vslip

transV

Y

A3

X

0 10 in/sec



1

in


sec


sec

5. The true velocity of point A on link 4 is Vtrans,

VA4 Vtrans VA4 12.95in

sec


From the linkage layout above: c 0.939 in and 132.661 deg

VA4

c 13.791

rad

sec CW

7. Determine the magnitude and sense of the vector VB using equation 6.7.

VB e VB 18.756in

sec

VA4 90 deg VA4 42.661 deg

8. Draw links 1, 4, 5, and 6 to a convenient scale. Indicate the directions of the velocity vectors of interest. (Seenext page.)


C

O 6

Direction of VC

Direction of VCB

O 2

2

3

A

X

B Direction of VB

4

Y

O 4


VC = VB + VCB

a. Choose a convenient velocity scale and layout the known vector VB.b. From the tip of VB, draw a construction line with the direction of VCB, magnitude unknown.c. From the tail of VB, draw a construction line with the direction of VC, magnitude unknown.d. Complete the vector triangle by drawing VCB from the tip of VB to the intersection of the VC constructionline and drawing VC from the tail of VB to the intersection of the VCB construction line.

V

2.088

Y

VB

C

X

VCB

0 10 in/sec



1

in VC 90 deg


sec VC 2.0 deg


PROBLEM 6-65

Statement: Figure P6-27 shows a cam and follower. Distances are given below. Find the velocities of points Aand B, the velocity of transmission, velocity of slip, and 3 if 2 = 50 rad/sec (CW). Use agraphical method.

Given: 50 rad sec1

Distance from O2 to A: a 1.890 in

Distance from O3 to B: b 1.645 in

Assumptions: Roll-slide contact



Direction of VA

1.645

1.890

Axis of slip

B

O2

2

A

3

O3

Direction of VB



VA a VA 94.500in

sec

3. Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A. The equationto be solved graphically is

VA = Vtrans + VAslip

a. Choose a convenient velocity scale and layout the known vector VA.b. From the tip of VA, draw a construction line with the direction of Vslip, magnitude unknown.c. From the tail of VA, draw a construction line with the direction of Vtrans, magnitude unknown.d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA and drawing Vtrans fromthe tail of VA to the intersection of the Vslip construction line.


1.890

2.304

1.317

Y

3.256

VA

transV

A2slipV

X

V

V

B

Bslip

0 50 in/sec



1

in

VA 1.890 in kv VA 94.5in

sec


sec


sec


sec


VB

b 70.0

rad

sec CW


PROBLEM 6-66

Statement: Figure P6-28 shows a quick-return mechanism with dimensions. Use a graphical method to calculatethe velocities of points A, B, and C and the velocity of slip for the position shown.

Given: Link lengths and angles:

Link 1 (O2O4) d 1.69 in Angle O2O4 makes with X axis 15.5 deg

Link 2 (L2) a 1.00 in Angle link 2 makes with X axis 99 deg

Link 4 (L4) e 4.76 in

Link 5 (L5) f 4.55 in

Offset (O2C) g 2.86 in

Angular velocity of link 2 10 rad sec1

CCW



2.068

Direction of VA3

44.228°

O 4

2

O2

3

A

X


Y

B

Axis of slip

4

2. Use equation 6.7 to calculate the magnitude of the velocity at point A on links 2 and 3.

VA2 a VA2 10.000in

sec VA2 90 deg VA2 189.0deg

3. Use equation 6.5 to (graphically) determine the magnitude of the velocity components at point A on link 3. Theequation to be solved graphically is

VA3 = Vtrans + Vslip


a. Choose a convenient velocity scale and layout the known vector VA3.b. From the tip of VA3, draw a construction line with the direction of Vslip, magnitude unknown.c. From the tail of VA3, draw a construction line with the direction of Vtrans, magnitude unknown.d. Complete the vector triangle by drawing Vslip from the tip of Vtrans to the tip of VA3 and drawing Vtrans fromthe tail of VA3 to the intersection of the Vslip construction line.

1.1541.634Vslip

V

transV

A3

Y

X

0 5 in/sec



1

in


sec


sec

5. The true velocity of point A on link 4 is Vtrans,

VA4 Vtrans VA4 5.77in

sec


From the linkage layout above: c 2.068 in and 44.228 deg

VA4

c 2.790

rad

sec CCW

7. Determine the magnitude and sense of the vector VB using equation 6.7.

VB e VB 13.281in

sec


8. Draw links 1, 4, 5, and 6 to a convenient scale. Indicate the directions of the velocity vectors of interest. (Seenext page.)


Direction of VCB

Direction of VC

6C

44.228°

O4

2

O 2

3

A

X

Direction of VB

5.805°

Y

5

4

B


VC = VB + VCB



Y

1.659

V

V

CB

C

VB0

X

5 in/sec


1

in


sec

VC 180 deg


PROBLEM 6-67

Statement: Figure P6-29 shows a drum pedal mechanism . For the dimensions given below, find and plot themechanical advantage and the velocity ratio of the linkage over its range of motion. If the inputvelocity Vin is a constant and Fin is constant, find the output velocity, output force, and power inover the range of motion.


Link 2 (O2A) a 100 mm Link 3 (AB) b 28 mm



Link 2 rin 48 mm Link 3 (AP) rout 124 mm

Input force and velocity: Fin 50 N Vin 3 m sec1

Range of positions of link 2: 162 deg 171 deg



y

B

x

2

34

A

2

1O4

Fin

inV

3

O2

P


Rotation angle of local xy system to global XY system: 180 deg

1 deg


K1d

a K2

d

c

K1 0.5600 K2 0.8750

K3a

2b

2 c

2 d

2

2 ac K3 1.2850

A cos K1 K2 cos K3


B 2 sin

C K1 K2 1 cos K3


2 atan2 2 A B B 2 4 A C


K4d

b K5

c2

d2

a2

b2

2 a b K4 2.0000

K5 1.7543

D cos K1 K4 cos K5

E 2 sin

F K1 K4 1 cos K5


2 atan2 2 D E E 2 4 D F

7. Using equations 6.13d and 6.18a, where out is 3, calculate and plot the mechanical advantage of the linkageover the given range.

mA b sin

a sin

rin

rout

162 164 166 168 170 1720.1

0.11

0.12

0.13

MECHANICAL ADVANTAGE

Pedal Angle, deg

Mec

hani

calA

dvan

tage

mA

deg


8. Calculate and plot the velocity ratio using equation 6.13d,

mV 1

mA

162 164 166 168 170 1720

2

4

6

8

10VELOCITY RATIO

Pedal Angle, deg

Velo

city

Ratio

mV

deg

9. Calculate and plot the output velocity using equation 6.13a.

Vout Vin mV

162 164 166 168 170 1720

10

20

30OUTPUT VELOCITY

Pedal Angle, deg

Velo

city

,m/s

ec

Vout sec

m

deg


10. Calculate and plot the output force using equation 6.13a.

Fout Fin mA

162 164 166 168 170 1720

2

4

6

8OUTPUT FORCE

Pedal Angle, deg

Forc

e,N Fout

N

deg


PROBLEM 6-68

Statement: Figure 3-33 shows a sixbar slider crank linkage. Find all of its instant centers in the position shown:

Given: Number of links n 6

Solution: See Figure 3-33 and Mathcad file P0668.


Cn n 1( )

2 C 15

2. Draw the linkage to scale and identify those ICs that can be found by inspection (8).

1,6 at infinity

O

1,42O

2

1,2

4

X

4

5,6

5

6

2,3

Y

3 3,4; 3,5; 4,5

1,5

1,6 at infinity

2,5

3,4; 3,5; 4,5; 2,43,6

O

1,42O

2

1,2

1,34

X

4

5,6

5

6

2,6

2,3

4,6

Y

3

4

5 3

1

6 2

2. Use Kennedy's Rule and a lineargraph to find the remaining 7 ICs:I1,3; I1,5; I2,4; I2,5; I2,6; I3,6; and I4,6

I1,5: I1,6-I5,6 and I1,4-I4,5

I2,5: I1,2-I1,5 and I2,3-I3,5

I1,3: I1,2-I2,3 and I1,5-I3,5

I3,6: I1,6-I1,3 and I3,5-I5,6

I2,4: I2,3-I3,4 and I2,5-I4,5

I2,6: I1,2-I1,6 and I2,5-I5,6

I4,6: I1,4-I1,6 and I4,5-I5,6


PROBLEM 6-69

Statement: Calculate and plot the centrodes of instant center I24 of the linkage in Figure 3-33 so that a pair ofnoncircular gears can be made to replace the driver dyad 23.


Input crank (L2) L2 2.170 Fourbar coupler (L3) L3 2.067

Output crank (L4) L4 2.310 Fourbar ground link (L1) L1 1.000

Two argument inverse tangent: atan2 x y( ) 0.5 return x 0= y 0if


atany

x

return x 0if

atany

x

otherwise


1. Invert the linkage, grounding link 2 such that the input link is 3, the coupler is 4, and the output link is 1.

a L3 b L4 c L1 d L2

2. Define the input crank motion for this inversion: 47 deg 47.5 deg 102.5 deg

3. Use equations 4.8a and 4.10 to calculate4 as a function of 2 (for the open circuit).

K1d

a K2

d

c K3

a2

b2

c2

d2

2 ac

K1 1.0498 K2 2.1700 K3 1.1237

A cos K1 K2 cos K3

B 2 sin C K1 K2 1 cos K3

2 atan2 2 A B B 2 4 A C

if 2 2

5. Calculate the coordinates of the intersection of links 1 and 3 in the xy coordinate system.

x242 d tan

tan tan y242 x242 tan

6. Invert the linkage, grounding link 4 such that the input link is 1, the coupler is 2, and the output link is 3.

a L1 b L2 c L3 d L4

8. Use equations 4.8a and 4.10 to calculate4 as a function of 2 (for the open circuit).

K1d

a K2

d

c K3

a2

b2

c2

d2

2 ac

K1 2.3100 K2 1.1176 K3 1.4271

A cos K1 K2 cos K3



2 atan2 2 A B B 2 4 A C

if 2 2

5. Calculate the coordinates of the intersection of links 1 and 3 in the xy coordinate system.

x244 d tan

tan tan y244 x244 tan

4 3 2 1 0 15

4

3

2

1

0

1

2

3

4

5LINK 2 GROUNDED

y242

x242

7. Define the input crank motion for this inversion: 41 deg 42 deg 241 deg

10 8 6 4 2 0 2 4 6 8 1010

8

6

4

2

0

2

4

6

8

10LINK 4 GROUNDED

y244

x244


PROBLEM 6-70a

Statement: Find the velocity of the slider in Figure 3-33 for2 = 110 deg with respect to the global X axis

assuming 2 = 1 rad/sec CW. Use a graphical method.



Link 4 (O4 to B) c 2.310 in Link 1 (O2 to O4) d 1.000 in

Link 5 (B to C) e 5.400

Coordinate angle 102 deg Crank angle: 2 110 deg


CW



B

147.635°

Direction of VB

Direction of VC

Direction of VCB

O4

2O

2

58.950°

4

X

C

158.818°5

6

Direction of VA

Direction of VBAA

Y

3


VA a VA 2.170in

sec

VA 2 90 deg VA 20.000 deg


VB = VA + VBA



1.325

1 in/sec0Y

X

VB

VBA

VA



1

in


sec VB 31.050 deg

5. Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C, the magnitude of therelative velocity VCB, and the angular velocity of link 3. The equation to be solved graphically is

VC = VB + VCB


V

1.4001 in/sec0

Y

X

VB CB

VC



1

in


sec VC 0.0 deg


PROBLEM 6-70b


assuming 2 = 1 rad/sec CW. Use the method of instant centers.




Link 5 (B to C) e 5.400

Coordinate angle 102 deg Crank angle: 2 110 deg


CW

Solution: See Figure 3-33 and Mathcad file P0670b.

1. Draw the linkage to scale in the position given, find instant centers I1,3 and I1,5, and the distances from the pinjoints to the instant centers. See Problem 6-68 for the determination of IC locations.

2 B

1,3O4

2O X

4

Y

A

3

C

5

6

1,5


AI13 2.609 in BI13 1.641 in BI15 9.406 in CI15 9.896 in



VA a VA 2.170in

sec



VA

AI13 0.832

rad

sec CW

4. Determine the magnitude of the velocity at point B using equation 6.9b. The direction of VB is down and to theright

VB BI13 VB 1.365in

sec


VB

BI15 0.145

rad

sec CCW

6. Determine the magnitude of the velocity at point C using equation 6.9b.

VC CI15 VC 1.436in

sec to the right


PROBLEM 6-70c


assuming 2 = 1 rad/sec CW. Use an analytical method.




Link 5 (B to C) e 5.400 in

Coordinate angle 102 deg Crank angle: 2XY 110 deg

Input crank angular velocity 2 1 rad sec1

CW



atany

x

return x 0if

atany

x

otherwise

Solution: See Figure P6-33 and Mathcad file P0670c.


B

102°

O4

x

2O

2

Xy

4

C

5

6

A

Y

3


Transform crank angle to the local xy coordinate system:

2 2XY 2 212.000 deg

K1d

a K1 0.4608 K2

d

c K2 0.4329

K3a

2b

2 c

2 d

2

2 ac K3 0.6755

A cos 2 K1 K2 cos 2 K3


B 2 sin 2

C K1 K2 1 cos 2 K3

A 0.2662 B 1.0598 C 2.3515


4xy 2 atan2 2 A B B2

4 A C 2 4xy 159.635 deg


K4d

b K5

c2

d2

a2

b2

2 a b K4 0.4838

K5 0.5178

D cos 2 K1 K4 cos 2 K5 D 2.2370

E 2 sin 2 E 1.0598

F K1 K4 1 cos 2 K5 F 0.3808


3xy 2 atan2 2 D E E2

4 D F 2 3xy 70.950deg

6. Determine the angular velocity of link 4 for the open circuit using equations 6.18.

4a 2

c

sin 2 3xy sin 4xy 3xy 4 0.591

rad

sec

7. Transform4 back to the global XY system.

4 4xy 4 57.635 deg

5. Determine5 and d, with respect to O4, using equation 4.17.

Offset: cc 0 in

5 asinc sin 4 cc

e

5 158.818 deg

dd c cos 4 e cos 5 dd 6.272 in


5c

e

cos 4 cos 5 4 5 0.145

rad

sec

7. Determine the velocity of pin C using equation 6.22b:

VC c 4 sin 4 e 5 sin 5

VC 1.436in

sec VC 1.436

in

sec arg VC 0.000deg


PROBLEM 6-71

Statement: Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver tocalculate and plot the angular velocity of link 4 and the linear velocity of slider 6 in the sixbarslider-crank linkage of Figure 3-33 as a function of the angle of input link 2 for a constant2 = 1

rad/sec CW. Plot Vc both as a function of 2 and separately as a function of slider position as

shown in the figure. What is the percent deviation from constant velocity over 240 deg < 2 < 270

deg and over 190 < 2 < 315 deg?


Input crank (L2) a 2.170 Fourbar coupler (L3) b 2.067

Output crank (L4) c 2.310 Sllider coupler (L5) e 5.40

Fourbar ground link (L1) d 1.000



atany

x

return x 0if

atany

x

otherwise

Crank velocity: 1rad

sec


1. This sixbar drag-link mechanism can be analyzed as a fourbar Grashof double crank in series with a slider-crankmechanism using the output of the fourbar, link 4, as the input to the slider-crank.

2. Define one revolution of the input crank: 0 deg 1 deg 360 deg

3. Use equations 4.8a and 4.10 to calculate4 as a function of 2 (for the open circuit) in the global XY coordinatesystem.

K3a

2b

2 c

2 d

2

2 ac

K1d

a K2

d

c

K1 0.4608 K2 0.4329 K3 0.6755

A cos K1 K2 cos K3


2 atan2 2 A B B 2 4 A C

102 deg

4. If the calculated value of4 is greater than 2, subtract 2from it and if it is negative, make it positive.

if 2 2

if 0 2

5. Determine the slider-crank motion using equations 4.16 and 4.17 with 4 as the input angle.

asinc sin

e

f c cos e cos



K4d

b K5

c2

d2

a2

b2

2 a b K4 0.4838

K5 0.5178

D cos K1 K4 cos K5

E 2 sin

F K1 K4 1 cos K5


2 atan2 2 D E E 2 4 D F


a

c

sin sin 102 deg

0 45 90 135 180 225 270 315 3602

1.75

1.5

1.25

1

0.75

0.5

deg


c

e

cos cos

9. Determine the velocity of pin C using equation 6.22b:

VC c sin e sin


0 45 90 135 180 225 270 315 3604

3

2

1

0

1

2

VC

deg

3 4 5 6 7 84

3

2

1

0

1

2

VC

f


PROBLEM 6-72

Statement: Figure 3-34 shows a Stephenson's sixbar mechanism. Find all of its instant centers in the positionshown:




Cn n 1( )

2 C 15

a. In part (a) of the figure. ___________________________________________________________________


5,6

4,5

53

O2 2

2,3

1,2

5 3,54

O4

O61,4

61,6

5,6

4,5

4,6

5

4

353

6

1,2; 2,5; 2,4; 2,6

1

2

O2 2

2,3

3,5; 1,3; 3,4; 3,65

4O4

1,4

1,561,6

O6

2. Use Kennedy's Rule and a linear graph to find the remaining 8 ICs:I1,3; I1,5; I2,4; I2,5; I2,6; I3,4; I3,6; and I4,6

I1,5: I1,6-I5,6 and I1,4-I4,5

I2,5: I1,2-I1,5 and I2,3-I3,5

I1,3: I1,2-I2,3 and I1,5-I3,5

I3,4: I1,4-I1,3 and I4,5-I3,5

I2,8: I2,3-I3,4 and I2,5-I4,5

I2,6: I1,2-I1,6 and I2,5-I5,6

I3,6: I1,3-I1,6 and I3,5-I5,6

I4,6: I1,4-I1,6 and I4,5-I5,6


b. In part (b) of the figure. ___________________________________________________________________


4,5

3

2,32 O2

1,2

3,5

45

4O

O6

1,4

65

1,6

5,6

2. Use Kennedy's Rule and a linear graph to find the remaining 8 ICs: I1,3; I1,5; I2,4; I2,5; I2,6; I3,4; I3,6; and I4,6

4,5

2,5; 2,4; 2,8

5

4

3

To 1,3

6

3

2,32 O2

1,2 3,6

2

1

3,5; 3,4

45

4O

O6

1,4; 1,52,6

65

1,6

5,6; 4,6

I1,5: I1,6-I5,6 and I1,4-I4,5

I2,5: I1,2-I1,5 and I2,3-I3,5

I1,3: I1,2-I2,3 and I1,5-I3,5

I3,4: I1,4-I1,3 and I4,5-I3,5

I2,4: I2,3-I3,4 and I2,5-I4,5

I2,6: I1,2-I1,6 and I2,5-I5,6

I3,6: I1,3-I1,6 and I3,5-I5,6

I4,6: I1,4-I1,6 and I4,5-I5,6


c. In part (c) of the figure. ___________________________________________________________________


3,5

2,3

1,2

3

O2

2

64

O4

O6

1,4

1,6

4,5

55

5,6






1,5

1,2; 2,5; 2,4: 2,6

4

3,5; 1,3; 3,4; 3,6

5 3

6

1

2

2,3

3

O2

2

64

O4

O6

1,4

1,6

4,6

4,5

55

5,6


PROBLEM 6-73a

Statement: Find the angular velocity of link 6 in Figure 3-34b for6 = 90 deg with respect to the global X axis

assuming 2 = 10 rad/sec CW. Use a graphical method.


Link 2 (O2 to A) g 1.556 in Link 3 (A to B) f 4.248 in

Link 4 (O4 to C) c 2.125 in Link 5 (C to D) b 2.158 in

Link 6 (O6 to D) a 1.542 in Link 5 (B to D) p 3.274 in

Link 1 X-offset dX 3.259 in Link 1 Y-offset dY 2.905 in

Angle CDB 5 36.0 deg

Output rocker angle: 90 deg Global XY system


CW

Solution: See Figure 3-34b and Mathcad file P0673a.


Direction of VC

Direction of VA

Direction of VAB

Direction of VCD

Direction of VD

Direction of VBD

3 4

B

5

2A

O2

C

5

O4

D

O

6

6

X

Y

2. Since this linkage is a Stephenson's II sixbar, we will have to start at link 6 and work back to link 2. We willassume a value for 6 and eventually find a value for2. We will then multiply the magnitudes of all velocities

by the ratio of the actual 2 to the found 2. Use equation 6.7 to calculate the magnitude of the velocity at pointD.

Assume: 1 rad sec1

CW

VD a VD 1.542in

sec VD 0 deg

3. Use equation 6.5 to (graphically) determine the magnitude of the velocity at point C, the magnitude of therelative velocity VCD, and the angular velocity of link 5. The equation to be solved graphically is

VC = VD + VCD

a. Choose a convenient velocity scale and layout the known vector VD.b. From the tip of VD, draw a construction line with the direction of VCD, magnitude unknown.


c. From the tail of VD, draw a construction line with the direction of VC, magnitude unknown.d. Complete the vector triangle by drawing VCD from the tip of VD to the intersection of the VC constructionline and drawing VC from the tail of VD to the intersection of the VCD construction line.

