MACHINE DESIGN (PART 2,GEARS)

GEARS

Types of years:

Depending on relation between the axes, shape, of solid on which teeth are developed, curvature of tooth – trace any other special feature, gears are categorized in to following types.

1.       Spur gear

2.       Parallel helical gears

3.       Herringbone gear

4.       St. bevel gears

5.       Spiral bevel gears

6.       Few bevel gears

7.       Hypoid gears

8.       Grossed helical gears

9.       Worm and worm wheel

Spur gears  are those gears in which the teeth are parallel to the axis of the gear. This is the simplest and least expensive type of gear. Spur gears can only be meshed if their axes are parallel.

Helical gears have their teeth at a helix angle, ----- with respect to the axis of the gear. Helical gears are either right-left-handed. When their axes are parallel, opposite hand helical gears mesh. When the axes are at an angle same hand helical gears mesh to form crossed helical gears. The helix angles can be designed to accommodate any skew angle between the non-intersecting shafts.

Helical gears are more expensive than super gears. But, they run quieter than spur gears because of the smoother and more gradual contact between the teeth as they come into mesh. Thus the sudden impact and consequent vibrations of spur gears are absent in helical gears. Also, for the same gear diameter and diametral pitch, the helical gear is stronger due to the thicker tooth form in a plane perpendicular to the axis of rotation.

Herringbone gears are formed by joining two helical gears of identical pitch and diameter but of opposite hand on the same shaft. These two sets of teeth are often cut on the same gear blank. The advantage compared to a helical gear is the internal cancellation of its axial thrust loads. Thus no thrust bearings are needed other than to locate the shaft axially. This type of gear is more expensive than a helical gear and is used in large, high power applications such as ship drives.

A spur gear set can be 98 to 99% efficient. A parallel helical gear set will be about 96- 98% efficient due to sliding friction along the helix angle. A crossed helical gear set will be only 50 to 90% efficient, because of the larger sliding of the meshing teeth.

If the helix angle of a helical gear is increased, it gives a spiral appearance. Such helical gears are called spiral gears.

If the helix angle is increased sufficiently, the result will be a worm which has only one tooth (in most cases) wrapped continuously around its circumference a number of items similar to a screw thread. The teeth are not involute over their entire face and therefore the centre distance must be maintained accurately to ensure conjugate action. Worms and gears are made as matched sets.

The worm gear sets have the advantage of very high gear ratios in a small package and can carry very high loads, because of the large surface area of contact between worm and gear. The disadvantage is very high sliding and thrust loads reducing the efficiency of the worm set to 40 to 85%

Another major advantage of the wormset is that it can be designed to eliminate back drive. A spur or helical gear set can be driven from either shaft. So, if the load being driven must be held in place after the power is shut off, they are not useful as they will backdrive, or a brake has to be provided. But the worm set can only be driven from the worm. The friction can be large enough to prevent it being backdriven from the worm gear.

If the diameter of the base circle of  a gear is increased to infinity, the base circle will become a straight line and the involute generating taut string will be pivoted at infinity thereby generating an involute profile that is a straight line. This linear gear is called a rack. Meshing a gear with the rack produces a rack and pinion arrangement that can convert rotary to linear motion or vice versa.

Bevel gears are used when the shaft angles are at any angle, including 9^o. Just as spur gears are based on rolling cylinders, bevel gears are based on rolling cones. The angle between the axes of the cones and the included angles of the cones can be any compatible values as long as the apices of the cones intersect. Otherwise, there would be a mismatch of velocity at the interface, as the periferal velocity of a cone is proportional to the distance of the periferal point from the apex.

If the teeth are parallel to the axis of the gear, it will be a straight level gear. If the teeth are angled with respect to the axis, it will be a spiral bevel gear similar to a helical gear.

Terminology and definitions

Terminology:

Pressure Angle: If a tangent is drawn to the involute profile of a tooth at any point on the curve and if a radical line in drawn through the point of tangency, connecting this point with the centre of the year, then the acute angle included between this tangent and the radial in defined an the pressure angle at that point.

Module: It is defined as the ratio of the pitch diameter to the number of teeth of a year. The value of module is expressed in millimeters.

         z – no. of leeth

Face: Surface above the pitch circle.

Flank: Surface below the pitch circle.

Circular pitch: It is the distance measured along the pitch circle between two similar points on adjacent teeth.

Diameteral pitch: This is the term used in gear technology in the FPS system.

