Final Exam(April2010)

Page 66 ce 470 final exam 12 april 2010 structural

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: (8 marks)- Explain your understanding of NBC’s and S16’s rationale for “seismic design” of steel structures. (page 6/6) CE 470: Final Exam (12 April 2010) –Structural Steel Design Some relevant information from Class Notes (this is not a question): Resultant force vector at centroids of connections: < Px, Py, Pz, Mxx(centroid), Myy(centroid), Mzz(centroid) >, where, Mxx(centroid)= Mxx(load point) _ (Py) (ez) + (Pz) (ey) --- twisting moment about x-x. Myy(centroid)= Myy(load point) + (Px) (ez) _ (Pz) (ex) --- flexural moment about y-y. Mzz(centroid)= Mzz(load point) _ (Px) (ey) + (Py) (ex) --- flexural moment about z-z. ___________________________________________________ - in an x-y plane the < x , y >components at any point of torsion-induced shears are: and, qx torque = y . P . e / [(Ixx + Iyy) / w] qy torque = x . P . e / [(Ixx + Iyy) / w] The resultant shear at any point along the weld then becomes: 2 qresultant = (q + q ) + (q + q )2 xd yto xto yd 2 ● inclined welds: Ixx= (sin ) (w L3)/12 , or page 7-54 of the CISC Handbook. ___________________________________________________ ● for a W-shape (thin-walled open section with double symmetry): Mu= ( /L) √ E Iy G J + ( /L)2 E2 Iy Cw - if: then, 1≡ E Iy G J , ( /L)2= {- 2≡ and, 1 ±√ [ 2 1 +4 E2 Iy Cw 2 Mu2 ] } / (2 2) - for a W-shape: Cw ≈ (Iflange h2 / 2) ≈ (Iyy-Beam h2 / 4) , and J ≈ ⅓ b t3 ~ The End & Good Luck as Engineers! ~...
View Full Document

Ask a homework question - tutors are online