Simplified Construction Estimate PDF

# SIMPLIFIED CONSTRUCTION ESTIMATE ## TABLE OF CONTENTS - CHAPTER-10 PAINTING - 10-1 Paint - 10-2 Ingredient of Paint - 10-3 Essential and Specific Properties of Good Quality Paint - 10-4 Elements of a Good Painting Job - 10-5 Surface Preparation - 10-6 Kinds of Paint, Uses...

# SIMPLIFIED CONSTRUCTION ESTIMATE ## TABLE OF CONTENTS - CHAPTER-10 PAINTING - 10-1 Paint - 10-2 Ingredient of Paint - 10-3 Essential and Specific Properties of Good Quality Paint - 10-4 Elements of a Good Painting Job - 10-5 Surface Preparation - 10-6 Kinds of Paint, Uses and Area Coverage - 10-7 Estimating Your Paint - 10-8 Paint Failures and Remedy - 10-9 Wall Papering - CHAPTER-11 AUXILIARY TOPICS - 11-1 Accordion Door Cover - 11-2 Glass Jalousie - 11-3 Water Tank - 11-4 Wood Piles - 11-5 Bituminous Surface Treatment - 11-6 Filling Materials - 11-7 Nipa Shingle Roofing - 11-8 Anahaw Roofing - CHAPTER 1 - CONCRETE - 1-1 PLAIN AND REINFORCED CONCRETE - 1-2 THE PRINCIPLES OF CONCRETE MIXING - 1-3 THE UNIT OF MEASURE - 1-4 CONCRETE PROPORTION - 1-5 CONCRETE SLAB - 1-6 ESTIMATING CONCRETE SLAB BY THE AREA METHOD - 1-7 SQUARE CONCRETE COLUMN - 1-8 ESTIMATING SQUARE CONCRETE COLUMN BY LINEAR METER METHOD - 1-9 POST AND FOOTING - 1-10 RECTANGULAR COLUMN - 1-11 RECTANGULAR BEAM AND GIRDER - 1-12 CIRCULAR COLUMN. - 1-13 CONCRETE PIPE - 1-14 QUANTITY OF CEMENT, SAND AND GRAVEL PER CONCRETE PIPE. - 1-15 ESTIMATING THE MATERIALS FOR CONCRETE PIPE - CHAPTER 2 - MASONRY - 2-1 CONCRETE HOLLOW BLOCK - 2-2 ESTIMATING CEMENT MORTAR - 2-3 CEMENT MORTAR FOR PLASTERING - 2-4 CONCRETE HOLLOW BLOCK FOOTING - 2-5 OTHER TYPES OF CONCRETE BLOCKS - 2-6 DECORATIVE BLOCKS - 2-7 ADOBE STONE - 2-8 RETAINING WALL - 2-9 RIP-RAP AND GROUTED RIP-RAP - 2-10 CONCRETE RETAINING WALL - 2-11 GABIONS AND MATTRESS - CHAPTER 3 - METAL REINFORCEMENT # CHAPTER 1 ## CONCRETE ### 1-1 PLAIN AND REINFORCED CONCRETE Concrete is either Plain or Reinforced. By definition, Plain Concrete is an artificial stone as a result of mixing cement, fine aggregates, coarse aggregates and water. The conglomeration of these materials producing a solid mass is called plain concrete. Reinforced Concrete on the other hand, is a concrete with reinforcement properly embedded in such a manner that the two materials act together in resisting forces. The Different Types of Concrete Used in Construction are: 1. The Ordinary Portland cement. 2. The Rapid Hardening Portland Cement which is preferred when high early strength concrete is desired. 3. The Blast Furnace or Sulfate Cement used on concrete structures designed to resist chemical attack. 4. The Low Heat Portland Cement used for massive sections designed to reduce the heat of hydration. 5. The Portland Pozzolan Cement with a low hardening characteristic concrete. 6. The High Alumina Cement. The High Alumina Cement is sometimes called aluminous cement or cement fundu. Its chemical composition is different from that of Portland cement for having predominant alumina oxide content of at least 32% by weight. The alumina lime is within the limit of 0.85% to 1.3%. ### SIMPLIFIED CONSTRUCTION ESTIMATE This type of cement has a very high rate of strength development compared with the ordinary Portland cement. Aside from its rapid hardening properties, it can resist chemical attack by sulfate and weak acids including sea water. It can also withstand prolonged exposure to high temperature of more than 1,000°C. Alumina cement however, is not advisable for mixing with any other types of cement. The Main Composition of Cement are: - 1.6 to 65% Lime - 18.0 25% Silica - 3.0 8% Alumina - 3.0 5% Iron oxide - 2.0 5% Magnesia - 1.0 5% Sulfur trioxide ### AGGREGATES Aggregates for concrete work are classified into two: 1. **Coarse Aggregate** such as crushed stone, crushed gravel or natural gravel with particles retained on a 5 mm sieve. 