trabajo de analisis estructural ii

UNIVERSIDAD NACIONAL DE CAJAMARCA

TRABAJO DE ANALISIS ESTRUCTURAL II

Se tiene un pórtico de 04 niveles

METODO KANY

SE PIDE:

DIAGRAMA DE MOMENTOS FLECTORES

DIAGRAMA DE FUERZAS CORTANTES

CONSIDERAR DIMENSIONAMIENTO DE VIGAS (h=L/11 ; b=25)

CONSIDERAR COLUMNAS :

1er. Y 2do nivel de 30x35

3er. Y 4to nivel de 30x30

ING. LINO CANCINO COLICHON ANALISIS ESTRUCTURAL II


DESARROLLO

Para la atura de la vigas se ha considerado la luz de mayor longitud (L=6.75)

Entonces h=l11

=6.7511

=0.614m

Consideramos h=0.65m, por lo tanto tenemos las vigas de .25x.65 m.

1) ENUMERANDO NUDOS



2) HALLANDO LOS MEP EN CADA TRAMO

M 1−2=−W L2

12=−7.4 Tn/m∗¿¿

Donde

M 1−2=M 5−6=M 9−10=M 13−14=M 3−4=M 7−8=M 11−12=M 15−16=−20.389Tn−m

M 2−1=−W L2

12=7.4Tn /m∗¿¿

M 2−1=M 6−5=M 10−9=M 14−13=M 4−3=M 8−7=M 12−11=M 16−15=20.389Tn−m

M 2−3=−W L2

12− PL8

M 2−3=M 6−7=M 10−11=M 14−15=−7.4Tnm

∗¿¿



M 3−2=W L2

12+PL8

M 3−2=M 7−6=M 11−10=M 15−14=7.4Tnm

∗¿¿

3) CALCULANDO LOS “ K”

DETERMINADO LAS INERCIAS

INERCIAVIGA=b h3

12=25∗(65)3

12=572135.417 cm4

INERCIAc olumna1 ro,2do piso=b h3

12=30∗(35)3

12=107187.5cm4

INERCIAcolumna3 ro ,4¿ piso ¿=bh3

12=30∗(30)3

12=67500cm4

PARA VIGAS DEL PORTICO “17-18” Y “19-20”

K=572135.417 c m4

575cm=995.018cm3

PARA VIGAS DEL PORTICO “ 18-19”

K=572135.417cm4

675cm=847.608cm3

PARA LAS COLUMNAS DEL 3ER. Y 4TO. PISO


K= IL


K=67500cm4

280cm=241.071cm3

PARA LAS COLUMNAS DEL 2DO. PISO

K=107187.5cm4

280cm=382.813cm3

PARA LAS COLUMNAS DEL 1ER. PISO

K=107187.5cm4

400cm=267.969cm3

4) DETERMINANDO LOS COEFICIENTES DE REPARTICION



C IJ=

−12

∗(K IJ)

