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Gears

Gears are used to transmit torque, rotary motion, and power from one

shaft to another. Compared to various other means of power transmission

(e.g., belts and chains), gears are the most rugged and durable. They have

transmission efficiency as high as 98%. However, gears are generally

more costly than belts and chains. Two modes of failure affect gear teeth:

fatigue fracture owing to fluctuating bending stress at the root of the tooth

and fatigue (wear) of the tooth surface. Both must be checked when

designing the gears. Selection of the proper materials to obtain

satisfactory strength, fatigue, and wear properties is important.

The shapes and sizes of the teeth are standardized by the American Gear

Manufacturers Association (AGMA). The methods of AGMA are widely

employed in design and analysis of gearing. There many are many types

of gears as discussed below.

1. Spur Gear This is a cylindrical shaped gear in which the teeth are parallel to the axis

as shown in figure(1). It has the largest applications and, also, it is the

easiest to manufacture.

Fig.(1) Spur gear

2. Helical Gear This is a cylindrical shaped gear with helicoid teeth as shown in figure

(2). Helical gears can bear more load than spur gears, and work more

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quietly. They are widely used in industry. A disadvantage is the axial

thrust force the helix form causes.

Fig.(2) Helical gears

3. Double Helical Gear This is a gear with both left-hand and right-hand helical teeth as shown in

figure (3). The double helical form balances the inherent thrust forces.

Fig.(3) Double Helical Gear

4. Spur and Helical Racks

This is a linear shaped gear which can mesh with a spur gear and helical

gears respectively with any number of teeth as shown in figure (4). The

spur rack is a portion of a spur gear with an infinite radius.

(a) Spur Rack (b) Helical Rack

Fig.(4) Spur and Helical Racks

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5. Straight Bevel Gear This is a gear in which the teeth have tapered conical elements that have

the same direction as the pitch cone base line as shown in figure (5).The

straight bevel gear is both the simplest to produce and the most widely

applied in the bevel gear family.

Fig.(5) Straight Bevel Gear

6. Worm and Worm Gear

Worm set is the name for a meshed worm and worm gear. The worm

resembles a screw thread; and the mating worm gear a helical gear,

except that it is made to envelope the worm as seen along the worm’s axis

as shown in figure (6).The outstanding feature is that the worm offers

a very large gear ratio in a single mesh. However, transmission

efficiency is very poor due to a great amount of sliding as the worm tooth

engages with its mating worm gear tooth and forces rotation by pushing

and sliding. With proper choices of materials and lubrication, wear can be

contained and noise is reduced.

Fig.(6) Worm gear

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7. Internal Gear This is a cylindrical shaped gear but with the teeth inside the circular ring

as shown in figure (7). It can mesh with a spur gear. Internal gears are

often used in planetary gear systems.

Fig.(7) internal gear

8. Spiral Bevel Gear

This is a bevel gear with a helical angle of spiral teeth as shown in figure

(8). It is much more complex to manufacture, but offers a higher strength

and lower noise.

Fig.(8) Spiral Bevel Gear

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Spur Gears

Geometry and Nomenclature

Consider two virtual friction cylinders (or disks) having no slip at the

point of contact, represented by the circles in Figure (9-a). A friction

cylinder can be transformed into spur gear by placing teeth on it that run

parallel to the axis of the cylinder. The surfaces of the rolling cylinders,

shown by the dashed lines in the figures, then become the pitch circles.

The diameters are the pitch diameters, and the cylinders represent the

pitch cylinders. The teeth, which lie in axial paths on the cylinder, are

arranged to extend both outside and inside the pitch circles (Figure9-

b).

Fig.(9) Spur gears are used to connect parallel shafts: (a) friction

cylinders and (b) an external gear set.

All calculations are based on the pitch circle. Note that spur gears are

used to transmit rotary motion between parallel shafts. A pinion is the

smaller of the two mating gears, which is also referred to as a pair of

gears or gear set. The larger is often called the gear. In most

applications, the pinion is the driving element, whereas the gear is the

driven element. This reduces speed, but increases torque, from the power

source (engine, motor, turbine): Machinery being driven runs slower. In

some cases, gears with teeth cut on the inside of the rim are needed. Such

a gear is known as an internal gear or an annulus (Figure 10-a). A rack

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(Figure 10-b) can be thought of as a segment of an internal gear of

infinite diameter

Fig.(10) Gear sets: (a) internal gear and pinion and (b) rack and pinion.

Properties of Gear Tooth

The face and flank portion of the tooth surface are divided by the pitch

cylinder. The circular pitch p is the distance, on the pitch circle, from a

point on one tooth to a corresponding on the next. This leads to the

definition

(1)

where

= the circular pitch, in.

d = the pitch diameter, in.

N = the number of teeth

The diametral pitch P is defined as the number of teeth in the gear per

inch of pitch diameter. Therefore,

(2)

The units of P are teeth/in. or in.−1

.

Both circular and diametral pitches prescribe the tooth size. The latter is a

more convenient definition. Combining Equations (1) and (2) yields the

useful relationship

(3)

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For two gears to mesh, they must have the same pitch.

In SI units, the size of teeth is specified by the module (denoted by m)

measured in millimeters. It can be defined as:

(4)

Where:

= pitch diameter (mm)

N= is the number of teeth.

Substituting equation (4) into equation (1) to get the circular pitch in

millimeters:

(5)

The diametral pitch, using Equation (3), is then

(6)

It is measured in teeth/mm or mm−1

.

Note that metric gears are not interchangeable with U.S. gears, as the

standards for tooth size are different.

Nomenclature of the spur gear teeth

Referring to figure (11)

Fig.(11) Nomenclature of spur gear

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a= The addendum is the radial distance between the top land and the

pitch circle

bd= The dedendum represents the radial distance from the bottom land to

the pitch circle.

b= The face width of the tooth is measured along the axis of the gear.

h=The whole depth which is the sum of the addendum and dedendum.

Clearance circle= represents a circle tangent to the addendum circle of the

mating gear.

f=The clearance represents the amount by which the dedendum in a

given gear exceeds the addendum of the mating gear. Clearance is

required to prevent the end of the tooth of one gear from riding on the

bottom of the mating gear. The difference between the whole depth and

clearance represents the working depth hk.

c= The distance between the centers of the two gears in mesh . Using

Equation (2) with d = 2r,

Or

(7)

Where subscripts l and 2 refer to driver and driven gears, respectively

Backlash

The width of space between teeth must be made slightly larger than the

gear tooth thickness t, both measured on the pitch circle. Otherwise, the

gears cannot mesh without jamming. The difference between the

foregoing dimensions is known as backlash. That is, the backlash is the

gap between mating teeth measured along the circumference of the

pitch circle, as schematically shown in Figure (12).

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Fig(12) Depiction of backlash in meshing gears

The amount of backlash must be limited to the minimum amount

necessary to ensure satisfactory meshing of gears. Excessive backlash

increases noise and impact loading whenever torque reversals occur.

Example :

For the gear set shown in figure (13) find the pitch diameters, module,

circular pitch, and center distance. N1=19, N2=124, P=16in-1

Note that in SI units, the module is

( )

( )

The circular pitch can be calculated as

The center distance can be calculated as

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( )

( )

Basic Law of Gearing

The main requirement of gear tooth geometry is the provision that

angular ratios are exactly constant. For quiet, vibrationless operation,

the velocities of two mating gears must be the same at all times. This

condition is satisfied when the pitch circle of the driver is moving with

constant velocity and the velocity of the pitch circle of the driven gear

neither increases nor decreases at any instant while the two teeth are

touching.

The basic law of gearing states that as the gears rotate, the common

normal at the point of contact between the teeth must always pass through

a fixed point on the line of centers. The fixed point is called the pitch

point P (Figure 9-b).

V = r1ω1 = r2ω2 (8)

Several useful relations for determining the speed ratio may be written as

follows:

(9)

where

rs = the speed or velocity ratio

ω = the angular velocity, rad/s

n = the speed, rpm

N = the number of teeth

d = the pitch circle diameter

Subscripts 1 and 2 refer to the driver and driven gears, respectively.

Figure (13) shows two involutes, on separate cylinders in mesh,

representing the gear teeth.

