25
EX 2.1
In a four bar chain ABCD , AD is fixed and is 150 mm
long. The crank AB is 40 mm long and rotate at 120
rpm (cw) while link CD is 80 mm oscillate about D.
BC and AD are of equal length. Find the angular
velocity of link CD when angle BAD is 60o.
Solution:
߱஺஻
ʹߨܰ
ʹߨͳʹͲ
ൌ
ൌ
ൌ ͳʹǤͷ͸ͺ‫݀ܽݎ‬ǤȀ‫ݏ‬
͸Ͳ
͸Ͳ
AB=40 mm= 0.04 m
vB= ߱஺஻ ‫ ܤܣݔ‬ൌ ͳʹǤͷ͸ͺ‫Ͳݔ‬ǤͲͶ ൌ ͲǤͷͲ͵݉Ȁ‫ݏ‬
1. Since the link AD is fixed, therefore points a, and d
are taken as one part in the velocity diagram .
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2. Draw vector ab ༗ AB (to some suitable scale in
this case) the scale is 1 m= 100 mm, so to represent
vB=0.503 m/s the vector should be (5.03 cm).
3. From point b draw bc ༗ CB.
4. From point d draw dc ༗ CD, so bc and dc will
intersect at point c.
5. By measuring dc is = 38.5 mm
ଷ଼Ǥହ
So, vc=
ଵ଴଴
ൌ ͲǤ͵ͺͷ݉Ȁ‫ݏ‬
And
߱஼஽ =
௩೎
஼஽
ൌ
଴Ǥଷ଼ହ
଴Ǥ଴଼
= 4.8 rad/s
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2.3 Slider Crank Mechanism:
1. Draw the space diagram.
Space diagram
2. Velocity diagram:
- Draw vector (ob) perpendicular ༗ OB =vB from
fixed point (o) (known)
- Draw vector (oa) parallel // to piston (A) known
direction.
- Draw vector from a to b from point (b) ab ༗ AB (
with known direction only.
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- The intersect of these to line gives point(a).
Then vA can be measured.
Ex 2.2 The crank and connecting rod of a theoretical steam
engine are 0.5 m and 2 m long respectively. The crank
makes 180 rpm in the cw direction when it has turned 45o
from the inner dead center position . determine the velocity
of the piston and the angular velocity of the connecting rod.
Space diagram
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Solution:
ʹߨܰ ʹߨ‫ͳݔ‬ͺͲ
‫݀ܽݎ‬
߱ை஻ ൌ
ൌ
ൌ ͳͺǤͺͷʹ
͸Ͳ
͸Ͳ
‫ݏ‬
VB=߱ை஻ x OB = 18.852 x 0.5 =9.426 m/s
Scale 1 cm = 2 m/s
Follow the above procedure and when we measure the
vector ob it gives 4.075 cm it mean that
vp= 2 x 4.075 = 8.15 m/s
And the vector bp gives 3.4 cm
vbp = 2 x 3.4 = 6.8 m/s
So
߱௉஻
‫ݒ‬௕௣ ͸Ǥͺ
ൌ
ൌ
ൌ ͵ǤͶ‫݀ܽݎ‬Ȁ‫ݏ‬
ʹ
‫ܲܤ‬
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2.4 Rubbing Velocity at a pin Joints:
The links in mechanism are mostly connected by means
of pin joints. The rubbing velocity is defined as “ The
algebraic sum between the angular velocities of the two
links which are connected by pin joint multiplied by the
radius of the pin”.
As shown below link oA , and oB connected by a pin
joint o .
Links connected by pin joint
Where
߱ଵ ൌ ܽ݊݃‫ܣ݋݈݂݇݊݅݋ݕݐ݅ܿ݋݈݁ݒݎ݈ܽݑ‬
߱ଶ ൌ ܽ݊݃‫ܤ݋݈݂݇݊݅݋ݕݐ݅ܿ݋݈݁ݒݎ݈ܽݑ‬
r = Radius of the pin o
so,
The rubbing velocity at the pin joint o .
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vRub= (߱ଵ െ ߱ଶ ሻ‫݁݉ܽݏ݊݅ݏ݁ݒ݋݉݁ݎܽݏ݈݄݇݊݅݁ݐ݂݅ݎ‬
vRub= (߱ଵ ൅ ߱ଶ ሻ‫݄݁ݐ݊݅ݏ݁ݒ݋݉݁ݎܽݏ݈݄݇݊݅݁ݐ݂݅ݎ‬
opposite direction.
Note: When the pin connects one sliding member and the
other turning member then the angular velocity of the
sliding member is Zero in such case
vRub= ߱‫ݎ‬
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Ex 2.3: For Ex 2.2 find the rubbing velocities at the pins of
the crank shaft, crank, and cross head when the pins
diameters are 50 mm, 60 mm, and 30 mm respectively.
Solution:
2.4 Velocity diagram for a block sliding on a rotating link:
Let ߱ be the angular velocity of the link OB, A is a block
sliding on the link and has absolute velocity va as shown in
figure below where va assumed to be known in magnitude
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and direction Ᾱ is the point on the link coincident with the
block.
‫ݒ‬௔ư = ߱‫ ݀݊ܽݎ‬༗ O‫ܣ‬ሗ
The block A has velocity relative ‫ܣ‬ሗ to and is parallel to
O‫ܣ‬ሗ
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3.Acceleration in Mechanism:
We have discussed the velocity diagram of various points
in the previous chapter . The acceleration will be discuss in
this chapter.
3.1 Acceleration Diagram for a link:
Consider two points A and B on a rigid link as
shown in figure below in which point B moves
with respect to A, with an angular
velocity߱‫݀ܽݎ‬Ȁ‫ ݏ‬, and let ߙ‫݀ܽݎ‬Ȁ‫ ݏ‬ଶ be the
angular acceleration of the link AB.
To Draw the acceleration diagram the following
steps to be follow:
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ሖ ̸ ̸ to AB represent the
1.From point ܾሗ draw vector ܾ‫ݔ‬
radial component of acceleration(ܽ஻௥ ) with known
magnitude and direction.
2.From point x draw vector xܽư ༗ AB to represent the
tangential component of acceleration (ܽ஻௧ ).
ሗ , the vector ܾܽư known acceleration image of
3. Join ܾܽư
link AB or total acceleration of point B i.e (aAB).
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Note that:
1. Tangential component at
ܽ௧ ൌ ‫ݎ‬Ǥ ‫ ן‬and always ༗ r
2. Radial component ar
௥
ଶ
ܽ ൌ ‫ݎ‬Ǥ ߱ ൌ
௩మ
௥
is always ̸ ̸ r
r.. length of link AB
ar is drawn first then at where at ༗ ar
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3.2 Acceleration diagram of a point on a link:
A two point A,B on link AB as shown in the
figure below. Let the acceleration of point A
i.e aA is known in magnitude and direction if
the path of point B is known . to determine the
acceleration of point
B magnitude and
direction the following procedure can be
follows:
1. From any point ‫݋‬ư draw vector ‫݋‬ư ܽư ̸ ̸ to the
direction of point A (i.e aA) to some suitable
scale.
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2. Draw vector ܽư x ̸ ̸ to AB which represent
(ܽ஻௥ )
and
ܽ஻௥
=
మ
௩ಳ
஺஻
ଶ
or ܽ஻௥ ൌ ߱஺஻
‫ܤܣݔ‬
With the same scale.
3. From point x draw vector xܾሗ ༗ to ܽư x
4. From ‫݋‬ư draw line ̸ ̸ to path B
5. Now the vectors xܾሗ and ‫݋‬ư ܾሗ will intersect at
point ܾሗ , and the values of aB and ܽ஻௧ can be
measured to the scale.
6. Join ܽư and ܾሗ the acceleration image of link
AB will represented by vector ܽư ܾሗ.
7. The acceleration of any point (c) on AB
may be obtained by
௔௖ư
ሗሗ
௔௕
=
஺஼
஺஻
֜ ܽܿư ൌaCA and ‫݋‬ư c = ac
8. The value of angular acceleration ‫ ן‬then
can be obtained by ‫= ן‬
௔೟
஺஻
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3.3 Acceleration in the Slider Crank Mechanism:
A slider crank mechanism shown in the figure below.
We know that ‫ݒ‬஻ = ߱஻ை ‫ܱܤݔ‬, and
௥
=
ܽ஻ை
ଶ
߱஻ை
‫ܤܱݔ‬
ൌ
మ
௩ಳೀ
ை஻
The acceleration diagram can be drawn as:
1. Draw vector ‫݋‬ư ܾሗ ӏӏӏ–‘ BO in magnitude of ܽ஻௥ With suitable scale . since point B moves with a
constant ߱ therefore will be no tangential component
acceleration.
2. From ܾሗ draw vector ܾሗx ӏӏӏ–‘ BA
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Where
௥
=
ܽ஺஻
ଶ
߱஺஻
‫ܤܣݔ‬
ൌ
మ
௩ಲಳ
஺஻
Note: VAB obtained from velocity diagram or given.
3.from point x draw vector xܽư ༗ ܾሗx (or AB) which is
௧
represent ܽ஺஻
.
4. Since the point A reciprocates along AO, therefore the
acceleration must be ӏӏӏ–‘velocity.Draw from ‫݋‬ưa vector
‫݋‬ưܽư ӏӏӏ–‘ AO, will intersect the vector xܽư at point ܽư
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௧
can be
NOW: aA … acceleration of piston A, and ܽ஺஻
measured to the scale.
5.Now aAB is represented by ܽư ܾሗ the connecting rod AB
acceleration , which is the sum of the vectors xܽư and ܾሗx.
6. The acceleration of any point on AB such as E may
be obtained as :
௔ư ௘ư
ሗሗ
௔௕
ൌ
஺ா
஺஻
or by measuring ‫݋‬ư ݁ư to the scale.
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7. The angular acceleration of the connecting rod AB
(‫ן‬஺஻ ) can be obtained from:
‫ן‬஺஻ ൌ ೟
௔ಲಳ
஺஻
(c.w ) about B.
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44
45
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3.3 Coriolis Component of Acceleration:
When a point on one link is sliding along another
rotating link, such as in quick return motion mechanism
as shown in the figure below, then the coriolis
component can be calculated as:
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ܽ௖ ൌ ʹ߱‫ݒ‬
Where
߱ ….Angular velocity of OA
‫…ݒ‬..Velocity of slider B
Always ܽ௖ ༗ OA
Direction of coriolis acceleration;
Note: The direction of ߱ and ‫ ݒ‬are given.