The floor framing plan shown in the figure below is subject

The floor framing plan shown in the figure below is subject to uniform distributed loads of: D = 30 psf, L = 70 psf. Determine the factored loading for each beam (B_), girder (G_), and column (C_) for a steel frame assuming one-way slab behavior.

https://imgur.com/a/PkHUe

Solution

In one way slab behaviour, the slab spans in the shorter direction

Deal load = 30 psf

Live load = 70 psf

Factored uniform load on floor per load combination 1.2D+1.6L = (1.2*30)+(1.6*70)=148 psf

loads on beams,girders and columns:

Beam B1:

Tributray width of B1=8/2=4\'

Uniform load on B1=148*4=592 plf=0.592 kip/ft

Beam B2:

Tributary width of B2 = (8/2)+(12/2)=10\'

uniform load on B2 = 148*10=1480 plf=1.48 kip/ft

Beam B3:

Tributary width of B3 = 12/2=6\'

uniform load on B3 = 148*6=888 plf=0.888 kip/ft

Gider G1 and G2:

Girders G1 and G2 are subjected to concentrated force from B2

Force from B2 on both G1 and G2 =reaction force at ends of B2

Reaction force at ends of B2 = 1.48*25/2=18.5 kips

Therefore, both girders G1 and G2 are subjected to concentrated forces of 18.5 kips

Column loads:

C1:

Load on C1 from G1 = (18.5*12/20)=11.1 kips

load on C1 from B1 = 0.592*25/2=7.4 kips

Therefore, total load on C1=11.1+7.4=18.5 kips

C2 is subjected to exactly equal loads from beam B2 and girder G2 as on column C1 from beam B2 and girder G1

Therefore, load on column C2=18.5 kips

C3:

Load on C3 from girder G2 = 18.5*8/20=7.4 kips

load on C3 from beam B3 = 0.888*25/2=11.1 kips

Total load on C3=7.4+11.1=18.5 kips

C4 is subjected to exactly equal loads from beam B3 and girder G1 as on column C3 from beam B3 and girder G2

Therefore, load on column C4=18.5 kips