Kinematic Analysis of Mechanisms(2)

Properties of the Instantaneous Centre

• A rigid link rotates instantaneously relative to another link at the instantaneous centre for configuration of that mechanism considered.
• The two rigid links have no linear velocity relative to each other at instantaneous centre. At this point i.e. instantaneous centre the two rigid links have the same linear velocity relative to the third rigid link.

Number of Instantaneous Centres in a Mechanism

The number of instantaneous centres in constrained kinematic chain is equal to number of possible combinations of two links. The number of pairs of links or number of instantaneous centres is the number of combinations of n links taken two at a time. Mathematically, number of instantaneous centres.

\[N = \frac{{n(n - 1)}}{2},\;\;\;\;\;\;\;\;\;Where\;\;\;n = number\;\;\;of\;\;\;link\]

Types of Instantaneous Centres

• Fixed instantaneous centres,
• Permanent instantaneous centres,
• Neither fixed nor permanent
  instantaneous centres.

Fig:Types of instantaneous centres

The instantaneous centres I₁₂ and I₁₄ are called fixed instantaneous centres as they remain in the same place for all type of configurations of the mechanism. The instantaneous centres I₂₃ and I₃₄ are permanent instantaneous centres as they move when the mechanism moves but the joints are of permanent nature. The instantaneous centres type I₁₃ and I₂₄ are the neither fixed nor permanent instantaneous centres as they vary with the configuration of the mechanism. The instantaneous centre of two links such as link 1 and link 2 is usually represented by I₁₂ and so on other.

Location of the Instantaneous Centres

Fig:Location of instantaneous centres

1. When two links are connected by a pin joint or pivot joint then the instantaneous centre lies on the centre of the pin as shown in fig (a). Such an instantaneous centre is of permanent nature but if one of that links is fixed then the instantaneous centre will be of fixed type.
2. When two links have a pure rolling contact mechanism i.e. link 2 rolls without slipping upon fixed link 1 which may be straight or curved the instantaneous centre lies on their point of contact as shown in fig (b). The velocity of any point A on the link 2 relatives to fixed link 1 will be normal to I₁₂ A and is proportional to I₁₂ A.
3. When the two links have a sliding contact the instantaneous centre lies on common normal at the point of contact. We shall consider the following three cases.

(a) When the link 2 slider moves on fixed link 1 which having straight surface which is shown in fig (c). The instantaneous centre lies at infinity and each point on slider have the same velocity.

(b) When the link 2 slidermoves on fixed link 1 having curved surface which is shown in fig (d). The instantaneous centre lies on the centre of curvature of a curvilinear path in the configuration at that instant.

(c) When link 2 slider moves on fixed link 1 having constant radius of curvature which is shown in fig (e). The instantaneous centre lies at centre of curvature i.e. the centre of the circle for all configuration of the links.

Kennedy’s theorem

The Kennedy’s theorem states that if the three bodies move relatively to each other. They have three instantaneous centres lie on a straight line.

Fig:Aronhold Kennedy’s theorem

Consider three kinematic links A, B and C having the relative plane motion. The number of instantaneous centres (N) is given by \[N = \frac{{n(n - 1)}}{2},\;\;\;\;\;\;\;\;\;Where\;\;\;n = number\;\;\;of\;\;\;link\]

The two instantaneous centres at pin joints of B with A and C with A i.e. I_ab and I_ac are permanent instantaneous centres. According to Aronhold Kennedy’s theorem, the third instantaneous centre I_bc must lie on the line joining I_ab and I_ac. In order to prove this let us consider that the instantaneous centre I_bc lies outside the line joining I_ab and I_ac which is shown in fig. The point I_bc belongs to both links B and C. Let us consider the point I_bc on link B. Its velocity v_BC must be perpendicular to line joining I_ab and I_bc. Now consider the point I_bc on link C. Its velocity v_BC must be perpendicular to line joining I_ac and I_bc.

Relative Motion Velocity Analysis; Velocity Polygons; Graphical or Vector algebra solutions

Relative Velocity of Two Bodies Moving in Straight Lines

The application of vectors for relative velocity of two bodies moving along parallel lines and inclined lines, as shown in fig (a) and (a) respectively.

Consider two bodies A and B moving along parallel lines in the same direction with absolute velocities v_A and v_B such that v_A > v_B which as shown in fig (a) . The relative velocity of A with respect to B v_AB = Vector difference of v_A and v_B = v_A −v_B

From fig (b), the relative velocity of A with respect to B i.e. v_AB may be written in the vector form

\[\overline {ba} = \overline {oa} - \overline {ob} \]

Similarly

The relative velocity of B with respect to A,

v_BA = Vector difference of v_B and v_A = v_B – v_A ...(ii)

\[\overline {ab} = \overline {ob} - \overline {oa} \]

Fig: Relative velocity of two bodies moving along parallel lines.

Now consider the body B moving in an inclined direction as shown in fig(a) . The relative velocity of A with respect to B may be obtained by law of parallelogram of velocities or triangle law of velocities. Take any fixed point o and draw vector oa to represent the v_A in magnitude and direction to some suitable scale. Similarly, draw vector ob to represent v_B in magnitude and direction to same scale. Then the vector ba represents the relative velocity of A with respect to B as shown in fig (b). In the similar way as discussed above, the relative velocity of A with respect to B,

v_AB = Vector difference of v_A and v_B = v_A – v_B

\[\overline {ba} = \overline {oa} - \overline {ob} \]

Fig: Relative velocity of two bodies moving along inclined lines

Similarly, the relative velocity of B with respect to A i.e.

V_BA = Vector difference of v_B and v_A = v_B – v_A

\[\overline {ab} = \overline {ob} - \overline {oa} \]

So that the relative velocity of point A with respect to B (v_AB) and the relative velocity of point B with respect A (v_BA) are equal in magnitude but opposite in direction, i.e.

v_AB = −v_BA and \[\overline {ba} = - \overline {ab} \]

It should be known that to find v_AB, start from point b towards a and for v_BA, start from point a towards b.

References:
1. H.H. Mabie and C. F. Reinholtz, “Mechanism and Dynamics of Machinery”, Wiley.
2. J.S. Rao & R.V. Dukkipati Mechanisms and Machine Theory, New Age International (P) Limited..
3. J.E. Shigley and J.J. Uicker, Jr., “ Theory of Machines and Mechanisms”, McGraw Hill.
4. B. Paul, “Kinematics and Dynamics of Planar Machinery”, Prentice Hall.
5. C. E. Wilson, J.P. Sadler and W.J. Michels, “Kinematics and Dynamics of Machinery”, Harper Row.