SIX-BAR LINKAGE MECHANISM THROUGH THE CONTOUR … · Cite this Article: Rame Likaj, Xhevahir...

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International Journal of Mechanical Engineering and Technology (IJMET) Volume 8, Issue 9, September 2017, pp. 442–453, Article ID: IJMET_08_09_048

Available online at http://www.iaeme.com/IJMET/issues.asp?JType=IJMET&VType=8&IType=9

ISSN Print: 0976-6340 and ISSN Online: 0976-6359

© IAEME Publication Scopus Indexed

SIX-BAR LINKAGE MECHANISM THROUGH

THE CONTOUR METHOD

Rame Likaj, Xhevahir Bajrami*, Astrit Sallauka

Faculty of Mechanical Engineering

University of Prishtina, 1000 Prishtina, Kosovo * Corresponding author: [email protected]

ABSTRACT

In this paper is presented a six-bar linkage mechanism of the pump for oil

extrusion. In this mechanism are introduced even higher kinematic pairs. Dimensions

and other incoming links will be adopted as necessary. For the six-bar linkage

mechanism will be carried out the kinematic analysis and for all linkages will be

presented the displacement, velocity and acceleration. The analysis will be performed

by Math Cad software, while kinetostatic analysis will be carried out using Contour

Method, comparing results of two different software‘s Math CAD and Working Model.

The simulation parameters will be computed for all points of the contours of

mechanism.

Keywords: Kinematics, Dynamic, Simulation.

Cite this Article: Rame Likaj, Xhevahir Bajrami and Astrit Sallauka, six-bar linkage

mechanism through the contour method, International Journal of Mechanical

Engineering and Technology 8(9), 2017, pp. 442–453.

http://www.iaeme.com/IJMET/issues.asp?JType=IJMET&VType=8&IType=9

1. INTRODUCTION

This paper has been realized with two application software’s: Math Cad and Working Model.

In this paper is presented a four-bar linkage ABCD and a simple crank mechanism DEF as

shown in the figure below. In this mechanism firstly is determined DFy and 5θ . Derivative

DFy represents the speed of the slider F. From the given picture o3601.0 2 ≤≤ θ should be

determined. The masses are adopted since the moments of inertia need to calculate. The

kinematic part of the paper will be completed by finding the velocities and accelerations of

each point A, B, C, D, E, and F for the centers from C1 to C6. In this way are determined the

angular accelerations and velocities of the linkages 2, 3, 4 and 5. Whereas for the kinetostatic

part will be determined the reaction forces of the points: XA, YA, XB,YB, XC,YC, XD,YD,

XE,YE, XF,YF, N=XF6 and Mtr which are acting on the leading link 2. In the picture are

shown 5 bodies: AB, BC right triangle, rod EF, and slider bar F.

Rame Likaj, Xhevahir Bajrami and Astrit Sallauka


Figure 1 Six bar-linkage mechanism ABCDEF

Given data:

56.725 ,4.62.4 ,2.4:,9.6

7.1,2[deg],360...deg5.0

2243

222

=+===

===

α

ωθ

Da rrr

r

(1)

Determination of angleα of Dr with 1x : Linkage I

→→→→

=++ Da rrrr 432

(2)

From then designed outline at x and y we have the following initial conditions:

)sin()sin()sin()sin(

)cos()cos()cos()cos(

443322

443322

αθθθ

αθθθ

⋅=⋅+⋅+⋅

⋅=⋅+⋅+⋅

Da

Da

rrrr

rrrr

(3)

Figure 2 Four-bar linkage ABCD and crank mechanism DEF

As shown in the Figure 2, there are 14 unknown sizes, 12 equations for the first four

bodies, ),( yxdy for the slider 14 equations.

Six-Bar Linkage Mechanism through the Contour Method


2. THE ANALYSIS OF THE POSITIONS, VELOCITIES AND

ACCELERATIONS OF SIX-BAR LINKAGE MECHANISM

In figure 3 is presented the velocity plan for the points A, B, C and D. Also are presented

angular velocities and angular accelerations for the points A, B and C. After finding the

velocities and accelerations of these points, the displacements, velocities and accelerations of

these points with angles 3θ and 4θ in function of 2θ are shown graphically. The outline is

shown in this form:

[deg]90[deg],180 43 == θθ (3)

The vector equation without the contour I is:

x

y

r4

rp

r2

r3

Figure 3 Velocity plan for the points A, B and C

=

+

+

0

)sin(

)cos(

0

)sin(

)cos(

0

)sin(

)cos(

0

)sin(

)cos(

44

44

33

33

22

22

α

α

θ

θ

θ

θ

θ

θ

p

p

r

r

r

r

r

r

r

r

(4)

