Question 10.4: Two solid steel shafts are connected by the gears shown. Kno...
Two solid steel shafts are connected by the gears shown. Knowing that for each shaft G = 11.2 \times 10^6 \text{ psi} and the allowable shearing stress is 8 ksi, determine (a) the largest torque \pmb{\text{T}}_0 that may be applied to end A of shaft AB and (b) the corresponding angle through which end A of shaft AB rotates.
STRATEGY: Use the free-body diagrams and kinematics to determine the relation between the torques and twist in each shaft segment, AB and CD. Then use the allowable stress to determine the torque that can be applied and Eq. (10.15) to determine the angle of twist at end A.
\phi =\frac{TL}{JG} (10.15)
MODELING: Denoting by F the magnitude of the tangential force between gear teeth (Fig. 1), we have
\pmb{Gear \ B.} \quad \quad \sum{M_B=0:} \quad \quad \quad F(0.875 \ \text{in.})-T_0=0 \quad \quad \quad T_{CD}=2.8T_0 \quad \quad \quad \pmb{(1)} \\ \pmb{Gear \ C.} \quad \quad \sum{M_C=0:} \quad \quad \quad F(2.45 \text{ in.})-T_{CD}=0Using kinematics with Fig. 2, we see that the peripheral motions of the gears are equal and write
r_B\phi_B=r_C\phi_C \quad \quad \quad \phi_B=\phi_C\frac{r_C}{r_B}=\phi_C\frac{2.45 \text{ in.}}{0.875 \text{ in.}}=2.8 \phi_C \quad \quad \quad \pmb{(2)}
ANALYSIS:
a. Torque T0. For shaft AB, T_{AB} = T_0 and c = 0.375 in. (Fig. 3); considering maximum permissible shearing stress, we write
\tau=\frac{T_{AB}c}{J} \quad \quad \quad 8000 \text{ psi}=\frac{T_0(0.975 \text{ in.})}{\frac{1}{2}\pi(0.375 \text{ in.})^4 } \quad \quad \quad T_0=663 \text{ lb.in.}
For shaft CD using Eq. (1) we have T_{CD} = 2.8T_0 (Fig. 4). With c = 0.5 in. and \tau_{\text{all}}=8000 \text{ psi}, we write
\tau=\frac{T_{CD}C}{J} \quad \quad \quad 8000 \text{ psi}=\frac{2.8T_0(0.5 \text{ in.})}{\frac{1}{2}\pi(0.5 \text{ in.})^4 } \quad \quad \quad T_0=561 \text{ lb.in.}
The maximum permissible torque is the smaller value obtained for T_0.
T_0=561 \ \text{lb.in.}
b. Angle of Rotation at End A. We first compute the angle of twist for each shaft.
Shaft AB. For T_{AB}=T_0=561 \text{ lb.in.}, we have
\phi_{A/B}=\frac{T_{AB}L}{JG}=\frac{(561 \text{ lb.in.})(24 \text{ in.})}{\frac{1}{2}\pi(0.375 \text{ in.})^4(11.2 \times 10^6 \text{ psi}) } =0.0387 \text{ rad}=2.22 ^\circ
Shaft CD. T_{CD}=2.8 T_0=2.8(561 \text{ lb.in.})
\phi_{C/D}=\frac{T_{CD}L}{JG}=\frac{2.8(561 \text{ lb.in.})(36 \text{ in.})}{\frac{1}{2}\pi(0.5 \text{ in.})^4 (11.2 \times 10^6 \text{ psi})}=0.0514 \text{ rad} =2.95 ^\circ
Since end D of shaft CD is fixed, we have \phi_C=\phi_{C/D}=2.95^\circ. Using Eq. (2) with Fig. 5, we find the angle of rotation of gear B is
\phi_B=2.8 \phi_C=2.8(2.95^\circ)=8.26^\circ
For end A of shaft AB, we have
\phi_A=\phi_B+\phi_{A/B}=8.26^\circ+2.22^\circ \quad \quad \phi_A=10.48^\circ
