Question

the bolts is in inches not Ft

need help asap

hw help please my case number is #3 Problem 1 Select for the conditions described, using F, - 50 ksi and F.-65ksi. Show calculations to select trial sizes and check for Gross Section Yielding and Tensile Rupture. Assume the member is to have two lines of bolts in each flange. Design of Tension members Method Select the lightest Po (kips) Case P(kips) Member Bolt diameter Length (ft) Section (ft) 7/8 ASD W10 100 75 75 125 25 25 25 2 W10 100 7/8 LRFD ASD 175 W10

Answer 1

GIVEN DATA Member Details A 992 Grade of Steel 25 ft Length of member, Connection Details Diameter of bolt in di Loads 175 kips P p Dead load, 125 kips P Live load, CALCULATIONS For grade of steel Yield stress A 992 50 ksi 65 ksi Ultimate stress, d. 1 + Diamter of bolt hole 8 1.125 in - Design load, 300 kips Yielding in the gross section Design strength of gross section is obtained as: T = ,F,A x Ag /1.67 50 29.94 Ag 300/ 29.9 A. Therefore, in 10.02 Fracture in the net section 0.85 (assumed) U Shear lag Factor 0.85 xA A Effective area, 0.85 A Design strength of net section is obtained as: T, = F A 0.85 A, /2 65 27.625 A 300 27.6 A. Therefore,

in 2 10.86 Therefore, required area is: 10.86 in Select W 10x45

GIVEN DATA Member Details Grade of Steel A 992 :W 10x45 Selected member Gross area of section, Ag 13.3 in Thickness of flange 0.62 in Breadth of flange, bf 8.02 in - Depth of section, Radius of gyration, d 10.1 in - 2.01 in ry 25 ft Length of member, Connection Details Member connected to web/two flanges two flanges 1 in Diameter of bolt, - Number of bolts across a section, 4 3 plus (assumed) Number of bolts in line m Pitch of the bolts 3 in S End Distance (assumed) 2 in Edge distance Distance of centroid from edge 1.26 in (WT5X22.5) 0.907 in CALCULATIONS A 992, For grade of steel, Yield stress, Fy 50 ksi Ultimate stress F, 65 ksi d h Diamter of bolt hole, 1+ 1 1.125 in DESIGN STRENGTH Yielding in the gross section Design strength of gross section is obtained as /1.67 F,A 13.3 /1.67 50 x 398.2 kips - Fracture in the net section A Ag -n (dt) Net area, 13.3 0.620 4 x 1.125 x 10.51 in Length of connection in the direction of loading, 6 in Alternative value of U [Case- 7(a)] Since the member is connected to two flanges

1-L bf 2/3d. U= Shear lag Factor 0.900 0.849 Thus, U = 0.900 (Maximum of Case-2 & Case-7(a) UA Effective area, A 0.900 x 10.51 9.459 in Design strength of net section is obtained as F yA /2 9.459 /2 65 x 307.42 kips Thus, design strength of the member is the smallest of the two values, that is 300 kips P 307.42 kips Hence, SAFE r 12/ 2.01 Slenderness ratio, 25 x 149.25 < 300 - Hence, SAFE