The clearance C is the amount by which the dedendum of a gear exceeds the addendum of the mating gear.

The backlash is the amount by which the width of a tooth space exceeds the thickness of the engaging tooth measured along the pitch circles.

Pressure Angle

The pressure angle in a gear set is defined as the angle between the axis of a transmission or line of action (common normal of the mating profiles) and the direction of velocity at the pitch point. Pressure angles of gear sets are standardized at a few values by the gear manufacturers. The standard values are 14.5^o , 20^o, and 25^o, with 20^o being the most commonly used and 14.5^o now being considered obsolete. Gears to be run together must be cut to the same nominal pressure angle.

Law of Gearing:

The common normal to the tooth profile at the point of contact should always part through a fixed point called pitch point, in order to obtain a constant velocity ratio, from Fig 1.0, O₁ and O₂ are the centers of the two gears rotating with angular velocities w₁ and w₂ respectively C is the point of contact between the teeth of the two gears and NN is the common normal at the point of contact.

CA in the velocity of point C, when it is considered on gear 1, while CB in velocity of point C, when it is considered on gear2. Also

The projections of the two vectors   and , i.e. CD along the common normal NN must be same. Otherwise the teeth will not remain in contact and there will be a slip.

       (a)

      (b)

Since  and  are similar,

              (c)

Similarly   and  CDB are similar

              (d)

From (c) and (d)

    (e)

From (a) (e)

               (f)

Since  and  are similar, therefore,

              (g)

From (f) & (g)

  

Spur Gear Design

The pitch circle is the curve of intersection of the pitch cylinder and a plane, which is involute to the axis of shaft. The pitch circle diameter is denoted by d’’. The base circle in the imaginary circle from which the involute curve of the teeth profile in generated.

Beam Strength of Gear Teeth: Lewis Equation:

The beam strength of year teeth is determined from an equation known as Lewis equation and load carrying ability of the toothed gear as determined by this equation.

Lewis assumed that as the load is being transmitted from one gear to another, it is all given and taken by one tooth, because it is not always safe to assume that the load is distributed among several teeth. When contact begins, the load is assumed to be at the end of the driven teeth and a contact ceases, it is at the end of the driving teeth this may not be true when the number of teeth in a pair of making gear is large, because the load maybe distributed among several teeth. But it is almost certain that at same-time during the contact of teeth, the proper distribution of load doesn’t exist and that one tooth must transmit full load.

Consider each tooth as a cantilever beam loaded by a normal load (W_N) an in fig (b). It is resolved into two components, i.e. tangential component (W_T) and radial components (W_R) acting  and parallel to the centerline of the tooth respectively. The tangential component (W_T) induces a compressive stern, which tends to break the tooth. The radical component (W_R) induces a compressive stern of relatively small magnitude; therefore its effect on the tooth may be neglected.

M = Maximum bending moment at critical section

W_T = Tangential load acting at the tooth

  = Length of tooth

Y = Half the thicken of the tooth (t) at critical section

Let  and

where x and k are constants

  

Substituting ; we have.

The quantity y is known as Lewis form factor.

Permissible Working Stress for Gear Teeth in the Lewis equation.

The permissible working strem in the Lewis equn depends up on the material for which an allowable static strem may be determined. The allowable static strem in the stern at the elastic limit of material. It is also called the basic stern. In order to account for the dynamic effects, which become more severe on the pitch line velocity increases, the value of  is reduced. According to Barth formula, the permissible   working strem,

 Allowable static strem

 Velocity Factor

The values of the velocity factor (CV) are given as follows.

 for ordinary cutting gears operating velocities upto 12.5 m/s.

 for carefully cut gears operating at velocities up to 12.5 m/s.

 for very accurate cut and ground metallic gears operating at velocities up to 20 m/s.

   for precision years cut which high accuracy and operations at velocities up to 20 m/s.

   for non-metallic gears.

Dynamic Tooth Load:

The dynamic loads are due to the following reasons:

1.       Inaccuracies of tooth spacing.

2.       Irregularities in tooth profiles.

3.       Deflections of teeth under load.

       

who C in the deformation factor:

K=A factor depending up on the form of the teeth

= 0.107 for 14½^o full depth involute system

= 0.111 for 20^o full depth involute system

= 0.115 for 20^o stub system.

Static Tooth load:

The static tooth load in obtained by Lewis formula by sustaining flexural endurance limit or elastic limit strem ( ) in place of permenable working system.

