Shaft Design Criteria In Planetary System

The best way is to make a computation about the fatigue capability of the shaft. This has to consider :

- how torque is transferred from shaft to the "user"?

- what kind of load will be applied to the shaft?

- is it a risk of dynamic resonances and stiffness is the guide value?

- is it an intensive use? high number of cycles? speed? MTBF? required degree of safety?

Only the nominal torque is an indicative value but not enough for the right computation of the shaft diameter.

If you send the answers I can verify and indicate the best choice.

- how torque is transferred from shaft to the "user"?

The planetary is mounted on an intermediary jackshaft that reduces speed 4 times, that speed is then conveyed to a a 17T sprocket at one end of the shaft and again reduced by 1.57 times with a 27T sprocket at the wheel axle. The 17T-27T are connected by a chain drive.

- what kind of load will be applied to the shaft?

Input shaft is being supported by bearings and retainers at the ends. A secondary clutch mounts on the input side of the shaft before going into the planetary. I have not gone as deep as to calculate the bending moments and bearing loads, have only calculated the torsional shear stress acting in the shaft, which comes out to about 136 MPa with the given dimensions. Shaft is hollow. What loading/loadings do you feel we should account for other than torque?

- is it a risk of dynamic resonances and stiffness is the guide value?

I'm not sure how to answer that.

- is it an intensive use? high number of cycles? speed? MTBF? required degree of safety?

The shafts are used for transmission in an off-road vehicle so it might see shock loads. We're going to use the vehicle about 5-6 hours per day so operation is limited. Engine produces a max RPM of 3600. We want a factor of safety of 5-10 at the minimum but please advise if that's reasonable.

Thank you.

1- what I meant was do you have a key? which type? or how is the sproket fastened on the shaft? This is required for the estimation of local notch effect which increases stresses even more than 4x depending on type of "connection".

2- if you have a chain drive on the shaft this introduces a radial variable force due to the chain and limited number of sprocket' s teeth. Non uniformity especially in low number of teeth sprockets is dangerous introducing torsional fatigue stresses.

3- the radial force mentioned above leads to a bending which is to be taken into consideration as well, its effect could be the major one depending on the shaft design. Some shaft parts can be less sensitive in torsion and much more sensitive in bending.

4- if shaft is hollow which is the ID? Both bending and torsional resistance depend on both diameters. According to your indicated stress value and my computations it is 10.6mm, is that true?

5- the off-road application leads to high dynamic coefficients, the presence of chains will reduce the amplitude due to their compliance but never the less a dynamic coefficient has to be considered. Its magnitude will depend on the vehicle mass and speed. It could be estimated knowing the chains type and length + pitch.

6- the link for the steel properties did not work could you send the values you found?

7- what you understand by safety? with respect to what should the factor be considered?

8- what MTBF do you expect ? how many years without a high risk of failure?

If you only consider the stress you computed and forget any notch factor (which I shall not recommend) then the elastic limit of the shaft material should be over 1178 N/mm² for a safety factor of 5 only and without any bending either !

Give the answers and I shall give you the values. The best would be you add a sketch of the shaft output with the sprocket on so that the bending can be estimated.

1. The secondary sits on the shatft via a key, that's the intention. The sprocket on the end on the output shaft from the planetary will probably have to be set using a keyway and a setscrew, easiest solution.

2. I have no experience with calculation of radial variable forces due to the chain. I would appreciate it if you clarify how I could go about calculating it? Is there a guide? 17T sprocket is not that low a number, we had to bump it up from 12 T because of chance of chordal action.

3. Input and output shafts are same size and hollow - 25 mm OD, 12.5 mm ID. Those are the constraints we have right now, but if we are led to a calculation that tells us that the output shaft sizing is not enough for the torque and bending loads, we probably have to size it up and look for different bearings.

4. Chain type is a D.I.D 402 V roller chain with a pitch of 1/2 inch I believe but slightly higher in width than a standard ANSI 40 chain, this stuff is used in ATV's and motorcycles. I estimated the length we would need, around 50 inches.

5. I'm sorry. The 4340 steel values are here and here. Some relevant data :

Tensile Strength, Ultimate 185900 Psi, 1282 Mpa
Tensile Strength, Yield 125000 Psi 862 Mpa
Shear Strength, Ultimate 139425 Psi 948.09 Mpa (75% of Ultimate Tensile Stength)

Modulus of Elasticity In Tension 30E6 psi 204000 Mpa

Modulus of Elasticity In Shear 114E6 psi 77520 Mpa (38% of Mod. of Elasticity in tension)

7. Factor of safety of 10 at the minimum with respect to torsional shear stresses, but if its a combined loading, maybe more?

8. Car is going to be used a handful of times for 32 hours total, but periodic use, and 4-5 hours max during a single use. Not more.

