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INTRODUCTION to VOLUME MEASUREMENTS Volume INTRODUCTION TO VOLUME MEASUREMENTS Volume measurements are needed for three different categories of pay items: • Earthwork --items such as borrow excavation, and subsoil excavation • Concrete -- the various classes of concrete used in bridges and other structures; and • Truck Measurements -- material delivered by trucks and measured by volume. Each of these categories is handled differently in the field. UNITS OF MEASUREMENTS The pay item unit of measurement for volume usually is the cubic yard. Keep this relationship in mind: One cubic foot = 1728 cubic inches; One cubic yard = 27 cubic feet; and One cubic foot = 7.48 gallons. METHODS FOR COMPUTING VOLUMES Volumes of earthwork usually are computed from cross sections taken before and after construction. The volume of concrete in structures is most often computed by formulas for the geometric shapes involved. And, when items are paid for by truck quantity, use the manufacturer’s certification or permanent decal showing the truck’s capacity and then simply count the number of loads delivered. CROSS SECTIONS In Chapter Two we talked about cross section notes and how they shall be recorded in field notebooks. In this chapter we want to learn how to use cross sections for measuring volumes of earthwork. Figure 4-1 below, illustrates typical earthwork cross sections. The example shows all fill sections probably roadway embankment. Sometimes the sections may be all cut, such as borrow areas or ditches. Quite often, however, both cut and fill areas are shown on the same section. A common method of determining volumes from cross sections is that of average end areas. It assumes that the volume between successive cross sections is the average of their end areas multiplied by the distance between them. This is expressed in the formula: V (Ft.3) = [A1 (FT2) + A2 (Ft2)] X L (Ft) 2 In which V is the volume in cubic feet, A1 and A2 are the end areas in square feet of successive sections and L is the length in feet between the sections. (The end areas will be divided by 2 to come up with the average end area). This formula is exact when A1 equals A2 in both area and configuration, but is only approximate when the cross sections have different areas. It is generally accepted as close enough for computing earthwork volumes. Later in this chapter we will look at a more accurate approach called the prismoidal formula. It is used for computing volumes of concrete, a much more expensive item. Using the examples of cross sections shown in Figure 4-1, let's see if we can compute the volume of earthwork by applying the above formula. The end areas shown on the sections were computed by a method described in the Volume Measurements section. We must compute the volume between each pair of cross sections and add the individual volumes to obtain the total volume of earthwork between station (9 + 75) and station (12 + 20). See Figure 4-2 below. FIGURE 4-1 Begin Sta. 9+75 Area = 0 FIGURE 4-1 End Sta. 12+20 Area 7 = 0 To make the answer come out in cubic yards, we must divide by 27. (Remember that the end areas were divided by 2 originally). So the formula will now read: (Below are two equations of how to achieve this). Volume (V) = (A1 + A2 ) L or V = (A1 + A2) L 2 X 27 54 Figure 4.2 4 VOLUME MEASUREMENTS Volume measurements are needed for which three different categories of pay items? A. Slope Pavement, Pile Driving and Fencing. B. Concrete, Truck measurement and Earthwork. C. Concrete, Pile Driving and Earthwork. D. Fencing, Truck Measurement and Slope Pavement. E. None of the above. 4 VOLUME MEASUREMENTS How many Cubic Feet are in 350 Gallons? Note: 1 CF = 7.48 Gallons. Round your answer to the tenth of a cubic foot. A. 46.8 cubic feet B. 95.4 cubic feet C. 25.6 cubic feet D. 12.9 cubic feet E. None of the above. 