A deflected beam is the shape of aparabola x2 = 4ay. The beam is supportedat its ends which are 12m apart. The
deflection in the centre is 9cm only. Findthe points on the beam where deflection
is 5cm.
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Answers
Answer:
2 m far from the center is the deflection 1 cm.
To find : How far from the center is the deflection 1 cm?
Given :
A beam is supported at its ends by supports, 12 m apart.
Load is concentrated at its center.
Deflection occurs at the center of 3 cm.
Deflected beam is in the shape of parabola.
Note :
With the given data graph has drawn which is attached below, refer it for better understanding.
From the drawn graph it shows that parabola is formed in y-axis :
So, the parabola equation is x² = 4ay.
Here, "R" is the midpoint of P and Q.
PQ = 12 cm.
OR = 3 cm.
RQ = = = 6 cm.
Converting 6 m to 6 cm :
1 m = 100 cm
6 m = 6 × 100 = 600 cm.
Co-ordinates of Q = (600, 3)
Where, x = 600 ; y = 3
Applying the values of "x" and "y" in the equation of parabola we get,
x² = 4ay.
(600)² = 4 × a × 3
120000 = 12 a
a = = 30000
a = 30000.
Applying the values of "a" in the equation of parabola we get,
x² = 4ay.
x² = 4 × 30000 × y.
x² = 120000y.
Finding deflection :
AB = 1 cm.
AC = 3 cm.
OC = x cm.
BC = ?
BC = AC - AB
= 3 - 1 = 2
BC = 2 cm.
B(x, 2) lies on the parabola.
Applying the values of "B" in the equation of parabola, we get
B(x, 2)
x² = 4ay.
x² = 4 × 30000 × y.
x² = 4 × 30000 × 2.
x² = 240000
x =
x = 200 cm ⇒ 2 m.
x = 2 m.
Therefore, the distance is 2 m.
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Hence RQ=1cm
QR=3cm−1cm=2cm=
100
2
m
let OP=x
⇒Q is (x,
100
2
) lies on parabola
x
2
=4(300)((
100
2
)=24
⇒x=
24
=2
6
m
∴ Required distance is 2
6
m.