What is the volume of earthwork for constructing a tank that is excavated in the level ground to a depth of 4 m ? The top of the tank is rectangular in shape having an area of 50 m × 40 m and the side slope of the tank is 2: 1 (horizontal: vertical).

Option 2 : 6688 m^{3}

__Concept____:__

The prismoidal formula is also known as Simpson’s rule.

a) Trapezoidal Formula:

Volume (v) of earthwork between a number of sections having areas A1, A2,…, An spaced at a constant distance d.

\({\rm{V}} = {\rm{d}}\left[ {\frac{{{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}}}{2} + {{\rm{A}}_2} + {{\rm{A}}_3} + \ldots + {{\rm{A}}_{{\rm{n}} - 1{\rm{\;}}}}} \right]\)

b) Simpson’s formulae:

Volume (v) of the earthwork between a number of sections having area A1, A2 … An spaced at constant distance d apart is

\({\rm{V}} = \frac{{\rm{d}}}{3}[\left( {{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}} \right) + 4\left( {{{\rm{A}}_2} + {{\rm{A}}_4} + \ldots + {{\rm{A}}_{{\rm{n}} - 1}}} \right) + 2\left( {{{\rm{A}}_3} + {{\rm{A}}_5} + \ldots + {{\rm{A}}_{{\rm{n}} - 2}}} \right)\)

c) Simple prismoidal rule:

\(V = \frac{D}{3}\left[ {({A_1} + 4Am + {A_2})} \right]\)

**Calculation:**

Given, A_{1} = 50 × 40 = 2000 m^{2}, As the side slope is given 2:1 i.e. H:V

So for a depth of 1 m, there is a change of 2 m in a Horizontal Direction.

So at 4 m vertical depth

Bottom Dimension is ( 50 - 8 ) = 42 m & (40 - 8) = 32 m

**∴ The bottom area is 42 m × 32 m = 1344 m ^{2}**

Mean area (A m) = (A_{1 }+ A_{2})/2 = (2000 + 1344) /2 = 1672 m^{2}

According to simple Trapezoidal rule for volume,

^{\(V = \frac{D}{2}\left[ {({A_1} + 2Am + {A_2})} \right]\)}

∴\(V = \frac{2}{2}\left[ {(2000 + 2\times (1672) + 1344)} \right]\) = **6688 m ^{3}**

__ Note:__ For calculating A

Option 3 : Subtractive

__Explanation:__

The volume of earthwork by trapezoidal method = V_{1}

V_{1} = \(common\: distance\left \{ \frac{First\: area + Last\: area}{2}+the \:sum \:of\: remaining \:area \right \}\)

The volume of earthwork by prismoidal formula = V_{2}

V_{2} = \(=\frac{Common\: distance}{3}\left \{ First\: area+ Last\: area + 2(Sum\: of odd\: area) + 4(Sum\: of even\: area)\right \}\)

**Prismoidal correction:**

- The volume by the prismoidal formula is more accurate than any other method
- But the trapezoidal method is more often used for calculating the volume of earthwork in the field.
- The difference between the volume computed by the trapezoidal formula and the prismoidal formula is known as a
**prismoidal correction.** **Since the trapezoidal formula always overestimates the volume, the prismoidal correction is always subtractive in nature is usually more than calculated by the prismoidal formula, therefore the prismoidal correction is generally subtractive.**- Volume by prismoidal formula = volume by the trapezoidal formula - prismoidal correction

**Prismoidal correction (C _{P})**

**\(C_{P}=\frac{DS}{6}\left \{ d- d_{1}\right \}^{2}\)**

Where, D = Distance between the sections, S (Horizontal) : 1 (Vertical) = Side slope, d and d_{1} are the depth of earthwork at the centerline

The areas enclosed by the contours in a lake are as follows:

Contour (m) |
270 |
275 |
280 |
285 |
290 |

Area (m |
50 |
200 |
400 |
600 |
750 |

The volume of water between the contours 270 m and 290 m by trapezoidal formula is _______.

Option 2 : 8000 m^{3}

__Concept:__

**Trapezoidal Formula:**

Volume (v) of earthwork between a number of sections having areas A_{1}, A_{2}, …, A_{n} spaced at a constant distance d.

