Theory of Machines and Mechanism

Lecture 3

Łukasz Jedliński, Ph.D., Eng.

The following points are to be considered while solving velocities

problems.

1. Draw the configuration design to a suitable scale.

2. Locate all fixed point in a mechanism as a common point in velocity

diagram.

3. Choose a suitable scale for the vector diagram velocity.

4. The velocity vector of each rotating link is ⊥ to the link.

5. Velocity of each link in mechanism has both magnitude and direction.

Start from a point whose magnitude and direction is known.

6. The points of the velocity diagram are indicated by small letters.

Kinematic analysis by graphical methods

Velocity diagram of the link

The link BCM is in plane motion. Velocities of three points vB, vC, vM are

known. To obtain velocity diagram we need to start drawing vectors from a

point (polar) πv in suitable scale κv.

Kinematic analysis by graphical methods

=mm

smv

/

drawingin value

velocityof valueκ

Velocity diagram of the link

Figure bcm is similar to figure BCM and rotated 90° in the same direction asangular velocity ω.We can write equations:

Kinematic analysis by graphical methods

CBBC vvvrrr

+=

MBBM vvvrrr

+=

MCCM vvvrrr

+=

Direction of relative velocity

vector vMC is ⊥ to CM

Acceleration diagram of the link

In similar way we construct an acceleration diagram.

Figure bcm is similar to figure BCM and rotated about angle ϕ - 180° in thedirection of angular acceleration ε.

Kinematic analysis by graphical methods

2ωε

ϕ arctg=

t

MC

n

MCMC aaarrr

+=

We can write equation about

relative acceleration:

MCCM aaarrr

+=

Acceleration of point M

could be express:

=mm

sma

2/

drawingin value

naccelerato of valueκ

Acceleration diagram of the link

Kinematic analysis by graphical methods

t

MC

n

MCMC aaarrr

+=

Direction of relative normal acceleration aMC is || to CM and sense

(indicated by the arrowhead) is from point M to C.

Direction of relative tangential acceleration aMC is ⊥ to CM.

Calculate the velocity and acceleration of the points C2 (part of link 2) and E

of the mechanism presented in figure using the diagram method.

Dimensions and positions of the links are known, as is the fact that link 1

rotates with the constant rotational speed:

n1 = 120 rev/min,

lAB = 86 mm, lBC = 180 mm, lCD = 115 mm, lDE = 70 mm, lBD = 218,45 mm,

j1 = 130°, j2 = 14,17°, j3 = 7,33°, j4 = 114°

Kinematic analysis by graphical methods - example

Determining velocities

Links 1 and 3 perform rotational motion, whereas link 2 moves in general

plane motion. Firstly, calculate the angular velocity of the driving link:

Kinematic analysis by graphical methods - example

rad/s 430

120

30

11 π

ππω =⋅== n

and the linear velocity of the point B:

m/s 0807,1086,041 =⋅== πω ABB lv

Obviously, in a rotational pair, the velocities of

the connected ends of the elements are equal,

thus: vB1 = vB2 = vB.

Determining velocities

Since there is not enough information on the movement of links 2 and 3, in order to

specify the velocity of the point C2, the velocity of link 3 in the point B3 has to be

determined first. It should be mentioned that no part of link 3 is located in that point.

Still, it is assumed that a point, which is rigidly connected with link 3, is situated in this

position. The movement of the point B3 will be considered as relative motion. The rise

velocity is vB and the relative velocity is vBB3, thus the vB3 velocity equals:

Kinematic analysis by graphical methods - example

CD

BB

AB

B

BD

B vvv||

33 +=⊥⊥

Determining velocities

Due to the construction of link 2, which includes the slider, the distance of the points B

and B3 in relation to the line determined by the points C and D is always constant during

the movement of the mechanism. It means that the direction of the relative velocity

vBB3 is parallel to the segment CD (hence, single underlining in the formula). The

direction of the velocity vB is perpendicular to the segment AB; additionally, the velocity

value is known, therefore, double underlining was used.

Kinematic analysis by graphical methods - example

CD

BB

AB

B

BD

B vvv||

33 +=⊥⊥

Direction of the velocity vB3is also known. Because link 3

moves in rotational motion, its velocity is perpendicular

to the segment BD. To determine the vB3 velocity from

the plot, the

=mm

1

s

m 01,0vκ

velocity scale

has been adopted

Determining velocities

The length of the velocity vector vB3, read from the fig., is 175,18 mm, thus, the velocity

value is:

Kinematic analysis by graphical methods - example

( )s

m 7518,101,018,17533 =⋅== vBB vv κ

CD

BB

AB

B

BD

B vvv||

33 +=⊥⊥

Determining velocities

The length of the velocity vector vB3, read from the fig., is 175,18 mm, thus, the velocity

value is:

Kinematic analysis by graphical methods - example

( )s

m 7518,101,018,17533 =⋅== vBB vv κ

CD

BB

AB

B

BD

B vvv||

33 +=⊥⊥

( )s

m 0205,101,005,10233 =⋅== vBBBB vv κ

and

Determining velocities

hence the angular velocity of link 3 equals:

