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Chapter 9—Excavation LEARNING OBJECTIVES Learn how to demine excavation quantit ies using the cross-sectional method, the average-end-area method, and the average width and length; and when to use each method. Explain how shrink and swell affect the volume of earthen mater ials that need to be excavated, hauled, and compacted. Learn how to estimate the quantities of landscaping, granular base, and asphalt. Learn how to determine the cost of each ofthese items.

HOMEWORK QUESTIONS 1.

What type of information a bout the excavation can the estimator learn from the specifications?

The extent of work, excess excavated material, the persons who responsible on clearing, grubbing, trees, and trenching. 2.

How does the type of so il to be excavated affect the estimate?

The estimator should investigate the soil conditions at the job site, and also need to check the drawings and the specifications of the project. 3.

What is the unit of measure in excavation?

The unit of measure in excavation is cubic yard. 4.

How will the type of so il, shape of the excavation, and amount of work to be done affect the equipment selection?

Different phase of excavation requires different equipments. For instances, if material must be hauled some distances, then the excavated material should be hauled out or the fill material hauled in, such as trucks or tractor-pulled wagons. 5.

What is the difference between a bank cubic yard, a loose cubic yard, and a compacted cubic yard? Why are they importa nt in estimating excavation?

Bank cubic yards is the material in its natural state. The loose cubic yards are the uncompacted excavated materials. The compacted cubic yards are the compacted materials. They are important because as estimator, it is essential to understand the swell and shrinkage factors for various types of soils. 6.

What is meant by cut and fill?

Cutting means to bring the ground to a lower level by removing earth. Filling means to bring soil in to build the land to a higher elevation. 7.

What does the estimator have to consider if there is a substantial amount of cut on the job? What if there is a substa ntial amo unt of fill?

The estimator needs quantities to estimate the amount of effort required to convert the cut material into the fill material. If there is a substantial amount of fill, then it must be hauled to where it will be used and then compacted. 8.

How can the estimator get an estimate of the depth of topsoil on the project?

The estimator needs to determine the depth of the topsoil, where it will be stockpiled, and what equipment should be used to strip the topsoil and move it

Chapter 9 Excav atio n

2 to the stockpile area.

9.

What type of excavation is co nsidered to be general excavation?

It is the removal of all types of soil that can be handled in fairly large quantities, and it is required for basement, mat footing, or a cut for a highway or parking area. 10.

How does general excavation differ from special excavation?

The special excavation is the portion of the work that requires hand excavation, and requires special equipment to be used for a particular portion. Most of the work includes footing holes, small trenches, and the trench-out areas. 11.

What is excess and borrow, and how are each considered in the estimate?

It the determination on whether there will be an excess of materials that must be discarded, or whether there is a shortage of materials and some must be brought in. The estimator needs to know how many cubic yards are required, and then the estimator must find a place to haul the material. The equipments are also need to be considered. 12.

How will the possibility of a high water table or underground stream affect the bid?

It depends on the projects; some projects require constant pumping to keep the excavation dry, and the variation in costs is extreme. Also the problem of water will depend the season of the year, location, type of work, general topography, and weather. 13.

What are piles, and under what conditions might they be requ ired on the project?

Piles are used to support loads are referred to as bearing piles. The conditions of use might be depended on the types of soil, and depth to be driven make accurate cost details difficult to determine unless the details of each particular job are known. 14.

For a project, your company needs to haul away 1,450 bank cubic yards (bcy) of so il. If the soil has a swell of 25 percent, how many loose cubic yards ( lcy) of so il will need to be hauled? How many truckloads are requ ired to haul the soil o ffs ite if each truck can haul nine loose cubic yards? Cubic yards of haul = 1,450 x (1 + 0.25) = 1812.5 lcy Loads = 1812.5 / 9 = 202 loads

15.

For a project, your company needs to haul away 350 bank cubic yard s of sand. If the sand has a swell of 12 percent, how many loose cubic yards of sand will need to be hauled? How many truckloads are required to haul the sand offsite if each truck can haul 12 loose cubic yards? Cubic yards of haul = 350 x (1 + 0.12) = 392 lcy Loads = 392/12 = 33 loads

16.

If 750 compacted cubic yards (ccy) of in-place so il is requ ired for a project, how many loads of import will be required? The import mater ial has a swell of 30 percent and shrinkage of 90 percent. The trucks can haul 10 loose cubic yards. Required bcy = 500/0.9 = 556 bcy Loose cubic yards = 556 x (1 + 0.3) = 573 lcy Loads = 573 / 10 = 58 loads

Chapter 9 3 Excav atio n 17. If 1,490 compacted cubic yards of in-place soil is required for a project, how many loads of import will be requ ired? The import mater ial has a swell of 14 percent and shrinkage of 95 percent. The trucks can haul 12 loose cubic yard s. Determine the volume of the haul and the number of loads as follows: Required bcy = 1,490 / 0.95 = 1569 bcy Loose cubic yard s = 1569 bcy x (1 + 0.14) = 1789 lcy Loads = 1789 / 12 = 149 loads 18.

Using the cross-sectional method, determine the cuts and fill for the project shown in Figure 9.47. The grids are 50 feet apart in both directio ns. The existing grade appears above the proposed grade. Express your cuts in bank cubic yards (bcy) and your fills in compacted cubic yards (ccy). The cuts and fills are shown in Figure 9-1.

Figure 9-1

Chapter 9 Excav atio n

4

The dividing line between the cuts and fills and the distances from the grid intersections to this line are shown in Figure 9-2.

