The maximum transmitted power throughout the solid circular rod is approximately 86.066 horse power.
Let consider that the solid circular rod experiments an uniform torsion and that the material represents a ductile material.
The solid circular rod rotates at constant angular speed ([tex]\omega[/tex]), in radians per second, and all losses throughout the system are negligible. Net power ([tex]\dot W[/tex]), in horse power, is calculated as the product of the torque ([tex]T[/tex]), in kilopound-force-inches, and the angular speed. That is to say:
[tex]\dot W = T \cdot \omega[/tex] (1)
In addition, we add a formula that relates the torque with the shear stress experimented by the solid circular rod:
[tex]\tau = \frac{T \cdot D}{2 \cdot J}[/tex] (2)
Where [tex]J[/tex] is the torsion module, in quartic inches, whose formula for a solid circular cross section is described below:
[tex]J = \frac{\pi \cdot D^{4}}{32}[/tex] (3)
The torsion module is calculated herein: ([tex]D = 2.5\,in[/tex])
[tex]J = 3.835\, in^{4}[/tex]
The maximum allowable torsion is found by isolating the corresponding variable in (2): ([tex]J = 3.835 in^{4}[/tex], [tex]D = 2.5\,in[/tex], )
[tex]T_{max} = \frac{2 \cdot J \cdot \tau_{max}}{D}[/tex]
[tex]T_{max} = 31.907\, kip \cdot in\,(2.659 \,kip \cdot ft)[/tex]
Then, the maximum transmitted power is determined by (1):
[tex]\dot W = (2.659 kip \cdot ft) \cdot (2)\cdot (\pi) \cdot (170 rpm)\\\dot W \approx 2840.188 \,\frac{kip \cdot ft}{min} \\\dot W \approx 86.066 \,h.p.[/tex]
The maximum transmitted power throughout the solid circular rod is approximately 86.066 horse power. [tex]\blacksquare[/tex]
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