## Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

## Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

(OP)

I'm working on a "simple" wood beam design tool. The beam can be rotated about it's longitudinal axis, with a rotation of 0 meaning that it's oriented in the strong direction for gravity loading, 90 degrees is a flatwise orientation, etc. The loads are always acting vertically (in the gravity direction) and for the sake of this discussion are assumed to act at the centroid/neutral axis (no torsion). The sketch below should clarify the basic conditions. I'm trying to calculate the total deflection of the beam under a uniform load for any rotation of the section. Despite scouring my resources and the internet, I can't find much guidance on this.

With the beam oriented either vertically or flat, the calculation is simple. I use either Ix (strong axis MOI) or Iy (weak axis MOI) of the section and the equation d = 5wL^4/384EI. For the rotated section, though, I can think of 2 different ways to calculate the total deflection, which give very different results:

Method 1:

With the beam oriented either vertically or flat, the calculation is simple. I use either Ix (strong axis MOI) or Iy (weak axis MOI) of the section and the equation d = 5wL^4/384EI. For the rotated section, though, I can think of 2 different ways to calculate the total deflection, which give very different results:

Method 1:

- Calculate the component of the load acting in the direction of the member's local y-axis. Then using this load (we'll call wy) and Ix, calculate the deflection, dy, which is the deflection in the direction of the member's local y-axis (strong axis).
- Using a similar procedure, calculate the deflection dx, in the direction of the member's local x-axis (weak axis).
- Calculate the total deflection: d = sqrt(dy*dy + dx*dx)

- Given Ix and Iy, calculate I_rotated, which is the MOI with respect to the Global X-Axis, which is a horizontal line going through the centroid of the section (I should have included that in the sketch above). From my strength of materials text: I_rotated = (Ix+Iy)/2 + cos(2*angle)(Ix-Iy)/2
- With I_rotated calculated, calculate d from the equation above (using I_rotated in place of I).

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

A few observations:

- For rotations of 0 and 90 degrees, the results are the same for both methods.
- At a rotation of 45 degrees, the total deflection is 10.87" for Method 1 and 0.79" for Method 2!
- At only a small rotation of 5 degrees, Method 1 gives a deflection nearly 1" greater than Method 2. Intuitively, this seems incorrect.

I initially based my analysis on Method 2, but in checking my results against Woodworks software, I was rather shocked. After some digging, it turns out that they use Method 1. I was hoping to come across some literature discussing this, but found nothing. If anybody is aware of anything or has insights into this, I'd be much appreciative! If all else fails, I might have to construct a model to test.## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

You need to combine the vertical component of each dy and dx.

the resultant you have calculated could be at an angle to the vertical

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

anydirection, so it should be an apples to apples comparison. But you have a good point seeing that we're typically most concerned with vertical deflection. See the attached updated spreadsheet which also includes vertical deflection for Method 1.## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

Don't think I understand what exactly you are doing.

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

Method 2 directly calculates the moment of inertia of the rotated section (I_rotated) with respect to the Global X axis. From there, deflection is calculated as: d = 5wL^4/384E(I_rotated).

For example, at a 45 degree angle, the 2x10 (1.5"x9.25") section has I_rotated = 50.77 in^4.

This is calculated from the equation: I_rotated = (Ix+Iy)/2 + cos(2*angle)(Ix-Iy)/2

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

In Method 2, I calculated I_rotated with respect to the Global X-axis, which I assumed to be the neutral axis of the section. The problem is that for all rotations except 0, having the neutral axis in this location results in there being internal moments acting about the Global Y-axis due to the resultant tension and compression forces (to either side of the n.a.) not being vertically aligned. Since there are no applied moments acting in this direction, this means the section is not in equilibrium ("an unsymmetrical bending problem" as mentioned by Celt83), and therefore this is not a valid solution.

I believe I could modify Method 2 to be a valid approach by determining the neutral axis angle of the rotated section corresponding with zero moment about the Global Y-axis. Using this n.a. I_rotated would be calculated. Graphically, for a section rotated at 45 degrees, it seems like the neutral axis might be somewhere between the local y-axis and the section diagonal. It's obvious that this would result in a much lower value for I-rotated than what I erroneously calculated above, with the value probably being rather close to Iy.

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

Your method 2 could work it just requires much more computation effort up front you'll need to compute Iy, Iz, and Iyz and all three components will show up in the final deflection equations. The cross-section stresses still have to obey static equilibrium and resolve to the applied moment. You may have computed stresses as just My/I, this formula is not correct for unsymmetric bending the more detailed formula, with 0 axial loading, is sigma = (-My Iyz - Mz Iy) y + (My Iz + Mz Iyz) z / Iy Iz - Iyz^2 , for any arbitrary y,z axis with origin at the cross-section centroid.

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

Have you considered constructing a Mohr circle to obtain the the I value of the beam at different rotations? If not see this link:-

https://www.researchgate.net/figure/Mohrs-circle-f...

