Question

1.100 kips Problem 5: The frame with the factored loads in 10 kips Figure 5 is fixed at the base. The columns are oriented so that in-plane bending will be about the x-axis. The columns are continuously braced in the transverse direction such that they can only buckle in-plane. Use A992 steel and select the lightest W14 for the column. The beam size is W24x76 15 f 20 ft

Answer 1

GIVEN DATA Grade of steel Factored axial Load, A 992 P1100 kips Restrain Unsupported Length of column Effective length factor 15 ft (assumed for preliminary design) CALCULATIONS Yield stress of steel Modulus of Elasticity of steel 50 ksi E29000 ksi KL/ry 40 Assuming 0.529< 1.5 Note λe <1.5, Fcr : (0.65842 )F, For 0.877 Fa44.47 ksi Therefore, critical stress, Therefore, required area of cross-section, -1100/0.9 x 44.47 27.48 in2 Select W 14x99 CHECK CAPACITY erties of Steel section Grade of steel Column Section Gross area of cross section, Radii of gyration, A 992 W 14x99 Ag29.1 in Tx= 6.17 in ry3.71 in K = 1.25 (from frame analysis) 1.25 15 x 12 in) / 6.17 36.467

0.482 1.5 Note FO = (0.6584" )F, λ, < 1.5, For >1.5, Fe= .877 Ae Therefore, critical stress, Fa45.37 ksi Hence, Design axial strength of column, 0.9x 45.37 x 29.1 1188 kips1100 kipsP SAFE Hence,

MOMENTS OF INERTIA Columns 1110 in W14x99 - Beams/ Girders W24x76- 2100 in W24x76 15 W14x99 1110 15 x 12 20 2100 20 x 12 0.70 From alignement chart, for sway not prevented 1.25 [for unbraced frame]