2Earthmoving Materialsand Operations

2-1 INTRODUCTION TO EARTHMOVING

TheEarthmoving Process

Earthmoving (Figure 2-1) is the process of moving soil or rock from one loca-tion to another and processing it so that it meets construction requirementsoflocation, elevation, density, moisture content, and so on. Activities involvedin this process include excavating, loading, hauling, placing (dumping andspreading), compacting, grading, and finishing. The construction proceduresand equipment involved in earthmoving are described in Chapters 3 to 6. Effi-cientmanagement of the earthmoving process requires accurate estimating ofworkquantities and job conditions, proper selection of equipment, and com-petent job management.

Equipment Selection

Thechoice of equipment to be used on a construction project has a majorinfluenceon the efficiency and profitability of the construction operation.Althoughthere are a number of factors that should be considered in selectingequipmentfor a project, the most important criterion is the ability of theequipmentto perform the required work. Among those items of equipmentcapableof performing the job, the principal criterion for selection should bemaximizingthe profit or return on the investment produced by the equip-ment.Usually, but not always, profit is maximized when the lowest cost perunitofproduction is achieved. (Chapter 17 provides a discussion of construc-tion economics.) Other factors that should be considered when selectingequipmentfor a project include possible future use of the equipment, its

21

22 Chapter 2

r

FIGURE 2-1 Earthmoving: an elevating scraper self-loads. (Courtesy of ClarkEquipment Company)

availability, the availability of parts and service, and the effect of equipmentdowntime on other construction equipment and operations.

After the equipment has been selected for a project, a plan must bedeveloped for efficient utilization of the equipment. The final phase of theprocess is, of course, competent job management to assure compliance withthe operating plan and to make adjustments for unexpected conditions.

Production of Earthmoving Equipment

The basic relationship for estimating the production of all earthmoving equip-ment is:

Production =Volume per cycle x Cycles per hour (2-1)

The term "volume per cycle" should represent the average volume of materialmoved per equipment cycle. Thus the nominal capacity of the excavator orhaul unit must be modified by an appropriate fill factor based on the type ofmaterial and equipment involved. The term "cycles per hour" must includeany appropriate efficiency factors, so that it represents the number of cyclesactually achieved (or expected to be achieved) per hour. In addition to thisbasic production relationship, specific procedures for estimating the produc-tion of major types of earthmoving equipment are presented in the chapterswhich follow.

2-2 Earthmoving Materials 23

The cost per unit of production may be calculated as follows:

. . Equipment cost per hourCost per umt of productIOn = E

. d. h

(2-2)qUlpment pro uctIOn per our

Methods for determining the hourly cost of equipment operations areexplainedin Chapter 17.

There are two principal approaches to estimating job efficiency in deter-miningthe number of cycles per hour to be used in Equation 2-1. One methodis to use the number of effective working minutes per hour to calculate thenumber of cycles achieved per hour. This is equivalent to using an efficiencyfactor equal to the number of working minutes per hour divided by 60. Theother approach is to multiply the number of theoretical cycles per 60-minhour by a numerical efficiency factor. A table of efficiency factors based on acombination of job conditions and management conditions is presented inTable2-1. Both methods are illustrated in the example problems.

2-2 EARTHMOVING MATERIALS

Soil and Rock

Soiland rock are the materials that make up the crust of the earth and are,therefore,the materials of interest to the constructor. In the remainder of thischapter,we will consider those characteristics of soil and rock that affect theirconstruction use, including their volume-change characteristics, methods ofclassification,and field identification.

TABLE2-1.Job efficiency factors for earthmoving operations

Management Conditions*

Poor

0.700.650.600.52

*Management conditions include:Skill, training, and motivation of workers.Selection, operation, and maintenance of equipment.Planning, job layout, supervision, and coordination of work.

**Job conditions are the physical conditions of a job that affect the production rate (not includ-ing the type of material involved). They include:

Topography and work dimensions.Surface and weather conditions.Specification requirements for work methods or sequence.

Job Conditions** Excellent Good Fair

Excellent 0.84 0.81 0.76Good 0.78 0.75 0.71Fair 0.72 0.69 0.65Poor 0.63 0.61 0.57

24 Chapter 2

General Soil Characteristics

Several terms relating to a soil's behavior in the construction environmentshould be understood. Trafficability is the ability of a soil to support theweight of vehicles under repeated traffic. In construction, trafficability con-trols the amount and type of traffic that can use unimproved access roads, aswell as the operation of earthmoving equipment within the construction area.Trafficability is usually expressed qualitatively, although devices are avail-able for quantitative measurement. Trafficability is primarily a function ofsoil type and moisture conditions. Drainage, stabilization of haul routes, orthe use of low-ground-pressure construction equipment may be requiredwhen poor trafficability conditions exist. Soil drainage characteristics areimportant to trafficability and affect the ease with which soils may be driedout. Loadability is a measure of the difficulty in excavating and loading a soil.Loose granular soils are highly loadable, whereas compacted cohesive soilsand rock have low loadability.

