Rotary Polishing Tool 2

7/30/2019 Rotary Polishing Tool 2

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Can I Get a Shiner?MEMS1029 Dr. Ludwick

By: Robert Leslie Cairns


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I. Introduction

Mechanically powered machinery has allowed for human beings to accomplish almost anything

imaginable. Machines can be designed to a wealth of different actions including lifting heavy objects,

hurling projectiles, and cutting grass. As our grasp of the mechanics involved with these machines has

grown we have come to find new applications for instance polishing brass. Though polishing brass maybe a seemingly simple albeit tedious task to do by hand, the complexity of a machine capable of this

action is rather astounding. Although there are many different mechanical means which may

accomplish the task, the following details specifically the design of a sufficient belt and gear system.

II. Functional Requirements

For consideration of this design, a successful rotary polishing machine tool must achieve the

following goals. The device must include a belt drive and gear train. The tool should be driven by a

nominal input power of 1,500 Watts on a shaft rotating at 2,000 rotations per minute. To accomplish a

reasonable brassy shine the polishing fixture must attain a speed of at least 10,000 rotations per minute.

The design should be capable of accomplishing one hundred million (108) cycles with a system reliability

of 95%. A safety factor for the gears of at least three is desired with a minimum gear safety factor of

two. The overall size of the design happens to be unimportant as it will be a stationary unit inside of a

large warehouse and the cost should be preferably $250 dollars or less. Additionally, an efficiency of 80-

95% will be acceptable. If all of these conditions are met the design will be successful and those

decrepit brass badges will be as shiny as new.

III. Design Concepts

There is an almost endless supply of potential belt drives coupled with gear trains. Some factors

to consider at this stage are what overall torque will be required and distance from the initial powersource. Three slightly different designs are depicted in Figures 1-3 below, corresponding to Design A,

Design B, and Design C.

Figure 1 Design A Figure 2 Design B Figure 3 Design C

Flat belt without twist V-belt Flat belt with twist


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For a rotary polishing tool our final torque value can be relatively low, however we do want to

have a high rotational velocity for polishing. Therefore flat belt drive as depicted in Figure 1 and Figure 3

can be considered while the V-belt in Figure 2 can be eliminated as V-belts transmit much higher

torques but at slightly lower efficiency and also limit the distance to the power source. While Design C

has better belt wear characteristics as a result of the uniform wear from the crossover, this could

potentially require additional tensioning devices, or guides for the belt. Design A on the other hand

might be capable of catenaryinduced tensioning to provide the initial tension and maintain operational

tension after elongation of the belt. Based upon these very basic assumptions, Design A (Figure 1) will

be detailed.

IV. Design Analysis

Using the basic conceptual configuration presented in Design A (Figure 1), a set of free body

diagrams can be developed for use in the analysis of the belt drive and gear train. Figure 4 below

illustrates the free body diagram for the belt drive where the larger pulley is mounted on the input

power shaft.

Figure 4 Free body diagram of pulley configuration

The important factors to note in Figure 4 are that the driving pulley is larger than the driven

pulley, the slack side is on the top, and the pulleys will rotate in the same direction. The slack side has

been configured to be on the top so that the dip will contribute to increasing the tension as a result of a

larger contact angle. Since the driving pulley is larger than the driven pulley a higher rotational speed

should be expected on the driven pulley, which is important towards reaching the design goal of 10,000

RPM.

The gear train configuration represented in Figure 5 below illustrates the free body diagram for

the gearing from Design A, the gears have been drawn with space between them for force indications

when in reality they would be in contact. Note that the larger driving gear is attached to the same shaft

as the driven pulley, thus it will have the same rotational speed as that pulley.


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Figure 5 Free body diagram of gear train configuration

Some important factors to note from Figure 5 are that the tool will be mounted on the same

shaft as the pinion thus having the same rotational speed and the pinion is driven by the larger gear thuswill have a higher rotational speed. Another important factor is that the tangential force of the gear will

be the same as that of the pinion but in the opposite direction thus resulting in the pinion rotating in the

opposite direction of the gear.

V. Belt Analysis

This design has many potential variables that can be manipulated to produce different results.

