VI. EARTHWORK AND EXCAVATION
VI.1. Excavating by hand – Numerous types and sizes of excavating equipment is generally used in construction except in very small projects where using equipment will be more expensive as shown in
VI.2.1. Trenching machines‐ There are two types of trenching machines: (1) Wheel‐type and (2) Ladder‐
Example VI.2 illustrates estimating the cost of earth excavation by hand.
Example VI.2:
A trench in a confined area in a refinery has to be excavated by hand. The trench is 15 ft long, 3 ft wide and 4 ft deep and the soil is sandy loam. Calculate the cost of excavation. Cost of labor is $15.56/hr.
Solution:
Quantity of work:
Volume of earth = (3 x 4 x 15)/27 = 6.7cy Production rate:
Using data from Table VI.1, Loosening earth = 6.7 x 2 hrs = 13.4 hrs Shoveling loose earth from trench = 6.7 x 1hr = 6.7 hrs Shoveling back to the trench = 6.7 x 0.5 hr =3.4 hrs Backfilling trench = 6.7 x 0.5 hr = 3.4 hrs Total labor hours = 26.9 hrs Cost:
Total cost = 26.9 hrs x $15.56 = $418.56 Cost/cy = $418.56/6.7 = $62.47
Cost/ft = $418.56/15 = $27.90
VI.2. Excavating by machine – For larger jobs, the cost of excavation by machine is considerably less than the cost by hand once the machine is transported to the job site. The savings in excavating cost must be sufficient to offset the cost of transporting the machine to the job and back to storage after the job is completed. Otherwise, hand labor is more economical.
VI.2.1. Trenching machines‐ There are two types of trenching machines: (1) Wheel‐type and (2) Ladder‐
type.
Wheel‐type machine is frequently used for water mains, gas lines, and oil pipe lines. The wheel rotates at the rear of the machine, which is mounted on crawler tracks. A combination of teeth and buckets attached to the wheel loosens and removes the earth from the trench as the machine advances. The
in. wide and depths up to 6 ft. Table VI.4 provides data on this type of machines.
Table VI.4 Data on wheel‐type trenching machines
Ladder‐type machine is used for deeper trenches such as those required for sewer pipes and other utilities. Inclined or vertical booms are mounted at the rear of the machine. Cutter teeth and buckets are attached to endless chains that travel along the boom. The depth of cut is adjusted by raising or lowering the boom. By adding side cutters, the width of the trench can be increased. This machine is used to excavate trenches from 16 to 36 in. wide and depths up to 12 ft. Table VI.5 provides data on this type of machines.
Table VI.5 Data on ladder‐type trenching machines
Example VI.3 illustrates earth excavation using a wheel‐type trenching machine.
Example VI.3:
Estimate the total cost and cost/linear foot for excavating 2,940 ft of trench in common earth using a ladder‐type trenching machine. The trench will be 30 in. wide and the average depth will be 7 ft. The machine will work for 45 min. in an hour. A machine operator, a laborer and a foreman will be employed on the job with a pickup truck. The rate for trenching machine is $87.50/hr. The machine operator is paid at $21.67/hr, the laborer is paid at $25.56/hr and the foreman is paid at $25.00/hr. The rate for pickup truck is $12.00/hr. In addition, transporting the machine to and from the job cost a lump sum of $1,500.00
Solution:
From Table VI.5, the digging speed is found to vary between 25 and 50 ft/hr for a width of 30 in and depth of 7 ft. Use the average digging speed of 37.5 ft/hr.
Time for digging trench = 2,940/37.5 = 78.4 hrs
Adjusting for 45 min/hr, time required = 78.4 x (60/45) = 104.5 hrs Cost:
Trenching machine : 104.5 hrs @ $87.50/hr = $9,143.75 Machine operator 104.5 hrs @ $21.67 = 2,264.50 Laborer 104.5 hrs @ $15.56 = 1,626.02 Foreman 104.5 hrs @ $25.00 = 2,612.50 Pickup truck 104.5 hrs @ $12.00 = 1,254.00 Transporting machine to and from job = 1,500.00 Total cost = $18,400.77 Cost/linear foot = $18,400.77/2,940 = $6.26
VI.2.2 Drag line – Drag lines are used for excavating earth for drainage channels and building levees where water is present. It can operate on wet ground and can dig earth out of water‐logged pits because they do not have go into pits or hole to excavate. They cannot excavate rock. Frequently they are used with a long boom to dispose of the material along the canal or near the pit. This eliminates the need for hauling units. The size of the drag line is indicated by the size of the bucket expressed in cubic yards. Most drag lines can handle more than one size bucket depending on the length of the boom and class and weight of the material excavated. Greatest output can be achieved if the job is planned to permit excavation at the optimum depth of cut. Table VI.6 gives ideal output, in bank measure, Table VI.6 Ideal output of draglines, bank measure, cy/hr
on a 60‐min hour. The upper figure is the optimum depth in feet and the lower number is the ideal output in cubic yards. Adjustments have to be made for outputs at other depths and angle of swing.
