Stress Analysis of a Helical-Gear Train Problem Redesign the spur-gear train of Examples 12-4 through 12-7 (pp. 707 to 730) using helical gears and compare their safety factors. Given The referenced examples address, respectively, the kinematics, bending stresses, surface stresses, and safety

Question

Stress Analysis of a Helical-Gear Train

Problem Redesign the spur-gear train of Examples 12-4 through , 12-5, 12-6, 12-7 (pp. 707 to 730) using helical gears and compare their safety factors.

Given The referenced examples address, respectively, the kinematics, bending stresses, surface stresses, and safety factors for a 3-gear train with the following data: [latex]W_{t}[/latex] = 432.17 lb, [latex]N_{p }[/latex] = 14, [latex]N_{idler}[/latex] = 17, [latex]N_{g}[/latex] = 49, [latex]\phi [/latex] = 25°, [latex]p_{d}[/latex] = 6, [latex]F[/latex] = 2.667 in, pinion speed = 2 500 rpm, and 20 hp. The velocity factor [latex]K_{ u}[/latex] = 0.66 from previous calculations.

Assumptions The teeth are standard AGMA full-depth profiles. The load and source are both uniform in nature. A gear-quality index of 6 will be used. All gears are steel with [latex] u[/latex] = 0.28. The service life required is 5 years of oneshift operation. Operating temperature is 200°F. Based on the assumption of uniform load and source, the application factor [latex]K_{a}[/latex] can be set to 1. The load distribution factor can be estimated from Table 12-16 (p. 715) based on the assumed face width: [latex]K_{m}[/latex] = 1.6. The idler factor [latex]K_{I}[/latex] = 1 for the pinion and gear and [latex]K_{I} [/latex] = 1.42 for the idler gear. The size factor [latex]K_{s}[/latex] = 1 for all three gears. [latex]C_{f}[/latex] = 1. [latex]K_{B}[/latex] = 1. Keep the same [latex]\phi[/latex] and [latex]p_{d}[/latex] as the previous examples’ spur gears and try a 20° helix angle.

Table 12-16 Load Distribution Factors [latex]K_{m}[/latex]
Face in	Width (mm)	[latex]K_{m}[/latex]
<2	(50)	1.6
6	(150)	1.7
9	(250)	1.8
≥20	(500)	2.0

Accepted Answer

1 The bending geometry factor [latex]J[/latex] for a 25° pressure angle, 20° helix angle, 14-tooth pinion in mesh with the 17-tooth idler is found from Table 13-5 to be [latex]J_{pinion}[/latex] = 0.51. The pinion-tooth bending stress is then

[latex] \sigma_{b_p}=\frac{W_t p_d}{F J} \frac{K_a K_m}{K_{ u}} K_s K_B K_I \cong \frac{432.17(6)}{2.667(0.51)} \frac{1(1.6)}{0.66}(1)(1)(1) \cong 4620 psi [/latex] (a)

2 The bending geometry factor [latex]J[/latex] for the 25° pressure angle, 20° helix angle, 17-tooth idler in mesh with the 14-tooth pinion from Table 13-5 is [latex]J_{idler}[/latex] = 0.54. The idler-tooth bending stress is then

[latex] \sigma_{b_i}=\frac{W_t p_d}{F J} \frac{K_a K_m}{K_{ u}} K_s K_B K_I \cong \frac{432.17(6)}{2.667(0.54)} \frac{1(1.6)}{0.66}(1)(1)(1.42) \cong 6200 psi [/latex] (b)

Note in Table 13-5 that the idler has a different [latex]J[/latex] factor when considered to be the “gear” in mesh with the smaller pinion than when considered to be the “pinion” in mesh with the larger gear. The smaller value of the two is used because it gives the higher stress.

