Question

Problem No. 1: (25 points) Compute the capacity of the connection shown in Fig. 1 under tensile loading. The connection is in between a C10x25 (channel section) and aL6x4x 3/4 angle with 3/4" @bolts at the longer side of the angle C10x25, A36 Steel L6 x4x 3/4, A36 Steel 1.5" 3" 3/4 ibolts 1.5 1.5 1.5" 1.5" C10x25, A36 Steel 1.5 3" 339 Fig.1 15:

Answer 1

GIVEN DATA Member Details Grade of Steel 36 : 10x25 Member designation 7.34 in Gross area of section, A, Thickness fweb 0.526 in t Connection Details Member connected to web/flange, web Diameter of bolt, d b in 4 Pitch of the bolts, 1.5 in (along axis of member) in (along transverse direction) 7.5 in (in the direction Gage of the bolts g loading) Length of connection, 0.617 in Distance of centroid from edge, CALCULATIONS For grade of steel А 36 Yield stress, 36 ksi Ultimate stress, 58 ksi Diamter of bolt hole, d i 3 + 1 л 8 0.875 in DESIGN STRENGTH Yielding in the gross section Design strength of gross section is obtained as: tTn Fy Ag 0.9 x 36 x 7.34 237.8 kips Fracture in the net section i) Fracture path: 1 hole 0 staggen = Ag-n (dht) Net area, 0.875 0.526 7.34 1 x 6.88 in i Fracture path: 2 holes 1 stagger = Ag-n (dt)+m(s'/4g x t) A Net area. 0.875 x 7.34 2 x 0.526 = 1.524x 1 x 3 ) х 0.526 6,518 in 6.518 in Hence, net area, A

NOTE: Shear lag Factor when connected to entire section 1.0 when connected to part of section =1-x/Ls0.9 is distance of centroid from edge and Lis the where Y length of connection Since the member is connected to web = 1-x/L Shear lag Factor 0.9 = UA Effective area, A 0.9 x 6,518 5.866 in2 - Design strength of net section is obtained as: Tn= Fy A 58 x 5.866 0.75 x 255.2 kips The design strength of the member is the smallest of the design strength of gross section design strength of net section. Thus, ,T= 237.8 kips

GIVEN DATA Member Details Grade of Steel Member designation A 36 L 6x4x3/4 in 2 Gross area of section, A c 6.94 Thickness of member, 0.75 in Connection Details Member connected to one/two legs one leg d. Diameter of bo lt, 3 in 4 Number of bolts across a section, 2 n Number of bolts in line 3 m Pitch of the bolts 1.5 in S Gage of bolts 3 in 2.75 in Length of connection, (in the direction of loading) 6 in 1.07 in Distance of centroid from edge CALCULATIONS For grade of steel, A 36 Yield stress, F. 36 ksi F.. Ultimate stress, 58 ksi Diamter of bolt hole, d a 3 1 + 4 0.875 in i) Fracture path: 2 hole 0 stagger - А,-n (d,t) Net area, A. 6,94 2 x 0.875 x 0.750 5.628 in ii) Fracture path: 3 holes 2 stagger Ag-n(dt)+m(s/4gx t) Net area, A, = 6.94 3х 0.875 x 0.750 1.5 /4x 1.5 4x 2.75 )x 0.750 1 x 1 x 3 ) x 0.750 5.265 in2 5.265 in A . Hence, net area,

NOTE Shear lag Factor, when connected to two legs 1,0 when connected to one leg = 1-x/L0,9 is distance of centroid from edge and L is the where x length of connection Since the member is connected to one leg, U= 0.6 (with three fasteners-Case 8) 1-x/L Shear lag Factor = 0.822 Thus, 0.822 (Maximum of Case-2 & Case-8) U : UA Effective area, A 0.822 x 5.265 4,328 in2 DESIGN STRENGTH Yielding in the gross section Design strength of gross section is obtained as ,T,= .FyAg 0,9 x 36 6.94 224.9 kips Fracture in the net section Design strength of net section is obtained as: T FyA 0.75 x 58 4.328 X 188.3 kips Thus, design strength of the member is the smaller of the two values, that is dTn188.3 kips Hence, the design strength of the connection is the smaller of the two values. = 188.3 kips Therefore,