How to Determine Resultant Force in Vertical Space Frame

A picture is worth a thousand words.

Bruce,

I just received an answer from the CR4 Administrator. Complex drawings cannot be attached using this website. They suggested to swap emails to get this information to all of the respondents, is that okay with you? Please advise.

Thank you for your time,

Scott

Hi, mach1gtx!

PDF's will not be deliverable on CR4's message centers because they use "Slash", a low-grade form of "Slashpot", (or some such thing). It's somethin g quick and easy the originators of the CR4 site thought to use at the time.

Here's how to send an image:

1. Draw the image in an art program. (I use 'microsoft PAINT').
2. Save the image as a JPG image with a name.
3. Click on the green camera icon at the top of the CR4 dialogue box. An image will appear that looks something like:
4. Browse to your saved JPG image.
5. Click on the "OK" button on your image browser to choose that image.
6. Click on the 'Submit' button to put your image into the dialogue box.
7. Treat the rest of the dialogue box in an ordinary manner.

Looking forward to seeing your images.

Mark

Scott,

It isn't a very good image, but here is your PDF in GIF format.

Bruce,

Thanks for placing the drawing on the website! I just got a reply from MarktheHandyman, and he recommended a procedure to do the same, but it was converting it to a .GIF, which I think is about the same. But, in any case, you got it done!

Thanks!

Scott

Scott,

I should have included the text from your email. Here it is:

Hi Bruce,

Attached is the sketch of what I trying to calculate. In short, I would like to miniaturize a so-called space frame, as found in modern building construction. The 'frame' is composed of two outer wythes of wire grid. The two grids are tied together with interconnecting, sloped wire chords. When looking at the space frame in section and elevation, you would see a typical module composed of a pyramidal type structure, similar to a real space frame. At this point in time, I am just trying to use the correct method to figure the tension and/or compression in a typical joint. If I can determine that, then I could roughly size the wire gauge necessary, then the necessary 'module' size. In the attached sketch, I think the details show an assumed 3 to 4 inch (76 to 100mm) square module.

The calculated uniform axial loading would be in the range of 1,000 to 1,500 lbs./lineal foot or 454 to 680 kg/0.3048m. The anticipated lateral loading is negligible, since the vertical edges, top and bottom connections are basically pinned.

Please let me know if you have any problems reading the attachment, etc. I certainly appreciate your time on this!

Thanks,

Scott <<ResultForce.pdf>>

A drawing or sketch will help us help you.

Willyap06,

Thanks for responding. I was attempting to send a .PDF drawing yesterday to illustrate my question. I just received word from the CR4 Administrator that the only way to provide a rather complex drawing is if we exchange email addresses. I can send it to you, if that is not a problem?

Thanks for your time!

Scott

Hi, mach1qtx!

Please make a drawing and include it in your next blog by

1. Saving it under a name on your computer
2. Clicking on the little camera icon at the top of the answer frame
3. browsing to your sketch under its saved name
4. entering the sketch to be included in the blog

In the meantime, a great big round of welcome to the CR4 blogsits. Looking forward to helping you out, and glad to have you aboard.

Thanks,

Mark

If the space frame is statically determinate, I would think that you could solve for all of the member forces by the Method of Joints.

Consider a tripod, for example with a force in any direction acting at the top. There are three equations available and three unknown forces. The applied force F in an arbitrary direction may be expressed in three components, namely Fx, Fy and Fz. Knowing the geometry of the tripod, each of the member forces can be resolved into X, Y and Z components. Say the unknown forces are A, B and C.

The three equations would be:

Ax + Bx + Cx + Fx = 0

Ay + By + Cy + Fy = 0

Az + Bz + Cz + Fz = 0

A set of three equations and three unknowns can be solved uniquely.

If the structure is a quadrapod instead of a tripod, it cannot be solved by the Method of Joints because there are still only three equations available but four unknowns. To solve it, one would need to consider strain compatibility.

Scott,

If a vertical line load is applied along one edge of the wall as it appears on your sketch, then no load will be transferred to the opposite edge. You would need a transfer beam to deliver the line load to each vertical. If the line load is 1500#/' and the spacing of verticals is 4" and if all joints are hinged, then each vertical member on the loaded side carries 500 pounds over its full height. The vertical members on the unloaded side carry 0 pounds.

In this way, it does not differ from a truss shaped as you have shown on "SIDE VIEW MATRIX". A truss with compression applied at each end of one chord places that chord in pure compression. The rest of the members are not loaded.

If, on the other hand, you were to stipulate that one chord was loaded at the top and the other chord was loaded at the bottom, then the diagonal members must come into play and a lateral reaction of 500*1.791/96 = 9.3 pounds per vertical would be required to maintain equilibrium. In that case, the compression in the loaded chords would start at 500# each at the top and taper down to 0 at the bottom while the opposite chords would have 500# at the base and 0 at the top.

Each apex in the wall would transfer a vertical shear of 500/n where n is the number of apexes in the height of the wall. For a height of 8' and spacing of 4" there would be 23 apexes, so each would transfer 500/23 = 21.7# from the loaded chord to the unloaded chord. From symmetry, each diagonal member meeting at an apex would carry a vertical component of 21.7/4 = 5.4#. The member force would then be 5.4*(1.791²+2²+2²)^1/2 * 1/2 = 9# per diagonal. For each apex, two members would be in tension and two in compression.

Incidentally, the structure is indeterminate but because of symmetry, a solution was possible without resorting to FEA (finite element analysis).

Bruce,

There will be a continuous 2 1/2" x 1 1/2" x 0.135" (63.5mm x 38.1mm x 3.5mm) 'U'-channel along the top and bottom of the steel panel. The 'U'-channels will be welded to the 12 gauge metal skins, as the 'U'-channel will encompass the top and bottom of the panel. There will be a uniform load bearing on top of the panel. The bottom of the panel will bear on concrete slab. Regarding the 2 outer wythes of metal grid; they will be welded to the interior faces of the 12 gauge metal skins. The grid matrix will be a welded assembly. No lateral loading will be incurred on the design. What I am hoping to achieve is a contiguous assembly, where all of the parts are restrained. The upper uniform load would transfer to the top edges of the 12 gauge metal skins. Since the skins are welded to the interior grid matrix, hopefully I would achieve a smooth load transfer to the grid matrix all of the way to the bottom 'U'-channel. The bottom 'U'-channel would then distribute its load uniformly as well.

I apologize for not providing this additional information. In my rush to get the drawing out, I realized this morning that I left out the top and bottom 'U'-channels.

Thank you,

Scott

From your latest description, the structure will be statically indeterminate to a high degree with rotational restraint at every joint. You cannot use the method of joints to analyze this structure.

The 12 gauge skin has a thickness of 0.1046" so it has much more area than the 1/8" diameter chords of the space frame. The skin will take the brunt of the load.

If the load is centrally applied at the top of the panel and centrally supported at the bottom, then it seems to me that the vertical chords and skin will all be stressed uniformly in compression while the diagonals will carry no load.

Under lateral load, the diagonals would be stressed but you have indicated that the lateral load is negligible, so that is not at issue.

Bruce,

Back to the drawing board. Thanks Bruce!!

Scott