Question

Question .2 A structural system that supports a reading room is being considered for a new library. The floor consists of a 125 mm thick slab of normal density reinforced concrete supported by steel beams and columns. The steel beams weigh 90 kg/m while the steel columns weigh 150 kg/m. In addition to self weight, dead loads must include a 1.0 kPa allotment for partitions and a 0.5 kPa allotment for flooring/ceiling/services. Determine, (a) the specified (i.e. unfactored) distributed load on a typical interior beam (in kN/m). (b)the factored distributed load on a typical interior beam (in kN/m)

Answer 1

DEAD WAD DUE TO SLAB = 25 Kw/m^3 times 0.125 m = 3.125 Kw/m^2 DEAD WAD DUE TO PARTITIONS = 1.0 Kw/m^2 DEAD WAD FOR FLOORING/CEILING SERVICES = 0.5/4.625 Kw/m^2 LIVE WAD = 2.87 Kw/m^2. a) TRIB WIDTH = 4/2 + 4/2 = 4 m DISTRIBUTED WAD(DEAD) = 4.625 Kw/m^2 times 4 m = 18.5 Kw/m. (LIVE) = 2.87 Kw/m^2 times 4 m = 11.48 Kw/m b) FACTORED WAD = 1.2 D + 1.6 L = 1.2 times 18.5 + 1.6 times 11.48 = 40.568 Kw/m.