solution6_2136

solution6_2136 - DEPARTMENT OF CIVIL ENGINEERING, LAKEHEAD...

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Unformatted text preview: DEPARTMENT OF CIVIL ENGINEERING, LAKEHEAD UNIVERSITY 1 Engineering 2136 Steel Design Solutions for Assignment No.6, Fall 2006 Instructor: Dr. Y Gong 1. (a) A trail location of the instantaneous center is chosen as r o =80 mm At ultimate limit state, at bolts 1 & 4, R 1 =R 4 =74 kips, and their corresponding deformation is 0.34 inches. At bolts 2 & 3, the amount of deformation is 20 . 34 . 157 8 . 91 34 . 135 80 45 80 2 2 2 2 max max 2 2 = = + + = = r r in. From the load-deformation curve, it is estimated the force for bolts 2 & 3 is 68 kips The vertical components of these bolt forces are: At bolts 1 & 3, 7 . 37 74 157 80 1 1 , 1 = = = R r r R o v kips At bolt 2, 2 . 59 68 87 . 8 . 91 80 2 , 3 , 2 = = = = R R R v v kips Then 8 . 193 2 2 . 59 2 7 . 37 1 , = + = = = m n v n R P kips Check: ( ) ( ) 1 = + = m n n n o R r r e P ( ) 5682 2 68 8 . 91 2 74 157 80 75 8 . 193 = + kip- mm. (b) A trail location of the instantaneous center is chosen as r o =120 mm At ultimate limit state, at bolts 1 & 4, R 1 =R 4 =74 kips, and their corresponding deformation is 0.34 inches. At bolts 2 & 3, the amount of deformation is 24 . 34 . 6 . 180 2 . 128 34 . 135 120 45 120 2 2 2 2 max max 2 2 = = + + = = r r in. From the load-deformation curve, it is estimated the force for bolts 2 & 3 is 70.2 kips The vertical components of these bolt forces are: At bolts 1 & 3, 2 . 49 74 6 . 180 120 1 1 , 1 = = = R r r R o v kips At bolt 2, 7 . 65 2 . 70 936 . 2 . 128 120 2 , 3 , 2 = = = = R R R v v kips Then 230 2 7 . 65 2 2 . 49 1 , = + = = = m n v n R P kips 3@90 mm P u 75 mm i.c R 1 R 2 R 3 R 4 DEPARTMENT OF CIVIL ENGINEERING, LAKEHEAD UNIVERSITY 2 Check: ( ) ( ) 1 = + = m n n n o R r r e P ( ) 122 2 2 . 70 2 . 128 2 74 6 . 180 120 75 230 = + kip-mm. Conclusion : resistance is 230 kips. 230/74=3.1 (see Page 3-29) 2. Solution The plates have a cross-sectional area of 4000 mm2. The tensile resistance of the member is governed by gross sectional yielding, i.e., T r = A g F y = 0.9(4000)(300) = 1080,000 N = 1080 kN. In order to fully utilize the full tension strength of the plates, the fillet welds are required to have a factored shear capacity not smaller than 1080 kN....
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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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solution6_2136 - DEPARTMENT OF CIVIL ENGINEERING, LAKEHEAD...

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