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 Np = 20, Ng = 35, ϕ = 25°,

Question

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 [latex]N_{p} = 20, N_{g} = 35, \phi [/latex]= 25°, and [latex]p_{d}[/latex] = 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: [latex]K_{a} = C_{a} = K_{s} = C_{s} = C_{f} = C_{H} = C_{R} = C_{T} = 1, K_{m} = C_{m} = 1.6, K_{ u} = C_{ u} = 0.652, C_{L} = 0.890[/latex], and [latex]C_{p}[/latex] = 2 276. From this section assume: [latex]C_{xc} = K_{x} = 1, C_{b} = 0.634[/latex], and [latex]C_{md}[/latex] = 1.5.

Accepted Answer

1 Determine the pinion torque from the given power and speed.

[latex] T_p=\frac{P}{\omega_p}=\frac{10 hp \left\lgroup 6600 \frac{ in - lb }{ sec } / hp ight group}{2500 rpm (2 \pi / 60) \frac{ rad }{ sec } / rpm }=252.1 lb - in [/latex] (a)

2 Find the pitch diameters of pinion and gear.

[latex] d_p=\frac{N_p}{p_d}=\frac{20}{8}=2.50 in , \quad d_g=\frac{35}{8}=4.375 ext { in } [/latex] (b)

3 Find the pitch cone angles from equation 13.7b (p. 762):

[latex] m_G=\frac{\omega_p}{\omega_g}=\frac{N_g}{N_p}=\frac{d_g}{d_p}= an \alpha_g=\cot \alpha_p [/latex] (13.7b)

[latex] \begin{array}{l} \alpha_g= an ^{-1}\left\lgroup \frac{N_g}{N_p} ight group = an ^{-1}\left\lgroup \frac{35}{20} ight group =60.26^{\circ} \ \alpha_p=90-\alpha_g=90-60.26=29.74^{\circ} \end{array} [/latex] (c)

4 Find the pitch cone length [latex]L[/latex] from equation 13.7a (p. 762):

[latex] L=\frac{r_p}{\sin \alpha_p}=\frac{d_p}{2 \sin \alpha_p}=\frac{d_g}{2 \sin \alpha_g} [/latex] (13.7a)

[latex] L=\frac{d_p}{2 \sin \alpha_p}=\frac{2.50}{2 \sin 29.74}=2.519 ext { in } [/latex] (d)

5 Use the pitch cone length [latex]L[/latex] to find a suitable face width, set to the maximum recommended value.

[latex] F=\frac{L}{3}=\frac{2.519}{3}=0.840 ext { in } [/latex] (e)

6 Look up the bending geometry factors for pinion and gear in Figure 13-5 (p. 752) to find [latex]J_{p}[/latex] = 0.237 and [latex]J_{g}[/latex] = 0.201.

7 Find the bending stress in the pinion from equation 13.9 (p. 763) using [latex]J_{p}[/latex].

[latex] \sigma_b=\frac{2 T_p}{d} \frac{p_d}{F J} \frac{K_a K_m K_s}{K_{ u} K_x} \quad psi [/latex] (13.9us)

[latex] \sigma_b=\frac{2000 T_p}{d} \frac{1}{F m J} \frac{K_a K_m K_s}{K_{ u} K_x} [/latex] (13.9si)

[latex] \sigma_{b_{ ext {pinion }}}=\frac{2 T_p}{d} \frac{p_d}{F J} \frac{K_a K_m K_s}{K_{ u} 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 [/latex] (f)

8 Find the bending stress in the gear from equation 13.9 using [latex]J_{g}[/latex].

[latex] \sigma_{b_{ ext {gear }}}=\frac{2 T_p}{d} \frac{p_d}{F J} \frac{K_a K_m K_s}{K_{ u} 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 [/latex] (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 [latex]I[/latex] = 0.076. Use this in equation 13.11 (p. 764) to find [latex]T_{D}[/latex]

[latex] 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} ight group ^2 \quad lb - in [/latex] (13.11us)

[latex] 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} ight group^2 \quad N \cdot m [/latex] (13.11si)

[latex] \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} ight group ^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)} ight group ^2 \cong 347.5 lb - in \end{aligned} [/latex] (h)

10 Since [latex]T_{D} > T_{p}, z[/latex] = 0.667. Use these data to find the surface stress with equation 13.10.

[latex] \sigma_c=C_p C_b \sqrt{\frac{2 T_D}{F I d^2}\left(\frac{T_p}{T_D} ight)^z \frac{C_a C_m}{C_{ u}} C_s C_f C_{x c}} [/latex] (13.10)

[latex] \begin{aligned} \sigma_c &=C_p C_b \sqrt{\frac{2 T_D}{F I d^2}\left\lgroup \frac{T_p}{T_D} ight group ^z \frac{C_a C_m}{C_{ u}} 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} ight group ^{0.667} \frac{1(1.6)}{0.652}(1)(1)(1)} \ & \cong 84753 psi \end{aligned} [/latex] (i)

11 The safety factors can now be found as

[latex] N_{b_{ ext {pinion }}}=\frac{S_{f b}}{\sigma_{b_{ ext {pinion }}}}=\frac{38937}{19880} \cong 2.0[/latex] (j)

[latex] N_{b_{ ext {gear }}}=\frac{S_{f b}}{\sigma_{b_{ ext {gear }}}}=\frac{38937}{23440} \cong 1.7 [/latex] (k)

[latex] N_c=\left\lgroup \frac{S_{f c}}{\sigma_c} ight group^2=\left\lgroup \frac{105063}{84753} ight group^2 \cong 1.5 [/latex] (l)

12 These are acceptable safety factors. The files EX13-02 are on the CD-ROM.

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