Uniformly distributed torsion in HSS beam
Uniformly distributed torsion in HSS beam
(OP)
Hi all, I'm designing an HSS member that has closely spaced 4' outriggers along one side of its 21' span (fixed ends). No other loads. I want to determine the deflection by figuring out the rotation of the section. I'm using AISC Design Guide 9 an the graphs in the appendix and assuming a uniformly distributed torque.
They all reference this torsional property "a" (lowercase), and I don't know what this is. They list it for WF shapes but don't for HSS. What is this value and how do I determine it for an HSS shape?
Thanks!
They all reference this torsional property "a" (lowercase), and I don't know what this is. They list it for WF shapes but don't for HSS. What is this value and how do I determine it for an HSS shape?
Thanks!






RE: Uniformly distributed torsion in HSS beam
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RE: Uniformly distributed torsion in HSS beam
RE: Uniformly distributed torsion in HSS beam
RE: Uniformly distributed torsion in HSS beam
I'm not sure where in the Manual it says Cw = 0 for HSS shapes but this document (http://www.cisc-icca.ca/files/technical/techdocs/u...) from CISC indicates it on page 4.
RE: Uniformly distributed torsion in HSS beam
RE: Uniformly distributed torsion in HSS beam
It may just be a semantics thing but it's important to recognize that you probably don't have uniform torsion. What you likely have is:
1) a uniform distributed load applied eccentrically and;
2) torsion varying linearly from zero at the middle to maximums at the ends.
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
RE: Uniformly distributed torsion in HSS beam
W X e X L/2 X L / 4
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
RE: Uniformly distributed torsion in HSS beam
My problem is actually calculating rotation. You need "a" to determine the Torsional Function from the AISC DG 9 appendix graphs.
RE: Uniformly distributed torsion in HSS beam
RE: Uniformly distributed torsion in HSS beam
For a quick and dirty solution, check "Design of Welded Structures" by Blodgett. He provides a value of R = 2tb2d2/(b+d) in which b and d are the sides of the rectangle formed by the centerline of the HSS walls and t is the thickness.
Then θ = TL/EsR
where T is torque and Es is shear modulus, usually called G and L is the length of tube subjected to a uniform torque. θ is the angle of twist expressed in radians.
BA
RE: Uniformly distributed torsion in HSS beam
Is the "T" a single torque (k-ft) or uniform (k-ft/ft)? I have that text but not with me at the moment.
RE: Uniformly distributed torsion in HSS beam
BA
RE: Uniformly distributed torsion in HSS beam
I'm still confused about how to use the AISC DG 9 method if you have a tube. It works for WF shapes but many of high-torsion applications want to be tubes. You'd think they would indicate the limitation. Or maybe I've just missed something. I sent a note to the AISC solutions center so we'll see what they come back with. I'll update if anyone is interested.
RE: Uniformly distributed torsion in HSS beam
BA
RE: Uniformly distributed torsion in HSS beam
Sounds like you're on your way with Blodgett but I'll answer this anyhow for good measure. I should have divided my previous equation by GJ and that would represent an approximation of the angle of twist at midspan assuming only St. Venant torsion (no warping). Essentially the area under the torsion/stiffness diagram between the supports and midspan.
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
RE: Uniformly distributed torsion in HSS beam
Thanks everyone for the help.
RE: Uniformly distributed torsion in HSS beam
I reviewed Blodgett's most relevant example (#5) of which a portion is shown below. With the following substitutions, it's identical to the formula that I provided above:
W X e --> T
R --> J
Es --> G
Blodgett's R is essentially a single factor accounting for both J and Cw. For open sections the difference between using J and R is huge. For closed sections, it's modest. Notably, for closed sections, Blodgett indicates that tested rotations for closed sections were too small to measure. As such, it's not clear to me why Blodgett assumed R to be an improvement over J for closed sections. To some degree, it depends how one defines J. The formulation included in this document is very simple to calculate and is what I'd probably use: Link
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
RE: Uniformly distributed torsion in HSS beam
It assumes a fixed torsional boundary condition. With a closed section, there's really no such thing as torsionally pinned. We assume closed sections to be restrained against warping which, by definition makes them torsionally fixed.
You don't use it. At least not the case chart business which is intended for use with sections that will be meaningfully affected by section warping behaviour.
If you review all of the places where "a" is defined in DG9, you will find that it takes on a value of zero wherever Cw would take on a value of zero (closed sections).
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
RE: Uniformly distributed torsion in HSS beam
I believe you are misinterpreting what Blodgett intended. The problem is with the symbol 'J' which Blodgett is taking as the polar moment of inertia of a section. The J value in my Steel Handbook is the torsional constant which is quite different than polar moment of inertia. Actually, Blodgett's R is the same (or nearly the same) as the J value for a section.
For open sections such as tees, angles and WFs, J is the sum of d.t3/3 for each flange or web. Blodgett's R value is precisely the same as he lists in Article 2.2-8.
For closed sections such as HSS, the torsional constant J, listed in the Steel Handbook is slightly less than Blodgett's R value which I believe is due to the rounded corners on an HSS. The value of the warping constant, Cw for an HSS is actually zero so Blodgett's R value is intended as the torsional constant which we label J.
BA
RE: Uniformly distributed torsion in HSS beam
Yeah, I had not clued into the fact that Blodgett was drawing comparisons between R and polar moment of inertia J for closed sections rather than R and the St. Venant torsion constant J. Regardless, I stand by my proposed solution and my previous comments that did not pertain to Blodgett's stuff.
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
RE: Uniformly distributed torsion in HSS beam
Max rotation for a simple span beam with uniform torque is defined as: rot = t*L^2 / (8*Es*R)
Max rotation for a simple span beam with a single torque at mid span is: rot = T*L / (4*Es*R)
Max rotation for a cantilevered beam is: rot = T*L / (Es*R) [this is the formula you gave me above]
rot = angle of twist (theta)
T = torque in force-dist units (e.g. kip-in)
t = torque in force-dist/dist units (e.g. kip-in/in)
L = span
Es = Shear Modulus
R = as you defined above for a tube section
Looks like for the same total torque, rotation is half if it is uniformly distributed rather than isolated in midspan. It also appears the twist for my case is 8 times less than I previously thought, which I'm ok with!
RE: Uniformly distributed torsion in HSS beam
RE: Uniformly distributed torsion in HSS beam
RE: Uniformly distributed torsion in HSS beam
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
RE: Uniformly distributed torsion in HSS beam
RE: Uniformly distributed torsion in HSS beam
RE: Uniformly distributed torsion in HSS beam
It may not be stated clearly in the document but AISC Design Guide 9 (Torsional Analysis of Structural Steel Members) is largely intended for open members such as wide-flange shapes. This was more clearly the case in the AISC publication that preceded Design Guide 9.
RE: Uniformly distributed torsion in HSS beam
If Mt is defined as a torsional moment, it is incorrect to state that "The units for Mt are k-ft2 (or k-in2, lb-ft2, etc.)". The units for a moment, torsional or otherwise, are k-ft, k-in, lb-ft etc.
As well, it is incorrect to state that rotation is defined by the equation Mt/GJ unless you mean unit rotation, i.e. rotation per unit length of the member. Rotation of a member of length L under a constant torque of Mt is Mt.L/GJ. When the principle is clear, the expression for a simple span beam becomes apparent.
BA
RE: Uniformly distributed torsion in HSS beam
RE: Uniformly distributed torsion in HSS beam
The expression for rotation of a member due to torsion is Mt.L/GJ where L is the length of member under consideration. To combine Mt and L into one unit is simply wrong.
BA
RE: Uniformly distributed torsion in HSS beam
BA - in the equation M.tL/GJ how do you derive the value of M.t? Meaning for the case of uniform torsion M.t = 1/8*w*e*L, right? that would get us to the 1/8*w*e*L^2 part once multiplied by L. However (for me) the 1/8*w*e*L is an odd expression, but I'm probably missing something.
EIT
www.HowToEngineer.com
RE: Uniformly distributed torsion in HSS beam
Correction: Torsional rotation at midspan should be weL2/8 divided by GJ
BA
RE: Uniformly distributed torsion in HSS beam
But it is not, it would be weL/4 * L/2 = weL^2/8 Right?
EIT
www.HowToEngineer.com
RE: Uniformly distributed torsion in HSS beam
BA
RE: Uniformly distributed torsion in HSS beam
This statement now helps me understand the M.t*L/GJ equation, I think...
M.t is the average torsional moment = (w*e*L/2) * (1/2)
L = is actually L/2 because this is our point of interest.
Now lets say we were looking for the rotation at any point, X, along the beam.
M.t = the average torsional moment up to point X ?
L = X ?
Close?
EIT
www.HowToEngineer.com
RE: Uniformly distributed torsion in HSS beam
That's the average in some cases but not in general.
I like to debate structural engineering theory -- a lot. If I challenge you on something, know that I'm doing so because I respect your opinion enough to either change it or adopt it.
RE: Uniformly distributed torsion in HSS beam
RE: Uniformly distributed torsion in HSS beam
EIT
www.HowToEngineer.com
RE: Uniformly distributed torsion in HSS beam
Similarly, the change in torsional rotation between any two points on a beam subjected to torsional load is the area under the Mt/GJ diagram.
BA