LECTURE 10

Earthworks

Theory and Practice of Surveying Review Innovations

 Problem 10.1

Compute the area of the following cross-sectional data given below.

25+100

7.85 0 8.45
+1.90 +3.20 +2.30

Theory and Practice of Surveying: Lecture 10

 Problem 10.1
25+100 𝑾idth of the road
left side centerline right side bisected by the centerline
x 7.85 0 8.45 𝑺ide slope
y 𝟎 run over rise ratio
+1.90 +3.20 +2.30
expressed as H:V
+𝟑. 𝟐𝟎 𝟖. 𝟒𝟓
𝟕. 𝟖𝟓
+𝟐. 𝟑𝟎
+𝟏. 𝟗𝟎

+𝟑. 𝟐𝟎
+𝟏. 𝟗𝟎 𝑺 +𝟐. 𝟑𝟎
𝑺

𝟏. 𝟗𝑺 𝑾/𝟐 𝑾/𝟐 𝟐. 𝟑𝑺
𝟕. 𝟖𝟓 𝟖. 𝟒𝟓
 Problem 10.1
𝟎
+𝟑. 𝟐𝟎 𝟖. 𝟒𝟓
𝟕. 𝟖𝟓
+𝟐. 𝟑𝟎
+𝟏. 𝟗𝟎

+𝟑. 𝟐𝟎
+𝟏. 𝟗𝟎 𝑺 +𝟐. 𝟑𝟎
𝑺

𝟏. 𝟗𝑺 𝑾/𝟐 𝑾/𝟐 𝟐. 𝟑𝑺
𝟕. 𝟖𝟓 𝟖. 𝟒𝟓
𝟕. 𝟖𝟓 = 𝟏. 𝟗𝑺 + 𝟎. 𝟓𝑾 𝑺 = 𝟏. 𝟓 = 𝟏. 𝟓: 𝟏
𝟖. 𝟒𝟓 = 𝟐. 𝟑𝑺 + 𝟎. 𝟓𝑾 𝑾 = 𝟏𝟎
Theory and Practice of Surveying: Lecture 10
 Problem 10.1

• Computation using triangles and trapezoids

• Computation using DMD/DPD – DPA method
• Computation using coordinate method

Theory and Practice of Surveying: Lecture 10

 Problem 10.1
𝟎
𝟖. 𝟒𝟓
𝟕. 𝟖𝟓 +𝟑. 𝟐𝟎
+𝟐. 𝟑𝟎
+𝟏. 𝟗𝟎

𝑨𝟐 𝑨𝟑
+𝟑. 𝟐𝟎
+𝟏. 𝟗𝟎 𝑺 +𝟐. 𝟑𝟎
𝑺 𝑨𝟒
𝑨𝟏

𝟐. 𝟖𝟓 𝟓 𝟓 𝟑. 𝟒𝟓
𝟕. 𝟖𝟓 𝟖. 𝟒𝟓

𝑨 = 𝑨𝟏 + 𝑨𝟐 + 𝑨𝟑 + 𝑨𝟒

𝟓(𝟏. 𝟗) 𝟑. 𝟐(𝟕. 𝟖𝟓) 𝟑. 𝟐(𝟖. 𝟒𝟓) 𝟓(𝟐. 𝟑𝟎)

