Mathematics for Construction

Mathematics for Construction
1

Contents
Contents...........................................................................................................................................2
TASK 1............................................................................................................................................3
Scenario 1........................................................................................................................................3
1...................................................................................................................................................3
2...................................................................................................................................................3
Scenario 2........................................................................................................................................4
Scenario 3........................................................................................................................................5
1...................................................................................................................................................5
2...................................................................................................................................................6
3...................................................................................................................................................7
TASK 2............................................................................................................................................8
Scenario 1........................................................................................................................................8
(a).................................................................................................................................................8
(b).................................................................................................................................................8
(c)...............................................................................................................................................10
Scenario 2......................................................................................................................................12
TASK 3..........................................................................................................................................12
Scenario 1......................................................................................................................................12
Scenario 2......................................................................................................................................14
TASK 4..........................................................................................................................................15
Scenario 1..................................................................................................................................15
Scenario 2..................................................................................................................................18
Scenario 3..................................................................................................................................19
2

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 = meter square = [M0L2T0]
Cd = No dimension
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