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# 226-Lect 12 - Balancing Balancing We have considered forces...

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1 Week 12 MECH226 1 Balancing Chapter 12 Norton Additional reading in chapters 13 and 14 for engine balancing Phil Commins Week 12 MECH226 2 Balancing We have considered forces applied to links in motion What about link 1? Link 1 was only used to restrain the mechanism and has never entered the calculations. 1 1 Week 12 MECH226 3 Balance Any link that is pure rotation can, theoretically, be balanced Usually all mechanisms are designed to be balanced. When are they not? Both static and dynamic balancing is usually required. Balancing is achieved by placing masses at a desired location Week 12 MECH226 4 Balance Variations in the external loads or the loads due to inertia forces are resisted by the grounding forces. Balancing for external loads – Use energy storage devices, such as flywheel to moderate the resulting affect Balancing for inertia loads

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2 Week 12 MECH226 5 Balance The magnitude of the inertia forces and torques reflect the balance of the mechanism. Often called “shaking forces” Usually undesirable and causes fatigue failure May excite natural mechanical frequencies causing further damage. Week 12 MECH226 6 Balance Balancing of rotors Balancing of reciprocating mechanisms Balancing the general four bar linkage Balancing an inverted pendulum??? Week 12 MECH226 7 Rotors In most cases, balancing requires a static and dynamic analysis. Basic requirement is the same as any system in equilibrium Sum of Forces = zero Sum of Moments = zero In this case, it is the inertia forces and torques/moments to be in equilibrium. Week 12 MECH226 8 Rotors 3 cases will be analysed Unbalanced masses in a single transverse plane Unbalanced masses in a single axial plane General case – masses in various axial and transverse planes
3 Week 12 MECH226 9 Examples Aeroplane propeller Single gear or pulley Bicycle tyre Car tyre – reasonably thick compared to radius. Suited for a dynamic balance Week 12 MECH226 10 Rotors – single transverse plane The inertia force resulting from an eccentric mass at constant angular velocity is • F o = - ma cg = m R ω 2 directed away from the axis Where m is the mass, R is the radial distance of point mass from axis of rotation and ω is the speed. It may be balanced by placing a balance mass directly opposite, have the same “mR” term. Week 12 MECH226 11 Single transverse plane - m 1 R 1 ω 2 - m 2 R 2 ω 2 - m b R b ω 2 = 0 The shaft speed is a common factor and can be factored out.

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