𝑨= + + + = 𝟑𝟔. 𝟓𝟖𝒎𝟐
𝟐 𝟐 𝟐 𝟐
Computation using triangles and trapezoids

𝐿𝑒𝑣𝑒𝑙 𝑠𝑒𝑐𝑡𝑖𝑜𝑛

𝑪
𝒅𝑳
𝐹𝑖𝑣𝑒 𝐿𝑒𝑣𝑒𝑙 𝑆𝑒𝑐𝑡𝑖𝑜𝑛
𝑺

𝑩 𝒅𝑹
𝑨= 𝑩𝑪 + 𝑺𝑪𝟐 𝑪
𝒇𝑳 𝒇𝑹

𝑇ℎ𝑟𝑒𝑒 𝐿𝑒𝑣𝑒𝑙 𝑆𝑒𝑐𝑡𝑖𝑜𝑛 𝒅𝑹

𝒅𝑳 𝑩
𝒉𝑳 𝒉𝑹 𝟏
𝑨 = 𝑩𝑪 + 𝒇𝑹 𝒅𝑹 + 𝒇𝑳 𝒅𝑳
𝑪 𝟐

𝑩
𝟏 𝑩
𝑨 = 𝑪 𝒅𝑹 + 𝒅𝑳 + 𝒉𝑹 + 𝒉𝑳
𝟐 𝟐
Computation using triangles and trapezoids

𝑇ℎ𝑟𝑒𝑒 𝐿𝑒𝑣𝑒𝑙 𝑆𝑒𝑐𝑡𝑖𝑜𝑛 𝒅𝑹 𝐹𝑖𝑣𝑒 𝐿𝑒𝑣𝑒𝑙 𝑆𝑒𝑐𝑡𝑖𝑜𝑛

𝒅𝑳 𝒅𝑳
𝒉𝑳 𝒉𝑹

𝑪 𝒅𝑹
𝑪
𝒇𝑳 𝒇𝑹
𝑩
𝟏 𝑩 𝑩
𝑨 = 𝑪 𝒅 𝑹 + 𝒅𝑳 + 𝒉𝑹 + 𝒉𝑳
𝟐 𝟐 𝟏
𝑨 = 𝑩𝑪 + 𝒇𝑹 𝒅𝑹 + 𝒇𝑳 𝒅𝑳
𝟐
Computation using triangles and trapezoids

𝑇ℎ𝑟𝑒𝑒 𝐿𝑒𝑣𝑒𝑙 𝑆𝑒𝑐𝑡𝑖𝑜𝑛 𝐹𝑖𝑣𝑒 𝐿𝑒𝑣𝑒𝑙 𝑆𝑒𝑐𝑡𝑖𝑜𝑛

+ + +
+ 𝐴=
𝐴=
2 2
 Problem 10.1

Using the same side slope and road width data, compute the area of
the following cross-sectional data given below.

25+150
𝑥 0 10.70
+2.90 +2.60 𝑦

Theory and Practice of Surveying: Lecture 10

 Problem 10.1
𝒙 = 𝟗. 𝟑𝟓
𝟎 𝟏𝟎. 𝟕𝟎
+𝟐. 𝟗𝟎
+𝟐. 𝟔𝟎 𝒚 = 𝟑. 𝟖

+𝟐. 𝟗𝟎 +𝟐. 𝟔𝟎 𝟓. 𝟕𝟎
𝟓 𝟓 𝟏. 𝟓: 𝟏 𝒚=
𝟏. 𝟓: 𝟏 𝑺
𝟎 𝟎

𝟐. 𝟗𝑺 𝟓 𝟓 𝟓. 𝟕𝟎
𝟒. 𝟑𝟓 𝟏𝟎. 𝟕𝟎
𝒙

0 +10.70 +5 -5 -9.35 0
+2.60 +3.80 0 0 +2.90 +2.60
𝟏
𝑨 = | − 𝟓(𝟐. 𝟗) −𝟗. 𝟑𝟓(𝟐. 𝟔) −𝟐. 𝟔(𝟏𝟎. 𝟕) −𝟑. 𝟖(𝟓)| = 𝟒𝟐. 𝟖𝟏𝟓 𝒎𝟐
𝟐
Computation using triangles and trapezoids

𝑥 0 10.70
+2.90 +2.60 𝑦

+ +
𝐴=
2
S = 1.5
10.70+9.35 2.6 +2.90 5 +3.8(5)
W =10 A= = 42.815
2
x = 5 + 1.5(2.9) = 9.35
1
y= 10.70 − 5 = 3.8
1.5
Volume Computation

• End area method

𝑨𝟐
𝑳
𝑳
𝑽𝒆 = 𝑨𝟏 + 𝑨𝟐
𝟐

𝑨𝟏

𝑨𝟏 + 𝑨𝒏
𝑽𝒆 = 𝑳 + 𝑨𝟐 + 𝑨𝟑 + ⋯ + 𝑨𝒏−𝟏
𝟐

Theory and Practice of Surveying: Lecture 10

Volume Computation
𝑨𝟐
𝑳
• Prismoidal Formula
𝑨𝒎

𝑳 𝑨𝟏
𝑽𝒑 = 𝑨𝟏 + 𝟒𝑨𝒎 + 𝑨𝟐
𝟔

NOTE: 𝐴𝑚 is determined by averaging the corresponding linear dimensions of the

end sections and NOT by averaging the end areas 𝐴1 and 𝐴2 .
𝑳
𝑽𝒑 = 𝑨𝟏 + 𝑨𝒏 + 𝟐𝚺𝑨𝒐𝒅𝒅 + 𝟒𝚺𝑨𝒆𝒗𝒆𝒏
𝟑
Theory and Practice of Surveying: Lecture 10
Volume Computation

• Prismoidal Correction 𝒄𝟏

𝑪𝒑 = 𝑽𝒆 − 𝑽𝑷 𝒅𝟏

𝑳
𝑪𝒑 = 𝒄𝟏 − 𝒄𝟐 𝒅𝟏 − 𝒅𝟐
𝟏𝟐
𝒄𝟐

𝒅𝟐
Theory and Practice of Surveying: Lecture 10
 Problem 10.1

Find the volume of excavation between the two stations using:

a. End Area Method
b. Prismoidal formula
c. Calculate the Prismoidal Correction

Theory and Practice of Surveying: Lecture 10

 Problem 10.1

Find the volume of excavation between the two stations using End area
method.
𝑳
𝑽𝒆 = 𝑨𝟏 + 𝑨𝟐
𝟐

𝑳 = 𝟓𝟎

𝑨𝟏 = 𝟑𝟔. 𝟓𝟖

𝑨𝟐 = 𝟒𝟐. 𝟖𝟏𝟓

𝑽𝒆 = 𝟏𝟗𝟖𝟒. 𝟖𝟕𝟓 𝒎𝟑

Theory and Practice of Surveying: Lecture 10

 Problem 10.1
Find the volume of excavation between the two stations using
Prismoidal formula.

𝑳
𝑽𝒑 = 𝑨𝟏 + 𝟒𝑨𝒎 + 𝑨𝟐
𝟔
𝑳 = 𝟓𝟎

𝑨𝟏 = 𝟑𝟔. 𝟓𝟖 𝑨𝒎 : 𝟎
𝟗. 𝟓𝟕𝟓
𝑨𝟐 = 𝟒𝟐. 𝟖𝟏𝟓 𝟖. 𝟔𝟎 +𝟐. 𝟗𝟎
+𝟑. 𝟎𝟓
+𝟐. 𝟒𝟎
𝑨𝒎 = 𝟑𝟗. 𝟗𝟕𝟖𝟕𝟓 𝒎𝟐

𝑽𝒑 = 𝟏𝟗𝟗𝟒. 𝟐𝟓 𝒎𝟑
𝟓 𝟓
 Problem 10.1
Compute the Prismoidal correction.
𝟎
𝟖. 𝟒𝟓
𝟕. 𝟖𝟓 +𝟑. 𝟐𝟎
+𝟐. 𝟑𝟎
𝑪𝒑 = 𝑽𝒆 − 𝑽𝒑 +𝟏. 𝟗𝟎
𝟑. 𝟐𝟎
𝑪𝒑 = −𝟗. 𝟑𝟕𝟓 𝒎𝟑
𝟏𝟔. 𝟑𝟎
𝑳
𝑪𝒑 = 𝒄𝟏 − 𝒄𝟐 𝒅𝟏 − 𝒅𝟐
𝟏𝟐 𝟗. 𝟑𝟓 𝟎 𝟏𝟎. 𝟕𝟎
+𝟐. 𝟗𝟎 +𝟐. 𝟔𝟎 𝟑. 𝟖𝟎
𝑳 = 𝟓𝟎

𝑪𝒑 = −𝟗. 𝟑𝟕𝟓 𝒎𝟑 𝟐. 𝟔𝟎

𝟐𝟎. 𝟎𝟓
 Problem 10.2

Given the following section of an earthworks for a proposed road

construction on a hilly portion of the route. The width of the road base
for cut is 6m for allowance of drainage canals and 5m for fill. Side
slopes for cut is 1:1 and for fill is 1.5:1.
3.7 0 𝑥
-0.8 0 +1.8
a. Compute the value of x.
b. Compute the area in fill.
c. Compute the area in cut.
Theory and Practice of Surveying: Lecture 10
 Problem 10.2
Given the following section of an earthworks for a proposed road construction on a hilly portion of
the route. The width of the road base for cut is 6m for allowance of drainage canals and 5m for fill.
Side slopes for cut is 1:1 and for fill is 1.5:1.
a. Compute the value of x.
3.7 0 𝑥
-0.8 0 +1.8 𝒙 = 𝟒. 𝟖
+𝟏. 𝟖
𝟎
𝟏. 𝟐 𝟐. 𝟓 𝟎 𝟏: 𝟏 𝟏. 𝟖

𝟎. 𝟖
𝟑. 𝟕 𝟑 𝟏. 𝟖
−𝟎. 𝟖
Theory and Practice of Surveying: Lecture 10
 Problem 10.2
Given the following section of an earthworks for a proposed road construction on a hilly portion of
the route. The width of the road base for cut is 6m for allowance of drainage canals and 5m for fill.
Side slopes for cut is 1:1 and for fill is 1.5:1.
𝟒. 𝟖
b. Compute the area in fill.
+𝟏. 𝟖
𝟎
𝟏. 𝟐 𝟐. 𝟓 𝟎 𝟏: 𝟏 𝟏. 𝟖

𝟎. 𝟖
𝟑. 𝟕 𝟑 𝟏. 𝟖
−𝟎. 𝟖 𝟏
𝑨𝒇𝒊𝒍𝒍 = × 𝟐. 𝟓 × 𝟎. 𝟖
𝟐

𝑨𝒇𝒊𝒍𝒍 = 𝟏. 𝟎 𝒎𝟐
Theory and Practice of Surveying: Lecture 10
 Problem 10.2
Given the following section of an earthworks for a proposed road construction on a hilly portion of
the route. The width of the road base for cut is 6m for allowance of drainage canals and 5m for fill.
Side slopes for cut is 1:1 and for fill is 1.5:1.
𝟒. 𝟖
b. Compute the area in cut.
+𝟏. 𝟖
𝟎
𝟏. 𝟐 𝟐. 𝟓 𝟎 𝟏. 𝟖

𝟎. 𝟖
𝟑. 𝟕 𝟑 𝟏. 𝟖
−𝟎. 𝟖 𝟏
𝑨𝒇𝒊𝒍𝒍 = × 𝟑 × 𝟏. 𝟖
𝟐

𝑨𝒇𝒊𝒍𝒍 = 𝟐. 𝟕 𝒎𝟐
Theory and Practice of Surveying: Lecture 10
 Problem 10.3

From station 10+100 with center height of 3.4 m in cut has a ground
line which makes uniform slope of -4% to station 10+180 whose center
height in fill is 1.2 m. Assume both sections to be trapezoidal having a
width of roadway of 8.0 and side slope of 1:1 for both cut and fill. If the
shrinkage factor is 1.3, compute the difference in volume of cut and fill
in cubic m.

Theory and Practice of Surveying: Lecture 10

 Problem 10.3
From station 10+100 with center height of 3.4 m in cut has a ground line which makes uniform slope of -4% to
station 10+180 whose center height in fill is 1.2 m. Assume both sections to be trapezoidal having a width of
roadway of 8.0 and side slope of 1:1 for both cut and fill. If the shrinkage factor is 1.3, compute the difference in
volume of cut and fill in cubic m.
𝟏𝟎 + 𝟏𝟎𝟎 𝟏𝟎 + 𝟏𝟖𝟎
𝟑. 𝟒 𝟑. 𝟒 𝟖 𝟑. 𝟒 𝟖

