Question

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.

Answer 1

Answer:

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

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

            $= 502.82 N/mm^2$

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

           $= 502.82 MPa$