# Wing weight

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#### pictsidhe

##### Well-Known Member
I really don't want to watch 9+ 15 minute videos, thanks. Your images won't load on my slow connection.

#### Dave Hodges

##### Well-Known Member
I really don't want to watch 9+ 15 minute videos, thanks. Your images won't load on my slow connection.
Bottom line: assakkaf.com formula said expect 953 pounds. Euler's equation says expect 2,650 pounds. Actual buckling load was 1,022 pounds. You want me to email it to you?

#### wsimpso1

##### Super Moderator
Staff member
Log Member
OK, I looked up Professor Assakaf, he is a systems reliability engineer. Structural Mechanics is NOT his area... Like most professors, he has to teach some undergraduate courses, and that included a structures course where he presents long and short column theory with examples within the area of reinforced concrete pillars. He did not invent any of this. Calling an equation picked from an undergraduate lecture "the Assakaf.com equation" is at least as inappropriate as deliberately pronouncing Euler's name differently than Euler did.

Our standard aero and mechanical engineering texts cover column buckling. Euler first described a theoretical approach to buckling of long columns and is definitely the starting point for all discussions on the topic. The general field is Elastic Instability, with Timoshenko doing a bunch of good teaching in Mechanical Engineering texts. In the Aero structures field, Peery & Azar and Bruhn cover it under "crippling". They all describe global and local buckling and the theory all comes back to the basic theory presented by Euler, modified and improved by Johnson for short columns and then further refined as the Elastic Instability and Crippling.

In the end, steel and aluminum tubing has multiple buckling modes, so can be tedious to check them all. This is necessary in fabricated tubes and monocoque structures. Metal tubing is usually bought in standard sizes, these tubes have been well tested and the information collated and summarized by a number of steel tube companies. I have Summerill Company's plots where they show the compressive strength for each AN standard size tube in 4130 and in 2024 as curves based upon length. They also published torsional instability plots for steel tubes. This information is reproduced in, among other places, the Evans Lightplane Design Handbook. These plots are VERY useful when designing things like fuselages and tail booms and pushrods.

The empirical based curves could be used to predict buckling of Hodge's tubes. It may over estimate the compression load some, as his boom was also carrying bending which may have added to the stresses in the tubes and the particular tube may have had other defects, such as curvature, dents, etc.

Has someone made note of the length, OD, wall thickness, out of straight and calculated compression load on the example tube that buckled? If someone knows the numbers, I will check and report the predicated strengths for comparison.

Billski

#### Dave Hodges

##### Well-Known Member
It's 0.041 inch thick. The distance between the top flange and the centerline through the bottom flanges is 6 inches. The distance between the 168-pound load and the failure point is 73 inches.

#### Dave Hodges

##### Well-Known Member
I calculated the moment of inertia and radius of gyration of an L-shaped half of the U-shaped failed member to be:
I=0.000189 inches∆4
k=0.0995 inch

#### BBerson

##### Well-Known Member
HBA Supporter
Is that homemade "U" section compression members?
Hat section might be much better.

#### pictsidhe

##### Well-Known Member
Bottom line: assakkaf.com formula said expect 953 pounds. Euler's equation says expect 2,650 pounds. Actual buckling load was 1,022 pounds. You want me to email it to you?
If the pages are a lot smaller than 3.4MB each

#### Dave Hodges

##### Well-Known Member
BBerson, yes, it's half of 1/2-inch electric conduit, squeezed onto the flange of a 4x4 beam, which is 5/16 inch thick.

#### Dave Hodges

##### Well-Known Member
If the pages are a lot smaller than 3.4MB each

#### pictsidhe

##### Well-Known Member
Combined compression and bending needs more than a simple Euler analysis.

#### pictsidhe

##### Well-Known Member
My connection is too slow to see uncompressed images...

#### Dave Hodges

##### Well-Known Member
My connection is too slow to see uncompressed images...
I don't guess you want to watch the videos. That would take a while. What if I text you pictures of the pages?

#### wsimpso1

##### Super Moderator
Staff member
Log Member
Not round, but I decided to analyze it as a half ellipse with minor and major radii of .178" and .334". I struck an ellipse, and using numerical integration I got a cross sectional area of 0.0337 in^2, the neutral axis at 0.205 from the open end, got the smaller I of 0.00056 In^4.

Fixity - the half tubes are welded, but no big deal was made to rigidly connect the parts. Assuming pinned is common even in welded joints in airplane fuselages.

Pinned - n=1 and Pcr = 981 lb.

Firmly cantilevered - n=4 (generally not done in airplane structures - it really needs to be STURDY to qualify for n=4) and we get Pcr = 3925 lb.

Hmmm, looks like actual parts somewhat exceeded the usual prediction of failure used in airplane parts.

I also did a quick check of the tube charts. Looking at it and saying it is approximately the same dispersion of material as 3/8 tube, I have charts for 0.035 and 0.049 wall. 0.035 wall is good to about 1000 pounds, while 0.049 wall is good to about 1600 pounds.

Given that the half-ellipse part is between the tubes on wall thickness, we would expect the value for an 0.041 wall to be between 1000 and 1600 pounds. Given that the half-ellipse part has free edges, we also know that the example half-ellipse will not be as elastically stable as the tubes, and will fail sooner at somewhat lower load than the tube would. I would guess at the low end of the range, somewhere around 1000 pounds.

From these outputs, standard design practice would have also predicted the failure at about the load levels seen with real parts.

I do not see anything wrong with applying standard design processes to these parts. These methods were "in the ballpark". The claims that the standard methods are wrong appears to be in error...

Billski

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#### BBerson

##### Well-Known Member
HBA Supporter
BBerson, yes, it's half of 1/2-inch electric conduit, squeezed onto the flange of a 4x4 beam, which is 5/16 inch thick.
Conduit is really low yield, to make it easy to bend. I don't know if accurate yield numbers are available or if it meets any standard.

#### wsimpso1

##### Super Moderator
Staff member
Log Member
Conduit. What do you figure for yield, 25,000 psi? It matters in the Johnson criteria for short columns.