Registered users: Bing [Bot], Google [Bot], Google Adsense [Bot], Yahoo [Bot]
 
User Information


Site Menu


Sponsored


Who is online
In total there are 69 users online :: 4 registered, 0 hidden and 65 guests Most users ever online was 218 on Wed Dec 07, 2016 6:58 pm Registered users: Bing [Bot], Google [Bot], Google Adsense [Bot], Yahoo [Bot] based on users active over the past 5 minutes 

The Team
Administrators
Global Moderators


Sponsored


Design engineering discussionI want to open up with my engineering techniques for the Striker to receive feedback, criticism and advice. I posted this as its own topic because the discussion could prove valuable to anyone designing a high pressure cannon. I will go over my choice of materials and the way I determined that the design is structurally sound to withstand the pressures of 5x combustion.
Chamber material choice is 3 1/2" x 10" SCH40 6061T6 aluminum pipe and 5/8" thick 6061T6 end caps. Six full length studs hold the end caps to the chamber. The studs are made of grade B7 studs, commonly used for pressure vessel applications. Stress analysis was conducted using Solidworks to determine that the chamber and end caps were strong enough to withstand a maximum theoretical pressure 600psi and an operating pressure of 375psi according to HGDT using a golf ball. The safety factor in the design for these components is as follows: Chamber pipe @ 600psi : 3 End caps @ 600psi : 2.9 Chamber pipe @ 375psi : 4.6 End caps @ 375psi : 4.9 Next, I had to make sure the studs would be strong enough. This took a different approach to find the safety factor. I used a thread stress area calculator that multiplies the area of a given thread by the materials tensile strength. Since the studs are being pulled in 2 directions, the strength of each stud had to be divided by 2 in order to arrive at the correct value. Using 5/1624 studs, the min pitch diameter is 0.2806" and at 24 threads per inch, this results in a tensile stress area of 0.05602. It is this value that is multiplied by the materials tensile strength of a min. of 125,000 psi. This results in an ultimate tensile strength of each stud being 7002.5lbs. The studs are torqued to about 80%, so 80% of this strength value equals 5602lbs per stud. This is then divided in half since each stud is being pulled in two directions, equalling 2801lbs of strength per end cap. Six studs equals 16,806lbs of strength per end cap. Each end cap is being pushed by 7,200lbs of force @600psi and by 4,500lbs of force @375psi. This results in a stud safety factor of 2.3 and 3.73 respectively. These values are conservative because the 125,000 of tensile strength per stud is a min value, and even if the 80% torque is exceeded by the combustion force, it doesn't mean the studs will break. They will simply extend briefly and then retract again. If this is repeated many times the studs can fatigue and eventually fail. So all of the safety factors appear to be inline and appropriate with each other. The nuts holding the studs are grade 8, which are high strength nuts. Each nut is stronger than than the stud itself, and since the stud's effective strength is divided by 2, then that means that the nuts themselves would be the last of all things to fail. Side note: This cannon uses pure O2 and MAPP. A modelling program predicts a max pressure of 440psi with pure O2 and MAPP. I used the higher value of 600psi predicted by HGDT assuming the use of air and MAPP to be on the safer side. I am not a math wizz by any means, so I have to rely in these modeling programs to do these chemistry oriented calculations for me. The 1 1/2 SCH10 6061T6 barrel is not of much concern as there are no holes in it, unlike the chamber and thus results in a safety factor of 5.5 @ 600psi. All safety factors are yield based and not based on ultimate failure. Lets stop here and begin the discussion.
Unless I'm missing something, this statement resembles one linebacker pulling on a rope with 300 lbs of pull, or needing a stronger rope with a pair of linebackers pulling against each other playing tug of war. I don't think the divided in half is correct unless I'm missing something. Calculate the strength to hold on one cap and you are done, even with through bolts (studs) and a cap on the other end.
*high five* you deserve way more then 5 spud bux for that post
everything seems perfectly calculated, but i'm missing why you have to divide the tensile strength of the bolts by to. "because there being pulled two ways" But the tensile strength would be measured in the first place by testing them in the application they were designed for A.K.A by pulling them, meaning that this has already been accounted for? EDIT: ops cross posting
It is confusing to me. It seems that the force is on both sides of the stud so the force is doubled. If the force only applies once then the studs are twice as strong as thought. Erroring on the side of caution is great sometimes.
The line backer analogy is what I was thinking. Two endcaps so two linebackers right? Then again, if there was only one linebacker then he would take off running with the rope and there would be no stress on it. I am confused. If each end cap has force on it, doesn't that double the force that the stud experiences? If not why not?
This is correct. The tension force on the stud will be equivalent to the force exerted on one cap. In a tension system with no net motion, a reaction force equivalent to the tension force must exist to oppose the motion of the system. In this particular case, the force applied to the other cap by combustion pressure will act as the opposing force. It is not factored into the calculation.
Consider the rope is tied to a tree. The linebacker pulls on this rope with a force of 100 pounds. However, he nor the rope are going anywhere  because there's no net force  the tree is providing 100 pounds of force in the opposite direction. Replace that tree with another linebacker, also pulling with 100 pounds of force. The two situations result in equal stress on the rope, of 100 pounds of force  not 200 pounds. To explain, imagine, if you will, a 100 pound weight hanging off a rope tied to the ceiling. What is the stress in the rope? 100 pounds, an easy enough question. This is despite the fact that the 100 lb weight is pulling down with 100 pounds, and the ceiling is pulling up with 100 pounds. You only need to consider the load on one end.
Does that thing kinda look like a big cat to you?
This did it. Thanks. With that being said, the design could use 1/4 studs and save a lot of weight!
Now that the linebacker, ropes, and trees are out of the way, are you doubling the force being applied(inside the chamber)???
My Cannons can be found by clicking the following link.
http://www.spudfiles.com/forums/viewtop ... tml#256896
No, because the rest of the cannon is designed for the current pressure.
I meant, the whole TieBolt thing, the pressure being applied to one end cap is also being applied to the opposite end cap, so the force being applied to the tiebolts is doubled, because the pressure applied to each endcap is 7200lbs, the force acting on the studs is 14,400, so in essence I can see why you were halving the stud strength, but you should've been doupling the load being applied not dividing...
My Cannons can be found by clicking the following link.
http://www.spudfiles.com/forums/viewtop ... tml#256896
I divided the strength of the studs, not the force. EDIT: Either way would end up with the same result. Studs being half as strong or force being twice as strong. I just chose to divide the stud strength.
I sometimes get confused by stacked forces like this too. Think of two air cylinders placed back to back, each pushing with a force of 100lb. What is the combined force exerted by the two cylinders? 100lb.
Now take those two cylinders and place them side by side, both pushing on the same object, the combined force is 200lb. Series and parallel.
Blah Blah Blah LobLab: http://propanetennisballcannon.blogspot.com/
Ahh, thats interesting.
 
Who is onlineRegistered users: Bing [Bot], Google [Bot], Google Adsense [Bot], Yahoo [Bot] 
