Simultaneous Equations Practical Construction
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Simultaneous Equations Practical Construction

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Simultaneous equations for practical construction problems

Simultaneous equations for practical construction problems

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Simultaneous Equations Practical Construction Simultaneous Equations Practical Construction Presentation Transcript

  • Tools: For now you will need….
    • Pen
    • Paper
    • Calculator
    Context…
  • Mathematics in Construction and the Built environment
    • Construction, civil engineering and building services engineering are technical disciplines which require the collection, processing and use of numerical data.
    • Construction team members will include ….
    • Designer – architect – engineer - cost controller - quantity surveyor
    • In complex situations, design engineers use formulae to….
    • calculate bending moments in beams and sizing for structural elements.
    • “ It is therefore essential that learners develop an understanding of the mathematical methods and techniques required for key activities, and of how to apply them correctly”.
    • The unit explores the rules for manipulation of formulae and equations
    Aims….
  • Aims for this topic:
    • You will be Learn how to solve real problems by forming and solving simultaneous equations using an algebraic method
    • You will learn how to solve simultaneous equations using graphs
    • On completion of this unit a learner should:
    • Know the basic underpinning mathematical techniques- methods used to manipulate and/or solve formulae, equations and algebraic expressions
    Recap….
  • Simultaneous Equations recap How do you go about solving simultaneous equations?...
  • A few hints . . . (1)Scale up each term in one, or both equations to make the same number in front of either the x terms or the y terms. (2)Subtract if the signs in front if the signs are the same. (3) Add if the signs in front of the numbers are different. Lets work through the first one together – have your NO MESS handout ready…..
  • 5x + y = 20 2x + y = 11 … (1) … (2) Number the Equations 3x = 9 Subtract (to get rid of a letter) Divide (to find x) x = 3 Substitute in (2) (to find y) 2 x 3 + y = 11 6 + y = 11 y = 5 1 You try the next one…
  • 7x + y = 43 3x + y = 23 … (1) … (2) Number the Equations 4x = 20 Subtract (to get rid of a letter) Divide (to find x) x = 5 Substitute in (2) (to find y) 3 x 5 + y = 23 15 + y = 23 y = 8 2 A nice System….
  • when using the elimination method to solve equations FOR NO MESS Remember
    • FOR M your equations
    • N eatly
    • O rganise your equations a x + b y = number
    • M ake the number of x’s or y’s the same if you have to
    • E liminate a letter
    • S olve the equation
    • S ubstitute
    • C heck your answer!
    Next stage - Form your own…
  • Forming equations
    • Next we will look at forming our own equations from given information-turning word problems into equations
    • We will do the first task
    • together…
  • Task 1: Financial Penalty clause
    • A penalty clause states that a contractor will forfeit a certain sum of money for each day that they are late in completing a contract
    • (i.e. the contractor gets paid the value of the original contract less any forfeit).
    • If they are 6 days late they receive £5000
    • if they are 14 days late they receive £3000.
    • (a) Find the amount of the daily forfeit
    • (b) Determine the value of the original contract.
    How?…
  • Task 1 Solution
    • To solve this problem - there are 2 unknowns…..
    • the original contract amount C
    • the Daily forfeit D
    • For 6 Days late C+6D = 5000 (1)
    • For 14 Days late C+14D = 3000 (2) Then use (2) - (1)
    • 8D = -2000
    • D = - 250
    • Sub D into (1) gives C + 6 (-250) = 5000
    • C-1500 = 5000
    • C = £6500
    You need a Strategy – see NO MESS handout….
  • Strategy Remember to use the phrase…
    • FOR NO MESS
    -Now lets try one…
  • Task 2: General Office equipment
    • The total cost of equipping 2 offices, A and B is £30,000.
    • Office B costs £2,000 more than office A.
    • find the cost of equipment for each of them.
  • Task 2: Solution
    • A+B = 30,000 (1)
    • A-2000 = B (2)
    • Re-arrange (2) giving
    • A-B = 2000 (3)
    • (1) - (3) leaves B- -B = 28000
    • 2B = 28000
    • B = £14000
    • Then A = £16000
  • Task 3: Office heating: costs
    • A heating installation for one office consists of 5 radiators and 4 convector heaters and the cost including labour is £2080.
    • In a second type of office 6 radiators and 7 convector heaters are used and the cost, including labour is £3076.
    • In each office, installation costs are £400.
    • Find the cost of a radiator and the cost of a convector heater.
  • Task 3: Solution
    • 5r + 4c = 2080 - 400 (1)
    • 6r + 7c = 3076 - 400 (2)
    • 5r +4c = 1680 (1)
    • 6r + 7c = 2676 (2)
    • (1) x 6/5 gives 6r + 4.8c = 2016 (3)
    • (2) –(3) 2.2c = 660
    • C = £300
    • Sub c into (1)
    • 5r + 4 x 300 = 1680
    • r = £480
  • Task 4: Plumbing costs
    • 100m of tubing and 8 elbow fittings cost £100.00.
    • 150m of tubing and 10 elbow fittings cost £147.50.
    • Find the cost of 1m of tubing and the cost of an elbow fitting.
  • Task 4: Solution
    • 100m + 8e = 100 (1)
    • 150m + 10e = 147.50 (2)
    • (1) x 1.5 gives 150m + 12e = 150 (3)
    • (3) – (2) gives 2e = 2.50
    • e = £1.25
    • Sub e into (1) 100m + 8x£1.25 = 100
    • 100m + 10 = 100
    • 100m = 90
    • 90/100 = m
    • m= 90p
  • Task 5: Orders-Pricing-inflation
    • The cost of grade B bolts is £0.60p
    • The cost of grade A bolts is £0.90p.
    • A combination of several samples of each costs £9.00.
    • After a year into the construction, inflation raises the cost of each sample by 15p and the cost of the combination rises to £ 10.65.
    • Find the number of each sample ordered.
  • Task 5: Solution
    • 0.60B + 0.90A = 9.00 (1)
    • 0.75B + 1.05A = 10.65 (2)
    • (1) x 0.15 gives 0.75B + 1.125A = 11.25 (3)
    • (3) – (2) gives 0.075A = 0.60
    • 0.60/0.075 = 8 = A
    • Sub A into (1) gives 0.60B + 0.90 x 8 = 9
    • 0.60B + 7.2 = 9
    • 0.60B = 1.8
    • 1.8/0.60 = 3 = B
  • Task 6: Construction plant
    • For the crane which lifts the items it is found that the effort E Newtons and the load W Newtons are connected by the equation E =aW + b.
    • An effort of 90N lifts a load of 100N
    • An effort of 130N lifts a load of 200 N.
    • Find the values of a and b
    • Determine the effort required to lift a load of 300N
  • Task 6: Solution
    • E = aW + b
    • 90 = 100a + b (1)
    • 130 = 200a + b (2)
    • (2) - (1) gives: 40 = 100a
    • a=0.4
    • Sub a into (1) gives:90 = 100x0.4 + b
    • 90 = 40 + b
    • b = 50
    • Now use E = aW + b
    • E = 0.4 x 300 + 50
    • E = 170N