Question 13.2: Stress Analysis of a Bevel-Gear Train Problem Determine the ...
Stress Analysis of a Bevel-Gear Train
Problem Determine the bending and surface stresses and safety factors in a straight bevel gearset made of the same steel materials, and operating under the same conditions for the same 5-year life as in Example 12-7 (p. 730).
Given N_{p} = 20, N_{g} = 35, \phi = 25°, and p_{d} = 8, passing 10 hp at 2 500 rpm. From Example 12-7: the corrected bending strength is 38 937 psi, and the surface strength is 118 000 psi uncorrected and 105 063 psi corrected.
Assumptions From Example 12-7: K_{a} = C_{a} = K_{s} = C_{s} = C_{f} = C_{H} = C_{R} = C_{T} = 1, K_{m} = C_{m} = 1.6, K_{\nu} = C_{\nu} = 0.652, C_{L} = 0.890, and C_{p} = 2 276. From this section assume: C_{xc} = K_{x} = 1, C_{b} = 0.634, and C_{md} = 1.5.
1 Determine the pinion torque from the given power and speed.
T_p=\frac{P}{\omega_p}=\frac{10 hp \left\lgroup 6600 \frac{ in – lb }{ sec } / hp \right\rgroup}{2500 rpm (2 \pi / 60) \frac{ rad }{ sec } / rpm }=252.1 lb – in (a)
2 Find the pitch diameters of pinion and gear.
d_p=\frac{N_p}{p_d}=\frac{20}{8}=2.50 in , \quad d_g=\frac{35}{8}=4.375 \text { in } (b)
3 Find the pitch cone angles from equation 13.7b (p. 762):
m_G=\frac{\omega_p}{\omega_g}=\frac{N_g}{N_p}=\frac{d_g}{d_p}=\tan \alpha_g=\cot \alpha_p (13.7b)
\begin{array}{l} \alpha_g=\tan ^{-1}\left\lgroup \frac{N_g}{N_p} \right\rgroup =\tan ^{-1}\left\lgroup \frac{35}{20} \right\rgroup =60.26^{\circ} \\ \alpha_p=90-\alpha_g=90-60.26=29.74^{\circ} \end{array} (c)
4 Find the pitch cone length L from equation 13.7a (p. 762):
L=\frac{r_p}{\sin \alpha_p}=\frac{d_p}{2 \sin \alpha_p}=\frac{d_g}{2 \sin \alpha_g} (13.7a)
L=\frac{d_p}{2 \sin \alpha_p}=\frac{2.50}{2 \sin 29.74}=2.519 \text { in } (d)
5 Use the pitch cone length L to find a suitable face width, set to the maximum recommended value.
F=\frac{L}{3}=\frac{2.519}{3}=0.840 \text { in } (e)
6 Look up the bending geometry factors for pinion and gear in Figure 13-5 (p. 752) to find J_{p} = 0.237 and J_{g} = 0.201.
7 Find the bending stress in the pinion from equation 13.9 (p. 763) using J_{p}.
\sigma_b=\frac{2 T_p}{d} \frac{p_d}{F J} \frac{K_a K_m K_s}{K_{\nu} K_x} \quad psi (13.9us)
\sigma_b=\frac{2000 T_p}{d} \frac{1}{F m J} \frac{K_a K_m K_s}{K_{\nu} K_x} (13.9si)
\sigma_{b_{\text {pinion }}}=\frac{2 T_p}{d} \frac{p_d}{F J} \frac{K_a K_m K_s}{K_{\nu} K_x}=\frac{2(252.1)}{2.5} \frac{8}{0.840(0.237)} \frac{1(1.6)(1)}{0.652(1)} \cong 19880 psi (f)
8 Find the bending stress in the gear from equation 13.9 using J_{g}.
\sigma_{b_{\text {gear }}}=\frac{2 T_p}{d} \frac{p_d}{F J} \frac{K_a K_m K_s}{K_{\nu} K_x}=\frac{2(252.1)}{2.5} \frac{8}{0.840(0.201)} \frac{1(1.6)(1)}{0.652(1)} \cong 23440 psi (g)
Note that the gear tooth is more highly stressed than the pinion tooth, because the long addendum on the pinion makes it stronger at the expense of the short-addendum gear tooth.
9 Look up the surface geometry factor for this combination of pinion and gear in Figure 13-6 (p. 767) to find I = 0.076. Use this in equation 13.11 (p. 764) to find T_{D}
T_D=\frac{F}{2} \frac{I C_v}{C_s C_{m d} C_f C_a C_{x c}}\left\lgroup \frac{S_{f c}^{\prime} d}{C_p C_b} \frac{0.774 C_H}{C_T C_R} \right\rgroup ^2 \quad lb – in (13.11us)
T_D=\frac{F}{2000} \frac{I C_v}{C_s C_{m d} C_f C_a C_{x c}}\left\lgroup \frac{S_{f c}^{\prime} d}{C_p C_b} \frac{0.774 C_H}{C_T C_R} \right\rgroup^2 \quad N \cdot m (13.11si)
\begin{aligned} T_D &=\frac{F}{2} \frac{I C_v}{C_s C_{m d} C_f C_a C_{x c}}\left\lgroup \frac{S_{f c}^{\prime} d}{C_p C_b} \frac{0.774 C_H}{C_T C_R} \right\rgroup ^2 \\ &=\frac{0.840}{2} \frac{0.076(0.652)}{1(1.5)(1)(1)(1)}\left\lgroup \frac{118000(2.5)}{2276(0.634)} \frac{0.774(1)}{1(1)} \right\rgroup ^2 \cong 347.5 lb – in \end{aligned} (h)
10 Since T_{D} > T_{p}, z = 0.667. Use these data to find the surface stress with equation 13.10.
\sigma_c=C_p C_b \sqrt{\frac{2 T_D}{F I d^2}\left(\frac{T_p}{T_D}\right)^z \frac{C_a C_m}{C_{\nu}} C_s C_f C_{x c}} (13.10)
\begin{aligned} \sigma_c &=C_p C_b \sqrt{\frac{2 T_D}{F I d^2}\left\lgroup \frac{T_p}{T_D}\right\rgroup ^z \frac{C_a C_m}{C_{\nu}} C_s C_f C_{x c}} \\ &=2276(0.634) \sqrt{\frac{2(347.5)}{0.840(0.076)(2.5)^2}\left\lgroup \frac{252.1}{347.5}\right\rgroup ^{0.667} \frac{1(1.6)}{0.652}(1)(1)(1)} \\ & \cong 84753 psi \end{aligned} (i)
11 The safety factors can now be found as
N_{b_{\text {pinion }}}=\frac{S_{f b}}{\sigma_{b_{\text {pinion }}}}=\frac{38937}{19880} \cong 2.0 (j)
N_{b_{\text {gear }}}=\frac{S_{f b}}{\sigma_{b_{\text {gear }}}}=\frac{38937}{23440} \cong 1.7 (k)
N_c=\left\lgroup \frac{S_{f c}}{\sigma_c}\right\rgroup^2=\left\lgroup \frac{105063}{84753}\right\rgroup^2 \cong 1.5 (l)
12 These are acceptable safety factors. The files EX13-02 are on the CD-ROM.
