Design of Transmission Systems Using Spur and Helical Gears

(302051) Design of Transmission
Systems
Unit 1 Spur and Helical Gears
Dr. S. G. Kolgiri
Associate Professor, Mech. Department
S B Patil College of Engineering, Indapur
SPPU Pune University

Gears
.
• Classification of gears
• Desirable properties and Selection of
material gear
• Standard system of gear tooth
• Basic mode of gear tooth failures and type
of gear tooth failure
• Gear Lubrication Methods
• Terminologies or nomenclatures
• Number of teeth and face width
• Forces analysis
• Beam strength (Lewis) equation
• Velocity factor
Design of transmission System
(Dr.S G Kolgiri)

Gears Purpose
• Transmit rotating motion
• Normally mounted on shaft
• Transmits rotating motion from one parallel
shaft to another

Type of Gears
• Spurs
• Helical
• Bevel
• And Worm Gears

Types of Gears
• Spur Gear
– Cut Straight across
– Spur gears only have
one tooth in contact at
a time
– Minimizes chance of
popping out of gear
– Handles torque well
– Used for Reverse

Helical Gear
• Are quieter than spur
gears
• Two teeth at a time
contact
• Has a tendency to move
shaft for and aft
• Are left and right handed
• Opposites on parallel
shafts

Helical Gears
• Initial contact is a point
• Changes into a line as teeth
come into engagement

Spur Gears vs. Helical Gears
• Spur gears are
straight cut parallel
to shaft
• Helical gears are
angle cut to gears
axis of rotation

Bevel Gears
• Change the direction of
rotation
• Spider gears are
straight cut bevel gears
• Transfer motion
between two shafts at
an angle to each other

Worm Gear
Note the axis of the shafts

Worm Gear
• Drives a spur gear
• Provides maximum
tooth contact
• Used in recirculating
ball steering boxes
• Speedometer cable
drive mechanisms

External Gears
• When two external
gears mesh, they
rotate in opposite
directions

Internal Gear Teeth
• Most gears are
external. (teeth on the
outside)
• Internal gear used with
pinion gears
• External gear rotates
same direction of
internal gear
• External gear rotates
opposite with another
external gear

Idler Gears
• Located between drive and driven gear
• Used in reverse gear trains
• Transfers motion without changing
rotational direction

Desirable properties and Selection of
material gear
• Load carrying capacity of the gear tooth depends upon the
ultimate tensile(yield)strength of material.
• Wear rating- The resistance of wear depends upon alloying
element, grain size, percentage of carbon-sufficient surface
endurance strength to avoid failure due to destructive pitting.
• For high speed transmission, the sliding velocity are very high
– material should have low coefficient of friction due to
scoring.
• The amount of heat distortion or warping –load gets
concentrated at one corner of tooth- alloy steel are superior to
plane carbon
• Cast iron FG 200,260, steel, bronze and phenolic resins and
45C8, 40Ni3Cr65Mo55

Basic mode of gear tooth failures
and type of gear tooth failure
• Breakage of tooth due to
static and dynamic loads
• Surface destruction
– Abrasive wear
– Corrosive wear
– Initial pitting
– Destructive pitting
– Scoring

Gear Ratios
• Can multiply torque and speed
• Can reduce torque and speed
• Same size and number of teeth = no change in
output
• Equal size gears create equal output
• Small drive gear to larger driven gear = driven
gear speed decreases
• Larger drive gear to smaller driven gear =
driven gear speed increases

Gear Ratio Calculations
• Gear Ratio can be found by dividing
number of teeth on driven gear by the
number of teeth on the driving gear.
• Ex: 75 driven teeth : 45 drive teeth = 1.66
gear ratio
Page 740

Gear Reduction
• Speed decreases
• Torque Increases
• When speed is
halved, torque
doubles
• Most manual
transmissions are
speed reducing,
torque increasing

Gear Overdrive
• Speed Increases
• Torque decreases
• When speed is doubled,
torque is halved
• Used for fuel milage
• Factory overdrives vs.
Aftermarket overdrives

Gear Nomenclature
• Pitch circle is a theoretical circle upon which all
calculations are based. Pitch circles of mating gears are
tangent to each other.
• Pinion is the smaller of two mating gears. Gear is the
larger.
• Circular pitch is equal to the sum of the tooth thickness
and the width of space measured on the pitch circle.
• Diametral pitch is the ratio of the number of teeth to the
pitch diameter.
• Module is the ratio of the pitch diameter to the number
of teeth (SI).
• Addendum is the radial distance from the top land to the
pitch circle.

Gear Nomenclature
• Dedendum is the radial distance from the bottom
land to the pitch circle.
• Whole depth is the sum of the addendum and
dedendum
• Clearance circle is a circle that is tangent to the
addendum circle of the mating gear.
• Clearance is the amount by which the dedendum
in a given gear exceeds the addendum of its
mating gear.
• Backlash is the amount by which the width of a
tooth space exceeds the thickness of the engaging
tooth measured on the pitch circles.
.

