Fiber Orientation questions

[Skid]

I know this is slightly off topic from RC planes but for one of my AOE classes I have to design an aircraft tire (simplified to exclude sidewall, tread design, etc.) but I do have to figure out which reinforcing fibers I want to use and such. We have stress and strain specs to meet for the overall tire and so my question is this - what are the equations relating longitudal and transverse strength of a composite with biased fibers? I have searched my books and so far haven't been able to find anything other than longitudal or transverse loading of a unidirectional fiber composite. Oh and if anyone has good industrial suppliers for Kevlar, nylon, etc., other than CST and fellows, feel free to post Thanks for the help guys.

Lee Skidmore

[Ollie]

Don't try to use only one fiber direction! For each direction of load calculated, use the required amount of kevlar unidirectional fibers oriented to that load direction. The problem is to account for all the loads and their directions. The rest is straight forward. The rubber is so much weaker and has so much lower modulus of elasticity that it can be ignored from a load point of view. It is there to hold air, to abrade and to keep the unidirectional fibers in place while they carry the loads.

[Skid]

Right, I understand that the rubber is just hte matrix material and that my strength is limited by the strength of the fiber. I guess what my question was then is there a single set of equations that relates the multiple layers to the total strength given the angle between them. Otherwise, I take it that I can use my formula for longitudal loading and just apply it to two seperate cases, one for the radial direction and one for the biased direction? Does this take in account the transverse strength of each layer though or do the transverse strength of one layer add with the longitudal strength of the next? (assuming alternating 90 degree layers) Thanks.

Lee

[Ollie]

There is no transverse strength due to the fibers. In that dierction, perpendicular to the fiber length, the rubber takes all the load. It's just like the links in a chain. The weakest link takes the load until failure.

When the fibers are in tension the layers are considered seperately and independently. However, if some of the fibers are in compression, then the transverse fibers help to hold the compressed fibers in column and delay buckling failure. The closer the orthoganal layers are the better they support each other to prevent buckling under compression.

The main reason that fiberous materials are stronger in tension than in compression is buckling at the individual fiber level because the matrix isn't stiff enough to hold the fiber in column.

[Skid]

None of the fibers will be in much compression so I am not really worreid about buckling. I also understand what you are saying about in transverse loading the matrix takes the load. However, if I need to find the total strength perpendicular to the circumference, do the strength of the matrix in the the transverse (transverse in one layer) and that of the fibers and matrix (longitudual in the other layer)? Sorry if I seem like I am going in circles but it seems that somewhere along the way I've seen equations relating strengths in the transverse and longitudal (longitudal and transverse with respect to the longest dimenision) direction for fiber layup of varying angles. Thank you for the help as always.

Lee

[Weisse Luft]

A tire is a torroid in basic geometry. If you have a balloon tire (unlike the current trend street low profile, belted tires), the primary hoop stress equation from Barlow applies. Fiber stress is roughly the internal pressure multiplied by the diameter divided by twice the cross sectional area of the fibers. Primary hoop stress runs the outside diameter of the tire and is known as the meridonal stress. Secondary hoop stress arises from the profile diameter of the tire and is calculated the same way. The relationship of these stresses gives rise to the principal stress and its angle respective to the tire. Additional plys are laid orthogonally to the first. This is known as a bias ply tire construction and has been the standard in bicycle tires for a hundred or so years. Find an old bicycle tire and tear it apart to see a graphical explaination.

Radial tire construction uses the first ply to constrain the secondary membrane stress arising from the profile of the tire. Cords run from bead to bead in a radial pattern. Belts on the circumfrence constrain the primary stress and stabilize the radial plies. Belts of smaller diameter constrain the section profile to a less than round geometry but still follow the PD/2t rule for stress.

[Skid]

I understand how a tire is contrustructed and it's make up, both bias and radial. I also understand the stress induced in a tire from inflation, however, what I am after is the failure strength of a composite material in both the transverse and the longitudual (again, the longest axis I'm calling longitudual since this will actually be transverse for some of the layers). The composite is essentially loaded while it is a flat sheet. I have a set of strength and strain requirements that I must meet with this compositie material that will ultimately be used in a tire application. Does this clear up what I'm asking for any?

Lee

[Ollie]

You have to ask yourself what the failure mode is in the transvers direction. If there are no transverse fibers the rubber or rubber to fiber joint will fail first. Those strengths and the rubber's modulus determine when failure occurs. The presence of fibers at right angles to the load just reduces the crossectional area of the rubber and lowers the transverse strength a wee bit.

If there is a load applied that is not in line with either the longitudinal fibers or the radial fibers then that load vector has to be analyzed into two component vectors which are in line with the fibers. This is the last time I am going to tell you that the strengths of orthagonal fibers are independent of each other as long as the loads are tension loads.

[Skid]

Okay, thanks, I see what you're saying now Ollie. I wasn't reading close enough and missed the line that said 'When the fibers are in tension the layers are considered seperately and independently.'

Lee