2016 optimisation a rear wing endplate in a rotating domain
PROJECT
1. I. Header
November 29, 2015
To: ProfessorJohn Harvey
University of California, Davis
One Shields Ave. Davis, CA 95616
From: Samuel Ray 999576030
Subject: Pavement Analysis and Design,
Bogota International Airport
I am submitting a report on the design of a newly reconstructed takeoff taxiway. The
taxiway will only carry the loads of the followingaircraft:
Boeing 777-300 ER(B777)
Airbus A350-900 (WV001) (A350)
2. II. Purpose:
The Bogota International Airport is considering expanding a terminal in order to
accommodate new flights by the above mentioned aircraft B777 and A350. The Mechanistic-
Empirical design method was used to create alternate pavement designs. Both asphalt concrete
and Portland cement concretewere used. The results were acquired by using both openpave and
EverFefor each new design. Openpave was used forthe design of the asphalt concrete (AC) and
EverFewas used for the design of the Portlandcement concrete(PCC).
III.Summary of Data:
Flexible:
It is expected that 3 B777 and 4 A350 will use the terminal every day. The main speed on the
taxiways is 25 km/hour. It will be used 365 days per year and will be evenly divided between
winter and summer, and between day and night. Expectedpavement temperatures at the site are
20ºC during summer days, 15ºC during summer nights and winter days, and 4ºC during winter
nights, all at 1/3 depth in the asphalt concrete.
The followingtwo tables were provided by Consultant John Harvey.
Table S1:ProposedPavementStructure:
Existing Layer Thickness Description
Asphalt Concrete 150 mm Dense graded AC, badly
cracked
Aggregate Base 300 mm Imported aggregate base,
compacted to 98 percent
modified Proctor
Subgrade Very thickfill. Original
design CBR values of 4-6
Table S2:Deflectionsof existing taxiways
Sensor location
(mm)
0 200 300 600 900 1200 1500
777 704 649 499 385 302 242
Table 1 specifies the existing layers (Asphalt Concrete, Aggregate base, and subgrade) and their
thicknesses. The shear strengths of the unbound layers are 100 kPafor the subgrade, and 420 kPa
or 350 kPafor the aggregate base. The design life for this pavement is 20 years. A Heavyweight
Deflectometerwas used to test the deflectionsin the existing runway.The load was 100kN on the
150 mm radius plate. The results are shown above in Table 2.
PCC Design:
3. For the alternative concretedesign we assumed a doweled (50 mm diameter) plain slab with
dimensions of 4m by 4m, with the PCC having a stiffness of 28 GPa and a MR of 5 MPa. The
nighttime temperature gradient is -5ºC forsummer and -3 ºC for winter, and daytime
temperature gradients of +4 ºC forboth seasons. The slab wouldalso have a constant equivalent
temperature of -2 ºC from moisture.
IV. Summary of Methods
Asphalt Concrete
Using the deflections in table 2, the stiffness of the subgrade and aggregate base were
obtained by back calculation in openpave. Given values of viscosity and penetration (Viscosity at
60 C is 18,000 poises and penetration at 25 C is 45 dmm) The Shell method was used to calculate
the Asphalt Stiffness. Intermediate steps of finding loading time, diameter at 1/3 depth, percent by
volume of aggregate and bitumen were needed to complete the shell method. Stiffness was found
for all 4 season conditions forboth air craft.
Given dimensions/positions of tires and their given loads and pressures were put into
openpave. Critical points were determined and analyzed to find max tensile strains and vertical
stresses. 4 open paves for each season condition were created. With the max tensile strains, for
each plane, for each season condition, fatigue life (Nf) was found foreach pass of each wheel.
Expected number of repetitions for each plane foreach season condition (n) was found with the
given information. Miner’s law was used to checkif the pavement failed due to fatigue. The
pavement does fail due to fatigue under the tires of both the B777 and the A350.
Simultaneously at the criticalpoints determined, vertical stress was obtained from each
openpave created. The thickness of the aggregate base (AB) was to be designed to protect the
subgrade (SG) from rutting. Iterations in open pave were used to find this thickness based on the
criteria of .4saturated shear strength >.5 verticalstress. The thickness foundthat satisfies this
criteria is 1225 mm. Same procedure was done for rutting of the AB but, it seems that the AB will
rut regardless of any changes made.
PCC
k-value was found using the funky chart and previously backcalculated subgrade stiffness
modulus. Overall temperature gradient was found for each season temperature. The dimensions of
the wheels were placed into EverFe along with the configurations of the wheel with its
corresponding loads.
