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6556K377 on McMaster was selected as the piston. The relevant dimensions and properties of 6556K377 are enumerated in the table below:

Bore Diameter, Inner (in)Bore MaterialStroke Length (in)Rod MaterialRod Diameter (in)Total Length (baseplate to baseplate; in)
1-1/4  Aluminum5.5Steel13/8-168.4

Given a 4.5" coupling section, this gives us a 1" margin on separation distance.

1 Specs for 303 stainless steel used below as estimate.

Buckling Calculation

We need to ensure that the rod of the piston will not buckle when it transfers load to the diaphragm. To perform these calculations we know that the tensile modulus of steel is 28000 ksi:

Mathinline
bodyP_{cr} = \frac{4 \pi^2 E I}{L^2}

Given the area moment of inertia for a circular cross section is:

Mathinline
bodyI = \frac{1}{4}\pi r^4

Pcr,steel = 12,789 lbs

As you can see, there is approximately a 35x factor of safety on rod buckling.

Burst Factor of Safety

Here we calculate the precise burst factor of safety on our piston. Some details:

    • Aluminum 6061-T6 has a tensile yield strength of approximately 276 MPa = 40.03 ksi
    • The inner radius of the piston bore is 0.625"
    • The wall thickness is approximately 0.125"

 

Mathinline
body\sigma_{hoop} = \frac{pR}{t} = \frac{p*0.625in}{0.125 in} = 40030 \frac{lb}{in^2}

Pburst = 8006 lb/in2

The necessary pressure for a separation of 360 pounds is calculated as follows:

Mathinline
bodyP_{sep} = \frac{F_{sep}}{A_{inner}} = \frac{360 lb}{\pi*(0.5625 in)^2}

Psep = 362.17 lb/in2

This gives a ~22x factor of safety on burst.

Premature Separation Factor of Safety

Next we check the factor of safety on premature separation, knowing that as previously calculated the pressure difference for flight will be approximately 12.56 psi (it will actually be much less due to a much lower expected altitude as of 2/2/2018):

The Design Space

Our redesigned piston must have the same form factor as a McMaster piston to allow for easy descope. Given the team's Fall 2017 semester experience with tie-rod pistons, we elect to continue using this style.

DTEG Requirements

As of 1/4/2018, the latest edition of the Spaceport America Cup's Design Test and Evaluation Guide has the following requirements for SRAD pressure vessels. These requirements can be read in more detail here (add link).

4.2.2 DESIGNED BURST PRESSURE FOR METALLIC PRESSURE VESSELS

4.2.4.1 PROOF PRESSURE TESTING

4.2.4.2 OPTIONAL BURST PRESSURE TESTING 

You can find a complete list of DTEG requirements that affect the Recovery system on the Hermes Recovery System page.

Desired Performance

The piston must be able to supply enough force at its operating pressure to break the shear pins with a 2x safety factor, which is the safety-critical guideline for parachute components presented by NASA (Section 3.3.1.5). As of January 4, 2018, we are designing for 180lbs of shear pins, and thus the piston must supply 360lbs. The same source (Section 3.3.1.6) dictates a design burst pressure factor of 2x the maximum design pressure, which aligns with DTEG requirement 4.2.2. [5] Thus, we expect the piston to burst when it supplies 720 lbs. Here, we make use of thin-walled pressure vessel theory [2], paraphrased below:

Neglecting end effects, the limiting factor will be the hoop stress in the piston bore:

 

Mathinline
body\sigma_{hoop} = \frac{pR}{t}

Given Aluminum 6061-T6 as the material, which typically has a tensile yield strength of approximately 276 MPA (this analysis neglects the internal temperature of the piston due to the gas produced by the combustion of black powder. A transient thermal spike could degrade material properties when the piston is pressurizing, but we assume that the magnitude of energy released is negligible compared to the thermal mass of the aluminum). The tensile yield strength can be used to calculate the design burst pressure. For this preliminary analysis, the wall thickness is chosen to be a 0.25x reduction in that of the previously-qualified piston bore (part 6491K254 on McMaster):

Mathinline
bodyt_{new} = \frac{1}{4}*t_{6491K254} = \frac{1}{4}*0.25in = 0.0625 in

 

Mathinline
body\sigma_{burst} = \sigma_{tensile \ yield} = 276*10^6 Pa \approx 40030 psi

Applying P = F/A where F is 720 lbs at burst and A is the area of bore, we find:

Mathinline
body\sigma_{burst} = \frac{F*r}{A*t}

Then, assuming a circular bore, area takes the form A = πr2 and we can solve for the radius of the cylinder.

Mathinline
bodyr = \frac{F}{t*\sigma_{burst}*\pi}

Plugging in numbers, we find the minimum radius of the piston bore:

Mathinline
bodyr = \frac{720 lb }{0.0625 in * 40030 psi * \pi}

rbore, min = 0.092 in

Now, we seek to find an upper bound on the possible piston radius. Another requirement of the piston is that it cannot break the shear pins prematurely due to an internal build-up of pressure caused by the altitude difference. Between 4,245 ft (the altitude of Truth or Consequences, NM) and 152,945 ft ASL (a simulated upper bound on performance as of January 4, 2018), the pressure difference is approximately (given by the 1976 Standard Atmospheric Calculator using no temperature offset) -86600 Pa ≈ 12.56 psi.

Applying a 2x safety factor to premature separation (again in accordance with safety-critical recovery components as dictated by NASA), we calculate the maximum allowable radius of the piston [5]:

...

Mathinline
bodyF_{sep} = P_{diff}*\pi*r_{bore}^2

Mathinline
body\sqrt\frac{F_{sep}}{P_{diff}*\pi} = r_{bore}
Mathinline
body\sqrt\frac{0.5 * 180lb}{ 12.56 psi*\pi} = r_{bore}

rbore, max = 1.51 in

Of course, an additional constraint on piston radius is the allowable space inside the Avionics Bay Coupler. The Team previously found that 6491K254, which had a 1in radius, was large and provided little room for Avionics to house its hardware, especially the batteries. Thus, a logical conclusion is to restrict the new piston geometry to radii below 1 in, which will provide an even larger safety factor on premature separation due to a pressure differential.

Geometry

Given the allowable bore-radius range of 0.092 in 1.51 in (and more accurately, a 1 in upper limit), we opt to initially select a radius of 0.50 in, which is roughly in the middle. This is also a logical decision because 0.5 in is the minimum typical bore diameter for tie rod air cylinders.

Tolerances

To create the bore, we plan to take an existing piece of hollow aluminum tube and turn it down to the proper outer radius (the inner radius can be achieved by drill and then reamer). The chosen wall thickness of 0.0625 is achievable within the 0.5 thousandths diameter tolerance of lathes on campus.

COTS Solutions

First, it is necessary to determine an appropriate COTS solution for the piston. Due to timeline constraints associated with the difficult task of engineering base plates (most notably, all of the required seals), it is logical to take an existing piston and modify it to meet our needs (i.e. changing the throw on the piston, making mass saving cuts, etc).

Solution No. 1: 1691T104

...

*(0.625 in)^2 = 15.4 lb

Given that we plan to use 180lb of shear pins, this provides a minimum of ~11.7 factor of safety on premature separation due to internal pressure buildup.

 

Resources:

The following resources are useful materials for learning about pressure vessel and piston theory:

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Note that this standard was cancelled in July, 2002.

[4] Aerospace Corporation, Operational Guidelines for Spaceflight Pressure Vessels

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