15 replies to this topic

[pheyden1448]

I am currently using a 4130 steel pushrod, 5/16th tube - 0.035 wall. These are made for me by Manton Racing. The steel is processed with a salt solution heat treat bath. The pushrod has a 7mm diameter cup and tip at the respective ends, with an effective length of 7.240 inch.

I have changed the type of cam follower that I am using and it has resulted in shortening the pushrod to 5.5 inches (a reduction of weight of 6 gram), but in all other respects the push rod is the same.

I presume that as the pushrod is shorter, that it is less likely to deflect. What I would like to compute is what the increase in stiffness and/or lack of deflection might be as compared to the original push rod.

Further info: Cam Flat tappet 300 degree duration, .332 lift at the lobe, 108 L/C. Rocker ratio is 1.45:1. Spring seat pressure 65 lbs. Over the nose 180 lbs.

Any help would be greatly appreciated.

[Greg Locock]

I am currently using a 4130 steel pushrod, 5/16th tube - 0.035 wall. These are made for me by Manton Racing. The steel is processed with a salt solution heat treat bath. The pushrod has a 7mm diameter cup and tip at the respective ends, with an effective length of 7.240 inch.

I have changed the type of cam follower that I am using and it has resulted in shortening the pushrod to 5.5 inches (a reduction of weight of 6 gram), but in all other respects the push rod is the same.

I presume that as the pushrod is shorter, that it is less likely to deflect. What I would like to compute is what the increase in stiffness and/or lack of deflection might be as compared to the original push rod.

Further info: Cam Flat tappet 300 degree duration, .332 lift at the lobe, 108 L/C. Rocker ratio is 1.45:1. Spring seat pressure 65 lbs. Over the nose 180 lbs.

Any help would be greatly appreciated.

roughly speaking the end deflection of a cantilever is proportional to L^3. In your case you haven't actually got a cantilever and you don't really know what your effective L is but deflection is likely to be about half of the original.

[Wolf]

Pheyden, I hope this image will help- on the left You'll find types of bending (below each drawing is formula for l0, which You use to calculate lambda*). Then on the bottom of the right side, You'll find a table containing values of omega (bending coefficient) as a function of lambda and strength of material- in this case values of omega510 is what You're looking for, seeing the yield strength of 4130 is IIRC 510N/mm2. Then use the values in formula above the table (F- force, S- area of cross-section, and sigma- resulting stress, which should be kept below yield strength to prevent buckling)

http://www.imagebam..../966f9c73406245

Hope it helps.

* i in that formula is radius of inertia of the cross section and the expression is found a bit below the pic

[McGuire]

A push rod is a free beam in compression but that's not what bends them. You can fold a piece of notebook paper so it will support a hard-cover book, but only in an extremely stable state. If any dynamic conditions are introduced (say, a vibration in the table) it will buckle like... a piece of notebook paper. If you look at a push rod the deflections are angular, eccentric, elliptical, and resonant. Personally, I would be skittish about .035 wall. Known to bend just on startup, like when a valve momentarily sticks in the guide. The VW Beetle racers seem to get away with it, but...

Since the pushrod is on the front side of the rocker arm's motion ratio, its mass is not as critical as it might appear. Meanwhile, for a pushrod of conventional length (unlike VW, nearly a foot long) the difference between .035 and .065 or .085 is not terribly significant in total pushrod mass -- the weight is in the ends and adjuster. The same is true of pushrod diameter -- in most cases, you want the largest that will fit. That said, a 5.5-inch pushrod will be stiffer and lighter than a 7+ in. pushrod, but then all else being equal, the longer cam follower will weigh more too.

[pheyden1448]

A push rod is a free beam in compression but that's not what bends them. You can fold a piece of notebook paper so it will support a hard-cover book, but only in an extremely stable state. If any dynamic conditions are introduced (say, a vibration in the table) it will buckle like... a piece of notebook paper. If you look at a push rod the deflections are angular, eccentric, elliptical, and resonant. Personally, I would be skittish about .035 wall. Known to bend just on startup, like when a valve momentarily sticks in the guide. The VW Beetle racers seem to get away with it, but...

Since the pushrod is on the front side of the rocker arm's motion ratio, its mass is not as critical as it might appear. Meanwhile, for a pushrod of conventional length (unlike VW, nearly a foot long) the difference between .035 and .065 or .085 is not terribly significant in total pushrod mass -- the weight is in the ends and adjuster. The same is true of pushrod diameter -- in most cases, you want the largest that will fit. That said, a 5.5-inch pushrod will be stiffer and lighter than a 7+ in. pushrod, but then all else being equal, the longer cam follower will weigh more too.

Thanks all for the formulas and advice.

McGuire,

Thanks for the insight. The cam follower is in fact the standard length, only I have added a cup insert at the top of the stem. So, there is some small additional 5 gram of weight. The pressure point is slightly higher and MAY cause some additional friction as the follower goes up an down. For this reason I plan on putting in bronze inserts in the block.

