Unidirectional Fiber-Reinforced Rubber

Abstract

The mystery why a tire, just one part of a vehicle, realizes many functions simultaneously can be explained by the fact that a tire has a composite structure made from composite material. Unidirectional fiber-reinforced rubber (UFRR) is the main composite material of a tire. The elastic and viscoelastic properties of UFRR are defined in principal directions, which are the direction of the fiber or cord and the direction perpendicular to the fiber or cord. Applying the rotation matrix to the properties of UFRR in the principal directions, the properties of UFRR can then be calculated in an arbitrary direction. Because UFRR has nonlinear properties with respect to the direction of the applied external force, the angle and width of UFRR used for a tire belt need to be carefully determined in tire design.

Appendices

Appendix: Viscoelasticity

Using Boltzmann’s superposition principle, the stress response σ(t) with respect to the loading history of the strain ε(t) is expressed by

$$\sigma (t) = \int\limits_{ - \infty }^{t} {E(t - \hat{\tau })\frac{{{\text{d}}\varepsilon (\hat{\tau })}}{{{\text{d}}\hat{\tau }}}} {\text{d}}\hat{\tau }.$$

(1.148)

The nondimensional relaxation function ϕ(t) is introduced as

$$\phi (t) = E(t)/E_{\infty } - 1,$$

(1.149)

where E_∞ is the elastic modulus at t = ∞. The substitution of Eq. (1.148) into Eq. (1.149) yields

$$\sigma (t) = \sigma_{\infty } (t) + \int\limits_{ - \infty }^{t} {\phi (t - \hat{\tau })\frac{{{\text{d}}\sigma_{\infty } (\hat{\tau })}}{{{\text{d}}\hat{\tau }}}} {\text{d}}\hat{\tau } = \sigma_{\infty } (t) - \int\limits_{0}^{\infty } {\phi (\tau )\frac{{{\text{d}}\sigma_{\infty } (t - \tau )}}{{{\text{d}}\tau }}} {\text{d}}\tau ,$$

(1.150)

where σ_∞ = E_∞ε(t) and ${\tau } = {t} - \hat{\tau }.$

Suppose that the strain ε harmonically oscillates with the angular velocity ω and ε is expressed by

$$\varepsilon (t) = \varepsilon^{*} {\text{e}}^{{{\text{i}}\omega t}} .$$

(1.151)

The substitution of Eq. (1.151) into Eq. (1.150) yields

$$\sigma (t) = E_{\infty } \varepsilon (t) - \int\limits_{0}^{\infty } {\phi (\tau )\frac{\text{d}}{{{\text{d}}\tau }}} \left( {E_{\infty } \varepsilon^{*} {\text{e}}^{{{\text{i}}\omega \left( {t - \tau } \right)}} } \right){\text{d}}\tau = E_{\infty } \left( {1 + {\text{i}}\omega \int\limits_{0}^{\infty } {\phi (\tau ){\text{e}}^{{ - {\text{i}}\omega \tau }} } {\text{d}}\tau } \right)\varepsilon (t).$$

(1.152)

Considering that ϕ(τ) = 0 is satisfied when τ < 0, the Fourier transformation of ϕ(t) is expressed as

$$\phi^{*} (\omega ) = \int\limits_{0}^{\infty } {\phi (\tau ){\text{e}}^{{ - {\text{i}}\omega \tau }} } {\text{d}}\tau .$$

(1.153)

Equation (1.152) is rewritten as

$$\sigma (t) = E^{*} (\omega )\varepsilon (t),$$

(1.154)

where $E^{*} \left( \omega \right)$ is given by

$$E^{*} (\omega ) = E_{\infty } \left( {1 + {\text{i}}\omega \phi^{*} (\omega )} \right).$$

(1.155)

Expressing ϕ^*(ω) in complex form, Eq. (1.155) can be rewritten as

$$E^{*} (\omega ) = E_{\infty } \left( {1 - \omega \phi^{\prime \prime } } \right) + {\text{i}}E_{\infty } \omega \phi^{{\prime }} = E^{\prime} + {\text{i}}E^{\prime\prime},$$

(1.156)

where $\phi^{\prime}$ and $\phi^{\prime\prime}$ are, respectively, real and imaginary parts of ϕ^*(ω). $E^{\prime}$ and $E^{\prime\prime}$ are, respectively, referred to as the storage modulus and loss modulus. The loss tangent tan δ is defined by

$$\tan \delta = E^{\prime\prime}/E^{\prime}.$$

(1.157)

Identification of the Damping Matrix [45]

The vibration problem of linear elastic materials can be solved by following one of two approaches of the finite element method. One is the direct approach, which is an exact method, but the computational cost is high and a method of identifying the damping matrix has not been established. The other approach is the modal superposition method, which is a kind of approximation method that has low computational cost. The damping coefficient can be determined from experimental modal analysis. Yoshida [45] proposed identifying the damping matrix for the direct method using modal damping coefficients.

