Question

Question 6 refers to a frame in an unbraced multi-story building shown below. The column on gridline C is a W12x72 oriented such that strong axis buckling would occur in the plane of the frame. B W16x26 W1835 15 ft W16x26 W18X35 15 ft 30 ft 40 ft 6. (5 points) Determine the effective length, Lc (ft) and available strength (k) of the column. You may use either LRFD or ASD.

Answer 1

MOMENTS OF INERTIA Columns W12x72 = W12x72 = 597 in 597 in Beams/ Girders W18x35 = 510 in W12x72 W18x35 kt* t W12x72 GAEIL 13/L3 597 15 x 12 + 597 15 x 12 20510 510 40 x 12 = 6.24 G = EI/L 1,1L = 1 From alignement chart, for sway not prevented, Ky = 1.75 [for unbraced frame) Ky = 1 {for pinned ends]

GIVEN DATA Properties of Steel section Grade of steel Column Section Gross area of cross section, Radii of gyration, A 992 w12x72 A, = 21.1 in 2 x = 5.31 in ry = 3.04 in Lx = Restraints Along x direction Unsupported Lengths of column, Effective length factor, Along y direction Unsupported Lengths of column, Effective length factor, 15 ft 1.75 L, = K = 15 ft 1 CALCULATIONS Yield stress of steel Modulus of Elasticity of steel Fy = E = 50 ksi 29000 ksi A 1.75 x 15 x 12 in ) / 5.31 = 59.322 K, L, 1x ( 15 x 12 in )/ 3.04 = 59.211 Since KxLx/x > K, Lyly KxLx/rx controls. KL F VE = 0.784 < 1.5 Note: For 2. <1.5, F, = (0.6584)F, 4, >1.5. F. -[0,??) Therefore, critical stress, F = 38.658 ksi Hence, Design axial strength of column, 6.P, = 0 F.,A = 0.9 x 38.7 X = 734 kips 21.1