solution5_2136

solution5_2136 - DEPARTMENT OF CIVIL ENGINEERING, LAKEHEAD...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
DEPARTMENT OF CIVIL ENGINEERING, LAKEHEAD UNIVERSITY 1 Engineering 2136 – Steel Design Solutions for Assignment No.5, Fall 2006 Instructor: Dr. Y Gong 1. Solution (1) Considering double channel sections C200 × 21 section: F y =300 MPa, F u = 450 MPa A=2600 mm 2 , w =7.7 mm, t = 9.9 mm Yielding of gross section: T r = φ A g F y = 0.9(2 × 2600)(300) = 1404,000 N = 1404 kN (a) Pattern 1b Fracture of net section 1-1: The section is normal to the tensile force. Shear lag shall be considered, and C sl =0.75 A ne = 0.75A n =0.75 [2 × 2600 –2 × 2 × 24 × 7.7] = 3345 mm 2 T r = 0.85 φ A ne F u = 0.85(0.9)(3345)(450) = 1151516 N = 1152 kN Tension and shear block failure, section 2-2: T r + V r = φ A nt F u + 0.6 φ A gv F y = 0.9[(80-24) × 7.7 × 2]450+0.6(0.9)(120 × 2 × 7.7 × 2)300 = 349272 + 598752 = 948024 N = 948 kN (governs) T r + V r φ A nt F u + 0.6 φ A nv F u =349272 + 0.6(0.9) [(120 × 2-3 × 24) × 7.7 × 2]450 = 349272 + 628689 = 977961 N = 978 kN (b) Pattern 1c Fracture of net section 1-1: The section is normal to the tensile force. Shear lag shall be considered, and C sl =0.85 A ne = 0.85A n =0.85 [2 × 2600 –2 × 1 × 24 × 7.7] = 4105 mm 2 T r = 0.85 φ A ne F u = 0.85(0.9)(4105)(450) = 1413146 N = 1413 kN Tension and shear block failure, section 2-2: T r + V r = φ A nt F u + 0.6 φ A gv F y = 0+0.6(0.9)(280 × 2 × 7.7 × 2)300 = 0 + 1397088 = 1397 kN (governs) T r + V r φ A nt F u + 0.6 φ A nv F u =0 + 0.6(0.9) [(280-3.5 × 24) × 2 × 7.7 × 2]450 = 0 + 1466942 = 1467 kN (c) Pattern 1d Fracture of net section 1-1: The section is normal to the tensile force. Shear lag shall be considered and C sl =0.85 203 59 7.7 9.9 C200 × 21 40 80 40 80 1 1 2 2 [email protected] 40 1 1 2 2
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
DEPARTMENT OF CIVIL ENGINEERING, LAKEHEAD UNIVERSITY 2 A ne = 0.85A n =0.85 [2 × 2600 –2 × 1 × 24 × 7.7] = 4105 mm 2 T r = 0.85 φ A ne F u = 0.85(0.9)(4105)(450) = 1413146 N = 1413 kN Fracture of net section 2-2: Shear lag shall be considered and C sl =0.85 A ne = 0.85A n =0.85 [2 × 2600 –2 × 2 × 24 × 7.7+2
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/23/2009 for the course ENGI 2136 taught by Professor Gong during the Fall '06 term at Lakehead.

Page1 / 5

solution5_2136 - DEPARTMENT OF CIVIL ENGINEERING, LAKEHEAD...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online