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## Q. 3.SP.6

The stepped shaft shown is to rotate at 900 rpm as it transmits power from a turbine to a generator. The grade of steel specified in the design has an allowable shearing stress of 8 ksi. (a) For the preliminary design shown, determine the maximum power that can be transmitted. (b) If in the final design the radius of the fillet is increased so that $r=\frac{15}{16} in.$, what will be the percent change, relative to the preliminary design, in the power that can be transmitted?

## Verified Solution

a. Preliminary Design. Using the notation of Fig. 3.32, we have:
$D=7.50 \text { in., } d=3.75 \text { in., } r=\frac{9}{16} \text { in. }=0.5625 \text { in. }$

$\frac{D}{d}=\frac{7.50 in. }{3.75 in .}=2 \quad \frac{r}{d}=\frac{0.5625 in .}{3.75 in .}=0.15$

A stress-concentration factor K = 1.33 is found from Fig. 3.29.

Torque. Recalling Eq. (3.25), we write

$\tau_{\max }=K \frac{T c}{J} \quad T=\frac{J}{c} \frac{\tau_{\max }}{K}$                          (1)

where J/c refers to the smaller-diameter shaft:

$J / c=\frac{1}{2} \pi c^{3}=\frac{1}{2} \pi(1.875 \text { in. })^{3}=10.35 in ^{3}$

and where            $\frac{\tau_{\max }}{K}=\frac{8 ksi }{1.33}=6.02 ksi$

Substituting into Eq. (1), we find $T=\left(10.35 \text { in }^{3}\right)(6.02 ksi )=62.3 kip \cdot \text { in. }$

Power. Since $f=(900 rpm ) \frac{1 Hz }{60 rpm }=15 Hz =15 s ^{-1}$, we write

$\begin{gathered}P_{a}=2 \pi f T=2 \pi\left(15 s ^{-1}\right)(62.3 kip \cdot \text { in. })=5.87 \times 10^{6} in . \cdot lb / s \\P_{a}=\left(5.87 \times 10^{6} in. \cdot lb / s \right)(1 hp / 6600 in . \cdot lb / s ) \quad P_{a}=890 hp\end{gathered}$

b. Final Design. For $r=\frac{15}{16} \text { in. }=0.9375 \text { in. },$

$\frac{D}{d}=2 \quad \frac{r}{d}=\frac{0.9375 in .}{3.75 in .}=0.250 \quad K=1.20$

Following the procedure used above, we write

$\frac{\tau_{\max }}{K}=\frac{8 ksi }{1.20}=6.67 ksi$

$T=\frac{J}{c} \frac{\tau_{\max }}{K}=\left(10.35 in ^{3}\right)(6.67 ksi )=69.0 kip \cdot in.$

$\begin{gathered}P_{b}=2 \pi f T=2 \pi\left(15 s ^{-1}\right)(69.0 kip \cdot \text { in. })=6.50 \times 10^{6} in. \cdot lb / s \\P_{b}=\left(6.50 \times 10^{6} in. \cdot lb / s \right)(1 hp / 6600 in . \cdot lb / s )=985 hp\end{gathered}$

Percent Change in Power

$\text { Percent change }=100 \frac{P_{b}-P_{a}}{P_{a}}=100 \frac{985-890}{890}=+11 \%$