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(Fig.13)

Gear Tooth Action and Systems of Gearing

To illustrate the action occurring when two gears are in mesh, consider

Figure (13). The pitch radii r1 and r2 are mutually tangent along the line

of centers O1O2, at the pitch point P. Line ab is the common tangent

through the pitch point. Note that line cd is normal to the teeth that are in

contact and always passes through P at an angle ϕ to ab. Line cd is also

tangent to both base circles. This line, called line of action or pressure

line, represents (neglecting the sliding friction) the direction in which the

resultant force acts between the gears. The angle ϕ is known as the

pressure angle, which is measured in a direction opposite to the direction

of rotation of the driver. The involute is the only geometric profile

satisfying the basic law of gearing that maintains a constant-pressure

angle as the gears rotate. Gears to be run together must be cut to the same

nominal pressure angle. As pointed out, the base circle is tangent to the

pressure line. Referring to Figure (13), the radius of the base circle is

then

rb = r cos ϕ (10)

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where r represents the pitch circle radius. The base pitch pb refers to the

distance measured on the base circle between corresponding adjacent

teeth:

pb = p cos ϕ (11)

where p is the circular pitch.

Standard Gear Teeth

Most gears are cut to operate with standard pressure angles of 20° or 25°.

The tooth proportions for some involute, spur gear teeth are given in

Table (1) in terms of the diametral pitch P.

Table(1) Commonly Used Standard Tooth systems

For spur gears

Figure (14) depicts the actual sizes of 20° pressure angle, standard, full-

depth teeth, for several standard pitches from P = 4 to P = 80. Note the

inverse relationship between P and tooth size.

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Fig.(14) Actual size gear teeth of various diametral pitches.

Transmitted Load With a pair of gears or gearset, power is transmitted by the load that the

tooth of one gear exerts on the tooth of the other. The transmitted load Fn

is normal to the tooth surface; therefore, it acts along the pressure line or

the line of action (Figure 15). This force between teeth can be resolved

into tangential force and radial force components, respectively:

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Fig.(15) Gear tooth force Fn, shown resolved at pitch point P.

And (12)

Φ is the pressure angle in degree.

The tangential component Ft when multiplied by the pitch-line

velocity, accounts for the power transmitted while radial component

Fr does no work but tends to push the gears apart.

The velocity along the pressure line is equal to the tangential velocity of

the base circles. The tangential velocity of the pitch circle (in feet per

minute, fpm) is given by

(13)

where

d represents the pitch diameter in in.

n is the speed in rpm

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In design, we assume that the tangential force remains constant as the

contact between two teeth moves from the top of the tooth to the bottom

of the tooth. The applied torque and the transmitted load are related by

(14)

The horsepower is defined by

or

(15)

in which the torque T is in pounds-inch and n is in rpm. The tangential

load transmitted can be obtained as:

(16)

where V is given by Equation (13). Since 1 hp equals 0.7457 kW. In SI

units, the preceding equations are given by the relationships

(17)

(18)

(19)

In the foregoing, we have

Ft = the transmitted tangential load (N)

d = the gear pitch-diameter

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n = the speed (rpm)

T = the torque (N · m)

V = πdn/60 = pitch-line velocity (in meters per second, m/s)

Dynamic Effects

The dynamic load Fd or total gear tooth load, in U.S. customary units, is

estimated using one of the following formulas:

( )

(20)

( )

(21)

√

( )

(22)

where V is the pitch-line velocity in fpm. To convert to m/s, divide the

given values in these equations by 196.8. Clearly, the dynamic load

occurs in the time while a tooth goes through mesh. Note that the

preceding relations form the basis of the AGMA dynamic factors.

Example ( Gear Force Analysis)

The three meshing gears shown in Figure (16-a) have a module of 5 mm

and a 20°pressure angle. Driving gear 1 transmits 40 kW at 2000 rpm to

idler gear 2 on shaft B. Output gear 3 is mounted to shaft C, which drives

a machine. Determine and show, on a free-body diagram,

a. The tangential and radial forces acting on gear 2

b. The reaction on shaft B

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Note: The intermediate gears, called idler gears, do not influence the

overall speed ratio. Hence, it affects only the direction of rotation of gear

3.

Solution

The pitch diameters of gears 1 and 3, can be calculated as

d1 = N1m = 20(5) = 100 mm

and d3 = N3m = 30(5) = 150 mm.

The free body diagram can be shown in the figure below:

Fig.(16) (a) a gear set and (b) free-body diagram of the forces acting on

gear 2 and reaction on shaft B. Bearing reaction can be calculated by

knowing that sum of the forces in x and y directions at the bearing are:

Hence, √

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Bending Strength of a Gear Tooth: The Lewis Formula

Wilfred Lewis was the first to present the application of the bending

equation to a gear tooth. The formula was announced in 1892, and it still

serves the basis for gear tooth bending stress analysis. Simplifying

assumptions in the Lewis approach are as follows :

1. A full load is applied to the tip of a single tooth.

2. The radial load component is negligible.

3. The load is distributed uniformly across the full-face width.

4. The forces owing to tooth sliding friction are negligible.

5. The stress concentration in the tooth fillet is negligible.

To develop the basic Lewis equation, consider a cantilever subjected to a

load Ft, uniformly distributed across its width b (Figure15-a). We have

the section modulus I/c=bt2/c = bt

2/6. So the maximum bending stress is

(23)

Fig.(15) Beam strength of a gear tooth: (a) cantilever beam and (b) gear

tooth as cantilever. This flexure formula yields results of acceptable

accuracy at cross sections away from the point of load application . We

now treat the tooth as a cantilever fixed at BD (Figure15). It was noted

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already that the normal force Fn is considered as acting through the corner

tip of the tooth along the pressure line. The radial component Fr causes a

uniform compressive stress over the cross section. This compressive

stress is small enough compared to the bending stress, due to the

tangential load Ft , to be ignored in determining the strength of the tooth.

Clearly, the compressive stress increases the bending stress on the

compressive side of the tooth and decreases the resultant stress on the

tensile side. Therefore, for many materials that are stronger in

compression than in tension, the assumption made results in a stronger

tooth design. Also note that because gear teeth are subjected to fatigue

failures that start onthe tension side of the tooth, the compressive stress

reduces the resultant tensile stress and thus strengthens the tooth.

U niform Strength Gear Tooth

In a gear tooth of constant strength, the stress is uniform; hence,

b/6Ft = constant = C

and Equation (12) then leads to

L = Ct2.

The foregoing expression represents a parabola inscribed through point A,

as shown by the dashed lines in Figure (15.b). This parabola is

tangent to the tooth profile at points B and D, where the maximum

compressive and tensile stresses occur, respectively. The tensile stress is

the cause of fatigue failure in a gear tooth.

Referring to the figure, by similar triangles ABE and BCE, we write

(t/2)/x = L/(t/2)

or

L = t2/4x.

Carrying this into Equation (12) and multiplying the numerator and

denominator by the circular pitch p, we have

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( )

(24)

The Lewis form factor is defined as

Substitution of the preceding into Equation (13) results in the original

Lewis formula:

(25)

Because the diametral pitch rather than circular pitch is often used to

designate gears, the following substitution may be made:

p = π/P and Y = πy.

Then the Lewis form factor is expressed as

(26)

Similarly, the Lewis formula becomes

(27)

When using SI units, in terms of module m = 1/P,

(28)

Both Y and y are the functions of tooth shape (but not size) and thus vary

with the number of teeth in the gear. Some values of Y determined from

Equation (26) are listed in Table ( 2 ).

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Table(2) Values of the Lewis Form Factor for Some Common

Full-Depth Teeth

Let bending stress (σ) in Equation (16) be designated by the allowable

static bending stress (σo) and so tangential load (Ft ) by the allowable

bending load (Fb). Then this equation becomes

(29)

or, in SI units,

(30)

The values of σo for some materials of different hardness are listed in

Table (3).

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Table(3) Allowable Bending Stress For Use in the Lewis Equation

The stress in a gear tooth is greatly influenced by size of the fillet radius

rf

(31)

(32)

As a reasonable approximation, Kf = 1.5 may be used in these equations.

The load capacity of a pair of gears is based on either the bending or wear

capacity, whichever is smaller. For satisfactory gear performance, it is

necessary that the dynamic load should not exceed the allowable load

capacity. That is,

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in which the dynamic load ( Fd ) is given by Equations(20,21,22). Note

that this dynamic load approach can be used for all gear types. When a

gear set is to be designed to transmit a load Fb, the gear material should

be chosen so that the values of the product σo Y are approximately the

same for both gears.

Ex:

A 25° pressure angle, 25-tooth spur gear having a module of 2 mm, and

45 mm face width are to operate at 900 rpm. Determine

a. The allowable bending load applying the Lewis formula.

b. The maximum tangential load and power that the gear can transmit.

The gear is made of SAE 1040 steel. A fatigue stress-concentration

factor of 1.5 is used.

Solution

From table (2) for gear with 25teeth We have Y = 0.402 and from table

(3)

σo= 172 MPa

The pitch diameter is

d = mN = 2(25) = 50 mm and V = πdn /60= π (0.05)(900/60) = 2.356

m/s = 463.7 fpm.

