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No it will not. This is a common misconception people have with lugs. As long as the applied loads to the lug are less than the clamping force, nothing really happens.
Are you sure? I know that the objects being clamped will not see any additional load until the fastener force is exceeded, but right now I can't visualize how a separating force wouldn't add more load to the fastener. I'll have to dig out my textbooks some time. The only reference I found on a quick search is in an ASME abstract. It is for a pressure vessel application, but I think it translates well to this topic:
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The amount of clamp load due to an externally applied separating force is determined for a boiled assembly in which the fastener is elongated past its proportional limit, while the clamped joint remained within its elastic range. After the initial tightening of the fastener, the joint is subsequently subjected to a tensile separating force, which further increases the fastener tensile stress into the nonlinear range. Such separating force will simultaneously reduce the clamping force in the bolted joint. Upon the removal of the separating service load, the bolted joint system reaches a new equilibrium point between the fastener tension and the joint clamping force. At the new equilibrium point, the fastener tension is reduced from its value at initial assembly, due to the plastic elongation of the fastener. The reduction in fastener tension translates into a partial—yet permanent—loss of the clamping load that may lead to joint leakage, loosening, or fatigue failure.
Abstract from ASME Journal of Pressure Technology
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Also, the steel will become stronger (higher tensile strength) as it is cold worked by the plastic deformation (but it will become more brittle too).
Oops. That should say "yield strength" instead of "tensile strength". It will still break at the same stress, but won't plastically deform again until the new yield stress is reached.