Download presentation

Presentation is loading. Please wait.

Published byCarl Cordrey Modified over 2 years ago

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 = 137.33’ 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 = 10.833’+40.833’= 81.67’ 2.Width = 27.833’+27.833’= 55.67’ 3.Perimeter = 81.67’+55.67’= 137.34 Total Linear ft 4.Volume = 137.34’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" = 1 200 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' + 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 1. 67 lbs of rebar per 1.56 SQ FT of floor area. 1.67 / 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.

Similar presentations

OK

THE NORTHBROOK CORPORATE CENTER Redesign of the Lateral Load Resisting System.

THE NORTHBROOK CORPORATE CENTER Redesign of the Lateral Load Resisting System.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on election in india 2012 Ppt on film industry bollywood dance Mis ppt on hospital waste Ppt on layer 3 switching explained Ppt on area related to circle of class 10 Ppt on wildlife tourism in india Ppt on mars orbiter mission Free pdf converter to ppt online Run ppt on html editor Ppt on main idea and supporting detail