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Module 1. Introduction to Theory of Machine

Module 2. Planar Mechanism

Module 3. Velocity and Acceleration Analysis

15 March - 21 March

22 March - 28 March

29 March - 4 April

5 April - 11 April

12 April - 18 April

19 April - 25 April

26 April - 2 May

## Lesson 6.

**6.1** **SLIDER CRANK MECHANISM ANALYSIS BY RELATIVE VELOCITY METHOD**

Angular velocity of link OA is given ω and velocity of point A can be given by, V_{a }= ω×OA = V_{AO} in the clockwise direction about O. V_{a} is perpendicular to OA. So we know the direction and magnitude of V_{a}. The velocity of slider B is along OB. The configuration and velocity diagram for this problem is given below.

**Fig. 3.5**

**Step 1.** Take any point **o** and draw vector oa such that** oa**= ω × OA=V_{AO}, perpendicular to OA in some suitable scale as shown in the velocity diagram.

**Step 2.** The velocity of point B with respect to A (V_{BA}) is perpendicular to AB. From **a** draw a vector **ab** perpendicular to the line AB.

**Step 3.** The velocity of slider B is along the line of stroke OB, from **o **draw a line parallel to OB which will intersect vector **ab** at **b**. The vector **ob **represent the velocity of slider B (V_{B}).

The velocity of any point C on the connecting rod can be determined by the help of relation:

**\[\frac{{{\mathbf{ac}}}}{{{\mathbf{ab}}}} = \frac{{{\text{AC}}}}{{{\text{AB}}}}{\text{ or }}{\mathbf{ac}} = {\mathbf{ab}} \times \frac{{{\text{AC}}}}{{{\text{AB}}}}\] **

The point **c **can be located on the velocity diagram. Join **o** with** c. **vector **oc **represents the absolute velocity of point C with respect to O.

**Example 3.1 **In the mechanism as shown below, the crank O_{2}A rotates at 600 r.p.m. in the anticlockwise direction. The length of link O_{2}A= 12 cm and of link O_{1}B= 60 cm. Find

a. Angular velocity of link O_{1}A.

b. Velocity of slider at B.

**Fig. 3.6**

**Solution: **

Given: N=600 r.p.m

\[\omega={\text{ }}\frac{{{\text{2}}\pi{\text{N}}}}{{{\text{6}}0}} = \frac{{{\text{2}}\pi\times {\text{6}}00}}{{{\text{6}}0}} = {\text{2}}0{\text{ }}\pi {\text{ rad}}/{\text{sec}}\]

The velocity of A with respect to O_{2}

\[{{\text{V}}_{{\text{AO2}}}}= \omega\times{{\text{O}}_{\text{2}}}{\text{A}}={\text{2}}0\pi\times \frac{{{\text{12}}}}{{{\text{1}}00}} = {\text{7}}.{\text{53 m}}/{\text{s}}\]

In the configuration diagram let us take a point C on the link O_{1}B. The velocity diagram can be drawn as explain below.

**Step 1. **O_{1 }and O_{2 }are fixed points on the configuration diagram so they are taken as single point **(o _{1,}o_{2})** on the velocity diagram. The velocity of point A with respect to O

_{2}, V

_{AO2}= 7.53 m/s is drawn as vector

**o**in some suitable scale perpendicular to O

_{2}a_{2}A.

**Step 2.** The velocity of point C with respect to A, V_{CA} along the path of slider, O_{1}B. From **a** draw a vector **ac **representing V_{CA }along O_{1}A. This will contain point **c**.

**Step 3.** The velocity of point C with respect to O_{1} is V_{CO1} and it will be perpendicular to O_{1}A.Then from **o _{1}** draw a vector o

_{1}c representing V

_{CO1}this will intersect

**ac**at point

**c**.

**Step 4.** Locate the point **b **corresponding to point B such that

\[\frac{{{{\mathbf{o}}_{\mathbf{1}}}{\mathbf{b}}}}{{{{\mathbf{o}}_{\mathbf{1}}}{\mathbf{c}}}} =\frac{{{{\text{O}}_{\text{1}}}{\text{B}}}}{{{{\text{O}}_{\text{1}}}{\text{C}}}}\]

** o _{1}b= **5.55 m/s by measurement

**o _{1}b=**V

_{BO1}

Angular velocity of link O_{1}A

\[{\omega_{{\text{O1A}}}}={\omega _{{\text{O1B}}}}=\frac{{{{\text{V}}_{{\text{BO1}}}}}}{{{{\text{O}}_{\text{1}}}{\text{B}}}}= \frac{{{\text{5}}.{\text{55}}}}{{0.{\text{6}}0}}\]

= 9.25 rad/sec

(anticlockwise direction)

**6.2 INSTANTANEOUS CENTRE OF ROTATION**

A rigid body undergoing in plane motion, there always exist a point in the plane of motion at which the velocity is zero at that particular instant. This point is called instantaneous centre of rotation or I-centre. It may or may not lie on the body. If the location of this point is determined then we can simplify the velocity analysis. To locate the I-centre, we use the fact that the velocity of a point on a body is always perpendicular to the position vector from I-centre to that point. If the velocity at two points A and B are known, I-centre will lie at the intersection of the perpendiculars to the velocity vectors through A and B.

**Fig. 3.7 I-centre of a body**

If the velocity vectors at A and B are perpendicular to the line AB, I-centre will lie at the intersection of the line AB with the line joining the extremities of the velocity vectors at A and B.If the velocity vectors are equal & parallel, I-centre will lie at infinity and the angular velocity is zero (pure translation).

**6.3 NUMBER OF I-CENTRE**

The number of I-centre ‘N’ in a mechanism is given by

\[{\text{N}} = \frac{{n{\text{ }}(n - 1)}}{{\text{2}}}\]

Where *n* = No. of links

The number of I-centre in a four bar mechanism will be \[\frac{{{\text{4}} \times \left( {{\text{4}} - {\text{1}}} \right)}}{{\text{2}}} = {\text{6}}\] .

**6.4 HOW TO LOCATE I-CENTRE**

There are two methods to locate the I-centre

1. by using rules to locate I-centre by inspection

2. by applying Aronhold-Kennedy Theorem

**Rules to locate I-centre**

- The centre of the pivot is the I-centre for two links in the pivoted joint.

**Fig. 3.8**

- The I-centre lies at infinity perpendicular to the path of motion of the slider.

**Fig. 3.9**

- In pure rolling or in case of no slip between the two links. The I-centre will lie at the point contact between the two links.

**Fig. 3.10**