Appendix A Position Kinematic Analysis. Trigonometric Method978-3-319-31970... · 2017. 8. 27. ·...

Appendix APosition Kinematic Analysis.Trigonometric Method

Chapter 3 shows the kinematic analysis of several mechanisms by using Raven’smethod. Writing the equations that solve the problem when using this method iseasy. However, finding the solution to such equations can be a complicated task.For that reason, we introduce the trigonometric method in this appendix, which ismuch simpler to write and solve.

A.1 Position Analysis of a Four-Bar Mechanism

Consider the mechanism shown in Fig. A.1 in which O2O4, O2A, AB and O4A arethe lengths of links 1, 2, 3 and 4 respectively. On the other hand, angles h2, h3 andh4 define the angular position of links 2, 3 and 4 considering the counterclockwiserotations positive.

In order to determine angles h3 and h4, we need to find the value of distance O4A(Eq. A.1) as well as angles b (Eq. A.3), d (Eq. A.7) and / (Eq. A.5). The value ofdistance O4A can be determined in triangle DO2AO4:

O4A ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiO2O4

2 þO2A2 � 2O2O4O2A cos h2q

ðA:1Þ

The same triangle verifies (Eq. A.2):

O4A sin b ¼ O2A sin h2 ðA:2Þ

where:

b ¼ arcsin O2AO4A

sin h2

� �ðA:3Þ

Angles / and d between bars 3 and 4 and diagonal O4A respectively, can beworked out from triangle DABO4. It verifies (Eq. A.4)

© Springer International Publishing Switzerland 2016A. Simón Mata et al., Fundamentals of Machine Theory and Mechanisms,Mechanisms and Machine Science 40, DOI 10.1007/978-3-319-31970-4

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O4B2 ¼ AB2 þO4A2 � 2ABO4A cos/ ðA:4Þ

We can clear / from Eq. (A.4):

/ ¼ arccosAB2 þO4A2 � O4B2

2ABO4AðA:5Þ

In the same triangle its verified (Eq. A.6):

O4B sin d ¼ AB sin/ ðA:6Þ

Thus:

d ¼ arcsin ABO4B

sin/� �

ðA:7Þ

Once the values of b, d and / have been determined, we can obtain h3 (Eq. A.8)and h4 (Eq. A.9) in the mechanism (Fig. A.1):

h3 ¼ /� b ðA:8Þ

h4 ¼ �ðbþ dÞ ðA:9Þ

When angle h2 takes values between 180° and 360°, angle b has a negative valueand Eqs. (A.8) and (A.9) are also applicable (Fig. A.2).

2O 4O

A

B

2θ

3θ

4θ

β δφ

Fig. A.2 Open four-barmechanism with link 2 in aposition between 180° and360°

2O 4O

A

B

2θ

3θ

4θ

β

δ

φ

Fig. A.1 Parametersinvolved in the calculation ofthe link positions in a four-barmechanism by means of thetrigonometric method

368 Appendix A: Position Kinematic Analysis. Trigonometric Method

For a crossed four-bar mechanism (Fig. A.3) we will use Eqs. (A.10) and (A.11):

h3 ¼ �ð/þ bÞ ðA:10Þ

h4 ¼ d� b ðA:11Þ

Again, when angle h2 takes values between 180° and 360°, angle b has anegative value and Eqs. (A.10) and (A.11) are also applicable (Fig. A.4).

A.2 Position Analysis of a Crank-Shaft Mechanism

Figure A.5 shows a crank-shaft mechanism. xB and yB are the Cartesian coordinatesof point B with respect a system centered on point O2 with its X-axis parallel to thepiston trajectory. xB is positive while yB is negative.

2O 4O

A

B

2θ

3θ

4θ

βδ

φ

Fig. A.3 Calculation of theposition of links 3 and 4 in acrossed four-bar mechanismby means of the trigonometricmethod

2O 4O

AB

2θ

3θ

4θ

βδ

φ

Fig. A.4 Crossed four-barmechanism with link 2 in aposition between 180° and360°

2O

A

B

23

4

2θ

3θ

μ

Bx

By

Fig. A.5 Crank-shaft mecha-nism and the parametersinvolved in the position anal-ysis with the trigonometricmethod

Appendix A: Position Kinematic Analysis. Trigonometric Method 369

Position of links 3 and 4 can be worked out using Eqs. (A.12)–(A.15):

AB sin l ¼ O2A sin h2 � yB ðA:12Þ

l ¼ arcsinO2A sin h2 � yBAB

ðA:13Þ

h3 ¼ �l ðA:14Þ

The x position of point B will be given by Eq. (A.15):

xB ¼ O2A cos h2 þAB cos h3 ðA:15Þ

It can easily be verified that Eq. (A.15) works for any position of input link 2.When the trajectory of point B is above O2, the sign of yB is positive and theseequations are also applicable.

A.3 Position Analysis of a Slider Mechanism

Consider the slider mechanism in Fig. A.6, where link 3 describes a straight tra-jectory along link 4 that rotates about O4 with offset O4B.

Similarly as in previous problems, O2O4 and O2A are the lengths of links 1 and 2respectively while angles h2 and h4 define the angular position of links 2 and 4.Assuming that we know O2O4, O2A and h2, we can obtain unknown values AB(Eq. A.17) and h4 (Eq. A.20). To do so, we start by obtaining the value of O4A(Eq. A.16):

O4A ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiO2O4

2 þO2A2 � 2O2O4O2A cos h2q

ðA:16Þ

We can calculate AB as:

AB ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiO4B

2 � O4A2q

ðA:17Þ

2O

A

B

2

3

42θ

4θ

4Oβ

δ

Fig. A.6 Position analysis ofa slider-mechanism by meansof the trigonometric method


The value of angle d (Eq. A.18) is:

d ¼ arctan O2A sin h2O2O4 þO2A cos h2

ðA:18Þ

Finally, h4 can be determined after first computing the value of b (Eq. A.19):

b ¼ arctan ABO4B

ðA:19Þ

h4 ¼ dþð90� � bÞ ðA:20Þ

If the offset is opposite, point B is below the X-axis and Eq. (A.20) changes toEq. (A.21):

h4 ¼ d� ð90� � bÞ ðA:21Þ

A.4 Two Generic Bars of a Mechanism

Let us consider that we have carried out the kinematic analysis of links 2, 3 and 4 ofthe mechanism shown in Fig. A.7. We will continue the position analysis of links 5and 6 considering that the position of point C of link 3 is known.

To find the position of links 5 and 6 we have to define triangle DCDO6 first(Fig. A.8).

The length of side O6C (Eq. A.22) can be calculated by means of the x andy coordinates of points C and O6:

O6C ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðxC � xO6Þ2 þðyC � yO6Þ2

qðA:22Þ

2O 4O

A

B

C

D

6O

2

3

4

5

6

Fig. A.7 Six-bar mechanism

Appendix A: Position Kinematic Analysis. Trigonometric Method 371

And its angle c (Eq. A.23) is:

c ¼ arctan yC � yO6xC � xO6

ðA:23Þ

Angle / (Eq. A.25) can be computed by using the law of cosines (Eq. A.24):

DO62 ¼ CD2 þO6D2 � 2CDO6D cos/ ðA:24Þ

/ ¼ arccosCD2 þO6D2 � DO62CDO6D

ðA:25Þ

Finally, angle d (Eq. A.26) is determined by using the law of sines:

d ¼ arcsin CDO6D

sin/� �

ðA:26Þ

Therefore, angles h5 and h6 are (Eqs. A.27 and A.28):

h5 ¼ /� ð180� � cÞ ðA:27Þ

h6 ¼ 180� þ c� d ðA:28Þ

6O

C

γ

5θ

6θ

δ

φ

DFig. A.8 Position analysis ofbars 5 and 6 by means of thetrigonometric method


Appendix BFreudenstein’s Method to Solvethe Position Equations in a Four-BarMechanism

In Chap. 3 we developed the position analysis of a four-bar mechanism by means ofRaven’s method. In this appendix we explain Freudenstein’s method to solve theobtained equations and calculate the value of angles h3 and h4.

