617 Assuming axial loads only select W10 sections for the in

6-17. Assuming axial loads only, select W10 sections for the interior column of the lat- erally braced frame shown in the accompanying illustration. Use F = 50 ksi and the LRFD method only. A column splice will be provided just above point B; select a column section for column AB and a second different column section for column BC and CD. Miscellaneous data: Concrete weighs 150 lb/ft3

Solution

Let us determine the design loads on the columns:

Column BC:

tributary area of column BC from each floor = (25/2+25/2)*(35/2+35/2)=875 sq ft

self weight of 2 in conc slab = (2/12)*150=25 psf

self weight of 6 in conc slab = (6/12)*150=75 psf

roof dead load = 10 psf

roof live load=30 psf

floor dead load=12 psf

partitional load=15 psf

floor live load=75 psf

Total dead load on column BC=(25+75+10+12)*875/1000=106.75 kips

Total live load on column BC=(30+75+15)*875/1000=105 kips

Factored load on column BC=1.2*106.75 + 1.6*105=296.1 kips

unbraced length of BC=12 ft=effective length

for effective length of 12 ft,the lightest W10 section that has capapcity greater than 296.1 kips is W10x39

Column AB:

tributary area of column AB from each floor = (25/2+25/2)*(35/2+35/2)=875 sq ft

self weight of 2 in conc slab = (2/12)*150=25 psf

self weight of 6 in conc slab = (6/12)*150=75 psf

roof dead load = 10 psf

roof live load=30 psf

floor dead load=12 psf

partitional load=15 psf

floor live load=75 psf

Total dead load on column AB=(25+2*75+10+2*12)*875/1000=182.875 kips

Total live load on column AB=(30+2*75+2*15)*875/1000=183.75 kips

Factored load on column AB=1.2*182.875 + 1.6*183.75=513.45 kips

unbraced length of AB=15 ft=effective length

for effective length of 15 ft,the lightest W10 section that has capapcity greater than 513.45 kips is W10x60