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Helical gear: Theory Q&A

Submitted by admin on Mon, 05/22/2017 - 01:42

Q.1. Compare the contact between mating of spur and helical gears teeth? Which is better?

Ans : Helical gears are manufactured by cutting teeth in the form of a helix on the pitch cylinder. Due to which they appear as inclined to the axix of the gear. In spur gears the contact between meshing teeth occurs along the entire face width of the tooth, which causes the sudden application of load and impact like situation, whereas in case of helical gears the contact begins at the leading edge of the toot and then it gradually extends along the diagonal line across the tooth. This results in gradually increasing load and smooth engagement and quiet running even at higher speeds.So the Helical gears are better in this prespective of contact between mating gear teeth.

Q.2. State the advantages of Helical gears over spur gears?

Ans : Following are the advantages of Helical Gears

1. Quiet Operation : The teeth engage in small area at a time rather than the entire face contact, which results in noiseless operation.

2. Strength of Gear teeth : For the same module and equivalent width, the helical gears can take more load because they are diagonally placed.

3. Connecting Non parallel Shaft : This is one of the exclusive advantage of helical gear over spur gear that one can connect two non parallel shafts also with helical gearing. There will be reduction in the efficiency in this case.

4.Efficiency : Helical gears are more efficient in transmission than spur gears.

5.Less Vibration : In spur gearing the load is transmitted by one or two teeth at a time , so the elastic flexibility is continuously changing as load is transferred from single-tooth to dual-tooth contact and back. Whereas in helical gearing the load is shared between sufficient teeth, which allows smoother transference and nearly constant elastic flexibility,  as a result helical gearing has less noise and vibrations as compared to spur gearing.

6.Less Wear and tear : Since the load is distributed between several teeth at a time, the wear and tear is very less as compared to spur gears.

Disadvantages of Helical gearing

  1. Axial thrust :   Helical gearing generates an axial thrust which is not present in spur gearing. {this problem is overcome by using double helical/ Herringbone gear}
  2. More heat generation : There is more sliding action between mating gear teeth of helical gears resulting in more heat generation as compared to spur gears. Hence helical gearing system needs proper lubrication and cooling arrangement.
  3. Manufacturing cost : As compared to spur gears helical gears need special tools for manufacture as well as inspection and testing. It also increase its manufacturing costs
  4. Power loss in transmission :Power lost in transmission is more in case of helical gearing than that of spur gearing due to more rubbing action between meshing teeth.


Q.3. Where do you use Helical Gears?

Ans : Helical gears are found in several machines and equipment where Higher power transmission in smaller space is required as well where noise/vibration is prime concern.Also helical gears are favored in the situation of high speed transmission. Following are the some applications,

  1. Automobile gear boxes
  2. Printing machinery.
  3. Rolling mills
  4. Textile machinery
  5. Conveyors,elevators etc
  6. Crushers, Feeders and Extruders.
  7. Coal pulverizing mills

Q.4. What is the relationship between actual and virtual number of teeth and the helix angle for Helical gear?

Ans : In design of helical gears, an imaginary spur gear is considered having a pitch clirle radius r’ and module Mn. The number of teeth z’ on this imaginary spur gear is called virtual number of teeth. It is given by relation

\LARGE z' =\frac{z}{cos^3 \psi }

Where z’= virtual number of teeth

          z= actual number of teeth

        \psi= Helix angle

Q.5. What is Double helical and Herringbone helical gear? What advantages they offer?

Ans : In helical gears power transmission both input and output shafts are subjected to ‘Thrust Load’. These forces cause the reactions on the bearings, which ultimately causes increase in size and cost of bearing. These thrust forces can be eliminated by using ‘Double Helical’ or ‘Herringbone’ gear. In both these gears two sets of helical gears are used which nulify the thrusts with each other.


Numerical Problems on Helical Gears

1) Design a  pair of parallel helical gears consist of 24 teeth pinion, which is  rotating at 5000 rpm and is supplying 2500 W power. The speed reduction is 4:1. The normal pressure angle and helix angle are 20° and 25° respectively. Both gears are made up of hardened steel having ultimate tensile strength = 750 N/mm 2 . The service factor is 1.5 and factor of safety is 2. The gears are machined to grade of 4. Deformation factor is 11400 N/mm2 .

