Junior Engineer Seeking Help Designing Gearbox for 25:1 Reduction

In summary: I think.In summary, the client requires a 25:1 reduction from 1500rpm to 60rpm with a power throughput of 25kW. The engineer is required to design all aspects of the gearbox, including the strength of material. The selected motor has a shaft diameter of 48mm, and the engineer is considering using spur gears or helical gears. The engineer is having difficulty finding the correct centre distance (CD) for the worm gear because the worm shaft is too small.
  • #1
MinhD
3
0
Hello good people,
I'm a junior engineer currently studying in an Adv.Diploma in Mechanical Engineering.
I have been tasked with the project of designing a gearbox and I'm having a bit of difficulty getting started as I'm not that experienced with gears.

The client requires a 25:1 reduction from 1500rpm to 60rpm with a power throughput of 25kW
I am required to design all aspects of the gearbox including that strength of material is ensured.
The motor I have found has a shaft diameter of 48mm

The design I proposed is a two stage reduction gear box using spur gears. This will use 2 pairs of 20 teeth gears and 100 teeth gears on parallel shafts. I'm consider using helical gears instead or rather than that, opt for a worm drive but I can't seem to find the relevant standards (Australian) or design guide to assist me concerning this.

Is there a procedure anyone could suggest I follow?
Any information will be much appreciated.
 
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  • #2
Remember that torque * RPM = power. Since it is a reduction box, it will be the output shaft size and the weight of the bull gear that will decide the size. The input shaft will be smallest. It will have a small pinion.

One specification will be quantity to manufacture, one will be simple, 10,000 will be built to a price and reliability spec. For a one off design, keep it simple but list possible optimisations.

An induction motor will probably have 5% slip under power. Find out if you can have a ratio close to 25:1 rather than exactly 25:1. The problem is that 25 is 5x5. The stages have an integer ratio of 5. Consider 19 : 95 as your tooth count, (avoid 20:100). Since 19 and 96 share no common factors, it makes a hunting tooth system that will evenly wear all teeth against all others.

Your pinions could use less than 20 teeth if they used stub teeth. Design for straight cut spur gears, then redesign with helical for silence and check end thrust on bearings.

Do an advanced search of PF with keyword= gear and username= baluncore. That will save me some repetition. Ask more specific questions when you have a better idea of the required configuration. I am in TAS, not that far away.
 
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  • #3
Thank you for the reply,

I have decided on a worm drive composed of a 3 start worm and a 75 teeth worm gear
The motor I have selected is this one:
http://www.inverterdrive.com/group/Motors-DC/25kW-DC-Motor-TT-Electric-1500RPM-33HP-shunt-B3/

My task does not consist of a use, it is to just design a gearbox.
With the 48mm motor shaft, I think I should have the worm shaft about ∅100mm bored to fit
The worm gear is ∅154mm; I'm unsure what size hole should be on my worm gear for output shaft (maybe make it ∅48mm?).
I've been following the initial sizing of the worm gear according to this but haven't made much progress:
http://www.roymech.co.uk/Useful_Tables/Drive/Worm_Gears.html

I can provide the information I have calculated so far that it will have:

Torque:
P = T.ω
25000W=T.(1500∏/30)
T=159.15N.m
Output Torque = 3978.87N.m

Materials:
Worm: Hardened steel 500 BHN
Worm gear: Bronze

I am having difficulty finding the correct centre distance (CD) because my worm shaft is too small. According to the procedure here, I'm supposed to have a worm shaft of 16.18mm which seems too small for the input shaft:
http://www.brighthubengineering.com...gn-calculations-using-agma-empirical-formula/

Let D1 be pitch diameter of worm shaft
D1 = 2.4 x 2∏ + 1.1


Module: 2
Axial pitch: 2∏
Worm gear pitch diameter: ∅150mm
Pressure angle: 20°

So according to the information I currently have, my questions are:
-What do you think of my current specifications?
-What should my worm shaft size be for the rated motor?
-What is my worm shaft pitch diameter?
-How would I go about designing the worm gear if I wanted it to have grooves cut into the teeth to better fit the turning worm shaft?

Please let me know if I'm missing anything or I'm incorrect anywhere.
I apologize for the long post.
 
