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ELEMENTOS DE MÁQUINAS II EQE A S UM
• Frenos (Guía 13)

1. Freno de Yugo de Zapata circular 2. Frenos de tambor con zapatas internas largas

187

ELEMENTOS DE MÁQUINAS II F ns re o

PresiónUniforme

P = Pmáx dFa = Pmáx = r ⋅ dθ ⋅ dr Fa =
ro θ 2 ri θ 1

∫ ∫ Pmáx ⋅ r ⋅ dθ ⋅ dr = Pmáx ⋅ (θ 2 − θ1 ) ∫ r ⋅ dr = Pmáx (θ 2 − θ1 )
ri ro θ 2

ro

(ro2 − ri 2 ) 2

dT f = µ ⋅ Pmáx ⋅ r⋅ dθ ⋅ dr ⋅ r → T f = T f = µ ⋅ Pmáx ⋅ (θ 2 − θ1 )
Desgaste Uniforme
W P

P ⋅ V P ⋅ r Px ⋅ m á ⋅ r = P m á x ⋅ r → P = i r Px ⋅ r m á i d F = ⋅ r ⋅⋅ d d r = Px a m á r F = Px ⋅ r ( − ( r − r ) am á i 2 1 ) o i d T
f

T

f

= ⋅ ⋅ Px r ⋅ ( m á i 2

αα θ θθ θ µθ θ µ
ri 1

∫ θ∫ µ ⋅ P

máx

⋅ r 2 ⋅ dθ ⋅ dr

(r − ri ) 3
3 o 3

P m á x = ⋅ r

⋅ r i ⋅ r ⋅⋅ d d r

(r 2 o − 1)

1. Freno de Yugo de Zapata circular

Fa = Pprom ⋅ π ⋅ R 2 → T f = Fa ⋅ µ ⋅ re = µ ⋅ Pprom ⋅ π ⋅ R 2 ⋅ re

188

ELEMENTOS DE MÁQUINAS II

R/e 0 0,1 0,2 0,3 0,4 0,5

δ=re/e 1 0,983 0,9690,957 0,947 0,938

Pmáx/Pprom 1 1,083 1,212 1,367 1,578 1,875

2. Frenos de tambor con zapatas internas largas

Suposiciones 1) La presión en cualquier punto de la Zapata se supone proporcionala la distancia desde el pasador de la articulación siendo nula en el talón. 2) Se ignora el efecto de la fuerza centrifuga.

189

ELEMENTOS DE MÁQUINAS II
3) Se supone una zapata rígida. 4)Todo el análisis tendrá como base un coeficiente de fricción que no varia con la presión.

p = K ⋅ sen θ ⇒ K =

p sen θ

p máx p = , donde θ máx = mín( θ 2 ,90 0 ) sen θ sen θ máx

pmáx ⋅ senθ ⋅r ⋅ b ⋅ dθ senθ máx p ⋅ senθ dF f = µ ⋅ dN = µ ⋅ máx ⋅ r ⋅ b ⋅ dθ senθ máx dN = p ⋅ dA = p ⋅ r ⋅ dθ ⋅ b =
p máx ⋅ r ⋅ b ⋅ a p ⋅ r ⋅ b ⋅ a θ2 ⋅ sen 2θ ⋅ dθ ⇒ M N = máx ⋅ ∫ sen 2θ ⋅ dθ θ1 senθ máx senθmáx
θ

dM N = dN ⋅ a ⋅ senθ =
MN =

p máx ⋅ r ⋅ b ⋅ a  θ sen 2θ  2 ⋅ −  sen θ máx 4 θ1 2
= dF f (r − a ⋅ cos θ)

dM

F

dM F = µ ⋅

pmáx ⋅ r ⋅ b ⋅ ( r ⋅ senθ − a ⋅ cosθ ⋅ senθ...
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