## Torsional Stress

A hard-drawn wire with a diameter of 5 cm will be used as a spring material in a static assembly. Approximate the max allowable torsional stress for the wire.

##
__
__**Hint**

**Hint**

The max allowable torsional stress for cold-drawn carbon steel in static applications can be approximated as:

$$$S_{sy}=\tau = 0.45\times S_{ut}$$$

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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 cold-drawn carbon steel (ASTM A227, A228, & A229) in static applications can be approximated as:

$$$S_{sy}=\tau = 0.45\times S_{ut}$$$

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 hard-drawn wire:

$$$S_{ut}=\frac{1510}{(50mm)^{0.201}}=\frac{1510}{2.195}=687.83\: MPa$$$

Thus, the max allowable torsional stress for a hard-drawn wire is approximately:

$$$\tau = 0.45\times (687.83MPa)=310\:MPa$$$

310 MPa