Shock load force? How can I calculate it ?

OK, more information... the pin is in the exact center of the axle, or 14.036 inches from the center to the wheel. (28.072" across).

Labyguy,

I dont understand the szenario you want to evaluate. A shock load is understood to be a sudden event - e.g. if your tractor hits a mine or crashing with max speed against a massive rock. In those cases the shock force is

F = m a (m=mass, a = accelleration).

But I suppose, this is not what you are looking after. Looking after the breaks efficiency, you may calculate the foce required to get the truck speed reduced from max. speed (=v1) to a lower speed (v2). For this szenario you need the time t1 and t2 for beginn and end of the period. Then you can use:

F = m (v2-v1)/(t2-t1)

This formula can be used for the accellaration as well, as far as accellaration is of constant value (wich probably is not the case for a diesel engine. It probably has a poor torque for speed near zero - up to a max. torque for a certain engine rotation speed).

With the wheel diameter you can transform the translative movement of your truck in a torque for the axles. You then have to decide whether the torque fully becomes effective for one wheel, unique for both wheels (or a certain share per wheel).

The torque is T= F r (T= torque, r = radius of the wheel)

From max torque per wheel you are able to claculate the torsional tension in the axles diameter by

TAU = T / Wp ; (Wp = PI d**3 / 32; d=diameter, PI=3.1415)

The litertature gives "admissible tension" data for certain materials, where the calculated torsion tension TAU needs to be compared with - appying a safety (better: uncertainity) factor, to get a long term reliable design for the shaft or pin.

There are some other constrains determining the stress sensitivity of a given design, those like grooves, keyways an bores in the shaft, recess design between two shaft diameters, and those things more, who are effective for the "local flow of forces".

But possibly your intention is a different one...

Albert

AS far as I am aware (and anyone correct me if I am wrong) but if you are are travelling in a forward motion of 12mph then a lot of teh resulting forces from impact will be absorbed by the tractor body itself.

What is the reversing speed? This may have a better guide on the impact force.

If the rear wheels were lifted then dropped this could be a good way of working out the impact force. Say, tractor at maximum speed going over a bump which lifts the rear wheels off the ground. Here you would have the speed of the tractor as a factor and the mass with gravity also a factor.

What material are the axles made from? Youngs modulus etc will have an effect on the strength on the axle.

Kev,

you said:

If the rear wheels were lifted then dropped this could be a good way of working out the impact force. Say, tractor at maximum speed going over a bump which lifts the rear wheels off the ground. Here you would have the speed of the tractor as a factor and the mass with gravity also a factor.

That is it, that is the problem, how do I figure this in with a margin of safety?

It seems to me that you have a beam axle supported on a single pivot. If this pivot pin is in double-shear then the stress developed within the pin is the applied load divided by the area resisting the load.

The area = 5.493 in²

and the load = 2750 lb so the static shear stress = 500.6 lb/in²

For a suddenly applied load this is doubled = 1001.3 lb/in²

If the pin material is a Chrome/Moly steel with a proof stress of about 92098 lb/in² (SEA/AISI 4340)

then you could conceivably allow a 90g shock load but you may have some imposed factor of safety which would bring this down somewhat.

This may not be what you had in mind but I hope it helps.

Drew

I think Ad016 got the question. Our friend here asked about how to evaluate the impact effect over the material resistance. As a standard, we double the stress in the material when you have impact, because all the tabulated data for material resistance is based in static load tests, i.e., very slow load application, controlled environment, etc.

Now, a little mental exercise, just to improve the discussion.

I also would mention that, in the case of the steels, remember to use as a general rule a fatigue stress limit of 1/3 of the yeld stress (roughly, of course). As you don't know how many times in the part life it will suffer impact. Then, as a tractor, I'd also expect a safety margin of 2 for in service loads, because people loves to do that! (hey, let's see what it can do...).

Until now, we got to a safety margin required of 2x3x2=12! (yes, it's a product, nto a sum...) or, around 12000 psi shear stress. Shear stress being 1/3 (approx.) of tensile stress, your material used goes to a 36 ksi proof stress. Still safe in the order or around 2.5, or maybe the manufacturer had another considerations, or less, and the material was right available (assuming that it's 4340, of course).

I have a question regarding your theory on dynamic vs. static loading. You state that the stresses are doubled for a sudden impact load. Is this a general rule of thumb? How would a load applied for (0.0005) seconds to a pin in double shear, compare to the same load slowly applied in the static state?

Basically, I'm trying to determine what the force is. I know the yield strengh and cross sectional area of the shear pin, and I want to derive the approximate load. What assumptions should I make for such a quick impact load, to achieve an accurate force?

Any information you could provide would be appreciated.

-Marcus

Drew,

Your the closest to what I'm looking for. I'm redesigning the axle, I don't want it to break. I want to use ductile iron, so what forces will the axle see? this is what I got, but I don't know if it's correct.

Shocks from bodies in motion. A body of the weight W moving horizontally with the velocity of v feet per second, has a stored up energy Ek:

Given g is gravity 32.18504

We have

Ek = ½ (Wv^2/g) = Foot pounds

So

W = 1465 lbs per axle.

v^2 = ((12 mph * 5280 ft pr mile/60 for feet/min. )/ 60 for feet per sec) = 17.6

Given:

Ek = 7,090.6 ft. lbs of force applied to the axle.

Thoughts? Anyone?