SPUR GEAR
DESIGN
(Dr. S G Kolgiri)

Gear failure can occur in various modes. In this chapter details of failure are given. If care is
taken during the design stage itself to prevent each of these failure a sound gear design can
be evolved. The gear failure is explained by means of flow diagram in Fig.
Gear Tooth Failure
Different modes of failure

Abrasive wear is the principal reason for the failure of open gearing and closed gearing of
machinery operating in media polluted by abrasive materials. Examples are mining
machinery; cement mills; road laying, building construction, agricultural and transportation
machinery, and certain other machines. In all these cases, depending on the size, shape and
concentration of the abrasives, the wear will change. Abrasive wear is classified as mild and
severe.
MILD ABRASION
Mild abrasion is noticed when there is ingress of fine dust particles in lubricating oil which
are abrasive in nature. Since abrasive is very fine, the rate of metal removal is slow. It takes
a long time for p erceptible wear.
Fig (a) Mild abrasion
Abrasive Wear

The surface appears as though it is polished. A spiral bevel pinion with mild abrasion is
shown in Fig. (a) Mild abrasive wear is faced in cement mills, ore grinding mills. Fine
dust particles entering the lubricating medium cause three body abrasions. The prior
machining marks disappear and surface appears highly polished as shown in Fig.(b).
Noticeable wear occurs only over a long time. Sealing improvement and slight
pressurization of the gear box with air can reduce the entry of dust particles and
decrease this wear.
Fig. (b) Mild abrasion
Abrasive Wear

Cause:
Foreign particles in the lubricant like dirt, rust, weld spatter or
metallic debris can scratch the tooth surface.
Remedy:
Provision of oil filters.
Increasing surface hardness.
Use of high viscosity oils.
Abrasive Wear

SEVERE ABRASION
This wear occurs due to ingress of larger abrasive particles in the lubricating medium
and higher concentration of the particles. The particles will plough a series of groove on
the surface in the direction of sliding on the gear tooth as seen in the Fig. (c). High rate
of wear in this case will quickly reduce the tooth thickness. Thinned tooth may later on
fracture leading to total failure.
Fig. (c) Severe abrasion
Abrasive Wear

Corrosive wear is due to the chemical action of the lubricating oil or the additives.
Tooth is roughened due to wear and can be seen in the Fig.(d1). Chemical wear of
flank of internal gear caused by acidic lubricant is shown in Fig. (d2)
(d1)
(d2)
Fig. d Corrosive wear
Corrosive Wear

Cause: Corrosive of the tooth is caused by corrosive
elements such as
Extreme pressure additives present in lubricating oils.
Foreign materials.
Remedies:
Providing complete enclosure for the gears free from external
contamination.
Selecting proper additives.
Replacing the lubricating oil at regular intervals.
Corrosive Wear

Pitting is a surface fatigue failure of the gear tooth. It
occurs due to repeated loading of tooth surface and the
contact stress exceeding the surface fatigue strength of
the material. Material in the fatigue region gets removed
and a pit is formed. The pit itself will cause stress
concentration and soon the pitting spreads to adjacent
region till the whole surface is covered. Subsequently,
higher impact load resulting from pitting may cause
fracture of already weakened tooth. However, the failure
process takes place over millions of cycles of running.
There are two types of pitting, initial and progressive.
Pitting of Gears

INITIAL / INCIPIENT PITTING
Initial pitting occurs during running-in
period wherein oversized peaks on the
surface get dislodged and small pits of
25 to 50 μm deep are formed just below
pitch line region. Later on, the load gets
distributed over a larger surface area
and the stress comes down which may
stop the progress of pitting.
Fig. (e) Initial pitting
Pitting of Gears

In the helical gear shown in Fig. (e). pitting started as a
local overload due to slight misalignment and progressed
across the tooth in the dedendum portion to mid face.
Here, the pitting stopped and the pitted surfaces began to
polish up and burnish over. This phenomenon is common
with medium hard gears. On gears of materials that run
in well, pitting may cease after running in, and it has
practically no effect on the performance of the drive since
the pits that are formed gradually become smoothed over
from the rolling action. The initial pitting is non-
progressive.
Pitting of Gears

INITIAL PITTING :
Cause:
Errors in tooth profile.
Surface irregularities.
Misalignment.
Remedies:
Precise machining of gears.
Correct alignment of gears.
Reducing the dynamic loads.
Pitting of Gears

PROGRESSIVE OR DESTRUCTIVE PITTING
During initial pitting, if the loads are high and the corrective action of initial pitting is unable
to suppress the pitting progress, then destructive pitting sets in. Pitting
spreads all over the tooth length. Pitting leads to higher pressure on the unpitted surface,
squeezing the lubricant into the pits and finally to seizing of surfaces.
Fig. (f) Tooth surface destroyed by extensive pitting
Pitting of Gears

Pitting begins on the tooth flanks near the line along the tooth passing through the
pitch point where there are high friction forces due to the low sliding velocity. Then it
spreads to the whole surface of the flank. Tooth faces are subjected to pitting only in
rare cases.
Fig. (f) shows how in destructive pitting, pitting has spread over the whole tooth and
weakened tooth has fractured at the tip leading to total failure
Fig. (g) Whole tooth is destroyed by extensive pitting
Pitting of Gears

DESTRUCTIVE PITTING:
Cause:
Load on the gear tooth exceeds the surface endurance
strength of the material.
Remedies:
Designing the gears in such a way that the wear strength of the gear
tooth is more than the sum of static and dynamic
loads.
The surface endurance strength can be improved by increasing the
surface hardness.
Pitting of Gears

Scoring is due to combination of two distinct activities: First, lubrication failure in the
contact region and second, establishment of metal to metal contact. Later on, welding and
tearing action resulting from metallic contact removes the metal rapidly and continuously
so far the load, speed and oil temperature remain at the same level. The scoring is
classified into initial, moderate and destructive.
INITIAL SCORING
Initial scoring occurs at the high
spots left by previous machining.
Lubrication failure at these spots
leads to initial scoring or scuffing as
shown in Fig. (h). Once these high
spots are removed, the stress
comes down as the load is
distributed over a larger area. The
scoring will then stop if the load,
speed and temperature of oil remain
unchanged or reduced. Initial
scoring is non-progressive and has
corrective action associated with it. Fig. (h) Initial scoring
Scoring

MODERATE SCORING
After initial scoring if the
load, speed or oil
temperature increases,
the scoring will spread
over to a larger area. The
Scoring progresses at
tolerable rate. This is
called moderate scoring
as shown in Fig. (i).
Fig. (i) Moderate scoring
Scoring

