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.

Answers

Answer 1
Answer:

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 * (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 = (130)/(1000) = 0.130m

The pitch line velocity is mathematically represented as

                     V_t = (d_p )/(2) (2 \pi \omega_p)/(60)

Substituting values

                          = (0.130)/(2) * (2 * 3.142 * 1800 )/(60)

                          = 12.25\  m/s

Generally the dynamic factor is mathematically represented as

                      K_v = [(A)/(A +√(200V_t) ) ]^B

Now B is a constant that is mathematically represented as

                B = ((12 -Q_r )^(2/3))/(4)

substituting values

                  = ((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 = [(83.779)/(83.779 + √(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 = (1)/(M)

Substituting values

            p_d  = (1)/(5)

                =0.2mm^(-1)

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

          \sigma_(bg) = (F_T \cdot p_d )/(W_f * J_g)(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) = (22*10^3 *0.2)/(62 * 0.41) * (1 * 1.7 )/(0.831)  * 1 *1 *1.42

            = 502.82  N/mm^2

            = 502.82 * 1000 * (N)/(m^2)

           = 502.82 MPa

           

           

       

 

               

                 


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Answer:

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Answers

Answer:

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Answers

Answer:

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Answers

Answer:

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L=9.2mm

Explanation:

Heat transfer consists of the propagation of energy in the form of heat in different ways, these can be convection if it is through a fluid, radiation through electromagnetic waves and conduction through solid solids.

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part b

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Answers

Answer:

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Explanation:

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