Mathematical Calculation on Constructional Site Question and Answers

Task 1

Question number A.

In this question, it has to find out the value of the height of the aerial. Let the height of the aerial is h.

The surveyor is due to the east of the aerial, creating a triangle where one angle is 900 with the height of the aerial, the angle of elevation becomes about 480.

Let the horizontal distance from the surveyor to the base of the aerial is d.

So, using these it can be said that, tan (480) =h/d.

After the moving of the surveyor due south by 30 meters, the new angle of the elevation transforms to 440.

So, now a new equation is formed, that is tan (440) =h/ (d+30).

tan (480) =h/d

or, d =h/ tan 480

so, tan 440= h/( h/ tan 480) +30

or, h/ tan 480+30 = h/ tan 440

or, h {(1/ tan 480)- (1/ tan 440)= -30

or, h( 1/1.11-1/0.966) = -30

or, h(0.9-1.035)= -30

or, h= -30/-0.135

or, h= 222.23 (approx.)

So, the height of the aerial is 222.23 (approximately).

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Question number B.

For estimating the total volume of soil which is essential to be transported from the site the formula of the trapezoidal rule is necessary to calculate the volume of each portion and at last, to sum up the values.

The formula of the volume of the trapezoidal prism is,

V= (h/2). (A1+A2).L

In this formula, the h represents the height of the trapezoidal prism. In this case, it is 10 meters. A1 and A2 values denote the two areas of the bases of the trapezoidal prism and L represents the length of the trapezoidal prism, in this case, this value also is 10 meters.

In this context, the five areas are A1, A2, A3, A4 and A5. The values of these areas are 1.44 m2, 1.62 m2, 1.83 m2, 2.0 m2 and 2.1 m2 respectively.

Here the bulk factor for the excavated soil is 1.5.

Now,

V=V? +V2+V3+V4

or, Vi=(h/2) (Ai+Ai+1)?L [i= 1,2,3,4]

or, V = 10/2{ (A1+A2)+(A,+ A3) + (A3+ A4) + (A4+A5)}*10

= 50 {(1.44+1.62)+(1.62+1.83) + (1.83+2.0) +(2.0+2.1)}

Or, V=50{1.44+2( 1.62 + 1.83 + 2.0)+2.1} .

= 50{1.44 + 2(5.45) + 2.1}

= 50{1.44 + 10. 9 + 2.1}

=50 *14.44

=722m³.

The bulk factor is 1.5.

so, the increased volume due to excavation is = 722 * 1.5 = 1083m3.

So, the total volume of soil which will need to be transported from the site is 1083 m3.

Question number C.

To find these three lengths of FB, BD, and EB first it has to observe the triangles that are AFB, EFB, EBD, and DBC. Here, the values of AB and BC are 5m, AF = 2.5cm, the value of angle A is 500, EF= ED = 4 cm, and the value of the angle EBD is 500,

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Since, in the triangle ABF, one angle is 500, one angle is the right angle, and angle FBA is {1800-(500+900)} = 400 so, the angle EBF is 900-400=500.

In the triangle ABF, by utilizing the law of cosines it is evaluated,

BF2= AB2+ AF2-2. AB.AF.cos (A)

In the triangle BFE, by utilizing the Pythagorean Theorem it is evaluated,

BE2= BF2+ EF2

In the triangle EBD, by utilizing the law of sines it is evaluated,

(EB / sin B)=(ED / sin E)

In the triangle DBC the values of AB and BC are 5 cm and the angle C is 500.

So here is an equation, DB/sin C =BC/ sin B.

Now by performing these equations, the values can be extracted properly.

EB/sin B = ED/sin E

or, 5.583 / sin 50° = 4/sin E

Or, sin E= 4sin 50°/5.583= 4* 0.766/ 5.583 = 0.549 (approx.).

Next, DB /sin C = BC/ sin B

Or, DB/ sin 50° = 5/ sin 50°

Or , DB= 5

The values of FB, BD, and EB are 3.896 cm, 5 cm, and, 5.583 cm approximately.

Question number D.

By observing this formula it can be said here that w denotes the permed work by the water pump, w0 is a constant and P1 and P2 are the beginning and final pressures respectively.

In this context, it has to find out the value of P2 and here it is mentioned that the value of the w= 3w0 and P1 = 7 kPa.

So, the final Pressure P2 evaluates approximately 0.3485 kPa.

Question number E.

Question number F.

First in this context in the first portion the first part is discussed on the basis of the first T section.

1. i) On the reference x-axis the centroid of the T section can be evaluated easily. Since the T section is symmetrical about the Y axis, the process is quite easy.

Here the location of centroidal axes x-x is in the middle of the above rectangle

So, the xc = 240/2 mm= 120 mm. And the yc = 15/2 = 7.5 mm. (c denotes a short form of centroidal)

1. ii) As the T shape is symmetrical about the Y axis then the neutral axis is lies between the centroidal axes x so, the xneutral is 120 mm.

iii) Since the same logic is applicable for the neutral axis y then the yneutral is 7.5 mm.

1. iv)

1. v)

Now for the below figure these calculations are again performed one by one.

1. i)
2. ii) and iii)
3. iv)
4. v)

Question number G.

Question number H.

In measuring the radius of the Sample, where the radius of the Circular cross section is 1% large, that means if the actual measurement is 100% the measured radius is 101%

Let's denote the actual measurement is ra, and, the measured measurement of radius is rm.

so, mathematically,

rm = 1.01*ra

Now, when these values are used in determining the area of a circle,

Am = πrm2

and, Aa = πra2

So, Am= π * (1.01 * ra)2

=π /1.0201* ra2

= Aa *1.0201

So, the error in the area= Am- Aa

=1.0201Aa- Aa

=Aa(0.0201)

So the estimating error in the area is (0.0201* π * ra2).

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