Download Drive Mechanisms

ENES100_0702
Prof. R. J. Phaneuf
Fall 2002
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Drive Mechanisms
What comes between the source of power, in this case the hands or feet of the operator,
and the moving parts of the machine, in this case a water pump, is the drive mechanism.
In designing the drive mechanism there are a number of considerations:
1. The nature of the motion of the pump: reciprocating vs. rotary
2. The required vs. available forces and rotational rates
3. Assuring adequate strength of the various connecting members and the support
structure.
4. Alignment of the various parts of the drive with adequate precision.
5. Minimizing power losses to work done against frictional and other dissipative
forces.
6. Simplicity. Simpler is usually better. Try to use the minimum number of parts
possible to accomplish all of the above so as to minimize the number of potential
failure mechanisms.
The simplest way of transmitting power to the shaft of a reciprocating pump is merely to
add a handle. Providing the force required is within the range that the operator can
supply repetitively over an extended period this might be a viable option, since it has the
virtue of simplicity, both in manufacture, and in operation.
The next level of sophistication beyond this is to add a lever to the drive mechanism
which provides a means of force amplification. A lever is a rod or bar which pivots about
a fixed point called the fulcrum. (If the fulcrum is between the load applied force, it is a
1st class lever. If the fulcrum is at one end and the load ia between the fulcrum and
applied force it is a 2nd class lever. If the applied force is between the load and fulcrum it
is a 3rd class lever.) The ratio of the applied force to the load force for the case of
constant angular velocity about the fulcrum is simply given by equating the torques
associated with each. This follows from Newton’s second law for angular motion for this
case:


  I  0
(1)
Recall that the torque associated with an applied force is simply given by the cross
product of the displacement from the rotation axis to the point of application, times the
force:



 r F
(2)
and that the magnitude is simply
  rF sin  ,
(3)
where  is the angle between the displacement and the force. In the simplest case, for
=90º, this gives a torque of magnitude:
  rF
(4)
If the two vectors are not at right angles, then the component of the force perpendicular to
the displacement should be used in this last equation. Now equating the magnitudes of
the torque associated with the load and the applied force gives the simple relation, for
uniform angular velocity:
rload Fload  rappl Fappl
(5)
which can be solved for the ratio of the forces:
MA  Fappl / Fload  rappl / rload
(6)
This ratio is referred to as the mechanical advantage.
For a rotary drive, it is also possible to use a simple handle, like that on a screw driver.
Again, it is often advantageous to add a force amplifier, in this case called a crank. As
for the lever, the ratio of the applied force to the load force for the case of uniform
angular motion is simply given by the inverse ratio of the displacements, where the
displacement for the shaft is simply its radius.
Step up/step down transmissions
More complex power transmissions are also possible for rotary motion to allow an
amplification or reduction in the angular velocity from that provided by the operator.
The most basic types fall into three classes: 1) Gear trains, 2) belt and pulley assemblies,
3) chains and sprockets.
The analysis of the angular velocity ratio is similar for the three cases. Consider first the
case of a chain and two sprockets of different radii (taken to be measured to a radial point
midway between the peak of a tooth, and valley between successive teeth), r1 and r2. By
necessity, the angular spacing between teeth times the radius of the sprocket must be
equal to the spacing between links in the chain, . Thus if the number of teeth in the
sprockets are N1 and N2,
N1  2r1
(7)
N 2   2r2
(8)
  2r1 / N1  2r2 / N2
(9)
Assuming the chain neither stretches nor slips, the magnitude of the velocity of a point at
the outer circumference of each sprocket is the same as that of the chain, vchain. Now
recall that the angular velocity is defined as the cross product of the displacement from
the rotation axis to the point at which the velocity is measured, times the velocity:


 
r v
r2
(10)
v sin 
,
r
(11)
and that the magnitude is simply

in this case the velocity is tangential, so that =90º, so that the angular velocity has
magnitude

v
r
(12)
or
v  r
(13)
In this case:
vchain  1r1  2 r2
So that
1 / 2  r2 / r1 ,
(14)
i.e. the ratio of angular velocities goes inversely with the ratio of radii. From equation (9)
and (14) it is easy to show that, in addition:
1 / 2  N1 / N2
(15)
meaning that the ratio of angular velocities is also the reciprocal of the ratio of the
number of teeth.
We can derive one more useful relationship for the case where the loss of power to
frictional forces is zero (this is nearly true for a well-oiled chain if it does not slip). In
this case the power delivered by the first sprocket, P1, is equal to the power transferred to
the second sprocket, P2.
P1=P2
(16)
Recall the relationship between power and torque is analogous to that between power and
velocity:
 
P  
(17)
and, taking the torque to be parallel to the angular velocity for each of the sprockets, the
dot products both yield cos =1:
P1  P2  11   22
(18)
 1 /  2  2 / 1
(19)
so that
and combining equation (19) with equation (14) yields
1 /  2  2 / 1  r1 / r2
(20)
so that the ratio of the torques is equal to the ratio of radii. Note that if the sprockets do
not slip about their shafts, the torque at the shaft is equal to the torque at the radius where
the chain engages (see the discussion of the crank, above)
Alignment
You first need to decide how you will position the operator: standing or sitting, perhaps
with his/her back supported. The pump should be fixed in an orientation compatible with
the position of the operator.
The pump shaft, whether reciprocating or rotary must also be located and aligned,
generally though the use of bearings to constrain the motion to be of the desired type, i.e.
without unwanted motion in other directions.
The alignment of the pump and shaft support bearings is important in minimizing
unwanted frictional forces, or in the case of extreme misalignments, forces or torques
which can damage the parts of the pump or transmission. Generally, the Pump location
and orientation should be adjustable.
If you decide to include a mechanism to increase or decrease the angular velocity, the
various shafts must also be positioned by bearings. Some adjustability of the relative
position must be included to assure adequate tension in the chain or belt, or to assure
adequate radial load force for a gear train to prevent slipping, which wastes power, and
can even bring the system to a halt.
Friction
As we’ve discussed earlier, friction dissipates power in the form of heat. It originates
from the formation of bonds between points of contact between moving surfaces.
Friction is taken to be proportional to the force normal to the surface of contact, N, with a
constant of proportionality which depends on the materials in contact.
f  N ,
(21)
where f is the frictional force, N is the normal force, and  is the coefficient of friction.
Frictional power losses can be decreased by choosing material combinations with a low
coefficient of friction, and by lubricating joints appropriately. Note that not all lubricants
work well for all materials. Frictional forces can also be reduced by tuning the normal
forces to allow adequate constraint, without making the frictional forces too large.
Generally, it is better to have dissimilar materials in contact to reduce their
tendency to bond together.
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