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: Ceramic tile ½ inch mortar Ceiling: Wood furring Gypsum board ¼ inch thick Mechanical Duct Allowance (MDA) Frame Partition: Wood, studs with gypsum board h=2.8m 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=3m (no interior walls) Live Load Residential
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: Ceramic tile ½ inch mortar Ceiling: Wood furring Gypsum board ¼ inch thick Mechanical Duct Allowance (MDA) Frame Partition: Wood, studs with gypsum board h=2.8m 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=3m (no interior walls) Live Load Residential
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 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:
Ceramic tile ½ inch mortar
Ceiling:
Wood furring
Gypsum board ¼ inch thick
Mechanical Duct Allowance (MDA)
Frame Partition:
Wood, studs with gypsum board h=2.8m
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=3m (no interior walls)
Live Load
Residential
Transcribed Image Text: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/m. Always simplify your load distributions (distributed loads acting on
the same points/locations should be added up).
Monolithic Floor
2 5m
Zm ..
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:
Sm
Ceramic tile 2 inch mortar
Ceiling:
Wood furring
Gypsum board Va inch thick
Mechanical Duct Allowance (MDA)
Frame Partition:
Wood, studs with gypsum board h=2.8m
Utilitles: 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=3m (no interior walls)
Live Load
Residential
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