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In beams when we are converting uniform disturbed load to point load, the point load is assumed to acting at the centre of length why is it so? And converting uniform varying load we assume it to act 1/3 or 2/3 of the total length, What's the logic behind it?

Thejeswini S
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1 Answers1

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Without getting into the technical details of the nature of supports for the particular beam, I would assume that you are referring to a cantilever beam for simplicity.

Assume that a load per unit length $w$ acts over a length $L$ of the beam causing it to deflect. The main idea behind converting distributed loads to point loads is to ease calculations, while still retaining the physical constraints of the problem intact. In this respect, the net force and moment created by the equivalent point load should be the same as the distributed load.

For equivalence of forces, the process is trivial.

$$ P = w.L $$

For equivalence of moments,

$$ P.d = \int w.x\, dx $$

where, $d$ is the distance at which the equivalent point load should be acting in order to ensure the external moment should remain the same.

Assuming an uniformly distributed load, we get,

$$ d = \frac{wL^2}{2wL} $$

which gives the distance of acting of equivalent load as, $d = \frac{L}2$

This process can be repeated for different distributed loads such as uniformly varying loads, sinusoidally varying loads, and so on.

Hope that helps !

Alucard Nosferatu
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  • Can you help me understand what does the term( int w.x dx) means? – Thejeswini S Mar 19 '24 at 04:33
  • If $w$ is taken to be the force per unit length, then $(w).dx$ simply refers to the force acting on an infinitesimally small length of the beam. By extension,$w.x.dx$ refers to the moment acting on the infinitesimally small length (w.r.t to some chosen reference point). Integration of the quantity results in the total moment acting on the beam w.r.t the given reference point.

    Hope that helps !

     – Alucard Nosferatu Mar 25 '24 at 06:41