Qualification - HND

Unit number and title - N5E08 Mathematics for Construction

Assignment - Mathematics for Construction

The purpose of this assignment is to:

1. Identify the relevance of mathematical methods to a variety of conceptualized construction examples

2. Investigate applications of statistical techniques to interpret, organize and present data by using appropriate computer software package

3. Use analytical and computational methods for solving problems by relating sinusoidal wave and vector functions to their respective construction applications

4. Illustrate the wide-ranging uses of calculus within different construction disciplines by solving problems of differential and integral calculus

Scenario:

An unnamed island in polar region is approximately 50,000 mile2, 75 percent of which is covered by ice with an average thickness of 1300 meters. Estimate the mass of the ice in this island (assume two significant figures). The density of ice is 0.917 g/mL (1cm3 = 1mL)

i. Clearly specify the units (dimensions) of the desired product

ii. Specify all known values with their associated units

iii. Specify relevant formulas and all conversion factors

iv. Develop an equation such that the units of the left side are equivalent to the units of the right side

v. Perform the calculation

A constant force of magnitude F acts over a distance d on an object of mass m. Find the final speed v of the body.

vi. List the quantities that might be involved to calculate the speed v and propose a generic equation [M1]

vii. Using a dimensional analysis on both left and right-hand side and follow the algebraic relations [M1]

viii. Derive the equation of final speed v [M1]

i. What is arithmetic progression [P2]

ii. Write down the general expression for n'th term of an arithmetic progression

iii. Determine the 8th term of series 2, 7, 12, 17...

iv. What is geometric progression [P2]

v. Write down the general expression for the n'th term of a geometric progression [P2]

vi. A drilling machine is to have 6 speeds ranging from 30 rev/min to 960 rev/min. If the speed form a geometric progression, determine the value of each speed. [P2]

i. RC Discharge

When a capacitor C discharge through a resistor R the voltage V at any time t after the start is related to time t by V = 12 e-t/Rc Volts, Make t the subject of the formula. Given RC = 2 seconds, evaluate the time when V - 6 Volts. [P3]

ii.

The angle of a tapered groove is checked using a 20mm diameter roller shown above. If the roller lies 2mm below the top of the groove, determine the value of angle θ. [P3]

iii.

A chain hangs in the form give by y = 40 ch (x/40). Determine, correct to 4 significant figures,

(a) The value of y when x is 30 [P3]
(b) The value of x when y = 50 [P3]

Scenario:
A sample of five tests was taken determine the unconfined compression strength (in tons/foot2) of soil, with the test results shown in table below

Soil strength of sampled soil

 Sample number Strength of tested soil (Tons/foot2) 1 2.8 2 3.5 3 6.9 4 4.6 5 4,3

i. Calculate the mean of samples [P4]
ii. Calculate the variance and standard deviation of those samples [P4]
iii. Use coefficient of variation to determine the safety of the soil when it is used in a construction site. [P4]

Scenario:

This is the simplest control chart.

Sample size - the number of selected sample from a batch =k

Number of measurements on each sample test = n

After computing the mean of the k x n measurement values (it) and standard deviation (a), the upper and lower control limits for quality control can be determined by

LCLx = μ - 3σ/√n UCLx = μ + 3σ/√n

Quality control on the 1C chip by measuring the output voltage using 3 σ control chart.

5 samples randomly picked up from a storage bin with IC chip mass produced from a process.

3 measurements of voltage output (mV) from each sample, recorded as follows

 Sample 1: 2.25 3.16 1.80 Sample 2: 2,60 1.95 3.22 Sample 3: 1.75 3.06 2.45 Sample 4: 2.15 2.80 1.85 Sample 5: 3.15 2.76 2.20

iv. Calculate the mean of k x n measurement values (μ) and standard deviation (σ) [D1]

v. Calculate the upper and lower control limits for quality control [D1 ]

vi. Draw the quality control chart and explain how to use the chart to control manufacturing process [D1]

Analyze the probability distributions for discrete and continuous data with the following situations.

i. Concrete blocks are tested and it is found that, on average, 7% fail to meet the required specification. To pick up 9 concrete blocks from a batch of concrete blocks in a construction site, determine the probabilities that less than three blocks out from the chosen 9 concrete blocks will fail to meet the specification. [P5]

ii. A construction company wants to recruit a technical assistant to work in a new site and the offered salary is HK\$11,500. Current research has indicated that the average monthly income for a technical assistant in construction industry in above 13K Hong Kong dollars, It is also known that the standard deviation of income is \$9000. A random sample of 30 technical assistants' income is shown in Table 2.2, Use the sample data from the table to predict whether the construction company is able to recruit the assistant? Use the level of confidence to support your claim on this recruitment matter.

a) Use z value to predict the chance of hiring a technical assistant using an offered salary of HK\$11,500, [P5]

b) Applying the z value theorem, determine how much monthly income would be able to recrui a new technical assistance, by the construction company, to work in the new site with a change of 90% successful. [P5]

The formula for find the z value is given as follows:

z = (X‾ - μ)/(σ/√N)

X -
where μ is the population mean. σ is the population standard deviation. and N is the sample size.