“Do not worry about your problems with mathematics, I assure you mine are far greater.” Albert Einstein

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

desertfox, I (very) briefly looked at a Mohrs circle approach but I was thinking it wouldn't apply to a rectangular section because I incorrectly assumed that the product of inertia would be zero for a double symmetric section. I see now that this is incorrect. Well, rather, it's correct if the section is oriented either vertically or horizontally but not for rotated sections. Good call on that. I'll probably look at that approach as well, which it appears might be the same as the equation suggested above by Celt83.

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

In the AISC Manual (p.17-43 of the 15th ed.):

I3 (your global Ix) = Ix*sin^2(theta)+Iy*cos^2(theta).

How would that compare with your method 1 and method 2 for vertical deflection?

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

----------------------------------------------------------------------

Why yes, I do in fact have no idea what I'm talking about

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

I have run two analyses using Strand7:

1) Loads were specified in the vertical and horizontal directions, and incrementally factored so that the resultant load rotated through 90 degrees in 5 degree increments. The resulting deflections were then rotated to find the resultant deflections parallel and perpendicular to the resultant applied load.

2) The load was applied in the vertical direction only and the beam was rotated in 5 degree increments; i.e the model replicated the actual loading as closely as possible. For this case 2nd degree geometric non-linearity was included.

The results were:

... so close, but significant differences between the two runs. This waas because geometric non-linear effects were significant, and were only included in the the second run, so I reduced the load to 10% and found:

... near exact agreement between the two runs.

In summary "Method 1" from the OP gives accurate results for the maximum resultant deflection, with the vertical deflection being a little lower.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

IDS, Thanks for running that analysis. I would have liked to check my results against those from a frame analysis program, but don't currently have a license for one. As you mentioned, my Method 1 results appear to match your plotted results above. The "Total Deflection" and "Vertical Deflection" lines from mine look very close to the "DYZ" and "DY'/DY2" lines from yours.## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

It was actually pretty easy to set up this problem in the spreadsheet because you can specify a rotation angle for any beam:

For each change of angle you need to recalculate the frame:

Then you can extract the midspan deflections for each angle:

The results are in near exact agreement with my Strand7 results where I rotated the loads, and didn't include non-linear geometric effects.

The download spreadsheet is set up with this model and the results summary table shown above.

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

IDS, Your spreadsheet looks impressive. I'll definitely take a look. I started writing a 3D Frame analysis tool awhile ago, but have yet to finish. Definitely not simple stuff. I imagine there being a lot of VBA code running behind the scenes.DoubleStud, I believe a square shape will have the same moment of inertia no matter how it's rotated, meaning that it will deflect vertically by the same amount for any rotation (like IDS mentioned). I was a little surprised to learn this. I figured the deflection would change slightly, especially if turned 45 degrees to a diamond orientation.For my "simple" beam analysis tool, I was hoping to not have to resort to a 3D Frame analysis to account for the member rotation (but perhaps it's inevitable). I did derive an expression for moment of inertia which seems to give total deflection (not vertical) of the beam without needing to split up the applied loads into individual components relative to the beams local x and y axes. That expression is:

I = 1 / sqrt[sin

^{2}(alpha)/Iy^{2}+ cos^{2}(alpha)/Ix^{2}]. If this leads to the total deflection (relative to the neutral axis), then if the neutral axis angle is known, the vertical deflection can be found.I'm not fully confident this approach is correct, but it seems to work. I'll need to review mechanics of materials a bit more. It's crazy how complicated the problem becomes when the section is not perfectly vertical or horizontal.

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

https://newtonexcelbach.com/2009/01/31/frame-analy...

https://newtonexcelbach.com/2009/02/04/frame-analy...

https://newtonexcelbach.com/2009/02/10/frame-analy...

https://newtonexcelbach.com/2009/02/15/frame-analy...

https://newtonexcelbach.com/2009/03/05/frame-analy...

https://newtonexcelbach.com/2009/03/29/frame-analy...

https://newtonexcelbach.com/2009/05/11/frame-analy...

Or for links to all the above + others (including 3D frames) see:

https://newtonexcelbach.com/2014/01/22/frame-analy...

or to also include all the stuff since 2014 just search the blog for "Frame Analysis"

Doug Jenkins

Interactive Design Services

http://newtonexcelbach.wordpress.com/

## RE: Deflection of Simply Supported, Uniformly Loaded Beam, with Rotated Section

The simple beam program that I'm currently working on uses the 2D (planar) stiffness method. While this obviously isn't as versatile as a 3D frame analysis, I figure it will cover about 95 percent of the member analysis that I typically do for a residential project (mostly wood framed). When/if I eventually create a 3D frame program, I'll also be able to use this to verify some of the results. I suppose I'm trying to create something similar to Woodworks or Forte.

Concerning the original problem I was having calculating the beam deflection, the method I described above works, at least for the example problem. I get identical results to those from your spreadsheet as well as Woodworks software: