Respuesta :
Answer:
The bending stress of the face tooth is [tex]\sigma _{bg} = 502.82 MPa[/tex]
Explanation:
From the question we are told that
The number of tooth of the pinion is [tex]N_t = 26 \ tooth[/tex]
The velocity of rotation is given as [tex]\omega_p = 1800 rpm[/tex]
The number of tooth is of the gear is [tex]N_g = 55 \ tooth[/tex]
The quality level is [tex]Q_r = 10[/tex]
The transmitted tangential load is [tex]F_T = 22\ kN[/tex] = [tex]22 KN * \frac{1000N}{1KN} = 22*10^3 N[/tex]
[tex]k_m = 1.7[/tex]
The angle of the teeth is [tex]\theta_t = 20^o[/tex]
The module is [tex]M= 5[/tex]
The face width is [tex]W_f = 62mm[/tex]
The diameter of the pinion is mathematically represented as
[tex]d_p = M * N_t[/tex]
Substituting the values
[tex]d_p = 5 *26[/tex]
[tex]= 130 mm = \frac{130}{1000} = 0.130m[/tex]
The pitch line velocity is mathematically represented as
[tex]V_t = \frac{d_p }{2} \frac{2 \pi \omega_p}{60}[/tex]
Substituting values
[tex]= \frac{0.130}{2} * \frac{2 * 3.142 * 1800 }{60}[/tex]
[tex]= 12.25\ m/s[/tex]
Generally the dynamic factor is mathematically represented as
[tex]K_v = [\frac{A}{A +\sqrt{200V_t} } ]^B[/tex]
Now B is a constant that is mathematically represented as
[tex]B = \frac{(12 -Q_r )^{2/3}}{4}[/tex]
substituting values
[tex]= \frac{(12- 10 )^{2/3}}{4}[/tex]
[tex]=0.3968[/tex]
A is also a constant that is mathematically represented as
[tex]A = 50 + 56(1 -B)[/tex]
Substituting values
[tex]= 50 +56 (1- 0.3968)[/tex]
[tex]= 83.779[/tex]
Substituting these value into the equation for dynamic factor we have
[tex]K_v = [\frac{83.779}{83.779 + \sqrt{200 * 12.25} } ]^{0.3968}[/tex]
[tex]= 0.831[/tex]
The geometric bending factor for a 20° profile from table
"AGMA Bending Geometry Factor J for 20°, Full -Depth Teeth with HPSTC Loading , Table 2-9"
That corresponds to 55 tooth gear meshing with 26 pinion is
[tex]J_g = 0.41[/tex]
the diameter pitch can be mathematically represented as
[tex]p_d = \frac{1}{M}[/tex]
Substituting values
[tex]p_d = \frac{1}{5}[/tex]
[tex]=0.2mm^{-1}[/tex]
The mathematically representation for gear tooth bending stress in the teeth face is as follows
[tex]\sigma_{bg} = \frac{F_T \cdot p_d }{W_f * J_g}\frac{K_a K_{dt} }{K_v} K_s K_B K_t ----(1)[/tex]
Where [tex]W_t[/tex] is the tangential load
[tex]W_f[/tex] is the face width
[tex]K_a[/tex] is the application factor this is obtained from table "Application Factors, Table 12-17 " and the value is [tex]K_a[/tex] = 1
[tex]K_{dt}[/tex] is the load distributed factor
[tex]K_s[/tex] is the size factor
[tex]K_B[/tex] is the rim thickness factor which is obtained for M which has a value 1
[tex]K_t[/tex] is the idler
Substituting values into equation 1
[tex]\sigma_{bg} = \frac{22*10^3 *0.2}{62 * 0.41} * \frac{1 * 1.7 }{0.831} * 1 *1 *1.42[/tex]
[tex]= 502.82 N/mm^2[/tex]
[tex]= 502.82 * 1000 * \frac{N}{m^2}[/tex]
[tex]= 502.82 MPa[/tex]
