Grua

Cálculo de la caja reductora:

MTrueda = 71620 × Pot / rpm
MTrueda = 71620 × 10,244 / 47,7
MTrueda = 15381,033[Kg·cm]
α = 7°

i = ηt / ηr
i = 2930[rpm] / 47,7[rpm]
i = 61,42
i = 60

Zrueda = 60
m = 1

Mtornillo = 71620 × Pot / rpm
Mtornillo = 71620 × 10,244 / 2930
Mtornillo = 250,40[Kg·cm]

Mn rueda = 8 × 3√[Mt rueda / (Z × σadm)]
Mn rueda = 8 × 3√[15381 / (60 × 1600)]
Mnrueda = 4,3[mm]
Mn rueda = 4,5[mm]

Mn tornillo = 8 × 3√[Mt tornillo / (Z × σadm)]
Mn tornillo = 8 × 3√[250,40 / (1 × 2200)]
Mn tornillo = 3,8[mm]
Se adopta el módulo mayor, por lo tanto tomamos 4,5[mm] al módulo de la relación de transmisión.

Dp = M × Z
Dp = 4,5[mm] × 60
Dp = 270[mm]

Dext = Dp + 2[mm]
Dext = 270 + (4,5 × 2)
Dext = 279[mm]

Dpt = (Mnt × m) / senα
Dpt = (4,5[mm] ×1) / sen7°
Dpt = 36,924[mm]

De = Dpt + 2 × Mnt
De = 36,924[mm] + 2 × 4,5[mm]
De = 45,924[mm]

Cálculo del carro transportador:

Carga sobre cada rodamiento:
Referencias:
Q = carga útil
G = Peso del aparato elevador
n = cantidad de ruedas
k = coeficiente 0,3[Kg/mm2]
Rv = radio de la rueda

q = (Q + G) / n
q = (4000[Kg] + 30[Kg]) / 4
q = 1007,5[Kg]

bv = 39,4 → 35[mm] = Anchode contacto

Rv = q / (2 × bv × k)
Rv = 1007,5[Kg] / (2 × 35[mm] × 0,3[Kg/mm2]
Rv = 47,98[mm]




Peso del cable:
Peso por metro lineal = 0,21[Kg]
Peso total = 0,21 × 24
Peso total = 5,04[Kg]

Peso de la polea = 1,5[Kg] c/u → 1,5 × 6 = 9[Kg] Peso de las 6 poleas

Gancho = 5[Kg]
σadm = F / A
σadm = 5[Kg] / 1,5

A = 11173,036[Kg] / 12000[Kg/cm2]
A = 9,311[cm2]

Calculo de lapotencia del motor.

Datos:
M = 11000[Kg]
Vf = 10[Km/h] ≡ 2,7[m/s]
V0 = 0
∆x = 15[m]
∆L/∆t = ?

F × ∆x = ½ × m × Vf2 – ½ × m × V02
F = (½ × m × Vf2 – ½ × m × V02) / ∆x
F = (½ × 11000[Kg] × (2,7[m/s])2 – ½ ×11000[Kg] × 0) / 15[m]
F = 2829,218[N]

F × ∆t = m × Vf - m × V0
∆t = (m × Vf - m × V0) / F
∆t = (11000[Kg] × 2,7[m/s] - 11000[Kg] × 0) / 2829,218[N]
∆t = 10,799[seg]




W =L / ∆t
W = (F ×) / ∆t
W = (2829,218[N] × 15[m]) / 10,799[seg]
W = 3929,469[watt]
a [cv] → 3929,469[watt] / 736 = 5,339[cv]

a = F / m
a = 2829,218[Kg·m/s2] / 11000[Kg]
a = 0,257[m/s]

Dimensionamiento del gancho.
(d2 × π) / 4 = Q / Kz
d = √((Q × 4) / (Kz × π))
d = √((5000[Kg] × 4) / (7[Kg/mm2] × π))
d = 30[mm]

d1 = 1,25·d
d1 = 1,25 × 30[mm]
d1 = 37,5[mm]

Anilla.
(∆02 × π)/ 4 = Q / (2 × Kz)
∆ = √((Q × 4) / (2 × Kz × π))
∆ = √((5000[Kg] × 4) / (2 × 7[Kg/mm2] × π))
∆ = 21,3[mm]

Radio del gancho.
W = 0,75∆0 a ∆0 adoptando ancho del cable
W = ∆0 = 21,3[mm]

Dimensiones por tabla.
a = 90[mm]
h = 100[mm]
B = 80[mm]
b = 30[mm]

h1 = ((B + 2b) / (B + b)) × h/3
h1 = ((80[mm] + 2 × 30[mm]) / (80[mm] + 30[mm])) × 100[mm]/3
h1 = 42,42[mm]

h = h1 + h2
h2 =h - h1
h2 = 100[mm] – 42,42[mm]
h2 = 57,57[mm]

J = (((6 × b2) + 6 × b × (B – b) + (B – b)2) / (36 × (2 × b + B – b)) × h3
J = (((6 × 302 + 6 × 30 × (80 – 30) + (80 – 30)2) / (36 × (2 × 30 + 80 – 30)) × 1003
J = 4267676,77[mm4]

W1 = J / h1
W1 = 4267676,77[mm4] / 42,42[mm]
W1 = 100595,23[mm3]

W2 = J / h2
W2 = 4267676,77[mm4] / 57,57[mm]
W2 = 74122,8[mm3]

F = (B + b × h) / 2
F =(80[mm] + 30[mm] × 100) / 2
F = 5500[mm2]

h = 2,6 √((5000[Kg] / 70[Kg/mm2]) × (5/4 + 2 × 21,3[mm] / 100))
h = 89,95[mm]

Tornillo sin fin y corona:
Datos:
Øtambor = 240[mm]
i = 60
m = 1

Mn = 8 × 3√[Mt rueda / (Z × σadm)]
Mn = 8 × 3√[9276,432 / (60 × 900)]
Mn = 4,5[mm]

Mnr = Mnt

Momento torsor en el Tornillo:

Mn = 8 × 3√[Mt rueda / (n × σadm)]
Mt = n × σadm × (Mn / 8)3Mt = 1 × 1400 × (4,5 / 8)3
Mt = 249[Kg·cm]
Mt = 2,49[Kg·m]

Según la tabla del motor: Mn = 30[N·m] ≡ 3,06[Kg·m] → 3,06 > 2,49

Cálculo del aparejo: (Se eligió una trócola de 6 poleas iguales)

Tensión del cable:
P = Q / (N × η)
P = 4000[Kg] / (6 × 0,862)
P = 773,395[Kg]

Cable.
Pr = Q × Ks
Pr = 773,395 × 5
Pr = 3866,976[Kg]
Por tabla: cable 6×19+1 180[Kg/mm2] → δ = 0,5[mm]

Ø...