Design and Thermomechanical Finite Element Analysis of Frictional Contact Mechanism on Automotive Disc Brake Assembly

Abstract

The braking phenomenon is an aspect of vehicle stopping performance where with kinetic energy due to speed of 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 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 static structural analysis was 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 analysis helped us in the calculation of the values of the heat transfer coefficients (h) that have been exploited in 3D transient evolution of the brake disc temperatures. Two different rotor designs and 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 von Mises stress of the disc, as well as the contact pressure of the brake pads. Following the analysis of the results obtained, we drew several conclusions from this investigation. The choice allowed us to deliver the rotor design excellence to ensure and guarantee the good braking performance of the vehicles.

Appendices

Appendix 1

Calculating Heat Flux Entering the Disc

The Forces Acting on the Wheels During Braking

By observing the situation described in Fig. 40, the longitudinal and transverse equilibrium of the vehicle can be written along the local axes x, y of the car.

$$\sum {F_{X} = 0}$$

$$F_{\text{RRV}} + F_{\text{FV}} + F_{\text{RRH}} + F_{\text{FH}} + F_{\text{RA}} - (F_{\text{RP}} + F_{\text{D}} ) = 0$$

(28)

$$F_{\text{FV}} + F_{\text{FH}} = F_{\text{RP}} + F_{\text{D}} - F_{\text{RRV}} - F_{\text{RA}} - F_{\text{RRH}}$$

(29)

$$F_{\text{F}} = F_{\text{RP}} + F_{\text{RF}} - F_{\text{RR}} - F_{\text{RA}}$$

(30)

with $F_{\text{F}} = F_{\text{FV}} + F_{\text{FH}}$, $F_{\text{RR}} = F_{\text{RRV}} + F_{\text{RRH}}$

$$\sum {F_{Y} = 0} \Rightarrow {\kern 1pt} F_{\text{G}} \cos \alpha - (F_{\text{QV}} + F_{\text{QH}} ) = 0$$

(31)

$$F_{\text{QH}} = F_{\text{G}} \cos \alpha - F_{\text{QV}}$$

(32)

$$\sum {M_{\text{B}} = 0} \Rightarrow {\kern 1pt} F_{\text{QV}} L + F_{\text{RA}} h - h(F_{\text{RF}} + F_{\text{RP}} ) - hF_{\text{G}} \sin \alpha$$

(33)

$$F_{\text{QV}} = \frac{{\left[ {(F_{\text{RF}} + F_{\text{RP}} )h + F_{\text{G}} L_{\text{H}} - F_{\text{RA}} h} \right]}}{L}$$

(34)

Fig. 40

Definition of the forces acting on an automobile during braking

On a road vehicle, the rolling force F_RR = F_Gf_r cos α is due to the flat formed by a tire on the road f_r is the rolling resistance coefficient. For a high pressure tire (f_r = 0.015).

$$F_{\text{RP}} = F_{\text{G}} \sin \alpha$$

(35)

The aerodynamic force is given by:

$$F_{\text{RA}} = C_{X} A_{\text{F}} \frac{{\rho_{\text{a}} }}{2}v^{2}$$

(36)

With C_X is the coefficient of form, equal to: 0.3 to 0.4 on the car, A_F (m²) is the frontal surface; on the approach, for a road passenger vehicle, we can take: A_F = 0.8 × height × width S and ρ_a is the air density.

Total Braking Power

$$P_{\text{tot}} = P_{\text{R}} + P_{\text{F}}$$

(37)

By

$$P_{\text{F}} = \sum {F_{\text{F}} v = (F_{\text{FV}} + F_{\text{FH}} )v}$$

(38)

$$P_{\text{R}} = \sum {F_{\text{R}} v = (F_{\text{RR}} + F_{\text{RP}} + F_{\text{RA}} )v}$$

(39)

In the case of flat braking (Fig. 41), the resistances due to rolling and the slope are neglected (F_RR = 0 and F_RP = 0), the penetration into the air is generally negligible, for this reason (F_RA = 0)

$$P_{\text{R}} = \sum {F_{\text{R}} v = (F_{\text{RR}} + F_{\text{RP}} + F_{\text{RA}} )v = 0}$$

(40)

$$P_{\text{F}} = \sum {F_{\text{F}} v = (F_{\text{FV}} + F_{\text{FH}} )v}$$

(41)

$$(F_{\text{FV}} + F_{\text{FH}} ) = F_{\text{D}} = ma$$

(42)

$$P_{\text{tot}} = P_{\text{F}} = mav$$

(43)

If we define, Let ϕ the factor of the ratio of the braking power with respect to the rear wheels P_FH = ϕmav then, P_FV = (1 − ϕ)mav.

