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Exercicio resolvido de estruturas metalicas
~1.:::;-r~IVI~~ E3-,C>_ T_E:_ - 2007UFPR2a. parte: Ao: valor 5 pontos4. uesto : (2,5 pontos}~:~~~~
eixos dos perfis, de modo Ia tomar iguais asresistncias nas direes x .e y.
b) A carga de clculo (NRd -compresso) que pode ser aplicada nessa condio. I 148 m
y
900 emx
Dados:
Ao 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...""""",,,,,,,,,,,,,~.,,.,,,,,,",,,,,-------------------------------------------~_________ ~- h -~- I
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