Question

advanced design metal structures

PROBLEM: AN ANGLE WITH STAGGERED FASTENERS IN EACH LEG IS SHOWN. A-36 STEEL IS USED AND HOLES ARE FOR 7/8" DIAMETER BOLTS. a} DETERMINE THE DESIGN STRENGTH USING LRFD APPROACH. b} WHAT WILL BE THE ALLOWABLE STRENGTH USING ASD METHOD? 24" 2 4 a 16 O ob 26 1 - oC 1 46") F--------- 0 0 0 o 311 o od L8 x 6 x 8:

Answer 1

GIVEN DATA Member Details A 36 Grade of Steel : 1 х Member dimension 13 ( Length = 1.25+2.5+4.25+3+2 Thickness = 1/2 Connection Details 7 in Diameter of bolt, do 2 /4 Number of bolts across a section, (in the direction of loading) Number of bolts in line Pitch of the bolts 1.5 in (in the direction of loading) 2.5 in (along transverse direction) Gage of the bolts, = 4.25 in 3 in %3D CALCULATIONS A 36 , For grade of steel, 36 ksi Yield stress, Fy Ultimate stress, 58 ksi in 2 Gross area of section, Ag 6.5 Thickness of member, in 0.5 Diamter of bolt hole, in %3D DESIGN STRENGTH i) Yielding in the gross section Design strength of gross section is obtained as: LRFD ASD F, -Ag O. T, = 0:FyAg 0.9 х 36 х 6.5 6.5 / 1.67 36 х 210.6 kips %3D 140.1 kips ii) Fracture in the net section i) Fracture path: a-b-d-f 2 holes O staggers A,= A,-n (d ,t) Net area, 2 х 1.000 x 6.5 0.500 5.5 in? a-b-c-d-e-g ii) Fracture path: 3 staggers 4 holes A,= A,-n (d ,t )+m(s°/4g xt) Net area, - 4 x 1.000 x 6.5 0.500 1 x ( 1.5 /4x 2.5 ) x 0.500

1 x ( 1.5 /4x 4.25 ) x 1x( 1.5 /4x 0.500 3 )х 0.500 4.772 in? A,= 4,772 in? Hence, net area, A, = UA , Effective area, 1 x 4.772 in? 4.772 Design strength of net section is obtained as: LRFD ASD т, 3 O, T, = 0, FuA, A. 0.75 x 58 x 4.772 207.6 kips 4.772 / 2.0 58 x 138.4 kips The design strength of the member is the smaller of the design strength of gross section, design strength of net section. Thus, LRFD ASD 207.6 kips 138.4 kips Тл 3