9. The input and output shafts have to be splined at the ends that the connect with the planetary. Yes, in my elementary analysis, I neglected the notch factors due to those splines, only because I'm not sure how to design a transmission shaft with allowance for splines. Honestly, I was looking at a quick and dirty way to know whether the said diameters of the hollow output shaft would be suitable for our application, where the maximum theoretical transmitted torque is about 290-300 ft-lb, not more.

Thank you for your help and guidance, waiting to hear more from you.

From: Tim Hawley Master Mech.

G4340 Alloy Steel can reach a hardness of 30 Rockwell C, unless Oil quenched and non nitrated will bring up to 37 Rockwell C. With a 6. mm wall thickness will reduce its strength by approx. 4 times according to the information below. You may want to go to 50 mm O.D. shaft with 12.5 mm I.D. This will give you a thicker wall of 18.75mm.

Do the calculations: and consider length of the shaft, Circular Shaft and Polar Moment of Inertia-as well as wall strength.

If you require additional information let me know.

Best Regards, Tim

http://www.engineeringtoolbox.com/torsion-shafts-d_947.html

Shear Stress in the Shaft

When a shaft is subjected to a torque or twisting, a shearing stress is produced in the shaft. The shear stress varies from zero in the axis to a maximum at the outside surface of the shaft.

The shear stress in a solid circular shaft in a given position can be expressed as:

σ = T r / I_p (1)

where

σ = shear stress (MPa, psi)

T = twisting moment (Nmm, in lb)

r = distance from center to stressed surface in the given position (mm, in)

I_p = "polar moment of inertia" of cross section (mm⁴, in⁴)

The "polar moment of inertia" is a measure of an object's ability to resist torsion.

Circular Shaft and Maximum Moment

Maximum moment in a circular shaft can be expressed as:

T_max = σ_max I_p / R (2)

where

T_max = maximum twisting moment (Nmm, in lb)

σ_max = maximum shear stress (MPa, psi)

R = radius of shaft (mm, in)

Combining (2) and (3) for a solid shaft

T_max = (π/16) σ_max D³ (2b)

Combining (2) and (3b) for a hollow shaft

T_max = (π/16) σ_max (D⁴ - d⁴) / D (2c)

Circular Shaft and Polar Moment of Inertia

Polar moment of inertia of a circular solid shaft can be expressed as

I_p = π R⁴/2 = π D⁴/32 (3)

where

D = shaft outside diameter (mm, in)

Polar moment of inertia of a circular hollow shaft can be expressed as

I_p = π (D⁴ - d⁴) /32 (3b)

where

d = shaft inside diameter (mm, in)

Diameter of a Solid Shaft

Diameter of a solid shaft can calculated by the formula

D = 1.72 (T_max/σ_max)^1/3 (4)

Torsional Deflection of Shaft

The angular deflection of a torsion solid shaft can be expressed as

θ = 584 L T / (G D⁴) (5)

where

θ = angular shaft deflection (degrees)

L = length of shaft (mm, in)

G = modulus of rigidity (Mpa, psi)

The angular deflection of a torsion hollow shaft can be expressed as

θ = 584 L T / (G (D⁴- d⁴)) (5b)

Torsion Resisting Moments of Shafts of Various Cross Sections

Thanks Tim,

Incidentally, these are the same calculations I also used in my elementary analysis. So I know given the torque of 290 ft-lbs acting on the output shaft of the planetary, the maximum torsional shear stress is going on happen on the outer surface of the hollow shaft, and my calculation comes out to be round about 136 Mpa. Now given that number, what do I compare it to , in order to make a judgement that this size shaft is no good? Here's what I did :

Factor of safety = Ultimate Shear Strength/Allowable Shear Stress

I can do some back calculation. If my factor of safety were to be 21.07, my allowable shear stress comes out to 45.59 Mpa. If I use that value to design my shaft, using the equations for diameter, my shaft OD and ID comes out to exactly 25mm and 12.5 mm respectively, which are the specifications of the shaft I actually have originally.

So I was wondering is 45.59 MPa good?

The ASME says that the maximum permissible shear stress may be taken as :

1) 56 Mpa for shafts without allowance for keyways

2) 42 Mpa for shafts with allowance for keyways

They also say that for shafts purchased under definite physical specifications, the permissible shear stress may be taken as 30% of elastic limit in tension or 18% of the ultimate tensile strength, whichever is less.

Let's say I choose the para above for definite physical specs. For 4340 steel, 18% of ultimate tensile is lesser, which is around 230 Mpa. My calculation of allowable shear stress is lesser than that by 40%. Question is, is this good enough? And what about the 42 MPa in 2) above, which the ASME says may not be exceeded (my design will make allowance for splines)?

I'm confused.

Keep in mind throughout this discussion, we just looked at the shaft acting in pure torsion and nothing else, which is not a complete analysis either. :)

Hi,

I apologize for the delay but I had some other things to do.