B.1 Position Analysis of a Four-Bar Mechanism by UsingRaven’s Method

We will apply Raven’s method to the four-bar mechanism shown in Fig. B.1.The vector loop equation (Eq. B.1) for the position analysis of the mechanism is:

r1eih1 ¼ r2eih2 þ r3eih3 þ r4eih4 ðB:1Þ

By converting this equation into its trigonometric form (Eq. B.2):

r1ðcos h1 þ i sin h1Þ ¼ r2ðcos h2 þ i sin h2Þþ r3ðcos h3 þ i sin h3Þþ r4ðcos h4 þ i sin h4Þ

ðB:2Þ

And by separating its real and imaginary parts, we obtain the system (Eq. B.3)with two unknowns (h3 and h4):

r1 cos h1 ¼ r2 cos h2 þ r3 cos h3 þ r4 cos h4r1 sin h1 ¼ r2 sin h2 þ r3 sin h3 þ r4 sin h4

)ðB:3Þ

B.2 Freudenstein’s Method

We substitute h1 ¼ 0 in Eq. (B.3) and isolate h3 (Eq. B.4):


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r1 � r2 cos h2 � r4 cos h4 ¼ r3 cos h3�r2 sin h2 � r4 sin h4 ¼ r3 sin h3

)ðB:4Þ

We raise each equation to the second power and add them term by term(Eq. B.5):

r21 þ r22 þ r24 � 2r1r2 cos h2 � 2r1r4 cos h4 þ 2r2r4ðcos h2 cos h4 þ sin h2 sin h4Þ ¼ r23ðB:5Þ

By dividing all terms by the coefficient of term cos h2 cos h4 þ sin h2 sin h4,2r2r4, it yields Eq. (B.6):

r21 þ r22 � r23 þ r242r2r4

� r1r4cos h2 � r1r2 cos h4 þðcos h2 cos h4 þ sin h2 sin h4Þ ¼ 0 ðB:6Þ

In order to simplify Eq. (B.6), we use the following coefficients (Eq. B.7):

k1 ¼ r1r2k2 ¼ r1r4k3 ¼ r

21 þ r22 � r23 þ r24

2r2r4

9>>>>>>=>>>>>>;

ðB:7Þ

Thus, Eq. (B.6) remains Eq. (B.8):

k3 � k2 cos h2 � k1 cos h4 þðcos h2 cos h4 þ sin h2 sin h4Þ ¼ 0 ðB:8Þ

We substitute cos h4 and sin h4 for their expressions in terms of the half angletangent (Eq. B.9):

2O 4O

2r

3r

4r

A

B

1r2θ

3θ

4θFig. B.1 Position analysis ofa four-bar mechanism bymeans of Raven’s method

374 Appendix B: Freudenstein’s Method to Solve …

k3 � k2 cos h2 � k11� tan2 h421þ tan2 h42

þ cos h21� tan2 h421þ tan2 h42

þ sin h22 tan h42

1þ tan2 h42

!¼ 0

ðB:9Þ

Next, we remove the denominators and group the terms for tan, tan2 and theindependent term (Eq. B.10), all in the same member.

ðk3 � k2 cos h2 � k1 � cos h2Þ tan2 h42 þ 2 sin h2 tanh42

þðk3 � k2 cos h2 � k1 þ cos h2Þ ¼ 0ðB:10Þ

Again, we rename the different coefficients (Eq. B.11) of the second degreeequation:

A ¼ k3 � k2 cos h2 � k1 � cos h2B ¼ 2 sin h2C ¼ k3 � k2 cos h2 � k1 þ cos h2

9=; ðB:11Þ

Thus, Eq. (B.10) can be written as Eq. (B.12):

A tan2h42

þB tan h42

þC ¼ 0 ðB:12Þ

Hence, h4, which is the unknown that defines the angular position of link 4, is(Eq. B.13):

h4 ¼ 2 arctan�B�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiB2 � 4AC

p

2AðB:13Þ

where the + and − signs indicate two possible solutions for the open and crossedconfigurations of the four-bar mechanism respectively.

Similarly, but in this case isolating h4 in one of the members, we reach(Eq. B.14) for h3, which defines the angular position of link 3. Again, there are twopossible solutions depending on the configuration of the four-bar mechanism:

h3 ¼ 2 arctan�E �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiE2 � 4DF

p

2DðB:14Þ

where the different coefficients (Eq. B.15) of the second degree equation (Eq. B.14)are:

Appendix B: Freudenstein’s Method to Solve … 375

D ¼ k1 � k4 cos h2 þ k5 � cos h2E ¼ 2 sin h2F ¼ �k1 � k4 cos h2 þ k5 þ cos h2

9=; ðB:15Þ

And k4 and k5 (Eq. B.16) are:

k1 ¼ r1r2k4 ¼ r1r3k5 ¼ r

21 þ r22 þ r23 � r24

2r2r3

9>>>>>>=>>>>>>;

ðB:16Þ

376 Appendix B: Freudenstein’s Method to Solve …

Appendix CKinematic and Dynamic Analysisof a Mechanism

The conveyor transfer mechanism shown in Fig. C.1 pushes boxes with a mass of8 kg from one conveyor belt to another. The motor link turns at a constant speed of40 rpm in counter clockwise direction.

In order to make a complete kinematic and dynamic analysis of the mechanism,we will use all the analysis methods described in this book. We will carry out theanalysis at a given position. In general, the most interesting one for dynamicanalysis is the position at which the acceleration of the piston is maximum. Thisway we can determine the forces that act on the links in extreme conditions. Theposition chosen for this study is h2 ¼ 350�.

This analysis includes the following sections:

• Kinematic chain. Study and identification of the kinematic pairs. Number ofD.O.F of the mechanism. Kinematic inversion that results from fixing link 4.

• Kinematic graph of slider displacement versus crank rotation.• Velocity analysis by means of the relative velocity method.• Velocity analysis by means of the method of Instantaneous Centers of Rotation.• Acceleration analysis by means of the relative acceleration method.• Velocity and acceleration analysis by means of Raven’s method.• Calculation of the inertial force and inertial torque of each of the links in the

mechanism.• Dynamic analysis by means of the graphical method.• Dynamic analysis by means of the matrix method.

C.1 Kinematic Chain

We begin the study of the mechanism by drawing its kinematic diagram as shownin Fig. C.2. This figure also shows the nomenclature that will be used along thisstudy.

Table C.1 shows the different types of kinematic pairs in the mechanism and thedegrees of freedom of each pair.


377

4m

1m

2m

2m

1m6m

3.2m

3m

3.7m

6.5m

Fig. C.1 Conveyor transfer mechanism

23

4

A

B

2O

4O

C

D5

6

Y

X

2θ

3θ

4θ

4ϕ

5θ

2DOx

Fig. C.2 Kinematic diagram of the mechanism

Table C.1 Type of kinematic pairs in the mechanism

PAIR Type Number of D.O.F.

1–2 Rotation 1

2–3 Rotation 1

3–4 Rotation 1

1–4 Rotation 1

4–5 Rotation 1

5–6 Rotation 1

1–6 Prismatic 1

378 Appendix C: Kinematic and Dynamic Analysis of a Mechanism

We use Kutzbach’s equation to calculate the number of degrees of freedom ofthe mechanism (Eq. C.1):

• Number of links: N = 6• Kinematic pairs with 1 DOF: J1 ¼ 7• Kinematic pairs with 2 DOF: J2 ¼ 0

DOF ¼ 3ðN � 1Þ � 2J1 � J2 ¼ 3ð6� 1Þ � 2 � 7� 0 ¼ 1 ðC:1Þ

To better understand the mechanism, we will draw the kinematic diagram of oneof its inversions. In this case we will consider link 4 as the frame. This is shown inFig. C.3.

C.2 Slider Displacement Versus Crank Rotation

We will draw the kinematic graph of point D displacement versus crank rotation bymeans of the graphical method. To do so, we divide the whole turn of the crank in12 positions starting from position 0�. This way, we find the 12 positions of point Awhich correspond to 12 angular positions of the crank in steps of 30°. Knowing thelength of the links, we can find the equivalent 12 positions for points B, C and D(Fig. C.4).

