Question

Design the lightest available W14 (A992) tension member by the 2016 AISC LRFD for a truss that will carry service DL = 100 kips and service LL = 320 kips. The member is 30 ft. long. The flanges will be bolted to the connection plates (A36) with -in bolts so that four bolts will appear on the cross-section Assume there are at least three bolts per line, standard holes, and the connection is adequate. Hint: Consider all applicable limit states.

Answer 1

GIVEN DATA Member Details A 992 Grade of Steel Length of member, 30 ft Connection Details Diameter of bolt dh 3 in 4 Loads 100 kips Dead Load, Pr Live Load, PL 320 kips CALCULATIONS A 992 For grade of steel, F y Yield stress 50 ksi - Ultimate stress. 65 ksi d h Diamter of bolt hole, 3 1 + 4 8 0.875 in Design load, Р, 1.2PD+1.6P = 632 kips Yielding in the gross section Design strength of gross section is obtained as: FyAg ФТ, 3 х А, 0.9 x 50 45 A - 9 632 Therefore, A g 45 in 14.04 Fracture in the net section shear lag Factor 0.85 (assumed) U = 0.85 XA А, Effective area, = 0.85 A Design strength of net section is obtained as: T F A ф,т, - 0.85 A 0.75 x 65 x 41.438 A IT Therefore A 632 41.4 2 in 15.25 2 Therefore, required area is: in A 15.25 Select W 14x68

GIVEN DATA Member Details Grade of Steel A 992 W 14x68 Selected member Gross area of section 20 in 2 A g Thickness of flange 0.722 in t Breadth of flange Depth of section, Radius of gyration, Length of member bf 10 in d 14 in 2.46 in ry 30 ft Connection Details Member connected to web/two flanges two flanges Diameter of bolt, 3 in h 4 Number of bolts across a section, 4 Number of bolts in line 3 plus 3 in (assumed) 1.5 in (assumed) m Pitch of the bolts S End Distance Edge distance 2.25 in 1.29 in (WT7x34) Distance of centroid from edge - CALCULATIONS A 992 For grade of steel, Yield stress Fr 50 ksi F Ultimate stress, 65 ksi Diamter of bolt hole, d 3 1 + 4 8 0.875 in DESIGN STRENGTH Yielding in the gross section Design strength of gross section is obtained as: T FyAg ф.Т, - x 20.000 0.9 x 50 900 kips Fracture in the net section Ag-n (dt) Net area, А, 20 4 x 0.875 x 0.722 17.47 in2 Length of connection in the direction of loading, L 6 in

Alternative value of U [Case- 7(a)] bf 2/3d U Since the member is connected to two flanges 1-L Shear lag Factor 0.900 U 0.785 Thus U = 0.900 (Maximum of Case-2 & Case-7(a)) UA Effective area, А. 0.900 x 17.473 15.73 in Design strength of net section is obtained as: ф.Т, FA 0.75 x 65 15.726 х 766.6 kips Block Shear 4 [Length (a-b) xt Gross and net sections of shear, A gv — 4х 7.5 x 0.722 21.66 in А у 4 ILength (a-b)] minus [(m-1)+0.5]d,}xt nv 2.5 x 0.875 ) x = 4 ( 7.5 0.722 15.34 in 4 length (b-c) xt Gross and net sections of tension, A gt x 0.722 = 4x 2.25 6.498 in2 А t 4 length (b-c] minus [0.5xdA} x t = 4 ( 0.875 x 2.25 0.5 x 0.722 5.235 in Comparing 0.6F,Anv and FAt 0.6FAV0.6 x F A 598.38 kips 65 15.34 х nv 340.28 kips 65 x 5.235 = Therefore 0.6F,Av nt F A Hence shear fracture-tension yielding controls (0.6F,AnFyA 9t ] R bs Design rupture strength, + 0.75 x[ 598.38 692.5 kips 6.50 50 x + Thus, design strength of the member is the smallest of the three values, that is 632 kips Pu 692.5 kips Tn > SAFE Hence, Ir Slenderness ratio, 30 х 12 2.46 146.3 300 SAFE Hence,