Causes of Gear Tooth Failure:

1.       Bending Failure

2.       Pitting

3.       Scoring

4.       Abrasive wear

5.       Corrosive wear

Design Procedure for Spur Gear

Step-1:

Select Material for gear and pinion.

Pg 186 table 12.7 → Data book (Mahadevan)

Step-2:

Find tangential tooth load (Ft) from power transmelled; equation (12.20(a))

 when p is kw.

Step3:

Determination of Lewis form factor for gear and pinion (y) e.g: 12.27

Form factor

Step4:

Determination of strength factor for gear and pinion.

Step5:

Apply Lewis equation to the weaker of two wheels. Determine tangential load (Ft) equation; 12.5

Step6: Determination of module ‘m’ in mm. Equate tangential load obtained from power transmission; equation 12.20(a) to the tangential load obtained from equation.

Step7:

Calculation of Dynamic Load (Fd) using bucking hums equation (12.33)

 Step8:

Check the design for wear Condition:

                       equation 12.36

Step9:

To check dynamic strength Fs;’

                      

Fs is determined from 12.7

Step10:

Determination of Gear proportion

Table 12.1

Causes of Gear Tooth Failure:

1.         Bending failure

Every gear tooth acts as a cantilever. If the total respective dynamic load acting on the gear tooth is greater than the beam strength of gear tooth fact in Bending.

2.         Pitting

It is the surface of fatigue failure, which occurs due to many repetitions of Hertz contact strens.

3.         Scoring

The extensive heat is generated when there is a extensive surface pressure high speed or supply of lubricant facts.

4.         Abrasive Wear

The foreign particle in the lubricant such as dirt, dust, or burr enters between tooth and damage the form of teeth.

5.         Corrosive Wear

The corrosion of the tooth surface in is mainly caused due to the presence of corrosive environments such as additives present in the lubricating acts.

Spur Gear Construction

The gear construction may have different designs depending upon the size and its application. When the dedendum circle dia is slightly greater than the shaft dia, the pinion teeth are cut integral with the shaft.

Small gears up to 250 mm pitch circle dia are built with a web; thicknen is generally equal to half the circular pitch on it may be taken as 1.6 mm to 1.9 m  where m is the module.

S.No.	Pitch circle dia	Number of Armo
1	upto 0.5 m	4 or 5
2	0.5 – 1.5 m	6
3	1.5 – 2 m	8
4	Above 2.0 m	10

Design of Shaft for Spur Gear:

Step-1:          Calculation of normal Load (F_N)

Equation 12.24a

Step-2:          To find Wt. of the gear;

Step-3:          To find resultant F_N and Fg

Step-4:          Calculation of bending moment;

x → Distance between the centre of gear and the centre of bearing

Step-5:          Calculation of twisting moment;

Step-6:          To calculate equalent torque;

Step-7:          To determine dia of shaft;

→ shear stren for the selected material.

Design for Gear Arms:

Step-1:          To determine hub dia for year and pinion

Hub dia = 1.8 * bore

Bore= Dia of shaft

Step-2:          Length of the hub;

Step-3:          To determine no. of arms;

Table 12.18 page: 194

Step-4:          To find stalling load and section modules;

Equation 12.40 12.41

by m

Step-5:          Equation 12.42 to find out major dia.

Step-6:          Design of rim

Qn. 1. A compressor running at 300 rev/min is driven by a 15 kW, 12000 rev/min motor through a 14½^o full depth gears. The centre distance is 0.375 m.  The motor pinion is to be of C-30 forged s hardened and tempered, and the driven gear is to be of cast steel. Assuming medium shock condition hardened and tempered, and the driven gear is to be of cast steel. Assuming medium shock condition

a) Determine the module, the face width, and the number of teeth on each gear.

b) Check the gears for dynamic load and wear

c) Design the drive completely

Solution: Now centre distance

Hence D_p = 0.15 m and D_g = 0.60 m since the speed ratio is

Now pitch line velocity V =

Now design power = rated power × C₀

From table 8, C₀ = 1.25

 Design Power = 1.5 × 1.25 = 18.75 kW

Now design transmitted load

Now velocity factor or dynamic factor C_v is given as

 for carefully cut gears with V upto 12.5 m/s

(a) Lewis Equation: We shall apply the Lewis equation to gear for preliminary calculations. For trial purposes, let us take

f = 10 m for Y = 0.25 as listed in the design procedure.

Now,

From table 7

f_b, design stress is 140 MPa, for Cast Steel

Now K_f =  stress concentration factor

 = 1.6 for full depth 14½^o teeth

m = 0.0053 m

Now from Table 2 let us take the standard module of 5mm

z_p = number of teeth on pinion

z_g = number of teeth on gear

Now the exact value of Lewis form factor Y can be determined.

For pinion,

For gear,

Now applying the Lewis strength equation for pinion and gear to find the tooth face width or pinion:

= 0.0277 m, (f_b =  224 MPa (Table 7)

Let us take the value ‘f’ as 45 mm, both for pinion and gear, since it should be between 9.5 m to 12.5 m.

Gear Tooth Proportions:

Addendum, a = m = 5 mm

Dedundum, d = 1.25 m = 6.25 mm

Clearance = 0.25 m = 1.25 mm

Total depth = a + d = 11.25 mm

Working depth = 11.25 – 1.25 = 10 mm

Circular pitch, p_c = m = 15.71 mm.

Now from Table 22.4, for m = 5 mm and V > 8 m/s, backlash = 0.28 mm.

Now tooth thickness + width of space = circular pitch

 (Width of space – backlash) + width of space = circular pitch

Width of space =

Tooth thickness = 7.995 – 0.28 = 7.715 mm.

Diameter of Addenum circle for pinion = D_p = 2a = 0.15 + 2 × 0.05 = 0.16 m

Diameter of Addendum circle for gear = D_g + 2a = 0.60 + 2 × 0.005 = 0.61 m

Diameter of Dedundum circle for pinion =

         

Diameter of dedendum circle for gear

(b) Check for dynamic Load: By Buckingham’s equation

Now for V = 9.42 m/s, the permissible error from Table 10 is approximately 0.040 mm.  For module equal to 5 mm, a class 2 gear would have an error of 0.028 mm and class 1 gear of 0.056 mm.  Since the permissible error is 0.040 mm, therefore, a class 2 gear must be used with error of 0.028 mm.

Now

Now beam strength for Gear, (gear is weaker than pinion, compare f_b.Y)

Now from Table 6, f_ef = 225 MPa with BHN = 160.

which is > 1.5 F_d  safe

Now for pinion, let us take

 BHN of pinion material  =

Check for Wear: The limiting wear load is,

Now D_p =  0.15 m . f = 0.045 m

f_es = surface endurance limit = 2.8 × BHN  - 70 MPa

Now average value of BHN for gear pair = 155

Now E_p = E_g = 210 GPa

Q-ratio factor

 which is quite less than F_d,

 The greater pair will fail in wear. Therefore, it would be necessary to heat treat the gear pair to an average BHN of 315.

; hence O.K., also   

(c) Design of Drive:

(i) Pinion: Now since the diameter of the pinion is 0.15 m and the limit of diameter for the provision of web, is

mm or 0.2075 m

 The pinion will be provided with a web and not arms.

web thickness = 1.6 m to 1.9 m.

ie, 8 mm to 9.5 mm, let us take 9 mm

Rim thickness: t_r = 1.6 m.i.e, 8 mm

Shaft: The shaft should be designed for the maximum torsional moment to be transmitted and not for the useful torsional moment. Maximum torsional moment will be induced in the shaft when the driver is started from stationary position.  The tooth load acting at the pitch line at that moment will be, corresponding to the stalling load.

Maximum Tangential load =

Torsional moment on the pinion shaft =

Now

Neglecting the weight of the pinion, if we consider the pinion to be over hanging.

Bending moment, M_b = F_n × over hang.

Let us take overhang, distance from the centre of pinion to centre of bearing 0.10 m.

Equivalent torsional moment,                                  

Now the shaft material is usually C – 45 and C – 55 steel with ultimate tensile strength of 450 MPa and 550 MPa. Let us choose C – 55 steel.

Now shear stress (ultimate) = 60% of 550 = 330 MPa.

Allowing for key way,

Now as per procedure given in Chapter 16,

Let us take

, say 40 mm

Hub diameter = 40 × 1.65 = 66 mm

Hub length = 1.25 × 40 = 50 mm

(ii) Gear: Since the diameter of gear is 0.60 m, six arms will be provided.

Arms Stalling Load F₀ = 6161 N

Bending moment, per arm,

Taking elliptical section for the arm. Let us take F.O.S. as 5 as already discussed.

