CHAPTER 6 RESULTS AND NUMERICAL SIMULATION OF VARIABLE BRAKING FORCE SYSTEM

141 CHAPTER 6 RESULTS AND NUMERICAL SIMULATION OF VARIABLE BRAKING FORCE SYSTEM 6.1 TWO-WHEELER TESTING After mounting all the rear brake system hardwares, the motor cycle is tested for unladen (191 kg) and laden (261 kg) conditions that are applied on the motor cycle. It is concluded that stopping distance of the motor cycle is reduced if the effective disc radius is changed based on the pillion load on the motor cycle. Stopping distance is measured for different vehicle speed, when the rear brake is only applied. The sudden deceleration test is conducted for various speeds of the vehicle, effective disc radius as well as loads (unladen and laden). 6.2 TEST PROCEDURE Tests are conducted in morning, midday and evening for all the load conditions. The tests are conducted ten times for various loads (unladen and laden) and various motorcycle speeds for finding the stopping distance of the motorcycle. The average value of stopping distance is calculated. The stopping distance is measured by using a tape from a point where the brake is applied to a point where the linear velocity of the motorcycle becomes zero. A rider (5 kg) alone sits on the motorcycle while testing the motorcycle for unladen condition. A person who weighs 7 kg sits on the rear seat of the motorcycle for laden condition while testing. The effective disc radius is 11 mm when the C clamp is located in the top most position in the guide. Also, each revolution of stepper motor rotation makes C clamp move 1.75 mm linearly. Hence the number of revolutions is calculated for the required

142 effective disc radius. The coefficient of friction between the tyre and the ground is calculated from the stopping distance details which are given by the manufacturer. The stopping distance of the same motor cycle is 19.13m (6 kmph kmph) as per technical specification given by the manufacturer. 2 V Stopping distance (SD) = (6.1) 2 a1 Where, a 1= Motorcycle deceleration, m/s 2 a 1 = 7.26 m/s 2 V = Motorcycle velocity, m/s a 1 g =.74 Where, = coefficient of friction between the tyre and the road surface. The manufacturer considers that the coefficient of friction between the tire and the ground is.74 and designs the braking system for the motorcycle. Hence, the coefficient friction between the tyre and the ground is assumed as.7 and variable braking force system is designed in this research work. Moreover the coefficient of friction mainly depends on the types of mating surfaces (tyre and ground). In Figure 6.1, if the motorcycle is in unladen condition (191 kg), the stopping distance is less when it is compared with loaded case (261 kg) for an effective radius of 55mm. From Figures 6.2, 6.3 and 6.4 stopping distance of the vehicle at unladen condition is more for an effective disc radius of 7mm, 84mm and 11mm respectively. It is determined that the effective disc radius is 55mm for the motorcycle with the total weight of 191 kg (Kerb weight 141kg and rider weight 5 kg) for the same vehicle geometry =.6283 and =.4678). In other words it may be concluded that the optimum disc radius of the motor cycle (Kerb weight 141kg and rider weight 5 kg) is 55mm without pillion rider. This is due to the braking force

143 developed between the tire and the ground being nearly equal to maximum frictional force between the tire and the ground for an effective disc radius of 55mm when there is no pillion rider on the motor cycle (141 kg) with a rider of 5 kg. 6 4 2 Unladen (191kg) loaded (261kg) 3 4 5 6 7 Vehicle speed in 'kmph' Figure 6.1 Stopping distance (Vs) vehicle speed for an effective radius of 55 mm 6 4 2 Unladen (191kg) Loaded (261kg)) 3 4 5 6 7 Vehicle Speed in 'kmph' Figure 6.2 Stopping distance (Vs) vehicle speed for an effective radius of 7 mm

144 If the effective disc radius is increased for the motorcycle under unladen (191 kg) condition with a rider of 5 kg, the braking force developed between the tire and the ground is more than the maximum frictional force between the tire and the ground for an effective disc radius of 7, 84 and 11 mm. Hence the wheel gets locked (Premature lock in the following order - 11mm, 84mm,7mm i.e., For 11mm, wheel lock very earlier.) that results in an increase in the vehicle stopping distance. If the disc radius is increased the stopping distance increases as given in Table 6.1 for same vehicle speed and load for unladen case. The braking force (621.72N) developed for the effective radius of 11mm very fast (Braking system takes less time to develop 621.74 N of force) which leads earlier wheel lock-up, compared with 7mm, 84mm, 11mm lock-up. In Figure 6.3, if the motor cycle is in loaded condition (261kg), the stopping distance is less when it is compared with unladen (191kg) condition for an effective radius of 84mm. Stopping distance is more for an effective radius of 55mm, 7mm and 11mm if the motor cycle is in loaded condition. 6 4 Unladen (191kg) Loaded (261kg) 2 3 4 5 6 7 Vehicle Speed in 'kmph' Figure 6.3 Stopping distance (Vs) vehicle speed for an effective radius of 84 mm

145 It is determined that the effective disc radius is 84mm for the motor cycle with the total weight of 261kg (Kerb weight 141kg, rider weight 5kg and pillion load 7kg) for the same vehicle geometry ( =.7145and =.4678). In other words, it may be concluded that the optimum disc radius of the motor cycle (Kerb weight 141kg, rider weight 5kg and pillion load7kg) is 85mm with a 7 kg pillion rider. This is due to the braking force developed between the tire and the ground being approximately equal to the maximum frictional force between the tyre and the ground for an effective disc radius of 85mm when there is pillion rider (7kg) on the motor cycle (141kg) with a rider of 5kg. If the effective disc radius is increased, the braking force developed between the tire and the ground may be more than the maximum frictional force between tire and ground for an effective disc radius of 11mm when there is a pillion rider (7kg) on the motor cycle (141kg) with a rider of 5kg. Hence the wheel gets locked that results in an increase in the vehicle stopping distance. Model calculations are given in Appendix 9. Relationship between tracive (or braking) force Vs wheel slip is discussed in Appendix 1. 6 4 2 Unladen (191kg) Loaded (261kg) 3 4 5 6 7 Vehicle Speed in 'kmph' Figure 6.4 Stopping distance (Vs) vehicle speed for an effective radius of 11 mm

146 If the effective disc radius is decreased, the braking force developed between the tyre and the ground may be less than the maximum frictional force between the tire and the ground for an effective disc radius of 55 and 7 mm when there is a pillion rider (7 kg) on the motor cycle (141kg) with a rider of 5 kg. In other words, the maximum traction capacity between the tyre-road system is not utilized. The stopping distance for various vehicle speeds is tabulated in Table 6.2 for loaded case. In Figure 6.4, the stopping distance is more for unladen (191kg) and loaded (261kg) vehicle for an effective radius of 11mm. This is due to braking force being more than maximum frictional force developed between the tyre and the ground by brake hardware system. The stopping distance (Vs) vehicle speed is plotted and given in Figures 6.5 and 6.6 for various effective radii when the vehicle is in unladen and loaded condition respectively. Table 6.1 Stopping distance at unladen condition (191kg) Sl. No. Motor Cycle Speed (kmph) Effective Disc Radius (mm) Stopping Distance (m) 1 4 55 23.32 2 4 7 25.1 3 4 84 27.45 4 4 11 31. 5 5 55 36.45 6 5 7 39.1 7 5 84 4. 8 5 11 41. 9 6 55 52.5 1 6 7 54.3 11 6 84 56.2 12 6 11 6.