Four iterations were made foreach plane and for each season temperature making a total of
12 iterations. The max tensile stresses and the deflections at the corners were recorded foreach
iteration. Flexural Strength (MR) was equal to 5 meaning the acceptable max tensile stress would
be 2.75.
The design thickness was found to be 605.5 mm with a deflection of 2.848 mm and a
maximum stress approximately 2.75 MPa. The criticalplane was A350 during summer/winter day
because this combination had the largest max stresses. Plots were then made to show the
relationship between thickness of PCC and deflection as wellas max tensile stress.
*All calculations, tables and figures are in the Attachments section with step by step methods.
4. V.Final Recommendation
Flexible
My final recommendation would be to design the AB to a thickness of 1225 mm to protect
the SB from rutting. Due to the large loads and the thin required AC thickness, the AB willrut
regardless of thickness. I’d recommend using the AB with a saturated shear strength of 420 kPa
because it will rut less than the alternative. The pavement also fatigues due to the large loads of the
tires and the slow velocity as well. Because minors law forthe max criticalpoint was around 6 the
pavement cannot be expected to last 20 years but around 6 times less than 20 years. The pavement
will fatigue after 3-4 years so preventive maintenance before these years could be an option. The
thin asphalt, heavy loads and slow speeds of the planes willcause the AB to rut and the AC to
fatigue. Changing one of those variables could greatly increase the life of this pavement.
Rigid
My final recommendation would be to design the PCC to a thickness of 605.5 mm to achieve
allowable tensile stresses. Rigid pavement design wouldbe better to handle the heavy loads of the
aircraft however,it is more expensive and a considerable large thickness would be required. If the
money is there I’d suggest using PCC instead of the thin asphalt.
VI. Closure
There are many issues withthe Asphalt pavement such as fatigue and rutting of the AB. The PC
howeverwhen designed correctly did not fail. The PCC however,costs more and more of it is
used. The PCC would be a better option because the AC will fail and a reliable expensive
pavement is better than a failing inexpensive one.
5. VII. Attachments
Part 1
Table 1
Sensor
location
(mm)
0 200 300 600 900 1200 1500
Deflectio 777 704 649 499 385 302 242
Using the sensor and deflectionmeasurements stiffness can be foundusing open pave.
Stiffness of asphalt concreteis given for this backcalculation as 10500 MPa. Thickness of
aggregate base (AB), radius and load were also given. When you enter in the load and radius
openpave calculates the pressure.
Table 2
Location X (mm) Y (mm)
Load
(kN)
Pressure
(kPa)
Radius
(mm)
1 0 0 100 1414.7106 150
Using the given information and iterations the stiffness of the AB and the subgrade (SG) can
be found.
Table 3
Layer
Thickness
(mm)
Elastic
Modulus
(MPa)
Poisson's
Ratio
Friction
1 150 10500 0.35 1
2 300 280 0.35 1
3 0 80 0.35 1
*Poisson’s ratio and Friction are assumed
Table 4
Point 1 2 3 4 5 6 7
X (mm) 0 200 300 600 900 1200 1500
Y (mm) 0 0 0 0 0 0 0
Z(mm) 0 0 0 0 0 0 0
Layer 1 1 1 1 1 1 1
6. Table 5
z
(mm) 0.7776996 0.7035079 0.6491723 0.4994464 0.3846662 0.301618 0.2421815
Therefore the elastic modulus of AB=280 MPa and SG=80 MPa
Step 2: Find stiffness of AC
We must find the Stiffness of the asphalt concrete(AC). First Plot
Viscosity at 60 C: 18,000 poises
Penetration at 25 C: 45 dmm
On the Bitumen Test Data Chart
Figure 1
from chart
Tr&b = 62°C
T1 = 25 °C
Penetration @ T1 = 45 dmm
PI calculation
20 − PI
10 + PI
=
log(800) − log(Penetration @T1)
Tr&b − T1
7. 20 − PI
10 + PI
=
log(800) − log(45)
62 − 25
PI = 1.16
Find area of tire cross-section on the surface
LoadB777 = 288120 N
PressureB777 = 1520550 Pa
A =
Load
Pressure
=
288
1521
= .189 m2
Find the Diamter of tire cross-section on the surface
A =
dsurface
2
π
4
. 189 = dsurface
2
∗
π
4
dsurface = .49 m
Find Diameter at 1/3 depth
1
3
thickess = .058 m
d1
3
depth
= dsurface + 2 ∗
1
3
thickness
d1
3
depth
= .49 + 2 ∗ .058 = .608 m
Find loading time (t)
velocity (v) = 6.9
m
s
t =
d1
3
depth
v
t =
. 608
6.9
= .09 sec
Table 6
max tire Mass (kg) P/g (kg/m^2) Area (m^2) diameter (m) thickness (m)
B777 29370 155000 0.18948387 0.491180575 0.175
A350 34112.5 169381.9735 0.20139392 0.506381976 0.175
9. Table 9
E1 (Pa)
9000000 day/summer
20000000 night/summer
20000000 day/winter
70000000 night/winter
Find percent volume of bitumen and aggregate
Figure 3
From Sample Calculations
Table 10
Vb % Vg %
10.7 84
With calculated values of stiffness modulus of bitumen, volume of bitumen and volume of aggregate
we can find stiffness modulus of bituminous mix.
Plotthese values on nomograph formixed stiffness.
12. Table 12
Stiffness of AC (Pa)
day/summer 1900000000
night/summer 3200000000
day/winter 3200000000
night/winter 8000000000
Step 3: Find position of wheels on open pave
Determine verticalStress
Set up openpave withloads forB777 and A350
Only one side of the plane is needed to be evaluated due to symmetry.
Loads and Pressures were calculated before.
Table 13
Load (N)
Pressure
(Pa)
288119.7 1520550 B777
334643.6 1661637.16 A350
Positions of wheels and criticalpoints with loads were calculated and put in openpave for the A350
and B777
Table 14: Wheel Loads
A350
Location X (mm) Y (mm) Load (kN)
Pressure
(kPa)
Radius
(mm)
1 4432.5 0 335 1662 253.29811
2 4432.5 2040 335 1662 253.29811
3 6167.5 0 335 1662 253.29811
4 6167.5 2040 335 1662 253.29811
13. Table 15: Wheel Loads
B777
Location X (mm) Y (mm) Load (kN)
Pressure
(kPa)
Radius
(mm)
1 4790 0 288 1521 245.50294
2 4790 1450 288 1521 245.50294
3 4790 2930 288 1521 245.50294
4 6190 0 288 1521 245.50294
5 6190 1450 288 1521 245.50294
6 6190 2930 288 1521 245.50294
Table 16: Critical Points
Table 17: Critical Points
Figure 5
Step 4: Find Thickness of AB to protect SG from rutting.
*we only have to checkforrutting for Summer Day because that is the time that is most susceptible
to rutting.
find desired maxvalue for σzz (vertical stress)
A350
Point 1 2 3 4 5 6 7 8 9 10 11 12
X (mm) 4790 5490 4432.5 5300 4432.5 5300 4432.5 5300 4432.5 5300 4432.5 5300
Y (mm) 0 0 0 0 0 0 1020 1020 0 0 1020 1020
Z(mm) 175 175 175 175 1400 1400 1400 1400 175 175 175 175
Layer 1 1 1 1 2 2 2 2 2 2 2 2
B777
Point 1 2 3 4 5 6 7 8 9 10 11 12
X (mm) 4790 5490 4432.5 5300 4790 5490 4790 5490 4790 5490 4790 5490
Y (mm) 1450 1450 1450 1450 1450 1450 725 725 1450 1450 725 725
Z(mm) 175 175 175 175 1400 1400 1400 1400 175 175 175 175
Layer 2 2 2 2 3 3 3 3 2 2 2 2
0
500
1000
1500
2000
2500
3000
3500
0 2000 4000 6000 8000
y(mm)
x (mm)
CoordinateAxis for B777 and A350
A350 load
B777 Load
Critical points fatigue
Critical Points Rutting
14. .4 ∗ shear strength > .5σzz
ssSG = 100 kPa
σzz < 80 kPa
Varying the depths wegot these values
*note through observation using openpave the B777 caused more verticalstress therefore this was
the plane used to determine if there will be rutting in the SG.
Table 18
Looking at AB 1200 mm and AB 1300mm we have just above and below 80 kPa respectively,when
looking at the max verticalstress.
Figure 6: AB thickness vs rutting in subgrade A350
SG SG SG SG AB AB AB AB
point 5 6 7 8 9 10 11 12
AB 300 mm szz (kPa) -224.96 -120.33 -106.76 -92.067 -733.281 -55.4613 -47.90628 -31.0746
AB 1200 mm szz (kPa) -77.337 -81.076 -71.932 -76.151 -799.741 -39.2644 -32.40267 -0.77226
AB 1300 mm szz (kPa) -72.208 -76.598 -67.948 -72.508 -799.998 -39.449 -32.64279 -0.94547
0
50
100
150
200
250
0 200 400 600 800 1000 1200
verticalStress(kPa)
Thickness of AB (mm)
A350 AB thickness to protect Subgrade
Stress vs Thickness
Design thickness
15. Figure 7: AB thickness vs rutting in subgrade B777
Through interpolation we get a design thickness of 1225 mm
Table 19
B) Determine if Subgrade ruts
Find desired σzz
same method as before
σzz < 336 kPa
By looking at point 9, on table 19, it shows a vertical stress of
800 kPa therefore, the AB will rut. Changing any thickness of the AB
will not protect the AB.
0
20
40
60
80
100
120
0 200 400 600 800 1000 1200 1400
(verticalstress)/2kPa
thickness of AB (mm)
B777 Actual Design Thickness
(vertical stress)/2 vs depth
Design thickness
SG SG SG SG AB AB AB AB
point 5 6 7 8 9 10 11 12
AB 1225 mm szz (kPa) -75.987 -79.928 -70.912 -75.237 -799.811 -39.3083 -32.46416 -0.80931
16. Figure 8: Rutting in AB vs Thickness of AB
C) Determine if the AC fails due to fatigue
Step 1: find the tensile strains due to passing of planes
Using open Pavethe followingmicro strains were calculated at the bottom of the
asphalt layer.
0
100
200
300
400
500
600
700
800
900
1000
0 200 400 600 800 1000 1200 1400
VerticalStress(kPa)
Thickness (mm)
Check for Rutting in AB1 & AB2
A-350
B777
Saturated Shear
Stress for AB1
Saturated Shear
Stress for AB2
17. Table 20
B777 wheel
A350
wheel
critcalpoint A B C D
A350 summer day (micro
strain) 358 124 797 125
A350 summer night (micro
strain) 327 90.4 659 95.9
A350 winter day (micro
strain) 327 90.4 659 95.9
A350 winter night (micro
strain) 244 102 421 83.9
B777 summer day (micro
strain) 677 168 269 137
B777 summer night (micro
strain) 568 76.9 479 125
B777 winter day (micro
strain) 568 76.9 479 125
B777 winter night (micro
strain) 370 88.2 364 116
Step 2: find allowed number of repetitions
for A350 summer day point A
ϵtension = 358 ∗ 10−6
Nfield = SF ∗ e−23.63−4.2913∗ln(ϵtension)
Nfield = 5 ∗ e−23.63−4.2913ln(358∗10−6)
= 167847 allowed repititions
Table 21
18. Step 3: find actual repetitions
nactual for each season for B777 = 365 ∗ 20 ∗ 3 ∗
3
2
2
= 16425
*n willonly change between the different planes
Table 22
n
(repetitions)
A350 14600
B777 16425
Step 3: use minors law tofind whether pavement fails due to fatigue
n
N
for A350 summer day at critical point A
n
N
=
14600
167847
= .087
*same process foreach minor law calculation
Table 23
Now sum up n/N at each criticalpoint
wheel B777 wheel A350
critcal point NA NB NC ND
A350 summer day (micro strain) 167846.8382 15881478.1 5412.06872 15343394.73
A350 summer night (micro strain) 247578.3282 61643290.2 12237.9428 47842427.57
A350 winter day (micro strain) 247578.3282 61643290.2 12237.9428 47842427.57
A350 winter night (micro strain) 869731.9717 36718546.4 83716.2255 84908602.8
B777 summer day (micro strain) 10901.4996 4314412.61 572262.698 10353386.51
B777 summer night (micro strain) 23155.6184 123400152 48113.5729 15343394.73
B777 winter day (micro strain) 23155.6184 123400152 48113.5729 15343394.73
B777 winter night (micro strain) 145702.6124 68517586.6 156292.101 21143789.56
wheel B777 wheel A350
critcal point minors A minors B minors C minors D
A350 summer day (micro strain) 0.086984063 0.00091931 2.69767454 0.00095155
A350 summer night (micro strain) 0.058971236 0.00023685 1.19301097 0.000305168
A350 winter day (micro strain) 0.058971236 0.00023685 1.19301097 0.000305168
A350 winter night (micro strain) 0.016786781 0.00039762 0.17439869 0.00017195
B777 summer day (micro strain) 1.506673449 0.00380701 0.02870185 0.001586437
B777 summer night (micro strain) 0.709331088 0.0001331 0.34137976 0.001070493
B777 winter day (micro strain) 0.709331088 0.0001331 0.34137976 0.001070493
B777 winter night (micro strain) 0.112729619 0.00023972 0.10509168 0.000776824
19. AC fails due to fatigue at A and C
Figure 9
Figure 10: Minors Law vs AB thickness
sum A sum B sum C sum D
3.259778561 0.006104 6.074648 0.006238
0
1
2
3
4
5
6
7
0 1 2 3 4 5
n/Nsum
Path
Minors Law for design thickness
Path A
Minors Law Failure line
Path B
Path C
Path D
0
2
4
6
8
10
12
0 500 1000 1500 2000 2500
Σn/N
Thickness (mm)
Minors Law vs AB thickness for A350 & B777
1225mm
300mm
2000mm
Max Strain before
Fatigue
20. As youcan see here the AC will rut no matter what even if youchange the thickness of the AB. It
levels off. AB cannot protect the AC.
Part 2
A) Find k-value
from funky chart
Figure 11: funky chart
@stiffness = 80 MPa
Bearing value = 42 psi
deflection = .2 in
k − value =
42
.2
= 210
psi
in
Using the funky chart and the back calculated stiffness modulus of SG, we can find the k-
value. K-value=210 psi/in
210
psi
in
= .0543
MPa
mm
= k − value
B) Find design thickness of asphalt
Step 1: Find wheel dimensions
Area B777 = .189m2
From B777 info website we found the width W of the wheel to be .53 m
L =
A
W
=.
189
.53
= .357 m
*same procedure forA350
Table 24:wheel dimentions
21. W (m) L (m)
B777 0.53 0.357516738
A350 0.53 0.379988533
Step 2: set up wheel configuration.
Figure 12:B777 configuration in EverFe
Figure13: A350 configurationin EverFe
Step 3:Find Changes of Temperature
22. Figure 14
Step 4: iterate to find design thickness
Do 4 iterations for each temperature gradient and for each plane.
Check the deflections and the max stress by using visualize for points
Here is a table of all the iterations of PCC thickness and corresponding
plane/season/thickness/deflection and Max tensile stress.
23. Table25:Thickness and respective stresses and deflections
Looking at chart we can see that the design thickness wouldbe forthe A350 summer day between
600 and 615
Step 5: find design thickness
Here is the criteria for Max stress
stress < .55 ∗ MR = .55 ∗ 5 MPa = 2.75 MPa
Through interpolation of A350 between 600 and 615 mmm.
Design thickness will be 605.5mm
Table 26 Design Thickness
thickness (mm) Max (Mpa) Displacement (mm)
B777 summerday 250 7.25 4.203
500 2.77 3.608
600 2.72 3.482
400 3.74 3.893
summernight 250 6.32 3.907
500 2.346 3.58
400 3.3469 3.665
450 2.787 3.618
winternight 250 6.468 4.055
500 2.4 3.651
400 3.425 3.753
450 2.84 3.713
A350 winternight 250 7.23 3.652
500 2.5454 3.057
400 3.63 3.231
450 3.01 3.134
Summernight 250 6.997 3.628
400 3.497 3.201
450 2.9 3.113
500 2.655 3.077
summerday 250 7.9565 3.868
500 3.1277 3.065
600 2.768 2.868
615 2.718 2.83
thickness (mm) Max (Mpa) Displacement (mm)
605.5 2.75 2.848
24. Figure 15: results for points
Figure 16: results for stresses
C) Plots of deflections and Max stress vs thickness for each season and each plane.
25. Figure 17
Figure 18
0
1
2
3
4
5
6
7
8
0 200 400 600 800
stress(MPa)
thickness (mm)
B777 Stress vs Thickness
B777 Max Stress Summer/Winter
Day
B777 Max Stress Summer Night
B777 Max Stress Winter Night
Max allowable stress
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 100 200 300 400 500 600 700
Deflections(mm)
thickness (mm)
B777 Deflections vs Thickness
B777 Defliction Summer/Winter
Day
B777 Deflection Summer Night
B777 Deflection Winter Night
26. Figure 19
Figure 20
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
0 100 200 300 400 500 600 700
Deflections(mm)
thickness (mm)
A350 Deflections vs Thickness
A350 Deflections Winter Night
A350 Deflections Summer Night
A350 Deflections Summer/Winter
Day
0
1
2
3
4
5
6
7
8
9
0 200 400 600 800 1000
MaxStress(MPa)
Thickess (mm)
A350 MaxStress vs Thickness
A350 Max Stress Winter Night
A350 Max Stress Summer Night
A350 Max Stresses Summer/Winter
Night
Max Allowable Stress
