(i)
In the case that
l
h >
l/ 2,
$$\begin{array}{*{20}l} {S_{y} = 0\quad 0 \le x < l_{h} } \hfill \\ {S_{y} = x\tan \alpha - S_{h} \quad l_{h} \le x < x_{h} } \hfill \\ {S_{y} = x\tan \alpha - \mu q_{z} (x)/C_{y}^{\text{down}} \quad x_{h} \le x < l}. \hfill \\ \end{array}$$
(14.190)
In the case that
l
h >
l/ 2,
$$\begin{array}{*{20}l} {S_{y} = 0\quad 0 \le x < l_{h} } \hfill \\ {S_{y} = x\tan \alpha - S_{h} \quad l_{h} \le x < x_{h} } \hfill \\ {S_{y} = x\tan \alpha - \mu q_{z} (x)/C_{y}^{\text{down}} \quad x_{h} \le x < l}. \hfill \\ \end{array}$$
(14.190)
The force equilibrium at the sliding point gives
$$C_{y}^{\text{up}} \tan \alpha \cdot l_{h} = \mu q_{z} (l_{h} ) = 4\mu p_{m} \frac{{l_{h} }}{l}(1 - \frac{{l_{h} }}{l}).$$
(14.191)
We introduce a new parameter:
$$\zeta_{y} = C_{y}^{\text{up}} l\tan \alpha /(4\mu p_{m} ).$$
(14.192)
l
h is expressed by
$$\begin{array}{*{20}l} {l_{h} = l\left( {1 - \zeta_{y} } \right)\quad 0 \le \zeta_{y} \le 1} \hfill \\ {l_{h} = 0\quad 1 < \zeta_{y} }. \hfill \\ \end{array}$$
(14.193)
x
h is expressed by
$$S_{h} = \tan \alpha \cdot l_{h} = \mu q_{z} (l_{h} )/C_{y}^{\text{up}} = \mu q_{z} (x_{h} )/C_{y}^{\text{down}} .$$
(14.194)
It follows that
$$q_{z} (x_{h} ) = C_{y}^{\text{down}} /C_{y}^{\text{up}} \cdot q_{z} (l_{h} ) = \rho q_{z} (l_{h} ),$$
(14.195)
where
$$\rho = C_{y}^{\text{down}} /C_{y}^{\text{up}} \le 1.$$
(14.196)
Wear energy
\(E_{y}^{\text{w}}\) is given by
$$\begin{aligned} E_{y}^{\text{w}} & = \int\limits_{0}^{2\pi r} {\mu q_{z} (x){\text{d}}y} = \int\limits_{{l_{h} }}^{l} {\mu q_{z} (x)\frac{{{\text{d}}y}}{{{\text{d}}x}}{\text{d}}x} = \int\limits_{{l_{h} }}^{{x_{h} }} {\mu q_{z} (x)\tan \alpha {\text{d}}x} \\ & \quad + \int\limits_{{x_{h} }}^{l} {\mu q_{z} (x)\left( {\tan \alpha - \frac{{\mu q_{z}^{\prime } (x)}}{{C_{y}^{\text{down}} }}} \right){\text{d}}x} \\ & = \mu \tan \alpha \int\limits_{{l_{h} }}^{l} {q_{z} (x){\text{d}}x} - \frac{1}{2}\frac{{\mu^{2} }}{{C_{y}^{\text{down}} }}\int\limits_{{x_{h} }}^{l} {\frac{{{\text{d}}q_{z}^{2} (x)}}{{{\text{d}}x}}{\text{d}}x} \\ & = \mu \tan \alpha \int\limits_{{l_{h} }}^{l} {q_{z} (x){\text{d}}x} + \frac{1}{2}\frac{{\mu^{2} }}{{C_{y}^{\text{down}} }}q_{z}^{2} (x_{h} ). \\ \end{aligned}$$
(14.197)
The substitution of Eq. (
14.195 ) into Eq. (
14.197 ) yields
$$\begin{array}{*{20}l} {E_{y}^{\text{w}} = \mu \tan \alpha \int\limits_{{l_{h} }}^{l} {q_{z} (x){\text{d}}x} + \frac{1}{2}\frac{{\mu^{2} }}{{C_{y}^{\text{up}} }}q_{z}^{2} (l_{h} ) - \frac{1}{2}\frac{{\mu^{2} }}{{C_{y}^{\text{up}} }}q_{z}^{2} (l_{h} ) + \frac{1}{2}\rho \frac{{\mu^{2} }}{{C_{y}^{\text{up}} }}q_{z}^{2} (l_{h} )} \hfill \\ { = \left\{ {\mu \tan \alpha \int\limits_{{l_{h} }}^{l} {q_{z} (x){\text{d}}x} + \frac{1}{2}\frac{{\mu^{2} }}{{C_{y}^{\text{up}} }}q_{z}^{2} (l_{h} )} \right\} - \left\{ {\frac{1}{2}\frac{{\mu^{2} }}{{C_{y}^{\text{up}} }}q_{z}^{2} (l_{h} )(1 - \rho )} \right\}}. \hfill \\ \end{array}$$
(14.198)
The first term of Eq. (
14.198 ) is the wear energy without hysteresis loss while the second term is the wear energy due to hysteresis loss. Because (1 −
ρ ) ≥ 0 is satisfied, the wear energy is reduced by hysteresis loss.