\({\rm{V}} = {\rm{d}}\left[ {\frac{{{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}}}{2} + {{\rm{A}}_2} + {{\rm{A}}_3} + \ldots + {{\rm{A}}_{{\rm{n}} - 1}}} \right]\)

**Simpson’s Formula:**

Volume (v) of the earthwork between a number of sections having area A_{1}, A_{2}, …, A_{n} spaced at constant distance d apart is

\({\rm{V}} = \frac{{\rm{d}}}{3}\left[ {\left( {{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}} \right) + 4\left( {{{\rm{A}}_2} + {{\rm{A}}_4} + \ldots {{\rm{A}}_{{\rm{n}} - 1}}} \right) + 2\left( {{{\rm{A}}_3} + {{\rm{A}}_5} + \ldots + {{\rm{A}}_{{\rm{n}} - 2}}} \right)} \right]\)

__Calculation:__

The areas enclosed by the contours in a lake are as follows:

Contour (m) |
270 |
275 |
280 |
285 |
290 |

Area (m |
50 |
200 |
400 |
600 |
750 |

Given contour interval (d) = 275-270 = 5 m

**So using trapezoidal formula:**

\({\rm{V}} = 5\left( {\left( {\frac{{50 + 750}}{2}} \right) + 200 + 400 + 600} \right) = 5 \times \left( {400 + 200 + 400 + 600} \right) = 8000{\rm{\;}}{{\rm{m}}^3}\)

Option 4 : odd

**Explanation:**

**Simpson's rule:**

This rule is based on the assumption that the figures are **trapezoids.**

In order to apply Simpson's rule, the area must be divided in even number i.e., the **number of offsets must be odd** i.e., n term in the last offset 'O_{n}' should be odd.

The area is given by Simpson's rule:

\(Area = \frac{d}{3}\left[ {({O_1} + {O_n}) + 4({O_2} + {O_4} + ........ + {O_{n - 1}}) + 2({O_3} + {O_5} + ......{O_{n - 2}})} \right]\)

where O_{1}, O_{2}, O_{3}, .........O_{n }is the offset

__Important Points__

- In case

A road embankment 10 m wide at the formation level with side slopes 2:1 and with an average height of 5 m is constructed with an average gradient of 1:40 from the contour 220 m to 280 m. Find the volume of earthwork.

Option 4 : 2,40,000 m^{3}

**Concept:**

**Gradient: **

A gradient is the rate of rise or falls along the length of the road with respect to horizontal. It is expressed as ‘1' vertical unit to 'N' horizontal units.

\(Tan\ \theta \ =\frac{{\bf{h}}}{{\bf{l}}}\)

\(\frac{{\bf{h}}}{{\bf{l}}} = \frac{1}{{N}}\)

**Area of trapezoidal:**

According to the trapezoid area formula, the area of a trapezoid is equal to half the product of the height and the sum of the two bases.

Area = ½ x (Sum of parallel sides) x (perpendicular distance between the parallel sides).

**Calculation:**

Road embankment = 10 m

Average height = 5 m

Difference in elevation(h) = 280 - 220 = 60 m

Average gradient = \( \frac{1}{{40}}\)

\(\frac{{\bf{h}}}{{\bf{l}}} = \frac{1}{{40}}\)

\(\frac{{\bf{60}}}{{\bf{l}}} = \frac{1}{{40}}\)

L = 2400 m

Average cross-sectional area(A) = \(% MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qadaWcaaWdaeaapeGaaGymaaWdaeaapeGaaGOmaaaacqGHxdaTdaqa % daWdaeaapeGaaGymaiaaicdacqGHRaWkcaaIZaGaaGimaaGaayjkai % aawMcaaiabgEna0kaaiwdaaaa!423F! \frac{1}{2} \times \left( {10 + 30} \right) \times 5\)\(\frac{1}{2} \times \left( {10 + 30} \right) \times 5\)

A = 100 m^{2}

Volume of earthwork = A × L

Volume of earthwork = 100 × 2400

**∴ Volume of earthwork = 2,40,000 m ^{3}**

Option 4 : 2085 m3

**Concept:**

**Volume of Sloped earthwork (V) = (bd + sd2) × L **

Where, B = Width, d = Depth and S = Side slope of the cross-section

__Calculation:__

Given,

slope = s: 1 = 3: 1

b = 10 m

L = 120 m

d_{1} = 1.5 m and d_{2} = 1 m

Volume of Sloped earthwork (V) = (bd + sd2) × L

\(V = \;\frac{1}{2}\left[ {(b{d_1}\; + {\rm{ }}s{d_1}^2) + (b{d_2}\; + {\rm{ }}s{d_2}^2)} \right] \times 120\)

^{\(V = \;\frac{1}{2}\left[ {(10 \times 1.5\; + {\rm{ }}3 \times {1.5^2}) + (10 \times 1\; + {\rm{ }}3 \times {{1}^2})} \right] \times 120 = 2085{m^3}\)}