We can graph the position of point D versus the crank position. This is shown inFig. C.5. We can see that the stroke end positions of the piston are close to posi-tions h2 ¼ 10� and h2 ¼ 195�. In these positions, the velocity of the piston has to benull. As the velocity is the first time-derivative of displacement, this can be verifiedby tracing a line tangent to the curve at the end-of-stroke position. If the line ishorizontal, the velocity is null.

23

4

5

61

Fig. C.3 Kinematic inver-sion of the mechanism whenlink 4 is fixed

Appendix C: Kinematic and Dynamic Analysis of a Mechanism 379

C.3 Velocity Analysis by Relative Velocity Method

Before starting the velocity analysis, the positions of the links have to be deter-mined. To do so, we will use the trigonometric method explained in Appendix A.Figure C.6 shows the angles and distances used to solve the position problem.

We start with the four-bar mechanism formed by links 1, 2, 3 and 4. DistanceO2O4 (Eq. C.2) and angles h1 (Eq. C.3) and c (Eq. C.4) can be calculated as:

23

4

A

B

2O

4O

C

D5

6

Fig. C.4 Kinematic diagram of the mechanism in a complete turn of the crank in steps of 30°

2DOx

2θ0

5−

4−

360°

3−

2−

90° 180° 270°

1−

Fig. C.5 Kinematic graph of the slider displacement versus the crank rotation


O2O4 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiðxO4 � xO2Þ2 þðyO4 � yO2Þ2

q¼ 4:206m ðC:2Þ

h1 ¼ 270� þ arctan 3:72 ¼ 331:6� ðC:3Þ

c ¼ 180� � 90� � arctan 3:72

¼ 28:4� ðC:4Þ

The application of the cosine rule to triangle DO2AO4 yields (Eq. C.5):

O4A ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiO2O4

2 þO2A2 � 2O2O4O2A cosðh2 � h1Þq

¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi4:2062 þ 12 � 2 � 4:206 cos 18:4�

p¼ 3:272m

ðC:5Þ

And the sine rule on the same triangle yields (Eqs. C.6–B.7):

O4A sinðb� cÞ ¼ O2A sinðh2 � h1Þ ðC:6Þ

b ¼ cþ arcsinO2A sinðh2 � h1ÞO4A

¼ 28:4� þ arcsin sin 18:43:272

¼ 33:9� ðC:7Þ

The application of the cosine rule to triangle DABO4 yields (Eqs. C.8 and C.9):

O4B2 ¼ AB2 þO4A2 � 2ABO4A cos/ ðC:8Þ

23

4

A

B

2O

4O

C

D5

6

1θφ 4

ϕ

4DOx

β

γ

δ

4DOy

2 4O Ox

2DOx

Fig. C.6 Calculation of theposition of links 3, 4, 5 an 6by means of the trigonometricmethod


/ ¼ arccosAB2 þO4A2 � O4B2

2ABO4A

¼ arccos 42 þ 3:2722 � 322 � 4 � 3:272 ¼ 47:4

�ðC:9Þ

Thus, the positions of the link 3 (Eq. C.10) is:

h3 ¼ /� b ¼ 47:4� � 33:9� ¼ 13:5� ðC:10Þ

The application of the sine rule to triangle DABO4 yields (Eqs. C.11 and C.12):

O4B sin d ¼ AB sin/ ðC:11Þ

d ¼ arcsin ABO4B

sin/� �

¼ arcsin 4 sin 47:4�

3

� �¼ 79:12�

ðC:12Þ

Therefore, the position of link 4 (Eq. C.13) is:

h4 ¼ 180� � b� d ¼ 180� � 33:9� � 79:12� ¼ 67� ðC:13Þ

We continue with the position analysis of the crank-shaft mechanism formed bylinks 4, 5 and 6 in Fig. C.6.

Although angle u4 formed by O4B and O4C has a fixed value and it could bepart of the data of the mechanism, in this case we have the length of the sides oftriangle DO4BC instead of angle u4 itself. We can easily obtain its value (Eq. C.15)by means of the rule of cosines (Eq. C.14):

BC ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiO4B

2 þO4C2 � 2O4BO4C cosu4q

ðC:14Þ

u4 ¼ arccosO4B

2 þO4C2 � BC22O4BO4C

!

¼ arccos 32 þ 62 � 3:222 � 3 � 6

� �¼ 15:1�

ðC:15Þ

The projection of triangle DO4CD over a direction perpendicular to the trajectoryof the piston yields the trigonometric equation (Eq. C.16):

yDO4 þCD sin h5 ¼ O4C sinðh4 þu4Þ ðC:16Þ


And clearing h5 (Eq. C.17), we obtain its value:

h5 ¼ arcsinO4C sinðh4 þu4Þ � yDO4CD

¼ arcsin 6 sinð67� þ 15:1�Þ � 56:5

¼ 8:34�ðC:17Þ

The projection of the sides of triangle DO4CD over de piston trajectory yields(Eq. C.18):

xDO4 ¼ O4C cosðh4 þu4Þ � CD cos h5¼ 6 cosð67� þ 15:1�Þ � 6:5 cos 8:34� ¼ �5:607m ðC:18Þ

Hence, the horizontal component of the distance between D and O2 (Eq. C.19)is:

xDO2 ¼ xDO4 � xO2O4 ¼ �5:607m� ð�3:7mÞ ¼ �1:907m ðC:19Þ

Therefore, the positions of the links (Eq. C.20) corresponding to crank positionh2 ¼ 350� are:

h3 ¼ 13:5�h4 ¼ 67�h5 ¼ 8:34�

xDO2 ¼ �1:907m

9>>>=>>>;

ðC:20Þ

The following step is to find the velocity of the links when link 2 rotates at anangular speed of 40 rpm counterclockwise. We have to use the velocity of link 2 inradians per second: 4.19 rad/s.

The velocity of point A (Eq. C.21) can be calculated as:

vA ¼ x2 ^ rAO2 ¼î ĵ k̂

0 0 4:19

1 cos 350� 1 sin 350� 0

�� ¼ 0:73̂iþ 4:13̂j

¼ 4:19 cm=s\80�ðC:21Þ

To calculate the angular velocity of links 3 and 4 we have to use the relativevelocity vector equation: vB ¼ vA þ vBA.

Vectors vB and vBA can be obtained the following way (Eqs. C.22 and C.23):


vBA ¼ x3 ^ rBA ¼î ĵ k̂0 0 x3

4 cos 13:5� 4 sin 13:5� 0

��

¼ �4x3ðsin 13:5�̂i� cos 13:5� ĵÞ ðC:22Þ

vB ¼ x4 ^ rBO4 ¼î ĵ k̂0 0 x4

3 cos 67� 3 sin 67� 0

�� ¼ �3x4ðsin 67�̂i� cos 67� ĵÞ

ðC:23Þ

By introducing the three velocity vectors, vA, vB and vBA, in the relative velocityequation and projecting them on the X and Y Cartesian axles, we reach the system ofequations (Eq. C.24):

0:73� 4x3 sin 13:5� ¼ �3x4 sin 67�4:13þ 4x3 cos 13:5� ¼ 3x4 cos 67�

)ðC:24Þ

The solution to the system of equations (Eq. C.24) yields the velocities of links 3and 4 (Eq. C.25):

x3 ¼ �1:27 rad=sx4 ¼ �0:69 rad=s

�ðC:25Þ

Using these values, we can calculate velocities vB (Eq. C.26) and vBA(Eq. C.27):

vB ¼ 1:91̂i� 0:81̂j ¼ 2:08m=s\336:96�13:5 ðC:26Þ

vBA ¼ 1:185̂i� 4:939̂j ðC:27Þ

The velocity of point C (Eq. C.28) can be determined by using the value of x4:

vC ¼ x4 ^ rCO4 ¼î ĵ k̂

0 0 x46 cosð67� þ 15:1�Þ 6 sinð67� þ 15:1�Þ 0

��

¼ 4:12̂i� 0:58̂j ¼ 4:16m=s\352:1�ðC:28Þ

We use vector equation (C.29) to calculate the angular velocity of link 5 and thelinear velocity of link 6.


