Question

Question 12 A column ABC is supported at ends A and Cand compressed by an axial load P (figure a). The column is pin-supported at the ends (support A and C). Lateral support is provided at point B but only in the plane ZX. The column is constructed of two channel sections (C6 x 8.2) back to back (see figure). (Given: A=4.78 in?, 1x=26.2 in*, Ly=2.627 in^, E = 29,500 ksi and o, = 36 ksi) 1. Determine the largest vertical force P that can be applied without causing buckling 2. Compared the critical stress to the yield stress. 3. If the FS -2.5, what is the allowable value of load P. N P C MacBook Air 30 3 DOD DOO F4 FS 1 # 3 * $ 4 % 5 & 7 6 8 9

Answer 1

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Given Ix By Ix = 26.2 in 4 Iy = 2.697 int 26.2 4.72 -2.341 in 2.527 -0.741 is 4.78 Yy = 115ft deffected A shape. 30ft E = 29500 ksi A=4.78 in 2 k=1, pinned-pinned klx 360 234 :153.85 - 242.9 ry= 36 ksi 15ft we have, k=1 / pinned-fixed. fa-citetin) y direchim) Effective length in x- di rechin, Lex = k be = 30 = 3044 = 360 in effective length in y-direction, Ley= k Ly = 1x15 = 1544 = 180 in (i) Slender den ratio ina-direction (SR), a li) Slenduren ratio in y-direchom (Srly = key ingebry (SR)y> ERIye y-direction governs the design (bucklung) : critical buckling load, MER &2 x 295008 9.627 Per = (80) (1) Largest value of p; Now, de = : sc>1.5 (No in elastic buckling) = 23.5 kips (kly2 KL by 36 229500 180 0.741X* = 2.7 ) sheen, for Por A 23.6 4.78 - 4.94 ksi 0.877*X36 but critical storns in given by, ter = 0.877 64.. 2.3 Siren :4.33 os bey = 4.33. ksi < fer : buckling :: Largest jurce applied , p = 4.33x Axle = 4.33X4.78 = 20.7 Kips <2) Critical stren, tér = 4.33 > $or (1.94), [cribeat stress > yieldsnen] (3) of F.S=3,5, P= = 8.98 kips >> Allowed, P = 8.28 kips 20.7 2.5