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minimum radius of gyration of L5x3-1/2x3/4 = 0.744 in
length of angle section = 10 ft
assuming pinned condition, effective length = 10*12=120 in
slenderness ratio = 120/0.744 = 161.3
Euler buckling stress = pi2E/(KL/r)2 = pi2*29000/(161.32)=11 ksi
critical stress=0.658(36/11)*36=9.1 ksi
area of cross section=5.85 in2
allowable load on member = 5.85*9.1=53.2 kips
PLEASE, I WANT THE ANSWER WITH GOOD HANDWRITING PLEASE, AND CLEAN AND CLEAR. THANK YOU 3. An L 5x 3 %x % angle is used...
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