Exercício resolvido

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2a. parte: Aço: valor 5 pontos
4. uestão : (2,5 pontos}~:~~~~
eixos dos perfis, de modo I
a tomar iguais as
resistências nas direções x .
e y.
b) A carga de cálculo (NRd -
compressão) que pode ser í
aplicada nessa condição. I 148 m
y
900 em
x
Dados:
Aço AR 345 fy = 345 MPa;
fu = 485 MPa.
Perfis:
A = 16 60 crrr': Ix = 635 00 em":9 , , , ,
rx = 6,18 cm; Iy = 82,00 em";
ry = 2,22 cm ; br= 100 mm;
tf = 4,9 mm; tw = 4,~' rnrn.
a
adote O = 1;
perfil Laminado rr 'T7
d = a/2 -:~a= 2.5,7718 = 11,54 clll...""""",,,,,,,,,,,,,~••••••.•,,.,•••,,,,,",,,,,
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À =1,47 rr. =35,83 > 32,13 _ OK!! I Icom ~= 0,251, na curva c, parâmetro de esbeltez = 1,76 i
p ~ fy ~aproximadamente. •
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Àp=0,55 rr.fE=13,41 > 10,20 - OK!! IÀ = 176.11:. 205000 = 134 78 .~~ I' 345 '
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i i, = ix = 900/134,78 = 6,68 cm; .I-------------------L------------------------
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