The question and Image are shown below:

What I tried: Using the conjugate beam method to find deflection and hence stiffness of the beam, but I am unable to find reactions in the conjugate beam.


Let us ignore the mass of the beam.

Then by inspection of symmetry and the rigidity of the two beams, we note that the deflection of beam under mass m, Yb is equal to deflection of the point D, Yd, because when the beam vibrates it will form an isosceles triangle whit the point c vibrating from maximum to minimum and points A and E stationary.

So we stablish $ \quad Y_{B} = Y{_ d} \quad $ meaning the deflection of spring is exactly as if it were under mass m, so the natural frequency is:

$ F = m*\alpha = k*y $

$ \omega_{ n} = \sqrt{\frac{ k}{m}} $

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  • $\begingroup$ It is given that the beam is rigid, so there must be an additional force due to the stiffness of beam( I think). $\endgroup$ – Shyamal Verma Jan 28 '19 at 8:03
  • $\begingroup$ @ShyamalVerma, I did assume the beam as rigid, otherwise the system would be three springs and one mass. But because the mass of the beam is not denoted I left it out. Or else one would need to add 2*Mb*Lb^2/3 in the denominator under the radical. If your instructor asks for that I will modify my answer. $\endgroup$ – kamran Jan 28 '19 at 8:40
  • $\begingroup$ I'd suggest mentioning the fact that "inspection of symmetry" must also take into consideration the fact that the beam is rigid. If that weren't the case, the deflection at B and D wouldn't be the same since there's a spring at D but not at B. $\endgroup$ – Wasabi Jan 29 '19 at 14:36
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    $\begingroup$ @Wasabi, I modified my answer to include your comment. Thank you. $\endgroup$ – kamran Jan 29 '19 at 17:00

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