Chrome Silicon Wire

A chrome silicon wire with a diameter of 10 cm will be used as a spring material in a static assembly. Approximate the max allowable torsional stress for the wire.

Hint
The max allowable torsional stress for hardened and tempered carbon and low-alloy steels in static applications can be approximated as:
$$$S_{sy}=\tau = 0.50\times S_{ut}$$$
Hint 2
For spring materials, the minimum tensile strength for common spring steels can be determined from:
$$$S_{ut}=\frac{A}{d^{m}}$$$
where $$S_{ut}$$ is the tensile strength in MPa, and $$d$$ is the wire diameter in millimeters.
The max allowable torsional stress for hardened and tempered carbon and low-alloy steels (ASTM A232 & A401) in static applications can be approximated as:
$$$S_{sy}=\tau = 0.50\times S_{ut}$$$
For spring materials, the minimum tensile strength for common spring steels can be determined from:
$$$S_{ut}=\frac{A}{d^{m}}$$$
where $$S_{ut}$$ is the tensile strength in MPa, $$d$$ is the wire diameter in millimeters, and $$A$$ and $$m$$ can be found in the below table:
Since the problem is asking for a chrome silicon wire:
$$$S_{ut}=\frac{1960}{(100mm)^{0.091}}=\frac{1960}{1.521}=1,289\: MPa$$$
Thus, the max allowable torsional stress for a chrome silicon wire is approximately:
$$$\tau = 0.50\times (1,289MPa)=645\:MPa$$$
645 MPa