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For a simply supported beam with a UDL, with boundary conditions of X=0, V=0 & X=L, V=0. Using a cubic trial function.

$$V=a+bx+cx^2+dx^3$$

The following curve is obtained (Red Line). It is shown in comparison to the simple beam theory curve (Blue line).

RR Graph

Is this the level of accuracy to be expected from a Rayleigh-Ritz method for this scenario using these conditions? I don't have an intuitive sense of how accurate the approximation should be so am unsure if I have made a mistake in my calculations or if this is an expected result.

FEA42
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  • See https://engineering.stackexchange.com/q/35235/10902 – Solar Mike Apr 18 '20 at 19:31
  • We have no idea how you actually made the approximation, so the question isn't answerable. But IMO either you made a mistake, or you used an inappropriate method. – alephzero Apr 18 '20 at 22:54
  • You have four unknowns. You list two boundary conditions that yield one equation each, and virtual work can provide a third. What is the forth? Please show the system of four equations and explain why you chose the ones you did. Show the calculation of the coefficients. The system is overconstrained, so you are relaxing some physical constraints. – Phil Sweet Apr 19 '20 at 03:28
  • http://www.mathcs.emory.edu/~haber/math315/chap4.pdf for more insight into methods, selection of orthagonal bases, problem conditioning, and error prediction. – Phil Sweet Apr 19 '20 at 03:46
  • @phil sweet The third equation is total energy and one of the unknowns resulted in zero leaving only 3 unknowns – FEA42 Apr 19 '20 at 07:25
  • Something is wrong there. Please show all your work. We know a = 0. Now you have three unknowns. We know from x(L) = 0 that bL + cL*2 +dL^3 = 0. And there is an integral equation for the work (the integral of the deflection) which equals the stored energy in the beam (the integral of the bending moment). You are still short a constraint to have a single expression. You may choose to set the slope at L/2 = 0, or set the slope at 0 and L to be equal and opposite. – Phil Sweet Apr 19 '20 at 13:25
  • I want you to get a handle on this, so that you understand why the problem can be conditioned to get a much better answer. By inspection, you know the shape should be symmetrical, so just model half the deflection curve. Normalize the known 4th order solution for length L to the domain from x = -1 to x = +1. Find the 3rd order polynomial over 0 to 1 that models half the curve. Deflection at x(0) is unknown, but slope is zero. Deflection is 0 at X(1). Match the energy balance (a family of curves) and find the one that minimizes the deviation from the original (Lagrangian optimization). – Phil Sweet Apr 19 '20 at 13:41

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