4/8. ESTIMATING ROAD CONSTRUCTION UNIT
COSTS
4.1 Introduction
4.2 Surveying
4.3 Clearing and Piling
4.4 Earthwork
4.5 Finish Grading
4.6 Surfacing
4.7 Drainage
4.2 Surveying
4.3 Clearing and Piling
4.4 Earthwork
4.5 Finish Grading
4.6 Surfacing
4.7 Drainage
4.1 Introduction
The unit cost of road construction in dollars
per kilometer is the sum of the subunit costs of the road construction
activities. Road construction unit costs are estimated by dividing the machine
rates by the production rates for the various activities involved in road
construction. The road construction activities considered here are surveying,
clearing and grubbing, excavation, surfacing, and drainage.
4.2 Surveying
Surveying and staking costs vary considerably
depending on type and size of the job, access, terrain, and job location. One
method of estimating production is to estimate the number of stakes which can
be set per hour and the number of stakes which must be set per kilometer. For
example, assume about 15 stakes can be set per hour with a two-man crew with
the preliminary survey line already in place. A typical five-point section
consists of two reference stakes, two slope stakes, and one final centerline
stake.
The surveying production rate in km per hour is
equal to the number of stakes the crew sets per hour divided by the number of
stakes required per km.
Example:
A survey crew is setting 300 stakes per km at a
rate of 15 stakes per hour. The cost of a survey crew including transport is
$10 per hr.
P = 15/300 = .05 km/hr
UC = 10/.05 = $200/km
4.3 Clearing and Piling
The clearing and piling cost can be calculated
by estimating the number of hectares of right-of way to be cleared and piled
per kilometer of road. The clearing and piling production rate in km/hr is the
hectares per hour which can be cleared and piled per hour divided by the number
of hectares per km to be cleared and piled. Clearing can be accomplished in a
number of ways, including men with axes or power saws. Merchantable logs may be
removed by skidder or tractor and the remainder piled by tractor for burning or
decay. Felling rates and skidding rates for logging can be used for determining
the cost of the removal of merchantable logs.
On gentle terrain, if a wide right-of-way is
being cleared to permit sunlight to dry the road surface after frequent rains,
the project might be estimated as a land clearing project. A method for
estimating the total time per hectare required to clear, grub, and pile on
gentle terrain with a tractor and shearing blade is shown below. Additional
details can be found in the Caterpillar Performance Handbook No. 21, Caterpillar,
Inc.
4.3.1 Mechanized Clearing
The clearing time will depend upon the size of
tractor and the number and size of the trees. The clearing time, Tc, in machine
hours per hectare is
Tc = (X/60) (AB + M1N1 +
M2N2 + M3N3 + M4N4 +
DF)
where X is the hardwood density factor, A is the
vine density factor, B is the base minutes per hectare, M is the minutes per
tree in each diameter range, N is the number of trees per hectare in each
diameter range, D is the sum of the diameters of all trees per hectare larger than
180 cm, and F is the minutes per cm of diameter to cut trees with diameters
greater than 180 cm.
TABLE 4.1. Production
factors for felling with Rome KG blade.
Tractor
|
Factors
|
Diameter
Range, cm
|
Min
per cm of diameter for trees
|
|||
GHP
|
30-60
|
61-90
|
91-120
|
121-180
|
>
180 cm
|
|
B
|
M1
|
M2
|
M3
|
M4
|
F
|
|
140
|
100
|
0.8
|
4.0
|
9.0
|
-
|
-
|
200
|
62
|
0.5
|
1.8
|
3.6
|
11
|
0.110
|
335
|
45
|
0.2
|
1.3
|
2.2
|
6
|
0.060
|
460
|
39
|
0.1
|
0.4
|
1.3
|
3
|
0.033
|
X = 1.3 if the percentage of hardwoods > 75
and X = 0.7 if percentage of hardwood is < 25, X = 1 otherwise.
A = 2.0 if number of trees/ha > 1500 and A =
0.7 if number of trees/ha < 1000, A = 1.0 otherwise. Increase value of A by
1.0 if there are heavy vines, and by 2.0 for very heavy vines.
For hectares which must be cleared and where
stumps must be removed (grubbed), multiply the total time for clearing by a
factor of 1.25.
4.3.2 Mechanized Piling
To compute piling time, when a rake or angled
shearing blade is used, an equation to calculate the piling time per hectare,
Tp, is
Tp = (1/60) (B + M1N1 +
M2N2 + M3N3 + M4N4 +
DF)
where the variables are defined as above. Table
4.2 shows the coefficients for piling when stumps have not been removed.
TABLE 4.2. Production
factors for piling in windrows.
