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. Assumeunit weight of concrete to be 23.6 kN/m3. Always simplify your load distributions (distributed loads acting onthe same points/locations should be added up).Monolithic Floor3m2m2.5mAll concrete slabs, t = 100 mmAll concrete beams, b = 250 mm ; h = 500 mmAll concrete columns, 300 mm x 300mmBCSuperimposed LoadsFloor Finish:Ceramic tile 2 inch mortar5mCeiling:Wood furringGypsum board 4 inch thickMechanical Duct Allowance (MDA)GFrame Partition:Wood, studs with gypsum board h=2.8mUtilities: twice the weight of MDAExterior WallsNormal Unit Weight Concrete Masonry/4 inch thick with plastering on both faces, full groutheight=3m (no interior walls)ncrete Hollow Blocks, CHBLive LoadResidentialNOTE: REFER TO NSCP 2015 CHAPTER 2 FOR THE MINIMUM DESIGN LOADS.USE LRFD LOAD FACTORSS: 1.2 DL + 1.6 LL

The Ultimate Load is generally the Limit Load that is multiplied by a subjected to safety Factor of…

etermine the ultimate load distribution acting on all beams and girders in the given floor framing below ssume that the supports are simply supported. Compute for the reactions and loads in each beam. Assume nit weight of concrete to be 23.6 kN/m³. Always simplify your load distributions (distributed loads acting or ne same points/locations should be added up). onolithic Floor 3m 2m 2.5m All concrete slabs, t 100 mm All concrete beams, b 250 mm ; h = 500 mm All concrete columns, 300 mm x 300mm uperimposed Loads Floor Finish: C D Ceramic tile 2 inch mortar 5m Ceiling: Wood furring Gypsum board 4 inch thick Mechanical Duct Allowance (MDA) G Frame Partition: Wood, studs with gypsum board h=2.8m

etermine the ultimate load distribution acting on all beams and girders in the given floor framing below ssume that the supports are simply supported. Compute for the reactions and loads in each beam. Assume nit weight of concrete to be 23.6 kN/m³. Always simplify your load distributions (distributed loads acting or ne same points/locations should be added up). onolithic Floor 3m 2m 2.5m All concrete slabs, t 100 mm All concrete beams, b 250 mm ; h = 500 mm All concrete columns, 300 mm x 300mm uperimposed Loads Floor Finish: C D Ceramic tile 2 inch mortar 5m Ceiling: Wood furring Gypsum board 4 inch thick Mechanical Duct Allowance (MDA) G Frame Partition: Wood, studs with gypsum board h=2.8m

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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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 23.6 kN/m3. Always simplify your load distributions (distributed loads acting on the same points/locations should be added up). Monolithic FloorAll concrete slabs, t = 100 mmAll concrete beams, b = 250 mm ; h = 500 mmAll concrete columns, 300 mm x 300mm Superimposed LoadsFloor Finish:Ceramic tile ½ inch mortar Ceiling:Wood furringGypsum board ¼ inch thickMechanical Duct Allowance (MDA) Frame Partition:Wood, studs with gypsum board h=2.8m Utilities: twice the weight of MDA Exterior WallsNormal Unit Weight Concrete Masonry/ Concrete Hollow Blocks, CHB4 inch thick with plastering on both faces, full groutheight=3m (no interior walls) Live LoadResidential
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