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 Floor All concrete slabs, t = 100 mm All concrete beams, b = 250 mm; h = 500 mm All concrete columns, 300 mm x 300mm Superimposed Loads Floor Finish: Ceiling: Ceramic tile ½ inch mortar Frame Partition: Exterior Walls Wood furring Gypsum board 14 inch thick Mechanical Duct Allowance (MDA) Utilities: twice the weight of MDA Live Load Wood, studs with gypsum board h=2.8m 5m A H 3m Normal Unit Weight Concrete Masonry/ Concrete Hollow Blocks, CHB 4 inch thick with plastering on both faces, full grout height=3m (no interior walls) Residential NOTE: REFER TO NSCP 2015 CHAPTER 2 FOR THE MINIMUM DESIGN LOADS. USE LRFD LOAD FACTORS: 1.2 DL + 1.6 LL B G 2m C F 2.5m D

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 Floor All concrete slabs, t = 100 mm All concrete beams, b = 250 mm; h = 500 mm All concrete columns, 300 mm x 300mm Superimposed Loads Floor Finish: Ceiling: Ceramic tile ½ inch mortar Frame Partition: Exterior Walls Wood furring Gypsum board 14 inch thick Mechanical Duct Allowance (MDA) Utilities: twice the weight of MDA Live Load Wood, studs with gypsum board h=2.8m 5m A H 3m Normal Unit Weight Concrete Masonry/ Concrete Hollow Blocks, CHB 4 inch thick with plastering on both faces, full grout height=3m (no interior walls) Residential NOTE: REFER TO NSCP 2015 CHAPTER 2 FOR THE MINIMUM DESIGN LOADS. USE LRFD LOAD FACTORS: 1.2 DL + 1.6 LL B G 2m C F 2.5m D

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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Please answer correctly and box the final answer.
The floor of a classroom is made of 150-mm thick lightweight concrete. If the floor is a slab having a length of 8m and width of 6m, determine the resultant force (in KN) caused by the dead load, the live load, and the total load. Note: lightweight concrete is 0.015KN/m2 per mm thickness & classroom is 1.92KN/m2
Minimum of 3 decimal places in all computations.
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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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