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Worked Examples · Example 16

Q.A beam is supported at its ends by supports which are 12 metres apart. Since the load is concentrated at the centre of the beam 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?

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Model the deflected beam as a parabola with vertex at the point of maximum (3 cm) deflection; use the support points (zero deflection) to fix the parabola, then solve for the position where the deflection is 1 cm.

Parabola with vertex at origin, axis vertical, opening upward: x2=4ayx^2=4ay.

  1. Place the origin at the centre of the beam, at the point of maximum deflection (3 cm). Let yy measure how much the beam has risen back toward the support level as we move away from the centre (so y=0y=0 at the centre, and y=3y=3 cm at each support, where deflection is 0).
  2. The supports are 12 m apart, so each support is at horizontal distance x=6x=6 m from the centre. At the support, deflection is 0, i.e. the beam has risen the full 3 cm, so y=3y=3 there.
  3. Substitute the point (x,y)=(6,3)(x,y)=(6,3) (working in metres for xx and cm for yy, consistent with the given data) into x2=4ayx^2=4ay:

62=4a(3)⇒36=12a⇒a=36^2=4a(3)\Rightarrow36=12a\Rightarrow a=3

  1. Parabola: x2=12yx^2=12y.
  2. We need the point where the deflection is 1 cm, i.e. the beam has risen 3−1=23-1=2 cm from the centre, so y=2y=2:

x2=12(2)=24⇒x=24=26≈4.899 mx^2=12(2)=24\Rightarrow x=\sqrt{24}=2\sqrt6\approx4.899\text{ m}

  1. Self-check: at x=26x=2\sqrt6, y=(26)212=2412=2y=\dfrac{(2\sqrt6)^2}{12}=\dfrac{24}{12}=2 cm risen ⇒\Rightarrow deflection =3−2=1=3-2=1 cm ✓, matching the requirement.
✓Final answer

The deflection is 1 cm at a horizontal distance x=26≈4.90x=2\sqrt6\approx4.90 m from the centre of the beam.

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