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Orientation Distributions Galore!
A] Perfectly Arbitrary Fiber Orientation
Known
=
E1 200MPa
=
E2 20MPa
=.
ν12 33
=.
ν21 033
=
.
G12 13 33 MPa
Determining Values
=

=
.
Q11 E11 ν12ν21 202 202
=

=
= .
Q12 ν12E21 ν12ν21 Q21 6 673
=

=
.
Q22 E21 ν12ν21 20 22
=
=
.
Q66 G12 13 33
=
.
(
) +
.
+ *
.
+
.
(
)
Q11 202 202 cosθ 4 26 672 2 13 33sinθ2cosθ2 20 22 sinθ 4
=
.
(
) +
.
+ *
.
+
.
(
)
Q22 202 202 sinθ 4 26 672 2 13 33sinθ2cosθ2 20 22 cosθ 4
Solving for Effective Moduli
=
Ex
π2π2Q11Pθdθ
=
Ey
π2π2Q22Pθdθ
Maple Output
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=
=
.
Ex Ey 91 74
.
This solution makes sense, because if the fibers are oriented perfectly in
random sequence, then all constitutive relationships experienced in any direction will be exactly the same.
The probability distribution of such a material would look like as shown below.
B] Various Locations along Ligament
The equation for the normal distribution is as follows:
=
* (( 
)
px 1σ2π e x μ22σ2
Where μ is the mean, which will always be 0 in this case.
σ is the standard deviation, which is what I will
be varying in order to get different shapes of normal distribution curves (leptokurtic and platykurtic).
X
is the angle in degrees that I am at, which in my case I broke my curve down into discrete blocks of every
15 degrees … So x will vary from the radian conversions of 0 to 90 degrees.
i] The Middle
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 Spring '10
 WEINER

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