mechanics of machines assignment

1

Graphical Position Analysis

2

Graphical Velocity Analysis

1 2

and

. . . 30

i i i i

i i i i

i i i i

i i i i

i i i i i i i i

i i i i i i i i

A A

B A B A

D C DC

C A C A

C B C B

AB ab AB ab AC ac BC bc

V OA OAn ms

V V V

V V V

V V V

V V V

π ω −

= =

= =

= +

= +

= +

= +

3

Graphical Acceleration Analysis

4

Velocity Analysis Using Instant Centres

1

Analytical Kinematic Analysis (Assignment Guidelines)

The Assur-Artobolevsky Method

The aim of the analytical solution is to determine the position, velocity and acceleration function for each joint (A, B, C and D) as well as the angular velocity and angular acceleration functions for the coupler, the slider and each linkage. The resulting functions are to be evaluated at a number of points and the output compared with that from the graphical method.

When highly accurate results are necessary or the kinematic analysis has to be repeated for a large number of configurations the analytical approach is best. The Assur-Artobolevsky method is based on the principle of analysing the mechanism as a set of kinematic pairs. This is essential for our example because it includes a coupler and a slider. This is a complex mechanism in terms of its motion.

In order to solve the complex six-bar linkage we have to decompose the mechanism into three kinematic pairs:

Part 1 Crank (point A)

Since the underlying four-bar linkage is Grashof, the crank OA will complete a full revolution. Hence the position of the endpoint A is given by:

2 2

2 2

2

cos sin

A

A

x y

OA

θ θ

= =

=

2

It follows that the components of velocity are:

2 2 2 2 2 2

2 2 2 2

2

2 2

sin and cos

(velocity angle) 2

(angular velocity)

A

A A

A A A

v

x y

v x y

θ θ θ θ

θ

π δ θ

θ ω

= − =

⇒ = + =

= +

=

Similarly, the acceleration at the point A can be found:

( )

2 2 2 2 2 2 2

2 2 2 2 2 2 2

2 2 2 2

2 2 2 2 2

2 2 2 2 2

2 2

sin cos

cos sin

cos sin arctan (acceleration angle)

sin cos

A

A

A

A

v

x

y

a x y

θ θ θ θ

θ θ θ θ

θ θ θ θ δ

θ θ θ θ

θ α

= − −

= −

⇒ = +

⎛ ⎞−⎜ ⎟= ⎜ ⎟− +⎝ ⎠

= (angular acceleration)

Part 2 Coupler and Rocker (points B and C)

Because the coupler undergoes complex motion we will represent the links as position vectors. The planar vectors for position, velocity and acceleration can then be represented in complex number form using Euler’s identity:

cos sinie iθ θ θ= +

Thus the position vectors in the four-bar linkage consisting of crank, rocker and coupler can be represented graphically as:

In order to obtain expressions for the angles θ3 and θ4 we use the above diagram to write:

( )

( ) ( ) ( )

1 2

3 4 3 4

3 4 4

1 2

1 1 2 2 1 1 2 2

and

now cos cos sin sin

real component imaginary component

A E AE

E A E A

i i AE E A

AE AE

R R R R R R R R R R R R R R

R R R e e i

i

θ θ

θ θ θ θ

+ = + = +

⇒ = + − = − +

= − = −

= − + −

= +

For convenience we will now define the real and imaginary components:

1 1 2 2

1 1 2 2

Re = Real component cos cos Im = Imaginary component sin sin

θ θ θ θ

= − = −

This then leads to:

( ) ( ) ( ) ( )

3 34 4 3 4 3 4

2 2 2 3 4 4 4 4 4 4

2 4 4 4 4 4 4 4

2 2 2 2 4 4 4 4 4 3

Re Im and Re Im

Re Im Re cos sin Im cos sin

Re cos sin Im cos sin

Re Im 2 Re cos 2 Im sin 0

i ii ie i e e i e

i i i

i i i

θ θθ θ

θ θ θ θ

θ θ θ θ

θ θ

−= + + = − +

⇒ = + + + − + +

− + + +

⇒ + + + + − =

From this we can obtain:

24 4

4 4 2 24 4

2 tan 1 tan 2 2sin and cos

1 tan 1 tan 2 2

θ θ

θ θ θ θ

− = =

+ +

If we now let: 2 2 2 2

4 3

4

Re Im 2

C + + −

=

then we obtain:

2 2 2 4

4 1,2

2 2 2

4

Im Re Im tan (where 0 )

2 Re

Im Re Im 2 arctan

Re

C C

C C

θ θ π

θ

− ± + −⎛ ⎞ = < <⎜ ⎟ −⎝ ⎠

⎛ ⎞− ± + − ⇒ = ⎜ ⎟

⎜ ⎟−⎝ ⎠

3

Since we want an open configuration we will take the positive value. To obtain an expression for θ3, we require the following diagram:

