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8/17/2019 Ejemplo Pandeo Local
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EJEMPLO PANDEO LOCAL
Global KL
r y
44.274=
Alas bf
2tf
10.938=
Alma h d 2tf −:= h
tw
78= (Despreciando filetes de soldadura)
Limites de esbeltez
Alas k c4
h
tw
:= k c 0.453= 0.64k c E⋅
Fy
9.099= < bf
2tf
10.938=
=> alas esbeltas
Alma 1.49 E
Fy
31.476= < h
tw
78=
=> alma esbelta
MPa 1 N
mm2
⋅:= kN 1000N:= tonf 1000kgf := ksi 1000psi:=
IN 50x110:
d 500mm:= tf 16mm:= Fy 65ksi:= Fy 4.57 tonf
cm2= E 200000MPa:=
bf 350mm:= tw 6mm:= KL 4m:=t
t
d
bf
tw
(i) Propiedades de la seccion
Ag d 2tf −( )tw 2bf tf ⋅+:= Ag 1.401 104× mm2=
Iy d 2tf −( )tw
3
122
bf 3
12tf ⋅+:= Iy 1.143 10
8× mm
4=
r y
Iy
Ag:= r y 90.347 mm=
(ii) Relaciones de esbeltez
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8/17/2019 Ejemplo Pandeo Local
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EJEMPLO PANDEO LOCAL
Qa 0.902=QaAeff
Ag
:=
Aeff 1.264 104× mm2=Aeff 2bf t f ⋅ be tw⋅+:=
h 468mm=h
tw
78=
f Fcr :=Fcr 371.995 MPa=Fcr 0.658
Fy
Fe
Fy⋅:=
Fe 1007.021MPa=Feπ2E
KL
r y
2:=
4.71 E
Fy
99.499=<KL
r y
44.274=
Necesitamos determinar f = Fcr (Q=1)
- Elementos atiesados (alma), Qa
Qs 0.915=Qs 1.415 0.65 bf
2tf
Fy
k c E⋅−:=
1.17k c E⋅
Fy
16.634=< bf
2tf
10.938=
8/17/2019 Ejemplo Pandeo Local
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EJEMPLO PANDEO LOCAL
- Tension critica:
Q Qs Qa⋅:= Q 0.826=
KL
r y
44.274= < 4.71 E
Q Fy⋅ 109.503=
Fcr Q 0.658
Q Fy⋅
Fe
⋅ Fy⋅:= Fcr 317.269 MPa=
- Resistencia nominal
Pn Fcr Ag⋅:= Pn 4444.307kN= Pn 453.193 tonf =
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