Construction estimate by max fajardo pdf free download

Find the number of 10 cm. Refer to Table , multiply: Find the cement and sand for block laying. Cement: x. Solving for CHB cause we do not plaster the wall below the ground line.

Find the Perimeter of the hollow block ,fence. For plastering, refer to Table Subtract the length occupied by the posts. Using a 10 x 40 footing class" S" mixture; multiply: Cement: Find the volume of the concrete posts. Find the volume of the footing. Total Volume of posts and footing. Using class US" mixture; multiply: Cement: 3.

Summary: 2, pieces 10 x 20 x 40 cm. CHS 2. Ji;W;i-- 1. From the following figures, find the following materia'is re- quired for its construction. There 19 em 19 are however various type of blocks especially designed for ar- chitectural and structural purposes such as; the stretcher,block, the L-comer block, the single end block, the half block and the beam block.

The purpose of making these types of blocks is to create a wider core to accommodate concrete and steel reinforcement. The thickness is only used in computing the cement mortar for block laying. The wall is de- signed as a continuous interlocking support of the roof framing. Comer blocks combined with stretcher blocKs are used on cor- ners provided with reinforcement instead of reinforced concrete post. With this method, the costs of forms and labor were felt as a substantial savings.

Half Block In block laying work, concrete being a mixture of cement, sand and gravel should be used to fill the hollow core of the blocks for economical reasons. If cement mortar will be used to fill the core, cost will be substantially high because of the. Cell ent in bag 1. Itemized the blocks according to its category and indicate Size in Class Mixture the number of pieces.

Centimeters B C Sand Gravel 1, pieces 2-core 15 cm. Block 80 L - Corner Block 2- core ' 20 cm. Gravel : x. The mortar for block laying is a mixture of cement and sand c. The hollow core is filled Cement : x. And to fill the core Sand: x ,. How to use the table, the fol,lowing example is pre- d. A masonry wall 15 cm. Find the cement Sand: 80 x. These types of construction materials had ' been widely used for ventilation and decorative purposes. Lately however, the used of adobe stone was no longer limited to the ordinary zocalo and fencing work but also extensiv.!

The use of adobe stone fOl' fences, buttresses, cross- foot- ings, and stairs minimizes the use of mortar filler unlike in work- ing with concrete hollow blocks. Plastering is sometimes not applied speCially when the design calls for exposure of the natural texture of the stone. Sand Buttress Cross No. Find the total length of the post A. Determine the length of the fen,ce minus the space oc- cupied by the buttresses. Refer t6 Table , using class.

For a 45 x 60 buttress footing refer to Table , the number of course ,is 4. Add the results of 2 and 3. Class of Mixture Sand B C 0 cu. Solving fo[ Wall Footing One Face. Length of the fence minus the space occupied by the Two Faces.

Find the total surface area of the wa ll and the but- tresses to be plastered. Multiply by 6 stones ' per meter length. The adobe stones are laid crosswise the wall, see figure The height is only 1. Solving for the Cement Mortar wall below the ground line. Refer to. Table 1. Using class "B" mixture, multiply:. The area of the fence is Us- Cement : This is for one side plaster only. Tf two sides will be Sand: 28;20 x. Refer Summary of the Materials to Table Along 30 x 45 buttress class "B" mixture multiply: pieces.

Indeed, the required wall cross section in- the fence using a Class A mixture; b Class C mixture. The creases with heights due to the effect of the triangular soil pres- plan specify plastering both sides of the wall using class 8 mix- sure distributed behind the retaining wall. If the ground water level rises into the backfill behind a re- taining wall due to either changing ground water condition or percolating water surface, the lateral pressure against the wall is also changed. The combined effect of soil and water pressure causes overturning moments and sliding forces.

This is greater than the condition where there is 'no water. To avoid the rise of water that is building up behind the re- -f- FI Weep holes should be at least 3 inches in diameter provided with a granular ,. The hori- 90 zontal spacing ranges from to centimeters apart. The wall must be safe against overturning. The wall structure must be safe against sliding -, - I I 4.

