05-05ChapGere.0002 - F (per inch of length of weld) must be...

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Solution 5.11-2 Welded steel girder SECTION 5.11 Built-Up Beams 351 All dimensions in millimeters. Allowable load in shear for one weld is 900 kN/m. [ f allow 5 2(900) 5 1800 kN/m 5 1638 3 10 6 mm 4 Q 5 Q flange 5 A f d f 5 (280)(25)(312.5) 5 2.1875 3 10 6 mm 3 5 1.35 MN V max 5 f allow I Q 5 (1800 kN / m)(1638 3 10 6 mm 4 ) 2.1875 3 10 6 mm 3 I 5 bh 3 12 2 ( b 2 t ) h 1 3 12 5 1 12 (280)(650) 3 2 1 12 (265)(600) 3 f 5 VQ I Ê V max 5 f allow I Q z y O b = 280 t = 15 25 25 h 1 = 600 weld h = 650 Problem 5.11-3 A welded steel girder having the cross section shown in the figure is fabricated of two 18 in. 3 1 in. flange plates and a 64 in. 3 3/8 in. web plate. The plates are joined by four longitudinal fillet welds that run continuously throughout the length of the girder. If the girder is subjected to a shear force of 300 kips, what force
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Unformatted text preview: F (per inch of length of weld) must be resisted by each weld? Solution 5.11-3 Welded steel girder z y O 18 in. in. 1 in. 1 in. 64 in. 3 8 z y O b = 18 t = 0.375 1.0 1.0 h 1 = 64 weld h = 66 All dimensions in inches. V 5 300 k F 5 force per inch of length of one weld f 5 shear flow 5 46,220 in. 4 Q 5 Q flange 5 A f d f 5 (18)(1.0)(32.5) 5 585 in. 3 F 5 VQ 2 I 5 (300 k)(585 in. 3 ) 2(46,220 in. 4 ) 5 1900 lb / in. I 5 bh 3 12 2 ( b 2 t ) h 1 3 12 5 1 12 (18)(66) 3 2 1 12 (17.625)(64) 3 F 5 VQ 2 I f 5 2 F 5 VQ I A-PDF Split DEMO : Purchase from www.A-PDF.com to remove the watermark...
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This note was uploaded on 09/20/2009 for the course COE 3001 taught by Professor Armanios during the Spring '08 term at Georgia Institute of Technology.

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