1.289

0

1.309

127.003°

VVC

VD

CD

54.195°

1 in/sec



1

in


sec VC 54.195 deg

VCD 1.309 in kv VCD 1.309in

sec VCD 127.003 deg


VCD

b 0.607

rad

sec

VC

c 0.607

rad

sec

6. Determine the magnitude and sense of the vector VBD using equation 6.7.

VBD p VBD 1.986in

sec

VBD VCD 5 VBD 163.003deg


VB = VD + VBD

a. Choose a convenient velocity scale and layout the known vector VD.b. From the tip of VD, layout the (now) known vector VBD.c. Complete the vector triangle by drawing VB from the tail of VD to the tip of the VBD vector.

0.682VD

121.607°

VB

VBD 0 1 in/sec




1

in


sec VB 121.607 deg

9. Use equation 6.5 to (graphically) determine the magnitude of the velocity at point A, the magnitude of the relativevelocity VAB, and the angular velocity of link 2. The equation to be solved graphically is

VA = VB + VAB


0.645

ABV 0VB

131.690°

VA

1 in/sec



1

in

VA 0.645 in kv VA 0.645in

sec VA 131.690 deg

11. Determine the angular velocity of link 2 with respect to the assumed value of 6 using equation 6.7.

VA

g 0.415

rad

sec CW

12. Calculate the actual value of the angular velocity of link 6.

rad

sec 24.124

rad

sec


PROBLEM 6-73b


assuming 2 = 10 rad/sec CW. Use the method of instant centers.






Angle CDB 5 36.0 deg

Output rocker angle: 90 deg Global XY system


CW

Solution: See Figure 3-34b and Mathcad file P0673b.

1. Draw the linkage to scale in the position given, find instant centers I1,3 and I1,5, and the distances from the pinjoints to the instant centers. See Problem 6-72b for determination of IC locations.

To 1,3

3

2 O2A

O

O1,5

B

45

4

6

D

65C


AI13 22.334 in BI13 23.650 in BI15 1.124 in DI15 2.542 in


2. Start from point D with an assumed value for 6 and work to find 2. Then, use the ratio of the actual vale of 2

to the found value to calculate the actual value of 6. Use equation 6.7 and inspection of the layout to determinethe magnitude and direction of the velocity at point D.

1rad

sec

VD a VD 1.542in

sec

VD 90 deg VD 0.0 deg


VD

DI15 0.607

rad

sec CW

4. Determine the magnitude of the velocity at point B using equation 6.9b.

VB BI15 VB 0.682in

sec


VB

BI13 0.029

rad

sec CCW

6. Determine the magnitude of the velocity at point A using equation 6.9b.

VA AI13 VA 0.644in

sec to the left

7. Use equation 6.9c to determine the angular velocity of link 2 based on the assumed value of6.

VA

g 0.414

rad

sec CW

8. Multiply the assumed vaue of 6 by the ratio of 21 over22 to get the value of 6 for 2 = 10 rad/sec.

rad

sec 24.166

rad

sec CW


PROBLEM 6-73c


assuming 2 = 10 rad/sec CW. Use an analytic method.






Angle CDB 5 36.0 deg Link 1 (O4 to O6) d 1.000 in

Output rocker angle: 6XY 90 deg Global XY system


CW

Coordinate rotationm angle 90 deg



atany

x

return x 0if

atany

x

otherwise

Solution: See Figure 3-34b and Mathcad file P0673c.

1. Transform the crank angle to the local coordinate system. Draw the linkage to scale and label it.

6 6XY 6 180.000 deg

C

3

A2 O2

Y

y

x

B

45

4O

O6

65

DX


K1d

a K2

d

c K3

a2

b2

c2

d2

2 ac

K1 0.6485 K2 0.4706 K3 0.4938


A cos 6 K1 K2 cos 6 K3 A 0.6841

B 2 sin 6 B 0.0000

C K1 K2 1 cos 6 K3 C 2.6129


4 2 atan2 2 A B B2

4 A C 4 234.195 deg


K4d

b K5

c2

d2

a2

b2

2 a b K4 0.4634

K5 0.5288

D cos 6 K1 K4 cos 6 K5 D 2.6407

E 2 sin 6 E 0.0000

F K1 K4 1 cos 6 K5 F 0.6563


51 2 atan2 2 D E E2

4 D F 51 307.003 deg

6. Determine the angular velocity of links 4 and 5 for the open circuit using equations 6.18. Initially assume 6 = 1

rad/sec. then by trial and error, change it to make2 = 10 rad/ sec CW.

6 1 rad sec1

5a 6

b

sin 4 6 sin 51 4 5 0.607

rad

sec

4a 6

c

sin 6 51 sin 4 51 4 0.607

rad

sec


PROBLEM 6-74

Statement: Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver tocalculate and plot the angular velocity of link 6 in the sixbar linkage of Figure 3-34 as a function of2 for a constant2 = 1 rad/sec CW.






Angle CDB 5 36.0 deg Link 1 (O4 to O6) d 1.000 in

Output rocker angle: 6XY 90 deg Global XY system


CW

Coordinate rotationm angle 90 deg


1. This problem is long and may be more appropriate for a project assignment. The solution involves definingvector loops and solving the resulting equations using a method such as Newton-Raphson.


PROBLEM 6-75

Statement: Figure 3-35 shows a Stephenson's sixbar mechanism. Find all of its instant centers in the positionshown:




Cn n 1( )

2 C 15

a. In part (a) of the figure. ___________________________________________________________________


1,2

O2 2,3

21,6

3O4

3,4

O6

5

1,4

4,5

46

5,6


2,31,2; 2,5; 2,4; 2,6

O23

21,6 4,6

1,3; 3,5; 3,6

1,5

5

O4

3,4

6

O6

4

3

2

1

6

5

4 1,4

4,5

5,6

I1,5: I1,6-I5,6 and I1,4-I4,5

I1,3: I1,2-I2,3 and I1,4-I3,4

I3,5: I1,5-I1,3 and I3,4-I4,5

I2,5: I1,2-I1,5 and I2,3-I3,5

I2,4: I2,3-I3,4 and I2,5-I4,5

I2,6: I1,2-I1,6 and I2,5-I5,6

I3,6: I1,3-I1,6 and I3,5-I5,6

I4,6: I1,4-I1,6 and I4,5-I5,6


b. In part (b) of the figure. ___________________________________________________________________


2

O

2,3

2

31,2

1,6

5,6

1,4

O4

3,4

45

64,5






5

6

3

4

2

12

O

1,2; 2,5; 2,4; 2,6

2,3

2

3

1,2

3,4; 1,3; 3,5; 3,6

4,6

1,5

5,6

1,4

O4

45 1,6

6

4,5


PROBLEM 6-76a

Statement: Use a compass and straightedge to draw the linkage in Figure 3-35 with link 2 at 90 deg and find theangular velocity of link 6 assuming 2 = 10 rad/sec CCW. Use a graphical method.




Link 4 (O4 to D) e 1.429 in Link 5 (D to E) f 1.286

Link 6 (O6 to E) g 0.771 in Crank angle: 2 90 deg


CCW

Solution: See Figure 3-35 and Mathcad file P0676a.


Direction of VBA

Direction of VB

Direction of VA

33.359°

O2

A

23

Y

X

122.085° 100.938°

Direction of VED

Direction of VEB

O44

O6

6

35.228°E

Direction of VD

D

5


VA a VA 10.000in

sec

VA 2 90 deg VA 180.000 deg


VB = VA + VBA



1.063 VBA5 in/sec0

VA

1.671

X

Y

VB



1

in


sec VB 147.915 deg


sec VBA 56.641 deg

VB

c 6.497

rad

sec CW

5. Calculate the magnitude and direction of VD.

VD e VD 9.284in

sec VD 55.085 deg

6. Use equation 6.5 to (graphically) determine the magnitude of the velocity at point E, the magnitude of the relativevelocity VED, and the angular velocity of link 3. The equation to be solved graphically is

VE = VD + VED

a. Choose a convenient velocity scale and layout the known vector VD.b. From the tip of VD, draw a construction line with the direction of VED, magnitude unknown.c. From the tail of VD, draw a construction line with the direction of VE, magnitude unknown.d. Complete the vector triangle by drawing VED from the tip of VD to the intersection of the VE constructionline and drawing VE from the tail of VD to the intersection of the VED construction line.

V

2.450

X

VDE

E

VD

5 in/secY 04. From the velocity triangle we have:


1

in

VE 2.450 in kv VE 12.250in

sec

VE

g

15.888rad

sec CW


PROBLEM 6-76b

Statement: Use a compass and straightedge to draw the linkage in Figure 3-35 with link 2 at 90 deg and find theangular velocity of link 6 assuming 2 = 10 rad/sec CCW. Use the method of instant centers.




Link 4 (O4 to D) e 1.429 in Link 5 (D to E) f 1.286

Link 6 (O6 to E) g 0.771 in Crank angle: 2 90 deg


CCW

Solution: See Figure 3-35 and Mathcad file P0676a.

1. Draw the linkage to scale in the position given, find instant centers I1,3 and I1,5, and the distances from the pinjoints to the instant centers. See Problem 6-68 for the determination of IC locations.

1,3

O2

A

23

B

O44

O6

6

E

D

1,5

5


AI13 7.152 in BI13 5.975 in BI15 1.740 in DI15 0.947 in EI15 1.249 in


VA a VA 10.000in

sec



VA

AI13 1.398

rad

sec CCW



VB BI13 VB 8.354in

sec


VB

c 6.496

rad

sec CW

6. Determine the magnitude of the velocity at point D using equation 6.9b.

VD e VD 9.283in

sec


VD

DI15 9.803

rad

sec CW

8. Determine the magnitude of the velocity at point E using equation 6.9b.

VE EI15 VE 12.244in

sec


VE

g 15.88

rad

sec CW


PROBLEM 6-76c

Statement: Use a compass and straightedge to draw the linkage in Figure 3-35 with link 2 at 90 deg and find theangular velocity of link 6 assuming 2 = 10 rad/sec CCW. Use an analytical method.




Link 4 (O4 to D) e 1.429 in Link 5 (D to E) f 1.286 in

Link 6 (O6 to E) g 0.771 in Link 1 (O4 to O6) h 0.786 in

Crank angle: 2 90 deg


CCW

Solution: See Figure 3-35 and Mathcad file P0676c.


33.359°

O2

A

23

Y

X

122.085° 100.938°

B

O44

O6

6

35.228°E

D

5


K1d

a K2

d

c K1 3.8570 K2 2.9992

K3a

2b

2 c

2 d

2

2 ac K3 1.2015


B 2 sin 2

C K1 K2 1 cos 2 K3

A 2.6555 B 2.0000 C 5.0585


41 2 atan2 2 A B B2

4 A C 41 237.915 deg


K4d

b K5

c2

d2

a2

b2

2 a b K4 1.0150 K5 3.7714

D cos 2 K1 K4 cos 2 K5 D 7.6284


E 2 sin 2 E 2.0000

F K1 K4 1 cos 2 K5 F 0.0856


3 2 atan2 2 D E E2

4 D F 3 326.641 deg


3a 2

b

sin 41 2 sin 3 41 3 1.398

rad

sec

4a 2

c

sin 2 3 sin 41 3 4 6.496

rad

sec

7. Link 4 will be the input to the second fourbar (links 4, 5, 6, and 1). Let the angle that link 4 makes in the secondfourbar be

42 41 157 deg 360 deg 42 34.915deg


K1h

e K2

h

g K1 0.5500 K2 1.0195

K3e

2f

2 g

2 h

2

2 eg K3 0.7263


B 2 sin 42

C K1 K2 1 cos 42 K3

A 0.1603 B 1.1447 C 0.3796


6 2 atan2 2 A B B2

4 A C 6 35.228 deg


K4h

f K5

g2

h2

e2

f2

2 e f K4 0.6112 K5 1.0119

D cos 42 K1 K4 cos 42 K5 D 0.2408

E 2 sin 42 E 1.1447

F K1 K4 1 cos 42 K5 F 0.7807


5 2 atan2 2 D E E2

4 D F 5 280.938 deg

12. Determine the angular velocity of link6 for the open circuit using equations 6.18.

6e 4

g

sin 42 5 sin 6 5

6 15.885rad

sec


PROBLEM 6-77

Statement: Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver tocalculate and plot the angular velocity of link 6 in the sixbar linkage of Figure 3-35 as a function of2 for a constant2 = 1 rad/sec CCW.




Link 4 (O4 to D) e 1.429 in Link 5 (D to E) f 1.286 in

Link 6 (O6 to E) g 0.771 in Link 1 (O4 to O6) h 0.786 in


CCW


1. Define the range of the input angle: 0 deg 1 deg 360 deg


K1d

a K2

d

c K1 3.8570 K2 2.9992

K3a

2b

2 c

2 d

2

2 ac K3 1.2015


C K1 K2 1 cos K3


2 atan2 2 A B B 2 4 A C


K4d

b K5

c2

d2

a2

b2

2 a b K4 1.0150 K5 3.7714


F K1 K4 1 cos K5


2 atan2 2 D E E 2 4 D F


a

b

sin sin

a

c

sin sin

7. Link 4 will be the input to the second fourbar (links 4, 5, 6, and 1). Let the angle that link 4 makes in the secondfourbar be


157 deg 360 deg


K1h

e K2

h

g K1 0.5500 K2 1.0195

K3e

2f

2 g

2 h

2

2 eg K3 0.7263

A' cos K1 K2 cos K3 B' 2 sin

C' K1 K2 1 cos K3


2 atan2 2 A' B' B' 2 4 A' C'


K4h

f K5

g2

h2

e2

f2

2 e f K4 0.6112 K5 1.0119

D' cos K1 K4 cos K5 E' 2 sin

F' K1 K4 1 cos K5


2 atan2 2 D' E' E' 2 4 D' F'

12. Determine the angular velocity of link6 for the open circuit using equations 6.18.

e

g

sin sin

0 45 90 135 180 225 270 315 3602

1

0

1

2

3ANGULAR VELOCITY - LINK 6

deg


PROBLEM 6-78

Statement: Figure 3-36 shows an eightbar mechanism. Find all of its instant centers in the position shown inpart (a) of the figure:


Solution: See Figure 3-36a and Mathcad file P0678.


Cn n 1( )

2 C 28


7,8

5,7

1,4; 1,8; 4,8

O2 O6O 4

1,2

34

6

82

2,3

3,4 4,5

1,65

7

5,6


I1,3: I1,2-I2,3 and I1,4-I3,4

7,8

5,72,6; 2,8; 6,8; 2,7 at infinity

4,71

2

3

45

6

7

8

4,5; 3,5

1,4; 1,8; 4,8; 5,8; 1,5

O6O 4

34

6

82

2,3

1,6O 2

1,2; 2,4

2,5

1,7

3,4; 1,3

5

7

5,6; 4,6

6,73,6; 3,7; 3,8

I4,6: I1,4-I1,6 and I4,5-I5,6

I2,4: I1,2-I1,4 and I2,3-I3,4

I2,8: I2,4-I4,8 and I1,2-I1,8

I2,6: I1,2-I1,6 and I2,4-I4,6

I6,8: I2,8-I2,6 and I1,8-I1,6

I6,7: I5,6-I5,7 and I6,8-I7,8

I2,5: I2,6-I5,6 and I2,3-I4,5

I2,7: I2,8-I7,8 and I2,6-I6,7

I3,5: I3,4-I4,5 and I2,3-I2,5

I3,6: I2,3-I2,6 and I3,5-I5,6





PROBLEM 6-79

Statement: Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver tocalculate and plot the angular velocity of link 8 in the linkage of Figure 3-36 as a function of2 for a

constant 2 = 1 rad/sec CCW.


Input crank (L2) a 0.450 First coupler (L3) b 0.990

Common rocker (O4B) c 0.590 First ground link (O2O4) d 1.000

Common rocker (O4C) a' 0.590 Second coupler (CD) b' 0.325

Output rocker (L6) c' 0.325 Second ground link (O4O6) d' 0.419

Link 7 (L7) e 0.938 Link 8 (L8) f 0.572

Link 5 extension (DE) p 0.823 Angle DCE 7.0 deg

Angle BO4C 128.6 deg


CCW


1. See problem 4-43 for the position solution. The velocity solution will use the same vector loop equations forlinks 5, 6, 7, and 8, differentiated with respect to time. This problem is suitable for a project assignment and isprobably too long for an overnight assignment.


PROBLEM 6-80

Statement: Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver tocalculate and plot the magnitude and direction of the velocity of point P in Figure 3-37a as afunction of2 for a constant2 = 1 rad/sec CCW. Also calculate and plot the velocity of point P

versus point A.








atany

x

return x 0if

atany

x

otherwise



0 deg 2 deg 360 deg


K1d

a K2

d

c

K1 10.3971 K2 1.4140

K3a

2b

2 c

2 d

2

2 ac K3 7.4187

A cos K1 K2 cos K3

B 2 sin

C K1 K2 1 cos K3


2 atan2 2 A B B 2 4 A C


K4d

b K5

c2

d2

a2

b2

2 a b K4 1.4140 K5 7.4187

D cos K1 K4 cos K5

E 2 sin

F K1 K4 1 cos K5



2 atan2 2 D E E 2 4 D F


a

b

sin sin

a

c

sin sin




VPA Rpa sin j cos

VP VA VPA

9. Plot the magnitude and direction of the velocity at coupler point P.

Magnitude: VP VP

Direction: VP1 arg VP

VP if VP1 0 VP1 2 VP1

0 45 90 135 180 225 270 315 3600.1

0.12

0.14

0.16

0.18MAGNITUDE

Crank Angle, deg

Velo

city

,mm

/sec

VP VA

deg


0 45 90 135 180 225 270 315 3600

100

200

300

DIRECTION

Crank Angle, deg

Vect

orAn

gle,

deg

VP deg

deg


PROBLEM 6-81

Statement: Calculate the percent error of the deviation from constant velocity magnitude of point P in Figure3-37a.








atany

x

return x 0if

atany

x

otherwise



0 deg 2 deg 360 deg


K1d

a K2

d

c

K1 10.3971 K2 1.4140

K3a

2b

2 c

2 d

2

2 ac K3 7.4187

A cos K1 K2 cos K3

B 2 sin

C K1 K2 1 cos K3


2 atan2 2 A B B 2 4 A C


K4d

b K5

c2

d2

a2

b2

2 a b K4 1.4140 K5 7.4187

D cos K1 K4 cos K5

E 2 sin

F K1 K4 1 cos K5



2 atan2 2 D E E 2 4 D F


a

b

sin sin

a

c

sin sin




VPA Rpa sin j cos

VP VA VPA

9. Calculate the magnitude of the velocity at coupler point P.

Magnitude: VP VP

Deviation from constant speed: err VP VA

VA

0 45 90 135 180 225 270 315 36020

10

0

10

20

30DEVIATION FROM CONSTANT SPEED

Crank Angle, deg

Perc

entE

rror

err %

deg


PROBLEM 6-82

Statement: Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver tocalculate and plot the magnitude and the direction of the velocity of point P in Figure 3-37b as afunction of2. Also calculate and plot the velocity of point P versus point A.


Input crank (L2) a 0.50 First coupler (AB) b 1.00

Rocker 4 (O4B) c 1.00 Rocker 5 (L5) c' 1.00

Ground link (O2O4) d 0.75 Second coupler 6 (CD) b' 1.00

Coupler point (DP) p 1.00 Distance to OP (O2OP) d' 1.50



atany

x

return x 0if

atany

x

otherwise

Crank speed: 1rad

sec

Solution: See Figure 3-37b and Mathcad file P0682.

1. Links 4, 5, BC, and CD form a parallelogram whose opposite sides remain parallel throughout the motion of thefourbar 1, 2, AB, 4. Define a position vector whose tail is at point D and whose tip is at point P and anotherwhose tail is at O4 and whose tip is at point D. Then, since R5 = RAB and RDP = -R4, the position vector from O2to P is P = R1 + RAB - R4. Separating this vector equation into real and imaginary parts gives the equations forthe X and Y coordinates of the coupler point P.

XP d b cos c cos = YP b sin c sin =

2. Define one revolution of the input crank: 0 deg 0.5 deg 360 deg

3. Use equations 4.8a and 4.10 to calculate4 as a function of 2 (for the crossed circuit).

K1d

a K2

d

c K3

a2

b2

c2

d2

2 ac

K1 1.5000 K2 0.7500 K3 0.8125

A cos K1 K2 cos K3


2 atan2 2 A B B 2 4 A C

2


K4d

b K5

c2

d2

a2

b2

2 a b K4 0.7500 K5 0.8125


F K1 K4 1 cos K5



2 atan2 2 D E E 2 4 D F

2

6. Define a local xy coordinate system with origin at OP and with the positive x axis to the right. The coordinates ofP are transformed to xP = XP - d ', yP = YP.

xP d b cos c cos d'

yP b sin c sin

7. Use equations 6.18 to calculate3 and 4.

a

b

sin sin

a

c

sin sin

8. Differentiate the position equations with respect to time to get the velocity components.

VPx b sin c sin

VPy b cos c cos

VP VPx 2 VPy 2

0 45 90 135 180 225 270 315 3600.6

0.4

0.2

0

0.2

0.4

0.6MAGNITUDE

Crank Angle, deg

Velo

city

,mm

/sec

VPx VPy

deg


PROBLEM 6-83

Statement: Find all instant centers of the linkage in Figure P6-30 in the position shown.




