Chief Delphi - Drive-train shaft size

How big should a drive shaft be on a FIRST Robot?

I think that this is an excellent question, since every robot I can think of is using wheels of some sort to carry it’s load. Unfortunately, of all the answers I’ve seen, I haven’t seen any math involved. Some students get inspired by seeing numbers (I know I did) so... here are some numbers.

Assumptions

To first quantify this question, you need to make a number of assumptions. I’ll try to go ahead and make some assumptions, and hopefully we’ll see if we can understand why some of these sizes that have been mentioned are okay, why some are too large, and why some might be a bit too small.

There are a number of considerations that need to be addressed, some that immediately come to mind are the following:

Will fatigue be a factor?
What is the typical load on the shaft?
What is the maximum torque transmitted by the shaft?
What will the maximum load on the shaft be?

The above questions all need to be addressed to answer the quetsion.

Will fatigue be a factor?

This is an interesting question. I’m going to say that from experience, a FIRST robot won’t see more than 10 hours of continuous run time. Assuming that the robot runs around on average at 2 ft/s with 6” diameter wheels, that gives us a life time of 72,000 feet traveled, or about 46,000 cycles (revolutions). In the scheme of steel shaft, I’m personally not too terribly worried about that, but it’s not negiligble. Whereas you should begin to start designing for infinite life around a million completely reversed bending cycles, depending on the assumptions you make above, you could possibly reach 100,000 cycles pretty quickly. Since there is actually a logarithmic relationship, 100,000 cycles is a lot closer to a million than to a static loading case. Whereas for inifinite life (million cycles+) you’d treat the allowable bending strength around 50.4% of the ultimate strength, I’d say in a situation like this, you’re going to not give a lot more leverage than say 70% the ultimate strength. What this means, in short, is that fatigue should probably be somewhat considered in situations where robots are heavily practiced with and demonstrated.

What are the forces on the shaft?

One should probably consider the average load of a full capacity robot, (read: holding maximum number of playing field objects).

Let’s assume that on the average game this is a nice round 150 pounds, and there’s a slight bias to one particular wheel of 33% of the weight of the robot. This would put a load of roughly 50 pounds on one shaft.

One often neglected factor is the fact that something is driving the shaft, and often times, this is a sprocket. For the sake of this sample calculation, we can probably assume that the sprocket is pulling in the orthogonal direction. This may actually be slightly less than conservative, but I’ll hope you’ll forgive me.

In a full fledged pushing match, assuming some super sticky tires with a coefficient of friction around 1.2, the chain will be pulling with roughly 60 lbs of force.

Also, you’ll have some torque about the shaft, equal to the 60 lbs times the radius of the wheel, and assuming a 6 inch diameter wheel, that’s 180 inch lbs of torque.

To sort of simplify some things, the critical elements, because the weight and chain pulling forces are orthogonal, are at the extreme fibers in perpendicular directions. Hence, you wouldn’t want to combine these force vectors, or you would get a critical element with bending stress that is higher than it actually is.

Because these stresses are in directions that are orthogonal (that is, the torsion is causing stress in the direction perpendicular to that of the robot weight on the shaft and the force the sprocket is pulling on the shaft) you need to combine these forces to get the total stress. Fortunately, you can combine these using von Mises or von Mises-Hencky theory, which is essentially just using distortion-energy theory.

It should be noted that this is not a particularly conservative estimate, so it should bring some answers to our question that can be somewhat reliable.

For those of you unfamiliar with von Mises, for a plane stress, it is essentially:

TotalStress = SQRT(stressX² + stressX*stressY + stressY² + 3*Torsion²)

In our case, we only have 1 stress in the bending direction, it simplifies to:

TotalStress = SQRT(stressX² + 3*Torsion²)

Now there are a few more questions to ask:

Where is the shaft loaded?
How long is the shaft?
How is the shaft supported?
Is the shaft notched?
What is the shaft material?
How does that whole fatigue thing come into play?

I’ll answer these with the following assumptions:

I’ll assume that the shaft is loaded in the center, which is a conservative assumption.

I’m going to assume a typical FIRST robot has an inside shafts length of 5” long.

I’ll assume the shaft is supported by fixed bearings on either end.

I’ll also say that the ends will be turned down to fit into some bearing blocks, but with some pretty sharp edges because a lot of teams don’t have the tooling to put good round corners in. (Stress concentration factor ~2)

Let’s do a steel shaft, say just like McMaster-Carr keyed shaft, of ANSI 1045 with a Rockwell Hardness (B Scale) of about 90. That hardness translates over to about 91 kpsi from a chart that I found here: http://www.gordonengland.co.uk/hardn...version_1c.htm

In a textbook I have (Mechanical Engineering Design, 8th edition, Shigley et al) it actually places the ultimate tensile strength of 1045 steel at exactly 91 kpsi, so I feel comfortable also using it’s yield strength of 77 kpsi for calculations involving static (or nearly static) loading.

