 |
|
|
1985-1998 [Subscribe to Daily Digest] |
Nathan,
The coefficient of friction, is 0.39 (for design purposes). And is not much affected by pressure or by velocity - which should not exceed 18 m/s. The maximum allowable temperature is 400oC. Below are some calc's on a prob from a year ago (or so)....(maybe help you?)
Prob#6)
A brake consists of two identical short shoes arranged and actuated as shown. The distance between drum and hinge centres is 240 mm.
Assuming a friction coefficient of 0.5, determine :
the total braking torque on the drum, To
the sensitivity S of each shoe
the load RH on each shoe hinge
the load supported by the drum shaft, RO
the brake's practicability given that the contact area of each lining is 100 mm2.
common shoe geometry - refer to short shoe sketch ( a) above
r = 150 mm ; a = 240 mm ; e = 150 mm ; qm = 70o ; so from ( 6a)
m = 1/0.5 x 240/150 x cos 70o = 3.007 ; and n = 1 - 240/150 x sin 70o = 0.4528
The "significant figures" quoted here for checking purposes are anything but - the friction coefficient accuracy is two sigfigs at the very most.
individual shoe parameters
LEFT SHOE RIGHT SHOE
Shoes are external so D = +1 leading so d = -1 trailing so
d = +1
P (N) 800 600
M = Pe (Nm) 0.15x800 = 120 0.15x600 = 90
(6a) N = M/mr( m -dDn) (N) 120/0.5x0.15( 3.007+0.453)
= 462 90/0.5x0.15( 3.007-0.453)
= 470
(vii) h = 1/( m -dDn) 1/( 3.007+0.453) = 0.289 1/( 3.007-0.453) = 0.392
(viii) S = mh 3.007x0.289 = 0.869 3.007x0.392 = 1.177
(5a) Fx = N( D cosqm -dm sinqm ) (N) 462(0.342+0.5x0.940)=375 470 ( 0.342-0.5x0.940)= -60
(5a) Fy = N(dm cosqm+ D sinqm ) (N) 462(-0.5x0.342+0.940)=355 470(0.5x0.342+0.940)=522
(5a) T = mNr or = hM (Nm) 0.5x462x0.15 = 34.7 0.5x470x0.15 = 35.2
So for the brake as a whole,
To = TL + TR = 34.7 + 35.2 = 69.9 Nm
Shoe free bodies with actuation, drum contact and hinge reaction components
RHx from S Fx = 0 (N) 800 cos40o -375 = 238 600 cos40o +60 = 520
RHy from S Fy = 0 (N) 800 sin40o -355 = 159 600 sin40o -522 =-136
RH by Pythagoras (N) 286 537
Drum free body with reactions to above shoe-drum contacts at drum centre, and horizontal & vertical components of drum shaft support
ROh from S Fhoriz = 0 (N) (522-355) cos30o - (60+375) sin30o = -73
ROv from S Fvert = 0 (N) (522+355) cos60o + (60-375) sin60o = 166
RO by Pythagoras (N) 181
If the brake is to be practicable then the average lining pressure pm = N/A ( 5a) cannot exceed a value which experience has shown to be feasible - around 700 kPa for a drum brake ( see earlier Rp table). The pressure resultant N is higher for the right shoe here so the highest average lining pressure in the brake is pm = 470/100 = 4.7 >> 0.7 MPa. The brake is clearly not practicable unless actuations and resulting braking torque are reduced to 0.7/4.7 = 15% of their values above.
Some brakes just don't have enough lining area to give useful braking effects with sustainable lining pressures, however they may be applicable to one-off emergency situations.
The left shoe is counter-actuating hence requires a larger actuation than the self-actuating right shoe for approximately the same braking torque.
If the left shoe's LH coordinate system causes difficulty in comprehension, then the brake should be viewed from the rear - the left shoe in the front view becomes the right in the rear view. This illustrates how the two opposed linings tend to balance one another so that the net load on drum shaft and bearings is small
posted by 24.48.24...
No Site Registration is Required to Post - Site Membership is optional (Member Features List), but helps to keep the site online
for all Saabers. If the site helps you, please consider helping the site by becoming a member.
 |
 |
 |
 |
 |
