topic 8 - rigid pavement stress analysis
TRANSCRIPT
Rigid Pavement Stress Analysis
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 t
YXα
==
• 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
LY
X
Y
)2(1∆TEC
)2(1∆TEC
σ 2Y
2X
X ναν
να tt
−+
−=
)C(C)2(1
∆TΕσ YX2X νν
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∆TE
σ 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(112Eh
⎥⎦
⎤⎢⎣
⎡−⋅
=ν
l
σXσY
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 νν
α t +−
=
)C(C)2(1
∆TEσ XY2Yint νν
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 StressesThree 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
kP
δ
2a1
h3P
σ
2c
0.6
2c
Where:k = modulus of subgrade reactionl = radius of relative stiffnessa = load contact radiusP = load
2.1.4 Interior Loading
⎪⎩
⎪⎨⎧
⎪⎭
⎪⎬⎫
⎟⎠⎞
⎜⎝⎛⎥⎦
⎤⎢⎣
⎡−⎟
⎠⎞
⎜⎝⎛+=
⎥⎦
⎤⎢⎣
⎡+⎟
⎠⎞
⎜⎝⎛=
2
2i
2i
a0.673
2a
21
18kP
δ
1.069b
4h
0.316Pσ
lll
l
lnπ
log 0.675hh1.6ab
ab22 −+=
= when a≥1.724h
when a<1.724h
Topic 8 – Rigid Pavement Stress Analysis
2.1.5 Edge Loading
⎥⎦
⎤⎢⎣
⎡⎟⎠⎞
⎜⎝⎛−=
⎥⎦
⎤⎢⎣
⎡−⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=
ll
ll
a0.821
k0.431P
δ
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
0.5227qPS
qP0.8521
a
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
2c2a
1h3P
σl
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:
⎥⎦
⎤⎢⎣
⎡−⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛= 0.034
a0.666
a4
h0.803P
σ 2e lllog
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 FabricWhat does (6 x 12 – W8 x W6) mean?
Transverse
Long
itudi
nal
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
s fhLγf
A =
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 =