Re: How thin can it be and still hold pressure? #3

How thin can it be? That depends upon a number of factors.
Note: The following contains many simplifications in order to illustrate the problem.
First let's look at allowable stresses, minimum tensile stresses and loads.
Generally the ASME B&PV Code allows a maximum allowable stress of 25% of minimum tensile strength for a pressure containing part.
For a staybolt constructed of material with a 40,000 psi minimum tensile strength, The maximum allowable tensile stress on the staybolt will be 10,000 psi. Under a hydrotest of 1.5 x MAWP, the stress on the staybolt will be 15,000 psi.
A staybolt could be reduced in area by 62.5% and still not fail under the hydrotests. (I.e., the staybolt has only 37.5% of its original area.)
It is important to realize that materials often have a tensile strength well above the minimum tensile strength.
In the above staybolt example, if the tensile strength of the material were 60,000 psi rather than 40,000 psi, then the staybolt could have a reduced area of 25% of the original area and conceivably not fail under the hydrotest.
As far as reduction in thickness of plates, the problem is quite a bit more complicated. The remaining strength of a plate depends upon the principle loading and the dimensions of the reduced thickness area.
In evaluating these conditions, the critical dimensions are the length and depth of the wasted area. The critical length dimension is perpendicular to the principal stress axis. Generally the shorter the critical length dimension, the greater the depth of the wasted area can be.
In the case that you pointed out in your previous post of the wasted area adjacent to a staybolt, this area is principally loaded in shear. Thus it is quite conceivable that the material could be as thin as you described and still be able to withstand operating stresses without failure.
An additional point to be made is that allowances for corrosion are often added to the size of the material determined by the design process. This corrosion allowance is not used in determining the strength of the part. For example, a plate thickness is determined to be 1/2" based upon pressure loadings. If a corrosion allowance of 1/8" is added, the wasteage that can occur to reduce the pressure strength to below 1.5 x MAWP will be considerable.
To illustrate, the plate with 1/2" thickness will operate at 25% of the tensile strength of the material. With the added 1/8" corrosion allowance, at MAWP the plate will be operating at 20% of the tensile strength of the material. At hydrotest pressures, the material will have a stress of only 30% of the tensile strength. It follows that the plate could be uniformly ( and I stress uniformly ) corroded to 30% of its original thickness and still pass the hydrotest. For the 5/8" original thickness, this would be aproximately 3/16".
Hope this helps.