X‾ is sample mean

The z value table of normal sampling distribution is shown in Appendix I.

Table 2.2 Sample of monthly income of technical assistance

 13800 16500 18500 12500 12000 11800 11200 9800 14500 14300 11800 16500 10050 13600 11500 12800 15500 10900 9600 13900 10800 8900 20000 15000 11300 11600 11800 12800 13500 12800

Scenario:

Population mean of IQ = 100, population standard deviation = 16

The IQ of some people were sampled and measured.

The hypothesis was that the mean IQ of those measured samples was significantly higher than the population mean.
The sample measurement was below]
Sample size N = 16
Sample mean IQ = 108
In social sciences, typically 5% probability of Type I error is acceptable.
tit Describe Type I and Type II error in significant test [M2]
iv. Calculate the z value [M2]
v. Find the probability p value from the z value table [M21
vi. Interpret the statistical result and comment on the hypothesis. [M21

Scenario:

A support beam, within an industrial building, is subjected to vibrations along its length; emanating from two machines situated at opposite ends of the beam. The displacement caused by the vibrations can be modelled by the following equations,
x1 =3.75 sin (100Πt+ 2Π/9) mm
x2= 4.42 sin (100Πt - 2Π/5) MM

i. State the amplitude, phase, frequency and periodic time of each of these waves. [P6]

ii. When both machines are switch on, how many seconds does it take for each machine to produce its maximum displacement? [P6]

iii. At what time does each vibration first reach a displacement of -2mm? [P6]

iv. Use the compound angle formulae to expand x1 and x2 into the form A sin 100Πt ± B cos 100Πt, where A and B are numbers to be found. [M3]

v. Using your answers from part iv, express x1 + x2 in a similar form. Convert this expression into the equivalent form R sin ((100Πt+ α).[M3]

vi. Using appropriate spread sheet software, copy and complete the following table of values: [D2]

 t 0.000 0.002 ' 0.004 0.006 0.008 0.010 0.012 0.014 0.016 0.018 0.020 x1 x2

vii. Plot the graphs of x1 and x2 on the same axes using any suitable computer package. Extend your table to include x1 + x2 and plot this graph on the same axes as the previous two. State the amplitude and frequency of the new wave. [D2]

viii. Using your answers from parts v and vii, what conclusions can be drawn about x1 + x2 and the two methods that were used to obtain this information? [D2]

Scenario:

A pipeline is to be fitted under a road and can be represented on 3D Cartesian axes as below, with the x-axis pointing East, the y-axis North, and the z-axis vertical. The pipeline is to consist of a straight section AB directly under the road, are in meters.

i. Calculate the distance AB.
The section BC is to he drilled in the direction of the vector 3i+4j+k.
ii. Find the angle between the sections AB and BC direction.
The section of pipe reaches ground level at the point C(a,b,0).
iii. Write down a vector equation of the line BC. Hence find a and b. [P7]

The distance x meters moved by a car in a time t seconds is give by x = 4t3 - t2 + 5t + 1.
1. Determine the velocity and acceleration when [P8]
a) t = 0s
b) t = 2s

Newtons laws of cooling is given by θ = θoe-kt, where temperature at zero time is θo°C and at time t seconds is θ°C.

ii. Determine the rate of change of temperature after 30s, given that θo = 20°C and k = -0.03 [P8]
iii. Using a general curve to describe the similarity and difference of maximum and minimum points in a curve [D3]
iv. Find the maximum and minimum value of the curve y = x3- 12x +8 by [D3]

a ). Examining the gradient on either side of the turning points, and

b) Determining the sign of the second derivative

v. Determine the height and radius of a cylinder of volume 300cm3 which has the least surface area [D3]

Scenario:

In a construction project, a slope excavation is shown by change of slope shape. Use calculus to find the area enclosed by curves represented the prior and final slope shape as shown in Figure 4.2
The two slopes shapes were represented by two curves y=x2 and y2-8x
i. Use calculus to the area shown by the shaded lines in Figure 4.2. [P9]

ii. In this area is rotated 360° about the x-axis, determine the volume of the soil of revolution produced. The volume of revolution , obtained by rotating area A through one revolution about the x-axis is given by the following expression: [P9]

iii. Consider the rate of change equation is dx/dt =kx [M4]
where t and x are variable and k is a constant with k ≠ 0

The number k is called the continuous growth rate if it is positive, or the continuous decay rate if it is negative.
Derive the exponential growth and decay formula by using into ration methods.