149 Table 6.2 Stopping distance at laden condition (261 kg) Sl No Motor Cycle Speed (kmph) Effective Disc Radius (mm) Stopping Distance (m) 1 4 55 24.5 2 4 7 23.5 3 4 84 22.66 4 4 11 26.6 5 5 55 4. 6 5 7 38.2 7 5 84 35.4 8 5 11 4.3 9 6 55 56.2 1 6 7 53.1 11 6 84 51. 12 6 11 57. 6.3 NUMERICAL SIMULATION OF VARIABLE BRAKING FORCE SYSTEM 6.3.1 Matlab Programe using M-File A computer program is written using MATLAB (M-file) for finding dynamics rear axle and front axle forces (Normal reaction) with respect to vehicle acceleration in g units. A plot which is given in Figure 6.7 is obtained for rear axle and front axle forces (Vs) motorcycle deceleration. In this computer program, vehicle load and other vehicle geometry are entered as input, the magnitude of above said forces and the plots are obtained. The MATLAB program is as follows:

15 w=191; %Enter vehicle mass in kg a=12; %static rear axle load in kg b=.62218; % enter vehicle vertical CG in m Lf=.8356; % enter horizontal distance between front axle and CG in m Lr=.49439; %enter horizontal distance between rear axle and CG in m l=1.33; %enter the wheelbase in m c=a/w; d=b/l; mu=.7; i=:.2:1; fzf=(1-c+d*i)*w*9.81; fzr=(c-d*i)*w*9.81; subplot(311); plot(i,fzf,i,fzr) axis([ 1 2 16]) xlabel('deceleration in g uints'); ylabel('fzf and Fzr'); 16 14 12 1 Dynamic front axle normal force Dynamic rear axle normal force 8 6 4 2.1.2.3.4.5.6.7.8.9 1 Deceleration in 'g' uints Figure 6.7 Dynamic rear axle and front axle normal forces (191kg)

151 Another computer program is written using MATLAB (M-file) for finding dynamic braking force at front and rear wheel, developed during sudden braking. In this program, if vehicle load, vehicle geometry and braking system hardware data are entered, the magnitude of the above said forces and effective disc radius are obtained for the particular load condition. A plot which is given in Figure 6.8 is obtained between dynamic rear and front axle braking force. A subplot, in which the coefficient of friction between the road surface and the tire is assumed as.7, is also obtained between dynamic rear axle force and front axle force. The MATLAB program is as follows: w=191; %Enter vehicle mass in kg a=12; %static rear axle load in kg b=.62218; % enter vehicle vertical CG in m Lf=.8356; % enter horizontal distance betweem front axle and CG in m Lr=.49439; %enter horizontal distance between rear axle and CG in m l=1.33; %enter the wheelbase in m PL=6.2;%Enter the hydraulic brake line pressure in N/mm^2 Po=.5;%Enter the pushout pressure in N/mm^2 Awc=84.28;%enter the value of caliper cylinder area in mm^2 c1=.98;%enter value of wheel cylinder efficiency BF=.7;%Enter the value of brake factor R=3;%Enter tyre radius in mm c=a/w; d=b/l; mu=.7; i=:.2:1; fzf=(1-c+d*i)*w*9.81; fzr=(c-d*i)*w*9.81; fxf=(i-c*i+d*(i.^2))*w*9.81; fxr=(c*i-d*(i.^2))*w*9.81;

152 subplot(311); plot(fxr,fxf) xlabel('dynamic rear axle force in N'); ylabel(' Dynamic front axle force in N'); fxf=fzf*mu; fxr=fzr*mu; subplot(312); plot(fxr,fxf) axis([2 5 16]) xlabel('dynamic rear axle force in N (mu=.7)'); ylabel(' Dynamic front axle force in N(mu=.7)'); d1=c*mu; d2=1+d*mu; decele=d1/d2; bt=(pl-po)*awc*c1*bf; WT=(decele*9.81)*w*R; r=wt/bt; disp(r) 2 15 1 5 5 1 15 2 25 3 35 4 Dynamic rear axle force in N 1 2 25 3 35 4 45 5 Dynamic rear axle force in N (mu=.7) Figure 6.8 Dynamic rear and front axle braking force

153 6.3.2 Simulation for Finding Vehicle deceleration, Velocity and Stopping distance Vs time In this simulation, normal reaction on wheel, vehicle deceleration, and effective disc radius are calculated using simulink model as shown in Figure 6.9. The Longitudinal Vehicle Dynamics block models a two-axle vehicle, with two equally sized wheels, moving forward or backward along its longitudinal axis. The dynamic front and rear braking forces F xf, F xr are applied at the front and rear wheel contact points, as well as the incline angle beta, as a set of simulink input signals. The block computes the vehicle velocity V and the dynamic front and rear axle normal forces F zf,dyn, F zr,dyn on the vehicle as a set of Simulink output signals. But in this simulation work F xf and beta are set to zero as the rear wheel is only braked at level road. The output F zr,dyn from the longitudinal vehicle dynamics block models is multiplied with mu at the end of third second of the simulation time and given that product as input for the vehicle dynamic block. The same magnitude of F zr,dyn is taken for stopping distance calculation. For calculating stopping distance, F zr,dyn is divided by the total mass of the motorcycle, and its value is motorcycle s deceleration. If the motorcycle s deceleration is integrated twice, the derived value is stopping distance. For finding the effective disc radius the braking system hardware data like brake line pressure, push out pressure, caliper piston cross section area, wheel cylinder efficiency and brake factor and effective radius of tyre are given as input. The minimum stopping distance is calculated and tabulated in Table 6.3.The following plots which are shown in Figures 6.1 to 6.26 are obtained from the simulation for the different load conditions and vehicle geometry. Normal reaction on rear wheel and front wheel Vs simulation time Acceleration, velocity and stopping distance Vs Simulation time

154 Figure 6.9 Simulink model for finding stopping distance and effective disc radius Table 6.3 Minimum stopping distance using Simulink model (191kg) Sl. No. Motor Cycle Speed (kmph) Effective Disc Radius (mm) Stopping Distance (m) 1 4 55 18.65 2 4 7 --- 3 4 84 --- 5 5 55 29.21 6 5 7 --- 7 5 84 --- 9 6 55 42.18 1 6 7 --- 11 6 84 ---

155 Table 6.4 Minimum stopping distance using Simulink model (261 kg) Sl No Motor Cycle Speed (kmph) Effective Disc Radius (mm) Stopping Distance (m) 1 4 55 25.9 2 4 7 21.92 3 4 84 16.96 5 5 55 4.43 6 5 7 38.2 7 5 84 26.47 9 6 55 58.25 1 6 7 53.1 11 6 84 38.14 12 11 Normal reaction on rear wheel (191kg) Normal reaction on front wheel (191kg) 1 9 8 7 6 1 2 3 4 5 6 7 8 9 Time in secs Figure 6.1 Load transfer effect during braking (191kg)

156 5 4 3 2 Stopping distance (191kg and 6kmph) Velocity (191kg and 6kmph) Deceleration (191kg and 6kmph) 1-1 1 2 3 4 5 6 7 8 9 Time in secs Figure 6.11 Performance of variable braking system (191kg) 13 12 Normal reaction on rear wheel (21kg) Normal reaction on front wheel (21kg) 11 1 9 8 7 1 2 3 4 5 6 7 8 9 Time in secs Figure 6.12 Load transfer effect during braking (21kg)

157 5 4 3 2 Stopping distance (21kg and 6kmph) Velocity (21kg and 6kmph) Deceleration (21kg and 6kmph) 1-1 1 2 3 4 5 6 7 8 9 Time in secs Figure 6.13 Performance of variable braking system (21kg) 14 13 12 Normal reaction on rear wheel (211kg) Normal reaction on front wheel (211kg) 11 1 9 8 7 1 2 3 4 5 6 7 8 Time in secs Figure 6.14 Load transfer effect during braking (211kg)

158 5 4 3 2 Stopping distance (211kg and 6 kmph) Velocity (211kg and 6 kmph) Deceleration (211kg and 6 kmph) 1-1 1 2 3 4 5 6 7 8 9 Time in secs Figure 6.15 Performance of variable braking system (211kg) 15 14 13 Normal reaction on rear wheel (221kg) Normal reaction on front wheel (221kg) 12 11 1 9 8 7 1 2 3 4 5 6 7 8 Time in secs Figure 6.16 Load transfer effect during braking (221kg)

159 5 4 3 2 Stopping distance (221kg and 6kmph) Velocity (221kg and 6kmph) Deceleration (221kg and 6kmph) 1-1 1 2 3 4 5 6 7 8 Time in secs Figure 6.17 Performance of variable braking system (221kg) 16 15 Normal reaction on rear wheel (231kg) Normal reaction on front wheel (231kg) 14 13 12 11 1 9 8 7 1 2 3 4 5 6 7 8 Time in secs Figure 6.18 Load transfer effect during braking (231kg)