22242

2

2

42

22322

2

3

2242

42232

3

43

432

)(,)(

)(,)(

[deg]48.59[deg],96.3[deg],18.104

),,()(

ωθθθ

εωθθθ

ε

ωθθθ

ωωθθθ

ω

βθθ

θθθ

⋅=⋅=

⋅=⋅=

=−==

=

d

d

d

d

d

d

d

d

FindZgj

(5)

After calculation of the values by Math Cad software, in the following are shown

graphically the values of positions, velocities, and accelerations respectively for 333 ,, εωθ and

444 ,, εωθ :



Figure 4 Diagrams for positions, velocities and accelerations

The position of the point B is calculated in direction of x and y :

]/[03.10)()(

)cos(,)sin(

)sin(),cos(

22

22

222222

2222

scmvvv

rvrv

ryrx

ByBxB

ByBx

BB

=+=

⋅⋅=⋅⋅−=

⋅=⋅=

θθ

ωθωθ

θθ

(6)

Figure 5 The position diagrams for the position equations of the point B in direction of x and y

]/[)()(

)sin(,)cos(

222

22

2222

2222

scmaaa

rvra

ByBxB

ByBx

θθ

ωθωθ

+=

⋅⋅−=⋅⋅−=

(7)

Angular velocity and acceleration for the point B:

][deg45.0[deg],87.0

,

244

2242

2

2

4242

4

=−=

⋅=⋅=

εω

ωθθ

εωθθ

ωd

d

d

d

(8)



3. THE EQUATIONS FOR POSITION (DISPLACEMENT), VELOCITY

AND ACCELERATION OF THE POINT C

In the following is shown the displacement of point C, along with velocities and accelerations

which are presented graphically by the diagrams via Math Cad program:

)(

)))(cos()sin((

0)();(

))(cos()sin(,0

))(sin()cos(0

2

223322

22

23322

23322

θ

ωθθθ

θθ

θθθ

θθθ

cyc

cy

cxccyc

cycx

cc

aa

rra

avv

rrvv

rryx

=

⋅⋅−⋅−=

==

⋅+⋅==

⋅+⋅==

(9)

Figure 6 Velocities and acceleration of point C

3.1. The expression for the middle point of the bars AB, BC and CD

In the following are extracted the displacement for the middle position of the bars AB, BC

and CD and another displacement which belong to these bars. Are also presented the

velocities and accelerations of these bars and their diagrams separately:

22

22

22

22

22

222

2

)cos(2

;)sin(2

)sin(2

);cos(2

:

ωθωθ

θθ

⋅=⋅−

=

==

ra

ra

ry

rx

AB

ycxc

cc

(9)

2223

33

2223

32

233

3233

3

))(cos(2

;))(sin(2

))(sin(2

));(cos(2

:

ωθθωθθ

θθθθ

⋅=⋅−

=

==

ra

ra

ry

rx

BC

ycxc

cc

(10)

2224

44

2224

44

243

4244

4

))(cos(2

;))(sin(2

))(sin(2

));(cos(2

:

ωθθωθθ

θθθθ

⋅=⋅−

=

==

ra

ra

ry

rx

CD

ycxc

cc

(11)



x

y R4a =42

50

R4a =40

R5 =92

Fy

E

Fig. 7 Crank mechanism

3.2. The equations for position (displacement), velocity and acceleration of the

point E

For the point, E of Crank Mechanism determined the displacement, and in the following

pictures are shown velocities and accelerations in the direction x and y:

22244

22244

2442244

244244

))(sin(;))(cos(

2))(cos(;))(sin(

))(sin());(cos(

ωθθωθθ

ωθθωθθ

θθθθ

⋅−=⋅−=

⋅=⋅−=

==

bEybEx

bEbE

bEbE

rara

ryrx

ryrx

(12)

Figure 8 Diagrams of the velocities and acceleration for the E

4. KINETOSTATIC ANALYSIS OF THE SIX-BAR LINKAGE

MECHANISM

For the kinetostatic part will be presented the kinetostatic analysis of six-bar linkage

mechanism by Math Cad software. Moments of inertia, masses of bodies are used in the

following:

2555

2244

2333

2222

65432

12

1);2.14.2(

12

1

2

1;

2

1

807.9;5;15;20;15;10

rmJmJ

rmJrmJ

gmmmmm

CC

CC

⋅⋅=−⋅=

⋅⋅=⋅⋅=

======

(13)



Linkage I:

The equilibrium conditions for the point A are equal to zero. Six linkages are used for the

kinetostatic analysis. For the first linkage are given the following equilibrium conditions

XA,YA and XB,YB for the middle points of the linkage AB, point C2 and for the body mass

m2.

