Question

PROBLEM 3.154 \( 3.154 \) In the bevel-gear system shown \( \alpha=18.43^{\circ} \). Knowing that the allowable shearing stress is \( 55 \mathrm{MPa} \) in each shaft, determine the largest torque \( \mathbf{T}_{A} \) which may be applied at \( A \). SOLUTION Shaft \( A \) : \[ \begin{aligned} \tau & =55 \mathrm{MPa} \quad c=\frac{1}{2} d=6 \mathrm{~mm} \\ T_{A} & =\frac{J \tau}{c}=\frac{\pi}{2} c^{3} \tau=\frac{\pi}{2}(0.006)^{3}\left(55 \times 10^{6}\right)=18.66 \mathrm{~N} \cdot \mathrm{m} \end{aligned} \] Shaft \( B: \quad \tau=55 \mathrm{MPa} \quad c=\frac{1}{2} d=8 \mathrm{~mm} \) \[ T_{B}=\frac{J \tau}{c}=\frac{\pi}{2} c^{3} \tau=\frac{\pi}{2}(0.008)^{3}\left(55 \times 10^{6}\right)=44.23 \mathrm{~N} \cdot \mathrm{m} \] From Statics: \( T_{A}=\frac{r_{A}}{r_{B}} T_{B}=(\tan \alpha) T_{B}=\left(\tan 18.43^{\circ}\right)(44.23) \) \[ =14.74 \mathrm{~N} \cdot \mathrm{m} \] Allowable value of \( T_{A} \) is the smaller \[ T_{A}=14.74 \mathrm{~N} \cdot \mathrm{m} \]

          PROBLEM 3.154
\( 3.154 \) In the bevel-gear system shown \( \alpha=18.43^{\circ} \). Knowing that the allowable shearing stress is \( 55 \mathrm{MPa} \) in each shaft, determine the largest torque \( \mathbf{T}_{A} \) which may be applied at \( A \).
SOLUTION
Shaft \( A \) :
\[
\begin{aligned}
\tau & =55 \mathrm{MPa} \quad c=\frac{1}{2} d=6 \mathrm{~mm} \\
T_{A} & =\frac{J \tau}{c}=\frac{\pi}{2} c^{3} \tau=\frac{\pi}{2}(0.006)^{3}\left(55 \times 10^{6}\right)=18.66 \mathrm{~N} \cdot \mathrm{m}
\end{aligned}
\]
Shaft \( B: \quad \tau=55 \mathrm{MPa} \quad c=\frac{1}{2} d=8 \mathrm{~mm} \)
\[
T_{B}=\frac{J \tau}{c}=\frac{\pi}{2} c^{3} \tau=\frac{\pi}{2}(0.008)^{3}\left(55 \times 10^{6}\right)=44.23 \mathrm{~N} \cdot \mathrm{m}
\]
From Statics: \( T_{A}=\frac{r_{A}}{r_{B}} T_{B}=(\tan \alpha) T_{B}=\left(\tan 18.43^{\circ}\right)(44.23) \)
\[
=14.74 \mathrm{~N} \cdot \mathrm{m}
\]
Allowable value of \( T_{A} \) is the smaller
\[
T_{A}=14.74 \mathrm{~N} \cdot \mathrm{m}
\]

PROBLEM 3.154
3.154 In the bevel-gear system shown α=18.43^∘. Knowing that the allowable shearing stress is 55 MPa in each shaft, determine the largest torque 𝐓A which may be applied at A.
SOLUTION
Shaft A :

τ =55 MPa c=(1)/(2) d=6 mm

TA =(J τ)/(c)=(π)/(2) c^3τ=(π)/(2)(0.006)^3(55 × 10^6)=18.66 N·m

Shaft B: τ=55 MPa c=(1)/(2) d=8 mm

TB=(J τ)/(c)=(π)/(2) c^3τ=(π)/(2)(0.008)^3(55 × 10^6)=44.23 N·m

From Statics: TA=(rA)/(rB) TB=(tanα) TB=(tan 18.43^∘)(44.23)

=14.74 N·m

Allowable value of TA is the smaller

TA=14.74 N·m

Added by Ibrahim A.

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