16 5 4 3 2 Stopping distance (231kg and 6kmph) Velocity (231kg and 6kmph) Deceleration (231kg and 6kmph) 1-1 1 2 3 4 5 6 7 8 Time in secs Figure 6.19 Performance of variable braking system (231kg) 17 16 Normal reaction on rear wheel (241kg) Normal reaction on front wheel (241kg) 15 14 13 12 11 1 9 8 7 1 2 3 4 5 6 7 8 Time in secs Figure 6.2 Load transfer effect during braking (241kg)

161 5 4 3 2 Stopping distance (241kg and 6kmph) Velocity (241 kg and 6kmph) Deceleration(241kg and 6kmph) 1-1 1 2 3 4 5 6 7 8 Time in secs Figure 6.21 Performance of variable braking system (241kg) 18 16 Normal reaction on rear wheel (251kg) Normal reaction on front wheel (251kg) 14 12 1 8 6 1 2 3 4 5 6 7 8 Time in secs Figure 6.22 Load transfer effect during braking (251kg)

162 5 4 3 2 Stopping distance (251kg and 6kmph) Velocity (251kg and 6kmph) Deceleration (251kg and 6kmph) 1-1 1 2 3 4 5 6 7 8 Time in secs Figure 6.23 Performance of variable braking system (251kg) 2 18 16 Normal reaction on rear wheel (261kg) Normal reaction on front wheel (261kg) 14 12 1 8 6 1 2 3 4 5 6 7 8 Time in secs Figure 6.24 Load transfer effect during braking (261kg)

163 5 4 3 2 Stopping distance (261kg and 6kmph) Velocity (261kg and 6kmph) Deceleration (261kg and 6kmph) 1-1 1 2 3 4 5 6 7 8 Time in secs Figure 6.25 Performance of variable braking system (261kg) 2 18 16 14 12 1 NRR191 NRF191 NRR21 NRF21 NRR211 NRF211 NRR221 NRF221 NRR231 NRF231 NRR241 NRF241 NRR251 NRF251 NRR261 NRF261 8 6 1 2 3 4 5 6 7 8 9 Time in Secs NRR-Normal reaction on rear wheel NRF-Normal reaction on front wheel Figure 6.26 Load transfer effect during braking (191kg -261kg) The vertical center of gravity for the results shown in Figure 6.26 varies from 622.18 mm (unladen-191kg) to 662.1 mm (laden-261kg) from the ground. It is also assumed that the vertical center of gravity is constant throughout the simulation period for all load conditions. Example: vertical

164 center of gravity is 622.18mm (constant) throughout the simulation period for unladen condition only. 6.3.3 Effect of Effective disc radius on Suspension due to Sudden Braking (MATLAB Model) In this simulation work, bicycle model of the motorcycle is considered. When the vehicle is suddenly parked a pitching moment is induced because of braking force that is acting at the tyre and the ground. The line of action of the braking force is separated by vehicles centre of gravity. Hence a pitching moment is developed. Mathematical modeling of motor cycle (Two Degree of Freedom Mass coupled systems) is as follows: In the spring mass systems of either single degree or two degree of freedom considered so far, either a linear translatory motion or angular motion only is allowed. Thus it is assumed that a rigid body can be taken as a point mass or inertia performing translatory or rotary motions respectively. In practice there are several examples where the translatory and rotary motions take place simultaneously. A common example is an automobile when suddenly parked, executing vertical translatory and pitching motions simultaneously. This is because the centre of gravity does not coincide with the centre of rotation. Consider a total motorcycle mass m, whose centre of gravity is G, supported by two springs K f and K r at distances L f and L r as shown in Figure 6.27. The points A and B are the front and the rear suspension mounting points respectively with sprung mass. The angled line which is passing through CG of the sprung mass, front and rear suspension mounting point, signify the position of sprung mass during vibration at any instant of time. It

165 is assumed that the body is making a pure translatory motion given by y, only when the moment of the restoring force at G is zero. i.e K f L f = K r L r. If such is not a special case, the body which is shown in Figure 6.27 experiences not only translatory motion y but also an angular motion simultaneously. Let r G be the radius of gyration of the body about the centroid G. Measuring y downwards and clockwise negative, the equilibrium diagrams of body for translatory and angular motions are as shown in Figure 6.27. Figure 6.27 Two degree of freedom Mass coupled systems Equation for only pitching moment is as follows: my C f (y - Lf ) C r(y Lr ) Kf (y - Lf ) K r(y Lr ) (6.1)

166 I C f (y - Lf )Lf C r(y Lr )Lr - K f (y - Lf )Lf K r(y Lr )Lr Mz (6.2) Equations for pitching moment and vertical bounce are as follows: m y C f (y - Lf ) Cr (y Lr ) K f (y - Lf ) K r (y Lr ) Fy (6.3) I C f (y - Lf )Lf C r(y Lr )Lr - K f (y - Lf )Lf K r(y Lr )Lr Mz (6.4) Where, K f, K r = Front and rear suspension spring constant, N/m C f, C r = Front and rear suspension damping rate, N-s/m, = Rotation (pitch) angle of mass m and its rate change, rad, rad/s respectively. y, y = Vertical displacement of mass m and its rate of change, m, m/s respectively. I = Mass moment of inertia about lateral axis of the motorcycle passing through CG, kg-m 2 y = Vertical acceleration of mass m in m/s2 F y = Vertical force due to road curb in N M z = Pitch moment induced due to vehicle deceleration in N-m = Angular acceleration of mass m in rad/s 2 F y (Kf K r)* R h (6.5) Where, R h = Road curb in m

167 Here y and are coupled in equation through the stiffness terms and damping and the coupling is thus said to be static coupling. K f L f = K r L r and as it is discussed above the motion can be purely translational or rotational. The above Equations (6.1) to (6.5) are solved for obtaining vehicle responses i.e. the displacement (y) and the rotational angle ( ) of mass m are calculated for a pitching moment induced due to sudden parking of the motorcycle and graphs are plotted for the same responses using the softwares MATLAB and ANSYS. 6.3.3.1 Simulink model (MATLAB) (Two Degree of Freedom Mass coupled systems of motorcycle) Figure 6.28 illustrates the modeled characteristics of the bicycle model of the motorcycle for the variable braking force system. The front and the rear suspension are modeled as spring/damper systems. The vehicle body has pitch and bounce degrees of freedom. They are represented in the model by four states: vertical displacement, vertical velocity, pitch angular displacement, and pitch angular velocity. The front suspension influences the bounce (i.e. vertical degree of freedom) according to Equation (6.6). Ffront 2K f(lf y) 2C f(lf y) (6.6) F front, F rear = upward force on motorcycle front and rear suspension Equation (6.7) The pitch contribution to the front suspension is given by M front Lf * Ffront (Pitch moment due to front suspension) (6.7)

168 Where, M front = Pitch moment due to front suspension Frear 2K r(y Lr ) 2C r(y Lr ) (6.8) Equation (6.8) The pitch contribution to the rear suspension is given by Mrear L r * Frear (6.9) Where, M rear = Pitch moment due to rear suspension my Ffront Frear mg (6.1) I Ffront * Lf Frear *Lr Mz (6.11) Pitching moment equation for variable braking force system is as follows Mz a )W * * h (6.12) R Pitching moment equation for conventional braking system is as follows Mz r (PL P ) A c BF * h R (6.13)

169 Figure 6.28 Simulink model of variable braking force system The suspension model shown in Figure 6.28 has an input (pitch moment) due to sudden parking of the vehicle. This input appears only as a moment about the pitch axis because the longitudinal body motion is not modeled. Figure 6.29 The Spring/damper model used in front Suspension subsystems

17 Figure 6.3 The Spring/Damper model used in Rear Suspension subsystems The spring/damper subsystem that models the front and the rear suspensions is shown in Figures 6.29 and 6.3 respectively. Right click on the Front/Rear Suspension block and select "Look Under Mask" to see the front/rear suspension subsystem. The suspension subsystems are used to model Equations (6.6) to (6.11). The equations are implemented directly in the Simulink diagram through the straightforward use of Gain and Summation blocks. As the subsystem is a masked block, a different data set can be entered for each instance. Furthermore, L is thought of as the Cartesian coordinate x, being negative or positive with respect to the origin, or center of gravity. Thus, K f, C f, and -L f are used for the front suspension block whereas K r, C r, and L r are used for the rear suspension block. The simulink model for the conventional braking system is shown in Figure 6.31. The difference between variable braking system and conventional system is only pitching moment whose values are given in Equation (6.12) and Equation (6.13) respectively.

