Engineering, 24.03.2020 21:30 arielpraczko1

A 26-tooth pinion rotating at a uniform 1800 rpm meshes with a 55-tooth gear in a spur gear reducer. Both pinion and gear are manufactured to a quality level of 10. The transmitted tangential load is 22 kN. Conditions are such that Km = 1.7. The teeth are standard 20-degree, full-depth. The module is 5 and the face width 62 mm. Determine the bending stress when the mesh is at the highest point of single tooth contact.

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Answer from: ayoismeisalex

The bending stress of the face tooth is $\sigma _{bg} = 502.82 MPa$

Explanation:

From the question we are told that

The number of tooth of the pinion is $N_t = 26 \ tooth$

The velocity of rotation is given as $\omega_p = 1800 rpm$

The number of tooth is of the gear is $N_g = 55 \ tooth$

The quality level is Q_r = 10

The transmitted tangential load is $F_T = 22\ kN$ = $22 KN * \frac{1000N}{1KN} = 22*10^3 N$

k_m = 1.7

The angle of the teeth is $\theta_t = 20^o$

The module is M= 5

The face width is W_f = 62mm

The diameter of the pinion is mathematically represented as

d_p = M * N_t

Substituting the values

d_p = 5 *26

$= 130 mm = \frac{130}{1000} = 0.130m$

The pitch line velocity is mathematically represented as

$V_t = \frac{d_p }{2} \frac{2 \pi \omega_p}{60}$

Substituting values

$= \frac{0.130}{2} * \frac{2 * 3.142 * 1800 }{60}$

$= 12.25\ m/s$

Generally the dynamic factor is mathematically represented as

$K_v = [\frac{A}{A +\sqrt{200V_t} } ]^B$

Now B is a constant that is mathematically represented as

$B = \frac{(12 -Q_r )^{2/3}}{4}$

substituting values

$= \frac{(12- 10 )^{2/3}}{4}$

=0.3968

A is also a constant that is mathematically represented as

A = 50 + 56(1 -B)

Substituting values

= 50 +56 (1- 0.3968)

= 83.779

Substituting these value into the equation for dynamic factor we have

$K_v = [\frac{83.779}{83.779 + \sqrt{200 * 12.25} } ]^{0.3968}$

= 0.831

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

J_g = 0.41

the diameter pitch can be mathematically represented as

$p_d = \frac{1}{M}$

Substituting values

$p_d = \frac{1}{5}$

$=0.2mm^{-1}$

The mathematically representation for gear tooth bending stress in the teeth face is as follows

$\sigma_{bg} = \frac{F_T \cdot p_d }{W_f * J_g}\frac{K_a K_{dt} }{K_v} K_s K_B K_t ----(1)$

Where W_t is the tangential load

W_f is the face width

K_a is the application factor this is obtained from table "Application Factors, Table 12-17 " and the value is K_a = 1

$K_{dt}$ is the load distributed factor

K_s is the size factor

K_B is the rim thickness factor which is obtained for M which has a value 1

K_t is the idler

Substituting values into equation 1

$\sigma_{bg} = \frac{22*10^3 *0.2}{62 * 0.41} * \frac{1 * 1.7 }{0.831} * 1 *1 *1.42$

= 502.82 N/mm^2

$= 502.82 * 1000 * \frac{N}{m^2}$

= 502.82 MPa

Answer from: Quest

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Answer from: Quest

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A 26-tooth pinion rotating at a uniform 1800 rpm meshes with a 55-tooth gear in a spur gear reducer....