DESTRUCTIVE SCORING
After the initial scoring, if
the load, speed or oil
temperature increases
appreciably, then severe
scoring sets in with heavy
metal torn regions
spreading quickly
throughout as shown in
Fig.(j). Scoring is normally
predominant over the pitch
line region since
elastohydrodynamic
lubrication is the least at
that region. In dry running
surfaces may seize.
Fig. (j) Destructive scoring
Scoring

Cause:
Excessive surface pressure.
High Surface Speed.
Inadequate supply of lubricant result in the breakdown of the oil
film.
Remedies:
Selecting the parameters such as surface speed, surface
pressure and the flow of lubricant in such a way that the
resulting
temperature at the contacting surfaces is within permissible limits.
T he bulk temperature of the lubricant can be reduced by
providing fins on the outside surface of the gear box and a fan
for forced
circulation of air over the fins.
Scoring

Module 2 –
SPUR GEAR
DESIGN
Beam Strength Of
Gear Tooth
Wear Strength Of
Gear Tooth
(Dr. S G Kolgiri)

FORCE ANALYSIS: In gears, power is transmitted by means of a force
exerted by the tooth of the driving gear on the meshing tooth of the driven
gear. As shown in Fig
•The tooth of the driving pinion exerting a force PN on the tooth of the
driven gear
•The resultant force PN can be resolved into two components— tangential
component Pt and radial component Pr at the pitch point as shown in Fig.
•The normal force PN can be resolved into two components;
•a tangential force Pt which does transmit the power and
•radial component Pr which does no work but tends to push the gears
apart. They can hence be written as,
Pt = PN cos α Pr = PN sin α
Pr = Pt tan α
The torque transmitted by the gears is given by,
Tooth Force Analysis

The above analysis of the gear tooth force is based on
the following assumptions:
•As the point of contact moves, the magnitude of the resultant
force PN changes. This effect is neglected in the above
analysis.
•It is assumed that only one pair of teeth takes the entire load.
At times there are two pairs, which are simultaneously in contact
and share the load. This aspect is neglected in the analysis.
•The analysis is valid under static conditions, i.e., when the
gears are running at very low velocities.
• In practice, there is dynamic force in addition to force due to
power transmission. The effect of this dynamic force is
neglected in the analysis.
Tooth Force Analysis

Photo-elastic Model of gear tooth
Stresses developed by Normal force in a photo-elastic model of gear
tooth as per Dolan and Broghammer are shown in the Fig. The
highest stresses exist at regions where the lines are bunched closest
together. The highest stress occurs at two locations:
A. At contact point where the force F acts
B. At the fillet region near the base of the tooth.
Gear - Tooth Stresses

The analysis of bending stresses in gear tooth was done by WILFRES LEWIS. The
lewis equation is considered as the basic equation in the design of gears.
In the lewis analysis, the gear tooth is consider as a cantilever beam
The accurate stress analysis of a gear tooth for a particular application is a complex
problem because of the following reasons:
•There is continuous change in the point of application of load on the tooth profile.
•The magnitude and direction of applied load also change.
•In addition to static load, the dynamic load due to inaccuracy of the tooth profile,
error in machining and mounting, tooth deflection, acceleration and stress
concentration also act on the tooth which are all difficult to model mathematically.
Beam Strength of Gear Tooth

Wilfred Lewis, in a paper titled, “The investigation of the strength of gear tooth”
published in 1892, derived an equation for determining the approximate stress in a
gear tooth by treating it as a cantilever beam of uniform strength. The beam
strength calculation is based upon the following assumptions:
Gear tooth as cantilever beam

Lewis considered gear tooth as a cantilever beam with static normal force F applied at the
tip.
Assumptions made in the derivation are:
1.The full load is applied to the tip of a single tooth in static condition.
2.The radial component is negligible.
3.The load is distributed uniformly across the full face width.
4.Forces due to tooth sliding friction are negligible.
5.Stress concentration in the tooth fillet is negligible.
Here force is denoted my P

The cross-section of the tooth varies from free end to the fixed end. Therefore , A
parabola is constructed within the tooth profile. Parabola is beam of uniform
strength. The stress at any cross-section of beam is uniform or same.
The weakest section of the gear tooth is at the section XX, where parabola is
tangent to the tooth profile.

At the section XX,
Mb= Pt x h
I= (1/12) b x t3
Y= t / 2
The bending stresses are given by, σ b = Mb x y / I
= (pt x h)(t/2) / [(1/12)bt3]
Rearranging the terms,
Pt = b σ b (t2/6h)
Multiplying the numerator and denominator of the right-hand side by m,
Pt =m b σ b (t2/6hm)

Defining a factor Y
,
Y=(t2/6hm)
The equation is rewritten as,
Pt= mbσ b Y ................ (i)
Y is called the lewis form factor.
Eq.(i) gives relationship between the tangential force(Pt) and the corresponding stress σ b.
When stresses reaches the permissible magnitude of bending stresses, the corresponding
force(Pt) is called the beam strength.
Therefore the beam strength(Sb) is maximum value of the tangential force that the tooth can
transmit without bending failure. Replacing Pt by Sb,
Sb = mbσbY ................ (ii) The minimum number of teeth to avoid interference is
given by,
Sb = beam strength of gear tooth(N)
σ b = permissible bending stress (N/mm2) The permissible bending stress is given by,

Eq.(Ii) is known as lewis equation. Table for
values of lewis form factor is given above
table
In order to avoid the breakage of gear tooth due to
bending, the bending strength should be more than
the effective force between the meshing teeth.
S b >= P eff

• A pair of spur gears with 20° full-depth involute teeth consists of a 20
teeth pinion meshing with a 41 teeth gear. The module is 3 mm while
the face width is 40 mm. The material for pinion as well as gear is steel
with an ultimate tensile strength of 600 N/mm2. The pinion rotates at
1450 rpm and the service factor for the application is 1.75. Assume that
velocity factor accounts for the dynamic load and the factor of safety is
1.5. Determine the rated power that the gears can transmit. the Lewis
form factor is 0.32 for 20 teeth
Since the same material is used for the pinion and the gear, the
pinion is weaker than the gear so design for pinion.