𝟏𝟎 + 𝟏𝟖𝟎 𝟑. 𝟒 𝟏. 𝟐
𝟏𝟎 + 𝟏𝟎𝟎
−𝟒%
𝑳 𝟖 𝟏. 𝟐 𝟖 𝟏. 𝟐
𝟏. 𝟐
𝑨 = 𝟑𝟖. 𝟕𝟔 𝒎𝟐 𝑨 = 𝟏𝟏. 𝟎𝟒 𝒎𝟐

𝑳 𝟖𝟎 − 𝑳 𝟓𝟗. 𝟏𝟑𝟎
= 𝑽𝒄𝒖𝒕 = (𝟑𝟖. 𝟕𝟔 + 𝟎) = 𝟏𝟏𝟒𝟓. 𝟗𝟒𝟖 𝒎𝟑
𝟑. 𝟒 𝟏. 𝟐 𝟐
𝑳 = 𝟓𝟗. 𝟏𝟑𝟎 𝟐𝟎. 𝟖𝟕𝟎
𝑽𝒇𝒊𝒍𝒍 = (𝟏𝟏. 𝟎𝟒 + 𝟎) = 𝟏𝟏𝟓. 𝟐𝟎𝟎 𝒎𝟑
𝟐
 Problem 10.3
From station 10+100 with center height of 3.4 m in cut has a ground line which makes uniform slope of -4% to
station 10+180 whose center height in fill is 1.2 m. Assume both sections to be trapezoidal having a width of
roadway of 8.0 and side slope of 1:1 for both cut and fill. If the shrinkage factor is 1.3, compute the difference in
volume of cut and fill in cubic m.
𝟏𝟎 + 𝟏𝟎𝟎 𝟏𝟎 + 𝟏𝟖𝟎
𝟑. 𝟒 𝟑. 𝟒 𝟖 𝟑. 𝟒 𝟖

𝟏𝟎 + 𝟏𝟖𝟎 𝟑. 𝟒 𝟏. 𝟐
𝟏𝟎 + 𝟏𝟎𝟎
−𝟒%
𝑳 𝟖 𝟏. 𝟐 𝟖 𝟏. 𝟐
𝟏. 𝟐
𝑨 = 𝟑𝟖. 𝟕𝟔 𝒎𝟐 𝑨 = 𝟏𝟏. 𝟎𝟒 𝒎𝟐

𝑳 𝟖𝟎 − 𝑳 𝑽𝒄𝒖𝒕 = 𝟏𝟏𝟒𝟓. 𝟗𝟒𝟖 𝒎𝟑 𝑽′𝒇𝒊𝒍𝒍 = 𝟏𝟏𝟓. 𝟐𝟎𝟎 × 𝟏. 𝟑

=
𝟑. 𝟒 𝟏. 𝟐
𝑽𝒇𝒊𝒍𝒍 = 𝟏𝟏𝟓. 𝟐𝟎𝟎 𝒎𝟑 𝑽′𝒇𝒊𝒍𝒍 = 𝟏𝟒𝟗. 𝟕𝟔𝟎 𝒎𝟑
𝑳 = 𝟓𝟗. 𝟏𝟑𝟎 ∆𝑽 = 𝟗𝟗𝟔. 𝟏𝟖𝟖 𝒎𝟑
𝑺𝒉𝒓𝒊𝒏𝒌𝒂𝒈𝒆 = 𝟏. 𝟑
 Problem 10.4

A square piece of land 60 m x 60 m is to be leveled down to 5 m above

MSL. To determine the volume of earth to be removed by the Borrow
Pit method, the land is divided into 9 squares as shown in the figure.
The grid elevations are as given. Find the volume of cut.
29.8 27.3 25.2 28.3

26.5 24.3 26.9 23.3

24.2 21.3 22.9 20.5

21.2 18.5 17.8 16.5

Theory and Practice of Surveying: Lecture 10

 Problem 10.4
A square piece of land 60 m x 60 m is to be leveled down to 5 m above MSL. To determine the
volume of earth to be removed by the Borrow Pit method, the land is divided into 9 squares as shown
in the figure. The grid elevations are as given. Find the volume of cut.
29.8 27.3 25.2 28.3
𝑨
24.8 22.3 20.2 23.3
𝑽= 𝚺𝒉𝟏 + 𝟐𝚺𝒉𝟐 + 𝟑𝚺𝒉𝟑 + 𝟒𝚺𝒉𝟒
𝟒

𝑨 = 𝟐𝟎 × 𝟐𝟎 = 𝟒𝟎𝟎 𝒎𝟐
26.5 24.3 26.9 23.3

𝚺𝒉𝟏 = 𝟐𝟒. 𝟖 + 𝟐𝟑. 𝟑 + 𝟏𝟔. 𝟐 + 𝟏𝟏. 𝟓

21.5 19.3 21.9 18.3
= 𝟕𝟓. 𝟖
24.2 21.3 22.9 20.5
𝚺𝒉𝟐 = 𝟐𝟐. 𝟑 + 𝟐𝟎. 𝟐 + 𝟏𝟖. 𝟑 + 𝟏𝟓. 𝟓
19.2 16.3 17.9 15.5 +𝟏𝟐. 𝟖 + 𝟏𝟑. 𝟓 + 𝟏𝟗. 𝟐 + 𝟐𝟏. 𝟓
= 𝟏𝟒𝟑. 𝟑
21.2 18.5 17.8 16.5

16.2 13.5 12.8 11.5

 Problem 10.4
A square piece of land 60 m x 60 m is to be leveled down to 5 m above MSL. To determine the volume of earth
to be removed by the Borrow Pit method, the land is divided into 9 squares as shown in the figure. The grid
elevations are as given. Find the volume of cut.
29.8 27.3 25.2 28.3 𝑨
𝑽= 𝚺𝒉𝟏 + 𝟐𝚺𝒉𝟐 + 𝟑𝚺𝒉𝟑 + 𝟒𝚺𝒉𝟒
24.8 22.3 20.2 23.3 𝟒
𝑨 = 𝟐𝟎 × 𝟐𝟎 = 𝟒𝟎𝟎 𝒎𝟐
26.5 24.3 26.9 23.3 𝚺𝒉𝟏 = 𝟕𝟓. 𝟖
21.5 19.3 21.9 18.3 𝚺𝒉𝟐 = 𝟏𝟒𝟑. 𝟑
𝚺𝒉𝟑 = 𝟎
24.2 21.3 22.9 20.5
𝚺𝒉𝟒 = 𝟏𝟗. 𝟑 + 𝟐𝟏. 𝟗 + 𝟏𝟔. 𝟑 + 𝟏𝟕. 𝟗
19.2 16.3 17.9 15.5 = 𝟕𝟓. 𝟒

21.2 18.5 17.8 16.5 𝑽 = 𝟔𝟔𝟒𝟎𝟎 𝒎𝟑

16.2 13.5 12.8 11.5
Distribution Analysis

• Excavation refers to the removal of volume of earth from the grade.

• Embankment refers to the addition of volume of earth from the
grade.

These terms are similar to cut and fill, which are used for cross-
sectional areas.

Theory and Practice of Surveying: Lecture 10

Mass-Haul Diagram

grade line

mass ordinate

distance

Theory and Practice of Surveying: Lecture 10

Mass-Haul Diagram

grade line

excavation

distance

embankment

• Mass-haul diagram refers to the accumulation of excavation or

embankment. Helpful for determining costs of operations.
• x axis is meter stations, y axis is cumulative volume.
 grade line

excavation

distance

embankment

• Upward slope represents excavation, downward slope represents

embankment.
• Extrema of the curve are the transition points of operations.
 grade line

excavation

distance

embankment

• An ordinate of 0 represents the balancing point, where the value of

cut is equal to the value of fill.
• Areas between balancing lines represents a haul.
 grade line

excavation

distance

borrow
embankment

• An excess of embankment is called a borrow.

• An excess of excavation is called a waste.
𝑳𝒐𝑶
• Free haul distance is a fixed distance
where hauling has no cost.
• Overhaul is the distance beyond the
free haul distance.
• The Limit of Economical Haul is the
𝑭𝑯𝑫 distance where the cost of
excavation and the cost of
𝑳𝑬𝑯 embankment are the same.
• Anything beyond the limit of
economical haul is just
waste/borrow.
𝑪𝒃 : cost of borrow per unit volume
𝑪𝒃 𝑪𝒉 : cost of haul per station
𝑳𝑬𝑯 = 𝑪 + 𝑭𝑯𝑫
𝑪𝒉
𝑪: length per station
 Problem 10.5

The earthworks data of a proposed highway is shown on the tabulated

data. Assume the ground surface to be uniformly sloping.
Stationing of limits of economical haul = 2+498.03 to 2+948.03
Stationing of limits of free haul = 2+713.12 to 2+763.12

STATION AREA
a. Compute the overhaul volume. CUT FILL
b. Compute the volume of waste. 2+440 50
c. Compute the volume of borrow. 2+740 Balancing point
3+040 70

