Design of Gears

SHARMA Prnav MCG 4322/5170

GUICHARD Cyril

DESIGN REPORT

ACKNOWLEDGEMENT

We are thankful to Dr. Atef Fahim for his kind help and guidance without which I could not have completed this project.We are also thankful to our T.A, Mr. Albert Mazzawi for his help.

ABSTRACT

We were asked to design an electric powered gear system that can add to an existing manual powered wheel chair and thus give an option to the user of motor powered wheel chair.

TABLE OF CONTENTS

INTRODUCTIONPage 3

FUNCTIONNAL ANALYSISPage 4

1./ Functional analysisPage 4

2./ Requirements concerning the systemPage 4

OUR CONCEPTSPage 6

1./ Concept #1Page 6

2./ Concept #2Page 6

3./ Concept #3Page 6

4./ Other – MiscellaneousPage 8

5./ Decision analysis

WATTAGE REQUIRED FROM THE MOTORPage 11

DESIGN OF GEAR BOXpage 12

1./ Selection of type of gearspage 12

2./ Selection of gear ratiospage 13

3./ Design of gearspage 14

4./ Design of gears to prevent failurespage 17

(a)./ Bending strengthpage 18

(b)./ Pitting strengthpage 19

DESIGN OF SHAFTpage 20

(1)./ Solid shaft in bending and Torsionpage 22

(2)./ Deflection of the shaftpage 24

(3)./ Torsional deflection of the shaftpage 25

(4)./ Selection of key for the shaftpage 25

DESIGNING PARTS FOR EXPECTED LIFEpage 26

CONCLUSIONpage 27

LITERATURE SEARCHPage 28

INTRODUCTION

We often see handicapped persons waiting to get some breath of air before moving her manual wheelchair again.

Some of them decide to afford an electric one. But it becomes a difficult situation here. Powered wheelchairs are very expensive – it costs around $1500 to $5500 to buy an electric powered wheelchair. Thus it is out of reach of many people. A manual powered wheelchair comes around $85 to $300 but has very restricted use.

Hence are we required, as CAD/CAM project, to design an electric propulsion package adaptable on manual powered wheelchairs, as a new alternative to help handicapped persons to feel more free.

FUNCTIONNAL ANALYSIS

1./ Functionnal analysis

We can write the following diagram :

TThis diagram represents our system in its environment. We can determine two main functions it as to fill (red) :

MF1 : Enable the user to drive his wheelchair electrically.
MF2 : Move the wheelchair on different types of roads.

We can sort three complementary functions (blue) :

CF1 : Be universally adaptable.
CF2 : Guarantee a range large enough to the wheelchair.
CF3 : Use standard batteries as power supply.

2./ Requirements concerning the system

We had to seek for what such a system needs to be. We found some specifications that our assembly will have to follow :

F1 : It should weight light.
F2 : It should be compact.
F3 : It should be universally adaptable.
F4 : It should have a large range.
F5 : It should be reliable.
F6 : It should be easily fixable and removable.
F7 : It should cost as less as possible.

Moreover, our subject imposes us to create a two-speed system.

We had to class these function in order to weight them. We concluded this with the following table :

F2 / F3 / F4 / F5 / F6 / F7
F1 / F2 / F3 / F4 / F5 / F1 / F7
F2 / F3 / F4 / F5 / F6 / F7
F3 / F3 / F5 / F3 / F7
F4 / F5 / F4 / F7
F5 / F5 / F5
F6 / F7

So we weighted our functions using their representative percentages:

% / Weight
F1 / 4.76 / 5
F2 / 4.76 / 5
F3 / 19.05 / 20
F4 / 14.29 / 15
F5 / 28.57 / 30
F6 / 4.76 / 5
F7 / 23.81 / 25

OUR CONCEPTS

1./ Concept #1

The first option that we considered uses a mechanism similar to these used in automotives.

It consists of an electrically driven motor coupled to a 2 speed gearbox. This gearbox is one of the most common system used. It is composed of a main shaft linked to the motor (we call it motor shaft). Two gears are connected to this shaft, indenting on two other gears. We use a compound gear train in order to have good gear ratios, like in cars. The last shaft is connected to a differential which is linked to the wheels. The gearbox is manual.

2./ Concept #2

It is almost the same as the first. We just changed the gearbox.

In this concept, we tried to have a type of automatic gearbox, using a centrifugal rod. We use the speed of rotation of the exiting shaft to change automatically the gear ratio thanks to the centrifugal rod which drives the gear selector on the motor shaft.

3./ Concept #3

It uses a basic cycling system of pinions. The main advantages are the large amount of gear ratios that are available, every parts are standard and cheep. The main drawback is that this system can’t be disengaged if the user want to use his wheelchair manually. Moreover, such a system, using a chain, is not really efficient because of the chain itself.

