Question

11. (10 points) Check to determine if a W14x159 used as a column can support an axial dead load of 250k and an axial live load of 1000k if the length of the column is 14 feet and the value of G for Joint A has been determined to be 0.8. Use the LRFD Design Method and assume that the columns are bending about their y axis as shown. (Hint: The answer is in the Manual) W14x159 Fixed Base at a steel W14x145 section used as a column can 12. (10 points) Determine the maximum compressive load th support if its length, L - 12 feet. Assume both ends are pinned and Fy-Soksi. height against buckling about the y-axis. Use the LRFD Design Method. The column is braced at mid 13. (10 points) Calculate the net area A, for the tension member shown below if %-in bolts are used: w12 × 16 14.(10 points) Determine the value of the shear lag factor, U, for the tension member below: 4 in 3in 3 in

Answer 1

GIVEN DATA Properties of Steel section Column Section Gross area of cross section, Radii of gyration, W 14x159 A 46.7 in rx6.38 in 4 in Restraint Alona x direction Unsupported Lengths of column, 14 ft Along y direction Unsupported Lengths of column, Bottom End Top end 14 ft : pinned : pinned ads Dead Load Live Load D250 kips L1000 kips CALCULATIONS Yield stress of steel Modulus of Elasticity of steel Factored Load 50 ksi E29000 ksi Pu1.2D+1.6 1900 kips K0.76 (for GA 0.8 in braced frame) Effective length factors :"1 -0.76 x ( 14 x 12 in )/ 6.38 - 20.013 K,L 1x( 14 x 12 in) 4 42 Since K,L/rx <KyLyfry, KyLy/ry controls K,L, F, 0.555 1.5 Note A 1.5, Fo-(0.658F For 0.877

F cr43.952 ksi Therefore, critical stress, Hence, Design axial strength of column, 0.9 x 44 x 46.7 1847 kips NOT SAFE 1900 kipsP Hence,

GIVEN DATA Properties of Steel section Column Section Gross area of cross section, Radii of gyration, W 14x145 A42.7 in2 rx6.33 in ry3.98 in Restraints Along x direction Unsupported Length of column, Bottom End Top end 12 ft : pinned : pinned Along y direction Unsupported Length of column Bottom End Top end 6 ft pinned : pinned CALCULATIONS Yield stress of steel Modulus of Elasticity of steel Effective length factors, 50 ksi E29000 ksi K,L 1 x 12 x 12 in6.33 - 22.749 K,L 1 x 6 x 12 in 3.98 18.09 Since KL/rr > KyLyyKxLx/rx controls - 0.3011.5 Note For 1.5, F (0.6581 )A, 0.877 Ae

Fcr= 48.139 ksi Therefore, critical stress Hence, Design axial strength of column, 0.9 x 48.14 x 42.7 = 1850 kips

3) FOR w12x16, 4-71-3 (る) , 22 4.1325 7in 11.98