tf tf d bf tw - U

EJEMPLO PANDEO LOCAL
N
MPa := 1⋅
mm
kN := 1000N
2
tonf := 1000kgf
ksi := 1000psi
H 500x350x110:
d := 500mm
t f := 16mm
Fy := 65ksi
b f := 350mm
t w := 6mm
KL := 4m
Fy = 4.57
tonf
cm
E := 200000MPa
2
tf
(i) Propiedades de la seccion
(
)
Ag := d − 2tf t w + 2b f ⋅ tf
3
(
)
tw
bf
Ag = 14008 mm
3
8
Iy := d − 2tf
+2
⋅t
12
12 f
ry :=
2
tw
d
4
Iy = 1.14 × 10 mm
tf
Iy
ry = 90 mm
Ag
bf
(ii) Relaciones de esbeltez
Global
Alas
Alma
KL
= 44.3
ry
bf
= 10.9
2t f
h
h := d − 2tf
tw
= 78
(Despreciando filetes de soldadura)
Limites de esbeltez
Alas
kc :=
4
kc = 0.45
h
0.64
kc⋅ E
Fy
= 9.1
<
bf
2t f
= 10.9
tw
=> alas esbeltas
Alma
1.49
E
Fy
= 31.5
<
h
tw
= 78
=> alma esbelta
1/3
EJEMPLO PANDEO LOCAL
(iii) Resistencia nominal
- Elementos no atiesados (alas), Qs
0.64
kc⋅ E
Fy
= 9.1
bf
<
= 10.9
2t f
<
kc⋅ E
1.17
 bf  Fy

 2t f  kc⋅ E
Q s := 1.415 − 0.65 
Fy
= 16.6
Q s = 0.91
- Elementos atiesados (alma), Qa
Necesitamos determinar f = Fcr (Q=1)
KL
ry
= 44.3
E
< 4.71
Fy
= 99.5
2
Fe :=
π E
 KL 
r 
 y
2
Fe = 1007 MPa
Fy 



Fe 

Fcr :=  0.658  ⋅ Fy
h
tw
= 78
Fcr = 372 MPa
> 1.49
E
f
= 34.5
E 
0.34
⋅ 1−
b e := 1.92⋅ t w⋅
f 
h


t 
 w
Aeff := 2b f ⋅ t f + b e⋅ t w
Q a :=
Aeff
Ag
f := Fcr
E
f



b e = 240 mm
Aeff = 12641 mm
<
h = 468 mm
2
Q a = 0.9
2/3
EJEMPLO PANDEO LOCAL
- Tension critica:
Q := Q s⋅ Q a
KL
ry
= 44.3
Q = 0.83
< 4.71
E
Q ⋅ Fy
= 109.5
Q⋅ Fy 



Fe 

Fcr := Q ⋅  0.658
 ⋅ Fy
Fcr = 317 MPa
- Resistencia nominal
Pn := Fcr⋅ Ag
Pn = 4444 kN
Pn = 453 tonf
3/3