Question 6.27: A schematic of a clutch-testing machine is shown. The steel ......
A schematic of a clutch-testing machine is shown. The steel shaft rotates at a constant speed ω. An axial load is applied to the shaft and is cycled from zero to P. The torque T induced by the clutch face onto the shaft is given by
T={\frac{f P(D+d)}{4}}where D and d are defined in the figure and f is the coefficient of friction of the clutch face. The shaft is machined with S_{{y}}=800\,\mathrm{MPa} and S_{{ut}}=1000\,\mathrm{MPa}. The theoretical stress concentration factors for the fillet are 3.0 and 1.8 for the axial and torsional loading, respectively.
(a) Assume the load variation P is synchronous with shaft rotation. With f = 0.3, find the maximum allowable load P such that the shaft will survive a minimum of 10^{6} cycles with a factor of safety of 3. Use the modified Goodman criterion. Determine the corresponding factor of safety guarding against yielding.
(b) Suppose the shaft is not rotating, but the load P is cycled as shown. With f = 0.3, find the maximum allowable load P so that the shaft will survive a minimum of 10^{6} cycles with a factor of safety of 3. Use the modified Goodman criterion. Determine the corresponding factor of safety guarding against yielding.
(a) From Fig. 6-20, for a notch radius of 3 mm and S_{u t}=1\;\mathrm{GPa},q\doteq0.92.
K_{f}=1+q(K_{t}-1)=1+0.92(3-1)=2.84\sigma_{\mathrm{max}}=-K_{f}{\frac{4P}{\pi d^{2}}}=-{\frac{2.84(4)P}{\pi(0.030)^{2}}}=-4018P
\sigma_{m}=-\sigma_{a}={\frac{1}{2}}(-4018P)=-2009P
T=f P\left({\frac{D+d}{4}}\right)
T_{\mathrm{max}}=0.3P\left({\frac{0.150+0.03}{4}}\right)=0.0135P
From Fig. 6-21, q_{s}\doteq0.95. Also, K_{t s} is given as 1.8. Thus,
K_{f s}=1+q_{s}(K_{t s}-1)=1+0.95(1.8-1)=1.76\tau_{\mathrm{max}}={\frac{16K_{f s}T}{\pi d^{3}}}={\frac{16(1.76)(0.0135P)}{\pi(0.03)^{3}}}=4482P
\tau_{a}=\tau_{m}={\frac{1}{2}}(4482P)=2241P
Eqs. (6-55) and (6-56):
\sigma_{a}^{\prime}=\left\{\left[(K_{f})_{\mathrm{bending}}(\sigma_{a})_{\mathrm{bending}}+(K_{f})_{\mathrm{axial}}{\frac{(\sigma_{a})_{\mathrm{axial}}}{0.85}}\right]^{2}+3\left[(K_{f s})_{\mathrm{torsion}}(\tau_{a})_{\mathrm{torsion}}\right]^{2}\right\}^{1/2} (6-55)
\sigma_{m}^{\prime}=\left\{\left[(K_{f})_{\mathrm{bending}}(\sigma_{m})_{\mathrm{bending}}+(K_{f})_{\mathrm{axial}}{{(\sigma_{m})_{\mathrm{axial}}}}\right]^{2}+3\left[(K_{f s})_{\mathrm{torsion}}(\tau_{m})_{\mathrm{torsion}}\right]^{2}\right\}^{1/2} (6-56)
\sigma_{a}^{\prime}=\sigma_{m}^{\prime}=\left[(\sigma_{a}/0.85)^{2}+3\tau_{a}^{2}\right]^{1/2}=\left[(-2009P/0.85)^{2}+3(2241P)^{2}\right]^{1/2}=4545PS_{e}^{\prime}=0.5(1000)=500\,\mathrm{MPa}
k_{a}=4.51(1000)^{-0.265}=0.723
k_{b}=\left({\frac{30}{7.62}}\right)^{-0.107}=0.864
S_{e}=0.723(0.864)(500)=312.3{\mathrm{~MPa}}
Modified Goodman: {\frac{\sigma_{a}^{\prime}}{S_{e}}}+{\frac{\sigma_{m}^{\prime}}{S_{u t}}}={\frac{1}{n}}
{\frac{4545P}{312.3(10^{6})}}+{\frac{4545P}{1000(10^{6})}}={\frac{1}{3}}\quad\Rightarrow\quad P=17.5(10^{3})\,\mathrm{N}=16.1\,\mathrm{kN}Yield (conservative): n_{y}={\frac{S_{y}}{\sigma_{a}^{\prime}+\sigma_{m}^{\prime}}}
n_{y}={\frac{800(10^{6})}{2(4545)(17.5)(10^{3})}}=5.03(actual): \sigma_{\mathrm{max}}^{\prime}=\left(\sigma_{\mathrm{max}}^{2}+3\tau_{\mathrm{max}}^{2}\right)^{1/2}=\left[(-4018P)^{2}+3(4482P)^{2}\right]^{1/2}=8741\,P
n_{y}={\frac{S_{v}}{\sigma_{\mathrm{max}}^{\prime}}}={\frac{800(10^{6})}{8741(17.5)10^{3}}}=5.22(b) If the shaft is not rotating, \tau_{m}=\tau_{a}=0.
\sigma_{m}=\sigma_{a}=-2009P
k_{b}=1 (axial)
k_{c}=0.85 (Since there is no tension, k_{c}=1 might be more appropriate.)
S_{e}=0.723(1)(0.85)(500)=307.3\mathrm{~MPa}n_{f}=\frac{307.3(10^{6})}{2009P}\quad\Rightarrow\quad P=\frac{307.3(10^{6})}{3(2009)}=51.0(10^{3})\,\mathrm{N}=51.0\,kN
Yield: n_{y}={\frac{800(10^{6})}{2(2009)(51.0)(10^{3})}}=3.90
