This document provides study material for Mathematics for Construction. It includes solved assignments, essays, and dissertations related to various topics in the subject. The document covers scenarios and equations related to length, width, velocity, power, lift, geometric progression, logarithms, and more.
Contribute Materials
Your contribution can guide someone’s learning journey. Share your
documents today.
Mathematics for Construction 1
Secure Best Marks with AI Grader
Need help grading? Try our AI Grader for instant feedback on your assignments.
TASK 1 Scenario 1 1. Let x be the length, then: l * w = 26.5 thus, x * (x-3.2) = 26.5 x2– 3.2x = 26.5 x2– 3.2x - 26.5 = 0 Solving quadratic equation applying quadratic yield formula: x = {- (-3.2) +/-√ [(-3.2)2– 4 * 1 * (- 26.5)]} / 2 * 1 x = 6.99 m Length = 6.99 m Width = 6.99 – 3.2 = 3.79 m 2. If she is 5 days late she receives = £4250 if she is 12 days late she receives = £2120 Difference in amount she receives = £4250 - £2120 = £2130 Difference in days late = 12 days – 5 days = 7 days Then per day pay is = £ 2130 / 7 = £ 304.28 So, constant pay is = 5771.4 Penalty per day is = 304.28 3
Scenario 2 A i. Fd= 0.5 ** u2* Cd* A Solution Fd= Force thus its dimension is: [M1L1T-2] Similarly, on Right hand side = density = kg / meter cube = [M1L-3T0] A = metersquare = [M0L2T0] Cd =Nodimension u2= velocity =u2 =[meter/ second]2= [M0L2T-2] 0.5 ** u2* Cd* A =[M1L-3T0] * [M0L2T-2] * [M0L2T0] =[M1L1T-2] Thus LHS = RHS and equation can be said to be dimensionally correct. ii) Power (P) = [M1L2T-3] Weight (W) = [M1L-2T-2] Height (h) = [M0L1T0] Fluid flow rate (q) = [M0L3T-1] Solution Power is proportional to W; h and q thus : Power = flow rate * work done P = q * Work Work = mass * height * gravity Work = (Weight /G) * h * g G and g can be considered as constant Thus dimension of work = dimension of weight * dimension of height dimension of work =[M1L-2T-2] * [M0L1T0] Power =flow rate * work done = [M0L3T-1] * [M1L-2T-2] * [M0L1T0] 4
Paraphrase This Document
Need a fresh take? Get an instant paraphrase of this document with our AI Paraphraser
Power = [M1L2T-3] Hence dimensions are verified Thus equation of power is: P = q * (W /G) * h * g Where Q = fluid flow rate w = weight h = height g = 9.8 m/s2 G = 6.67x10-11(N-m2/kg2) b. Lift = k** V2 *A Solution k= no dimension = kg / meter cube = [M1L-3T0] A = metersquare = [M0L2T0] V2 =[meter/ second]2= [M0L2T-2] Lift = [M1L-3T0] * [M0L2T-2] * [M0L2T0] Lift = [M1L1T-2] Thus dimensions of lift are:(Kilogram*meter) per second square = Newton Scenario 3 1. (i)-2b/3 5
(ii)b(4 –n)/3 (iii)-45/16 2. Consider the geometric progression 1, 1/2, 1/4 , 1/8 , 1/16 , . . . . or 1, ½, 1/22,1/23,1/24…... Thus, The 20th term of the progression = 1 / 219 We have a = 1 and r = 1/2 , and so we can calculate some sums. We get S1 = 1 S2 = 1 + 1/2 = 3/2 S3 = 1 + 1/2 + 1/4 = 7/4 S4 = 1 + 1/2 + 1/4 + 1/8 = 15/8 S4 = and there seems to be a pattern because 1 = 2 – 1 3/2 = 2 – 1/2 7/4 = 2 – 1/4 15/8 = 2 – 1/8 In each such case, should subtract small quantity out of 2, as well as take the successive total quantity getting smaller than smaller. When one capable to add the infinitely terms, then response ‘ought-to-be’ 2 or nearly 2. 6
