Cantilevered beam with variating height.

[Thorvald]

Hi.

I am new here (I am from Denmark), so I hope I found the right place... :-)

Well, I have a situation where I have a cantilevered beam and first this beam was straight, but then the architect found out about the size of the beam... Then I suggested to make the beam with variating height - so yes, it is my own fault that I put myself in an difficult enginnering problem. But problems are to be solved, right? Well at attached file you get an idea of the issue. I have calculated this beam as follows:
1) To determine that the strength is ok, I calculated the momentum and the cross sectional properties in a number of sections, and derived the bending stress and shear stress. This seemed ok.
2) To get an idea if the stiffness of the beam is ok, I calculated the deflection for a normal cantilevered beam, using the area moment of inertia in a section in the middle of the height-variating part of the beam - and added a factor 1,2 to have some margin.
The question is now: Does there exist formulas for correct calculation of the deflection of such kind of beam?
Maybe the moment and shear stress curves are even not completely correct for this kind of beam?
Does there exist (free, low price or demo) engineering / FEM software that can calculate this beam?

 

[nvn]

Thorvald: Can you give us the cross-sectional dimensions of a typical cross section of your beam?

 

[Thorvald]

It is a UNP320. See also attached detailed drawing.

 

[nvn]

Assuming this beam is mild structural steel (E = 200 GPa), and assuming the top flange is firmly attached to something at the 95 kN load that prevents torsional rotation, I am currently getting a tip deflection on your given beam (at section 1 in your first pdf) of u = -3,96 mm. For the moment of inertia between your two supports, I used I = 136,257e6 mm^4, because I think the average flange thickness for a UNP 320 beam is tf = 17,5 mm.

We see, therefore, that your approximate check in item 2 of post 1, which gives -6,02 mm at section 1, overestimates the deflection. Your moment and shear diagrams are correct.

 

[Thorvald]

It looks like you are doing more or less the same as I do - calculating the deflection of the beam at the middle. - Is that the correct method? I mean there are no formulas for calculating deflection of a tapered beam?

The steel quality is S235, and the top flange is attached to a concrete balcony. So your assumptions are quite ok. When I calculate the deflection in section 7 (see my first PDF), I get 4,05 mm (and then I have added factor 1,2 for extra security - because I was not sure if this method was correct). This section is slightly to the right of where you calculate it. See also attached calculation. For some reason I cannot print page 2 nicely to PDF.

 

[nvn]

No, the deflection values I posted in post 4 are vertical deflection at the cantilever tip (section 1), not section 7. The maximum deflection is at the cantilever tip. There are no easy formulas for tapered I-beam deflection.

 

[Bob S]

To get the correct deflection, you should calculate the angular deflection at every point x in the beam based on the applied torque, including the total distributed weight on the beam including the beam itself, then integrate dx from the wall to the tip of the beam to get the linear (vertical) deflection.

 

[nvn]

Nice post by Bob_S. I would strongly recommend neglecting the beam self weight in this problem, as I did. If you include the beam self weight, it changes the cantilever tip deflection in post 4 from -3,96 mm to -3,97 mm.

 

[Thorvald]

All right nvn, I see. So you have actually calculated the deflection at the tip of the beam. How did you do that? With a computer program? But it is interesting that our results are close to each other - so maybe my assumption to calculate the deflection in princip based on a average value of I, is not so far from correct? Or maybe I was just lucky in this case... :-)

Thank you also to Bob_S for the answer - but it is very long time ago I have calculated integration. I used to be very good at mathematics, but if you don't keep it running and use it, you forget it. The same as with language - you forget it if you don't use it. But of course I could probably find out if I studied it more deeply. But I didn't completely understand this: "the angular deflection at every point x in the beam based on the applied torque".

 

[nvn]

Thorvald: If you pretend your simply-supported beam plus cantilever is only a normal cantilever of length 1100 mm, with a rigid support at x = 1100 mm, exactly as shown in the second diagram of your first pdf, and assign to this cantilever a constant moment of inertia, along its entire length, equal to the moment of inertia of your real beam at x = 524 mm, then, without multiplying by 1,2, you will get the same tip deflection that occurs in your real beam. I.e., u = -F5*(L^3)/(3*E*I) = -95 000(1100^3)/(3*200 000*53,224e6) = -3,96 mm. Therefore, yes, your approximate method is quite accurate, but only in this particular case. This is luck, not a general rule.

 

[Thorvald]

Why not at x = 550 = L*½ ?? And why in this particular case it works with this simple method and why not in more general?

Why? Maybe if you make a cantilevered beam with rigid support, that is tapered from full height at rigid support to height = 0 at the tip - maybe in this case to calculate the deflection at the tip based on I at ½*L, is a correct method??

In my case I have a height of 100 mm at the tip, and the tapered part starts appr. 100 mm from the rigid support, as far as I remember. Maybe - if above is correct - this is why I get very close to the correct result (without multiplying with 1,2).

But still I would like to ask - how did you get your result? Did you use Bob_S' method or did you use a computer program, and in the latter case, what kind of computer program?

By the way: This beam is not intended to be a simple supported beam with cantilever, but just a cantilevered beam with rigid support. I have assumed that the 2x6 bolts can work as a rigid support, and I have calculated the 6 bolts to take the upward reaction and the other 6 bolts to take the downwards reaction from the momentum in the rigid support - by checking the shear strength of the bolts. I think maybe this solution is more economical regarding the bolts than to calculate as simple supported beam with cantilever?

 

[nvn]

I previously assumed your real beam had a simply-supported portion, as shown in your first pdf. If we say the real beam is the tapered cantilever shown in your first pdf, with a rigid support at x = 1100 mm, then the tip deflection of your real cantilever is -2,48 mm. Therefore, an equivalent nontapered cantilever of length 1100 mm would need to have a constant moment of inertia equal to the moment of inertia of your real cantilever at x = 737 mm to get the same tip deflection that occurs in your real cantilever.

You cannot taper the cantilever to a point, because you then get an additional 4,3 mm of deflection near the sharp point, which is an invalid design, and would be an irrelevant comparison. There is no general answer; the answer depends on the particular cross-sectional geometry.

I used a lot of numerical integration, similar to what Bob_S described, performing the integrations in a typical mathematics application, such as Octave[/color] (free). Refer to your favorite textbooks if you want to learn advanced stress analysis. If you want to use finite element analysis, you could perhaps try Calculix[/color] (free).

 

[Thorvald]

Hi nvn. Thank's for your answers - very interesting. I almost feel like studying mathematics again. But I am quite busy with my engineering, so I don't know when the time will be right... ;-) Everything indicates that the beam is having sufficient strength and stiffness. Calculix looks quite interesting - I will try that one day soon, I think. It is a little bit surprising that there are no "normal" engineering statics programs, that can make a calculation of these problem. A relative simple 2D beam program, like WinBeam at a little more advanced level.