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Mathematics for Construction Assignment Solved

   

Added on  2021-02-20

12 Pages2085 Words124 Views
Calculus(Mathematics forConstruction)

TASK 1(i) The project has a series of double cantilevered beams with the Bending Moment equationdetailed belowBM = ax2 -bx +cFor your delegate student number and the details given on sheet 1 calculate the point ofmaximum bending using calculusSolution: The data given for different values of a, b and c as per number of students are givenbelow:-For Student 8: a)25 b) 190.0338 c) 205.2365So, the point of bending can be calculated by solving quadratic equation as –25x2 -190.0338x +205.2365Using, Calculus method : x = -b ± √b2 - 4ac ÷ 2ax = (190.0338 ± √ (190.0338)2 - 4* 25 * 205.2365) ÷ 2*25= ( 190.0338 ± √36112.84 - 20523.65) ÷ 50= ( 190.0338 ± 124.85) ÷ 50 = 1.303 or 6.297 (approx.) therefore, maximum point of bending will be gained at x = 6.297(ii) A formula for the height of suspension bridge across the river Lune main cables has beencalculated as Y = 0.00455x2 – 0.2275x +10Calculate the lowest point of the suspension cable in distance (x) and height (y). Also calculatethe length of the suspension wires to the main road deck which is at 5.120mSolution: The lowest point of suspension cable can be calculated by -Y = 0.00455x2 – 0.2275x +10differentiating both sides with respect to x, dY/dx = d/dx (0.00455x2 – 0.2275x +10)= 0.0091x – 0.2275 at maximum or minimum point, dY/dx = 0which implies that, 0.0091x – 0.2275 = 0

x = 0.2275 / .0091= 25 so, lowest point of suspension level in distance x = 25m and y = 0.00455 (25)2 – 0.2275 (25) +10= 2.84 – 5.68 + 10= 7.16mLength of the suspension wires to the main road deck which is at 5.120m will be, Y = 0.00455(5.120)2 – 0.2275 (5.120) +10= 0.119 – 1.164 + 10= 8.955m (iii) Find the locations when the gradient of the voltage curve v given by the expression v =3(sin4θ) equals zeroSolution: Expression of voltage v is given by,v = 3 (sin 4θ)differentiating the equation with respect to θ, dv/dθ = d/dθ [3 (sin 4θ)] v' = 3 cos 4θ . 4= 12 cos 4θ The locations can be calculated by taking first derivative at 0,dv/dθ = 012 cos 4θ = 0cos 4θ = 04θ = 0° or πθ = 0° and π/4 or 45°

Task 2(i) Calculate using calculus the location of the centre of gravity along the x axis for the profile ofthe flood defence wall detailed on sheet 2.Solutions: Total area of following figure = area of triangle + area of big rectangle + area of smallRectangle= ½ x 3.2 x 7 + 3.8 x 7 + 3.6 x 3.3= 11.2 + 25.6 + 11.88= 48.68 square unit. Now, centre of gravity can be determined by, Centre of gravity for triangle = h/3= 7/3 = 2.33Centre of gravity for rectangle = h/2= 7/2 = 3.5 Centre of gravity for small rectangle = h/2= 3.3/2 = 1.65 Therefore, total centre of gravity = 2.3 x 11.2 + 3.5 x 25.6 + 1.65 x 11.8811.2 + 25.6 + 11.88= 135.962 / 48.68 = 2.796Therefore, centre of gravity of given figure is at a distance of 2.8 units from left corner of figure.

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