WEAR STRENGTH
Due to rolling and sliding actions of the gear teeth, the following types of surface
destructions (wear) may occur:
Abrasive wear. Scratching of the tooth surface due to the presence of foreign materials
in the lubricant is called abrasive wear.
Corrosive wear. Chemical reactions on the surface of a gear cause corrosive wear.
Pitting. Repeated application of the stress cycle, known as pitting, cause fatigue failure.
Scoring. Inadequate lubrication between metal-to metal contact cause scoring.
It is observed from the results of various experiments that abrasion, corrosive, the
scoring are caused by improper lubrication, whereas pitting usually occurs because of
repeated application of Hertz contact stress on the portion of a gear tooth which has
relatively little sliding motion compared to rolling motion. Clearly, the spur gear will have
pitting near the pitch lien where motion is almost all of rolling.
Wear Strength of Gear Tooth

The failure of the gear tooth due to pitting occurs when the contact stress between two
meshing teeth exceed the surface endurance strength of the material.
Pitting is a surface fatigue failure, characterized by small pits on the surface of the gear
tooth.
To avoid this type of failure, properties of gear tooth and surface properties should be
selected such a way that the wear strength of the gear tooth is more than the effective
load between the meshing teeth.
The analysis of wear strength of gear tooth is done by Earle Buckingham. Buckingham
’s equation gives the wear strength of the gear tooth. which is based on Hertz theory of
contact stresses.

When two cylinders are pressed together as
shown in fig.(a), the contact stress is given
by,
σ c= 2*P/π*b*l ................(1.1)
and b= { [ 2P(1-μ2)(1/E1 + 1/E2)] / π l(1/d1 + 1/d2) }1/2
................(1.2)
where,
σ= maximum value of compressive stress (N/mm2) P=force pressing the two
cylinders together (N)
b= half width of deformation (mm) l= axial length of the cylinder (mm)
d1,d2= diameter of two cylinders (mm)
E1,E2= moduli of elasticity of 2 cylinders (N/mm2) μ = Poisson’s ratio
fig. contact stresses

Due to deformation under the action of load P, a rectangular surface of width (2b) and
length (l) is formed between the two cylinders. The elliptical stress distribution across the
width (2b) is shown in fig.
Substituting eq.(ii) into(i) and squaring both sides,
σc
2 =[1/π(1-μ2)]* (P/l)* {(1/r1+ 1/r2) / (1/E1+ 1/E2)} ................(1.3)
Where, r1,r2= radii of two cylinders
σc
2 =0.35*(P/l)* {(1/r1+ 1/r2) / (1/E1+ 1/E2)} ................(1.4)
The above equation of the contact stress is based on the following assumptions:
•The cylinders are made of isotropic materials.
•The elastic limit of the material is not exceeded.
•The dimensions r1,r2 are very large when compared to the width (2b) of the
deformation.

Fig. shows the contact between
two meshing teeth at the pitch
point
Fig. contact between two meshing teeth at pitch point

r1,r2 are to be replaced by radii of curvature at the pitch point. Therefore,
r1= dp’ * sinα /2 r2= dg’ * sinα /2
There are some reasons for taking the radii of curvature at the pitch point.
•The wear on the gear tooth generally occurs at or near the pitch line.
•When only one pair of teeth carries the entire load, contact occurs at the pitch point.
•The dynamic load is imposed on the gear tooth near the pitch line area.
(1/r1 + 1/r2)= 2/sinα [1/dp’ + 1/dg’] ............(1.5) A ratio factor Q is defined as,
Q= 2*zg/(zg+zp) ..............(1.6)

Substituting, dp’=m zp and dg’= m zg,
Q= 2 dg’/(dg’ + dp’)
Therefore,
[1/dp’ + 1/dg’] = (dg’ + dp’)/dg’ *dp’
= 2/ Q*dp’ ...............(1.7)
From eq.(v) and (vii),
..............(1.8)
(1/r1 + 1/r2)= 2/sinα*Q*dp’
The force acting along the pitch line in fig.(b) isPN,
P=PN= Pt / cosα .............(1.9)
The axial length of the gears is the face width b,i.e. l=b
Therefore,
σc = 1.4*Pt / b*Q*dp’*sinα*cosα* (1/E1+1/E2)
2 ............(1.10)

a load stress factor K is defined as,
K= σc
2* sinα*cosα* (1/E1+ 1/E2) /1.4 ...............(1.11)
Substituting the above eq. Into (1.1),
Pt= b*Q*dp’*k ...............(1.12)
This equation gives a relationship between the tangential force Pt and the
corresponding contact stress σc .
The tangential force is increase with the increasing contact stress. Pitting occurs when
the contact stress reaches the magnitude of the surface endurance strength. The
corresponding value of Pt is called wear strength.

Wear strength is maximum value of the tangential force that tooth can transmit without
pitting failure.
Replacing Pt by Sw,
Sw= b*Q*dp’*k .............(1.13)
Where,
Sw= wear strength of the gear tooth (N)
σc= surface endurance strength of the material (N/mm2)
equation (1.13) is known as Buckingham’s equation for wear
The ratio factor for internal gears is defined as,
Q= 2*zg/(zg-zp) .............(1.14)

Example for find out load-stress factor K,
The expression for the load-stress factor K can be simplified when both the gears are
made of steel with a 20° pressure angle.
In the special case,
(for steel)
E1=E2= 206000 N/mm2
α= 20°
according to G Niemann,
σc= 0.27 (BHN) kgf/m2
= 0.27(9.81) (BHN) N/mm2
Where BHN is Brinell Hardness Number.

K = σc
2*sinα*cosα*(1/E1+1/E2) / 1.4
= (0.27*9.81)2(BHN)2*sin(20) * cos(20) *(2/206000) /1.4
=0.156(BHN/100)2
Or
K=0.16(BHN/100)2
The above equation is applicable only when both the gears are made of steel with 20°
pressure angle.
In order to avoid failure of the gear tooth due to pitting, the wear strength should be
more than the effective force between the meshing teeth.

1. A pair of spur gears with 20° full-depth involute teeth consists of a 19
teeth pinion meshing with a 40 teeth gear. The pinion is mounted on a
crankshaft of 7.5 kW single cylinder diesel engine running at 1500 rpm.
The driven shaft is connected to a two-stage compressor. Assume the
service factor as 1.5. The pinion as well as the gear is made of steel
40C8 (Sut = 600 N/mm2). The module and face width of the gears are 4
and 40 mm respectively. A) Using the velocity factor to account for the
dynamic load, determine the factor of safety B) If the factor of safety is
two for pitting failure, find out (recommend) surface hardness for the
gears.
Step I Factor of safety based on dynamic load by
velocity factor

In helical gears, the two meshing gears may be
mounted on parallel or intersecting shafts. The
teeth on helical gear are cut at an angle (helix
angle) to the gear axis.
The helix angle usually ranges between 15º and
30º.
8
0
(Dr. S G Kolgiri)
Helical Gears

Spur gears, are merely helical gears with a zero
helix angle.
Carry more load than equivalent-sized spur gears
8
1
Helical Gears

Helical Gears» Helix Angle
The helix angle of helical gears is generally
selected from the range 6,8,10,12,15, 20, 30
degrees.
The larger the angle the smoother the motion and
speed possible however the
on the supporting bearings
thrust
also
the higher
loadings
increases.
8
2

Helical gears are classified as
Parallel helical gears Crossed helical gears and
Herringbone gears
All the helical gears generate thrust loads on the shafts
because of inclined teeth; hence, these must be taken
care while designing the machines.
Helical Gears» Classification
8
3

Helical Gears» Thrust & Radial Load
THRUST load is load parallel to the shaft of the gear. It is
produced by helical gears because the helix angle, not the
pressure angle. It is not produced by spur gears, which have
straight teeth that are parallel to the shaft axis.
84
RADIAL load is the load that tends to separate the gears. It
acts perpendicular to the shaft. This is what is produced by
the pressure angle. Both spur gears and helical gears produce
this kind of load.

Helical Gears» Parallel Helical Gears
85
• Refers to when the shafts are parallel to each other.
• Note that mating helical gears (on parallel shafts)
must have the same helix angle but the opposite
hand.

Helical Gears» Crossed Helical Gears
The shafts are non-parallel, and in this configuration the
gears are sometimes known as "skew gears"
86

Helical Gears» Herringbone Gears
Double helical gears overcome the problem of axial
thrust presented by single helical gears by using two
87
sets of teeth that are set in a V shape. This arrangement
cancels out the net axial thrust, since each half of the
gear thrusts in the opposite direction, resulting in a net
axial force of zero.
However, double helical gears are more difficult to
manufacture due to their more complicated shape.

Helical Gears» Herringbone Gears
88
To avoid axial thrust, two
helical gears of opposite hand
can be mounted side by side,
to cancel resulting thrust
forces.
Herringbone gears are mostly
used on heavy machinery.

Helical Gears» Left Hand & Right Hand Gear
89

Helical Gears» Advantages
90
The angled teeth engage more gradually than do spur
gear teeth causing them to run more smoothly and
quietly.
Helical gears are highly durable and are ideal for
high load applications.
Helical gears can be used on non parallel and even
perpendicular shafts, and can carry higher loads than
can spur gears.
At any given time their load is distributed over
several teeth, resulting in less wear.

Helical Gears» Drawbacks
One of the disadvantages of these gears is the thrust
which results along the gear axis, which needs to be
accommodated by using adequate thrust bearings.
There is a greater degree of sliding friction between
the teeth. This produces greater wear during
operation, and the need for lubrication systems.
The helical gear efficiency is lower due to the contact
between its teeth, which produces axial thrust and
generates heat. A greater loss of energy reduces
efficiency.
15

Helical Gears» Efficiency
92
Spur gears are the gears in which the teeth are parallel
to the gear axis, whereas in helical gear, the teeth are
not parallel to the gear axis.
Because of this arrangement of teeth in helical gears,
they have sliding contact between the teeth.
Due to the sliding contact between the teeth of helical
gear, axial trust of gear shaft also produces and also
produce more heat. Thus the efficiency of helical gear
is less than that of the spur gear.

Spur Gears Helical Gears
Teeth are cut parallel to the axis of the
shaft
Teeth are cut in the form of helix on
the pitch cylinder between
meshing
gears
Contact between meshing teeth occurs
along the entire face width of the tooth
Contact between meshing gears begins
with a point on the leading edge of the
tooth and gradually extends along the
diagonal line across the tooth
Load application is sudden resulting
into impact conditions and generating
noise in high speed applications
Pick up of load by the tooth is gradual,
resulting in smooth engagement and
quiet operation even at high speeds
Helical Gears» Helical vs Spur Gear
93

Helical Gears» Geometry
94
Spur gears have the diametral and circular pitches. Helical gear
geometry requires additional pitches. Figure shows a portion of the
top view of a helical rack.

