Control of the nonlinear dynamics of a truck and trailer combination

Abstract

The directional instability problem of a truck and trailer combination in a forward motion has been investigated by developing a dynamic model of the planar rigid-body dynamics of the system. Two control strategies have been devised based on the reference model following controller configuration. The main objective of the controllers is to ensure that the nonlinear model follows the desired response generated by a linearized version of the truck and trailer model. The first RFMC uses an integral-plus-state feedback controller for its compensator, which was designed based on the eigenstructure assignment methodology. The second RFMC relies on a sliding mode compensator. The control inputs were applied through the actuation of differential wheel torques. Different driving scenarios and road conditions, such as braking and accelerating maneuver with a flat tire on a dry road, and a double-lane change maneuver on a slippery road, were considered. Overall, the simulation results demonstrated the superiority of the sliding mode controller over the integral-plus-state feedback controller in ensuring better tracking of the truck and trailer combination to a target path under different driving maneuvers.

Appendix

1.1 Appendix A: Truck and trailer lateral tire forces

The formulation used in determining the lateral tire force of the truck is [30, 31]

$$\begin{aligned} F_{y} =F_{yV}^{\max } \sin [b_{{}V} \tanh \left( {c{}_{V} \alpha } \right) ] \end{aligned}$$

(A-1)

where

$$\begin{aligned} F_{yV}^{\max }= & {} 4359.62f_{z} -112.786f_{z}^{2} -14.3349f_{z}^{3}\\&\quad +\,1.27521f_{z}^{4} -5.59463\times 10^{-2}f_{z}^{5} \\ b_{V}= & {} 2.57769-9.42715\times 10^{-3}f_{z} \\&\quad -\,3.77976\times 10^{-4}f_{z}^{2} -1.68722\times 10^{-4}f_{z}^{3} \\&+\,2.29386\times 10^{-5}f_{z}^{4} -1.02129\times 10^{-6}f_{z}^{5} \\ c_{V}= & {} 0.10517+5.03038\times 10^{-2}f_{z} \\&\quad -\,4.44333\times 10^{-2}f_{z}^{2} +1.16291\times 10^{-2}f_{z}^{3} \\&\quad -\,2.73167\times 10^{-3}f_{z}^{4} +1.7083\times 10^{-4}f_{z}^{5} \\ f_{z}= & {} F_{z} /F_{z}^{\mathrm{nom}} \end{aligned}$$

Similarly, the formulation used in determining the lateral tire force of the trailer is:

$$\begin{aligned} F_{y} =F_{yT}^{\max } \sin \left( {b_\mathrm{T} \tanh \left( {c_\mathrm{T} \alpha } \right) } \right) \end{aligned}$$

(A-2)

where

$$\begin{aligned}&F_{yT}^{\max } =4471.74f_{z} -81.061f_{z}^{2} -8.45943f_{z}^{3}\\&\quad +\,0.62684f_{z}^{4} -2.32964\times 10^{-2}f_{z}^{5} \\&b_\mathrm{T} =2.69816-9.03815\times 10^{-3}f_{z} \\&\quad +\,8.11047\times 10^{-4}f_{z}^{2} -3.68044\times 10^{-4}f_{z}^{3} \\&\quad +\,4.61300\times 10^{-5}f_{z}^{4} -2.12143\times 10^{-6}f_{z}^{5} \\&c_\mathrm{T} = 0.15247-3.92322\times 10^{-2}f_{z} \\&\quad +\,1.97822\times 10^{-2}f_{z}^{2} -4.4911\times 10^{-3}f_{z}^{3} \\&+\,4.59281\times 10^{-4}f_{z}^{4} -1.7325\times 10^{-5}f_{z}^{5} \\&f_{z} =F_{z} /F_{z}^\mathrm{nom} \end{aligned}$$

Note that $\alpha$ in both Eqs. (A-1) and (A-2) is expressed in degrees.

1.2 Appendix B: Linearized differential equation of motion

$$\begin{aligned}&A_{1} =\left[ {{\begin{array}{ccc} {a_{11} } &{}\quad {a_{12} } &{}\quad {a_{13} } \\ {a_{21} } &{}\quad {a_{22} } &{}\quad {a_{23} } \\ {a_{31} } &{}\quad {a_{32} } &{}\quad {a_{33} } \\ \end{array} }} \right] \nonumber \\&\quad \quad =\left[ {{\begin{array}{ccc} 1 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad {J_{zz} +h(c+h)\mu } &{}\quad {-ch\mu } \\ 0 &{}\quad {J_{zz} -I_{zz} +\left( {h^{2}-c^{2}} \right) \mu } &{}\quad {I_{zz} +c\left( {c-h} \right) \mu } \\ \end{array} }} \right] \nonumber \\ \end{aligned}$$

(B-1)

$$\begin{aligned}&B_{1} =\left[ {{\begin{array}{ccc} 0 &{}\quad 0 &{}\quad {-1} \\ {b_{21} } &{}\quad {b_{22} } &{}\quad {b_{23} } \\ {b_{31} } &{}\quad {b_{32} } &{}\quad {b_{33} } \\ \end{array} }} \right] \end{aligned}$$

(B-2)

where

$$\begin{aligned}&\mu =\frac{Mm}{m+M} \nonumber \\&b_{21} =-h\left( {\frac{\mu }{m}C_{\alpha _\mathrm{T} } +F_{x_\mathrm{T} } } \right) \nonumber \\&b_{22} =\frac{a^{2}C_{\alpha _\mathrm{f} } +b^{2}C_{\alpha _\mathrm{r} } +h(c+d+h)C_{\alpha _\mathrm{T} } }{V_{x} }\nonumber \\&\quad +\left( {\frac{aC_{\alpha _\mathrm{f} } -bC_{\alpha _\mathrm{r} } -(c+d+h)C_{\alpha _\mathrm{T} } }{V_{x} }} \right) \frac{\mu h}{M} \nonumber \\&b_{23} =-\frac{h(c+d)C_{\alpha _\mathrm{T} } }{V_{x} }\frac{\mu }{m} \nonumber \\&b_{31} =C_{\alpha _\mathrm{T} } \left( {c+d-h} \right) \nonumber \\&\quad -[C_{\alpha _\mathrm{T} } \left( {c-h} \right) +cF_{x_\mathrm{f} } ]\frac{\mu }{M}-hF_{x_\mathrm{T} } \nonumber \\&b_{32} =\frac{a^{2}C_{\alpha _\mathrm{f} } +b^{2}C_{\alpha _\mathrm{r} } +[h^{2}-\left( {c+d} \right) ^{2}]C_{\alpha _\mathrm{T} } }{V_{x} }\nonumber \\&\quad +\frac{aC_{\alpha _\mathrm{f} } -bC_{\alpha _\mathrm{r} } -\left( {c+d+h} \right) C_{\alpha _\mathrm{T} } }{V_{x} }\frac{\left( {h-c} \right) \mu }{M} \nonumber \\&b_{33} =\left( {\left( {h-c} \right) \frac{\mu }{M}+\left( {c+d-h} \right) } \right) \frac{(c+d)C_{\alpha _\mathrm{T} } }{V_{x} } \nonumber \\&C_{1} =\left[ {{\begin{array}{cc} 0 &{}\quad 0 \\ 1 &{}\quad 0 \\ 1 &{}\quad {-1} \\ \end{array} }} \right] \end{aligned}$$

(B-3)

$$\begin{aligned}&F_{1} =\left[ {{\begin{array}{c} 0 \\ {f_{21} } \\ {f_{31} } \\ \end{array} }} \right] \end{aligned}$$

(B-4)

where

$$\begin{aligned} f_{21}= & {} -\left( {aC_{\alpha _\mathrm{f} } +aF_{x_\mathrm{f} } +(C_{\alpha _\mathrm{f} } +F_{x_\mathrm{f} } )\frac{h\mu }{M}} \right) \delta _\mathrm{f} \\&+\left( aC_{\alpha _\mathrm{f} } -bC_{\alpha _\mathrm{r} } -hC_{\alpha _\mathrm{T} }\right. \\&\left. +(C_{\alpha _\mathrm{f} } +C_{\alpha _\mathrm{r} } +C_{\alpha _\mathrm{T} } )\frac{h\mu }{M} \right) \frac{V_{y} }{V_{x} } \\ f_{31}= & {} \left( {C_{\alpha _\mathrm{f} } \frac{\left( {c-h} \right) \mu }{M}-a} \right) (C_{\alpha _\mathrm{f} } +F_{x_\mathrm{f} } )\delta _\mathrm{f} \\&+\left( aC_{\alpha _\mathrm{f} } -bC_{\alpha _\mathrm{r} } +C_{\alpha _\mathrm{T} } \left( {c+d-h} \right) \right. \\&\left. -\left( {C_{\alpha _\mathrm{f} } +C_{\alpha _\mathrm{r} } +C_{\alpha _\mathrm{T} } } \right) \frac{\left( {c-h} \right) \mu }{M} \right) \frac{V_{y} }{V_{x} } \end{aligned}$$