# MTO OVERLAY DESIGN - spin.mohawkc.on.ca

```MTO OVERLAY DESIGN
The following presentation contains references to Figures 6.02 and
6.03, Tables 6.01 and 6.05, “Overlay Design Model” and “Overlay
Example Problem”, all of which are posted under subsection 2.6 of the
course notes on the instructor’s website. Viewer discretion is advised
as some scenes contain material of a graphic nature.
The basic problem is to test the pavement’s
strength and compare it to the strength
required to carry an estimated number of
Instead of coring surveys and guesstimates
of strength, its more economical to measure
deflections under standard loads as with…
 THE BENKELMAN BEAM
or
 THE DYNAFLECT
Deflection Tests
The Dynaflect is more efficient for mass
inventory (many tests) but the original
strengthening technology is based on
Benkelman Beam Rebounds (BBR’s)
 The Dynflect Sensor 1, (DYN) readings must
first be converted into BBR’s
 in Ontario this conversion is made using the
following equation:
BB  0.0024 0.019DYN  0.0095DYN
2
Statistical Criteria
1 i n
m ean X   X i
n i 1
X  2sx
0+210
0+180
Chainage
0+150
0+060
X  sx
0+030
0+000
X
0+120
X  sx
X  2sx
0+090
Deflection
Plotting deflections along the road alignment
gives a deflection profile:
i n
Standard Deviation = s x 
X
i 1
2
i
 nX
n 1
2
Design Deflection, DD
 Assuming the variation in deflections is
random and therefore normally distributed,
using the mean plus two standard deviations
will ensure that the maximum deflection is
taken into consideration 95% of the time
 Engineers always like to consider the worst
case scenario in design
 The pavement is at its weakest in the early
spring, when the frost has just come out of
the ground...
TABLE 6.05
Design Deflection,
DD
(Continued)
SEASONAL VARIATION
OF STRENGTH
RATIO
(After: Chong and Stott, Dept. of T ransportation and Communication, Ontario, Report No. IR42, October, 1971)
of
Type of
Spring/Fall
 The testsType
are
usually
taken
some
time
later,
Soil
Construction
Ratio
Remarks
quite often
inClay
theStandard
autumn
when the pavement
Heavy
gravel or crushed
When estimating the maximum
Silt
rock base with maximum 3.5"
spring deflection for heavy clay
structure has
stabilized
asphalt concrete surface
2.0
of high plasticity in standard
of 2.0 reflect
to 2.5 should be used
Light to
For all standard construction
the pavement’s
strength
in
the
early
spring
1.8
Medium Clay
including heavy clay and silt
construction, a spring/fall ratio
To Summarize: Design Deflection isgenerally occurs during the
spring-thaw period (April).
Rockthe mean plus two1.5standard
calculated as
Deep strength construction
Peak deflection generally
All
Types
1.4
deviations of thewithBenkelman
Beam Rebounds
thin granular base layer
occurs in June or July rather
Full depth construction - more
than in April. Peak-to-peak
on a pavement
to spring
by
than 7" of asphalt concrete
curvature, unlike standard
1.4 conditions
directly on subsoil
factoring by
a spring/fall
Soil cement-stabilized baseratio
All Types
1.4
Sand and
Gravel
1.6
with asphalt concrete surface
above, the peak deflection
Maximum
DD
= 0.067”
Tolerable
Deflection, MTD
DTN
DTN==48.9
125
If
the
MTD
was
how
much
traffic
Say,
The
Studies
for
MTD
example,
in
for
Ontario
a0.067”
DTN
that (and
of
itthen
is
125
elsewhere)
desired
is 0.05287”.
to know
have
could
it carry?
whether
correlated
or the
not design
a pavement
deflection
with aof
DD
a pavement
of
0.067”
withwill
itsbe
ability
able to carry another
10 years
as
Since
DD
>
MTD,
this
pavement
can’t
handle
It
could
carry
a
DTN
of
48.9.
of traffic characterized
as reflected by
bya the
DTN
DTN.
of 125.
this much traffic.
For the portion of this relationship for DTN >
10, the following equation can be used instead
of the graph:
0.1787907
MTD 
DTN
0.2523450
 0.10"
Required Strengthening
 Through extensive research, the Asphalt
Institute and MTO have developed design
curves relating Design Deflection to
Strengthening Requirements
 The
EachStrengthening
curve is for a different
is in termsMTD
of inches
value of
which(new
GBE
range
Granular
from 0.02”
A) up to 0.10”
Interpolation
 What happens when the MTD is between the
curves?
Example: DD = 0.045” & MTD = 0.033”
Proportion 
0.033"0.030"  0.3
(0.040"0.030")
T0.033"  6.913"(6.913"2.026")  (0.3)  5.447"
T0.033” = 5.447”
T0.03” = 6.913”
T0.04” = 2.026”
DD = 0.045”
Overlay Thickness Equations
Each curve on these graphs
has an equation with two
coefficients, a and b in the
a( DD  MTD)
form…
T
WARNING
Only the MTD
values
tabulated can
be used in this
equation!!!!!!
( DD  MTD)  b
MTD
a
b
MTD
a
b
0.020
33.880888
0.0250955
0.070
14.948930
0.0892606
0.030
26.179140
0.0418036
0.075
11.960133
0.0723573
0.040
19.330437
0.0427002
0.080
11.927564
0.0832181
0.050
14.308789
0.0414468
0.090
9.600747
0.0728982
0.060
14.342670
0.0740099
0.100
12.565574
0.1565992
OVERLAY EXAMPLE PROBLEM
The following Dynaflect Sensor 1 readings were recorded along a 4-lane
section of highway in October. If the present AADT is 16000 vpd with
20% heavy vehicles and an annual growth rate of 2.0%/year, design an
asphalt concrete overlay for a design period of 12 years if the subgrade
in the area is a light clay.
Then
Finally,
First calculate
Next
step
calculate
multiply
is tothe
mean
convert
by spring/fall
mean
andDynaflect
plus
standard
two
ratio
standard
Sensor
to find1
TABLE 6.05
STA
Sensor
1
BBR
deviations:
Design
deviation:
Deflection:
to Benkelman Beam rebounds:
SEASONAL VARIATION OF STRENGTH RATIO
0+030
0.013964
0.65
(After: Chong and Stott, Dept. of Transportation and Communication, Ontario, Report No. IR42, October, 1971)
Type of
0.60 Spring/Fall0.012420
Soil
Construction
Ratio
Remarks
Heavy
Clay
Standard
gravel
or
crushed
When
estimating
the maximum
0.013032
0+090
0.62
Silt
rock base with maximum 3.5"
spring deflection for heavy clay
0+120asphalt concrete surface
0.57 2.0 0.011517
of high plasticity in standard
Type of
0+060
0.013341
of 2.0 to 2.5 should be used
0+150
0.63
Light to
0+180
Medium Clay
Sand and
0+210
Gravel
0.61
1.8
0.69
1.6
above, the peak deflection
0.015233
0.82
1.5
0.019568
All
Types with thin granular base
layer
0+270
0.75
1.4
0.017194
occurs in June or July rather
than in April. Peak-to-peak
curvature, unlike standard
Rock
0+240
Deep strength construction
Full depth construction - more
than 7" of asphalt concrete
directly on subsoil
Soil cement-stabilized base
with asphalt concrete surface
construction, a spring/fall ratio
0.012725
including heavy clay and silt
For all standard construction
generally occurs during the
spring-thaw period (April).
Peak deflection generally
All
Types
0+300
0.68
1.4
0.014913
All
Types
0+330
0.70
1.4
0.015555
in  0.0024 0.019DYN  0.0095DYN 2
1 BB
1
2
0.159462
XBB
  
X

0.0144965
i  0.019


0.0024
0.65
 0.00950.
65
 0.013964
60
60
012420
n i1
11
in
sx 
X
i1
i
2
 nX
n 1
2

0.00236657- 110.0144965
11 - 1
2
 0.0023439
  0.019184
x  2s  0.0144965
 20.0023439
Therefore,
DD
= 0.034531”
x
DD  0.019184  1.8  0.034531
For a light clay, S/F = 1.8 according to
Table 6.05.
OVERLAY EXAMPLE (Continued)
TABLE 6.01
MTO LANE DISTRIBUTION FACTORS
TFi = 0.90 TFf = 0.96
Truck Factor, TF
Now to find the DTN from the traffic data given:
12
20,292
16,000
Highway Type
2.0


/lane
5073
  1  4000
 16,000

20291.8687

2 lanes
44 100 
Since AADTi =16,000, LDFi = 0.75
Since AADTf =20,292, LDFf = 0.75
Ti = 0.20; Tf = 0.20
4 lanes
(After: "Pavement Design and Rehabilitation Manual", 1990)
Lane Distribution
Factor, LDF
All
1.00
< 5000
0.85
5000 - 15000
0.80
15000 - 25000
0.75
> 25000
0.70
40% H
1.5
30% H
ESALi/day  16,000 0.5 0.20 0.75 0.90  1080
< 15000
20% H
10% H
0.60
ESALf/day  20,291.868
7  0.5 0.20 0.7515000
 0.96
 1461.0145
0.96
- 25000
0.55
1.0
0.90
6 lanes


1080  1461.0145
>
40000
0.45
DTN 
 1270.5073
25000 - 40000
Then find the DTN:
and finally the MTD…
0.50
2
0.5
0.1787907
MTD 
 0.02945"
0.2523450
1270.5073
0.0
2000
3000
4000
5000
6000
7000
Traffic/Lane (= AADT/4 for 4-Lane Highways)
FIGURE 6.03b - M TO TRUCK FACTORS: 4-LANE HIGHWAYS
8000
OVERLAY EXAMPLE (Continued)
Since MTD is between the 0.02” and 0.03” curves,
evaluate T0.02” and T0.03” and interpolate
For T0.02”:
For T0.03”:
a = 33.880888
b = 0.0250955
a = 26.179140
b = 0.0418036
T0.02" 
33.880888(0.034531"0.02")
 12.424"
(0.034531"0.02")  0.0250955
T0.03" 
26.179140(0.034531"0.03")
 2.560"
(0.034531"0.03")  0.0418036
The proportion that 0.02945”
0.02945"0.020"  0.945
divides the interval from 0.02” Proportion 
(0.030"0.020")
to 0.03”:
The interpolated
thickness of extra
GBE required:
MTD
a
b
MTD
a
b
T0.02945"  12.424"
(12.424"
2.560"
)  (0.945)
33.880888
0.0250955
14.948930  3.103"
0.0892606
0.020
0.070
And finally,the overlay
thickness in millimetres
of new AC:
0.030
26.179140
0.0418036
0.075
11.960133
0.0723573
3.103"
 25.4 0.080
mm/" 11.927564 0.0832181
0.0427002
0.040
T(mm
AC)19.330437

 39.4  40mm
2.0
14.308789
0.0414468
9.600747
0.0728982
0.050
0.090
0.060
14.342670
0.0740099
0.100
12.565574
0.1565992
```