( )

Since V=463.7fpm then

( )

The limiting value of the transmitted load is

Or

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The corresponding gear power transmitted:

Wear Strength of a Gear Tooth: The Buckingham Formula

The failure of the surfaces of gear teeth is called wear. Wear is a broad

term, which encompasses a number of kinds of surface failures. So it is

evident that gear tooth surface durability is a more complex matter than

the capacity to withstand gear tooth bending failure. Tests have shown

that pitting, a surface fatigue failure due to repeated high contact

stress, occurs on those portions of a gear tooth that have relatively

little sliding compared with rolling. Clearly, spur gears and helical

gears have pitting near the pitch line, where the motion is almost pure

rolling. The allowable wear load can be evaluated as:

(33)

Where:

(

)

=Surface endurance limit.

Equation (33) called Buckingham formula

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Table(4) Surface Endurance Limit Se and Wear Load Factor K for Use in

the Buckingham Equation

For satisfactory gear performance, the usual requirement is that

Where Fd is the dynamic load.

To prevent too much pinion wear, particularly for high-speed gearing, a

medium-hard pinion with low hardness gear is often used. This has the

advantage of giving some increase in load capacity and slightly lower

coefficient of friction on the teeth.

Ex

A 25° pressure angle, 25-tooth spur gear having a module of 2 mm, and

a 45 mm face width are to operate at 900 rpm mesh with a gear of 60

tooth made of cast iron. Determine

a. The allowable load for wear for the gear set using the Buckingham

formula.

b. The maximum load that can be transmitted

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Given: Np = 25, dp = 50 mm, b = 45 mm, Ng = 60.

Solution

The wear load can be calculated as:

Using table (4) for gear with 200Bhn to get

K=1.014MPa, Hence

( )( ) (

) ( )

Since

V = πdn /60= π (0.05)(900/60) = 2.356 m/s = 463.7 fpm.

( )

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Supplementary Problems

(Q1)A gear set of steel with average Bhn of pinion and gear of 200 has a

5 mm module and a width of 40 mm. A 25-tooth pinion rotates at 120

rpm and drives a 50-tooth gear. The gears are cut using a pressure angle

of 20°. What is the maximum horsepower that can be transmitted on the

basis of wear strength by applying the Buckingham formula?

Solution

The allowable wear load can be evaluated as:

The pitch circle diameter can evaluated as

From table(4) for both gear and pinion steel with average hardness

200Bhn

Substitute the above values in the wear load equation to get:

Or

The pitch line velocity can be evaluated as

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Since 1m=3.3ft

Then the pitch line velocity in fpm can be evaluated as

The dynamic load can be evaluated as

( )

For satisfactory gear performance, the usual requirement is that

Or

The power transmitted can be calculated as:

Or

( )

(Q2) A pair of gears has a 20o pressure angle and a diametral pitch of 6

teeth/in. Determine the maximum horsepower that can be transmitted,

based on bending strength and applying the Lewis equation and Kf = 1.4.

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The gear is made of phosphor bronze, has 60 teeth, and rotates at 240

rpm. The pinion is made of SAE 1040 steel and rotates at 600 rpm.Both

gears have a width of 3.5 in.

Solution

The bending load can be calculated from Lewis equation as:

Since the pinion is made of SAE1040 steel, then from table (3)

The number of the teeth for the pinion can be calculated as

Or

From table(2) The Lewis form factor for the pinion with 24 teeth

Substitute the above values in the bending load equation to get

To evaluate the pitch line velocity of the pinion, the pitch diametr of the

pinion must be calculated:

Or

So

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The pitch line velocity is calculated as:

( )

(Q3) Two meshing gears have face widths of 2 in. and diametral pitches

of 5 teeth/in. The pinion is rotating at 600 rpm and has 28 teeth, and the

velocity ratio is to be 4/5. Determine the horsepower that can be

transmitted, based on wear strength and using the Buckingham equation.

The gears are made both of steel hardened to a 350 Bhn and have a 20°

pressure angle.

Q4) The gears shown in Figure (16) have a module of 10 mm, a 20°

pressure angle, and a tooth width of 15 mm. Determine

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a. The allowable bending load, using the Lewis equation and Kf = 1.5, for

the tooth of gear 4 when the gear and pinion made of cast iron ASTM50.

b. The allowable load for wear, applying the Buckingham equation, for

gears 1 and 2 when the gears are made of hardened steel (200 Bhn), and

gears 2 and 3 are mounted on shaft B.

Fig.(16)