Question

For Problems 3-4 through 3.8, compute the values requested using the basic floor framing plan shown in the accompanying figure. Consider floor live-load reductions following the provisions of ASCE 7-16 as appropriate. For purposes of these problems, assume that this is the uppermost floor in a multistory office building. Also assume that it has a one-way slab behavior. Basic Building Layout 2 -40 ft -40 ft-*-40 ft- (3) (4) 40 ft 35 ft 4-35 ft 35 ft- Problems 3.4–3.8 P34 Load on column B3 contributed by this floor if the floor live load is 150 psf. P3.7 Load on an interior floor beam if the floor live load is 75 psf. (Ans: w = 740 plf)

Answer 1

ANSWERl:

GIVEN THAT

CONSIDER FLOOR LIVE LOAD REDUCTIONS FOLLOWING THE PROVISIONS OF ASCE 7 AS APPROPRIATE .

ASSUME THIS IS THE UPMOST FLOOR IN AMULTISTORY OFFICE BUILDING ALSO ASSUME THAT IT HAS A ONE WAY SLAB BEHAVIOUR.

A) LOAD ON COLOUMN B3 CONTRIBUTED BY THIS FLOOR LIVE LOAD IS 150 PSF

B) LOAD ON AN INTERIOR FLOOR BEAM IF THE FLOOR LIVE LOAD IS 75 PSF

Baste Peitling Soyout. a) Lead on colourn Be Contributed by this flood of the floor live load is 150 put Tributary Area Art Greco e ] [ 2015 + 3a"] [20+20] [ 17.5 +17,5] - 40 / 35] | ATE 1400ft? 10 = 150 psf Simce to > loopst, it should not be reduced .: Load On Cobouwin B3 = 150 pst X1400H12 = 210000 ob 1) load on an Interior flooo beam at the 210 Kipe of the Hoor live load u 75 psf Tributary Area do= 40 * 15' - 13.344 35 1 = 466067 ft2 Lo=75pst Ke=2 Are kee to => 2x466.67 - 933.33 %? **osoa e Final, que un Load on Interio become = 55.57X40 717602.9379