TASK 2 Scenario 1 (a) (b) Upper Class January (f) Cumulative Frequency 8
boundary 42727 93865 1440105 1922127 2913140 394144 Median = 144/2 = 72 Thus as per graph median would be = 10 lower quartilen/4 = 144/4 = 36 upper quartile = 36 * 4 = 144 Upper Class boundary JulyCumulative Frequency 42222 93961 1469130 1941171 2920191 9
395196 Median = 196/2 = 98 Thus as per graph median would be = 12 lower quartile = n/4 = 196/4 = 49 upper quartile = 49 * 4 =196 (c) RevenueMid- Point January (f)Cumulative Frequency m*fm2f*m2 Less than 522727544108 5andless than 10 73865266491862 10andless than 15 12401054801445760 15andless than 20 17221273742896358 20andless than 30 24.513140318.5600.257803.25 30andless34.541441381190.254761 10
Paraphrase This Document
Need a fresh take? Get an instant paraphrase of this document with our AI Paraphraser
than 40 Sum of the product of the midpoints and frequencies = 1630.5∑f⋅\M2 =26652.25 Mean= 1630.5/144 = 11.32292 Range= 30-0 =30 Standard Deviation =(26652.25-144*11.32* 11.32)/144-1 = 47.27 RevenueMid- Point JulyCumulative Frequencym*fm2f*m2 Less than 52222244488 5 and less than 10 73961273491911 10 and less than 15 12691308281449936 15 and less than 20 174117169728911849 20 and less than 30 24.520191490600.2512005 30 and less than 40 34.55196172.51190.255951.25 Sum of the product of themidpointsand frequencies = 2504.5∑f⋅\M2 = 41740.25 11
Mean= 2504.5/196 = 12.77806 Range= 30-0 =30 Standard Deviation =(41740.25-144*12.78* 12.78)/196-1 = 49.94 Scenario 2 (a). z for the 10th percentile of failure is -1.282 length of life is z=(x-mean)/sd so -1.282*60=x-360 -76.92=x-360 x=283 days (b). Here, Practicality implies to amount of works needed--replacing might be timing intensive. The costs of the replacing all such bulbs have to balance against the issue is one burn-out. (c). B bulbs which lasts 450 days, 90 days longer, will require all to be replaced (same policy) in 373 days. (d). If B lasts 25% longer, by the means, it is exactly 25% longer. But because the sd is less per the mean, the 10th percentile for B is 31% longer than it is for A costing 25% more. It's a close call slightly in favour of B. (e). Type C would be -1.282*45=x-432 so x=374.31. It is slightly more than B, and has less variability, generally a good thing, its costs should be similar to B although slightly less would be reasonable (0.3%). TASK 3 Scenario 1 (i)The displacement caused by the vibrations can be modelled by the following equations : x1=3.75 (sin100πt cos(2π/9 ) + cos100πt sin(2π/9)) dx1/dt = 3.75 (100 π cos100πt cos(2π/9 )-100π sin100 πt(2π/9)) dx1/dt=375π cos(100 πt+2π/9 ) = 0 12
cos(100 πt+2π/9 ) = 0 100πt+2π/9 = π/2 100πt = 5π/18 100 t = 5/18 t = 5/1800 seconds (ii) x2=4.42 (sin100πt cos(2π/5 ) - cos100πt sin(2π/5)) dx2/dt = 4.42(100 π cos100πt cos(2π/5 )+100π sin100 πt(2π/5)) dx2/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 sin (100 πt+2π/9 ) sin (100 πt+2π/9 ) = -0.533333 100πt+2π/9 = arc sin(-0.533333) t=1.8809/100π=0.00599 second (iv) -2=442 sin (100 πt-2π/5 ) sin (100 πt-2π/5 ) = -0.45249 100πt=0.78708 t=0.00251 seconds R Sin(100πt+a) x1=3.75 sin (100 πt+2π/9 ) x2=4.42 sin (100 πt-2π/5 ) 13
Secure Best Marks with AI Grader
Need help grading? Try our AI Grader for instant feedback on your assignments.