171 Figure 6.31 Simulink model for the conventional braking system Simulation results of two Degree of Freedom Mass coupled systems are as follows: The analyses are carried out for variable braking force and conventional system. In these analyses the following parameters are determined when a pitching moment and other parameters shown in the Table 6.5, are given as input for two Degree of Freedom mass coupled systems of the motorcycle for different load conditions: Normal reaction on front and rear wheel. Rotation (pitch) angle and vertical displacement of the motorcycle mass.

172 The analyses results are shown in Figures 6.32 to 6.36. Table 6.5 Assumed input values (191kg) for simulink model Sl No Input Parameter Value 1.6283 2.4678 3 L f 835.6mm 4 L r 494.39mm 5 h 622.18mm 6 m 191kg 7 I 335 kg-m 2 8 K f 6686N/m 9 K r 113N/m 1 C f 319 N-s/m 11 C r 398 N-s/m 12 Initial displacement of mass due to own weight of motorcycle mass.142 m 13 Road height.853m 12 11 NRR (191kg) NRF (19 1kg) NRF with suspension (191kg) NRR with suspension (191kg) 1 9 8 7 6 1 2 3 4 5 6 7 8 9 1 Time in secs Figure 6.32 Normal reactions on tyre with and without suspension system (191 kg)

173-5 -1 1 2 3 4 5 6 7 8 9 1 2 1 Conventional ( 191kg, r =11mm) 1 2 3 4 5 6 7 8 9 1 2 VBF ( 191kg, r = 55mm) 1 1 2 3 4 5 6 7 8 9 1.2.1 1 2 3 4 5 6 7 8 9 1-3 x 1 5-5 1 2 3 4 5 6 7 8 9 1 Time in secs Figure 6.33 Motorcycle responses due to pitching moment at CG (Variable braking force system with 191 kg) 2 18 16 NRR (261kg) NRF (261kg) NRF with suspension (261kg) NRR with suapension (261kg) 14 12 1 8 6 1 2 3 4 5 6 7 8 9 1 Time in secs Figure 6.34 Normal reactions on tyre with and without suspension system (261 kg)

174-5 -1 1 2 3 4 5 6 7 8 9 1 2 1 Conventional ( 261kg, r = 11mm 1 2 3 4 5 6 7 8 9 1 2 VBF (261kg, r = 84mm) 1 1 2 3 4 5 6 7 8 9 1.2.1.1 1 2 3 4 5 6 7 8 9 1 -.1 1 2 3 4 5 6 7 8 9 1 Time in secs Figure 6.35 Motorcycle responses due to pitching moment at CG (261kg) -5-1 1 2 3 4 5 6 7 8 9 1 2 1 r =11mm, m=191kg r = 55mm, m=191kg 1 2 3 4 5 6 7 r = 11mm, 8 m=261kg 9 1 2 r = 84mm,m=261kg 1 1 2 3 4 5 6 7 8 9 1.2.1.1 1 2 3 4 5 6 7 8 9 1 -.1 1 2 3 4 5 6 7 8 9 1 Time in secs Figure 6.36 Motorcycle responses due to pitching moment at CG (Conventional and Variable braking force system- 191kg and 261kg)

175 6.3.4 Comparison between Conventional and Variable Braking Force System using MATLAB and ANSYS The motorcycle response due to pitching moment developed during hard braking is compared between conventional and variable braking system using MATLAB and ANSYS. The effect of pitching moment on suspension is analyzed for conventional braking system using simulink model. In this analysis, it is assumed that sufficient amount of frictional force is available between the tire and the ground. The only difference between this simulation work and the previous simulation (6.2.3) is pitching moment magnitude. In conventional braking system pitching moment is equal to the product of braking force developed between the tyre and the road surface due to braking system hardware and vertical centre of gravity of the vehicle. The following formula is used to calculate the pitching moment: Mz r (PL P ) A c BF * h (6.14) R The same analyses are carried out using ANSYS. The step by step procedure for analysis the response of the motorcycle for unladen condition (191kg) is as follows: Selection of analysis type- structural Selection of element types- mass 21-option-2D (w rot inert) - combin14 - beam3 Assigning real constant value for the selected elements Mass 21 - Enter mass and mass moment of inertia of the motorcycle

176 Combin12 - Enter the stiffness and damping coefficient of front and rear suspension Beam3 - Enter the cross sectional area and area moment of inertia Selection of material properties of the beam element Modeling of suspension system Creating nodes, as shown in Figure 6.37 at the corresponding position of front and rear suspension spring and damper and mass center. Node 1 (, ) Node 2 (,.622) Node 3 (.835,.622) Node 4 (1.33,.622) Node 5 (1.33,) Figure 6.37 Creating nodes

177 Creating element as shown in Figure 6.38 using defined nodes Figure 6.38 Creating Elements Apply boundary condition as shown Figure 6.39 Figure 6.39 Apply boundary condition

178 Type of analysis Solution-analysis type-new analysis-transient-reduced-mater DOFs-Define-UY and ROT Z at node 3 as shown in Figure 6.4 Figure 6.4 Locating MDOF Apply load (Based on analysis apply pitching moment /vertical force or both at CG of the motorcycle) Solution- load step opts-output ctrls-db/results file-select Every N th substep-ok Time/frequency- Time-Time step-[deltim] time step sizeenter.1 Stepped or ramped b.c-select stepped-ok Write LS file load step file number n Enter 1 Time/frequency- Time-Time step-time at end of load step-enter 3 Write LS file load step file number n Enter 2

179 Solution-apply-loads-Force/moment- on nodes-select node number 3-select Mz and enter its value at value Time/frequency- Time-Time step-time at end of load step-enter 7 Write LS file load step file number n Enter 3 Solution-apply-loads-Force/moment- on nodes-select node number 3-select Fy and enter its value at node 3 which is shown in Figure 6.41 Figure 6.41 Applying load Write LS file load step file number n Enter 4 Solution-solve-From LS Files-starting LS File number-enter 1- Ending LS File number enter 4-ok Time history post processing-define variables-pick node 3-select nodal displacement UY and ROTZ.-Plot graph-select variable 2(UY) and 3(ROTZ).

18 The results of the analyses using MATLAB and ANSYS are given in Figures 6.42 to 6.55..2.15 VBF ( m = 191kg, r = 55mm) Conventional (m = 191kg, r = 11mm).1.5 1 2 3 4 5 6 7 8 9 1 Time in secs -3 x 1 4 3 2 VBF ( m =191kg, r =55mm) Conventional (m = 191kg, r = 11mm) 1-1 -2 1 2 3 4 5 6 7 8 9 1 Time in secs Figure 6.42 Comparison of motorcycle responses between conventional and variable braking force (VBF) system due to pitching moment at CG (191kg) Figure 6.43 Rotation (or) pitch angle (radian) of motorcycle mass due to pitching (variable braking force system- 191kg-ANSYS)

181 Figure 6.44 Rotation (or) pitch angle (radian) of motorcycle due to pitching (Conventional braking system- 191kg-ANSYS) Figure 6.45 Displacement (y) of motorcycle due to pitching (Variable braking force system -191kg-ANSYS)

182 Figure 6.46 Displacement (y) of motorcycle due to pitching (Conventional braking force system -191kg-ANSYS).18.16.14.12.1.8.6.4.2 -.2 2 4 6 8 1 12 Time in secs Mat Lab ( m = 191kg, r = 11mm) Mat Lab ( m = 191kg, r = 55mm) Ansys ( m = 191kg, r = 11mm) Ansys ( m = 191kg, r = 55mm) Figure 6.47 Motorcycle responses for conventional (r=11mm) and VBF (r=55mm) system under unladen condition (m =191kg)

183.4.3.2.1 -.1 2 4 6 8 1 12 -.2 -.3 Time in secs Mat Lab ( m = 191kg, r = 11mm) Mat Lab ( m = 191kg, r = 55mm) Ansys ( m = 191kg, r=11mm) Ansys ( m = 191kg, r = 55mm) Figure 6.48 Motorcycle responses for conventional (r = 11mm) and VBF (r=55mm) system under unladen condition (m =191kg).2.15 VBF (m = 261kg, r = 84mm) Conventional (m = 261kg, r = 11mm).1.5 1 2 3 4 5 6 7 8 9 1 Time in secs -3 x 1 6 4 2 VBF (m = 261kg, r = 84mm) Conventional (m = 261kg, r = 11mm) -2-4 1 2 3 4 5 6 7 8 9 1 Time in secs Figure 6.49 Comparison of motorcycle responses between conventional and variable braking force (VBF) system due to pitching moment at CG (261kg)

184 Figure 6.5 Rotation (or) pitch angle (radian) of motorcycle due to pitching (variable braking system - 261kg-ANSYS) Figure 6.51 Rotation (or) pitch angle (radian) of motorcycle due to pitching (Conventional braking system - 261kg)

185 Figure 6.52 Displacement (y) of motorcycle due to pitching (Variable braking force system - 261kg) Figure 6.53 Displacement (y) of motorcycle due to pitching (Conventional braking force system -261kg)

186.6.4.2 -.2 -.4 2 4 6 8 1 12 Time in secs Mat Lab (m = 261kg, r = 11mm) Mat Lab (m = 261kg, r = 84mm) Ansys (m = 261kg, r = 11mm) Ansys (m = 261kg, r = 84mm) Figure 6.54 Motorcycle responses for conventional (r = 11mm) and VBF (r = 84mm) system under laden condition (m =261kg).2.15.1.5 -.5 2 4 6 8 1 12 Time in secs Mat Lab (m = 261kg, r = 11mm) Mat Lab ( m = 261kg, r = 84mm) Ansys (m = 261kg, r = 11mm) Ansys (m = 261kg, r = 84mm) Figure 6.55 Motorcycle responses for conventional (r = 11mm) and VBF system under laden condition (m =261kg) From the above analyses, it is determined that the displacement and the rotational angle for conventional braking system are more compared with variable braking force system.

187 6.3.5 Effect of Effective disc radius on Suspension during Real time Braking (MATLAB Model and ANSYS) If the motorcycle rider notices a road curb on his way, he will apply brakes. This activity is simulated using MATLAB. In this work, simulation time runs for 1secs. First -3 secs the rider is riding the vehicle at constant speed and at the end of 3sec., he starts to apply brakes and slows down the vehicle as he notices the road curb (.853m) on his way. At the end of 7 th sec the vehicle comes across the road curb and simulation ends at 1 th sec. In this simulation work road height shown in Figure 6.56, is given as second input in addition to the pitching moment due to sudden braking. UY and ROTZ denote the vertical displacement (y) and rotation (or) pitch angle (theta) of mass m due to braking and road curb. CG Vehicle motion m K r C r C f K f R h (a) Figure 6.56 (Continued)

188 (b) Figure 6.56 Vehicle and Simulink model of motorcycle suspension system with road curb The same simulation work is also done using ANSYS. The results of the analyses are given below in the Figures 6.57 to 6.64. -5 VBF ( m = 191kg, r = 55mm) Conventional ( m = 191kg, r = 11mm) -1 1 2 3 4 5 6 7 8 9 1.2 Road curb (Rh).1 1 2 3 4 5 6 7 8 9 1.2.1 1 2 3 4 5 6 7 8 9 1.2 -.2 1 2 3 4 5 6 7 8 9 1 Time in secs Figure 6.57 Motorcycle (191kg) responses due to pitching (Mz) at 3 rd sec and road curb (R h ) at 7 th sec -MATLAB

189 Figure 6.58 Motorcycle (m = 191kg, r = 11mm) responses due to pitching (Mz) at 3 rd sec and road curb (R h ) at 7 th sec -ANSYS Figure 6.59 Motorcycle (m = 191kg, r = 55mm) responses due to pitching (Mz) at 3 rd sec and road curb (R h ) at 7 th sec -ANSYS

19-5 VBF (m = 261kg, r = 84mm) Conventional (m = 261kg, r =11mm) -1 1 2 3 4 5 6 7 8 9 1.1.5 Road curb (Rh) 1 2 3 4 5 6 7 8 9 1.2.1 1 2 3 4 5 6 7 8 9 1.2 -.2 1 2 3 4 5 6 7 8 9 1 Time in secs Figure 6.6 Motorcycle (261kg) responses due to pitching (Mz) at 3 rd sec and road curb (R h ) at 7 th sec -MATLAB Figure 6.61 Motorcycle (m = 261kg, r = 84mm) responses due to pitching (Mz) at 3 rd sec and road curb (R h ) at 7 th sec -ANSYS

191 Figure 6.62 Motorcycle (m = 261kg, r = 11mm) responses due to pitching (Mz) at 3 rd sec and road curb (R h ) at 7 th sec -ANSYS.16.14.12.1.8.6.4.2. -.2 2 4 6 8 1 12 Time in secs ROTZ - MATLAB (m =191kg, r =55mm) UY - ANSYS (m = 191kg, r =55mm) ROTZ - ANSYS (m = 191kg, r = 55mm) UY- MATLAB (m=191kg, r =55mm) Figure 6.63 Motorcycle responses for VBF system under unladen condition

192.14.12.1.8.6.4.2 -.2 2 4 6 8 1 12 Time in secs ROTZ - ANSYS (m = 261kg, r =84mm) ROTZ - MATLAB (m = 261kg, r =84mm) UY - ANSYS (m = 261kg, r =84mm) UY - MATLAB (m = 261kg, r =84mm) Figure 6.64 Motorcycle responses for VBF system under laden condition 6.4 THERMAL ANALYSIS OF DISC BRAKE 6.4.1 Analysis using Two Dimensional Models A simple two dimensional (2D) model is constructed to model the brake disc to capture the major physics of the brake disc without the venting holes. A time-varying heat flux was applied on the model to simulate the braking action. Detailed procedures are as follows: 6.4.1.1 Modeling A schematic diagram of the brake disc is shown in Figure 6.65.

193 Figure 6.65 Schematic diagram of brake disc As the temperature distribution in the circumferential direction is assumed to be insignificant, an infinitely thin strip of disc (in the axial direction, or x-axis) is used to model the above mentioned problem. The 2D models of disc brake are shown in Figures 6.66 and 6.67 for unladen condition and loaded condition respectively. Figure 6.66 2D model of the brake disc under unladen condition Due to symmetrical loading conditions, only half thickness is used (3mm). To simulate the action of the brake pads on the brake disc, a

194 time-varying heat flux q(t) is applied on the contact surface. The actual model used in ANSYS is given in Figure 6.68. Figure 6.67 2D model of the brake disc under laden condition Figure 6.68 Actual model of brake disc

195 6.4.1.2 Meshing 87 second order (quadratic) quadrilateral element is used in the meshing of the model. The reasons for prefering quadratic elements rather than the linear elements are that simulation results have been compared and that in essence, second order elements give a more accurate answer. As the thickness of the model is very small (only 3mm), and the temperature distribution in that direction is going to be insignificant, two elements in the thickness direction are going to give a good approximation of the actual results. The meshed model in ANSYS is given in Figure 6.69. Figure 6.69 Meshed model of brake disc

196 6.4.1.3 Material properties Cast iron and stainless steel are used in the construction of the brake disc. Its material properties can be found in Table 6.6. The material properties are entered into the ANSYS pre-processor directly. Table 6.6 Material properties of brake disc Sl No Material Properties CAST IRON STAINLESS STEEL 1 Density 7228 kg/m 3 78 kg/m 3 2 Specific Heat 419 J/kg.ºC 46 J/kg.ºC 3 Thermal Conductivity 48.46 W/m.ºC 25 W/m.ºC 4 Young s Modulus 1e11 N/m 2 2e11 N/m 2 5 Poission ratio.25.3 6 Thermal expansion coefficient 9e-6 11e-6 6.4.1.4 Boundary conditions A time-varying heat flux q(t) is applied to the model as shown in Figure 6.7 to simulate the action of the brake pad on the disc. The procedure to determine the time varying heat flux is as follows: Braking energy, Eb = mv 2 1 /2 + Iw 12 /2 N-m (6.15) Where, Eb = Braking energy, N-m I w = Mass moment of inertia of rotating parts, kg-m 2 m = Total mass of motorcycle, kg V 1 = Motorcycle velocity at the beginning of braking, m/s 1 = Angular velocity of rotating parts at beginning of braking, 1/s Using V 1 = R 1, then Eb = (m/2)(1 + I/R 2 m)v 1 2 = (K1 mv 1 2 )/2 (6.16)

