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To solve for the pin diameter with a safety factor of F/2 (safety factor of two, but designed for the pin to break, therefore the force is divided by 2):
D=sqrt(2*F/π*τallow)
From a research paper found on google books (reference), the force required to cause damage to a human chest was 8800N, therefore, a pin diameter was found where F=8800N with the materials nylon, steel and aluminum. Shear strength reference steel and aluminum, shear strength reference for nylon.
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Previous on MSXII, the shear pin used was (believed) to be 4-5mm nylon pin, therefore, it would be suitable for MSXIV to also use a nylon pin, from calculations above, the nylon pin should be between 2.68 to 6.33mm 4 to 6mm in diameter.