4./ Other – Miscenalleous

We considered designing a fourth concept, using a semi-automatic or an automatic gearbox. Since we didn’t find enough documentation about this kind of gearbox, we got rid of this idea, which, moreover, is very complicated.

5./ Decision analysis

We used the functions studied before, and we weighted them for each concept.

Concept #1 / Concept #2 / Concept #3
Criteria / Weight / Score / Weighted score / Score / Weighted score / Score / Weighted score
F1 / 5 / 30 / 150 / 25 / 125 / 45 / 225
F2 / 5 / 40 / 200 / 30 / 150 / 30 / 150
F3 / 20 / 45 / 900 / 35 / 700 / 20 / 400
F4 / 15 / 55 / 825 / 35 / 525 / 10 / 150
F5 / 30 / 40 / 1200 / 35 / 1050 / 25 / 750
F6 / 5 / 30 / 150 / 30 / 150 / 40 / 200
F7 / 25 / 30 / 750 / 20 / 500 / 50 / 1250

Total 4175 3200 3125

So , our final winning concept is concept no. 1.It is all teh characteristics we need.Though It may be not very light and easy to fix and remove but it has very good reliability and long life.Also it is universally adaptable.

WATTAGE REQUIRED FROM THE MOTOR:

Prior to our calculations We make the following assumptions:

(1)We assume the efficiency of the system to be 85%.

(2)The weight of the wheel chair including batteries be 75 Kg.

(3)Let us consider the weight of the person sitting on the chair to be 150 kg.

(4)We assume the acceleration of the chair to constant and about 3m/s2.

(5)We design our motor for 10 to 5 kmph speed.

Now the total weight that the motor has to drive be

75 + 150 = 225 kg.

Now the energy ( E ) required to move this chair can be derived as follows,

E = mv2/2,

Here, E is energy of motor.

M is total mass of wheel chair

V is the velocity of the chair

i.e, E= 225. 3.3/2 = 1012.5 kgm2/s2,

Now lets assume it achieves this speed in 5 seconds.

Then,

E = 1012.5/5 =202.5 watt.

Now we have assumed the system to have 85% efficiency so the power that the motor should be in order of

E = 202.5 .100/85 = 238.23 watts.

Thus, the wattage of the motor should be 238.23 watts or 0.4 H.P.

DESIGN OF GEARS BOX

1./ Selection of type of gears.

The most common pressure angles used for spur gears are 14.5,20,25.In general ,the 14.5 pressure angle is not used for new designs( and has, in fact been withdrawn as an AGMA standard tooth form); However it is still used for special designs and for some replacement gears.Lower pressure angles have the advantage of smoother and quiter tooth action because of large profile contact ratio.In addition,lower loads are imposed on the support bearings because of decreased radial load component;however,the tangential load components remains unchanged with pressure angle.The problem of undercutting associated with small numbers of pinion teeth is more severe with lower pressure angles.Lower pressure angles gears also have lower bending strength and surface durability ratings and operate with higher sliding velocities( which contributes to their relatively poor scoring and wear performance characteristics) than their higher pressure angle counterparts.

Higher pressure angles have the advantage of better load-carrying capacity,with respect to both strength and durability,and lower sliding velocities( thus better scoring and wear performance characteristics).In some cases,very high pressure angles,28,30 and in few cases as highas 45 are employed.

We selected spur gear for for gear box.

2./ SELECTION OF GEAR REDUCTION RATIO

(1). In the first step of design of gears we try to determine the desired reduction ratio and the input speed by the motor.We shall denote this speed ( say p).We also determine the main function of the gear, i.e, to transmit motion or power.

In our design cost and adaptibility were very important factor’s of consideration so we have selected spur gears.As they are cheaper and are most widely used gears.

After selecting the type of gear to be used we select the desired gear.Its is generally undesirable to use external spur gears of ratio greater than 1: 5.We can compute the gear ratio as follows.

First find out the RPM input through the motor,We have selected a 0.5 H.P,D.C motor with iron core,1750 rpm.Now we determine that the motor will have the maximum speed of the chair to be 2.76 m/s.

Now Since we are working on as existing wheel chair of wheel diameter of 0.80 m.Thus, the required angular acceleration will be

V =  .r

Here, V is the desired velocity of the chair

R is the radius of the wheel(r = 40 cm)

 is the angular acceleration of the wheel

Thus, we have

2.76 =  .(0.40)

or,  = 6.9 rad/s

or, = 66 RPM

Now at this point we have our input speed and the desired output speed.So we will try to divide our ratio in such a way that non of the gears have reduction more than 1: 5.

