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Example 18 A beam is supported at its ends by supports which are 12 metres apart. Since the load is concentrated at its centre, there is a deflection of 3 cm at the centre and the deflected beam is in the shape of a parabola. How far from the centre is the deflection 1 cm? Beam is always facing upwards with the axis vertical Since, the axis is positive y-axis, its equation is x2 = 4ay First we find coordinates of point B Given Width of beam = 12 m Hence, AB = 12 m So, BC = 𝐴𝐵/2 = 12/2 = 6 m Also, there is a deflection of 3 cm from centre So, OC = BD = 3 cm OC = BD = 3 cm = 3/100 m Hence point B is B(6, 𝟑/𝟏𝟎𝟎) Now, Since point B(6, 3/100) lies on the parabola Putting x = 6, y = 3/100 in equation x2 = 4ay (6)2 = 4a (3/100) 36 = (3𝑎 )/25 3/25 a = 36 a = (36 × 25)/3 a = 12 × 25 a = 300 m Now, we need to find how far from the centre is the deflection 1 cm Hence RQ = 1 cm, We need to find OP QP = 3 cm – 1 cm = 2cm = 2/100 m Let OP = x So, coordinates of point Q is Q (x, 𝟐/𝟏𝟎𝟎) Since point Q lies on parabola it will satisfy the equation of parabola Equation of parabola is x2 = 4ay Putting x = x & y = 2/100 m & a = 300 m x2 = 4 (300) (2/100) x2 = 1200 × 2/100 x2 = 24 x = √24 = √(6 ×4) = 2√6 𝑚 Thus, the required distance is 2√𝟔 𝒎

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo