LOAD DISTRIBUTION $(50$ PTS$)$

Determine the ultimate load distribution acting on all beams and girders in the given floor framing below.

Assume that the supports are simply supported. Compute for the reactions and loads in each beam. Assume

unit weight of concrete to be $23\mathrm{.}6k\frac{N}{m^3}.$ Always simplify your load distributions $($distributed loads acting on

the same points$/$locations should be added up$).$

Monolithic Floor

All concrete slabs, $t=100mm$

All concrete beams, $b=250mm$;$h=500mm$

All concrete columns, $300mm300mm$

Superimposed Loads

Floor Finish:

Ceramic tile $\frac{1}{2}$ inch mortar

Ceiling:

Wood furring

Gypsum board $\frac{1}{4}$ inch thick

Mechanical Duct Allowance $($MDA$)$

Frame Partition:

Wood, studs with gypsum board $h=2\mathrm{.}8m$

Utilities: twice the weight of MDA

Exterior Walls

Normal Unit Weight Concrete Masonry$/$ Concrete Hollow Blocks, CHB

$4$ inch thick with plastering on both faces, full grout

height$=3$m $($no interior walls$)$

Live Load

Residential

NOTE: REFER TO NSCP $2015$ CHAPTER $2$ FOR THE MINIMUM DESIGN LOADS.

USE LRFD LOAD FACTORS: $1\mathrm{.}2DL+1\mathrm{.}6LL$