Castigliano's method for the deflection of a cantilevered beam

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SUMMARY

The discussion centers on the application of Castigliano's method for calculating the deflection of a cantilevered beam. The user initially derived a formula for the deflection at the midpoint, \(y_{B}=\frac{3L^{2}M}{8EI}\), but later discovered an error in the bending moment function \(M(x_{2})\). After correcting this to \(M(x_{2})=-M-P \left( \frac{L}{2} + x_{2} \right)\), the user arrived at the correct deflection formula \(y_{B}=\frac{ML^{2}}{8EI}\), which matched the results obtained from Finite Element Analysis (FEA). This exchange highlights the importance of accurate moment functions in applying Castigliano's theorem.

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FEAnalyst
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Hi,

it may look like a homework but believe me that it's not. Castigliano's method was omitted when I was attending mechanics of materials course at my university and now I'm catching up. Another reason why I want to solve this is that I'm just curious what's the formula for the deflection of such beam as it can't be found in the literature.

Anyway, here's a beam that I want to solve for deflection of middle point (B) using dummy force P:

cantilever 2.png


And here's my solution:
$$M(x_{1})=-M$$ $$\frac{\partial M(x_{1})}{\partial P}=0$$ $$M(x_{2})=-M-P \left( \frac{L}{2}+x_{2} \right) $$ $$\frac{\partial M(x_{2})}{\partial P}=-\frac{L}{2}-x_{2}$$ $$y_{B}=\frac{1}{EI}\int_{0}^{\frac{L}{2}} M(x) \cdot \frac{\partial M(x)}{\partial P}dx=\frac{1}{EI}\int_{0}^{\frac{L}{2}} \left( -M-P \left( \frac{L}{2}+x_{2} \right) \right) \cdot \left( - \frac{L}{2} - x_{2} \right) dx_{2} = \frac{1}{EI} \int_{0}^{\frac{L}{2}} -M \cdot \left( - \frac{L}{2}-x_{2} \right) dx_{2}=\frac{3L^{2}M}{8EI}$$

It seems fine but I solved an exemplary case and used FEA to find reference solution. This way I found out that my formula obtained from Castigliano's method is not correct. Do you have an idea what's wrong here ?

Thanks in advance for your help
 

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  • cantilever 2.png
    cantilever 2.png
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This is an interesting problem, I was involved in a thread some time ago where the goal was to derive a similarly complex beam's analytical equations here: Thread: Harder beam equation

The general integration method followed this form which seems to be about what you're doing as well:
Mech_Engineer said:
The last attachment I posted is basically a fully symbolic derviation of the beam bending formula using the integration procedure. MathCAD did all the heavy lifting for me in terms of symbolic manipulation, but it can at least give you an idea of what you're in for (a lot of work).

You'll end up having to split the beam into three sections, integrate three times for each section, and then solve a system of 9 equations with 9 unknowns at the end. It's not pretty. A pdf of the MathCAD sheet you'll be most interested in is attached to the following post:

https://www.physicsforums.com/showpost.php?p=1600375&postcount=19

The integration tree you'll need is as follows:
\nu''''=\frac{q(x)}{EI}
\nu'''=\frac{V(x)}{EI}
\nu''=\frac{M(x)}{EI}
\nu'=\theta(x)
\nu=\delta(x)

With this in mind, I do find solving this sort of problem is easier if you lay out the bending moment and shear force diagrams for the beam to help define boundary values. Do you have any diagrams you could post, maybe that will help us find a potential discrepancy?

MIT Open Courseware: Bending Moment and Shear Force Diagrams
 
Thanks for reply. I think that the bending moment and shear force diagrams for this beam should look like that:

cantilever 2 diagrams.png


To be honest, I suspect that the error is somewhere in these two bending moment functions ##M(x_{1})## and ##M(x_{2})##. I omitted support reactions but I guess it's correct. Anyway, something may be wrong in the second function.

I ignore shear forces in Castigliano's method as I only want to account for flexural strain energy.

P.S. In the thread that you've cited different method was used. I want to utilize Castigliano's theorem as this is what I'm trying to learn but of course another method may provide correct solution which would be fairly helpful.
 

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  • cantilever 2 diagrams.png
    cantilever 2 diagrams.png
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Ok, it seems that I've found an error - as I suspected that second bending moment function was incorrect. It should be ##M(x_{2})=-M-P \left( \frac{L}{2} + x_{2} \right)##. When you solve the rest with this corrected term the result will be ##y_{B}=\frac{0.125M L^{2}}{EI}=\frac{ML^{2}}{8EI}##. Now the results agree with FEA solution.
 
Nice catch and well done seeing it through to the finish line!
 

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