5 Special Fluid Medium Lubrication
5.3.2 Deformation Analysis of Liquid Crystal Lubrication
Because the liquid crystal has a layered structure, its lubrication analysis method is different from the usual lubrication theory. So far, there has been no good LC lubrication theory. Rhim et al. set a simple model of liquid crystal lubrication for numerical deformation analysis [3]. The following discussion is limited to the situation that the orientation of liquid crystal is aligned to the normal direction of the layer as shown in Figure 5.12a.
As shown in Figure 5.14, the layers of the liquid crystal before entering the wedge-shaped gap are parallel to each other. When they enter the wedge zone, the liquid crystal layers inside Figure 5.14 Schematic diagram of liquid crystal lubrication.
k k Figure 5.15 Rhim’s rheological model of LC.
the two surfaces are still maintained in parallel. Therefore, at the smaller end, LC will inevitably be squeezed. If the wedge angle is 2𝜃, the wedge-shaped length isl, the larger film thickness is 2h, the smaller thickness is 2h1,Δais the average thickness of liquid crystal layer, andbis the Burgers vector of values (i.e. LC the atomic spacing), the displacement at the boundary is L=bΔa/𝜃.
Figure 5.15 is a liquid crystal rheological model proposed by Rhim et al. [3]. They idealized a liquid crystal plane as a porous layer, connected by springs between the liquid crystal layers. LC can be thought of as the same as a Newtonian fluid flowing parallel to the plane. In the thickness direction, the flow of liquid crystal will be resisted by the porous layer in the same way as the penetration. The resistance is proportional to the pressure gradient. With the isothermal and incompressible assumptions, and without regard to the body force, the elastic deformation of the liquid crystal layers can be described only by one displacement componentw(x,y,z), where wis the local displacement of a liquid crystal layer, and it obeys the differential equation
e𝜕2w
𝜕z2 −k [𝜕4w
𝜕x4 +2 𝜕4w
𝜕x2𝜕z2 + 𝜕4w
𝜕z4 ]
=0, (5.25)
wherezis the coordinate perpendicular to the local liquid crystal layer;xandyare the coordi- nates along the local layer, and perpendicular to each other;eandkare the compression elastic modulus and the separation modulus of liquid crystal material, whileehas a stress dimension, andkhas a power dimension.
For simplicity, the following discussions are only limited to the two-dimensional infinite slid- ing problem. The dimensionless coordinates and displacements are
X=x∕l;Z=z∕h0;W =w∕h0, (5.26)
wherelis slider width;h0is the half thickness at the larger end;Wis the displacement of defor- mation.
Equation 5.25 can be written as
𝜕2W
𝜕Z2 −𝜆2𝜕4W
𝜕X4 =0, (5.27)
where𝜆is the liquid crystal permeability coefficient, for unit width,𝜆=
√ kh20∕el2.
If liquid crystal only has one boundary dislocation, the local deformationWiof Equation 5.27 has the solution
Wi= −Δa(Z−Z0)
|Z−Z0| [
erf
( X−X0
√4𝜆|Z−Z0| )
+erf
( X0
√4𝜆|Z−Z0| )]
, (5.28)
k k whereX0andZ0are the boundary coordinates ofXandZ;Δais the average thickness of liquid
crystal layer; erf() is the error function.
When there are multiple internal dislocations, and as long as they are far enough apart from each other, the liquid crystal layer displacement solution can be obtained by superimposing all the single dislocation solutions. If in Figure 5.14, LC is limited to a thin, wedge-shaped area in the rigid boundaries, the deformation of liquid crystal layer can be obtained from Equation 5.28:
W =
∑n
i=1
[Wi(X,Z,X0i,Z0i,) +𝛿W(X,Z)]. (5.29) With the assumption that liquid crystal layers are parallel to the wedge surface, the corre- sponding boundary conditions are
𝛿W[(X,−H(X)] =1−H(X) −
∑n
i=1
Wi[X,−H(X)] (5.30)
𝛿W[(X,H(X)] =1+H(X) −
∑n
i=1
Wi[X,H(X)] (5.31)
𝛿W(0,Z) =0 𝛿2W(0,Z)
𝛿X2 =0
𝛿W(1,Z) = −[1−H1]sign(Z) −
∑n
i=1
Wi[1,Z]
𝛿2W(1,Z)
𝛿X2 =0, (5.32)
where sign(Z) is the function. WhileZ≥0, it takes a positive sign, and whileZ<0 a negative sign;His the dimensionless thickness,H=h/h1.
In fact, under these boundary conditions, an analytical solution of Equation 5.27 still cannot be directly obtained, such as Equation 5.28. Therefore, numerical methods must be used to solve𝛿W(X,Z). The usual approach is to use the finite element method or the finite difference method. Substitute the deformation of the nodes as the unknown quantities into Equation 5.27, and then solve them according to the above boundary conditions. For example,
𝛿W(X,Z) =Ni(𝜉, 𝜂)𝛿Wi, (5.33)
whereNi(𝜉,𝜂) is the element function, known as the shape function;𝛿Wiis the unknown defor- mation on nodei.
Rhim et al. gave the analysis below on the finite element method. The parameters used in the calculation are:l=0.01 m,h=10−4m,h0=10−9/2m,e=107Pa,𝜆2=hh02/el4=10−18.
Figure 5.16 shows that if there is no dislocation, while the liquid crystal layer enters a wedge-shaped region, its shape is twisted to be a quadrilateral. In order to adapt to the wedge-shaped area, it gradually reduces its thickness so that the liquid crystal layer becomes thinner as a spring is gradually pressed. Clearly, the deformation would be impossible if the amount of compression was greater than the thickness of the layer.
Figure 5.17 shows three sub-images. They are the total displacement, the node displacements, and the side shape. The boundary dislocation isL=bΔa=0.1 (H1=0.8). In Equation 5.20, there exists a singular point, which can be clearly seen in Figure 5.17. Near the solid surface, LC is
k k Figure 5.16 Wdistribution without dislocation (H1=0.8); (a) displacement of liquid crystal layer; (b) side shape.
Figure 5.17 Wdistribution with dislocations (H1=0.8,L=0.1); (a) total displacement; (b) node displacements;
(c) side shape.
almost parallel to the surface, keeping a constant. However, the dislocations near the center occur abruptly rather than gradually (Figure 5.17a). From the node displacements (Figure 5.17b) there exist many small wrinkles, which force (or cause) the liquid crystal layer distorted, which is quite different from Figure 5.16.
Figure 5.18 shows the displacementWof the small Burgers vectors with dislocations. Unlike Figure 5.17, the figure has very small Burgers vectors, or the distance between the liquid crystal layers are larger,L=bΔa=0.02. It can be seen from Figure 5.18 that the minimum thickness H1is equal to 0.8 and there are 20 dislocations per unit length at the boundary. In addition to dislocations, tiny wrinkles can also be seen near the dislocations, but wrinkles are hardly seen
k k Figure 5.18 Wdistribution with dislocation of (H1=0.8,L=0.02).
a little further away from the dislocations. This is mainly because the Burgers vectors are too small. In addition, the amplitudes also reduce in the central area.