Question

13-13. Determine the maximum load P the frame can support without buckling member AB. Assume that AB is made of steel and is pinned at its ends for y-y axis buckling and fixed at its ends for x-x axis buckling. Ex = 200 GPa, y = 360 MPa. 3 m А P T 50 mm H-X 50 mm 4 m 50 mm - 6 m

Answer 1

• Moment of Suchow_puopaties :- Tueetia Ixx bd3 sox1003 = 4.167 X106 mmt 12 Pyy = b3d 563x100 =1.04167 X 106 mmt 12 12 4h I 12 bm • Auea Cross section. of С A= bd=50x100 = 5000 mm2 Radius of gination B ly har Ian A 4.167x 106 Sooo 28.867 mm Somm f 2- rey Tyy А omm 1.04167x lot 5000 14.434 mm ہوا 15ohm *Effective length • Tu x-n dinebian [Both ends fined] (KL)an = LX0.5 = LX0.5 = 6x0.5 = 3m = 3000mm • The 4-y-diretion [ Both end (KL)yy = L=6m = 6oooman hinged

From the stuuture, take point A, A i. From Equilibrium of point A. P Efy = 0 7 Faco Coso + FAB=0 e FAC { fx=07 FAC - Sino =P FAL р Sino FAB 3 SP tand = 3 FAB = FAC .coso Sino = 3 mit colch th cos - - 50 x 4 = -4p 4 [c] [negative sign indicate that AB is under compression ]

•Frane Eule's buckling lead formula, Par = π²EL (KL)2 i. For an anis Pai= T2X (2008 103 N/mm2) (4.1678106 mm 4) (3000)? Par 9.1392 x 105 N Pai= 913.925 KN & For yy-anis Prz= TC2 X (200 X 103 N/mm2) (1.04167 x 106 mm) (6000) 2 - 5.716 8 104 N Perz 570116 KN i. Per=Minimum of [Poy , Paz]=570116 IN Here, FAB < Per 4P AB < 57.116 IN P < 42.837KN . Maximum load p=42.837 IN