The equation of motion for the structure with damping is given by

$$\left[ M \right]\left\{ {\ddot{u}} \right\} + \left[ C \right]\left\{ {\dot{u}} \right\} + \left[ K \right]\left\{ u \right\} = \left\{ F \right\},$$

(1.158)

where {u} is the displacement vector, the dot indicates the time derivative, and [M], [C], [K] and {F} are the mass matrix, damping matrix, stiffness matrix and external force. When the damping and external force are zero in Eq. (1.158), the structure freely vibrates. The eigenvalue problem is thus defined, where the r-th real eigenvector is expressed by {X_r}. The solution {u} to Eq. (1.158) can be expressed as the superposition of orthogonal vectors {X_r}:

$$\left\{ u \right\} = \left[ {\left\{ {X_{1} } \right\}\left\{ {X_{2} } \right\} \ldots \left\{ {X_{r} } \right\} \ldots \left\{ {X_{n} } \right\}} \right]\left\{ q \right\} = \left[ \xi \right]\left\{ q \right\},$$

(1.159)

where {q} is the modal displacement, [ξ] is the matrix for which the eigenvector {X_r} is in the column. Substituting Eq. (1.159) into Eq. (1.158) and multiplying by [ξ]^T from the left yields

$$\begin{aligned} & \left[ m \right]\left\{ {\ddot{u}} \right\} + \left[ c \right]\left\{ {\dot{u}} \right\} + \left[ k \right]\left\{ u \right\} = \left[ \xi \right]^{\text{T}} \left\{ F \right\} \\ & \left[ m \right] = \left[ \xi \right]^{\text{T}} \left[ M \right]\left[ \xi \right] \\ & \left[ c \right] = \left[ \xi \right]^{\text{T}} \left[ C \right]\left[ \xi \right] \\ & \left[ k \right] = \left[ \xi \right]^{\text{T}} \left[ K \right]\left[ \xi \right], \\ \end{aligned}$$

(1.160)

where mass matrix [m] and stiffness matrix [k] are diagonal matrices because of the orthogonality of eigenvectors. The r-th diagonal element m_r and k_r are referred to as the modal mass and modal stiffness and are expressed as

$$\begin{aligned} & \left[ m \right]\left\{ {\ddot{u}} \right\} + \left[ c \right]\left\{ {\dot{u}} \right\} + \left[ k \right]\left\{ u \right\} = \left[ \xi \right]^{\text{T}} \left\{ F \right\} \\ & m_{r} = \left\{ {X_{r} } \right\}^{\text{T}} \left[ M \right]\left\{ {X_{r} } \right\} \\ & k_{r} = \left\{ {X_{r} } \right\}^{\text{T}} \left[ K \right]\left\{ {X_{r} } \right\}. \\ \end{aligned}$$

(1.161)

Meanwhile, the eigenvector {X_r} does not satisfy orthogonality with the damping matrix [C], and the third equation [c] of Eq. (1.160) is not a diagonal matrix in general. If [C] is expressed by the linear combination of [K] and [M], it is referred to as Rayleigh damping and the matrix [c] becomes a diagonal matrix. However, because Rayleigh damping has a damping matrix with two parameters, the properties of the damping matrix cannot be expressed satisfactory. Yoshida assumed that [c] is a diagonal matrix in which the diagonal element is the modal damping c_r and then expressed [C] by $c_{r}$. This assumption gives a relationship between c_r and [C]:

$$c_{r} = \left\{ {X_{r} } \right\}^{\text{T}} \left[ C \right]\left\{ {X_{r} } \right\}.$$

(1.162)

Here, c_r is expressed by

$$c_{r} = 2\omega_{r} \zeta_{r} m_{r} ,$$

(1.163)

where ζ_r is the modal damping ratio.

The substitution of Eq. (1.163) into Eq. (1.162) yields

$$\left[ {2\omega_{r} \zeta_{r} m_{r} } \right] = \left[ \xi \right]^{\text{T}} \left[ C \right]\left[ \xi \right],$$

(1.164)

where [2ω_rζ_rm_r] is a diagonal matrix in which the r-th diagonal element is 2ω_rζ_rm_r. Using the second equation of Eq. (1.160), Eq. (1.164) can be transformed to

$$\left[ C \right] = \left[ M \right]\left[ \xi \right]\left[ {2\omega_{r} \zeta_{r} /m_{r} } \right]\left[ \xi \right]^{\text{T}} \left[ M \right],$$

(1.165)

where [2ω_rζ_r/m_r] is a diagonal matrix in which the r-th diagonal element is 2ω_rζ_r/m_r . Note that [ξ]⁻¹ cannot be calculated because [ξ] is not a square matrix.

Physical Meaning of Modal Damping [28]

When [c] is a diagonal matrix, the equations given as Eq. (1.161) can be rewritten as n one-dimensional equations of motion:

$$m_{r} \ddot{q}_{r} + 2\omega_{r} \zeta_{r} m_{r} \dot{q}_{r} + k_{r} q_{r} = f_{r} ,$$

(1.166)

where the undamped natural frequency ω_r and generalized external force f_r are

$$\begin{aligned} \omega_{r} & = \sqrt {k_{r} /m_{r} } \\ f_{r} & = \left\{ {X_{r} } \right\}^{\text{T}} \left\{ F \right\}. \\ \end{aligned}$$

(1.167)

When f_r is zero in Eq. (1.166) and the condition ζ_r < 1 is satisfied, the solution is damped vibration. Suppose that the period is T. Multiplying both sides of Eq. (1.166) by dq_r and integrating over one period of time, we obtain

$$\begin{aligned} & \int\limits_{0}^{T} {\frac{{{\text{d}}K_{\text{en}} }}{{{\text{d}}t}}{\text{d}}t} + \int\limits_{0}^{T} {\frac{{{\text{d}}U}}{{{\text{d}}t}}{\text{d}}t} = - 2\omega_{r} \zeta_{r} m_{r} \int\limits_{0}^{T} {\dot{q}_{r}^{2} {\text{d}}t} \\ & K_{\text{en}} = \frac{1}{2}m_{r} \dot{q}_{r}^{2} \\ & U = \frac{1}{2}k_{r} q_{r}^{2} , \\ \end{aligned}$$

(1.168)

where K_en and U are, respectively, the kinetic energy and strain energy. Equation (1.168) gives the energy conservation in a period. Assuming the initial displacement is q_r0 at t = 0, the solution to (1.166) is given by

$$\begin{aligned} q_{r} & = q_{r0} {\text{e}}^{{ - \omega_{r} \zeta t}} \cos \omega_{rd} t \\ \omega_{{r{\text{d}}}} & = \omega_{r} \sqrt {1 - \zeta_{r}^{2} } , \\ \end{aligned}$$

(1.169)

where ω_rd is the natural frequency of the damped vibration of the r-th mode. Substituting Eq. (1.169) into the right side of Eq. (1.168), applying a Taylor series expansion with respect to ζ_r and then neglecting more than the quadratic term of ζ_r, we obtain

$$2\omega_{r} \zeta_{r} m_{r} \int\limits_{0}^{T} {\dot{q}_{r}^{2} {\text{d}}t} \approx 4\pi \zeta_{r} \left( {\frac{1}{2}k_{r} q_{r0}^{2} } \right) = 4\pi \zeta_{r} E_{0} ,$$

(1.170)

where E₀ is the initial strain energy. Because the first term on the left side of Eq. (1.168) is the energy −ΔE to lose in a period, from Eq. (1.168) and (1.170), we obtain

$$\Delta E = 4\pi \zeta_{r} E_{0} .$$

(1.171)

Equation (1.171) shows that the dissipation energy in a period, ΔE, is proportional to the initial strain energy E₀ and the modal damping ratio ζ_r is equivalent to the ratio of the two energies. When the material is linear viscoelastic, the constitutive equation is Eq. (1.154). When the strain is assumed to be harmonic vibration expressed by Eq. (1.151), and the dissipation energy Δe in a period per unit volume is the same as the work of internal stress, this is expressed by

$$\Delta e = \oint {\text{Re} \left( {\left\{ {\sigma_{r} } \right\}} \right)^{\text{T}} \text{Re} \left( {\left\{ {{\text{d}}\varepsilon_{r} } \right\}} \right)} = \oint {\text{Re} \left( {\left[ {E^{*} } \right]\left\{ {\sigma_{r} } \right\}} \right)^{\text{T}} \cdot \text{Re} \left( {\left\{ {\dot{\varepsilon }_{r} } \right\}} \right)} {\text{d}}t = \pi \left\{ {\varepsilon_{r} } \right\}^{\text{T}} \left[ {E^{\prime\prime}} \right]\left\{ {\varepsilon_{r} } \right\},$$

(1.172)

where Re indicates the real part of the complex number. Integrating Δe in the volume of the structure, the energy loss in a period ΔE and the initial strain energy E₀ are expressed by

$$\begin{aligned} \Delta E & = \pi \int\limits_{V} {\left\{ {\varepsilon_{r} } \right\}^{\text{T}} \left[ {E^{\prime\prime}} \right]\left\{ {\varepsilon_{r} } \right\}{\text{d}}V} \\ E_{0} & = \frac{1}{2}\int\limits_{V} {\left\{ {\varepsilon_{r} } \right\}^{\text{T}} \left[ {E^{\prime}} \right]\left\{ {\varepsilon_{r} } \right\}{\text{d}}V} . \\ \end{aligned}$$

(1.173)

Equations (1.171) and (1.173) give the modal damping ratio ζ_r expressed by

$$\zeta_{r} = \frac{1}{2}\frac{{\int_{V} {\left\{ {\varepsilon_{r} } \right\}^{\text{T}} \left[ {E^{\prime\prime}} \right]\left\{ {\varepsilon_{r} } \right\}{\text{d}}V} }}{{\int_{V} {\left\{ {\varepsilon_{r} } \right\}^{\text{T}} \left[ {E^{\prime}} \right]\left\{ {\varepsilon_{r} } \right\}{\text{d}}V} }}.$$

(1.174)

Hence, if the complex elastic matrix $[E^{*} ]$ is known, ζ_r can be calculated from the strain energy distribution of the mode corresponding to the real eigenvector.

Notes

Note 1.1 Incompressibility of Rubber

Rubber has specific properties that the volume is maintained under external forces. Referring to Fig. 1.35a, the incompressibility is expressed by

$$\left( {1 + \varepsilon_{x} } \right)l_{x} \left( {1 + \varepsilon_{y} } \right)l_{y} \left( {1 + \varepsilon_{z} } \right)l_{z} = l_{x} l_{y} l_{z} .$$

The low-strain assumptions ε_x, ε_y, ε_z ≪ 1 yield

$$\varepsilon_{x} + \varepsilon_{y} + \varepsilon_{z} = 0.$$

Figure 1.35b shows the deformation of the element where the external force is applied in the z-direction. When the strain along the z-axis is ε, strains along the x- and y-axes can be expressed by ε and Poisson’s ratio ν. Referring to Fig. 1.35b, the incompressibility can be expressed by

$$\left( {1 - \nu \varepsilon } \right)l_{x} \left( {1 - \nu \varepsilon } \right)l_{y} \left( {1 + \varepsilon } \right)l_{z} = l_{x} l_{y} l_{z} .$$

The assumption of low strain ε ≪ 1 yields

$$\left( {1 - \nu \varepsilon } \right)^{2} \left( {1 + \varepsilon } \right) \cong \left( {1 - 2\nu \varepsilon } \right)\left( {1 + \varepsilon } \right) \cong 1 + \left( {1 - 2\nu } \right)\varepsilon = 1.$$

Poisson’s ratio for incompressible materials is ν = 1/2 from the above equation.

Fig. 1.35

Incompressibility of rubber

Note 1.2

When −1 < cos 2θ < 1 is satisfied, the extreme value is located away from θ = 0° and θ = 90°. Because $\frac{{E_{T} - E_{L} }}{{E_{L} E_{T} /G_{LT} - \left( {E_{L} + E_{T} + 2\nu_{L} E_{T} } \right)}} < 1$ is satisfied, G_LT > E_L/{2(1 + ν_L)} is obtained considering the relation E_L − E_T > 0.

In the case of FRR, from Eq. (1.98), the elastic properties can be expressed as

$$\begin{aligned} & G_{LT} = \frac{{E_{\text{m}} }}{3} \\ & \frac{{E_{L} }}{{2(1 + \nu_{L} )}} = \frac{{E_{\text{f}} V_{\text{f}} }}{3} \\ & \frac{{E_{T} }}{{2(1 + \nu_{T} )}} = \frac{2}{3}E_{\text{m}} . \\ \end{aligned}$$

Because the condition G_LT < E_T/{2(1 + ν_T)} is satisfied for FRR, E_x for FRR has an extreme value within 0 < θ < 90°.