Question

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The following question is based on the theoretical Euler Buckling that you covered in Mechanics of Solids. For the Column shown below, calculate: i. The controlling (ie, maximum) theoretical slenderness ratio, L . ii. The critical theoretical Euler Buckling load, Por L1 = m 6.6 L = 11 m Radius of Gyration Typ 236 mm y 242 mm Known Properties: E = 2.60E+05 MPa box = 2.08E+07 mm ly = 2.80E+05 mm L= 11 m L1 = 6.6 m L2 = 4.4 m L2 = 4.4 m End condition for major axis (xx) buckling End condition for major axis (y) buckling

Answer 1

AS Both ends are hinged The Cowmo Buckles in this manner Les effective length} =kor e given length in Major Aris Xx-Buckling this case, since Both kezel ends are hinges Lex = (kex) = 1x1 = il mo Lex 11 - 56.61 0.236 and 6.6m in this case; Both ends are hinges there is another kirge in between The Buckling follows the pattern i drown n 4-4 Major Axis yox Buckling Since; Longer the length of Colum More choso chance of Buckling, the Upper and buckles first key =) Ley = 1x6-6 = 6.6m Ley 2 27.27 = 6. 6 0-242 Controlling cre Maximum) slenderness ratio Less- greater of (46.6), 27.27) = 466) We know Per = T² E Imin Le

E = 2.6x105 MPa Imin = 2.8x105 mm le = anem z Effective length corresponding Slenderness ratios to Controlling Per = TEII - ² x 2.6x105 x 2.8x105 lez inte (11000) 2 avenly = 5938:09 N = 5.938 KN