The stone Excavation and Placing shall be tough, durable and dense. It shall be resistance to the action of air and water and suitable in all aspects for the pur- The bed for 'rip-rap is excavated dow! Adobe stone shall not be used unless specified. The rip-rap is thor- oughly rammed into place and the finished stones are laid to an even tight surface. Intersections between stones are filled with small broken fragments firmly rammed into place.

H The stones are placed by hand or individually laid by ma- chine. Spaces between stones are then filled with cement mor- 30x30x30 tar sufficient enough to completely fill all the voids except the Gravel backfill face surface of the stones left exposed.

After grouting, the surface is cured like structural concrete for a period of at least 3 days after the installation. Class A Stone ranging from 15 to 25 kilograms with at least Class - A 2.

Class C Stones ranging from 60 to kg. Class 0. Stones weighing from kg. Summary ofthe materials cubic meters Class. Solve for the volume of the proposed riprap. This pattern of fail- 2 Total volume.. A gravity wall 50 meters long has the following dimensions Sand: x.

Length of the riprap divided by the weep holes at 2. Find the volume of stem. Find the volume of backfill materials.

Summary of the Materials bags cement 64 cu. Find the volume of the footing.. Total volume of 1 and 2. Gabions are box shaped containers made of tough 4. Using class A mixture, multiply: woven hexagonal netting strengthened by selvedges of heavier wires. To further strengthen the container, diaphragms are Cement: Recommended Uses 5. For Reinforcement, see plan and refer to Chapter-3 When filled with quarried stpne on site, Gabions and Mat- tresses prove to be excellent materials for ccnstruction of 6.

For Weep Hole pipe, divide length by 2. Slope stabilization for embankment and cutting 2 2. Prevention of erosion in river embankments. Tensile reinforcement of a beam b. Classification of the reinforcement is under tension.

In estimating the quantity of steel reinforcing bars, one has. Length of splice for compression bars. Concrete Hollow Block Remforcement. This type of reinforcement is Insta"ed and spliced pro- gressively with the rise of the concmte blocks.

Foeting Reinforcements 1. Dowels, cut bars, anchor, and bend bars 3. Main vertical reinforcement a. Single or b. Bundled bars 2. Lateral Ties a. Outer ties bf Inner ties. Straight ties [ 3. Spiral Ties for circular column 4. Dowels for partitions and other future attachments. Main Reinforcement a. Straight bars 1.

The main reinforcement of post, column, beam, girder and b. Bend bars the like structures, are determined by the simple Direct c. Additional cut bars for tension and compression Counting Method, that Is" by countlrig the number of main d.

Dowel bars for future attachment. Open stirrups 2. Closed stirrups and splices for end lapping joints. The additional length Is c. Straight stirrups. In ordering steel bars, always think of minimizing the 3. Cut Bars splices If it cannot be totally avoided. Over and across the support ' b. Between supports 3. For other reinforcement parts such as lateral ties, stirrups, c. Dowels and hangers for ceiling and partition ' spirals, dowels, cut bars and the like 'should be treated or estimated separately one at a time taking Into consideration the additional length for the hook, bend, and spliCing length.

Floor Slab Reinforcement 1. Main reinforcements. After knowing the length of the lateral ties, stirrups and other similar parts, select the steel bars from the various s. Stralght 'main reinforcing bars extending from one commercial length of 8. Alternate main reinforcing bars with bend between unnecessary extra cuts.

Main alternate bars over support beam or girder , '5. The length of each tie 2. Additional altemate cuf barS over the support beam. Not until after Steel bar as reinforcement is a requirement for all types 'of. The National Building Code has ment could one make a sound and reliable estimate. Concrete hollow block reinforcement could be determined In Hortzontal Rein', at everv 3 layera three different ways: 1. By the Direct Counting Method 2. By the Area Method Natural Ground 3.