Tractor
|
Factors
|
Diameter
Range, cm
|
Min
per cm of diameter for trees
|
|||
GHP
|
30-60
|
61-90
|
91-120
|
121-180
|
>
180 cm
|
|
B
|
M1
|
M2
|
M3
|
M4
|
F
|
|
140
|
185
|
0.6
|
1.2
|
5.0
|
-
|
-
|
200
|
135
|
0.4
|
0.7
|
2.7
|
5.4
|
-
|
335
|
111
|
0.1
|
0.5
|
1.8
|
3.6
|
0.03
|
460
|
97
|
0.08
|
0.1
|
1.2
|
2.1
|
0.01
|
When piling is to include piling of stumps,
increase the total piling time by 25 percent.
EXAMPLE:
Five hectares per km of right-of-way in
hardwoods are being cleared for a road (extra width is being used to help the
road dry after rains). Of the five hectares, 1.2 hectares per km will need to
have the stumps removed. Tractor machine rate is $80 per hour. All material
will be piled for burning. Work is being done by a 335 HP bulldozer. The
average number of trees per hectare less than 180 cm diameter are in Table 4.3.
There is also one tree per hectare with a diameter of approximately 185 cm.
TABLE 4.3 Data for
clearing, grubbing and piling example.
Number
of trees
|
Diameter
Range, cm
|
Sum
of tree diameters for trees
|
|||
<30
cm
|
30-60
|
61-90
|
91-120
|
121-180
|
>
180 cm
|
N1
|
N2
|
N3
|
N4
|
D
|
|
1100
|
35
|
6
|
6
|
4
|
185
|
Tc = (X/60) (AB + M1N1 +
M2N2 + M3N3 + M4N4 +
DF)
Tc = (1.3/60) [(1) (45) + (.2) (35) + (1.3) (6)
+ (2.2) (6) + (6) (4) + (185) (0.06)] = 2.34 hr/ha
Tp = (1/60) (B + M1N1 +
M2N2 + M3N3 + M4N4 +
DF)
Tp = (1/60) [111 + (.1) (35) + (.5) (6) + (1.8)
(6) + (3.6) (4) + (185) (0.03) ] = 2.47 hr/ha
Total tractor time/km = 3.8 (2.34 + 2.47) +
1.2(1.25) (2.34 + 2.47) = 25.5 hr/km
P = 1/25.5 = .039 km/hr
UC = 80 × 25.5 = $ 2039/km
UC = 80 × 25.5 = $ 2039/km
4.4 Earthwork
The earthwork cost is calculated by estimating
the number of cubic meters of common material and rock which must be moved to
construct the road. The earthwork production rate is calculated as the cubic
meters per hour which can be excavated and placed divided by the number of
cubic meters per km to be excavated.
Road construction superintendents can often
estimate the number of meters per hour that their equipment can build road
based upon local experience after looking at the topography. The engineer's
method is to calculate the number of cubic meters to be excavated using
formulas or tables for calculating earthwork quantities as a function of
sideslope, road width, cut and fill slope ratios. Production rates for
bulldozers and hydraulic excavators are available.
For example, a 6.0 meter subgrade on a 30
percent slope with a 1.5:1 fill slope and 0.5:1 cut slope with a one foot ditch
and a 20 percent shrinkage factor would be approximately 2100 bank cubic meters
per km for a balanced section.
An average production rate in common material
(no rock) from an equipment performance handbook might be 150 bank cubic meters
per hour for a 300 hp power-shift tractor with ripper. The tractor cost is
$80/hr. The rate of excavation would be
P = (150 m3/hr)/(2100 m3/km) = .07 km/hr
UC = 80/.07 = $1143/km
If the earthwork is not being placed or sidecast
within 50 meters of the cut, the production rate for pushing the material to
the placement location must be made. Scrapers or excavators and dump trucks may
be used.
Excavation rates in rock vary with the size of
job, hardness of rock and other local conditions. Often there is a local market
price for blasting. Estimates of blasting production can be made by knowing the
size of equipment and the type of job. For example, a 10 cm track-mounted drill
and 25 cubic meter per minute air-compressor may prepare 40 cubic meters per
hour for small, shallow blasts and 140 cubic meters per hour for larger, deeper
blasts including quarry development to produce rock surfacing. A major cost
will be explosives. For example, 0.8 kg of explosive such as Tovex might be
used per cubic meter of rock at a cost of approximately $2 per kg.
4.5 Finish Grading
Finish grading of the subgrade can be estimated
by determining the number of passes a grader must make for a certain width
subgrade and the speed of the grader. This number can be converted to the
number of hours per hectare of subgrade. For example, a 120 hp grader may
require about 10 hours of productive machine time without delays per hectare of
subgrade or 0.1 hectares per hour. The production rate for final grading of a
6.0 meter subgrade would then be,
P = (0.1 ha/hr)/(0.6
ha/km) = .17 km/hr
If the grader cost is $30/hr, the unit cost of
grading is
UC = 30/.17 = $176/km
Similarly, the rate of pulling ditches per
kilometer can be estimated.
