Shear Center of a J section

[FJCV]

Dear fellow engineers,
I am requesting your help figuring out the best way to calculate the shear center of an unsymmetrical cross section as shown in the attachment. I am familiar with the theoretical solutions of Timoshenko for cross sections with at least one axis of symmetry.
I appreciate your help finding solutions for cross sections with multiple flanges without a symmetry axis.

tiff image:

http://files.engineering.com/getfil...94-491e-af7e-adb400b69ec8&file=J-Section.tiff

jpg image:

http://files.engineering.com/getfil...51f-479a-b289-00d38c7d6e60&file=J-Section.jpg

 

[ukbridge]

Theres no attachement as far as im aware. If you have a decent FE package like Ansys or LUSAS there will be some kind of a section property calculator based on Prandtls soap film analogy, which will calculate shear centre for you. Failing that there should be something simple in Roark...

As a guess for an open section I might work out the shear centre by taking the first moment of area of all the bt3/3s, or very least as a validation of some famcy method, dont just use this approach though!

 

[FJCV]

Hello Ukbridge,
I added the picture attachment. Can you give an the example calculation using the first moment of area of all the bt3/3s?
I have been able to find a solution with FEM but like to understand the theory better. Thank you for your help.

 

[ukbridge]

I cant read .tff files unfortunately...if you post it as a jpeg or gif ill take a look.

Why not try the analysis i suggested for a simpler shape and compare it to those from standard tables? Have a look through your softwares manual on section property calculations to see if theres anything useful in there.

Also, not going to do your work for you sorry!

 

[KootK]

This document explains some of the theory:Link. It's by a fellow aerospace guy no less. This might be hulpful as well: Link.

If you want to go really deep, this is the most in depth torsion reference that I know of: Link. I had to wait a good while for one to show up on eBay.



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.

 

[ukbridge]

I must have misread your post sorry. For open sections ONLY, Simply divide the cross section into rectangles. Calculate the J of each rectangle i.e. Bt^3/3 where t is always the smaller dimension. If you have shapes with an aspect ratio less than 10 you will need to modify the 1/3 coefficient as it is no longer a good approximation of J. Also, dont ask me how to do this as 10minutes googling will tell you!

Then calculate the "centroid" of the section in a similar manner to working out the elastic neutral axis, but using J instead of A i.e. Shear centre = sum of J* distance to centroid/ sum of J.

Disclaimer: For lititgious reasons, I don't officially endorse using this exact method for actual design purposes. No doubt theres a winning ticket to willy wonkas fudge factory lurking within its "derivation" but rather the method i would first compare for known shapes with tabulated values!

I cant see the section, but maybe the solution is in your load path. Are there any clever ways you can avoid torsion in your system? If not, why not use a RHS section or similar which has a much better torsional resistance. Alternatively if i needed a quick solution you could try taking the torsion as a warping couple in the top and bottom flanges.

 

[KootK]

Qualitatively, based on "structural senses", I would say this:

1) SC vertical location would be closer to the upper flange about in proportion to the weighted average of the upper and lower flange lengths.

2) SC horizontal location would be left of the vertical web as it's on the web for a zee and left of the web for a cee.

I've not heard of UK's method but it sounds as though it would generate a consistent result.

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.

 

[BAretired]

Calculate the shear flow distribution for each element using VQ/I. Then find the shear force in each element. The shear center is the point where all shear forces multiplied by their distance to the shear center equals zero.

You should have four forces, two for the top flange in opposite directions, one for the web (shear in the web is V) and one for the bottom flange. The shear center is the point where ∑F.d = 0 where F is the shear in each section and d is distance to shear center.



BA

 

[FJCV]

Thank you all for your responses.

BAretired,
I tried the method you described and even rotated the cross section to its principal axis first but was not able to match the solution I got from my FEM (x= -0.32743, y= -0.19332 if a=1.000" and t=0.100"). I used Timoshenko's integration method where one expresses the static area moment and shear stress as a function of dimension s (distance along the flange centerlines). With Timoshenko's method I was able to match the known solutions for a an I and C section but mentioned J section remains a project in work....
Note that I applied shear loads in the FEM and observed no rotation, confirming mentioned shear center location. Not sure what I am missing.

Ukbridge,
I will try to add a jpg file to the original post and will try your method shortly. Thx.

 

[BAretired]

I found x = -0.392", about 20% more than your FEM value, but I haven't checked my work so I may have made a mistake. I found I to be 0.357 in[sup]4[/sup].

I did not calculate a y value.



BA

 

[FJCV]

BAretired,
I sppreciate it if you could post a scan of your analysis. Thx.

 

[ukbridge]

If your FEM analysis correctly calculates every other parameter like Mpl, J, shear area etc I dont see any reason why you cant hang your hat on its answer for the shear center. As an additional validation you could use the same analysis for something like an equal angle section and see if you get the same answer as those given in tabulated values for rolled sections.

 

[jimstructures]

FJCV,

IES Software out of Bozeman, MT has a program called Shapebuilder. You can download a working copy for 30 days and run your shape in it and get all the values as well as the Shear Center in either Metric or Imperial dimensions. This would allow you to check you numbers.

http://www.iesweb.com

Jim

 

[Teguci]

Just to add to the confusion, I'm getting x0 = -0.48 and y0 = -0.62.

I follow this stepwise procedure with appropriate corrections (in particular, pg 11 equations are wrong)

http://www.cranerepairengineer.com/Torsional Warping Constant (Cw)/OpenWarpingConst.pdf

 

[BAretired]

My scanner is not working this morning for some reason.
I now find I of the section to be 0.34769 in[sup]4[/sup]
The c.g. is 0.6724 below the X axis.
Draw a line diagram representing the center of the three elements.

Label top flange A, B and C
Label bot. flange D, E

Q[sub]B[/sub] = 3(0.1)0.6224 = 0.18672in[sup]3[/sup]
Q[sub]D[/sub] = 0.95(0.1)1.2776 = 0.12137in[sup]3[/sup]

Assume V = 100#
q[sub]B[/sub] = VQ/I = 100*0.18672/0.34769 = 53.7 psi

F[sub]BA[/sub] = 53.7/3 * 1/2 = -8.95#
F[sub]BC[/sub] = 2*53.7/3 *2/2 = 35.8

F[sub]top flg[/sub] = 35.8 - 8.95 = 26.85#

q[sub]D[/sub] = 100 * 1.2137/034769 = 34.9 psi
F[sub]bot. flg[/sub] = 34.9 * 0.95/2 = -16.58#

F[sub]web[/sub] = 100#

Solving for x,
x = [26.85(0.6224) + 16.58(1.2776)]/100 = 0.379" left of Y axis, or -0.379"

A value for y could be obtained in a similar way with V acting from left to right.


BA

 

[BAretired]

Scanner is working now. Please see attached two page file.

BA

 

[Archie264]

Teguci, I love following behind your calculations; they are things of beauty. As for that link you provided, I'm not going through that guy's magnum opus. I'm glad you were able to and that you caught his error. Come to think of it, it would be amazing if he had gone through all of that without an error. I'm not entirely sure why he did all that; isn't that why we have computers?

 

[Teguci]

If I ignore Ixy and the rotation required to get to the principal axis, I get x0 = -0.377 and y0 = -0.104. I believe V/Q is a simplification based on an assumption of symmetry (Ixy = 0). For an unsymmetric condition page 9 of the enclosed should be used

http://soliton.ae.gatech.edu/people/jcraig/classes/ae3145/Lab7/shear-center-theory.pdf

The long form - [ Vx ( 1/(Ixx Iyy - Ixy^2) x INT (Ixx X - Ixy Y ) dA ] becomes Vx INT (X dA) / Iyy only when Ixy = 0 (bending about the principal axis) INT X dA = Qyy, INT Y dA = Qxx

 

[Teguci]

@Archie264:
I only put the pretty stuff up. The actual calculations are a disaster and a half long Excel Spreadsheet following the procedure in my first link, which I am still fixing (important but not critical). Someday when I'm happy with it(ie never), I will get it into C++

 

[Archie264]

Well, it's impressive stuff in any case.