## 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.

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__
__**Hint**

**Hint**

The max allowable torsional stress for hardened and tempered carbon and low-alloy steels in static applications can be approximated as:

$$$S_{sy}=\tau = 0.50\times S_{ut}$$$

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__**Hint 2**

**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