Unit soil weight is normally expressed in pounds per cubic yard or kilo-grams per cubic meter. Unit weight depends on soil type, moisture content,and degree of compaction. For a specific soil, there is a relationship betweenthe soil's unit weight and its bearing capacity. Thus soil unit weight is com-monly used as a measure of compaction, as described in Chapter 5. Soil unitweight is also a factor in determining the capacity of a haul unit, as explainedin Chapter 4.

In their natural state, all soils contain some moisture. The moisture con-tent of a soil is expressed as a percentage that represents the weight of waterin the soil divided by the dry weight of the soil:

M. t t t (n! )

Moist weight - Dry weight 100O1Sure con en -/0 =D . h

xry weig t

If, for example, a soil sample weighed 120 lb (54.4 kg) in the natural stateand 100 lb (45.3 kg) after drying, the weight of water in the sample would be20 lb (9.1 kg) and the soil moisture content would be 20%. Using Equation2-3, this is calculated as follows:

(2-3)

. 120 - 100MOIsture content = 100 x 100 = 20%

[= 54.4 - 45.3 x 100 = 20%

]45.3

2-3 SOIL IDENTIFICATION AND CLASSIFICATION

Soil is considered to consist of five fundamental material types: gravel, sand,silt, clay, and organic material. Gravel is composed of individual particleslarger than about Ytin. (6 mm) in diameter but smaller than 3 in. (76 mm) indiameter. Rock particles larger than 3 in. (76 mm) in diameter are called cob-

2-3 Soil Identification and Classification 25

bles or boulders. Sand is material smaller than gravel but larger than theNo. 200 sieve opening (0.7 mm). Silt particles pass the No. 200 sieve but arelarger than 0.002 mm. Clay is composed of particles less than 0.002 mm indiameter. Organic soils contain partially decomposed vegetable matter. Peatis a highly organic soil having a fibrous texture. It is normally readily identi-fied by its dark color, odor, and spongy feel. It is generally considered to beunsuitable for any construction use.

Because a soil's characteristics are largely determined by the amountand type of each of the five basic materials present, these factors are used forthe identification and classification procedures described in the remainder ofthis section.

Soil Classification Systems

There are two principal soil classification systems used for design and con-struction in the United States. These are the Unified System and theAASHTO [American Association of State Highway and Transportation Offi-cials, formerly known as the American Association of State Highway Officials(AASHO)]System. In both systems soil particles 3 in. or larger in diameterare removedbefore performing classification tests.

The liquid limit (LL) of a soil is the water content (expressed in per-centage of dry weight) at which the soil will just start to flow when subjectedto a standard shaking test. The plastic limit (PL) of a soil is the moisture con-tent in percent at which the soil just begins to crumble when rolled into athread ~ in. (0.3 cm) in diameter. The plasticity index (PI) is the numericaldifferencebetween the liquid and plastic limits and represents the range inmoisture content over which the soil remains plastic.

The Unified System assigns a two-letter symbol to identify each soiltype. Field classification procedures are given in Table 2-2. Soils that haveless than 50% by weight passing the No. 200 sieve are further classified ascoarse-grainedsoils, whereas soils that have more than 50% by weight pass-ing the No. 200 sieve are fine-grained soils. Gradation curves for well-gradedand poorlygraded sand and gravel are illustrated in Figure 2-2.

Under the AASHTO System, soils are classified as types A-l throughA-7,corresponding to their relative value as subgrade material. Classificationproceduresfor the AASHTO System are given in Table 2-3.

Field Identification of Soil (Unified System)

When identifying soil in connection with construction operations, adequatetime and laboratory facilities are frequently not available for complete soilclassification.The use of the procedures described here together with Table2-2 should permit a reasonably accurate soil classification to be made in aminimum of time.

~=

TABLE 2-2Unified system of soil classification-field identification

Coarse-Grained Soils(Less Than 50% PassNo. 200 Sieve)

Symbol

GWGPSWSPGMGCSMSC

Name

Well-graded gravelPoorly graded gravelWell-graded sandPoorly graded sandSilty gravelClayey gravelSilty sandClayey sand

Percent ofCoarse FractionLess Than 1f4in.

Percent ofSample Smaller

Than No. 200 Sieve Comments

Wide range of grain sizes with all intermediate sizesPredominantly one size or some sizes missingWide range of grain sizes with all intermediate sizesPredominantly one size or some sizes missingLow-plasticity fines (see ML below)Plastic fines (see CL below)Low-plasticity fines (see ML below)Plastic fines (see CL below)

50 max.50 max.51 min.51 min.50 max.50 max.51 min.51 min.

<W<W<W<W~W~W~W~W

Tests on Fraction Passing No. 40 Sieve (Approx. 1f64in. or 0.4 mm)*

Fine-Grained Soils(50% or More Pass No. 200 Sieve)

Symbol

MLCLOLMHCHOHPt

Name

Low-plasticity siltLow-plasticity clayLow-plasticity organicHigh-plasticity siltHigh-plasticity clayHigh-plasticity organicPeat

Dry Strength Shaking Other

Low Medium to quickLow to medium None to slowLow to medium Slow Color and odorMedium to high None to slowHigh NoneMedium to high None to slow Color and odorIdentified by dull brown to black color, odor, spongy feel,

and fibrous texture

*Laboratory classification based on liquid limit and plasticity index values.

0.01 0:001

Grain size in millimeters

Silt or clayFine

FIGURE 2-2 Typical gradation curves for coarse-grained soils. (U.S. Army Engineer School)

~'I

100

90

80

70

II>>II> 60'f;;a>c:

'gj501..

C. I...c:II>40f:!

II>11.

30

201

10

01000

ICobbles

~00

TABLE 2-3AASHTO system of soil classification

Group Number

A-I A-2

A-I-a A-I-b A-2-4 A-2-5 A-2-6 A-2-7 A-3 A-4 A-5 A-6 A-7

Percent passingNo. 10 sieveNo. 40 sieveNo. 200 sieve

Fraction passing No.40Liquid limitPlasticity index

.Typical material

.

50 max.30 max. 50 max. 51 min.15 max. 25 max. 35 max. 35 max. 35 max. 35 max. 10 max. 36 min. 36 min. 36 min. 36 min.

6max. 6 max. 40 max. 41 min. 40 max. 41 min. 40 max. 41 min. 40 max. 41 min.10 max. 10 max. 11 min. 11 min. 10 max. 10 max. 11 min. 11 min.

Gravel and sand Silty or clayey sand or gravel Fine Silt Silt Clay Claysand

2-4 Soil Volume-Change Characteristics 29

All particles over 3 in. (76 mm) in diameter are first removed. The soilparticles are then separated visually at the No. 200 sieve size: this correspondsto the smallest particles that can be seen by the naked eye. If more than 50%of the soil by weight is larger than the No. 200 sieve, it is a coarse-grained soil.The coarse particles are then divided into particles larger and smaller than *in. (6 mm) in diameter. If over 50% of the coarse fraction (by weight) is largerthan ~ in. (6 mm) in diameter, the soil is classified as gravel; otherwise, it issand. Ifless than 10% by weight of the total sample is smaller than the No. 200sieve, the second letter is assigned based on grain size distribution. That is, itis either well graded (W) or poorly graded (P). If more than 10% of the sampleis smaller than the No. 200 sieve, the second classification letter is based onthe plasticity of the fines (L or H), as shown in the table.

If the sample is fine-grained (more than 50% by weight smaller than theNo. 200 sieve), classification is based on dry strength and shaking tests of thematerial smaller than ~4 in. (0.4 mm) in diameter.

Dry Strength Test. Mold a sample into a ball about the size of a golf ballto the consistency of putty, adding water as needed. Allow the sample to drycompletely.Attempt to break the sample using the thumb and forefinger ofboth hands. If the sample cannot be broken, the soil is highly plastic. If thesample breaks, attempt to powder it by rubbing it between the thumb andforefingerof one hand. If the sample is difficult to break and powder, it hasmedium plasticity. Samples of low plasticity will break and powder easily.

Shaking Test. Form the material into a ball about % in. (19 mm) in diam-eter, adding water until the sample does not stick to the fingers as it ismolded. Put the sample in the palm of the hand and shake vigorously.Observe the speed with which water comes to the surface of the sample toproducea shiny surface. A rapid reaction indicates a nonplastic silt.

Construction Characteristics of Soils

Some important construction characteristics of soils as classified under theUnified System are summarized in Table 2-4.

2-4 SOIL VOLUME-CHANGE CHARACTERISTICS

Soil Conditions

There are three principal conditions or states in which earthmoving materialmay exist: bank, loose, and compacted. The meanings of these terms are asfollows:

·Bank: Material in its natural state before disturbance. Often referredto as "in-place"or "in situ." A unit volume is identified as a bank cubicyard (BCY)or a bank cubic meter (BCM).

