Estatica

30 Klb
30 Klb
50 Klb
50 Klb

4.5)
4Klb/Ft
3klb/ft
3klb/ft

AB
15 m 10m 15m


∑Fx =0Ax =0
∑Fy =0 Ay + By – 50 -30 – 140 =0
Ay + By =220

CENTROIDE DE LA CARGA TOTAL
AREA | X promedio | Y promedio | A*Xprom. | A*Y prom. |
120 | 20 |1.5 | 2400 | 180 |
20 | 13.3 | 4.3 | 266 | 86 |
∑A=140 | | | ∑=2666 | ∑=266 |


Centroide: X=19.04 Ft
Y=1.9 Ft

(Horario +)∑MA=0 140(19.04) + 50(15) + 30(25) –By (40) = 0
BY = 104.14 Klb AY = 115.86 Klb
Distancias verticales: se hace por triángulossemejantes y da:
Y= 5/8 d= 3/8
1) Área(1)

AREA | X promedio | Y promedio | A*Xprom | A*Y prom |
30 | 5 | 1.5 | 150 | 67.5 |
6,25 | 5 | 3.31 | 31.25 | 52.5 |
-1,25 | 6,66 | 3.54 | -8,33| -10,88 |
∑A=35 | | | ∑=172.92 | ∑=109.12 |

Centroide: X=7.37 Klb
Y=1.9 Klb
2) AREA (2)

AREA | X promedio | Y promedio | A*Xprom | A*Y prom |
30 | 5 | 1.5 |150 | 67.5 |
6,25 | 5 | 3.31 | 31.25 | 52.5 |
-1,25 | 6,66 | 3.54 | -8,33 | -10,88 |
∑A=35 | | | ∑=172.92 | ∑=109.12 |

Alturas: Y2 = 3/8 ft d2=1/4 ft
Centroide: X= 4.94Y=1.75
3) AREA:
AREA | X prom. | Y prom. | A*X prom | A*Y prom. |
45 | 7.5 | 1.5 | 337.5 | 67.5 |
2,81 | 5 | 3.125 | 14.05 | 8.781 |
∑A=47.81 | | | ∑=351.55 | ∑=76.3 |Centroide: X=7.35 Klb
Y= 1.6 Klb

Entonces; mediante secciones se hayo:

Cortante:

a) V=115.86-3.7X por que: 115-V-3.7=0

b) V=67.75-(61/16)X por que:...