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