4 VOLUME MEASUREMENTS Based on the areas determined for the Stations shown below, what is the total volume of earthwork between Station 71+25 and 72+75? (solve to the nearest CY) A. 1,406 cubic yards B. 2,828 cubic yards C. 1,916 cubic yards D. 1,667 cubic yards Station Area Volume 71+25 308.0 Ft2 71+48 287.0 Ft2 71+81 291.5 Ft2 72+23 304.0 Ft2 72+75 315.3 Ft2 Total _______ 4 VOLUME MEASUREMENTS The table shows the end areas determined for the indicated cross sections. Compute the total volume of earthwork between stations 408+00 and 410+10. (Answer to the nearest Cubic Yard) A. 2,925 Cubic Yards B. 2,332 Cubic Yards C. 1,927 Cubic Yards D. 1,513 Cubic Yards Station Area (S.F.) Volume (CY) 408+00 244 408+62 263 409+25 212 409+81 259 410+10 303 Total = CURVATURE CORRECTION When volumes are computed along a curved base survey line, some error will be introduced unless the center of gravity of the cross sectional area lies approximately along the base line. This situation is illustrated in the example below: G = Center of Gravity R = Radius of base line Note: Multiline handles the curvature correction calculations. VOLUME FORMULAS Cross sections do not work well for computing the volumes of some pay items such as reinforced concrete and excavation. For these items it is much better to measure the dimensions of the construction and use conventional formulas to compute geometric shapes. The Construction Mathematics training course provides a good background in the use of formulas for calculating volumes. Let's take a quick look at some of the formulas we will be using. The simplest geometric shape of course, is the rectangular solid with opposite sides parallel. In both cases, V =LWH (volume equals length times width times height) Example: Find the volume of the concrete block seen here. First we must make sure all measurements have the same components. The width is 2 Ft and 3 inches. The inches will have to be converted into feet. 3 ÷12 = 0.25 Ft. The Width is 2' 3" = 2.25 Ft. V = LWH V = 8 X 2.25 X 5 V = 90 Cubic Foot. If the answer is in Cubic Yards, then we must divide by 27 Ft/CY, and the Volume will be: V = 90 ÷ 27 = 3.33 CY = 3 CY (If answer is to be to the nearest CY) When you studied areas you learned that a triangle is really half of a rectangle or parallelogram. The same concept applies to volumes. Example: Calculate the volume of the triangle seen below to the nearest cubic foot. Now how about trapezoidal solids? Do these formulas look familiar? When you stop to think about it, we are really computing an end area and then multiplying by a third dimension to find the volume. Example: Calculate the volume of the trapezoid seen below to the nearest cubic foot. First we must make sure all measurements have the same components (inches need to be converted to feet). a = 4' 4" and 4" ÷ 12 = 0.33; a= 4.33 Ft. b = 9' 4" and 4" ÷ 12 = 0.33; b = 9.33 Ft. H = 7' 2" and 2" ÷ 12 = 0.17; H = 7.17 Ft. W= 3' 9" and 9" ÷ 12 = 0.75’ W= 3.75 Ft V = [(a + b) ÷ 2] HW = [(4.33 + 9.33) ÷ 2] X 7.17 X 3.75 = V = 183.64 = 184 CF Let’s try it with a cylindrical solid and a cone: Why do we divide by 3 instead of 2 in the cone formula? The tip of the cone obviously has a zero area, and if we average the end areas the formula should be: V = A x H 2 But this is wrong. Remember, when we talked about cross sections we said that the average-end-area method was close enough for earthwork volumes but not accurate enough for items like concrete. So we need to use the cone formula or "prismoidal formula." The prismoidal formula is: Where: A1 and A2 are the two end areas and Am (m = mean) is the area at a point halfway between the two ends. Let us solve some examples together: Example 1 Calculate the volume of the cone using the cone formula. Calculate the answer to the nearest cubic foot. V = πR2 H 3 Note: the answer is to the nearest cubic foot H = 30 in We will first convert the inches to feet. 30 ÷ 12 = 2.5 Ft. 10 ÷ 12 = 0.83 Ft. V = 3.1416 X (0.83)2 X 2.5 3 R=10 in V = 1.8 = 2 CF Example 2 How about an example with a cylinder? Calculate the volume of the Cylinder to the nearest D= 3' Cubic Yard. 2 V = πR X H H= 8' R = (D ÷ 2) = 3 ÷ 2 = 1.5' V = 3.1416 X (1.5)2 X 8 V = 56.5 CF ÷ 27 CY/CF = 2.1 = 2 CY R= 6' All right, so when should the prismoidal formula be used? Let's look at a definition and see what it means to us: Prismoidal shape: A solid with parallel end areas and sloping sides, but whose mid-area (mean area) is not necessarily the average of the two end areas. How and when we apply the prismoidal formula to our work is straightforward. Since we usually calculate from vertical plane to vertical plane (i.e., cross sectional areas) the "end areas" in the above definition are self-explanatory.
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