Problemas De Embragues y Frenos Faires

SECTION 16 – BRAKES AND CLUTCHES

ENERGY TO BRAKES 881. A motor operates a hoist through a pair of spur gears, with a velocity ratio of 4. The drum on which the cable wraps is on the same shaft as the gear, and the torque cause by the weight of the load and hoist is 12,000 ft-lb. The pinion is on the motor shaft. Consider first on which shaft to mount the brake drum; in the process make trialcalculations, and try to think of pros and cons. Make a decision and determine the size of a drum that will not have a temperature rise greater than ∆t = 150o F when a 4000-lb. load moves down 200 ft. at a constant speed. Include a calculation for the frp/sq. in. of the drum’s surface.

Solution: Consider that brake drum is mounted on motor shaft that has lesser torque. 12,000 ft − lb T f= = 3000ft − lb = 36,000 in − lb 4 From Table AT 29, Assume f = 0.35 , p = 75 psi , max. vm = 5000 fpm
Tf = FD 2

F = fN = N= 2T f

2T f D

fD N p= A A = π Db p= 2T f N 2(36,000 ) = = = 75 2 π Db π D bf π D 2b(0.35)

D 2b = 873 use D 2b = 873 873 b= 2 D Then, U ft − lb ∆t o F = f Wm c Assume a cast-iron, ρ = 0.253 lb in3 c = 101 Wm = ρ V

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SECTION 16 – BRAKES AND CLUTCHES

D2   D 2t = π t  Db +  4 4    U f = (4000 )(200 ) = 800,000 ft − lb V = π Dbt +

π

∆t = 150o F

c∆t 800,000 0.253V = (150)(101) V = 208.7 in 3 But  D2   V = π t  Db +  4    873 b= 2 D  873 D 2  V = π t  D + 4     For minimum V : dV  − 873 D  = π t 2 +  = 0 dD 2  D 3 D = 2(873) D = 12 in For t :  873 (12 )2  V = 208.7 = π t  +  4   12 t = 0.611 in 5 say t =in 8
873 1 = 6.0625 in = 6 in 2 16 (12 ) 5 1 Therefore use D = 12 in , t = in , b = 6 in 8 16 b=

Wm = ρ V =

Uf

For fhp sq. in. = fhp = Fvm 33,000

fhp A

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SECTION 16 – BRAKES AND CLUTCHES 2(36,000) = 6000 lb D 12 vm = 5000 fpm (max.) (6000)(5000) = 909 hp fhp = 33,000  1 A = π Db = π (12 ) 6  in 2  16  fhp 909 fhp sq. in. = = = 3.98 (peak value) A 228.55 F= 2Tf = 882. A 3500-lb. automobile moving on level ground at 60 mph, is to be stopped in a distance of 260 ft. Tire diameter is 30 in.; all frictional energy except for the brake is to be neglected. (a) What total averaging braking torque must be applied? (b) What must be the minimum coefficient of friction between the tires and the road in order for the wheels not to skid if it is assumed that weightis equally distributed among the four wheels (not true)? (c) If the frictional energy is momentarily stored in 50 lb. of cast iron brake drums, what is the average temperature rise of the drums?

Solution: (a) Solving for the total braking torque. W 2 U f = − ∆KE = vs − vs22 2g 1 W = 3500 lb vs1 = 60 mph = 88 fps

(

)

vs2 = 0 mph = 0 fps

g = 32.2 fps 2 3500 Uf = (882 − 02 ) = 421,000ft − lb 2(32.2) (T ft − lb )ωm = (T f in − lb )n fhp = f 33,000 63,000 2 2 2 vs − vs1 0 − (88) a= 2 = = −14.892 fps 2 2s 2(260) vs − vs1 0 − 88 t= 2 = = 5.91 sec a − 14.892 U − ∆KE 421,000 fhp = = f = = 130 hp (t )(550) 550t 550(5.91)

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SECTION 16 – BRAKES AND CLUTCHES 1 (88 fps )(60 sec min ) vm 2 n= = = 336 rpm πD  30  π  ft   12  T n fhp = f 63,000 63,000(130 ) Tf = =24,375 in − lb 336 (b) f =
F N 3500 = 875 lb 4

for each wheel, N =
Tf =

24,375 = 6094 in − lb 4 2T 2(6094) F= f = = 406 in − lb D 30 F 406 f = = = 0.464 N 875

(c) ∆t =

Uf

Wm c U f = 421,000 ft − lb Wm = 50 lb c = 101 ft − lb lb − F for cast-iron 421,000 ∆t = = 83.4o F (50)(101)

884.

An overhead traveling crane weighs 160,000 lb. with its load and runs 253 fpm. It is driven bya 25-hp motor operating at 1750 rpm.The speed reduction from the motor to the 18-in. wheels is 32 to 1. Frictional energy other than at the brake is negligible. (a) How much energy must be absorbed by the brake to stop this crane in a distance of 18 ft.? (b) Determine the constant average braking torque that must be exerted on the motor shaft. (c) If all the energy is absorbed by the rim of the...