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22.322 Mechanical Design II

22.322 Mechanical Design II. Spring 2013. Lecture 20. Some Review. The requirement for static balance is that the sum of all forces on the moving system must be zero Unbalanced forces of concern are due to the accelerations of masses in the system

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22.322 Mechanical Design II

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  1. 22.322 Mechanical Design II Spring 2013

  2. Lecture 20 Some Review • The requirement for static balance is that the sum of all forces on the moving system must be zero • Unbalanced forces of concern are due to the accelerations of masses in the system • Another name for static balance is single-plane balance • The masses generating the inertia forces are in, or nearly in, the same plane • Essentially a 2-D problem • Devices that can be statically balanced are all short in the axial direction compared to the radial direction • Can be considered to exist in a single plane In Lecture 19, we figured out what mbRb needed to be to statically balance the links Note that for balancing, it does not matter what external forces are acting on the system  the only forces acting on this system are inertia forces External forces cannot be balanced by making any changes to the system’s internal geometry For balancing, it does not matter how fast the system is rotating, only that it is rotating.

  3. Lecture 20 Some Review Inertia forces form a couple which rotates with the masses about the shaft Rocking couple causes a moment causing the left and right ends of the shaft to lift and drop • Dynamic balance is sometimes called two-plane balance • Requires that the sum of the forces AND sum of the moments must be zero • Devices that can be dynamically balanced are all long in the axial direction compared to the radial direction • It’s possible for an object to be statically balanced but not dynamically balanced. Statically balanced Unbalanced moment

  4. Lecture 20 Some Review • To correct dynamic imbalance requires either adding or removing the right amount of mass at the proper angular locations in two correction planesseparated by some distance along the shaft. • Creates the necessary counterforces to statically balance the system and also provide a countercouple to cancel the unbalanced moment • Automobile tire  the two correction planes are the inner and outer edges of the wheel rim • Correction weights are added at the proper locations in each of these correction planes based on a measurement of the dynamic forces generated by the unbalanced, spinning wheel.

  5. Lecture 20 Balancing Linkages Essentially adding a “dummy” mechanism just to cancel dynamic effects • Complete balance of any mechanism can be obtained by creating a second “mirror image” mechanism connected to it so as to cancel all dynamic forces and moments. • This approach is expensive and only justified if the added mechanism serves some purpose (increasing power). Certain configurations of multicylinder internal combustion engines do this  pistons and cranks of some cylinders cancel inertial effects of others.

  6. Lecture 20 Balancing Linkages • Many methods have been devised to balance linkages • Some achieve a complete balance of one dynamic factor (shaking force), at the expense of other factors (shaking moment) • Shaking force – net unbalanced force acting on mechanism • Shaking moment – net unbalanced moment acting on mechanism • Others seek an arrangement that minimizes (but does not zero) shaking forces, moments, and torques for a best compromise • For a four bar linkage, the rotating links (crank and rocker) can be individually statically balanced using the balancing methods we described in lectures 18 and 19. • Note that the process of statically balancing a rotating link, in effect, forces its mass center (CG) to be at its fixed pivot and thus stationary. • But the coupler (which can have complex motion) has no fixed pivot and thus its mass center is, in general, always in motion.

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  11. Lecture 20 Max 462 lb at 15o Essentially zero (computational round-off errors)

  12. Lecture 20 Example m1 = 3.50 kg m2 = 2.64 kg m3 = 8.75 kg R1 = 2.65 m @ 100o R2 = 5.20 m @ -60o R3 = 1.25 m @ 30o l1= 4 m l2 = 9 m l3 = 11 m

  13. Lecture 20 Example mARA = 7.482 qA = -154.4o mBRB = 7.993 qB = 176.3o • For the instantaneous “freeze-frame” position shown, write down the x and y components of the position vectors: • R1x = r1(cos(q1)) --> do the same for R2 and R3 • R1y= r1(sin(q1)) --> do the same for R2 and R3 • The shaking forces and moments can be calculated using the angular velocity (100 rpm = 10.47 rad/s) • Sum the moments about point O • Find the position angle and mass-radius product required in plane B • Solve for x and y forces in plane A • Find the position angle and mass-radius product required in plane A

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