(ii)
In the case that
l
h ≤
l/ 2,
$$\begin{array}{*{20}l} {S_{y} = 0\quad 0 \le x < l_{h} } \hfill \\ {S_{y} = x\tan \alpha - \mu q_{z} (x)/C_{y}^{\text{up}} \quad l_{h} \le x < l/2} \hfill \\ {S_{y} = x\tan \alpha - S_{l/2} \quad l/2 \le x < x_{h} } \hfill \\ {S_{y} = x\tan \alpha - \mu q_{z} (x)/C_{y}^{\text{down}} \quad x_{h} \le x < l}, \hfill \\ \end{array}$$
(14.199)
where
l
h is given by Eq. (
14.191 ) and
x
h is given by
$$S_{l/2} = \mu p_{m} /C_{y}^{\text{up}} = \mu q_{z} (l/2)/C_{y}^{\text{up}} = \mu q_{z} (x_{h} )/C_{y}^{\text{down}} .$$
(14.200)
In the case that
l
h ≤
l/ 2,
$$\begin{array}{*{20}l} {S_{y} = 0\quad 0 \le x < l_{h} } \hfill \\ {S_{y} = x\tan \alpha - \mu q_{z} (x)/C_{y}^{\text{up}} \quad l_{h} \le x < l/2} \hfill \\ {S_{y} = x\tan \alpha - S_{l/2} \quad l/2 \le x < x_{h} } \hfill \\ {S_{y} = x\tan \alpha - \mu q_{z} (x)/C_{y}^{\text{down}} \quad x_{h} \le x < l}, \hfill \\ \end{array}$$
(14.199)
where
l
h is given by Eq. (
14.191 ) and
x
h is given by
$$S_{l/2} = \mu p_{m} /C_{y}^{\text{up}} = \mu q_{z} (l/2)/C_{y}^{\text{up}} = \mu q_{z} (x_{h} )/C_{y}^{\text{down}} .$$
(14.200)
It follows that
$$q_{z} (x_{h} ) = C_{y}^{\text{down}} /C_{y}^{\text{up}} \cdot q_{z} (l/2) = \rho q_{z} (l/2).$$
(14.201)
The wear energy
\(E_{y}^{\text{w}}\) is given by
$$\begin{aligned} E_{y}^{\text{w}} & = \int\limits_{{l_{h} }}^{l} {\mu q_{z} (x)\frac{{{\text{d}}y}}{{{\text{d}}x}}{\text{d}}x} = \int\limits_{{l_{h} }}^{l/2} {\mu q_{z} (x))\left( {\tan \alpha - \frac{{\mu q_{z}^{\prime } (x)}}{{C_{y}^{\text{up}} }}} \right){\text{d}}x} + \int\limits_{l/2}^{{x_{h} }} {\mu q_{z} (x)\tan \alpha {\text{d}}x} \\ & \quad + \int\limits_{{x_{h} }}^{l} {\mu q_{z} (x)\left( {\tan \alpha - \frac{{\mu q_{z}^{\prime } (x)}}{{C_{y}^{\text{down}} }}} \right){\text{d}}x} = \mu \tan \alpha \int\limits_{{l_{h} }}^{l} {q_{z} (x){\text{d}}x} - \frac{1}{2}\frac{{\mu^{2} }}{{C_{y}^{\text{up}} }}\int\limits_{{l_{h} }}^{l/2} {\frac{{{\text{d}}q_{z}^{2} (x)}}{{{\text{d}}x}}{\text{d}}x} \\ & \quad - \frac{1}{2}\frac{{\mu^{2} }}{{C_{y}^{\text{down}} }}\int\limits_{{x_{h} }}^{l} {\frac{{{\text{d}}q_{z}^{2} (x)}}{{{\text{d}}x}}{\text{d}}x} \\ & = \mu \tan \alpha \int\limits_{{l_{h} }}^{l} {q_{z} (x){\text{d}}x} - \frac{1}{2}\frac{{\mu^{2} }}{{C_{y}^{\text{up}} }}\left\{ {q_{z}^{2} (l/2) - q_{z}^{2} (l_{h} )} \right\} + \frac{1}{2}\frac{{\mu^{2} }}{{C_{y}^{\text{down}} }}\left\{ {q_{z}^{2} (x_{h} )} \right\} \\ & = \mu \tan \alpha \int\limits_{{l_{h} }}^{l} {q_{z} (x){\text{d}}x} + \frac{1}{2}\frac{{\mu^{2} q_{z}^{2} (l_{h} )}}{{C_{y}^{\text{up}} }} - \frac{1}{2}\frac{{\mu^{2} }}{{C_{y}^{\text{up}} }}q_{z}^{2} (l/2) + \frac{1}{2}\frac{{\mu^{2} }}{{C_{y}^{\text{down}} }}\left\{ {q_{z}^{2} (x_{h} )} \right\}. \\ \end{aligned}$$
(14.202)
The substitution of Eq. (
14.201 ) into Eq. (
14.202 ) yields
$$E_{y}^{\text{w}} = \left\{ {\mu \tan \alpha \int\limits_{{l_{h} }}^{l} {q_{z} (x){\text{d}}x} + \frac{1}{2}\frac{{\mu^{2} }}{{C_{y}^{\text{up}} }}q_{z}^{2} (l_{h} )} \right\} - \left\{ {\frac{1}{2}\frac{{\mu^{2} }}{{C_{y}^{\text{up}} }}q_{z}^{2} (l/2)(1 - \rho )} \right\}.$$
(14.203)
The first term of Eq. (
14.203 ) is the wear energy without hysteresis loss, and the second term is the wear energy due to hysteresis loss. Because (1 −
ρ ) ≥ 0 is satisfied, the wear energy is reduced by hysteresis loss.
Equation
(
14.45 )
Referring to Fig.
14.92 ,
F
y is given as
$$F_{y} = G\gamma S = G \cdot y/H \cdot ab = \left( {G/H \cdot ab} \right)y,$$
where
G and
γ are the shear modulus and shear strain of rubber and
y is shear displacement. Considering that
C
y is the shear spring rate per unit area of the tread rubber, we obtain
C
y =
G/H .
C
y cannot be expressed by the shear modulus of rubber when the tread pattern is considered. The block rigidity discussed in Sect.
7.1 is therefore used.
Fig. 14.92
Shear deformation of the tread pattern
×
Equation
(
14.46 )
From Eq. (
11.53 ), we obtain
$$C_{F\alpha } = \frac{{bl^{2} C_{y} }}{{2\left( {1 + \frac{{b\lambda^{3} l^{3} }}{{12k_{s} }}C_{y} } \right)}} = \frac{1}{{\frac{2}{{bl^{2} C_{y} }} + \frac{{b\lambda^{3} l^{3} }}{{6bl^{2} C_{y} k_{s} }}C_{y} }} = \frac{1}{{\frac{2H}{{bl^{2} G_{y} }} + \frac{{\lambda^{3} l}}{{6k_{s} }}}}.$$
Substituting the relation
\({\lambda} = \root 4 \of {{{k_{s} /\left( {4EI_{z} } \right)}}}\) into the above equation, we obtain
$$C_{F\alpha } = \frac{1}{{\frac{2H}{{bl^{2} G_{y} }} + \root 4 \of {{{\frac{{k_{s}^{3} }}{{4^{3} \left( {EI_{z} } \right)^{3} }}}}\frac{l}{{6k_{s} }}}}} = \frac{1}{{\frac{2H}{{bl^{2} G_{y} }} + \frac{l}{{12\sqrt 2 \root 4 \of {{{\left( {EI_{z} } \right)^{3} k_{s} }}}}}}}.$$
Referring to Fig.
14.93 , we have
d
2 =
PL
3 /(48
EI ) from formulae of the mechanics of materials. Using the relation
τ =
Gγ and noting that the in-plane bending problem of contact patch is approximated by the bending problem of cantilever beam with length
L/2 and load
P/2 , we obtain
$$\frac{P/2}{A} = G\frac{{d_{1} }}{L/2} \to d_{1} = \frac{PL}{4AG}.$$
Fig. 14.93
Bending and shear deformations of the tread block
×
The reason for the value of 2 in the third term is that when the variable of integration is changed from
L to
α , the integral is taken twice in the integration region from
α
0 −
α
m to
α
0 +
α
m .
$$\begin{aligned} \frac{1}{\pi }\int\limits_{0}^{{\alpha_{0} + \alpha_{m} }} {\frac{{\alpha^{2} }}{{\sqrt {\alpha_{m}^{2} - \left( {\alpha - \alpha_{0} } \right)^{2} } }}{\text{d}}\alpha } & = \frac{3}{2\pi }\alpha_{m} \alpha_{0} \sqrt {1 - \left( {\frac{{\alpha_{0} }}{{\alpha_{m} }}} \right)^{2} } + \frac{1}{2}\left( {\frac{{\alpha_{m}^{2} }}{2} + \alpha_{0}^{2} } \right) \\ & \quad + \frac{1}{\pi }\left( {\frac{{\alpha_{m}^{2} }}{2} + \alpha_{0}^{2} } \right)\sin^{ - 1} \left( {\frac{{\alpha_{0} }}{{\alpha_{m} }}} \right) \\ \frac{1}{\pi }\int\limits_{{\alpha_{0} - \alpha_{m} }}^{0} {\frac{{\alpha^{2} }}{{\sqrt {\alpha_{m}^{2} - \left( {\alpha - \alpha_{0} } \right)^{2} } }}{\text{d}}\alpha } & = \frac{1}{\pi }\int\limits_{0}^{{ - \alpha_{0} + \alpha_{m} }} {\frac{{\alpha^{2} }}{{\sqrt {\alpha_{m}^{2} - \left( {\alpha + \alpha_{0} } \right)^{2} } }}{\text{d}}\alpha } \\ \end{aligned}$$
Using the above relation, 〈
E
w 〉 is obtained as
$$\begin{array}{*{20}l} { \left\langle E^{\text{w}} \right\rangle = \frac{1}{\pi }\int\limits_{{\alpha_{0} - \alpha_{m} }}^{0} {a^{ - } \frac{{\alpha^{2} }}{{\sqrt {\alpha_{m}^{2} - \left( {\alpha - \alpha_{0} } \right)^{2} } }}{\text{d}}\alpha } + \frac{1}{\pi }\int\limits_{0}^{{\alpha_{0} + \alpha_{m} }} {a^{ + } \frac{{\alpha^{2} }}{{\sqrt {\alpha_{m}^{2} - \left( {\alpha - \alpha_{0} } \right)^{2} } }}{\text{d}}\alpha } } \hfill \\ {\quad = \left( {\frac{{\alpha_{m}^{2} }}{2} + \alpha_{0}^{2} } \right)a_{0} + \frac{2}{\pi }\left\{ {\left( {\frac{{\alpha_{m}^{2} }}{2} + \alpha_{0}^{2} } \right)\sin^{ - 1} \left( {\frac{{\alpha_{0} }}{{\alpha_{m} }}} \right) + \frac{3}{2}\alpha_{m} \alpha_{0} \sqrt {1 - \left( {\frac{{\alpha_{0} }}{{\alpha_{m} }}} \right)^{2} } } \right\}b \cdot y} . \hfill \\ \end{array}$$
When lateral forces on left and right sides have similar amplitudes,
\(\alpha_{m} \gg \alpha_{0}\) is satisfied. Applying Taylor series expansion to the above equation, we obtain
$$\left\langle E^{\text{w}} \right\rangle \cong \left( {\frac{{\alpha_{m}^{2} }}{2} + \alpha_{0}^{2} } \right)a_{0} + \frac{4}{\pi }\alpha_{m} \alpha_{0} b \cdot y.$$
$$\begin{aligned} & \int\limits_{0}^{{\alpha_{0} + \alpha_{m} }} {\frac{{\alpha^{2} }}{{\sqrt {2\pi } \sigma }}{\text{e}}^{{ - \frac{{\left( {\alpha - \alpha_{0} } \right)^{2} }}{{2\sigma^{2} }}}} {\text{d}}\alpha } \cong \int\limits_{0}^{\infty } {\frac{{\alpha^{2} }}{{\sqrt {2\pi } \sigma }}{\text{e}}^{{ - \frac{{\left( {\alpha - \alpha_{0} } \right)^{2} }}{{2\sigma^{2} }}}} {\text{d}}\alpha = \frac{1}{{\sqrt {2\pi } \sigma }}\int\limits_{{ - \alpha_{0} }}^{\infty } {x^{2} {\text{e}}^{{ - \frac{{x^{2} }}{{2\sigma^{2} }}}} {\text{d}}x} } \\ & + \frac{{2\alpha_{0} }}{{\sqrt {2\pi } \sigma }}\int\limits_{{ - \alpha_{0} }}^{\infty } {x{\text{e}}^{{ - \frac{{x^{2} }}{{2\sigma^{2} }}}} {\text{d}}x} + \frac{{\alpha_{0}^{2} }}{{\sqrt {2\pi } \sigma }}\int\limits_{{ - \alpha_{0} }}^{\infty } {\frac{{x^{2} }}{{\sqrt {2\pi } \sigma }}{\text{e}}^{{ - \frac{{x^{2} }}{{2\sigma^{2} }}}} {\text{d}}x} \\ & = - \frac{{\sigma \alpha_{0} }}{{\sqrt {2\pi } }}{\text{e}}^{{ - \frac{{\alpha_{0}^{2} }}{{2\sigma^{2} }}}} + \frac{{\sigma^{2} }}{2}erfc\left( { - \frac{{\alpha_{0} }}{\sqrt 2 \sigma }} \right) + \frac{{2\sigma \alpha_{0} }}{{\sqrt {2\pi } }}{\text{e}}^{{ - \frac{{\alpha_{0}^{2} }}{{2\sigma^{2} }}}} + \frac{{\alpha_{0}^{2} }}{2}erfc\left( { - \frac{{\alpha_{0} }}{\sqrt 2 \sigma }} \right) \\ & = \frac{{\sigma \alpha_{0} }}{{\sqrt {2\pi } }}{\text{e}}^{{ - \frac{{\alpha_{0}^{2} }}{{2\sigma^{2} }}}} + \frac{{\sigma^{2} + \alpha_{0}^{2} }}{2}erfc\left( { - \frac{{\alpha_{0} }}{\sqrt 2 \sigma }} \right), \\ \end{aligned}$$
where
$$\begin{array}{*{20}l} {erfc\left( x \right) = 1 - erf(x)} \hfill \\ {erf(x) = \frac{2}{\sqrt \pi }\int\limits_{0}^{x} {{\text{e}}^{{ - t^{2} }} } {\text{d}}t = \frac{2}{\sqrt \pi }\left( {x - \frac{{x^{3} }}{3} + \cdots } \right)} \hfill \\ {erf(x) = - erf( - x)} \hfill \\ \end{array} ,$$
$$\begin{aligned} & \int\limits_{{\alpha_{0} - \alpha_{m} }}^{0} {\frac{{\alpha^{2} }}{{\sqrt {2\pi } \sigma }}{\text{e}}^{{ - \frac{{\left( {\alpha - \alpha_{0} } \right)^{2} }}{{2\sigma^{2} }}}} {\text{d}}\alpha \cong } \int\limits_{ - \infty }^{0} {\frac{{\alpha^{2} }}{{\sqrt {2\pi } \sigma }}{\text{e}}^{{ - \frac{{\left( {\alpha - \alpha_{0} } \right)^{2} }}{{2\sigma^{2} }}}} {\text{d}}\alpha } \\ & = \frac{1}{{\sqrt {2\pi } \sigma }}\int\limits_{{\alpha_{0} }}^{\infty } {x^{2} {\text{e}}^{{ - \frac{{x^{2} }}{{2\sigma^{2} }}}} {\text{d}}x} - \frac{{2\alpha_{0} }}{{\sqrt {2\pi } \sigma }}\int\limits_{{\alpha_{0} }}^{\infty } {x{\text{e}}^{{ - \frac{{x^{2} }}{{2\sigma^{2} }}}} {\text{d}}x} + \frac{{\alpha_{0}^{2} }}{{\sqrt {2\pi } \sigma }}\int\limits_{{\alpha_{0} }}^{\infty } {\frac{{x^{2} }}{{\sqrt {2\pi } \sigma }}{\text{e}}^{{ - \frac{{x^{2} }}{{2\sigma^{2} }}}} {\text{d}}x} \\ & = - \frac{{\sigma \alpha_{0} }}{{\sqrt {2\pi } }}{\text{e}}^{{ - \frac{{\alpha_{0}^{2} }}{{2\sigma^{2} }}}} + \frac{{\sigma^{2} + \alpha_{0}^{2} }}{2}erfc\left( {\frac{{\alpha_{0} }}{\sqrt 2 \sigma }} \right), \\ \end{aligned}$$
$$\begin{aligned} \left\langle {E^{\text{w}} } \right\rangle & = \frac{1}{{L_{0} }}\int\limits_{{\alpha_{0} - \alpha_{m} }}^{{\alpha_{0} + \alpha_{m} }} {a(y)\frac{{\alpha^{2} }}{{\sqrt {2\pi } \sigma }}{\text{e}}^{{ - \frac{{\left( {\alpha - \alpha_{0} } \right)^{2} }}{{2\sigma^{2} }}}} {\text{d}}\alpha } \\ & = \frac{{\sigma^{2} + \alpha_{0}^{2} }}{2}\left\{ {erfc\left( {\frac{{\alpha_{0} }}{\sqrt 2 \sigma }} \right) + erfc\left( { - \frac{{\alpha_{0} }}{\sqrt 2 \sigma }} \right)} \right\}a_{0} \\ & \quad + \left[ {\frac{{2\sigma \alpha_{0} }}{{\sqrt {2\pi } }}{\text{e}}^{{ - \frac{{\alpha_{0}^{2} }}{{2\sigma^{2} }}}} + \frac{{\sigma^{2} + \alpha_{0}^{2} }}{2}\left\{ { - erfc\left( {\frac{{\alpha_{0} }}{\sqrt 2 \sigma }} \right) + erfc\left( { - \frac{{\alpha_{0} }}{\sqrt 2 \sigma }} \right)} \right\}} \right]b \cdot y \\ \end{aligned}$$
Using the relation
$$\begin{aligned} erfc\left( {\frac{{\alpha_{0} }}{\sqrt 2 \sigma }} \right) + erfc\left( { - \frac{{\alpha_{0} }}{\sqrt 2 \sigma }} \right) & = 2 - erf\left( {\frac{{\alpha_{0} }}{\sqrt 2 \sigma }} \right) - erf\left( { - \frac{{\alpha_{0} }}{\sqrt 2 \sigma }} \right) \\ & = 2 - erf\left( {\frac{{\alpha_{0} }}{\sqrt 2 \sigma }} \right) + erf\left( {\frac{{\alpha_{0} }}{\sqrt 2 \sigma }} \right) = 2 \\ \left\{ { - erfc\left( {\frac{{\alpha_{0} }}{\sqrt 2 \sigma }} \right) + erfc\left( { - \frac{{\alpha_{0} }}{\sqrt 2 \sigma }} \right)} \right\} & = erf\left( {\frac{{\alpha_{0} }}{\sqrt 2 \sigma }} \right) - erf\left( { - \frac{{\alpha_{0} }}{\sqrt 2 \sigma }} \right) \\ & = 2erf\left( {\frac{{\alpha_{0} }}{\sqrt 2 \sigma }} \right) \\ \end{aligned}$$
the above equation for 〈
E
w 〉 can be simplified as
$$\left\langle E^{\text{w}} \right\rangle = \left( {\sigma^{2} + \alpha_{0}^{2} } \right)a_{0} + \left[ {\frac{{\sqrt 2 \sigma \alpha_{0} }}{\sqrt \pi }{\text{e}}^{{ - \frac{{\alpha_{0}^{2} }}{{2\sigma^{2} }}}} + \left( {\sigma^{2} + \alpha_{0}^{2} } \right)erf\left( {\frac{{\alpha_{0} }}{\sqrt 2 \sigma }} \right)} \right]b \cdot y.$$
Assuming
\(\sigma \gg \alpha_{0}\) and applying Taylor series expansion to the second term of the above equation, we obtain
$$\frac{{\sqrt 2 \sigma \alpha_{0} }}{\sqrt \pi }{\text{e}}^{{ - \frac{{\alpha_{0}^{2} }}{{2\sigma^{2} }}}} + \left( {\sigma^{2} + \alpha_{0}^{2} } \right)erf\left( {\frac{{\alpha_{0} }}{\sqrt 2 \sigma }} \right) = \frac{{\sqrt 2 \sigma \alpha_{0} }}{\sqrt \pi } + \left( {\sigma^{2} + \alpha_{0}^{2} } \right)\frac{2}{\sqrt \pi }\frac{{\alpha_{0} }}{\sqrt 2 \sigma } \cong 2\sqrt {\frac{2}{\pi }} \sigma \alpha_{0} .$$
(i)
The case that 0 ≤
l
h l :
The case that 0 ≤
l
h ≤
l :
In braking (
s > 0),
$$\begin{aligned} E^{{{\text{w}}({\text{braking}})}} & = \int\limits_{{l_{h} }}^{l} {\mu_{d} q_{z} (x)\frac{{{\text{d}}y}}{{{\text{d}}x}}{\text{d}}x} \\ & = \int\limits_{{l_{h} }}^{l} {\mu_{d} q_{z} (x)\left( {\frac{{\sqrt {C_{x}^{2} s^{2} \cos^{2} \alpha + C_{y}^{2} \sin^{2} \alpha } }}{C} - \frac{{\mu_{d} q_{z}^{\prime } (x)}}{C}} \right){\text{d}}x} \\ & = \mu_{d} \frac{{\sqrt {C_{x}^{2} s^{2} \cos^{2} \alpha + C_{y}^{2} \sin^{2} \alpha } }}{C}\int\limits_{{l_{h} }}^{l} {q_{z} (x)dx} - \frac{1}{2}\frac{{\mu_{d}^{2} }}{C}\int\limits_{{l_{h} }}^{l} {\frac{{{\text{d}}q_{z}^{2} (x)}}{{{\text{d}}x}}{\text{d}}x} \\ & = \mu_{d} \frac{{\sqrt {C_{x}^{2} s^{2} \cos^{2} \alpha + C_{y}^{2} \sin^{2} \alpha } }}{C}\int\limits_{{l_{h} }}^{l} {q_{z} dx} - \left. {\frac{1}{2}\frac{{\mu_{d}^{2} }}{C}q_{z}^{2} (x)} \right|_{{l_{h} }}^{l} \\ & = \mu_{d} \frac{{\sqrt {C_{x}^{2} s^{2} \cos^{2} \alpha + C_{y}^{2} \sin^{2} \alpha } }}{C}\int\limits_{{l_{h} }}^{l} {q_{z} dx} + \frac{1}{2}\frac{{\mu_{d}^{2} }}{C}q_{z}^{2} (l_{h} ) \\ & = \frac{{2\mu_{d} p_{m} l}}{3C}\sqrt {C_{x}^{2} s^{2} \cos^{2} \alpha + C_{y}^{2} \sin^{2} \alpha } \left\{ {1 - 3\left( {\frac{{l_{h} }}{l}} \right)^{2} + 2\left( {\frac{{l_{h} }}{l}} \right)^{3} } \right\} \\ & \quad + \frac{{8\mu_{d}^{2} p_{m}^{2} }}{C}\left( {\frac{{l_{h} }}{l}} \right)^{2} \left( {1 - \frac{{l_{h} }}{l}} \right)^{2}. \\ \end{aligned}$$
In driving (
s < 0),
$$\begin{aligned} E^{{\text{w}}({\text{driving}})} & = \frac{{2\mu_{d} p_{m} l}}{3C}\sqrt {C_{x}^{2} s^{2} + C_{y}^{2} (1 + s)^{2} \tan^{2} \alpha } \left\{ {1 - 3\left( {\frac{{l_{h} }}{l}} \right)^{2} + 2\left( {\frac{{l_{h} }}{l}} \right)^{3} } \right\} \\ & \quad + \frac{{8\mu_{d}^{2} p_{m}^{2} }}{C}\left( {\frac{{l_{h} }}{l}} \right)^{2} \left( {1 - \frac{{l_{h} }}{l}} \right)^{2} . \\ \end{aligned}$$
For simplicity, assuming
C
x =
C
y =
C and considering the relation
p
m = 3
F
z / (2
lb ), we obtain
$$\begin{array}{*{20}l} \begin{aligned} E^{{{\text{w}}({\text{braking}})}} & = \frac{{\mu_{d} F_{z} }}{b}\sqrt {s^{2} \cos^{2} \alpha + \sin^{2} \alpha } \left\{ {1 - 3\left( {\frac{{l_{h} }}{l}} \right)^{2} + 2\left( {\frac{{l_{h} }}{l}} \right)^{3} } \right\} \\ & \quad + \frac{{18\mu_{d}^{2} F_{z}^{2} }}{{Cl^{2} b^{2} }}\left( {\frac{{l_{h} }}{l}} \right)^{2} \left( {1 - \frac{{l_{h} }}{l}} \right)^{2} \\ \end{aligned} \hfill \\ \begin{aligned} E^{{{\text{w}}({\text{driving}})}} & = \frac{{\mu_{d} F_{z} }}{b}\sqrt {s^{2} + (1 + s)^{2} \tan^{2} \alpha } \left\{ {1 - 3\left( {\frac{{l_{h} }}{l}} \right)^{2} + 2\left( {\frac{{l_{h} }}{l}} \right)^{3} } \right\} \\ & \quad + \frac{{18\mu_{d}^{2} F_{z}^{2} }}{{Cl^{2} b^{2} }}\left( {\frac{{l_{h} }}{l}} \right)^{2} \left( {1 - \frac{{l_{h} }}{l}} \right)^{2} \\ \end{aligned} \hfill \\ \end{array}$$
(ii)
In the case that
l
h
In the case that
l
h < 0:
In braking (s > 0),
$$\begin{aligned} E^{{{\text{w}}({\text{braking}})}} & = \int\limits_{0}^{l} {\mu_{d} q_{z} (x)\frac{{{\text{d}}S_{y} }}{{{\text{d}}x}}{\text{d}}x} = \frac{{2\mu_{d} p_{m} l}}{3C}\sqrt {C_{x}^{2} s^{2} \cos^{2} \alpha + C_{y}^{2} \sin^{2} \alpha } \\ & = \frac{{\mu_{d} F_{z} }}{bC}\sqrt {C_{x}^{2} s^{2} \cos^{2} \alpha + C_{y}^{2} \sin^{2} \alpha } . \\ \end{aligned}$$
In driving (
s < 0),
$$\begin{aligned} E^{{{\text{w}}({\text{driving}})}} & = \frac{{2\mu_{d} p_{m} l}}{3C}\sqrt {C_{x}^{2} s^{2} + C_{y}^{2} (1 + s)^{2} \tan^{2} \alpha } \\ & = \frac{{\mu_{d} F_{z} }}{bC}\sqrt {C_{x}^{2} s^{2} + C_{y}^{2} (1 + s)^{2} \tan^{2} \alpha } . \\ \end{aligned}$$
Assuming the relation
C
x =
C
y =
C , we obtain
$$\begin{array}{*{20}l} {E^{{{\text{w}}({\text{braking}})}} = \frac{{\mu_{d} F_{z} }}{b}\sqrt {s^{2} \cos^{2} \alpha + \sin^{2} \alpha } } \hfill \\ {E^{{{\text{w}}({\text{driving}})}} = \frac{{\mu_{d} F_{z} }}{b}\sqrt {s^{2} + (1 + s)^{2} \tan^{2} \alpha } } \hfill \\ \end{array} .$$
The substitution of Eq. (
14.85 ) into Eq. (
14.89 ) yields
$$\eta \dot{y} + k_{\text{tread}}\, y = \mu \left( {AVt + B} \right).$$
The solution to the above differential equation is
$$y = \xi {\text{e}}^{{ - \frac{{k_{\text{tread}} }}{\eta }t}} + \frac{\mu }{{k_{\text{tread}} }}\left\{ {AV\left( {t - \frac{\eta }{{k_{\text{tread}} }}} \right) + B} \right\}.$$
Considering initial conditions, we obtain
$$\begin{aligned} & V_{y} t_{c} = \xi {\text{e}}^{{ - \frac{{k_{\text{tread}} }}{\eta }t_{c} }} + \frac{\mu }{{k_{\text{tread}} }}\left\{ {AV\left( {t_{c} - \frac{\eta }{{k_{\text{tread}} }}} \right) + B} \right\}, \\ & \xi = - \left[ {\frac{\mu }{{k_{\text{tread}} }}\left\{ {AV\left( {t_{c} - \frac{\eta }{{k_{\text{tread}} }}} \right) + B} \right\} - V_{y} t_{c} } \right]{\text{e}}^{{\frac{{k_{\text{tread}} }}{\eta }t_{c} }} . \\ \end{aligned}$$
Using Eqs. (
14.85 ) and (
14.86 ), at
t =
t
c , Eq. (
14.89 ) can be rewritten as
$$\eta V\sin \alpha + k_{\text{tread}} Vt_{c} \sin \alpha = \mu \left( {AVt_{c} + B} \right).$$
The above equation can be rewritten as
$$\mu A - k_{\text{tread}} \sin \alpha = \left( {\eta V\sin \alpha - \mu B} \right)/(Vt_{c} ).$$
The substitution of the above equation into equation for
ξ yields
$$\xi = - \left[ {\frac{\mu }{{k_{\text{tread}} }}\left( { - \frac{\eta }{{k_{\text{tread}} }}AV + B} \right) + \frac{1}{{k_{\text{tread}} }}\left( { - \mu B + \eta V\sin \alpha } \right)} \right]{\text{e}}^{{\frac{{k_{\text{tread}} }}{\eta }t_{c} }} = \frac{\eta V}{{k_{\text{tread}} }}\left( {\frac{\mu A}{{k_{\text{tread}} }} - \sin \alpha } \right){\text{e}}^{{\frac{{k_{\text{tread}} }}{\eta }t_{c} }} .$$
The substitution of
ξ and
t =
T +
t
c into equation for
y yields
$$y = \frac{\mu }{{k_{\text{tread}} }}\left\{ {AV\left( {T + t_{c} - \frac{\eta }{{k_{\text{tread}} }}} \right) + B} \right\} + \frac{\eta V}{{k_{\text{tread}} }}\left( {\frac{\mu }{{k_{\text{tread}} }}A - \sin \alpha } \right){\text{e}}^{{ - \frac{{k_{\text{tread}} }}{\eta }T}} .$$
Fujikawa et al. [
23 ] estimated the intra-shear stiffness of a block. As shown Fig.
14.94 , the shear strain
γ at longitudinal position
x and vertical position
z is expressed by
$$\gamma (x,z) = \frac{z}{h}\left\{ {\frac{{\delta_{y} (x + l_{E} ) - \delta_{y} (x)}}{{l_{E} }}} \right\},$$
where
l
E is the distance between adjacent elements. The shear force
f (
x ) acting between the tread elements is therefore expressed by
$$f(x) = bG\int\limits_{0}^{h} {\gamma (x,z){\text{d}}z} = \frac{bhG}{2}\left\{ {\frac{{\delta_{y} (x + l_{E} ) - \delta_{y} (x)}}{{l_{E} }}} \right\} = \frac{bhG}{{2l_{E} }}{\Delta }\delta_{y} .$$
Fig. 14.94
Local slip model and shear deformation of tread (reproduced from Ref. [
21 ] with the permission of
JSAE )
×
From the above equation, the intra-shear stiffness
k
2 is obtained as
k
2 =
bhG/ (2
l
E ), where
b is the width of the block. Meanwhile, the shear stiffness of a block element
\(t_{\text{tread}}^{\prime }\) is expressed as
\(t_{\text{tread}}^{\prime } = bl_{E} G/h\) . For a block with dimensions of 20 mm (length) × 20 mm (width) × 8 mm (height), considering that
h = 8 mm,
l
E = 20/3 mm and
b = 20 mm, the relation between the intra-shear stiffness and shear stiffness of a block element is
\(k_{2} = 0.72t_{\text{tread}}^{\prime }\) . Therefore, Fujikawa’s assumption of
\(k_{2} = 2t_{\text{tread}}^{\prime }\) may not be appropriate for this block.
Using Eqs. (
14.136 ), (
14.139 ) and (
14.140 ), we obtain
$$\delta = \left( {1 + K_{r} /K_{f} } \right)\beta - \left( {C_{Pf} + C_{\Pr } } \right)/K_{f} .$$
Using Eqs. (
14.138 ), (
14.139 ) and the above equation, we obtain the equation for the side slip angle of the vehicle
β .
When there is a time delay in the spring term for the equation of motion with one degree of freedom, the governing equation is
\(m\ddot{x}(t) + c\dot{x}(t) + kx(t - \tau ) = 0\) . When the time delay is small, using the relation
\(x(t - \tau ) = x(t) - \tau \dot{x}(t)\) , the above equation can be rewritten as
\(m\ddot{x}(t) + (c - k\tau )\dot{x}(t) + kx = 0\) . Unstable vibration occurs when the relation
c −
kτ < 0 is satisfied. If the time delay is not small, by applying the Laplace transform to the original equation, we obtain the characteristic equation
ms
2 +
cs +
ke
−τs = 0. Because this characteristic equation contains the term
e
−τs , the number of solutions is infinite. This is because the relation
\({\text{e}}^{ - j(\theta + 2n\pi )} = {\text{e}}^{j\theta } (j = \sqrt { - 1} )\) is satisfied for an arbitrary
n . This is a feature of a time-delay system.
Equation
(
14.160 )
From the relations
W
1 (
t ) = 0 and
W
2 (
t ) = 0, we obtain
X
0 (
t ) =
V
0 (
t ) =
U
0 (
t −
T ). Using Eq. (
14.157 ), we obtain
$$U_{0} (t) = U_{0} (t - T) + \bar{k}\beta^{n} P_{0}^{n}$$
From the relation
\(\ddot{X}(t) = 0\) and the initial condition
X
0 (0) = 0, we obtain
X
0 (
t ) =
At , where
A is a constant. From the relation
X
0 (
t ) =
U
0 (
t −
T ),
U
0 (
t ) is found to be a linear function with respect to
t . Using the above equation,
U
0 (
t ) and
U
0 (
t −
T ) are given as
$$\begin{array}{*{20}l} {U_{0} (t) = \bar{k}\beta^{n} P_{0}^{n} (t + T)/T} \hfill \\ {U_{0} (t - T) = \bar{k}\beta^{n} P_{0}^{n} t/T}. \hfill \\ \end{array}$$
Equation
(
14.162 )
From Eq. (
14.152 ), we obtain
where
$$p(t) = \eta_{T} \dot{W}_{2} (t) + K_{T} W_{2} (t).$$
The substitution of Eqs. (
14.152 ) and (
14.161 ) into Eq. (
14.157 ) yields
$$\begin{array}{*{20}l} {U_{0} (t) + u(t) = U_{0} (t - T) + u(t - T) + \bar{k}\beta^{n} \left\{ {P_{0} + p(t)} \right\}^{n} } \hfill \\ { \cong U_{0} (t - T) + u(t - T) + \bar{k}\beta^{n} P_{0}^{n} (t) + \bar{k}\beta^{n} nP_{0}^{n - 1} (t)p(t)} \hfill \\ { = U_{0} (t - T) + u(t - T) + \bar{k}\beta^{n} P_{0}^{n} (t) + \bar{k}\beta^{n} nP_{0}^{n - 1} (t)\left\{ {\eta_{T} \dot{W}_{2} (t) + K_{T} W_{2} (t)} \right\}}, \hfill \\ \end{array}$$
where
$$\begin{array}{*{20}l} {U_{0} (t) = U(t - T) + W(t) = U_{0} (t - T) + \bar{k}\beta^{n} P_{0}^{n} (t)} \hfill \\ {W_{2} (t) = X_{0} (t) + x(t) - U_{0} (t - T) - u(t - T) = x(t) - u(t - T)}. \hfill \\ \end{array}$$
From the above equation, we obtain
$$u(t) = u(t - T) + \bar{k}\beta^{n} P_{0}^{n - 1} n\left[ {\eta_{T} \left\{ {\dot{x}(t) - \dot{u}(t - T)} \right\} + K_{T} \left\{ {x(t) - u(t - T)} \right\}} \right].$$
Schallamach’s equation, given as Eq. (
14.30 ), corresponds to
n = 2.
