Applied mathematics for CBE assignment

1. HND in Construction & Built Environment (Civil Engineering) BCAS DOHA QATAR
Unit 3: Applied Mathematics Page 1
Task 1 Estimate the cost of masonry water platform from the given drawings
specifications
1.1.Estimate the required quantity for the following specifications given.
i.Earth work in excavation in foundation
ii.Earth work in filling inside above ground level (G.L)
iii.Lime concrete in foundation
iv.First class brick work in lime mortar
v.12mm cement sand plastering 1:6 in walls outside
vi.2.5cm C.C. 1:2:4 floor over and including 7.5cm L.C.
vii.2.5cm C.C. 1:2:4 floor on top of walls
I. Volume of excavation = Length × Width × Height
= [(6x2)+(4.2x2)]×0.6×(0.2+0.2+0.1+0.2)
= 8.568m³
II. Filling inside above ground level = Length × Width × Height
= (6-0.8) × (5-0.8) × (2-0.075-0.025)
= 41.496m³
2.5cm c.c over 7.5cm l.c .floor

2. HND in Construction & Built Environment (Civil Engineering) BCAS DOHA QATAR
Unit 3: Applied Mathematics Page 2
III. 1st
concrete foundation volume = Length × Width × Height
= [(6x2)+(4.2x2)]×0.6 × 0.2
= 2.448m³
2nd
concrete foundation volume = Length × Width × Height
= [(6x2)+(4.2x2)] × 0.5 × 0.2
=2.04m³
3rd
concrete foundation volume = Length × Width × Height
= [(6x2)+(4.2x2)] × 0.45 × 0.1
= 0.918m³
Total foundation volume = 2.448m³+2.04m³+0.918m³
= 5.406m³
IV. First class brick work = Length × Width × Height
= (6x2+4.2x2) × 0.388 × (2+0.2-0.1)
= 16.62m³
V. 12mm cement sand plastering = Length × Width
= [(6x2)+(5x2)] × 2
=44m²
VI. 25mm C.C 1:2:4 floor over and = Length × Width × Height
including 75mm L.C
= 6×5×(0.025+0.075)
= 3m³
VII. 25mm C.C 1:2:4 floor = 6×5
=30m²

3. HND in Construction & Built Environment (Civil Engineering) BCAS DOHA QATAR
Unit 3: Applied Mathematics Page 3
1.2.Rates for the construction are given below
 Earthwork excavation Rs 1050.00 per cubic meter
 Lime Concrete in foundation and floor per cum is Rs 750.00
 1st class brickwork in 1:6 cement mortar Rs 650 per cum
 12mm cement plaster 1:2 with coarse sand Rs 85.00 per square meter.
 12mm cement plaster 1:4 with local sand Rs 75.00 per square meter
 5cm cement concrete 1:1 ½:3 floors is Rs. 110.00 per square meter.
Estimate the cost for the each construction given by considering,
i.3% for contingencies
ii.2% for work charged establishments from total expenses.(P1, M1.1, M1.3, D1.2)
Item Description Unit Qty
Unit
cost
SLR
Material
cost
3%
contingencies
2% work
charged
Total cost
Earthwork m³ 8.568 1050 8996.4 269.892 179.928 9446.22
Lime concrete in
foundation
m³ 5.406 750 4054.5 121.635 81.09 4257.225
1st class brick work m³ 16.62 650 10803 324.09 216.06 11343.15
12mm plastering coarse
sand
m² 44 85 3740 112.2 74.8 3927
12mm plastering local
sand
m² 44 85 3740 112.2 74.8 3927
Cement concrete for floor m² 30 110 3300 99 66 3465
Total expenses 36365.595
Total expenses = 36, 365.60 /= SLR

4. HND in Construction & Built Environment (Civil Engineering) BCAS DOHA QATAR
Unit 3: Applied Mathematics Page 4
Task 2 The bearings observed in traversing with a compass at a place where local
attraction was suspected are given below.
Line Fore bearing Back Bearing
AB S 450
30’ E N 450
30’ W
BC S 600
00’ E N 600
40’ W
CD N 030
20’ E S 050
30’ W
DA S 850
00’ W ` N 830
30’ E
2.1.Calculate the following
i. Convert the above fore and back bearings to W.C.B. (W.C.B. = Whole Circle Bearings)
ii. Draw the diagram
iii. At what stations do you suspect local attraction?
iv. Find the corrected bearings of the lines
v. Find the interior angle .
vi. BD length cannot measure due to a pond on the line BD. Let B D=∝ where cos ∝=0.6497. Find
the length of BD. Where BC=901m and CD=1100m
Line
R.B WCB
Different
Corrected WCB
FB BB FB BB FB FB
AB S 45.5 E N 45.5 W 134.5 314.5 180 134.5 314.5
BC S 60 E N 60.66 W 120 299.34 179.34 120 300
CD N 3.33 E S 5.5 W 3.33 185.5 182.17 4 184
DA S 85 W N 83.5 E 265 83.5 -181.5 263.5 83.5

