For the welded builtup section shown determine the maximum f

For the welded built-up section shown, determine the maximum factored axial load that can be applied. Use A572 Grade 50 steel. Pu PL 1x 10 6 ft Weak axis braces 6 ft PL 1 x 10 Section a-a

Solution

Moment of inertia of section about major axis = 2*10*1³/12+2*1*10*5.5²+0.75*10³/12=669.17 in⁴

Moment of inertia about minor axis = 2*1*10³/12 + 10*0.75³/12=167 in⁴

Area of cross section = 2*1*10 + 0.75*10=27.5 in²

radius of gyration about major axis = sqrt(669.17/27.5)=4.93 in

radius of gyration about minor axis = sqrt(167/27.5)=2.46 in

KL/r about major axis = 20*12/4.93=48.68

KL/r about minor axis = 8*12/2.46=39.0

Therefore, KL/r about major axis governs the design

4.71*sqrt(E/fy)=4.71*sqrt(29000/50)=113.4 > KL/r about major axis

Fe = pi²*E/ (KL/r)²= 120.78 ksi

F_cr = 0.658^Fy/Fe*Fy = 0.658^50/120.78*50 = 42 ksi

Design strength = 0.9*42*27.5=1039.5 kips