Geometry of five link mechanism with two degrees of freedom

David Tavkhelidze

Internal combustion engine

A-Crankshaft;B-Connecting rod;C-Slider (piston);D-Frame;E-Valve mechanism

E

Kinematic pairs

Degree of freedom Degree of freedom for spatial mechanism

W=6n-P1-2P2-3P3-4P4-5P5

Degree of freedom of planar mechanism

W=3n-2P5

3n=2P5

53

2Pn

53

2Pn

Kinematic chains

Four link mechanism

104233 w

Five link mechanism

25243 W

Mechanisms used in technological machines

Four link slider - crank mechanism

Four link mechanism with rotating kinematic pairs

Six link gear mechanism Five link mechanism with gear pair, reducing number degrees of freedom of the mechanical system

Mechanisms with two degrees of freedom

Kinematic scheme of five link mechanism with two degrees of freedom

Various scheme of mechanisms with two degrees of freedom

ETTTTT )4,0()3,4()2,3()1,2()0,1(

Straight geometrical task

The design diagram of five link mechanism

.

For the closed kinematic chain it is necessary that the product of matrices of transformation between coupled coordination systems connected with all incoming links has to be equal to unit matrix (1)

For simplification of calculation it would be written

(2))1,0()2,1()3,2()4,0()3,4( TTTTT

Straight geometrical task

The transformation matrix between of two sequential i-1 and i plain coordinate systems has the following general form:

(3)

.

)cos()sin(sin)sin(sin

)sin()cos()cos()cos(cos

001

231201231201231201023121232

231201231201231201023121232

lll

lll

)cos()sin(sin

)sin()cos(cos

001

40434043434

404340434343

l

ll

Taking in the account the previous equations will be obtained

(4)

Straight geometrical taskThe 4th matrix equation of five-link mechanism blockage contains full information about parameters of link motion characteristics. In order to determine relative and absolute displacement of links the respective elements of left and right parts of equation should be equated and receive system of algebraic equations the solution of which will enable to determine displacements of mechanism links.

(5)

, .

)cos(cos)cos(cos 2312010232231214343 lllll

231201023121232434 sinsinsinsin llll

2312014043 sinsin

2312014043 coscos

.

Besides these equations, in order to solve the problem the subsidiary condition should be added according to which the sum of internal angles of any five link is equal to 3π.

(6)

34034231201

Straight geometrical taskAfter transformations we get the following quadratic equation

(7) 03sincos2cos3sin 01

223434

2201

2 CAAEEC

,

Here: 0120303224

23

22

21

20 3cos222 lllllllllllA

4243 22 llllB 402 llC 013cos CBE

From the equation (7) we will obtain meanings of angelsϕ34 and ϕ23 that determines position of the point C of the mechanism.

(8)

(9)

1

403401034423

6sinsinarcsin

l

ll

. .

2 2 2 2 2 201 01

34 2 201

2 4 4 sin 3 sin 3arccos

2 sin 3

AE A E C E C A

C E

And hence, in case of differentiating on time the received values of obtaining equations, we shall receive values of speeds and acceleration of the links of the mechanism.

The inverse geometrical taskIn spite of the straight geometrical problem, here on the basis of the given angels of rotation of the actuators mounted on the frame of the mechanism the trajectory of the output link of the considered mechanical system is defined.

The formulation of inverse task of kinematics of five link mechanism is done in the following way: the location of C point of mechanism i.e. its coordinates in coordinate system connected with base, is given and it’s necessary to find generalized coordinates of the mechanism which provide the location of C point.