Table VI.7 provides the adjustment factors.
Table VI.7 Adjustment factors for angle of swing and other than ideal depth of cut for Drag lines
Example VI.4 illustrates calculation of excavation cost using a dragline.
Example VI.4:
A drainage ditch 10 ft wide at the bottom, 36 ft wide at the top, 12 ft deep, and 15,100 ft long is to be excavated using a dragline. The soil is good common earth and the excavated earth can be deposited on one or both sides of the ditch. A 1.5cy dragline with an average angle of swing of 1200 will be used. It will require 2 days to set up and remove the dragline. Two laborers will assist the operation which will be supervised by a foreman. The dragline will operate an average of 45 min/hr. The following cost data will apply: dragline $87.69/hr., machine operator $21.67/hr., laborer $15.56/hr. In addition, it will cost
$2,500 to move the dragline to and from the job. Estimate the total cost and cost/cy for excavating the ditch.
Solution:
Quantity of material:
Volume of earth = (10 + 36)/2 x 12 x 15,110 = 4,170,360cf = 4,170,360/27 = 154,458cy Production rate:
From Table VI.6, the ideal output = 190 cy/hr, bank measure, and the optimum depth of cut = 9.0 ft Percent of optimum depth= 12/9 = 1.33 or 133%
From Table VI.7, for 133% optimum depth and 1200 angle of swing, the depth‐swing adjustment factor = 0.89
For a 60‐min/hr operation, production rate = 0.89 x 190 = 169.1 cy/hr
For a 45‐min/hr operation, production rate = 169.1 x (45/60) = 126.8 cy/hr bank measure.
Time required for full excavation = 154,458/126.8 = 1,218 hrs Cost:
Dragline 1,218 hrs@ $87.69/hr = $106,806.42 Operator 1,218 hrs @ $21.67/hr = 26,394.06 Laborers 1,218 hrs x 2 @ $15.56/hr = 37,904.16 Foreman 1,218 hrs @ $25.00/hr = 30,450.00 Cost to set up and remove the dragline:
Operator 16 hrs @ $21.67/hr = 346.72 Laborers 16 hrs x 2 @ $15.56/hr = 497.92 Cost to move the dragline to and from the job = 2,500.00 Total cost =$204,999.28 Cost/cy = $204,999.28/154,458 = $1.33
VI.2.3. Shovel – A shovel is a hydraulic excavator. There are two kinds of shovels: one having the digging action in an upward direction and the other having the digging action in a downward direction.
The former is called front shovel or simply shovel and the latter is called hoe or backhoe or trackhoe.
Shovels are mostly used for pit excavation where the bucket load is obtained from the vertical face of the excavation pit above and in front of the excavator. These machines can handle all classes of earth without loosening. The output of a shovel depends upon the class of earth to be excavated, the height of the cut, the ease with which hauling equipment can approach the shovel, the angle of swing from digging to emptying the bucket, and the size of the bucket. The size of the shovel is designated by the size of the bucket, expressed in cubic yards, loose measure.
A bucket can be rated as struck capacity or heaped capacity. The struck capacity is defined as the volume in the bucket when it is filled even with, but not above, the sides. The heaped capacity is defined as the volume that a bucket will hold when the earth is piled above the sides. The heaped capacity will depend upon the depth of the earth above the sides and the base area of the bucket. The heaped slope usually is 1:1 or 2:1 above the sides of the bucket. The equipment has a fill factor which is a percentage to be used to multiply the heaped capacity to obtain the average payload of the bucket.
Table VI.8 provides fill factors for shovels.
