Question 5.5: Find the required fillet radius for the juncture of a 6-in-d......
Find the required fillet radius for the juncture of a 6-in-diameter shaft with a 4-indiameter segment if the shaft transmits 110 hp at 100 rpm and the maximum shear stress is limited to 8000 psi.
Given: Dimensions of and requirements for shaft performance.
Find: Fillet radius for connecting two segments of shaft.
Assume: Hooke’s law applies. Transition between segments is only stress–concentration.
We make use of the relationship between applied torque, power output, and rotational frequency (see also Example 5.4) to find the applied torque:
T=\frac{63,000 \cdot 110 \mathrm{hp}}{100 \mathrm{rpm}}=69,300 \mathrm{in}-\mathrm{lb} .The shear stress in the shaft cannot exceed 8000 psi. We obtain the maximum allowable stress concentration factor, using the smaller segment’s radius for c and in J :
K=\frac{\tau_{\max } J}{T c}=\frac{8000 \mathrm{psi}\left[\frac{\pi}{2}(2 \mathrm{in})^4\right]}{69,300 \mathrm{in}-\mathrm{lb}(2 \mathrm{in})}=1.45and
\frac{D}{d}=\frac{\text { big shaft diameter }}{\text { small shaft diameter }}=\frac{6}{4}=1.5 \text {. }From Figure 5.5, we find that this K and this D/d correspond to an r/d ratio of 0.085. This means that the allowable fillet radius is
r=(0.085)(4 \mathrm{in})=0.34 \mathrm{in} .