95
The angle Ψ is the helix angle.
The transverse circular pitch (p) or circular pitch is measured on a plane
normal to the shaft axis (A-A plane).
The normal circular pitch (pn)is the distance between corresponding points of
adjacent teeth, measured on a plane perpendicular to the helix (B-B plane).
The axial pitch (pa) is the distance between corresponding points of adjacent
teeth, measured on a plane parallel to the shaft axis. For smooth transfer of
load, the face width of helical gear (b) is usually made at least 20% longer
than the axial pitch (pa).

96
So,
Diametral pitch spur gear Circular pitch spur gear
p
cos
pn pn  pcos
No.of teeth (z)
P 
per pitch circle diameter (d) No.of teeth (z)

d
p 
Pp 

97
So, normal and transverse diametral pitches are
and normal (mn) and transverse (m) module
Pp  Pn pn 
Pn P
n
p  pcos
 
cos
P
cos
Pn 
1 1
mn mcos
 mn  mcos

The Module of a Gear
is defined as inverse of diametral pitch
98

99
For axial pitch (pa)
The relation between the normal pressure angle (αn) in the
normal plane and transverse pressure angle (α) is
pa
p
tan
p
tan
pa 
tan
cos
tan
n

100
z  P cos
p 
d
 d 
zp

z
 zm 
zmn
d 
zmn
cos
The pitch circle diameter of the helical gear (d), is related to the
number of teeth (z), normal module (mn) and helix angle (ψ) as

101
Centre distance between the mating gears is
a 
d1 
d2 
z1mn 
z2mn
2 2 2cos 2cos
a 
mn z1  z2 
2cos

102
Centre distance between the mating gears is
a 
d1 
d2 
z1mn 
z2mn
2 2 2cos 2cos
a 
mn z1  z2 
2cos

Helical Gears» Virtual Number of Teeth
103
The virtual number of teeth for a helical gear may be defined as
the number of teeth that can be generated on the surface of a
cylinder having a radius equal to the radius of curvature at a point
at the tip of the minor axis of an ellipse obtained by taking a
section of the gear in the normal plane.

104
b
The shape of the tooth in
the normal plane is nearly
the same as the shape of a
spur gear tooth having a
pitch radius equal to radius
of curvature r’ at point P of
the ellipses is
a 2
r 


105
The equation of an ellipse with its centre at origin of an xy system
with a and b as the semi-major and semi-minor axes is
Also the formula for radius of curvature is

y 2

1
a2 b2
x2 3
d 2 y
dx2
By these two equations and by putting x = 0 and y = b, we get

dx 
2  2
  
dy
1 



a2
b

106

107
b
a2
r 

2cos
d
a 
Semi  major axis of the ellipse
b 
d
 Semi  minor axis 2
2cos2
d
r 

108
cos2
d  2r 
d
In the design of helical gears, an
imaginary spur gear is
considered in the plane A-A with
a pitch circle radius r’ and
module mn. It is called virtual
spur gear.

109
pn
d
mn
2r
z 

 
  
 2cos2
2
The number of teeth z’ on this
imaginary spur gear is called the
virtual number of teeth.
mn cos2
d
d 

 cos2


z    
mn

110
cos3
The design of helical gears is based on
virtual number of teeth (z’) and z is the
actual number of teeth..
cos
d
mn cos2
zmn
mn cos2
z


z 

Helical Gears» Tooth Proportions
111
As a general guide, tooth proportions should be based on a normal
pressure angle of 20°. In helical gears, the normal module mn
should be selected from standard values.
mn (in mm) = 1, 1.25, 1.5, 2, 2.5, 3, 4, 5, 6, 8 and 10
The standard proportions of the addendum and the dedendum are,
addendum (ha)
dedendum (hf)
clearance (c)
= mn
= 1.25 mn
= 0.25 mn (Tip and root clearance)

112
The addendum circle diameter da is given by
Similarly, dedendum circle diameter df is
da  d  2ha 
zmn  2mn
cos

 
 2

cos
z
da  mn

 2.5mn

zmn
d f  d  2hf
cos



 2.5

cos
d f  mn
 z

113
Trailing edge of
the tooth
Leading edge of the tooth
Leading edge of the tooth should be advanced ahead of the
trailing edge by a distance greater than the circular pitch
x  p
From theΔA1A2C
So,
 x  b tan
tan
A2C

x
A1C b
tan
b tan p or b 
p

114
And
Therefore,
This is the minimum face width
Many authorities recommended that the face width be at least
twice the axial pitch to obtain true helical-gear action.
sin
tancos
tan
tan
mn 
mn

p

m

sin
b 
mn

Problem 2.1: A pair of parallel helical gears consists of a 20 teeth
pinion meshing with a 40 teeth gear. The helix angle is 25° and the
normal pressure angle is 20°. The normal module is 3 mm. Calculate:
(a)Transverse module
(b)Transverse pressure angle
(c)Axial pitch
(d)Pitch circle diameter of the pinion and the gear
(e)Centre distance
(f)Addendum and Dedendum circle diamters of the pinion
Helical Gears» Numerical Problem» 2.1
115

Helical Gears» Numerical Problem»
Prob. 2.1 solution
116
Transverse Module
Transverse PressureAngle
mn 3
 3.31mm
m  
cos cos25
cos25
cos
tan
tan
n 
tan 20
 21.88
Axial Pitch
p m 3
pa 
tan

tan

tan 25
 22.3 mm

Prob. 2.1 solution
117
Pitch circle diameter of pinion and gear
Centre distance
z pmn 203
d p 
cos

cos 25
 66.2 mm
2

66.2 132.4
 99.3mm
2
d p  dg
a 
zg mn 403
dg 
cos

cos25
 132.4 mm

Prob. 2.1 solution
118
Addendum and Dedendum circle diameters of pinion
z 20


cos25
 2  72.2 mm



 2

 3
da  mn

cos
z 20


cos25
 2.5  58.7 mm



 2.5

 3
d f  mn

cos

Problem 2.2: A pair of helical gears consists of a 25 teeth pinion
meshing with a 50 teeth gear. The normal module is 4 mm. Find the
required value of helix angle, if the centre distance is exactly 165 mm.
Helical Gears» Numerical Problem 2.2
119