Table VI.8 Fill factors for front shovel buckets
There are four elements in the production cycle of a shovel: load bucket, swing with load, dump load, and return swing empty. Adding times for these elements provides the cycle time for the shovel. Table VI.9 provides the range of element times for shovels with bucket sizes ranging from 3 to 5cy.
Table VI.9 Time for elements in a shovel production cycle
The optimum height of cut for a shovel is that depth at which the bucket comes to the surface of the ground with a full load without overcrowding or under crowding the bucket. The optimum depth varies with the class of soil and the size of the bucket and varies from 30 to 50 percent of the maximum digging height. The output has to be adjusted for the percentage of optimum heights of cut and angles of swing. Table VI.10 provides the adjustment factors.
Table VI.10 Adjustment factors for depth of cut and angle of swing for a shovel
Example VI.5 illustrates the calculation of cost of excavation using a shovel.
Example VI.5:
A 4‐cy shovel will be used to excavate and haul 58,640cy, bank measure, of common earth with a swell factor of 0.25. The maximum digging height for the shovel is 34 ft and the average height of cut is 15 ft.
The shovel swings 1200 to load the haul units. The earth will be hauled 4 mi by 20cy, loose measure trucks, at an average speed of 30 mph. The expected time at the dump is 4 min. The truck time waiting at the shovel to move into loading position will average 3 min and the trucks will work 45 min in an hr.
In addition to the shovel operator and truck drivers, 2 laborers and a foreman will be employed on the job. The rate of renting a shovel will be $135.00/hr and that for the truck is $55.00/hr. The wage rates for the workers will be as follows: shovel operator @ $21.67/hr, truck driver @ $18.17/hr, laborer @
$15.56/hr, and foreman @ $25.00. Calculate the total cost and cost/cy based on using sufficient trucks to balance the production rate of the shovel.
Solution:
Volume of earth = 58,640cy, bank measure
Volume of truck = 20/(1 + 0.25) = 16cy, bank measure Cycle time of shovel:
From Table VI.9 using average values, Load bucket = 8 sec Swing with load = 5 sec Dump bucket = 3 sec Return swing = 4 sec Total time = 20 sec Cycles/min = 60/20 = 3
Production rate of shovel:
Bucket fill factor for common earth from Table VI.7 = 105%
Average bucket payload = 4 x 105% = 4.2cy Ideal production rate = 4.2 x 3 = 12.6cy/min Optimum height of cut = 40% x 34 = 13.6 ft Average height of cut = 15 ft
Percent optimum height of cut = 15/13.6 = 1.1 or 110%
Adjustment factor for 110% optimum depth of cut and 1200 angle of swing from Table VI.9 = 0.87 Probable output = 12.6 x 0.87 x 45 = 493 cy/hr, loose measure
Swell factor for common earth is 25%
Probable production rate = 493/(1 + 0.25) = 395 cy/hr, bank measure Cycle time for trucks:
Loading truck: 16/395 = 0.04 hr Traveling: 8/30 = 0.26 hr Dump time: 4/60 = 0.07 hr Waiting to load: 3/60 = 0.05 hr Total cycle time = 0.42 hr/trip Production rate of trucks:
Number of trips/hr = 1/0.42 = 2.4
Number of trips in 45 min = 2.4 x 45/60 = 1.8 trips/hr Volume hauled/trip = 1.8 x 16 = 28.8 cy/hr, bank measure
Time to complete the job:
Use 14 trucks – there will be more trucks than required and hence the shovel production rate of 395 cy/hr will govern the time required = 58,640/395 =148.5 hrs
Cost:
Shovel 148.5 hrs @ $135.00/hr = $20,047.50 Trucks 145.5 hrs x 14 x $55.00/hr = 112,035.00 Shovel operator 145.5 hrs @ $21.67/hr = 3,152.99 Truck drivers 145.5 hrs x 14 x $18.17/hr = 37,012.29 Laborers 145.5 hrs x 2 x $15.56/hr = 4,527.96 Foreman 145.5 hrs @ $25.00/hr = 3,637.50 Transporting the shovel to and from the job = 2,850.00 Total cost = $183,263.24
Cost/cy = $183,263.24/58,640 = $3.13/cy bank measure
VI.2.4. Backhoe –Backhoes are used below natural surface such as for trenches and basements requiring precise control of depths. It may be a wheel type or track type. The latter is called trackhoe. A backhoe is superior to the machines already described due to its convenient loading into dump trucks, its greater tooth pressure and its capacity for digging utility trenches requiring no shoring. The production rate depends upon bucket payload, average cycle time, and job efficiency. The cycle time depends upon the difficulty in loosening the soil, the angle of swing, the size of the truck the backhoe must load, and the skill of the operator. Table VI.11 provides representative cycle time for backhoes under average conditions and Table VI.12 provides representative values of the bucket fill factor for backhoes.
Table VI.11 Representative cycle time for backhoes
Table VI.12 Representative values of bucket fill factors
VII. FOUNDATIONS
Foundations that support structures include footings, piles, and drilled shafts. Footings are shallow while piles and drilled shafts are deep.
Footings are placed at shallow depths, usually less than 5 ft. Footings may be isolated or continuous.
Isolated footings are placed at one location to support a column. Continuous footing provides support for a wall of a building. Construction of a footing includes excavating the soil to the required depth, erecting formwork, setting reinforcing steel, placing concrete, removing formwork, and backfilling soil above the footing to the surface of the ground. A trench is excavated into the soil to construct a continuous footing. If the sides of the excavated soil are stable enough to support itself without caving in, there is no need for providing formwork. If the soil is unstable, it will be necessary to install a system of shores, braces, and solid sheeting along the excavated walls to hold the earth in position. The
sheeting is typically constructed with 2‐,3‐ or 4‐in thick lumber, placed side by side or overlapping along the whole length of the trench. Wales consisting of 4 x 6‐in thick lumber can be placed in a horizontal direction to provide additional support for the sheeting. Depending upon the stability of the earth, it may be necessary to install braces at suitable intervals. Figure VII.1 shows a cross section of a trench with sheeting, wales and braces.
Figure VII.1 Cross section of a trench with sheeting, wales and braces
Example VII.1 illustrates the cost estimation for a trench with sheeting, wales and braces.
Example VII.1
Estimate the cost of installing and removing solid sheeting and bracing for a trench 100 ft long and 7 ft deep. The sheeting will be 2 x 12‐in lumber, 8 ft long. Two horizontal rows of 4 x 6‐in wales will be placed on each side of the trench for the full length of the trench. Trench braces will be placed 4 ft apart along each row of wales. The sheeting will be driven one plank at a time by using an air compressor and a pneumatic hammer. Two laborers combined will install the sheeting at 4/hr and wales and braces at 3/hr.
The following material cost will apply: sheeting $0.73/bf, wales $0.92/bf and braces $3.50 each. The following equipment cost will apply: air compressor $8.75/hr and air hammer $1.75/hr. Labor costs are
$21.67/hr for driving sheet piles and $18.17 for installing wales and braces.
Solution:
Quantity of materials:
Sheeting, using actual dimension of 2 x 12 lumber as 1.5 in x 11.25 in Number of pieces for two sides= 100/(11.25/12) x 2 = 214 pieces. = Quantity of sheeting = 214 x 2 x (12/12) x 8 = 3,424 bf
Wales: Number of pieces for one row = 100/8 12.5., use 13 pieces = Quantity of lumber = 13 x 2 x 2 x 4 x (6/12) x 8 = 832 bf
Trench braces = (100/4) x 2= 50
Time to perform the job:
2 laborers installing sheeting = 214/4 = 53.5 hrs
2 laborers installing wales and braces = (50 + 13)/3 = 21 hrs Labor cost:
Laborers driving piles 53.5 hrs x 2 x $21.67 = $2,318.69 Laborers installing wales and braces = 21hrs x 2 x $18.17 = $763.14 Total labor cost = $3,081.83 Material cost:
Sheeting 3,424 bf @ $0.73 = $2,499.52 Wales 832 bf @ $0.92 = 765.44 Braces 50 @ $3.50 = 175.00 Total material cost = $3,439.96
Equipment cost:
Air compressor 53.5 hrs x $8.75 = $468.13 Air hammer, hose etc 53.5 hrs x $1.75 = 93.63 Total equipment cost = $561.76 Summary of costs:
Labor = $3,081.83 Material = 3,439.96 Equipment = 561.76 Total cost = $7,083.55
Cost/linear foot of trench = $7,083.55/100 = $70.84 Cost/sf of sheeting = $7,083.55/(100 x 7 x 2) = $5.06 VII.1 Pile foundation – The following type of piles are in use: