XS60C PCP Repeater project

[rsterne]

Providing the plug in the tube is a relatively close fit compared to it's length, and the screws are far enough apart, it doesn't matter whether the screws are evenly spaced around the tube or not, or whether you have 1, 2, or 3 (of course if 3 you have 3 times the strength of 1).... However, if the screws are of different sizes, or the holes in the tube not accurate, one can fail first, leading to a "zipper" failure because the others are then under more stress.... Proper design would have a safety margin of at least 3:1 to failure, better is 3:1 to yield....

Beyond that, all the details are needed to do such a calculation....

Bob

[Buldawg76]

Stalwart
Its been quite awhile since I have been on here but I am wondering if you have all the details for the mods to the 60C to use the Marauder magazines I it and if so what is the cost for you to do so on a 22 caliber 60C. I had issues with my valve stem seal being cut early on after getting my 60c but since have modified the crosman 22xx and 13xx series old brass valves to work with a delrin seat in the 60c by pressing the stem out of the crosman valve head and pressing in a new longer stem that is also a much tighter fit in the valve end so there is less loss of air past the stem when shooting.

The stock 60C stem is .118' in diameter with the hole in the valve body being right at .125" and the stem that I pressed into the crosman valve head is .123' so it just right for a smooth but tight fit.

If you have the Mrod mag fitment and production worked out I am interested in the costs to have it done.

Buldawg76

[Buldawg76]

Stalwart
What kind of progress have you made in getting these into the production stage as you were saying in the beginning of this thread. it has been awhile since I have been on the site but was wondering iof you are making the mad in a production run that can be purchased by the members of the forum and if so what is the cost of the breech and barrel mods and do you have exchanges or do you do the parts as they are sent to you.

Any further info on your progress would be welcomed

BD.

[ali]

Hi rsterne, thanks for the response. I really appreciate your great help for us – airgun enthusiasts.
 
Way back to my inquiry, your answer is pointing to the strength of the screws that are in shear stress.
This is the way I understand your reply ofcourse. I think what you are pointing is the same like the
FIRST SITUATION shown in the picture here?

What I’m referring is the tensile strength of the tube/cylinder in relation to the way the screws are
placed –SECOND SITUATION.



For Details, let’s assume these:
Tube: 2024-T3
Yield Strength: 50 ksi
Ultimate Tensile Strength: 70 ksi
Tube I.D.:  25mm
Tube O.D:  30mm
Thickness: 2.5mm
Allen head screw’s Yield strength: 120 ksi
Allen head O.D: 8mm
Pressure inside the tube: 3000 psi


FIRST SITUATION: All four pieces allen screws are placed on the same plane; 10 mm from the tip of the tube (as shown in the picture).

SECOND SITUATION: Two allen screws are on the same plane; 10mm from the tip of the tube
                                      Another two allen screws on the other stress plane; 28mm from the tip of the tube

Question: Does the tube in the SECOND SITUATION has more strength than the FIRST SITUATION?
               Does placing the screws like the one in the SECOND SITUATION gives higher safety factor for the tube?

[rsterne]

I am not an engineer, but my gut reaction is this.... The loading on the screw heads and the tube wall is exactly the same in both cases, so in terms of the shear strength of the screw heads and the yield strength of the pockets in the tube wall there would be no difference.... I assume that the O-ring sealing the tube from the air pressure is at the front end of the valve, so there is no radial load on the tube near the screws....

The only difference, IMO, would be the strength of the remaining tubing wall between the four holes in terms of the tensile load applied.... and the tearout strength of the holes due to their distance from the end of the tube.... A quick back-of-the-napkin calculation:

Area of end of valve = 25^2 x PI/4 = 491 sq.mm. = 0.760 sq.in. x 3000 psi = 2281 lb. end force @ 3000 psi....

