Question

Q2. Determine the effective net area of a channel section C15 x 50 (A36 steel) with two lines of 4-in bolts in its web (four in a line 4-in on center) as shown in Fig. Q2. The properties of the channel are as follows: A 14.7 in2, d 0.65 in, I 404 in, Iy 11.0 in, = 0.799 in. = 15.0 in, tw = 0.716 in, br = 3.72 in, t - (25%) ELEVATION $4 in 2 in 2 in 2 in ELEVATION Fig. 2b p Fig. 2a

Answer 1

GIVEN DATA Member Details Grade of Steel 36 : 15x50 Member designation 14.7 in Gross area of section, A. Thickness fweb 0.716 in Connection Details Member connected to web/flange, web Diameter of bolt, d b in 4 in (along axis of member) in (along transverse direction) 14 in (in the direction Pitch of the bolts, Gage of the bolts. 9 loading) Length of connection, 0.799 in Distance of centroid from edge, - CALCULATIONS For grade of steel А 36 Yield stress, 36 ksi Ultimate stress, 58 ksi Diameter 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: Ф.Т, 3 фFА, 0.9 x 36 x 14.7 476.3 kips Fracture in the net section i) Fracture path 1 hole 0 stagger = Ag-n (dht) Net area, 0.875 14.7 1 x 0.716 14.074 in i Fracture path: 2 holes 1 stagger = A,-n (dt)+m(s/4g x t) A. Net area. 14.7 2 x 0.875 0.716 22/4x 1 x 4) x 0.716 =13.626 2 =13.626 in Hence, net area, A, NOTE:

Shear lag Factor when connected to entire section 1.0 when connected to part of section F1-x/Ls0.9 where is distance of centroid from edge and Lis the length of connection Since the member is connected to web = 1-x/L Shear lag Factor 0.9 - Effective area, A. = UA 0.9 x 13.626 =12.263 in2