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
Top Related