# Highest point of stress in a multistage gear reducer

Assume a multistage gear reducer (used in forklifts or bed lifts) like the one presented in the following image. Assume that the motor is connected to shaft 1 and the load on shaft 3. Which gear pair has the highest stress?

1. Is it the slowest moving gear towards the lifting end, or
2. is the stress equal throughout the mechanism? An internet search has been fruitless.

My intuition is #1. Since the mechanism converts velocity to torque, it has high velocity and low torque on the motor side, but low velocity and high torque on the lifting side. So while the strength of individual gear teeth is the same throughout the mechanism, they're subjected to higher forces on the lifting side, and this is where they'll break first.

I'm been presented with the counter argument that it's simply an energy converter, and the same amount of total energy is present throughout the mechanism. So the stress is the same on the motor side as it is on the lifting side. This feels counterintuitive, but I'm unable to refute it.

TO CLARIFY: The question is about multiple gears in succession, e.g. a series of 5 gears, and whether the output stage is under more stress than the input stage, or whether all teeth experience the same stress throughout. I do readily accept that a simple two gear reducer would have the same stress between the teeth of each gear.

Example would be like this, with motor input on the right and load output on the right, and whether the stress is higher on the output stage teeth (left). • In multistage gear trains you can change tooth geometry. Sep 8 at 15:19
• @Per are you talking about gear teeth or the shaft that the gear is on? Sep 8 at 15:35
• The gear teeth.
– Per
Sep 8 at 15:45
• could you also share a picture showing the two gears at the back share the same shaft? Sep 8 at 15:58
• The assumption would be that they wouldn't share a shaft, so the picture I pasted isn't ideal.
– Per
Sep 8 at 16:08

You are basically correct. The output gear since it is the one applying the most torque and will experience the most force on the gear teeth.

That said, you asked about stress which means something very specific. Stress is measured as force over area. You can always make the final gear thicker which spreads the force on the gear teeth over a wider area thereby reducing the stress. In that way you can vary the stress on each gear's teeth however you want. But it certainly does not need to remain the same. However, optimal design when all gears are made of the same material would probably dictate that you choose gear thicknesses (tooth geometry in general) such that the stress on the gear teeth are all the same all throughout the gear train to minimize the use of material.

The counter claim is wrong simply because energy and power are not force or stress. Where did you hear this claim?

• Thanks! I upvoted, but I apparently got a new StackExchange account (couldn't log in with my StackOverflow account), so my upvote doesn't count. But thanks! Discussion was with my dad. Suddenly we got into a brain loop.
– Per
Sep 8 at 15:23
• @Per I revised my answer using less sloppy definitons. Please see. Sep 8 at 15:58
• Thanks! I agree that there are many variables, which I'm trying to not consider. For the sake of argument, it's only about a multi-stage reducer with identical strength teeth throughout, and whether there's more stress on the teeth on the slow-but-strong side. It sounds like you agree, since this is exactly the kind of problem you'd want to solve by changing the geometry where the stress is highest.
– Per
Sep 8 at 16:04
• @Per Ahhh, but with identical tooth strength I now am not sure if you mean same material for same specific strength or same absolute strength. But one thing is certain: the final output gear teeth experience the most FORCE, regardless of how it is distributed over the gear tooth. It is the one most liable to break, all other things being equal. The resulting STRESS (pressure) from the FORCE can be manipulated via geometry. Sep 8 at 16:06
• Thanks, I really appreciate the high quality discussion!
– Per
Sep 8 at 16:09

Update: Gears are designed to be manufactured from material. As such they are designed so that the stresses are such that the material can withstand the stresses.

So although the last gear pair of your multistage gear system will experience larger forces -if the gear pairs have identical diameters (see below) - (btw the forces on each gear of the gear set are equal), the engineer has the opportunity of reducing the stresses simply by changing the width.

So for a multistage reducer,

• the power on each gear set P_i = constant\$

where $$i$$, refers to each shaft.

• the torque on each shaft will increase (as the angular velocity decreases).

$$P_i = M_i \cdot \omega_i$$

where: P is the power, M is the torque and $$\omega$$ is the angular velocity

• forces on the contact of each shaft are depended on the diameter of each gear.

$$M_i = F_{i, i+1}\cdot d_i$$

where $$M_i$$ refers to the torque on the shaft, $$F_{i,i+1}$$ refers the contact force between the two gears situated on shaft i, and i+1, and $$d_i$$ is the gear on shaft i.

So if the torque on the input shaft is 1Nm and the diameter is 0.2m for the the input gear, then the force will be 5 N. While, if you had a 1m diameter the force would only be 1N.

• the stresses (there are many like contact, bending etc) on the gear teeth are linearly depended on the width. So the designer can easily increase the width and set the stresses at the required level (high or low).

• But let's assume that gear material and size and geometry are the same throughout the mechanism. Is your contention then that the stress is the same for all teeth in the mechanism, and that the input stage is as likely to break as the output stage?
– Per
Sep 8 at 15:45
• Also crucially, the assumption is multiple gears, not simply two gears against each other. I would accept that two gears experience the same tooth stress. The question is whether a succession of e.g. 5 reducing gear experience the same stress on the output stage as the input stage.
– Per
Sep 8 at 15:46
• Although the stresses are approximately the same, the reason that the smaller gear would wear out first, is that (assuming a gear ratio of 2, so twice the number of teeth), for every turn on the large wheel there are two turns on the smaller wheel. So, the load is more frequent. Sep 8 at 15:47
• @per I've updated the answer to put some images for clarification. Are you talking about gears in line in figure 2 or Figure 3? Sep 8 at 15:49
• @Per Technically, the output gear teeth take the most LOAD (experience the most force). That said, I believe with the technical definition of stress, you can just make the output gear wider which would distribute the force over a larger gear tooth thereby reducing the stress. Sep 8 at 15:52

As per the other answers, when teeth are uniform width, and are engaged, then pressure and shear-stress should generally be highest on the slowest moving gear.

There are also impact loads as they engage and disengage -- in particular with spur gears, as opposed to helical gears (?! Or so I thought?? the reference below doesn't reflect this). Analyzing this is way beyond my experience, but here's a reference (RoyMech) that has proved reliable in other cases:

https://roymech.org/Useful_Tables/Drive/Gears.html

There's a formula for stress in there, with a "K_v" factor accounting for impact loads.

If the overall reduction ratio is high, and the velocity on the motor side is high, and the output torque is low, maybe the impact stress on the input side would be dominant as the OP question suspects. I think it would be unusual, but do the calculation and see!