Spring Properties of Tires

Abstract

The spring properties of tires are the most fundamental characteristics of tires and are closely related to the four fundamental functions discussed in Sect. 1.1.

Notes

Note 6.1

Equation (6.13)

Because the length of the sidewall does not change after loading, it follows that

$$r\phi_{\text{s}} = const.$$

(6.178)

Differentiating Eq. (6.178) with respect to ϕ_s, we obtain

$${\text{d}}r/{\text{d}}\phi_{\text{s}} = - r/\phi_{\text{s}} .$$

(6.179)

Using Eqs. (6.9) to (6.12), we obtain

$$K_{\text{p}} = 2\bar{K}_{\text{p}} = 2\frac{\partial P}{\partial x} = 2\frac{{\partial (pr\cos \phi_{\text{s}} )}}{{\partial (2r\sin \phi_{\text{s}} )}} = p\frac{{\frac{{\partial r\cos \phi_{\text{s}} }}{{\partial \phi_{\text{s}} }}}}{{\frac{{\partial r\sin \phi_{\text{s}} }}{{\partial \phi_{\text{s}} }}}} = p\frac{{\frac{\partial r}{{\partial \phi_{\text{s}} }}\cos \phi_{\text{s}} - r\sin \phi_{\text{s}} }}{{\frac{\partial r}{{\partial \phi_{\text{s}} }}\sin \phi_{\text{s}} + r\cos \phi_{\text{s}} }}.$$

(6.180)

The substitution of Eq. (6.179) into Eq. (6.180) yields

$$K_{\text{p}} = p\frac{{\cos \phi_{\text{s}} + \phi_{\text{s}} \sin \phi_{\text{s}} }}{{\sin \phi_{\text{s}} - \phi_{\text{s}} \cos \phi_{\text{s}} }}.$$

(6.181)

Pacejka [13] derived Eq. (6.13) taking the following approach. Referring to Fig. 6.64, the external vertical force acting on the segment is defined as

Fig. 6.64

Force equilibrium of a tire

$$q_{z} = 2\int\limits_{0}^{b} {p_{z}\; {\text{d}}y} .$$

(6.182)

The force equilibrium in Fig. 6.64 yields

$$q_{z} /2 + T_{\text{s}} = p(r_{\text{s}} + b_{e} ).$$

(6.183)

Substituting the relation T_s = pr_s into Eq. (6.183), we obtain

$$q_{z} = 2pb_{e} ,$$

(6.184)

where 2b_e denotes the effective width indicated in Fig. 6.9. When the radial deflection is −w, it follows from Fig. 6.9 that

$$h_{s} = h_{s0} + w = l\sin \phi_{\text{s}} /\phi_{\text{s}} ,$$

(6.185)

$$2\left( {b - b_{e} } \right) = l\cos \phi_{\text{s}} /\phi_{\text{s}} .$$

(6.186)

Using Eqs. (6.184), (6.185) and (6.186), we obtain

$$K_{\text{P}} = - \frac{{{\text{d}}q_{z} }}{{{\text{d}}w}} = - 2p\left( {\frac{{{\text{d}}b_{e} }}{{{\text{d}}w}}} \right) = - 2p\left( {\frac{{{\text{d}}b_{e} /{\text{d}}\phi_{\text{s}} }}{{{\text{d}}w/{\text{d}}\phi_{\text{s}} }}} \right) = p\frac{{\cos \phi_{\text{s}} + \phi_{\text{s}} \sin \phi_{\text{s}} }}{{\sin \phi_{\text{s}} - \phi_{\text{s}} \cos \phi_{\text{s}} }}.$$

(6.187)

Note 6.2

Equation (6.19)

When the circumferential displacement at the tread is v_D in Fig. 6.45, the shear strain at the sidewall is γ = v_D/l, where l is the length of the sidewall. The reaction force per unit length in the circumferential direction at the tread is F = Gγh = Gv_Dh/l. Hence, the structural component of the circumferential fundamental spring per unit length in the circumferential direction due to the sidewall rubber is given by

$$F/v_{\text{D}} = Gh/l.$$

(6.188)

Referring to Fig. 6.45, the tensile component of the circumferential fundamental spring is given by N_ϕ(c)γ, where N_ϕ(c) and γ are, respectively, given by Eqs. (6.122) and (6.126). N_ϕ(c)γ is expressed as

$$N_{\phi } (c)\gamma = \frac{{p\left( {r_{\text{D}}^{2} - r_{\text{c}}^{2} } \right)}}{{2r\sin \phi_{\text{D}} }}\left( {\frac{v}{r} - \frac{{{\text{d}}v}}{{{\text{d}}r}}} \right)\cos \phi .$$

(6.189)

Neglecting the term dv/dr, Eq. (6.189) can be rewritten for r = r_D as

$$N_{\phi } (c)\gamma = \frac{{p\left( {r_{\text{D}}^{2} - r_{\text{c}}^{2} } \right)v_{\text{D}} }}{{2r_{\text{D}}^{2} \tan \phi_{\text{D}} }}.$$

(6.190)

The tensile spring component is given by

$$\frac{{N_{\phi } (c)\gamma }}{{v_{\text{D}} }} = \frac{p}{{2\tan \phi_{\text{D}} }} - \left( {\frac{{r_{\text{c}} }}{{r_{\text{D}} }}} \right)^{2} \frac{p}{{2\tan \phi_{\text{D}} }}.$$

(6.191)

ϕ_D in Fig. 6.45 corresponds to ϕ_s in Fig. 6.9. If (r_C/r_D)² is neglected, adding Eqs. (6.188) and (6.191) and considering two sidewalls, we obtain

$$k_{\text{t}} = 2Gh/l + p/\tan \phi_{\text{s}} .$$

(6.192)

Note 6.3

Equation (6.108)

When the neutral plane is located at the middle of the bead filler with thickness h, the flexural rigidity D_ϕ is equal to ${{E_{\text{m}} h^{3} } \mathord{\left/ {\vphantom {{E_{\text{m}} h^{3} } {\left\{ {12\left( {1 - \nu_{\text{m}}^{2} } \right)} \right\}}}} \right. \kern-0pt} {\left\{ {12\left( {1 - \nu_{\text{m}}^{2} } \right)} \right\}}}$. Meanwhile, when the neutral plane is located on the carcass with thickness h, the flexural rigidity D_ϕ is equal to ${{E_{\text{m}} h^{3} } \mathord{\left/ {\vphantom {{E_{\text{m}} h^{3} } {\left\{ {3\left( {1 - \nu_{\text{m}}^{2} } \right)} \right\}}}} \right. \kern-0pt} {\left\{ {3\left( {1 - \nu_{\text{m}}^{2} } \right)} \right\}}}$.