[SOLVED] Thermodynamic and Mechanical Problems

Mathematics for construction

Table of Contents
TASK 1............................................................................................................................................3
TASK 2............................................................................................................................................9
Scenario 1.........................................................................................................................................9
a. Producing histogram with frequency density.....................................................................9
b. Producing cumulative frequency curve and extracting median and interquartile range. .11
c. Extracting mean, range and standard deviation................................................................13
Scenario 2.......................................................................................................................................13
A.a Normal distribution test.................................................................................................13
A.b Deciding interval allowed for replacement of not more than 10% prior to replacement
Type A..................................................................................................................................13
A.c. Practical consideration for replacement policy.............................................................14
A.d Replacement time affected Type B...............................................................................14
A.e Increment of 25% cost is preferable or not....................................................................15
A.f. Type C..........................................................................................................................15
B Testing hypothesis with 5% level of significance with both two and one tailed test.......15
TASK 3 .........................................................................................................................................18
TASK 4 .........................................................................................................................................22

TASK 1
Question – 1 A building services engineer is to design a water tank for a project. The tank has a
rectangular area of 26.5m2. With the design specifics of the width being 3.2m shorter than the
length, calculate the length and width to 3 significant figures for resource requirements
Ans :
width= 3.2 m
length = 3.2 m
area = 26.5 meter square
area of rectangle= l X b
222 5 = l x 3.2
l= 26.5 / 3.2
l = 8.28
where
width = 3 m
area = l x b
26.5 = lx 3
l= 26.5 / 3
l= 8.83 m

Question – 2 As an employee of company JR construction you have received a letter regarding
the project your company is working on. It has a penalty clause that states the contactor will
forfeit a certain some of money each day for late completion. (i.e. the contractor gets paid the
value of the original contract less any sum forfeit). If she is 5 days late she receives £4250 and if
she is 12 days late she receives £2120. Calculate the daily forfeit and determine the original
contract.
Ans :
Let the original contract amount be x and per day forfeit be y.
When work is delayed by 5 days:
x – 5y = 4250 (1)
Similarly when work is delayed by 12 days:
x – 12y = 2120 (2)
On solving equations 1 and 2 we get
y = 304.28 and
x = 4250 +(5*304.28) = 5771.4
Thus
Daily forfeit = £304.28
Original contract amount = £5771.4
Question – 3 A car driving at an average speed 65 miles/hour
Ans :
A)
1). Average speed= 65 miles/ hour [1 mile = 1760 yards]
65 miles per hour = 65 x 1.609 = 104.585 yard/ second
65 miles per hour = 29.05746 meter per second
2). Time = distance / speed
Time = 100/ 29
Time = 3.44 hour/ second

3). 30 miles / gallon
(miles/100 km ) x (liters/gallon) / (miles/gallon)
(3.785 * 62.14)/30 = 235.19/30 = 7.84 liters/100 km
4). 100 km/ 7.84 liter = 12.755 km per litre
= 7.84 l x 100 /100
= 7. 84 litre for the journey.
B). Formula → Lift =k x ρ x V^2 x A
lift = meter2 x Kg/meter3 x in2
Lift = Kg. m/s2 = newtons or slug
1. An arithmetic sequence is given by b, , , 0....... 3 3 2b b  Determine the sixth term  State
the kth term  If the 20th term has value of 15 find the value of b and the sum of the first 20
terms
Ans :
a). a = a+(n-1)d
a = b(6-1)b/3
a = bx 5b/3
a = 8b/3 6 th terms
b). The k term is given by :
a(k) = b + (k-1)* (b/3)
c). a(20)= b +(20-1)*(b/3)
15 = (3b +19b)/3

45= 22b
b= 2.04
Sum of n terms of A.P. = (n/2)*[2a + (n-1)*d]
n = 20 , a= b= 2.04, d =b/3 =0.68
On substituting the values in equation we get
S(20) = 10 * [4.08 + 12.92]
= 170
Thus the sum of first 20 terms of given AP is 170
2. For the following geometric progression 1, ½, 1/4 ........ determine
a) The 20th term of the progression = a(1- r^n) / 1-r
= (1-(1/2)^20) / 1- ½
= 39 x 2 / 40
= 39 / 20
b). Sn = ∑ (n=0) ar^n
Sn = 1/2(1-r^∞) / 1-r
when the value of r is infinity it means the entire system it
Solve the following Equations for x :
(a) 2Log (3x) + Log (18x) = 27
Ans : log ((3x^2)) + log (18x) = 27 
log ((3x)^2 . 18X ) = 27
log ((3x)^2 . 18X ) = log (10^27) 
(3x)^2.18x = 10^27