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Cantilever Beam Experiencing Bending & Torsion

Cantilever Beam Experiencing Bending & Torsion

Cantilever Beam Experiencing Bending & Torsion

(OP)
thread507-180666: Allowable Torsion Strees on Steel

I have a cantilever beam on which a load is creating bending and torsion. I am looking at using 5" x 5" x 3/8" square HHS ASTM A500 Gr B material. This material has a yield of 45,700 psi and ultimate tensile of 58,000 psi. FEA is resulting a 33,690 psi torsional stress. According to Blodgett in Design of Welded Structures, in the absence of test data the ultimate shear stress is assumed to be 75% of the materials ultimate tensile stress (resulting a 43,500 psi ultimate shear stress). This is almost a 1.3 safety factor. Any concerns with this approach? Should I look at using a thicker/larger member?

RE: Cantilever Beam Experiencing Bending & Torsion

I have concerns with your approach. Are you combining torsional shear with beam shear? 34,000 psi seems too high for working stress.

What type of load produces these results? Can you provide a loading diagram?

BA

RE: Cantilever Beam Experiencing Bending & Torsion

designcoeng,

At a quick glance, it doesn't seem too bad to me. I can't understand how your software came up with a torsional stress of 33,690 psi. That seems much too high, but I don't have time to check it right now. I will look at it again tomorrow or Monday.

BA

RE: Cantilever Beam Experiencing Bending & Torsion

check AISC design guides for torsion of steel beams. I have a pdf version somewhere in my old design files....if you need it, let me know and I will email it to you.

What about deflection?

RE: Cantilever Beam Experiencing Bending & Torsion

Bending
V = 7500#
v = 7500/(2*5*3/8) = 2000 psi
M = 7500*34 = 255,000"#
S = 9.23in3
fb = M/S = 27,600 psi


Torsion
Mt = 7500*3.75 = 28,125"#
vt = Mt/(2At) = 28125/(2*(5 - 0.375)2*3/8) = 1753 psi

Combined shear (max. in wall nearest load)
v + vt = 2000 + 1753 = 3750 psi (okay)

theta = Mt*s/(4*A2*G*t) = 28,125*4(5-0.375)/(4*4.6254*11,165,000*0.375) = 0.0000679 rad/in or a total of 0.0023 rads in 34 inches (negligible rotation).

The above calculation is approximate as I have not considered the rounded corners of the HSS, but it demonstrates that the torsional shear stress in the OP is much too high.

BA

RE: Cantilever Beam Experiencing Bending & Torsion

(OP)
Thank you for your responses. Jim, the max deflection according to FEA is 0.0075 inches which is comparable to my calculations. I have access to AISC and will take a look at that. Thanks for the tip. BA, my hand calculations agree with yours which lead me to scratch my head at the FEA results. 3750 psi is significantly better that 34 ksi.

Let me explain the whole picture a little better and let you see if something in my thinking is wrong. This 5" square tube will slide back and forth inside a larger tube. It will be fully supported inside the larger tube with spacers. The maximum projection is the 9" as illustrated in my earlier attachment. I have attached another document displaying the way I constrained the 5" tube in FEA.

My model is generated in Autodesk Inventor, which allows me to draw a sketch to reflect a beam length (the green light in the first image of the attachment). I then use "Frame Generator" to select the beam shape/size (represented by the transparent tube). With their frame analysis tools, I apply the constraints and loads accordingly. I might try taking it into the other Stress Analysis environment and create a mesh and so on.

I wonder why I am getting such a large value in FEA.

Thanks for your time!!!

RE: Cantilever Beam Experiencing Bending & Torsion

What happens if you make the left support fixed in all directions for displacement and rotation and remove the right support entirely?

BA

RE: Cantilever Beam Experiencing Bending & Torsion

Designcoeng:
The devil’s going to be in the details of the schematic idea you are trying to show. Goodness knows what you are really trying to do, or how. You are keeping that a secret, and trying to cover it up with a bunch of fancy FEA calcs. The type of calcs. BA has done are really all you need until you start to hone in on the details. He apparently didn’t notice your right support (pin support, your first sketch), so he is using 34" as a cantilever length. I agree with the 5.7ksi normal bending stress. And, I suspect that the large (33.69ksi) torsional stress is some small area of stress concentration immediately around the load application point, an anomaly having to do with the exact way the software assumes you have applied the eccentric load, or the way you have imported the model to the FEA. And, I would bet it is on the side of the load application too, where the pin shearing stress will be greatest.

You would do well to do these types of hand calcs. and hand sketching, rather than CAD and FEA when you start a problem. Although, your second CAD sketch is fine, but missing a bunch of important detail. Think about how it works, how it is manufactured, and goes together while you are doing your hand calcs. and sketches. You can just draw and analyze the hell out of things which can’t be practically built when you rely totally on CAD and FEA to start your analysis. The fact that you have fixed so many of your boundary conditions at the two supports is likely making the FEA software think that warping and parallelograming can’t take place, and this could lead to some very high localized stresses.

I don’t really think you have fixity in any direction at either of your reaction points, and you don’t explain what those reactions are or how they work to cause any fixity. You do limit the y & z displacements to some amount. But, even that is a funny number because you have a telescoping cantilever within an outer supporting tube member which will bend and deflect. And, now all of a sudden, you show a new support from above and about 18" left of your right support. It’s really difficult to find two perfectly square and straight tubes which will slide, one inside the other. So, you’ll end up with 4" wide (5" less the corners) x 2 or 3" long bearing pads; one set of 4 pads at the back end of the sliding tube; and the other set of 4 pads at the front end of the support tube. Thus, it is only about 24" btwn. the reaction points. Now, the question is, should there be a wear pad on top of the inner tube in the region of that new top support point? If you want the sliding/telescoping action to be smooth and easy, you might consider cam rollers as your support points. How are you getting the 7.5k eccentric load into the tube wall nearest the load, what’s the pin through the tube for? What is that whole elliptical device at the load point? You might want to put end plates on each end of the inner tube as stiffeners. What is the actual support structure and what is this thing actually supposed to do?

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