1 Select the leastweight W12 A992 column to carry a live loa

1. Select the least-weight W12 A992 column to carry a live load of 130 kips and a dead load 10 pointsof 100 kips with an effective length about both axes of 14 ft. Use Table 4-1 for design, and then apply Table 4-14 and Specification Chapter E to verify the factored axial force capacity for the column. After determining the appropriate section for flexural buckling use the Specification to determine the factored torsional buckling capacity (as you would have to include that limit state in your analysis for design) 2. 10 points Repeat problem 1 using A913 Grade 65 steel. Do not consider torsional buckling. 3. A column with an effective length of 21 ft must support a dead load of 120 kips, a live 0 points load of 175 kips, and a wind load of 84 kips. Select the lightest W14 A992 member to support the load. Use Table 4-1 for design, and then apply Table 4-14 and Specification Chapter E to verify the factored axial force capacity for the column. Do not consider torsional buckling

Solution

1) live load=130 kips

dead load=100 kips

factored load per LRFD = 1.2*100 + 1.6*130=328 kips

effective length of column=14 ft

Lightest W12 that has capacity greater than 328 kips for effective length of 14 ft = W12x45

2) grade of steel = 65 ksi

Let us consoider W12x40 and check its axial load capapcity

radius oif gyration of section about minor axis = 1.93 in

slenderness ratio = 14*12/1.93=87

4.71*sqrt(E/fy)=4.71*sqrt(29000/65)=99.5>87

Euler buckling stress = pi²*29000/(87²)=37.8 ksi

area of cross section=11.8 in²

crtical stress = 0.658^65/37.8*65=31.6 ksi

Design strength of section=0.9*31.6*11.8=336 kips >328kips

Required section is W12x40

3)dead load=120 kips

live load=175 kips

wind load=84 kips

the governing load combination that will give maximum demand on column = 1.2D+1.6L

Factored load = 1.2*120 + 1.6*175=424 kips

effective length =21 ft

lightest W14 section that has capacity greater than 424 kips for effective length of 21 ft is W14x74