E. M.. PART XV
CHAPTER 3
October 1954
pavement thickness requirements.) The lower portion of the base (at least 4 inches) should be designed as a filter.
Surface freezing index = (air freezing index) x (correction factor)
= 5500 x (0.7)
= 3850 degree days (F)
By figure 9, for 24-in. base, depth of freeze below base = 58 in.
By figure 9, for 48-in. base, depth of freeze below base = 45 in.
Interpolating for 32-in. base depth of freeze below
32 - 24
base = 58-
(58-45) = 54 in.
48 - 24
Depth of thaw:
Surface thawing index = 4800 degree days (F)
By figure 8, for 24-in. base, depth of thaw below base = 52 in.
By figure 8, for 48-in. base, depth of thaw below base = 44 in.
32 - 24
Interpolating, for 32-in. base 52 =
(52-44) = 49 in.
48 - 24
This would indicate that the permafrost probably would not degrade, since the depth of freeze is, by computation, 5
inches greater than the depth of thaw.
The soils should be investigated to a depth of at least 4 feet beneath the proposed subgrade elevation for nonuniformity
and isolated pockets of high frost-susceptibility.
(2) Rigid payment. Although the use of rigid pavements is not recommended over an F4 soil, assume for the
purpose of this example that special conditions dictate its use.
(a) Design to restrict annual thaw to pavement and base depth. It would be impractical to design a base
depth to keep the subgrade from thawing, as was shown for the flexible pavement.
(b) Design based on reduced strength of subgrade. A base thickness equal to one-half the depth of the
subgrade that will be subject to freezing and thawing is required. In no case should the base thickness be less than 24
inches nor greater than 48 inches.
Surface freezing index = 3850 degree days (F)
By figure 9, for 24-in. base, depth of freeze below base = 58 in.
By figure 9, for 48-in. base, depth of freeze below base = 45 in.
Depth of thaw:
Surface thawing index = 4800 degree days (F)
By figure 8, for 24-in. base, depth of thaw below base = 52 in.
By figure 8, for 48-in. base, depth of thaw below base = 44 in.
From the above, with a depth of thaw for the minimum base thickness (24 inches) less than the depth of freeze for that
thickness, it is evident that only 52 inches of subgrade is subjected to annual freeze and thaw. Therefore, a base
thickness one-half the depth of thaw or 26 inches would be used. At least 4 inches of the lower portion of this base
should be designed as a filter.
The slab thickness may be determined by use of figure 10 and the pavement design curves in part XII, chapter 3. By
figure 10, a subgrade modulus of 150 lbs. per sq. in. per in. is obtained. By the curve of part XII, chapter 3 a slab
thickness of 11 inches is obtained.
b. Example 2.
Design an airfield taxiway for both flexible and rigid pavements, to withstand a 100,000 pound dual wheel load,
contact area each wheel 267 sq. in., for the following conditions:
Mean air thawing index, 300 degree days (F)
Mean air freezing index, 9000 degree days (F)
Subgrade = silty sand, 10 percent by weight of grains finer than 0.02 mm., moisture content=25 percent group
F2
Subgrade modulus, k (test made on top of gravel base) =300 lbs./sq. in./in.
Base CBR=80 percent
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