Following are the calculation to the diameter of the brass:
wall reactions are identical:
[tex]\to T_{BC} =T_{AB} =T= 680\ N.m[/tex]
Determine the polar moment of inertia of the steel shaft:
[tex]\to J_{st}=\frac{\pi }{32} d^4_{st} \\\\[/tex]
[tex]=\frac{\pi}{32} \times 0.03^4\\\\=7.95 \times 10^{-8}\ m^4\\\\[/tex]
Determine the polar moment of inertia of the brass shaft.
[tex]\to J_{brass}=\frac{\pi }{32} d^4_{br} \\\\[/tex]
Determine the size of the brass via equating both angles of twist of the brass rod and the steel rod.
[tex]\to (\phi_{st})_{AB}=(\phi_{br})_{BC}\\\\\to \frac{T_{st} L_{st}}{J_{st} G_{st}}= \frac{T_{br} L_{br}}{J_{br} G_{br}}\\\\\to \frac{680\times 0.75}{7.95\times 10^{-8} \times 75 \times 10^{9}}= \frac{680 \times 1.6}{\frac{\pi}{32} d^{4}_{br}\times 39 \times 10^9}\\\\\to 0.086\ d^{4}_{br}=2.84\times 10^{-7}\\\\\to d^{4}_{br} =3.304 \times 10^{-6}\\\\\to d_{br}=0.0426\ m\\\\\to d_{br}=42.6\ mm \\\\[/tex]
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