Question

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 ?​

Answer 1

2$\sqrt{6}$ 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 = $\frac{PQ}{2}$ = $\frac{12}{2}$ = 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 = $\frac{360000}{12}$ = 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 = $\sqrt{240000}$

x = 200$\sqrt{6}$  cm  ⇒ 2$\sqrt{6}$ m.

x = 2$\sqrt{6}$ m.

Therefore, the distance is  2$\sqrt{6}$ m.

To learn more...

1. The length of the intercept made by parabola x^2 - 7x + 4y + 12 = 0 on x-axis is

(A) 4

(B) 3

(C) 1

D)2​

brainly.in/question/13789722

2. The focus and vertex of a parabola are (4, 5) and (3, 6) equation of Axis is

brainly.in/question/1170545