Tube circumference (inner) = 25 x PI = 78.5 mm (using outer diameter increases safety margin)
Minus holes = 4 x 8 mm = 32 mm.... so remaining total "width" between the holes is 78.5 - 32 = 46.5 mm
Tube thickness = 2.5 mm, so the cross-sectional area between the holes is 46.5 x 2.5 = 116.25 sq.mm..... which is 0.180 sq.in....
Maximum tensile load on the portion of the tube between the holes is 70,000 x 0.180 = 12,613 lbs.... Safety margin is 5.53:1 based on tensile strength....
Using yield strength.... 50,000 x 0.180 = 9000 lbs.... Safety margin is 3.95:1 based on yield strength.... The strength of the 2nd situation would be greater because there is more area to provide tensile strength between the holes....

The area of the pockets subject to the bearing load is.... 8 mm x 2.5 mm x 4 = 80 sq.mm. = 0.124 sq.in. x 50,000 = 6,200 lbs.... Safety Margin of the bearing load is 2.72:1 based on yield strength....

The "tearout" (shear) area between the pockets and the end of the tube is.... 10 mm x 2 x 2.5 mm x 4 = 200 sq.mm. = 0.310sq.in. x 70,000 x 60% = 13020 lbs.... Safety Margin is 5.7:1 based on shear strength.... The strength of the 2nd situation would be greater because two of the screws are further from the end of the tube....

Since the lowest safety margin is the load on the pockets, IMO there is no practical difference between the strengths of the two situations.... Please remember, I am NOT an engineer, just a guy trying to understand the safety aspects involved.... I am not approving or suggesting any design....

Bob

[ali]

rsterne, thanks for the time in sharing your numbers here.

I assume that the O-ring sealing the tube from the air pressure is at the front end of the valve, so there is no radial load on the tube near the screws....

Yes, you're right. I'm thinking the same way.

The "tearout" (shear) area between the pockets and the end of the tube is.... 10 mm x 2 x 2.5 mm x 4 = 200 sq.mm. = 0.310sq.in. x 70,000 x 60% = 13020 lbs....

Is it not 10mm x 8mm(diameter of the hole) x 4(number of holes)= 320 sq. mm =0.496 sq. in. ( Based from the FIRST SITUATION)?

Regards.

[rsterne]

The calculation for tearout strength, as I understand it, is based on the two shear planes in the direction of force from the edges of the holes to the end of the tube.... I used the distance from the center of the hole to the end of the tube (10mm) instead of 14mm, for increased safety (in case it doesn't tear straight from the edge of the hole).... The way you are suggesting is to use all the area between those two shear planes, which is not my understanding of the way the calculation is done.... and you neglected the thickness of the tube (2.5mm)....

The bearing area is, however, the width of the hole times the thickness of the tube, times the number of holes, and THAT is the limiting factor, and in fact it usually is....

Bob

[stalwart]

Hey Eric, anything new going on with the project?  Did everyone go to the Bahama's? 

The Bahamas sound pretty good right now...

Just real busy with non-AG life right now. All good, but hectic. The planned production will happen, but I have customs and lots of R&D to do before that. I'll let everyone know what the schedule will be... as soon as I do. I'm looking for some competent help... no luck yet.

I've been running into some pretty non-concentric barrels lately. That can decide whether or not I can make them removable... fixed barrels are no problem. I'm looking into solving that with custom barrels. If I can, I'd like to come up with something at a good price. Any suggestions are welcome...

[silent_airman]

Hey Eric, anything new going on with the project?  Did everyone go to the Bahama's? 

The Bahamas sound pretty good right now...

Just real busy with non-AG life right now. All good, but hectic. The planned production will happen, but I have customs and lots of R&D to do before that. I'll let everyone know what the schedule will be... as soon as I do. I'm looking for some competent help... no luck yet.

I've been running into some pretty non-concentric barrels lately. That can decide whether or not I can make them removable... fixed barrels are no problem. I'm looking into solving that with custom barrels. If I can, I'd like to come up with something at a good price. Any suggestions are welcome...

Darn right the Bahamas sound good! -14F here this morning. Heck, your neck of the woods sounds pretty good to me right now.

[RMM]

Hey Eric, anything new going on with the project?  Did everyone go to the Bahama's? 

The Bahamas sound pretty good right now...

