Answer:
The axial force is [tex]P = 15.93 k N[/tex]
Explanation:
From the question we are told that
The diameter of the shaft steel is [tex]d = 50mm[/tex]
The length of the cylindrical bushing [tex]L =100mm[/tex]
The outer diameter of the cylindrical bushing is [tex]D = 70 \ mm[/tex]
The diametral interference is [tex]\delta _d = 0.005 mm[/tex]
The coefficient of friction is [tex]\mu = 0.2[/tex]
The Young modulus of steel is [tex]207 *10^{3} MPa (N/mm^2)[/tex]
The diametral interference is mathematically represented as
[tex]\delta_d = \frac{2 *d * P_B * D^2}{E (D^2 -d^2)}[/tex]
Where [tex]P_B[/tex] is the pressure (stress) on the two object held together
So making [tex]P_B[/tex] the subject
[tex]P_B = \frac{\delta _d E (D^2 - d^2)}{2 * d* D^2}[/tex]
Substituting values
[tex]P_B = \frac{(0.005) (207 *10^{3} ) * (70^2 - 50^2))}{2 * (50) (70) ^2 }[/tex]
[tex]P_B = 5.069 MPa[/tex]
Now he axial force required is
[tex]P = \mu * P_B * A[/tex]
Where A is the area which is mathematically evaluated as
[tex]\pi d l[/tex]
So [tex]P = \mu P_B \pi d l[/tex]
Substituting values
[tex]P = 0.2 * 5.069 * 3.142 * 50 * 100[/tex]
[tex]P = 15.93 k N[/tex]
