Select salient dimensions for a drum brake whose lining temperature is estimated to be 200^oC. The lining wear constants are   R_wo = 8.0 mm³/MJ ;   T_o = 185 ;   n = 1.78.
The normal intermittent duty is :-

energy dissipated	:	7.5 kJ /application (which includes a design factor)
average speed	:	240 rpm   (during braking)   ( rpm ≡ rev/min )
braking time	:	0.25 s / application
required lining life	:	1.6 E6 applications
For one application	:	mean power dissipation rate ( 3)	:	7.5/0.25 = 30 kW
		Rp from the above table	:	600 kW/m2
		so, minimum req'd lining area ( 4)	:	30/600 = 0.05 m2
Over the whole life	:	total energy dissipated	:	7.5 E-3 ∗ 1.6 E6 = 12 E3 MJ
		Rw from (ii)	:	8.0 exp((200/185)1.78 ) = 25 mm3 /MJ
		so volume lost over the life	:	12 E3 ∗ 25 = 300 E3 mm3
		and thickness = volume/area	:	300 E3/0.05 E6 = 6.1 mm
This is increased by 100% to allow for possible uneven wear and fixing with rivets - so required thickness is 12 mm. Note that if the temperature were to increase by only 60oC, then the thickness would have to be doubled.
Over the braking period	:	displacement of drum from (1)	:	(240/60) ∗ 0.25 = 1 rev
		braking torque from (2)	:	7.5/2π = 1.2 kNm

The drum diameter   D and lining width   w may be estimated as above. Thus with two 100^o extent shoes :
      A   =   w ( radius ∗ lining extent )   =   w D ∗ 5π/9       and eliminating w
      3   ≤   D/w   =   5π D² / 9 A   ≤   10       leads to     0.29   ≤   D   ≤   0.53     say   D = 400 mm

The mean pressure   p_m may be estimated from ( i), assuming   m = 0.35 say.
      v_m   =   ( π D N )_m   =   π 0.4 (240/60)   =   5.0 m/s        as always, keep track of the units !
      p_m   =   600 / 5.0 ∗ 0.35   =   340 kPa   which is acceptable by the above table.