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 equation is x=A sin(2πRt+P) x=Rsin (100πt+a) Where R is the amplitude t is the time tx1x2x1+x2 02.411261-4.20436-1.7931 0.0023.639093-2.598921.040178 0.0043.4758339.81E-163.475833 0.0061.983892.5989154.582805 0.008-0.266424.204363.937938 0.01-2.414894.2026321.787742 0.012-3.640232.594392-1.04584 0.014-3.47405-0.00559-3.47964 0.016-1.97986-2.60343-4.5833 0.0180.271152-4.20608-3.93493 0.022.418516-4.2009-1.78238 14
Scenario 2 (i). From the figure, Distance AB = √((40-0)^2+(0-(-40)^2+(-20-0)^2 ) = √((40)^2+ (40)^2+ (-20)^2 ) = √(1600+1600+400) = 60 meters (Distance AB) (ii) 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 > = cosθABBC (iii). Denotes the norm of vector cosθ=((40 i+40 j-20k)(60 i+80j+20k))/(√((40)^2+(40)^2+(20)^2 ) √((60)^2+(40)^2+(20)^2 )) cosθ=(2400+3200-400)/(60 √10400)=52/61.14 cosθ=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 TASK 4 Scenario 1 (a.) Bending moment M = 3000 – 550 x – 20 x2 Solution M = 3000 – 550 x – 20 x2 dm/dx = -550 – 40 x 15
equating dm/dx = 0 we have: -550 – 40 x = 0 so: 40x = - 550 x = -13.75 d2M /dx2= -40 It is negative thus the given momentum function haslocal maxima at x = -13.75 The maximum value of function is given by: M (x = -13.75) =3000 – 550 x – 20 x2 M (x = -13.75) = 6781.25 The bending moment is zero at x = 4.664 . (aii) The bending moment (b.) Ө = 300 + 100 e^ (-0.1 t) Ө is temperature in degree Celsius and t is time in minutes Solution At t = 0 Ө = 300 + 100 e^ (0) = 400 At t = 1 Ө = 300 + 100 e^ (-0.1*1) = 390 At t = 2 16
Paraphrase This Document
Need a fresh take? Get an instant paraphrase of this document with our AI Paraphraser
Ө = 300 + 100 e^ (-0.1*2) = 381 At t = 5 Ө = 300 + 100 e^ (-0.1*5) = 360 At t = ∞ Ө = 300 + 100 e^ (-0.1*∞) = 300 At t = -∞ Ө = 300 + 100 e^ (∞) =∞ Thus value of temperature is positive for all values of t between -∞ and +∞ (C) log P + n log V – log C Solution P stands for pressure; V for volume and C is constant and n is known as index log P + n log V = log C By using the identity: a log x = log xa log P + log Vn= log C By identity: log a + log b = log ab we can write the above equation: Log P Vn= log C P Vn= CHence proved Also log P =log C - n log V Differentiating both sides with respect to P; we have: 1/P = -n (1/V) dV/d P dV/d P = - (V/ n*P) For n = 2 we have dV/d P = - (V/ 2P) 17
Scenario 2 C = 16 t2+ 2t Solution (a. ) T12345678 C18681502644105887981040 (b) The calculus can be used to provide the analytical solutions for the given condition. Using concept of differentiation, it is possible to find the values of production time at which maximum or minimum cost is obtained. The information helps in optimising the production cost by making adjustments to the production time. (C) C = 16 t2+ 2t Differentiating cost function with respect to time. 18
dC/dt = 32t + 2 Equating first derivative to zero 32t + 2 = 0 t = -0.062 Second derivative of cost function is: d2c/dt2= 32 Which is greater than zero. It means at t = -0.06 minimum production cost is obtained which is £32. (D). The turning point is at t= -0.06 at which cost is minimum and beyond that time the cost function becomes negative. Scenario 3 Solution Heat flow H = e3t Where t is the temperature difference outside the building. At t = -20H = 8.75651E-27 At t = 0H = 1 At t = 20H = 1.14…. E26 19