Integrated Composite Analyzer (ICAN/JAVA)
Fiber Slicing Computation

Fiber Slicing

ICAN/JAVA may be used to analyze a composite material where fibers are embedded in a matrix. If the fibers are not sliced, the analysis for a cylindrical fiber in an equivalent cube is used. Further information is available about computing volume ratios for a cylindrical fiber. However, if the fibers are sliced, there was no analysis that exactly fit that situation.

Because the fibers are assumed to be evenly spaced, it is sufficient to consider one fiber within a volume that is repeated as required.
Fiber Sliced D
The size of the cube containing the fiber may be computed from the diameter of the fiber, df , and the fiber volume ratio, kf . The fiber volume ratio may be expressed as the ratio of the volume of the fiber to the volume of the cube that may be repeated.
(1.1)   kf = (pi/4) (df )² / D²
where D is length of the side of the cube over which the fibers are evenly spaced.

Solving for D, one obtains:
(1.2)   D = df [(pi/4)/ kf ](½)

Consequently, one need consider a horizontal slice, possibly containing a piece of the fiber, only within a cube over which the fibers are evenly spaced. One such slice is shown below:
Fiber Sliced D
The fiber slice volume ratio may be expressed as the ratio of the volume of the fiber within the horizontal slice to the volume of the horizontal slice. Further information is available about computing volume ratios for a slice of a cylindrical fiber. The analysis below uses the volumes and volume ratios for the slice.

To analyze further, the horizontal slices may be sliced in the vertical direction; for example, along the indicated lines above. The round edges are approximated by square edges, as shown below:
Fiber Subsliced D

Fiber Subslicing

In fiber subslicing, the horizontal slices are also sliced in the vertical direction to form boxes. To perform the calculations, one must compute or approximate the amount of each material within the subslice. The amount of the fiber within the subslice is the volume of the intersection of the fiber and the subslice. The fiber is considered to be a cylinder of diameter df . The amount of the interphase within the subslice is the intersection of the interphase and the subslice. The interphase is considered to be the space between two concentric cylinders of diameter df (fiber) and larger diameter di (interphase).

Since the fibers are aligned with the 1 (X) axis, the subslice may be moved anywhere along the 1 (X) axis and the amount of overlap will not be affected. Thus the distance in the 1 (X) direction may be ignored and the problem reduced to the two-dimensional problem of computing the amount of overlap between a circle and a rectangle.
Intersection of Fiber + Subslice D
Let the origin of the cylindrical fiber be denoted by O. The radius, R, is ½ df , then ½ di , to compute the required intersections. The center point of the subslice, C, and the size of the subslice, (dX,dY,dZ), are specified. The farthest point of the subslice, F, may be computed as F = C + ½ D. The nearest point of the subslice, N, may be computed as N = C - ½ D. Where the size vector, D, is computed as: D = [ 0 , dY sign([C - O]Y), dZ sign([C - O]Z) ].

  1. If the subslice is entirely inside the cylinder, i.e. the cylinder contains the entire subslice, there is obviously an intersection. The entire subslice is inside the cylinder if the farthest point of the subslice is within a radius from the center of the cylinder, i.e. ||F - O|| < R.

  2. If the subslice contains the entire cylinder, i.e. the cylinder is entirely inside the subslice, there is obviously an intersection.
    Intersection of Fiber + Subslice D
    For each axis, let T be a unit vector defining the direction to test. The 1 (X) axis is omitted. The cylinder is within the subslice if, for all such directions, T, |(C-O)·T| < ½ |D·T| - R.

  3. If the subslice is entirely outside the cylinder, i.e. the cylinder is entirely outside the subslice, there is obviously no intersection.
    Intersection of Fiber + Subslice D
    For each axis, let T be unit vector defining the direction to test. The 1 (X) axis is omitted. The cylinder is outside the subslice if, for any such direction, T, |(C-O)·T| > ½ |D·T| + R. This test determines whether the cylinder is outside of the box enlarged by the radius of the cylinder on each side. The "green" areas may be described as the area outside the locus of points made by the center of the cylinder rolling around on the outside of the box and inside a box enlarged by the radius of the cylinder on each side. The "green" areas in the figure, near each corner or edge, are left for the following analysis.

  4. If the center of the cylinder is outside the subslice by an amount greater than R, the cylinder does not intersect the subslice. For each axis, let T be unit vector defining the direction to test. The 1 (X) axis is omitted. The distance that the center of the cylinder is outside the subslice in the direction specified by T, is given by:
    dT = |(C-O)·T| - ½ |D·T|
    If dT < 0, use zero for this term in the next equation, because one only needs the distance outside the corresponding face of the subslice.
    Let d = SQRT( dTy² + dTz² ).
    There is no intersection when d > R.

  5. Otherwise, there is an intersection.

The above analysis may be used as follows:

The volume of the intersection may be computed by integration or approximation. The problem with integration is computing the limits of integration, which occur at either the cylindrical interface or some edge of the subslice, and integrating differently within the different regions (difficult). Approximation allows evaluating the limits along a single line, computing the length of the line within the intersection, and using the weighted sum of many such lengths to represent the integral (easy).

To approximate the integral, sum the vertical lengths over all y (2) values. The z location(s) where the vertical line at y intersects the cylinder may be computed as follows:
z = OZ ± SQRT[ R² - (y - OY)² ]
If the computation gives imaginary results, there is no intersection. If real z value(s) can be computed, there will be two solutions z1 and z2, corresponding to the negative and positive signs, respectively. The length inside the cylinder at y is the length of the line from (x,y,z1) to (x,y,z2). The amount of this line to use, i.e. the amount inside the subslice, is computed by limiting the (maximum) z2 value to the maximum for the slice and limiting the (minimum) z1 value to the minimum for the slice. If the resulting (z2 - z1) is positive, sum the corresponding volume: (z2 - z1) * dA.

Normal Moduli

Assuming the elements are connected in series, from the force equilibrium,
(2.1)   sigmac11 = sigmam11 = sigmai11 = sigmaf11
From the geometric compatibility, the total displacement equals the sum of the displacements in the component materials. This may be written:
(2.2)   epsilonc11 Lc = epsilonm11 Lm + epsiloni11 Li + epsilonf11 Lf
where L is the distance over which the strain is applicable. Using a simplified Hooke's Law:
(2.3)   epsilon = sigma / E
Substituting, one may write:
(2.4)   Lc sigmac11 / Ec11 = Lm sigmam11 / Em11 + Li sigmai11 / Ei11 + Lf sigmaf11 / Ef11
Using (2.1), one obtains:
(2.5)   Lc / Ec11 = Lm / Em11 + Li / Ei11 + Lf / Ef11
The moduli are constant for a given material type over the subslices. The lengths of each type of material may vary over the subslices. The integral of the lengths, over the volume of the slice, will be the corresponding volumes of the material type. Thus, one obtains:
(2.6)   Vc / Ec11 = Vm / Em11 + Vi / Ei11 + Vf / Ef11
Dividing by the total volume of the slice, Vc, the equation may be expressed in terms of volume ratios for the slice:
(2.7)   1 / Ec11 = km / Em11 + ki / Ei11 + kf / Ef11
For a cylindrical fiber, there is no difference in the equations for longitudinal or transverse directions within the slice. Since the 11 and 22 directions are equivalent because of symmetry,
(2.8)   Ec22 = Ec11
The 33 direction does not work the same way because it is sliced. Also, it may not be exactly symmetric because of the MFIM and loads that change through the thickness. Consider that the regions are connected in parallel. From the force equilibrium:
(2.9)   sigmac33 Lc = sigmam33 Lm + sigmai33 Li + sigmaf33 Lf
where L is the distance over which the strain is applicable. From the geometric compatibility:
(2.10)   epsilonc33 = epsilonm33 = epsiloni33 = epsilonf33
Using a simplified Hooke's Law (2.3) and substituting:
(2.11)   Ec33 Lc = Em33 Lm + Ei33 Li + Ef33 Lf
As previously noted, one may integrate the lengths to obtain volumes and divide to obtain the equation in terms of volume ratios:
(2.12)   Ec33 = km Em33 + ki Ei33 + kf Ef33

Shear Moduli

Assuming the elements are connected in series, from the force equilibrium,
(3.1)   sigmaSc = sigmaSm = sigmaSi = sigmaSf
From the geometric compatibility, the total displacement equals the sum of the displacements in the component materials. This may be written:
(3.2)   epsilonSc Lc = epsilonSm Lm + epsilonSi Li + epsilonSf Lf
where L is the distance over which the strain is applicable. Using a simplified Hooke's Law:
(3.3)   epsilonS = sigmaS / G
Substituting, one may write:
(3.4)   Lc sigmac / Gc = Lm sigmam / Gm + Li sigmai / Gi + Lf sigmaf / Gf
Using (3.1), one obtains:
(3.5)   Lc / Gc = Lm / Gm + Li / Gi + Lf / Gf
The moduli are constant for a given material type over the subslices. The lengths of each type of material may vary over the subslices. The integral of the lengths, over the volume of the slice, will be the corresponding volumes of the material type. Thus, one obtains:
(3.6)   Vc / Gc = Vm / Gm + Vi / Gi + Vf / Gf
Dividing by the total volume, Vc, the equation may be expressed in terms of the volume ratios for the slice:
(3.7)   1 / Gc = km / Gm + ki / Gi + kf / Gf

Poisson's Ratio

Using the rule of mixtures, Poisson's ratio may be written as:
(4.1)   nuc = km num + ki nui + kf nuf

Thermal Expansion Coefficients

From the thermal expansion displacement compatibility, the total displacement equals the sum of the displacements in the component materials. This may be written:
(5.1)   alphac Lc = alpham Lm + alphai Li + alphaf Lf
This is similar to the Normal moduli. The result being:
(5.2)   alphac = km alpham + ki alphai + kf alphaf

Thermal Heat Conductivities

Similar to the derivation for the normal moduli, one obtains:
(6.1)   1 / Kc = km / Km + ki / Ki + kf / Kf

Heat Capacity

Heat capacities are related as follows:
(7.1)   Vc rhoc Cc = Vm rhom Cm + Vi rhoi Ci + Vf rhof Cf
Dividing by Vc and using the appropriate volume ratios, one obtains:
(7.2)   rhoc Cc = km rhom Cm + ki rhoi Ci + kf rhof Cf
The masses add as follows:
(7.3)   Vc rhoc = Vm rhom + Vi rhoi + Vf rhof
Dividing by Vc and using the appropriate volume ratios, one obtains:
(7.4)   rhoc = km rhom + ki rhoi + kf rhof
Substituting 7.4 int 7.2 and solving for Cc, one obtains:
                              km rhom Cm + ki rhoi Ci + kf rhof Cf
(7.5)   Cc = ------------------------------------------------------------
                              km rhom + ki rhoi + kf rhof

Moisture Expansion Coefficient

Similar to the derivation for the Thermal Expansion Coefficient, one obtains:
(8.1)   betac = km betam + ki betai + kf betaf

Moisture Diffusivities

Similar to the derivation for the Thermal Heat Conductivities, one obtains:
(9.1)   1 / Dc = km / Dm + ki / Di + kf / Df


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Last updated on Mon Nov 8 11:28:52 EST 2004
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