For each specimen draw two curves on the same graph showing the torque $($N$-$m$)$ versus the rate of twist $($rad$/$m$)$: one curve from zero to the yield point and the other one $($on a different scale for the rate of twist$)$ from zero to rupture. Using two curves will allow both the elastic and the plastic range of behaviour to be viewed at once. If you are using a computer to prepare the graphs, you may simply plot two graphs for each specimen. Determine for each of the specimens: $1.$ The elastic limit $($proportional limit$)$ for shear stress. $2.$ The yield strength at an offset of $0\mathrm{.}5$ rad$/$m of gauge length $($the concept of an $$offset$$ to define a property is shown in Hibbeler Section $8\mathrm{.}3).3.$ Modulus of rigidity, G; $4.$ The Poisson's ratio if Est $=210$ GPa, Eal $=68$ GPa, and Ebr $=97$ GPa $($see Hibbeler Section $8\mathrm{.}6).5.$ Fully plastic torque is given by the equation: $32$ c T $=$ y u $3$ Calculate the plastic torque for both materials using the yield stress calculated above with a $0\mathrm{.}5$ rad$/$m offset and comment on any difference between this value and the torque value observed at failure. $36.$ The elastic energy up to the elastic limit$.$ Useful equations include: shear strain is given by: L c$=$ bar rotation is given by: GJ TL $=$ and the maximum shear stress $($for a given value of T$)$ is given by: J Tc $$ max $=$ for a circular shaft with radius c$.$