Obras civiles

∑ m = + ; -

6000*7 – Ay*5 + 9000*2,5 + 5000*2 = 0
Ay = 6000*7 – Ay*5 + 9000*2,5 + 5000*2
Ay = 14900Kgf

∑Fy = 0
-6000 + 14900 – 9000 - 5000 + By = 0
By = 6000 + 9000 + 5000 – 14900
By = 5100Kgf
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0 ≤ x ≤ 2

∑ Fy = 0
-6000 – V1 = 0
V1 = 6000Kgf

∑ m1 = + ; -
6000x + m1 = 0M1 = -6000x

2 ≤ x ≤ 5

∑ Fy = 0
-6000 + 14900 – 1800(x-2) – V2 = 0
V2 = -6000 + 14900 – 1800(x-2)
V2 = 8900 – 1800(x-2)

∑ m2 = + ; -
6000x – 14900(x-2) + 1800x-2(x-2)2 + m2 = 0
m2 = -6000x + 14900(x-2) – 900(x-2)2

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2 ≤ x ≤ 5

∑ Fy = 0
-6000 + 14900 – 1800(x-2) – 5000 – V3 = 0
V3 =-6000 + 14900 – 1800(x-2) – 5000
V3 = 3900 – 1800(x-2)

∑ m3 = + ; -

6000x – 14900(x-2) + 1800x-2(x-2)2 + 5000(x-5) + m3 = 0
m3 = -6000x + 14900(x-2) – 900(x-2)2 - 5000(x-5)

Tabla

X = 0 V1 = -6000 M1 = 0
X = 2 V1 = -6000 M1 = -12000
X = 2 V1 = 8900 M1 = -12000
X = 5V1 = 3500 M1 = 6600
X = 5 V1 = -1500 M1 = 6600
X = 7 V1 = 0 M1 = 0

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Acá van los gráficos

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T adm = 1500 Kgf/cm²

Acá falta lo que hay que pedirle a laKathleen .

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T máx = 1200000*1522500 = 800

X = L2 = 72 = 3,5 cm
Y = 8 cm

M2 = -6000*3,5 + 14900(3,5 – 2) – 900(3,5 – 2)²
M2 = -675Kgf
675Kgf * 100cm² = 67500Kgf/cm²

T = 67500*822500
T = 24Kgf

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Elección de las vigas
IN A37-24
T adm= 0,6*2400
T adm = 1440Kgf/cm²

W = 12000001440
W = 833,33cm3

IN A42-27
T adm = 0,6*2700
T adm = 1620Kgf/cm²

W = 12000001620
W = 740,74cm3

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6)

IN1 40x49,3 I = 18000cm4

T máx 1 = 1200000*2018000
T máx 1 = 1333,33Kgf/cm2IN1 35x46,9 I = 13400cm4

T máx 1 = 1200000*17,513400
T máx 1 = 1567,16Kgf/cm2












7)

X = 1,3 de un apoyo M2 = 3,3

M2 = -6000*3,3 + 14900(3,3 - 2) – 900(3,3 – 2)²
M2 = -1951Kgf

T1 = 195000*2018000
T1 = 216,67Kgf

T2 = 195000*17,513400
T2 = 254,66Kgf


Θ 2 = 1EI M2 dx
= 1EI -6000x+14900(x-2)dx
= 1EI -3000x²+7450x-22- 300x-23+ C


X = 5 Θ 2 = 0

0 = -3000*5² + 7450(5 – 2)² - 300(5 – 2)³ + c
C = 3000*5² - 7450(5 – 2)² + 300(5 – 2)³
C = 16050
Ecuación de la pendiente
2 ≤ x ≤ 5

Θ 2 = 1EI -3000x²+7450x-22- 300x-23+ 16050

X = 2

Θ 2*EI = -3000*2²+745002- 30003+ 16050

Θ 2*EI = Θ 1*EI = 4050



Θ 1 = 1EI M1 dx
=1EI -600x
= 1EI -3000x²+ C

4050 = -3000x² + c
C = 4050 + 3000*2²
= 16050

Θ 1 = 1EI -3000x²+ 16050


Θ 3 = 1EI M3 dx
= 1EI -6000x+14900x-2- 900x-22- 5000x-5dx
= 1EI -3000x²+7450x-22- 300x-52- 2500x-52+ C

X = 5 Θ 3 = 0

0 = -3000*5² + 7450*3² - 300*3³ - 2500*0² + c
C = 3000*5² 7450*3² - 300*3³
C = 16050Ecuación de la pendiente
2 ≤ x ≤ 5


Θ 3 = 1EI -3000x²+7450x-22- 300x-52- 2500(x-5)²+ 16050









9)

E=2,1*1010Kgf/cm² I = Icm41004

IN 40x41,3 I = 18000 / 1004 = 1,8*10-4 m4
IN 35x46,9 I = 13400 / 1004 = 1,34*10-4 m4

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Primer tramo
0 ≤ x ≤ 2

IN 1
X = 0 Θ 1 =...