197 Where, R = Effective rolling radius of tyre. In this work, K 1 (correction factor for rotating masses (1 + I/R 2 m)) is estimated to be about 1.1. Mass equivalent may be around 2% for a motorcycle. But the mass equivalent is considered for the rear wheel only because only the rear wheel is braked. Moreover, the mass of the rear wheel is 12 kg. If it is assumed that the entire mass of the wheel is located at the radial position of 3mm (R- Effective rolling radius of tyre) then the mass moment of inertia of rotating components is 1.8kg-m 2 ( 12*(.3) 2 ). K 1 = (1+I w / mr 2 ) (6.17) For unladen condition (m = 191kg) K 1 = 1.6 Approximately 6% if the mass moment of inertia of rotating gears (gear box) may be negligible because of less weight. The mass moment of inertia of other rotating parts like flywheel magneto, crankshaft etc., may be neglected because the clutch is disengaged during braking operation. For laden condition (m = 261kg) K 1 = 1.45 Approximately 5% if the mass moment of inertia of rotating gears (gear box) may be negligible because of less weight. The mass moment of inertia of other rotating parts like flywheel magneto, crankshaft etc., may be neglected because the clutch is disengaged during braking operation.

198 Braking power, Pb = d(eb)/dt (6.18) For constant deceleration a, V(t) = V 1 a 1 t (6.19) Combining Equations (6.16), (6.17), (6.18) and (6.19) leads to Pb = K 1 ma 1 (V 1 a 1 t) (6.2) Heat flux, Q(t) is obtained by dividing the braking power by the swept area of the brake rotor. i.e. Q(t) = K 1 ma 1 (V 1 a 1 t)/a p (6.21) where, a 1 = Motorcycle deceleration taken from Table 3.3, m/s 2 K 1 = Correction factor for rotating masses. t = Stopping time=v 1 /a, Secs A p = Pad swept area=8639 mm 2 (unladen condition-191kg) A p = 13195 mm 2 (loaded condition-261kg) m = Total mass of motorcycle, kg V 1 = Motorcycle velocity at the beginning of braking, m/s This heat flux is the averaged heat flux generated during the braking cycle. In the 2D model, it is assumed that the temperature variation in the circumferential direction is insignificant and thus it is modeled to have infinitely small thickness in the x direction. As such, if the brake disc is to be divided into infinitely many such small strips, each strip would experience the same averaged heat flux.

199 Since only half of the disc thickness is modeled, the heat flux entering the disc should also be halved. Furthermore, not all of the heat enters the brake disc. A proportion of heat entering into the brake disc is called heat distribution factor. 1 p d 1 c c p d k k P d 1/ 2 (6.22) Where, is heat distribution factor. p, c p, c d d = Density of pad and disc respectively. = Specific heat of pad and disc respectively. k P, k d = Thermal conductivity of pad and disc respectively. Pad : p = 2592 kg/m 3, Cast iron disc: cd = 7228 kg/m3, c p = 1465 N-m/kg-K, Stainless steel disc: sd = 78 kg/m3, k P = 1.61 W/m-k c cd = 419 N-m/kg-K, k cd = 48.46 W/m-k c sd = 46 N-m/kg-K, k sd = 25 W/m-k 1 cd p 1 c c p cd k k P cd 1/ 2 =.83, For cast iron 1 p sd 1 c c p sd k k P sd 1/ 2 =.8, For stainless steel Therefore, the heat flux function should be modified to be: Q(t) =.5* *K 1 ma 1 (V 1 a 1 t)/a p (6.23)

2 Heat flux is set to zero on the right side of the model to simulate the symmetric loading conditions. Also, convection is set at all other lines as they are exposed to free air. The heat transfer coefficient is set to 12 W/m 2 -k and the bulk temperature set at 3K. Figure 6.7 Heat flux applied at pads and disc contact area for the effective disc radius of.55m 6.4.2 Results of the Thermal Analysis The 2D simulation contour plots can be found for various load conditions and vehicle speeds. The rise in disc surface temperature in these plots is measured in degree centigrade. The Figure 6.71 shows how the disc surface temperature is rising with respect to time when heat flux is applied between.45m and.65m (effective disc radius of.55m-unladen condition). Figure 6.72 shows the temperature along the disc thickness for various times of braking operation. The results of thermal analyses given in Figures 6.71 to 6.79 for unladen condition (191kg) are carried out for various velocities. The same above said thermal analyses are carried out for the effective disc radius of.85m (loaded condition) for various vehicle velocities and the results are shown in Figures 6.8 to 6.88. The thermal stress values on the brake disc of the motorcycle at laden condition (261kg) are

21 plotted in Figures 6.89 to 6.91 for various speeds. The 2D simulation results can be a quick and accurate way to find out what the general temperature of the brake disc will be during braking. In order to obtain more accurate analysis, 3D models of the brake discs with venting holes with a diameter 6 mm are constructed and solved using ANSYS. Figure 6.71 Temperature distributions on cast iron brake disc at unladen condition (CI-191kg &4 kmph) Figure 6.72 Disc surface temperature Vs braking time at various axial distances (CI-191kg &4kmph)

22 Figure 6.73 Temperature distributions along the various axial distances and braking times (CI-191kg &4kmph) Figure 6.74 Temperature distributions on brake disc at unladen condition (CI-191kg &5 kmph)

23 Figure 6.75 Disc surface temperature Vs braking time at various axial distances (CI-191kg &5kmph) Figure 6.76 Temperature distributions along the various axial distances and braking times (CI-191kg &5kmph)

24 Figure 6.77 Temperature distributions on brake disc at unladen condition (CI-191kg &6 kmph) Figure 6.78 Disc surface temperature Vs braking time at various axial distances (CI-191kg &6kmph)

25 Figure 6.79 Temperature distributions along the various axial distances and braking times (CI-191kg &6kmph) Figure 6.8 Temperature distributions on brake disc at loaded condition (CI-261kg &4 kmph)

26 Figure 6.81 Surface temperature Vs braking time at various axial distances (CI-261kg &4kmph) Figure 6.82 Temperature distributions along the various axial distances and braking times (CI-261kg &4kmph)

27 Figure 6.83 Temperature distributions on brake disc at loaded condition (CI-261kg &5 kmph) Figure 6.84 Surface temperature Vs braking time at various axial distances (CI-261kg &5kmph)

28 Figure 6.85 Temperature distributions along the various axial distances and braking times (CI-261kg &5kmph) Figure 6.86 Temperature distributions on brake disc at loaded condition (CI-261kg &6 kmph)

29 Figure 6.87 Surface temperature Vs braking time at various axial distances (261kg &6kmph) Figure 6.88 Temperature distributions along the various axial distances and braking times (CI-261kg &6kmph)

21 Figure 6.89 Thermal stress distributions (261 kg & 4kmph) Figure 6.9 Thermal stress distributions (261 kg & 5kmph)

211 Figure 6.91 Thermal stress distributions (261 kg & 6kmph) 7 6 5 4 3 2 1..5 1. 1.5 2. 2.5 3. 3.5 4. Braking time, secs CI ( Effective disc radius 55mm-191kg) CI ( Effective disc radius 84mm-261kg) CI ( Effective disc radius 11mm-191kg) CI ( Effective disc radius 11mm-261kg) SS ( Effective disc radius 55mm-191kg) SS ( Effective disc radius 84mm-261kg) SS ( Effective disc radius 11mm-191kg) SS ( Effective disc radius 11mm-261kg) Figure 6.92 Motorcycle velocity 4kmph - Rise in disc surface temperature at the axial distance (x=m)

