Katarzyna Topczewska

Technical Issues
3/2016 pp. 113-119
ISSN 2392-3954

THE TEMPERATURE OF A BRAKE DISC DURING FRICTIONAL HEATING

WITH LINEAR DISTRIBUTION OF THE CONTACT PRESSURE

Katarzyna Topczewska
Bialystok University of Technology
Faculty of Mechanical Engineering
Department of Mechanics and Applied Computer Science
45A Wiejska Street
15-351 Białystok
e-mail: k.topczewska@o2.pl

Abstract: In this paper the analytical solution of the boundary–value heat conduction problem for a brake rotor was
developed. A solid brake disc is heated by frictional heat flux during braking with constant deceleration. Intensity of the
heat flux affecting friction surface of the disc is proportional to the specific power of friction. It was assumed that
contact pressure between the pad and the disc increases linearly, from zero in the initial moment of the braking process
to the maximum value in standstill. Calculations were carried out on variables and parameters in the dimensionless
form. The obtained results were compared with adequate resultsduring braking with constant deceleration, with an
assumption of pressure constant in time.

Keywords: contact pressure, frictional heating, heat conduction equation, temperature, friction, brake disc.

Introduction temperature on the pad/disc interface is formulation and

solving of friction heat problems. These are boundary–
Nomenclature: value heat conduction problems with defined heat flux
a – effective depth of the heat penetra-tion [m]; on the friction surface. The intensity of frictional heat
erf(x) – Gauss error function; flux is proportional to the specific power of friction
erfc(x)=1 – erf(x) – complementary error function; forces. It is equal to product of friction coefficient,
ierfc ( x )   1 2 exp(  x 2 )  xerfc( x ) – integral of the contact pressure and sliding speed [3]. In this study a
mathematical model of transient temperature field in a
complementary error function;
brake disc was developed using this method. It was
f – friction coefficient;
assumed that, the braking process proceeds with linear
K – thermal conductivity [W K-1 m-1];
distribution of pressure and constant deceleration.
k – thermal diffusivity [m2 s-1];
p – contact pressure [Pa];
q – intensity of the frictional heat flux[W m-2]; Statement to the problem
t – time [s];
ts – braking time[s]; Frictional heating of the contact surface of a brake disc is
T – temperature [K]; considered. The assumptions were applied:
T* – dimensionless temperature; - the material of the disc is homogeneous and isotropic;
T0 – temperature scaling factor [K]; - spatial distribution of the contact pressure on pad/disc
Ta – ambient temperature [K]; interface increases linearly:
V – velocity sliding [m s-1 ]; p(t )  p0 p  (t ), p  (t )  t t s , 0  t  t ;
s
(1)
τ – Fourier number; - braking proceeds with constant deceleration i.e.
ζ – dimensionless spatial coordinate. velocity of the vehicle decreases linearly from the
maximal value V0 in the initial time t = 0 , to zero in the
In the mechanical brakes large amount of heat is retention time t = ts:
generated due to friction. High temperature on the
V (t )  V0V * (t ), V  (t )  1  t / t s , 0  t  t s ; (2)
contact surface causes changes of thermophysical
properties and deterioration of frictional couple, which - the sum of heat fluxes intensities directed
results in reduction of the braking efficiency [8]. The perpendicularly from the friction surface to the inside of
temperature is a main parameter for the assessment of the brake disc, is equal to the specific power of friction
abrasive wear and thermal stability of friction systems. forces:
Therefore, knowledge of the temperature field is a q(t )  q0 q * (t ), 0  t  t s , (3)
priority in the brake system design process. One of the where:
effective methods used to predict the maximal (4)
q 0  fp 0V0 ,

113 Technical Issues 3/2016

 THE TEMPERATURE OF A BRAKE DISC DURING FRICTIONAL HEATING WITH LINEAR …

q * (t )  p * (t ) V * (t ), 0  t  t s ; (5) semi–space in the Cartesian coordinate system Oxyz

(Fig. 1).
- gradient of temperature in radial and circumferential
 2T ( z , t ) 1 T ( z , t ) 0  z   , 0  t  t , (6)
directions are negligible;  , s
- the free surface of the brake rotor is adiabatic; z 2 k t
- in the initial time of the braking process the T ( z , t )
K  q (t ), 0  t  t s , (7)
temperature of the disc is constant, equal to Ta. z z 0
According to the above assumptions the transient T ( z, t )  0, z  , 0  t  t s , (8)
temperature field of the disc is one–dimensional. To find
T ( z ,0)  Ta , 0  z  , , (9)
the distribution of the temperature, we had the following
parabolic heat conduction problem boundary–value for where intensity of the heat flux q(t) has form (3–5).

Fig. 1. Scheme of the problem.

