Test: Measurements of Area & Volume - 2 - Civil Engineering (CE) MCQ

Concept:

Simpson’s Rule

Calculation:

By Simpson’s rule

A = d/3 × [(O₁ + O_n) + 4 (O₂ + O₄ + … + O_n-2) + 2 (O₃ + O₅… + O_n-1)]

d = 20 m,

O₁ + O_n = 2.35 + 5.65 = 8 m,

4 (O₂ + O₄ + …. O_n-2) = 4(4 6.5 + 6.65 + 6.2 + 6.5) = 103.4

2 (O₃ + O₅ + …. O_n-1) = 2 (5.4 + 7.85 + 2.35) = 31.2

∴ A = 20/3 × (8 + 103.4 + 31.20) = 950.66 sq. metres

So, the correct answer will be 950 m²

Application of Simpson rule related to shipping:

• The underwater (displaced) volume can be calculated in many ways, for example, if the immersed area, is divided into a number of sections throughout the length of a ship and then the area is calculated by Simpson's rule, by integrating the area of a particular section, within the given range of the length of the ship
• ​The longitudinal position of the center of buoyancy with respect to any reference point on the ship is called the longitudinal center of buoyancy (LCB).
  • Usually, the reference point for locating the LCG is either of the forward or aft perpendiculars. Calculation of the longitudinal center of buoyancy of the hull is the application of Simpson's rule.
• Now, to obtain the extreme displacement, it is necessary to add this to the shell displacement, and with regard to the displacement of the shell, this is determined by first of all calculating the wetted surface area.
  • This area when multiplied by the mean thickness of the shell planting gives the volume displaced by the shell. The wetted surface area is not easy to calculate since the outside surface of a ship has double curvature.
  • It can be approximated by girths around the various sections and then applying Simpson's rule to find the area

Concept:
Volume of prismoid

Where A₁ = Area of one section (or side) of prismoid

A₂ = Area of the second section (or side) of prismoid

Calculation:

Given
A₁ = 76.32 m²
A₂ = 24.56 m²
d = 4 m

Volume (V) = 201.76 m³

Concept:

Trapezoidal Formula:

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

Simpson’s Formula:

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

Calculation:

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

 

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

So using trapezoidal formula:
V = 5 ((50+750/2) + 200 + 400 +600) = 5 x (400 + 200 + 400 + 600) = 8000 m³

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:

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

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.


 

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 = 1/40
 

L = 2400 m

 

Average cross-sectional area(A) = 12×(10+30)×5

A = 100 m²

Volume of earthwork = A × L

Volume of earthwork = 100 × 2400

∴ Volume of earthwork = 2,40,000 m³

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

This method is based on the assumption that the mid-area is the mean of the end areas, which make it the Average end area method

Concept: 

Correction for Erroneous Length of Chain/Tape: 

• The chain surveying depends only on linear measurement of distances. For traversing only the errors in distance measurements are of importance and significance.
• Measuring devices either chain or tape can either be longer or shorter than the designated length.
• The measured distance will be smaller than the actual if the length of the chain is longer than the designated length.
• It will be larger than the actual if the chain is shorter than the designated length.
• The actual measured distance can be corrected by the following formula :   

 True or Correct Distance = L'/L x measured Distance
Where L' = Actual incorrect length of chain, and L = Designated length of chain
∴ True or correct Volume of ebankment  = (L'/L)3 x Measured Volume of embankment

Given, L' = Actual incorrect length of chain = 20.1 m (as chain is 10 cm too long),  L = Designated length of chain = 20 m,  measured/calculated volume of embankment = 368.85 m³

∴ True or correct volume of embankment = (20.1/20)3 x 368.85 = 374.41 m³

he cube is open from the top, hence total surface area of the cube will be equal to 6l² – l² = 5l²
= 5(20)²
= 5(400)
= 2000.

A cube has six equal sides.
Hence, total surface area of a cube = sum of areas of 6 sides (as shown in the figure below)

Area of one side of a cube having side equal to ‘l’ = l²
Total surface area of the cube = l² + l² + l²+l²+ l² + l²
= 6l².