Page 140 - DISEÑO DE ELEM MAQUINAS I

CAPITULO VII: ELEMENTOS MECÁNICOS FLEXIBLES [131]

dN = F ⋅ dθ − ds (a)

∑ Ft = fdN + F Cos dθ − (F + dF ) Cos dθ = 0 (Ec. 7.7)

22

f (F.dθ − ds) + F (1) − (F + dF ).(1) = 0

f ⋅ F ⋅ dθ − f ⋅ ds − dF = 0 (b)

La fuerza centrífuga (F = m.a) sobre el elemento.

  = m / s2
 Vs
 
an = Vs2 /(r /100)  r = cm

  g 0 = 9.81 m
  S2

 Vs = f . p.s
 r = pu lg.
an = Vs2 /(r /12) g0 = 32.2 fps 2




El volumen del elemento para un espesor de correa de b (cm ó pulg)

btdL = bt rdθ (cm3 o pu lg3) (Ec. 7.8)

• El peso = ρ ⋅ v = ρ ⋅ b ⋅ t ⋅ r ⋅ dθ (Ec. 7.9)

• La masa :  unidadesmétricas: ρ b ⋅ t ⋅r ⋅ dθ (kg _ masa)
Para 9.81 ⋅r ⋅ dθ (slugs)

Para unidadesinglesas: ρ b ⋅ t
32.2

ds = dm ⋅ an = ρbtr dθ Vs2 = 100 ρbt Vs2 dθ = K dθ (Ec. 7.10)
g0 (r 100) g0

ds = dm ⋅ an = ρbtr dθ Vs 2 = 12 ρbt Vs 2 dθ = K´dθ (Ec. 7.11)
g0 (r 12) g0 (Ec. 7.10)

Reemplazando en (b)

fFdθ − fkdθ − dF = 0

fdθ (F − K ) = dF → 1 dF = fdθ
F−K

DISEÑO DE ELEMENTOS DE MÁQUINAS I