From the angle AB’B we obtain:

4 4 3

3 4 4

Im sin 2 arctan

Re cos θ

θ θ

⎛ ⎞± = ⎜ ⎟

+ +⎝ ⎠

Finally we obtain the following expression for XB and XE:

4 4

4 4

cos sin

B E

B E

X X Y Y

θ θ

= + = +

Similarly we can analyse the position of the point C. Clearly: ( )32

2 (+ when C is above AB) ii

C A AC ACR R R e e θ βθ β+= + = +

This leads to expressions for XC and YC:

( ) ( )

2 2 3

2 2 3

cos cos

sin sin C AC

C AC

X

Y

θ θ β

θ θ β

= + +

= + +

Differentiation then yields expressions for the velocity and acceleration components.

4

Part 3 Slider (point D)

The final part of the analysis concerns the position of the slider (point D). You will need this diagram:

From the diagram it follows that:

( ) / 25 5 5 5 5 5 5

5 5

5 5

cos sin cos and sin

sin

arcsin

i C C

C C

C

C

X iY i He S iH S X S Y H

H Y

H Y

πθ θ

θ θ

θ

θ

+ + + = + = +

⇒ + = + = −

⇒ =

⎛ ⎞− ⇒ = ⎜ ⎟

⎝ ⎠

Again, velocities and accelerations can be found by differentiating the appropriate functions. Finally:

5 5cos and D C DX S X Y Hθ= = + = −

5

Dynamic Force Analysis

Analysing link 2

( ) ( )

12 32 2 12 12 32 32 12

12 12 12 32 32 2

12 32

Summing the forces yields, Unknowns , , , ,

Summing the Torque yields:

Representing Force as X-components:

G X Y X Y

G

X X

F F F mg F F F F T

T T R F R F I

F F mg

α

= + =

= + × + × =

+ =

∑

∑

( ) ( )

2

12 32 2

12 12 12 12 12 32 32 32 32 2

Representing Force as Y-components:

Representing Torque as:

G X

Y Y G Y

X Y Y X X Y Y X G

F F mg

T R F R F R F R F I α

+ =

+ × − × + × − × =

1

Analysing link 3:

( ) ( ) ( )

43 32 53 3 43 43 32 32 53 53

23 23 53 53 43 43 3

Summing the forces yields, Unknowns , , , , , ,

Summing the Torque yields:

Representing Force as X-componen

G X Y X Y X Y

G

F F F F mg F F F F F F

T R F R F R F I α

= − + =

= × + × + × =

∑

∑

( ) ( ) ( )

( )

43 32 53 3

43 32 53 3

23 32

23 23 23 23 53 53 53 53

43 43 43 43 3

23 32 23 32 53

ts:

Representing Force as Y-components:

Representing Torque as: ( )

X X X G X

Y Y Y G Y

X Y Y X X Y Y X

X Y Y X G

X Y Y X

F F F mg

F F F mg F F

R F R F R F R F

R F R F I

R F R F R

α

− + =

− + =

= −

× − × + × − × +

× − × =

− × − × + ( ) ( )

53 53 53

43 43 43 43 3

X Y Y X

X Y Y X G

F R F

R F R F I α

× − × +

× − × =

2

Analysing link 4:

( ) ( )

14 43 4 14 14 43 43

34 34 14 14 4

14 43 4

Summing the forces yields, Unknowns , , ,

Summing the Torque yields:

Representing Force as X-components:

Repre

G X Y X Y

G

X X G X

F F F mg F F F F

T R F R F I

F F mg

α

= − =

= × + × =

− =

∑

∑

( ) ( ) ( ) ( )

14 43 4

34 43

34 34 34 34 14 14 14 14 4

23 43 23 43 14 14 14 14

senting Force as Y-components:

Representing Torque as: ( )

4

Y Y G Y

X Y Y X X Y Y X G

X Y Y X X Y Y X

F F mg F F

R F R F R F R F I

R F R F R F R F I

α

α

− = = −

× − × + × − × =

− × − × + × − × =

3

Analysing link 5:

( ) ( )

65 53 5 65 65 53 53

35 35 65 65 5

65 53 5

Summing the forces yields, Unknowns , , ,

Summing the Torque yields:

Representing Force as X-components:

Repre

G X Y X Y

G

X X G X

F F F mg F F F F

T R F R F I

F F mg

α

= − =

= × + × =

− =

∑

∑

( ) ( ) ( ) ( )

65 53 5

35 53

35 35 35 35 65 65 65 65 5

35 53 35 53 65 65 65 65 5

senting Force as Y-components:

Representing Torque as: ( ) Y Y G Y

X Y Y X X Y Y X G

X Y Y X X Y Y X G

F F mg F F

R F R F R F R F I

R F R F R F R F I

α

α

− = = −

× − × + × − × =

− × − × + × − × =

4

Analysing link 6:

16 65 6 16 65 65

16

Summing the forces yields, Unknowns , ,

Summing the Torque yields: 0 (all forces through centre of mass)

Representing Force as X-components:

G X Y X

X

F F F mg F F F

T

F F

= − =

=

−

∑

∑

65 6 16 16

16 65 6

16 65

knowing 0.2

0.2 Representing Force as Y-components:

0

X G X X Y

Y X G X

Y Y

mg F N F F

F F mg

F F

µ µ= = ± ⇒ =

− =

− =

Obtaining a Complete Matrix

Once the accelerations about the centre of gravity for each link have been obtained, the masses of each link, together with the angular acceleration, moment of inertia and the position vector values can be inserted into the matrix.

Note: The coefficient of friction for the slider has been taken to be µ = 0.3

5

Use the program MATRIX to solve the matrix for the unknown dynamic forces and driving torque.

6

Shaking Force and Moment

Shaking Forces

The net effect of all the dynamic forces experienced by the ground of the mechanism, translate to vibrations in the structure which supports the mechanism. The sum of all the forces acting on the ground plane is referred to as the shaking force, which for this particular case is represented as:

21 41 6 12 21 41 14 61 16

21

21

41

41

61

61

given , , where

1990.37 999.27 4.139

566.15 76.16

253.97

S

X

Y

X

Y

X

Y

F F F F F F F F F F

F N F N F N F N F N F N

= + + = − = − = −

= = = − = − = =

In order to determine the result force experienced by the ground you need to sum the x and y components of the force respectively.

Finding:

21 41 61

1990.37 4.139 76.16 2062.4

RX X X XF F F F

N

= + +

= − + =

∑

21 41 61

999.27 566.15 253.97 687.09

RY Y Y YF F F F

N

= + +

= − + =

∑

Therefore FR = 2173.84 N

1

The shaking force will tend to move the structure (ground) in the transverse direction, but due to the influence of the driving link with respect to the entire mechanism shaking torque is induced to the system. The shaking torque is simply the reaction torque felt by the ground that tends to rock the structure about its driveline axis. The shaking torque is represented as:

21 12ST T T= = −

with –T12 representing the negative of the source torque which is delivered by the driving link.

Therefore:

TS = -143.491 Nm

Shaking Moments

The shaking moment of a linkage mechanism is the summation of the reaction torque T21 and the shaking couples introduced by the connection to the ground, which is represented as:

( ) ( )21 1 41 2 61SM T R F R F= + × + ×

with:

T21 = the negative of the driving torque

R1 = the position vector from ground of driver (O) to ground of link 4

F41 = force of link 4 onto ground

R2 = the position vector from ground of driver (O) to ground of slider

F61 = force of slider onto ground

2

Therefore:

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

21 1 41 1 41

2 61 2 61

143.491 0.5478 566.135

0.1164 4.139 0.035 76.16

0.951 253.87 209.04

S X Y Y X

X Y Y X

M T R F R F

R F R F

Nm

Nm

= + × − ×

− × − ×

= + ×

− × − ×

− ×

=

The effect of the shaking moment is basically a rocking effect about the axis of the motor’s drive shaft.

3

• Specify a frequency of 1000 Hz in the World → Accuracy → Animation Step.

• The parameters of each link can be customised in the Window → Properties/Appearance menus.

1

Computational Model with Working Model 2D Configuring the Software

• Turn on the grid, axes and ruler in the View → Workspace menu.

• Customise the units in the View → Numbers and Units menu. Set the rotational velocity to revs/min and the frequency to Hertz.

Model Construction

• Know each link's dimension to its COG (X, Y, θ, L).

• Sketch three rectangles using the Rectangle tool (around 10 mm thick) in the positive X, Y plane. Specify the required dimensions, ensuring each is the correct position.

• Place Point Elements on both ends of each link with the exception of the ground end of link 4 and the end of link 5 that connects to the slider. Place Pin Joints at these ends.

2

• Sketch the main body using the Polygon tool. Click on the appropriate ends of the three links.

• Drag the body to the left and place Point Elements on each edge. Reposition the body and connect it to the links using the Join tool, selecting an element on the body and link.

3

• Sketch the slider using the Rectangle tool. Place a pin joint at its COG. Connect the slider to the bottom edge of link 5.

• Create a Slot Joint near the slider. Grab its pin joint and place it over the sliders pin joint (drop it when an ‘x’ appears).

• Place a motor at the appropriate location using the Motor tool.

4

• Rotate the main body to check if the model is connected.

• Customise each link using the Appearance menu. Uncheck Track outline and check Show centre of mass.

• Specify the correct masses and motor speed using the Properties menu.

5

Monitoring the Links

Link velocities and accelerations:

• Select the appropriate link and choose the appropriate output variable to monitor using the Measure menu (i.e. Centre of mass acceleration).

• Each output box can be configured as desired.

Joint forces and motor torque:

• Select the appropriate joint and check Measure → Force.

• Select the motor and check Measure → Torque.

6

Running the Simulation

• Check each frame in the World → Tracking tab.

• Specify a frequency of 1000 Hz in the World → Accuracy → Animation Step.

Exporting Results:

• Select File → Export, choosing an appropriate last frame.

• Open the file in Excel as Tab delimited.

You can use the following URL to run the simulation. You will need to logon with your student number and password to do this.

http://blackboard.rmit.edu.au/bin/common/content.pl?action=LIST &render_type=EDITABLE&mode=&content_id=_384906_1

Example of Working Model Results

The model represents the mechanism consisting of a motor, linkages, pin joints, slider and a slot for the slider to travel along. In order to achieve one revolution the time incrementation was set for one frame of sixteen using the ‘Accuracy’ command, and then by setting the ‘Pause Control’ for sixteen frames the desirable stopping location was obtained.

Working Model

7

Plot of Path

Working Model: Plot of paths for joint A, B, C and D

Velocity/Acceleration of Joints In order to determine the correlation between the analytical and computational results for velocity at θ = 60° a computational analysis needs to be done for the joints of the mechanism. The results obtained from the software are shown in figures.

Working Model: Velocity and Acceleration for joints A (28) and B (24)

8

Working Model: Velocity and Acceleration for joints C (25) and D (26)

The results were obtained by running the mechanism for the velocity and acceleration at joints A, B, C and D. The results were exported from Working Model into Excel while the model was running.

Note: The charts below were completed for one revolution of the crank (link OA).

9

10

Force Analysis

In order to determine the forces present in the model, the forces at the linkages need to be investigated using Working Model, with the result displayed in the figure for θ = 60°.

Working Model: Forces results for joints A, B, C and D

The results of the force analysis are displayed below using the Working Model.

Computational Method Software Generated (N)

F A

= 1929.351

F B = 747.243

F C = 739.126

F E = 709.716

F S = 267.186

T 12

= 156.354 Nm

11

Finally, don’t forget to provide a comprehensive summary of all results obtained using graphical, analytical and computational approaches.

Graphical Method Analytical Method

Computational Method

Linear Velocity

Vector (m/s)

I.C. (m/s)

θ (°)

Vector (m/s)

θ (°)

Vector (m/s)

V A

7.1834 7.1834 150 7.183 150 7.183

V B 4.983 4.983 164 4.983 164 4.958

V C 6.103 6.103 160.2 6.103 160 6.113

V D

6.932 6.938 180 6.956 180 6.926

Angular Velocity

Vector (rad/s)

Vector (rad/s)

Vector (rad/s)

ω 2 51.31 51.31

ω 3 3.7357 3.77045

ω 4 12.4575 12.4569

ω 5 4.2714 4.2823

Linear Acceleration

Vector (m/s2)

θ (°)

Vector (m/s2)

θ (°)

Vector (m/s2)

A A

368.58 240 368.58 240 368.635

A B

446.60 171.98 446.60 172 448.912

A C 280.598 195.84 280.598 196

.8 275.657

A D

252.1 180 251.89 180 251.98

Angular Acceleration

Vector (rad/s2) Vector (rad/s2)

Vector (rad/s2)

α A

0 0

α B 600.6 600.5972

α C 1105.65 1105.646

α D

109.6 109.5921

Table of Velocity and Acceleration results for θ2 = 60°

Summary of All Results

1

Analytical Method Computational Method Raw Data (N) Joint Forces (N) Software Generated (N) F

12X = -1990.307

F 12Y

= -999.273 F

12 = 2227.07

F 32X

= 1806.007 F

32Y = 680.073

F 32

= 1929.81 F A

= 1929.351

F 43X

= 335.819 F

43Y = 519.405

F 43

= 618.51 F B = 747.243

F 53X

= 624.438 F

53Y = -201.632

F 53

= 656.18 F C = 739.126

F 14X

= 4.139 F

14Y = 566.135

F 14

= 566.15 F E = 709.716

F 65X

= 680.073 F

65Y = -253.872

F 65

= 295.15

F 16Y

= -253.872 F

16X = 76.16

F 16

= 265.05 F S = 267.186

T 12

= 143.491 Nm T 12

= 156.354 Nm

Table of Force Results for θ2 = 60°

The stroke distance is determined to be:

• graphical method: 266.64 mm

• analytical method: 266.1 mm

The shaking force, shaking torque and shaking moment at θ2 = 60 ° are

respectively:

• FS = 2173.84 N

• TS = -143.491 Nm

• MS = 209.04 Nm

Please Note: You are expected to propose and describe a method for balancing of this linkage mechanism based on the obtained values of shaking forces and moments.

2