Cn n 1( )

2 C 6


1,2

O

P

P2

14

Y

41,4

3,4

O2

2

2,3

X

3

1,3

1,2

O

P

P2

14

Y

41,4

3,4

O2

2,4

2

2,3

X

3




PROBLEM 6-84a

Statement: Find the angular velocities of links 3 and 4 and the linear velocity of points A, B and P1 in the XY

coordinate system for the linkage in Figure P6-30 in the position shown. Assume that2 = 45 deg

in the XY coordinate system and 2 = 10 rad/sec. The coordinates of the point P1 on link 4 are(114.68, 33.19) with respect to the xy coordinate system. Use a graphical method.



Link 4 (O4 to B) c 51.26 in

Link 1 X-offset dX 47.5 in Link 1 Y-offset dY 76.00 in 12.00 in

Coupler point x-offset px 114.68 in Coupler point y-offset py 33.19 in

Crank angle: 45 deg


CCW

Solution: See Figure P6-30 and Mathcad file P0684a.1. Draw the linkage to a convenient scale. Indicate the directions of the velocity vectors of interest.

B

Direction of VB

y

O

P

P2

1

x

4

29.063°

Y

4

45.000°

Direction of VA

99.055°

O2

2

A

X

Direction of VBA

3


VA a VA 140.000in

sec VA 45 deg 90 deg


VB = VA + VBA



119.063°

135.000°

0.409"

1.206"

9.055°

VA

VBA

0 100 in/sec

VB

X

Y



1

in

VB 1.206 in kv VB 120.600in

sec VB 119.063 deg


sec


VBA

b 0.511

rad

sec

VB

c 2.353

rad

sec

6. Transform the xy coordinates of point P1 into the XY system using equations 4.0b.

Coordinate transformation angle atan2 dX d Y 126.582deg

PX px cos p y sin PX 94.998 in

PY px sin py cos PY 72.308 in

7. Calculate the distance from O4 to P1 and use equation 6.7 to calculate the velocity at P1.

Distance from O4 to P1: e PX dX 2 PY dY 2 e 48.219 in

VP1 e VP1 113.446in

sec


PROBLEM 6-84b



in the XY coordinate system and 2 = 10 rad/sec. The coordinates of the point P1 on link 4 are(114.68, 33.19) with respect to the xy coordinate system. Use the method of instant centers.






Crank angle: 45 deg


CCW


1. Draw the linkage to scale in the position given, findinstant center I1,3 and the distance from the pin jointsto the instant center.

1,3

O

P

P2

14

Y

4

B

O2

2

A

X

3


AI13 273.768 in BI13 235.874 in


VA a VA 140.000in

sec




VA

AI13 0.511

rad

sec CW


VB BI13 VB 120.622in

sec


VB

c 2.353

rad

sec CCW

6. Transform the xy coordinates of point P1 into the XY system using equations 4.0b.

Coordinate transformation angle atan2 dX d Y 126.582deg

PX px cos p y sin PX 94.998 in

PY px sin py cos PY 72.308 in

7. Calculate the distance from O4 to P1 and use equation 6.7 to calculate the velocity at P1.

Distance from O4 to P1: e PX dX 2 PY dY 2 e 48.219 in

VP1 e VP1 113.467in

sec


PROBLEM 6-84c



in the XY coordinate system and 2 = 10 rad/sec. The coordinates of the point P1 on link 4 are(114.68, 33.19) with respect to the xy coordinate system. Use an analytical method.






Crank angle: 2XY 45 deg XY coord system

Coordinate transformation angle: 126.582 deg


CCW



B

81.582°

126.582°

y

O

97.519°

P

P2

1

x

4

4

Y

27.528°

45.000°

O2

2

A

X

3

Transform the crank angle from global to local coordinate system:

2 2XY 2 81.582 deg

Distance O2O4: d d X2

dY2

d 79.701 in


K1d

a K2

d

c K1 5.6929 K2 1.5548

K3a

2b

2 c

2 d

2

2 ac K3 1.9340

A cos 2 K1 K2 cos 2 K3 B 2 sin 2

C K1 K2 1 cos 2 K3

A 3.8401 B 1.9785 C 7.2529



4 2 atan2 2 A B B2

4 A C 4 262.482 deg


K4d

b K5

c2

d2

a2

b2

2 a b K4 0.9963 K5 4.6074

D cos 2 K1 K4 cos 2 K5 D 10.0081

E 2 sin 2 E 1.9785

F K1 K4 1 cos 2 K5 F 1.0849


3 2 atan2 2 D E E2

4 D F 3 332.475 deg


3a 2

b

sin 4 2 sin 3 4 3 0.511

rad

sec

4a 2

c

sin 2 3 sin 4 3 4 2.353

rad

sec


VA a 2 sin 2 j cos 2

VA 138.492 20.495j( )in

sec VA 140.000

in

sec

VAXY arg VA VAXY 135.000 deg

VB c 4 sin 4 j cos 4

VB 119.588 15.781j( )in

sec VB 120.624

in

sec

VBXY arg VB VBXY 119.064 deg

8. Calculate the distance from O4 to P1 and the angle BO4P1.

Distance from O4 to P1: e px d 2 py2

e 48.219 in

4 180 deg atanpy

px d

4 136.503 deg

9. Determine the velocity of point P1 using equations 6.35.

VP1 e 4 sin 4 j cos 4

VP1 55.122 99.181j( )in

sec VP1 113.469

in

sec

VPXY arg VP1 VPXY 245.646 deg


PROBLEM 6-85

Statement: Write a computer program or use an equation solver such as Mathcad, Matlab, or TKSolver tocalculate and plot the absolute velocity of point P1 in Figure P6-30 as a function of 2 for2 = 10rad/sec. The coordinates of the point P1 on link 4 are (114.68, 33.19) with respect to the xycoordinate system.






Crank angle: 2XY 0 deg 1 deg 360 deg XY coord system



CCW



B

81.582°

126.582°

y

O

97.519°

P

P2

1

x

4

4

Y

27.528°

45.000°

O2

2

A

X

3

Transform the crank angle from global to local coordinate system:

2XY 2XY

Distance O2O4: d d X2

dY2

d 79.701 in


K1d

a K2

d

c K1 5.6929 K2 1.5548

K3a

2b

2 c

2 d

2

2 ac K3 1.9340

A 2XY cos 2XY K1 K2 cos 2XY K3

B 2XY 2 sin 2XY

C 2XY K1 K2 1 cos 2XY K3



2XY 2 atan2 2 A 2XY B 2XY B 2XY 2 4 A 2XY C 2XY


K4d

b K5

c2

d2

a2

b2

2 a b K4 0.9963 K5 4.6074

D 2XY cos 2XY K1 K4 cos 2XY K5

E 2XY 2 sin 2XY

F 2XY K1 K4 1 cos 2XY K5


2XY 2 atan2 2 D 2XY E 2XY E 2XY 2 4 D 2XY F 2XY

6. Determine the angular velocity of link 4 for the crossed circuit using equations 6.18.

2XY a

c

sin 2XY 2XY sin 2XY 2XY

7. Calculate the distance from O4 to P1 and the angle BO4P1.

Distance from O4 to P1: e px d 2 py2

e 48.219 in

180 deg atanpy

px d

136.503deg

8. Determine and plot the velocity of point P1 using equations 6.35.

VP1 2XY e 2XY sin 2XY j cos 2XY

VP1 2XY VP1 2XY VPXY 2XY arg VP1 2XY

0 45 90 135 180 225 270 315 3600

50

100

150

200P1 VELOCITY MAGNITUDE

Crank Angle - Global XY System

Velo

city

-in/

sec

VP1 2XY sec

in

2XY

deg


0 45 90 135 180 225 270 315 3600

50

100

150

200

250

300P1 VELOCITY DIRECTION

Crank Angle - Global XY System

Vel

ocity

angl

e-

Glo

balX

YSy

stem

VPXY 2XY deg

2XY

deg


PROBLEM 6-86

Statement: Find all instant centers of the linkage in Figure P6-31 in the position shown.




Cn n 1( )

2 C 6


4

X

2,3

Y

2

1

O

yO21,2

1,4

x

3

4P

3,4



4

X

1,3

2,4

2,3

Y

2

1

O

yO21,2

1,4

x

3

4P

3,4


PROBLEM 6-87a

Statement: Find the angular velocities of links 3 and 4 and the linear velocity of point P in the XY coordinatesystem for the linkage in Figure P6-31 in the position shown. Assume that2 = -94.121 deg in the

XY coordinate system and 2 = 1 rad/sec. The position of the coupler point P on link 3 with

respect to point A is: p = 15.00, 3 = 0 deg. Use a graphical method.





Coupler point data: p 15.00 in 0 deg



CW



P

60.472°

x

Direction VBA

Direction VB

Direction VPA

y

26.000°

Direction VA A

2

1

O2

Y

68.121°

72.224°3

44O

B

X


VA a VA 9.170in

sec



VB = VA + VBA



1.783"

Y

1.798"

V VBA B

25 in/sec0

VAX



1

in

VB 1.783 in kv VB 44.575in

sec VB 262.352 deg


sec VBA 274.103 deg


VBA

b 3.466

rad

sec

VB

c 4.658

rad

sec

6. Determine the magnitude and sense of the vector VPA using equation 6.7.

VPA p VPA 51.985in

sec

VPA VBA VPA 274.103 deg

7. Use equation 6.5 to (graphically) determine the magnitude of the velocity at point P. The equation to be solvedgraphically is

VP = VA + VPA

a. Choose a convenient velocity scale and layout the known vector VA.b. From the tip of VA, layout the (now) known vector VPA.c. Complete the vector triangle by drawing VP from the tail of VA to the tip of the VPA vector.

P

96.051°

2.059"

Y

VPA V

25 in/sec0

VAX



1

in

VP 2.059 in kv VP 51.475in

sec

P 96.051 deg


PROBLEM 6-87b



respect to point A is: p = 15.00, 3 = 0 deg. Use the method of instant centers.








CW


1. Draw the linkage to scale in the position given, find instant center I1,3 and the distance from the pin joints to theinstant center.

P

y

1,3

A

2

1

O2

Y

x

3

44O

B

X


AI13 2.647 in BI13 12.862 in PI13 14.825 in


VA a VA 9.174in

sec



VA

AI13 3.466

rad

sec CW


VB BI13 VB 44.577in

sec



VB

c 4.657

rad

sec CW

6. Determine the magnitude of the velocity at point P using equation 6.9b.

VP PI13 VP 51.381in

sec


PROBLEM 6-87c



respect to point A is: p = 15.00, 3 = 0 deg. Use an analytical method.






