7/18/2019 Emparrillado Plano 3
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MATRIZ DE RIGIDEZ PARA EMPARRILLADOS
A= C= E= G=
B= D= F= H=
A B A B
A C D -A G F
K = B D E - B F H
- A - B - A - B
A G F - A C DB F H - B D E
BARRA PARALELA A "X"
0 0
0 K 0 0 G F
K = 0 0
0 00 0 0 0
0 0
BARRA PARALELA A "Y"
0 0
0 0
K = 0 0 0 0
0 0
0 0
0 0 0 0
=/ _1=12 ^3 /
_2 _2 _1^2 + 〖 _2^2 〗 _3
〖− 〗 _1 _2 〖 _1 〗 _2 (−
_1
_1
____________
〖−〗 ____________
−
〖−〗 ____________
〖−〗 ____________
7/18/2019 Emparrillado Plano 3
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_2=6 /
_3 )
〖−〗
7/18/2019 Emparrillado Plano 3
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METODO DE RIGIDEZ
Dado ! #a$$%!!ado d !a &'($a) * #%d+
Determinar los desplazamienos y giros del extremo del voladizo
Determinar los diagramas de es!"erzos
, Mod!o d C.!/(!o
#e en"meran nodos$ tal %omo se indi%a en la &g y se determina los grados de li'er
,, Ma$%1 d $%'%d1
((((((21*(((
= =1,22,((( 1(
16((.( 32
1(6(((( ((((( -16((.((( -1(6((((
((((( 1,22,(( ((((( (((((
-16((.((( ((((( 32(1(((( 16((.(((
-1(6(((( ((((( 16((.((( 1(6((((
((((( -1,22,(( ((((( (((((
-16((.((( ((((( 16((.((( 16((.(((
=/ _1=12 ^3 / _2=6 /
= _1=
_2= _3=
=
_=___________
7/18/2019 Emparrillado Plano 3
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16*2 m 61,3((2
11...1(2(*1 m 2*31*2
61,3(( 11.6 ((((( -61,3((
11.6 2*31*3 ((((( -11.6
((((( ((((( 16*2(( (((((
-61,3(( -11.6 ((((( 61,3((
11.6 131.1 ((((( -11.6
((((( ((((( -16*2(( (((((
, MATRIZ GLOBAL
13,3(( -11.6 -16((.(((
K= -11.6 2,36((*3 (((((-16((.((( ((((( 336.2((
,2 E*ado I#e introd"%en !"erzas de empotramiento a &nes de 'lo/"ear los desplazamientos n
_
_2= _3=
_=___________
7/18/2019 Emparrillado Plano 3
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-1
(
- = (,
-1
(
-(,
n este estado no se prod"%en desplazamientos pero s 4ay re!"erzos
,3 E*ado IIl ve%tor de %argas nodales es ig"al y op"esto a las !"erzas de empotramiento per!
3,
5 = -122.
-(,
-21
-122.
- = (
-21 m
122.
(
2 CALC4LO DE DESPLAZAMIENTOS
os desplazamientos en ese estadoo se %al%"lan mediante la matriz de rigidez y el
.,-(3 777
8 = 1,-(32.-(3 777
7/18/2019 Emparrillado Plano 3
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777
3 CALC4LO DE ESF4ERZOS EN BARRAS
9 &nes de determinar los diagramas$ se determinan las !"erzas de 'orde$ s"perponi
(*22,.,
(321(,2
-(*163-*(22,6
-(321(,
-626.(1**
7/18/2019 Emparrillado Plano 3
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-32(2*
-.*1*.
-(*163
-(*22,6
-(321(,
(*163*
DIAGRAMAS FINALES
7/18/2019 Emparrillado Plano 3
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EMPARRILLADOS PLANOS
3 m
3. m
12 t:m
= 3(((((( t:m2
(1
'= (1. m
4= (* m
ad del pro'lema
2(.(((((( 5a ((((((2.
((
1((
((((( -16((.(((
-1,22,(( (((((
((((( 16((.(((
((((( 16((.((( BARRA 2-,
1,22,(( (((((
((((( 32(1((((
_1=
_2=
/=
=
^2 _3=* /
= =
7/18/2019 Emparrillado Plano 3
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,1.*.1, m
.1*2, m
11.6 (((((
131.1 (((((
((((( -16*2((
-11.6 ((((( BARRA 3-2
2*31*3 (((((
((((( 16*2((
dales0
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%to
ve%tor de %argas
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endo stados y
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