# Calculate deflection of a beam with an arbitrary shape

I have a beam with rectangular cross section, but rather an strange shape in the length wise direction as seen below:

So I have taken out the symmetry part of it:

And it actually has some nice radius's due to metal forming:

the beam is fixed at 0 and symmetric about 5. The Beam sees a Distributed load at s4. The distributed load is actually a gasket so when my beam deflect I actually need to iterate to make sure I reach a steady condition.

The reason why I would like analytic solution is for understanding of whats going on. The problem is a coupled problem, below the beam is a fluid flowing and also the beam has electric current flowing through it, so its all coupled and I would like to experiment with different materials.

I have tried calculating the moment along the curve by saying that the axial force can be neglected and that the fixed(welded end) can rotate, it might not be 100% accurate but I assumed this will be close enough to the real world:

and the moment seems to make sense:

this also fits with FEM

The problem is that I can not figure out how to calculate the deflection:

Can you not simply take the derivative of the moment or something to get the deflection?

I have tried to solve this but it gives me completely wrong answer and not a nice symmetric deflection curve, I also need to guess the slope at the beginning of the beam and make sure it's the same as the end (6.03225deg). So I have the curve of the beam (inside/outside i.e outside -inside= thickness) as well as the moment and the Young's modulus and inertia of the cross section, so I should be able to calculate the deflection right?

• Could you please clarify what the actual question is? Of course, it is possible to calculate the deflection, but if what you are asking is an explanation for the deviation from a FEM-model is, you should include results from that model. I will note that your calculation of the section forces isn't right: You'll need to include the axial force as well and note that you've described a statically indeterminate structure, so it is somewhat more work than this. – ingenørd Apr 6 '20 at 16:36
• Also, your bending moment diagram is wrong: you say the ends are fixed, yet you have zero bending moment at the ends. – Wasabi Apr 6 '20 at 18:45
• It's a bit cheating, but you can always use programs to give you an idea of what your bending moment diagram should look like. See here for the result I got from a very rough model (uniform vertical loading throughout the beam). – Wasabi Apr 6 '20 at 18:55
• Thanks for the feedback, I have updated the question with more details – skatun Apr 7 '20 at 11:29
• @Wasabi which program did you use? Did you apply the load only on the flat top part? – skatun Apr 15 '20 at 14:19

• The axial force is essential to calculating the bending moment and is important for more than the axial deformation, which is likely to be small in this case. If a beam makes a 90 degrees bend, the axial force and shear force "switch places", i. e. the axial force immediately before the bend is equal to the shear force immediately after the bend. Therefore, as you ignore the axial force, you are calculating the shear force and bending moment for a straight beam without any bends.

• As Wasabi has noted in a comment, you are missing the bending moment at the ends. (The integration constant from integrating the shear force.) Your FEM-model shows a global maximum here, which is the expected result.

• You don't give complete equations for both deflection and angular rotation, both the one you give for the deflection isn't correct. The curvature is $$\kappa(x) = \frac{M(x)}{EI}$$ and the angular rotation is $$\alpha(x) = \int_0^x \kappa(x) dx + C$$ with C being an integration constant. You seem to be using this, so so far so good. To calculate the deflection (ignoring the axial deformations for simplicity), you would use $$\frac{d\delta}{dx} = \alpha$$. For a straight beam, you could use $$\delta(x) = \int_0^x \alpha(x) dx + C_2$$ with $$C_2$$ being another integration constant, which I will note is not what you are doing. (You need to integrate the curvature twice to get a deflection.) As the beam isn't straight, the calculation becomes more complicated and you will need to use a vectorized version: $$\delta$$ is a deformation perpendicular to the beam and will change direction around the bends. In orders words, the underlying assumption of the integration, that $$\delta(x+dx) = \delta(x) + \alpha dx$$, is simply not true (in the non-vector version), because $$\delta(x)$$ and $$\alpha dx$$ are deformations in two different directions.

In conclusion, the current approach of your calculations isn't more accurate than approximating the beam with a straight beam that is fixed in both ends. You should be able to look up a complete solution for such a regular, straight beam in a decent textbook of your choice, so even though you almost certainly have considered this already, I'll just add that it might be worth the effort to compare this crude approximation to your FEM-model just to check just how crude it is.

• So one approximation would be to use a straight beam with length=( L_inside+L_Outside)/2 and the put my distributed load on it, compute the deflection and then subtract the deflection along the midcurve of the beam of original shape. I will try to look more into decomposition of integral in y and x axis – skatun Apr 8 '20 at 9:38