Force couple acting on rigid frame 2

[310toumad]

How would I go about solving for the moment(s) at the base where this structure is fixed? From looking at examples of frames, a few assumptions are typically made to reduce the indeterminacy of the problem. For this case, can I assume the reactions acting opposite to the applied 1430 forces are also 1430? The problem I run into with this is that, if I assume that the horizontal member has a point of inflection at the midway point, how do I resolve the internal forces when "cutting it in half" and doing an force equilibrium analysis to find the moment at the base? If I already have assumed the applied force and the reaction are equal, then any internal force in the beam would have to be zero in a force balance, in which case your moment is just 1430*25.5. I don't think that is correct, the horizontal member connecting the two should 'stiffen' and lessen the bending moment logically. What am I assuming incorrectly? Wouldn't there also be a torsional component as well?

 

[XR250]

The horizontal member adds a bunch of degrees of freedom to the system. It will stiffen the system by inducing torsion in the columns.
Unless you have a 3D frame package, I would just do 1430 lbs x 25.5 ft .

 

[310toumad]

So basically the analytical solution would be too time consuming and convoluted to even attempt? I'd be interested to see the method, although I'm sure it involves creating an enormous stiffness matrix. Unfortunately I don't have access to FEA tools.

 

[Shotzie]

Why don't you request a trial of Risa3d from their website? Link

 

[KootK]

So basically the analytical solution would be too time consuming and convoluted to even attempt?

I would argue that XR's solution basically is the analytical solution. Unless your columns are monstrous HSS, the response of the frame should be very accurately represented by 1430 lbs x 25.5 ft. From a global frame torsion perspective, flexure in the the columns is analogous to warping torsion and torque in the columns is analogous to St.Venant. As usual, warping dominates for open sections.

As with all problems, you can always drill deeper analytically. There's no need here for practical design purposes however.

I don't think that is correct, the horizontal member connecting the two should 'stiffen' and lessen the bending moment logically. What am I assuming incorrectly?

For the loads shown, I don't think that the horizontal member would stiffen the system much really. That said, it may have some other benefits from a stability perspective etc.


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.

 

[cal91]

Here's the method:

You'd need to do a 3D stiffness matrix, having 6 DOFs at each of the two nodes. By symmetry you could do 6 DOFs.

I'm assuming you know how to get shear, flexural, and axial stiffnesses.

Torsional stiffness is GJ/L if torsionally pinned (warping is unrestrained).

If warping is restrained, torsional stiffness is:

et1 and et2 are numbers from 0 to 1, 0 being a 100% warping unrestrained end connection and 1 being 100% restrained.

lambda is dimensionless: Length * sqrt(GJ/ECw)

 

[cal91]

I realized that this is actually a "grid", so you only need 4 DOFs. There will be no x or y translations (no translations in the plane of the frame). So you have z translation, and three rotational DOFs. There will be no axial forces.

But, as KootK said, accounting for this is not going to change much unless the columns have huge torsional stiffnesses. The solution will show you that the bending moment is almost exactly 1430 lbs * 25'6" at the column bases.

 

[rb1957]

the couple Ax and Bx will be slightly smaller than the applied 1430 lbs, as some of the applied torsion is reaction by the in-plane moments (Ma and Mb, along the y-axis)

Ay and By should be very small, and will probably induce small fixed end moments about the z-axis)

there will also be another fixed end moment, along the zed axis (driven by the applied x-load) ... moments at a and b will be equal and opposite.

The problem (in my mind) with applied torsion couples is that they induce equal and opposite reactions at other places in the structure that you can't see from a simple FBD ... the simple reaction to your applied load is Ax = Bx = 1430 lbs ... but then you ask yourself "what is happening in the vertical members ? they look (taken as individuals) like cantilever beams ...

another day in paradise, or is paradise one day closer ?

 

[cal91]

I ran the analysis for fun, assuming all joints fixed flexurally and torsionally.

1. -All members W10x30

Column:
Moment at top = 11.96 k-in
Moment at base = 402 k-in
Torsion = 7.28 k-in
Shear = 1.35 kip

Beam:
End Moment = 7.28 k-in
Torsion = 11.96 k-in
Shear = 0.08 kip

2. - All members HSS6x6x3/8

Column:
Moment at top = 73.1 k-in
Moment at base = 143 k-in
Torsion = 69.5 k-in
Shear = 0.71 kip

Beam:
End Moment = 69.5 k-in
Torsion = 73.1 k-in
Shear = 0.72 kip

As expected, the torsional stiffness plays a large part for closed sections, and is negligible for open sections.

 

[310toumad]

I think I should have clarified, the dimensions for lengths are in inches so it would be 25.5 inches. I was thinking of using HSS 2.5x2.5x.25 Wall.. what software are you using?

 

[cal91]

Sorry, I was reading it in feet.

Just did some matrix operations with Excel.

Here's what I get Using HSS2.5x2.5x.25 and dimensions in inches...

Column:
Moment at top = 6.09 k-in
Moment at base = 11.6 k-in
Torsion = 5.87 k-in
Shear = 0.70 kip

Beam:
End Moment = 5.86 k-in
Torsion = 6.09 k-in
Shear = 0.73 kip

 

[rb1957]

"HSS 6x6x3/8" is a closed tube section, yes?? ... why would that section make the beam much more effective in bending (the beam reacts much more shear in 2, reducing significantly the bending in the vertical members) ?

"W10x30" is an I-beam, yes? ... it would matter which axes was bending in the "beam", strong or weak axes.

the center of the beam (and the foot of the vertical) remain fixed in location (by symmetry), so the lateral deflection of the beam matches the lateral deflection of the vertical post. so you could solve it with a very large beam, and a wimpy little post ... but it doesn't sound like a very robust solution ...

another day in paradise, or is paradise one day closer ?

 

[cal91]

What in the world is this structure?

 

[rb1957]

in the OP's world ?

another day in paradise, or is paradise one day closer ?

 

[310toumad]

Its part of a carriage assembly for a lift to raise up golf carts. I don't have it drawn exactly how it would be oriented in actual use, just the way that seemed a little easier to understand the concept. Think of something similar to this:

http://www.heftee.com/products/product_details.cfm?ProductID=1

with the lifting fork adjusted all the way to the end, it generates a force couple on the carriage (tubed structure we've discussed here), and I'm trying to discern the stress at which point it attaches to the vertical column.

 

[cal91]

Yes to both of your questions.

1. The torsional stiffness of the beam allows the column to behave more like a cantilever with it's free end restrained against rotation.

If the beam had no torsional stiffness, the column would act as a normal cantilevered beam.



2. The torsional stiffness of the columns allows the beam to act more like a beam with end rotations fixed when it's ends are being translated in opposite transverse directions.

If the columns had no torsional stiffness the beam would be a straight line, but it has reverse curvature when there is torsional stiffness.

 

[rb1957]

i think you're answering my questions ?

1) in isolation I see the beam working like a, ummm, beam ... bending under the applied load, cantilevered at the CL ... I won't call that torsion on the beam. but my question was why would one shape be more effective than the other in reacting shear ?

my question 2) was more about how do you orient the W section in the beam ? is the applied load bending about the strong axes or weak ?

we agree, if the beam was a "noodle", the the vertical posts would behave like cantilevers.



another day in paradise, or is paradise one day closer ?

 

[310toumad]

Would you mind attaching your spreadsheet? I've never really done a problem like this, mostly just stuff involving your standard beam formulas and Mc/I stuff. I would like to see the math to learn more. I appreciate the insight guys.

 

[cal91]

i think you're answering my questions ?

1) in isolation I see the beam working like a, ummm, beam ... bending under the applied load, cantilevered at the CL ... I won't call that torsion on the beam. but my question was why would one shape be more effective than the other in reacting shear ?

my question 2) was more about how do you orient the W section in the beam ? is the applied load bending about the strong axes or weak ?

we agree, if the beam was a "noodle", the the vertical posts would behave like cantilevers.

It's not the beam's weak axis flexural stiffness that causes the shear to be dumped into the beam, it's the two points I explained above. Let me try to be more clear.

1. First imagine the beam has high torsional stiffness (i.e. the tip of the cantilevered column resists rotation).
As the cantilever column deflects, the tip rotates, inducing torsion on the beam. Because of the beams torsional stiffness, it attracts more load and the cantilevered column attracts less load as well as developing a reverse moment at it's tip.

2. Imagine the column has no torsional stiffness. As the beam ends translate transversely, the column freely twists and the beam stays in a straight line.
Now add torsional stiffness to the column, and it resists the twisting, which puts the beam into an S bend reverse curvature.

It's not that the beam is more effective in shear, it's that the torsional resistance of both the beam and the column stiffens up the joints.

 

[cal91]

Here's my spreadsheet. Uses the stiffness method for a grid system. Not very pretty but gets the job done.