Theory and Practice of Surveying: Lecture 10

 The earthworks data of a proposed highway is shown on the tabulated data. Assume the ground
surface to be uniformly sloping
Stationing of limits of economical haul = 2+498.03 to 2+948.03
Stationing of limits of free haul = 2+713.12 to 2+763.12
𝒉𝟏 𝟓𝟎
a. Compute the overhaul volume. =
𝟐𝟒𝟏. 𝟗𝟕 𝟑𝟎𝟎
𝒉𝟏 2+713.12 2+763.12
𝒉𝟏 = 𝟒𝟎. 𝟑𝟑
𝟓𝟎 𝑭𝑯𝑫
𝑽𝒐𝒗𝒉 𝒉𝟐 𝟓𝟎
=
𝒉𝟐 𝟐𝟔. 𝟖𝟖 𝟑𝟎𝟎
𝟕𝟎
𝒉𝟐 = 𝟒. 𝟒𝟖
𝑳𝑬𝑯 𝟐𝟏𝟓. 𝟎𝟗(𝟒𝟎. 𝟑𝟑 + 𝟒. 𝟒𝟖)
𝑽𝒐𝒗𝒉 =
𝟐
2+440 2+498.03 2+740 2+948.03 3+040
𝑽𝒐𝒗𝒉 = 𝟒𝟖𝟏𝟖. 𝟗𝟏 𝒎𝟑
 The earthworks data of a proposed highway is shown on the tabulated data. Assume the ground
surface to be uniformly sloping
Stationing of limits of economical haul = 2+498.03 to 2+948.03
Stationing of limits of free haul = 2+713.12 to 2+763.12

b. Compute the volume of waste.

𝟓𝟖. 𝟎𝟑(𝟒𝟎. 𝟑𝟑 + 𝟓𝟎)
𝒉𝟏 2+713.12 2+763.12 𝑽𝒘𝒂𝒔𝒕𝒆 =
𝟐
𝟓𝟎 𝑽 𝑭𝑯𝑫
𝒘𝒂𝒔𝒕𝒆

𝑽𝒘𝒂𝒔𝒕𝒆 = 𝟐𝟔𝟐𝟎. 𝟖𝟖 𝒎𝟑
𝟕𝟎

𝑳𝑬𝑯

2+440 2+498.03 2+740 2+948.03 3+040

 The earthworks data of a proposed highway is shown on the tabulated data. Assume the ground
surface to be uniformly sloping
Stationing of limits of economical haul = 2+498.03 to 2+948.03
Stationing of limits of free haul = 2+713.12 to 2+763.12

c. Compute the volume of borrow.

2+713.12 2+763.12 𝒉𝟑 𝟕𝟎
=
𝟓𝟎 𝑭𝑯𝑫 𝟐𝟎𝟖. 𝟎𝟑 𝟑𝟎𝟎
𝑽𝒃𝒐𝒓𝒓𝒐𝒘
𝒉𝟑 = 𝟒𝟖. 𝟓𝟒
𝒉𝟑
𝟕𝟎
𝟗𝟏. 𝟗𝟕(𝟒𝟖. 𝟓𝟒 + 𝟕𝟎)
𝑽𝒃𝒐𝒓𝒓𝒐𝒘 =
𝑳𝑬𝑯 𝟐

2+440 2+498.03 2+740 2+948.03 3+040 𝑽𝒃𝒐𝒓𝒓𝒐𝒘 = 𝟓𝟒𝟓𝟏. 𝟎𝟖 𝒎𝟑

 Problem 10.6

The cost of borrow per cubic m is P500 and the cost of haul per meter
station is P25. Cost of excavation is approximately P650 per cubic m.
The free haul distance is 50 m long, and the length of overhaul is equal
to 201.40 m. If the mass ordinate of the initial point of the free haul
distance is +800 cubic m and the mass ordinates of the summit mass
diagram from 10+000 to 10+600 are -60 cubic m and -140 cubic m,
respectively.
a. Compute the length of economical haul.
b. Compute the mass ordinate of the initial point of the limit of
economical haul is the total cost of hauling is P171190.
c. Compute the total cost of waste.

Theory and Practice of Surveying: Lecture 10

 Problem 10.6
The cost of borrow per cubic m is P500 and the cost of haul per meter station is P25. Cost of
excavation is approximately P650 per cubic m. The free haul distance is 50 m long, and the length of
overhaul is equal to 201.40 m. If the mass ordinate of the initial point of the free haul distance is
+800 cubic m and the mass ordinates of the summit mass diagram from 10+000 to 10+600 are -60
cubic m and -140 cubic m, respectively.
a. Compute the length of economical haul.
𝑪𝒃 𝟓𝟎𝟎
𝑳𝑬𝑯 = 𝑪 + 𝑭𝑯𝑫 𝑳𝑬𝑯 = 𝟐𝟎 + 𝟓𝟎
𝑪𝒉 𝟐𝟓
𝑪𝒃 = 𝟓𝟎𝟎 𝑳𝑬𝑯 = 𝟒𝟓𝟎 𝒎