2. **Fine Aggregate** such as crushed stone, crushed gravel, sand or natural sand with particles passing on a 5 mm sieve. **Size of Aggregates.** For coarse aggregate (gravel), the maximum nominal size varies from 40, 20, 14 or 10 mm diameter. The choice from the above sizes depends upon the dimensions of the concrete member more particularly, the spacing of the steel bars reinforcement or as specified. Good practice demand that the maximum size of coarse aggregate (gravel) should not exceed 25% of the minimum thickness of the member structure nor exceed the clear distance between the reinforcing bars and the form. ### CONCRETE The coarse aggregate should be small enough for the concrete mixture to flow smoothly around the reinforcement. This is referred to as workability of concrete. ### 1-2 THE PRINCIPLES OF CONCRETE MIXING The purpose in mixing concrete is to select an optimum proportion of cement, water and aggregates, to produce a concrete mixture that will meet the following requirements: 1. Workability 2. Strength 3. Durability 4. Economy The proportion that will be finally adopted in concrete mixing has to be established by actual trial and adjustment processes to attain the desired strength and quality of concrete required under the following procedures: 1. The water cement ratio is first determined at the very first hour of mixing to meet the requirements of strength and durability. 2. The cement-aggregate ratio is then chosen and established to satisfy the workability requirements. Workability, means the ability of the fresh concrete to fill all the voids between the steel bars and the forms without necessarily exerting much effort in tamping. Laboratory tests showed that the water-cement content ratio is the most important consideration in mixing because it determines not only the strength and durability of the concrete but also the workability of the mixture. Concrete mixtures in a paste form, is preferred than those mixtures which are flowing with water. ### The ACI Requirements for Concrete are as follows: 1. Fresh concrete shall be workable, meaning that fresh concrete could freely flow around the reinforcements to fill all the voids inside the form. 2. That, the hardened concrete shall be strong enough to carry the design load. 3. That, hardened concrete could withstand the conditions to which it is expected to perform. 4. That, concrete should be economically produced. ### Concrete Mixture may be classified as either: - **a. Designed Mixture** - **b. Prescribed Mixture** **Designed Mixture.** Where the contractor is responsible in establishing the mixture proportion that will achieve the required strength and workability as specified in the plan. **Prescribed Mixture.** Where the designing engineer specify the mixture proportion. The contractor's responsibility is only to provide a properly mixed concrete containing the right proportions as prescribed in the plan. ### 1-3 THE UNIT OF MEASURE Prior to the worldwide acceptance of Metrication, otherwise known as 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 galions per bag respectively. Lately however, under the Sl measures, the 94 pounds per bag cement equivalent to 42.72 kilograms was changed and fixed at 40 kilograms per bag. The traditional wooden bex used to measure the sand and gravel is 12 inches wide