**Answer:**

0.0806 mm

735.588 MPa

**Explanation:**

Given data:

Esteel = 200 GPa

vsteel = 0.3,

EAI = 70 GPa, vAI = 1/3

diameter of steel ball = 20-mm

Applied force = 1 kN force

density of steel ( ρ ) = 8050 kg/m^3

weight of steel = ρvg = 0.33 N

**<u>Calculate the size of </u>**

<u><em>i) contact patch</em></u>

size of contact patch = ----------- ( 1 )

where: F = 1000 + 0.33 = 1000.33 N , r = 0.01 m,

E = 57.26 * 10^9 ( calculated value )

**Input values into equation 1 **

size of contact patch ( a **) = 0.0806 mm**

<u><em>ii) maximum contact pressure </em></u>

P = 3F / 2πa^2

= ( 3 * 1000.33 ) / 2π(0.0806)^2

**= 735.588 MPa**