I analysed your problem and here are the results.

First important parameter is the fact the shaft works in a fatigue range since at 3600/4=900rpm and 40h of life the number of cycles will be 2.2E6. The loads have to be thus analysed with care and not only as "nominal value".

Torque: the torque will vary depending on the adherence to ground but will stay in a range which I assume between 25% and 125%. The number of cycles is difficult to estimate since it depends on speed and ground, let us assume that it is such that we can consider 1.5E5 macro-cycles. There is also a variation due to the chain transmission but this one is low being proportional to 1-cos(π/z) ≈0.02.

Bending is in phase with torque since bending is induced by the chain force and this one is directly proportional to the torque. For the most loaded section (in the plane where the key is) bending is an alternative symmetrical load with 1 cycle/rotation because the chain force direction is the same and the shaft rotates.

The second step is to estimate as good as possible the stress level at the maximal moment.

It is thus necessary to determine the resistance module of the shaft in the most sensitive section where the key grove is milled.

The presence of the key grove makes the use of the classical equations for a round section impossible! Those equations are for circular sections and NOT for sections with variable wall thickness with a ratio as high as 2.78. I am sorry to say that the error to use equations without looking at the way they were build up and their validity domain is a common error and lead to too many problems.

The simplest approach based on the fact that the torque generated stresses "flow" in the section is to define the biggest diameter inscriptible in the section as in the right part of the figure.

For the resulting section it is possible to use (with some errors of course) the relations for the stress computations in closed thin profiles which is based on the circulation integral. The highest stress occurs where the wall is the thinnest and will be with communicated data 399.3 N/mm²

The chain force is 1.15E4 N and assuming it is at 30mm the bending moment will be 3.44E5 Nmm. The bending stress with the minimal module of shaft with respect to the (in the figure) horizontal axis will be 125.1 at the edge of grove and 89.3 at bottom of grove.

The computation of the equivalent stress according to von Mises will consider thus σ=89.3 and τ=399.3 N/mm². Resulting value is 697.3 N/mm² in presence of bending and 691.5 N/mm² with only shear. This shows first that bending in this configuration has no influence on the result since the ratio σmin/σmax= 0.99. The problem is related thus only to the torque magnitude which according to the 1^st assumption will vary between 174.3 and 871.6 N/mm². The ratio will be 0.2 which is low enough to assume a pulsating cycle. The value for this kind of load estimated with the mechanical data for the steel is 853.8 N/mm². The comparison between obtained value of 694.4 and 853.8 leads to a safety margin which is only 1.23. The dispersion of mechanical properties is such that this margin is too low and even at a lower number of cycles the crack risk is quite high.

Conclusion: The 25 mm shaft is not safe enough for the applications, the failure risk being high if a key is used.

There is a solution which is to transmit torque between shaft and sprocket with a different connection as for instance a double conical bushing with axial pressure (Ringspan makes such bushings) and thus use full capacity. The advantage is that it is with this "fastener" no screw for axial positioning will be any more needed.

I have to look deeper into every line you have written here. It seems to coincide with my fears that the output wouldn't work, but I obviously looked at the wrong equations and didn't consider the stress concentrations of the splines, keyways etc. Well, the output and input shaft are connected to the planetary by splines and the sprocket at the end of the output will also have some sort of spline connection with the shaft. The only keyway being where the secondary clutch is mounted to the input shaft.

Here's another thing. This shaft is "through hardened" 4340, so properties are bound to be different than the normalized 25mm OD samples found on Matweb. But the key to knowing how different they are is knowing what kind of heat treatment they have gone through and that's one thing my team is working on to find from the supplier. However, my doubts still remain.

Thank you for correcting my blunder. I'll have to print this out and read deeper into it, and expect me to get back to you with some questions on the double conical bushing idea.

Meanwhile, what do you think about the possibility of perhaps using a stepped down shaft for the output, 38 mm near the planetary and turned down to something lesser, ofcourse, with a proper fillet radius to reduce stress concentration? Just another question I had in my mind...

Thanks!!

Nick

What is your feeling about a 38 mm OD for that output shaft?

That's our next possible size we can go up by but it will be a challenge trying to fit it in our design. Constraints are the specifications of bearings, clearance holes of the end connection of the output shaft and such.

Here are some answers to your questions:

The picture shows how the tempering temperature modifies the mechanical properties.

Type RLK 402 25x50 transmits 577 Nm if the maximal possible torque is less you can use it. manufacturer is RINGSPANN as you see the shaft does not need a key or splines. The only limitation can be the fact that you have a hollow shaft. Do yoy need it or can a full one satisfy the function?

I shall have a look for the 38 mm and tell you what I think. If the pictures are not ggod enough then send via the private channel a mail address so that I can mail you the files in jpg.