So, final reduction ratio is

66/1750

We will try to divide this ratio in set of intermediate gears,

i.e,

66/1750 = 2.3.11/2.7.5.5.5

= (20/50) (30/70)(22/20) )(20/100

Similarly for low speed

40/1750 = 2.2.5.2/2.7.5.5.5

= (20/20)(20/50)(20/70)(20/100)

Hence we have the divided ratio in set of small imtermediate gears.Care should be taken that atleast one ratio should be common so that the transfer of power from final shaft to diffrential takes place.

3./ DESIGN OF GEARS:

We determine the gear ratioof the first pair i.e,

N g / N p

Where, N g is number of teeth in gears that is being driven.

N p is the number of teeth in pinion that drives the motor.

Now, say

N g / N p= Mg.

(2). In the next step we determine the pressure angle.The selection of pressure angle( say ) is based on the operating requirements.If its possible we try to use the standard system.We have selected a pressure angle of 20.

(3). Pick up the approximate number of pinion teeth (say N p).

(a). consult table 4.2 for general information.

(b). Check table 4.3 and 4.4 for undercut limitations.

(c). If power gearing ,check table 4.5 for data on balancing strength and wear capacity.

As  p /a = N a / N p

From above calculation’s we selected Na = 50 and N p= 20

(4). Having approximate the number of pinion teeth,we determine appropriate center distance and face width.If power gearing,estimate size.

We take the face width (F) of gears to be one third of the pitch diameter of the gears (Dp).

i.e, Fd = F/Dp,

We also define a pinion torque (say Tp)[1]

Tp = 9549.3 P[kw]/np[r/min].

Then the necessary pitch diameter is determined

Dp[mm] = {2000 Tp[N.m] (1 + S2 Np /Ng)/ FdK[N/mm2]}1/3[2]

Here,

Tp is Pinion torque.

P is the wattage of motor in kw.

S2 is 1 for external gears and –1 for internal gears.

K is synthetic surface loading factor for RH.The value of K may be selected from the table 11.1(attached).Thus in the above calculation we calculate the Pitch diameter of the gear (Dp).

Now we have an empirical relation between the pitch diameter andthe center distance between the gears.

Dp = 2C/Mg + S2

Dg = Mg.Dp

Where,

C is the center distance between two gears.

Mg is the gear ratio i.e, N g / N p.

(5) Based on number of pinion teeth and center distance,determine approximate pitch.Check the table 4.6 for standard pitches,and if possible ,use standard pitch.Readjust pinion tooth numbers,center distance,and ratio to agree with pitch chosen.

(6). Determine the whole depth of pinion and gear.Usetables 4.7 and 4.1.

(7). Determine addendum of pinion and gear.Consult tables4.9 and 4.1.

Follow these rules:

Use enough long addendum to avoid undercut.
If a critical power job ,speed decreasing,balance addendum for strength (table 4.7).
If no undercut or power problems,use standard addendum (table 4.1).

(8). Determine operating circular pitch .If standard pitch is used,consult table 4.12.If enlarged center distance is used,determine operating circular pitch from operating pitch diameters..

(9). Determine design tooth thickness.Decide first on how much to thin teeth for backlash.Table 4.11 gives recommended amount for power gearing having in mind normal accuracy of center distance and normal operating temperature variations between gear wheel and castings.

T = a.2tan n

After the theoretical tooth thickness are obtained,subtract one-half teh amount the teeth are to be thinned for backlash from the pinion and gear theoretical tooth thickness.This is the maximun design tooth thickness.Obtain the minimum design tooth thickness by subtracting a reasonable tolerance for machining from the maximum design tooth thickness.

(10). Recheck load capacity using design proportions just obtained.The check has been explained in the next page.

(11). Certain general dimensions must be calculated and toleranced.

a. Outside diameter, D + 2(a  a).

b. Root diameter, Do – 2 ht

Face width.
Chordal addendum and chordal thickness.

3./ DESIGN OF GEARS TO PREVENT FAILURES

The three most common types of gear tooth failure are tooth breakage, surface pitting ,and scoring.

(a)Tooth breakage may be caused by an unexpected heavy load imposed on the teeth .A more common type of failure is due to bending fatigue, which results from the large number of repetitions of load imposed on the tooth as the gear rotates. A small value of the radius of the fillet may accentuate the bending fatigue effects.

[1] Dudley’s gear handbook,Dennis P.Townsend,Chap 11,Sec 11.3

[2] Dudley’s gear handbook,Dennis P.Townsend,Chap 11,Sec 11.4.2

(b) Pitting is a surface phenomenon caused by stresses exceeding the endurance limit of the surface material. After a sufficient

number of repetitions of the loading the bits of metal on the surface will fatigue and drop out. The process sometimes continues at an increasing rate since the remaining unpitted areas are less able to carry the load .Lubrication difficulties may contribute to pitting failures.

(c) Scoring can occur under heavy loads and inadequate lubrication. The oil film breaks down and metal to metal contact occurs. High temperature results and the high spots of the two surfaces weld together. The welds are immediately broken, but the surface of the teeth go rapid wear. Gearboxes should be broken in by preliminary operation at lower loads and smaller speeds till the tooth surface are highly polished. Also the misalignment of the shafts may shift the entire load to one edge of the tooth resulting excessive stress and the probability of scoring.

A properly designed set of gears should never fail by bending fatigue since the design can be made strong enough to limit the bending stress well below the fatigue endurance limit. On the other hand it is generally not possible to design a set of gears that will have infinite life with respect to the surface contact stresses .Thus the gears will eventually fail and most commonly by pitting unless lubrication is inadequate. The equation for bending stress in gears is based on the notion that gear teeth are cantilever beams that are subjected to end loads. The bending in the gear teeth were first studied by W.Lewis.

The AGMA Bending Stress Equation, as defined in AGMA standard 2001-C95,is based on the following assumptions about the teeth and the gear geometry:

(1). None of the teeth are damaged.

(2). The transverse contact ratio is between 1and 2.

(3). No interference exists between the tips of the teeth and root filets and there is no undercutting of teeth above the theoretical start of the active profile of the tooth.

(4). The teeth is not pointed.

(5). The backlash is zero.

(6). The fillets of the roots are standard ,smooth and are produced by a process of generation.

3.a./ BENDING STRENGTH

Bending strength is the strength of teeth, that resists the teeth against the bending when subjected to loads.

The fundamental formula for the bending stress in a gear tooth is given by:

S = WtKoKvKsPKmKb/F J[3]

Where S = bending stress( psi );

Wt = the transmitted tangential load(lbf)

Ko = the overload factor;

Kv= the dynamic factor;

Ks = the size factor;

Km = the load distribution factor;

Kb = the rim thickness;

J = the geometry factor for the bending strength;

For proper design ,this bending stress must not exceed the design stress value given by

S = Sat Yn/SfKtKr

Where:

Sat = the allowable bending stress(psi)

Yn= the stress cycle factor for bending strength

Sf = the factor of safety for bending strength

Kt = the temperature factor

Kr= the reliability factor

[3] Design of machine elements, M.F.Spoutts, T.E.Spoutts, Chap 10,Sec 10-22

3.b./ PITTING STRENGTH:

Themode of stress application for pitting failure is contact stress. The fundamental formula for the pitting stress in a gear tooth is:

Sc = Cp (Wt * Ko * Kv * Km*Cf´ d* F *I) [4]

Where:

Sc = contact stress (psi).

Cp = the elastic coefficient (lbf/in2).

Wt = the transmitted tangential load (lbf).

Ko = the overload factor.

Kv = the dynamic factor.

Km = the load distribution factor.

Cf´ = the surface condition factor for pitting resistance.

d = the operating pitch diameter of the pinion.

F = the net face width of the narrowest member.

I = the geometry factor for pitting resistance.

For proper design the calculated value must be smaller than

Sc = Sac Zn Ch/Sh Kt Kr

Where:

Sac = the allowable contact stress for the material

Zn = the stress cycle factor for pitting resistance.

Ch = the hardness ratio factor for pitting resistance.

Sh = the safety factor for pitting.

Kt = the temperature factor.

Kr = the reliability factor.

[4] Design of machine elements, M.F.Spoutts, T.E.Spoutts, Chap 10,Sec 10-22

Design of Shaft

Shaft are classified in two broad categories:

(a)Machine shaft.

(b)Transmission shaft.

The machine shaft is used when motion has to be transferred between two elements.The transmissoin is used where the power has to be transmitted between two elements.In our case we have both the machine shaft and transmission shaft in use.We have three machine shaft and one transmission shaft in our drawing.

The design of shaft will depend on the torque transmitted by the motor,the revolution of he motor itself.The shaft is subjected to bending or shear stresses or the combination of both.The shaft may also be subjected to axial loads due to the meshing of the gears.Lets denote the axial load on the shaft be Fa and we proceed on our design.

Let the Torque on the shaft from the motor be denoted by T.Now the force developed by output surface i.e, tooth of gear will be,

Fg = T/R.P.M,

Taking moments about Rr .Now since the shaft has uniform weight per foot of length so we can assume that total weight of shaft is concentrated at the bottom of the shaft under consideration.

So,

L  Lr _- Wg  L/2 -Fg  a = 0,

Or we can write this equation as

Lr = Fg  a  Wg  L/2/L

Wg is the concentrated weight

Similarly ,

Rr  L – Fg (L - a) – Wg  L/2 = 0

Or we can write this equation as