30 Chapter 2

TABLE 2-4Construction characteristics of soils (Unified System)

·Loose: Material that has been excavated or loaded. A unit volume isidentified as a loose cubic yard (LCY) or loose cubic meter (LCM).·Compacted: Material after compaction. A unit volume is identified asa compacted cubic yard (CCY) or compacted cubic meter (CCM).

Swell

A soil increases in volume when it is excavated because the soil grains areloosened during excavation and air fills the void spaces created. As a result, aunit volume of soil in the bank condition will occupy more than one unit vol-

1.0 cubic yard innatural condition

(in-place yards)..1.25 cubic yards

after digging

(loose yards)

~

0.90 cubic yard

after compaction(compacted yards)..

FIGURE 2-3 Typical soil volume change during earthmoving.

Suitabilityfor Subgrade Suitability

Construction (No Frost forSoil Type Symbol Drainage Workability Action) Surfacing

Well-graded gravel GW Excellent Excellent Good GoodPoorly graded gravel GP Excellent Good Good to excellent PoorSilty gravel GM Poor to fair Good Good to excellent FairClayey gravel GC Poor Good Good ExcellentWell-graded sand SW Excellent Excellent Good GoodPoorly graded sand SP Excellent Fair Fair to good PoorSilty sand SM Poor to fair Fair Fair to good FairClayey sand SC Poor Good Poor to fair ExcellentLow-plasticity silt ML Poor to fair Fair Poor to fair PoorLow-plasticity clay CL Poor Fair to good Poor to fair FairLow-plasticity organic OL Poor Fair Poor PoorHigh-plasticity silt MH Poor to fair Poor Poor PoorHigh-plasticity clay CH Very poor Poor Poor to fair PoorHigh-plasticity organic OH Very poor Poor Very poor to poor PoorPeat Pt Poor to fair Unsuitable Unsuitable Unsuitable

2-4 Soil Volume-Change Characteristics

ume after excavation. This phenomenon is called swell. Swell may be calcu-lated as follows:

(Weightlbank volume _ 1)x 100 (2-4)Swell (%) = WeightJloose volume

EXAMPLE 2-1

Find the swell of a soil that weighs 2800 lb/cu yd (1661 kg/m3) in its naturalstate and 2000 lb/cu yd (1186 kg/m3) after excavation.

Solution

(2800

)Swell = 2000 - 1 x 100 =40%

[= (:~:~ - 1) x 100= 40%]That is, 1 bank cubic yard (meter) of material will expand to 1.4 loose cubicyards (meters) after excavation.

(2-4)

Shrinkage

When a soil is compacted, some of the air is forced out of the soil's void spaces.As a result, the soil will occupy less volume than it did under either the bankor looseconditions. This phenomenon, which is the reverse of the swell phe-nomenon, is called shrinkage. The value of shrinkage may be determined asfollows:

Shrinkage (%) =(1- ~eightlbank volume )x 100 (2-5)WeIght/compacted volume

Soil volume change due to excavation and compaction is illustrated in Figure2-3. Note that both swell and shrinkage are calculated from the bank (or nat-ural) condition.

EXAMPLE 2-2

Find the shrinkage of a soil that weighs 2800 lb/cu yd (1661 kg/m3) in its nat-ural state and 3500 lb/cu yd (2077 kg/m3) after compaction.

Solution

.(

2800

)Shnnkage = 1 - 3500 x 100 x 20%

[

1661

]=(1- 2077) x 100 =20%

(2-5)

31

32 Chapter 2

Hence 1 bank. cubic yard (meter) of material will shrink to 0.8 compactedcubic yard (meter) as a result of compaction.

Load and Shrinkage Factors

In performing earthmoving calculations, it is important to convert all mater-ial volumes to a common unit of measure. Although the bank. cubic yard (ormeter) is most commonly used for this purpose, any of the three volume unitsmay be used. Apay yard (or meter) is the volume unit specified as the basisfor payment in an earthmoving contract. It may be any of the three volumeunits.

Because haul unit and spoil bank volume are commonly expressed inloose measure, it is convenient to have a conversion factor to simplify theconversion of loose volume to bank. volume. The factor used for this purposeis called a load factor. A soil's load factor may be calculated by use of Equa-tion 2-6 or 2-7. Loose volume is multiplied by the load factor to obtain bankvolume.

L d f: t _ Weight/loose unit volumeoa ac or -W

.ht/bank.

.IeIg umt vo ume

(2-6)

or

1Load factor =~we

(2-7)

A factor used for the conversion of bank volume to compacted volume issometimes referred to as a shrinkage factor. The shrinkage factor may be cal-culated by use of Equation 2-8 or 2-9. Bank volume may be multiplied by theshrinkage factor to obtain compacted volume or compacted volume may bedivided by the shrinkage factor to obtain bank volume.

Shrinka e factor = ~eight/bank unit ~olume (2-8)g WeIght/compacted umt volume

or

Shrinkage factor = 1 - shrinkage (2-9)

EXAMPLE 2-3

A soil weighs 1960 lb/LCY(1163kg/LCM),2800lb/BCY(1661kg/BCM),and3500 lb/CCY (2077 kg/CCM). (a) Find the load factor and shrinkage factorfor the soil. (b) How many bank cubic yards (BCY)or meters (BCM) andcompacted cubic yards (CCY) or meters (CCM) are contained in 1 millionloose cubic yards (593,300 LCM) of this soil?

2-5 Spoil Banks 33

Solution

1960(a) Load factor =2800 =0.70 (2-6)

[

1163

]= 1661 =0.70

. 2800Shrinkagefactor =3500 =0.80 (2-8)

[

1661

]= 2077 =0.80

(b) Bank volume = 1,000,000x 0.70 = 700,000BCY[= 593300 x 0.70 = 415310 BCM]

Compacted volume = 700,000x 0.80 = 560,000CCY[= 415310 x 0.80 = 332248 CCM]

Typicalvalues of unit weight, swell, shrinkage, load factor, and shrink-age factor for some common earthmoving materials are given in Table 2-5.

2-5 SPOIL BANKS

Whenplanning and estimating earthwork, it is frequently necessary to deter-mine the size of the pile of material that will be created by the materialremovedfrom the excavation. If the pile of material is long in relation to itswidth, it is referred to as a spoil bank. Spoil banks are characterized by a tri-.angular cross section. If the material is dumped from a fixed position, a spoilpile is created which has a conical shape. To determine the dimensions ofspoilbanks or piles, it is first necessary to convert the volume of excavationfrom in-place conditions (BCY or BCM) to loose conditions (LCY or LCM).

TABLE 2-5

Typical soil weight and volume change characteristics*

Unit Weight [lb/cu yd (kglm3)]Swell Shrinkage Load Shrinkage

Loose Bank Compacted (%) (%) Factor Factor

Clay 2310 (1370) 3000 (1780) 3750 (2225) 30 20 0.77 0.80Commonearth 2480 (1471) 3100 (1839) 3450 (2047) 25 10 0.80 0.90Rock(blasted) 3060 (1815) 4600 (2729) 3550 (2106) 50 -30** 0.67 1.30**Sand and

gravel 2860 (1697) 3200 (1899) 3650 (2166) 12 12 0.89 0.88

*Exactvalues vary with grain size distribution, moisture, compaction, and other factors. Tests are required todetermineexact values for a specific soil.**Compactedrock is less dense than is in-place rock.

34 Chapter 2

Bank or pile dimensions may then be calculated using Equations 2-10 to2-13 if the soil's angle of repose is known.

A soil's angle of repose is the angle that the sides of a spoil bank or pilenaturally form with the horizontal when the excavated soil is dumped ontothe pile. The angle of repose (which represents the equilibrium position of thesoil) varies with the soil's physical characteristics and its moisture content.Typical values of angle of repose for common soils are given in Table 2-6.

Triangular Spoil Bank

Volume =Section area x Length

(4V

)1/2

B= LxtanR

H=BxtanR2

(2-10)

(2-11)

where B =base width (ft or m)H =pile height (ft or m)L =pile length (ft or m)R = angle of repose (deg)V =pile volume (cu ft or m3)

Conical Spoil Pile

Volume = ~ x Base area x Height

D =(7.64V

)1/3

tanR

DH=-xtanR

2

(2-12)

(2-13)

where D is the diameter of the pile base (ft or m).

TABLE 2-6Typical values of angle of repose ofexcavated soil

Material

ClayCommon earth, dryCommon earth, moistGravelSand, drySand, moist

Angle of Repose (deg)

353237352537

2-6 Estimating Earthwork Volume 35

EXAMPLE 2-4

Find the base width and height of a triangular spoil bank containing 100BCY(76.5 BCM) if the pile length is 30 ft (9.14 m), the soil's angle of reposeis 37°, and its swell is 25%.

SolutionLoose volume =27 x 100 x 1.25 =3375 cu ft

[= 76.5 x 1.25 = 95.6 m3]. 4 x 3375 1/2

Base WIdth =( 30 x tan 370) = 24.4 ft

[=( 4 x 95.6 )1/2= 7.45 m

]9.14 x tan 37°

Height = 2;.4 x tan 37° =9.2 ft

[= 7.:5 x tan 37° = 2.80 m]

(2-10)

(2-11)

EXAMPLE 2-5

Find the base diameter and height of a conical spoil pile that will contain100 BCY (76.5 BCM) of excavation if the soil's angle of repose is 32° and itsswell is 12%.

SolutionLoose volume = 27 x 100 x 1.12 = 3024 cu ft

[= 76.5 x 1.12 = 85.7 m3]

. 7.64 x 3024 1/3Base dIameter = ( M~) =33.3 fttan

[=(

7.64 x 85.7

)1/3 = 10.16 m]tan 32°

Height = 3;.3 x tan 32° = 10.4 ft

[

10.16

]=~xtan32°=3.17m

(2-12)

(2-13)

2-6 ESTIMATING EARTHWORK VOLUME

When planning or estimating an earthmoving project it is often necessary toestimate the volume of material to be excavated or placed as fill. The proce-dures to be followedcan be divided into three principal categories: 1) pit exca-

36 Chapter 2

vations (small, relatively deep excavations such as those required for base-ments and foundations), 2) trench excavation for utility lines, and 3) excavat-ing or grading relatively large areas. Procedures suggested for each of thesethree cases will now be described.

The estimation of the earthwork volume involved in the construction ofroads and airfields is customarily performed by the design engineer. Theusual method is to calculate the cross-sectional area of cut or fill at regularintervals (such as stations (l00 ft or 33 m» along the centerline. The volumeof cut or fill between stations is then calculated, accumulated, and plotted asa mass diagram. While the construction of a mass diagram is beyond thescope of this book, some construction uses of the mass diagram are describedin Section 2-7.

When making earthwork volume calculations, keep in mind that cut vol-ume is normally calculated in bank measure while the volume of compactedfill is calculated in compacted measure. Both cut and fill must be expressed inthe same volume units before being added.

Pit Excavations

For these cases simply multiply the horizontal area of excavation by the aver-age depth of excavation (Equation 2-14).

Volume=Horizontal area x Average depth (2-14)

To perform these calculations, first divide the horizontal area into a conve-nient set of rectangles, triangles, or circular segments. After the area ofeach segment has been calculated, the total area is found as the sum of thesegment areas. The average depth is then calculated. For simple rectangu-lar excavations, the average depth can be taken as simply the average ofthe four corner depths. For more complex areas, measure the depth atadditional points along the perimeter of the excavation and average alldepths.

EXAMPLE 2-6

Estimate the volume of excavation required (bank measure) for the basementshown in Figure 2-4. Values shown at each corner are depths of excavation.All values are in feet (m).

SolutionArea =25 x 30 =750 sq ft

[= 7.63 x 9.15 = 69.8 m2)

A d th 6.0 + 8.2 + 7.6 + 5.86 9 ftverage ep = 4 = .

2-6 Estimating Earthwork Volume 37

FIGURE 2-4 Figure forExample2-6. 6.0

(1.8)

30.0 ft(9.15 m) 8.2

(2.5)

25.0 ft(7.63 m)

5.8(1.8)

7.6(2.3)

[= 1.8 + 2.5 : 2.3 + 1.8 =2.1 m ]

Volume = 750;76.9 191.7 BCY

[= 69.8 x 2.1 = 146.6 BCM ]

Trench Excavations

Thevolume of excavation required for a trench can be calculated as the prod-uct of the trench cross-sectional area and the linear distance along the trenchline (Equation 2-15). For rectangular trench sections where the trench depthand width are relatively constant, trench volume can be found as simply theproduct of trench width, depth, and length. When trench sides are sloped andvary in width and/or depth, cross sections should be taken at frequent linearintervals and the volumes between locations computed. These volumes arethen added to find total trench volume.

Volume = Cross-sectional area x Length (2-15)

EXAMPLE 2-7

Find the volume (bank measure) of excavation required for a trench 3 ft (0.92m) wide, 6 it (1.83 m) deep, and 500 it (152 m) long. Assume that the trenchsideswill be approximately vertical.

38 Chapter 2

SolutionCross-sectional area =3 x 6 =18 sq ft

[= 0.92 x 1.83 = 1.68 m2 ]

Volume = 18 ~7500 333 BCY

[= 1.68 x 152 = 255 BCM ]

Large Areas

To estimate the earthwork volume involved in large or complex areas, onemethod is to divide the area into a grid indicating the depth of excavation orfill at each grid intersection. Assign the depth at each corner or segmentintersection a weight according to its location (number of segment lines inter-secting at the point). Thus, interior points (intersection of four segments) areassigned a weight of four, exterior points at the intersection of two segmentsare assigned a weight of two, and corner points are assigned a weight of one.Average depth is then computed using Equation 2-16 and multiplied by thehorizontal area to obtain the volume of excavation. Note, however, that thiscalculation yields the net volume of excavation for the area. Any balancing ofcut and fill within the area is not identified in the result.

