In Example $6\mathrm{.}1,$ the basic mechanical properties of a $2024-$T$81$ aluminum are calculated based on its stress$-$strain curve $($Figure $6\mathrm{.}3).$ Given at the top of page $164$ are load$-$elongation data for a type $304$ stainless steel similar to that presented in Figure $6\mathrm{.}2.$ This steel is similar to alloy $3($a$)$ in Table $6\mathrm{.}1$ except that it has a different thermomechanical history, giving it slightly higher strength with lower ductility. $($a$)$ Plot these data in a manner comparable to the plot shown in Figure $6\mathrm{.}2.($b$)$ Replot these data as a stress$-$strain curve similar to that shown in Figure $6\mathrm{.}3.($c$)$ Replot the initial strain data on an expanded scale, similar to that used for Figure $6\mathrm{.}4.$ Using the results of parts $($a$)-($c$),$ calculate $($d$)E,($e$)$ Y$.$S$.,($f$)$ T$.$S$.,$ and $($g$)$ percent elongation at failure for this $304$ stainless steel. For parts $($d$)-($f$),$ express answers in both Pa and psi units.

$\backslash$table$[[$Load $($N$),$Gage length $($mm$),$Load $($N$),$Gage length $($mm$)],[0,50\mathrm{.}8000,35,220,50\mathrm{.}9778],[4,890,50\mathrm{.}8102,35,720,51\mathrm{.}0032],[9,779,50\mathrm{.}8203,40,540,51\mathrm{.}816],[14,670,50\mathrm{.}8305,48,390,53\mathrm{.}340],[19,560,50\mathrm{.}8406,59,030,55\mathrm{.}880],[24,450,50\mathrm{.}8508,65,870,58\mathrm{.}420],[27,620,50\mathrm{.}8610,69,420,60\mathrm{.}960],[29,390,50\mathrm{.}8711,69,670($maximum$),61\mathrm{.}468],[32,680,50\mathrm{.}9016,68,150,63\mathrm{.}500],[33,950,50\mathrm{.}9270,60,810($fracture$),66\mathrm{.}040($after fracture$)],[34,580,50\mathrm{.}9524,,],[$Original$,$ specimen diameter: $12\mathrm{.}7$ mm$.,]]$