Prob. 2.2 solution
Helix angle from equation
a 
mn z1  z2 
2 cos
120
425 50
2cos
165 
2cos
425 50 1.818
165
cos 0.909
 24.619

Problem 2.3: A pair of parallel helical gears consists of a 20 teeth
pinion and velocity ratio 3:1. The helix angle is 15° and the normal
module is 5 mm. Calculate:
(i)Pitch circle diameter of the pinion and the gear
(ii)Centre distance
Helical Gears» Numerical Problem
2.3
121

Prob. 2.3 solution
122
(i) Pitch circle diameters
z pmn 205
d p 
cos

cos15
 103.53 mm
zg mn 605
dg 
cos

cos15
 310.58 mm

Prob. 2.3 solution
(ii) Centre distance
a 
mn z1  z2 
2 cos
123
520  60
2cos15
a 
a  207.06 mm

Problem 2.4: Twenty degree full depth involute profiled 19-tooth
pinion and 37-tooth gear are in mesh. If the module is 5 mm, the centre
distance between the gear pair will be
(a)140 mm.
(b)150 mm
(c)280 mm
(d)300 mm
2.4
124

Problem 2.5: If α = helix angle, and pc = circular pitch; then which one
of the following correctly expresses the axial pitch of a helical gear?
(a.) (c) (d)
2.5
125
(b) pc cos
tan cos
pc pc pc sin

Problem 2.6:
Assertion (A): Shafts supporting helical gears must
have only deep groove ball-bearings.
Reason (R): Helical gears produce axial thrusts.
(a)Both A and R are individually true and R is the correct explanation
of A.
(b)BothA and R are individually true but
R is not the correct explanation of A
(c)A is true but R is false
(d)A is false but R is true
2.6
126

Problem 4.7:
Assertion (A): While transmitting power between two parallel shafts,
the noise generated by a pair of helical gears is less than that of an
equivalent pair of spur gears.
Reason(R): A pair of helical gears has fewer teeth in contact as
compared to an equivalent pair of spur gears.
(a)Both A and R are individually true and R is the correct explanation
of A
(b)BothA and R are individually true but
R is not the correct explanation of A
(c)A is true but R is false.
(d)A is false but R is true
4.7
127

Fa

F
 n
Fr

Ft

Helical Gears» Force Analysis

Resultant force F acting on the
tooth of helical gear
Ft = tangential component
Fr = radial component
Fa = axial component
A
B
D
C
FromΔABC
Fr  F sinn
BC  F cosn
E

A
B
D
C
FromΔBDC
Fa  BC sin
 F cosn sin
Ft  BC cos
 F cosn cos
Dividing Fa and Ft
Fa 
F cosn sin
 tan
Ft F cosn cos
Fa  Ft tan
E

A
B
D
C
Dividing Fr and Ft
cos
Ft F cosn cos

tann
Fr 
F sinn

 cos 
tan 
n
r t
F  F


E

A
B
D
C
Ft
FromΔEDC
tan
Fr
E
and
cos
Ft
tan
n 
Fr cos tancos
Ft
Fr
BC
tann 
Fr 
tan
cos
tan
n

133
The tangential component is calculated from the relationship
2
d
Mt
t
F 
Mt = transmitted torque (N-mm)
d = pitch circle diameter (mm)
The following information is required in order to decide the direction
of three components
(i)Which is the driving element? Which is driven element ?
(ii)Is pinion rotating in clockwise or anticlockwise ?
(iii)What is hand of helix? Is it right handed or left handed ?

134

135

Problem 2.8: A pair of parallel helical gears, 5 kW power
at 720 rpm is supplied to the pinion A through its shaft. The
normal module is 5 mm and the normal pressure angle is
20°. The pinion has right-hand teeth, while the gear has left-
hand teeth. The helix angle is 30°. The pinion rotates in the
clockwise direction when seen from the left side of the
page. Determine the components of the tooth force and
draw a free-body diagram showing the forces acting on the
pinion and the gear.
2.8
136

Prob. 2.8 solution
137
Power (kW) = 5
nA = 720 rpm
zA = 20
zB = 30
mn = 5 mm
ψ = 30°
αn = 20°

Prob. 2.8 solution
138
Components of tooth force
Ft = tangential component
Fr = radial component
Fa = axial component
As we know
 
 
2
Ft  A
 dA 
Mt  
A
2n
60106  Power(kW )
Mt A 
cos
ZAmn
dA 

Prob. 2.8 solution
So,
266314.56)
115.47
1148.6 N
Ft 
Mt A   66314.56 N-mm
60106 5
2720
A
139
cos30
d 
Z Amn 
205 115.47 mm
cos
Design of Machine Elements II
(Dr. S G Kolgiri)

Prob. 2.8 solution
So,
266314.5611)
(5.47
1148.6 N
Ft 
Fa  Ft tan
Fa  Ft tan 1148.6tan30 663.14 N



 cos
 tann 
Fr  Ft
 482.73 N
140
cos  
 
cos30
tan 20
  1148.6



Fr  Ft
 tann 

Prob. 2.8 solution
ts II MEC306
7
Free-body diagram of forces
The direction of three components on
the gear at the point-2 will be opposite
to that of the pinion.