Crank angle: 2XY 94.121 deg Global XY coord system



CW


P

60.472°

x

y

26.000°

A

2

1

O2

Y

68.121°

72.224°3

44O

B

X


Transform crank angle into local xy coordinate system:

2 2XY 2 26.000 deg

Calculate distance O2O4: d d X2

dY2

d 7.487 in


K1d

a K2

d

c K1 0.8161 K2 0.7821

K3a

2b

2 c

2 d

2

2 ac K3 0.3622


C K1 K2 1 cos 2 K3

A 0.2581 B 0.8767 C 0.4234



4 2 atan2 2 A B B2

4 A C 2 4 60.488 deg


K4d

b K5

c2

d2

a2

b2

2 a b K4 0.5772 K5 0.9111

D cos 2 K1 K4 cos 2 K5 D 0.3096

E 2 sin 2 E 0.8767

F K1 K4 1 cos 2 K5 F 0.4749


3 2 atan2 2 D E E2

4 D F 2 3 72.236 deg


3a 2

b

sin 4 2 sin 3 4 3 3.467

rad

sec

4a 2

c

sin 2 3 sin 4 3 4 4.658

rad

sec

7. Determine the velocity of points A and B using equations 6.19.

VA a 2 sin 2 j cos 2

VA 4.022 8.246i( )in

sec VA 9.174

in

sec


VB c 4 sin 4 j cos 4

VB 38.807 21.967i( )in

sec VB 44.593

in

sec


8. Determine the velocity of the coupler point P for the crossed circuit using equations 6.36.

VPA p 3 sin 3 j cos 3

VPA 4.127 1.322i( )ft

s

VP VA VPA VP 3.792 2.009i( )ft

s VP 51.501

in

sec

VPXY arg VP VPXY 96.039 deg


PROBLEM 6-88

Statement: Figure P6-32 shows a fourbar double slider known as an elliptical trammel. F ind all its instantcenters in the position shown.




Cn n 1( )

2 C 6


1

3,4

2

To 1,2 at infinity

2,3

Y

3

4

X

To 1,4 at infinity



1

3,4

2

To 2,4 at infinity

To 1,2 at infinity

2,3

Y

1,33

4

X

To 1,4 at infinity


PROBLEM 6-89

Statement: The elliptical trammel in Figure P6-32 must be driven by rotating link 3 in a full circle. Points on lineAB describe ellipses. Find and draw (mannually or with a computer) the fixed and movingcentrodes of instant center I13. (Hint: These are called the Cardan circles.)


1. The instant center I13 is shown as the point P in the diagram below. The length of link 3, c, is constant and it is adiagonal of the rectangle OBPA. Therefore, the other diagonal, OP, has a constant length, c, regardless of thecurrent lengths a and b . Thus, the point P (I13) travels on a circle of radius c. This circle is the fixed centrode.

2

a

A1

c

4

3BP

Ob

2. Now, invert the mechanism by holding link 3 fixed and allowing the slots to move, which can only be in a circleabout O. The locus of points P will then be a circle of radius 0.5c with center at O'. This is the moving centrode.

2

a

A1

c

4

3BP

OO' b


PROBLEM 6-90

Statement: Derive analytical expressions for the velocities of points A and B in Figure P6-32 as a function of3, 3 and the length AB of link 3. Use a vector loop equation.


1. Establish the global XY system such that the coordinates of the intersection of the slot centerlines is at (dX,dY).Then, define position vectors R1X, R1Y, R2, R3, and R4 as shown below.

1

2 R4

R

1XR

1Y

A

Y

R2

R3

B3C

4

X

3

2. Write the vector loop equation: R1Y + R2 + R3 - R1X - R4 = 0 then substitute the complex number notation foreach position vector. The equation then becomes:

dY ej

2 a e

j 0( ) c e

j 3 dX e

j 0( ) b e

j

2 0=

3. Differentiate this equation with respect to time.

tad

dj c e

j

tb e

j

2

d

d 0=

4. Substituting the Euler identity into this equation gives:

Va jc cos 3 j sin 3 Vb j 0=

5. Separate this equation into its real (x component) and imaginary (y component) parts, setting each equal to zero.

Va c sin 3 0= c cos 3 Vb 0=

6. Solve for the two unknowns Va and Vb in terms of the independent variables 3 and3

Va c sin 3 = Vb c cos 3 =


PROBLEM 6-91

Statement: The linkage in Figure P6-33a has link 2 at 120 deg in the global XY coordinate system. Find 6 and

VD in the global XY coordinate system for the position shown if 2 = 10 rad/sec CCW. Use thevelocity difference graphical method.



Link 4 (O4 to B) c 3.00 in Link 5 (C to D) e 5.60 in

Link 3 (A to C) p 2.25 in Link 4 (O4 to D) f 3.382 in


Angle ACB 3 0.0 deg Angle BO4D 4 110.0 deg

Input rocker angle: 120 deg Global XY system


CW



Direction of VBA

113.057°

11.709°

110°

Direction of VDC

Direction of VB

4

4OB

87.972°

Axis of slip


D

Direction of VA

Direction of VCA

5 C

3

A

120°

XO2

2

Y


VA a VA 62.000in

sec VA 120 deg 90 deg


VB = VA + VBA



VBA

BV

1.433

1.206

78.291°

AV

0 50 in/sec

Y

X



1

in

VB 1.206 in kv VB 60.300in

sec VB 78.291 deg


sec VBA 23.057 deg


VBA

b 15.922

rad

sec CCW

VB

c 20.100

rad

sec CW


VCA p VCA 35.825in

sec

VCA VBA 3 VCA 23.057 deg

7. Determine the magnitude and sense of the vector Vtrans using equation 6.7.

Vtrans f Vtrans 67.978in

sec

trans VB 4 trans 31.709 deg


VC = VA + VCA

a. Choose a convenient velocity scale and layout the known vector VA.b. From the tip of VA, layout the (now) known vector VCA.c. Complete the vector triangle by drawing VC from the tail of VA to the tip of the VCA vector.


VCA

CV0.991

114.720°

AV

0 50 in/sec

Y

X



1

in

VC 0.991 in kv VC 49.550in

sec VC 114.720 deg

9. Use equation 6.5 to (graphically) determine the magnitude of the velocity at point D, the magnitude of therelative velocity VDC. The equations to be solved (simultaneously) graphically are

VD = VC + VDC and VD = Vtrans + Vslip

a. Choose a convenient velocity scale and layout the known vector VC.b. From the tip of VC, draw a construction line with the direction of VDC, magnitude unknown.c. Repeat steps a and b with Vtrans and Vslip.c. From the tail of VC, draw a construction line to the intersection of the two construction lines.d. Complete the vector polygon by drawing VDC from the tip of VC to the intersection of the VD constructionline and drawing VD from the tail of VC to the intersection of the VDC construction line.

95.189°3.045

CV0

DCV Vslip

DV

Y

50 in/sec

V trans

X

10. From the velocity polygon we have:


kv50 in sec

1

in

VD 3.045 in kv VD 152.250in

sec

VD 95.189 deg

20.100rad

sec


PROBLEM 6-92


VD in the global XY coordinate system for the position shown if 2 = 10 rad/sec CCW. Use theinstant center graphical method.







Input rocker angle: 120 deg Global XY system


CW


1. Draw the linkage to scale in the position given, find instant centers I1,3 and I1,5, and the distances from the pinjoints to the instant centers.

3,5

4

4O 1,4

3,4

6

To 4,6 at infinity

D 55,6

B

C

31,5

2,3

XO2

1,2

3,62

1,3

1,6 Y

A


AI13 3.893 in BI13 3.788 in CI13 3.113 in CI15 1.411 in DI15 4.333 in DI16 7.573 in


VA a VA 62.000in

sec



VA

AI13 15.926

rad

sec CCW


VB BI13 VB 60.328in

sec


5. Use equation 6.9c to determine the angular velocity of links 4 and 6.

VB

c 20.109

rad

sec CW

w6 w6 20.109rad

sec CW


VC CI13 VC 49.578in

sec


VC

CI15 35.137

rad

sec CW

8. Determine the magnitude of the velocity at point D using equation 6.9b.

VD DI15 VD 152.247in

sec @ 95.185 deg


PROBLEM 6-93


VD in the global XY coordinate system for the position shown if 2 = 10 rad/sec CCW. Use ananalytical method.







Input rocker angle: 2XY 120 deg Global XY system



CCW



atany

x

return x 0if

atany

x

otherwise



120°

175.455°

O2

y

4

110°

O

x

D

4

5

6

55.455°B 16.254°

C

32

A

71.487° Y

X

Calculate the distance O2O4: d d X2

dY2

d 7.825 in

Transform2XY into the local coordinate system: 2 2XY 2 55.455 deg


K1d

a K2

d

c K1 1.2620 K2 2.6082


K3a

2b

2 c

2 d

2

2 ac K3 2.3767


C K1 K2 1 cos 2 K3

A 0.2028 B 1.6474 C 1.5927


41 2 atan2 2 A B B2

4 A C 41 163.746 deg


K4d

b K5

c2

d2

a2

b2

2 a b K4 1.7388 K5 1.9877

D cos 2 K1 K4 cos 2 K5 D 1.6967

E 2 sin 2 E 1.6474

F K1 K4 1 cos 2 K5 F 0.3067


3 2 atan2 2 D E E2

4 D F 2 3 71.488 deg

6. At this point write vector loop equations for links 3, 4, 5, and 6 to solve for the position and velocty of link 6.


PROBLEM 6-94

Statement: The linkage in Figure P6-33b has link 3 perpendicular to the X-axis and links 2 and 4 are parallel toeach other. Find 3, VA, VB, and VP if 2 = 15 rad/sec CW. Use the velocity difference graphicalmethod.



Link 4 (O4 to B) c 2.75 in Link 1 (O4 to B) d 4.43 in



CW



36.351°

A

Direction of VBA

Direction of VB

Direction of VA

O2

B

O4

36.352 deg


VA a VA 41.250in

sec



VB = VA + VBA


4. Without actually drawing the vector polygon, we see that VA and VB have the same angle with respect to the Xaxis. Therefore,


VB VA and VBA 0in

sec VB VA VBA 0 deg


VBA

b 0.000

rad

sec

VB

c 15.000

rad

sec

6. Determine the magnitude and sense of the vector VPA using equation 6.7.

VPA p VPA 0.000in

sec

VPA VBA VPA 0.000deg

7. Use equation 6.5 to (graphically) determine the magnitude of the velocity at point P. The equation to be solvedgraphically is

VP = VA + VPA

a. Choose a convenient velocity scale and layout the known vector VA.b. From the tip of VA, layout the (now) known vector VPA.c. Complete the vector triangle by drawing VP from the tail of VA to the tip of the VPA vector.

