Question

The uniformly distributed live load on the floor plan in the figure given below is 65 Ib/ft2. Consider the live load reduction if permitted by the ASCE standard. .9 6 @ 6.67' 40 C4 GI B3 1182 1182 2 @ 10 20 B4 C2 iA G3 31 C3 CI 20' 40'

find :

1) Loading (W) for floor beam B1 in Kips/ft

2) Loading (W) for floor beam B2 in Kips/ft

3) Loading (P) for Grider G1 in Kips

4) Loading (W) for Grider G2 in Kips

Answer 1

1) floor load =65 psf

Tributry width of B1=8 ft

Uniform load on B1=65*8/1000=0.52 kip/ft

2) tributray width of B2=6.67/2=3.335 ft

Uniform load on B2=3.335*65/1000=0.217 kip/ft

3)loading P on G1 is basically equal to reaction from B3

Tributtary width of B3=10 ft

Load on B3=10*65=650 plf=0.65 kip/ft

Span if B3=20 ft

Reaction at G1=0.65*20/2=6.5 kips

Therefore P on G1 is 6.5 kips

4) tributray width of G2=(40/2)+(20/2)=30 ft

Uniform load on G2=65*30/1000=1.95 kip/ft