by 12 inches long and 12 inches high, having a riet volume of 1 cubic foot. Today, instead of the traditional measuring wooden box, the empty plastic bag of cement is popularly used to measure the volume of sand and gravel for convenience in handling aggregates during the mixing operations. ### TABLE 1-1 CONVERSION FROM INCHES TO METER | Number | Accurate Value | Approximate Value | Number | Accurate Value | Approximate Value | |---|---|---|---|---|---| | 1 | 0254 | .025 | 21 | 5334 | 525 | | 2 | 0508 | 050 | 22 | 5588 | 550 | | 3 | 0762 | .075 | 23 | 5842 | .575 | | 4 | 1016 | .100 | 24 | 6096 | .600 | | 5 | 1270 | .125 | 25 | 6350 | 625 | | 6 | 1524 | .150 | 26 | 6604 | .650 | | 7 | 1778 | .175 | 27 | 6858 | .675 | | 8 | 2032 | .200 | 28 | 7112 | 700 | | 9 | 2286 | .225 | 29 | 7366 | 725 | | 10 | 2540 | .250 | 30 | 7620 | 750 | | 11 | 2794 | .275 | 31 | 7874 | 775 | | 12 | 3048 | .300 | 32 | 8128 | 800 | | 13 | .3302 | .325 | 33 | 8382 | 825 | | 14 | 3556 | .350 | 34 | 8636 | 850 | | 15 | 3810 | .375 | 35 | 8890 | 875 | | 16 | .4064 | .400 | 36 | 9144 | 900 | | 17 | 4316 | .425 | 37 | 9398 | 925 | | 18 | 4572 | .450 | 38 | 9652 | 950 | | 19 | 4826 | .475 | 39 | 9906 | 975 | | 20 | 5080 | .500 | 40 | 1.016 | 1.00 | The values presented in Table 1-1 could be useful in: 1. Finding the accurate conversion of length from English to Metric measure. 2. Determining the approximate value to be used generally in our simplified methods of estimating. ### For Instance: - A) In solving problems, the probability of committing error is substantially high when severa! digit numbers are being used. - B) To memorize the values given in Table 1-1 is a waste of time and not a practical approach in estimating. A simple guide will be adopted so that one could easily determine the equivalent values from English to Metric or vise versa. ### Example: 1. To convert Meter to Feet: Divide the length by .30 Say 6.00 meters / .30 = 20 ft. 2. To convert Feet to Meters: Multiply by .30 Say, 30 feet x .30 = 9.00 meters 3. To convert Inches to Meter, just remember the following values of equivalent. - 1 inch = 2.5 cm. or .025 meters - 2 Inches = 5.0 cm. or .050 meters - 3 Inches = 7.5 cm. or .075 meters - 4 Inches = 10.0 cm. or .10 / meters - 5 Inches = 12.5 cm. or .125 meters ### CONCRETE Take note that all length in inches is divisible by one or any combination of these five numbers. Thus, it could be easily converted to meters by summing up their quotient equivalent. ### Example: - a. What is the meter length equivalent of 7 inches ? By simple analysis, 7 inches could be the sum of 4 and 3, therefore: - 4 inches = .10 meter - 3 Inches = 075 meter - Answer =.175 meter - b. How about 21 inches ? - 5 x 4 inches = 20 + 1 = 21 inches - since 4" = .10 m. and 1" = .025; multiply - 5 x .10 m. = .50 + .025 = .525 m. ### Problem Exercise Using the foregoing simple guide, convert the following numbers from inches to meters or vise versa. | Inches | to | Meters | Meters | to | Inches | |---|---|---|---|---|---| | 66 | | 2.42 | 2.42 | | 66 | | 99 | | 3.35 | 3.35 | | 99 | | 113 | | 4.27 | 4.27 | | 113 | | 178 | | 4.88 | 4.88 | | 178 | | 233 | | 5.19 | 5.19 | | 233 | ### 1-4 CONCRETE PROPORTION Proportioning concrete mixture is done in two different ways: by weight or by volume method. The most common and convenient way is by the volume method using the empty plastic bag of cement, or by a measuring box for sand and gravel as explained in Section 1-3. Measuring the aggregates and water by weight is sometimes used in a concrete batching plant for ready-mix concrete or as specified in the