Starts with the hoop stress calculation. Since hoop stress is tensile, you can likely assume loading perpendicular to the long (tube) axis. The woven fibres will cross the tensile stress direction at some angle theta. The strength of the composite increases as teh angle approaches zero. The following equations are from notes from a class taken (6-7_years ago) the equations may not be valid.
(below results valid for composite loaded
parallel to fibre direction)
modulus of a continuous reinforced composite as a function of Fibre fraction (volume) and respective moduli.:
Ec = f(Ef)+(1-f)Em
where: Ec - modulus of the composite
Ef - modulus of the fibre
Em - modulus of the matrix
f - volume fraction of fibre
Ok now to bring strength into the equation.
Basic strength equations:
Sc = f(Sf) ; f > f{c
Sc = f(Sm)[Ef/Em] + (1-f)Sm ; f < f{c
where: Sc-(Sigma sub C)-Stress to failure of the composite
Sf-(Sigma sub f)-Stress to failure of the fibre
Sm-(Sigma sub m)-Stress to failure of the matrix
The modifiers after the equations above are used to determine failure mode. f{c is the crossover of change over between matrix controlled failure and fibre controlled failure. If you plot the above Sc equations together on a graph of Sigma (yield or tensile strenght) verses fibre fraction the two equations cross at f{c. For all fibre vol%< f{c the failure is completely controlled by the matrix. When it failes so does the composite. (The load x-fers (completely to the fibre) when the matrix fails causing the stress in the fibres to increse over their ultimate strength.
The matrix equations governing the effects of orientation on long fibre composites are too complex to include here. Due to the angle however some weaves are better suited to certain types of forces, obviously continuous rings are good for hoop stress while a perfect 45deg braid is best for torque. You might want to try mtu.edu site and go to the Materials Science and Engineering pages, look for Dr. John Pilling's web-page, IIRC he had a Math-Cad applet that could calc the different vector directions for angular loadings.
Good luck, Hope this Helps.
Nick
I love materials science!