Design stress in bending

Now section modulus, b = major axis at hub

 × 110 × 10⁶

 b = 0.385 m that is 40 mm

 minor axis at hub = 20.0 mm

Now major axis at rim end = b – taper =

 Minor axis at the rim end = 10.25 mm

Rim: The rim thickness is given as

Now, z = 120, and z_a = 6

Depth of circumferential rib, h = t_r = 11 mm

Thickness of rib, w = thickness of arm at this end = 10.25 mm

Shaft: Torsional moment on the shaft may be considered as the stalling torsional moment to be transmitted. Therefore,

Now

Neglecting the bending moment due to the weight of gear, bending moment on the shaft,

, say 50 mm

Hub diameter = 1.8 d = 1.8 × 50 = 90 mm

Length of Hub = 1.25 d = 1.25 × 50 = 62.5 mm

Qn. 2. Determine the proper pitch, module, face, number of teeth, and outside diameters of a pair of 20° involute full depth spur gears to transmit 112.5 k W, from a pinion running at 750 rev/min to a gear running at 140 rev/min. The service is intermittent with light shocks.

Solution: Since nothing is mentioned about diameters, centre distance, materials and pitch line velocity, suitable assumptions will have to be made for these and then the drive calculated and if need be suitable changes made afterwards.

Pitch line velocity : Since the speed of pinion and of gear is not very high, a moderate value of pitch line velocity may be selected. As the pitch line velocity for medium velocity drives ranges from 3 to 15 m/s let us choose a velocity of 8 m/s.

Materials: Suitable materials for pinion and gear may be taken as : Forged C-30 steel with ultimate tensile strength of 500 MPa and Cast Iron, grade 35, heat treated, with ultimate tensile strength of 350 MPa respectively, from Table 22.6.

Now the approximate pitch diameter of pinion.

(1) Design load:

Now useful transmitted load,

Now,

If the power source is taken as electric motor, then since the service is intermittent,

= 1.25 × 0.8 = 1.00

Design tangential load, F_t =  1.00 × 14.0625 = 14.0625 kN

Gear Strength: The Lewis equation is:

Let us take Y = 0.29 and f = 10 m, K_f = 1.5

Let, us apply the gear strength equation for the pinion.

FS = 3

Design stress

m = 0.011 m, that is 11 mm.

Let us choose a standard value of ‘m’ as 10 mm (First choice, Table 2)

Now number of teeth on pinion = , minimum needed.   O.K.

Let us make z_p = 20

Exact,

 and

Exact value of Y, for pinion

m = 0.010 m, that is , 10 mm

Chosen standard value of m = 10 mm is O.K.

Now from the Lewis equation for pinion,

For Gear:

Making z_g =  107, will given n_g = 749 rev/min. ie, an error of only 0.133% which is negligible.

Now since f lies between 9.5 m to 12.5 m;   Let us take f = 110 mm.

Outside diameters:

(Da) p = m(z_p + 2) = 10(20 +2) = 220 mm.

(Da) g = m(z_g + 2) = 10(107 + 2) = 1090 mm

Check by Buckingham’s equation for dynamic Load;

Now

Now for V = 7.83 m/s, the maximum permissible error from Table 10 is approximately 0.047 mm. For module equal to 10 mm, a class 2 gear would have an error of about 0.045 mm.  Therefore, a class 2 gear must be used, with e = 0.045 mm.

Now

For 20^o full depth profile, k = 0.111.

Now strength factor for pinion,

and for gear,

 Gear is weaker.

Now Beam strength of gear = f_ef f . m . Y . N

Now f_ef for C.I. with BHN 160 = 84 MPa

F_b = 84 × 10⁶ × 0.11 × 0.01 × 0.457 = 42.27 kN

Now, for light shocks, , that is .

Now from Table 6, iron grade 35 heat treated has BHN = 300.

The material would have much higher value of endurance limit (in bending) and hence beam strength.  So, the design is safe for dynamic load.

Now f_ef for steel pinion =

 BHN for steel pinion (required) =

Check for Wear

Now Limiting wear Load,

D_p =  0.20 m . f = 110 mm

Now

Now from table 22.6 BHN value for the forged C-30 steel with f_ut = 500 MPa, is 143.  Now from Table 11 for steel pinion with BHN = 150 and C.I. gear, f_es = 350 MPa.

which is less than F_d,   The design is not safe for wear.

Let us make , say

Now from Table the required BHN for steel pinion will be about 225.

