Transcript Earthwork QTO

Stage 1 Preconstruction
Stage 2: Procurement
Conceptual Planning
Stage3: Construction
Design
Stage 4: Project Close-out
Earth Work QTO
• Soils that contain large amounts of clay
or organic materials or soils that are
contaminated are not suitable for fill.
• The ideal situation is that the amount
of usable cut equals the amount of fill.
Soil Volume Changes
• The soil will be in one of three conditions:
• Loose: after cut. Least unit mass and
maximum volume.
• Compacted: after the soil is dumped and
utilized as fill. Highest unit mass and least
volume.
• In situ or bank: In between unit mass and
volume.
Basic Terminology
• Volumetric Measures:
– Vl, Vb, Vc.
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In meters.
Density Measures:
Sw = Swell, density, percentage.
Sh = Shrinkage, density, percentage.
Wl = Weight per Loose Cubic Meter
Wb = Weight per bank Cubic Meter
Wc = Weight per compacted Cubic Meter
Common Formulation
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Wb = Wl ( 1 + Sw )
Wb = Wc ( 1- Sh )
Sw = (Wb / Wl ) – 1 = (Vl / Vb ) -1
Sh = 1- (Wb / Wc) {relationship between bank and compacted}
Swell Factor = Load Factor (LF) = Wl /Wb =
1/ (1+ Sw) = Vb / Vl
or
Vl = Vb (1+ Sw)
• Multiply the bank weight by the load factor to get loose wt.
Wl = LF * Wb
• Shrinkage Factor (SF) = Wb / Wc = 1 - Sh
• Vl = Vb / LF = Vc /(LF *SF)
• Vb = Vc / SF = Vc /(1 - Sh ) = (Wc * Vc ) / Wb
Then, Vl = Vb (1+ Sw) , Vc = Vb (1 - Sh ), and
LF = Vb / Vl = Wl /Wb = 1/(1+ Sw)
Example
A given material exhibits the following characteristics:
shrinkage = 10% swell = 15%. if 10,000 cubic meters (in
situ measure) of material are to be transported in trucks
capable of carrying 8 cubic meters, how many truck loads will
be required? How much volume will the material consume
when compacted?
• total number of truck loads = 10,000 (1 + 0.15) / 8 = 1438
• compacted volume = 10,000 (1 - 0.1) = 9000 cubic meters
If the material has an in situ (bank) unit mass of 1600
kilograms per cubic meter, how heavy is a single truck load?
•LF = Wl /Wb = 1/(1+ Sw) = 0.87, Wl = Wb (LF) then
•load per truck load = 8 X 1600 X 0.87 = 11,136 kilograms.
Grid (Borrow-Pit) Method
• Consists of overlaying the topographic
information of the site with a grid
• Grid size is determined according to the desired
level of accuracy.
• The volume on any part of the grid is equal to the
area of that part times the average height of cut or
fill over that part.
• Elevations at the corners of the grid are estimated
from the contour lines within a foot, usually.
40 ft grid “each square is 40 ft”
Before
After
• The height of cut or fill is computed at the corners
of the grid.
• A line at design elevation is drawn to separate the
cut from the fill. Points along that line are of zero
cut or fill (the daylight point).
• The volume of the cut or fill on the full squares is
computed first.
• The volume of the cut or fill on the rest of the
shapes is then computed individually.
• The total Volume of the cut and the total volume
of the fill are then computed
• Examples and homework
Double End-Area Method
• More Suitable for linear structures such as highways and
utility trenches.
• The method consists of cutting a series of parallel crosssections through the transverse direction of a given cut
and/or fill area.
• A minimum of two section cuts must be made on a
project. The number of the cross sections needed depends
on how rough the topography is.
• The existing and the proposed topography are drawn to
scale at each cross section. Section cuts could be cut, fill
or transition.
• A reference vertical line is chosen along the cross sections.
The distance of each inflection point from that reference
line is written on that point. Distances to the left of the
reference line are considered negative.
• A scale that represents the elevations is drawn to the right
of the cross sections.
• The area of the cross section is then computed using the
traverse area method. Assuming that the values along the X
axis are the distances from the reference line, and that the
values along the Y axis are the elevations, the area can be
computed as follows:
A = 1/2 { Xi (Yi+1 – Yi-1)}
• Volume between two end areas is then
computed. That volume is equal to the
average of the two areas times the distance
between them.
• The cut and fill volumes are computed and
summed up separately.
• Example
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