Finite Element Analysis of a Disc Brake Mounted on the Axle of a Train

Abstract

Axle-mounted disc brakes are generally employed in high-speed trains owing to the lesser weight and sustainable thermal properties. Most widely used material for manufacturing of the disc brake is grey cast iron (GCI), because of its availability. However, it being a brittle material has certain limitations. Hence, an aluminium composite material (AlSiC) is considered for analysis under different loading conditions. Three models of brake discs are developed on which both structural analysis and coupled analysis were done and results are compared.

Appendix

Structural load calculations: In this case, a railway vehicle travelling at a speed of 160 kmph on a horizontal track stops due to application of emergency brake was considered. Time of travel before stopping, deceleration, weight of the vehicle, clamping force on the brake disc, brake pad area, coefficient of friction, etc., for calculating the loads are taken from the railway specification. Value of Mass of railway vehicle—M = 64000 kg, No. of axles per vehicle = 4, Maximum load per axle = 16000 kg, no. of brake discs per axle = 2, Load on each wheel = 8000 kg, Start speed v₀ = 44.4 m/s, Deceleration a = 1.2 m/s², Braking time t_a = 36 s, Effective radius of the brake disc r_disc = 0.247 m, Radius of the wheel r_wheel = 0.458 m, Mean coefficient of friction brake pad µ = 0.35, Clamping force F_c = 42.1 kN, Surface area of brake pads A_c = 400 cm², Maximum temperature under sun = 70 ℃, Maximum temperature under shade = 45 ℃, Factor of Safety = 1.5 (Fig. 82.10).

Fig. 82.10

Representation of forces acting on wheel and disc brake

• Stopping distance

$$S = v_{o} t_{s} - \frac{1}{2}at_{s}^{2} = 822.24m$$

(82.1)

• Determination of pressure on disc

Pressure acting on the brake disc,

$$P = \frac{{F_{c} }}{{A_{c} \times \mu }} = \frac{42.1 \times 1000}{{800 \times (10)^{ - 4} \times 0.35}} = 1.5\,{\text{Mpa on each side}}$$

(82.2)

where

$F_{c}$ :

Clamping force (i.e. 42 kN)

A _c :

Contact area of brake pad on each side (i.e. 400 cm²)

μ :

Coefficient of friction (i.e. 0.35).

• Angular velocity

$$\omega = \frac{{{\text{Velocity }}}}{{{\text{radius}}}} = \frac{{v_{0} }}{{r_{{{\text{wheel}}}} }} = \frac{{44.44}}{{0.458}} = 97.12\,{\text{rad}}/{\text{s}}$$

(82.3)

• Thermal load

The kinetic energy for one wheel (disc brake) is equivalent to the energy balance

$$\begin{aligned} 0.125 * \frac{1}{2} * M * v^{2} & = \mathop \smallint \limits_{0}^{{t_{s} }} P(t){\text{d}}t \\ & = 2*F_{\text{disc}} \mathop \smallint \limits_{0}^{{t_{s} }} v_{\text{disc}} (t){\text{d}}t \\ \end{aligned}$$

(82.4)

The energy change at the moment is equal to the heat flux on the surface of the disc.

Equation (82.4) is valid in the case of constant braking deceleration. The braking force on the disc is equal to Eq. (82.7)

$$F_{\text{disc}} = \frac{{0.125 * \frac{1}{2 } * M * v_{0}^{2} }}{{2 * \frac{{r_{\text{disc}} }}{{r_{\text{wheel}} }} \left( {v_{0} * t_{s} - \frac{1}{2} * a * t_{s}^{2 } } \right) }} = 8940\,{\text{N}}$$

(82.5)

The heat flux at the moment, which affects one half of the disc, is calculated according to the

$$\begin{aligned} Q(t) & = F_{\text{disc}} *v_{\text{disc}} (t) = F_{\text{disc}} * \frac{{r_{\text{disc}} }}{{r_{\text{wheel}} }}\left( {v_{0} - a*t} \right) \\ & = 8940* \frac{0.247}{0.458}\left( {44.44{-}1.2*t} \right) = 214261 {-} 5786*t\,{\text{Watts}} \\ \end{aligned}$$

(82.6)

$${\text{Area of friction surface}} = \frac{\pi }{4}\left( {0.64^{2} {-}0.35^{2} } \right)*2 = 0.45\,{\text{m}}^{2}$$

$$Q(t) = \frac{{214261 {-} 5786*{\text{t}}}}{0.45}\,{\text{W/m}}^{2}$$

(82.7)

For the case of emergency braking on horizontal track from 160 kmph to stop, the analysis was carried out in 36 steps, each step being 1 s long.