CE414_SP2017 Stresses and Deflections in Rigid Pavements

•Temperature Stresses
Due to the temperature
differential between the top and
bottom of the slab, curling stresses
(similar to bending stresses) are
induced at the bottom or top of
the slab

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

Frictional stresses
Due to the contraction of slab
due to shrinkage or due to drop
in temperature tensile stresses
are induced at the middle
portion of the slab

Wheel Load Stresses

CC slab is subjected to
flexural stresses due to the
wheel loads

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

Temperature Stresses
• Temperature differential between the top and bottom of
the slab causes curling (warping)stress in the pavement
• If the temperature of the upper surface of the slab is higher
than the bottom surface then top surface tends to expand
and the bottom surface tends to contract resulting in
compressive stress at the top, tensile stress at bottom and vice
versa

Temperature Differential
Temperature at top = T
•Temperature differential = t
•Temperature at bottom = T – t
•Average Temperature (at mid height) = ( T+T-t)/2 = T - t/2
•Increase in temperature of top fiber above average
temperature = t/2
•Decrease in temperature of bottom fiber below average
temperature = t/2

Curling Stresses in infinite Slab

Bending of Infinite Plate

 The difference between a beam and a

plate is that the beam is stressed in only
one direction, the plate in two
directions
 For stresses in two directions, the strain ε
, in the x direction can be determined
by the generalized Hooke's law ,
𝜎𝑥 𝜎𝑦
𝜖𝑥 = −𝜇
𝐸 𝐸
𝜎𝑦 𝜎𝑥
𝜖𝑦 = −𝜇
𝐸 𝐸

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

Curling Stresses in infinite Slab

t = temperature differential between the top and the

bottom of the slab and
αt = coefficient of thermal expansion of concrete
∝𝑡 ∆𝑡
𝜖𝑥 = 𝜖𝑦 =
2
stress in the x direction due to bending in the x direction
𝐸 ∝𝑡 ∆𝑡
𝜎𝑥 =
2(1 − 𝜇 2 )

stress in the x direction due to

bending in the y direction
𝜇𝐸 ∝𝑡 ∆𝑡
𝜎𝑥 =
2(1 − 𝜇 2 )

Curling Stresses in infinite Slab

Total stress in the x direction :

𝐸∝ ∆
𝑡 𝑡 𝜇𝐸∝ ∆
𝑡 𝑡
𝜎𝑥 = 2(1−𝜇 2 ) + 2(1−𝜇 2 )

𝐸 ∝𝑡 ∆𝑡
𝜎𝑥 = (1 + μ)
2 1 − 𝜇2
Curling Stresses in finite Slab
Total stress in the x direction :
𝐸∝ ∆
𝑡 𝑡 𝐸∝ ∆
𝑡 𝑡
𝜎𝑥 = 𝐶𝑥 2(1−𝜇 2 ) +Cy 𝜇 2(1−𝜇 2 )

𝐸∝ ∆
𝑡 𝑡
𝜎𝑥 = 2(1−𝜇 2 ) (Cx +𝜇𝐶𝑦 )

Cx and Cy are correction factors for a finite slab

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

Curling Stresses in finite Slab

𝐸∝ ∆
𝑡 𝑡
𝜎𝑥 = 2(1−𝜇 2 ) (Cx +𝜇𝐶𝑦 )

𝐸∝ ∆
𝑡 𝑡
𝜎𝑦 = 2(1−𝜇 2 ) (Cy +𝜇𝐶𝑥 )

These equations give the

stress at interior due to
temperature differential.

The stresses at the edges is obtained by substituting

𝝁 = 0 in the previous equation:
𝐸 ∝𝑡 ∆𝑡
𝜎 =
2

Curling Stresses in infinite Slab

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

Example 4.1 : Curling Stresses in infinite Slab

A concrete slab, 7 .62 m long, 3 .66 m wide, and 203 mm thick , subjected to a
temperature differential of 11 .1°C. Assuming that k = 54 .2 MN/m3 and αt =9X10 -6
mm/mm/°C, determine the maximum curling stress in the interior and at the edge of the
slab. Take the radius of contact as a= 152 mm.
7.62 m

Solution ΔT= 11.1 °C

Determine Cx and Cy ?

