Question

Steel Design

2. (50 pts) Design an A36-steel tension member 16 ft long consisting of two channels, bolted through their webs to a single 1 inch gusset plate, to transmit 75 kips DL and 93 kips LL. The bolts are 7/8 in. Bolt spacing 4 inch and bolt edge distance 3 inch. In addition to checking block shear in the channels, design the thickness of the gusset plate to resist block shear.

Answer 1

GIVEN DATA Member Details А 36 Grade of Steel Span of member 16 ft = Loads PD 75 kips Service dead load, PL Service live load, 93 kips CALCULATIONS А 36 , For grade of steel, 36 Мра Yield stress, Fy Ultimate stress FM 58 Mpa = 1.2D 1.6Lu Ти Design Load 238.8 kips Yielding in the gross section Design strength of gross section is obtained as: F,Ag ф.Т. 0.9 x 36 хА, 32.4 Ag Therefore, 238.8 32.4 Ag 2 7.37 in - Fracture in the net section Shear lag Factor (assumed) 0.85 - UA Effective area, Ae = 0.85 A Design strength of net section is obtained as: .T 0.85 A 0.75 x 58 X 36.98 A Therefore 238.8 36.98 A. 2 6.46 in = Threfore, required area is 7.37 in A = Select 2C 9x13.4

GIVEN DATA Member Details Grade of Steel A 36 Member designation 2C 9x13.4 2 7,88 in Gross area of section, A g Thickness of web t 0.233 in Radius of gyration, 0.666 in rmin Connection Details Member connected to web/flange, web : Diameter of bolt, d b 7 in 8 Number of bolts across a section, 2 n Number of bolts in line, 3 m Pitch of the bolts (along axis of member) 4 in Gage of the bolts, End Distance 4 in (along transverse direction) 3 in , Distance of centroid from edge, 0.601 in CALCULATIONS A 36 For grade of steel Yield stress, 36 ksi F y Ultimate stress, 58 ksi FV Diamter of bolt hole, d h 7+ 1 8 8 1 in DESIGN STRENGTH Yielding in the gross section Design strength of gross section is obtained as Tn= 4.FyA 0.9 x 36 x 7.88 255.3 kips Fracture in the net section = Ag-n (dt) Net area A n 7.88 2 x 1.000 x 0.233 7.414 in NOTE Shear lag Factor when connected to entire section 1.0 when connected to part of section 1-x/Ls0.9 where is distance of centroid from edge and Lis the length of connection Length of connection in the direction of loading, 8 in

Since the member is connected to web =1-/L Shear lag Factor U 0.9 Effective area, UA А, 0.9 х 7.414 6.673 in Design strength of net section is obtained as: Tn= ,FA, x 6.673 0.75 x 58 290.3 kips Block Shear 4 x [Length (a-b )] x1 A gv Gross and net sections of shear, 4х 11 х 0.233 10.252 in2 = 4 x {[Length (a -b)] minus [(m-1)+0.5]d , } xt A nv 4 11 2.5 1.000x 0.233 X 7.922 in2 = 2x[length (b-c )] xt Gross and net sections of tension, A gt 2 x 4 x 0.233 1.864 in 2x{Ilength (b-c )] minus [(n-1)xd}xt 2 x4 1.0 nt 1.000x 0.233 X 1.398 in2 Comparing 0.6F,A n and FAt 0.6F Am0.6 FuAnt and FyA 275.69 kips 0,6 58 x 7.922 х X х 81.08 kips 58 x 1.4 Therefore 0.6F Am Fu A nt shear fracture-tension yielding controls Hence, R= [0.6FA nu FyA gr l Design rupture strength bs х 1.864 ] 0.75 x [ 275.69 36 257.1 kips The design strength of the member is the smallest of the design strength of gross section, design strength of net section and design rupture strength. Thus, Tn =255.3 kips 238.8 kips Pu > Hence SAFE CHECK FOR SLENDERNESS 16 x 12 0.666 288.29 300 SAFE < To resist block shear, the minimum required thickness must be greater than the sum of thicknesses of web of each channel. Therefore,

t 2 x 0.233 = 0.466 in Hence, provide thickness 0.5 in