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Parcial de diseño de máquinas 2 UIS

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Parcial resuelto de diseño de máquinas 2 de la universidad industrial de Santamder

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  • March 30, 2022
  • 10
  • 2021/2022
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Universidad industrial de Santander
Escuela de ingeniería mecánica.
Diseño de maquina II
Prof. Gilberto Parra R.
Agosto 2020.

El esquema nos muestra un cilindro de hierro fundido de diámetro
interior 800 mm, que tiene una presión que varía desde 20 Kgf/cm2
hasta 80 Kgf/cm2.
El cilindro está unido a la cabeza del mismo de acero de bajo
carbono, por medio de pernos de grado métrico 12.9, si el número
de pernos interiores son las 2/3 partes de los tornillos exteriores y
la fuerza sobre cada perno son inversamente proporcionales a su
radio y el empaque de espesor 5 mm y es de cobre asbesto
Halle el diámetro de los pernos y el número de ellos requerido si el
factor de seguridad es de 2,5.
Considere vida infinita y confiabilidad del 99%.


1400mm

1100 mm




30 mm

5 mm
35 mm




Cobre
asbesto
Fe
800 mm
fundido

, Hallaremos la fuerza máxima aplicada:
F max= p A
2 2
Kgf πx 80 cm
F max=80 2 X =402123,86 Kgf
cm 4

Kgf πx 802 cm2
F min=20 2 X =100530,96 Kgf
cm 4




Longitud de la circunferencia interior:
Li=π Di

Li=πx 110 cms=345,58 cms

Para seleccionar el paso de los tornillos
kgf
Pmax =80 2
=447 psi=7,85 Mpa
cm




Si Pres > 200 psi; 2,0 d≤ Paso ≤3,5 d


Asumo p=3 d; por tanto en el circulo interior habrán
Li 3455,8 mm 1151.9
N ti = = =
pi 3 xd d

Por tanto puedo calcular el N° de tornillos del exterior
2
Del enunciado tenemos N ti = x N te
3

3
N te = x N ti
2

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