solution4_2136

# solution4_2136 - DEPARTMENT OF CIVIL ENGINEERING LAKEHEAD...

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DEPARTMENT OF CIVIL ENGINEERING, LAKEHEAD UNIVERSITY Engineering 2136 – Steel Design Solutions for Assignment No.4, Fall 2006 Instructor: Dr. Y Gong 1. section area A=350 × 25 × 2 + (350-25 × 2) × 16 = 22300 mm 2 The moment of inertia () ( ) 6 3 3 10 499 2 25 350 16 350 12 1 350 350 12 1 × = × × × = x I mm 4 (a) 3 6 10 2852 2 / 350 10 499 2 / × = × = = d I S x mm 3 (b) Plastic neutral axis coincides with elastic neutral axis due to symmetry with respect to x-axis. and y x p F Z M = c c t t x y A y A Z + = where A c = the compressive zone of the cross-section at the plastic limit state; A t = the tensile zone at the plastic limit state; y t = the distance of the centriod of the tensile zone to the plastic neutral axis; y c = the distance of the centriod of the compressive zone to the neutral axis. 1 A c = A t = 0.5A = 0.5 × 22300 = 11150 mm 2 2 x x 7 . 143 11150 2 150 16 150 2 25 2 350 25 350 = × × + × × = = i i i c A y A y mm Z x = 11150 × 143.7 + 11150 × 143.7 = 3204 × 10 3 mm 3 (c) shape factor f = Z x / S x = 3204 × 10 3 /2852 × 10 3 = 1.12 2. Solution (1) Sectional properties: b = 209 mm, t =13.3 mm, d = 528 mm, h = d – 2 t = 501 mm, w = 9.5 mm, Z x = 2070 × 10 3 mm 3 I x = 478 × 10 6 mm 4 , F y =350 MPa (2) sectional classification Flange: 1 . 9 350 170 170 9 . 7 3 . 13 2 209 2 = = < = = y F t b , class 2 Web: 8 . 58 350

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

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solution4_2136 - DEPARTMENT OF CIVIL ENGINEERING LAKEHEAD...

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