Figure 9-2 Unless the division line between the cuts and fills passes through a cell, the cut or fill area is 2,500 square feet (50 ft × 50 ft). The cuts and fills are as fo llows: Cell 1: Cut1 = (0′ + 0′ + 0′ + 0.1′) / 4 × 2,500 sf = 62 bcf Fill 1 = 0 ccf Cell 2: Cut2 = (0′ + 0′ + 0.1′ + 0.3′) / 4 × 2,500 sf = 250 bcf Fill 2 = 0 ccf Cell 3: Cut3 = (0′ + 0′ + 0.3′ + 0.9′) / 4 × 2,500 sf = 750 bcf Fill 3 = 0 ccf Cell 4: Cut4 = (0′ + 0′ + 0.9′ + 1.1′) / 4 × 2,500 sf = 1,250 bcf Fill 4 = 0 ccf

Chapter 9 Excav atio n

Cell 5: Cut5 = (0′ + 0′ + 1.1′ + 1.5′) / 4 × 2,500 sf = 1,625 bcf Fill 5 = 0 ccf Cell 6: Cut6 = (0′ + 0′ + 1.5′ + 0′) / 4 × 2,500 sf = 938 bcf Fill 6 = 0 ccf Cell 7: Cut Area7 = (12.5′ × 50′) / 2 = 312 sf Cut7 = (0′ + 0′ + 0.1′) / 3 × 312 sf = 10 bcf Fill Area 7 = (37.5′ × 50′) / 2 = 938 sf Fill 7 = (0′ + 0′ + 0.3′) / 3 × 938 sf = 94 ccf Cell 8: Cut Area8 = (50′ × 50′) – (37.5′ × 30′) / 2 = 1,938 sf Cut8 = (0.1′ + 0.3′ + 0′ + 0′ + 0.2′) / 5 × 1,938 sf = 233 bcf Fill Area 8 = (37.5′ × 30′) / 2 = 562 sf Fill 8 = (0′ + 0.3′ + 0′) / 3 × 562 sf = 56 ccf Cell 9: Cut9 = (0.3′ + 0.9′ + 0.2′ + 0.4′) / 4 × 2,500 sf = 1,125 bcf Fill 9 = 0 ccf Cell 10: Cut10 = (0.9′ + 1.1′ + 0.4′ + 0.7′) / 4 × 2,500 sf = 1,937 bcf Fill 10 = 0 ccf Cell 11: Cut11 = (1.1′ + 1.5′ + 0.7′ + 0.9′) / 4 × 2,500 sf = 2,625 bcf Fill 11 = 0 ccf Cell 12: Cut12 = (1.5′ + 0′ + 0.9′ + 0′) / 4 × 2,500 sf = 1,500 bcf

5

Chapter 9 Excav atio n Fill 12 = 0 ccf Cell 13: Cut13 = 0 bcf Fill 13 = (0′ + 0.3′ + 0′ + 0.7′) / 4 × 2,500 sf = 625 ccf Cell 14: Cut Area14 = (20′ × 33.3′) / 2 = 333 sf Cut14 = (0′ + 0.2′ + 0′) / 3 × 333 sf = 22 bcf Fill Area 14 = (50′ × 50′) – 333 sf = 2,167 sf Fill 14 = (0.3′ + 0′ + 0′ + 0.7′ + 0.1′) / 5 × 2,167 sf = 477 ccf Cell 15: Cut Area15 = (50′ × 50′) – (16.7′ × 12.5′) / 2 = 2,396sf Cut 15 = (0.2′ + 0.4′ + 0′ + 0′ + 0.3′) / 5 × 2,396 sf = 431 bcf Fill Area 15 = (16.7′ × 12.5′) / 2 = 104 sf Fill 15 = (0′ + 0.1′ + 0′) / 3 × 104 sf = 3 ccf Cell 16: Cut16 = (0.4′ + 0.7′ + 0.3′ + 0.6′) / 4 × 2,500 sf = 1,250 bcf Fill 16 = 0 ccf Cell 17: Cut17 = (0.7′ + 0.9′ + 0.6′ + 0.6′) / 4 × 2,500 sf = 1,750 bcf Fill 17 = 0 ccf Cell 18: Cut18 = (0.9′ + 0′ + 0.6′ + 0′) / 4 × 2,500 sf = 938 b cf Fill 18 = 0 ccf Cell 19: Cut19 = 0 bcf Fill 19 = (0′ + 0.7′ + 0′ + 1.2′) / 4 × 2,500 sf = 1,188 ccf Cell 20: Cut20 = 0 bcf

6

Chapter 9 Excav atio n

Fill 20 = (0.7′ + 0.1′ + 1.2′ + 0.9′) / 4 × 2,500 sf = 1,812 ccf Cell 21: Cut Area21 = (37.5′ × 18.75′) / 2 = 352 sf Cut21 = (0′ + 0.3′ + 0′) / 3 × 352 sf = 35 bcf Fill Area 21 = (50′ × 50′) – 352 sf = 2,148 sf Fill 21 = (0.1′ + 0′ + 0′ + 0.9′ + 0.5′) / 5 × 2,148 sf = 644 ccf Cell 22: Cut Area22 = (50′ × 50′) – (31.25′ × 41.7′) / 2 = 1,848 sf Cut22 = (0.3′ + 0.6′ + 0′ + 0′ + 0.1′) / 5 × 1,848 sf = 370 bcf Fill Area 22 = (31.25′ × 41.7′) / 2 = 652 sf Fill 22 = (0′ + 0.5′ + 0′) / 3 × 652 sf = 109 ccf Cell 23: Cut23 = (0.6′ + 0.6′ + 0.1′ + 0.2′) / 4 × 2,500 sf = 938 bcf Fill 23 = 0 ccf Cell 24: Cut24 = (0.6′ + 0′ + 0.2′ + 0′) / 4 × 2,500 sf = 500 bcf Fill 24 = 0 ccf Cell 25: Cut25 = 0 bcf Fill 25 = (0′ + 1.2′ + 0′ + 1.3′) / 4 × 2,500 sf = 1,562 ccf Cell 26: Cut26 = 0 bcf Fill 26 = (1.2′ + 0.9′ + 1.3′ + 0.9′) / 4 × 2,500 sf = 2,688 ccf Cell 27: Cut27 = 0 bcf Fill 27 = (0.9′ + 0.5′ + 0.9′ + 0.6′) / 4 × 2,500 sf = 1,812 ccf Cell 28:

7

Chapter 9 Excav atio n Cut Area28 = (8.3′ × 8.3′) / 2 = 34 sf Cut28 = (0′ + 0.1′ + 0′) / 3 × 34 sf = 1 bcf Fill Area 28 = (50′ × 50′) – 34 sf = 2,466 sf Fill 28 = (0.5′ + 0′ + 0′ + 0.6′ + 0.5′) / 5 × 2,466 sf = 789 ccf Cell 29: Cut Area29 = 50′ × (8.3′ + 10′) / 2 = 458 sf Cut29 = (0.1′ + 0.2′ + 0′ + 0′) / 4 × 458 sf = 34 bcf Fill Area 29 = 50′ × (41.7′ + 40′) / 2 = 2,042 sf Fill 29 = (0′ + 0′ + 0.5′ + 0.8′) / 4 × 2,042 sf = 664 ccf Cell 30: Cut Area30 = (10′ × 50′) / 2 = 250 sf Cut30 = (0′ + 0′ + 0.2′) / 3 × 250 sf = 17 bcf Fill Area 30 = (40′ × 50′) / 2 = 1,000 sf Fill 30 = (0′ + 0.8′ + 0′) / 3 × 1,000 sf = 267 ccf Cell 31: Cut31 = 0 bcf Fill 31 = (0′ + 1.3′ + 0′ + 0′) / 4 × 2,500 sf = 812 ccf Cell 32: Cut32 = 0 bcf Fill 32 = (1.3′ + 0.9′ + 0′ + 0′) / 4 × 2,500 sf = 1,375 ccf Cell 33: Cut33 = 0 bcf Fill 33 = (0.9′ + 0.6′ + 0′ + 0′) / 4 × 2,500 sf = 938 ccf Cell 34: Cut34 = 0 bcf Fill 34 = (0.6′ + 0.5′ + 0′ + 0′) / 4 × 2,500 sf = 688 ccf Cell 35: Cut35 = 0 bcf

8

Chapter 10 Concrete

9

Fill 35 = (0.5′ + 0.8′ + 0′ + 0′) / 4 × 2,500 sf = 812 ccf Cell 36: Cut36 = 0 bcf Fill 36 = (0.8′ + 0′ + 0′ + 0′) / 4 × 2,500 sf = 500 ccf Total Cuts: Total cuts = 62 bcf + 250 bcf + 750 bcf + 1,250 bcf + 1,625 bcf + 938 bcf + 10 bcf + 233 bcf + 1,125 bcf + 1,937 bcf + 2,625 bcf + 1,500 bcf + 0 bcf + 22 bcf + 431 bcf + 1,250 bcf + 1,750 bcf + 938 bcf + 0 bcf + 0 bcf + 35 bcf + 370 bcf + 938 bcf + 500 bcf + 0 bcf + 0 bcf + 0 bcf + 1 bcf + 34 bcf + 17 bcf + 0 bcf + 0 bcf + 0 bcf + 0 bcf + 0 bcf + 0 bcf Total Cuts = 18,591 bcf / 27 cf per cy = 689 bcy Total Fills: Total Fills = + 0 ccf + 0 ccf + 0 ccf + 0 ccf + 0 ccf + 0 ccf + 94 ccf + 56 ccf + 0 ccf + 0 ccf + 0 ccf + 0 ccf + 625 ccf + 477 ccf + 3 ccf + 0 ccf + 0 ccf + 0 ccf + 1,188 ccf + 1,812 ccf + 644 ccf + 109 ccf + 0 ccf + 0 ccf + 1,562 ccf + 2,688 ccf + 1,812 ccf + 789 ccf + 664 ccf + 267 ccf + 812 ccf + 1,375 ccf + 938 ccf + 688 ccf + 812 ccf + 500 ccf Total Fills = 17,915 ccf / 27 cf per cy = 664 ccy

Chapter 10 Concrete

10

Chapter 10—Concrete LEARNING OBJECTIVES Learn how to estimate the quantity and cost of concrete including footings, foundation walls, columns and piers, grade beams, and sla bs.

HOMEWORK QUESTIONS 1.

What is the difference between plant ready-mixed concrete and job-mixed concrete?

The ready mixed concrete mostly used on commercial and residential work. The quality control, proper gradation, water, and design mixes are different from the mixed on the job. 2.

Under what circumstances might it be desirable to have a field batching plant for job-mixed concrete?

The estimator will need to compute the amounts of each material required, evaluate the local availability, and determine if these ingredients will be used or if materials will be shipped to the site from somewhere else. 3.

What is the unit of measure for concrete?

The unit cubic yard or cubic foot, and then converted into cubic yards. 4.

Why do es the estimator have to keep the different places that the concrete will be used separate in the estimate (e.g., concrete sidewalks, floor slabs)

The reason for that is the irregularly shaped projects are broken down into smaller areas for more accurate and convenient manipulation. 5.

Why shou ld the different strengths of concrete be kept separate?

It should be kept separate because it is more organized and accurate for the estimation. 6.

Where would the estimator most likely find the strength of the concrete required?

The estimator most likely find the strength of the concrete required under the specifications. 7.

How is rebar ta ken off? In what unit of measure are large quantit ies ordered?

The rebar are taken off by linear feet. The takeoff sheet should have included the number of the bars, pieces, lengths, and bends. 8.

How does lap affect the rebar quantit ies?

The lap splicing costs may range from 5 to 15 percent, and it depends on the size of the bar and yield strength of steel used. The waste may also range from less than 1 percent for precut and preformed bars to 10 percent when the bars are cut and bent on the job site. 9.

How is wire mesh taken off? How is it ordered?

Chapter 10 Concrete

11 The takeoff must be broken up into the various sizes required and the number of square feet required of each type.

10.

How is vapor barrier ta ken o ff? How do es the estimator determine the number of rolls requ ired?

It is placed between the gravel and the slab poured on it and is usually included in the concrete portion of the takeoff. The method of determining is to sketch a layout of how it might be placed on the job. 11.

What unit of measure is used when taking off expansion joint fillers?

The measurement of taking off expansion joint fillers is by the linear footage. 12.

What unit of measure is used when taking off concrete finishes?

The unit of measurement of taking off concrete finishes is square foot. 13.

Why shou ld each finish be listed separately on the est imate?

It should be separate, because each are requiring a different finish, such as footing, column, walls, beam, and girder areas are most commonly found in the form calculations. 14.

Where would the estimator look to determine if any curing of concr ete is requ ired on the project?

The type of curing and in what type of weather the concrete will be poured. 15.

Why must the estimator consider how the concrete will be transp orted to the job site?

The reason for that is because the equipment of transportation, the time of transportation, and the labor are different for the costing. 16.

What unit of measure is used when taking off concrete forms? How can reuse of forms affect the estimate? The unit is feet per hour, and the reuse forms can affect the estimate by repetitive pour item, or the form may have to be taken apart and reworked into a new form.

17.

What unit of measure is used for form liners? When might it be more economical to rent instead o f purchasing them?

The liners may be rented, and the cost of the forms required for the concrete work can be reduced substantially. 18.

What two methods of pricing might a subcontractor use for precast concrete?

Most precast concrete is priced by the square foot, linear foot, or in a lump sum. Most suppliers of precast concrete items price them delivered to the job site and installed. 19.