𝑪𝒉 = 𝟐𝟓
𝑭𝑯𝑫 = 𝟓𝟎
𝑪 = 𝟐𝟎
Theory and Practice of Surveying: Lecture 10
 Problem 10.6
The cost of borrow per cubic m is P500 and the cost of haul per meter station is P25. Cost of
excavation is approximately P650 per cubic m. The free haul distance is 50 m long, and the length of
overhaul is equal to 201.40 m. If the mass ordinate of the initial point of the free haul distance is
+800 cubic m and the mass ordinates of the summit mass diagram from 10+000 to 10+600 are -60
cubic m and -140 cubic m, respectively.
b. Compute the mass ordinate of the initial point of the limit of economical haul is
the total cost of hauling is P171190.
𝑪𝒉 × 𝑳𝒐𝒗𝒆𝒓𝒉𝒂𝒖𝒍 × 𝑽𝒐𝒗𝒆𝒓𝒉𝒂𝒖𝒍
𝑪𝒉,𝒕𝒐𝒕𝒂𝒍 =
𝑪
𝑪𝒉,𝒕𝒐𝒕𝒂𝒍 = 𝟏𝟕𝟏𝟏𝟗𝟎
𝟐𝟓 × 𝟐𝟎𝟏. 𝟒𝟎 × 𝑽𝒐𝒗𝒆𝒓𝒉𝒂𝒖𝒍
𝟏𝟕𝟏𝟏𝟗𝟎 =
𝑪𝒉 = 𝟐𝟓 𝟐𝟎
𝑳𝒐𝒗𝒆𝒓𝒉𝒂𝒖𝒍 = 𝟐𝟎𝟏. 𝟒𝟎 𝑽𝒐𝒗𝒆𝒓𝒉𝒂𝒖𝒍 = 𝟔𝟖𝟎 𝒎𝟑

𝑪 = 𝟐𝟎
Theory and Practice of Surveying: Lecture 10
 Problem 10.6
The cost of borrow per cubic m is P500 and the cost of haul per meter station is P25. Cost of excavation is
approximately P650 per cubic m. The free haul distance is 50 m long, and the length of overhaul is equal to
201.40 m. If the mass ordinate of the initial point of the free haul distance is +800 cubic m and the mass
ordinates of the summit mass diagram from 10+000 to 10+600 are -60 cubic m and -140 cubic m, respectively.
b. Compute the mass ordinate of the initial point of the limit of
economical haul is the total cost of hauling is P171190.

800 FHD 𝑽𝒐𝒗𝒆𝒓𝒉𝒂𝒖𝒍 = 𝟔𝟖𝟎 𝒎𝟑

680 𝒙 = 𝟖𝟎𝟎 − 𝟔𝟖𝟎

x
LEH 𝒙 = 𝟏𝟐𝟎 𝒎𝟑

-60 10+000

-140 10+600
 Problem 10.6
The cost of borrow per cubic m is P500 and the cost of haul per meter station is P25. Cost of
excavation is approximately P650 per cubic m. The free haul distance is 50 m long, and the length of
overhaul is equal to 201.40 m. If the mass ordinate of the initial point of the free haul distance is
+800 cubic m and the mass ordinates of the summit mass diagram from 10+000 to 10+600 are -60
cubic m and -140 cubic m, respectively.

c. Compute the total cost of waste.

800 FHD 𝒆𝒙𝒄𝒂𝒗𝒂𝒕𝒊𝒐𝒏 𝒗𝒐𝒍𝒖𝒎𝒆 = 𝟏𝟐𝟎 − (−𝟔𝟎)

𝒆𝒙𝒄𝒂𝒗𝒂𝒕𝒊𝒐𝒏 𝒗𝒐𝒍𝒖𝒎𝒆 = 𝟏𝟖𝟎

120 LEH
𝑪𝒘,𝒕𝒐𝒕𝒂𝒍 = 𝟔𝟓𝟎(𝟏𝟖𝟎)

-60 10+000 𝑪𝒘,𝒕𝒐𝒕𝒂𝒍 = 𝑷 𝟏𝟏𝟕𝟎𝟎𝟎

-140 10+600
 Lecture 10 DONE!

CONGRATS, Engineer!

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