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ESTIMATING BULK MATERIAL REQUIREMENTS All material estimates, including those for bulk materials, are used as a basis for material procurement and as a check to determine if sufficient materials are available to constructor complete a project. In general, the term bulk material refers to concrete, bituminous paving materials, and mineral products, such as sand, gravel, or rock. A few examples of estimating these materials are described below.Concrete Estimating the amount of concrete required for a project consists of determining the volume (in cubic yards or, in many locations outside the United States, in cubic meters) of the spaces that will be occupied by the concrete. As an example, lets look at figure 5-2. This figure shows the foundation plan and a typical foundation detail for a small 20 foot by 32 foot building. As shown in the detail, the foundation is continuous and the floor is a 4-inch-thick concrete slab. Our task is to determine the amount of concrete that will be required for the foundation and slab. Since in any concrete job a certain amount of concrete will be unavoidably lost during placement, we will include a 10-percent waste factor. You can find this waste factor listed in the P-405. Although we could proceed in various ways to estimate the amount of concrete that is required, an easy method is tabulated as follows:Foundation wall: (32.00 ft x .67 ft x 2 fi) x 2 = 85.76 cu ft (18.67 ft x .67 ft x 2 ft) x 2 = 50.04 cu ft Footing: (33.33 ft x 2 ft x 1 ft) x 2 = 133.032 cu ft (17.33 ft x 2 ft x 1 ft) x 2 = 69.32 cu ft Slab: From the above tabulation, you can see that separate estimates were prepared for the foundation wall, footing, and slab. You can also see that both the foundation wall and the footing were further subdivided based on the length and width of the building; for example, the foundation wall consists of two walls measuring 32 feet long and two measuring 18 feet 8 inches long (allowing for overlap at the foundation corners). Then, the separate estimates were added together, the waste factor was applied, and the final cubic feet of concrete was converted to cubic yards. Now lets see how much compacted fill will be required for this job. For this example, we will assume that the project specifications call for sand to be used as the fill material.Figure 5-2 shows that the fill material is to be 8 inches thick after compaction. Therefore, the volume of the compacted fill is 383.65 cubic feet. However, from your knowledge of soils, you know that compacted sand occupies less volume than loose sand. Since the sand for this project will be delivered to the jobsite in a loose condition, a compaction factor must be applied. NAVFAC P-405 lists compaction factors for various materials. For sand, the compaction factor listed is 1.17. So, the total amount of sand required for this project is 383.65 x 1.17=448.87 cubic feet, or 16.6 cubic yards. Again, however, we know that a certain amount of sand will be wasted. So, lets increase the total by 10 percent. Now we need approximately 18 cubic yards of sand for the job.Bituminous Paving Although not always, most bituminous paving projects that are accomplished by the Seabees use hot-mixed bituminous concrete that is purchased from a central plant. In this case, the job of estimating consists of determining the compacted volume, in cubic feet, of the pavement. This volume is then multiplied by the unit weight of the mix, in pounds per cubic foot (pcf), and the final result is converted to tons of mix required. An equation for determining the required tons of mix can be expressed as follows:Tons of mix = Where: L= Length of paved area in feet W= Width of paved area in feet T= Compacted thickness of the pavement in inches UW = Unit weight of the mix in pounds per cubic footTo illustrate the use of this formula, lets assume that we are estimating a 2-inch-thick hot-mix bituminous pavement on a 150-foot by 600-foot parking lot. The unit weight (which usually ranges from 140 to 160 pcf) should be determined from laboratory testing when possible; however, when the unit weight is not known, an estimated weight of 160 pcf maybe used. In this example, lets assume a unit weight of 147 pcf. From this, we can estimate the tons of plant mix required by substitution into the above formula as follows:Then if we include a loss factor of, lets say 5 percent, we will need 1,158 tons of plant mix for this parking lot. Now, lets assume that this same parking lot is to be laid over a compacted-soil subbase. In this case, we will need a prime coat also. The prime coat is a low-viscosity liquid bitumen that is sprayed on the subbase. It provides a seal and promotes adhesion between the subbase and the pavement. To estimate the amount of bitumen required for the prime coat, multiply the area to be treated by the rate of application The estimate should include enough bitumen for an additional width of 1 foot on each side of the pavement. A formula for estimating the number of gallons of primer needed is as follows:Gallons = Where: L= Length of paved area in feet W= Width of paved area in feet AR = Application rate of bitumens in gallons per square yardSo, if the project specs for the parking lot we have been discussing call for an application rate of 0.3 gallons of prime coat per square yard of surface and if we assume a 5 percent loss factor, how many gallons of primer will be required? You can try this one on your own. |
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