Shock load aka impact load is very difficult to calculate, normally people just add a large enough factor into the calculation to justify that. For a more detail study you need to work out the struture stiffness. look at the struture like a spring, when a load is suddenly stop, you loss the kinetic energy (E=1/2*mv^2), but where the kinetic energy goes? it go to the struture as a form of spring energy(E=1/2*kx^2). By equaling these two you can find the spring deflection (beam deflection), and work back from the spring deflection you can find the load that causes the deflection.

If I remember correctly that should be the way.

I will check it out and let you know later.

Zichau,

The plot thickens... Don't you love it. In the background, the potential supplier has been doing a FEA and on a part we have been talking about. The results are :

..........Steel Tube Design .............Casting Design................Casting Design

..........................................(Previous Version 4) ...........(Current Version 5)

Vertical Stiffness 22727 ..................8547 .................................10416

For-after Stiffness 11235 ................6993 ....................................8196

Now the new problem. Will the results of Version 5 suffice for the project I've outlined to you?

My thoughts are yes. I'm thinking that a ductile part should be able to withstand 7,090 pounds of force. But the units they presented are pressure, no lbs. so where do we go to start talking apples to apples??

Thanks for the help guys.

Hey, of course they are presenting pressure units. But don't say pressure, it's ugly in terms of material selection. Talk about stress.

They presented you with a simulation that indicates a requirement for 10.5 ksi X 8 ksi of material resistance in casting, 22.7 ksi X 11.2 ksi for a tube design. Are these results of the worst case load applied? Remember to add those (remember Mohr's circle?). Of course, I am assuming this is the worse case, and the data is correct, with all combinations of loads included.

Check the posts of the other people. My last consideration is to use a material to withstand 36 ksi. That's pretty close to the results you have now too.

You have all you need, in my oppinion. Check your suppliers for inexpensive materials. We are talking about a material with yeld stress around 250 MPa (or 36 ksi). SAE 1045 at a low hardness is a great candidate. 1020, if cold drawn and not annealed would also do the job (select the raw material spec with some care in both cases - it's cheaper to use the material as supplied and without need for heat treatment). I'm pretty sure you won't need to stick with expensive alloys or heat treatment.

I got some more information on this. They used 4,000 lbs of force on the pin in the FEA program. When they did, they got .000527 mm of deflection on the Ductile part where the steel part had .0000851mm of deflection. They changed the design a couple of time and improved the casting by 50%. I don't see any other numbers, but that would put it at .0002635 or better. But that still is not meeting the original steel part at .00008. What amount of deflection would be acceptable for a ductile axle?

I checked out your original post, and it calls for dimensioning in the shear pin, right?

So, do not worry about displacement, as it's really small comparing to the rest of the geometry (tractor dimensions). It doesn't make too much sense to speak in terms of deflection in a shear pin. You mention that they are using 4000 lbs of force, and returning displacements of 10E-5 milimeters... FEA software is very good, but be aware that everything that goes on gives a result, consistent or not. This may be correct, but the data is not so usefull. Is it a shear pin, or it's actually a drag brace or other long structure subject to load? The requirements are really different.

Concentrate on pin stress, i.e., ksi or MPa applied to the pin under load, then go for material. I would then calculate the "deflection", and then take a look if it's reasonable for the design. Just to make it clear, it's not the deformation that the pin would suffer under overload. The permanent deformation must be zero.

One thing occurred to me: could this pin be a shear pin, used to break in overload case, to protect the rest of the structure? If so, go reverse-engineering the original one, discover material resistance and dimensions, and just replicate it in a new design. In this case, and only in this case, a hollow pin (tube) is a great choice (high resistance in traction, low resistance in shear). And do not worry, the pin should deform and brake, this is what it is intended to do in case of overload.

bhrescobar,

Your input has been a great help. It's not a shear pin, I never said it was, I just said it was a pin. My fellow engineers here think the worst thing that could possiable happen is for the pin to break.

If they had used 8,000 lbs, I'd be all over that thing as a go project, but with only 4,000. I'm on the fence. But I do agree with you, deflection that does not stay deflected is acceptable.

Thanks for your help..

Ok, but they may be able to provide you some valid data.

If you feel comfortable with 8000 lbs, ask the fellows the higher stress in the material in the model they tested. Consider it half of your load, and then select material.

Displacement should be the double too, unless material yelds.

You'll end with a valid result. I think you have what you need.

I am intrigued by this post: the best FEA package in the world wont give you sensible results unless you enter realistic loadings. Where did the FEA guys get their data from? As it is your design, I would assume it was from you; if not where did the 4000lbs come from. None of the FEA software that I have used, or seen demonstrated, would allow you to enter just the KE of the attached part. Please would you give us some more detail of your process here.

Thanks,

Drew

Automotive applications are not my area of experience but I'll offer a few thoughts for what they're worth:

You need to ask yourself 'What is the worst-case loading condition?'
In order to answer that question, you need some more information. For example:
Is the axle a) Passive or b) does it have brakes or power?

If a) then the loading will come from external sources - road bumps etc.
If b) then it will have all the loads from a) plus axial torque loads from brakes (where your energy calculation above comes in) and/or motor.

These loads could/will occur in combination so some analysis is needed to decide the worst combination.

Also: the road bumps could affect both wheels simultaneously, e.g. speed humps, which will induce an acceleration upwards, or they could affect only one wheel, in which case there will also be another torque loading on the central mounting in the same plane as the road as if the axle was trying to steer.

There will also be bending forces working on the 'arms' of the axle in two planes at least.
There will be shear forces trying to tear the wheels off.

Apologies if these are a bit random, I just jotted them down as they occurred to me.
I hope they are of some help.

Drew