METODO DE VIGA CONJUGADA
6) Determinar P para que la flecha en B sea igual a 1cm. EI AC=500T−m2 EI AC=500T−m2
yB=1cm=0.01m
∑M A=0→MA=2 P (1 )+P (2 )→M A=4 P
∑ FY=0→R A=3P
→M1=MA+RA X1→M 1=−4 P+3 P X1
→M2=−P X2
yB=MB=−[ P250 (1 )( 12 )( 23 )]−[ 3P500 (1 )( 12 )( 23 +1)]−[ P500 (1)(12+1)]
MB=−P750
− P200
− 3P1000
=−7 P750
= 7 P750
↓
≫ yB=7 P750
=0.01→P=1514
=1.0714T
7) Calcular los giros en los apoyos de la viga (EI=55MN-m2)
SUPERPOSICION:
+
∑ F A=0→RC (10 )=35 (2 )+35(8)→RC=35 ∑M A=RC (10 )=RB (5 )→RC=RB2
RA=RC=35KN RA=RC=RB2
∑M A=0=−RC (10 )+12( 70EI
)(2)( 43)+6 ( 70EI ) (5 )+ 1
2 ( 70EI )(2 )( 23 +8)−5 RB2 EI(10)( 1
2)(5)
250RB4 EI
+10RC=2803EI
+ 2100EI
+ 1820EI
10MC=2800EI
−250 RB4 EI
→RC=RA=1120−25RB
EI
MB=0→−( 1120−25 RBEI ) (5 )−12 (5 RB2EI ) (5 )( 53 )+ 12 ( 70EI ) (2 )( 23 +3)+( 70EI ) (3 )( 32 )=0RB=
12068275
=43.884
RA=RC=1120−25( 12068
275)
EI=25211EI
RA=RC=θ A=θB=25211EI
=0.00042 rad
→θB=V B=12 ( 70EI )(2 )+( 70EI ) (3 )− 252
11EI−( 5RBEI )( 12 ) (5 )
θB=( 70EI )+(210EI )− 25211EI
−( 254 EI )(12068275 )=−0.00031 rad
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