3.66 m
4 𝐸ℎ3
𝑙= h= 203 mm
12 1 − 𝜇 2 𝑘
K= 54.2 MN/m3
4 27 .6 GPa6 ×(203𝑚𝑚)3 αt= 9x10-6 /°C
𝑙= = 776.4 𝑚𝑚
12 1−0.152 54.2MN/m3 a= 152 mm
Lx/l = 7.62×1000/776.4 = 9.81
Ly/ l = 3.66 x 1000/776.4 = 4.71

Example : Solution Curling Stresses in infinite Slab

Determine Cx and Cy ?

Lx/l = 9 .81
Given: Ly/ l = 4 .71

From Chart
Cx = 1 .07
Cy = 0.63

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

Example : Solution Curling Stresses in infinite Slab

Determine Cx and Cy ?

Lx/l = 9 .81
Given:
Ly/ l = 4 .71

From Table
Cx = 1 .07

Cy = 0.63

Curling Stresses in infinite Slab

Example : Solution

From Eq . 4 .9a, the maximum stress in the interior is in the

x direction ,
𝑬𝜶𝒕 ∆𝒕
𝝈𝒙 = 𝑪𝒙 + 𝝁𝑪𝒚
𝟐 𝟏 − 𝝁𝟐

27.6 𝐺𝑃𝑎 × 9x10−6 mm/mm/°C × 11.1

𝜎𝑥 = 1.07 + 0.15 × 0.63 = 𝟏. 𝟔𝟒 𝑀𝑃𝑎
2 1 − 𝜇0.152

From Eq. 4 .11, the maximum stress at an edge is also in the x direction,

𝐶𝐸𝛼𝑡 ∆𝑡 1.07 × 27.6 𝐺𝑃𝑎 × 9x10−6 mm/mm/°C × 11.1

σ= = = 𝟏. 𝟒𝟖 𝑀𝑃𝑎
2 2

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

𝐸 ∝ 𝑡 ∆𝑡
𝜎𝑐 = 𝑎 𝑙
3(1 − 𝜇)

c = corner warping stress (Pa)

E = modulus of elasticity of PCC (Pa)
t = thermal coefficient of PCC (~0.000005/F)
t = temperature differential between the top and bottom ofthe
slab (C)
μ = Poisson’s ratio for PCC
a = radius of wheel load distribution for corner load
l = radius of relative stiffness

Unless actual field measurements are made, it is reasonable to assume a maximum temperature
gradient of 0 .055 to 0 .077°C/mm during the day and about half the above values at night

 Three methods can be used to determine the stresses and deflections

in concrete pavements:
1. closed-form formulas
2. influence charts
3. finite-element computer programs

 These formulas are applicable only to a very large slab with a single-wheel load
applied near the corner, in the interior of a slab at a considerable distance from
any edge, and near the edge far from any corner .
 Corner Loading
 Interior Loading
 Edge Loading

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

 Corner Loading
The formula is based on a concentrated load P
applied at the slab corner,
𝑃𝑥 3𝑃
𝜎𝑐 = 1 =
6
(2𝑥)ℎ2 ℎ2

For circular load near the corner an approximate

solution is given by Westergaard as:

3𝑃 𝑎 2
0.6
𝜎𝑐 = 1−
ℎ2 𝑙

𝑃 𝑎 2
∆𝑐 = 1.1 − 0.88
𝑘𝑙 2 𝑙

 Corner Loading
For rectangular loading area near the corner , Ioannides et al. (1985) suggested the
use of the relationships :
3𝑃 0.72
𝜎𝑐 = 2 1 − 𝐶𝑙
ℎ
c = 1 .772a
𝑃 𝐶
∆𝑐 = 2 1.205 − 0.69 𝑙
𝑘𝑙
 Example:
A concrete slab subjected to a
P= 44.5 kN
corner loading, Given:
k =27 .2 MN/m3 ,
h = 254 mm,
a = 152 mm, a= 152 mm
P = 44.5 kN,
h= 254 mm
determine the maximum stress and K= 27.2 MN/m3
deflection due to corner loading .

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

 Solution:
Eh3 27 .6 GPa  (254 mm) 3
l4 l4  1.09 m
12(1   2 )k 12(1  0.15 2 )27.2 MN / m 3

 0.6

 From Eq. 4 .13,   3P 1   a 2  
c   h   l  
2


3  44.5kN   152 mm 2 0.6 

c  1      1.29 MPa
(254 mm) 2   1.09m  
 
 Eq. 4.14, P   a 2 
c  1.1  0.88 

kl 2   l 

44.5kN   152 mm 2 
c  1.1  0.88   1.27 mm

27.2MN / m3  (1.09m) 2   1.09m 

 Solution:
 From Eqs. 4 .15 and 4.17 [rectangular loading area]:
3P   c  
0.72
3  44.5kN   1.772 152 mm 0.72 
c  1     c  1      1.31MPa
h 2   l   (254 mm) 2   1.09m  

 which is 2% larger than the value obtained from Eq. 4.13.

 From Eqs. 4.16 and 4.17,
P   c 
c  1.205  0.69 l 
kl 2   
44.5kN   1.772 152mm 
c  1.205  0.69   1.42mm
27.2MN / m3 (1.09m) 2   1.09m 

 which is 11% greater than the value obtained from Eq. 4.14.

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

 Interior Loading
The earliest formula developed by Westergaard (1926b) for the stress in
the interior of a slab under a circular loaded area of radius a is
3(1   ) P l
i  (ln  0.6159)
2h 2
b

 For a Poisson ratio of 0.15 for concrete slab

0.316 P l
i  [ 4 log( )  1.069]
h2 b
P 
 1  a  a 
2


i  1  ln  2l   0.673 2l  
8kl 2  2
      


 Example:
Same as previous example except that the load is applied in the
interior,.
Given:
k =27 .2 MN/m3 , a= 152 mm
P= 44.5 kN
h = 254 mm,
a = 152 mm,
P = 44.5 kN,
h= 254 mm
K= 27.2 MN/m3

Determine the maximum stress and deflection due to

interior loading.

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

 Solution:
From equation 4.19b: P= 44.5 kN

a= 152 mm

h= 254 mm
1.724h = 1.724x254 = 438 mm > a K= 27.2 MN/m3
So,
b  1.6a  h  0.675h
2 2
when a  1.724h
b  1.6(152 ) 2  (254 ) 2  0.675(254 )  147 mm
From equation 4.20, 0.316 P l
i  2
[4 log( )  1.069]
h b
0.316  44.5kN 1.09m
i  2
[4 log( )  1.069]  992 kPa
(254 mm) 147 mm
From equation 4.21, P 
 1  a  a 
2


i  1  ln  2l   0.673 2l  
8kl 2 
 2      

44.5 
 1   152   152 
2


i  1  ln  2 1.09   0.673 2 1.09    0.17 mm
8  27.2 1.09 2 
 2      


 Edge Loading
generalized solutions for maximum stress and
deflection produced for a circular or semicircular
loaded area are given by the following
equations :

Edge
Stress

Edge
deflection

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

 Example:
 Same as previous example except that the load is
applied to the slab edge, Determine the maximum
stress and deflection under both circular and
semicircular loaded areas ..
Given: h= 254 mm
K= 27.2 MN/m3
k =27 .2 MN/m3 ,
P= 44.5 kN
h = 254 mm,
a = 152 mm,
P = 44.5 kN,
a= 152 mm

Determine the maximum stress and deflection under

both circular and semicircular loaded areas ..

Example: h= 254 mm
K= 27.2 MN/m3

P= 44.5 kN

 For circular area,

use equation 4.26 for edge stress: a= 152 mm

σe = 1.93 Mpa

 Use Eq. 4.28 for deflection:

Δe = 0.525 mm

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

Example:
h= 254 mm
K= 27.2 MN/m3

P= 44.5 kN

 For semicircular area,

use equation 4.27 for edge stress: a= 152 mm

σe = 2.28 Mpa

 Use Eq. 4.29 for deflection:

Δe = 0.564 mm

Dual Tires

 When a load is applied over a set of dual tires, it is

necessary to convert it into a circular area, so that the
equations based on a circular loaded area can be applied

Sd

P
Pd Pd

0.6L
2a 2a

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

Dual Tires

 When a load is applied over a set of dual tires, it is necessary to

convert it into a circular area, so that the equations based on a
circular loaded area can be applied
Sd

0.6L
2a

Dual Tires
• the circle has an area equal to the contact area of the duals plus the area
between the duals, as indicated by the hatched area shown in the figure .
• If Pd is the load on one tire and q is the contact pressure, the area of each tire is

𝑷𝒅 Sd
= 𝝅(𝟎. 𝟑𝑳)𝟐 +(0.4L)(0.6L) = 0.5227L2
𝒒

𝑷𝒅 0.3L
𝑳=
𝟎. 𝟓𝟐𝟐𝟕𝒒 L
0.4L
• The area of an equivalent circle is:
0.3L
𝜋𝑎2 = 2 × 0.5227𝐿2 + 𝑆𝑑 − 0.6L L
𝜋𝑎2 = 0.4454𝐿2 + 𝑆𝑑 L 0.6L Sd -0.6L 0.6L
0.8521𝑃𝑑 Pd
• Substituting for L: 𝜋𝑎2 = + 𝑆𝑑 𝑃𝑑
0.5227𝑞 2a
𝑞
q
1 2
0.8521𝑃𝑑 𝑆𝑑 𝑃𝑑
• So the radius of contact area is: 𝑎 = +𝜋
𝑞𝜋 0.5227𝑞

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

 Example:
Using Westergaard's formulas, determine the maximum stress in last
three Examples if the 44.5-kN load is applied on a set of duals spaced
at 356 mm on centers, as shown in Figure, instead of over a 152-mm
circular area .

22.25 kN
610 kPa i = ?
356 mm

22.25 kN
i= ?

h= 254 mm
K= 27.2 MN/m3

e = ? c = ?
e= ? c= ?

 Solution:
With Sd = 356 mm, q = 44.5kN/(152mm)2 = 613 kPa, and Pd = 22 .25 kN,
from Eq . 4 .31 ,
22.25 kN
1 2 356 mm i = ?
0.8521𝑃𝑑 𝑆𝑑 𝑃𝑑 610 kPa
22.25 kN
i= ?
𝑎= +𝜋
𝑞𝜋 0.5227𝑞 h= 254 mm
K= 27.2 MN/m 3

e = ? c = ?
e= ? c= ?
0.8521×22.25𝑘𝑁 22.25 𝑘𝑁 1 2
356𝑚𝑚
𝑎= + 𝜋
= 199 mm
(613 𝑘𝑃𝑎)𝜋 0.5227×613𝑘𝑃𝑎

3P   a 2  
0.6

 From Eq. 4 .13,  c  2 1    

h   l  
 

3  44.5kN   199mm 2  
0.6

c  1    
(199mm) 2   1.09m    1.15 MPa
   

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

 Solution:
22.25 kN
i = ?
From Eq. 4 .19b,
356 mm
610 kPa i= ?
 22.25 kN

h= 254 mm
K= 27.2 MN/m 3

b  1.6a  h  0.675h
2 2
when a  1.724h
e = ? c = ?
e= ? c= ?

b  1.6(199mm)  (254mm)  0.675  254mm 186mm

2 2

 From Eq. 4 .20,  i  0.316 P [4 log( l )  1.069]

h2 b
0.316  44.5kN 1.09m
i  [ 4 log( )  1.069]  902 kPa
(254mm) 2 186mm
 From Eq. 4 .26,  e  0.803P [4 log( l )  0.666( a )  0.034]
h2 a l
0.803  44.5kN 1.09m 199mm
e  2
[4 log( )  0.666( )  0.034] 1.68 MPa
(254mm) 199mm 1.09m

 Three methods can be used to determine the stresses

and deflections in concrete pavements:
1. closed-form formulas
2. influence charts
3. finite-element computer programs

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

• Influence charts based on liquid foundations (Pickett and Ray, 1951)

were used previously by the Portland Cement Association for rigid
pavement design
• The charts are based on Westergaard's theory with a Poisson ratio of
0.15 for the concrete slab .
• Only charts for interior and edge loadings are available
 interior loading being used for the design of airport pavements
(PCA, 1955)
 edge loading for the design of highway pavements (PCA, 1966)

Interior Loading

Figure 4.12 shows the applications of

influence charts for determining the
moment at the interior of slab.
𝒒𝒍𝟐 𝑵
𝑴=
𝟏𝟎, 𝟎𝟎𝟎

For example, if l is 1.45 m,

the scale on the top of Figure is 1.45 m
This scale should be used to draw the
configuration of the contact area .
If the actual length of tire imprint is 574
mm, the length to be drawn on the
influence chart is 0.574/1.45 or 39.6% n
of the length shown by the scale l

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

Interior Loading
𝒒𝒍𝟐 𝑵
𝑴=
𝟏𝟎, 𝟎𝟎𝟎

𝟔𝑴
𝝈𝒊 =
𝒉𝟐

Interior Loading

Deflection:
𝟎. 𝟎𝟎𝟎𝟓𝒒𝒍𝟒𝑵
∆𝒊 =
𝑫

Rigidity:
𝑬𝒉𝟑
𝑫=
𝟏𝟐(𝟏 − 𝝁𝟐 )

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

Example:
A concrete slab 254 mm thick is subjected to an interior loading
of 44.5 kN. Applied on a circular area of radius a=152 mm .Given
that k = 27 .2 MN/m3, determine the maximum stress and
deflection by the influence charts.

Solution:
l = 1.09 m
The scale on chart is 1.09 m
Radius of contact area a=
152 mm
a/l = 0.152/1.09 = 0.14
Radius of loading area
=0.14l

0 20 50 100
 L = 1.09 m 

Solution:
a= 0.14l
l = 1.09 m
The scale on chart is 1.09 m
Radius of contact area a=
152 mm
a/l = 0.152/1.09 = 0.14
Radius of loading area
=0.14l

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

37
blocks

N=37x4=148
blocks

Solution:

44.5 𝑘𝑁
𝑞 = 𝜋 (152)2 = 610 kPa

𝒒𝒍𝟐 𝑵 𝟔𝟏𝟎× 𝟏.𝟎𝟗 𝟐 (𝟏𝟒𝟖)

𝑴 = 𝟏𝟎,𝟎𝟎𝟎 = =10.8 m-kN/m
𝟏𝟎,𝟎𝟎𝟎

𝟔𝑴 𝟔×𝟏𝟎.𝟖
𝝈𝒊 = = = 𝟏. 𝟎 𝑴𝑷𝒂
𝒉𝟐 (𝟐𝟓𝟒)𝟐

From Figure 4.13 the number of blocks for

deflection could be determined

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

Solution:
From Figure 4.13 the
number of blocks
for deflection could
be determined a= 0.14l

Deflection:
𝑬𝒉𝟑
𝑫=
𝟏𝟐(𝟏 − 𝝁𝟐 )
(𝟐𝟕.𝟔 𝑮𝑷𝒂)(𝟐𝟓𝟒 𝒎𝒎)𝟑
𝑫= = 38.5 m-MN
𝟏𝟐(𝟏−𝟎.𝟏𝟓𝟐 )

𝟎. 𝟎𝟎𝟎𝟓𝒒𝒍𝟒𝑵
∆𝒊 =
𝑫
𝟎.𝟎𝟎𝟎𝟓×𝟔𝟏𝟎× 𝟏.𝟎𝟗 𝟒 (𝟏𝟔)
∆𝒊 = = 0.18 mm N = 16
𝟑𝟖.𝟓 𝒎−𝑴𝑵

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

Rigid Pavement Construction

Rigid Pavement Construction

23
CE414_SP2017 Stresses and Deflections in Rigid Pavements

Rigid Pavement Construction

Rigid Pavement Construction

24
CE414_SP2017 Stresses and Deflections in Rigid Pavements

 Rigid Pavement Construction

STRESSES DUE TO
FRICTION

• The friction between a concrete slab and its foundation causes tensile stresses
in the concrete, in the steel reinforcements, if any, and in the tie bars.
• For plain concrete pavements, the spacing between contraction joints must be
so chosen that the stresses due to friction will not cause the concrete to crack .
• For longer joint spacings, steel reinforcements must be provided to take care of
the stresses caused by friction. The number of tie bars required is also
controlled by the friction .

