---------------------------------------------------------- Pietenpol-List Digest Archive --- Total Messages Posted Sat 01/06/18: 7 ---------------------------------------------------------- Today's Message Index: ---------------------- 1. 10:07 AM - Re: Wing Strut Attachment (taildrags) 2. 10:41 AM - Re: Wing Strut Attachment (taildrags) 3. 10:50 AM - Re: Wing Strut Attachment (taildrags) 4. 02:04 PM - Re: Re: Wing Strut Attachment (helspersew@aol.com) 5. 03:45 PM - Re: Wing Strut Attachment (taildrags) 6. 04:05 PM - Re: Re: Wing Strut Attachment (John C Black) 7. 07:33 PM - Re: Re: Wing Strut Attachment (Clif Dawson) ________________________________ Message 1 _____________________________________ Time: 10:07:50 AM PST US Subject: Pietenpol-List: Re: Wing Strut Attachment From: "taildrags" John; I'm running some calculations to determine where the weakest link is in the string of things between the wing attachment fitting for the lift strut and the attachment to the fuselage at the bottom end. I'll post the results here when I'm done. In the meantime, I'm intrigued by your mention of building a test stand to load-test a sample to failure in tension. This is easily done if you want to "backyard" it. Simple hardware and lumber, but be prepared to be surprised at what it takes to fail the assembly. Let's just pick one of the numbers that you threw out there, the Heim joint at 14,000 lbs. If you have some nice heavy-wall steel pipe out there, you can develop that kind of load by using a 5 ft chunk of that pipe, mounting one end on something very sturdy using a bolt or other attachment that will allow the pipe to pivot, and then attaching your test piece 1 ft away from the pivot end of the pipe. You now have a 4:1 lever and you can get 14,000 lbs on the test piece by loading the long end of your pipe with 2800 lbs of "something". One such "something" is a fairly common 2-ton "come-along" (Maasdam Pow'r-Pull). I have one in my truck. Using a come-along, you can gradually load the test piece but since you don't know what's going to fail or when it's going to fail, you should use an extension on the come-along handle to get you some distance away and put up a plywood shield or something you can work behind when you conduct the test. If you have something stronger than pipe, you can make the lever arm longer to get more advantage. For example, if you have an 8 ft section of steel I-beam, you could develop a 7:1 advantage and only need to put 1,750 lbs on the long end, which you can now do with just a 1-ton come-along. If you wanted to know how much load you were putting on it as you loaded it, you could attach a 200 gallon container to the long end and start filling it with water (8.34 lbs/gal), with marks on the container to let you know how much load was accumulating. I know, this is getting ludicrous, but you get the point... you don't need a special load testing apparatus to run your test. I prefer to just run the calculations instead of conducting exciting load tests like this, especially since there are so many Air Campers flying around safely out there with all sorts of lift strut arrangements. -------- Oscar Zuniga Medford, OR Air Camper NX41CC "Scout" A75 power, 72x36 Culver prop Read this topic online here: http://forums.matronics.com/viewtopic.php?p=477122#477122 ________________________________ Message 2 _____________________________________ Time: 10:41:43 AM PST US Subject: Pietenpol-List: Re: Wing Strut Attachment From: "taildrags" John: also, in regard to thread engagement, there are some general rules of thumb as to how many threads need to be engaged in order for the full strength of the connection to be developed. The question you're asking has to do with a steel fastener threading into an aluminum bar, so the question has two answers... one for the steel fork end fitting and another for the bar it's threaded into. The *general* rule for a steel part is that it must be engaged at least 1x its diameter in order to develop full strength. The fork end in question has a 3/8-24 thread, so at least 3/8" has to be engaged in the threads of the mating part for the fork end to develop its full strength in tension. For a 24 threads-per-inch threading, that would be at least 9 threads engaged. However, for *aluminum*, the rule of thumb is 1.5x the diameter due to the lower shear strength of the material. Again for the same thread pitch and diameter, that would be about 11 threads engaged, or roughly 1/2" of full engagement for the aluminum bar to receive the full load being transferred to it by the fork end. More threads are better though ;o) Speaking of rules for threads in general, the first thread always takes the most load and it drops off with successive threads as you come out. Here's the theoretical breakdown as you add threads: 1st thread takes 34%; 1st and second combined take 57%; 1st, 2nd, and 3rd combined take 73%; 1st through 4th take 84%, 1st through 5th take 91%; 1st through 6th threads take 98%. It would seem, then, that all you would need is 6 or 7 threads engaged even in aluminum, but the problem is the shear strength because when a threaded fitting fails by stripping, once the first thread strips the following ones strip in rapid succession so it's important to prevent a pull-out failure by thread stripping. If you've worked in the shop any length of time, you know that sickening feeling when you're putting some muscle onto a ratchet on a nut and it strips. It all goes mushy and that fastener is done for. -------- Oscar Zuniga Medford, OR Air Camper NX41CC "Scout" A75 power, 72x36 Culver prop Read this topic online here: http://forums.matronics.com/viewtopic.php?p=477124#477124 ________________________________ Message 3 _____________________________________ Time: 10:50:33 AM PST US Subject: Pietenpol-List: Re: Wing Strut Attachment From: "taildrags" John: found just the thing for your test rig: Northern Tool has a 3-ton hydraulic jack for only $60! No problem developing a couple of thousand pounds of load on that test rig ;o) -------- Oscar Zuniga Medford, OR Air Camper NX41CC "Scout" A75 power, 72x36 Culver prop Read this topic online here: http://forums.matronics.com/viewtopic.php?p=477125#477125 ________________________________ Message 4 _____________________________________ Time: 02:04:07 PM PST US From: helspersew@aol.com Subject: Re: Pietenpol-List: Re: Wing Strut Attachment Here is an example that has been flying for 7 years. Upper strut attachment. Carlson "small" strut, and Mcmaster-Carr 7071 alloy insert, with two AN4 bolts. Dan Helsper Loensloe Airfield Puryear, TN ________________________________ Message 5 _____________________________________ Time: 03:45:46 PM PST US Subject: Pietenpol-List: Re: Wing Strut Attachment From: "taildrags" Ive taken a look at the forces in the wing lift struts of the Air Camper. To run the analysis, Ive used the dimensions given in the Orrin Hoopman (1933-34) plans to create the geometry, but theres not a lot of difference in any of the others that are commonly used. The main lift strut length is given as 89-3/4 between the bolt holes, the front cabane length is given as 21-1/4, and the distance from the cabane mounting point to the lift strut mounting point at the bottom of the fuselage side is 22-3/4, so that side of the triangle is 44. With those two dimensions, we find that the angle that the lift strut makes with the bottom of the wing is 29.4 degrees. Assuming an aircraft gross weight of 1,088 lb and a design loading of +3.8G (normal category aircraft), we get a total load to be supported by the wing of 4,134 lb, and each half of the wing thus has to support 2,067 lb. According to the source quoted in my spar strength article in the BPA Newsletter (Noel Becars analysis in a 1963 Sport Aviation), the main spar in a strut-braced monoplane at a high positive angle of attack carries about 84% of the total load, so that results in 1,737 lbs that each front lift strut carries. If the load on the spar where the lift strut attaches is 1,737 lbs and the strut is at an angle of 29.4 degrees to the wing, the tensile force in the strut will be 1737/(sin 29.4) = 3,538 lbs. This is the force that every element in the lift strut assembly must meet or exceed in order for the wing to sustain +3.8G loading. Lets start with the item that everybody is asking about, which is the small Carlson extruded aluminum strut itself. I have graphically checked the value that Carlson gives for the cross-sectional area of aluminum material in the strut, 0.431 sq.in., and my number comes very close to theirs so Ill use theirs. Assuming that there are no obvious stress concentrating aspects of the cross-section and that it all gets evenly loaded by transfer from the attachments at the ends, and that the tensile strength of 6061-T6 aluminum is 45,000 psi, the strut should be able to carry a tensile load of 19,395 lbs. The Carlson literature gives it as 18,012 lbs so well use their number. This is about 5 times more than what it will need to carry at +3.8 G of wing loading. The shape is important in the other regime of *negative* G loading (compression) and in bending, but thats a separate topic from just the loading in tension under wing lifting condition. I agree with the person who stated that the information that Carlson gives about round aluminum tubing is confusing and of no use in strength analysis. They compare their strut to a 12.75 diameter, 0.065 wall tube... absurd. The closest match to their strut and a round tube with that wall thickness is a 2-1/4 OD 6061-T6 tube, which has a metal area of 0.446 sq.in., a weight of 0.533 lb/lin.ft., and a tensile strength of 20,079 lb. It is obvious by inspection that the Carlson strut will have significantly lower aerodynamic drag than a round 2-1/4 tube, which along with cost is just about the only useful comparison between the two. Moving along the strut to the ends, the fork ends wont be a problem... with nominal AN6 ends (3/8) and 125,000 psi steel strength, theyre good for at least 13-14,000 pounds in tension... about 4 times the needed strength. The catalog touts them as high strength, but no strength value is given. The Heim ends appear to have the same strength in tension as the forks, so they shouldnt be a problem either. The trickiest part is the attachment between the fork (or Heim) ends and the lift strut. Looking at the 1 x 3/4 6061-T6 bar that Carlson uses, when undrilled it has a cross-sectional area of 0.75 sq.in. so its good for about 33,750 pounds in tension but at the spot where the first mounting hole is drilled through it to attach it to the strut, the cross-sectional area is reduced to about 0.5625 sq.in. and the capacity drops to 25,312 pounds. Still about 7 times whats needed. The aluminum bar is just fine; steel is not necessary. Where the fork ends are threaded into the bar stock, the edge distance of the hole in the end of the bar is about 3/16 on the short side, so that shouldnt be a limiting factor. If a sufficient number of threads of the fork end are engaged into the threaded hole in the bar, that wont be the weak point either. The problem then comes down to the end of the bar that slides up into the hollow end of the strut and how the two are connected. Since the bar is much thicker than the wall of the lift strut, the weakest point of the connection is at the mounting bolt holes through the sides of the strut. Although there are rigorous methods for determining the load distribution between the bolts in a multi-bolt connection, a fairly safe assumption in this case is that if the bolts are not placed too close to one another or too close to the edges of the material, at ultimate loading all of the bolts will be carrying about the same amount of load. Lets look at the load at the bolt holes through the strut wall nearest to the end of the strut. There is no need to look at shear strength of the AN4 connecting bolts.... they will be far stronger in shear than the thin walls of the strut. There are at least two possible modes of failure at the connecting bolts but Ill look only at shear out of the bolt, where the steel bolt pulls or tends to elongate the hole in the softer aluminum as it yields. In that mode, the force that the bolt can carry before the hole in the aluminum shears out is equal to the shear strength of the material (extruded 6061-T6 is about 24,000 psi) times the area that the bolt bears on. For AN4 bolts and a strut wall thickness of 0.119 at its thinnest point, with the first bolt in the row being about 3/4 from the end of the strut (such as in the image Piet_construction_449.jpg that Dan Helsper shared earlier), that works out to about 8,568 lbs for the first bolt in the row. That one alone should be able to carry twice the required load, so two bolts through each end of the strut should be more than adequate but if they give you three holes, use three bolts. The distance of the first bolt hole from the end of the strut is very important though... if the edge distance is reduced from 3/4 to 1/2 (for example), the pull-out load drops to about 5,712 lbs... only about 1.6 times the required capacity. With at least two attachment bolts, the connection should be adequate, but by all means keep the first bolt hole comfortably away from the end of the strut. -------- Oscar Zuniga Medford, OR Air Camper NX41CC "Scout" A75 power, 72x36 Culver prop Read this topic online here: http://forums.matronics.com/viewtopic.php?p=477134#477134 ________________________________ Message 6 _____________________________________ Time: 04:05:36 PM PST US From: John C Black Subject: Re: Pietenpol-List: Re: Wing Strut Attachment Oscar, WELL DONE !! Thank God you are interested in Pietenpols !! This really helps. John PS Regarding doing a destruction test to confirm I think I have borrowed 6 Jersey barriers weighing about 2 tons each to anchor my test piece to the ground. Several guys around here have cranes and other construction equip that can lift 10 tons. Creating a test piece is no problem. The problem is how to measure how much force is being applied as we lift. Some cranes estimate force. All this is crude. Im trying to find a universtity that has a machine set up to do this with some accuracy. On Jan 6, 2018, at 3:45 PM, taildrags wrote: Ive taken a look at the forces in the wing lift struts of the Air Camper. To run the analysis, Ive used the dimensions given in the Orrin Hoopman (1933-34) plans to create the geometry, but theres not a lot of difference in any of the others that are commonly used. The main lift strut length is given as 89-3/4 between