BRAKES
Brakes
5.Design of Braking System
5.1. Introduction:
Brakes are the mechanical device whose function is to slow and stop the vehicle. They are mandatory for the safe operation of the cars. When a vehicle is in motion, it has the kinetic energy or energy derived from this motion. To slow down the car, this energy must be decreased. This is accomplished by converting into another form. Thus the braking system transforms the kinetic energy into heat energy by creating friction between the parts. Various methods actuate the brakes. Hydraulic braking is the most common type of braking system in 4-wheelers because of its efficiency and simplicity in design.
5.2. Hydraulic brake system:
Hydraulic brake is an arrangement of braking mechanism which uses brake fluid to transfer pressure from the controlling unit, which is usually near the operator of the vehicle, to the actual brake mechanism, which is nearer to the wheel of the car.
5.2.1. Components:
Brake pedal or lever
A pushrod or actuating the rod
A master cylinder assembly containing piston assembly( made up of either one or two pistons, a return spring, a series of gaskets and a fluid reservoir)
Reinforced hydraulic lines
Brake caliper assembly consisting of one or two caliper pistons, an asset of thermally conductive brake pads and a rotor
5.3. Steps involved in brake calculations:
Step 1: Find the center of gravity of the vehicle
From chapter 1
Weight of the vehicle,W = 270 kg = 2648.7 N
Wheelbase, L=142 cm
Front wheel track, Bf =122 cm
Rear wheel track, Br=117 cm
Location of Centre of gravity(C.G):
The distance of C.G from the rear wheel, Lr=55.6 cm
The distance of C.G from the front wheel, Lf=86.4 cm
The height of location of C.G from the ground, H=75 cm
Static reaction force at front wheels,
Rf = W× Lr/L = 270×55.6/142
Rf =105.72 kg
Static reaction force at rear wheels,
Rr =W× Lf/L = 270×86.4/142
Rr =164.28 kg
Step 2: To find the maximum deceleration that can be achieved by the vehicle:
Let,
Rrdyn= Dynamic normal reaction acting on the the rear whee l, N
Rfdyn= Dynamic normal reaction acting on the front wheel, N
FR = Frictional force at rear wheels, N
FF = Frictional force at front wheels, N
a = Deceleration of the vehicle, g units
W×a = Inertial force, N
µ = coefficient of friction between the wheel and ground
W×a and ( FR+FF) acts in the opposite direction which creates a couple.
When the brake is applied, the couple will compress the front wheel and lifts the rear wheel.
FF+ FR– Wa=0 (5.1)
Σ FY=0
Rrdyn+Rfdyn– W=0 (5.2)
FR=µ×Rrdyn (5.3)
FF=µ×Rfdyn (5.4)
Substituting in (5.1)
FF+FR– (W×a)=0
µ×Rfdyn + µ×Rrdyn =W×a
µ(Rfdyn+Rrdyn ) =W×a
µ×W=W×a
Therefore , (5.5)
Maximum deceleration(a) is limited by the coefficient of friction between the wheel and ground(µ).
Step 3: To find the dynamic reaction loads :
Taking moment about front wheel,
Rrdyn×L+ (W×a×H) = W×Lf (5.6)
Rrdyn×L= (W×Lf)–(W×a×H) (5.7)
Considering Co-efficient of friction betwthe een the ttiresand surface to be 0.6
From equation 5.5 Deceleration =0.6 g units
Substituting the values from step 1
Rrdyn×142=(2648.7×86.4) –(2648.7×0.6×75)
(5.8)
TakTaking a moment about rear wheel
Rfdyn×L = W×Lr+Wa×H (5.9)
Rfdyn×142 = 2648.7×55.6 + 2648.7×0.6×75
(5.10)
For various values of deceleration, the dynamic loads are calculated and are shown in e 5.1 and visualized in the graph 5.1 as follows:
Deceleration
Robynn(in newtons)
Robyn(in newtons)
0.2
1316.89
1331.81
0.4
1596.68
1052.02
0.6
1876.47
772.23
0.8
2156.27
492.43
1.0
2436.05
212.64
Table 5.1: Variation of dynamic loads for various values of deceleration
Graph 5.1: Variation of deceleration with front and rear dynamic load
Step 4: To find the maximum Frictional (Braking) Force :
From equations (5.3) and (5.4),
FR=µ×Rrdyn (5.11)
=0.6×772.23
FF = µ×Rfdyn (5.12)
=0.6×1876.47
Step 5: Design of braking components:
It is assumed that FF & FR values induce the braking system
( KfW+KaAfV2)+FF+FR=W×a
Where,
Kr =Co-efficient of rolling resistance
Ka=Co-efficient of air drag resistance
Af=Frontal area in m2
V=Velocity of the vehicle in m/s
Neglecting air and rolling resistance, we get
FF+FR=W×a
Disc:
A simple diagram of the drivers shown in figure 5.3
Braking torque = Fr×R=F1×r×2
µ×Rrdyn × R=F1×r×2
µ×Rrdyn× R=µ1×Pcal×Acal ×r×2
where,
R =Radius of the wheel, m
r =Effective radius of the is, m
F1 =Normal force acting on the disc, N
µ1=Co-efficient of friction between disc and brake pad
Pcal=Pressure developed in the caliper, N/m2
Acal=Area of the piston in the caliper, m2
Step 6: Calculation of braking torque (TB)
Radius of the wheel(R) = 11” = 0.2794 m
TB(r) = FR × R
= 463.34 × 0.2794
= 129.46 N-m
TB(f) = 1125.88 × 0.2794
= 314.57 N-m
Considering uniform wear rate theory,
R= (r1+r2)/2
Where
r1 is the inner radius in ‘m’
r2 is the outer radius in ‘m’
Braking condition is given by
µ×Rrdyn× R ≥ µ1×Pcal×Acal×r×2
If Pcal=0, then deceleration is zero.
When the vehicle is slowed down then
µ×Rrdyn× R – µ1×Pcal×Acal ×r×2 = I×α
I×α is the residual torque produced due to rotating parts
While braking I×α tends to zero as there is no angular acceleration.
Therefore, µ×Rrdyn× R – µ1×Pcal×Acal ×r×2 = 0
Considering caliper of bore diameter 1.12" with double piston :
Brake pad width = 25.4 mm
Disc diameter(d1) = 200 mm
r2 = disc radius - caliper width
= 74.6 mm
r= (r1+r2)/2 = (100+74.6)/2 = 87.3 mm
For the rear wheel, the normal braking force is given by
F1(r) = TB(r)/( µ1×r×2)
= 129.46/ (0.3×0.0872×2)
= 2474.39 N
Similarly for the front wheel,
F1(f) = 314.57/ (0.3×0.0872×2)
= 6012.42 N
Choosing master cylinder with bore diameter 1"
Area of the master cylinder = 506.71 mm2
Caliper bore diameter = 1.12" = 28.45 mm
Area of the caliper Acal = 635.7 mm2
Pressure developed in the caliper, Pcal = F1(f)/ (n× Acal)
where ‘n’ represents the number of pistons in the caliper
Pcal = 6012.42/(2×635.7)
= 4.73 N/mm2
Since Pmaster cylinder = Pcal
Force at the master cylinder/ Area of the master cylinder = 4.73 N/mm2
Force at the master cylinder = 4.73 × Area of the master cylinder
= 4.73 × 506.71
= 2396.74 N
From the
P1×l1=P2×l2
where
P1 =Pedal force, N
P2=Force exerted on master cylinder, N
l1&l2=Distance of the pivot from pedal and master cylinder respectively, m
Assuming the pedal force to be 250 N,
Pedal force(P1) × Mechanical leverage(l1/l2) = Force at the master cylinder(P2)
250 × (l1/l2) = 2396.74
= 2396.74/250
Mechanical leverage (l1/l2)= 9.5
Step6: Stopping distance ' S '
S=V2/2f
f =deceleration of the vehicle in 'm/s2'
V=Velocity of the vehicle in 'm/s'
If the vehicle moves at a speed of 55 kmph, then
S = V2/ (2 × 0.6g)
v = 55× (5/18) = 15.27 m/s
S = (15.27)2/ (2 × 0.6g)
S = 19.81 m
5.4. Summary:
In this chapter, the basic introduction about the braking system is discussed. The hydraulic braking system and its components were also discussed. Further the calculations that are essential for a braking system is derived and results are made. Graphs are also plotted for various decelerations (in g units) with corresponding dynamic loads for visual interpretation.
