Question

Problem 3: A W 410 x 46.1 is connected at its ends with 12.5 mm plates at both flanges with 6 M 22 bolts at each flange, as shown in figure. If the member is made of steel A36, determine its factored tensile resistance based on; 1- Yielding on Ag 2- Fracture on Ae 3- Block shear rupture. 丰丰丰 丰丰丰 W 410 x 46.1 Given that for W 410 x 46.1 Ag = 5890 mm2, t f= 11.2 mm, bf= 140 mm d= 403 mm, tw 7.0 mm 田田田 t- *:田田田 50 75 75 50 mm

steel design

Answer 1

Effective area eUA 0.850 x 4814.8 4093 in Design strength of net section is obtained as 400 0.75 x 1228 kips x 4092.580 Block Shear Gross and net sections of shear, Agy4 Length (a-b)]xt - 4x 200 x 11.2 8960 in Anv4 Length (a-b)] minus(m-1)+0.5]dhxt 4( 200-2.5 x 24.000)x 11.2 6272 in Gross and net sections of tension, Agt4 [length (b-c)] xt - 4x 30x 11.2 1344 in A nt4 length (b-c)] minus [0.5xdhl xt 4( 30-0.5 x 24.000) x 11.2 806.4 in2 Comparing 0.6F An ad FAnt 0.6F,AnF 0.6 x 400 x 6272 1505.28 KN 322.56 kN FAF400 x 806.4 Therefore Hence, shear fracture-tension yielding controls Design rupture strength 0.75 x 1505.28 250 x 1344.0 /1000] 1381 kN Thus, design strength of the member is the smallest of the three values, that is 1228 kN