A 50 mm diameter steel shaft and a 100 mm long steel cylindrical bushing with an outer diameter of 70 mm have been incorrectly shrink fit to

Question

A 50 mm diameter steel shaft and a 100 mm long steel cylindrical bushing with an outer diameter of 70 mm have been incorrectly shrink fit together and have to be separated. What axial force, Pa, is needed for this if the diametral interference is 0.005 mm and the coefficient of friction is 0.2? E (steel) = 207 x 103 MPa (N/mm2)

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niczorrrr 4 years 2021-07-19T03:30:53+00:00 1 Answers 13 views 0

Answers ( )

  1. thienthi
    0
    2021-07-19T03:32:08+00:00
    Reply

    Answer:

    The axial force is  P = 15.93 k N

    Explanation:

    From the question we are told that

         The diameter of the shaft steel is  d = 50mm

          The length of the cylindrical bushing  L =100mm

         The outer diameter of the cylindrical bushing  is  D = 70 \ mm

           The diametral interference is \delta _d = 0.005 mm

           The coefficient of friction is  \mu = 0.2

           The Young modulus of  steel is  207 *10^{3} MPa (N/mm^2)

    The diametral interference is mathematically represented as

               \delta_d = \frac{2 *d * P_B * D^2}{E (D^2 -d^2)}

    Where P_B is the pressure (stress) on the two object held together  

         So making P_B the subject

                P_B = \frac{\delta _d E (D^2 - d^2)}{2 * d* D^2}

    Substituting values

                    P_B = \frac{(0.005) (207 *10^{3} ) * (70^2 - 50^2))}{2 * (50) (70) ^2 }

                     P_B = 5.069 MPa

    Now he axial force required is

                 P = \mu * P_B * A

    Where A is the area which is mathematically evaluated as

                   \pi d l

    So   P = \mu P_B \pi d l

    Substituting values

          P = 0.2 * 5.069 * 3.142 * 50 * 100

           P = 15.93 k N

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