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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Concept explainers

Floor Plan Details

Floor plans are the scale drawings that show the interplay of rooms, spaces, and physical objects. They are an integral part of the house plan. They let you see how people will move around the room. Floor plans determine the suitability of space for its intended use, work through any potential issues, and redesign before moving on to more comprehensive planning or constructing phases. It shows the different alternatives and perspectives related to the location and dimension of rooms, windows, doors, kitchen and all other things as per the requirement by the user. Many consultants give their own floor plans as reference to the clients for a better idea and to develop good floor plans.

Question

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

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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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