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Prof Awad S. Hanna Concrete Structures Concrete Structures.

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Presentation on theme: "Prof Awad S. Hanna Concrete Structures Concrete Structures."— Presentation transcript:

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2 Prof Awad S. Hanna Concrete Structures Concrete Structures

3 Prof Awad S. Hanna Units of Measure Concrete:Cubic Yard Steel:Lb or tons Formwork:Square feet per contact area

4 Prof Awad S. Hanna Types of Concrete construction 1.Foundations 2.Columns 3.Slabs and beams

5 Prof Awad S. Hanna

6 Shallow Foundation Isolate d Footing Combined Footing Continuous Wall Footing Strap Footing Raft Foundatio n

7 Prof Awad S. Hanna Deep Foundation Caissons Piles

8 Prof Awad S. Hanna Columns  Shapes  Square - Rectangle - Circular  Others  Reinforcement  Ordinary - Spiral

9 Prof Awad S. Hanna Slabs and Beams: One-Way slabs One-Way Slab, Beam, and Girder One-Way Slab, Supported by Beams or Walls One-Way Joist Slab

10 Prof Awad S. Hanna Slabs and Beams: Two-Way Slabs Flat Plate Two-Way Slab Supported by Beams Flat Slab Waffle Slab

11 Prof Awad S. Hanna Estimating Issues 1.Formwork is the largest cost item that represent 40% to 60% of concrete structure cost. Distribution of costs for cast-in-place concrete slab wall:

12 Prof Awad S. Hanna 2.Waste allowance from 2% to 10% should be added for concrete and concrete formwork.

13 Prof Awad S. Hanna 1. Center Line Method Calculations: 1.Length = 30’10” - 1’6”= 29’4” 2.Width = 40’10” - 1’6”= 39’4” 3.Perimeter = 29’4” + 29’4” + 39’4” +39’4”= 137”4” 4.Volume of concrete = ’ x 1’ x 1.5’= 206 ft 3 30’10” 30’ 40’0” 40’10” Building size: 30’ x 40’

14 Prof Awad S. Hanna 2. Sectional Method Calculations: 1.Length = ’ ’= 81.67’ 2.Width = ’ ’= 55.67’ 3.Perimeter = 81.67’+55.67’= Total Linear ft 4.Volume = ’x1’x1.5’= 206 Cubic feet 5.volume in CY = 206/27= 7.6 Cubic Yards 30’10” 27’10” 40’10”

15 Prof Awad S. Hanna COST OF OTHER COMPONENTS OF CONCRETE WORK 1.Reinforcement Steel 2.Forms 3.Concrete Mix 4.Finishing 5.Curing

16 Prof Awad S. Hanna Table 1. DIMENSIONS & WEIGHT OF WELDED WIRE FABRIC

17 Prof Awad S. Hanna Table 2 SIZES & WEIGHTS OF REINFORCING BAR

18 Prof Awad S. Hanna EXAMPLE: How many pounds of rebar are required for the concrete floor area below? 6” Concrete slab with #4 rebar 15” on center each way 20’ 60’ 40’ 30’

19 Prof Awad S. Hanna Step #1: Calculate the total SF of area involved. 20'0" X 30'0" = 600 SF 30'0" X 40'0" = SF Total= 1,800 SF floor area Step #2: Calculate the pounds of rebar per SF of floor based on the unit spacing of the rebar at any point of intersection. Unit area: 15" X 15" = 1.25' X 1.25' = 1.56 SF Unit rebar length: 1.25' ' = 2.5 LF Lbs. of rebar/unit:= 2.5 LF X.668 lbs/ LF = 1.67 lbs  It is now known that there are lbs of rebar per 1.56 SQ FT of floor area / 1.56 = 1.07 lbs/SF of floor area. SO: 1.07 lbs/ SF X 1,800 SF = 1,926 lbs of #4 rebar required.


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