Just real busy with non-AG life right now. All good, but hectic. The planned production will happen, but I have customs and lots of R&D to do before that. I'll let everyone know what the schedule will be... as soon as I do. I'm looking for some competent help... no luck yet.

I've been running into some pretty non-concentric barrels lately. That can decide whether or not I can make them removable... fixed barrels are no problem. I'm looking into solving that with custom barrels. If I can, I'd like to come up with something at a good price. Any suggestions are welcome...

-14 F outside this morning, 50 F in the house, furnace repair man here right now...  Bahamas sound real real good right now.   

[ali]

I used the distance from the center of the hole to the end of the tube (10mm) instead of 14mm,..

You're right, and that's actually the only way.

The way you are suggesting is to use all the area between those two shear planes, which is not my understanding of the way the calculation is done.... and you neglected the thickness of the tube (2.5mm)....

Based from the FIRST SITUATION, using thickness (2.5mm) is already 'perpendicular' to the direction of the force. Tearing (shear) of the tube actually rsterne should be 'parallel' to the direction of the force. That's why I used 10mm(tearing distance) x 8mm(hole diameter for screw) cause its the only dimensions that are parallel to the direction of the force.

BTW, from my first post, I was just asking if the strength of the tube is strengthened if the holes for screws are fashioned like the one in the SECOND SITUATION of my drawing. And your calculations add up to the way i see them. Honestly, thanks.

As for the tear ( shear), and bearing strength, this is just what I've learned. Peace.

[rsterne]

If the 2.5mm tube thickness makes no difference to the tearing strength, then I don't understand what is going on.... good thing I'm not an engineer, eh?.... I was taught that the thickness of the tube was paramount in the tearout calculations....

As I see it, the 8mm width of the hole is perpendicular to the force, just as much as the tube thickness is.... only the 10mm length is parallel to it.... To develop a force in lbs. required for failure requires a pressure (lb/sq.in) times an area (sq.in), just look at the units.... One of those dimensions it the length of the area being sheared (the 10 mm, agree on that).... That leaves only two other possibilities for calculating the area, the width of the hole or the thickness of the tube (x 2).... It we reduce the width of the hole to zero (a point load) then there is only one shear plane, and the area being sheared is the length (10mm) times the tube thickness (2.5 mm).... This is just the same as shearing a bolt, where the area of the bolt is used in the calculation.... Double the tube thickness, double the area, double the force to shear it....

When we make the hole wider, we develop two shear planes, one at either edge of the hole.... Think about using a square peg in a square hole, that is 8mm wide.... The center of the peg is being supported by the bearing load of the tube, and the shearing stress is concentrated at the corners, developing two shear planes, 8mm apart.... If we double the thickness of the tube, we double the area to be sheared, but either way, the area to be sheared is the length (10 mm) times the tube thickness (2.5 mm) times 2.... Doubling the width of the square bolt only moves the shear planes further apart.... At least that is what I was taught....

I think perhaps if the distance between the hole and the end of the tube is large enough (and I don't know how large, maybe 10 times the hole diameter?) then the situation approaches the idea of a point load, with only one (wide) shear plane, but the thickness of the tube still must a governing factor in the strength.... Basically the tube cannot tearout, because the failure mode becomes the bearing load exceeding the tensile strength of the tube.... Surely at that point, we are dealing with a bearing load failure, not a tearout failure, and of course the bearing load calculation uses the width of the bolt hole and the thickness of the tube....

The fact remains that in nearly every case in a PCP, the bearing load is the one that fails first, unless the holes are WAY too close to the end of the tube.... In our example, that doesn't occur until the edge of the hole is less than 5mm from the end of the tube.... at which point tearout becomes the likely failure mode instead of (or combined with) a bearing yield failure....

Bob

[ali]

good thing I'm not an engineer, eh?....

It doesn't matter whether you are an Engineer or not. What is important is that we come up with the right analysis for our safety purposes right?  You actually think and see things like an engineer.

As I see it, the 8mm width of the hole is perpendicular to the force, just as much as the tube thickness is.... only the 10mm length is parallel to it....

You're right, 8mm width of the hole looks like perpendicular same as 2mm thickness. But, only if you look at it like a line. What I'm referring here is the plane area that is in stress. Please see my drawing below so that we can picture it out;



Now you may say that it's the same as 2mm thickness right? because you projected the area same as like that. Though it may seem like that, 2mm (which is the thickness) is already perpendicular with respect to the force. Let's see the next drwaing:



The upper drawing is what I'm talking about. The stress area (Side view of the tube) is the shaded one cause by screw to the tube. If you will see it from the 'Force A'  to  'Force B' direction, the area that is under stress is in parallel to the direction of the force.

The Lower drawing is what I understood with respect to your calculations   " 10mm(tearout/shear distance) x 2( two planes) x 2.5mm (thickness) ". Again, as you can see, 2mm thickness is already in perpendicular plane. Moreover, you have missed the area between the two planes.

Same principles is applicable to the bolt in shear.

Though your calculation is safe, imagine if you're in a project calculating huge area. You will surely spend more, 

Peace.

[rsterne]

So you are arguing that the tube thickness is irrelevant?.... I still don't get it....

In your upper diagram, while the entire shaded area is under compression, only the two thin bands in my diagram are in shear.... IMO....

Bob

[ali]

Tube thickness is relevant only in determining how much pressure the tube can contain ( Barlow's principle) . And then followed by tensile, then tear/shear stress.

For the tear/shear, yes it is irrelevant. I'm not arguing bro. It just so happen that i happen to see how you calculate, that's why I asked and got curious. Peace.

[rsterne]

Tube thickness is relevant only in determining how much pressure the tube can contain ( Barlow's principle).
For the tear/shear, yes it is irrelevant

OK, if you say so.... I guess you have the qualifications to back it up.... I just got my methods from guys who do it for a living in the oil patch, like I said, I'm not an engineer.... So the tube thickness is irrelevant for the bearing load of the screw head on the tube wall, too, I guess?.... It could be 1" thick , or 0.001" and still support the same load?.... for the tearout too?.... Wow !

Bob

[ali]

I would welcome your friends' methods with my open heart. If somewhere I missed something, I would admit it.

Remember, we where talking about 'shear' that can contain by the tube with 10 mm tear distance. You would not design a pressure vessel with 0.001" thickness, 1 inch ID and pressurize it with 3000 psi. It will surely burst.

As I said, the way I see it, how much pressure the tube can contain is the first thing to be calculated before proceeding to the tensile, and shear cause by the plug. If it's safe, that's the time you will proceed with the tensile and shear.

As for the 1 inch thick or more, if you are familiar with the principle of punching stress, you will know what I mean.



------------------------------
P.S.

This discussion is for educational purpose only.

[rsterne]

The methodology is either valid or not, the tube does not have to be a pressure vessel, it could simply be a vertical tube, supporting a weight vertically, hanging from the plug that is bolted into the tube by the four 8mm screw heads, as per your diagrams.... Somehow I don't think a 0.001" thick tube wall could support 2281 lbs. but I'm darn sure a 2.5 mm one could.... I am certain that the wall thickness is a vital part of the calculation, but as I say, I'm not an engineer, I just use what I was taught, and common sense....

Bob

[QVTom]

The upper drawing is what I'm talking about. The stress area (Side view of the tube) is the shaded one cause by screw to the tube. If you will see it from the 'Force A'  to  'Force B' direction, the area that is under stress is in parallel to the direction of the force.

The Lower drawing is what I understood with respect to your calculations   " 10mm(tearout/shear distance) x 2( two planes) x 2.5mm (thickness) ". Again, as you can see, 2mm thickness is already in perpendicular plane. Moreover, you have missed the area between the two planes.

I agree 100% with Bob.  His analysis of the tear out and bearing load of the screw heads on the tube was spot on.

Tear out is calculated using the areas in the second picture.    Two lines that are tangent to the hole and parallel to the load are the tube thickness x the length from the tangent point to the end of the tube.

I don't understand how the area in the fist picture is of any value calculating tear out.

Tom

[rsterne]

Tom, thanks for the confirmation.... I guess I'm being conservative by using the distance from the center of the hole to the end of the tube, nice to know....

Bob