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Chapter 3

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?

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.

Step-by-Step

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 \%

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.
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.
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.
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.