A solid circular shaft has factor of safety 2 under a given torque. A concentric hole of radius R/2 is made, leaving the same outer radius R and the same torque. Find the resulting factor of safety (one decimal place).
Maximum torsional shear is TR/J. The hollow-to-solid polar-moment ratio is $(R^4-(R/2)^4)/R^4=15/16$. With outer radius, torque and material strength unchanged, factor of safety scales with J. Thus the new value is $2(15/16)=1.875$, or 1.9.
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