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You can find a complete list of DTEG requirements that affect the Recovery system on the Hermes 1 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:

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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.

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EDIT: After first deciding to use McMaster 6453K153, it became apparent that we cannot purchase a companion eye-nut. Although there are some workarounds, none are worthwhile in comparison to switching to 6556K377, which has a thread size of 3/8"-16. The bore size is only marginally larger .

The relevant dimensions and properties of 6556K377 are enumerated in the table below:

and although the overall piston will be larger and heavier, there are actually some significant advantages to this. Notably, we will get a larger factor of safety on burst, which would only be ~5.8x with the 6453K153. This isn't a bad factor of safety, but it doesn't include end cap effects. Also, the previously selected piston only had 2 bolt holes on the payload bulkhead side, versus this one has 4. Lastly, having a slightly longer total length, it will be less of a drastic change for Avionics to make.

The relevant dimensions and properties of 6556K377 are enumerated in the table below:

Bore Diameter, Inner (in)Bore MaterialStroke Length (in)Bore Diameter, Inner (in)Bore MaterialStroke Length (in)Rod MaterialRod Diameter (in)Total Length (baseplate to baseplate; in)
1-1/4  Aluminum5.5303 Stainless Steel3/87.75

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

Rod Material Improvement

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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 and the elastic modulus of aluminum is 10000 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 = 15,765 lb and Pcr,aluminum = 5,630 lb12,789 lbs

As you can see, both steel and aluminum rods have large factors there is approximately a 35x factor of safety on rod buckling due to piston force.

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 0.5625in625"
    • The outer diameter of the piston bore is 1.18125wall thickness is approximately 0.125"

 

Mathinline
body\sigma_{hoop} = \frac{pR}{t} = \frac{p*0.5625in625in}{0.02972 125 in} = 40030 \frac{lb}{in^2}

Pburst = 2115.007 8006 lb/in2

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

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Psep = 362.17 lb/in2

This gives a ~5.8x ~22x factor of safety on burst.

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Mathinline
bodyF_{sep} = 12.56 psi*\pi*(0.5625 625 in)^2 = 1215.48 4 lb

Given that we plan to use 180lb of shear pins, this provides a minimum of 14~11.4 7 factor of safety on premature separation due to internal pressure buildup.

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