Question

Question #2 (30 points) A solid rectangular column has the dimensions and properties below. Determine if it is a Johnson or an Euler column and find the critical load: (a) If its boundary conditions are pinned- pinned. (b) If its boundary conditions are fixed-free. Length of column L - 120 mm Thickness = 5 mm Height = 10 mm Material Steel Yield strength Sy = 305 MPa Modulus of elasticity E = 207 GPa 2 * 20 7

Answer 1

Question #2 9. Both ends are pinned. moment in Joumn -py. we know the deflection a 22 El dy op The E-ME-Py. idmatan yo ☺ The Genreel sol of the above differenticel egn.. y= A cos pr. ] + B in [ Boundry condition. 1 = 0 ; y no UL) 90' put in eq? ② one by one. o = A Coscolt B Sin (0) and OF A Cool2.) + sin (son VEL LIP-NT VEL

P = n E 1772 mm for critical load. I = Imin, of (Ixx and Iyy). - a2 = effective length of cloumn. a = length of fixity Coreff?. P=pe - Buckling load A t e Modulus of elasticity. . le= Thelman mitibo para Imin = I x 10 x 5² = 1250 mmt. h = 120mm E = 207x109 Pa Per r2 x 207x10 x 1250x1012 = 14.778704X10N 1202 x 10 6

@ Ple= 14.7787 KN Ans. by one end & fixed. then effective length = 2XL = 2x120= 240mm. le - T2 202x18x1280 x 1612 12 2402 x 156 - 3.69467X13 N Pe = 3:694 67 kn Ans Caushing load: p. P = Syc XARE here. byc = yield Strength A/C = Cs Arca N = factors of softy. P = 305 4160 x 5x10x186 P= 15.25X10N Crushing load in higher then of Bucking I so that we find load. crushing load.