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Chapter 6

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sameerg edited this page May 26, 2011 · 2 revisions
  • Determining the Velocity of a Point on a Link
      1. 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. cir

  •       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

    1. 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

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