**The approximate quantity of earthwork = 2085 m ^{3}**

Option 3 : B × d + S × d2

**Explanation:**

Given,

B = Formation width

d = Height of the embankment

S:1 = Side slope of the embankment (H: V)

The area of embankment = The area of the trapezium

Area of Trapezium = \(\frac{1}{2}\) × (Sum of Lengths of parallel sides) × Height

Area of Embankment \(= \frac{1}{2}\left[ {B + \left( {B + 2sd} \right)} \right] \times d\)

Option 2 : 122.6 m^{3}

Volume of tank by prismoidal formula is given by:

\(V = \frac{d}{3}\left( {{A_1} + 4{A_m} + {A_2}} \right)\)

d = step size = 4 m

A_{m} = Mean area of the tank

A_{1} = 6 × 4 = 24 m^{2}

A_{2} = 4 × 2 = 8 m^{2}

Mean area is

length = (6 + 4 )/2 = 5m ,

width = (4+2)/2 = 3

Mean area A_{m} = 5 × 3 = 15 m^{2}

\({A_m} = 15\;{m^2}\)

\(V = \;\frac{4}{3}\left( {24 + 4 × 15 + 8} \right)\)

\(V = 122.6\;{m^3}\)

**Note:**

According to prismoidal formula, d=h/2

Option 1 : 5300 m^{3}

**Prismoidal formula** can also be said as Simpson’s rule applied for two intervals.

**Trapezoidal Formula: **

a) Volume (v) of earthwork between a number of sections having areas A1, A2,…,An spaced at a constant distance d.

\({\rm{V}} = {\rm{d}}\left[ {\frac{{{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}}}{2} + {{\rm{A}}_2} + {{\rm{A}}_3} + \ldots + {{\rm{A}}_{{\rm{n}} - 1\:}}} \right]\)

b) Simpson’s formulae:

Volume (v) of the earthwork between a number of sections having area A1, A2 … An spaced at constant distance d apart is

\({\rm{V}} = \frac{{\rm{d}}}{3}[\left( {{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}} \right) + 4\left( {{{\rm{A}}_2} + {{\rm{A}}_4} + \ldots + {{\rm{A}}_{{\rm{n}} - 1}}} \right) + 2\left( {{{\rm{A}}_3} + {{\rm{A}}_5} + \ldots + {{\rm{A}}_{{\rm{n}} - 2}}} \right)\)

__Calculation:__

The given data can be depicted as below:

∴ V = (30/3) × (20 + 30 + 4 × (40 + 50) + 2 × (60)) = 5300 m^{3}

Option 1 : 92 m^{3}

__Concept____:__

Prismoidal formula is also known as Simpson’s rule.

Trapezoidal Formula:

a) Volume (v) of earthwork between a number of sections having areas A_{1}, A_{2},…,A_{n} spaced at a constant distance d.

\({\rm{V}} = {\rm{d}}\left[ {\frac{{{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}}}{2} + {{\rm{A}}_2} + {{\rm{A}}_3} + \ldots + {{\rm{A}}_{{\rm{n}} - 1{\rm{\;}}}}} \right]\)

b) Simpson’s formulae:

Volume (v) of the earthwork between a number of sections having area A_{1}, A_{2} … A_{n} spaced at constant distance d apart is

\({\rm{V}} = \frac{{\rm{d}}}{3}[\left( {{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}} \right) + 4\left( {{{\rm{A}}_2} + {{\rm{A}}_4} + \ldots + {{\rm{A}}_{{\rm{n}} - 1}}} \right) + 2\left( {{{\rm{A}}_3} + {{\rm{A}}_5} + \ldots + {{\rm{A}}_{{\rm{n}} - 2}}} \right)\)

__Calculation:__

**Given: **2L = 6m ⇒ L = 3m

Top area = 6 × 4 = 24 m^{2} = A_{1}

Bottom area = 4 × 2 = 8 m^{2} = A_{2}

\({{\rm{A}}_{\rm{m}}} = \left( {\frac{{6 + 4}}{2}} \right) \times \left( {\frac{{4 + 2}}{2}} \right) = 15{\rm{\;}}{{\rm{m}}^2}\)

\({\rm{Volume\;}} = \frac{{\rm{L}}}{3}\left( {{{\rm{A}}_1} + 4{{\rm{A}}_{\rm{m}}} + {{\rm{A}}_2}} \right)\)

\({\rm{Volume}} = \frac{3}{3}\left( {24 + 4 \times 15 + 8} \right) = 92{\rm{\;}}{{\rm{m}}^3}\)

Option 1 : Simpson’s

__Explanation__

(i) For irregular boundaries or curve boundary, Simpson’s rule is preferred over the trapezoidal rule to calculate the given area.