212 From Figure 6.92, the rise in disc surface temperature ( T) increases when the effective disc radius decreases because the disc heating mass is decreased as the radius decreases (KE = m d c d, = KE/ m d c d m d decreases - increases). Hence the disc surface temperature increases as effective disc radius decreases for cast iron (CI) as well as for stainless steel (SS) for the same velocity and mass of the motorcycle. Furthemore, the stainless steel has less rise in disc surface temperature, when compared with cast iron because the density ( d ) and specific heat (c d ) of steel are high when compared with cast iron. The magnitude of d c d for steel is 3588 J/m 3 K and that of cast iron is 328532 J/m 3 K. Further steel disc has less heat distribution factor ( =.8) compared with cast iron disc ( =.83). However, the steel has more rises in disc surface temperature during the starting of braking operation because of its low thermal diffusivity and heat flux. Furthermore, the stainless steel disc retains the heat at the disc surface for a particular period of time because of low thermal diffusivity (Cast iron diffuses the heat faster). 2 Ld The heat flux penetration time (t b ) is given by t, hr (6.24) b 5 a Where, L d = Half the brake disc thickness, m a td = Where d a td = Thermal diffusivity, m 2 /h k d, (6.25) cd c, k = Density, Specific heat, Thermal conductivity of brake d, d a =.57, For Cast Iron. td d disc respectively. a td =.25, For Stainless Steel. t b = 3.1578 x 1-5 hr, td

213 t b =.1137 s, For Cast Iron t = 7.2 x 1-5 hr, b t b =.2592 s, For Stainless Steel As the heat flux penetration time for the stainless steel disc is more compared with cast iron, the stainless steel disc retains its heat at the disc s outer surface. Moreover the heat influx into the CI disc is more during the starting of braking operation. Hence there is greater rise in disc outer surface temperature ( T) during the starting of braking for stainless steel than in cast iron. The thermal analyses are carried out for various motorcycle velocities, axial distances of brake disc and their results are given in Figures 6.93 to 6.1. 7 6 5 4 3 2 1..5 1. 1.5 2. 2.5 3. 3.5 4. Time in secs CI ( Effective disc radius 55mm-191kg) CI ( Effective disc radius 84mm-261kg) CI ( Effective disc radius 11mm-191kg) CI ( Effective disc radius 11mm-261kg) SS ( Effective disc radius 55mm-191kg) SS ( Effective disc radius 84mm-261kg) SS ( Effective disc radius 11mm-191kg) SS ( Effective disc radius 11mm-261kg) Figure 6.93 Motorcycle velocity 4kmph - Rise in disc surface temperature at the axial distance (x=.15m)

214 7 6 5 4 3 2 1..5 1. 1.5 2. 2.5 3. 3.5 4. Braking time, secs CI ( Effective disc radius 55mm-191kg) CI ( Effective disc radius 84mm-261kg) CI ( Effective disc radius 11mm-191kg) CI ( Effective disc radius 11mm-261kg) SS ( Effective disc radius 55mm-191kg) SS ( Effective disc radius 84mm-261kg) SS ( Effective disc radius 11mm-191kg) SS ( Effective disc radius 11mm-261kg) Figure 6.94 Motorcycle velocity 4kmph - Rise in disc surface temperature at the axial distance (x=.3m) 1 9 8 7 6 5 4 3 2 1.5 1 1.5 2 2.5 3 3.5 4 4.5 Braking time, secs CI ( Effective disc radius 55mm-191kg) CI ( Effective disc radius 84mm-261kg) CI ( Effective disc radius 11mm-191kg) CI ( Effective disc radius 11mm-261kg) SS ( Effective disc radius 55mm-191kg) SS ( Effective disc radius 84mm-261kg) SS ( Effective disc radius 11mm-191kg) SS ( Effective disc radius 11mm-261kg) Figure 6.95 Motorcycle velocity 5kmph - Rise in disc surface temperature at the axial distance (x=m)

215 1 9 8 7 6 5 4 3 2 1.5 1 1.5 2 2.5 3 3.5 4 4.5 Braking time, secs CI ( Effective disc radius 55mm-191kg) CI ( Effective disc radius 84mm-191kg) CI ( Effective disc radius 11mm-191kg) CI ( Effective disc radius 11mm-261kg) SS ( Effective disc radius 55mm-191kg) SS ( Effective disc radius 84mm-261kg) SS ( Effective disc radius 11mm-191kg) SS ( Effective disc radius 11mm-261kg) Figure 6.96 Motorcycle velocity 5kmph - Rise in disc surface temperature at the axial distance (x=.15m) 1 9 8 7 6 5 4 3 2 1.5 1 1.5 2 2.5 3 3.5 4 4.5 Braking time, secs CI ( Effective disc radius 55mm-191kg) CI ( Effective disc radius 84mm-261kg) CI ( Effective disc radius 11mm-191kg) CI ( Effective disc radius 11mm-261kg) SS ( Effective disc radius 55mm-191kg) SS ( Effective disc radius 84mm-261kg) SS ( Effective disc radius 11mm-191kg) SS ( Effective disc radius 11mm-261kg) Figure 6.97 Motorcycle velocity 5kmph - Rise in disc surface temperature at the axial distance (x=.3m)

216 14 12 1 8 6 4 2 1 2 3 4 5 6 Braking time, secs CI ( Effective disc radius 55mm-191kg) CI ( Effective disc radius 84mm-261kg) CI ( Effective disc radius 11mm-191kg) CI ( Effective disc radius 11mm-261kg) SS ( Effective disc radius 55mm-191kg) SS ( Effective disc radius 84mm-261kg) SS ( Effective disc radius 11mm-191kg) SS ( Effective disc radius 11mm-261kg) Figure 6.98 Motorcycle velocity 6kmph - Rise in disc surface temperature at the axial distance (x=m) 14 12 1 8 6 4 2.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 Braking time, secs CI ( Effective disc radius 55mm-191kg) CI ( Effective disc radius 84mm-261kg) CI ( Effective disc radius 11mm-191kg) CI ( Effective disc radius 11mm-261kg) SS ( Effective disc radius 55mm-191kg) SS ( Effective disc radius 84mm-261kg) SS ( Effective disc radius 11mm-191kg) SS ( Effective disc radius 11mm-261kg) Figure 6.99 Motorcycle velocity 6kmph - Rise in disc surface temperature at the axial distance (x=.15m)

217 14 12 1 8 6 4 2.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 Braking time, secs CI ( Effective disc radius 55mm-191kg) CI ( Effective dsic radius 84mm-261kg) CI ( Effective disc radius 11mm-191kg) CI ( Effective disc radius 11mm-261kg) SS ( Effective disc radius 55mm-191kg) SS ( Effective disc radius 84mm-261kg) SS ( Effective disc radius 11mm-191kg) SS ( Effective disc radius 11mm-261kg) Figure 6.1 Motorcycle velocity 6kmph - Rise in disc surface temperature at the axial distance (x=.3m) 14 12 1 8 6 4 2 1 2 3 4 5 6 Braking time, secs CI ( Vehicle velocity 4kmph-191kg) CI ( Vehicle velocity 5kmph-191kg) CI ( Vehicle velocity 6kmph-191kg) SS (Vehicle velocity 4kmph-191kg) SS ( Vehicle velocity 5kmph-191kg) SS ( Vehicle velocity 6kmph-191kg) Figure 6.11 Rise in disc outer surface temperature for various vehicle velocities ( m=191kg, r = 54mm, x =mm)

218 Figure 6.11 shows that when the kinetic enegy of the motorcycle is increased, there will be more rise in disc outer surface temperature ( T). Figures 12-14 show the rise in brake disc surface temperature for various motorcycle velocities, mass, and effective radius of disc. 12 1 8 6 4 2.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Braking time, secs CI ( Vehicle velocity 4kmph-261kg) CI ( Vehicle velocity 5kmph-261kg) CI ( Vehicle velocity 6kmph-261kg) SS ( Vehicle velocity 4kmph-261kg) SS ( Vehicle velocity 5kmph-261kg) SS ( Vehicle velocity 6kmph-261kg) Figure 6.12 Rise in disc outer surface temperature for various vehicle velocities (m=261kg, r = 84mm, x =mm) 1 9 8 7 6 5 4 3 2 1.5 1 1.5 2 2.5 3 3.5 4 4.5 5 Braking time, secs CI ( Vehicle velocity 4kmph-261kg) CI ( Vehicle velocity 5kmph-261kg) CI ( Vehicle velocity 6kmph-261kg) SS ( Vehicle velocity 4kmph-261kg) SS ( Vehicle velocity 5kmph-261kg) SS ( Vehicle velocity 6kmph-261kg) Figure 6.13 Rise in disc outer surface temperature for various vehicle velocities (m=261kg, r = 11mm, x =mm)

219 7 6 5 4 3 2 1.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 Braking time, secs CI ( Vehicle velocity 4kmph-191kg) CI ( Vehicle velocity 5kmph-191kg) CI ( Vehicle velocity 6kmph-191kg SS ( Vehicle velocity 4kmph-191kg) SS ( Vehicle velocity 5kmph-191kg) SS ( Vehicle velocity 6kmph-191kg) Figure 6.14 Rise in disc outer surface temperature for various vehicle velocities (m=191kg, r = 11mm, x =mm) The analysis results are validated by calculating the average rise in disc surface temperature ( T avg ) as follows: Consider a graph (Rise in disc outer surface temperature Vs braking time) for the following condition: Disc material is cast iron m = 191kg r = 11 mm V = 4 kmph Axial distance of brake disc(x) = mm It may be assumed that the profile of the Figure 6.15 between the rise in disc surface temperature and braking time is a parabola. 2 Area under the parabola is = 3 *3. 43 3

22 Figure 6.15 Rise in disc surface temperature Vs braking time Area under the parabola = Area of the retangle. Length of x axis in the Figure 6.15 is equal for parabola and rectangle 2 3 3 *3.43 = T avg *3.43 T avg = 2 1 4 Initial kinetic energy of motorcycle = *191* *1. 1 2 3.6 = 12969 Nm 2 (6.26) Final kinetic energy of motorcycle is zero because vehicle velocity becomes zero. So average kinetic energy (KE avg ) during braking process is equal to average value of initial and final kinetic energy. 12969 KE avg = 2 6484.5Nm Average kinetic energy (KE avg ) = m c T (6.27) cd cd avg Where m cd = Heating mass of cast iron (Density of cast iron X heating volume of disc) = 7228 X 17279 X 1-6 X.6

221 c cd = Specific heat of cast iron disc T avg = Average rise in disc surface temperature. T avg = 2.65 (This value may be checked in the Table 6.7.) 6.4.3 Calculation of Rise in Disc Average Surface Temperature KE avg = m d c d T avg (6.28) Where, m d c d KE avg T avg = Heating mass of disc. = Specific heat of disc. = Average kinetic energy of motorcycle. = Average rise in disc surface temperature. T avg is calculated from the Equation (6.28) and given in Tables 6.7 and 6.8 for cast iron and stainless steel respectively. Table 6.9 shows the percentage rise in average disc temperature for various load conditions and disc materials. Table 6.7 Average rise in disc surface temperature for cast iron Sl No Loading condition Vehicle velocity (kmph) Effective disc radius (mm) Average energy during braking (N-m) 1 Unladen (191kg) 4 55 6484.56 41.3 2 Unladen (191kg) 4 11 6484.56 2.7 3 Unladen (191kg) 5 55 1132.14 64.5 4 Unladen (191kg) 5 11 1132.14 32.3 5 Unladen (191kg) 6 55 1459.28 92.9 6 Unladen (191kg) 6 11 1459.28 46.5 7 Laden (261kg) 4 84 8861.11 37. 8 Laden (261kg) 4 11 8861.11 28.2 9 Laden (261kg) 5 84 13845.49 57.7 1 Laden (261kg) 5 11 13845.49 44.1 11 Laden (261kg) 6 84 19937.5 83.2 12 Laden (261kg) 6 11 19937.5 63.5 avg

222 Table 6.8 Average rise in disc surface temperature for stainless steel Sl No Loading condition Vehicle velocity (kmph) Effective disc radius (mm) Average energy during braking (N-m) 1 Unladen (191kg) 4 55 6484.56 34.9 2 Unladen (191kg) 4 11 6484.56 17.4 3 Unladen (191kg) 5 55 1132.14 54.5 4 Unladen (191kg) 5 11 1132.14 27.2 5 Unladen (191kg) 6 55 1459.28 78.5 6 Unladen (191kg) 6 11 1459.28 39.2 7 Laden (261kg) 4 84 8861.11 31.2 8 Laden (261kg) 4 11 8861.11 23.8 9 Laden (261kg) 5 84 13845.49 48.7 1 Laden (261kg) 5 11 13845.49 37.2 11 Laden (261kg) 6 84 19937.5 7.2 12 Laden (261kg) 6 11 19937.5 53.6 The rise in disc temperature ( T) is high for VBF heating compared with conventional heating ( r = 11mm) because the magnitude of disc mass is less for VBF system as disc radius is varied based on pillion load on the motorcycle. avg 1 9 8 7 6 5 4 3 2 1 35 4 45 5 55 6 65 Vehicle velocity (kmph) CI (m =191kg), r = 54mm) CI ( m = 261kg, r = 84mm) CI - Conventional ( m = 191kg, r = 11mm) CI - Conventional ( m = 261kg, r = 11mm) SS ( m = 191 kg, r = 55mm) SS ( m = 261kg, r = 84mm) SS - Conventional ( m = 191kg, r = 11mm) SS - Conventional ( m = 261kg, r =11mm) Figure 6.16 Average rise in disc surface temperature (Vs) vehicle velocity

223 6.5 DISC ANALYSIS USING THREE DIMENSIONAL MODEL Since the 2D simulation results are unable to show the general temperature distribution with reasonable accuracy, the 3D models of brake disc are built to investigate the temperature distribution. 6.5.1 3D Modeling Taking advantage of the symmetry of the brake disc, only 1/6 of it is modeled. Each model is divided into 3 areas to facilitate the ease of application of the time- varying heat flux. A schematic diagram of the 3D model is shown in Figure 6.17. The pink area represents the brake pad contact area. Figure 6.17 Schematic diagram of 3D model of brake disc

225 The braking action is simulated by the same averaged time-varying heat flux used in the 2D simulations, and is applied on the brake contact area. The actual 3D models of the brake disc designs in ANSYS can be found in Figure 6.18. Figure 6.18 3D Model of brake disc 6.5.2 3D Meshing 2-node quadratic brick elements are used in the meshing of the 3D models. Brick elements are chosen over tetrahedral elements because they give a more accurate answer. Similarly, quadratic elements are more accurate than linear elements. Furthermore, due to the complex shapes of the brake disc, quadratic elements will provide for a better mesh. In the thickness direction, only 1 element is used as the temperature variation is insignificant. Areas around piercing holes were given finer mesh in order to obtain a more

226 accurate answer around the region. The actual meshed models of the brake discs can be found in Figure 6.19. Figure 6.19 Meshed model of brake disc 6.5.3 Disc Material Properties Similar to the 2D simulations, the material properties of the disc are entered into the ANSYS pre-processor directly. 6.5.4 Boundary Conditions Similar to the 2D simulations, the same time-varying heat flux, convection parameters as well as symmetrical boundary conditions are used. The heat flux is applied as shown in Figure 6.11 on the elements in the brake-pad contact area, which has been earlier defined in the modeling section and convective heat transfer is considered in the remaining two areas of brake

227 disc. The reason for applying the heat flux in this manner is that since the brake disc rotates so many times in one braking cycle, it is reasonable to assume that the entire brake disc experiences the same heat flux and that the temperature variation in the circumferential direction is going to be negligible. Figure 6.11 Load and boundary condition applied on brake disc 6.5.5 3D Disc Thermal Analysis Results The 3D results look like they are just an extension of the 2D simulation results in the circumferential direction. Similarly, these 3D results predict the general temperature distribution with a high degree of accuracy. The temperature distribution of the brake disc for the load condition 191 kg (unladen condition) for various velocities are plotted in the Figures 6.111-6.113 and for loaded condition (261kg) in the Figures 6.114-6.116. The results of 2D models and 3D models are given in Table 6.1

228 Figure 6.111 3D-Temperature distribution (CI, m=191kg & V=4kmph) Figure 6.112 3D-Temperature distribution (CI, m=191kg & V=5kmph)

229 Figure 6.113 3D-Temperature distribution (CI, m=191kg &V=6kmph) Figure 6.114 3D-Temperature distribution (CI, m=261kg &V=4kmph)

23 Figure 6.115 3D-Temperature distribution (CI, m=261kg &V= 5kmph) Figure 6.116 3D-Temperature distribution (CI, m=261kg &V=6kmph)