By the Unit Block Method ::! The hook, bend and lap- ping splices are Imagine! Layer P,er Block Per Sq. The values presented in the table include the allowances required 1. Solve for the Area of the fence for standard bend, hook and lapping splices. From Figure , determine the number of: 2. For vertical reinforcement spaced at 80 centimeters, re- a 10 x 20 x 40 cm CHB fer to Table Under column length per sq. Add the results of a and b 3. Solve for the horizontal bars at every 3 layers.

From Ta-. Convert' thls length to the commercial size of steel bars, 12 sq. Tie wire is used to secure the steel bars 6. Ordering tie wire is not by feet nor by meter length but in kHograms or roll.

Solving the same problem of illustration , by the Unit Block Method, we have the following solution: The length of each tie '. Tie wire is cut into length ranging from 20 to 1.

For larger steel bars, the length of ties varies depending upon the conven- Area: I: 4. I: 12 sq. Determine the number of hollow blocks. The quantity is determined through a more or less calculation. The only thing that is certain is either it is over estimated or under esti- 3. The common size of steel bar reinforcement specified for concrete hollow block work is either; 10 mm, 12 mm or 16 mm.

For this Table Find the wall area. Along 60 em. The reinforcement , for this type of structure is detennined by direct counting from ' 1. Vertil" I t:li:a"S at SO centimeters and one. Solve for the wall area. Along 80 cm. Know the actual dimensions of the foo Consider the 7. If the plan does not specify hook or bend of the footing reinforcement, the length f the bar is equal to the lengttt. Know the spacing distance of the steel bars both ways to determine the exact number required.

As much as pos- 1. Find the net length of one cut reinforcing bar. One Important consideration in estimating steel bar rein- 2. Find the total number of these cut bars in one footing. For instance, if 5 pieces at 1. Find the total number of cut bars in 24 footings. Likewise, if the work requires 2. The result simply mean that in one 6 meters long steel bar we could get 4. We will not consider the fractional value of.

Instead, we will only use the whole value 4. Divide the result of dtep 3 by 4. Find the total cut bars for 30 footings. The following rules will help in making the right choice. Find the total length of pes. Determine the net length of one reinforcing cut bar x 1. Divide 6. Divide by 6. If the result in dividing the length of one steel bar by ,the length of one cut bar is a whole number exact value use m.

If the result has a fractional value, use the first method. The above solution is correct because in dividing 6. Estimating the Tie Wire was initially discussed in Section by 1. Applying the same principles of 5. Meaning, there is no extra cut in one 6. In step 5, dividing by 6. Therefore, I , 1. Looking at the plan of Figure there are: all cuts are exactly the same without excesses.

The net length of one cut bar is 1. Solve for the total ties of the 24 footings: number of bars in 30 footings is pieces. Select a commercial length of steel bar which is divisible 3.

Using a 30 centimeters long ties, multiply: H by 1. Try 6. Convert this leng h to kilogram. One kilogram jf No. Divide: 3. This simply mean that in a 6. Divide the lli. Looking at the plan of Figure There are 36 steel bar to the additional length I. I wire a. Additional length for beam depth and floor thickness if 2. Solve for the total ties of 30 footings. Distance from floor to footing slab. Provisions for splices of succeeding floors.

If one tie is 30 cm. From the following figure, list down the main reinforcement 4, Convert this length to kilograms. Using the value of 53 from the footing to the second floor using 20 mm if mere are 10 meters per kilogram, divide: columns in the plan. The m Ii or vertical reinforcement, b. The lateral ties or c. SpirAl tip. Identify the bars with hook and bend for adjustment of their order length.

Determine the total length of the main steel bar reinforce- ments. Slab Slab Beam Beam a. Bend at the base footing Height from ground line to beam Depth of beam Dowel for second floor 20 x 20 mm Select a 6. Multiply by number of bars in one post x'10 post. Order: 80 pes. Verify the plan if the span or distance of the column Tied column has reinforcement 'conSisting of vertical bars where the beam is resting indicates the following condi- held In a position by lateral reinforcement calied lateral ties.

Take note that "the lesser th splice the Ie.. Determine the spacing distance of the lateral ties. Ties should not be less than No. The lowest value is Therefore, adopt 30 centimeters shall not exceed the following: spacing of the lateral ties 1.

Designation Inches mm. The least dimension of the column 2 Y- '. Determine the spacing of- the lateral ties. TIES 16 x 20 mm. Find how many 1. By trial division we have: The least dimension of the column is 50 cm. Adopt 32 cm. Determine the number of lateral ties in one column. Iong steel bar. CommeBU: 32cm. Step 7 is very important treca use wtihout these trial divi- sions, cutting could be done on a 6. In cases where results of the trial divisions does not give an exad quotient, it becomes the estimator's choice to FIGURE decide which length to use that will not produce exces- sive waste.

TaKe note that this 22 is the spacing distance between lateral ties. What we are after is the number of ties in one '3. In this example, we have learned. Add one to get the number of lateral ties. Solve for the total lateral ties In the 26 cohllmns. Find the number of main. Detennlne the length of one Lateral Tie. By inspection, ties. Find the total intersections in the 26 columns: 1. Or 32 cm. This 4, is the total numbl:! Or 48 cm.

The least dimension of the column is 40 cm. Adopt 32 em. One kilogram of No. Find the total lateral ties in th e 30 columns.

By Inspection, the length of the lateral ties are : tie wire in kilograms. By trial divi- main reinf. The results dictate that we use 6.

Convert to kilograms. The x - en- Spacing per M. The table will guide the estimator in selecting rei nforcing 15 6. To use the table, consider the following example: 25 4. Find the 40 2. If the length of each tie is 40 cm. By Inspection, there are two types of lateral ties. Convert to kilograms, divide by 53 m. Find the total length of the 20 columns. Problem Exercise 7. From the following figure, find the number of 10 mm lateral ties and tie wire if there are 36 columns with cross sectional di- 3.

Refer to Tatile Under spacing of lateral ties at 30 cm. Multiply: m. Refer again to Table Under length of ties along centimeters, 4 and 6 pi'eces could pe derived from a 5. For the 80 cm. If there are 16 beams of the The procedures adopted in estimating the number of stirrups same design, find the materials required for the stirrups. By'direCt counting, there are 17 stimJps at 99 cm. For a 1. For easy handling use 6.

By inspection the length of one stinup is cm. Refer' to Table , along cm. If we chose 6. If we chose 7. The spiral reinforce. Find the number of 10 mm b'ilr spirals.

Refer to Table 3- spacers under the following considerations 9. For a 50 cm. That, the clear spacing between the spirals should not 3. Multiply: 3. Finding the Tie Wire 7. Find the number of vertical bars per column :: 12 pieces. Under 50 cm. Total Ties for 14 cplumn at 7. TotallengtlJ 9f the wire at.

Convert to' kilograms. Divide by 5j, , 5. The sup- tennlne the number of 12 mm. The principal reinforcement runs in one direction parallel to the slab span and perpendicular to the A. One method used in finding the number of steel bars for a one-way reinforced concrete slab is either by the direct Given Data: counting or by the area method. Find the number of main reinforcements at. This Find the number of cut bars in.

Using a 6. Find the main reinforcement; add step 2 and step 5. Across the main reinforcement, divide by. Along the 1. Since there are ,2 sides at 1. Add the results of step I and step 3. Detennine the floor area. Refer to Table , along b ,r spacing at 15 crn. I '"'! I "- T-rii -l 3. Bars 7. Solve for the Area of the floor: 7. Skin pybelle s60v3 x ice cream sandwich.

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