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]