5.Design of Braking System
5.1. Introduction:
Brakes are the mechanical device whose function is to slow and stop the vehicle. They are mandatory for the safe operation of the cars. When a vehicle is in motion, it has the kinetic energy or energy derived from this motion. To slow down the car, this energy must be decreased. This is accomplished by converting into another form. Thus the braking system transforms the kinetic energy into heat energy by creating friction between the parts. Various methods actuate the brakes. Hydraulic braking is the most common type of braking system in 4-wheelers because of its efficiency and simplicity in design.
5.2. Hydraulic brake system:
Hydraulic brake is an arrangement of braking mechanism which uses brake fluid to transfer pressure from the controlling unit, which is usually near the operator of the vehicle, to the actual brake mechanism, which is nearer to the wheel of the car.
5.2.1. Components:
Brake pedal or lever
A pushrod or actuating the rod
A master cylinder assembly containing piston assembly( made up of either one or two pistons, a return spring, a series of gaskets and a fluid reservoir)
Reinforced hydraulic lines
Brake caliper assembly consisting of one or two caliper pistons, an asset of thermally conductive brake pads and a rotor
5.3. Steps involved in brake calculations:
Step 1: Find the center of gravity of the vehicle
From chapter 1
Weight of the vehicle,W = 270 kg = 2648.7 N
Wheelbase, L=142 cm
Front wheel track, Bf =122 cm
Rear wheel track, Br=117 cm
Location of Centre of gravity(C.G):
The distance of C.G from the rear wheel, Lr=55.6 cm
The distance of C.G from the front wheel, Lf=86.4 cm
The height of location of C.G from the ground, H=75 cm
Static reaction force at front wheels,
Rf = W× Lr/L = 270×55.6/142
Rf =105.72 kg
Static reaction force at rear wheels,
Rr =W× Lf/L = 270×86.4/142
Rr =164.28 kg
Step 2: To find the maximum deceleration that can be achieved by the vehicle:
Let,
Rrdyn= Dynamic normal reaction acting on the the rear whee l, N
Rfdyn= Dynamic normal reaction acting on the front wheel, N
FR = Frictional force at rear wheels, N
FF = Frictional force at front wheels, N
a = Deceleration of the vehicle, g units
W×a = Inertial force, N
µ = coefficient of friction between the wheel and ground
W×a and ( FR+FF) acts in the opposite direction which creates a couple.
When the brake is applied, the couple will compress the front wheel and lifts the rear wheel.
FF+ FR– Wa=0 (5.1)
Σ FY=0
Rrdyn+Rfdyn– W=0 (5.2)
FR=µ×Rrdyn (5.3)
FF=µ×Rfdyn (5.4)
Substituting in (5.1)
FF+FR– (W×a)=0
µ×Rfdyn + µ×Rrdyn =W×a
µ(Rfdyn+Rrdyn ) =W×a
µ×W=W×a
Therefore , (5.5)
Maximum deceleration(a) is limited by the coefficient of friction between the wheel and ground(µ).
