Question

PROBLEM \( 3.85 \) \( 150 \mathrm{~mm} \) The stepped shaft shown rotates at \( 450 \mathrm{rpm} \). Knowing that \( r=5 \mathrm{~mm} \), determine the maximum power that can be transmitted without exceeding an allowable shearing stress of \( 50 \mathrm{MPa} \). SOLUTION \[ \begin{aligned} d & =125 \mathrm{~mm} \quad D=150 \mathrm{~mm} \quad r=5 \mathrm{~mm} \\ \frac{D}{d} & =\frac{150}{125}=1.20 \quad \frac{r}{d}=\frac{5}{125}=0.04 \end{aligned} \] From Fig. \( 3.32 \) \[ K=1.55 \] For smaller side \[ \begin{aligned} c & =\frac{1}{2} d=62.5 \mathrm{~mm} \quad \tau=\frac{K T c}{J}=\frac{2 K T}{\pi c^{3}} \\ T & =\frac{\pi c^{3} \tau}{2 K}=\frac{\pi(0.0625)^{3}\left(50 \cdot 10^{6}\right)}{(2)(1.55)}=12.4 \mathrm{kN} \cdot \mathrm{m} \\ f & =450 \mathrm{rpm}=7.5 \mathrm{~Hz} \end{aligned} \] Power \[ P=2 \pi f T=2 \pi(7.5)\left(12.4 \cdot 10^{3}\right)=584 \mathrm{~kW} \] \[ P=584 \mathrm{~kW} \]

          PROBLEM \( 3.85 \)
\( 150 \mathrm{~mm} \) The stepped shaft shown rotates at \( 450 \mathrm{rpm} \). Knowing that \( r=5 \mathrm{~mm} \), determine the maximum power that can be transmitted without exceeding an allowable shearing stress of \( 50 \mathrm{MPa} \).
SOLUTION
\[
\begin{aligned}
d & =125 \mathrm{~mm} \quad D=150 \mathrm{~mm} \quad r=5 \mathrm{~mm} \\
\frac{D}{d} & =\frac{150}{125}=1.20 \quad \frac{r}{d}=\frac{5}{125}=0.04
\end{aligned}
\]
From Fig. \( 3.32 \)
\[
K=1.55
\]
For smaller side
\[
\begin{aligned}
c & =\frac{1}{2} d=62.5 \mathrm{~mm} \quad \tau=\frac{K T c}{J}=\frac{2 K T}{\pi c^{3}} \\
T & =\frac{\pi c^{3} \tau}{2 K}=\frac{\pi(0.0625)^{3}\left(50 \cdot 10^{6}\right)}{(2)(1.55)}=12.4 \mathrm{kN} \cdot \mathrm{m} \\
f & =450 \mathrm{rpm}=7.5 \mathrm{~Hz}
\end{aligned}
\]
Power
\[
P=2 \pi f T=2 \pi(7.5)\left(12.4 \cdot 10^{3}\right)=584 \mathrm{~kW}
\]
\[
P=584 \mathrm{~kW}
\]

PROBLEM 3.85
150 mm The stepped shaft shown rotates at 450 rpm. Knowing that r=5 mm, determine the maximum power that can be transmitted without exceeding an allowable shearing stress of 50 MPa.
SOLUTION

d =125 mm D=150 mm r=5 mm
(D)/(d) =(150)/(125)=1.20 (r)/(d)=(5)/(125)=0.04

From Fig. 3.32

K=1.55

For smaller side

c =(1)/(2) d=62.5 mm τ=(K T c)/(J)=(2 K T)/(π c^3)

T =(π c^3τ)/(2 K)=(π(0.0625)^3(50 · 10^6))/((2)(1.55))=12.4 kN·m

f =450 rpm=7.5 Hz

Power

P=2 π f T=2 π(7.5)(12.4 · 10^3)=584 kW

P=584 kW

Added by Ibrahim A.

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