)sin(2

)cos(2

)cos(2

)cos(2

)();(

22

22

22

22

222222

θθθθ

θθ

rX

rY

rX

rYM

amYYamXX

BBAAtr

ycBAxcBA

−+−=

⋅=−⋅=−

(14)

x

y

XA

YA

C2 Cxc2

Cyc2

XB

YB

Figure 9 Linkage I–Equilibrium conditions for the point AB

Linkage II

Also, for the linkage II are written the equilibrium conditions BC, which are XB,YB , XC,YC

center C3 body mass BC and moment of inertia for the point C3.

))(sin(2

))(2

))(cos(2

))(sin(2

)(

)();(

233

233

233

233

23

233233

θθθθθθθθθε

θθ

⋅⋅−⋅⋅−⋅⋅−⋅⋅=⋅

⋅=−⋅=−

rX

rY

rY

rXJ

amYYamXX

BBCCC

ycBCxcBC

(15)

x

y

XB

YB

C3 Cxc3

Cyc3

XC

YC

Figure 10 Linkage II – Equilibrium conditions for the part BC

Linkage III

)()()(

)(

)(

2244244

244

244

θθθε

θ

θ

mDzbaC

yCEDC

xCEDC

rXrYEHrYrYJ

amYYY

amXXX

⋅+⋅−−−+⋅−=⋅

⋅=−+

⋅=−+

(16)



x

y

XD

YD

C4 Cxc4

Cyc4

XC

YC

XE

YE

Figure 11 Linkage III of the points C, D, E and center C4

))(sin(2

)())(sin(2

)()(

)(

)(

255

255

255

244

255

θθθθθε

θ

θ

⋅⋅+−⋅⋅+=⋅

⋅=−+

⋅=−

rYY

rXXJ

amYYY

amXX

FEFEC

yCEDC

xCFE

(17)

x

y

XF

YF

C5 Cxc5

Cyc5

XE

YE

Figure 12 Linkage IV of the part EF in direction of x and y

Linkage V

))(( 626 gmamY yF ⋅+⋅−= θ (18)

x

y

F

XF YF

m6

Figure 13 Linkage V of the point F in direction of x and y

All unknown parameters for the linkages in both directions x and y are calculated by Math

Cad. From point A to the point Find the Transmission moment Mtr:

trFFEEDDC

CBBAA

MYXYXYXY

XYXYXFindSolve

⋅⋅⋅⋅⋅⋅⋅

⋅⋅⋅⋅⋅= (:)2(θ

(18)

In the following diagrams are given the values of all unknown parameters for the positions

(0; 0.5; 1) by Math Cad software. This it was really a tough work since the working process

by Math Cad was very slow especially to follow the procedure which is given by kinetostatic

analysis.



Figure 14 Diagrams of kinetostatic analysis of the given mechanism for the points A, B, C, D, E, F

and transmission moment Mtr

5. SIMULATION FOR SIX-BAR MECHANISM BY WORKING MODEL

SOFTWARE

In the second part of this paper is carried out the simulation for all points A, B, C, D, E, F and

Mtr (time dependent)of the six bar linkage mechanism by Working Model, which is shown in

the following.

Figure 15 Diagrams in Working Model for angular velocity and acceleration 3, 4, 5.



Figure 17 Diagrams in Working Model for equilibrium points A, B, C, D, E, F and transmission

moment (Mtr) in direction y

Figure18 Diagrams for 3θ, 4θ

and 5θ(time dependent), under command RUN

6. CONCLUSIONS AND RECOMANDATIONS

By application of the software’s Math Cad or Matlab we have reached to carry out the full

analysis of this paper. In this paper are made the calculations of all positions (displacement)

for the whole mechanism, and also are determined the plans for velocities and accelerations

for each point.

However, in this paper are shown the outline planes of the mechanism same as the

diagrams for each linkage through Math Cad and Working Model software, Six-bar linkage

mechanism diagrams which are derived by Working Model are almost similar to the diagrams

derived by Math Cad, same as the derived results, Through Working Model software, are

derived the results of reactions from the equilibrium conditions of six bar linkage mechanism,

angular velocities and accelerations for the points 3, 4 and 5, for the angles 3θ , 4θ and 5θ in

time domain through the command run, as the general conclusion; the results derived by both

software, same as for their diagrams, for all points of the six bar linkage mechanism are

within the reasonable boundaries.

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SIX-BAR LINKAGE MECHANISM THROUGH THE CONTOUR … · Cite this Article: Rame Likaj, Xhevahir...

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