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Rigid Pavement Stress Analysis
Dr. Antonis Michael
Frederick University
Notes Courtesy of Dr. Christos Drakos
University of Florida
Topic 8 Rigid Pavement Stress Analysis
Curling
Load
Friction
Cause of Stresses in Rigid Pavements
Where is the tension zone?
1. Curling Stresses
Topic 8 Rigid Pavement Stress Analysis
1.1 Curling Because of Temperature
Topic 8 Rigid Pavement Stress Analysis
1.3 Curling Because of Shrinkage
1.2 Curling Because of Moisture
Topic 8 Rigid Pavement Stress Analysis
X due to curling in X-direction:
X due to curling in Y-direction:
1.4 Curling Stress of Infinite Plate
2
T tYX
==
Assume linear t = coefficient of thermal expansion
T
T+T
Topic 8 Rigid Pavement Stress Analysis
1.5 Bending Stress of Finite Slab
LX
LYX
Y
)2(1
TEC
)2(1
TEC
2Y
2X
X
tt
+
=
)C(C)2(1
T YX2X
t +
=
Topic 8 Rigid Pavement Stress Analysis
Correction Factor Chart
Topic 8 Rigid Pavement Stress Analysis
Maximum Interior Stress @ Center of Slab
)C(C)2(1
TE
)C(C)2(1
TE
XY2Y
YX2X
t
t
+
=
+
=
Edge Stress @ Midspan
C2
TE t
=
1.5 Bending Stress of Finite Slab (cont)
may be x or y, depending on whether C is taken as Cx or Cy
Topic 8 Rigid Pavement Stress Analysis
1.6 Temperature Curling Example
12
25
8k=200 pcit=5x10
-6 /oF
t=20oFEc=4,000,000 psi=0.15
Calculate Stresses
i. Radius of Relative Stiffness:
1/4
2
3
)k(112
Eh
=
l
X
Y
Topic 8 Rigid Pavement Stress Analysis
)C(C)2(1
TE
)C(C)2(1
TE
XY2Y
YX2X
t
t
+
=
+
=
ii. Maximum Interior Stress @ Center of Slab
Topic 8 Rigid Pavement Stress Analysis
)C(C)2(1
TE YX2Xint
t +
=
)C(C)2(1
TE XY2Yint
t +
=
1.6 Temperature Curling Example (cont)
Topic 8 Rigid Pavement Stress Analysis
iii. Edge Stress @ Midspan
XX C2
TE t
=
1.6 Temperature Curling Example (cont)
Topic 8 Rigid Pavement Stress Analysis
1.7 Combined Stresses
Joints and steel relieve and take care of curling stresses (as long as the cracks are held together by reinforcement and are still able to transfer load they will not affect performance)
Curling stresses add to load stresses during the day and subtract to load stresses during the night
Fatigue principle is based on # of repetitions; curling effect limited compared to load repetitions
Curling stresses are high, but usually not considered in the thickness design for the following reasons:
Topic 8 Rigid Pavement Stress Analysis
2. Loading Stresses
Three ways of determining & : Closed form solutions (Westergaard single-wheel) Influence charts (Picket & Ray, 1951 multiple-wheel) Finite Element (FE) solutions
2.1 Closed-form solutions Westergaard theory
2.1.1 Assumptions
All forces on the surface of the plate are perpendicular to the surface
Slab has uniform cross-section and constant thickness Slab length Slab placed
Topic 8 Rigid Pavement Stress Analysis
2.1.2 Limitations
Only corner loading/edge loading or mid-slab deformation and stresses can be calculated
No discontinuities or voids beneath the slab Developed for single wheel loads
Topic 8 Rigid Pavement Stress Analysis
2.1.3 Corner Loading
=
=
ll
l
2a0.881.1
k
P
2a1
h
3P
2c
0.6
2c
Where:k = modulus of subgrade reaction
l = radius of relative stiffness
a = load contact radiusP = load
2.1.4 Interior Loading
+=
+
=
2
2i
2i
a0.673
2
a
2
11
8k
P
1.069b
4h
0.316P
lll
l
ln
log 0.675hh1.6ab
ab
22 +=
= when a1.724h
when a
Topic 8 Rigid Pavement Stress Analysis
2.1.5 Edge Loading
=
+
=
ll
l
l
a0.821
k
0.431P
0.034a
0.666a
4h
0.803P
2e
2elog
Topic 8 Rigid Pavement Stress Analysis
2.1.6 Dual Tires
Assume that:
Then, area of the equivalent circle:
0.5227qL d
P
( )1/2
ddd
d22
0.5227q
PS
q
P0.8521a
L0.6LS0.5227L2a
+=
+=
Topic 8 Rigid Pavement Stress Analysis
2.1.7 Dual Tire Example
14
P=10000 lbq=88.42 psik=100pciSd=14Ec=4,000,000 psih=10
Calculate stresses.
Topic 8 Rigid Pavement Stress Analysis
iii. Corner Stress:
=
0.6
2c
2a1
h
3P
l
iv. Interior Stress:
+
= 1.069b
4h
0.316P
2i
llog
2.1.7 Dual Tire Example (cont)
Topic 8 Rigid Pavement Stress Analysis
v. Edge Stress:
+
= 0.034a0.666a
4h
0.803P
2el
llog
2.1.7 Dual Tire Example (cont)
Topic 8 Rigid Pavement Stress Analysis
3. Friction Stresses
L
L/2
h
Friction between concrete slab and its foundations induces internal tensile stresses in the concrete. If the slab is reinforced, these stresses are eventually carried by the steel reinforcement.
What happens to PCC w/ T?
Where:
c=Unit weight of PCC
fa=Average friction between slab & foundation
Topic 8 Rigid Pavement Stress Analysis
Steel Stresses: Reinforcing steel Tie bars Dowels
Wire fabric or Do Increase
L/2
ht
f
Where:As = Area of required steel per unit widthfs = Allowable stress in steel
3.1 Reinforcement
Topic 8 Rigid Pavement Stress Analysis
3.1.1 Welded Wire Fabric
What does (6 x 12 W8 x W6) mean?
Transverse
Longitudinal
Orientation
Minimum wires W4 or D4 (because wires are subjected to bending and tension)
Minimum spacing 4in (allow for PCC placement and vibration) Maximum 12x24
Wire fabric should have end and side laps: Longitudinal: 30*Diam. but no less than 12 Transverse: 20*Diam. but no less than 6
Fabric should extend to about 2in but no more than 6in from the slab edges
Wire Reinforcement Institute Guidelines:
Topic 8 Rigid Pavement Stress Analysis
Topic 8 Rigid Pavement Stress Analysis
3.2 Tie Bars
s
'ca
sf
hLfA =
L = distance from the longitudinal joint to
L
L
L
Length of tie bars
= allowable bond stressd = bar diameter
Many Agencies use
Placed along the
Spacing of tie bars
Topic 8 Rigid Pavement Stress Analysis
4. Joint Opening
Where: = Joint openingt = Coefficient of thermal contraction = Drying shrinkage coefficientL = Slab lengthC = adjustment factor for subgrade friction
Stabilized = Granular =
