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The above is a past exam question from an introductory structural analysis course, one in which although we have studied the Euler buckling load equation, we have just been given parameters for the equation based on the standard end support conditions (fixed/fixed, fixed/pinned, etc.). I don't think the above example fits any of the standard end conditions and am therefore finding it difficult to derive an equation for $P_{cr}$ I have tried to use the bending moment equation by including torque (that opposes bending) from B and the net load (due to spring C) as follows: $$ T_B = \beta_r \theta $$ Taking a small deflection x in the vertical direction and y in the horizontal: $$ tan\theta = \frac{dy}{dx} \approx \theta $$ $$ \therefore T_B = \beta_r \frac{dy}{dx} $$ Using the bending moment equation: $$ M=-EI\frac{d^2y}{dx^2} $$ $$ \therefore \sum M=-EI\frac{d^2y}{dx^2} + \beta_r \frac{dy}{dx} = Qy $$ The net load $Q$ can be expressed in term of the force P and the reaction force due to the spring at C as: $$ Q = P - \beta dx $$ Leaving the final second order equation: $$ EI\frac{d^2y}{dx^2} - \beta_r \frac{dy}{dx} + (P-\beta dx) y = 0 $$

This is beyond the scope of our course, so I have no idea if:
a) The above equation is correct or solvable (and if so how one goes about solving it)
b) There is a different approach using more basic techniques (which is more than likely the case)

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To answer the specific question: (a) No, that equation is irrelevant and (b) Yes. see below.

The question isn't about Euler buckling. It says the bar is rigid. Euler buckling only applies to a flexible bar - the Euler formulas include Young's modulus, and also the moment of inertia of the bar which is not mentioned in the question.

One way to solve the question is to use the principle of virtual work, and compare the strain energy in the springs with the work done by the external force P.

You might also notice that the condition $\beta_r = 3\beta L^2/2$ is related to the geometry of the triangle ABC.

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  • $\begingroup$ Thank you and apologies for the late reply, I have been busy with exams recently. Does that mean for virtual work that I take unit loads in the horizontal and vertical direction? Seeing as there is no real load in the horizontal, I do not know how to equate the vertical load to the energy in spring A. $\endgroup$ – user120568 Jan 10 '17 at 15:00
  • $\begingroup$ apologies, forgot to tag you in the previous comment $\endgroup$ – user120568 Jan 10 '17 at 23:30

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