vD ¼ vC þ vDC ðC:29Þ

Since points C and D are two points of the same link, their relative velocity isgiven by Eq. (C.30):

vDC ¼ x5 ^ rDC ¼î ĵ k̂

0 0 x56:5 cosðh5 þ 180�Þ 6:5 sinðh5 þ 180�Þ 0

��

¼ �6:5x5ðsinðh5 þ 180�Þ̂i� cosðh5 þ 180�Þ̂jÞ

ðC:30Þ

The velocity of point D (Eq. C.31) has the same direction as the trajectory.Therefore, its vertical component is null:

vD ¼ vD î ðC:31Þ

By substituting the three velocity vectors, vC, vD and vDC, in Eq. (C.29) weobtain the system of equations (Eq. C.32):

4:12� 6:5x5 sin 188:3� ¼ vD�0:58þ 6:5x5 cos 188:3� ¼ 0

)ðC:32Þ

Hence, the values of the velocities of links 5 and 6 (Eq. C.33) are:

x5 ¼ �0:09 rad=sv6 ¼ vD ¼ 4:04m=s

)ðC:33Þ

And the vector velocity of point D (Eq. C.34) is:

vD ¼ 4:04̂i ¼ 4:04m=s\0� ðC:34Þ

Figure C.7 shows the velocity polygon of the mechanism. We can see howabsolute velocities start at velocity pole O and relative velocities connect the endpoints of the absolute velocity vectors. It can also be seen that triangle Dobc in thepolygon is similar to DO4BC in the mechanism since their sides are perpendicular.

C.4 Instantaneous Center Method for Velocities

To calculate the ICRs in the mechanism, we start by identifying the ICRs whichcorrespond to real joints. In this case, the known ICRs are: I12, I23, I34, I14, I45, I16,and I56 (Fig. C.8).


Then we draw a polygon with as many vertexes as links in the mechanism. Eachof the sides or diagonals of the polygon represents one ICR. A solid line is used todraw those ICRs that are already known while those which are unknown are drawnas dotted lines.

In this case, to calculate the velocity of points B, C and D (Eqs. C.35–C.39), wehave to obtain ICRs I16, I24 and I46 by using Kennedy’s theorem.

26D I=v v

23

4

2 12O I=

45C I=

56D I=5

6

46I

24I

16 ( )I ∞

23I

B

4 14O I=

13I26I

23A I=v v

B′

B′v

Bv34I

C′C′v

Cv

A′A′v

Dv

A

36

1

5 4

2

Fig. C.8 Velocity calculation by means of the ICR method

o

Av

Dv

BvCv

BAv

CBvDCv

a

bc

d

AB⊥ 2O A⊥

4O B⊥4O C⊥

DC⊥

BC⊥

Fig. C.7 Velocity polygon


I26I16I12I24I46

�I24

I23I34I14I12

�I46

I45I56I14I16

�

Figure C.8 shows the graphical development of the method and the vectorobtained for each velocity.

v23 ¼ vAv23 ¼ I13I23x3 ¼ 3:33m � x3

�! x3 ¼ 1:26 rad=s ðC:35Þ

vB ¼ BI13x3 ¼ 2:08m=s ðC:36Þ

v24 ¼ I12I24x2 ¼ 0:58m � x2v24 ¼ I14I24x4 ¼ 3:52m � x4

�! x4 ¼ 0:69 rad=s ðC:37Þ

vC ¼ I14Cx4 ¼ 4:14m=s ðC:38Þ

v26 ¼ I12I26x2 ¼ 0:96m � x2v26 ¼ vD

�! vD ¼ 4:04m=s ðC:39Þ

C.5 Acceleration Analysis with the Relative AccelerationMethod

We know that the motor link turns at a constant rate of 40 rpm. Therefore, itsangular acceleration is null (a2 ¼ 0). In order to calculate the acceleration of thelinks, we start with the acceleration of point A (Eq. C.40). The tangential compo-nent will be zero as it depends on the angular acceleration value. Therefore, it willhave only one normal component:

aA ¼ anAO2 ¼ x2 ^ vA ¼î ĵ k̂

0 0 4:19

0:73 4:13 0

��

¼ �17:3̂iþ 3:06̂j ¼ 17:55m=s2\170�ðC:40Þ

To calculate the angular acceleration of links 3 and 4, we use the vectors(Eqs. C.41 and C.42):

aB ¼ aA þ aBA ðC:41Þ

anB þ atB ¼ anA þ atA þ anBA þ atBA ðC:42Þ


anB ¼ x4 ^ vB ¼î ĵ k̂0 0 �0:69

1:91 �0:81 0

�� ¼ �0:559̂i� 1:318̂j ðC:43Þ

atB ¼ a4 ^ rBO4 ¼î ĵ k̂0 0 a4

3 cos 67� 3 sin 67� 0

�� ¼ �3a4 sin 67�̂iþ 3a4 cos 67� ĵ

ðC:44Þ

anBA ¼ x3 ^ vBA ¼î ĵ k̂0 0 �1:27

1:185 �4:939 0

�� ¼ �6:272̂i� 1:506̂j ðC:45Þ

atBA ¼ a3 ^ rBA ¼î ĵ k̂0 0 a3

4 cos 13:5� 4 sin 13:5� 0

��

¼ �4a4 sin 13:5�̂iþ 4a3 cos 13:5� ĵ ðC:46Þ

Substituting these vectors (Eqs. C.43–C.46) in Eq. (C.42) and projecting themon the Cartesian axles, we reach the system of equations (Eq. C.47):

�0:559� 3a4 sin 67� ¼ �17:3� 6:272� 4a3 sin 13:5��1:311þ 3a4 cos 67� ¼ þ 3:06� 1:506þ 4a3 cos 13:5�

)ðC:47Þ

The solution yields the angular speed of links 3 and 4 (Eq. C.48).

a3 ¼ 1:98 rad=s2a4 ¼ 9 rad=s2

�ðC:48Þ

Once a4 is known, we can calculate the acceleration of point B (Eq. C.49):

aB ¼ �25:4̂iþ 9:24̂j ¼ 27:03m=s\160� ðC:49Þ

The acceleration of point C can be calculated by means of (Eq. C.50):

aC ¼ aB þ aCB ðC:50Þ

As points B and C belong to the same link, the components of the relativeacceleration (Eq. C.51) are:

anCB ¼ x4 ^ vCBatCB ¼ a4 ^ rCB

�ðC:51Þ


Substituting the values previously obtained in Eq. (C.51), we calculate theacceleration vector of point C (Eq. C.52):

aC ¼ �53:89̂iþ 4:59̂j ¼ 54:06m=s\175� ðC:52Þ

To determine the angular acceleration of link 5 and the linear acceleration of link6, we use the vector equation (C.53):

anD þ atD ¼ anC þ atC þ anDC þ atDC ðC:53Þ

where:

anD ¼ 0atD ¼ aD î

�ðC:54Þ

anDC ¼ x5 ^ vDCatDC ¼ a5 ^ rDC

�ðC:55Þ

Substituting vectors aC (Eq. C.52), aD (Eq. C.54) and aDC (Eq. C.55) inEq. (C.53) and projecting them onto the Cartesian axles, we reach to the system ofequations (Eq. C.56):

aD ¼ �53:89þ 0:052� 6:5a5 sin 188:3�0 ¼ 4:59þ 0:0076þ 6:5a5 cos 188:3�

)ðC:56Þ

The solution of the system yields the accelerations of links 5 and 6 (Eq. C.57):

a5 ¼ 0:72 rad=s2aD ¼ �53:14m=s2

)ðC:57Þ

Thus, the vector acceleration of point D (Eq. C.58) is:

aD ¼ �53:14̂i ¼ 53:14m=s2\180� ðC:58Þ

Figure C.9 shows the acceleration polygon of the mechanism. It can be noticedthat triangle Dobc of the acceleration polygon is similar to triangle DO4BC of themechanism. In the acceleration polygon, the sides of triangle Dobc are not per-pendicular to the sides of triangle DO4BC like in the velocity polygon. Angle /4(Eq. C.59) between the sides of both triangles can be calculated as:

/4 ¼ arctanatCanC

¼ arctan atB

anB¼ arctan a

tCB

anCB¼ arctan a4

x24ðC:59Þ


Therefore, triangle Dobc in the polygon is similar to triangle DO4BC in themechanism and rotated angle /4.

C.6 Raven’s Method

The number of needed vector loop equations depends on the number of unknowns.In this case, the position unknowns are h3, h4, h5 and r6. As each vector equationallows solving two unknowns and we have four, we will need 2 vector equations(Eq. C.60) (Fig. C.10).

r1 þ r4 ¼ r2 þ r3r10 þ r5 þ r6 ¼ r40

)ðC:60Þ

Using the complex exponential form for the vectors, vector equation (Eq. C.60)can be written as (Eq. C.61):

o

Da

a

b

c

d

AaBa

Ca BAa

CBa

DCa

4φ

4φ

4φ

Fig. C.9 Acceleration polygon

3

4

A

B

2O

4O

C

D5

6

1r

2r 3r

4r

4′r

5r

6r

1′r

2θ1θ

3θ

4θ

5θ

1θ ′6θ

4ϕ

Fig. C.10 Kinematic dia-gram of the mechanism withthe two vector loop equationsused to solve the problem


r1eih1 þ r4eih4 ¼ r2eih2 þ r3eih3r10eih10 þ r5eih5 þ r6eih6 ¼ r40eiðh4 þu4Þ

)ðC:61Þ

By separating the real and imaginary parts we obtain a system (Eq. C.62) withfour equations and four unknowns: h3, h4, h5 and r6:

r1 cos h1 þ r4 cos h4 ¼ r2 cos h2 þ r3 cos h3r1 sin h1 þ r4 sin h4 ¼ r2 sin h2 þ r3 sin h3

r10 cos h10 þ r5 cos h5 þ r6 cos h6 ¼ r40 cosðh4 þu4Þr10 sin h10 þ r5 sin h5 þ r6 sin h6 ¼ r40 sinðh4 þu4Þ

9>>>=>>>;

ðC:62Þ

Using Freudenstein’s equation, explained in Appendix B of this book, the firsttwo equations of the system yields (Eqs. C.63 and C.64):

h3 ¼ 2 arctan�B�ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiB2 � 4AC

p

2AðC:63Þ

h4 ¼ 2 arctan�E �ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiE2 � 4DF

p

2DðC:64Þ

where A, B, C, D, E and F coefficients (Eq. C.65) are:

A ¼ k3 cos h1 � k2 cosðh2 � h1Þþ k1 � cos h2B ¼ 2ðsin h2 � k3 sin h1ÞC ¼ k3 cos h1 � k2 cosðh2 � h1Þþ k1 þ cos h2

9>=>;

D ¼ k3 cos h1 � k5 cosðh2 � h1Þþ k4 þ cos h2E ¼ 2ð� sin h2 þ k3 sin h1ÞF ¼ k3 cos h1 � k5 cosðh2 � h1Þþ k4 � cos h2

9>=>;

ðC:65Þ

And where k1, k2, k3, k4 and k5 geometrical data (Eq. C.66) are:

k1 ¼ r21 þ r22 þ r23�r24

2r2r3

k2 ¼ r1r3k3 ¼ r1r2k4 ¼ r

21 þ r22 þ r24�r23

2r2r4

k5 ¼ r1r4

9>>>>>>>>>=>>>>>>>>>;

ðC:66Þ


Using Freudenstein’s equation, explained in Appendix B of this book, the lasttwo equations of the system (Eq. C.62) yields (Eqs. C.67 and C.68):

h5 ¼ arcsin r40 sinðh4 þu4Þ � r10

r5ðC:67Þ

r6 ¼ r5 cos h5 � r40 cosðh4 þu4Þ ðC:68Þ

Using Eqs. (C.63), (C.64)–(C.67), (C.68), we can plot the position of the linksrelative to the positions of link 2 along one full turn. Figure C.11a shows angles h3,h4 and h5 and Fig. C.11b shows distance r6.

These figures illustrate the benefits of mathematical methods over graphicalones. The latter would only yield the solution to one of the points in such curvesand the problem has to be solved again when there are any changes in the geometricparameters of the mechanism. Conversely, the expressions in Raven’s method yielda solution for all the points in the curve and they do not need to be modifiedwhenever geometrical data are modified.

The solution of the obtained equations for h2 ¼ 350� yield (Eq. C.69) the fol-lowing values for the position unknowns:

iθ

2θ0°

3θ25°

50°

360°

4θ

5θ

75°

100°

90° 180° 270°

6r

9

8

7

6

5

10

2θ360°90° 180° 270°0°

(a)

(b)

Fig. C.11 a Angular positionof links 3, 4 and 5 in terms ofh2, b Plot of the linear posi-tion of link 6 versus h2


h3 ¼ 13:5�h4 ¼ 67:03�h5 ¼ 8:3�r6 ¼ 5:607m

9>>=>>; ðC:69Þ

The position along the horizontal path of link 6 with respect to the coordinatesystem origin (Eq. C.71) will be given by the position of point D (Eq. C.70):

r6 ¼ �xDO4 ¼ �xDO2 � xO2O4 ¼ �xDO2 � ð�3:7mÞ ¼ 5:607m ðC:70Þ

So for h2 ¼ 350� the X coordinate of link 6 is:

xDO2 ¼ �1:907m ðC:71Þ

By differentiating with respect to time (Eq. C.61), we obtain (Eq. C.72):

ir2x2eih2 þ ir3x3eih3 ¼ ir4x4eih4v6eip þ ir5x5eih5 ¼ ir40x4eiðh4 þu4Þ

)ðC:72Þ

We separate the real and imaginary parts in Eq. (C.72), which yields the equationsystem (Eq. C.73) with four unknowns: x3, x4, x5 and v6:

�r2x2 sin h2 �r3x3 sin h3 ¼ �r4x4 sin h4r2x2 cos h2 þ r3x3 cos h3 ¼ r4x4 cos h4

�v6 � r5x5 sin h5 ¼ �r40x4 sinðh4 þu4Þr5x5 cos h5 ¼ r40x4 cosðh4 þu4Þ

9>>>=>>>;

ðC:73Þ

From the first two algebraic equations in Eq. (C.73) we can obtain the expres-sions for x3 (Eq. C.74) and x4 (Eq. C.75):

x3 ¼ r2r3sin h4 � h2ð Þsin h3 � h4ð Þx2 ðC:74Þ

x4 ¼ r2r4sin h3 � h2ð Þsin h3 � h4ð Þx2 ðC:75Þ

Finally, from the third and fourth algebraic equation we reach expressions for x5(Eq. C.76) and v6 (Eq. C.77):

x5 ¼ r40

r5

cosðh4 þu4Þcos h5

x4 ðC:76Þ

v6 ¼ �r5x5 sin h5 þ r40x4 sinðh4 þu4Þ ðC:77Þ


Using Eqs. (C.74) and (C.75), we can plot the kinematic curve of the linkvelocity versus the position of link 2. These curves are shown in Fig. C.12a, b.

Again, equations (Eqs. C.74–C.77) can be particularized for h2 ¼ 350� yieldingthe values for the velocity unknowns (Eq. C.78):

x3 ¼ �1:27 rad=sx4 ¼ �0:69 rad=sx5 ¼ �0:09 rad=sv6 ¼ �4:04m=s

9>>>=>>>;

ðC:78Þ

Once more, Eq. (C.72) can be time-differentiated again in order to find accel-erations (Eq. C.79):

ð�r2x22 þ ir2a2Þeih2 þð�r3x23 þ ir3a3Þeih3 ¼ ð�r4x24 þ ir4a4Þeih4ð�r5x25 þ ir5a5Þeih5 þ a6eih6 ¼ ð�r40x24 þ ir40a4Þeiðh4 þu4Þ

)ðC:79Þ

By separating real and imaginary parts we reach, once more, a system (Eq. C.80)with four equations and four unknowns: a3, a4, a5 and a6.