the bolt holes, the front cabane length is given as 21-1/4, and the distance from the cabane mounting point to the lift strut mounting point at the bottom of the fuselage side is 22-3/4, so that side of the triangle is 44. With those two dimensions, we find that the angle that the lift strut makes with the bottom of the wing is 29.4 degrees. Assuming an aircraft gross weight of 1,088 lb and a design loading of +3.8G (normal category aircraft), we get a total load to be supported by the wing of 4,134 lb, and each half of the wing thus has to support 2,067 lb. According to the source quoted in my spar strength article in the BPA Newsletter (Noel Becars analysis in a 1963 Sport Aviation), the main spar in a strut-braced monoplane at a high positive angle of attack carries about 84% of the total load, so that results in 1,737 lbs that each front lift strut carries. If the load on the spar where the lift strut attaches is 1,737 lbs and the strut is at an angle of 29.4 degrees to the wing, the tensile force in the strut will be 1737/(sin 29.4) = 3,538 lbs. This is the force that every element in the lift strut assembly must meet or exceed in order for the wing to sustain +3.8G loading. Lets start with the item that everybody is asking about, which is the small Carlson extruded aluminum strut itself. I have graphically checked the value that Carlson gives for the cross-sectional area of aluminum material in the strut, 0.431 sq.in., and my number comes very close to theirs so Ill use theirs. Assuming that there are no obvious stress concentrating aspects of the cross-section and that it all gets evenly loaded by transfer from the attachments at the ends, and that the tensile strength of 6061-T6 aluminum is 45,000 psi, the strut should be able to carry a tensile load of 19,395 lbs. The Carlson literature gives it as 18,012 lbs so well use their number. This is about 5 times more than what it will need to carry at +3.8 G of wing loading. The shape is important in the other regime of *negative* G loading (compression) and in bending, but thats a separate topic from just the loading in tension under wing lifting condition. I agree with th! e person who stated that the information that Carlson gives about round aluminum tubing is confusing and of no use in strength analysis. They compare their strut to a 12.75 diameter, 0.065 wall tube... absurd. The closest match to their strut and a round tube with that wall thickness is a 2-1/4 OD 6061-T6 tube, which has a metal area of 0.446 sq.in., a weight of 0.533 lb/lin.ft., and a tensile strength of 20,079 lb. It is obvious by inspection that the Carlson strut will have significantly lower aerodynamic drag than a round 2-1/4 tube, which along with cost is just about the only useful comparison between the two. Moving along the strut to the ends, the fork ends wont be a problem... with nominal AN6 ends (3/8) and 125,000 psi steel strength, theyre good for at least 13-14,000 pounds in tension... about 4 times the needed strength. The catalog touts them as high strength, but no strength value is given. The Heim ends appear to have the same strength in tension as the forks, so they shouldnt be a problem either. The trickiest part is the attachment between the fork (or Heim) ends and the lift strut. Looking at the 1 x 3/4 6061-T6 bar that Carlson uses, when undrilled it has a cross-sectional area of 0.75 sq.in. so its good for about 33,750 pounds in tension but at the spot where the first mounting hole is drilled through it to attach it to the strut, the cross-sectional area is reduced to about 0.5625 sq.in. and the capacity drops to 25,312 pounds. Still about 7 times whats needed. The aluminum bar is just fine; steel is not necessary. Where the fork ends are threaded into the bar stock, the edge distance of the hole in the end of the bar is about 3/16 on the short side, so that shouldnt be a limiting factor. If a sufficient number of threads of the fork end are engaged into the threaded hole in the bar, that wont be the weak point either. The problem then comes down to the end of the bar that slides up into the hollow end of the strut and how the two are connected. Since the bar is much thicker than the wall of the lift strut, the weakest point of the connection is at the mounting bolt holes through the sides of the strut. Although there are rigorous methods for determining the load distribution between the bolts in a multi-bolt connection, a fairly safe assumption in this case is that if the bolts are not placed too close to one another or too close to the edges of the material, at ultimate loading all of the bolts will be carrying about the same amount of load. Lets look at the load at ! the bolt holes through the strut wall nearest to the end of the strut. There is no need to look at shear strength of the AN4 connecting bolts.... they will be far stronger in shear