Outcome

P1 Apply dimensional analysis techniques to solve complex problems

Criteria - Using dimensional analysis to solve problem

P2 Generate answers from contextualized arithmetic and geometric progressions

Criteria - Understand and answer arithmetic and geometric progressions,

P3 Determine the solutions of equations using exponential, trigonometric and hyperbolic functions

Criteria - Able to solve the problem of exponential, trigonometric and hyperbolic functions,

P4 Summarize data by calculating mean and standard deviation, and simplify data into graphical form

Criteria - Able to calculate mean and standard deviation

P5 Calculate probabilities within both binomially distributed and normally distributed random variables

Criteria - Calculate the probability of both binomially and normally distributed random variables.

P6 Solve construction problems relating to sinusoidal functions

Criteria - Calculation of sinusoidal functions

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(ii) Given:

Area: 50,000 Sq.Mile

Covered by ice: 75% = 0.75(50,000)=37500 Sq.Miles

Avg.Thickness: 1300 meters

Densilty: 0.917 g/mL

To find:

Mass of ice, M.

Now,

(i)

The units of the Desired Produce is in Kilogram in case of MKS dimensional analysis.

(iii)

In case of Dimensionless units the Unit for mass can be derived as Below:

In general the Mass can be denoted by the Formula:

Mass,M= ρV

Where,

ρ is the Density

V is the Velocity

Now,

(iv)

In Dimesionless form:

Density can be denoted as: ML-3

Volume, V = L3

Now,

Mass,M=ML^3 L^(-3)

Where,

M=M

(V)

Calculation:

M= ρV

ρ = 0.917 g/mL

V = ?

To find Volume:

V = Ah

Where,

A is the Area

h is the Thickness of ice

V=1.29499e11*0.75* (1300)=1.261e14 m3

Now,

Mass,

M = 917 (1.261e14) = 1.15678e 17 Kg

The mass of the ice is found to be 1.1578 E17 Kilo Grams

(Vi) Force F acts over a Distance d on object of Mass m, find Final speed V of body.

Now,

We know that the force acting at a distance is torque, denoted by T

Now,

Torque, T = F.d (N.m)

Now,

(Vii) (Viii)The angular Speed can be given as:

v= a/r=F/mr (r/r)=Fr/?Mr?^2

Thus the equation for speed is found to be:

v= Fr/?Mr?^2 =T/I

Where,

T is torque

I is the moment of inertia.

i) Arithmetic progression (AP) : An Arithmetic Progression is also known as arithmetic sequence. It is defined as the sequence of number such that the difference between consecutive terms is constant.

ii) General expression for n'th term of an arithmetic progression :

The general form of an Arithmetic Progression is  a, a+d , a + 2d, a + 3d ...... so on

n'th term of an arithmetic progression (AP) Tn = a + (n-1) d

whereTn = nth term and a  = first term

(iii) T8 = 2 + (8-1) 5 = 37 (8th Term of series is 37)

(iv) Geometric Progression (GP) : It is defined as the sequence of number where each term after the first is found by multiplying the previous one by the fixed ,non-zero number called common ratio .This is known as Geometric progression or Sequence .

(v) The general form of an geometric sequence is ; a , ar , ar2, ar3 ........................

Where r ≠ 0

n'th term of an geometic progression (GP) with initial value a and common ratio r = an = a r n-1

(vi) The speeds are 30,A,B,C,D and 960 rev/min

Total 6 speeds .

In a geometric progression, each term is the product of the previous term and a common ratio (r).  So, the sequence is:

aar              ar2           ar3               ar4                           ar5

30         30 r           30r2        30 r3             30r4            960

So therefore ,

ar5  =  960

30 r5 = 960

r = 2

The speeds are

30 , 60 , 120 , 240 , 480 , 960 rpm

i) RC Discharge

Given that :When a capacitor C discharge through a resistor R the voltage V at any time t after the start is related totime t by V = 12 e-t/RC Volts.

RC = 2 seconds,

V = 6 Volts.

Taking log both sides,

Log V = 12 x -t/RC x log e

Log 6 = 12 x (-t/2) x 0.434

Time constant t = 0.298 sec

ii)

Length BC = 10 mm

AB = 30 - 10 - 2 = 18 mm

sin??(θ/2?)=10/18

(θ/( 2))=sin^(-1)??(10/18?)=33.748

θ=67.49°

iii) Given that :

y = 40 cos h (x/40)

(a)    x = 30

y = 40 cos h (30/40) = 40 x 1.294 = 51.76

(b)   y = 50

50 = 40 cos h (x/40)

Cos-1 h (5/4) = x/40

x = 27.72

(i) The Mean of the samples is

μ=(2.8+3.5+6.9+4.6+4.3)/5

μ = 4.42

(ii) The variance of those samples

σ^2=(?(2.8)?^2+?(3.5)?^2+?(6.9)?^2+?(4.6)?^2+?(4.3)?^2)/5

σ^2=21.47

Standard deviation

σ=√21.47

σ= 4.63

(iii) Coefficient of variation

CV=σ/μ

CV=1.047

 K1 2.25 3.16 1.8 K2 2.6 1.95 3.22 K3 1.75 3.06 2.45 K4 2.15 2.8 1.85 K5 3.15 2.76 2.2 MEAN 2.38 2.746 2.304

(iv) Mean-
μ=(2.38+2.746+2.304)/3=2.477
σ=√18.5133
σ=4.302

(v). Upper control Limit and Lower control Limit
Lcl=μ-3σ/√n
Lcl=-4.975
Ucl=μ+3σ/√n
Ucl=9.92

(vi).

Based on the upper and lower limits the graph is drawn. The following chart can be employed to control the manufacturing process by regulating the processes within the limits.

(i) Probability that less than three blocks will fail

=3/9 =1/3

(a).x ¯=13000
μ=12985
σ=9000
n=30
z=((x ¯-μ)/σ)/n=0.05

From the table the chance of hiring a technical assistant is

P=0.2797

(b).  90%

Z=100-90=10%=0.1

Therefore the value of p from the table is

P=0.557

x ¯=300(0.557)+12985

The monthly income is 13145

(iii). Type 1 error:

If the null hypothesis is true and it is considered as false it is termed as Type 1 Error

Type 2 error:

If the null hypothesis is false and it is considered as true it is termed as Type 2 Error.

(iv) The value of z is

z=((x ¯-μ)/σ)/n=(108-100)/16=0.5

(v) From the z table, we locate 0.5 along the z column. Since the value of z from (iv) is 0.5, we pick the value below the 0.00 column and 0.5 row. Thus. P = 0.6915.

(vi) Since the z value is positive from the result we can come to a conclusion that type 1 error is acceptable

Hypothesis

H0: Type one error is acceptable

H1: Type one error is not-acceptable

(i)The displacement caused by the vibrations can be modelled by the following equations :
?(iv)x?_1=3.75 (sin??100πt cos??(2π/9? ?) + cos?100πt sin?(2π/9))
?dx?_1/dt = 3.75 (100 π cos??100πt ?cos(???2π/9? ? )-100π sin??100 πt(2π/9?))
?dx?_1/dt=375π cos??(100? πt+2π/9 ) = 0
cos??(100? πt+2π/9 ) = 0
100πt+2π/9 = π/2
100πt = 5π/18
100 t = 5/18
t = 5/1800 seconds
x_2=4.42 (sin??100πt cos??(2π/5? ?) - cos?100πt sin?(2π/5))
?dx?_2/dt = 4.42(100 π cos??100πt ?cos(???2π/5? ? )+100π sin??100 πt(2π/5?))
?dx?_2/dt = 442π cos??(100? πt-2π/5 ) = 0
442 cos??(100? πt-2π/5 ) = 0
cos??(100? πt-2π/5 ) = 0
100πt-2π/5 = π/2
100πt = 9π/10
100 t = 9/10
t = 9/1000 seconds
(iii)
-2=3.75 s??in (100? πt+2π/9 )
s??in (100? πt+2π/9 ) = -0.533333
100πt+2π/9 = arc sin??(-0.533333)?
t=1.8809/100π=0.00599 second
(ii)
-2=442 s??in (100? πt-2π/5 )
s??in (100? πt-2π/5 ) = -0.45249
100πt=0.78708
t=0.00251 seconds

R Sin(100πt+a)
x_1=3.75 s??in (100? πt+2π/9 )
x_2=4.42 s??in (100? πt-2π/5 )

 Amplitude Phase Frequency Period x1 3.75 2π/9 (Sleading) 50 c/s 1/50 sec x2 4.42 -2π/5 (Slagging) 50 c/s 1/50 sec

x_1attains its max x_1 in 1/900 second after start.
x_2attains its max x_1 in 9/1000 second after start.
About 0.0095 second after start x_1 reaches -2mm
About 0.0025 second after start x_2 reaches -2mm
(v) A general wave equations is
x=A sin??(2πRt+P)?
x=Rsin (100πt+a)
Where R is the amplitude
t is the time

vi)

 t x1 x2 x1+x2 0 2.411261 -4.20436 -1.7931 0.002 3.639093 -2.59892 1.040178 0.004 3.475833 9.81E-16 3.475833 0.006 1.98389 2.598915 4.582805 0.008 -0.26642 4.20436 3.937938 0.01 -2.41489 4.202632 1.787742 0.012 -3.64023 2.594392 -1.04584 0.014 -3.47405 -0.00559 -3.47964 0.016 -1.97986 -2.60343 -4.5833 0.018 0.271152 -4.20608 -3.93493 0.02 2.418516 -4.2009 -1.78238

(vii)

From the figure,
Distance AB = √((40-?0)?^2+(0-(-4?0)?^2+(-20-?0)?^2 )
= √((40)^2+ (4?0)?^2+ (-2?0)?^2 )
= √(1600+1600+400) = 60 meters (Distance AB)

The given section BC is to be drilled in the direction of vector 3i+4j+k
The vector equation of BC is :
(a-40) i + (b - 0) j + (0-(-20)) k
(a-40) i + b j + 20k
Compare this equation by c (3i+4j+k )
a - 40 = 3c
b = 4c
20 = c
a = 60 + 40 = 100
b = 80 , c = 20
The vector equation of AB is
40 i + 40 j - 20 k
The vector equation of BC is
60 i + 80 j + 20 k
We know that,
<AB , BC > = cosθ?AB ? ? BC?
?-Denotes the norm of vector
cos??θ=((40 i+40 j-20k)(60 i+80j+20k))/(√((4?0)?^2+(40)^2+(20)^2 ) √((60)^2+(40)^2+(20)^2 ))?
cos?θ=(2400+3200-400)/(60 √10400)=52/61.14
cos?θ=0.85050
θ=cos^(-1)?( 0.85050)
θ=31.7339^0= 32^0
Angle between section AB and BC is 31.7339^0= 32^0

(i) The distance x meters moved by a car in a time t seconds is give by x = 4t3 - t2 + 5t + 1

t = 0 sec

Velocity :

a)      V = dx/dt = 12t2 - 2t + 5 = 5 m/s

t = 2 sec

b)      V = dx/dt = 12t2 - 2t + 5 = 49 m/s

Acceleration :

t = 0 sec

a = dv/dt = 24t - 2 = -2 m/s2

t = 2 sec

a = dv/dt = 24t - 2 = 46 m/s2

Cooling curve :

Newtons laws of cooling is given by ,

θ=θ_o e^(-kt)
dθ/dt = θ_o (-k)e^(-kt)

(ii) Rate of change of temperature = 20 (0.03) e^((0.03)30) = 1.4757

(iii)             The minimum and maximum points of a function the extrema of the function. So, relative extrema will refer to the relative minimums and maximums while absolute extrema refer to the absolute minimums and maximums.

(iv)             Maximum (-2,24) ,Minimum (2,-8)

max = c - (b2 / 4a) = 8 - (122/4) = -28 (Maximum point)

(V) V = 300 cm3

300 =300 = πr^2 h
h=300/(πr^2 )
Area=2πr^2+2πrh=2πr+2πr ( 300/(πr^2 ))
Area = 2πr^2+600/r
dA/dr= 0 for minimum area
= 4πr-600/r^2 =0
4πr=600/r^2
r^3= 600/4π=47.746
r=3.6278 cm
Area=2πr^2+2πrh=2π (3.62780)+2πr ( 300/(π?(3.62780?^2 ))
Area=248.08 ?cm?^2

(i)                 The two slopes shapes were represented by two curves y=x^2 and y^2=8x

= ∫_0^2?(x?^2 -2√2 √x ) dx
= 8/3
∫_a^b?πy^2 ? dx
∫_a^bπ8x dx=8π [ x^2/2 ]_a^b
= 8π/2[b^2-a^2] = 4π [b^2-a^2]

(iii)       dx/dt=kx

k≠0
k is called the continuous growth rate.

f (t) = e^kt
d/dt e^kt = k .e^kt
d/dt ?c.e?^kt = k .c.e^kt