3 The bending geometry factor [latex]J[/latex] for the 25° pressure angle, 20° helix angle, 49-tooth gear in mesh with the 17-tooth idler is found from Table 13-5 as [latex]J_{gear}[/latex] = 0.66. The gear-tooth bending stress is then

[latex] \sigma_{b_g}=\frac{W_t p_d}{F J} \frac{K_a K_m}{K_{ u}} K_s K_B K_I \cong \frac{432.17(6)}{2.667(0.66)} \frac{1(1.6)}{0.66}(1)(1)(1) \cong 3570 psi [/latex] (c)

4 We will need the pitch diameter and pitch radius of each gear for this calculation. From the data in Example 12-4 (p. 707):

[latex] \begin{aligned} d_p &=2.333 \quad d_i=2.833 \quad d_g=8.167 \ r_p &=1.167 \quad r_i=1.417 \quad r_g=4.083 \end{aligned} [/latex] (d)

5 The addenda, dedenda, and center distances of the meshes are:

[latex] \begin{array}{l} a_p=a_i=a_g=\frac{1.0}{p_d}=\frac{1}{6}=0.167 ext { in } \ C_{p i}=\frac{d_p+d_i}{2}=\frac{2.333+2.833}{2}=2.58 in \ C_{i d}=\frac{d_i+d_g}{2}=\frac{2.833+8.167}{2}=5.50 in \end{array} [/latex] (e)

6 Find the lengths of action [latex]Z_{pi}[/latex] and [latex]Z_{ig }[/latex] for the two meshes using equation 12.2 (p. 686).

[latex] Z=\sqrt{\left(r_p+a_p ight)^2-\left(r_p \cos \phi ight)^2}+\sqrt{\left(r_g+a_g ight)^2-\left(r_g \cos \phi ight)^2}-C \sin \phi [/latex] (2.12)

[latex] \begin{array}{l} Z_{p i}=\sqrt{\left(r_p+a_p ight)^2-\left(r_p \cos \phi ight)^2}+\sqrt{\left(r_i+a_i ight)^2-\left(r_i \cos \phi ight)^2}-C_{p i} \sin \phi\ \quad =\sqrt{(1.167+0.167)^2-\left(1.167 \cos 25^{\circ} ight)^2}\ \quad \quad +\sqrt{(1.417+0.167)^2-\left(1.417 \cos 25^{\circ} ight)^2}-2.58 \sin 25^{\circ}=0.647\ Z_{i g}=\sqrt{\left(r_p+a_p ight)^2-\left(r_p \cos \phi ight)^2}+\sqrt{\left(r_g+a_g ight)^2-\left(r_g \cos \phi ight)^2}-C_{i g} \sin \phi\ \quad =\sqrt{(1.167+0.167)^2-\left(1.167 \cos 25^{\circ} ight)^2}\ \quad \quad +\sqrt{(4.083+0.167)^2-\left(4.083 \cos 25^{\circ} ight)^2}-5.50 \sin 25^{\circ}=0.692 \end{array} [/latex] (f)

7 The transverse contact ratios for the two meshes are found from equation 12.7b (p. 694).

[latex] m_p=\frac{p_d Z}{\pi \cos \phi} [/latex] (12.7b)

[latex] \begin{array}{l} ext {for the pinion - idler mesh :} \quad m_{p_{p i}}=\frac{p_d Z_{p i}}{\pi \cos \phi}=\frac{6(0.647)}{\pi \cos 25}=1.36 \ ext {for the idler - gear mesh :} \quad m_{p_{i g}}=\frac{p_d Z_{i g}}{\pi \cos \phi}=\frac{6(0.692)}{\pi \cos 25}=1.46 \end{array} [/latex] (g)

8 The axial contact ratio [latex]m_{F}[/latex] is found from equation 13.5 (p. 752) and the axial pitch [latex]p_{x}[/latex] from equations 13.1a, b, and c (pp. 749–750).

[latex] m_F=\frac{F}{p_x}=\frac{F p_d an \psi}{\pi} [/latex] (13.5)

[latex] p_t=p_n / \cos \psi [/latex] (13.1a)

[latex] p_x=p_n / \sin \psi [/latex] (13.1b)

[latex] p_d=\frac{N}{d}=\frac{\pi}{p_c}=\frac{\pi}{p_t} [/latex] (13.1c)

[latex] \begin{array}{l} m_F=\frac{F p_d an \psi}{\pi}=\frac{2.667(6) an 20^{\circ}}{\pi}=1.85 \ p_x=p_n / \sin \psi=p_t \cos \psi / \sin \psi=\frac{\pi \cos \psi}{p_d \sin \psi}=\frac{\pi \cos 20^{\circ}}{6 \sin 20^{\circ}}=1.44 ext { in } \end{array} [/latex] (h)

9 Find the normal pressure angle [latex]\phi _{n}[/latex] and the base helix angle [latex]\psi _{b}[/latex] from equations 13.2 and 13.6f (pp. 750 and 753), respectively.

[latex] an \phi_t= an \phi= an \phi_n / \cos \psi [/latex] (13.2)

[latex] \psi_b=\cos ^{-1}\left\lgroup \cos \psi \frac{\cos \phi_n}{\cos \phi} ight group [/latex] (13.6f)

[latex] \begin{array}{l} \phi_n= an ^{-1}(\cos \psi an \phi)= an ^{-1}\left(\cos 20^{\circ} an 25^{\circ} ight)=23.66^{\circ} \ \psi_b=\cos ^{-1}\left\lgroup\cos \psi \frac{\cos \phi_n}{\cos \phi} ight group=\cos ^{-1}\left\lgroup\cos 20^{\circ} \frac{\cos 23.66^{\circ}}{\cos 25^{\circ}} ight group=18.25^{\circ} \end{array} [/latex] (i)

10 Find the minimum length of the lines of contact for each mesh using equations 13.6c and 13.6d or e (p. 753) and use it to find the load sharing ratio [latex]m_{N}[/latex] from equation 13.6b (p. 753).

[latex] n_r = ext {fractional part of} m_p \ n_a ext {fractional part of} m_F [/latex] (13.6c)

[latex] ext { if } n_a \leq 1-n_r ext { then } L_{\min }=\frac{m_p F-n_a n_r p_x}{\cos \psi_b} [/latex] (13.6d)

[latex] ext { if } n_a>1-n_r ext { then } L_{\min }=\frac{m_p F-\left(1-n_a ight)\left(1-n_r ight) p_x}{\cos \psi_b} [/latex] (13.6e)

[latex] \begin{aligned} ext {for the pinion - idler mesh :} \quad n_{r_{p i}} &= ext { fractional part of } m_{p_{p i}}=0.36 \ ext {for the idler - gear mesh :} \quad n_{r_{i g}} &= ext { fractional part of } m_{p_{i g}}=0.46 \ ext {for both meshes :} \quad n_a &= ext { fractional part of } m_F=0.85 \end{aligned} [/latex] (j)

for the pinion-idler mesh (equation 13.6e):

[latex] \begin{aligned} ext { if } n_a>1-n_{r_{p i}} ext { then : } L_{\min _{p i}} &=\frac{m_{p_{p i}} F-\left\lgroup1-n_a ight group \left\lgroup1-n_{r_{p i}} ight group p_x}{\cos \psi_b} \ L_{\min _{p i}} &=\frac{1.36(2.667)-(1-0.85)(1-0.36)(1.44)}{\cos \left(18.25^{\circ} ight)}=3.674 \end{aligned} [/latex] (k)

[latex] m_{N_{p i}}=\frac{F}{L_{\min p i}}=\frac{2.667}{3.674}=0.726 [/latex] (l)

for the idler-gear mesh (equation 13.6e):

[latex] \begin{array}{l} ext { if } n_a>1-n_{r_{i g}} ext { then: } \quad L_{m i n_{i g}}=\frac{m_{p_{i g}} F-\left\lgroup 1-n_a ight group \left\lgroup 1-n_{r_{i g}} ight group p_x}{\cos \psi_b}\ L_{\min _{i g}}=\frac{1.46(2.667)-(1-0.85)(1-0.46)(1.44)}{\cos \left(18.25^{\circ} ight)}=3.977 \end{array} [/latex] (m)

[latex] m_{N_{i g}}=\frac{F}{L_{\min _{i g}}}=\frac{2.667}{3.977}=0.671 [/latex] (n)

11 The radii of curvature of the gear teeth are:

[latex] \begin{aligned} ho_p &=\sqrt{\left\{0.5\left[\left(r_p+a_p ight) \pm\left(C_{p i}-r_i-a_i ight) ight] ight\}^2-\left(r_p \cos \phi ight)^2} \ &=\sqrt{\{0.5[(1.167+0.0167)+(2.58-1.417-0.0167)]\}^2-\left(1.167 \cos 25^{\circ} ight)^2} \ &=0.4931 ext { in } \ ho_i &=C_{p i} \sin \phi- ho_p=2.58 \sin 25^{\circ}-0.4931=0.5987 in \ ho_g &=C_{i g} \sin \phi- ho_i=5.5 \sin 25^{\circ}-0.5987=1.726 in \end{aligned} [/latex] (o)

12 The pitting geometry factor [latex]I[/latex] is calculated for a pair of gears in mesh. Since we have two meshes (pinion/idler and idler/gear), there will be two different values of [latex]I[/latex] to be calculated using equations 13.6.

[latex] I=\frac{\cos \phi}{\left\lgroup \frac{1}{ ho_p} \pm \frac{1}{ ho_g} ight group d_p m_N} [/latex] (13.6a)

[latex] m_N=\frac{F}{L_{\min }} [/latex] (13.6b)

[latex] n_r = ext {fractional part of} m_p \ n_a ext {fractional part of} m_F [/latex] (13.6c)

[latex] ext { if } n_a \leq 1-n_r ext { then } L_{\min }=\frac{m_p F-n_a n_r p_x}{\cos \psi_b} [/latex] (13.6d)

[latex] ext { if } n_a>1-n_r ext { then } L_{\min }=\frac{m_p F-\left(1-n_a ight)\left(1-n_r ight) p_x}{\cos \psi_b} [/latex] (13.6e)

[latex] \psi_b=\cos ^{-1}\left\lgroup\cos \psi \frac{\cos \phi_n}{\cos \phi} ight group [/latex] (13.6f)

[latex] \begin{array}{l} ho_p=\sqrt{\left\{0.5\left[\left(r_p+a_p ight) \pm\left(C-r_g-a_g ight) ight] ight\}^2-\left(r_p \cos \phi ight)^2} \ ho_g=C \sin \phi \mp ho_p \end{array} [/latex] (13.6g)

[latex] \begin{array}{l} I_{p i}=\frac{\cos \phi}{\left\lgroup \frac{1}{ ho_p}+\frac{1}{ ho_i} ight group d_p m_{N_{p i}}}=\frac{\cos 25^{\circ}}{\left\lgroup \frac{1}{0.4931}+\frac{1}{0.5987} ight group (2.333)(0.726)}=0.14 \ I_{i g}=\frac{\cos \phi}{\left\lgroup \frac{1}{ ho_i}+\frac{1}{ ho_g} ight group d_i m_{N_{i g}}}=\frac{\cos 25^{\circ}}{\left\lgroup \frac{1}{0.5987}+\frac{1}{1.726} ight group (2.833)(0.671)}=0.21 \end{array} [/latex] (p)

13 The elastic coefficient [latex]C_{p}[/latex] is found from equation 12.23 (p. 719) and, as before, is 2 276.

[latex] C_p=\sqrt{\frac{1}{\left[\left\lgroup \frac{1-v_p^2}{E_p} ight group +\left\lgroup \frac{1-v_g^2}{E_g} ight group ight]}} [/latex] (12.23)

14 The surface stress for the pinion-idler mesh is

[latex] \begin{aligned} \sigma_{c_p} &=C_p \sqrt{\frac{W_t}{F I_{p i} d_p} \frac{C_a C_m}{C_{ u}} C_s C_f} \ & \cong 2276 \sqrt{\frac{432.17}{2.667(0.14)(2.33)} \frac{1(1.6)}{0.66}(1)(1)} \cong 79 kpsi \end{aligned} [/latex] (q)

