1 gen
By considering the simmetry condition applied to the plate could be good idea considerer the crack as an half problem.
So the
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And
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So:
(*\[Beta]=Sqrt[Sec[Pi a /L]]*)
\[Beta] =
Sqrt[(4 L/(Pi a)) Tan[Pi a /(4 L)]];
(*\[Sigma]inf=\[Sigma]y\[LeftDoubleBracket]1,2\[RightDoubleBracket];*)
\
\[Sigma]inf = q;
Kfactor = \[Beta] \[Sigma]inf Sqrt[Pi a];
Kfactor2 = \[Sigma]inf*
Sqrt[Pi a]*((1 - (a/(2 L)) + 0.326*(a/L)^2)/Sqrt[1 - (a/L)]);
Print["Fattore di intensificazione degli sforzi teorico 1: ",
N[Kfactor]]
Print["Fattore di intensificazione degli sforzi teorico 2: ",
N[Kfactor2]]