UNIVERSIDAD FERMIN TORO

VICERRECTORADO ACADEMICO

FACULTAD DE INGENIERIA

Mecanica Estatica

Integrantes

Anthony Rojas c.i: 23537385

Ejercicios:

1)

∑F y=0 : R∆ y+RB=5kn (1 )

∑M∆=0 :−20 (5 )kn+25RB=0=¿ RB=100k .m

25m=4k Así R∆ y=5kn−4 kn=1kn

∑F y=0 1kn−V 1=0=¿V 1=1kn

∑M 1=0 :−1kn ( X )+M 1=0=¿ M 1=X (kn .m )

X=0 => M 1=0

X=0 =>M 1=20

∑F y=0 1kn−V 2−5kn=0=¿V 2=4kn

∑M 2=0 :−1knX+5 ( X−20 )+M 2=0=¿ M 2=X−5 X+100

M 2=−4 knX+100

2)

FT=area de triangulo=12

bh=12

(3 p ) 1200 ,FT=1800 lb

∑F y=0 : RBy+RDy=1800+4650=6450=¿ RBy+RDy=6450 lb

∑MB=0 1 (1800 )−3 (4650 )+6 RDy=0

RDy=12150

6lb=2025 lb

RBy=6450lb−2025lb=4425 lb

F1=12

bh=12

( 400x ) x ,F1=200 x2

∑F y=0

−V 1−200 x2=0 ,V 1=−200 x2 { x=0=¿V 1=0x=3=¿V 1=−1800 lb

∑M 1=0 :12

x F1+M 1=0=¿M 1=−13

x ¿

si x=3=¿ M 1=−1800 lbp

3<x<6

∑F y=0: −1800+4425−V 2=0 ;V 2=2625 lb

∑M 2=0 , ( x−2 ) FT−4425 ( x−3 )+M 2=0

¿>M 2=−1800 ( x−2 )+4425 ( x−3 )=2625x−9675 , {x=3=¿ M 2=−1800 lbx=6=¿ M 2=6075 lb

−1800 lb+4425 lb−4650−V 3=0=¿V 3=−2025lb

∑M 3=0

1800 ( x−2 )−4425 ( x−3 )+4650 ( x−6 )+M 3=0

M 3=−2025 x+18225 ( lbp )

X=6 => M 3=6075 lb

X=9 => M 3=0