Many suppliers take the respo nsibility of installing the precast units. Why might this be desirable?

It is because of doing your own precast may cost more than subcontracting the work to others. 20.

Under what conditions might it be desirable for a contractor to precast the concrete on the job site?

The precast concrete may be heavy weight or lightweight, and it should be noted that some types of lightweight concrete are not recommended by some

Chapter 10 Concrete

12 consultant engineers.

21.

Using a waste factor of 7 percent, determine the number of cubic yards of concrete needed to po ur the continuous footings shown in Figures 10.36 and 10.37. Assume that the wall is centered over the footing. The footings extend 11 inches beyond the wall on both sides. The lengths of the footings are shown in Figure 10-1.

Chapter 10 Concrete

13

Figure 10-1 Length of Footing (ft) = 22′10″ + 5′ + 15′ + 16′10″ + 37′10″ + 21′10″ = 119.33′ Cross-sectional area (sf) = 2 ′6″ × 1′ = 2.5 sf Volume of concrete (cy) = 2.5 sf × 119.33′ = 298 cf / 27 cf per cy = 11.04 cy Add 7% waste and round off. Volume of concrete (cy) = 11.04 cy × 1.07 = 11.81 cy – Use 12 cy

24.

Using a waste factor of 8 percent, determine the number of cubic yards of concrete needed to po ur the continuous footings shown in Figures 10.38 and 10.39. Assume that the wall is centered over the footing. The footings extend one foot beyond the wall on both sides. The lengths of the footings are shown in Figure 10-2.

Figure 10-2 Length of Footing (ft) = 26′ + 42′ + 26′ + 42′ + 12′ + 15′ = 163′

Chapter 10 Concrete Cross-sectional area (sf) = 3′ × 1′3″ = 3.75 sf Volume of concrete (cy) = 3.75 sf × 163′ = 611 cf / 27 cf per cy = 22.63 cy Add 8% waste and round off. Volume of concrete (cy) = 22.63 cy × 1.08 = 24.44 cy – Use 24.5 cy 25.

Determine the amount of rebar needed for the continuous footings shown in Figures 10.38 and 10.39. Allow for two inches of cover and extend the dowels 15 inches into the wall. Add 10 percent for lap and waste to the continuous bars. From P roblem 24, the length of the footing is 163 ft. The number of short bars and dowels is calculated as follows:

14

Chapter 10 Concrete

15

Number of spaces = 163′ / 1′ = 163 spaces Add 1 to get the nu mber of dowels or short bars – Use 164 dowels and short bars The short bars are 32″ (36″ – 2 × 2″) long. Total bar length (lf) = 164 short bars × 32″ per bar / 12 in per ft = 437.3 lf Total bar weight (pounds) = 437.3 lf × 0.668 pounds per lf = 292 pounds The dowels extend 15 inches into the wall, 13 inches (15″ – 2″) into the footing, and then turn and continue for 12 inches (12″ + 2″ – 2″). Their total length is 40 inches. There are two setsof dowels, one on each side of the wall. Total bar length (lf) = 2 × 164 dowel × 40″ per dowel / 12 in per ft = 1,093 lf Total bar weight (pounds) = 1,093 lf × 1.043 pounds per lf = 1,140 pounds The continuous bars are calcu lated as fo llows: Total length of long bars (lf) = 163′ × 5 bars = 815 lf Total weight of long bars (pounds) = 815 lf × 0.668 pounds per lf = 544 pounds Add for waste and lap. Total weight of long bars (pounds) = 544 pounds × 1.10 = 598 po unds Total weight of rebar (pounds) = 292 pounds + 1,140 pounds + 598 pounds = 2,030 pounds 26.

Determine the number of linear feet of forms needed for the continuous footings shown in Figures 10.38 and 10.39. From P roblem 24, the length of the footings is 163 ′. Length of forms (lf) = 2 × 163′ = 326′

27.

Using a waste factor of 8 percent, determine the number of cubic yards of concrete needed to po ur two of the spread footings in Figure 10.40. Find the volume of concrete as fo llows: Volume of concrete (cy) = 2 × 30″ × 30″ × 10″ /( 12 in per ft)3 = 10.42 cf / 27 cf per cy = 0.39 cy Add 8% waste and round off. Volume of concrete (cy) = 0.39 cy × 1.08 = 0.42 cy – Use 0.5 cy

28.

Determine the amount of rebar needed for two of the spread footings shown in Figure 10.40. Allow for two inches of cover. The dowels extend 34 inches (36″ – 2″) into the columns, 8 inches (10″ – 2″) into the footing, and turn and continue for 9 inches (30″ / 2 – 12″ / 2). Their total length is 51 inches.

Chapter 10 Concrete

16

Total bar length (lf) = 2 × 4 bars × 51″ per bar / 12 in per ft = 34 lf Total bar weight (pounds) = 34 lf × 1.043 pounds per lf = 35.5 pounds The horizontal bars are 26 inches long (30″ – 2 × 2″) and there are eight bars per footing, four in each direction. Total length of horizontal bars (lf) = 2 × 8 × 26″ / 12 in per ft = 34.7 lf Total weight of horizontal bars (pounds) = 34.7 lf × 0.668 pounds per lf = 23.2 pounds Total weight (pounds) = 35.5 pounds + 23.2 pounds = 58.7 pounds 29.

Using a waste factor of 10 percent, determine the number of cubic yards of concrete needed to pour the spread footing in Figure 10.41. Find the volume of concrete as fo llows: Volume of concrete (cy) = 8′ × 17′ × 2.5′ = 340 cf / 27 cf per cy = 12.6 cy Add 10% waste and round off. Volume of concrete (cy) = 12.6 cy × 1.1 = 13.86 cy – Use 14 cy

30.

Determine the amount of rebar needed for the spread footing Figure 10.41. The dowels extend 24 inches into the column. Allow for three inches of cover. The dowels extend 24 inches into the column, 27 inches (30″ – 3″ ) into the footing, and turn and continue for 30 inches (96″/2 – 36″/2). Their total length is 81 inches. Total bar length (lf) = 28 bars × 81″ per bar / 12 in per ft = 189 lf Total bar weight (pounds) = 189 lf × 4.303 pounds per lf = 813 pounds The short horizontal bars are 90 inches long (96″ – 3 × 2″). Spacing = (17′ × 12 in per ft – 3 × 2″) / 9″ = 22 spaces Add 1 to get the nu mber of short horizontal bars – Use 23 bars per side Total length of short horizontal bars (lf) = 2 × 23 × 90″ / 12 in per ft = 345 lf Total weight of short horizontal bars (pounds) = 345 lf × 2.67 pounds per lf = 921 pounds The long horizontal bars are 198 inches long (17′ × 12 in per ft – 3 × 2″) Spacing = (96″ – 3 × 2″) / 9″ = 10 spaces Add 1 to get the nu mber of long horizontal bars – Use 11 bars per side Total length of long horizontal bars (lf) = 2 × 11 × 198″ / 12 in per ft = 363 lf Total weight of long horizontal bars (pounds) = 363 lf × 2.67 pounds per lf = 969 pounds Total weight (pounds) = 813 pounds + 921 pounds + 969 pounds = 2,703 pounds

Chapter 10 Concrete

31.

17

Using a waste factor of 5 percent, determine the number of cubic yards of concrete needed to po ur two of the piers in Figure 10.40. First, find the cross-sectional area: Cross-sectional area (sf) = 1′ × 1′ = 1 sf Volume in piers (cy) = 2 ea × 1 sf × 3′ = 6 cf / 27 cf per cy = 0.22 cy Add 5% waste and round off. Volume in piers (cy) = 0.22 cy × 1.05 = 0.23 cy – Use 0.25 cy

32.

Determine the amount of rebar needed for two of the p iers shown in Figure 10.40. Allow for two inches of cover. The dowels are part of the footing rebar. The ties are 8 inches (12″ – 2 × 2″) square. The tie length is 32 inches (4 × 8″). Number of spaces = 3′ / 3″ = 12 spaces Add 1 to get the nu mber of ties – Use 13 per column, a total of 26 (13 × 2) Total length of ties (lf) = 26 ties × 32″ / 12 in per ft = 69.3 lf Total weight of ties (pounds) = 69.3 lf × 0.376 pounds per lf = 26 pounds

33.

Using a waste factor of 4 percent, determine the number of cubic yards of concrete needed to po ur the column in Figure 10.41. First, find the cross-sectional area: Cross-sectional area (sf) = 3′ × 12′ = 36 sf Volume in piers (cy) = 36 sf × 10′ = 360 cf / 27 cf per cy = 13.33 cy Add 4% waste and round off. Volume in piers (cy) = 13.33 cy × 1.04 = 13.86 cy – Use 14 cy

34.

Determine the amount of rebar needed for the column shown in Figure 10.41. Allow for three inches of cover. The dowels are part of the footing rebar. The ties are 30 inches (36″ – 2 × 3″) by 138 inches (144″ – 2 × 3″). Length = 30″ + 138″ + 30″ + 138″ = 336″ / 12 in per ft = 28′ Number of spaces = 120″ / 6″ = 20 spaces Add 1 to get the nu mber of ties – Use 21 Total length of ties (lf) = 21 ties × 28′ = 588 lf

Chapter 10 Concrete

18

Total weight of ties (pounds) = 588 lf × 1.502 pounds per lf = 883 pounds The vertical bars are 117 inches long (120″ – 3″). Total bar length (lf) = 28 bars × 117″ per bar / 12 in per ft = 273 lf Total bar weight (pounds) = 273 lf × 4.303 pounds per lf = 1,175 pounds Total weight (pounds) = 883 pounds + 1,175 pounds = 2,058 pounds 35.

A building requ ires four drilled p iers with a shaft diameter of 24 inches, a bell diameter of 48 inches, and a bell angle of 50 degrees. The height from the bottom of the bell to the top of the pier is 10 feet. Using a waste factor of 12 percent, determine the number of cubic yards of concrete needed to pour the drilled p iers. Find the height of the bell using Equation 10-2 or Figure 10-6 from the textbook: B − S′ H = ta n( A) ×----2

4′ − 2′ H = ta n(50°) × -------------2 H = 1.192′

Find the bell volume using Equation 10-4 or Figure 10-6: Volume of bell =

Volume of bell =

πH 12

(S 2 + B 2 + SB)

π ×1.192′ 12

(2′2 + 4′2 + 2′ × 4′)

Volume of bell = 8.74 cf Length of shaft (ft) = 10′ – 1.192′ = 8.808′ Shaft volume (cf) =

π S2 4

× (Length of Shaft)

π 2 '2 Shaft volume (cf) = × (8.808 ' ) = 27.67 cf 4 Volume of concrete in piers (cy) = Count × (Volume in bell + Volume in shaft) Volume of concrete in piers (cy) = 4 × (8.74 cf + 27.67 cf) = 146 cf / 27 cf per cy = 5.4 cy Add 12% waste and round off. Volume of concrete (cy) = 5.4 cy × 1.12 = 6.0 cy – Use 6 cy

Chapter 10 Concrete

36.

19

A building requ ires 10 dr illed piers with a shaft dia meter of 30 inches, a bell d iameter of 54 inches, and a bell angle of 60 degrees. The height from the bottom of the bell to the top of the pier is 15 feet. Using a waste factor of 15 percent, determine the number of cubic yards of concrete needed to pour the drilled p iers. Find the height of the bell using Equation 10-2 or Figure 10-6 from the textbook: B − S′ H = ta n( A) × 2 4.5′ − 2.5′ H = ta n(60°) × 2 H = 1.732′ Find the bell volume using Equation 10-4 or Figure 10-6: Volume of bell =

Volume of bell =

πH 12

(S 2 + B 2 + SB)

π ×1.732′ 12

(2.5′2 + 4.5′2 + 2.5′ × 4.5′)

Volume of bell = 17.12 cf Length of shaft (ft) = 15′ – 1.732′ = 13.268′ π S2 Shaft volume (cf) = 4

Shaft volume (cf) =

× (Length of Shaft)

π 2.5 '2 × (13.268 ' ) = 65.13 cf 4

Volume of concrete in piers (cy) = Count × (Volume in bell + Volume in shaft) Volume of concrete in piers (cy) = 10 × (17.12 cf + 65.13 cf) = 822 cf / 27 cf per cy = 30.4 cy Add 15% waste and round off. Volume of concrete (cy) = 30.4 cy × 1.15 = 35.0 cy – Use 35 cy 37.

Using a waste factor of 5 percent, determine the number of cubic yards of concrete needed to po ur the foundation walls shown in Figures 10.36 and 10.37. The wall lengths are shown in Figure 10-3.

Chapter 10 Concrete

20

Figure 10-3 Length of wall (lf) = 21′ + 4′4 ″ + 15′8″ + 18′8″ + 36′ + 23′8″ = 119.33′ Foundation wall concrete (cy) = 119.33′ × 10′ × 8″ / 12 in per ft = 796 cf / 27 cf per cy = 29.48 cy Add 5% waste and round off. Foundation wall concrete (cy) = 29.48 cy × 1.05 = 30.95 cy – Use 31 cy 38.

Determine the amount of rebar needed for the foundation walls shown in Figures 10.36 and10 .37 . Allow for two inches of cover. Add 10 percent for lap and waste to the continuous bars. From P roblem 37, the length of the wall is 119.33′. The vertical bars are calculated as fo llows: Number of spaces = 119.33′ / 1′ = 120 spaces Add 1 to get the nu mber of vertical bars – Use 121 bars The vertical bars are 9′10″ long (10 ′ – 2″). Total length of vertical bars (lf) = 121 bars × 9 ′10″ per bar = 1,190 lf Total weight of vertical bars (pounds) = 1,190 lf × 0.668 pounds per lf = 795 pounds The horizontal bars are calculated as fo llows: Number of spaces = 10′ / 1′ = 10 spaces Add 1 to get the nu mber of horizontal bars that are needed to meet the spacing requirements plu s an additional top and bottom bar (10 + 1 + 1 + 1) – Use 13 bars Total length of horizontal bars (lf) = 119.33′ × 13 bars = 1,551 lf Total weight of horizontal bars (pounds) = 1,551 lf × 0.668 pounds per lf = 1,036 pounds Add for waste and lap.

Chapter 10 Concrete

21

Total weight of horizontal bars (pounds) = 1,036 pounds × 1.10 = 1,140 pounds Total weight of rebar (pounds) = 795 pounds + 1,140 pounds = 1,935 pounds 39.

Determine the cost to install the rebar for the foundation walls in 10.36 and 10.37 using a productivity of 10.75 labo r hours per ton and an average labor rate of $25.62 per labo r hour. From P roblem 38, the weight of the rebar is 1,935 pounds. Labo r hour = 10.75 lhr per ton × 1,935 pounds / 2,000 pounds per ton = 10.4 lhr Labo r cost = 10.4 lhr × $25.62/lhr = $266.45

40.

Determine the square foot contact area (SFCA) of the forms needed for the foundation walls shown in Figures 10.36 and 10.37. From P roblem 37, the length of the foundation wall is 119.33 ′. Area of forms (SFCA) = 2 sides × 119.33′ × 10′ = 2,387 SFCA

41.

Determine the cost to form the foundation walls in Figures 10.36 and 10.37 using a produ ctivity of 0.125 labo r hours per SFCA and an average labo r rate of $25.62 per labor hour. From P roblem 40, the area of the forms is 2,387 SFCA. Labo r hour = 0.125 lhr per SFCA × 2,387 SFCA = 298 lhr Labo r cost = 298 lhr × $25.62/lhr = $7,634.76

42.

Using a waste factor of 6 percent, determine the number of cubic yards of concrete needed to po ur the foundation walls shown in Figures 10.38 and 10.39. The wall lengths are shown in Figure 10-4.

Figure 10-4

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73

Length of wall (lf) = 28′ + 40′ + 28′ + 40′ + 12′ + 17′ = 165′ Foundation wall concrete (cy) = 165′ × 12′ × 1′ = 1,980 cf / 27 cf per cy = 73.33 cy Add 6% waste and round off. Foundation wall concrete (cy) = 73.33 cy × 1.06 = 77.73 cy – Use 77.75 cy 43.

Determine the amount of rebar needed for the foundation walls shown in Figures 10.38 and10 .39 . Allow for two inches of cover. Add 10 percent for lap and waste to the continuous bars. From P roblem 42, the length of the wall is 165 ′. The vertical bars are calcu lated as follows: Number of spaces = 165′ / 1′ = 165 spaces Add 1 to get the nu mber of vertical bars – Use 166 bars per side The vertical bars are 11′10″ long (12 ′ – 2″). Total length of vertical bars (lf) = 2 sides × 166 bars per side × 11′10″ per bar = 3,929 lf Total weight of vertical bars (pounds) = 3,929 lf × 0.668 pounds per lf = 2,625 pounds The horizontal bars are calculated as fo llows: Number of spaces = 12′ / 1′ = 12 spaces Add 1 to get the nu mber of horizontal bars that are needed to meet the spacing requ irements for one side – Use 13 bars per side Total length of horizontal bars (lf) = 165′ × 13 bars per side × 2 sides = 4,290 lf Total weight of horizontal bars (pounds) = 4,290 lf × 0.668 pounds per lf = 2,866 pounds Add for waste and lap. Total weight of horizontal bars (pounds) = 2,866 pounds × 1.10 = 3,153 pounds Total weight of rebar (pounds) = 2,625 pounds + 3,153 pounds = 5,778 pounds

44.

Determine the cost to install the rebar for the foundation walls in Figures 10.38 and 10.39 using a productivity of 10.3 labor hours per ton and an average labor rate of $31.26 per la bor hour. From P roblem 43, the weight of the rebar is 5,778 pounds. Labo r hour = 10.3 lhr per ton × 5,778 pounds / 2,000 pounds per ton = 29.8 lhr Labo r cost = 29.8 lhr × $31.26/lhr = $931.55

45.

Determine the square foot contact area (SFCA) of the forms needed for the foundation walls shown in Figures 10.38 and 10.39. From P roblem 42, the length of the foundation wall is 165 ′. Area of forms (SFCA) = 2 sides × 165′ × 12′ = 3,960 SFCA

Chapter 10 Concrete

46.

74

Determine the cost to form the foundation walls in Figures 10.38 and 10.39 using a produ ctivity of 0.11 labo r hours per SFCA and an average labo r rate of $32.17 per labo r hour. From P roblem 9, t he area of the forms is 3,960 SFCA. Labo r hour = 0.11 lhr per SFCA × 3,960 SFCA = 436 lhr Labo r cost = 436 lhr × $32.17/lhr = $14,026.12

47.

A building requ ires 20 feet of the grade beam whose cross section is shown in Figure 10.42. Using a waste factor of 8 percent, determine the number of cubic yards of concrete needed to pour the grade beam. First, find the cross-sectional area: Cross-sectional area (sf) = 12″ × 14″ /(12 in per ft)2 = 1.167 sf Volume in grade beam (cy) = 20′ × 1.167 sf = 23.34 cf / 27 cf per cy = 0.86 cy Add 8% waste and round off. Volume in piers (cy) = 0.86 cy × 1.08 = 0.93 cy – Use 1 cy

48.

Determine the amount of rebar needed for the grade beam in Problem 47. Allow for two inches of cover and ignore waste. The ties are 8 inches (12″ – 2 × 2″) by 10 inches (14″ – 2 × 2″). The length of the ties is 36 inches (8″ + 10″ + 8″ + 10″). The number of ties is calcu lated as fo llows: Number of spaces = 20′ × 12 in per ft / 8″ = 30 spaces Add 1 to get the nu mber of ties – Use 31 ties Total length of ties (lf) = 31 ties × 3′ = 93 lf Total weight of ties (pounds) = 93 lf × 0.376 pounds per lf = 35 pounds The horizontal bars are 20 feet long calculated as fo llows: Total length of horizontal bars (lf) = 4 × 20′ = 80 lf Total weight of horizontal bars (pounds) = 80 lf × 1.043 pounds per lf = 83 pounds Total weight of rebar (pounds) = 35 pounds + 83 pounds = 118 pounds

49.

A building requ ires 35 feet of the grade beam whose cross section is shown in Figure 10.43. Using a waste factor of 10 percent, determine the number of cubic yards of concrete needed to pour the grade beam. First, find the cross-sectional area: Cross-sectional area (sf) = 12″ × 16″ = 1.333 sf Volume in grade beam (cy) = 35′ × 1.333 sf = 46.66 cf / 27 cf per cy = 1.73 cy

Chapter 10 Concrete

75

Add 10% waste and round off. Volume in piers (cy) = 1.73 cy × 1.10 = 1.90 cy – Use 2 cy 50.

Determine the amount of rebar needed for the grade beam in Problem 49. Allow for three inches of cover. Add 10 percent for lap and waste to the continuous bars. The ties are 6 inches (12″ – 2 × 3″) by 10 inches (16″ – 2 × 3″). The length of the ties is 32 inches (6″ + 10″ + 6″ + 10″). The number of ties is calcu lated as fo llows: Number of spaces = 35′ × 12 in per ft / 6″ = 70 spaces Add 1 to get the nu mber of ties – Use 71 ties Total length of ties (lf) = 71 ties × 32″ / 12 in per ft = 189 lf Total weight of ties (pounds) = 189 lf × 0.376 pounds per lf = 71 pounds The continuous bars are calcu lated as fo llows: Total length of continuous bars (lf) = 4 × 35′ = 140 lf Total weight of continuous bars (pounds) = 140 lf × 0.668 pounds per lf = 93.5 pounds Add for waste and lap. Total weight of horizontal bars (pounds) = 93.5 pounds × 1.10 = 103 pounds Total weight of rebar (pounds) = 71 pounds + 103 pounds = 174 pounds

51.

A building requ ires a 100-foot-long by 83-foot-wide by 5-inch-thick concrete sla b. Using a waste factor of 10 percent, determine the number of cubic ya rds of concrete needed to po ur the slab. Find the quantity of concrete as follows: Quantity of concrete (cy) = 100′ × 83′ × 5″ / 12 in per ft = 3,458 cf / 27 cf per cy = 128 cy Add 10% waste. Quantity of concrete (cy) = 128 cy × 1.10 = 141 cy

52.

Determine the number of rolls of wire mesh needed for the slab in P roblem 51. There are 750 square feet of wire mesh per roll. Add 12 percent for lap and waste. Find the area of the slab as follows: Area of slab (sf) = 100′ × 83′ = 8,300 sf Rolls of mesh required = 8,300 sf / 750 sf per roll = 11.1 rolls Add for waste and lap. Rolls of mesh required = 11.1 rolls × 1.12 = 12.43 rolls – Order 13 rolls

Chapter 10 Concrete 53.

76

Determine the amount of rebar needed for the slab in P roblem 51. The slab is reinforced with #4 rebar at 12 inches on center both ways. Assume two inches of cover and add 10 percent for lap and waste. The long bars are 99′8″ (100 ′ – 2 × 2) long. Determine the number of long bars: Long bar spaces = 83′ / 1′ = 83 spaces. Add 1 to get the nu mber of long bars – Use 84 bars Total length of long bars (lf) = 84 bars × 99′8″ per bar = 8,372 lf Total weight of long bars (pounds) = 8,372 lf × 0.668 pounds per lf = 5,592 pounds The short bars are 82′8″ (83′ – 2 × 2) long. Determine the number of short bars: Short bar spaces = 100′ / 1′ = 100 spaces. Add 1 to get the nu mber of short bars – Use 101 bars Total length of short bars (lf) = 101 bars × 82′8″ per bar = 8,349 lf Total weight of short bars (pounds) = 8,349 lf × 0.668 pounds per lf = 5,577 pounds Total weight of rebar (pounds) = 5,592 pounds + 5,577 pounds = 11,169pounds Add for waste and lap. Total weight of rebar (pounds) = 11,169 pounds × 1.10 = 12,286 pounds

54.

Determine the cost to finish the slab in P roblem 51 using a productivity of 0.011 labor hours per square foot and an average labor rate of $29.92 per labo r hour. From P roblem 52, the area of the slab is 8,300 sf. Labor hour = 0.011 lhr per sf × 8,300 sf = 91.3 lhr Labor cost = 91.3 lhr × $29.92/lhr = $2,731.70

55.

A vapor barrier is to be placed below the slab in Problem 51. Determine the number of 10-footby 100-foot rolls of vapor barrier ne eded. The vapor barrier needs to be lapped one foot. Hint: Subtract the lap from the width of the roll. Run the rolls the 100′ direction; therefore, one roll will cover the length of the sla b. Number of rolls wide = 83′ / 9 ′ = 9.2 – Use 10 rolls 100′ long

56.

Using a waste factor of 12 percent, determine the amo unt of concrete needed to pour the slab shown in Figure 10.44. The slab is 4 inches thick. Find the area of the slab as follows: Area of slab (sf) = 25′ × 30′ + 15′ × 20′ = 1,050 sf Quantity of concrete (cy) = 1,050 sf × 4″ / 12 in per ft = 350 cf / 27 cf per cy = 13.0 cy

Chapter 10 Concrete

77

Add 12% waste. Quantity of concrete (cy) = 13.0 cy × 1.12 = 14.6 cy – Use 14.75 cy 57.

The slab in Problem 56 is reinforced with wire mesh. Determine the number of rolls of wire mesh needed for the sla b. There are 750 square feet of wire mesh per roll. Add 11 percent for lap and waste. From P roblem 56, the area of the slab is 1,050 sf. Rolls of mesh required = 1,050 sf / 750 sf per roll = 1.4 rolls Add for waste and lap. Rolls of mesh required = 1.4 rolls × 1.11 = 1.6 rolls – Order 2 rolls

58.

Determine the amount of rebar needed for the slab in P roblem 56. The slab is reinforced with #4 rebar at 18 inches on center both ways. Assume two inches of cover and add 10 percent for lap and waste. Start with the 25′ by 30 ′ area. The long bars are 29 ′8″ (30′ – 2 × 2) long. Determine the number of long bars: Long bar spaces = 25′ / 1.5′ = 17 spaces. Add 1 to get the nu mber of long bars – Use 18 bars Total length of long bars (lf) = 18 bars × 29′8″ per bar = 534 lf The short bars are 24′8″ (25′ – 2 × 2) long. Determine the number of short bars: Short bar spaces = 30′ / 1.5′ = 20 spaces. Add 1 to get the nu mber of short bars – Use 21 bars Total length of short bars (lf) = 21 bars × 24′8″ per bar = 518 lf Calculate the rebar for the 15′ by 20 ′ area. The long bars are 19′8″ (20′ – 2 × 2) lo ng. Determine the number of long bars: Long bar spaces = 15′ / 1.5′ = 10 spaces. Add 1 to get the nu mber of long bars – Use 11 bars Total length of long bars (lf) = 11 bars × 19 ′8″ per bar = 216 lf The short bars are 14′8″ (15′ – 2 × 2) long. Determine the number of short bars: Short bar spaces = 20′ / 1.5′ = 14 spaces. Add 1 to get the nu mber of short bars – Use 15 bars Total length of short bars (lf) = 15 bars × 14′8″ per bar = 220 lf

Chapter 10 Concrete

78

Total length of rebar (lf) = 534 lf + 518 lf + 216 lf + 220 lf = 1,488 lf Total weight of rebar (pounds) = 1,488 lf × 0.668 pounds per lf = 994 pounds Add for waste and lap. Total weight of bars (pounds) = 994 pounds × 1.10 = 1,093 pounds 59.

Determine the cost to finish the slab in P roblem 56 using a productivity of 0.013 labor hours per square foot and an average labor rate of $31.17 per labo r hour. From P roblem 56, the area of the slab is 1,050 sf. Labor hour = 0.013 lhr per sf × 1,050 sf = 13.65 lhr Labor cost = 13.65 lhr × $31.17/lhr = $425.47

60.

A vapor barrier is to be placed below the slab in P roblem 56. Determine the number of 10-foot by 50-foot rolls of vapor barrier ne eded. The vapor barrier needs to be lapped one foot. Run the rolls the 30′ direction. The 25′ by 30′ area will requ ire three 10′ by 30′ p ieces and the 15 ′ by 20′ area will require two 10′ by 20′ pieces. The length of the plastic is calculated as fo llows: Length = 3 × 30′ + 2 × 20′ = 130′ Number of ro lls = 130′ / 50′ = 2.6 – Use 3 ro lls

61.

Determine the amount of concrete, reinforcing, forms, and acce ssories required for the Commercial Building Pro ject in Append ix C.

Chapter 10 Concrete

62.

79

Using the drawings of Billy’s Convenience Store found in Appendix F, determine the quantity of concrete and reinforcing required for the spread footings.

Chapter 10 Concrete

80

63.

Using the drawings of Billy’s Convenience Store found in Appendix F, determine the quantity of concrete, reinforcing, and formwork in the slab on grade.

64.

Determine the amount of concrete, reinforcing, forms, and accessories required the Real Estate Office in Appendix G. The fo llowing quantit ies can be extracted from the model.

Chapter 10 Concrete

81

The foundations are found under walls. Woo d framed walls must be filtered out and the walls sorted by width. The 8-inch walls repr esent the foundation wall and the 2-foot walls repr esent the footings. The footings must be ta ken off as fo llows: Length of perimeter footing = 73′4′′ + 47′4′′ + 73 ′4′′ + 47′4′′ = 241 ′4′′ Volume of perimeter footing = 241 ′4′′ × 1′ × 2′ = 483 ft3 = 17.88 yd3 Length of interior footings = 70 ′8′′ + 70′8′′ = 141′4′′ Volume of interior footings = 141′4′′ × 8′′ × 2′ = 188 ft3 = 6.98 yd3 Concrete volume = 17.88 yd 3 + 6.98 yd3 = 24.86 yd3 You need 483 feet (241′4′′ × 2) of 12-inch-high forms and 283 feet (141′4′′ × 2) of 8-inch-high forms. Horizontal #4 rebar = 3 × (241′4′′ + 141 ′4′′) = 1,148′ × 1.1 = 1,263′ Dowels length (assume 2′′ cover) = 10′′ + 10′′ + 18′′ = 38′′ Number of #4 dowels = 241′4′′ / 16 ′′ = 181 + 1 = 182 #4 rebar for dowels = 182 × 38′′ = 576′ Anchor bolts = 141′4′′ / 16 ′′ =106 + 1 = 107 The foundation is ta ken off as follows: Length of foundation = 72′ + 48′8′′ + 72′ + 48′8 ′′ = 241 ′4′′ Concrete volume = 241 ′4′′ × 8′′ × 1′8′′ = 268 ft3 = 9.93 cy3 Forms = 241′4′′ × 1 ′8′′ × 2 sides = 805 sfca Horizontal #4 rebar = 241′4 ′′ × 3 = 724 ′ × 1.1 = 796′ Slab dowels = 241 ′4′′ / 16 ′′ = 181 + 1 = 182

Chapter 10 Concrete #4 rebar for slab dowels = 182 × 32′′ = 486′ Anchor bolts = 241′4′′ / 16 ′′ = 181 + 1 = 182 #4 rebar = 1,263′ + 576′ + 796′ + 486′ = 3,121′ × 0.668 pounds per lf = 2,085 pounds Concrete slab = 72′ × 50′ × 4′′ = 1,200 ft3 = 44.44 yd3

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Chapter 10 Concrete

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