2θ360°90° 180° 270°

iω

−1.5

3ω

−1

0

4ω

5ω

1

1.5

2θ360°90° 180° 270°

6v

−10

−5

0

5

10

0.5

−0.5

(a)

(b)

Fig. C.12 a Angular veloci-ties of links 3, 4 and 5 interms of h2, b Plot of thelinear velocity of link 6versus h2


�r2x22 cos h2 � r2a2 sin h2 � r3x23 cos h3 � r3a3 sin h3 ¼ �r4x24 cos h4 � r4a4 sin h4�r2x22 sin h2 þ r2a2 cos h2 � r3x23 sin h3 þ r3a3 cos h3 ¼ �r4x24 sin h4 þ r4a4 cos h4

�r5x25 cos h5 � r5a5 sin h5þ a6 cos h6 ¼ �r40x24 cos h40 � r40a4 sinðh4 þu4Þ�r5x25 sin h5þ r5a5 cos h5 þ a6 sin h6 ¼ �r40x24 sin h40 þ r40a4 cosðh4þu4Þ

9>>>>=>>>>;

ðC:80Þ

Again, we start by considering the first two algebraic equations in the system(Eq. C.80), which yield the angular accelerations of links 3 and 4 (Eq. C.81).

a3 ¼ �r2a2 sin h2 þ r4a4 sin h4 � r2x22 cos h2 � r3x23 cos h3 þ r4x24 cos h4

r3 sin h3

a4 ¼ �r2a2 sinðh3 � h2Þþ r2x22 cosðh3 � h2Þþ r3x23 � r4x24 sinðh4 � h3Þr4 sinðh4 � h3Þ

9>>=>>;ðC:81Þ

Finally, a6 and a5 (Eq. C.82) are obtained from the last two algebraic equationsin the system (Eq. C.80):

a5 ¼ r40a4 cosðh4 þu4Þ � r40x24 sinðh4 þu4Þþ r5x25 sin h5

r5 cos h5a6 ¼ r40a4 sinðh4 þu4Þþ r40x24 cosðh4 þu4Þ � r5a5 sin h5 � r5x25 cos h5

9=;ðC:82Þ

These expressions (Eqs. C.81 and C.82) can be particularized for h2 ¼ 350�yielding the values for the unknowns (Eq. C.83):

a3 ¼ 1:98 rad=s2a4 ¼ 9 rad=s2a5 ¼ 0:72 rad=s2a6 ¼ 53:14m=s2

9>>=>>; ðC:83Þ

C.7 Mass, Inertia Moments, Inertia Forces and Inertia Pairs

We assume that we know the value of the mass and the moment of inertia of thelinks. Their values are included in Table C.2.

Figure C.13 shows the center of mass of each link. Their position (Eq. C.84) isgiven by the following distances:


O2G2 ¼ 0:5mAG3 ¼ 2mBG4 ¼ 0:52m

\O4BG4 ¼ 75:4�DG5 ¼ 3:25m

9>>>>>>>=>>>>>>>;

ðC:84Þ

The acceleration of the center of mass of each link (Eq. C.85) has been deter-mined by Raven’s Method yielding the following results:

aG2 ¼ �8:64̂iþ 1:52̂j ¼ 8:78m=s2\170�aG3 ¼ �21:35̂iþ 6:15̂j ¼ 22:22m=s2\163:94�aG4 ¼ �25:84̂iþ 4:57̂j ¼ 26:24m=s2\170�aG5 ¼ �53:55̂iþ 2:16̂j ¼ 53:59m=s2\177:69�aG6 ¼ �53:14̂i ¼ 53:14m=s2\180�

9>>>>=>>>>;

ðC:85Þ

Once the masses, moments of inertia and accelerations of each center of masshave been determined, we can calculate the forces (Eq. C.86) and moments(Eq. C.87) due to inertia:

Table C.2 Mass and moment of inertia of the links

Link 2 3 4 5 6

Mk (kg) 15.31 61.26 154.75 99.54 85

Ik (kg m2) 1.278 81.68 495.76 358.479 –

23

4

A

B

2O

4O

C

6D G=5

6 5G

4G

3G

2G4 4O BG∠

Fig. C.13 Position of thecenters of mass of the links


FIn2 ¼ 15:31 � 8:78 ¼ 134:42NFIn3 ¼ 61:26 � 22:22 ¼ 1361:2NFIn4 ¼ 154:75 � 26:24 ¼ 4060:64NFIn5 ¼ 99:54 � 53:59 ¼ 5334:35NFIn6 ¼ 85 � 53:14 ¼ 4516:9N

9>>>>>>=>>>>>>;

ðC:86Þ

MIn2 ¼ 1;278 � 0 ¼ 0MIn3 ¼ 81:68 � 1:98 ¼ 161:73NmMIn4 ¼ 495:76 � 9 ¼ 4461:84NmMIn5 ¼ 358:48 � 0:72 ¼ 258:1NmMIn6 ¼ 0

9>>>>>>=>>>>>>;

ðC:87Þ

Figure C.14 shows the force and moment that acts on each link due to inertia.We can see that the force is opposite to the linear acceleration of the center of massand the moment is opposite to the angular acceleration of the link.

C.8 Force Analysis. Graphical Method

In order to calculate the torque that is acting on link 2 to equilibrate themechanism,wewill consider the inertia of the links as well as the force needed tomove the 80 kg box.We will consider a friction coefficient of 0.4 and will neglect the inertia force of thebox. Obviously, in a real problem the inertia of the box would have to be considered.

3

4

A

B

2O

4O

C

6D G=

5

65G

4G

3G

2G2InF

3InF

4InF

5InF

6InF

6Ga

5Ga

4Ga

3Ga

2Ga

5InM

4InM

3InM

5α

4α

3α

Fig. C.14 Forces and moments due to inertia in the mechanism


So, in this example the force that has to be exerted by link 6 to move the box(Eq. C.88) is:

FR ¼ lN ¼ 0:4ð80 kg � 9:81m=s2Þ ¼ 314:2N ðC:88Þ

We study the forces on the mechanism starting with link 6 (Eq. C.89).

F56 þF16 þFR ¼ 0 ðC:89Þ

In order to simplify the problem, we will consider that force FR acts on point D.This will affect the position of reaction force F16 but as force FR is quite smallcompared to Fi6 and the distance from point D to the base of the piston is also smallcompared to the mechanism dimensions, the error is very small.

Since the direction of force F56 is unknown, it will be broken into a vertical andhorizontal component, FV56 and F

H56. We know that the direction of force F16 is

perpendicular to the slider trajectory. So, this force will be equilibrated by FV56. Thevalue of FH56 (Eq. C.91) can be calculated by means of the force equilibrium of thehorizontal components of the forces acting on the link (Eq. C.90) as shown inFig. C.15b.

0 ¼ RFx ¼ FR þFIn6 þFH560 ¼ RFy ¼ F16 þFV56

�ðC:90Þ

FH56 ¼ 4831:14N\180� ðC:91Þ

Figure C.16 shows the free body diagram of link 5. As we already know, forceFH56 is equal to force F

H65 but with opposite direction.

C

D

5 5G

5InF

5InM65

HF

45F5Inh

65Hh

65VF 65

VhFig. C.16 Free body dia-gram of link 5

6D G=6

6InF

56F

16F

RFRF 6

InF

56HF

(a) (b)Fig. C.15 a Free body dia-gram of link 6, b horizontalcomponents acting on link 6


Next, we analyze the moment equilibrium at point C (Eqs. C.92–C.94) assumingthe direction of FV65 to be upwards:X

j

MCjz ¼ 0 ! hIn5 FIn5 � hV65FV65 þ hH65FH65 �MIn5 ¼ 0 ðC:92Þ

0:6m � 5334:35N� 6:43m � FV65 þ 0:94m � 4831:14N� 258:1Nm ¼ 0 ðC:93Þ

FV56 ¼ 1163:88N\90� ðC:94Þ

Then, F65 will be (Eq. C.95):

F65 ¼ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiFV65� 2 þ FH65� 2

q¼

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi4831:142 þ 1163:882

p¼ 4969:32N

hF65 ¼ arctanFV65FH65

¼ arctan 1163:884831:14

¼ 13:5�

9>>>>>>>>>=>>>>>>>>>;

ðC:95Þ

where distances hIn5 , hV65 and h

H65 (Eq. C.96) are:

hIn5 ¼ G5C sinð360� � hFIn5 þ h5Þ ¼ 3:25 sin 10:65�

hV65 ¼ DC cos h5 ¼ 6:5 cos 8:34�

hH65 ¼ DC sin h5 ¼ 6:5 sin 8:34�

9>>=>>; ðC:96Þ

Back to link 6, we know the vertical forces acting on it, as FV65 ¼ �FV56 ¼ F16.The equilibrium of forces acting on link 5 (Eq. C.97) yields the value of force

F45 (Eq. C.98) as shown in Fig. C.17.

FH56 þFV56 þFIn5 þF45 ¼ 0 ðC:97Þ

F45 ¼ 10;207:9N\185:36� ðC:98Þ

Similarly, the equilibrium equations of links 3 and 4 yield the value of the forcestransmitted by the links.

5InF

65HF

65VF

65F45Fo

Fig. C.17 Polygon of forcesacting on link 5


In link 3, we break force F43 into components FN343 and FN443 (Eq. C.101), with

directions AB and O4B respectively (Fig. C.18). We consider the equilibrium ofmoments about point A (Eqs. C.99 and C.100) assuming that FN443 (Eq. C.101) goesupwards. X

j

MAjz ¼ 0 ! �hIn3 FIn3 þ hN443 FN443 �MIn3 ¼ 0 ðC:99Þ

�0:986m � 1361:2Nþ 3:215 � FN443 � 161:73Nm ¼ 0 ðC:100Þ

FN443 ¼ 467:77N ðC:101Þ

Where distances hIn3 and hN443 can be measured on the drawing of the mechanism

or be determined as (Eq. C.102):

hIn3 ¼ G3A cosð90� þ hFIn3 � h3Þ ¼ 2 cos 60:44�

hN443 ¼ BA sinðh4 � h3Þ ¼ 4 sin 53:5�

)ðC:102Þ

3A

B3G

3InF

3InM

23F

443NF

343NF

3Inh

443Nh

Fig. C.18 Free body dia-gram of link 3

4

B

4O

C

4G4InF

4InM

334NF

54F

14F

4Inh3

34Nh

54h4

34NF

Fig. C.19 Free bodydiagram of link 4


Figure C.19 shows the free body diagram of link 4. We can calculate FN343(Eq. C.105) by means of the equilibrium equation of moments with respect to pointO4 (Eqs. C.103 and C.104). We assume the direction of FN334 to be oriented to theleft. X

j

MO4jz ¼ 0 ! �hIn4 FIn4 þ hN334 FN334 � h54F54 �MIn4 ¼ 0 ðC:103Þ

0 ¼ �2:923m � 4060:64Nþ 2:412m � FN334� 5:84m � 10;207:9N� 4461:84Nm ðC:104Þ

FN334 ¼ 31;486N ðC:105Þ

where distances hIn4 , hN334 and h54 (Eq. C.106) can be determined as:

hIn4 ¼ O4B cosð90� þ hFIn4 � h4Þ ¼ 3 cos 13�

h54 ¼ O4C cosð90� þ hF54 � h40 Þ ¼ 6 cos 13:26�hN334 ¼ O4B sinðh4 � h3Þ

9>=>; ðC:106Þ

The force equilibrium analysis of the forces acting on link 4 (Eq. C.107) yieldsF14 (Eq. C.108). Figure C.20 shows the force polygon.

F54 þFIn4 þFN434 þFN334 þF14 ¼ 0 ðC:107Þ

F14 ¼ 18;454N\24� ðC:108Þ

Also, the analysis of the force equilibrium of link 3 (Eq. C.109) yields force F23(Eq. C.110). The force polygon of forces acting on link 3 is shown in Fig. C.21.

FIn3 þF43 þF23 ¼ 0 ðC:109Þ

F23 ¼ 31;520N\193:5� ðC:110Þ

4InF

334NF

54F

14F4

34NF

o

Fig. C.20 Polygon of forces acting on link 4


Finally, Fig. C.22 shows the equilibrium analysis of link 2, from which we findthe value of the equilibrating torque (Eq. C.114) as well as force F12 (Eq. C.112).Since the value of FIn2 is very small compared to F32, we will neglect it inEq. (C.111). Therefore:

F32 þF12 ¼ 0 ðC:111Þ

F12 ¼ �F32 ¼ F23 ðC:112Þ

Figure C.22 shows the free body diagram of link 2.Torque M0, which acts on link 2 to equilibrate the mechanism, can be obtained

with (Eq. C.113): Xj

MO2jz ¼ 0 ! h32F32 þM0 ¼ 0 ðC:113Þ

M0 ¼ �0:398m � 31;520N ¼ �12;568:6Nm ðC:114Þ

where distance h32 (Eq. C.115) can be calculated as follows:

h32 ¼ O2A sinð360� � hF32 � h2Þ ¼ 1 � sin 23:5� ðC:115Þ

43F

23F

3InFo

Fig. C.21 Polygon of forces acting on link 3

A2O

2G2InF

32F

0M

32h

Fig. C.22 Free body dia-gram of link 2


C.9 Dynamic Analysis. Matrix Method

We start the dynamic analysis of the mechanism by writing the equations of theforce and moment equilibrium of each link (Eqs. C.116–C.20). Figure C.23 showsradius vectors pi, qi and ri used in the moment equations:

• Link 2:

F32 � F21 ¼ m2aG2p2 ^ F32 � q2 ^ F21 þM0 ¼ IG2a2

)ðC:116Þ

• Link 3:

F43 � F32 ¼ m3aG3p3 ^ F43 � q3 ^ F32 ¼ IG3a3

)ðC:117Þ

• Link 4:

F54 � F43 þF14 ¼ m4aG4p4 ^ F54 � q4 ^ F43 þ r4 ^ F14 ¼ IG4a4

)ðC:118Þ

3

4

A

B

2O

4O

C

6D G=

5

65G

4G

3G

2G

5p5q

2p2q 3q

3p

4p

4q

4r4φ

Fig. C.23 Radius vectorsused in the moment equilib-rium equations


• Link 5:

F65 � F54 ¼ m5aG5p5 ^ F65 � q5 ^ F54 ¼ IG5a5

)ðC:119Þ

• Link 6:

F16 � F65 þFR ¼ m6aG6 ðC:120Þ

The moment equilibrium equation of link 6 is not necessary as we suppose theforces are concurrent at point D and a6 ¼ 0.

Equations C.116–C.20 yield a system of 14 algebraic equations (Eq. C.121) and14 unknowns where F16x ¼ 0 as the direction of force F16 has to be perpendicular tothe sliding trajectory. So, in this example it will only have a vertical component.

Projecting each force equation on the X and Y axles and finding the vectorproducts in the torque equations, we reach the following system of equations:

F32x � F21x ¼ m2aG2xF32y � F21y ¼ m2aG2y

ðp2xF32y � p2yF32xÞ � ðq2xF21y � q2yF21xÞþM0 ¼ IG2a2F43x � F32x ¼ m3aG3xF43y � F32y ¼ m3aG3y

ðp3xF43y � p3yF43xÞ � ðq3xF32y � q3yF32xÞ ¼ IG3a3F54x � F43x þF14x ¼ m4aG4xF54y � F43y þF14y ¼ m4aG4y

ðp4xF54y � p4yF54xÞ � ðq4xF43y � q4yF43xÞþ ðr4xF14y � r4yF14xÞ ¼IG4a4F65x � F54x ¼ m5aG5xF65y � F54y ¼ m5aG5y

ðp5xF65y � p5yF65xÞ � ðq5xF54y � q5yF54xÞ ¼ IG5a5F16x � F65x ¼ m6aG6x � FRF16y � F65y ¼ m6aG6y

9>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>=>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>;

ðC:121Þ

The system (Eq. C.121) can be written in matrix form (Eq. C.122):

½L�q ¼ F ðC:122Þ


where:

½L� ¼

�1 0 1 0 0 0 0 0 0 0 0 0 0 00 �1 0 1 0 0 0 0 0 0 0 0 0 0q2y �q2x �p2y p2x 0 0 0 0 0 0 0 0 0 10 0 �1 0 1 0 0 0 0 0 0 0 0 00 0 0 �1 0 1 0 0 0 0 0 0 0 00 0 q3y �q3x �p3y p3x 0 0 0 0 0 0 0 00 0 0 0 �1 0 1 0 1 0 0 0 0 00 0 0 0 0 �1 0 1 0 1 0 0 0 00 0 0 0 q4y �q4x �p4y p4x �r4y r4x 0 0 0 00 0 0 0 0 0 �1 0 0 0 1 0 0 00 0 0 0 0 0 0 �1 0 0 0 1 0 00 0 0 0 0 0 q5y �q5x 0 0 �p5y p5x 0 00 0 0 0 0 0 0 0 0 0 �1 0 0 00 0 0 0 0 0 0 0 0 0 0 �1 1 0

0BBBBBBBBBBBBBBBBBBBBBB@

1CCCCCCCCCCCCCCCCCCCCCCA

ðC:123Þ

q ¼

F21xF21yF32xF32yF43xF43yF54xF54yF14xF14yF65xF65yF16yM0



ðC:124Þ

F ¼

m2aG2xm2aG2yIG2a2m3aG3xm3aG3yIG3a3m4aG4xm4aG4yIG4a4m5aG5xm5aG5yIG5a5

m6aG6x � FRm6aG6y



ðC:125Þ


The analytical expressions of the radius vectors in matrix ½L� are defined in Eqs.(C.126)–(C.129). Angles h2, h3, h4, u4 and h5 are the ones defined in Sect. 3.6 forthe analysis with Raven’s Method. Angle /4 is defined in Fig. C.23.

• Link 2:

p2 ¼ AG2ðcos h2̂iþ sin h2 ĵÞq2 ¼ �O2G2ðcos h2̂iþ sin h2 ĵÞ

�ðC:126Þ

• Link 3:

p3 ¼ BG3ðcos h3̂iþ sin h3 ĵÞq3 ¼ �AG3ðcos h3̂iþ sin h3 ĵÞ

�ðC:127Þ

• Link 4:

p4 ¼ CO4ðcosðh4 þu4Þ̂iþ sinðh4 þu4Þ̂jÞ � O4G4ðcosðh4 þ/4Þ̂iþ sinðh4 þ/4Þ̂jÞq4 ¼ BO4ðcos h4̂iþ sin h4 ĵÞ � O4G4ðcosðh4 þ/4Þ̂iþ sinðh4 þ/4Þ̂jÞr4 ¼ �O4G4ðcosðh4 þ/4Þ̂iþ sinðh4 þ/4Þ̂jÞ

9=;

ðC:128Þ

• Link 5:

p5 ¼ �DG5ðcos h5̂iþ sin h5 ĵÞq5 ¼ CG5ðcos h5̂iþ sin h5 ĵÞ

�ðC:129Þ

If we find the values of these vectors (Eqs. C.126–C.129) for h2 ¼ 350� weobtain (Eq. C.130):

p2 ¼ 0:4924̂i� 0:0868̂jmq2 ¼ �p2p3 ¼ 1:9447̂iþ 0:4672̂jmq3 ¼ �p3p4 ¼ 0:1692̂iþ 3:1052̂jmq4 ¼ 0:5144̂i� 0:0766̂jmr4 ¼ �0:6598̂i� 2:8373̂jmp5 ¼ �3:2157̂i� 0:4712̂jmq5 ¼ �p5

9>>>>>>>>>>>>>>>>>>=>>>>>>>>>>>>>>>>>>;

ðC:130Þ


The analytical expressions of the acceleration vector of the center of mass ofeach link (Eqs. C.131–C.34) are:

aG2 ¼ O2G2a2ð� sin h2̂iþ cos h2ĵÞ � O2G2x22ðcos h2̂iþ sin h2 ĵÞ ðC:131Þ

aG3 ¼ O2Aa2ð� sin h2̂iþ cos h2 ĵÞ � O2Ax22ðcos h2̂iþ sin h2 ĵÞþG3Aa3ð� sin h3̂iþ cos h3 ĵÞ � G3Ax23ðcos h3̂iþ sin h3 ĵÞ

ðC:132Þ

aG4 ¼ O4G4a4ð� sinðh4 þ/4Þ̂iþ cosðh4 þ/4Þ̂jÞ� O4G4x24ðcosðh4 þ/4Þ̂iþ sinðh4 þ/4Þ̂jÞ

ðC:133Þ

aG5 ¼ CO4a4ð� sinðh4 þu4Þ̂iþ cosðh4 þu4Þ̂jÞ� CO4x24ðcosðh4 þu4Þ̂iþ sinðh4 þu4Þ̂jÞþG5Ca5ð� sinðh5 þ 180�Þ̂iþ cosðh5 þ 180�Þ̂jÞ� G5Cx25ðcosðh5 þ 180�Þ̂iþ sinðh5 þ 180�Þ̂jÞ

ðC:134Þ

aG6 ¼ CO4a4ð� sinðh4 þu4Þ̂iþ cosðh4 þu4Þ̂jÞ� CO4x24ðcosðh4 þu4Þ̂iþ sinðh4 þu4Þ̂jÞþDCa5ð� sinðh5 þ 180�Þ̂iþ cosðh5 þ 180�Þ̂jÞ� DCx25ðcosðh5 þ 180�Þ̂iþ sinðh5 þ 180�Þ̂jÞ

ðC:135Þ

If we find the values of all the elements in the vector F for h2 ¼ 350� and solvethe system (Eq. C.122), we obtain the values (Eq. C.136) for the unknowns(Eq. C.124):

F21 ¼ 32;239̂iþ 7386̂jNF32 ¼ 32;106̂iþ 7409̂jNF14 ¼ 16;636̂iþ 7550̂jNF54 ¼ 10;161̂iþ 944̂jNF65 ¼ 4833̂iþ 1173̂jNF16 ¼ 1173̂jNM0 ¼ �12;872Nm

9>>>>>>>>>>>>>=>>>>>>>>>>>>>;

ðC:136Þ


One of the main advantages of the Matrix Method for the dynamic analysis whencompared to the graphical method is the ability to calculate the value of theunknowns along a complete cycle.

With the latter we can only find one solution for one position of the crank in thecurve and the expressions cannot be used again if there are any changes in thegeometrical data of the mechanism.

Figure C.24 shows a curve with the value of the instantaneous motor torque forthe different positions of the crank, h2. The value of the motor torque is given byM0and it is the torque that is necessary to apply to motor link 2 in order to obtain thedesired speed and acceleration. In this case, x2 ¼ 4:19 rad=s and a2 ¼ 0.

2θ360°90° 180° 270°

0M

−10,000

−5,000

0

5,000

10,000

−15,000

−20,000

Fig. C.24 Instantaneousmotor torque M0 versus crankangle h2

2θ360°90° 180° 270°

SF

0

5,000

10,000

15,000

2,500

7,500

12,500

Fig. C.25 Shaking forcecurve versus crank angle h2


Moreover, we can obtain the curve of the magnitude of the shaking force versusthe positions of the crank (Fig. C.25). The shaking force is given by Eq. (C.137):

FS ¼ F21 þF41 þF61 ðC:137Þ

In Fig. C.25 we can see that the maximum value for the shaking force is at aposition close to h2 ¼ 350�. In this case, this position coincides with the maximumacceleration of link 6.


Appendix A Position Kinematic Analysis.Trigonometric MethodAppendix B Freudenstein’s Method to Solvethe Position Equations in a Four-BarMechanismAppendix C Kinematic and Dynamic Analysisof a Mechanism

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Appendix A Position Kinematic Analysis. Trigonometric Method978-3-319-31970... · 2017. 8. 27. ·...

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