4.6 Surfacing
Surfacing costs are a function of the type of surfacing
material, the quantity of surfacing material per square meter, and the length
of haul. Local information is the best guide in constructing surfacing costs
due to the wide range of conditions that can be encountered.
Natural gravel from streams may require only
loading with front-end loaders directly to dump trucks, transporting,
spreading, and may or may not be compacted.
Laterite may be ripped by crawler tractor,
loaded by front-end loader, transported, spread and grid-rolled with a
sheeps-foot roller to produce a sealed running surface.
Rock may have to be blasted, loaded into one or
more crusher(s), stockpiled, reloaded, transported, spread, and compacted.
The costs for each of these operations can be
developed by estimating the equipment production rates and machine rates.
EXAMPLE:
A relatively complex surfacing operation
requires developing a 20,000 cubic meter solid rock source (26,400 cubic meters
in the road prism) to surface 26.4 km of road including shooting and crushing
rock, loading, transporting, and spreading rock as follows.
To open up rock source, use data from clearing
and common excavation:
(a) To clear and excavate to rock:
Equipment
|
Machine
Hours
|
Machine
Rate
|
Cost
|
Tractor
|
27
|
72.00
|
1944.00
|
Cost per cubic meter solid rock = $0.10
(b) To drill and blast at a production rate of
140 cubic meters per hour
Equipment
|
Machine
Hours
|
Machine
Rate
|
Cost
|
Drills
|
1.0
|
60.00
|
60.00
|
Compressor
|
1.0
|
55.00
|
55.00
|
Explosives
|
0.8
kg × $2.0/kg × 140 m3
|
224.00
|
|
339.00
|
Cost per cubic meter solid rock = $2.42
(c) To crush 225 tons per hour (2.6 tons/solid
cubic meter):
Equipment
|
Machine
Hours
|
Machine
Rate
|
Cost
|
Tractor
|
0.5
|
72.00
|
36.00
|
Loader
|
1.0
|
90.00
|
90.00
|
Crusher
|
1.0
|
90.00
|
90.00
|
Stacker
|
1.0
|
15.00
|
15.00
|
Generator
|
1.0
|
20.00
|
20.00
|
251.00
|
Cost per cubic meter solid rock = $2.90
(d) To load, transport, spread 20,000 cubic
meters of rock.
1 truck × 3 loads/hr × 20 tons/ld × m3/2.6 ton = 23 m3/hr
If 4 trucks are used:
Equipment
|
Machine
Hours
|
Machine
Rate
|
Cost
|
4 trucks
|
870
|
50.00
|
43,500
|
Loader
|
218
|
90.00
|
19,600
|
Tractor
|
218
|
72.00
|
15,700
|
Grader
|
30
|
60.00
|
1,800
|
80,600
|
Cost per cubic meter solid rock = $4.03
The total unit cost of per cubic meter of rock
spread on the road is
Activity
|
$/m3
solid |
$/m3
prism |
$/km |
Develop pit
|
0.10
|
0.08
|
74
|
Drill and blast
|
2.42
|
1.83
|
1833
|
Crush
|
2.90
|
2.20
|
2197
|
Load, transport, and spread
|
4.03
|
3.05
|
3053
|
9.45
|
7.16
|
7157
|
Equipment balancing plays an important role in
obtaining the minimum cost per cubic meter for surfacing. In some areas, market
prices for various types of surfacing may exist and tradeoffs between aggregate
cost, aggregate quality, and hauling distance will have to be evaluated. Since
surfacing is often expensive, a surveying crew is sometimes added to stake and
monitor the surfacing operation.
4.7 Drainage
Drainage costs vary widely with the type of
drainage being installed. The costs of drainage dips (water bars), culverts,
and bridges are often expressed as a cost per lineal foot which can then be
easily applied in road estimating. Local values for cost per lineal foot for
culverts and different types of bridges are generally available. If not,
constructed costs can be made by using time study data.
EXAMPLE:
A 45 cm culvert, 10 meters long, is being
installed. Experience indicates that a small backhoe and operator, and two
laborers can install 3 culverts per day. The culvert crew uses a flat-bed truck
to transport themselves and the pipe each day.
To install 3 culverts:
Equipment
|
Machine
Hours
|
Machine
Rate
|
Cost
|
Backhoe
|
6
|
60.00
|
360.00
|
Truck
|
9
|
12.00
|
108.88
|
Pipe Cost
|
30 meters × $15/meter
|
450.00
|
|
918.00
|
Cost per lineal meter of culvert = $30.60 per meter
Alternatively the cost could be stated as $306
per culvert or if there were an average of 4 culverts per km, then $1224 per
km.
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