Question

PROBLEM \( 3.78 \) The shaft-disk-belt arrangement shown is used to transmit \( 2 \mathrm{~kW} \) from point \( A \) to point \( D \). (a) Using an allowable shearing stress of \( 66 \mathrm{MPa} \), determine the required speed of shaft \( A B \). (b) Solve Part \( a \), assuming that the diameters of shafts \( A B \) and \( C D \) are, respectively, \( 18 \mathrm{~mm} \) and \( 15 \mathrm{~mm} \). SOLUTION \[ \begin{aligned} \tau & =66 \mathrm{MPa}, \quad P=2 \mathrm{~kW} / \mathrm{s} \\ \tau & =\frac{T c}{J}=\frac{2 T}{\pi c^{3}} \quad T=\frac{\pi}{2} c^{3} \tau \end{aligned} \] Allowable torques \( 16 \mathrm{~mm} \) diameter shaft \( \quad c=8 \mathrm{~mm}, \quad T_{\text {all }}=\frac{\pi}{2}(0.008)^{3}\left(66 \cdot 10^{6}\right)=53 \mathrm{~N} \cdot \mathrm{m} \) \( 18 \mathrm{~mm} \) diameter shaft \( \quad c=9 \mathrm{~mm}, \quad T_{\text {all }}=\frac{\pi}{2}(0.009)^{3}\left(66 \cdot 10^{6}\right)=75.6 \mathrm{~N} \cdot \mathrm{m} \) Statics: (a) Allowable torques \( \quad T_{B, \text { all }}=53 \mathrm{~N} \cdot \mathrm{m}, T_{C, \text { all }}=75.6 \mathrm{~N} \cdot \mathrm{m} \) Assume Then \[ \begin{aligned} T_{C}=75.6 \mathrm{~N} \cdot \mathrm{m} & \\ T_{B}=(0.2727)(75.6) & =20.6 \mathrm{~N} \cdot \mathrm{m} \\ & <53 \mathrm{~N} \cdot \mathrm{m} \quad \text { (okay) } \\ P & =2 \pi F T \quad f_{A B}=\frac{P}{2 \pi T_{B}}=\frac{2000}{2 \pi(20.6)} \quad f_{A B}=15.45 \mathrm{~Hz} \end{aligned} \] (b) Allowable torques Assume Then \[ \begin{array}{l} T_{B, \text { all }}=75.6 \mathrm{~N} \cdot \mathrm{m}, \quad T_{C, \text { all }}=53 \mathrm{~N} \cdot \mathrm{m} \\ T_{C}=53 \mathrm{~N} \cdot \mathrm{m} \\ T_{B}=(0.2727)(53)=14.5 \mathrm{~N} \cdot \mathrm{m} \\ <75.6 \mathrm{~N} \cdot \mathrm{m} \\ P=2 \pi F T \quad f_{A B}=\frac{P}{2 \pi T_{B}}=\frac{2000}{2 \pi(14.5)} \\ f_{A B}=21.95 \mathrm{~Hz} \\ \end{array} \]

          PROBLEM \( 3.78 \)
The shaft-disk-belt arrangement shown is used to transmit \( 2 \mathrm{~kW} \) from point \( A \) to point \( D \). (a) Using an allowable shearing stress of \( 66 \mathrm{MPa} \), determine the required speed of shaft \( A B \). (b) Solve Part \( a \), assuming that the diameters of shafts \( A B \) and \( C D \) are, respectively, \( 18 \mathrm{~mm} \) and \( 15 \mathrm{~mm} \).
SOLUTION
\[
\begin{aligned}
\tau & =66 \mathrm{MPa}, \quad P=2 \mathrm{~kW} / \mathrm{s} \\
\tau & =\frac{T c}{J}=\frac{2 T}{\pi c^{3}} \quad T=\frac{\pi}{2} c^{3} \tau
\end{aligned}
\]
Allowable torques
\( 16 \mathrm{~mm} \) diameter shaft \( \quad c=8 \mathrm{~mm}, \quad T_{\text {all }}=\frac{\pi}{2}(0.008)^{3}\left(66 \cdot 10^{6}\right)=53 \mathrm{~N} \cdot \mathrm{m} \)
\( 18 \mathrm{~mm} \) diameter shaft \( \quad c=9 \mathrm{~mm}, \quad T_{\text {all }}=\frac{\pi}{2}(0.009)^{3}\left(66 \cdot 10^{6}\right)=75.6 \mathrm{~N} \cdot \mathrm{m} \)
Statics:
(a) Allowable torques \( \quad T_{B, \text { all }}=53 \mathrm{~N} \cdot \mathrm{m}, T_{C, \text { all }}=75.6 \mathrm{~N} \cdot \mathrm{m} \)
Assume
Then
\[
\begin{aligned}
T_{C}=75.6 \mathrm{~N} \cdot \mathrm{m} & \\
T_{B}=(0.2727)(75.6) & =20.6 \mathrm{~N} \cdot \mathrm{m} \\
& <53 \mathrm{~N} \cdot \mathrm{m} \quad \text { (okay) } \\
P & =2 \pi F T \quad f_{A B}=\frac{P}{2 \pi T_{B}}=\frac{2000}{2 \pi(20.6)} \quad f_{A B}=15.45 \mathrm{~Hz}
\end{aligned}
\]
(b) Allowable torques
Assume
Then
\[
\begin{array}{l}
T_{B, \text { all }}=75.6 \mathrm{~N} \cdot \mathrm{m}, \quad T_{C, \text { all }}=53 \mathrm{~N} \cdot \mathrm{m} \\
T_{C}=53 \mathrm{~N} \cdot \mathrm{m} \\
T_{B}=(0.2727)(53)=14.5 \mathrm{~N} \cdot \mathrm{m} \\
<75.6 \mathrm{~N} \cdot \mathrm{m} \\
P=2 \pi F T \quad f_{A B}=\frac{P}{2 \pi T_{B}}=\frac{2000}{2 \pi(14.5)} \\
f_{A B}=21.95 \mathrm{~Hz} \\
\end{array}
\]

PROBLEM 3.78
The shaft-disk-belt arrangement shown is used to transmit 2 kW from point A to point D. (a) Using an allowable shearing stress of 66 MPa, determine the required speed of shaft A B. (b) Solve Part a, assuming that the diameters of shafts A B and C D are, respectively, 18 mm and 15 mm.
SOLUTION

τ =66 MPa, P=2 kW / s
τ =(T c)/(J)=(2 T)/(π c^3) T=(π)/(2) c^3τ

Allowable torques
16 mm diameter shaft c=8 mm, Tall =(π)/(2)(0.008)^3(66 · 10^6)=53 N·m
18 mm diameter shaft c=9 mm, Tall =(π)/(2)(0.009)^3(66 · 10^6)=75.6 N·m
Statics:
(a) Allowable torques TB, all =53 N·m, TC, all =75.6 N·m
Assume
Then

TC=75.6 N·m

TB=(0.2727)(75.6) =20.6 N·m
<53 N·m (okay)

P =2 π F T fA B=(P)/(2 π TB)=(2000)/(2 π(20.6)) fA B=15.45 Hz

(b) Allowable torques
Assume
Then

TB, all =75.6 N·m, TC, all =53 N·m

TC=53 N·m

TB=(0.2727)(53)=14.5 N·m

<75.6 N·m

P=2 π F T fA B=(P)/(2 π TB)=(2000)/(2 π(14.5))

fA B=21.95 Hz

Added by Ibrahim A.

Question

Please give Ace some feedback