HELICAL GEARS

Axial pitch: Distance measured between teeth

Normal pitch: Perpendicular distance between two teeth

For helical gears, face width

b=12.5 mm - 20 mm

Virtual number of gears

In helical gears the plane normal to gear teeth intersects pitch cylinder to form an ellipse. In the design of helical gear, imaginary spur gear is considered in the plane of ellipse. This is known as formative or virtual spur gear. No. of teeth on this imaginary spur gear is known as virtual no. of teeth.

Forces Acting on a helical Gear:

(1)     Tangential Load

                                 -F_t

(2)     Radial Load

                                 -F_n

(3)     Axial Load

                                 -F_a

→ Angle made by the resultant on horizontal plane angle.

Design procedure for Helical and Herringbone gear:

Step-1:          Selection of material for gear and pinion

Table 12.7 page186

Step-2:          Tangential tooth Load Ft, from power transmitted

Equation: 12.20(a)

Step-3:          Determination of virtual no. of teeth (Ze)

Equation: 12.52(a)

Step-4:          Determination of Lewis form factor for Gear and pinion.

Equation: 12.17

Step-5:          Determination of strength Factor

Step-6:          Apply Lewis equation for weaker of 2 wheels and determine tangential Load F_t.

Equation 12.59

→ Velocity Factor – Equation 12.61.

→ Wear and Lubrication factor [Table 12.21 page 196]

b = 12.58 a for helical.

Step-7:          Determination of module (m)

Equation (S₂) and equation (S₆).

Step-8:          Calculation of dynamic Load

Equation 12.62

Step-9:          To check limiting load for wear.

Equation 12.64

Step-10:        To check dynamic strength (F_s)

Equation 12.63.

Step-11:        To Determine axial thrust (F_a)

Equation 12.51

Step-12:        To Determine Helical tooth proportion          

Equation 12.20

Design For Shaft:

Step-1:          Determination of F₁

Step-2:          Determination of F_a

Step-3:          Bending Moment due to F_t

M₁=Ft×*

Where is the overhang distance.

Step-4:          Bending Moment due to axial Load

Step-5: Resultant bending Moment

Step-6:          Twisting moment

Step-7:          Equalent torque

Step-8:          Diameter of shaft

 [Page 43 3.5b]

K=o for solid shaft.

Helical Gear Construction:

The procedure of helical gear construction is similar to that of spur gear construction.

(1)     Hub dia.

(2)     Length of Hub

(3)     Design for web if pitch dia is < 250 mm

(4)     No. of arms (Table 12.8) Page 194

(5)     Staking Load F (12.40)

(6)     Section modules 12.41

(7)     Width of elliptical arm (D) 12.42

(8)     Thickmen of rim J (page 168)

Example 1 :  Two helical gears are used in a speed reducer that is to be driven by an internal combustion engine.  The rated power of the speed reducer is 75 kW at a pinion speed of 1200 rev/min. The speed ratio is 3 to 1.  Assume medium shock conditions and 24 hr operation.

(a) Determine the module, face, number of teeth in each gear, and the material and heat treatment if the teeth are 20^o full depth in the normal plane.

(b) Design the arms and shaft  for the gear.

Solution: The power to be transmitted is 75 kW at a pinion speed of 1200 rev./min. both the values are neither low nor high, so let us assume a peripheral pitch line velocity of 10 m/s.  Therefore, the pitch diameter of the pinion is:

  Exact

Tangential Design Load:

Now service factor, form Table 22.8 C₀ = 1.50 × 1.25 = 1.875

Now

Lewis Equation:

         

Let the number of teeth in the pinion be 18.

Now ‘m’ ie, module in the plane of rotation

Since the teeth are to be hobbed, a standard value of ‘m_n’ should be chosen.

Let m_n = 8 mm

or 

 

Let  m_n = 7 mm,

The values of  for single helical gear lies between 20^o and 35^o. So both the values are correct. Let us take y = 29^o.

Now,

Pressure angle, in the plane of rotation , which is within the limits of  and . Now formative number of teeth in the pinion.

Material for the pinion may be taken as C-30 forged steel, from table 22.7. f_b = 175 MPa. BHN = 150.

\ From the Lewis equation, the active width of the pinion,

Now minimum face width is given as:

say 66 mm is OK.