than the thin walls of the strut. There are at least two possible modes of failure at the connecting bolts but Ill look only at shear out of the bolt, where the steel bolt pulls or tends to elongate the hole in the softer aluminum as it yields. In that mode, the force that the bolt can carry before the hole in the aluminum shears out is equal to the shear strength of the material (extruded 6061-T6 is about 24,000 psi) times the area that the bolt bears on. For AN4 bolts and a strut wall thickness of 0.119 at its thinnest point, with the first bolt in the row being about 3/4 from the end of the strut (such as in the image Piet_construction_449.jpg that Dan Helsper shared earlier), that works out to about 8,568 lbs for the first bolt in the row. That one alone should be able to carry twice the required load, so two bolts through each end of the strut should be more than adequate but if they give you three holes, use three bolts. The distance of the first bolt hole from the end of the strut i! s very important though... if the edge distance is reduced from 3/4 to 1/2 (for example), the pull-out load drops to about 5,712 lbs... only about 1.6 times the required capacity. With at least two attachment bolts, the connection should be adequate, but by all means keep the first bolt hole comfortably away from the end of the strut. -------- Oscar Zuniga Medford, OR Air Camper NX41CC "Scout" A75 power, 72x36 Culver prop Read this topic online here: http://forums.matronics.com/viewtopic.php?p=477134#477134 ________________________________ Message 7 _____________________________________ Time: 07:33:03 PM PST US From: "Clif Dawson" Subject: Re: Pietenpol-List: Re: Wing Strut Attachment Now if the rest of our beloved beast matched these figures ----- goodby Sukhoi. :-) Attached are pics of my strut ends and the jury struts on my Hemlock struts. Just for comparison, you understand. Clif "Hell, there are no rules here; we're trying to accomplish something." (Thomas Edison) Handle every Stressful situation like a dog. If you can't eat it or play with it, Piss on it and walk away. I?Tve taken a look at the forces in the wing lift struts of the Air Camper. To run the analysis, I?Tve used the dimensions given in the Orrin Hoopman (1933-34) plans to create the geometry, but there?Ts not a lot of difference in any of the others that are commonly used. The main lift strut length is given as 89-3/4? between the bolt holes, the front cabane length is given as 21-1/4?, and the distance from the cabane mounting point to the lift strut mounting point at the bottom of the fuselage side is 22-3/4?, so that side of the triangle is 44?. With those two dimensions, we find that the angle that the lift strut makes with the bottom of the wing is 29.4 degrees. Assuming an aircraft gross weight of 1,088 lb and a design loading of +3.8G (normal category aircraft), we get a total load to be supported by the wing of 4,134 lb, and each half of the wing thus has to support 2,067 lb. According to the source quoted in my spar strength article in the BPA Newsletter (Noel Becar?Ts analysis in a 1963 Sport Aviation), the main spar in a strut-braced monoplane at a high positive angle of attack carries about 84% of the total load, so that results in 1,737 lbs that each front lift strut carries. If the load on the spar where the lift strut attaches is 1,737 lbs and the strut is at an angle of 29.4 degrees to the wing, the tensile force in the strut will be 1737/(sin 29.4) = 3,538 lbs. This is the force that every element in the lift strut assembly must meet or exceed in order for the wing to sustain +3.8G loading. -------- Oscar Zuniga Medford, OR Air Camper NX41CC "Scout" A75 power, 72x36 Culver prop ------------------------------------------------------------------------------------- Other Matronics Email List Services ------------------------------------------------------------------------------------- Post A New Message pietenpol-list@matronics.com UN/SUBSCRIBE http://www.matronics.com/subscription List FAQ http://www.matronics.com/FAQ/Pietenpol-List.htm Web Forum Interface To Lists http://forums.matronics.com Matronics List Wiki http://wiki.matronics.com Full Archive Search Engine http://www.matronics.com/search 7-Day List Browse http://www.matronics.com/browse/pietenpol-list Browse Digests http://www.matronics.com/digest/pietenpol-list Browse Other Lists http://www.matronics.com/browse Live Online Chat! http://www.matronics.com/chat Archive Downloading http://www.matronics.com/archives Photo Share http://www.matronics.com/photoshare Other Email Lists http://www.matronics.com/emaillists Contributions http://www.matronics.com/contribution ------------------------------------------------------------------------------------- These Email List Services are sponsored solely by Matronics and through the generous Contributions of its members.