8. Again, without drawing the vector polygon we see that,

VP VA

At this instant, link 3 is in pure translation with every point on it having the same velocity.


PROBLEM 6-95

Statement: The linkage in Figure P6-33b has link 3 perpendicular to the X-axis and links 2 and 4 are parallel toeach other. Find 3, VA, VB, and VP if 2 = 15 rad/sec CW. Use the instant center graphicalmethod.






CW

Solution: See Figure P6-33b and Mathcad file P0695.1. Draw the linkage to scale in the position given, find instant center I1,3 and the distance from the pin joints to the

instant center.

36.351°

2

A

2 3

Y

OP

B

XO4

4

Since O2A and O4B are parallel, I1,3 is at infinity and link 3 does not rotate for this instantaneous position of thelinkage. Thus, link 3 is in pure translation and all points on it will have the same velocity.

2. Use equation 6.7 and inspection of the layout to determine the magnitude and direction of the velocity at pointsA, B, and P, which will be the same.

VA a VA 41.250in

sec

36.351 deg


The velocity vectors for points B and P are identical to VA.


PROBLEM 6-96

Statement: The linkage in Figure P6-33b has link 3 perpendicular to the X-axis and links 2 and 4 are parallel toeach other. Find 3, VA, VB, and VP if 2 = 15 rad/sec CW. Use an analytical method.





Input crank angular velocity w2 15 rad sec1

CW


1. Draw the linkage to a convenient scale and label it.

36.351°

A

2OP

O4

B

36.351 deg


K1d

a K2

d

c K1 1.6109 K2 1.6109

K3a

2b

2 c

2 d

2

2 ac K3 1.5949

A cos K1 K2 cos K3 A 0.5081

B 2 sin B 1.1855

C K1 K2 1 cos K3 C 1.1029


2 atan2 2 A B B2

4 A C 2 143.649 deg


K4d

b K5

c2

d2

a2

b2

2 a b K4 1.3589 K5 1.6873


E 2 sin E 1.1855

F K1 K4 1 cos K5 F 0.2127



2 atan2 2 D E E2

4 D F 2 89.995 deg


a w2

b

sin sin 0.000

rad

sec

a w2

c

sin sin 15.000

rad

sec

7. Determine the velocity of points A and B using equations 6.19.

VA a w2 sin j cos

VA 24.450 33.223i( )in

sec VA 41.250

in

sec


VB c sin j cos

VB 24.450 33.223i( )in

sec VB 41.250

in

sec



VPA p sin j cos

VPA 1.936 104

1.676i 108

in

sec

VP VA VPA VP 41.250in

sec

VPXY arg VP VPXY 126.351 deg


PROBLEM 6-97

Statement: The crosshead linkage shown in Figure P6-33c has 2 DOF with inputs at crossheads 2 and 5. Findinstant centers I1,3 and I1,4.




Cn n 1( )

2 C 10


1

2

31,2 at infinity

2,3

2,5 at infinity

4,5

1,5 at infinity

3,4

45

2. Use Kennedy's Rule and a linear graph to find the remaining ICs required to find I1,3 and I1,4.

1

2

3

2,3; 2,4

1,2 at infinity

2,5 at infinity

4,5; 3,5

1,5 at infinity

3,4

45

1,3; 1,4

I2,4: I1,2-I1,4 and I2,3-I3,4

I3,5: I2,3-I2,5 and I4,5-I3,4

I1,4: I1,2-I2,4 and I1,5-I4,5

I1,3: I1,2-I2,3 and I1,4-I3,4


PROBLEM 6-98

Statement: The crosshead linkage shown in Figure P6-33c has 2 DOF with inputs at cross heads 2 and 5. FindVB, VP3, and VP4 if the crossheads are each moving toward the origin of the XY coordinate systemwith a speed of 20 in/sec. Use a graphical method.


Link 3 (A to B) b 34.32 in Link 4 (B to C) c 50.4 in

Link 2 Y-offset a 59.5 in Link 5 X-offset d 57 in

Coupler point data: AP3 31.5 in BP3 22.2 in

BP4 41.52 in CP4 27.0 in

Input velocities: V2Y 20 in sec1

V5X 20 in sec1



2

3

Direction of VC

Direction of VA

A

Y

XC

Direction of VP3A

Direction of VBCDirection of VBA

Direction of VP4C

P4 4

5

B

P3

2. Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relativevelocities VBA, VBC and the angular velocities of links 3 and 4. The equations to be solved (simultaneously)graphically are

VB = VA + VBA VC = VB + VCB

a. Choose a convenient velocity scale and layout the known vectors VA and VCb. From the tip of VA, draw a construction line with the direction of VBA, magnitude unknown.c. From the tip of VC, draw a construction line with the direction of VCB, magnitude unknown.d. From the origin, draw a construction line to the intersection of the two construction lines in steps b and c.d. Complete the vector triangle by drawing VBA from the tip of VA to the intersection of the VB constructionline and drawing VB from the tail of VA to the intersection of the VBA construction line. And then do the samewith the VCB vector

(See next page)


6.004

5.932

4. 385

32.718°

VC

AV

0 10 in/sec

Y

60.123°46. 990°

X

BAV

VBC

VB

3. From the velocity triangles we have:


1

in

VB 4.385 in kv VB 43.850in

sec VB 46.990 deg


sec VBA 60.123 deg

VBC 5.932 in kv VBC 59.320in

sec VBC 32.718 deg


VBA

b 1.749

rad

sec CCW

VBC

c 1.177

rad

sec CW

6. Determine the magnitudes of the vectors VP3A and VP4C using equation 6.7.

VP3A AP3 VP3A 55.107in

sec


VP4C CP4 VP4C 47.234in

sec

7. Use equation 6.5 to (graphically) determine the magnitude of the velocities at points P3 and P4. The equations tobe solved graphically are

VP3 = VA + VP3A VP4 = VC + VP4C

a. Choose a convenient velocity scale and layout the known vector VA.b. From the tip of VA, layout the (now) known vector VP3A.c. Complete the vector triangle by drawing VP3 from the tail of VA to the tip of the VP3A vector.d. Repeat for VP4.

104.444°

164.011°

3.537

6.598

VC

AV

VP4C

VP4

0 10 in/sec

Y

VP3AVP3

X

8. From the velocity triangles we have:


1

in

VP3 3.537 in kv VP3 35.370in

sec P3 104.444 deg

VP4 6.598 in kv VP4 65.980in

sec P4 164.011 deg


PROBLEM 6-99

Statement: The linkage in Figure P6-33d has the path of slider 6 perpendicular to the global X-axis and link 2aligned with the global X-axis. Find VA in the position shown if the velocity of the slider is 20in/sec downward. Use the velocity difference graphical method.


Link 2 (O2 to A) a 12 in Link 3 (A to B) b 24 in

Link 4 (O4 to B) c 18 in Link 5 (C to D) f 24 in

Link 4 (O4 to C) e 18 in

Link 1 X-offset dX 19 in Link 1 Y-offset dY 28 in

Angle BO4C 4 0.0 deg

Output crank angle: 0 deg Global XY system

Input slider velocity VD 20 in sec1

VD 90.0 deg



B

AX

Y

4

2

6

D

O2

C

Direction of VD

Direction of VCDDirection of VC

Direction of VB

Direction of VA

Direction of VAB

114.410°

19.963°

O4

116.161°

5

3


VC = VD + VCD

a. Choose a convenient velocity scale and layout the known vector VD.b. From the tip of VD, draw a construction line with the direction of VCD, magnitude unknown.c. From the tail of VD, draw a construction line with the direction of VC, magnitude unknown.d. Complete the vector triangle by drawing VCD from the tip of VD to the intersection of the VC constructionline and drawing VC from the tail of VD to the intersection of the VCD construction line.


V

CD

70.037°

1.806

V

0

VD

10 in/sec

Y

C

X



1

in

VC 1.806 in kv VC 18.060in

sec VC 70.037 deg

4. Determine the magnitude and sense of the vector VB .

VB VC VB 18.060in

sec

VB VC 180 deg VB 109.963 deg

5. Use equation 6.5 to (graphically) determine the magnitude of the velocity at point A. The equation to be solvedgraphically is

VA = VB + VAB


1.977

VB

Y

V

V

A

AB

X

0 10 in/sec



kv10 in sec

1

in

VA 1.977 in kv

VA 19.770in

sec

VA 90.0 deg


PROBLEM 6-100

Statement: The linkage in Figure P6-33a has the path of slider 6 perpendicular to the global X-axis and link 2aligned with the global X-axis. Find VA in the position shown if the velocity of the slider is 20in/sec downward. Use the instant center graphical method.


Link 2 (O2 to A) a 12 in Link 3 (A to B) b 24 in

Link 4 (O4 to B) c 18 in Link 5 (C to D) f 24 in

Link 3 (A to C) p 2.25 in Link 4 (O4 to C) e 18 in

Link 1 X-offset dX 19 in Link 1 Y-offset dY 28 in

Angle BO4C 4 0.0 deg

Output crank angle: 0 deg Global XY system

Input slider velocity VD 20 in sec1

VD 90.0 deg


1. Draw the linkage to scale in the position given, find instant centers I1,3 and I1,5, and the distances from the pinjoints to the instant centers.

C

B

A

O44

2

5

6

O21,3

D1,5

3


AI13 70.085 in BI13 64.013 in CI15 63.095 in DI15 69.886 in


VD

DI15 0.286

rad

sec CW


VC CI15 VC 18.057in

sec


VC

e 1.003

rad

sec CW


VB c VB 18.057in

sec



VB

BI13 0.282

rad

sec CCW

7. Determine the magnitude of the velocity at point A using equation 6.9b.

VA AI13 VA 19.769in

sec Upward


PROBLEM 6-101

Statement: For the linkage of Figure P6-33e, write a computer program or use an equation solver to find andplot VD in the global coordinate system for one revolution of link 2 if2 = 10 rad/sec CW.


Input crank (L2) a 5.00 Fourbar coupler (L3) b 5.00

Output crank (O4B) c 6.00 Slider coupler (L5) e 15.00

Fourbar ground link (L1) d 2.500 Angle BO4C 83.621 deg



atany

x

return x 0if

atany

x

otherwise

Crank velocity: 10rad

sec

Solution: See Figure P6-33e and Mathcad file P06101.

1. This sixbar drag-link mechanism can be analyzed as a fourbar Grashof double crank in series with a slider-crankmechanism using the output of the fourbar, link 4, as the input to the slider-crank.


