Uniformly distributed torsion in HSS beam 1

[DETstru]

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!

 

[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.

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.

 

[DETstru]

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!

 

[SAIL3]

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

 

[RWW0002]

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.

 

[KootK]

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.

 

[DETstru]

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".

 

[BUGGAR]

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.

 

[Hokie93]

The general equation for determining rotation of a closed shaped is M_t/GJ, where M_t 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, M_t equals weL²/8, as pointed out by KootK. The units for M_t 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.

 

[BAretired]

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

 

[Hokie93]

The equation M_t/GJ returns the maximum rotation for the loading under consideration and the units of M_t 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.

 

[BAretired]

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

 

[RFreund]

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

http://www.HowToEngineer.com

 

[BAretired]

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

 

[RFreund]

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

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

http://www.HowToEngineer.com

 

[BAretired]

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

 

[RFreund]

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.

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

http://www.HowToEngineer.com

 

[KootK]

This but applied between any two points of interest:

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.

 

[Hokie93]

My point in bringing up M_t/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, M_t/GJ provides a convenient check. If one is hung up on the numerator term, feel free to rename it X_t/GJ or any other variable that suits one's fancy.

 

[RFreund]

@Kootk - Right, good point.

EIT

http://www.HowToEngineer.com

 

[BAretired]

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