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1. Analice el marco de la siguiente figura. Datos EI es constante y existe empotramiento en A y D.
SOLUCION
AX=DX=20kN
θA=θD=0
θB=θC=θ
φ AB=φCD=φ
M AB=2EI14
(θB−3φ )= EI7
(θ−3φ)
MBA=2 EI14
(2θB−3φ )=EI7
(2θ−3φ) …. (1)
MBC=2 EI18
(2θB+θC )=EI9
(3θ)
NUDO B
MBA+M BC=0
EI7
(2θ−3φ )+ EI9
(3θ)=0
θ= 913φ ……. (2)
MBA+M AB−20 (14 )=0
EI7
(2θ−3φ )+ EI7
(θ−3φ )=280
Sustituyendo: θ=913φ ó φ=
139θ
7EI
∗[ EI7 (2θ−3[ 139 θ])+EI7 (θ−3[ 139 θ])]=280∗[ 7EI ]
[(2θ−3{139 θ})+(θ−3 {139 θ })]=280∗[ 7EI ] θ=
−345.882EI
…….. (3)
φ= 913θ=
−345.882∗(9)EI∗(13)
φ=−239.457
EI ……….. (4)
Sustituir ecu (2) y (3) en (1)
M AB=2EI14
(θB−3φ )= EI7
(θ−3φ)
M AB=EI7
(−345.882EI
−3[−239.457EI ])M AB=−152.028kN .m
MBA=2 EI14
(2θB−3φ )=EI7
(2θ−3φ)
MBA=EI7
(2θ−3φ)
MBA=EI7
(2[−345.882EI ]−3 [−239.457EI ]) MBA=3.801kN .m
MBC=2 EI18
(2θB+θC )=EI9
(3θ)
MBC=EI9
(3θ)
MBC=EI9
(3 [−345.882EI ])MBC=115.294 k N .m
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