A: d th _ Sum of products of depth x weightverage ep - S f.

hurn 0 welg ts(2-16)

EXAMPLE 2-8

Find the volume of excavation required for the area shown below. The figureat each grid intersection represents the depth of cut at that location.

Solution

Corner points = 6.0 + 3.4 + 2.0 + 4.0 = 15.4 ft[= 1.83 + 1.04 + 0.61 + 1.22 = 4.70 m ]

Border points = 5.8 + 5.2 + 4.6 + 3.0 + 2.8 + 3.0+ 3.5 + 4.8 + 4.8 + 5.5 =43.0 ft

[= 1.77 + 1.59 + 1.40 + 0.92 + 0.85 + 0.92+ 1.07 + 1.46 + 1.46 + 1.68 =13.12 m ]

Interior points =5.0 + 4.6 + 4.2 + 4.9 + 4.0 + 3.6 = 26.3 ft[= 1.52+ 1.40+ 1.28+ 1.49+ 1.22+ 1.10=8.01 m ]

A: d th - 15.4 + 2(43.0) + 4(26.3) - 3 97 fitVffa~ ~ - ~ - .

[= 4.70 + 2(135;2) + 4(8.01) = 1.21 m ]

2-7 Construction Use of the Mass Diagram 39

6.0 (1.83) 5.8 (1.77) 5.2 (1.59) 4.6 (1.40) 3.4 (1.04)

5.5

ftm)

4.8

4.0

400 ft(121.9 m)

(.j

FIGURE 2-5 Figure for Example 2-8.

Area =300 x 400 = 120,000sq ft[= 91.4 x 121.9 = 11,142 m2 ]

Volume = 120,00~7x 3.98 = 17,689 BCY

[= 11,142 x 1.21 = 13,482 BCM ]

2-7 CONSTRUCTION USE OF THE MASS DIAGRAM

Amass diagram is a continuous curve representing the accumulated volumeof earthwork plotted against the linear profile of a roadway or airfield. Massdiagrams are prepared by highway and airfield designers to assist in select-ing an alignment which minimizes the earthwork required to construct thefacility while meeting established limits of roadway grade and curvature.Sincethe mass diagram is intended as a design aid, it is not normally pro-videdto contractors as part of a construction bid package. However, the massdiagram can provide very useful information to the construction manager andit is usually available to the contractor upon request. A typical mass diagramand corresponding roadway profile are illustrated in Figure 2-6.

,-,-

(1.68) 5.0 (1.52) 4.6 (1.40) 4.2 (1.28) 3.0 (0.92

II

300"

(91.4

I

(1.46) 4.9 (1.49) 4.0 (1.22) 3.6 (1.10) 2.8 (0.85

,

(1.22) . 4.8 (1.46) I 3.5 (1.07) 3.0 (0.92) 2.0, '" .. >- -..

40 Chapter 2

20

10

5

CD

a!£ 0&-CD

"C~alE

&-

-10

-5100

Mass diagram

CDE 50:Io~>>-<DU.~m 01!i~:I....E-<3 -50

50 CDE:Io~>U

o ~m~~:I....E-:IU

-50

-100oIo

1000 2000I

500

3000 4000 5000 6000 7000I I I

1000 1500 2000

Distance along centerline

8000 9000 10000 ftI I

2500 3000 m

FIGURE 2-6 A mass diagram.

Characteristics of a Mass Diagram

Some of the principal characteristics of a mass diagram include the fol-lowing.

· The vertical coordinate of the mass diagram corresponding to anylocation on the roadway profile represents the cumulative earthworkvolume from the origin to that point.·Within a cut, the curve rises from left to right.·Within a fill, the curve falls from left to right.·A peak on the curve represents a point where the earthwork changesfrom cut to fill.·A valley (low point) on the curve represents a point where the earth-work changes from fill to cut.·When a horizontal line intersects the curve at two or more points, theaccumulated volumes at these points are equal. Thus, such a line rep-resents a balance line on the diagram.

Problems 41

Using the Mass Diagram

Someof the information which a mass diagram can provide a constructionmanagerincludes the following.

· The length and direction of haul within a balanced section.· The average length of haul for a balanced section.· The location and amount of borrow (material hauled in from a bor-

row pit) and waste (material hauled away to a waste area) for theproject.

The following explanation of methods for obtaining this informationfroma mass diagram will be illustrated using Figure 2-7.

1. For a balanced section (section 1 on the figure), project the end pointsof the section up to the profile (points A and B). These points identifythe limits of the balanced section.

2. Locatepoint C on the profile corresponding to the lowest point of themass diagram within section 1. This is the point at which the exca-vation changes from fill to cut. The areas of cut and fill can now beidentified on the profile.

3. The direction of haul within a balanced section is always from cutto fill.

4. Repeat this process for sections 2, 3, and 4 as shown.5. Sincethe mass diagram has a negative value from point D to the end,

the ordinate at point E (-50,000 BCY or -38,230 BCM) represents thevolume of material which must be brought in from a borrow pit tocompletethe roadway embankment.

6. The approximate average haul distance within a balanced section canbe taken as the length of a horizontal line located midway betweenthe balance line for the section and the peak or valley of the curve forthe section. Thus, the length of the line F-G represents the averagehaul distance for section 1, which is 1800 ft or 549 m.

PROBLEMS

1. Express the job efficiencyfactors shown in Table 2-1 as equivalent effec-tive workingminutes per hour. Based on these values, what is your eval-uation of the validity of the frequently used 50 working minutes per houras average job performance?

II I 2. Asampleofgravelfroma stockpileweighs 15 lb (6.80kg).After oven dry-ing, the sample weighs 14.2 lb (6.44 kg). Calculate the moisture contentof the sample.

42

cv~ 3. A soil weighs 2500 lb/cu yd (1483 kg/m3) loose, 3100 lb/cu yd (1839 kg/m3)in its natural state, and 3650 lb/cu yd (2166 kg/m3) compacted. Find thissoil's load factor and shrinkage factor.

4. Calculate the size of a conical spoil pile resulting from the excavation of500 BCY (382 BCM) of dry common earth.

5. A 1000 ft (305 m) long pipeline requires an excavation 4 ft (1.2 m) wide toan average depth of 5 ft (1.5 m). If the soil is dry common earth, what sizespoil bank will be created by the excavation?

6. In making a field identification of a soil you find that less than 50% byweight is smaller than the No. 200 sieve and less than 50% by weight ofthe coarse fraction is smaller than % in. (6 rom). The fine fraction of thesoil exhibits low dry strength and gives a medium speed reaction to theshaking test. How would you classify this soil under the Unified System?

7. How many truck loads of trucks hauling an average volume of6 LCY (4.6LCM) would be required to haul 1 million CCY (764,600 CCM) of the soilof Problem 3 to a dam site?

8. Calculate the volume of excavation in bank measure required for thebasement shown in Figure Problem 2-8. Excavation depths are in ft (m).

Chapter 2

20

CD,,~es.(!:j

100

CDE 50::3"Os:->()~aJ~C')<ISO

"3....E~

<3 -50

-100oIo

5000 6000 7000I I

1500 2000

8000 9000 10000 ftI I

2500 3000 m

1000 2000 3000 4000I I

500 1000

Distance along centerline

FIGURE 2-7 Construction use of a mass diagram.

)

'j~

I

c. "

5

CD,,~!!!E(!:j~

-5

50 CDE::3

"O~

>()o ~aJ+::C")<ISo"3....E~

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References 43

8.2 (2.50) 6.6 (2.01)

28.0'(8.54 m)

45.0'

(13.7m)

6.9 (2.10) 6.0

(1.83)

7.4 I (2.26)

L-- 26.0'1- (7.93 m)

6.8 I (2.07)

1 24.0'(7.32 m)

(jox/of r c ,

,! I 9. Use the profile and mass diagram of Figure 2-7 to find the following val- 5ues. $0 "ID\r., Y /' GO )iIO~g(u,.\y ? CO y 10

a. The total volume of 1) cut, 2) fill, 3) waste, -and 4) borrow~b. The average length of haul for section 3. 6:+ I?tJ - 5 20,/ = \ C;tJ D ..( b

10. Developa computer program to determine the height and base diameter(it or m) of the conical spoil pile that will result from a rectangular exca-vation. Input should include the width, length, and average depth of theexcavation, or the bank volume of the excavation, as well as the soil'sangle of repose and swell. Solve Problem 4 using your program.

REFERENCES

1. Ahlvin, Robert G., and Vernon A. Smoots. Construction Guide for Soils andFoundations, 2nd ed. New York: Wiley, 1988.

2. Bowles, Joseph E. Engineering Properties of Soils and Their Measure-ments, 3rd ed. New York: McGraw-Hill, 1985.

3. Caterpillar Performance Handbook. Caterpillar Inc., Peoria, IL.

44 Chapter 2

4. Hickerson, Thomas F. Route Location & Design, 5th ed. New York:McGraw-Hill, 1967.

5. Nunnally, S. W. Managing Construction Equipment. Upper Saddle River,NJ: Prentice Hall, 1977.

6. Production and Cost Estimating of Material Movement with EarthmovingEquipment. Terex Corporation, Hudson, OH.

7. Soil Mechanics. Compaction America, Kewanee, IL.8. Soils Manual for Design of Asphalt Pavement Structures (MS-10). The

Asphalt Institute, Lexington, KY.