∑K

NUDO 1

C1−2=

−12

∗995.018

(995.018+241.071 )=−0.40

C1−5=

−12

∗241.071

(995.018+241.071 )=−0.10

NUDO 2

C2−1=

−12

∗995.018

(995.018+241.071+847.608 )=−0.24

C2−6=

−12

∗241.071

(995.018+241.071+847.608 )=−0.06

C2−3=

−12

∗847.608

(995.018+241.071+847.608 )=−0.20

NUDO 3

C3−2=

−12

∗847.608

(995.018+241.071+847.608 )=−0.20

C3−7=

−12

∗241.071

(995.018+241.071+847.608 )=−0.06



C3−4=

−12

∗995.018

(995.018+241.071+847.608 )=−0.24

NUDO 4

C4−3=

−12

∗995.018

(995.018+241.07 )=−0.40

C4−8=

−12

∗241.071

(995.018+241.071 )=−0.10

NUDO 5

C5−1=

−12

∗241.071

(995.018+241.071+241.071 )=−0.08

C5−9=

−12

∗241.071

(995.018+241.071+241.071 )=−0.08

C5−6=

−12

∗995.018

(995.018+241.071+241.071 )=−0.34

NUDO 6

C6−2=

−12

∗241.071

(995.018+241.071+241.071+847.608 )=−0.05

C6−10=

−12

∗241.071

(995.018+241.071+241.071+847.608 )=−0.05

C6−5=

−12

∗995.018

(995.018+241.071+241.071+847.608 )=−0.22



C6−2=

−12

∗847.608

(995.018+241.071+241.071+847.608 )=−0.18

NUDO 7

C7−3=

−12

∗241.071

(995.018+241.071+241.071+847.608 )=−0.05

C7−11=

−12

∗241.071

(995.018+241.071+241.071+847.608 )=−0.05

C7−8=

−12

∗995.018

(995.018+241.071+241.071+847.608 )=−0.22

C7−6=

−12

∗847.608

(995.018+241.071+241.071+847.608 )=−0.18

NUDO 8

C8−4=

−12

∗241.071

(995.018+241.071+241.071 )=−0.08

C8−12=

−12

∗241.071

(995.018+241.071+241.071 )=−0.08

C8−7=

−12

∗995.018

(995.018+241.071+241.071 )=−0.34

NUDO 9

C9−5=

−12

∗241.071

(995.018+241.071+382.813 )=−0.07



C9−10=

−12

∗995.018

(995.018+241.071+382.813 )=−0.31

C9−13=

−12

∗382.813

(995.018+241.071+382.813 )=−0.12

NUDO 10

C10−6=

−12

∗241.071

(995.018+241.071+382.813+847.608 )=−0.05

C10−9=

−12

∗995.018

(995.018+241.071+382.813+847.608 )=−0.20

C10−11=

−12

∗847.608

(995.018+241.071+382.813+847.608 )=−0.17

C10−14=

−12

∗382.813

(995.018+241.071+382.813+847.608 )=−0.08

NUDO 11

C11−7=

−12

∗241.071

(995.018+241.071+382.813+847.608 )=−0.05

C11−12=

−12

∗995.018

(995.018+241.071+382.813+847.608 )=−0.20

C11−10=

−12

∗847.608

(995.018+241.071+382.813+847.608 )=−0.17

C11−15=

−12

∗382.813

(995.018+241.071+382.813+847.608 )=−0.08

NUDO 12



C12−8=

−12

∗241.071

(995.018+241.071+382.813 )=−0.07

C12−11=

−12

∗995.018

(995.018+241.071+382.813 )=−0.31

C12−16=

−12

∗382.813

(995.018+241.071+382.813 )=−0.12

NUDO 13

C13−9=

−12

∗382.813

(995.018+267.969+382.813 )=−0.12

C13−14=

−12

∗995.018

(995.018+267.969+382.813 )=−0.30

C13−17=

−12

∗267.969

(995.018+267.969+382.813 )=−0.08

NUDO 14

C14−10=

−12

∗382.813

(995.018+267.969+382.813+847.608 )=−0.08

C14−13=

−12

∗995.018

(995.018+267.969+382.813+847.608 )=−0.20

C14−15=

−12

∗847.608

(995.018+267.969+382.813+847.608 )=−0.17

C14−18=

−12

∗267.969

(995.018+267.969+382.813+847.608 )=−0.05

NUDO 15



C15−11=

−12

∗382.813

(995.018+267.969+382.813+847.608 )=−0.08

C15−16=

−12

∗995.018

(995.018+267.969+382.813+847.608 )=−0.20

C15−14=

−12

∗847.608

(995.018+267.969+382.813+847.608 )=−0.17

C15−19=

−12

∗267.969

(995.018+267.969+382.813+847.608 )=−0.05

NUDO 16

C16−12=

−12

∗382.813

(995.018+267.969+382.813 )=−0.12

C16−15=

−12

∗995.018

(995.018+267.969+382.813 )=−0.30

C16−20=

−12

∗267.969

(995.018+267.969+382.813 )=−0.08



DIAGRAMA DE LOS MOMENTOS FLECTORES



DIAGRAMA DE CORTANTES


trabajo de analisis estructural ii

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