plan. <br/> # CHAPTER 2 ## MASONRY ### 2-1 CONCRETE HOLLOW BLOCKS Concrete Hollow Block is popularly known as CHB. It is classified as load bearing and non-bearing blocks. Load bearing blocks are those whose thickness ranges from 15 to 20 centimeters and are used to carry load aside from its own weight. Non-bearing blocks on the other hand, are blocks intended for walls, partitions, fences, dividers and the like carrying its own weight whose thickness ranges from 7 to 10 centimeters The standard hollow blocks has three void cells and two half cells at both ends having a total of four. These hollow cells vary in sizes as there are different manufacturers using different types of mold. Hence, it is recommended that concrete hollow blocks with bigger cells be considered in estimating for a more realistic result. In this study, what we want to know is the quantity of the materials needed for a certain masonry work made of concrete. ### SIMPLIFIED CONSTRUCTION ESTIMATE hollow block which generally comprises of the following items. 1. Concrete hollow blocks. 2. Cement and sand for block laying. 3. Cement, sand and gravel filler for the hollow core or cell. 4. Cement and fine sand for plastering. 5. Cement sand and gravel for foundation or footing. 6. Reinforcing steel bars and 7. Tie wires. Item 1 to 5 will be discussed in this chapter. The reinforcing steel bars and Tie wires will be presented in Chapter 3 -Metal Reinforcement. ### Estimating the materials for masonry work using hollow blocks, could be done in either of the following methods: - By Fundamental methods - By the Area methods ### ILLUSTRATION 2-1 A concrete hollow block wall has a general dimension of 3.00 meters high by 4.00 meters long. Determine the number of CHB, cement and sand required to construct the wall. ### SOLUTION-1 (By Fundamental Method) 1. Divide the height of the fence by the height of one block. 3.00 / 20 = 15 layers 2. Divide the length of the fence by the length of one block 4.00 / 40 = 10 pieces 3. Multiply the result of step 1 by step 2 15 x 10 = 150 pieces ### SOLUTION-2 (By the Area Method) Let us examine first how many pieces of CHB can cover up one square meter area. ### ILLUSTRATION 2-2 From the following Figure 2-4, find the number of 4" x 8" x 16" concrete hollow blocks to construct the fence. ### SOLUTION-1 (By Fundamental Method) 1. Find the Perimeter of the Fence - P=4+4+12+20+20= 60.00 meters. 2. Divide this perimeter by the length of one block - 60.00 / 40 = 150 pieces 3. Divide the height of the wall by the height of one block - 2.60 /.20 = 13 layers 4. Multiply 2 and 3: 150 x 13 = 1,950 pieces. ### SOLUTION - 2 (By the Area Method) 1. Find the area of the wall. A= 2.60 x 60.00 m. = 156 square meters. 2. If there are 12.5 blocks in one square meter then, multiply by the area. 156 x 12.5 = 1,950 pieces. ### Comments 1. Comparing the results obtained by the two methods, the answers are practically the same, but for convenience, the solution by the area method is much favored for being simple and direct to the answer. 2. Take note that in the above example, we computed the number of hollow blocks without posts. Suppose that Figure 2-4 was provided with the necessary posts as indicated in Figure 2-5, in this case, the area covered by the post will be subtracted from the total area of the wall, then solve for the CHB adopting the area method for simplicity of the process. ### ILLUSTRATION 2-3 From Figure 2-5, using class B mixture find the number of: - a) 10 x 20 x 40 cm. concrete hollow blocks ### SOLUTION-1 (Finding the CHB) 1. Find the perimeter of the wall. - P=20+20+12+8 - P = 60.00 meters 2. Find the space length occupied by the posts - Along 20 m. (.20 x 6) 2 = 2.40 - Along 12 m. (.20 x 4) 2 = 1.60 - Total space occupied by posts = 4.00 3. Subtract: - 60.00 m. - 4.00 m. = 56.00 m. net length for CHB. 4. Multiply by the height of fence to get the Net Area. - Net Area = 56.00 x 2.60 ht. - A = 145.6 square meters 5. Multiply by 12.5 to get the total number of concrete hollow blocks required. - 145.6 x 12.5 = 1,820 pieces. 