Chapter 6
- Determining the Velocity of a Point on a Link
-
-
- Instantaneous centre method, and 2. Relative velocity method.
-
-
* 1. A rigid link rotates instantaneously relative to another link at the instantaneous centre for -
- the configuration of the mechanism considered.
-
-
2. The two rigid links have no linear velocity relative to each other at the instantaneous
-
centre.
-
Draw A I and BI perpendiculars to the directions v A and v B respectively. Let these lines intersect at I,which is known as instantaneous centre or virtual centre of the link. Now resolving the velocities along A B[fix length]
number of instantaneous centres = n[n-1]/2 n= #links
-
- When the two links are connected by a pin joint (or pivot joint), the instantaneous centre lies on the centre of the pin
- When the two links have a pure rolling contact the instantaneous centre lies on their point of contact, The velocity of any point A on the link 2 relative to fixed link 1 will be perpendicular to I12 A and is proportional to I12 A > When the two links have a sliding contact, the instantaneous centre lies on the common
normal at the point of contact.
The Aronhold Kennedy\u2019s theorem states that if three bodies move relatively to each other, they have three instantaneous centres and lie on a straight line.
Example 6.1. In a pin jointed four bar mecha-nism, as shown in Fig. 6.9, AB = 300 mm, BC = CD = 360 mm, and AD = 600 mm. The angle BAD = 60\u00b0. The crankAB rotates uniformly at 100 r.p.m. Locate all the instantaneous centres and find the angular velocity of the link BC. Solution. Given : Nab = 100 r.p.m ->Wab=10.47 rad/s Since the length of crank A B = 300 mm = 0.3 m, therefore velocity of point B on link A B, Vb = Wab* A B = 10.47* 0.3 = 3.141 m/s c other examples if time allows
* Example 6.2. Locate all the instantaneous centres of the slider crank mechanism as shown
- in Fig. 6.12. The lengths of crank OB and connecting rod AB are 100 mm and 400 mm respectively.
- If the crank rotates clockwise with an angular velocity of 10 rad/s, find: 1. Velocity of the slider A,
- and 2. Angular velocity of the connecting rod AB.
-
Fig. 6.12 -
\u03c9OB = 10 rad/ s; OB = 100 mm = 0.1 m -
Solution. Given : -
We know that linear velocity of the crank OB, -
vOB = v B = \u03c9OB \u00d7 OB = 10 \u00d7 0.1 = 1 m/s - Location of instantaneous centres
-
The instantaneous centres in a slider crank mechanism are located as discussed below: -
1. Since there are four links (i.e. n = 4), therefore the number of instantaneous centres, -
n ( n \u2013 1) 4 (4 \u2013 1) -
N = = =6 -
2 2 -
Bearing block -
Pin -
Slider -
Connecting -
rod -
Crank -
Slider crank mechanism. -
2. For a four link mechanism, the book keeping table may be drawn as discussed in Art. 6.10. -
3. Locate the fixed and permanent instantaneous centres by inspection. These centres are I12, - I23 and I34 as shown in Fig. 6.13. Since the slider (link 4) moves on a straight surface (link 1), there-
- fore the instantaneous centre I14 will be at infinity.
- Note: Since the slider crank mechanism has three turning pairs and one sliding pair, therefore there will be three
- primary (i.e. fixed and permanent) instantaneous centres.
-
129 -
l -
Chapter 6 : Velocity in Mechanisms -
4. Locate the other two remaining neither fixed nor permanent instantaneous centres, by - Aronhold Kennedy\u2019s theorem. This is done by circle diagram as shown in Fig. 6.14. Mark four points
- 1, 2, 3 and 4 (equal to the number of links in a mechanism) on the circle to indicate I12, I23, I34 and I14.
-
Fig. 6.13 Fig. 6.14 -
5. Join 1 to 3 to form two triangles 1 2 3 and 3 4 1 in the circle diagram. The side 1 3, - common to both triangles, is responsible for completing the two triangles. Therefore the centre I13
- will lie on the intersection of I12 I23 and I14 I34, produced if necessary. Thus centre I13 is located. Join
- 1 to 3 by a dotted line and mark number 5 on it.
-
6. Join 2 to 4 by a dotted line to form two triangles 2 3 4 and 1 2 4. The side 2 4, common - to both triangles, is responsible for completing the two triangles. Therefore the centre I24 lies on the
- intersection of I23 I34 and I12 I14. Join 2 to 4 by a dotted line on the circle diagram and mark number 6
- on it. Thus all the six instantaneous centres are located.
-
By measurement, we find that -
I13 A = 460 mm = 0.46 m ; and I13 B = 560 mm = 0.56 m -
- Velocity of the slider A
-
Let vA = Velocity of the slider A . -
vA v -
= B -
We know that -
I13 A I13 B -
I13 A 0.46 -
vA = vB \u00d7 =1\u00d7 = 0.82 m/s Ans. - or
-
0.56 -
I13 B -
- Angular velocity of the connecting rod AB
-
\u03c9AB = Angular velocity of the connecting rod A B. -
Let -
vA v -
= B = \u03c9AB -
We know that -
I13 A I13 B - 130 l Theory of Machines
-
Exhaust -
waste heat -
Engine -
Hydraulic -
rams -
Load -
The above picture shows a digging machine. -
Note : This picture is given as additional information and is not a direct example of the current chapter. -
vB 1 -
\u03c9 AB = = = 1.78 rad/s Ans. -
\u2234 -
I13 B 0.56 - Note: The velocity of the slider A and angular velocity of the connecting rod A B may also be determined as
- follows :
-
From similar triangles I13 I23 I34 and I12 I23 I24, -
I12 I 23 I 23 I 24 -
= ...(i) -
I13 I 23 I 23 I 34 -
I13 I34 I12 I 24 -
= ...(ii) - and I 34 I 23 I 23 I 24
-
\u03c9 \u00d7 OB -
vB -
\u03c9AB = = OB ...(\u2235 vB = \u03c9OB \u00d7 OB) -
We know that I13 B I13 B -
I12 I 23 I I -
= \u03c9OB \u00d7 = \u03c9OB \u00d7 23 24 ...[From equation (i)] ...(iii) -
I13 I 23 I 23 I 34 -
I 23 I 24 -
vA = \u03c9AB \u00d7 I13 A = \u03c9OB \u00d7 \u00d7 I13 I 34 . -
Also ...[From equation (iii)] -
I 23 I34 -
= \u03c9OB \u00d7 I12 I24 = \u03c9OB \u00d7 OD ...[From equation (ii)]