(ii) According to this rule the short length of boundaries between the two adjacent ordinates is a parabolic arch.

__Important Points__

Simpson's rule:

In order to apply Simpson's rule, the area must be divided in even number i.e., the number of offsets must be odd i.e., n term in the last offset 'On' should be odd.

The area is given by Simpson's rule:

\(Area = \frac{d}{3}\left[ {({O_1} + {O_n}) + 4({O_2} + {O_4} + ........ + {O_{n - 1}}) + 2({O_3} + {O_5} + ......{O_{n - 2}})} \right]\)

where O1, O2, O3, .........On is the offset

Note:

(i) In case of an even number of cross-sections, the end strip is treated separately and the area of the remaining strip is calculated by Simpson's rule. The area of the last strip can be calculated by either trapezoidal or Simpson's rule.

Option 3 : Both (1) and (2)

** Concept**:

The area can be calculated by following methods:

1. **Mid-ordinate rule:**

Area = h_{avg} × L = (h_{1} + h_{2} + …. + h_{n})d = d(Σh_{i})

Here,

h_{i} – Ordinate at the mid-point of each division and L – length of the baseline.

2. **Average ordinate rule:**

Here,

O_{i} – is the ordinate at regular intervals.

\(Area = \frac{{sum\;of\;the\;ordinates}}{{no.\;of\;the\;ordiantes}} \times length\;of\;the\;base\;line\)

3.** Trapezoidal rule:**

Here,

O_{i} – ordinates at equal interval

d – the common difference

\(Area = \frac{d}{2}\left[ {{O_o} + {O_n} + 2\left( {{O_1} + {O_2} + {O_3} + \; \ldots \ldots + {O_{n - 1}}} \right)} \right]\)

4. **Simpson’s rule:**

This method is used when:

- The
**boundaries are not straight**. - The number of
**area segments**should be**even in number**. - The
**number of ordinates should be odd**.

\(Area = \frac{h}{3}\left[ {{y_1} + {y_n} + 4\left( {{y_{even}}} \right) + 2\left( {{y_{odd}}} \right)} \right]\)

Here,

h = x_{2} – x_{1} = x_{3} – x_{2 }

Option 2 : mass haul diagram

__Explanation:__

Mass Haul Curve:

This is a curve representing the **cumulative volume of earthwork** at any point on the curve, the manner in which earth to be removed.

It is necessary to plan the movement of excavated soil of worksite from cuts to fill so that haul distance is minimum to reduce the cost of earthwork.

The mass haul diagram helps to determine the economy in a better way.

The mass haul diagram is a curve plotted on a distance base with the ordinate at any point on the curve representing the algebraic sum of the volume of earthwork up that point.

A haul refers to the transportation of your project’s excavated materials. The haul includes the movement of material from the position where you excavated it to the disposal area or a specified location. A haul is also sometimes referred to as an authorized haul.

Haul = Σ Volume of earthwork × Distance moved.

In a cross staff survey, the perpendicular offsets are taken on right and left of the chain line AD as shown in figure – all values are in ‘metres’. The area enclosed by ABCDEFA, computed by trapezoidal method is

Option 3 : 3475 m^{2}

**Explanation**

The area bounded by ABCDEFA

= Area bounded by ABCDA + Area bounded by AFEDA

\(= \left[ {\frac{1}{2} \times 30 \times 20 + \frac{1}{2} \times(20+30) \times \left( {45 - 30} \right) + \frac{1}{2} \times 30 \times \left( {90 - 45} \right)} \right] + \left[ {\frac{1}{2} \times 35 \times 40 + \frac{1}{2}\left( {40 + 30} \right) \times \left( {65 - 35} \right) + \frac{1}{2} \times 30 \times \left( {90 - 65} \right)} \right]\)

= 1350 + 2125

= **3475 m ^{2}**

Option 3 : Simpson’s rule

**Explanation**

For irregular boundaries, Simpson’s rule is preferred over the trapezoidal rule to calculate the given area.

According to this rule the short length of boundaries between the two adjacent ordinates is a parabolic arch.

__Important Points__

Simpson's rule:

In order to apply Simpson's rule, the area must be divided in even number i.e., the number of offsets must be odd i.e., n term in the last offset 'On' should be odd.

