H beam con-rods

[Turbo-Brown]

Can someone explain to me why H-Beam conrods are considered stronger than I-Beam rods?

I was going to buy them as part of a stroker kit but it seems to me that all of their stiffness is in the wrong direction!

If anyone could shed any light I'd be greatful!

[Yaninnya]

Alex,
When the crank is turning and piston is going up and down the forces "trying" to bend the rod to the left and right, not to the front and rear of the engine, so the extra stiffness is in right direction.
But if you want to stroke your M20, a big temptation is M54B30 crank or even american market S52B32. American Metrics Mechanics is boring and stroking M20 to 3.2 l. Temptating, isn't it?
Jan

[Turbo-Brown]

This is just it, the second moment of area side to side of an H-Beam rod is MASSIVELY smaller than that of an I-Beam rod and a high 2ndMA is what will stop anything from bending.

This has been bugging me for a while but I've only recently thought about running the numbers.

All of the meat on an H rod is concentrated around the neutral axis where it only experiences shear forces and is therefore of bugger all use.

With an I rod, all the meat is concentrated around the extremeties where it's effect is amplified by a factor of it's distance from the neutral axis squared.

[Yaninnya]

It is a little to complicated english for me but if I understand corectly:
In H rod you have two big flat surfaces conected by one in the middle. And try to imagine how hard is to bend these two to the side of surface. It is a lot stronger than standard I rod and usually a lot lighter as well.
Jan

[Turbo-Brown]

Here ya go, I've made the amount of material for the H rod more generous than that of the I rod, but you can see that, despite having a smaller x-sec area, the I rod is MUCH stiffer and lighter than the H rod.

OK I did pluck the dimensions out of the air, but you can see that for a lighter rod, you're using the material much more efficiently with the I rod and still getting greater stiffness.

The only thing which the I rod would concede to the H is that it's tensile strength would be less, which is obviously important, but so it a rod's resistance to bending.

[maxfield]

That makes the mind boggle!

[bazza93]

That makes the mind boggle!

very i know a little about the workings of an engine but that is mind boggling

[Mattoz1]

Your problem is you forgot to square the distance(from the neutral axis) in your I-beam calculations... had you done that you obtain a value for Ixx of around 1.8^-8 ie the two beams have the same stiffness

[Mattoz1]

If you then compare the flanges in the other bending axis (Iyy) you will see that it has around 50 times the stiffness, now admittedly that is not all that useful, as Yani quite rightly pointed out, most of the stress is applied in the bending perpendicular to the crank. Perhaps if I wasn't trying to think about this at 3am I could produce a more intelligent answer than the above

[oldroydsr4]

I wrote an assignment on this in my final year at uni.

If i can find it i will dig it out, as currently i can't remember the facts.

Off topic

but another think that baffles me is why people install uprated big end bolts when they are running standard engine speeds. (forces on the bigend bolts are still them same even with greater power. only on the compression stroke and exhaust stroke forces effect the bolts when the mass of the piston assembly is trying to 'stretch' them.)

If maximium engine speed is not changed the bolts are under no greater forces

[Turbo-Brown]

oh p1ss, you're right I did forget to square the distance!

What a plonker

Good job someone's paying attention Mattoz1

With that in mind, it comes out at 1.38*10^-8 m^4 for the I beam.

Don't suppose anybody's got any actual H and I beam rods to measure have they?

We've demonstrated that an I beam of half the weight (ignoring the big end and little end) is less stiff than an H, but I'm really interested to know what the actual dims of some rods designed for the same engine are.

Basically what I'm getting at is that H beam rods really don't use their material very efficiently!

[Yaninnya]

You show the force in wrong direction. To the sides the rod is "block" by gudgeon pin and big end on crank. But because of rotation of crank and piston move up and down, the forcrs trying to bend it to the sides of engine (not to the front or rear of it). That is why you is confuse.
Jan

[HairyScreech]

a better question is why does noone take into consideration the aerodynamics of the engine components as they are still moving in air at high speed.

[Turbo-Brown]

Jan, if you imagine I've shown the rods in my sketch on plan, and the front or the engine is to the right, the rear is to the left.

The rod is trying to bend up or down, so the direction in which it's bending isn't blocked by the gudgeon pin and crank pin.

The direction in which our rods in the sketches resist bending is perpendicular to the axis x-x.

That's also a good question hairyscreech, but one for another thread I think as this one's getting complicated enough as it is!

[Turbo-Brown]

Just to remove any ambiguity, the calcs above are for loadings in these directions with the H being on the left and the I being on the right.



To me, it's clear that the greater concentrations of metal at the extremities of the I beam are better able to resist any bending moments than the centralised distribution of the H rod. Reason being that by trying to bend each rod in the directions shown, you're trying to stretch one side and compress the other whilst the middle bit just resists shear.

Materials are good at resisting shear (try to shear a piece of aluminium sheet with your bare hands) but bad at resisting bending (try bending a sheet of aluminium) as bending is a very effective way of applying a great deal of stress to a material.

The way to make materials deal with applied loads (i.e. limiting the stress in them) is to give them a lot of meat (cross sectional area) in the loaded areas and, in the case of a con-rod, those loaded areas are in the outer sides of the beam.

Now the ability to stop the piston from flying off because the beam has snapped is entirely down to the cross sectional area of the beam (or perhaps stresses where it joins the big and little ends) and we've displayed above that an I rod with half the beam weight (the weight of the big and little ends is likely to be similar between the two rod types) has almost the same stiffness in bending as an H rod, so we can safely assume that an I rod of the same beam weight as an H rod could, given half decent material distribution, be stiffer than the H rod.

We can also assume that for the same weight, it could be not only stiffer, but also have the same resistance to tensile loadings (i.e. it could be revved just as high)

I'm liking H rods less and less!

[Turbo-Brown]

I wrote an assignment on this in my final year at uni.

If i can find it i will dig it out, as currently i can't remember the facts.

Off topic

but another think that baffles me is why people install uprated big end bolts when they are running standard engine speeds. (forces on the bigend bolts are still them same even with greater power. only on the compression stroke and exhaust stroke forces effect the bolts when the mass of the piston assembly is trying to 'stretch' them.)

If maximium engine speed is not changed the bolts are under no greater forces

Just read this properly, it's a very good point. I guess if you're increasing power you might be more likely to zing the limiter or force the engine into an over-rev by changing down at the wrong point (something the limiter can't save you from) Just a guess though.

Now uprated main bolts make a better arguement with standard rev-limits and more power..

[Mattoz1]

I reckon I have it Turbo-Brown.... Buckling loads, even though there is thrusting which could cause bending poerpendicular to the shaft, buckling will occur in the direction of least stiffness, so it is important to have a very similar 2nd moment of area in both xx and yy directions, H beam does, I beam does not. so if you're engineering to lower factors of safety and using high performance engines where the buckling loads will be very high (high comp turbo springs to mind), I suspect the H beam may be better suited. I haven't run the numbers or anything this is just an instinct here. HTH

[Turbo-Brown]

I see your point, but if we assumed that our beam was fixed by the gudgeon pin and crank pin, the effective length for the purposes of our buckling calc front-to-back in the engine is only half that of the effective length in buckling across the crank (side to side).

I really wanna gather some data on the rods available and see what's what with real numbers now!

[HairyScreech]

then surly a rod with a hollow centre and fully rectangular csa would be the best, and not impracticle to make as it could still be investment cast, as you could leave a small hole inside the big end bearing housing to remove your core from.

[Yaninnya]

I'm liking H rods less and less!

And here you are very close to one of my main rules: don't even think about modify, if it is working fine.
Most engine parts are designed with very high safety ratio. So when you know what you are doing, you can built very strong engine on mostly standard parts. I'm not a H rods fan at all. For nearly any fast road engine it is just bling factor nothing else. The YB Cosworth engine is the best example: 500 BHP can be done quite easy with almost 100% standard parts (ex. head gasket, cylinder head bolts, cams, crank bearings, spark plugs, turbo, etc.). That is why I'm using original parts everywhere it is possible. With quite good results I think.
Jan

[Demlotcrew]

A H rod is two I rods 'glued' together without the extra weight?

Here is a better illustration between four different rods i have, and you can see they are not quite like the illustration shown in this thread.

All for S14's, i can measure whart ever you need.

Arrow, Farndon, BMW S14 shot penned lightened and polished, stock S14 rod.

Wish i had good scales i could use to weight them all.





Andrew

[Turbo-Brown]

Aah, couldn't measure the width of the rods at roughly the mid-point, the dimensions of the webs and also the flanges could you?

I know it's a pain in the arse but it's be interesting to see which actually are the stronger rods.

[hoshy]

Interesting thread!

[Silverfang]

Interesting considering we just did some work on this in lectures!

And Maxfield, it doesn't boggle the mind so much when you break it all down

[Dan318-is]

Your problem is you forgot to square the distance(from the neutral axis) in your I-beam calculations... had you done that you obtain a value for Ixx of around 1.8^-8 ie the two beams have the same stiffness

You have to cube that not square it i thought.

Alex, according to one of our senior lecturers, the who allegedly designed the conrods in one of the F1 engines in the turbo era, these are the dimensions of one of the conrods he used. (excuse the dire drawing, obv. not to scale)



Max stress load equated to 5489.5 mm ^4 but that was working in mm not meters.

[maxfield]

Interesting considering we just did some work on this in lectures!

And Maxfield, it doesn't boggle the mind so much when you break it all down

You're studying it!

I'm studying the basics, which ain't exactly difficult when you've spent a few years on here!

[Turbo-Brown]

Your problem is you forgot to square the distance(from the neutral axis) in your I-beam calculations... had you done that you obtain a value for Ixx of around 1.8^-8 ie the two beams have the same stiffness

You have to cube that not square it i thought.

Alex, according to one of our senior lecturers, the who allegedly designed the conrods in one of the F1 engines in the turbo era, these are the dimensions of one of the conrods he used. (excuse the dire drawing, obv. not to scale)



Max stress load equated to 5489.5 mm ^4 but that was working in mm not meters.

Interesting, wide but not all that deep (ooh err)

He's right about the square. To do a parallel axis equation like I was trying to, you take the I value of the element and add it to it's area times the distance of the element's I value from the overall centre of area squared.

So moving a connected body of mass away from the neutral axis of a beam will dramatically increase it's 'I' value (or stiffeness) without necessarily increasing weight. It won't however necessarily increase it's strength in compression of tension.

[Dan318-is]

Well he told us that a rod with those sort of dimensions is good for 1500 brake.

[reggid]

no need to use parallel axis thereom for either configuration.....

look at as the superposition of rectangular sections using the basic equation of I=1/12*B*D^3

so
Ixx = (1/12)*29*14.75^3 - 2* (1/12)*13*10.15^3 = 5489.56mm^4
Iyy = 2*(1/12)*2.3*29^3 + (1/12)*10.15*3^3 = 9371.95mm^4

this is provided that each sub rectangle you use has its own centroid symmetric about the axis of interest which is always possible for thr I and H sections.

I would call it I Beam Vs Column rather than H beam since it seems to hint at the type of loading it can best resist.

Also stress is not purely a function of Ixx or Iyy you need section modulus so two sections with the same I cannot neccesarily resist the same load.

It would difficult to compare the two what is the reference? Mass? Space envelope? What if mass is not so much an issue this chnages everything!

If given two section sizes the type of loading that each can resist in tension, compression, bending and less importantly in torsion but determining which would be better would depend on the application.

Also IMO the hp capacity of a rod is meaningless as a rod rated at 1500hp@ 12000rpm may not last on at 800hp at 200rpm

If you are not building a high rpm engine so mass is not that importnat but will be making alot of power and torque what would be the best section to resist the large compressive forces?

[Turbo-Brown]

For the example I sketched out (and forgot to square the flange centroid offset ) I'm pretty sure you do need to use the parallel axis theorem.

What does your Ixx calc come out to in m^4 for the I-beam in my sketch out of interest if you don't use the parallel axis theorem?

[reggid]

For the example I sketched out (and forgot to square the flange centroid offset ) I'm pretty sure you do need to use the parallel axis theorem.

What does your Ixx calc come out to in m^4 for the I-beam in my sketch out of interest if you don't use the parallel axis theorem?

Ixx = (1/12)*30^3*10 - (1/12)*22^3*8 = 15,401.33 mm^4 = 1.540133E-8 m^4