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ING.
J.R.T.G. METODO DE CROSS
Calcular los Cortantes y Momentos de la estructura adjunta:
Datos:
W
W =
a 1 b 2 c L1 =
L2 =
H =
4 ! H I 1 =
I 2 =
I =
d e % I 4 =
I ! =
L1 L2
MOMENTO DE EMPOTRAMIENTO PERFECTO
M&ab = '2.(# )n'm. M&ba = 2.(# )n'm.
M&bc = '2.(# )n'm. M&cb = 2.(# )n'm.
RIGIDECES RELATIVAS K ik = I / L FACTORES DE DISTRIBUCION C ik = Kik / S
*ab = I 1 + L1 = 0.!,2!,2!,2( I Cab = *ab + *ab-*ad = '0.#0 Cba = *ba
*bc = I 2 + L2 = 0.!,2!,2!,2( I Cad = *ad + *ab-*ad = '0.2,# Cbc = *bc
*da = I + H = 0.2! I '1.000 Cbe = *be
*eb = I 4 + H = 0.2! I
*%c = I ! + H = 0.2! I Ccb = *cb + *cb-*c% = '0.#0
Cc% = *c% + *cb-*c% = '0.2,#
'1.000
.!"# ".$%&
'".$%& 0.000 '.!"# 0.000
0.000 0.000 0.000 0.000
0.001 0.001 0.000 0.000
'0.002 '0.001 0.001 0.001
0.00 0.00# '0.001 '0.002
'0.01# '0.00$ 0.00# 0.00
0.024 0.04$ '0.00$ '0.01#
'0.11( '0.0!$ 0.04$ 0.024
0.1(! 0.1 '0.0!$ '0.11(
0.1( 0.0($ 0.1 0.1(!
'0.1,4 '0.$# '0.$#0 '1.#,
1.$#! 0.,$ '0.$# '0.1,4
'2.((# 2.((# '2.((# 2.((#
'0.#0 '0.41 '0.41 '0.#0
" . $
% &
0 . 0
0 0
' 0 . 0
0
1
' 0 . 0
0
#
' 0 . 0
4
,
0 . 0
! #
0 . #
, 1
' 0 . 2
,
#a (
' 0 . 1
#
4
' 0 . 1
(
0 . 1
4 0
0 . 0
2 0
0 . 0
0
0 . 0
0 0
0 . 0
0 0
" . "
" "
)
' 0 . 2
,
#
' 0 . #
4
' 0 . 0
4
,
' 0 . 0
0
#
' 0 . 0
0
1
" .
% !
0 . 0
0 0
' 0 . 0
0 1
' 0 . 0
0 4
' 0 . 0
2 !
0 . 0
2 ,
0 .
, (
0 . 0
0 0
0 . 0
0 0
' 0 . 0
$ 2
0 . 0
# 0
0 . 0
1 0
0 . 0
0 1
0 . 0
0 0
0 . 0
0 0
" . "
" "
0 . 0
0 0
' 0 .
( #
' 0 . 0
2 !
' 0 . 0
0 4
' 0 . 0
0 1
* + ,
M-i = L0 /&0
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FUER1AS CORTANTES Vik = 2V-ik'3Mik2Mki4/ Lik Vki = 'V-ki'3Mki2Mik4/ Lki
IG/ CLMN/
ab = -&ab'3Mab-Mba+Lab = .0 )n. da = -&da'3Mda-Mad+Lda
ba = '&ba'3Mba-Mab+Lba = '4.#0 )n. ad = '&ad'3Mad-Mda+Lad
bc = -&bc'3Mbc-Mcb+Lbc = 4.#0 )n. eb = -&eb'3Meb-Mbe+Leb
cb = '&cb'3Mcb-Mbc+Lcb = '.0 )n. be = '&be'3Mbe-Meb+Lbe
%c = -&%c'3M%c-Mc%+L%c
c% = '&c%'3Mc%-M%c+Lc%
DIAGRAMA DE FUER1AS CORTANTES 3T5.4
4.#0
.0
'.
'4.#0
' 0 .
0
0 . 0
0
0 .
0
5
L
DIAGRAMA DE MOMENTOS FLECTORES 3T5'6.4
'.(0 '.(0
'0.#, '0.#
'0.#,
1., 1.,
6 = 0.2( m. 6 2.#$ m. 7 r 2.#$ s
7 = 0.,( m.
0.40 0.40
0.0 )n.
0.40 )n'm
.0 )n. ,.41 )n. .0
Mo6+5tos M78i6os Positi9os
/6l8camos un corte en el 6unto de 8n%le98n del cortante
en el tramo ab y bc
ab W 9
ad )
1
2
5 1
5 2
= V &. L / 3V
&2V
0 4
M -1
L1
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a 9
51
La determ8nac8n de los sent8dos de las %uer;as y Momentos
de la estructura" se obt8ene de los d8a<ramas de uer;as Los s8<nos de los momentos 7ue s
cortantes y Momentos. de las car<as d8str8bu8das y otras"
?ara el desarrollo del 6roblema se tomar>n los s8<nos" tal an>l8s8s.
como se 8nd8ca en los d8a<ramas de %uer;as cortantes y
Momentos lectores.
?or esta ra;n los dos 6r8meros monom8os de la ecuac8n
7ue se e96resa se @a cons8derado con s8<no 6os8t8Ao.
2
Bl c>lcul8sta 6uede ele<8r otras o6c8ones 6ara el c>lculo del ' momento m>98mo 6os8t8Ao.
= 1., )n'm.= 1., )n'm.
Es;+<o >+ ?o *is,<>t+5
M 2& = @ L
&0 /
Mab M -ab
L& = 0 30 M
& / @
4&/0
M 2a( =Va(. &2Ma('@.
&0 /0
M 2() =V(). 0 2M()'@.
0 0 /0
M -
abM -bc
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Lam8na
MC-02
2.00 )n'm 8<a
4.00 m. T<a6o ( (/&0
4.00 m. ab 0.00 . (0.00 !40000.00
4.00 m. bc 0.00 . (0.00 !40000.00
40"000.00 cm4 = 2.#0 I
40"000.00 cm4 = 2.#0 I Columna
2#"$12.!0 cm4 = 1 I T<a6o ( (/&0
2#"$12.!0 cm4 = 1 I ad 0.00 4!.00 22#$12.!0
2#"$12.!0 cm4 = 1 I be 0.00 4!.00 22#$12.!0
c% 0.00 4!.00 22#$12.!0
i
*ba-*bc-*be = '0.41
*ba-*bc-*be = '0.41
*ba-*bc-*be = '0.1#4
'1.000
0 . 0
0 0
0 . 0
0 0
' " . $
%
&
0 . 0
0 0
0 . 0
0 0
' " .
% !
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= '0.0 )n.
= '0.0 )n.
= 0.00 )n.
= 0.00 )n.
= 0.0 )n.
= 0.0 )n.
= 1.(! m.
= 2.! m.
'0.#,
r = 0.,( m. .
s = 0.2( m.
0.0 )n.
0.40 )n'm
)n.
5 1
5 2
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e obt8enen d8rectamente
se determ8nar>n 6or