Question

Q3. Determine the LRFD strengths for aW12 x 50 (A36 steel) with two lines of 4-in bolts in its flange (three in a line 4-in on center) as shown in Fig. Q3. Gross section yielding, tensile rupture and blockshear strengths must be investigated. Properties of the beam properties of the channel are as follows: A 14.6 in2, d = 12.2 in, tw 8.08 in, t 0.64 in, I are as follows: The 0.370 in, by 391 in, ly 56.3 in, x 1.17 in. for WT 6x 25. (25%) 2 in 4 in 4 in 6 in Fig. Q3

Answer 1

GIVEN DATA Member Details Grade of Steel A 36 Member designation W 12x50 Gross area of section, А, 14.6 in t. Thickness flange Breadth of flange, Depth of section, 0.64 in bt 8,08 in - 12.2 in - Connection Details Member connected to web/two flanges two flanges 3 in Diameter of bolt, d Number of bolts across a section A Number of bolts in line Pitch of the bolts 4 in S End Distance 2 in Edge distance 1,04 in - Distance of centroid from edge 1.17 in (WT6X25) Y CALCULATIONS * steel, A 36 , For grade Yield stress F. 36 ksi Ultimate stress, 58 ksi d h Diamter of :hole, 3 л 8 0.875 in DESIGN STRENGTH Yielding in the gross section Design strength of gross section is obtained as: T FyAg x 14.600 0.9 x 36 473 kips Fracture in the net section Net area ("p)u- 4 x A . 14.6 0.875 x 0.640 12.36 in2 Length of connection in the direction f loading, 8 in Since the member is connected to two flanges Alternative value of U [Case-7(a)] bf<2/3d U 0.850 1-/L Shear lag Factor 0.854 0.854 (Maximum of Case-2 & Case-7(a) Thus,

UA Effective area, A . 0.854 x 12.36 10.55 in Design strength of net section is obtained as ф,т, FuA 0.75 x 58 10.552 459 kips Block Shear Length (a-b) xt Gross and net sections of shear, A gr = 4 x х 0.64 10 25.6 in2 = 4 {[Length (a-b)] minus [(m-1)+0.5]d Anr xt = 4 ( 0.875 x 10 2.5 x 0.640 20 in2 length (b-c)] xt Gross and net sections of tension, A a - =4x 1.04 x 0.64 2.662 in2 4 length (b-c)] minus [0.5xd h xt At - 0.640 = 4 ( 0.875)x 1.04 0.5 x 1.542 in2 Comparing 0.6F y A ny and Fy A nt 0.6F A 696.00 kips 0.6 x 58 20 X 89.44 kips FuA 58 > 1.542 - Therefore, 0.6F A F Ant Hence, shear fracture-tension yielding controls [0.6FA mFy Design rupture strength Rb 0.75 x 696.00 36 2.66 1 593.9 kips Thus, design strength of the member is the smallest of the three values, that is ф, Т, - 459 kips