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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
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