Question

A horizontal beam AB is pin-supported at the end A and carries a uniformly distributed load with intensity 20 kN/m and a concentrated load F as shown in the figure. The beam is also supported at C by a pinned-end column: the column is restrained laterally at mid-height in the plane of the figure but it is free to deflect perpendicular to the plane of the figure. Assume the column may buckle in any direction. The column is a solid steel bar (E = 200 GPa) of rectangular cross section that has a width b and a height h; determine: • The allowable concentrated load Fallowable using a factor of safety n = 3 with respect to Euler buckling of the column, if b = 0.12 m and h = 0.18 m. • The ratio hlb so that the critical load is the same for buckling about the two principal axes 1-1 and 2-2.

Answer 1

Solution : @ FBD of hstigated beam ACB. W=20 kW/m ton = 20x6 = 120 KN 3m Ray oo old Á ZMA =o → Gy(1) — F(6) - Faq (3) =0 → 49 -6F – 3 (120) = : + y = 6F7360 19 = {F +90 (kN) Buckling analysis of membey CD :- → Since too = = 200 16 = 90+3 F) and both portions CE and T ED Qle hinged connections = 20 at ends, eqzcelive tengths E k * of pstlions CE and ED üle same. and le ce = (deen = Î = 2m Note :- gal hinged connectionesty te - l = 1

Nowy X²E Imm = te? this case, 7 12 F= 200 G1 Pa = 200X109 MPiel te = 2m = 2x100 mm Inmin = mins her heart a min s pogled min $5.832x107, 2-59217, I'min - 2592x107 mm P= 12( 250X17) (2.592X107) (2x103) 2 ☆ ( 2x 2-592.) x polo 4X106 : = 12.791 x104 . Pod = 127.91 kN T Now, 984 sage condition against buckling : = 12111 = 49.64 cy = 42.64 . F +90 3 42.64 → =-47.36 F3-31-58 (kN)

F = - 358 KN means, -FZ 31-58kN. Let's say -F = F". > pl must be greater than 31.58kN. flimust be applied oling the years Tey, vehically cup weil ds. So, minimum zeice srequired to be applied at extreme bcatim + ACB along the y-artis 45 31 55k4 to avoid buckling

© -> Ratio := to have same colitical load. 71 buckling about two principal axes . 1-4 and 2-23 Sam → one = ole → bih=17 > 1 b=I=1