Recitation #01 summary double angle table

Recitation #01 summary double angle table - it will be this...

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Trial Sizes Minimum Required r __1.2 in ___ Minimum Required A g (double angle) 5.24 in 2 ____ Minimum Required A e (double angle) _4.84 in 2 _ Member Angle thickness Double angle A g r y r x Double Angle A n Double Angle A e Ok? 2L4x3x1/2 ½ 6.51 1.32 1.24 5.51 4.40 NG 2L4x3-1/2x1/2 1/2 7.01 1.57 1.23 6.01 4.81 OK 2L4x4x7/16 7/16 6.61 1.81 1.22 5.74 4.59 NG 2L5x3x7/16 7/16 6.62 1.23 1.59 5.75 4.60 NG 2L6x3-1/2x3/8 3/8 6.88 1.38 1.93 6.13 4.90 OK FINAL SELECTION ___2L6x3-1/2x3/8 _________ 4c: Check to see if fillers are required Need L z /r z <300 (and < max (L/r x , L/r y )). Smallest of r x and r y is r y, so max(L/r x , L/r y ) will by L/r y r y = 1.38 (for LLBB and 3/8” gusset, so
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Unformatted text preview: it will be this or larger if we use angles LLBB [long legs back to back]). r z = 0.76. This is less than 1.2, so if we do nothing, we will not satisfy L z /r z <300. Fix this by adding fillers to reduce L z (note this doesn’t reduce L, just reduces the length over which a single angle acts by itself). If we add one filler L z = L/2 = 15’ Æ L z /r z = 237, so this satisfies code (is less than 300). Standard of practice is to have L z /r z < max (L/r x , L/r y ). Because r y is smaller than r x. L/r y will be bigger than L/r x . L/r y = 261. Thus one filler also satisfies L z /r z < max (L/r x , L/r y )....
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This note was uploaded on 06/09/2010 for the course CIVIL 4401 taught by Professor Shield during the Fall '09 term at University of Minnesota Crookston.

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