Question

3) Determine the tensile design strength. The member is a PL 5/8 x 10 and the steel is A36. All bolts are 7/8 inch in diameter. (Hint: determine the YLS and FLS and compare with Pu) оо ооо оо

Answer 1

GIVEN DATA Member Details Grade of Steel A 36 5 х Plate dimension 10 8 Connection Details Diameter of bolt db= 7 in 8 Number of bolts across a section 2 /3 Number of bolts in line (in the direction of loading) 2 in (in the direction of loading) 3 m Pitch of the bolts S 3 in 3 in (along transverse direction) Gage of the bolts, g End Distance, 2 in Edge Distance, 2 in CALCULATIONS For grade of steel, A 36 Yield stress FT 50 ksi Ultimate stress, F= 62 ksi Gross area of section, Ag= 6.25 in Thickness of member 0.625 in = Diamter of bolt hole d = 7 + 1 8 8 1 in DESIGN STRENGTH Yielding in the gross section Design strength of gross section is obtained as: Tn=,FyAg 9 х 50 х 6.25 0.9 = 281.3 kips ii) Fracture in the net section i) Fracture path: 3 holes 0 staggers An Ag-n (dt) Net area, 1.000 x 6.25 3 x 0.625 4.375 in i) Fracture path: 3 holes 2 staggers Ag-n (dht+m{s*/4g x t) Net area, An 1.000 x 6.25 3х 0.625 2x( 2.0%/4х )х 0.625 3 + 4.792 in ii) Fracture path: 3 holes 2 staggers A-n (dht)+m(s%/4g x t Net area, An

1.000 x 6.25 3 x 0.625 2 x ( 3.02/4x )х 0.625 5.313 in 4.375 in Hence, net area, A, Ae UA Effective area, 1 x 4.375 in 4,375 = Design strength of net section is obtained as: Tn= ,FuA, 62 x 4.375 0.75 x 203.4 kips The design strength of the member is the smallest of the design strength of gross section, design strength of net section. Thus, Ф,т, 203.4 kips