Applying the following dimensionless variables and   ( 0)

parameters: T  ( , )   q  ( s ) T ( ,  s )ds ,   0, 0     s , (16)
0 
z
  ,   kt2 ,  s  kt s , T0  q0 a , T *  T  Ta , (10) where [2]
a a a2 K T0   ,
T (0) ( , )  2  ierfc    0 , 0     s , (17)
( a  3kt s – effective depth of the heat penetration 2  
inside the brake disc [4, 5]), considered boundary–value is solution to the problem (11)–(15) with constant
problem (6–9) was written in dimensionless form: intensity of heat flux q*(τ) = 1 in boundary condition
 2T * ( , ) 1 T * ( , ) 0   , 0     s , (11) (12).
 ,
 2 k2  Substituting function q*(τ) (15) and solution (17) in
*
T ( , ) equation (16), we have:
 q * ( ), 0     s , (12)

 s  s       ,
 0 T * ( , )   1   2   s ierfc ds
* 0 s   s     2   s 
T ( , )  0,   , 0     s , (13)
0     s . (18)
*
T ( ,0)  0, 0   , (14)
Taking into account the value of the following derivative
where [1]:
    d 2x  x2 ,
q * ( )  1   , 0     s . (15) erfc( x)   e (19)
s  s  dx 
derivative of integral of the complementary error
Solution and verification to the problem function was counted:
d 2 x  x2 d
[ierfc( x )]   e  erfc( x)  x erfc( x)  erfc( x) .
Solution to the formulated boundary-value problem of dx  dx
heat conduction (11–15) was found based on Duhamel's (20)
theorem [6], which for the considered case can be Using relation (20), we achieved:
written in the form:
2
  
   
2   
     1  1  2 
  s          e  2  s  . (21)
 2   s ierfc
 
   e  erfc   erfc  
   2   s   s   2 τs  2 τ  s  2(  s) 2  s   (  s)
 

Technical Issues 3/2016 114

 Katarzyna Topczewska

Substituting the partial derivative (21) to the formula Using the method of substitution the formula (25) was
(18), we have: received:
1 1  1 
T * ( , )  I1 ( , )  2 I 2 ( , ) , (22)
x 
s s    s  2 (26)
J 1 ( , )    L2 ( , ),
where ds  2  
dx
  
2  x3 
  
1 s where
I1 ( , )   e  2  s  ds ,
 0  s  
2
   x 2
2 dx
 


 L2 ( ,  )   e 2 . (27)
1  s2 x2
I 2 ( , )   e  2  s  ds . (23) 1
 0  s 
Subsequently, function I1(ζ,τ) was presented in the Counting the integral (27) [7]:
following form: 
 1 (  x )2    
I1 ( , )  J1 ( , )  J 2 ( , ) , (24) L2 ( , )   e 2   erf  x  ,   0 , (28)
where  x 2  2 
  1
  
2 
 
  1
J1 ( , )   e  2  s  ds ,
 0  s from equation (26) it was found:
2
  
   
1
J 2 ( , )    se  2   s  ds . (25)
 0

 2
 2   
   
.
2  1  ( 2 x )    2   2         (29)
J1 ( , )   e   erf  x   e  2 erfc   2  ierfc 
  x 2  2  1  2 2   2  
 


Function J2(ζ,τ) (25) was found in the similar way: Exploiting the following recurrence relation [7]:
2 2 2
e ( ax ) e ( ax ) 2a 2 e ( ax )
 1  L p ( , )   dx     dx ,
(n  1) x n1 n  1 x n2
2
 x    2
  s  2   2  x dx 2
xn
J 2 ( , )     e 4
 L4 ( , ). a  0 , n  2,3,..., (31)
ds  2   1 x 
dx
 x3  

(30)

and relation (28), we achieved:

 
 2    
2 
  2   x 2
  2  x 2
 2
  2   2 2  (32)
 e 2     1  2  x      2       e 
L4 ( , )   3
     e     erf  x          x 1 ierfc x    .
 3 x 3  2  
x 2 2    3  2   2 2  3x 3 
    
 1
  1
 
Taking into account function L4(ζ,τ) (32) in formula (30),
we obtained:

  
2 
 x 
 2  2    2
 
(33)
2 2     1    e  2          
J 2 ( ,  )        ierfc  x         1  2    ierfc    erfc   .
 3  2 2 x  2  3x 3  3   2   
 2   2  2   
  1

Substituting function J1(ζ,τ) (29) and J2(ζ,τ) (33) into the
right side of equation (24), we received:
2       .
2
         (34)
I1 ( , )  2 1    ierfc    erfc 
3    2    2   2   2  
  

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 THE TEMPERATURE OF A BRAKE DISC DURING FRICTIONAL HEATING WITH LINEAR …

Function I2(ζ,τ) (23) was written as a difference of the   .

integrals: J 3  ,   2 2  ierfc 
(37)
2  
I 2 ( , )  J 3  ,   J 4  ,  , (35)
To find integral in equation (36), the method of
where substitution was used. As a result we obtained:
2
    1  2
 x 
 
2  1   s    s 
 
  x 2
J 3  ,    e 2 ds , J 4  ,   
2 
 2
 2
e  
dx (38)
 0  s   2  x .
ds  2   1 x4
2 dx
    x3  
1  (  s )  s    
J 4  ,    e  2   s  ds . (36) Function J4(ζ,τ) (38) was written in the form:
  s 4 2
0
J 4  ,   L4  ,   L6  , . (39)
 
Based on relation (29), we can write: Based on equations (31) and (32) it was found:


2
2  
 
  x 2   x 2
 e 2 2 2
e 2 
L6  ,    6
dx   5
   L4  ,  
1 x 5x 5 2 

1

   
2
2    2
  
2
       
1  (40)
 2  exp           1  2    ierfc   erfc  .
5   2  
  15  2    2    2   2  2   
Taking into account relations (32) and (40) in formula (38),
it was counted:
2  
4 2   1     2  2 
2    2
          
J 4  ,   1     1  2    ierfc   erfc    exp     . (41)
3  5  2     2    2   2  2    5   2   
     
Substituting functions J3(ζ,τ) (37) and J4(ζ,τ) (41) to the right
side of equation (35), we received:
           (42)
2 2 4 2
2            
I 2 (  ,  )   ierfc    8  18    4       erfc    7  2     .
15  2    2   2     2   2    2    
In regard to functions I1(ζ,τ) (34) , I2(ζ,τ) (42) and relation
(22), we found the dimensionless temperature field:
2      
2
         
 
T * ( ,  )   2 1     ierfc      erfc    
3 s    2    2   2   2   

2 2     
4

2 2
                  0 , 0     s . (43)
  8  18    4    ierfc       7  2   erfc   ,
15 s2   2   2   
 2   2   2   
 2   


To verify the correctness of the obtained solutions (43)     1   ,

ierfc    erfc 
the accordance of the boundary (12), (13) and initial (14)  2   2  2  
conditions with achieved temperature fields were 2
  
examined. First, the accordance of boundary condition   
d     
(12) was checked, which defines thermal load on the erfc    e 2   . (44)
d 2   
outer surface of the strip ζ = 0. For this purpose, the
derivative of complementary error function (19) and the Differentiating solution (43) with respect ζ, with regard
derivative of its integral (20) were taken into account, we to the derivatives (44), was found:
achieved:

    
2
   1 3    
2
T * ( , ) 2   2        2  2   
   ierfc       erfc  e 
 3  s    2   2     2  2     2   2  
 
   
2
2 2   1      2       2 
2 2
   8     
  15  18     6    erfc   7  2   e  (45)
        .
15 s2  2    2     2     2   2    2    
 

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 Katarzyna Topczewska

Substituting in equation (45) ζ = 0, we received: which shows the fulfillment of the condition (13).
T * ( , )  2     Substituting τ = 0 in the solution (43) we can easy
   2   1  , (46) compute:
 s s s  s 
 0
T * ( ,0)  0 , (50)
which provides the fulfillment of the boundary condition
in this way, conformity of the result (43) with the initial
(12), in regard to the form of function q*(τ) (15).
condition (14) has been proved.
Taking into account, that [1]:
lim erfc( x)  0 , (47)
x Numerical analysis
we obtained:
 e  x2  Numerical analysis was carried out based on the devised
(48) solution (43), which describes transient temperature field
lim ierfc( x)  lim   x erfc( x)  0.
x  x    
  in a brake disc heated by frictional heat flux with linearly
increasing contact pressure. The results were compared
The limit of the solution (43) as ζ approaches infinity with the following, adequate results, which were
(ζ→ ∞) in regard to value of the limits (47)–(48), was received in the article [9], with an assumption of
calculated: constant pressure:
lim T * ( , )  0 , (49)
x 

 2  2    2
 
 
T * ( , )  2  ierfc   

 1  


 
 ierfc 
 1 
  
  
 erfc    ,   0 , 0  s . (51)
2   s  3  2    2   3 2   2   
 

During braking with linear distribution of contact field in the case with constant pressure, the value p*=1/3
pressure (1) and linearly decreasing sliding velocity (2), was adopted.
dimensionless total amount of thermal energy directed to The input dimensionless conditions used to numerical
a brake disc is: analysis were: distance from friction surface ζ and
 2
s Fourier numbers (dimensionless time) τ and τs. It was
s s
     3  1
Q    q  ( ) d     1   d       . assumed that, dimensionless braking time is τs=1.
0 s s   2 2  s
0  s 3 s  0 6 Evolution of the dimensionless temperature T* in time,
(52) on few depth ζ was presented in Fig. 2. At the beginning
Whereas, during braking with constant pressure of the process the temperature increases, attains the
p*(τ) = p* > 0 and constant deceleration this value is maximum value and decreases until the moment of
equal: standstill. The maximum dimensionless temperatures
s T*=0,425 (Fig. 2a) and T*=0,177 (Fig. 2b) are reached
 
s s     2
1
   
Q   q ( ) d   p   1  d   p      p  s. on the contact surface ζ = 0, exactly in the half of
0   s  
 2 s  2
0
0 braking time (τ = 0,5). The maximum temperature
(53) achieved in the braking with linearly changing pressure
Taking into account (52) and (53), to maintain the same is 140% higher than in the braking with constant
total amount of heat, during calculations of temperature pressure.

a) b)
Fig. 2. The evolution of the dimensionless temperature T* versus time τ with respect to braking time τs, on several distances of
friction surface ζ: a) linearly increasing contact pressure; b) constant pressure.

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 THE TEMPERATURE OF A BRAKE DISC DURING FRICTIONAL HEATING WITH LINEAR …

Cooling of the outer surface of the disc after maximum number τ were shown in Fig. 3. The temperature
temperature has been reached, in the case with linear monotonically decreases with increasing of the distance
distribution of pressure is more intense than with from the heated surface. The largest gradient between
constant pressure. The temperature is lower and the time temperature on the friction surface ζ = 0 and at the
to reach maximum temperature value increases with distance ζ = 1,5, occurs in the half time of braking
increasing distance from the friction surface to the center process τ = 0,5 (Fig. 3). The mentioned gradient reaches
of the disc. In the braking with constant pressure minimum value in the stop moment τ = τs = 1. Effective
maximum temperatures on particular distances ζ are depth of the heat penetration, i.e. distance from friction
achieved with higher time offset in the standstill surface, on which the temperature achieved 5% of the
direction. Monotonically temperature increases during maximum value on heated surface. During braking with
the entire braking process: with linear distribution of the linearly increasing pressure, effective depth of the heat
pressure it takes place under the distance ζ = 1,5, while penetration is equal ζeff = 1 (Fig. 3a) and the temperature
with constant pressure – under the distance ζ = 1. decreases more rapidly (Fig. 3a). Whereas, during
Distribution of the dimensionless temperature T* versus braking with constant pressure the mentioned depth has a
dimensionless depth ζ in several values of Fourier higher value ζeff > 1,5 (Fig. 3b).

a) b)

Fig. 3. Distribution of the dimensionless temperature T* inside disc at few different dimensionless time moments τ: a) linearly
increasing contact pressure; b) constant pressure.

Conclusions - effective depth of the heat penetration is higher during

braking with constant pressure;
Conducted analysis demonstrates, that: - the time to reach maximum temperature value increases
-with equal value of total amount of heat, during braking with increasing distance from the friction surface to the
with constant pressure reached temperatures near the center of the disc. In the braking with constant pressure
working surface are significantly lower. Adjacent the the maximum temperatures on particular distances ζ are
distance ζ = 1, obtained temperatures in both cases are achieved with higher time offset in the standstill
equal. Below this depth, the temperatures achieved in direction;
case of constant pressure are higher; - the temperature increases monotonically during the en-
- cooling of the heated surface during braking with linear tire braking process with linear distribution of the
distribution of the pressure is more intensive; pressure which takes place on the distance   1,5 , while
with constant pressure – on the distance   1 .

References

1. Abramowitz, M., Stegun, I.A., Handbook of Mathematical Functions with Formulas, Graphs, and Tables, National
Bureau of Standards, Washington, 1972.
2. Carlslaw, H.S., Jaeger, J. C., Conduction of Heat in Solids, 2nd ed.Clarendon Press, Oxford, 1959.
3. Jewtuszenko, O., (red), Analityczne i numeryczne modelowanie procesu nieustalonej generacji ciepła w elementach
tarciowych układów hamulcowych, Oficyna Wydawnicza Politechniki Białostockiej, Białystok, 2014.

Technical Issues 3/2016 118

 Katarzyna Topczewska

4. Kuciej, M., Zagadnienia cieplne tarcia dla układu warstwa-podłoże, Rozprawa doktorska, Politechnika Białostocka,
Białystok, 2007.
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Białostockiej, Białystok, 2012.
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Breach: New York, 1986.
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Technologii Eksploatacji, Radom, 1998.
9. Topczewska, K., Influence of the protective strip properties on the distribution of the temperature in brake disc. II –
Braking with constant deceleration, Zagadnienia aktualne poruszane przez młodych naukowców 3, s. 415-420,
Creativetime, Kraków, 2015.

119 Technical Issues 3/2016