Yes, I consider there to be a trade-off, with the follower marginally heavier, and the pushrod shorter (thus stiffer) and slightly lighter. However as you say this is on the front side of the rocker arm's motion ratio this may be inconsequential.

I have gone to some lengths to reduce to weight on the back side (valve side), where we have reduced the mass of the items by 34% over what we used last year.

Again many thanks for the comments.

[gruntguru]

Pheyden, I hope this image will help- on the left You'll find types of bending (below each drawing is formula for l0, which You use to calculate lambda*).

The pushrod case is shown by the second drawing (lo = l). (These are not "bending" situations rather "column" loading. I'm sure you knew that Wolf - just correcting your typo.)

[Allan Lupton]

What you are looking at is the crippling (or buckling) of struts. Euler had something to say on that, long before pushrods!
The buckling load is proportional to EI/(L)^2 so just shortening a pushrod increases the buckling load as the square of the lengths.
In your case (7.24/5.5)^2 = 1.733.

For a given mass of metal it is helpful to increase the section modulus (I) which is why a thin-walled larger diameter tube is often used. Although E (Young's Modulus) is equally powerful, a change of material spec. may help - e.g. use a less dense metal with a lower E but use more of it for the same mass and give it a large value of I

Hope this helps

[gruntguru]

The buckling load is proportional to EI/(L)^2 so just shortening a pushrod increases the buckling load as the square of the lengths.

increases?? typo??

[Allan Lupton]

increases?? typo??

Why should that be a typo.

Your profile says you are a Mechanical Engineer, but even the non-technical, I would think, should understand that a short strut is less likely to buckle than a long one, and with an (L)^2 term as a divisor it quantifies it.

Edited by Allan Lupton, 29 March 2010 - 08:45.

[gruntguru]

Why should that be a typo.
Your profile says you are a Mechanical Engineer, but even the non-technical, I would think, should understand that a short strut is less likely to buckle than a long one, and with an (L)^2 term as a divisor it quantifies it.

Now that I understand.

I realise now, when you said "shortening a pushrod increases the buckling load"
you meant "shortening a pushrod increases its buckling load capacity".

My apologies - a simple misunderstanding.

[Allan Lupton]

Now that I understand.

I realise now, when you said "shortening a pushrod increases the buckling load"
you meant "shortening a pushrod increases its buckling load capacity".

My apologies - a simple misunderstanding.

Buckling load is the load at which it buckles: what else could it possibly mean? You need no more words.
Sorry if that sounds a bit tetchy, but I thought, as this is the so-called Technical Forum, a simple technical explanation would be acceptable.

[Canuck]

Excellent...welcome to the tech forum.;)

[gruntguru]

Buckling load is the load at which it buckles: what else could it possibly mean? You need no more words.

Ahhh . . but some of us are slower than you think and we will make fewer mistakes if you give us more words - in this case "applied" or "critical"

[gruntguru]

A push rod is a free beam in compression but that's not what bends them.

A pushrod is a "column" in compression and that is what bends them. If you apply sufficient axial compression load (the critical buckling load) and nothing else to a pushrod with a slenderness ratio (l/r) > 40, it will buckle. The failure (bent pushrod) looks similar to a transverse loading failure.

Of course there are also transverse loads acting on a pushrod. The rocker end suffers a small, oscillating, transverse acceleration and that introduces a dynamic transverse distributed loading. The ball and socket joints apply a small bending moment at each end. All of these transverse disturbances reduce the axial load capacitly of the pushrod, but it is nevertheless the axial compression load that bends them.

Edited by gruntguru, 31 March 2010 - 08:43.

[McGuire]

A pushrod is a "column" in compression and that is what bends them. If you apply sufficient axial compression load (the critical buckling load) and nothing else to a pushrod with a slenderness ratio (l/r) > 40, it will buckle. The failure (bent pushrod) looks similar to a transverse loading failure.

Of course there are also transverse loads acting on a pushrod. The rocker end suffers a small, oscillating, transverse acceleration and that introduces a dynamic transverse distributed loading. The ball and socket joints apply a small bending moment at each end. All of these transverse disturbances reduce the axial load capacitly of the pushrod, but it is nevertheless the axial compression load that bends them.

A "free beam in compression" and a "column in compression" are the same thing. I'm only saying that pushrods do not fail due to simple compression loadings in excess. This would appear to be another one of your arguments for the sake of argument. Wish I had your free time.

[Allan Lupton]

A "free beam in compression" and a "column in compression" are the same thing. I'm only saying that pushrods do not fail due to simple compression loadings in excess. This would appear to be another one of your arguments for the sake of argument. Wish I had your free time.

As I wrote a week or so ago (above) we are dealing with crippling or buckling of struts, and I offered Euler's formula.
Of course it is the presence of end load that converts any instability into buckling failure, but without some assymetrical input that won't happen. A high enough end load would cause a simple collapse, of course, but we can't be thinking of that when dealing with push-rods (except in the unlikely event that the valvespring becomes coilbound before the cam has reached its maximum).