We are going to assume that we have a 6 inch diameter pulley (D) as the driving pulley, a 2.5 inch

diameter driven pulley (d), and a center to center distance (C) of 72 inches. Some additional

assumptions must be made about the service factor (K s) and the design factor (d) before we can

calculate the design power of our tool (Hd), each factor will be assumed to be 1.1. To calculate the

design power the equation below is used, where the nominal power (Hnom) is the given input power of

1500W or 2.0115 horsepower.

Hd= Hnom/Ksd = 2.0115/(1.1*1.1) = 1.662 horsepower


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Knowing that our design power will be 1.662 horsepower and the initial rotational speed of the

driving pulley (nD) of 2000 RPM, we can proceed to calculate the initial torque to be transmitted () with

the equation below.

= 63025Hd/nD = 63025*1.662/2000 = 52.38 lbf in

Based upon the previously assumed and calculated data we can calculate the wrap angle (d) of

the smaller driven pulley where the highest forces will occur, and the belt speed (V) through the use of

the following equations.

d= -2*sin-1[(D-d)/(2C)] = -2*sin-1[(6-2.5)/(2*72)] = 3.093 rad

V = nDD/12 = *2000*6/12 = 3141.6 ft/min

With the belt speed known, the rotational speed of the driven pulley (nd) can easily be found

with the following equation.

nd = 12V/(d) = 12*3141.6/(*2.5) = 4800 RPM

With these calculations and assumed constraints in mind, the belt thickness is restricted by the

specification of a polyamide belt, which we will assume is an A-2c. Based upon this assumption we

immediately know that the velocity correction factor (Cv) will be 1 since it is a polyamide belt. Shigleys

Table 17-2 suggests that for the selected belt, the thickness (t) will be 0.11, a specific weight () of 0.037

lbf/in3, an allowable tension per unit width (Fa) of 60 lbf/in, and a coefficient of friction () of .8. To

calculate the belt width (b), we must first do several calculations in terms of b. The first of which will be

to find the mass per unit length (w) as shown in the following calculation in terms of belt width.

w = 12bt = 12*.037*b*0.11 = 0.04884b lbf/ft

The next step is to calculate the hoop tension due to centrifugal force (Fc), which incorporates

the gravitational constant (g) of 32.17 ft/s2, the calculation is shown below.

Fc = (w/g)(V/60)2 = (.04884b/32.17)(3141.6/60)2 = 4.16b lbf

Now we need to find the largest allowable tension (F1)a which requires the pulley correction

factor (Cp) from Shigleys Table 17-4 which is .73 for the selected belt material. The largest allowable

tension calculation is determined below.

(F1)a = bFaCpCv = b*60*.73*1 = 43.8b lbf

The difference between the largest allowable tension and the slack side tension (F2) is

equivalent to twice the transmitted torque divided by the driven diameter. Thus the slack side tension

can be found as a function of the belt width as shown below.

(F1)a-F2 = 2/d => F2 = (F1)a-(2)/d = 43.8b-(2*52.38)/6 = 43.8b-17.5 lbf

Next, the initial tension (Fi) can be calculated based off of the previous tension calculations as

the following depicts.


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Fi = [(F1)a+F2]/2 -Fc = (43.8b+43.8b-17.5)/2 4.16b = 39.64b 8.75 lbf

Finally, to solve for a belt width we should base it off of the friction development (f) to ensure

slipping does not occur. To do this we use Shigleys equation 17-7 solved for f as shown below and plug

in our previously established values and set f to be equivalent to the belt coefficient of friction. With

some fancy mathematical manipulation we can determine a minimal value for the belt width to transmitthe load without slippage.

f= =(1/)ln{[(F1)a-Fc]/[F2-Fc]} =>

.8 = (1/3.093)ln{[43.8b-4.16b]/[43.8b-17.5-4.16b]} =>

b = .481 in

Since belt width is based upon what is easily manufactured and available, a belt width of 0.5

inches has been selected. With this value we can find the mass per unit length and the actual tension

values by plugging in the belt width into the equations that were derived in terms of belt width, these

calculations are shown below.

w = 12bt = 12*.037*0.5*0.11 = 0.0244 lb/ft

Fc = 4.16b = 4.16*0.5 = 2.08 lbf

(F1)a = 43.8b = 43.8*0.5 = 21.9 lbf

F2 = 43.8b - 17.5 = 43.8*0.5 - 17.5 = 4.4 lbf

Fi = 39.64b - 8.75 = 39.64*0.5 8.75 = 11.1 lbf

To purchase a belt we would also need to know the required belt length (L) but to do that we

need the wrap angle of the driving pulley (D). Both calculations are shown below.

