DISEÑO DE ELEMENTOS MECANICOS.
Enviado por Rebecca • 29 de Mayo de 2018 • 530 Palabras (3 Páginas) • 197 Visitas
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Datos
Su = 130 KSI
Sy = 120 KSI
P = ±5000 lb
N = 1.5 D = ?
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D = 1” Tabla Kt = 2.65 Diagrama Peterson [pic 130]
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A = 0.71; B = ; C = 0.71; Kc = 0.814 para 99 %;[pic 135]
Kd = 1 T ˂ 450°C ; Se = 0.5Su = 0.5(130,000) = 65 KSI
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P=35000lb
Mt=2500lb*plg
Su=100ksi
Sy=60ksi
Por tensión
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[pic 149]De gráfica kt=1.9
Por torsión
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A = 0.71; B = ; C = 0.73; Kc = 0.814 para 99 %;[pic 154]
Kd = 1 T ˂ 450°C ; Se = 0.5Su = 0.5(100,000) = 50000PSI
Por torsión
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P=+-2500lb
Mf=1500lb
=+-5°F[pic 162]
Su=100ksi
Sy=60ksi
Por carga
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Por temperatura
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Por flexion
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C=2.5/2=1.25´´
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d=D-2r=2.5-2(.125)=2.25´´
D/d=2.5/2.25=1.11
r/d=0.125/2.25=0.05
Kftension=2.7
Kf flex=2.4
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Por flexion
A=1
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Kf=2.32
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P=+3500lb
Mf=+-2500lb
=-3°F[pic 181]
Su=125ksi
Sy=80ksi
Por carga
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Por temperatura
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Por cortante
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Concentradores de esfuerzos para el cortante
Kf=1.45
Kfs=1.18
D/d=2/1.2=1.66
r/d=0.4/1.2=0.33
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A = 0.71; B = ; C = 1; Kc = 0.702 para 99.99 %;[pic 192]
Kd = 1 T ˂ 450°C ; Se = 0.5Su = 0.5(12500) = 62500 PSI
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