Up above I said that we ought to use a working strength of about 70% of the ultimate strength when we do design for fatigue, which means I’d use about 63 kpsi.

What I haven’t done very well so far is explain that I’m going to be determining basically 3 calculations:

The size of the shaft to support maximum static loads
The size of the shaft to withstand the fatigue of reversible bending cycles.
The size of the keyway that is needed in the shaft to ensure the wheel or sprocket stays in place.

The first calculation for a maximum total stress due to chain pull (60 lbs) involves the following calculations:

Bending Stress:

D = Diameter of shaft
Z = Section Modulus (Moment of inertia I over extreme fiber distance y (y = D/2)
L = length of shaft
F = force on shaft

My Machinery’s handbook tells me:

Z for a round shaft = pi * D³ / 32

Stress at extremes ends, center = (F * L) / (8 * Z) = (4 * F * L) / (pi * D³)

Torsional Stress

T = Torque applied
R= Radius of shaft
J = polar moment of inertia (for a round shaft, D^4/32)

Torsion = T * R / J

Now, I started to cheat and did some work in excel.

Here’s the numbers:

Wheel Diameter 6.00 in
Shaft Length 5.00 in
Allowable Stress 77,000 lbs / in²
Weight of Robot (max) 150 lbs
% Weight on Wheel (max) 33.3%
Coefficient of Friction 1.2
Shaft Load 60.0 lbs
Moment of Inertia 0.000192 in^4
Section Modulus 0.00153 in³
Modulus of Elasticity 30,000,000 psi

Shaft Estimated Max Stress 24,444 psi
Shaft Estimated Max Torsion 58,665 psi
von Mises 177,684 psi

Shaft Diameter 1/4 in
Shaft Deflection 0.007 in
Deflection as percent of length 0.14%
Factor of Safety 0.43

Key Width 3/32 in
Key Length 1/2 in
Key Allowable Torsion 45,000 psi
Key Applied Stress 30,717 psi
Key Factor of Safety 1.46

Wheel Diameter 6.00 in
Shaft Length 5.00 in
Allowable Stress 77,000 lbs / in²
Weight of Robot (max) 150 lbs
% Weight on Wheel (max) 33.3%
Coefficient of Friction 1.2
Shaft Load 60.0 lbs
Moment of Inertia 0.000971 in^4
Section Modulus 0.00518 in³
Modulus of Elasticity 30,000,000 psi

Shaft Estimated Max Stress 7,243 psi
Shaft Estimated Max Torsion 17,382 psi
von Mises 52,647 psi
Shaft Diameter 3/8 in
Shaft Deflection 0.001 in
Deflection as percent of length 0.03%
Factor of Safety 1.46

Key Width 3/32 in
Key Length 1/2 in
Key Allowable Torsion 45,000 psi
Key Applied Stress 20,478 psi
Key Factor of Safety 2.20

Wheel Diameter 6.00 in
Shaft Length 5.00 in
Allowable Stress 77,000 lbs / in²
Weight of Robot (max) 150 lbs
% Weight on Wheel (max) 33.3%
Coefficient of Friction 1.2
Shaft Load 60.0 lbs
Moment of Inertia 0.003068 in^4
Section Modulus 0.01227 in³
Modulus of Elasticity 30,000,000 psi

Shaft Estimated Max Stress 3,055 psi
Shaft Estimated Max Torsion 7,333 psi
von Mises 22,211 psi
Shaft Diameter 1/2 in
Shaft Deflection 0.000 in
Deflection as percent of length 0.01%
Factor of Safety 3.47

Key Width 1/8 in
Key Length 1/2 in
Key Allowable Torsion 45,000 psi
Key Applied Stress 11,519 psi
Key Factor of Safety 3.91

Now, there were a lot of assumptions made, but I think that this sort of tells the story pretty well in terms of what the shaft diameter needs to be in order to be reliable.

I’m not going to paste the numbers for fatigue, but the short story is that when you be more realistic about the loading for fatigue conditions (less weight and more evenly distributed) it sort of cancels out the slightly lower allowable stress, .

When I look at these numbers,and then consider the real world factors of roughness that tends to occur (setting the robot down too rough, flipping, etc) my humble opinion is that half inch shaft is probably a pretty safe bet, and three eights isn’t terribly risky either.

Keyway Calculations:

It should also be noted that if you choose to use a keyway, calculating their strength is important as well. If you decide to use some super strong keys, they’ll have an ultimate tensile strength of around 90,000 psi, (and a shear strength of roughly half that). If you assume that you have a 3/32 keyway for less than a half inch shaft and a 1/8 inch keyway for over a half inch, you can do some calculations in this respect as well, which I did in the colored text above.

Basically, you find the torque on the wheel, divide it by the radius of the shaft and divide that force by the area of your keyway. That will be the shear stress on your key. If the shear stress is less than the shear strength, you’re good to go.

I hope that this helps answer your questions. I've attached the quick spreadsheet for those who want to play along or check my quick math.

Matt