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Uniformly distributed torsion in HSS beam

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!

RE: Uniformly distributed torsion in HSS beam

(OP)
Thanks, the problem with that equation is that Cw is generally taken as 0 for HSS sections (it's not even listed in the AISC tables). That makes a = 0 and the rotation becomes 0 rad.

RE: Uniformly distributed torsion in HSS beam

Cw is the torsional constant from the steel book that is situational with every beam size. Where doe sit assume Cw is zero?>

RE: Uniformly distributed torsion in HSS beam

(OP)
"C" (units of in^3) is the Torsional Constant. "Cw" (units of in^6) is the Warping Constant, which is listed for WF shapes but not HSS because Cw = 0 in^6 for HSS shapes.

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

Ah thanks

RE: Uniformly distributed torsion in HSS beam

With warping off the table, you can just apply TL/JG to your concentrated torques and superimpose the results. A quick excel sheet will make short work of it. No doubt there's a direct formula someplace but I wasn't able to locate it with Google in 30 seconds or less.

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

Perhaps:

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

(OP)
I have outriggers at 12" on center imparting a 1.5 k-ft torque each. If I could calculate the rotation at the first outrigger, then the second, then the third, etc I would add them until I get to mid span.

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

(OP)
KootK, I see your second reply. can you elaborate a bit more? Not sure what that formula is expressing.

RE: Uniformly distributed torsion in HSS beam

Torsion of a hollow section is described in detail in "Theory of Elasticity" by Timoshenko and Goodier. An approximate solution can be found using membrane analogy.

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

(OP)
Thanks BAretired, I'll try that.

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

T in the formula is a constant torque. In your case, you will have a constant torque for a length of 4', then a sudden change in torque for the next 4' and so on. Your torque diagram will be a step function with maximum values at the fixed supports.

BA

RE: Uniformly distributed torsion in HSS beam

(OP)
Thanks. I think Blodgett will do the trick (as he many times has). I'll have to look at the chapter to see if this is for fixed ends or pinned.

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

The formula is for twist of a rectangular tube of length L subjected to a torsional moment of magnitude T. There is no mention of whether the supports are fixed or pinned. If you want the total deflection of the outriggers, you would need to add the beam deflection to the torsional deflection.

BA

RE: Uniformly distributed torsion in HSS beam

Quote (OP)

KootK, I see your second reply. can you elaborate a bit more? Not sure what that formula is expressing

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

(OP)
KootK, I'm not sure that's correct but I like the number I'm getting from Blodgett so I'm sticking to it!

Thanks everyone for the help.

RE: Uniformly distributed torsion in HSS beam

Quote (DETstr)

KootK, I'm not sure that's correct but I like the number I'm getting from Blodgett so I'm sticking to it!

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

Quote (OP)

I'll have to look at the chapter to see if this is for fixed ends or pinned.

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.

Quote (OP)

I'm still confused about how to use the AISC DG 9 method if you have a tube.

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.

Quote (OP)

You need "a" to determine the Torsional Function from the AISC DG 9 appendix graphs.

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

@KootK,
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

Quote (BA)

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.

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

(OP)
BAretired- a quick follow up: in Section 8.2, Blodgett defines twist formulas for various span and loading conditions.

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

as Koot pointed out...no such thing as pinned connection w/o warping....should be a closed section or the fixed connection should activate all four sides of the HSS member

RE: Uniformly distributed torsion in HSS beam

For what it is worth, this is in line with AISC Design Guide 9 Example 5.2 in which rotation is calculated as TL/4JG (sim. to T*L /(4*Es*R) above) for a mid-span torque applied to a closed section.

RE: Uniformly distributed torsion in HSS beam

WeL^2/8GJ. At least humour me as a double check.

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

(OP)
KootK- I agree, WeL^2/8GJ matches Blodgett's t*L^2 / (8*Es*R) except for the sight discrepancy between Blodgett's "R" and AISC's "J".

RE: Uniformly distributed torsion in HSS beam

Blodgett explains how he gets his "R" and why it may vary. Keep your b/t and h/t ratios in the right range and buckling is not a problem.

RE: Uniformly distributed torsion in HSS beam

The general equation for determining rotation of a closed shaped is Mt/GJ, where Mt is the torsional moment in the member resulting from the application of the eccentric loads. In the case of a simply supported member with a uniform eccentric load, Mt equals weL2/8, as pointed out by KootK. The units for Mt are k-ft2 (or k-in2, lb-ft2, etc.). The rotation unit is radians. Blodgett provides the torsional shear diagrams on page 8.2-1 in "Design of Welded Structures" and the moment diagrams can be constructed accordingly.

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

Hokie93,
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

The equation Mt/GJ returns the maximum rotation for the loading under consideration and the units of Mt are k-ft2 (or k-in2, etc.). The moment term has an additional length component because the load is acting at an eccentricity (for example, weL2/8 for a uniformly eccentrically loaded member rather than wL2/8 for a uniformly loaded member). The units check out (rotation in radians) for moment equal to k-in2, G equal to ksi, and J equal to in4.

RE: Uniformly distributed torsion in HSS beam

If the moment term has an additional length term, it is not a moment! It is a moment multiplied by a length. To suggest that Mt carries units of k-ft2 or k-in2 is not only wrong, it is confusing to anyone trying to understand torsion.

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

Not to disrespect BA but I think Hokie93's presentation is somewhat helpful. Meaning that it is helpful to view the applied torsion in a similar way that we view any other load applied to a beam. If you draw a loading diagram, in the case of uniform torsion you have an imaginary uniform load applied to a beam, kip-in / ft. You can then draw a "shear" diagram, in the case of uniform torsion you would have max positive torsion on one end with a uniformly varying line decreasing to a max negative torsion on the opposite end (kip-in). Then you would draw your "moment diagram" which would be a parabolic shape with your maximum "torsional moment" located at midspan (kip-in^2). Based on the above it becomes easier (or maybe not) to see how the w*e*L^2/8 is derived. The rotation at any point is then the integral of this... atleast I hope I am on the right track here...

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

For a simple span beam with eccentric load w at eccentricity e, the total torsional load on the beam is weL, half of which goes to each support, namely weL/2 which is the torsional moment at each end and tapers down to zero at midspan. Torsional rotation can be found at any point, but maximum rotation occurs at midspan and is the average torsional moment times L/2 or weL/4 * L/2 = weL/8.
Correction: Torsional rotation at midspan should be weL2/8 divided by GJ

BA

RE: Uniformly distributed torsion in HSS beam

@BA - What you have described is what I was trying to describe, except -

Quote (BA)

and is the average torsional moment times L/2 or weL/4 * L/2 = weL/8

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

Yes RFreund, you are correct. I omitted the ^2 but the point is that the torsional moment at any point along the beam is in units of force*length, not force*length^2.

BA

RE: Uniformly distributed torsion in HSS beam

Quote (BA)

but the point is that the torsional moment at any point along the beam is in units of force*length, not force*length^2
Very true I can't argue that.

Quote (BA)

but maximum rotation occurs at midspan and is the average torsional moment times L/2
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

This but applied between any two points of interest:

Quote (KootK)

Essentially the area under the torsion/stiffness diagram between the supports and midspan

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

My point in bringing up Mt/GJ is that it is an easy and convenient equation for calculating the rotation of a closed shape without having to rely on published equations, such as those on page 8.2-1 in Blodgett. Or, even if the Blodgett equations are readily available, Mt/GJ provides a convenient check. If one is hung up on the numerator term, feel free to rename it Xt/GJ or any other variable that suits one's fancy.

RE: Uniformly distributed torsion in HSS beam

In flexural theory, the change in rotation between any two points on a beam with gravity load is the area under the M/EI diagram.

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

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