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
==
)2(1∆TE
σ 2X ναt−⋅⋅
=
)2(1∆TEν
σ 2X ναt−⋅⋅⋅
=
Bending occurs in both X & Y directions, solution must be superimposed
Sum of both directions:
)2(1∆TEσX ν
α t−⋅⋅
=
• 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 +−
=
CX & CY = correction factors for finite slab
Correction factors depend on LX/l & LY/l
l = radius of relative stiffnessh = slab thicknessk = modulus of subgrade reaction
1/4
2
3
)k(112Eh
⎥⎦
⎤⎢⎣
⎡−⋅
=ν
l
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 in 30.57)(200)(0.15)(112
)810(41/4
2
36
=⎥⎦
⎤⎢⎣
⎡−⋅×
=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
Have to calculate CX & CY first!
4.7130.57144L
9.8130.57300L
Y
X
==
==
l
l
CX=1.07
CY=0.63
LX/lLY/l
Topic 8 – Rigid Pavement Stress Analysis
)C(C)2(1
∆TEσ YX2Xint νν
α t +−
=
238.26psi)0.15(0.63)(1.07)0.152(1
)2010)(510(4σ 2
66
Xint =+−
××=
−
)C(C)2(1
∆TEσ XY2Yint νν
α t +−
=
161.74psi)0.15(1.07)(0.63)0.152(1
)2010)(510(4σ 2
66
Yint =+−
××=
−
σXint > σYint Was that expected? Why?
1.6 Temperature Curling Example (cont)
Topic 8 – Rigid Pavement Stress Analysis
Concrete Tensile Strength≈10% fc’ ≈ 400 psi
iii. Edge Stress @ Midspan Since the critical stress is associated with the longest dimension, omit σY
XX C2∆TEσ tα=
214psi1.072
)2010)(510(4σ
66
X =××
=−
So, a wheel load stress of only 186 psi (400-214) will exceed the tensile stress when thermal and load stresses are combined
25’ Slabs are almost always reinforced
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• Shear deformations are small• Slab length is infinite• Slab placed on subgrade consist of discrete springs
Topic 8 – Rigid Pavement Stress Analysis
2.1.2 Limitations
Interior EdgeCorner
Interior & Edge: Tension @ the bottom
Corner: Tension @ the top
• 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
Where is the tension zone?
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:
Pd = single tire loadPd = P/2
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.
i. Effective Radius:
7.85ina
42)0.5227(88.(5000)(14)
(88.42)0)0.8521(500a
1/2
=
⎟⎠⎞
⎜⎝⎛+=
ππ
ii. Radius of Relative Stiffness:
What do we need first?
1/4
2
3
)k(112Eh
⎥⎦
⎤⎢⎣
⎡−⋅
=ν
l 42.97in)(100)(0.15)(112
)1010(41/4
2
36
=⎥⎦
⎤⎢⎣
⎡−⋅×
=l
Topic 8 – Rigid Pavement Stress Analysis
iii. Corner Stress:
⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−=
0.6
2c2a
1h3P
σl
166.82psi42.97
27.851
103(10,000)
σ0.6
2c =⎥⎥⎦
⎤
⎢⎢⎣
⎡⎟⎟⎠
⎞⎜⎜⎝
⎛−=
iv. Interior Stress:
⎥⎦
⎤⎢⎣
⎡+⎟⎠⎞
⎜⎝⎛= 1.069b
4h
0.316Pσ 2i
llog a<1.724h 0.675hh1.6ab 22 −+=
7.34in0.675(10)101.6(7.85)b 22 =−+=
130.8psi1.0697.3442.974
1000)0.316(10,0σ 2i =⎥⎦
⎤⎢⎣⎡ +⎟
⎠⎞
⎜⎝⎛= log
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
244psiσ
0.03442.977.85
0.6667.8542.97
4log10
00)0.803(10,0σ
e
2e
=
⎥⎦
⎤⎢⎣
⎡−⎟⎠⎞
⎜⎝⎛+⎟
⎠⎞
⎜⎝⎛=
2.1.7 Dual Tire Example (cont)
Topic 8 – Rigid Pavement Stress Analysis
3. Friction Stresses
L
L/2
hσt
σf
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.
Stresses by subgrade reaction:
What happens to PCC w/ ∆T?
1h2L
γf1hσ cat ⋅⋅⋅⋅=⋅⋅
0ΣFX = Nµ1hσ t ⋅=⋅⋅;
2Lγf
σ cat =
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 bar mats used to control cracking• Do NOT increase structural capacity• Increase joint spacing (slab length)
L/2
hσt
σf
1h2L
γf1hσ cat ⋅⋅⋅⋅=⋅⋅
sst fA1hσ =⋅⋅ s
cas 2f
LhγfA =
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?
Spacing Size
6” 6” 6”
12”
12”
Transverse
Long
itudi
nal
Orientation
A=0.08 in2
A=0.06 in2
• 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
Recommended Spacing: 6x12
Topic 8 – Rigid Pavement Stress Analysis
3.2 Tie Bars
s
'ca
s fhLγf
A =
L’ = distance from the longitudinal joint to the free edge where no tie bar exists
L’
L’
L’Length of tie bars
µdf
21
t s= µ = allowable bond stressd = bar diameter
Many Agencies use the standard tie bar design to simplify construction
Tie Bars: d = 0.5” ; L = 36” ; Spaced @ 30-40” apart
• Placed along the longitudinal joint to tie the 2 slabs together
Spacing of tie bars
)s(A required Steel of Areaentreinforcembar ofArea
pacingS =
Topic 8 – Rigid Pavement Stress Analysis
4. Joint Opening
δ
δ ≤ 0.05” OR use dowels
ε)∆TCL(αδ t +=
Where:δ = Joint openingαt = Coefficient of thermal contractionε = Drying shrinkage coefficientL = Slab lengthC = adjustment factor for subgrade friction
• Stabilized = 0.65• Granular = 0.80
δ2
δ2 ≤ 0.25” to reduce potential for bearing failure
moment arm increases w/ increasing δ2
Topic 8 – Rigid Pavement Stress Analysis
δ
δ ≤ 0.05” OR use dowels
ε)∆TCL(αδ t +=
4.1 Joint Opening Example
0.8C60∆T
40L100.5ε
105α4
6t
==
=×=
×=−
−
( ) ( )( )0.13δ
100.56010512400.8δ 4-6
=×+×××××= −
Dowels required
/oF
oFft
in/in