Find the maximum aggregate can be used in the concrete wall

Find the maximum aggregate can be used in the concrete wall shown below? Cover between steel and forms, C=1.5 in. Spacing between bars, Distance between forms. B-9 in.

Solution

Solution:-

Given

B = 9 inch

Let height of wall(h) = 11.2 feet

                                = 134.4 inch

Let length of wall(L) = 16 feet = 192 inch

Total volume of wall = L*B*h                           

                           = 192 * 9*134.4 = 232.243*10³ inch³

Let # 9 bar is used.

Diameter of bar = 1.128 inch

Volume of bar = 2 */4*(1.128)² *192 = 383.74 inch³

Volume of concrete = Total volume of wall – volume of steel

                                    = 232.243 *10³ – 383.74

                                    = 231.859 *10³ inch³

Let use M 20 concrete

Ratio for M20 is 1:1.5:3

Volume of coarse aggregate = 3/(1+1.5 +3) *231.859*10³ = 126.468*10³ inch³

We know that

Unit weight of coarse aggregate = 150 lbs/feet³ = 0.0868 lbs/inch³

weight of coarse aggregate = 126.468*10³ *0.0868

                                                  = 10977.42 pound   Answer

Volume of fine aggregate = 1.5/(1+1.5 + 3) * 231.859*10³ = 63.234*10³ inch³

Weight of fine aggregate = 63.234*10³ *0.0868 = 5488.7 pound Answer