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### Transcript of calculo brazo hidraulico

MOTORES BRAZO MECANICO MOTOR GRIP 0.07 A 12 V 320 RPM P=Vi P=12 V 0.07 A P=0.84 Watts 1W=1.341x10-3hp= 0.001126 hp63000 ( hp ) 63000 (0.001126 hp ) = = 0.22 N m rpm 320 .8rpm = 70 % 0.00126 = pot .desalida pot .desalida = 0.00782 hp T =

MOTOR BANDA 0.05 A 20 V 100 RPM P=Vi P=20 V 0.05 A P=1 Watts 1W=1.341x10-3hp= 0.001341hp63000 ( hp ) 63000 (0.001341 hp ) = = 0.84 N m rpm 100 rpm = 70 % 0.001341 hp = pot .desalida pot .desalida = 0.0009387 hp T =

MOTOR GRANDE 1.25 A 5V 16.5 RPM P=Vi P=5 V 1.25 A P=6.25 Watts 1W=1.341x10-3hp= 0.00838125 hp63000 (hp ) 63000 (0.00838125 hp ) = = 32 .00018 N m rpm 16 .5rpm = 70 % 0.00838125 hp = pot .desalida pot .desalida = 0.005866875 hp T =

MOTOR V 1.00 A 12 V 42.85 RPM P=Vi P=12 V 1.00 A P=12 Watts 1W=1.341x10-3hp= 0.016092 hp63000 ( hp ) 63000 (0.016092 hp ) = = 23 .6591 N m rpm 42 .85 rpm = 70 % 0.016092 hp = pot .desalida pot .desalida = 0.0112644 hp T =

MOTOR P 0.3 A

12 V 42 RPM P=Vi P=12 V 0.3 A P=3.6 Watts 1W=1.341x10-3hp= 0.0048276 hpT = 63000 ( hp ) 63000 (0.0048276 hp ) = = 7.24 N m rpm 42 rpm

= 70 % =0.0048276 hp pot .desalida pot .desalida = 0.00337932

hp

CABLE Cable = T 0.84 lb max = maxf max 4 3.1281 cm = (2.44 ) 2 / 4

= 77 .0 max = max

L =b c l ae 2 .8" 4 L = .69 0 24 m

3.1281 cm + (0.6294 x 77 ) (2.44 ) 2 / 4 3.128 = + 48 .46 = 48 .62 N 18 .70

MOTOR DEL CABLE. (MOTOR V) T =23.659 Nm Eje 42.8 RPM 0.016092 Hp

0.0004023 Hp 42.8 RPM x 2 =268.92 rad/min W= 4.48 rad/seg Op=0.25X P=1.57 1.12in/seg 0.0015in/sHp = fv 0.01126 = f = 0.84 lb 0.0015

0.001126 Hp 0.03Hp 11.88 Rpm Relacin 27=1 0.81 2.223 FUERZA GRIP

63000 (0.03 ) 11 .88 T =159 .09 N m T =

= n38 k (0.0144 ) = 0.00532 K m gF

T=7.24 Nm =0.00337432 Hp V=wr w= rad seg 1 RPM = 2 rad

L L

= + " +++ (0 5.4 1 .6 1 4 .2) 1 0 .6 1 4 .2 3.2 = 1 9 .0 5 4 "

2( 42 ) = 263 .89 rad / m in

Relacin de Catarina

W = 4.39 rad / seg

H p V 0.0 3 7 3 0392 F = =0.0 6 lb 14 0.2 6 A / seg 0 t =0.0 6 5 3 1 5 kg F P n d o u to elm to r F =

1.64 a 42 RPM Nota: La velocidad aumenta pero se disminuye el torque.D = 1 .1 2 6 " R = 0 .5 6 3 " P = 2 r

P = 2(o.563 ) P = 3.5374

V =148 in / m in V = 2.476 in / seg

PUNTO B 0.036155 1= 0.036158 kg F 0.48X0.014472 Estticamente Fuerza Mxima a 0 o 180

F(d)F = 0.014472 Kgf m D = 0.48 m

=19 .054 " LL =0.48 m

TORQUE DE LOS EJES Eje motor P = T = 7.24 Nm

ESFUERZO DE CORTE DEL MOTOR P

T motor = 32.00018 RPM = 16.5 RPM Hps = 0.005866875 Hp Hp = fv

p = 2r p = 1.4(2 ) p = 8.7"

16 .5 =1.89 in / m in 8.7 V = 0.031 in / seg V = 0.00263 ft / seg V =

H p 0.005166 = v 0.00263 F = 2.22826 F =1.0092 KgF F =

TORQUE DEL PUNTO X Motor T= 32.00018 TR = 50.24 Relacin de Catarinas 1.57 a 1

x = (11.505 2 + (5.75) 2 ) x = 165.42 x = 12.86"

F =1.0092 kgF

f a ( ) + FB ( ) + C = 0 1.0092 = 0.03 31 .91

FUERZAS DE CORTE DEL PUNTO X

TORQUE CON REDUCCIONES DE LA BASE