Check by Buckingham’s equation for dynamic load:

Now for V = 10.05 m/s, the maximum permissible tooth action error from Table 22.10 is 0.038 mm. From Table 22.9, for module equal to 8 mm, the probable error for class 2 gear is 0.04 mm and for class 3 gears it is 0.020 mm. Since the maximum permissible error is 0.038 mm, therefore, use class 3 gears with probable error of 0.02 mm.

k = 0.111

Let the material for gear be also steel, its exact composition will be determined later.  Let It be cast steel with f_b = 140 MPa (Table 22.7) with BHN = 180.

Now for gear,

Now strength factor for pinion, f_b . Y_f = 175 × 0.388 = 67.9 and strength factor for gear,

Therefore, gear is weaker,

Now

 = 12.81 kN.

Now Beam strength of gear.

Now it is clear from the Table 22.7 that the BHN of gear material = 180.

Now Beam strength is greater that 1.5 × F_d, therefore, the design is safe for dynamic loading.

Check of Wear:

The average BHN of gear pair =

 of teh gear pair = 2.8 × 165 – 70 = 392 MPa.

which is the less than F_d, therefore, the teeth will fail in wear.

Heat treat the gear pair to an average BHN value of 300.

 will come out to be = 1381.3, kN/m².

and  which is      safe

Final values are:

m = 8 mm; m_n = 7 mm; f = 66 mm; z_p = 20; z_g = 60

(b) Arms: Gear diameter will be = 0.16 × 3 = 0.48 m.

Let us take number of arms = 4

Now stalling load,

 Bending moment per arm =

Taking elliptical section, with major axis = 2 minor axis

Let 

 b = major axis at the hub centre = 0.054 m = 54 mm.

a = minor axis at the hub centre = 27 mm.

Taper of arm = 65 mm per metre length = 15.6 mm.

\ Major axis at the rim section = 54 – 15.6 = 38.4 mm.

\ Minor axis at the rim section = 19.2 mm.

Gear Shaft: Maximum useful turning force,

F₀ = 23.25 kN

Since the bearing positions are not given, let us assume that gear is over hanging.

Axial force, F_a =  23.25 × tan29^o = 23.25 × 0.554 = 12.88 kN.

B.M. due to F₀ = 23.25 × overhang (let overhang = 0.12 m)

B.M. due to

Since these act at right angles to each other, therefore, resultant, B.M. is:

Torsional moment on gear shaft,

\  d  = 0.0755 m, say 76 mm.

Direct compressive or tensile stress on the shaft due to F_a.

Now actual

\  Principal shear stress

             \ safe

                                                *DATA BOOK

                                                                K MAHADEVAN

                                                                K BALAVEERA REDDY

BEVEL GEARS

Design Procedure for Bevel Gear:

Step-1:          Selection of material for gear and pinion.

Table 12.7

Step-2:          Tangential tooth Load (Ft).

Equation [12.91a] from power transmelled.

 Step-3:         Determination of pitch angle for gear and pinion.

For rt angle 12.72.

For obtuse angle 12.73

For acute angle 12.67 and 12.68

Step-4:          Determination of formative no. of gear for gear and pinion

Equation 12.81

Step-5:          Determination of Lewis form factor

Step-6:          Determination of strength Factor

Step-7:          Lewis equations to weaker of two wheels and determine Ft. equation 12.86

Velocity factor Cv.

Step-8:          Determination of module m and select value from 12.23 pages 197

Step-9:          Calculation of dynamic load [Fd]

Step-10:        To check wear strength Fw equation 12.89

Step-11:        To check endurance strength Fen

Fen > 1.25 d.

Step-12:        Determination of effective tooth load.

Fte, Fr, Fa.

Step-13:        Bevel gear teeth proportion.

Step-14:        Outside dia of gear and pinion.

Example . 1 Design a bevel gear drive between two shafts whose axes are at right angles.  Speed of pinion shaft is 240 rev/min and that of the gear shaft is 120 rev/min. Pinion is to have 21 teeth of involute profile with module of 20 mm and a pressure angle of 20^o and is to be of suitable material. Gear is of cast iron. Power at gear shaft = 75 kW.

Solution: Now

Since velocity ratio is 2 : 1

Now pitch cone angles may be determined as below:

Now cone distance,

Design Load: Pitch line velocity, V =

Assuming electric motor drive and moderate shock conditions, with 24 hours per day running, the service factor or the overload factor from Table 22.8, C₀ = 1.25 × 1.25 = 1.5625.

Design transmitted load,

Velocity factor,

Strength: Applying the Lewis strength equation for gear

To find Y, the formative number of teeth in the gear should be found out

f_b =  70 MPa for C.I. good grade, Table 7

Now L = 0.47 m.

 From the Lewis equation,

From here, f =  0.086 m.

Now f should lie between  and  ie, about 0.12 m to 0.16 m.

 Let us take, f = 0.12 m, ie, equal to

Check for dynamic Load:

Now  for gear = 70 × 0.453 = 31.71

Now

Y for pinion =

Let the material for pinion be cast steel, heat treated with f_b = 196 MPa and BHN = 250,

 for pinion = 196 × 0.36 = 70.56

 Check for dynamic load will be applied to gear.

Now k = 0.111

         

Now at V = 5.28 m/s, the maximum permissible error from Table 10 is about 0.059 mm. From table 9, with module of 20 mm the error for class 2 gear is about 0.0568 mm.  Since the maximum permissible error 0.059 mm, use gears of class 2.

e = 0.0568 mm

Now Beam strength of gear,

         

Now

Let

Now F_b F_d =  1.55, therefore, the design is very much safe for dynamic load.

Check for Wear:

Now

Now from Table 11, for steel pinion and C.I. gear, with BHN of 250 for Steel,

Since F_w > F_d   Design is safe for wear alos.

Tooth proportions:

Addendum = m = 20 mm

Dedendum = 1.2 m = 24 mm

Working depth = 2.0 m = 40 mm

Clearance = 0.2 m = 4 mm

pc =m = 20  mm

Tooth thickness = space width =

For the gear, the blank diameter.

For the pinion,  = 0.42 + 2 × 0.02 × 0.894 = 0.456 m.

Gear shaft: Turning force, F_­1 = 22.2 kN.

Radial force at the pitch diameter of gear, F_r = F_t tan  = 22.2 × tan20^o =  8.075 kN.

If in Fig. 25, the point ‘b’ is the centre of the gear shaft bearing, then bc = 0.23 m.

 Bending moment due to

The bending moment on shaft, due to

Since the BM due to F_t and F_r act in planes at right angles.   Their resultant is

         

Torsional moment on the shaft is,

The shaft is subjected to direct stress, shear stress and alternating bending stress.

In the tooth design, we have taken FS as 1.5 on elastic limit or about 3 on ultimate tensile strength. Now, while calculating equivalent torque above, we have not considered the design load which includes dynamic factor but only the design transmitted load.  Therefore, for shaft design, shock factor of about 1.5 (as discussed in chapter 4) can be taken.  Thus, the FS may be taken as 4.5 say 5 to account for direct stress also.  Therefore as before f_s may be taken as,

         

or  , after considering key way effect.

d = 0.0961 m, say 100 mm

Hub diameter = 1.8 d = 1.8 × 100 = 180 mm

Hub length = 1.25 d = 1.25 × 100 = 125 mm

Pinion Shaft: Pinion shaft can also be designed in the above manner.

WORM GEARS

Design of Worm Gears:

Design of Procedure:

1.       From the data (available) given note down the power transmitted, speed, speed ratio and other working condition if any.

2.       From the consideration of power to be transmitted and speed ratio, material for worm (steel) and wheel (Bronze) are selected and the number of starts an worm is fixed.

Number of teeth on worm wheel in selected between 25 and 85. For the unit to be compact select towards lower limit and for high efficiency choose towards higher limits for high power take number of teeth on wheel from 60-80.

Complete mini centre distance based on the beam strength using equation.

Determine accurately centre distance a.

Determine sliding velocity Vs and find the corresponding value from the table.

Check for bending stren using the equation.

Determine the length of worm.

Determine the face width of worm wheel b

Determine heat generated (Hg).

Determine heat decrepitated.

Strength of Worm Gear Teeth:

In finding tooth size and strength, it is safe to anume that the teeth of worm gear are always weaker than threads of the worm. In worm gearing, two or more teeth are usually in contact, but due to uncertainty of load distribution among themselves it is anumed that the load is transmitted by one tooth only.

Forces Acting on Worm Gears.

When the worm is transmitting power, the forces acting on the worm are similar to those on power screw. Figure shows the forces acting on the worm.

It can be muted that the forces on a worm gear are equal in magnitude to that of worm.

References

1.      Mechanical Engineering Design – Joseph Shigley

2.      Machine Design – Mubeen

3.      Machine Design – Black

4.      Principles of Lubrication – Cameron A.