4. Determine the size of the contact patch, relative displacement and the maximum contact pressure for a 20-mm-dia steel ball rolled against a flat aluminum plate with 1 kN force. Take Esteel = 200 GPa, νsteel = 0.3, EAl = 70 GPa and νAl = 1/3.

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

Calculate the size of

i) contact patch

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

ii) maximum contact pressure

P = 3F / 2πa^2

  = ( 3 * 1000.33 ) / 2π(0.0806)^2

  = 735.588 MPa

x+2 because it just doubles hon so the pattern can actually tell you the expression.

so for example,

if x=2 then 2+2 is 4

if x=4 then 4+4 is 8

and so on and so forth.

i really hope that this makes sense and you out in a way hon: )

answer:

try looking it up that is always

explanation: