A predictive tool to evaluate braking system performance using a fully coupled thermo-mechanical finite element model

Abstract

The braking phenomenon is an aspect of vehicle stopping performance where with kinetic energy due to speed of the vehicle is transformed to thermal energy via the friction between the brake disc and its pads. The heat must then be dissipated into the surrounding structure and into airflow around the brake system. The frictional thermal field during the braking phase between the disc and the brake pads can lead to excessive temperatures. In our work, we presented a numerical modeling using ANSYS software adapted in the finite element method, to follow the evolution of the global temperatures for the two types of brake discs, full and ventilated disc during braking scenario. Also, numerical simulation of the transient thermal and the static structural analysis were performed here sequentially, with coupled thermo-structural method. Numerical procedure of calculation relies on important steps such that the Computational Fluid Dynamics (CFD) and thermal analysis have been well illustrated in 3D, showing the effects of heat distribution over the brake disc. This CFD approach helped in the calculation of the values of the thermal coefficients (h) that have been exploited in the 3D transient evolution of the brake disc temperatures. Three different brake disc materials were tested and comparative analysis of the results was conducted in order, to derive the one with the best thermal behavior. Finally, the resolution of the coupled thermomechanical model allows us to visualize other important results of this research such as; the deformations, and the equivalent stresses of Von Mises of the disc, as well as the contact pressure of the brake pads. Following our analysis and results we draw from it, we derive several conclusions. The choice allowed us to deliver the rotor design excellence to ensure and guarantee the good braking performance of the vehicles.

Appendix A

1.1 A.1. Analysis of disc rotor force

A free body diagram of a front wheel-rotor system, Fig. 43 , is used to drive the equation of equilibrium. Since large amount of the braking load is born by the front brakes, that amount of kinetic energy and potential energy into a single disc is given by

$$ E_{dissipated} = {\kern 1pt} \frac{ 1}{ 2}{\kern 1pt} k{\kern 1pt} m{\kern 1pt} v_{0}^{2} + S_{b} {\kern 1pt} m{\kern 1pt} {\kern 1pt} g\,{\kern 1pt} \sin \,\alpha $$

(A.1)

But $S_{b} {\kern 1pt} = \frac{{v_{0}^{2} }}{2\,a}$

$$ E_{dissipated} = {\kern 1pt} \frac{ 1}{ 2}{\kern 1pt} k{\kern 1pt} m{\kern 1pt} v_{0}^{2} + \frac{{v_{0}^{2} }}{2\,a}{\kern 1pt} m{\kern 1pt} {\kern 1pt} g\,{\kern 1pt} \sin \,\alpha $$

(A.2)

Fig. 43

Free body diagram of a front wheel-rotor system

The power dissipated by each rotor face is equal to the heat flux into the rotor face.

$$ E_{dissipated} = {\kern 1pt} \int {P_{dissipated} \,\,t\;d{\kern 1pt} t} \;\, = \,\int {(2\,F_{rotor} )\,v_{rotor} (t){\kern 1pt} \,dt} $$

(A.3)

$$ {\kern 1pt} \frac{ 1}{ 2}{\kern 1pt} k{\kern 1pt} m{\kern 1pt} v_{0}^{2} + \frac{{v_{0}^{2} }}{2\,a}{\kern 1pt} m{\kern 1pt} {\kern 1pt} g\,{\kern 1pt} \sin \,\alpha = K\,E_{dissipated} = $$

$$ \int {P_{dissipated} \,\,t\;d{\kern 1pt} t} = 2\,\int {F_{rotor} \,v_{rotor} (t){\kern 1pt} \,dt} $$

(A.4)

But from kinematic relationships.

$$ v_{vehicle} (t){\kern 1pt} \, = v_{0} - a{\kern 1pt} {\kern 1pt} t $$

$$ a = \frac{{v_{0} }}{{t_{stop} }} $$

$$ \frac{{v_{vehicle} (t)}}{{R_{tire} }}\, = \omega {\kern 1pt} \,(t) = \frac{{v_{rotor} (t)}}{{R_{rotor} }} $$

$$ v_{rotor} (t) = \frac{{R_{rotor} }}{{R_{tire} }}\left( {v_{0} - \left\{ {\frac{{v_{0} }}{{t_{stop} }}} \right\}\,t} \right) $$

$F_{rotor} {\kern 1pt}$ is constant with respect to time, and $v_{rotor}$ varies only linearly with time so the energy balance equation becomes:

$$ {\kern 1pt} \frac{ 1}{ 2}{\kern 1pt} k{\kern 1pt} m{\kern 1pt} v_{0}^{2} + \frac{{v_{0}^{2} }}{2\,a}{\kern 1pt} m{\kern 1pt} {\kern 1pt} g\,{\kern 1pt} \sin \,\alpha = 2\,F_{rotor} \,\int\limits_{0}^{{t_{stop} }} {v_{rotor} (t){\kern 1pt} \,dt} $$

$$ = 2{\kern 1pt} {\kern 1pt} F_{rotor} {\kern 1pt} \frac{{R_{rotor} }}{{R_{tire} }}\left( {v_{0} t_{stop} - \frac{1}{2}\left\{ {\frac{{v_{0} }}{{t_{stop} }}} \right\}\,t_{stop}^{2} } \right) $$

(A.5)

$$ F_{rotor} = \frac{{{\kern 1pt} \frac{ 1}{ 2}{\kern 1pt} k{\kern 1pt} m{\kern 1pt} v_{0}^{2} + \frac{{v_{0}^{2} }}{2\,a}{\kern 1pt} m{\kern 1pt} {\kern 1pt} g\,{\kern 1pt} \sin \,\alpha }}{{2{\kern 1pt} \frac{{R_{rotor} }}{{R_{tire} }}{\kern 1pt} \left( {v_{0} {\kern 1pt} t_{stop} - \frac{1}{2}{\kern 1pt} \left\{ {\frac{{v_{0} }}{{t_{stop} }}} \right\}{\kern 1pt} t_{stop}^{2} } \right)}} $$

(A.6)

When braking on a straight/flat track (α = 0), k is estimated to be about 0.30. Therefore, equation (A.6) should be modified to be:

$$ F_{disc} = \frac{{(30{\text{\% )}}{\kern 1pt} \frac{ 1}{ 2}{\kern 1pt} m{\kern 1pt} v_{0}^{2} }}{{2{\kern 1pt} \frac{{R_{rotor} }}{{R_{tire} }}{\kern 1pt} \left( {v_{0} {\kern 1pt} t_{stop} - \frac{1}{2}{\kern 1pt} \left\{ {\frac{{v_{0} }}{{t_{stop} }}} \right\}{\kern 1pt} t_{stop}^{2} } \right)}} $$

(A.7)