  • #4
Worms are notoriously inefficient because they have a sliding contact rather than the rolling contact of spur gears. I would be deeply concerned about the temperature rise of a worm transmitting 25kW.

Take a look at this page for design of worms boxes;
http://www.qtcgears.com/QTC_Redo/Product-Technical-Information/Worm-Gears-Technical-Information.htm

For worm design I would refer to Section 9 of the QTC Gears Technical Manual. I would strongly recommend you download that document, (it is about 25Mbyte). Google{ qtc “ELEMENTS OF METRIC GEAR TECHNOLOGY” .pdf }to find a very useful reference work.

Your specs are similar to the final drive reduction box found on 7 to 10 tonne excavators. They use straight cut spur gears with three stages of reduction.
 
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  • #5
Also get a copy of the “Gear Technical Reference”, 20Mbyte. At the bottom of this link.
http://www.khkgears.co.jp/en/gear_technology/guide_info.html
This also contains a summary of the Japanese standards for gear design and material specs.
 
  • #6
Thanks for the tips :)

I believe the heat generation you've brought up is significant for such a design so thank you for letting me know, I wouldn't want to complicate things by being that oblivious.

Now that I think about it, if I were to use a 2-3 stage parallel shaft set up I would be conflicted between using traditional spurs gears and helical gears.
According to wikipedia, the spur gear teeth's sudden contact causes stress and noise whereas the helical teeth is less noisy but there is thrust and sliding friction.

The next thing that comes to mind is using herringbone gears but I'm clueless in terms of using the same calculations and such as the single helical gear or if there are separate ones (which I can't seem to find). Ignoring any misalignment (though cost MAY be an issue), would you recommend it in this case if I were to use it in a compound gear train of about ~3 stages or would a standard spur/helical gear combo do?

Aside from that, I will follow up on the information you've provided before in more detail.

In the mean time, much appreciated.
 
  • #7
MinhD said:
I would be conflicted between using traditional spurs gears and helical gears.
There need be no conflict. If you design with spur gears then it is trivial to skew them into helical gears.

According to wikipedia, the spur gear teeth's sudden contact causes stress and noise
If you design correctly profiled straight cut spur gears they will be in rolling contact. There is no problem helically skewing them by one tooth pitch to reduce contact noise if it is a problem.

The next thing that comes to mind is using herringbone gears
Herringbone gears are effectively a pair of opposite handed helical gears. But you do not need to use herringbone gears unless you are building a gearbox to go between a turbine and the propeller shaft of a ship or a 250kW alternator.

Again, design for spur gears, skew those into helical if noise is a problem, then mirror up to herringbone only if it is necessary to reduce end thrust on shafts.
 

1. How do I determine the gear ratios for a 25:1 reduction?

The gear ratio is calculated by dividing the number of teeth on the driven gear (output) by the number of teeth on the driving gear (input). In this case, the gear ratio would be 25:1, meaning the output gear would have 25 times more teeth than the input gear.

2. What type of gears should I use for a 25:1 reduction?

For a 25:1 reduction, you will need to use multiple gears in a gearbox. The most common types of gears used in gearboxes are spur gears, worm gears, and helical gears. The specific type of gear will depend on your specific application and design requirements.

3. How do I calculate the torque required for a 25:1 reduction?

To calculate the torque required for a 25:1 reduction, you will need to know the torque required at the output of the gearbox. Once you have this information, you can use the gear ratio to determine the torque required at the input. For example, if the required output torque is 100 Nm, the input torque for a 25:1 reduction would be 2500 Nm (100 x 25).

4. What factors should I consider when designing a gearbox for a 25:1 reduction?

There are several factors to consider when designing a gearbox for a 25:1 reduction, including the required torque, speed, and precision, as well as the type of gears, materials, and lubrication. It is also important to consider the size and weight limitations of your application.

5. What are some common challenges when designing a gearbox for a 25:1 reduction?

Some common challenges when designing a gearbox for a 25:1 reduction include managing the high torque and speed requirements, ensuring proper alignment and meshing of gears, and minimizing backlash and noise. It is also important to consider the durability and maintenance requirements for the gearbox.

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