Step 3: To find the dynamic reaction loads :
Taking moment about front wheel,
Rrdyn×L+ (W×a×H) = W×Lf (5.6)
Rrdyn×L= (W×Lf)–(W×a×H) (5.7)
Considering Co-efficient of friction betwthe een the ttiresand surface to be 0.6
From equation 5.5 Deceleration =0.6 g units
Substituting the values from step 1
Rrdyn×142=(2648.7×86.4) –(2648.7×0.6×75)
(5.8)
TakTaking a moment about rear wheel
Rfdyn×L = W×Lr+Wa×H (5.9)
Rfdyn×142 = 2648.7×55.6 + 2648.7×0.6×75
(5.10)
For various values of deceleration, the dynamic loads are calculated and are shown in e 5.1 and visualized in the graph 5.1 as follows:
Deceleration
Robynn(in newtons)
Robyn(in newtons)
0.2
1316.89
1331.81
0.4
1596.68
1052.02
0.6
1876.47
772.23
0.8
2156.27
492.43
1.0
2436.05
212.64
Table 5.1: Variation of dynamic loads for various values of deceleration
Graph 5.1: Variation of deceleration with front and rear dynamic load
Step 4: To find the maximum Frictional (Braking) Force :
From equations (5.3) and (5.4),
FR=µ×Rrdyn (5.11)
=0.6×772.23
FF = µ×Rfdyn (5.12)
=0.6×1876.47
Step 5: Design of braking components:
It is assumed that FF & FR values induce the braking system
( KfW+KaAfV2)+FF+FR=W×a
Where,
Kr =Co-efficient of rolling resistance
Ka=Co-efficient of air drag resistance
Af=Frontal area in m2
V=Velocity of the vehicle in m/s
Neglecting air and rolling resistance, we get
FF+FR=W×a
Disc:
A simple diagram of the drivers shown in figure 5.3
Braking torque = Fr×R=F1×r×2
µ×Rrdyn × R=F1×r×2
µ×Rrdyn× R=µ1×Pcal×Acal ×r×2
where,
R =Radius of the wheel, m
r =Effective radius of the is, m
F1 =Normal force acting on the disc, N
µ1=Co-efficient of friction between disc and brake pad
Pcal=Pressure developed in the caliper, N/m2
Acal=Area of the piston in the caliper, m2
Step 6: Calculation of braking torque (TB)
Radius of the wheel(R) = 11” = 0.2794 m
TB(r) = FR × R
= 463.34 × 0.2794
= 129.46 N-m
TB(f) = 1125.88 × 0.2794
= 314.57 N-m
Considering uniform wear rate theory,
R= (r1+r2)/2
Where
r1 is the inner radius in ‘m’
r2 is the outer radius in ‘m’
Braking condition is given by
µ×Rrdyn× R ≥ µ1×Pcal×Acal×r×2
If Pcal=0, then deceleration is zero.
When the vehicle is slowed down then
µ×Rrdyn× R – µ1×Pcal×Acal ×r×2 = I×α
I×α is the residual torque produced due to rotating parts
While braking I×α tends to zero as there is no angular acceleration.
Therefore, µ×Rrdyn× R – µ1×Pcal×Acal ×r×2 = 0
Considering caliper of bore diameter 1.12" with double piston :
Brake pad width = 25.4 mm
Disc diameter(d1) = 200 mm
r2 = disc radius - caliper width
= 74.6 mm
r= (r1+r2)/2 = (100+74.6)/2 = 87.3 mm
For the rear wheel, the normal braking force is given by
F1(r) = TB(r)/( µ1×r×2)
= 129.46/ (0.3×0.0872×2)
= 2474.39 N
Similarly for the front wheel,
F1(f) = 314.57/ (0.3×0.0872×2)
= 6012.42 N
Choosing master cylinder with bore diameter 1"
Area of the master cylinder = 506.71 mm2
Caliper bore diameter = 1.12" = 28.45 mm
Area of the caliper Acal = 635.7 mm2
Pressure developed in the caliper, Pcal = F1(f)/ (n× Acal)
where ‘n’ represents the number of pistons in the caliper
Pcal = 6012.42/(2×635.7)
= 4.73 N/mm2
Since Pmaster cylinder = Pcal
Force at the master cylinder/ Area of the master cylinder = 4.73 N/mm2
Force at the master cylinder = 4.73 × Area of the master cylinder
= 4.73 × 506.71
= 2396.74 N
From the
P1×l1=P2×l2
where
P1 =Pedal force, N
P2=Force exerted on master cylinder, N
l1&l2=Distance of the pivot from pedal and master cylinder respectively, m
Assuming the pedal force to be 250 N,
Pedal force(P1) × Mechanical leverage(l1/l2) = Force at the master cylinder(P2)
250 × (l1/l2) = 2396.74
= 2396.74/250
Mechanical leverage (l1/l2)= 9.5
Step6: Stopping distance ' S '
S=V2/2f
f =deceleration of the vehicle in 'm/s2'
V=Velocity of the vehicle in 'm/s'
If the vehicle moves at a speed of 55 kmph, then
S = V2/ (2 × 0.6g)
v = 55× (5/18) = 15.27 m/s
S = (15.27)2/ (2 × 0.6g)
S = 19.81 m
5.4. Summary:
In this chapter, the basic introduction about the braking system is discussed. The hydraulic braking system and its components were also discussed. Further the calculations that are essential for a braking system is derived and results are made. Graphs are also plotted for various decelerations (in g units) with corresponding dynamic loads for visual interpretation.
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