Versions Compared

Key

  • This line was added.
  • This line was removed.
  • Formatting was changed.
Comment: Migration of unmigrated content due to installation of a new plugin

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"

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.

Desired Performance

The piston must be able to supply enough force at its operating pressure to break the shear pins with a ?x safety factor. According to standards established by the Aerospace Corporation, there must be a minimum 1.5 design burst factor. [4]

As of 12/27/2017, we plan on using 180lbs of shear pins, making for a desired yield force of 720lbs. This analysis makes use of thin-walled pressure vessel theory [2], paraphrased below:

...

 

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):

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

Mathinline
bodyF_{sep} = 12.56 psi*\pi*(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:

...

Note that this standard was cancelled in July, 2002.

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

[5] NASA, Structural Design Requirements and Factors of Safety for Spaceflight Hardware