The area is given by Simpson's rule:

\(Area = \frac{d}{3}\left[ {({O_1} + {O_n}) + 4({O_2} + {O_4} + ........ + {O_{n - 1}}) + 2({O_3} + {O_5} + ......{O_{n - 2}})} \right]\)

where O1, O2, O3, .........On is the offset

**Note:**

- In case of an even number of cross-sections, the end strip is treated separately and the area of the remaining strip is calculated by Simpson's rule. The area of the last strip can be calculated by either trapezoidal or Simpson's rule.

Option 1 : \(\frac{{\rm{h}}}{3} \times {\rm{\;}}\left[ {\left( {{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}} \right) + 4\left( {{{\rm{A}}_2} + {{\rm{A}}_4} + \ldots } \right) + 2\left( {{{\rm{A}}_3} + {{\rm{A}}_5} + \ldots } \right)} \right]\)

__Formula:__

a) Trapezoidal Formula:

\({\rm{V}} = \frac{{\rm{h}}}{2} \times {\rm{\;}}\left[ {\left( {{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}} \right) + 2\left( {{{\rm{A}}_2} + {{\rm{A}}_4} + \ldots+A_{n-1} } \right) } \right]\)

b) Mid-ordinate Formula:

\({\rm{V}} = {\frac{d}{n}\rm{}}\left[ {{{{{\rm{A}}_1} + }} {{\rm{A}}_2} + {{\rm{A}}_3} + \ldots + {{\rm{A}}_{{\rm{n}} }}} \right]\)

c) Simpson’s Formula:

\({\rm{V}} = \frac{{\rm{d}}}{3}\left[ {\left( {{{\rm{A}}_1} + {{\rm{A}}_{\rm{n}}}} \right) + 4\left( {{{\rm{A}}_2} + {{\rm{A}}_4} + \ldots + {{\rm{A}}_{{\rm{n}} - 1}}} \right) + 2\left( {{{\rm{A}}_3} + {{\rm{A}}_5} + \ldots + {{\rm{A}}_{{\rm{n}} - 2}}} \right)} \right]\)

Where,

V is the volume of earthwork between a number of sections having areas A1, A2 …, An spaced at a constant distance d.

Option 2 : Planimeter

**Planimeter:**

Planimeter is an instrument used in surveying to compute the area of any given plan. Planimeter only needs plan drawn on the sheet to calculate area.

Generally, it is very difficult to determine the area of irregular plot. So, by using planimeter we can easily calculate the area of any shape.

**Pedometer:**

It is a portable device usually an electronic or electromagnetic that counts each steps of a person by detecting the motion of the person’s hip or hand.

**Passometer: **

It is an instrument shaped like a watch used to count the number of persons steps.

Option 1 : Sum of the product of each load by its distance

**Mass Haul Curve**

This is a curve representing the cumulative volume of earthwork at any point on the curve, the manner in which earth to be removed.

It is necessary to plan the movement of excavated soil of worksite from cuts to fill so that haul distance is minimum to reduce the cost of earthwork.

The mass haul diagram** helps to determine the economy in a better way**.

The mass haul diagram is a curve plotted on a distance base with the ordinate at any point on the curve representing the algebraic sum of the volume of earthwork up that point.

**A haul refers to the transportation of your project’s excavated materials**. The haul includes the movement of material from the position where you excavated it to the disposal area or a specified location. A haul is also sometimes referred to as an authorized haul.

**Haul = Σ Volume of earthwork × Distance moved.**

Option 1 : Area

**Concept:**

The areas between curved boundaries are also computed by planimeter from the map of the **area**. Most commonly used is Amsler polar Planimeter.

The formula for calculating areas of the map by using planimeter.

**Area (A) = M × (FR – IR ± 10 N + C)**

Where,

M = Multiplying constant

FR = Final reading of Planimeter

IR = Initial Reading

N = Number of times zero of dial pass is index mark

C = Constant marked above scale division on tracing arm.

± ⇒ **+ ve sign when zero is passed in clockwise direction**

**– ve for when zero is passed in an anticlockwise direction.**

∴ The planimeter is used to measure the area.

Option 1 : + ve

__Concept:__

The areas between curved boundaries are also computed by planimeter from the map of the area. Most commonly used is Amsler polar Planimeter.

The formula for calculating areas of the map by using planimeter.

Area (A) = M × (FR – IR ± 10 N + C)

Where,

M = Multiplying constant, FR = Final reading of Planimeter, IR = Initial Reading, N = Number of times zero of dial pass is index mark, and C = Constant marked above scale division on tracing arm.

± ⇒ **+ ve sign when zero is passed in clockwise direction** and – ve for when zero is passed in an anticlockwise direction.