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CHAPTER 1:. CONCRETE. SIMPLIFIED CONSTRUCTION ESTIMATES BY MAX FAJARDO.

1-1 PLAIN AND REINFORCED CONCRETE. Concrete is either plain or reinforced..

AGGREGATES. Classification of Aggregates:. Course Aggregate such as crushed sone, crushed gravel or natural gravel with particles retained on a 5 mm sieve. Fine Aggregate such as crushed stone, crushed gravel, sand or natural sand with particles passing on a 5 mm sieve..

1-2 THE PRINCIPLES OF CONCRETE MIXING. Requirements to meet in production of concrete mixture:.

Classification of Concrete Mixture: 1. Designed Mixture The contractor is responsible in establishing the mixture proportion that will achieve the required strength and workability as specified in the plan. 2. Prescribed Mixture The designing engineer specified the mixture proportion 1-3 THE UNIT OF MEASURE Prior to System International (SI), materials for concrete structures were estimated in terms of cubic meter although, the components thereof like; cement, sand, gravel, and water, are measured in pounds, cubic foot, and gallons per bag respectively..

TABLE 1-1 CONVERSION FROM INCHES TO METER. TABLE 1-1 CONVERSION FROM INCHES TO METER .Awtodn*e Ninet 10 11 12 13 14 15 16 17 18 20 .0254 .050B .0762 .1016 .1270 .1524 .1778 .2032 .228E .2540 .2794 3048 .3302 3556 3810 .4064 .431B 4572 .4826 .5080 02S 050 075 .100 .125 .150 .175 .200 225 .250 .275 .300 .325 .350 .375 400 .425 475 23 24 27 28 29 30 32 33 37 40 Acurü .5334 .5588 .5842 6350 .6858 .7112 .7620 .7874 .8128 8382 8636 .8890 .9144 .9398 9652 1.016 .525 .550 .575 .600 .625 .650 .675 .700 725 .750 77S .82S 875 .92S .975 1.00.

1-4 CONCRETE PROPORTION Two different ways in proportioning concrete mixture: By Weight By Volume Method.

In actual and masonry work, there are several factors that might affect the accuracy of the estimate. Some of which are enumerated as follows: Ordering of Course Aggregate must be Specific as to:.

1-5 CONCRETE SLAB. . to cm. FIGURE 1-2 CONCRETE PAVEMENT.

SOLUTION: 1. Determine the volume of the proposed concrete pavement. Volume = Thickness x Width x Length V = .10 x 3.00 x 5.00 V = 1.5 cubic meters 2. Refer to Table 1-2. Using 40 kg. cement class C mixture; multiply the volume by the corresponding values..

1-6 ESTIMATING CONCRETE SLAB BY THE AREA METHOD. TABLE 1-3 QUANTITY OF CEMENT, SAND AND GRAVEL FOR SLAB AND WALLS PER SQUARE METER AREA Mixture Class • Slab 5.0 7.5 10.0 12.5 15.0 17.5 20.0 22.5 25.0 27.5 30.0 40 kg. Cement SO kg. Cement A .45Q .675 .900 1.125 1.350 1.575 1.800 2.030 2.250 2.475 2700 B .375 .563 .750 .938 1.125 1.313 1.500 1.688 1.875 2.063 2.250 C .300 450 .600 .750 .900 1.050 i 200 1.350 1.500 1.6S0 1.800 .350 .525 .700 .87S I.oso 1.225 1.400 1.575 1.750 1.925 2.100 B .300 .450 .600 .750 .900 1.050 1.200 I .350 1.650 C .250 .375 .soo .625 .750 .875 1.00 1.125 1.250 1.375 1.500 Sand cu m. .0250 .0375 .osoo .0630 .0750 .0880 .1000 .112S .1250 .1380 .1500 Gravel ce. m. .050 .075 .100 .125 .150 .17S • .200 .225 .250 .275 .300.

SOLUTION: 1. Solve for the pavement area Area = Width x Length A = 3.00 x 5.00 A = 15 squared meters 2. Refer to Table 1-3. Along 10 cm. slab thickness under 40 kg. cement class “C” mixture, multiply; Cement: 15 x .600 = 9.0 bags Sand: 15 x .050 = .75 cu. m. Gravel: 15 x .100 = 1.50 cu. m. 3. Compare this quantity to that Illustration 1-1, the results are the same.

ELEVATION CROSS SECTION X-X 1-12 COCUMN. 1-7 SQUARE CONCRETE COLUMN Estimating the quantity of materials for concrete post or column can be done in two simple ways: By Volume Method and By Linear Meter Method.

TAE 1.1 FROM TO METER Acuü 2 8 9 10 11 12 13 14 15 16 18 20 .050B .0762 .1016 1270 .1524 .1778 .2032 .2286 .2540 .2794 304B .3302 .3556 3810 .4064 .431B 4572 4826 .5080 02S 050 075 .100 .125 .150 .175 .200 225 .250 .27S .300 .325 .350 .375 400 .425 475 24 30 37 40 .5588 .5842 .6096 .6804 .6858 .7112 .7366 .7620 .7874 .8128 8382 8636 .8890 .91" .939e 9652 1.016 .525 .575 .625 .675 .700 .750 77S .82s 875 .975 1.00.

1-8 ESTIMATE SQUARE CONCRETE COLUMN BY LINEAR METER METHOD.

A concrete column is 7.00 meters high with a cross sectional dimension of 20 by 20 inches. Determine the quantity of cement, sand, and gravel content of the column if there are 8 columns in a row using class "A" concrete. CROSS.

1-9 POST AND FOOTING. Estimating the quantity of materials for concrete post and footing could be done by: Volume Method Area and Linear Meter Method combined for post and footing.

SOLUTION: By Volume Method Find the volume of 12 posts Volume = Sectional Area x Height x No. of Posts V = (.40 x .40) x 4.00 m x 12 posts V = 7.68 cubic meters 2. Solve for the volume of 12 footing slab V = (1.20 x 1.20) x .20 m x 12 posts V = 3.456 cubic meters.

SOLUTION: By Area and Linear Meter Method 1. Find the length of 12 posts L =12 posts x 4.00 m. height L = 48 meters 2. Refers to Table 1-4. Along the 40 x 40 cm column size class “A” mixture, multiply; Cement:48 x 1.440 = 69.12 bags Sand: 48 x .080 = 3.84 cu. m. Gravel: 48 x .160 = 17.68 cu. m. 3. Find the area of the footing slab Area =12 posts (1.2 x 1.2) L = 17.28 square meters.

SOLUTION: By Area and Linear Meter Method 1. Find the length of 12 posts L =12 posts x 4.00 m. height L = 48 meters 2. Refers to Table 1-4. Along the 40 x 40 cm column size class “A” mixture, multiply; Cement:48 x 1.440 = 69.12 bags Sand: 48 x .080 = 3.84 cu. m. Gravel: 48 x .160 = 17.68 cu. m..

1-10 RECTANGULAR COLUMN. The procedure in estimating rectangular column is practically the same as that of the square column.

SOLUTION: By Area Method 1. Find the length of the 8 columns L =8 columns x 5.00 m. height L = 40 meters 2. Refers to Table 1-4. Along with 40 x 60 cm column size under the 40 kg. cement class “A” mixture, multiply; Cement: 40 x 2.16 = 86.4 bags Sand: 40 x .120 = 4.8 cu. m. Gravel: 40 x .240 = 9.6 cu. m..

1-11 RECTANGULAR BEAM AND GIRDER. Beam – strong horizontal piece of reinforced concrete for spanning and supporting weights. Girder – a beam that is carrying or supporting another beam..

SOLUTION: By Volume Method 1. Find the volume of the beam V = (.25 x .40) x 8 m. span x 5 pcs. V = 4 cubic meters 2. Solve for the volume of the girder V = (.40 x .60) x 16 m. span x 2 pcs. V = 7.68 cubic meters 3. Total volume of the beam and girder Total Volume = 4 + 7.68 V = 11. 68 cubic meters.

1-12 CIRCULAR COLUMN. Estimating the materials for circular column is a typically the same as the volume method with the aid of Table-2. However, Table 1-5 was also prepared for circular problems.

A circular concrete column has a diameter of 60 cm by 6.00 meters high. Find the concrete materials required if there are 5 columns in the same size in a row..

SOLUTION: By Volume Method 1. Solve for the cross sectional area of the beam A = ??^2 or A= 0.7845?^2 A = (O.7854 x .〖60〗^2) =.283 sq. m. 2. Find the volume of one column V =.283 x 6 m. ht. = 1.698 cu. m. 3. Solve for the volume of 5 columns Total Volume: (1.698 x 5) = 8.49 cu. m. 4. Refer to Table 1-2. Using 40 kg. cement class “A” concrete, multiply; Cement: 8.49 x 9.0 = 76.4 bags Sand: 8.49 x .50 = 4.24 cu. m. Gravel: 8.49 x 1.0= 8.49 cu. m..

1-12 CIRCULAR PIPE. Circular Pipe is much in demand for small and medium drainage construction. The use of drainage structure materially save cost, time, and labor cost as well..

A road construction requires 12 pieces of 90 cm. diameter concrete pipes for drainage purposes. Determine the quantity of cement, sand, and gravel needed to manufacture the said pipes using class “A” concrete. (Excluding reinforcement which will be discussed in Chapter 3).

SOLUTION: By Volume Method 1. Solve for the gross volume of the concrete pipe Formula: V= 0.7845?^2H V = 0.7845 x 〖1.10〗^2 x 1= .950 cu. m. 2. Solve for the volume of the hole V = 0.7845 x ?^2H V = 0.7845 x 〖.90〗^2 x 1. = .636 cu. m. 3. Subtract result of step 2 from step 1 to get the net volume of the concrete pipe. Vc = .950 - .636 = .314 4. Total volume of the 12 pipes Vt = .314 x 12 = 3.768 cu. m..

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