Helical Gears» Beam Strength
• For Beam strength, the helical gear
is considered to be equivalent to a formative
spur gear.
• The pitch circle diameter of this gear is d´,
the number of teeth is z´ and the module mn.
• The beam-strength of the spur gear is given
by,
• Sb  mbbY 142
Beam strength (Sb) is the maximum value of the
tangential force that the tooth can transmit without
bending failure.

• For Beam strength, the helical gear
is considered to be equivalent to a formative
spur gear.
• The pitch circle diameter of this gear is d´,
the number of teeth is z´ and the module mn.
• The beam-strength of the spur gear is given
by,
• Sb  mbbY 143
Beam strength (Sb) is the maximum value of the tangential
force that the tooth can transmit without bending failure.

144
beam strength
to the tooth
Sb = (Sb)n =
perpendicular
element
m = mn = normal module
The beam-strength of the spur
gear is given by,
Sb  mb bY
cos
b
b 
b
b
S  mb
m b Y
n b
cos
n
b
 Y  S  

lements II MEC306
Bhagi)
7
Sb  mnbbY
cos
n
b 
mnbbY
 S 
Y = Lewis form factor based on
virtual number of teeth z´.
Sb is the component of (Sb)n in
the plane of rotation. Thus,
Sb  Sb n cos
In order to avoid the failure of gear
tooth due to bending, the beam
strength should be more than the
effective force between the
meshing teeth.
eff
b
S  F

Helical Gears» Effective Load on Gear Tooth
146
But when gears rotate at high speed then it is necessary to
consider the dynamic force resulting from the impact between
mating teeth.
t t
t
M  2M


2

When gears rotate at very low speed the tangential load (Ft) can
be considered as actual force between two meshing teeth.
 d  d
F 

147
The dynamic force is induced due to:
(i) Inaccuracies of tooth profile
(ii) Errors in tooth spacing
(iii) Elasticity of parts
There are two methods to account for dynamic load:
(i)Approximate estimation by velocity factor (Cv) in the preliminary
stages of gear design and
(ii)Precise calculationby Buckingham’s equation in the final stages of
gear design

148
Cs = service factor and Cv = velocity factor
The velocity factor for helical gears is (v >20m/s)
Where, v is the pitch line velocity in m/s
Cv
Feff
In the preliminary stages, the effective load Feff between two
meshing teeth is given by

Cs Ft
Cv 
5.6
5.6  v

149
In the final stages of gear design, the dynamic load is calculated
by equation derived by Earle Buckingham. The dynamic load is
given by,
 
2 Ft cos
Fd 
21v Cebcos
21v  Ceb cos2 Ft 
Fd = dynamic load; v = pitch line velocity
e = sum of errors between two meshing teeth
b= face width of tooth; Ft = tangential force due to rated
torque

150
The effective load is given by
Where, Fd is the dynamic load or additional load due to dynamic
conditions between two meshing teeth.
Feff  Cs Ft  Fd 

Helical Gears» Wear Strength
151
To calculate wear strength, pair of helical gears is considered to
be equivalent to a formative pinion and a formative gear in a
plane perpendicular to the tooth element. The wear strength of
the spur gear is given by
Sw  bQd p K
Wear strength is the maximum value of the tangential force that
the tooth can transmit without pitting failure.

152

153
Equation (18.24) is known as Buckingham’s equation of wear strength.
Equation (18.24) gives wear strength in the plane of rotation. Therefore,
wear strength (Sw) indicates the maximum tangential force that the tooth
can transmit without pitting failure. It should be always more than the
effective force between the meshing teeth. The virtual number of teeth on
the pinion and gear are zp and zg respectively. The ratio factor Q for
external helical gears is given by,

154
The ratio factor Q for external helical gears is given by,

The pressure angle in a plane perpendicular to the tooth element
is an. The K factor in Eq. (18.24) is given by

Design of Machine Elements II
(Dr. S G Kolgiri)
Example .1 A pair of parallel helical gears consists of a 20 teeth
pinion meshing with a 100 teeth gear. The pinion rotates at 720 rpm.
The normal pressure angle is 20°, while the helix angle is 25°. The face
width is 40 mm and the normal module is 4 mm. The pinion as well as
the gear is made of steel 4OC8 (Sut = 600 N/mm2) and heat treated to
a surface hardness of 300 BHN. The service factor and the factor of
safety are 1.5 and 2 respectively. Assume that the velocity factor
accounts for the dynamic load and calculate the power transmitting
capacity of gears.

Assignment
A simple speed reducer is composed of 2 spur gears. The pinion gear has a
pitch diameter of 0.75” and 36 teeth while the driven gear has a pitch
diameter of 4.0” and 192 teeth.
1. What is the Diametral pitch of each gear?
2. If an electric motor rotating CCW at 3000 rpm is coupled to the pinion, what
is the rotational speed of the driven gear?
3. If the torque delivered to the pinion is 1 N-m, what is the torque on the
driven gear?
4. What is the power transmitted by the gear train?

Design of gearbox for Hoist. (Use AGMA approach)
Example:
A conveyor drive involving heavy-shock torsional loading is operated by an
electric motor, the speed ratio is 1:2 and the pinion has Diameteral pitch P=10
in-1, and number of teeth N=18 and face width of b=1.5 in. The gear has Brinnel
hardness of 300 Bhn. Find the maximum horspower that can be transmitted,
using AGMA formula.

References
• Design Data - P.S.G. College of
Technology, Coimbatore.
• . Bhandari, V. B. Machine Design data
book, Tata McGraw Hill Publication Co.
Ltd.
• C.S. Sharma and Kamlesh Purohit, Design
of Machine Elements, PHI Learning Pvt.
Ltd
• D.K. Aggarwal& P.C. Sharma, Machine
Design, S.K Kataria and Sons.

Design of Transmission Systems Using Spur and Helical Gears

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