Question

Select the lightest C section that will safely support the service tensile loads PD = 50 k and PL = 65 k. The member is to be 14 ft long and is assumed to have two lines of holes for 4-in bolts in the web. Assume that there are at least three holes in each line 3-in on center. Use A36 steel. Neglect block shear.

Answer 1

GIVEN DATA Member Details Grade of Steel Span of member A 36 14 ft Loads Service dead load, Service live load, P P = = 50 kips 65 kips CALCULATIONS For grade of steel, A 36, Yield stress, Ultimate stress, F F = = 36 Mpa 58 Mpa Design Load T = 1.2D + 1.6L 164 kips Yielding in the gross section Design strength of gross section is obtained as: 0:1, = 0F,A, = 0.9 X 36 32.4 A, Therefore, A = 164 / 32.4 = 5.06 in? XA, Fracture in the net section Shear lag Factor Effective area, (assumed) U A = 0.85 = UA = 0.85 A Design strength of net section is obtained as: 0:1, = 0F,A = 0.75 x = 36.98 A, 58 x 0.85 A, Therefore, = = 164 / 36.98 4.44 in 2 Threfore, required area is A = 5.06 in 2 Select C 8x18.75

GIVEN DATA Member Details Grade of Steel Span of member A 36 14 ft Loads Service dead load, Service live load, P P = = 50 kips 65 kips CALCULATIONS For grade of steel, A 36, Yield stress, Ultimate stress, F F = = 36 Mpa 58 Mpa Design Load T = 1.2D + 1.6L 164 kips Yielding in the gross section Design strength of gross section is obtained as: 0:1, = 0F,A, = 0.9 X 36 32.4 A, Therefore, A = 164 / 32.4 = 5.06 in? XA, Fracture in the net section Shear lag Factor Effective area, (assumed) U A = 0.85 = UA = 0.85 A Design strength of net section is obtained as: 0:1, = 0F,A = 0.75 x = 36.98 A, 58 x 0.85 A, Therefore, = = 164 / 36.98 4.44 in 2 Threfore, required area is A = 5.06 in 2 Select C 8x18.75

CAPACITY CHECK GIVEN DATA Load Applied Dead Load Live Load Span of member 50 kips 65 kips 14 ft 1 = Member Details Grade of Steel Selected section, Gross area of section, Thickness of web Radius of gyration, A, t r : A 36 C 8x18.75 = 5.51 in? = 0.487 in = 0.598 in Connection Details Member connected to web/flange, Diameter of bolt, web 3 in d = 4 m = Number of bolts across a section, Number of bolts in line, Pitch of the bolts End Distance, Distance of centroid from edge, 3 plus 3 in (along axis of member) 2 in (assumed) 0.565 in F = , CALCULATIONS For grade of steel, A 36 Yield stress, Ultimate stress, Diamter of bolt hole, F F = = dm = 36 ksi 58 ksi 3 + Ā 0.875 in 1 8 = DESIGN STRENGTH Yielding in the gross section Design strength of gross section is obtained as: “T, = %,FA = 0.9 X = 178.5 kips 36 X 5.51 Fracture in the net section Net area, A = A,-n(dot) = 5.51 - = 4.658 in? 2 x 0.875 x 0.487 NOTE:

Shear lag Factor U = 1.0 when connected to entire section =1-*/L<0.9 when connected to part of section where x is distance of centroid from edge and Lis the length of connection Length of connection in the direction of loading, L = 6 in Since the member is connected to web Shear lag Factor U = 1-XL = 0.9 Effective area, A = UA, = 0.9 x 4.658 = 4.192 in Design strength of net section is obtained as: %T, = %,F,A, = 0.75 x 58 x 4.192 = 182.4 kips The design strength of the member is the smallest of the design strength of gross section, design strength of net section. Thus, ,1,= 178.5 kips CHECK Factored Design Load TL = 1.2D + 1.62 = 164 kips < 0,1 = 178.5 kips Hence, the connection is SAFE SLENDERNESS CHECK 1/ry = 14 x = 280.94 12/ 0.598 < 300 Hence, the slenderness limit is OK