6. Comparing this result to that of illustration 2-2, with 1,950 pieces hollow blocks, there is a material difference of 130 pieces because we subtracted the space occupied by the concrete posts. ### SOLUTION-2 (Concrete Posts and its Footing) 1. Find the volume of one concrete footing slab. ### ILLUSTRATION 2-4 From the following Figure, determine the number of 15 x 20 x 40 cm. CHB required to construct the building firewall. ### SOLUTION (By Direct Counting) 1. Find the CHB at Area A: 4.00 / 40 = 10 pieces 2. Height of wall A divided by height of one block 3.4.0 / 20 = 17 pieces 3. Multiply: (1) and (2): 10 x 17 = 170 pieces 4. Find the CHB at Area B: 7.00 / 40 = 17.5 5. Average Height of Area B divided by .20 ht. of one block 6.90 / .20 = 34.50 6. Multiply (4) and (5): 17.5 x 34.50 = 603.75 pieces 7. Find the CHB at Area C: 5.00 / .40 = 12.50 8. Height of C divided by height of one block 1.90 / 20 = 9.50 9. Multiply (7) and (8) 12.50 x 9.50 = 118.75 pieces ### Add Total CHB for Area A, B and C. 170 + 603.75 + 118.75 = 893 pieces ### Comment Take note that in the preceding example solution, fundamental methods of determining the number of blocks were used. The methods had undergone a very long process of finding the quantity by area one at a time. The process must be simplified with the aid of Table 2-2, presented as follows: ### SOLUTION - 2 (By the Area Method) 1. Find the Area of A: 3.40 x 4.00 = 13.60 sq. m. 2. Find the Area of B: 7.00 x 6.90 = 48.30 sq. m. 3. Find the Area of C: 5.00 x 1.90 = 9,50 sq. m. Total Area 71.40 sq. m. 4. Rer to Table 2-2. Along 15 x 20 x 40 CHB under column number per square meter, multiply: 71.40 x 12.50 = 893 pieces. ### Reminder Before estimating the quantity of concrete hollow blocks, be sure to verify the plan specially the clear height of the wall which is very important in the process. The following questions should be given due consideration for these might affect the result of the estimate. 1. Does the elevation as indicated in the plan specify the height from the first floor to the second floor line, or is it from floor to ceiling? In either case, the depth of the beam has to be considered in the estimate, either added or subtracted. 2. Have you considered the CHB to be installed from the underground foundation to the floor line? This particular portion of the wall is often overlook in the process of estimating especially when there is no detailed plan or cross section detail. Don't ever commit the same mistake experienced by most estimators. 3. See to it that the concrete hollow blocks to be installed are uniform in sizes and in thickness. Have it ordered from one manufacturer or supplier only. Installing different sizes of CHB means additional expenses for cement plaster and labor. If several suppliers cannot be avoided, have their respective blocks installed in a particular phase of work. ### 2-2 ESTIMATING CEMENT MORTAR After knowing the number of blocks needed for a particular masonry work, the next step is to find its work partner called cement mortar. Cement mortar is a mixture of cement, sand and water. It is used as bonding materials in installing masonry blocks and other various plastering work. In estimating cement mortar, one has to consider the following items. - a. The mortar to be used in between the layer of CHB. - b. The mortar filler for the hollow core or cell of the blocks. This filler could be pure mortar or mortar with gravel for economy. - c. Fine screened sand for plastering. ### ILLUSTRATION 2-5 Continuing the problem of Illustration 2-1 Figure 2-2, determine how many bags of cement and sand needed to install the 150 pieces 10 x 20 x 40 cm. CHB using class "B" mortar. There are three solutions offered in finding the cement mortar for concrete hollow blocks installation. - 1. By volume method. - 2. By the Area Method. - 3. Per Hundred Block method. ### SOLUTION-1 (By Volume Method) 1. Determine the volume of mortar in between the layer of the blocks, adopting the 12 mm (1/2") or .012 meters uniform thickness of the mortar. Volume = Thickness x Width of CHB x Length - V = .012 x 10 x 4.00 m. - V = .0048 cubic meter 2. Take note that 3.00 meters high wall divided by .20 m. height of one block is = 15 layers. Thus, multiply: - V = 15 layers x .0048 - V = .072 cu. m. This is the total volume of the mortar in between the 15 layers of concrete hollow blocks. 3. Aside from the cement mortar used in between block layers, there are 4 hollow cores or cells per block to be filled up with mortar. Find the volume per block. Volume = .05 x .075 x 20 x 4 cores - V = .003 cu. m. ### SOLUTION - 2 (By the Area Method) 1. Find the area of the wall. Area = (3.00 x 4.00) = 12 square meters. 2. Refer to Table 2-2. Along 10 x 20 x 40 CHB under class "B" mixture; multiply: Cement: 12 x .522 = 6.26 bags Sand 12 x .0435 = .522 cu. m. ### SOLUTION - 3 (By the Hundred Block Method) The Hundred Block Method is the third solution offered for a more simpler approach with the aid of Table 2-3 1. Find the number of concrete hollow blocks. Area: 3.00 x 4.00 = 12 sq. m. 12 sq. m. x 12.5 = 150 pieces CHB 2. Convert to unit of 100: 150 / 100 = 1.5 3. Refer to Table 2-3. Under class "B" mixture for a 10 x 20 x 40 CHB, multiply: Cement: 1.5 x 4.176 = 6.26 bags Sand: 1.5 x .348 = .522 cu. m. ### Another way of finding the mortar for block laying is by the Area Method with the aid of Table 2-2. ### SOLUTION - 2 (By the Area Method) 1. Find the area of the wall. ### ILLUSTRATION 2-6 Going back to the problem of Illustration 2-2, Figure 2-4, find the quantity of hollow blocks, cement and sand, for mortar using the area method of estimating. ### SOLUTION: 1. Find the area of the wall. - Area = Perimeter x Height - A = 60.00 m. x 2.60 m. - A = 156 square meters 2. Find the number of CHB. Refer to Table 2-2, under column number per sq. m. multiply: - 156 x 12.5 = 1,950 pieces ### SOLUTION-1 (Finding the CHB) 1. Find the perimeter of the wall. - P=20+20+12+8 - P = 60.00 meters 2. Find the space length occupied by the posts - Along 20 m. (.20 x 6) 2 = 2.40 - Along 12 m. (.20 x 4) 2 = 1.60 - Total space occupied by posts = 4.00 3. Subtract: - 60.00 m. - 4.00 m. = 56.00 m. net length for CHB. 4. Multiply by the height of fence to get the Net Area. - Net Area = 56.00 x 2.60 ht. - A = 145.6 square meters 5. Multiply by 12.5 to get the total number of concrete hollow blocks required. - 145.6 x 12.5 = 1,820 pieces. 6. Comparing this result to that of illustration 2-2, with 1,950 pieces hollow blocks, there is a material difference of 130 pieces because we subtracted the space occupied by the concrete posts. ### SOLUTION-2 (Concrete Posts and its Footing) 1. Find the volume of one concrete footing slab. ### ILLUSTRATION 2-9 The owner of a commercial lot wants to fence the frontage of his lot with 15 x 20 x 40 cm. concrete hollow blocks. The fence is 3.50 meters high and 40 meters long provided with a 25 x 25 cm. reinforced concrete posts spaced at 4.00 meters distance. Using class B mixture list down the materials required. ### SOLUTION 1. Solve for the gross area of the fence. ### ILLUSTRATION 2-10 Continuing the problem of Illustration 2-9, Figure 2-10, if the wall footing is 15 centimeters thick and 50 centimeters wide, determine the quantity of cement, sand and gravel necessary using class "A" concrete. ### SOLUTION-1 (By Volume Method) 1. Find the volume of the wall footing. Length minus the space occupied by the posts to get the net length. 40.00 m. - (.25 x 11 posts) = 37.25 m. net length Volume = .15 x .50 x 37.25 m. = 2.79 cu. m. 2. Refer to Table 1-2. Using class "A" concrete; multiply: Cement: 2.79 x 9.0 = 25 bags Sand: 2.79 x .50 = 1.4 cu. m. Gravel: 2.79 x 1.0 = 2.8 cu. m. ### SOLUTION - 2 (By the Linear Meter Method) 1. Solve for the Net Length of the CHB wall. Net length = 40.00 m. - (.25 x 11 posts) = 37.25 m. 2. Refer to Table 2-5. Along 15 x 50 cm. footing dimension class "A" mixture, multiply: ### ILLUSTRATION 2-11 From the following figure, prepare the bill of materials using class "B" mixture for concrete and mortar. ### SOLUTION #### A. Solving for CHB 1. Find the Perimeter of the hollow block fence. P = 3 (25.00) + 4(5.00) P = 95.00 meters. 2. Subtract the length occupied by the posts. Length = .20 width x 20 posts (see figure) L = 4.00 meters ### ILLUSTRATION 2-12 A masonry wall 15 cm. thick requires 1,500 pieces of 2-core stretcher blocks, 100 pieces single end block, 120 half block, 200 corner blocks and 80 pieces beam block. Find the cement sand and gravel using class "B" mortar mixture. ### SOLUTION 1. Itemized the blocks according to its category and indicate the number of pieces. - 2-core 15 cm. Stretcher block = 1,500 pieces - Single end block = 100 - Half block = 120 - L-Corner Block = 200 - Beam Block = 80 2. Refer to Table 2-6. Under column class "B" mixture, multiply the number of blocks to each corresponding value in the table to get the cement, sand and gravel required. - **a.) 1,500 Stretcher Blocks** - Cement: 1,500 x .0623 = 93.45 bags - Sand: 1,500 x .0030 = 4.50 cu. m. - Gravel: 1,500 x .0045 = 6.75 cu. m. - **b.) 100 Single End Block** - Cement: 100 x .0612 = 6.12 bags - Sand: 100 x .0044 = 0.45 cu. m. - Gravel: 100 x .0075 = 0.75 cu. m. - **c.) 120- Half Block.** - Cement: 120 x .0270 = 3.24 bags - Sand: 120 x .0020 = 0.24 cu. m. - Gravel: 120 x .0035 = 0.42 cu. m. - **d.) 200L Corner Block.** - Cement: 200 x .0590 = 11.80 bags - Sand: 200 x .0041 = 0.82 cu. m. - Gravel: 200 x .0070 = 1.40 cu. m. - **e.) 80-Beam Block** - Cement: 80 x .0488 = 3.90 bags - Sand: 80 x .0040 = 0.32 cu m. - Gravel: 80 x 0070 = 0.56 cu m. 3. Summary of the Materials - 1,500 stretcher blocks - 120 half block - 80 beam block - 6.33 cu. m. sand - 100 single end block - 200 L-corner block - 119 bags cement - 9.88 cu. m. gravel ### 2-6 DECORATIVE BLOCKS Decorative hollow blocks are manufactured from either cement mortar or clay. These types of construction materials had been widely used for ventilation and decorative purposes. ### TABLE 2-7 QUANTITY OF CECORATIVE BLOCKS CEMENT AND SAND PER 100 BLOCKS. | Size in cm. | Number per sq. m. | Cement in Bag at 40 kg. Class Mixture | Sand per 100 blocks | |---|---|---|---| | 5 x 10 | 200 | 180 | 120 | 010 | | 5 x 15 | 133 | 270 | 180 | 015 | | 5 x 20 | 100 | 360 | 240 | 020 | | 5 x 25 | 80 | 450 | 300 |