Fig. 41

External forces acting on the vehicle during braking on a flat road

If a is constant, we have:

$$v(t) = v_{0} - at$$

(44)

$$P_{\text{F}} = (1 - \phi )ma(v_{0} - at)$$

(45)

The braking power delivered to the brake disc is equal to half the total power:

$$P_{\text{FVI}} = \frac{(1 - \phi )}{2}ma(v_{0} - at)$$

(46)

At time t = 0, we have

$$P_{\text{FVI}} = \frac{(1 - \phi )}{2}mav_{0}$$

(47)

The braking efficiency is then defined by the ratio between the deceleration (a) and the acceleration (g):

$$Z = {{a_{\text{d}} } \mathord{\left/ {\vphantom {{a_{\text{d}} } g}} \right. \kern-0pt} g}$$

(48)

$$P_{\text{FVI}} = \frac{(1 - \phi )}{2}mZgv_{0}$$

(49)

Assuming that the amount of heat generated by friction is completely absorbed by the disc.

$$Q_{\text{v}} = \frac{(1 - \phi )}{2}m_{\text{tot}} \,g\,v\quad [{\text{Nm/s}}] = [{\text{W}}]$$

(50)

The expression of the transformed friction power per unit area is thus:

$$Q_{v}^{'} = \frac{(1 - \phi )}{2}\frac{{m_{\text{tot}} \,g\,v}}{{2A_{\text{d}} }}\quad [{\text{N}}\,{\text{m}}/{\text{s}}\,{\text{m}}^{2} ] = [{\text{W}}/{\text{m}}^{2} ]$$

(51)

The quantity $Q_{v}^{\prime }$ indicates the heat flow absorbed by the disc, which must be housed only on the actual contact surface. Where A_d is the surface of the rotor to which a brake pad pivots.

By definition, the operating factor ε_p of the friction surface is given by the following formula:

$$\varepsilon_{p} = {{Q_{v}^{'} } \mathord{\left/ {\vphantom {{Q_{v}^{'} } {Q_{{v_{\hbox{max} } }}^{'} }}} \right. \kern-0pt} {Q_{{v_{\hbox{max} } }}^{'} }}$$

(52)

Thus, the equation of the initial thermal flow of friction entering the disc, which is calculated as follows:

$$Q_{{v_{\hbox{max} } }}^{'} = \frac{(1 - \phi )}{2}\frac{{m_{\text{tot}} gv}}{{2A_{d} }}\quad [{\text{N}}\,{\text{m/s}}\,{\text{m}}^{2} ]] = [{\text{W/m}}^{2} ]$$

(53)

Appendix 2

Analysis of Disc Rotor Force

A free body diagram of a front wheel-rotor system, Fig. 42, 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

$$E_{\text{dissipated}} = \frac{1}{2}kmv_{0}^{2} + S_{b} mg\sin \alpha$$

(54)

But $S_{b} = \frac{{v_{0}^{2} }}{2a}$

$$E_{\text{dissipated}} = \frac{1}{2}kmv_{0}^{2} + \frac{{v_{0}^{2} }}{2a}mg\sin \alpha$$

(55)

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

$$E_{\text{dissipated}} = \int {P_{\text{dissipated}} t\,{\text{d}}t} = \int {(2F_{\text{rotor}} )v_{\text{rotor}} (t){\text{d}}t}$$

(56)

$${\kern 1pt} \frac{1}{2}kmv_{0}^{2} + \frac{{v_{0}^{2} }}{2a}mg\sin \alpha = KE_{\text{dissipated}} = \int {P_{\text{dissipated}} t\,{\text{d}}t} = 2\int {F_{\text{rotor}} v_{\text{rotor}} (t){\text{d}}t}$$

(57)

But from kinematic relationships.

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

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

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

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

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

$$\begin{aligned} & {\kern 1pt} \frac{ 1}{ 2}kmv_{0}^{2} + \frac{{v_{0}^{2} }}{2a}mg\sin \alpha = 2F_{\text{rotor}} \int\limits_{0}^{{t_{\text{stop}} }} {v_{\text{rotor}} (t)\,{\text{d}}t} \\ & \quad = 2{\kern 1pt} F_{\text{rotor}} \frac{{R_{\text{rotor}} }}{{R_{\text{tire}} }}\left( {v_{0} t_{\text{stop}} - \frac{1}{2}\left\{ {\frac{{v_{0} }}{{t_{\text{stop}} }}} \right\}t_{\text{stop}}^{2} } \right) \\ \end{aligned}$$

(58)

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

(59)

When braking on a straight/flat track (α = 0), k is estimated to be about 0.30. Therefore, Eq 59 should be modified to be:

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

(60)

Fig. 42

Free body diagram of a front wheel-rotor system