Calculation of Area and Volume MCQ Quiz - Objective Question with Answer for Calculation of Area and Volume - Download Free PDF

Concept:

The rule described is used for approximating the area under a curve using straight-line segments (trapezoids) between data points. This is known as the Trapezoidal Rule.

Formula:

\( A = \frac{h}{2} \left[ y_0 + y_n + 2(y_1 + y_2 + \dots + y_{n-1}) \right] \)

Where:

• \( h \) is the spacing between ordinates
• \( y_0 \) and \( y_n \) are the first and last ordinates
• All others are intermediate ordinates

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

Concept:

The formula for the area of the mid-ordinate can be given as

Area = common distance × sum of mid-ordinates

Area of plot = h1 × d + h2 × d + … + hn × d = d (h1 + h2 + … hn)

A = d × ∑O.

Calculation:

Given that,

d = 2 m

Ordinate = 25 m, 44 m, and 20 m

On substitution,

A = 2 × (25 + 44 + 20)

A = 178 sq. m.

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 '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

Important Points 

• 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.

Concept:

To estimate the earthwork quantity between multiple cross-sections, we use the Trapezoidal Formula:

\( V = D \left[ \frac{A_0 + A_n}{2} + A_1 + A_2 + \dots + A_{n-1} \right] \)

Given:

Distance between sections, \( D = 30 \, \text{m} \)

Cross-sectional areas: \( A_0 = 5.5 \, \text{m}^2, \, A_1 = 11.5 \, \text{m}^2, \, A_2 = 17 \, \text{m}^2, \, A_3 = 24.5 \, \text{m}^2 \)

Calculation:

Apply the trapezoidal formula:

\(V = 30 \left[ \frac{5.5 + 24.5}{2} + 11.5 + 17 \right]\)

\(V = 30 \left[ \frac{30}{2} + 11.5 + 17 \right] = 30 \left[15 + 11.5 + 17 \right]\)

\(V = 30 \times 43.5 = 1305 \, \text{m}^3\)

Explanation:

The volume of earthwork by trapezoidal method  = V₁

V₁ =  \(common\: distance\left \{ \frac{First\: area + Last\: area}{2}+the \:sum \:of\: remaining \:area \right \}\)

The volume of earthwork by prismoidal formula = V₂

V₂ = \(=\frac{Common\: distance}{3}\left \{ First\: area+ Last\: area + 2(Sum\: of odd\: area) + 4(Sum\: of even\: area)\right \}\)

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 d₁ are the depth of earthwork at the centerline

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.

Concept:

a) Trapezoidal Formula:

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

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

Also can be written as 

\(V = \frac{D}{2}\left[ {({A_1} + A_n +2 \times (A2 +..+ A_{n-1} )} \right]\)

Calculation:

Given, A₁ = 50 × 40 = 2000 m², 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 - 16 ) = 34 m & (40 - 16) = 24 m

∴ The bottom area is 34 m × 24 m = 816 m²

Mean area (A_m) = (2000 + 816) / 2 = 1408 m²

According to simple Trapezoidal rule for volume,

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

∴\(V = \frac{2}{2}\left[ {(2000 + 2\times (1408) + 816)} \right]\) = 5632 m³

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 O₁, O₂, O₃, .........O_n is the offset

Important Points

• 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.

Concept:

i) Mid-Ordinate Rule

Area of plot = h1 × d + h2 × d + … + hn × d = d (h1 + h2 + … hn)

∴ Area = common distance × sum of mid-ordinates

ii) Average-Ordinate Rule

\(\text{Area}=\frac{{{\text{O}}_{1}}+{{\text{O}}_{2}}+\ldots +{{\text{O}}_{\text{n}}}}{{{\text{O}}_{\text{n}+1}}}\times \text{l}\)

\(\text{i}.\text{e}.\text{ }\!\!~\!\!\text{ Area}=\frac{\text{Sum }\;\!\!~\!\!\text{ of }\;\!\!~\!\!\text{ ordinates}}{\text{No}.\;\text{ }\!\!~\!\!\text{ of }\;\!\!~\!\!\text{ ordinates}}\times \text{length }\;\!\!~\!\!\text{ of }\;\!\!~\!\!\text{ base }\;\!\!~\!\!\text{ line}\)

(iii) Trapezoidal Rule

\(\text{Total }{area}=\frac{\text{d}}{2}\left\{ {{\text{O}}_{1}}+2{{\text{O}}_{1}}+2{{\text{O}}_{2}}+\ldots +2{{\text{O}}_{\text{n}-1}}+{{\text{O}}_{\text{n}}} \right\}\)

\(\text{Total Area}=\frac{\text{Common distance}}{2}\times \{\left( 1\text{st ordinate}+\text{last ordinate} \right)+2\left( \text{sum of other ordinate} \right))\}\)

(iv) Simpson’s Rule

\(\text{Total }{ area}=\frac{\text{d}}{3}\left( {{\text{O}}_{1}}+4{{\text{O}}_{2}}+2{{\text{O}}_{3}}+4{{\text{O}}_{4}}+\ldots +{{\text{O}}_{\text{n}}} \right)\)

\(\text{Total Area}=\frac{\text{Common Distance}}{3}\times \left\{ \left( 1\text{st ordinate}+\text{last ordinate} \right)+4\left( \text{sum of even ordinates} \right)+2\left( \text{sum of remaining odd ordinates} \right) \right\}\)

Explanation:

Methods for measurement of earthwork:

1. Mid Section method  2. Trapezoidal Method  3. Prismoidal Method  4. Simpson's 3/8th rule

Mid Section Method:

In this method the quantity of earthwork is computed with the help of size of mid section.

Volume of earthwork \(= Area\times Length=\left\{BD+SD^{2} \right\} \times L \)​

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

d₁ = 1.5 m and d₂ = 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 m³

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 m²

Volume of earthwork = A × L

Volume of earthwork = 100 × 2400

∴ Volume of earthwork = 2,40,000 m³

Concept:

Calculation:

Given: L = 6m 

Top area = 6 × 4 = 24 m2 = A1

Bottom area = 4 × 2 = 8 m2 = A2

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

\({\rm{Volume\;}} = \frac{{\rm{L}}}{6}\left( {{{\rm{A}}_1} + 4{{\rm{A}}_{\rm{m}}} + {{\rm{A}}_2}} \right)\)

\({\rm{Volume}} = \frac{6}{6}\left( {24 + 4 \times 15 + 8} \right) = 92{\rm{\;}}{{\rm{m}}^3}\)

Explanation: Simpson's rule:

1. Simpson's rule is also known as Parabolic Rule.
2. For calculation of volume, this formula is known as the Prismoidal rule 
3. This rule is used when the ends of ordinates or straight to form an arc of a parabola.
4. This rule is applicable only when the number of divisions is even, i.e. the number of ordinates is odd.  
5. This rule can be stated as follows: One-third of common distance multiplied by the sum of last and first ordinate, four times the sum of even ordinates, and twice the sum of the remaining odd ordinates are added.

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