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\sigma_H = \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.
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\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.