Dynamic Strength - Hybrid Anisotropic Materials for Wind Power Turbine Blades - page 144

C

X - A
B

Φ ( X,A,B,C, ) = 1 - exp -

. . (5.76)

Here,.parameters.are:
X.>.0;.B.>.0;.C.>.0;
X,.A,.B,.C.each.are.equal.to.discrete.symbols.

For.strength.distribution,.we.designate:
X.. =.σ bi. ;
A..=.σ bm ;
B.. =.Sj;
C.. =.N;
N-base.of.testing.

Therefore,.Equation.(5.76).for.fatigue.strength.represents.Equation.(5.77):

N

σ -

Sj

σ

bi bm

Φ ( σ , σ , Sj, N) = 1 - exp -

. . (5.77)

bi bm

If.we.consider.that:

N

σ -

Sj

σ

bi b

Φ ( σ , σ , Sj, N) = 1 - P(t) = exp -

.

.(5.78)

bi bm

We.get.the.logarithmic.equation,.see.Equation.(5.79):

. ln[1.-.P(t)].=.N.ln(σ bm. -.σ bi). -.N.lnSj.. (5.79)

A.straight.line.in.logarithmic.coordinates.shows.in.Equation.(5.79)..Base.of.
testing.N.can.determine.the.inclination.of.this.straight.line..Therefore,.the.
period.of.testing.N.shall.be.determined.in.Equation.(5.80):

1n 1 - P(t)

. N = . (5.80)

( )

1 n σ - σ - nSj 1

bm bi

If. we. propose. that. the. shear. stress. responsible. for. delaminating. of. the.
composite,.the.period.of.testing.N.shall.be.determined.as: