Question 14.5: A 17-tooth 20° normal pitch-angle helical pinion with a righ...

All of the parameters in this example are the same as in Ex. 14–4 with the exception that we are using helical gears. Thus, several terms will be the same as Ex. 14–4. The reader should verify that the following terms remain unchanged: K_o = 1, Y_P = 0.303,

For helical gears, the transverse diametral pitch, given by Eq. (13–18),

\sigma _{c,all}= \begin{cases} \frac{S_C}{S_H}\frac{Z_NC_H}{K_TK_R} & (U.S.  customary  units) \\ \frac{S_C}{S_H}\frac{Z_NZ_W}{Y_{\theta }Y_Z} & (SI  units) \end{cases}                      (13–18)

p. 684, is

P_t=P_n \cos \psi =10 \cos 30° = 8.660  teeth/in

Thus, the pitch diameters are d_P = N_P/P_t = 17/8.660 = 1.963 in and d_G = 52/8.660 = 6.005 in. The pitch-line velocity and transmitted force are

V=\frac{\pi d_pn_p}{12}=\frac{\pi (1.963)1800}{12}=925  ft/min

W^t=\frac{33  000H}{V}=\frac{33  000(4)}{925}=142.7  lbf

As in Ex. 14–4, for the dynamic factor, B = 0.8255 and A = 59.77. Thus, Eq. (14–27)

K_v=\begin{cases} \left(\frac{A+\sqrt{V} }{A} \right)^B & V  in  ft/min \\ \left(\frac{A+\sqrt{200V} }{A} \right)^A & V  in  m/s \end{cases}                     (14.27)

gives

K_v=\left(\frac{59.77+\sqrt{925} }{59.77} \right)^{0.8255}=1.404

The geometry factor I for helical gears requires a little work. First, the transverse pressure angle is given by Eq. (13–19) p. 684,

\phi _t=\tan^{-1}\left(\frac{\tan\phi _n}{\cos\psi } \right)=\tan^{-1}\left(\frac{\tan20°}{\cos 30°} \right)=22.80°

The radii of the pinion and gear are r_P = 1.963/2 = 0.9815 in and r_G = 6.004/2 = 3.002 in, respectively. The addendum is a = 1/P_n = 1/10 = 0.1, and the base-circle radii of the pinion and gear are given by Eq. (13–6), p. 672, with Φ = Φ_t:

(r_b)_P=r_P\cos\phi _t= 0.9815 cos 22.80° = 0.9048 in

(r_b)_G = 3.002 cos 22.80° = 2.767 in

From Eq. (14–25),

Z = [(r_P + a)^2 – r^2_{bP}]^{1/2} + [(r_G + a)^2 – r^2_{bG}]1/2 – (r_P + r_G) sin Φ_t                (14.25)

the surface strength geometry factor

Z=\sqrt{(0.9815 + 0.1)^2 – 0.9048^2}+\sqrt{(3.004 – 0.1)^2 – 2.769^2}

-(0.9815 + 3.004) sin 22.80°

= 0.5924 + 1.4027 – 1.544 4 = 0.4507 in

Since the first two terms are less than 1.544 4, the equation for Z stands. From Eq. (14–24) the normal circular pitch p_N is

p_N=p_n\cos\phi_n=\frac{\pi }{P_n}\cos20°=\frac{\pi }{10}\cos20°=0.2952  in

From Eq. (14–21), the load sharing ratio

m_N=\frac{p_N}{0.95Z}=\frac{0.2952}{0.95(0.4507)}=0.6895

Substituting in Eq. (14–23),

I= \begin{cases} \frac{\cos \phi _t\sin \phi _t }{2m_N} \frac{m_G}{m_G+1} & external\ gears \\ \frac{\cos \phi _t\sin \phi _t }{2m_N} \frac{m_G}{m_G-1} & internal\ gears \end{cases}                             (14.23)

the geometry factor I is

I+\frac{sin 22.80° cos 22.80°}{2(0.6895)}\frac{3.06}{3.06 + 1}=0.195

From Fig. 14–7, geometry factors J'_P = 0.45 and J'_G = 0.54. Also from Fig. 14–8 the J-factor multipliers are 0.94 and 0.98, correcting J'_P and J'_G to

J_P = 0.45(0.94) = 0.423
J_G = 0.54(0.98) = 0.529

The load-distribution factor K_m is estimated from Eq. (14–32):

C_{pf}= \begin{cases} \frac{F}{10d_P}-0.025 & F \leq 1 in \\ \frac{F}{10d_P}-0.0375+0.0125F & 1\lt F\leq 17 in\\ \frac{F}{10d_P}-0.1109+0.0207F-0.000 228F^2 & 17\lt F\leq 40 in \end{cases}                        (14.32)

C_{pf}=\frac{1.5}{10(1.963)}- 0.0375 + 0.0125(1.5) = 0.0577

with C_{mc} = 1, C_{pm} = 1, C_{ma} = 0.15 from Fig. 14–11, and C_e = 1. Therefore, from Eq. (14–30),

K_m = C_{mf} = 1 + C_{mc}(C_{pf} C_{pm} + C_{ma} C_e)                   (14.30)

K_m = 1 + (1) [0.0577(1) + 0.15(1) ] = 1.208

(a) Pinion tooth bending. Substituting the appropriate terms into Eq. (14–15) using P_t gives

(\sigma )_P=\left(W^tK_oK_vK_s\frac{P_t}{F}\frac{K_mK_B}{J} \right)_P= 142.7(1)1.404(1.043)\frac{8.66}{1.5}\frac{1.208(1)}{0.423}

= 3445 psi

Substituting the appropriate terms for the pinion into Eq. (14–41) gives

(S_F)_P=\left(\frac{S_tY_N/(K_TK_R)}{\sigma } \right)_P =\frac{31  350(0.977)/[1(0.85)]}{3445} = 10.5

Gear tooth bending. Substituting the appropriate terms for the gear into Eq. (14–15) gives

(\sigma )_G=142.7(1)1.404(1.052)\frac{8.66}{1.5}\frac{1.208(1)}{0.529}=2779  psi

Substituting the appropriate terms for the gear into Eq. (14–41) gives

(S_F)_G=\frac{28  260(0.996)/[1(0.85)]}{2779}=11.9

(b) Pinion tooth wear. Substituting the appropriate terms for the pinion into Eq. (14–16) gives

(\sigma )_P=\left(W^tK_oK_vK_s\frac{K_m}{d_PF}\frac{C_f}{I} \right)^{1/2}_P

=2300\left[142.7(1)1.404(1.043)\frac{1.208}{1.963(1.5)}\frac{1}{0.195} \right]^{1/2}=48  230  psi

Substituting the appropriate terms for the pinion into Eq. (14–42) gives

(S_H)_P=\left(\frac{S_CZ_N/(K_TK_R)}{\sigma _C} \right)_P=\frac{106 400(0.948)/[1(0.85)]}{48  230}=2.46

Gear tooth wear. The only term in Eq. (14–16) that changes for the gear is K_s. Thus,

(\sigma _c)_G=\left[\frac{(K_S)_G}{(K_S)_P} \right]^{1/2}                (\sigma _c)_P=\left(\frac{1.052}{1.043} \right)^{1/2} 48  230 = 48  440  psi

Substituting the appropriate terms for the gear into Eq. (14–42) with C_H = 1.005 gives

(S_H)_G=\frac{93 500(0.973)1.005/[1(0.85)]}{48 440}=2.22

(c) For the pinion we compare S_F with S^2_H , or 10.5 with 2.46^2 = 6.05, so the threat in the pinion is from wear. For the gear we compare S_F with S^2_H , or 11.9 with 2.22^2 = 4.93, so the threat is also from wear in the gear. For the meshing gearset wear controls.