Post on 22-Jul-2018
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
• σX due to curling in X-direction:
• σX due to curling in Y-direction:
1.4 Curling Stress of Infinite Plate
2
∆Tεε tYX
α==
• Assume linear ∆Τ• αt = coefficient of thermal expansion
T
T+∆T
ε
ε
Topic 8 – Rigid Pavement Stress Analysis
1.5 Bending Stress of Finite Slab
LX
LYX
Y
)2(1
∆TEC
)2(1
∆TECσ
2Y
2X
Xν
αν
ν
α tt
−+
−=
)C(C)2(1
∆TΕσ YX2X ν
ν
αt +−
=
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
∆TEσ
)C(C)2(1
∆TEσ
XY2Y
YX2X
νν
α
νν
α
t
t
+−
=
+−
=
• Edge Stress @ Midspan
C2
∆TEσ tα=
1.5 Bending Stress of Finite Slab (cont)
σ may be σx or σy, depending on whether C is taken as Cx or Cy
Topic 8 – Rigid Pavement Stress Analysis
1.6 Temperature Curling Example
12’
25’
8”k=200 pciαt=5x10
-6 /oF
∆t=20oFEc=4,000,000 psiν=0.15
Calculate Stresses
i. Radius of Relative Stiffness:
1/4
2
3
)k(112
Eh
−⋅=
νl
σX
σY
Topic 8 – Rigid Pavement Stress Analysis
)C(C)2(1
∆TEσ
)C(C)2(1
∆TEσ
XY2Y
YX2X
νν
α
νν
α
t
t
+−
=
+−
=
ii. Maximum Interior Stress @ Center of Slab
Topic 8 – Rigid Pavement Stress Analysis
)C(C)2(1
∆TEσ YX2Xint ν
ν
α t +−
=
)C(C)2(1
∆TEσ XY2Yint ν
ν
α t +−
=
1.6 Temperature Curling Example (cont)
Topic 8 – Rigid Pavement Stress Analysis
iii. Edge Stress @ Midspan
XX C2
∆TEσ tα=
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 σ & δ:– 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
Pδ
2a1
h
3Pσ
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
Pδ
1.069b
4h
0.316Pσ
lll
l
lnπ
log 0.675hh1.6ab
ab
22 −+=
= when a≥1.724h
when a<1.724h
Topic 8 – Rigid Pavement Stress Analysis
2.1.5 Edge Loading
−=
−
+
=
ll
l
l
a0.821
k
0.431Pδ
0.034a
0.666a
4h
0.803Pσ
2e
2e log
Topic 8 – Rigid Pavement Stress Analysis
2.1.6 Dual Tires
Assume that:
Then, area of the equivalent circle:
0.5227qL dP≈
( )1/2
ddd
d22
0.5227q
PS
q
P0.8521a
L0.6LS0.5227L2a
+×=
−+×=
ππ
π
Topic 8 – Rigid Pavement Stress Analysis
2.1.7 Dual Tire Example
14”
P=10000 lbq=88.42 psik=100pciSd=14”Ec=4,000,000 psih=10”
Calculate stresses.
Topic 8 – Rigid Pavement Stress Analysis
iii. Corner Stress:
−=
0.6
2c
2a1
h
3Pσ
l
iv. Interior Stress:
+
= 1.069b
4h
0.316Pσ
2i
llog
2.1.7 Dual Tire Example (cont)
Topic 8 – Rigid Pavement Stress Analysis
v. Edge Stress:
−
+
= 0.034a
0.666a
4h
0.803Pσ
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:
• γc=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
hσt
σf
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
hLγfA =
L’ = distance from the longitudinal joint to
L’
L’
L’
Length of tie bars
µ = allowable bond stressd = bar diameter
Many Agencies use
• Placed along the
Spacing of tie bars
Topic 8 – Rigid Pavement Stress Analysis
4. Joint Opening
δ
Where:δ = Joint openingαt = Coefficient of thermal contractionε = Drying shrinkage coefficientL = Slab lengthC = adjustment factor for subgrade friction
• Stabilized = • Granular =