Question

Select a single angle for the case given below to support a tensile load A single gage line of at least three bolts is to be used Q 5 Bolt diameter Length (in.) Steel Case DL LL (ft) (kips) (kips) 15 3/4 40 А36 15 3/4 15 A572 Gr 50 40 2 15

Answer 1

Angle Section care 1. DL = 15 LL = 40 A 36 Steel bolt dia = 3 in Length = 15ft Fy = 36 ksi Fu= 58 ksi Factored load Tu = 1.20 + 1.62 = 1.2 (15) + 1-6 (40) | Tu = 82 kips (i) Tensile Strength of member rusing section. gross фе Т, 2 ти Tu =0.900 Fy x Ag 82 = 0.4 X 36 X AG Ag = 2. 53 m² li) of using Net Tensile strength member section at Tn = 0.75 X Fu & Ae 82 = 0.75 X 56 X AL Ae = 1.88 in? Effective Area Net area An = Aero

U shear lag = 0.8 An = 1.88 Net area, = 2.35 in 2 0.8 Let Slenderness - 5 be 305 Minimum r = - = 15x12 300 = 0.6 in Select Single Angle member: Ag = Ant ncan) = 2.35 + :(+b) (Ag) = 2.623 in 2 Required Let select an Angle: [ 5x5x5 - Ag = 3.07.in? > Ag required q=0.990 in > r rerired Hanc me LS480g nu se

Case (2) D = 15 L: 40 L= 156€ A 572 Gr 50 Fy = 50 ksi Fu = 65 ksi Factoved load : Tu = 1.2 D + 1.62 -(1.2 x 15)+ 1.6 (40) = 82 kips. (i) Tensile strengen of member using Gross section de in = 0.900 Fy x Ag 82 = 0.qx 50 x Ag Ag = 1.82 in? Met section : Tensile li Strength of member Ot Tr = 0.75X Ae X Fu 82 = 0.75 x 65 x Ae Ae = 1.68 in 2 area an t Net sectional An = 2.101h?

- An tnda Ag. = 2.10 + (+) Ag = 2.373 in? Slenderness Ratio - 300 - -06. 1512 300 Let select an angle: L 4* 43 : A = 2.86 in? > Agresuired r = 0.779 in 7 r required. use angle [L 4X4X%.