Determine the designcapacity using ASD and NSCPspecifications for the connectionshown. Use A36 steel and D20diameter bolt on an oversized holediameter. Show complete solutionand round your final answer to thenearest thousandths.Given formula:●.●●Due GSY:OPn = ØAgFy; = 0.90Due TRSOPn = QUAnFu : = 0.75U = 1-Due to BSSØRn = Ø[0.60AnFu + UbsAntFu]≤ Ø[0.60AgvFy+UbsªntFu] ;0.75Slendemess Ratio≤ 300;LT'minFor section propertiesIx = 1ox + A(y²)ly=Loy + A(x²)r ==117125←M20M22M24278300. Nominal hole diameterBoltHole diameterStandard OversizeddiameterM16182022242428273075-x 14.35 mmC8x18.75A₁ = 3,555 mm²bf 64.2 mmtf = 9.9 mmtw 12.4 mmd = 203.2 mmly 18.31 x10 mm"lx = 0.82 x10 mm"75dnet+ 2mm+ 2mm+ 2mm+2mm15PuORPA

C 8 x 18.75(A = 3555 mm2 , tw = 12.4 mm )A 36 steel (Fy = 250 Mpa and Fu = 400 Mpa)Bolt diameter…

Determine the design capacity using ASD and NSCP specifications for the connection shown. Use A36 steel and D20 diameter bolt on an oversized hole diameter. Show complete solution and round your final answer to the nearest thousandths. 117 L 278 300 O 75 75 Pu OF P

Determine the design capacity using ASD and NSCP specifications for the connection shown. Use A36 steel and D20 diameter bolt on an oversized hole diameter. Show complete solution and round your final answer to the nearest thousandths. 117 L 278 300 O 75 75 Pu OF P

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Determine the design capacity using ASD and NSCP specifications for the connection shown. Use A36 steel and D20 diameter bolt on an oversized hole 117 75 75 diameter. Show complete solution and round your final answer to the Pu OR nearest thousandths. PA 125 Given formula: Nominal hole diameter Due GSY: Bolt Hole diameter ØPn = ØA,F, ;ø= 0.90 dnet diameter Standard Oversized Due TRS M16 18 20 + 2mm ØP, = ØUA„Fu ;ø= 0.75 M20 22 24 + 2mm M22 24 28 + 2mm U = 1- M24 27 30 + 2mm Due to BSS ØRn = Ø[0. 60A,nyFu + U psAntFu] < ø[0.60A9„F, + UbgAntFu] ; ¢ = y -X = 14.35 mm 0.75 Slendemess Ratio < 300 ; C8x18.75 I'min For section properties I = Iox + A(y²) ly = Ioy + A(x²) A, = 3,555 mm? bf = 64.2 mm tf = 9.9 mm tw = 12.4 mm d = 203.2 mm ly = 18.31 x106 mm* Ix = 0.82 x106 mm* r = 008 75 75
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