3. Use equations 4.8a and 4.10 to calculate4 as a function of 2 (for the open circuit) in the global XY coordinatesystem.

K1d

a K2

d

c K3

a2

b2

c2

d2

2 ac

K1 0.5000 K2 0.4167 K3 0.7042

A cos K1 K2 cos K3


2 atan2 2 A B B 2 4 A C

4. If the calculated value of4 is greater than 2, subtract 2from it and if it is negative, make it positive.

if 2 2

if 0 2

5. Determine the slider-crank motion using equations 4.16 and 4.17 with 4 as the input angle.

asinc sin

e

f c cos e cos


K4d

b K5

c2

d2

a2

b2

2 a b


K4 0.5000 K5 0.4050

D cos K1 K4 cos K5

E 2 sin F K1 K4 1 cos K5


2 atan2 2 D E E 2 4 D F


a

c

sin sin


c

e

cos cos


VD c sin e sin

0 45 90 135 180 225 270 315 360100

75

50

25

0

25

50

VD

deg


PROBLEM 6-102

Statement: For the linkage of Figure P6-33f, locate and identify all instant centers.



6

7

8 7,8

5,6

To 1,8 at infinity

3,5

O41,4

4

3,4

5

3

5,7

O21,2

2,3

2

O6

1,6


Cn n 1( )

2 C 28

1. Draw the linkage to scale and identify those ICsthat can be found by inspection (10).

2. Use Kennedy's Rule and a linear graph to findthe remaining 18 ICs.

I1,3: I1,2-I2,3 and I1,4-I3,4

I1,5: I1,6-I5,6 and I1,3-I3,5

I2,4: I1,2-I1,4 and I2,3-I3,4

I4,5: I1,4-I1,5 and I3,5-I3,4

I2,5: I1,2-I1,5 and I2,4-I4,5

I2,6: I1,2-I1,6 and I2,5-I5,6

I3,6: I2,3-I2,6 and I1,2-I1,6

6

7

8 7,8

5,6

To 1,8 at infinity

1,5

2,4

4,5

5,8

1,3; 3,8

6,7

4,6

4,7

1,7

3,5

4,8

O4

1,4 4 3,4

5

3

5,7

2,8

To 2,6 and 3,6

O21,2

6,8

2,73,7

2,5

2,3

2

O6

1,6

I5,8: I5,7-I7,8 and I1,5-I1,8

I4,8: I4,5-I5,8 and I1,8-I1,4

I2,8: I2,4-I4,8 and I1,2-I1,8

I3,8: I3,4-I4,8 and I1,3-I1,8

I6,8: I5,8-I5,6 and I2,8-I2,6

I2,7: I2,8-I7,8 and I2,5-I5,7

I6,7: I5,6-I5,7 and I6,8-I7,8

I3,7: I3,6-I6,7 and I2,3-I2,7

I4,6: I1,4-I1,6 and I4,5-I5,6

I4,7: I4,6-I6,7 and I3,4-I3,7

I1,7: I1,8-I7,8 and I1,6-I6,7


PROBLEM 6-103

Statement: The linkage of Figure P6-33f has link 2 at 130 deg in the global XY coordinate system. Find VD inthe global coordinate system for the position shown if 2 = 15 rad/sec CW. Use any graphicalmethod.



Link 4 (O4 to B) c 2.5 in Link 1 (O2 to O4) d1 12.5 in

Link 5 (C to E) e 8.9 in Link 5 (C to D) h 5.9 in

Link 6 (O6 to E) f 3.2 in Link 7 (D to F) k 6.4 in

Link 1 (O2 to O6) d2 10.5 in Link 3 (A to C) g 2.4 in

Crank angle: 2 130 deg

Slider axis offset and angle: s 11.7 in 150 deg


CW



YF

8

A

7E

6

Direction of VF

Direction of VFD Direction of VE

Direction of VEC

Direction of VB

Direction VBA

B

O4

4

Direction of VDC

Direction of VCA 5

C

3

D

O2X

Direction of VA

2

O6

Angles measured from layout:

VB 24.351 deg VBA 79.348 deg VCA VBA

VE 64.594 deg VEC 162.461 deg VDC VEC

VF 150.00 deg VFD 198.690 deg


VA a VA 75.000in

sec



3. Use equation 6.5 to (graphically) determine the magnitude of the velocity at point B, the magnitude of the relativevelocity VBA, and the angular velocity of link 3. The equations to be solved graphically are

VB = VA + VBA and VC = VA + VCA


3.192

2.668

0 25 in/sec

Y 3.30222.053°

V

BAV

B

X

0.943

VC

VA

VCA



1

in

VB 3.192 in kv VB 79.800in

sec VB 24.351 deg


sec VBA 79.348 deg

VBA

b 9.827

rad

sec CCW

5. Calculate the magnitude and direction of VCA and determine the magnitude and velocity of VC from the velocitypolygon above.

VCA g VCA 23.586in

sec VCA 79.348 deg

Length of VCA on velocity polygon: vCAVCA

kv vCA 0.943 in

VC 2.668 in kv VC 66.700in

sec VC 22.053deg


6. Use equation 6.5 to (graphically) determine the magnitude of the velocity at point E, the magnitude of the relativevelocity VED, and the angular velocity of link 5. The equations to be solved graphically are

VE = VC + VEC and VD = VC + VDC

a. Choose a convenient velocity scale and layout the known vector VC.b. From the tip of VC, draw a construction line with the direction of VEC, magnitude unknown.c. From the tail of VC, draw a construction line with the direction of VE, magnitude unknown.d. Complete the vector triangle by drawing VEC from the tip of VC to the intersection of the VE constructionline and drawing VE from the tail of VC to the intersection of the VEC construction line.

1.716

1.900

0 25 in/sec

V

Y

E

45.934°

X

1.821

1.207

VDCVD

VEC

VC



1

in

VE 1.716 in kv VE 42.900in

sec VE 64.594 deg

VEC 1.821 in kv VEC 45.525in

sec VEC 162.461deg

VEC

e 5.115

rad

sec CCW

8. Calculate the magnitude and direction of VDE and determine the magnitude and velocity of VD from the velocitypolygon above.

VDC h VDC 30.179in

sec VDC 162.461 deg

Length of VDC on velocity polygon: vDCVDC

kv vDC 1.207 in

VD 1.900 in kv VD 47.500in

sec VD 45.934deg

9. Use equation 6.5 to (graphically) determine the magnitude of the velocity at point F, the magnitude of the relativevelocity VFD, and the angular velocity of link 5. The equation to be solved graphically is

VF = VD + VFD


a. Choose a convenient velocity scale and layout the known vector VD.b. From the tip of VD, draw a construction line with the direction of VFD, magnitude unknown.c. From the tail of VD, draw a construction line with the direction of VF, magnitude unknown.d. Complete the vector triangle by drawing VFD from the tip of VD to the intersection of the VF constructionline and drawing VF from the tail of VD to the intersection of the VFD construction line.

150.000°

F

1.158

V

VFD

Y

45.934°

X

25 in/secVD0



1

in

VF 1.158 in kv VF 28.950in

sec VF 150.000 deg


PROBLEM 6-104

Statement: For the linkage of Figure P6-34, locate and identify all instant centers.




Cn n 1( )

2 C 15


O4 O2

AB

D

C

E

2

3

4

5

6

1

3

1,4

1,2

2,33,4

4,5

5,6

3,6



5

6

4

3

2

1I4,6: I3,4-I3,6 and I4,5-I5,6 I1,3: I1,2-I2,3 and I1,4-I3,4



O4 O2

A

B

D

C

E

2

3

4

5

6

1

3

1,4

1,2

2,3

3,4

4,5

5,6

3,6

2,4

4,6

2,6

2,5

3,51,31,5

1,6


PROBLEM 6-105

Statement: For the linkage of Figure P6-34, show that I1,6 is stationary for all positions of link 2.


Input crank (L2) a 1.75 First coupler (AB) b 1.00

First rocker (O4B) c 1.75 Ground link (O2O4) d 1.00

Second input (BC) e 1.00 Second coupler (L5) f 1.75

Output rocker (L6) g 1.00 Third coupler (BE) h 1.75

Ternary link (AE) k 2.60


1. Because the linkage is symmetrical and composed of two parallelograms the analysis can be done with simpletrigonometry. We will show that the path of point E on link 6 is a circle and that the center of the circle is theinstant center I1,6.

O4 O2

AB

D

C

E

2

3

4

5

6

1

325.396°

R E

R P

R 2

x

y

a

be

c

f

g

kh

d

=

2. Calculate the fixed angle that line AB (b) makes with line AE (k) using the law of cosines. This is the angle thatthe coupler point, E, makes with link 3 in the fourbar made up of links 1, 2, 3, and 4. Because links 1, 2, 3, and 4are a parallelogram, link 4 will have the same angle as link 2 and AB will always be parallel to O2O4.

acosb

2k

2 h

2

2 bk

25.396deg

3. Define the vectors R2, RP, and RE as shown above. Then, RE = R2 + RP or,

a cos jsin k cos jsin =

Separating into real and imaginary parts, the coordinates of point E for all positions of link 2 are:

xE a cos k cos

yE a sin k sin


This pair of equations represent the parametric equation of a circle with radius a and center at

xEC k cos xEC 2.349

yEC k sin yEC 1.115


5. Plot the position of point E as a function of the angle of input link 2.

0 1 2 3 4 51

0

1

2

3

4PATH OF POINT E

1.115

yE

2.349

xE

6. From Problem 6-104, the instant center I1,6 is located at the intersection of a line through O4 that is parallel to BE(angle ) with a line through E that is parallel to O2A (angle 2). (See figure on next page). The coordinates of the

intersection of these two lines for the position shown ( 64.036 deg ) are:

Angle BAE: acosb

2h

2 k

2

2 b h

39.582 deg

Coordinates of point E: xE a cos k cos xE 3.115

yE a sin k sin yE 0.458

Line through E with angle 2:y x( ) x xE tan yE

Coordinates of point O4: xO4 d yO4 0


Line through O4 with angle :

y x( ) x xO4 tan

Solving for the intersection (point F, the instant center I1,6) of these two lines:

xFxE tan yE d tan

tan tan xF 2.349

yF xF d tan yF 1.115

These are the coordinates of the instant center I1,6 and the center of the circle through which point E passes.Thus, the instant center I1,6 is stationary and is a hinge point that link 6 rotates about with respect to link 1.

O4 O2

A

B

D

C

E

2

3

4

5

6

1

3

1,4

1,2

2,33,4

4,5

5,6

3,6

2,4

4,6

2,6

2,5

3,51,31,5

1,6

39.582°

64.036°

1.115"

2.349"

F

1.750"

design of machinery solution manual chapter 6

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