Overview
The case needs to not break.
Define Operating Parameters
The a priori requirements for the design of a case usually come from vehicle level design choices, such as outer diameter and length. These are usually supplemented by system level requirements, such as operating pressure, safety margins, and integration constraints. A typical set of requirements could look like the table below:
| Requirement | Parameter |
|---|---|
| Outer Diameter | 6.00 in |
| Length | 87.00 in |
| MAWP | 900 psi |
| Burst Pressure | 1800 psi |
| Working Temperature | 400° F |
Material Selection
The selection of materials for a rocket motor case is an interesting endeavor. Typical constraints are availability, cost, lead time, availability of appropriate casting tubes and liners, compatibility with commercial standards, etc. Find the main article here.
Stress in a Pressure Vessel
Hoop stress in a thin-walled cylinder can be written:
| \sigma_H = \frac{Pd}{2t} |
Where the stress is equal to the Pressure times the diameter divided by two times the thickness of the cylinder.
Since the diameter and pressure are usually known, this leaves you with two knobs to turn, the thickness of the case, and the material properties which dictate the maximum stress. Another way to work through these equations would be to select a diameter, a thickness, and a maximum stress and then calculate the maximum pressure you could run a case at. This route is more likely if you're re-rating a commercial case or trying to decide if an existing case is appropriate for a given test.
The longitudinal stress in a thin walled cylinder can be written:
| \sigma_L = \frac{Pd}{4t} |
Bolt Hole Tear Out Strength
A critical factor in determining the maximum working pressure of a radially bolted case is the tear out strength of the case. The shear strength of the material must take up all of the load from the bolt. This load makes up the primary component of the longitudinal stress we looked at earlier. The tear out strength of a hole is a function of how close it is to the edge of the material and how thick the material is. Good analysis should be conservative unless extensive testing is planned, so for this case we'll take the area loaded in shear to be from the edge of the bolt hole to the edge of the case. Note that the entire ring of bolts could still be ejected from the case if the force on the bolts exceeds the tensile strength of the material inbetween the bolt holes. Also note that the second level of the bolt pattern must support the loading on it's bolts, as well as the bolt circle below it.
The following set of equations assumes that there is no pre-load on the bolt. A more accurate analysis would calculate the bearing stress of the hole before calculating shear stress on the hole. The following analysis is valid for the first row of holes, or a single row of holes in a case. It can be extrapolated to additional circles of bolts if necessary.
| \tau_{shear} = \frac{P((\frac{d}{2})^2\pi)}{N_{bolts}(2A_{shear})} |
Where the shear area is defined as the thickness of the material times the spacing to the edge of the material.
There are some practical considerations as well. Empirical results show that to avoid tear out the bolt should be at least 1 bolt diameter from the edge or any other bolt in the pattern. In other materials it can be common to require that ratio to be 1.5 or 2.0 diameters from an edge or bolt. For additional reading please investigate AISC J 3.4.
Example Calculation Using Phoenix Booster Forward Closure
Here is a side view of the phoenix forward closure. To make sure that the case will not fail under the motors pressure, we will calculate the shear stress of the forward closure. We will also calculate the tensile stress on the area in-between the bolts.
Distance 1 = 0.63 in
Distance 2 = 1.63 in
Distance 3 = 0.75 in
Using the equation from earlier, we can calculate what the shear stress due to the bolts will be. However since there are two distances from the holes to the edge of the case and we have an equal number of bolts on each row, we will average them to use in the equation.
Thickness of Wall = 0.25 in
A = 0.25*(1.63 + 0.63)/2 = 0.28 in^2
Now we can calculate the shear stress:
P = 826 PSI
d = 4 in
Num Bolts = 24
T_shear = (826*(4/2)^2*pi)/(24*2*0.28) = 772 PSI
This amount of shear is significantly lower than the shear modulus of aluminum, so this should not be a concern.