FWLM Midterm
Calculations and plots were created using Excel and Maple software. Integration done
using wolframalpha.com
1a)
question 1 was solved in maple file “1.mw”
The displacements of point loads are additive meaning they can be superimposed. A
single normal point load distribution of displacement is given in Johnson, (Eq. 3.18):
ux 

P  xz
x
 3  1  2 

4G  
   z  
uy 

P  yz
y
 3  1  2 

4G  
   z  
uz 
P  z 2 21   
 

4G   3
 
Where   x 2  y 2  z 2 and z is positive downward.
Superimposing was done by replacing x with (x-xo) including in the equalting for  . xo
is the x location of the point load. y and z were unchanged, since location of the loads
only varied in the x direction. The tangential load equations were also modified by
replacing x.
The tangential point load displacement distribution is given in Johnson, (Eq. 3.75):
 1

Q  1 x2
x2
u x  x   3  1  2 

2 
4G   
   z    z  
uy 
Qx  xy
xy 
 3  1  2 

4G  
   z 2 
uz 
Qx  xz

x
 1  2 

3
4G  
   z 
Where   x 2  y 2  z 2 and z is positive downward.
Strain was solved from the displacements using the euler equation for small
displacement:
1  u j u i 
 ij  

 where i and j are replaced with x, y, and z for the 9 components of
2  xi x j 
the strain tensor.
Stress was calculated from strain using Hooke’s law,   2G  tr( ) I where Lame’s
E
E
, the shear modulus G=
(incorrect in HW 2 #2
1   1  2 
21   
answer), tr() is the trace function which is the sum of the main diagonal in the strain
tensor, and I is the identity matrix which is 0 except for ones in the main diagonal.
first parameter  
Entering these equations into Maple resulted in the following plots with titles below the
plots. Also, if additional plots are required, the Maple file can be modified as necessary.
 x along line in x direction, 1 micrometer below the surface
 y along line in y direction, 1 micrometer below the surface
 z along line in x direction, 1 micrometer below the surface
 z along line in z direction, 1 to 10 micrometers below the surface
 xy 2d contour in the xy plane, 1 micrometer below the surface
 yz 2d contour in the yz plane, x=3 micrometer to avoid the point loads
 xz 2d contour in the xz plane, y=1 micrometer
 x along line in x direction, 1 micrometer below the surface
 y along line in y direction, 1 micrometer below the surface
 z along line in x direction, 1 micrometer below the surface
 z along line in z direction, 1 to 10 micrometers below the surface
 xy 2d contour in the xy plane, 1 micrometer below the surface
 yz 2d contour in the yz plane, x=3 micrometer to avoid the point loads
 xz 2d contour in the xz plane, y=1 micrometer
ux along line in x direction, 1 micrometer below the surface
uy along line in y direction, 1 micrometer below the surface
uz along line in x direction, 1 micrometer below the surface
uz along line in z direction, 1 to 10 micrometers below the surface
1b)
Maple computed values at (x,y,z)=(10-6, 10-6, 10-6) :
u x   2.73 
  

u  u y   0.468 in µm
u   3.46 
 z 

 x

   xy

 xz
 xy
y
 yz
 xz   0.165  0.464  0.398
 

 yz    0.464  0.288  0.947 
 z   0.398  0.947  0.488
 x

   xy

 xz
 xy  xz    45.1  71.4  61.3 
 

 y  yz    71.4  114.8  145.7 in GPa
 yz  z   61.3  145.7  145.7
2.
Hertzian contact between sphere and plate. The radius of a plate is >> than for the
sphere, therefore the effective radius is equal to that of the sphere, R=R1=0.1 m. The
4
4
load on the sphere is due to gravity P= R 3 g   0.13 * 7800 * 9.8066 =320.4 N.
3
3
1/ 3
1/ 3
 3PR 
 3 * 320.4 * 0.1 
a) The Hertz contact radius is given as a   *   

11 
 4E 
 4 * 1.099 *10 
0.6025*10^-3m=0.6025 mm. Where, since material properties are equal (E1=E2=2*1011

 2 1  2
and v1=v2=0.3), E  
E

*
 


1

 2 1  0.3 2
 
11
 2 *10
 
1
11
 =1.099*10 .

The equations for stress were taken from the contact mechanics lecture notes:
Stress as a function of r along the surface:



 p F    1   2 1 / 2
 0
r  

1  2 a 2
p0

3r 2




,r  a
 p  F    2 1   2
 0
  

1  2 a 2
 p0

3r 2


,r  a

1/ 2

,r  a
,r  a

1/ 2

 p 0 1   2
,r  a

0
,r  a

z  
Where
F   

1  2
1 1  2
2
3


3/ 2
 ,   ar , and p =3/2*p
0
mean=422
MPa
Stress as a function of depth and r=0:
2 1

 z
1   z   


1  a 
 r     p0  1   1  tan    1   



a
 z  2   a   




1
  z 2 
 z   p 0 1    
 a 


The plots of stress are in dimensionless coordinates with distance divided by the
P
320.4
 281 MPa..
contact radius a and stress divided by pmean= 2 
 0.0006025
a
Though not asked for, strain fields are provided in the Excel workbook solved by
Hooke’s law.
Stress along surface
0.4
0.2
0
-3
-2
-1
-0.2 0
1
2
3
-0.4
Sr/pm
-0.6
St/pm
-0.8
Sz/pm
-1
-1.2
-1.4
-1.6
stress downward
(displayed to the right)
1
0.5
0
0
0.5
1
1.5
2
2.5
3
3.5
z/a
-0.5
-1
Sr/pm, St/pm
Sz/pm
-1.5
Tau/pm
-2
The plot of the z height by the contact radius was from the following equation:



1   2 p 0

2a 2  r 2
,r  a

E
4
a

uz  
2
1   2 p 0  2
 a  
2
1  a 

2a  r sin    ra 1   
E 2a 

r
 r  








,r  a
uz
-3
-2
0
-1
0
-0.000001
r/a
1
2
3
-0.000002
uz
-0.000003
-0.000004
2b) Maximum stress is located slightly below the loading surface, and at the center of
contact. Since  r    in the vertical direction, Von Mises is reduced to
 vm   r    z . At z/a=0.481, the ratio of Von Mises stress over yield stress = 1.045.
The steel will yield according to the Von Mises Criterion.
3.
Gravity effects were assumed negligible. Derivision was done in Wolfram alpha and
checked using maple file “3.mw”
Derivation:
  3 p 
h
(5-43 in Shaw) the pressure
h
  6 U
x  x 
x
p
distribution can be solved integrating by x, knowing h=hi-a x , h 3
 6 U h  c1  or
x
Using Reynolds’ differential equation,

p
1
 6U 
 h a x
x
 i



 . Integrating again gives the pressure distribution
3 
hi  a x 
c1
 
2

 2h

c
2 ln h  c1 x
p  6 U  2 i 

 21  c 2 
2
2
a
ah
a h
a h

using boundary conditions, p(0)=0,
 2 2 ln hi 
c1 
1
   2 2hi  2 ln hi hi  c1 
c2   2 

2
2
a
a hi  
a hi
a



2hi
2 ln hi  a x
c1 hi
p  6U  2


2
a h a x
a
a 2 hi  a x
i





2


2c1 x
a hi  a x

2


1





2
h

2
ln
h
h

c
i
i
i
1
2

a hi

using boundary conditions, p(L)=0,
0

2hi

a hi  a L
2
c1 hi
a hi  a L
2

2






2 ln hi  a L
c1hi

2
a
a 2 hi  a L
2c1 L
a hi  a L


2





2c1 L
a hi  a L


2


1
2hi  2 ln hi hi  c1 
a hi
2
2hi
2 ln hi  a L
c1
2
 2

 2 1  ln hi 
2
2
a hi a hi  a L
a
a




2hi
2 ln hi  a L
2
c1   2

 2 1  ln hi 
2
a h a L

a
a
i




2


hi

 2
 a hi  a L


2


2 L
a hi  a L

2

1 
a 2 hi 

Knowing hi=20 μm, a=1e-4 m0.5, μ=0.1Pa.s, U=1 m/s and L=0.01 m gives c1= -1.23e-5
and c2=1.90e9
For specific values of p(x), see the excel workbook, sheet “3”.
h
h  y  U
1 p
The volumetric flow rate per unit depth is q=  udy where u=

y  hy 
0
2 x
h
Uh
1 p 3
Uh

h . And taking h at p’=0 gives q 
2
2 12 x
dp/dx is 0 at x=0.005969 m. h(0.005969)=12.3 μm, U=1 m/s,
volumetric flow rate per unit depth q=6.14*10-6 m3/m.s and is reasonable considering
that the gap height is in the scale of micrometers.
(5-35 in Shaw), integrating gives q=
The pressure distribution is:
p
2500000
2000000
1500000
p
1000000
500000
0
0
0.002
0.004
0.006
0.008
0.01
0.012
P max is 2.356 MPa at x=0.005969 m.
L
The force per unit depth is F=  pdx (modified 5-50 in Shaw) is P average (1.53 MPa)
0
times length (0.01 m). = 15.3 kN/m of depth.
4.
A journal bearing with rj=0.02500 m, rb=0.02505 m, e=10-5 m, N=600 RPM
Fluid properties μ =0.1 Pa.s
Pressure distribution (5-86 in Shaw) is p  p0 
6Ur  n2  n cos  sin  


c 2  2  n 2 1  n cos  2 


Note: the lecture 5 notes equation have an extra factor of 6 added which is a typo.
In the above equation, Po is take as 0, μ=0.1, U=Ω*r, Ω =500*2π/60=52.4 rad/s,
r=rj=0.025, c=rb-rj=5*10-5, n=e/c=0.2.
p
2000000
1500000
1000000
500000
0
-500000
p
0
1
2
3
4
5
6
7
-1000000
-1500000
-2000000
The max pressure is 1.61 MPa at θ=1.869 rad
An upper bound for resulting force from 0 to π: Pavg*π*rb=1.007e6*3.14*0.0250=79.1
kN/m depth. Where Pavg was calculated using difference in angles, and below by
integration. It is an upper bound because the direction and magnitude of the force
changes with θ.
Alternatively, integrating the pressure equation from 0 to π for the average pressure

1 
6Ur 
1
 ln n cos   1 1.008 MPa. Multiplying by
results in  pd  2

2
 0
c 2  n  n cos   1



length gives 1.008*106*π*0.0250=79.1 kN/m
Note: Taking the integral without dividing by π gives units of kN.rad/m 2. Since rj has units of
m/rad, multiplying the integral by rj and not dividing by π results in the correct value.
5.
See excel workbook, sheets “5” and “5b” for details.
a) Histogram:
Histogram
35
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
0
Frequency
30
25
20
15
10
5
Normal Distribution
2.18
2.305
2.43
2.555
2.68
2.805
2.93
3.055
3.18
3.305
3.43
0
Frequency
Bin
Normal distribution was used to fit the Data using Z mean=2.78 and σs=0.254
Bin count was chosen as
n ~ 10
b)
The mean asperity height Zm=
1
N
N
z
i 1
i
=2.78 μm.
The centerline average roughness Ra =
1 N
 zi  z m =0.200.
N i 1
The standard deviation of the roughness σ=
The root mean square roughness Rq=
1 N
zi  z m 2 =0.254.

N i 1
1 N
zi  z m 2 =0.254.

N i 1
Note that Rq is the same as the standard deviation. In class and class notes, It was incorrectly
presented as Rq=
1 N
zi 2 =2.79 which is refuted in a number of references:

N i 1
http://membranes.edu.au/wiki/index.php/Surface_Roughness,
http://en.wikipedia.org/wiki/Surface_roughness,
http://www.rubert.co.uk/Ra.htm,
https://engineering.purdue.edu/ME556/Documents/Surface%20Roughness.pdf, etc.
The confusion is the definition of zi, which if z=0 is at the mean, then
1 N
zi 2 is appropriate,

N i 1
but this definition is non sequitur since z=0 is 2.78 μm from the mean. Therefore the use of the
equation is not correct for this problem.
Skewness Sk =
Kurtosis =
1
 4N
1
 3N
N
 z
i 1
N
 z
i 1
 z m  =0.545
3
i
 z m  =2.78
4
i
Note: the kurtosis of the above formula for normal distribution is 3. In Excel, the KURT function
subtracts 3 to make the normal distribution kurtosis =0 and would result in K=-0.22.
c)
The Autocorrelation Function was plotted redefining the time lag as the distance lagged
between data points. This resulted in an extremely jagged plot. Better applications for
autocorrelation are signal processing and time based surface finishing.
Note that the lecture on surface mechanics left out the length in the denominator and
correcting for a mean not equal to 0. See excel workbook, sheet “5” column “L” for
example. The excel function CORREL has
ACF=
N lag
1
 zxi   z m zxi  lag   z m  where lag is the difference in
 2 ( N  lag ) i 1
horizontal distance in μm, and N is the total horizontal distance in μm.
The resulting values are:
lag
1
2
3
4
5
6
7
8
9
10
ACF
lag
0.019583
-0.11439
-0.25049
0.034212
0.047041
-0.05429
0.07215
-0.03686
0.071358
-0.01633
11
12
13
14
15
16
17
18
19
20
ACF
lag
0.054085
-0.0121
-0.06849
-0.10818
-0.02254
0.057953
-0.08927
0.210441
0.012668
-0.01389
21
22
23
24
25
26
27
28
29
30
ACF
lag
-0.07162
-0.01335
-0.04043
-0.02284
-0.00432
-0.09917
0.085879
-0.01829
0.166336
-0.04402
31
32
33
34
35
36
37
38
39
40
ACF
lag
-0.02559
-0.43282
0.011081
0.251892
0.157105
-0.08363
-0.23392
0.014785
0.103645
0.205372
41
42
43
44
45
46
47
48
49
50
ACF
-0.15365
-0.02532
0.155837
-0.07305
0.069925
-0.04271
0.10433
0.000153
0.21235
0.106785
ACF
0.3
0.2
0.1
0
-0.1
0
10
20
30
40
50
60
ACF
-0.2
-0.3
-0.4
-0.5
The graph shows random correlation. An ACF of +1 means the heights match
preceding heights or that the values have good “correlation”, and an ACF of -1 would
mean the height are opposite preceding heights or that the values have
“anti-correlation”.