D = +2*sin-1[(D-d)/(2C)] = +2*sin-1[(6-2.5)/(2*72)] = 3.190 rad

L = (4C2-(D-d)2).5 + 0.5(DD+dd) = (4*722-(6-2.5)2 + 0.5(6*3.190+2.5*3.093) = 157.4 in

Flat belts are available in lengths made to order so a 157.4 inch belt should be available. The

last step is to calculate the belt dip (dip) on the slack side of the belt. This final calculation is shown

below.

dip = (C2w)/(96*Fi) = (722*0.0244)/(96*11.1) = 0.119 in

This dip should be acceptable as it is not great enough to conflict with the lower side of the belt

and will allow for the belt to stretch increasing the dip slightly but preserving the tension.

VI. Gear Analysis

The speed of the driving gear (nG) is equivalent to the speed of the driven pulley which is 4,800

RPM. To get to the rotational speed of at least 10,000 RPM given the driven pulley speed, we can

deduce that a gear ratio of at least 2.083:1, for simplicity and the fact that the desired rotational speed

can be greater than 10,000 RPM, a ratio of 3:1 will be used. This means that the rotational speed of the

pinion or driven gear (np) will be 14,400 RPM. Based upon gears available from Ondrivesus.com a


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common pressure angle () is 20 and their AGMA quality standard (Qv) is 10. These gears are made

from 303 Stainless steel with a Brinell hardness of 160, and have been assumed to have a Grade of 1

although not specified by the manufacturer. Through the use of the American Gear Manufacturers

Association (AGMA) equations, we can calculate the bending stresses and safety factors for bending, as

well as the contact stresses and safety factors relative to these stresses. Given the transmitted power

based off of the previous belt calculations, we know that our input power is 1.662 Hp at a rotational

speed of 4,800 RPM. Since the driving gear is mounted on the same shaft as the driven pulley, they will

share the same rotational speed thus the speed of the driving gear (nG) is also 4,800 RPM. To begin the

calculations we must select some value of teeth to examine, a pinion tooth count (N p) of 20 was selected

based on available gear sizes. Therefore the gear tooth count (NG) based off of the desired gear ratio.

The manufacturers indicate that for 16-pitch gears with 20 pressure angles, a 60 tooth gear will have a

diameter (dG) of 3.75 inches or 95.25mm, and a 20 tooth pinion will have a diameter (d p) of 1.25 inches

or 31.75mm. These dimensions allow for a module (m) to be calculated, which turns out to be .0625

inches or 1.59mm. All of the remaining AGMA calculations will be done with SI units and equations.

The AGMA calculations are shown in Figure 6, and figure 7 shows the final results withimportant values drawn from data contained in the tables and figures of Shigleys chapter 14. Some of

the values that are not derived directly from formulas include the Lewis form factor for the pinion (Y p)

and gear (YG), the face width (F) which is available from the gear manufacturer and found to be

adequate at 12.7mm. The load distribution factor (KH) required several bits of information from the

manufacturer including teeth crown, pinion mounting, and assembly adjustment, each yielded a

respective value of 1. Based upon the idea that the final product will have the gearing enclosed,

commercial enclosed unit conditions were used in the calculations. The rim thickness factor (KB) was

found to be 1 based off of the tooth height and rim thickness provided by the manufacturer. The

dynamic factor (Kv) was found to be 1.27 based upon the AGMA quality of the gear and the velocity of

the pinion since it is the fastest spinning gearing component. Since the overall system desires a 95%reliability, an reliability of 0.983 was used to calculate the reliability factor (Yz) which was found to be

0.845. It is assumed that the overload factor (Ko) will be 1, and the hardness ratio (CH) will also be 1

since both gears are made of the same material.


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Figure 6 AGMA Calculations


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Figure 7 AGMA Results

Based upon the AGMA calculations we find that our bending stresses in the pinion (p) are 11.46

MPa with a safety factor of 17.5 and the bending stress in the gear (G) is 9.23 MPa with a safety factor

of 21.7. The contact stress in the pinion (Cp) are 239.3 MPa with a safety factor of 2.6 and the contact

stress in the gear (CG) is also 239.3 MPa but with a safety factor of 2.7. These contact safety factors are

in between the design minimums and preferred factors of safety and thus justified. Although many

assumptions were made in this selection of AGMA calculations, these fairly large factors of safety should

compensate for at least mildly incorrect assumptions.

VII. Efficiency

The overall efficiency (e) of the machine can be calculated by taking the final output power (Hf)

and dividing by the initial power provided. The final output power is determined from the speed anddiameter of the pinion gear and the transmitted load (W t) from the gear to the pinion, the transmitted

load calculation is shown in Figure 6 and is 51.8N. Both calculations are shown below.

Hf= dpnpWt/60000 = *31.75*14400*51.8/60000 = 1239 W

e = Hf/Hnom = 1239/1500 = .826 * 100% = 82.6%

The efficiency of the design turns out to be 82.6% which is fairly decent and within the range set

by the functional requirements.

VIII. Visualization

Based upon the design and calculated dimensions and lengths, Figure 8 illustrates the overall

configuration. Although not depicted on the shafts or shaft configuration, bearings would be required

to support the shaft and allow free spinning or the various components.


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Figure 8 Design Configuration

Figure 9 illustrates the driving pulley on the power input shaft, notice the key that is used totransfer the shaft rotation into the pulley and thus through the system.


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Figure 9 Large Pulley on Input Shaft

Figure 10, below shows how the tool side of the machine will work. The small pulley has a

similar key and keyway to transfer the motion to the shaft, and the driving and driven gears each have

set screws that allow the motion to be transferred through the gears to and from each shaft. The final

element on the tool shaft is the polishing tool to shine up that brass.

Figure 10 Tool Side Pulley and Gear Train


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The shafts have been specially designed to allow for shoulders for positioning, along with

keyways and keys to transfer the torque from the shaft to the pulley. There are snap ring grooves

located to hold the pulleys in place on each shaft. For the gears, which are manufactured with set

screws, additional slight holes were added for the set screws to engage in to help transfer the torque

from the shaft to the gears without any slipping. Bearings are not illustrated on the shafts nor are any

shaft features associated with bearings such as shoulders or additional snap ring grooves. Additionally,

considerations for filleting sharp corners and shoulders should be considered as well. Figure 11 below

highlights some of these key design features.

Figure 11 Shaft Features


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The pulleys are very similar in their design; each has a keyway cut into the bore diameter to be

used with a key to transfer the torque from the shaft. Each pulley has a crowned surface, sloped sides,

and raised edges to assist in the tracking of the belt. Figure 12 shows the key features present on each

of the pulleys.

Figure 12 Pulley Features

IX. Cost

The bill of materials for this project is rather extensive, especially when considering components

such as shafts, keys, snap rings, bearings and mounting components. However, strictly for the belt,

gears, and associated pulley, Figure 13 depicts the bill of materials for the project.

Figure 13 Bill of Selected Materials

The total cost of the basic components of this design is approximately $283.21, which is slightlyout of the ideal provided by the functional requirements but still reasonable.

X. Conclusion

Based upon the technical calculations for this design, the functional requirements seem to be

appropriately accounted for. The design has been successfully powered using a shaft that is inputting

1,500 Watts at 2,000 RPM. A fairly large margin over the design requirement of 10,000 RPM was


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achieved, with this device rotating at about 14,400 RPM, 44% faster than required and perfectly capable

of a nice shiny finish. The design was based off the use of a cycle count of one hundred million cycles

and a system reliability of 95%, yielding safety factors of 2.61 and 2.67 for contact stresses. These safety

factors are in excess of the minimum factors and sufficient per the requirements. The cost of this design

is slightly more than the functional requirements desire, but within about a 10% margin, although the

cost excludes bearings, shafts, and shaft components which would likely cost well over double the

predicted cost at the moment. However, the efficiency of the design has been found to be about 83%

which is comfortably into the range desired by the functional requirements. Based upon these findings,

this design should be reasonably successful in meeting the demands set forth and should be quite

capable of providing a nice brass shiner.

XI. Appendices

Documents

Rotary Polishing Tool 2