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

Effect of Volume Change on Concrete

The volume change caused by the variation of temperature and moisture has two important
effects on concrete.
L
•induces tensile stresses and causes
the concrete to crack.
c = ?
•causes the joint to open and L/2
decreases the efficiency of load
transfer.
c h
Concrete Stress h

c h = ? Frictional Stresses

Fully
c h = c . fa. h. L/2 mobilize
d friction
𝜸𝒄𝑳𝒇
𝝈𝒄 = 𝟐
𝒂

fa= 1.5

Example 4.8 :
Given a concrete pavement with a joint spacing of 7 .6 m and a
coefficient of friction of 1 .5, as shown in Figure, determine the stress in
concrete due to friction .
7.6 m

Solution c = ?

With c = 23 .6 kN/m', L = 7 .6 m, and fa =1.5,

𝜸𝒄𝑳𝒇
𝝈𝒄 = 𝟐
𝒂 =𝟐𝟑.𝟔 ×𝟕.𝟔×𝟏.𝟓
𝟐
=135 kPa

The tensile strength of concrete ranges from 3 𝑓𝑐′ to 5 𝑓𝑐′

where 𝑓𝑐′ is the compressive strength of concrete. If 𝑓𝑐′ = 13.8 MPa, the tensile strength
is from 1.13 to 1.89 MPa, which is much greater than the tensile stress of 135 kPa

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

Joint Opening

• The spacing of joints in plain concrete pavements depends

more on the shrinkage characteristics of the concrete rather
than on the stress in the concrete
• Longer joint spacings cause the joint to open wider and
decrease the efficiency of load transfer .
• The opening of a joint can be computed approximately by

∆𝐿 = 𝐶𝐿 𝛼𝑡 ∆𝑇 + 𝜖 (4.36)

Joint Opening

∆𝐿 = 𝐶𝐿 𝛼𝑡 ∆𝑇 + 𝜖 (4.36)

ΔL = joint opening caused by temperature change and drying shrinkage of

concrete
αt = coefficient of thermal expansion of concrete,
generally 9 to 10 .8 x10-6/°C
ε = drying shrinkage coefficient of concrete, approximately 0 .5 to 2.5x10-4;
L = joint spacing or slab length ;
ΔT = temperature range, which is the temperature at placement minus the
lowest mean monthly temperature
C = adjustment factor due to slab-subbase friction,
0 .65 for stabilizedbase
0 .8 for granular subbase.

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

Example 2 :
Given ΔT = 19°C, αt = 9.9x10-6/°C, ε = 1.0X10-4 , C = 0.65, andthe
allowable joint openings for undoweled and doweled joints are 1.3
and 6.4mm, respectively, determine the maximum allowable joint
spacing .

Solution:
∆𝐿 ∆𝐿 ∆𝐿
𝐿= = =
𝐶 𝛼𝑡 ∆𝑇 + 𝜖 0.65(9.9 × 10−6 × 19 + 0.0001) 0.00019

For the un-doweled joint, L = 1.3/0 .00019 = 6942mm = 6.9 m

For the doweled joint, L = 6.4/0.00019 = 24564mm = 24 .5 m

Steel Stress

The design of longitudinal and transverse reinforcements and of

the tie bars across longitudinal joints is based on the stresses due
to friction.

Reinforcements, Wire fabric or bar mats may be used in

concrete slabs for control of temperature cracking .

