Lamination Theory

Abstract

A laminate structure is used for the belt of radial tires and the ply of bias tires that consist of steel cords and rubber, or organic cords and rubber. Properties of the laminate, such as the extensional stiffness and bending stiffness of tire belt, can be estimated using classical lamination theory (CLT). Although interlaminar shear is not considered in CLT, the elastic properties predicted by CLT agree with measurements fairly well. The finite element method is used to predict the effect of the belt structure on the tire performance. Furthermore, the belt structure optimized for maneuverability or durability is obtained by combining the genetic algorithm (GA) with the finite element method.

Notes

Note 2.1 Eq. (2.9)

Referring to Fig. 2.28 below, we have ${\text{d}}x = - \rho {\text{d}}\phi ,\quad {\text{d}}\phi {\text{ = }}\frac{\partial }{{\partial x}}\frac{{\partial w}}{{\partial x}}{\text{ = }}\frac{{\partial ^{2} w}}{{\partial x^{2} }}{\text{d}}x$.

Fig. 2.28

Plate in bending deformation

From the above equations, we obtain $\kappa = \frac{1}{\rho } = - \frac{\partial\phi }{\partial x} = - \frac{{\partial^{2} w}}{{\partial x^{2} }}$.

When the Kirchhoff–Love hypothesis is not satisfied, the straight lines normal to the mid-surface do not remain normal to the mid-surface after bending and shear deformation is therefore generated. Theory including shear deformation in the cross section of a plate is proposed and the displacements are expressed by

$$\begin{aligned} u_{1} (x,y,z) & = u(x,y) - z\phi_{x} (x,y) \\ u_{2} (x,y,z) & = v(x,y) - z\phi_{y} (x,y) \\ u_{3} (x,y,z) & = w(x,y), \\ \end{aligned}$$

where $- \phi_{x} (x,y)$ and $- \phi_{y} (x,y)$ are, respectively, the rotation angles with respect to the y- and x-axes. Because the above equations are of first order with respect to z, the theory is referred to as first-order shear deformation theory.

Note 2.2 Eq. (2.70)

In the case of α = 20° and E_T = 3 MPa, we obtain E_x = 152 MPa by Eq. (2.70). When the tire radius r is 280 mm, the inflation pressure p is 0.2 MPa, and the belt thickness h is 2 mm, the stress of belt in the circumferential direction $\sigma_{x}$ is given by $\sigma_{x} = pr/h = 28\,{\text{MPa}}$. The value of E_x is too small to support $\sigma_{x}$. The reason why a tire can support $\sigma_{x}$ is that the stress in the lateral direction $\sigma_{y}$ is not zero in a tire. Using Eq. (2.50) under the FRR approximation and the assumption of $\sigma_{y} \ne 0$, we obtain

$$\frac{{\sigma_{x} }}{{\varepsilon_{x} }} = \overline{E}_{xx} - \nu_{x} \overline{E}_{xy} \cong E_{L} (1 - \nu_{x} \tan^{2}\upalpha)\cos^{4}\upalpha$$

where ν_x = − ε_y/ε_x. Because the above equation is a function of E_L, the belt can support $\sigma_{x}$. The value of ν_x is 7.5 by Eq. (2.70), but the value of a tire is from 1 to 2 in the center of a tire owing to the radial ply which angle is 90°.