Rettt
Enviado por jeffersonlr • 11 de Septiembre de 2015 • Apuntes • 2.333 Palabras (10 Páginas) • 169 Visitas
- PRIMERA SECCIÓN
Inercia:
I = b* h3
12[pic 1]
I = (50) (75)3
12[pic 2]
I = 1757812,5 cm4
Momento: Conversión:
m = 5.40 Ton * 1000 Kg = 5400 Kg[pic 4][pic 3]
M = (2.4) ( 0.5 * 0.75) 6 1 Ton
M = 5.40 Ton 5400 Kg (9.8) = 52920 N
5290 N * 1Kg F = 5400 Kg carga puntual [pic 5][pic 6]
9.8 N
Valor de carga distribuida:
5400 Kg = 900 Kg/m
6 m
Reacciones:
∑MA = 0 + ∑Fy = 0
-20000 Kg (3) – 900 (3) (6) + RB (6) = 0 RA -20000 + RB + 5400 = 0
-60000 – 16200 + 6 RB = 0 RA = 20000 -12700 -5400
RB = 76200 /6 RA = 12700 Kg
RB = 12700 Kg
∑Fy = 0 ∑MA = 0 +
RA -900 (X) – V = 0 -900 (X)2/2 + V (X) + M = 0
12700 -900 X –V = 0 M = 900 X2/2 + V (X)
V = 12700 +900 (0.1) M = 4.5 + 12610
V = 12610 Kg M = 126520 Kg * cm
Esfuerzos normales:
σ = M * C[pic 7]
I
σ1 y 7 = [126520 (37.5)] / 1757812.5 = + 2.69 Kg/cm2 (+) tensión
σ2 y 6 = [126520 (25)] / 1757812.5 = + 1.79 Kg/cm2 (-) compresión
σ3 y 5 = [126520 (12.5)] / 1757812.5 = + 0.89 Kg/cm2
σ4 = [126520 (0)] / 1757812.5 = 0
Esfuerzos cortantes:
Q1 = A * y Q2 = A * y Q3 = A * y[pic 8][pic 9]
Q1 = 0 Q2 = (625) (31.25) Q3 = (625) (18.75)
Q2 = 19531.25 cm3 Q3 = 11718.75 + Q2
Q3 = 31250 cm3
Q4 = A * y
Q4 = (625) (6.25)
Q4 = 3906.25 + Q3
Q4 = 35156.25 cm3
τ = V * Q [pic 10]
I * t
τ1 = [12610 (0)] / 1757812.5 (50) = 0
τ2 = [12610 (19531.25)] / 1757812.5 (50) = 2.80 Kg/cm2
τ3 = [12610 (31250)] / 1757812.5 (50) = 4.48 Kg/cm2
τ4 = [12610 (35156.25)] / 1757812.5 (50) = 5.044 Kg/cm2
Densidad: 2.4 Ton/m3 Conversión:
m = 5.40 Ton * 1000 Kg = 5400 Kg[pic 12][pic 11]
M = (2.4) ( 0.5 * 0.75) 6 1 Ton
M = 5.40 Ton 5400 Kg (9.8) = 52920 N
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