Question

PROBLEM 3.67 Using an allowable shearing stress of \( 50 \mathrm{MPa} \), design a solid steel shaft to transmit \( 15 \mathrm{~kW} \) at a frequency of (a) \( 30 \mathrm{~Hz},(b) 60 \mathrm{~Hz} \). SOLUTION \[ \tau_{\text {all }}=50 \mathrm{MPa}=50 \times 10^{6} \mathrm{~Pa} \quad P=15 \mathrm{~kW}=15 \times 10^{3} \mathrm{~W} \] \( (a) \) \[ \begin{aligned} f & =30 \mathrm{~Hz}: \\ T & =\frac{P}{2 \pi f}=\frac{15 \times 10^{3}}{2 \pi(30)}=79.577 \mathrm{~N} \cdot \mathrm{m} \\ \tau & =\frac{T c}{J}=\frac{2 T}{\pi c^{3}} \\ c & =\sqrt[3]{\frac{2 T}{\pi \tau}}=\sqrt[3]{\frac{(2)(79.577)}{\pi\left(50 \times 10^{6}\right)}}=10.0438 \times 10^{-3} \mathrm{~m}=10.0438 \mathrm{~mm} \end{aligned} \] \( (b) \) \[ d=2 c=20.1 \mathrm{~mm} \] \[ \begin{aligned} f & =60 \mathrm{~Hz} \\ T & =\frac{P}{2 \pi f}=\frac{15 \times 10^{3}}{2 \pi(60)}=39.789 \mathrm{~N} \cdot \mathrm{m} \\ c & =\sqrt[3]{\frac{(2)(39.789)}{\pi\left(50 \times 10^{6}\right)}}=7.9718 \times 10^{-3} \mathrm{~m}=7.9718 \mathrm{~mm} \end{aligned} \] \[ d=2 c=15.94 \mathrm{~mm} \]

          PROBLEM 3.67
Using an allowable shearing stress of \( 50 \mathrm{MPa} \), design a solid steel shaft to transmit \( 15 \mathrm{~kW} \) at a frequency of (a) \( 30 \mathrm{~Hz},(b) 60 \mathrm{~Hz} \).
SOLUTION
\[
\tau_{\text {all }}=50 \mathrm{MPa}=50 \times 10^{6} \mathrm{~Pa} \quad P=15 \mathrm{~kW}=15 \times 10^{3} \mathrm{~W}
\]
\( (a) \)
\[
\begin{aligned}
f & =30 \mathrm{~Hz}: \\
T & =\frac{P}{2 \pi f}=\frac{15 \times 10^{3}}{2 \pi(30)}=79.577 \mathrm{~N} \cdot \mathrm{m} \\
\tau & =\frac{T c}{J}=\frac{2 T}{\pi c^{3}} \\
c & =\sqrt[3]{\frac{2 T}{\pi \tau}}=\sqrt[3]{\frac{(2)(79.577)}{\pi\left(50 \times 10^{6}\right)}}=10.0438 \times 10^{-3} \mathrm{~m}=10.0438 \mathrm{~mm}
\end{aligned}
\]
\( (b) \)
\[
d=2 c=20.1 \mathrm{~mm}
\]
\[
\begin{aligned}
f & =60 \mathrm{~Hz} \\
T & =\frac{P}{2 \pi f}=\frac{15 \times 10^{3}}{2 \pi(60)}=39.789 \mathrm{~N} \cdot \mathrm{m} \\
c & =\sqrt[3]{\frac{(2)(39.789)}{\pi\left(50 \times 10^{6}\right)}}=7.9718 \times 10^{-3} \mathrm{~m}=7.9718 \mathrm{~mm}
\end{aligned}
\]
\[
d=2 c=15.94 \mathrm{~mm}
\]

PROBLEM 3.67
Using an allowable shearing stress of 50 MPa, design a solid steel shaft to transmit 15 kW at a frequency of (a) 30 Hz,(b) 60 Hz.
SOLUTION

τall =50 MPa=50 × 10^6 Pa P=15 kW=15 × 10^3 W

(a)

f =30 Hz:

T =(P)/(2 π f)=(15 × 10^3)/(2 π(30))=79.577 N·m
τ =(T c)/(J)=(2 T)/(π c^3)

c =√((2 T)/(πτ))=√(((2)(79.577))/(π(50 × 10^6)))=10.0438 × 10^-3 m=10.0438 mm

(b)

d=2 c=20.1 mm

f =60 Hz

T =(P)/(2 π f)=(15 × 10^3)/(2 π(60))=39.789 N·m

c =√(((2)(39.789))/(π(50 × 10^6)))=7.9718 × 10^-3 m=7.9718 mm

d=2 c=15.94 mm

Added by Ibrahim A.

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