Final Exam(April2010)

# Page 66 ce 470 final exam 12 april 2010 structural

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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! ~...
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## This document was uploaded on 03/30/2014 for the course CE 470 at University of Saskatchewan- Management Area.

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