EJERCICIO N°2

MOMENETOS DE EMPOTRAMIENTO

TRAMO AB

VA=OT

MA=0T∗m

VB=50T

M B=53.33T∗m

TRAMO BC

MB=Q∗L2

20=16T∗m

MC=Q∗L2

30=−10.67T∗m

TRAMO CD

VC=OT

MD=0T∗m

TRAMO DE

MD=(Q∗a2

12L2∗(2 L (3L−4 a )+3a2 ))+Q∗a∗b∗b

L2=63T∗m

ME=−(Q∗a3

12 L2∗( (4 L−3a ) ))−Q∗a∗b∗a

L2=−63T∗m

TRAMO EF

ME=(Q∗a3

12 L2∗( (4 L−3 a ) ))=1.67T∗m

MF=−(Q∗a2

12 L2∗(2 L (3 L−4a )+3a2 ))=−6.11T∗m

TRAMO FG

VF=40T

M B=80T∗m

MG=−40T∗m

ECUACIONES GIRO DEFLEXION

MAB=0

MBA=−53.33T∗m

MBC=MBCE+ EIL

∗(4 θB+2θC+6 ∆L )=16+ 2EI4 ∗(2θB+θC+0 )

MBC=16+2θB+θC

MCB=MCBE+ EIL

∗(4θC+2θB+ 6∆L )=−10.67+ 2 EI

4∗(4θC+2θB+0 )

MCB=−10.67+θB+2θC

MCD=MCDE+ EIL

∗(4θC+2θD+ 6∆L )=0+ 3 EI3 ∗(4θC+2θD+0 )

MCD=0+4θC+2θD

MDC=MDCE+ EIL

∗(4θD+2θC+ 6∆L )=0+ 3 EI3 ∗(4θD+2θC+0 )

MDC=0+4θD+2θC

MDE=MDEE+ EIL

∗(4θD+2θE+6 ∆L )=63+ 1.5EI5 ∗(4 θD+2θE+0 )

MDE=63+1.2θD+0.6θE

MED=MEDE+ EIL

∗(4θE+2θD+6 ∆L )=−63+ 1.5EI

5∗(4θE+2θD+0 )

MED=−63+1.2θE+0.6θD

MEF=MEFE+ EIL

∗(4θE+2θF+ 6∆L )=1.67+ 3 EI3 ∗(4θE+2θF+0 )

MEF=1.67+4θE+2θF

MFE=MFEE+EIL

∗(4θF+2θE+6 ∆L )=−6.11+ 3 EI

3∗(4θF+2θE+0 )

M FE=−6.11+4θF+2θ E

MFG=80T∗m

MGF=−40T∗m

ECUACIONES DE EQUILIBRIO

∑MB=0

16+2θB+θC−53.33=0

∑MC=0

−10.67+θB+2θC+4θC+2θD=0

∑MD=0

4 θD+2θC+63+1.2θD+0.6θE=0

∑ME=0

−63+1.2θE+0.6θD+1.67+4θE+2θF=0

∑MF=0

−6.11+1.2θF+0.6θF+80=0

REEMPLAZANDO LOS ANGULOS DE GIRO CALCULAMOS LOS MOMENTOS FINALES

MAB=0T∗m

MBA=−53.33T∗m

MBC=16+2θB+θC=−53.33T∗m

MCB=−10.67+θB+2θC=−9.46T∗m

MCD=0+4θC+2θD=−9.46T∗m

MDC=0+4θD+2θC=−25.74 T∗m

MDE=63+1.2θD+0.6θE=−25.74T∗m

MED=−63+1.2θE+0.6θD=−2.87T∗m

MEF=1.67+4θE+2θF=−2.87T∗m

MFE=−6.11+4θF+2θE=−80T∗m

MFG=−80T∗m

MGF=−40T∗m