15 The surface stress for the idler-gear mesh is

[latex] \begin{aligned} \sigma_{c_i} &=C_p \sqrt{\frac{W_t}{F I_{i g} d_i} \frac{C_a C_m}{C_{ u}} C_s C_f} \ & \cong 2276 \sqrt{\frac{432.17}{2.667(0.21)(2.83)} \frac{1(1.6)}{0.66}(1)(1)} \cong 59 kpsi \end{aligned} [/latex] (r)

16 The corrected bending-fatigue strength of the steel from Example 12-7 (p. 730) is 39 kpsi, and its corrected surface-fatigue strength is 105 kpsi. The safety factors against bending failure are found by comparing the corrected bending strength to the bending stress for each gear in the mesh:

[latex] N_{b_{ ext {pinion }}}=\frac{S_{f b}}{\sigma_{b_{ ext {pinion }}}}=\frac{39}{4.6} \cong 8.5 [/latex] (s)

[latex] N_{b_{ ext {idler }}}=\frac{S_{f b}}{\sigma_{b_{ ext {idler }}}}=\frac{39}{6.2}=6.3 [/latex] (t)

[latex] N_{b_{ ext {gear }}}=\frac{S_{f b}}{\sigma_{b_{ ext {gear }}}}=\frac{39}{3.6}=10.8 [/latex] (u)

17 The safety factors against surface failure are found by comparing the corrected surface strength to the surface stress for each gear in the mesh:*

[latex] N_{c_{ ext {pinion-idler }}} =\left\lgroup\frac{S_{f c}}{\sigma_{c_{ ext {pinion-idler }}}} ight group ^2=\left\lgroup\frac{105}{79} ight group ^2 \cong 1.8[/latex] (v)

[latex] N_{c_{ ext {idler-gear }}} =\left\lgroup\frac{S_{f c}}{\sigma_{c_{ ext {idler-gear }}}} ight group ^2=\left\lgroup\frac{105}{59} ight group ^2 \cong 3.2 [/latex] (w)

18 Compare these results to the safety factors for the spur-gear train in Example 12-7 (p. 730). The helical gears have significantly larger safety factors than same-pitch spur gears.

[latex] \begin{array}{c} N_{b_{ ext {pinion }}}=8.5 \quad N_{b_{ ext {idler }}}=6.3 \quad N_{b_{ ext {gear }}}=10.8 \ N_{c_{ ext {pinion-idler }}}=1.8 \quad N_{c_{ ext {idler }- ext { gear }}}=3.2 \end{array} [/latex] (x)

19 The files EX13-01 can be found on the CD-ROM.

* The safety factor against surface failure should be found by comparing the actual load to the load that would produce a stress equal to the material’s corrected surface strength. Because surface stress is related to the square root of the load, the surface-fatigue safety factor can be calculated as the quotient of the square of the corrected surface strength divided by the square of the surface stress for each gear in the mesh.

Table 13-5 AGMA Bending Geometry Factor J for[latex] \phi=25^{\circ}, \psi=20^{\circ} [/latex] Full-Depth Teeth with Tip Loading
	Pinion teeth
	12		14		17		21		26		35		55		135
Gear teeth	P	G	P	G	P	G	P	G	P	G	P	G	P	G	P	G
12	0.47	0.47
14	0.47	0.50	0.50	0.50
17	0.48	0.53	0.51	0.54	0.54	0.54
21	0.48	0.56	0.51	0.57	0.55	0.58	0.58	0.58
26	0.49	0.59	0.52	0.60	0.55	0.60	0.69	0.61	0.62	0.62
35	0.49	0.62	0.53	0.63	0.56	0.64	0.60	0.64	0.62	0.65	0.66	0.66
55	0.50	0.66	0.53	0.67	0.57	0.67	0.60	0.68	0.63	0.69	0.67	0.70	0.71	0.71
135	0.51	0.70	0.54	0.71	0.58	0.72	0.62	0.72	0.65	0.73	0.68	0.74	0.72	0.75	0.76	0.76

Question 13.1: Stress Analysis of a Helical-Gear Train Problem Redesign the...

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