1. EJERCICIO
¡ APROXIMADAMENTE !

2. 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

3. ( K + 3) x d (3.5 + 3) x 80 [mm]
2 2 I RECOMENDADA ≥ ≥
(SEGÚN PIRELLI)
= 260 [mm]
(D – d)
2 x I a = p – 2 x sen-1 = p – 2 x sen-1
(280 – 80)
2 x 260
= p – 2 x 0.395 [rad]
= 2.35 [rad]
dm
I = RC + dm+ rm
a ¡ SIEMPRE MEDIDO SOBRE LA POLEA MENOR !

4. I MÍNIMA =
(PEZZANO-KLEIN)
D + d 280 + 80
+ 50 [mm] = + 50 [mm] = 230 [mm]
2 2
50
I = RC + rm + 50

5. 75 x P 75 x 5 [C.V.]
Vm [m/s] 10.26 [m/s]
FT [kgf] = =
= 36.55 [kgf] ≈ 359 [N]
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)
TF = = =
= 116 [N]
mQ 0.6 x 2.35

6. DEL MISMO MODO, REEMPLAZANDO EN LA
EXPRESIÓN DE TT:
mQ
T= Tx e = 116 [N] x 4.1 = T F 475 [N]
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]

7. DEL MISMO MODO, REEMPLAZANDO EN LA
EXPRESIÓN DE TT:
mQ
T= Tx e = 116 [N] x 4.1 = T F 475 [N]
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]