The error is given by:

\phi = m + 0.25 \times \sqrt{d}   and     e= 3.20 +0.25 \phi 


 i) Module (taking pitch line velocity of 10 m/s for initial calculations).
ii) Buckingham dynamic load.
iii) Specify the hardness of gears.
iv) All other dimensions of the gears.

2) It is required to design a pair of parallel Helical gears which consists of 20 teeth pinion meshing with 100 teeth gear. The speed of  pinion is 900 rpm. The normal pressure angle is 20°. While the helix angle is 25°. The face width is 40mm and normal module is 4mm. The pinion is madeup of plane carbon steel 55 C 8 (S ut =720 N/mm 2 ). While the gear is made up of plane carbon steel 40 C 8 (S ut= 580 N/mm 2 ). The pinion and gear are heat treated to a surface hardness of 350 BHN and 300 BHN respectively. The service factor and factor of safety are 1.5 and 2.0 respectively. Assuming the velocity factor account for dynamic load. Calculate the power transmitting capacity of helical gear pair.

Use velocity factor K_v = \frac{5.6}{5.6+\sqrt v}

3) A pair of helical gear is used to transmit 10kW power from an automotive multi plate clutch to a constant mesh gear box. The clutch rotates at 7200 rpm which further reduces the speed of gear box input shaft to 2600 rpm. Taking number of teeth on pinion as 20, normal pressure angle 20°, helix angle 25°, design the gear pair. The material for gear may be taken as 40 C8 with ultimate tensile strength 600MPa and surface hardness 300BHN. Using factor of safety & service factor of 2 & 1.5 & considering velocity factor initially.

i) Design the gear pair
ii) Use Grade 4of manufacture & determine available factor of safety.
iii) Explain how the all types of gear tooth failure in this case be avoided?
Use the following data

4) A pair of helical gear consist of 24 teeth pinion rotating at 5000 rpm and supplying 2.5 kW power to a gear. The speed reduction is 4 : 1. The normal pressure angle and helix angle
are 20° and 23° respectively. Both the gears are made of hardened steel (S ut = 750 N/mm 2 ). The service factor, factor of safety and load concentration factor are 1.5, 2.0 and 1.0 respectively.
The gears are finish as per grade –4.
(i) In initial stage of gear design assume velocity factor accounts dynamic load and face width is 10 X module and assume pitch line velocity V = 10 m/s, for estimating normal module.
(ii) Select first preference module and calculate dimensions of gears. (iii) Determine the dynamic load by Buckingham’s equation also calculate factor of safety in bending.
(iv) Specify the surface hardness at factor of safety 2.0.

First preference module (mm) — 1.0, 1.25, 1.5, 2.0, 2.5, 3, 4, 5, 6, 8, 10, 12, 16 and 20

5) A helical pinion 14 teeth made of alloy steel with S ut - 800 N/mm 2 mesh with gear made of plain carbon steel with S ut - 720 N/mm 2 . The gear is required to transmit 30 kW power from an electric motor running at 720 rpm to machine at 225 rpm. The application factor and load concentration factor are 1.3 and 1.1 resp. While the factor of safety is 2.0. The face width is 10 × normal module & tooth is 20o full depth involute. The deformation factor is 11000 × e N/mm. Design the gear pair by using velocity factor and buckingham’s equation for dynamic load. Also suggest the surface hardness for gear pair. Use following data:


6) A helical gear pair 20° full depth tooth profile consists of 18 teeth pinion meshing with 36 teeth gear. The pinion & gear is made of with same material. With S ut - 600 N/mm 2 module - 5 mm, face width - 10×module. helix angle 23°. BHN for pinion and gear - 280, factor of safety - 2. Pinion speed - 1440 rpm. Calculate

i) Beam strength
ii) Wear strength
use following data -

7) A pair of parallel helical gear consisting of 18 teeth pinion meshing with 63 teeth gear. The pinion rotates at 1440 rpm. The normal pressure angle is 20° while the helix angle is 23°. The face width is 30 mm and the normal module is 3 mm. The pinion and gear are made of plain carbon steel 40C8 (S ut = 600 N/mm 2 ). The service factor and factor of safety are 1.5 and 2.0 respectively. Assume that the velocity factor accounts dynamic load. Calculate the power transmitting capacity of the gears,

Use velocity factor K_v = \frac{5.6}{5.6+\sqrt v}



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