The design diagram of five link mechanism for inverse task

The inverse geometrical taskFor this we take C point radius vector from the origin of coordinates and represent it as the sum of two vectors:

(10)21 llRC

Projections of these vectors in immovable coordinate system are expressed as:

(11) 011

0111 sin

cos

l

ll

2 01 12

22 01 12

cos

sin

ll

l

12012011 coscos llX C

In projections formula (10) will have the following form:

(12) 12012011 sinsin llYC

The inverse geometrical taskIn order to find two generalized and coordinates determining the location of BC kinematic chain the expressions (12) should be squared and summed up:

(13)

120122

2120101210122

12 coscoscos2cos llllX C

120122

2120101210122

12 sinsinsin2sin llllYC

221221

21

22 cos2 llllYX CC

The obtained expressions (13) allows to calculate values of angels

Hence, we will have:

(14)

(15)

01 and 12

2 12 2 12 101 2 2 2

2 12 1 2 12

sin coscos

cos sinC Cl Y X l l

l l l

2 2 2 21 2

121 2

cos2

C CX Y l l

l l

The inverse geometrical taskWe behave similarly when we determine the location of CDO kinematic chain. We present radius vector of C point in the form of the following vectors sum:

(16)

(17)

(18)

340 lllRC

And hence we can obtain:

43

24

23

220

34 2cos

ll

llYlX CC

4343

34403040 cos

sinsincos

ll

Xll C

Based on derivations of the given formulas the values of velocities and accelerations of the links of the investigated mechanism have obtained .

The inverse geometrical taskBased on usage of MATLAB software here are given curves of alternations of phase angles of the five bar mechanism, when the two link junction point C is performing movement along the circle.

The inverse geometrical task

0 2 4 6 8 10 12 14 16 18 201.5

2

2.5

3

t

fi01 kuTxeebi

0 2 4 6 8 10 12 14 16 18 20-2.5

-2

-1.5

-1

t

fi12

Curves of alternation of angles

The inverse geometrical task

0 2 4 6 8 10 12 14 16 18 201

1.5

2

2.5

t

fi34

0 2 4 6 8 10 12 14 16 18 201.5

2

2.5

3

t

fi40

The inverse geometrical task

Values of angular velocities

0 2 4 6 8 10 12 14 16 18 20-2

-1

0

1

t

ffi0

1 kuTxur i si Cqar eebi

0 2 4 6 8 10 12 14 16 18 20-0.5

0

0.5

t

ffi1

2

The inverse geometrical task

0 2 4 6 8 10 12 14 16 18 20-0.5

0

0.5

1

t

ffi3

4

0 2 4 6 8 10 12 14 16 18 20-0.4

-0.2

0

0.2

0.4

t

ffi4

0

The inverse geometrical task

Values of angular accelerations

0 2 4 6 8 10 12 14 16 18 20-10

0

10

20

30

t

fffi01

kuTxur i aCqar ebebi

0 2 4 6 8 10 12 14 16 18 20-30

-20

-10

0

10

t

fffi12

The inverse geometrical task

0 2 4 6 8 10 12 14 16 18 20-0.4

-0.2

0

0.2

0.4

t

fffi3

4

0 2 4 6 8 10 12 14 16 18 20-1

-0.5

0

0.5

t

fffi4

0

Kinetostatics of five bar planar mechanisms

On the links of mechanical system are acting two type of force factors - External forces and Internal forces.

The internal forces – forces of weight, reduction forces of inertia and moments of inertia of force couples

Forces of inertia-Moments of inertia of force couples-

slus WmF

2

zsus IM

Kinetostatics of five bar planar mechanism

Reduction forces and moments of inertia acting on the links of five bar mechanism

Determination of forces and torques

Lagrange equation relatively to generalized

coordinate

Lagrange equation relatively to generalized coordinate

111

DMTT

dt

d

444

DMTT

dt

d

4

4

Determination of forces and torques

Equitant for determination of torque acting on A kinematic pair.

124

1

44

4

14

414

1121

1

11414

111

2

1

2

1

DMII

II

dt

dI

dt

dI

Equitant for determination of torque acting on O kinematic pair.

421

4

11

1

14

411

4424

4

44114

441

2

1

2

1

DMII

II

dt

dI

dt

dI

Determination of force factors

0 2 4 6 8 10 12 14 16 18 20-10

0

10

20

30

40

50

60

70

80

90

t

M1p

Determination of torque acting on A kinematic pair.

Determination of force factors

0 2 4 6 8 10 12 14 16 18 20-3

-2

-1

0

1

2

3

t

M4p

Determination of torque acting on O kinematic pair.

Thank you