5. HND in Construction & Built Environment (Civil Engineering) BCAS DOHA QATAR
Unit 3: Applied Mathematics Page 5
III. Stations C and D are local attraction.
V. .=64˚
VI.
COSФ=0.6497
Ф=49˚28’52”
Cosine rule = DC²=BC²+CD²-2 × BC × CD × COS C
X²= 901² + 1100²- 2 × 901 × 1100 × 0.6497
= 733965.66
X = 733965.66
X =856.72m
2.2.A rectangular plot of land ABCD has the following dimensions.
 AB=50m, BC=80m, CD=50m.
 Side DA is a curved line.
The following left offsets to the curved boundary taken while chaining along DA were taken.
Distance (m) 0 10 20 30 40 50 60 70 80
Offset (m) 0 2.0 3.0 4.0 5.0 4.0 3.0 2.0 0
i. Estimate the area of the plot of land using Simpson’s rule and trapezoidal Rule.

6. HND in Construction & Built Environment (Civil Engineering) BCAS DOHA QATAR
Unit 3: Applied Mathematics Page 6
ii. If the exact area of ABCD is less than the area you calculated by above two methods then
express your comments about the area using inequality.(P2, M2.1, M 2.2, D 2.1)
Simpson’s rule area = 1/3 (width of interval )[(first +last ordinate)+4(sum of even ordinates)+2(sum
of remaining ordinates)
= 1/3×10[(50+50)+4(52+54+54+52)+2(53+55+53)
= 4233.33m²
Trapezoidal rule area = (width of interval )[1/2(first +last ordinate)+ (sum of remaining ordinates)]
= 10 × ½(50+50)+(52+53+54+55+54+53+52)
=4230m²
Reasons for exact area of ABCD is less than above two methods
 Both method are average basis method so it's not exact area it can be little changes.
 In this case between the internals we have no any offset dimensions.

7. HND in Construction & Built Environment (Civil Engineering) BCAS DOHA QATAR
Unit 3: Applied Mathematics Page 7
Task 3
3.1.The thicknesses of 20 samples of steel plate are measured and the results (in millimeters) to two
significant figures are as follows.
7.3 7.1 6.6 7.0 7.8 7.3 7.5 6.2 6.9 6.7
6.5 6.8 7.2 7.4 6.5 6.9 7.2 7.6 7.0 6.8
I. Complies a table showing the frequency distribution and the relative frequency distribution for
regular classes of 0.2mm from 6.2mm to 7.9mm.
3.2.Present the above data by using
I. Histogram
II. Frequency polygon
III. Cumulative frequency curve
3.2
I. Histogram
Class of
thickness
Mid
points
Frequency
Cumulative
Frequency
6.2 - 6.4 6.3 1 1
6.5 - 6.7 6.6 4 5
6.8 - 7.0 6.9 6 11
7.1 - 7.3 7.2 5 16
7.4 - 7.6 7.5 3 19
7.7 - 7.9 7.8 1 20

8. HND in Construction & Built Environment (Civil Engineering) BCAS DOHA QATAR
Unit 3: Applied Mathematics Page 8
II. Frequency polygon
III. Cumulative frequency curve

9. HND in Construction & Built Environment (Civil Engineering) BCAS DOHA QATAR
Unit 3: Applied Mathematics Page 9
3.3.The following table gives the values of two variables x and y.
X= Demand Y= Supply
X 42 44 58 55 89 98 66
Y 56 49 53 58 65 76 58
I. Estimate the coefficient of correlation between x and y
II. Obtained the regression line of y and x
III. Hence predict the value of Y when x=60(P3, M2.7, D1.2)
Coefficient of correlation (r) = n(Σxy)-( Σx) (Σy)
[n(Σx²)-(Σx) ²][n(Σy²)-(Σy) ²]
= 7(27833)-( 452) (415)
[7(31970)-(452) ²][7(25075)-(415) ²]
r = 0.904
an + bΣxy = Σyj
aΣxj+bΣx²j=Σxjyj
7a + 452b= 415 1
452a+31970b=27833 2
7 = 415 452
=
415 452
7
n X Y X² Y² XY
1 42 56 1764 3136 2352
2 44 49 1936 2401 2156
3 58 53 3364 2809 3074
4 55 58 3025 3364 3190
5 89 65 7921 4225 5785
6 98 76 9604 5776 7448
7 66 58 4356 3364 3828
Σ 452 415 31970 25075 27833

10. HND in Construction & Built Environment (Civil Engineering) BCAS DOHA QATAR
Unit 3: Applied Mathematics Page 10
452
415 452
7
+ 31970 = 27833
b= 0.372
=
415 452
7
=
415 452 × 0.372
7
= 35.26
So regression line
= 35.258 + 0.372
X=60
= 35.258 + 0.372 × 60
= 57.578

11. HND in Construction & Built Environment (Civil Engineering) BCAS DOHA QATAR
Unit 3: Applied Mathematics Page 11
Task 4
4.1.PS and PQ are uniform two rods freely joint at P and the whole system has keep on a smooth
horizontal table in a vertical plane.
To keep the above system in a vertical plane the point x on PS and point Q have connected by a
inextensible string such that PX= PS where PQ=PS=1.8m, and weight of PQ or PS in 800N
respectively.The angle between the PS or PQ and horizontal table in 600
.
I. Draw the suitable diagram.
II. The reaction at joint P and its direction.
III. Estimate the tension of the string.
=
2
3
=
2
3
1.8
= 1.2 ,
< ,
( ) = ( ) + ( ) 2 × ( )( ) × 60°
( ) = (0.6) + (1.8) 2 × 0.6 × 1.8 ×
( ) = 0.36 + 3.24 1.08

12. HND in Construction & Built Environment (Civil Engineering) BCAS DOHA QATAR
Unit 3: Applied Mathematics Page 12
( ) = 2.52
= √2.52
= 1.59 ,
sin60 ° = 1.59 sin
0.6
√3
2
= 1.59 sin
sin = 0.3 ×
√3
1.59
sin = 0.327
= sin (0.327)
= 19.07°
Q)
1 × 1.8 × cos60° + 800 × 0.9 × cos 60° = 2 × 1.8 × sin60°
1 × 1.8 × + 800 × 0.9 × = 2 × 1.8 × √
2 1 + 800 = 2√3 F2
400 = √3 2 1
P)
× 1.8 × cos 60° = 800 × 0.9 × cos 60° + × cos × 1.8 × sin60°
× 1.8 × = 800 × 0.9 × + × cos × 1.8 ×
√3
2
400 = √3 × cos
=
√
=
400
√3cos(19.07)
= . Tension in the strip
①
②

13. HND in Construction & Built Environment (Civil Engineering) BCAS DOHA QATAR
Unit 3: Applied Mathematics Page 13
S)
1 × 1.8 × cos60° + 2 × 1.8 × sin 60° = 800 × 0.9 × cos60° + sin(60 + ) 0.6
1 × 1.8 × + 2 × 1.8 ×
√
= 800 × 0.9 ×
1
2
+ sin(60 + 19.07) 0.6
1 × 0.9 + 2 × 0.9 × √3 = 400 × 0.9 + 44.35 sin(79.07) 0.6
1 + 2√3 = 400 × 0.9 + 159.94
1 + 2√3 = 559.94
400 = √3 2 1
③ + ① = 2√3 2 = 959.94
2√3 2 = 959.94
= .
400 = √3 227.12 1
= .
= 79.97 +277.12
= . For the directions refer the drawing.
4.2.
a) The diagram shows a window. ABCD is a rectangle and DCE is
equilateral triangle. The perimeter of this window is 20m. The width of
the window is . To get the maximum area estimate the value of .
③
①
E
A B
D C
600
600
y
x
x

14. HND in Construction & Built Environment (Civil Engineering) BCAS DOHA QATAR
Unit 3: Applied Mathematics Page 14
= 20
= + + + +
= 3 + 2
20 = 3 + 2
=
20 3
2
= +
= × + × × sin60°
= +
√
= +
√
=
20 3
2
+ √
= 10
3
2
+
√
= 10 3 +
√
= 0 ↔ 10 = 3 √
10 = 3
√3
2
=
10 × 2
6 √3
= . maximum area estimation value of X

15. HND in Construction & Built Environment (Civil Engineering) BCAS DOHA QATAR
Unit 3: Applied Mathematics Page 15
b) Radius of a cylinder is . Height is .
 + = is a constant.
 The volume of this cylinder is .
I. Show that = ( 2 + )
II. Hence estimate the value of in terms of for maximum or minimum volume of the cylinder.
Deduce maximum volume when = 1
=
=
=
= +
= ( ) ×
= ( + 2 ) ×
= ( + 2 )
= +
= ( + 2 )
= 0 ↔ 3 4KX + = 0
(3 )(X K) = 0
(X K) = 0
(3 ) = 0
= =
Maximum volume when k = 1
V = 2K + K π
V = K
3
3
+
K
27
2K × K π
V =
9K + K 6K
27
π
V =
4K π
27
= =