Question

b) A 150mm diameter shaft has a 25mm diameter ball rolling circumferentially around the outside with a normal load of 1200N. The ball is made of silicon nitride (E = 314GPa, v = 0.26) and the shaft is made of steel (E = 206 GPa, v = 0.30). Apply Hertzian contact analysis to determine the, Curvature sum (include a diagram(s) to define the contact geometry considered) [4 Marks] ii. Dimensions of the contact ellipse [6 Marks] Maximum elastic deformation in the contact [3 Marks] iv. Maximum pressure in the contact [2 Marks] i.

Answer 1

Given,

Diameter of the shaft (D1) = 150 mm

Youngs Modulus of shaft (E1) = 206 GPa = 206*103 N/mm2

Poissons ratio of shaft ($\upsilon$1) = 0.30

Diameter of the ball (D2) = 25 mm

Youngs Modulus of ball (E2) = 314 GPa = 314*103 N/mm2

Poissons ratio of ball ($\upsilon$2) = 0.26

Normal force applied (F) = 1200 N

Hertzian contact analysis relates to the stress which is near to the area of contact between two curvatures of different radii.

(i). Here the diameter of the two surfaces are very far apart each other D>>>d, So the surface of the shaft is assumed to be a flat surface and the ball of diameter d is pushed onto the flat surface with a normal force.

So the curvature sum between the two surfaces becomes equal to the formula$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$

where R is the curvature sum, R1 = radius of the shaft, R2 = radius of the ball

So, (1 / R) = (1 / 75) + (1 / 12.5)

R = 10.714 mm is the curvature sum

R2 1P K24

(ii). The contact modulus of the surface of contact (E*) is equal to $\frac{1}{E^{*}}= \frac{(1-\upsilon _{1}^{2})}{E_{1}}+\frac{(1-\upsilon _{2}^{2})}{E_{2}}$

  

1 E* (1 -0.32) 206 * 103 + (1 – 0.262) 314 * 103

(1 / E*) = 7.386*10-6

   E* = 135391.28 N/ mm2

From (iii). d = 0.016mm

The dimension of the semi ellipse is a = $\sqrt{Rd}$

  a = sqrt(10.714*0.016)

  a = 0.414 mm

So the diameter = 2a = 0.828mm

F

(iii). Since the normal force (F) is $F=\frac{4}{3}E^{*}R^{(1/2)}d^{(3/2)}$ , where d is the indentation depth or the elastic deformation at the contact.

Therefore, $1200=\frac{4}{3}135391.28*(10.714^{(1/2)})(d^{(3/2)})$

d(3/2) = 0.00203

d = 0.016mm is the max elastic deformation at the contact

(iv). Maximum pressure at the contact (Po) is $P_{o}= \frac{3F}{2\pi a^{2}}$

Po= (3*1200) / (2*3.14*0.4142)

Po = 3344.584 N/mm2