Kinematic analysis by graphical methods - example

rad/s 0192,821845,0

7518,133 ===

BD

B

l

vω

The velocity of the point E can be calculated from the relation:

s

m 56134,007,00192,83 =⋅== DEE lv ω

The remaining issue is the calculation of the velocity of the point C2. The equation on

the velocity of this point in the relative motion can be

designed analogously to the equation for the point B3:

CD

CC

CD

CC vvv||

3232 +=⊥

Determining velocities

The value of the velocity vC3 is calculated from the formula:

Kinematic analysis by graphical methods - example

CD

CC

CD

CC vvv||

3232 +=⊥

m/s 9222,0115,00192,833 =⋅== CDC lv ωThere is not enough data to graphically solve the equation on the velocity of the point

C2. It requires one more relation, which may be established once the planar motion of

link 2 is regarded as the combination of the translational and rotational motions. Then,

the connection is formed between the velocities of the points of the element:

BC

BC

AB

BC vvv⊥⊥

+= 2222

BC

BC

AB

BC vvv⊥⊥

+= 2222

Determining velocities

The length of the velocity vector vC2 is 137,52 mm, thus, the velocity value is:

Kinematic analysis by graphical methods - example

( ) m/s 3752,101,052,13722 =⋅== vCC vv κ( ) m/s 0204,101,001,1023232 =⋅== vCCCC vv κ( ) m/s 4430,101,030,1442222 =⋅== vBCBC vv κ

Determining velocities

Kinematic analysis by graphical methods - example

Δb3c3d ∼ Δ BCD

Determining accelerations

The order of determining the accelerations is identical to the one for velocities, since it

results from the structure of mechanism and the available data.

The driving link rotates with the constant angular velocity, therefore, the only present

acceleration is the normal one, which, in the point B, equals:

Kinematic analysis by graphical methods - example

( ) 2221 m/s 5806,13086,04 =⋅=== πω ABnBB laaTo determine the acceleration in the point B3, two equations have to be used. In the

first one, the movement of the point B3 is treated as a relative motion:

CD

c

BB

CD

t

BB

n

BB

AB

BBBBB aaaaaaa

⊥=

+++=+= 13||

130

13

||

11313

The normal accelerationnBBa 13

The normal acceleration aB3B1 equals zero, for the path of the point B3 is rectilinear in

relation to the point B1 during the relative motion. Coriolis acceleration is calculated

from the relation:2

13313 m/s 3671,160205,10192,822 =⋅⋅== BBc

BB va ωand the manner of determining the direction is

shown in the acceleration plot. The second equation

for the point B3 results from the rotational motion of

link 3:

BD

t

B

BD

n

BB aaa⊥

+= 3||

33

Determining accelerations

while:

Kinematic analysis by graphical methods - exampleThe normal accelerationn

BBa 13

222

33 m/s 0513,14

21845,0

7518,1 ===BD

Bn

Bl

va

The adopted acceleration graduation scale equals

=mm

1

s

m 5,0

2aκ

CD

c

BB

CD

t

BB

n

BB

AB

BBBBB aaaaaaa

⊥=

+++=+= 13||

130

13

||

11313

BD

t

B

BD

n

BB aaa⊥

+= 3||

33

Determining accelerations

On the basis of the length of vectors on the plot, the following values are calculated:

Kinematic analysis by graphical methods - example

( ) 233 m/s 6525,225,0305,45 =⋅== aBB aa κ( ) 233 m/s 7678,175,05357,35 =⋅== atBtB aa κ

Determining the acceleration of the point C2 requires two equations as well. The first

one describes the relation of the point C2 against C3 during the complex motion:

CD

c

CC

CD

t

CC

n

CC

CD

t

C

CD

n

CCCCC aaaaaaaa

⊥=⊥

++++=+= 32||

320

323

||

33232

where:

2

22

33

s

m 3952,7

115,0

9222,0 ===CD

Cn

Cl

va

2

333

s

m 3557,9115,0

218450

7678,17 ====,

ll

ala CD

BD

t

BCD

t

C ε

232332 s

m 3656,160204,10192,822 =⋅⋅== CC

c

CC va ω

Determining accelerations

Considering the planar motion of link 2 as a sum of translational and rotational motion

leads to obtaining of the second equation:

Kinematic analysis by graphical methods - example

where:BC

t

BC

BC

n

BC

AB

BBCBC aaaaaa⊥

++=+= 22||

22

||

22222

2

22

2222

s

m 5680,11

18,0

443,1 ===BC

BCn

BCl

va

CD

c

CC

CD

t

CC

n

CC

CD

t

C

CD

n

CCCCC aaaaaaaa

⊥=⊥

++++=+= 32||

320

323

||

33232

Determining accelerations

hence, the acceleration of the point C2 equals:

Kinematic analysis by graphical methods - example

The final calculation will be the acceleration

of the point E:

( ) ( )

2

222

2

3

22

22

s

m 2576,707,0

218450

7678,17

07,0

5613,0 =

+

=

+

=+=

,

ll

a

l

vaaa DE

BD

t

B

DE

Et

E

n

EE

( ) 222 m/s 5656,275,01312,55 =⋅== aCC aa κ

Kinematic analysis by graphical methods - example

Δb3c3d ∼ Δ BCD