Learning with Purpose March 25, 2013 Learning with Purpose March 25, 2013 22.322 Mechanical Design...
-
Upload
ireland-mincher -
Category
Documents
-
view
222 -
download
3
Transcript of Learning with Purpose March 25, 2013 Learning with Purpose March 25, 2013 22.322 Mechanical Design...
Learning with Purpose March 25, 2013Learning with Purpose March 25, 2013
22.322 Mechanical Design II
Spring 2013
Learning with Purpose March 25, 2013
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 systemAnother 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
Lecture 20
Some Review
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.
In Lecture 19, we figured out what
mbRb needed to be to statically
balance the links
Learning with Purpose March 25, 2013
Dynamic balance is sometimes called two-plane balance• Requires that the sum of the forces AND sum of the
moments must be zeroDevices that can be dynamically balanced are all long in the axial direction compared to the radial directionIt’s possible for an object to be statically balanced but not dynamically balanced.
Lecture 20
Some Review
Statically balanced Unbalanced moment
Inertia forces form a couple which rotates with the masses about the shaftRocking couple causes a moment causing the left and right ends of the shaft to lift and drop
Learning with Purpose March 25, 2013
To correct dynamic imbalance requires either adding or removing the right amount of mass at the proper angular locations in two correction planes separated 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.
Lecture 20
Some Review
Learning with Purpose March 25, 2013
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).
Lecture 20
Balancing LinkagesEssentially adding a “dummy” mechanism just to cancel dynamic effects
Certain configurations of multicylinder internal combustion engines do this pistons and cranks of some cylinders cancel inertial effects of others.
Learning with Purpose March 25, 2013
Many methods have been devised to balance linkagesSome 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 compromiseFor 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.
Lecture 20
Balancing Linkages
Learning with Purpose March 25, 2013
Lecture 20
Learning with Purpose March 25, 2013
Lecture 20
Learning with Purpose March 25, 2013
Lecture 20
Learning with Purpose March 25, 2013
Lecture 20
Learning with Purpose March 25, 2013
Lecture 20
Essentially zero (computational round-off errors)
Max 462 lb at 15o
Learning with Purpose March 25, 2013
Lecture 20
Example
m1 = 3.50 kgm2 = 2.64 kgm3 = 8.75 kgR1 = 2.65 m @ 100o
R2 = 5.20 m @ -60o
R3 = 1.25 m @ 30o
l1 = 4 ml2 = 9 ml3 = 11 m
Learning with Purpose March 25, 2013
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 OFind the position angle and mass-radius product required in plane BSolve for x and y forces in plane AFind the position angle and mass-radius product required in plane A
Lecture 20
Example
mARA = 7.482qA = -154.4o
mBRB = 7.993qB = 176.3o