ASTM 232
A chrome vanadium 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.
Expand 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 shown table:
Since the problem is asking for a chrome vanadium wire:
$$$S_{ut}=\frac{1790}{(50mm)^{0.155}}=\frac{1790}{1.834}=976\: MPa$$$
Thus, the max allowable torsional stress for a chrome vanadium wire is approximately:
$$$\tau = 0.50\times (976.14MPa)=488\:MPa$$$
488 MPa
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