EJERCICIO
¡ APROXIMADAMENTE !
K = 3.5 = = dm = DC / 3.5 = 280 / 3.5 = 80 [mm]
nC = nm / 3.5 = 2 450 / 3.5 = 700 [r.p.m.]
VTGC =
= =
= 10.26 [m/s]
DC nm
dm nC
p x dm [mm] x nm [r.p.m.] 60 x 1 000
p x dm [mm] x nm [r.p.m.] p x 80 x 2 450 60 x 1 000 60 000
( K + 3) x d (3.5 + 3) x 80 [mm]
2 2I RECOMENDADA ≥ ≥ (SEGÚN PIRELLI)
= 260 [mm]
(D – d)
2 x I a = p – 2 x sen-1 = p – 2 x sen-1
= p – 2 x 0.395 [rad]
= 2.35 [rad]
(280 – 80)
2 x 260
dm
I = RC + dm+ rm
a¡ SIEMPRE MEDIDO SOBRE LA POLEA MENOR !
I MÍNIMA = (PEZZANO-KLEIN)
D + d 280 + 80 + 50 [mm] = + 50 [mm] = 230 [mm] 2 2
50
I = RC + rm + 50
FT [kgf] = =
= 36.55 [kgf] ≈ 359 [N]
75 x P 75 x 5 [C.V.] Vm [m/s] 10.26 [m/s]
De la expresión de Prony, y sabiendo que FT = TT – TF :
e = TT = TF x e FT = TF x (e - 1)mQ TT
TF
mQ mQ
FT 359 [N] 359 [N]
(e - 1) (e - 1) (4.1 – 1)
= 116 [N]
TF = = = mQ 0.6 x 2.35
TT = TF x e = 116 [N] x 4.1 = 475 [N]mQ
DEL MISMO MODO, REEMPLAZANDO EN LAEXPRESIÓN DE TT:
SE VERIFICA QUE:
FT = TT – TF = 475 [N] – 116 [N] = 359 [N]
LA FUERZA DE PRETENSADO RESULTA:
2 x T0 = TT +TF = 475 [N] + 116 [N] = 591 [N]
Y LA RELACIÓN DE TIRO y :
y = = = 0.61 61 % DE EFICIENCIA FT 359 [N]
2 x T0 591 [N]