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 = π΄π΅/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, π/πππ)
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 = (3π )/25
3/25 a = 36
a = (36 Γ 25)/3
a = 12 Γ 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 β 1 cm = 2cm = 2/100 m
Let OP = x
So, coordinates of point Q is Q (x, π/πππ)
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 Γ 2/100
x2 = 24
x = β24 = β(6 Γ4) = 2β6 π
Thus, the required distance is 2βπ π

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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