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Shear flow examples

Shear flow examples - The beam is fabricated from two...

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Unformatted text preview: The beam is fabricated from two equivalent 7-structural tees and two plates. Each plate has a height of U6 in. and a thickness of 0.5 in. If the bolts are spaced at “‘5 = 8 in, determine the maximum shear force V that can We applied to the cross section. Each bolt can resist a shear force of 15 kip. Section Properties: 1 [m = Et3l(9’) ~Tli(2.5}(83) 1 7240.510”) +fim(o’) = 93.25 in‘ {2 =2? ' = 2.5mm) +4.2s(3) (0.5) = 10.125 in’ Shear FIcw:Sincetherea.1-emvoshearplanas ondtebolt. the l'fl'iJ. aflowableshear flow is g = 2‘ IS] = 3.75 kiplin. V9 4'7" _ mans) 93. 25 3.15 y =34-5kip Am Internal Slur For“: as $110er on shear anagram. v“ = 300 lb. Section Properties: Z'A O.5(3)(1)+2.5(1)[3} "E: 3(l)+l{3) =15” mu 1 I,“ = Eon t') ”(nus—0.5): 1 -Ann '89:) +130“ 3’) + 1(3)(2.5— 1.5)“ =3.50in‘ Q=i§A’=I(3)(IJ=3.mm’ .5ch Flow: Since men: an: M0 glue joints. hence 1(9'9] 1[800(3.00) q=- — — 3.50 ]=l41 {Mil}. Ans 0691.4 |,_ .4 p.053]. 21"5 The channel is subjected to a shear of V = 75 RN. Determine the shear flow developed at point A. ; = fl = 0.015(0.4)(0.03) + 2[0.l3(0.2)[0.03)] =0.0725In 2A 0.4(0.03) + 2(0.2)(0.03) I = 1—;(0.4)(0.033) + 0.4(0.03)(Ct0725—0.015)a + 2[l—'2(0.03}(0.2’) + 0.03(0.2)(0.13—0.0?25)’] = 0.12025(10")m‘ Q. = im’ =0.0575(0.2)(0.03) = 0.3450(10")m’ - 1’9: 9 I = 75(10’}(0.3450)(10") 0.12025(10-3) q; = 215 thn Ans ...
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