Castigliano's method for the deflection of a cantilevered beam

In summary, the beam's equation can be written as:$$M(x_{1})=-M$$ $$\frac{\partial M(x_{1})}{\partial P}=0$$ $$M(x_{2})=-M-P \left( \frac{L}{2}+x_{2} \right)$$ $$\frac{\partial M(x_{2})}{\partial P}=-\frac{L}{2}-x_{2}$$ $$y_{B}=\frac{1}{EI}\int_{0}^{\frac{L}{2}} M(x) \cdot \frac{\partial M(x)}{\partial P}dx=\
  • #1
FEAnalyst
336
137
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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  • #2
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:
[tex]\nu''''=\frac{q(x)}{EI}[/tex]
[tex]\nu'''=\frac{V(x)}{EI}[/tex]
[tex]\nu''=\frac{M(x)}{EI}[/tex]
[tex]\nu'=\theta(x)[/tex]
[tex]\nu=\delta(x)[/tex]

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
 
  • #3
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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  • #4
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.
 
  • #5
Nice catch and well done seeing it through to the finish line!
 

What is Castigliano's method for the deflection of a cantilevered beam?

Castigliano's method is a mathematical approach for calculating the deflection of a cantilevered beam under a specific load. It uses the principle of virtual work to determine the displacement at any point on the beam.

How does Castigliano's method work?

Castigliano's method works by calculating the partial derivative of the strain energy with respect to the load at a specific point on the beam. This derivative is then multiplied by the displacement of that point to determine the deflection at that point.

What are the advantages of using Castigliano's method?

One of the main advantages of Castigliano's method is that it can be used for any type of loading, including point loads, distributed loads, and combinations of loads. It also provides a simple and efficient way to calculate the deflection of a cantilevered beam without the need for complex equations.

What are the limitations of Castigliano's method?

Castigliano's method assumes that the beam is linearly elastic and that the material properties are constant throughout the beam. It also does not take into account any deformations caused by shear forces, which may result in less accurate results for beams with high shear forces.

How accurate is Castigliano's method?

The accuracy of Castigliano's method depends on the assumptions made and the complexity of the beam. For simple beams with small deflections, it can provide accurate results. However, for more complex beams with large deflections, the accuracy may decrease.

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