Question

A single compression member of 30 ft effective length is obtained by connecting two C6 x 13 steel channels with lacing bars as shown in Figure 1. Determine the distance d for this cross-section corresponding to the most efficient design against buckling (the most efficient design occurs when the resistance to buckling is equal in both planes of symmetry).

C6x 13 properties: A (in) d (in) bf (in) tf (in) tw (in) Ix (in“) rx (in) ly (in) ry (in) x (in) 3.81 6 2.16 0.343 0.437 17.3 2.13 1.05 0.524 0.514 Y d IN -T Y Figure 1 - column

Answer 1

Calculation of distance, d' between channel sections:- * Most efficient design occurs when the resistance ito buckling is equal in both planes Resistance to buckling is to buckling is based on the "Radius of Gynation' .. Radius of Guguration is to be equal in both the planes (Consider combined section) i. llax - agy T Radius of Gegnation (s) = 1 Txx Tyy A a Area will be same in a direction. any So, we get Ixx= Iyy I Use Parallel axis Theorem (x-axis will not change for Gmbined channel section because of symmetry. 2x Ix = 2 (Ty + Ax2) ܛ

2X 17.3 = 2(1.05+ 8-81(7+ ) 2X 17.3 = 2 (1-05+ 3.81(0.61+ + d )° ) A x 17.3=1.05 +3.81/0.514td a عم -81(0.514 + ) 17.3-1.05 = 3.81 31(0.514 + -)? 16.25 n (0.514+ d)² 3.81 14.265 0.514td old 2. 065-0-514 - 2 d=2x10551 ... d=3.102 in ņ 3.15 in The distance à between the channel sections placed back to back is 3.15 in
The distance 'd' for the given cross-section corresponding to the most efficient design against buckling is 3.15inch