4.4 Compaction of a single tow
4.4.4 Compaction test case
In order to check the model has been derived and implemented correctly, it is important to run a simulation where the result is known. A tow where the fibres are arranged as
CHAPTER4: MECHANICAL MODELLING OF TOWS
shown in Figure 4.17 was constructed. This is not a realistic tow, but is one where an analytical solution is easily obtained and which is sufficiently complicated to test the computational model. L w h L 6 L 2
Figure 4.17:Two orthogonal views of the fibre compaction test case
Analytical solution
Due to symmetry it can be seen that the four slanted fibres do not bend at all. The 24 straight fibres each bend due to a force applied to each one at a single point. This point is at L
2 for 8 of the fibres. For the other 16 fibres the point is at L6. These assumptions are only valid for small deflections; once the fibres start to bend the contact points and direction of the forces will change. The ends of the fibres are pinned, so that no deformation or strain energy is present before compaction. From Equation 4.12 or 4.13, the displacement v at point a of a fibre pinned at either end caused by a force P acting at point a is:
v(a) = P(L−a)a3
6EIL −
Pa 2L2−3La+a2
6EIL a (4.58)
which can be simplified and re-arranged to give P in terms of v: P(a) = − 3EILv
a2(a−L)2 (4.59)
CHAPTER4: MECHANICAL MODELLING OF TOWS
An arbitrary set of dimensions and properties have been chosen as shown in Table 4.1. Note that a consistent set of units has been chosen to avoid having to convert to SI units.
Table 4.1:Fibre compaction test case dimensions
Fibre radius r 0.01 mm Domain width w 0.16 mm Domain height h 0.16 mm
Domain length L 1 mm
Young’s modulus E 80×103 MPa Second moment of area I 7.85×10−9 mm4 For the fibres which have the force applied at their centre (i.e. a= L
2), the force is:
P=30.1×10−3v (4.60)
For the fibres which have the force applied at a= L
6), the force is:
P=97.7×10−3v (4.61)
Figure 4.18 illustrates eight layers in which contacts between fibres occur. Compaction
Contact layers
Figure 4.18:Fibre contact layers
forces applied to the top surface of the tow are propagated through the tow by these contact layers. In each of these layers there is one fibre with force applied at the ax- ial centre and two fibres with force applied at axial position L
6. This test case has been designed to contain easily identifiable contact layers, however in a randomly dis- tributed tow these layers do not exist, hence demonstrating the need for an energy based method.
The total force F acting on one layer would be: F=
30.1×10−3
+2 97.7×10−3 v=225.5×10−3v (4.62) The strain energy within the fibres will also be obtained analytically for comparison with the model. In order to do this the moment M from Equations 4.10 and 4.11 is
CHAPTER4: MECHANICAL MODELLING OF TOWS
substituted into Equation 4.34 to obtain the strain energy U: Ui = 1 2EI Z a 0 P(L−a)x L 2 dx+ Z L a Pa1− x L 2 dx ! (4.63) Ui = 1 2EI P(L−a) L 2 x3 3 a 0 + (Pa)2 x− x 2 L + x3 3L L a ! (4.64) Ui = P2a2 L4−2L2a+La2+ (1−L)a3 6EIL2 (4.65)
For the fibres which have the force applied at the centre of the fibre (i.e. a = L 2), the strain energy is:
Ui =15.0×10−3v2 (4.66) For the fibres which have the force applied one sixth of the way along the length (i.e. a = L
6), the force is:
Ui =48.9×10−3v2 (4.67) The total strain energy in the system is thus:
U =
8 15.0×10−3
+16 48.9×10−3 v2 =902×10−3v2 (4.68) In Section 4.4.3 it was shown that the force F could be related to the strain energy with Equation 4.53. Thus F can be expressed in terms of U as:
F = d
ds(902×10
−3v2) (4.69)
Before it can be evaluated s needs to be related to v. Initially when the fibres are just touching the distance between the centrelines is 2r where r is the radius of the fibres. As the yarn is compacted by compaction Q, the distance between the centrelines will reduce to 2rQy. Rather than the fibres being compressed, one is deflected by distance v while the other remains undisplaced. The distance v can be expressed as:
v=2r−2rQy (4.70)
It can be seen from Figure 4.17 that the height h is equal to 16r, hence: v= h(1−Qy)
8 (4.71)
From Equation 4.57 it can be shown that:
s=8v (4.72)
which can then be used to evaluate F: F= d
dv
(902×10−3v2)
8 =225.5×10
−3v (4.73)
which agrees exactly with Equation 4.62 thus proving that the energy method is valid. 100
CHAPTER4: MECHANICAL MODELLING OF TOWS
Computation model solution
A computational model of the same tow has been built and can be seen in Figure 4.19. As well as the dimensions specified in Table 4.1, some additional parameters are needed for the computational model and are specified in Table 4.2.
Figure 4.19:Computation test case model
Table 4.2:Parameters for fibre compaction test case
Coefficient of friction µ 0.3 Final compaction~Q (1, 0.75, 1)
Number of steps ns 50 Number of fibre length divisions nd 200 Intersection convergence tolerance td 10−6
Strain convergence tolerance tU 0.001 Strain convergence iterations ni 50
Contact coefficient K 10
The contact coefficient K shown in the table is used to control how much force is applied to the fibres when they intersect by a distance d. In Section 4.3.2, it is suggested that a suitable equation to calculate the magnitude of force between two fibres is:
k~P′k = k~Pk +cd′ (4.74)
The value of c affects the rate of convergence and should be chosen such that the con- vergence rate is maximised without causing model instabilities. The optimum value of cdepends on the fibre properties such as modulus, radius and length. It is convenient to define c in terms of a new constant K that is independant of the dimensions and
CHAPTER4: MECHANICAL MODELLING OF TOWS
properties of the fibres. The magnitude of the force between two fibres becomes:
k~P′k = k~Pk + ~PeK L
nd (4.75)
where P~e is the estimated force necessary to displace two fibres so that they are just
touching assuming they are the only two fibres in contact:
k ~Pek = − 3EILd ′
2a2(a−L)2 (4.76)
This is the same as Equation 4.59 except that v has been replaced by d′
2, since the de- flection of each fibre will be half the total intersection distance. The term L
nd, which
is length of the fibres divided by the number of fibre length divisions, represents the distance between two consecutive cross-sections. It is included since shorter distances between cross-sections will result in more contact points, thus the force on each contact force must be less.
After about 20 minutes, 9443 iterations and 50 steps, the simulation successfully com- pleted. Figure 4.20 shows the simulation at step number 25 with forces illustrated in yellow and moments in green. The plot of strain energy U versus distance s for the ana- lytical and computation model are shown in Figure 4.21. Similarly the plot of F versus s is shown in Figure 4.22. The computational model follows the analytical solution very closely for small compactions. However as the compaction increases, the analytical model becomes invalid. The contact forces no longer remain vertical as the fibres begin to bend, as can be seen in Figure 4.20. As expected, the more accurate computational model shows higher values of U and F for higher degrees of compaction.
The curve of the computational solution in Figure 4.22 is slightly jagged. This is due to the approximation created by the fibre length divisions. If a curve of strain energy U were plotted against contact force location a (see Section 4.3.1), the curve should have continuity C1. However in the computational model the strain energy varies linearly between fibre length divisions hence resulting in a curve with continuity C0only. Since the force F is proportional to U′, a curve of F plotted against a does not have continuity C0. The jaggedness of the curve in Figure 4.22 is a result of one or more contact force locations a crossing a fibre length division. The problem is exacerbated in this case due to the symmetry of the model, as all of the contact force locations move in unison. Appendix G contains graphs showing the effect of varying the number of divisions nd. Nevertheless, the high level of agreement for low compaction levels is very encourag- ing. It shows that the computational model has been correctly implemented and that the iterative technique to obtain a converged solution is valid.
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Figure 4.20:Computational test case model at step 25
0 1 · 10−7 2 · 10−7 3 · 10−7 0 0.002 0.004 0 1 · 10−5 2 · 10−5 3 · 10−5 4 · 10−5 Strain Energy U (mJ) Strain Energy U (mJ) 0 0.01 0.02 0.03 0.04 Compaction distance s (mm) Compaction distance s (mm) Analytical Solution Computational Solution
CHAPTER4: MECHANICAL MODELLING OF TOWS 0 5 · 10−5 1 · 10−4 0 0.002 0.004 0 0.001 0.002 0.003 0.004 F orce F (N) F orce F (N) 0 0.01 0.02 0.03 0.04 Compaction distance s (mm) Compaction distance s (mm) Analytical Solution Computational Solution
Figure 4.22:Fversus s for compaction test case