Question

I know its a long problem, any help would be appreciates. I will like

Mat Foundation System Specifications: Length 20'-0" x Width 20'-0" 12-inches Size: Thickness: Concrete Strength: 3,000 psi Waste Factor: 10% Reinforcing: #5 Spaced at 1'-0" Each Way Waste Factor: 15% to account for lapping requirements Unit Weight of Bar 1.04 lbs/ft . Reinforcing Accessories: Assume you will need 3" Plastic high chairs at a rate of 4 Chairs per every 16 SF of slab area. Assume you will need 10-inches of tie wire at every intersection of #5 reinforcing bar. Unit Weight of tie wire 0.01 lbs/lIf Concrete Finishing: Curing: Screed and Bull Float Sprayed Membrane/Curing Compound

Answer 1

Size of foundation = 20*20 = 400 sqft and thickness = 12" = 1'

Thus volume of dry concrete = 20*20*1 = 400 cuft

Assuming volume of wet concrete = 1.54 * volume of dry concrete thus volume of wet concrete = 1.54*400 = 616 cuft

Accounting for 10% wastage of concrete , quantity of wet concrete required = 616+61.6 = 677.6 cuft say 678 cuft.

Number of #5 bars each way = 21 and neglecting any clear cover , cutting length of each #5 bar = 20'

Thus total cutting length of all bars including 15% overlap = 1.15*(21*2*20) = 966'

Thus weight (quantity) of reinforcement required= 966*1.04 = 1004.64 lb

Number of plastic chairs required = 4*400/16 = 100

Number of intersections = 21*21 = 441 thus quantity of tie wire = 441*10" = 4410" =( 4410/12 )' thus weight = 4410*0.01/12 = 3.675 lb