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

These reinforcements do not increase the structural capacity of

the slab but are used for two purposes :
1. to increase the joint spacing
2. to tie the cracked concrete together and maintain load transfers
through aggregate interlock.
When steel reinforcements are used, it is assumed that all tensile
stresses are taken by the steel alone, so σch must be replaced by Asfs

𝛾𝑐 ℎ𝐿𝑓𝑎
𝐴𝑠 =
2𝑓𝑠

As = area of steel required per unit width

fs = allowable stress in steel .

the amount of steel required is proportional to the length of slab

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

Welded wire fabric is prefabricated reinforcement

consisting of parallel series of high-strength, cold-
drawn wires welded together in square or
rectangular grids .
The spacings and sizes of wires are identified by
"style." A typical style designation is
6x12 —W8xW6
size of
spacing of transverse
longitudina size of wires is W6
l wires is 6 the spacing longitudinal wire with a cross
in of is W8 with sectional area
transverse across-sectional of 0 .06 in.
wires is 12 in area of 0 .08 in.2

The typical style with deformed welded wire fabric

is
6x12— D8 xD6.

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

A typical style designation is

6x12 —W8xW6

Example 3

Determine the wire fabric required for a two-lane concrete pavement,

203 mm thick, 18 .3 m long, and 7 .3 m wide, with a longitudinal joint at
the center, as shown in Figure.

203 mm

7 .3 m

18.3 m

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

Solution:

γc= 23.6 kN/m3, hc =203 mm, L=18.3m, fa=1.5, and fs = 297 MPa, the
required longitudinal steel is
𝛾𝑐 ℎ𝐿𝑓𝑎 23.6 × 203 × 18300 × 1.5
𝐴𝑠 = = =
2𝑓𝑠 2 × 297

As = 222 mm2/m.

The required transverse steel is

As =88 .9 mm2/m.
From Table 4.3, use 6x12–W5.5 X W4.5 with cross sectional areas of
0.11 in.2 (71 mm2 ) for longitudinal wires and 0.045 in.2 (29 mm2 ) for
transverse wires

From Table 4.3, use 6x12–W5.5 X W4.5 with cross

sectional areas of 0.11 in.2 (71 mm2 ) for
longitudinal wires and 0.045 in.2 (29 mm2 ) for
transverse wires

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

Rigid Pavement Construction

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

 Rigid Pavement Construction

The design of dowels and joints is mostly based on experience.

The size of dowels to be used depends on the thickness of slab .

the diameter of dowels is equal to one-eighth of the slab thickness.

In a recent edition of joint design, PCA (1991) recommended the use of 1 .25 in . (32
mm) diameter dowels for highway pavements less than 10 in . (254 mm) thick and 1.5
in . (38 mm) diameter dowels for pavements 10 in. (254 mm) thick or greater . A
minimum dowel diameter of 1 .25 to 1 .5 in. (32 to 38 mm ) is needed to control faulting
by reducing the bearing stress in concrete.

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

Example 5 :

Figure below shows a concrete pavement 8 in . (203 mm) thick having a joint width
of 0.2 in. (5.1 mm), a modulus of subgrade reaction of 100 pci (27 kN/m3 ), and a
modulus of dowel support of 1.5 x 106 pci (407 GN/m3 ) . A load of 9000 lb (40 kN) is
applied over the outermost dowel at a distance of 6 in . (152 mm) from the edge.
The dowels are ¾ in. (19 mm) in diameter and 12 in. (305 mm) on centers .
Determine the maximum bearing stress between dowel and concrete .

Example 5 :

Figure below shows a concrete pavement 8 in . (203 mm) thick

having a joint width of 0.2 in. (5.1 mm), a modulus of subgrade
reaction of 100 pci (27 kN/m3 ), and a modulus of dowel
support of 1.5 x 106 pci (407 GN/m3 ) . A load of 9000 lb (40 kN)
is applied over the outermost dowel at a distance of 6 in . (152
mm) from the edge. The dowels are ¾ in. (19 mm) in diameter
and 12 in. (305 mm) on centers . Determine the maximum
bearing stress between dowel and concrete .

Solution:
l = [4 x 106 x 512/(12 x 0.9775 x 100)] 025 = 36.35 in.
(427mm).
If the dowel directly under the load is subjected to a
shear force Pt , the forces on the dowels within a distance
of 1.8l, or 66 in . (1.68 m), can be determined by assuming
a straight-line variation, as shown in Figure.

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

The sum of the forces on all dowels is 3 .27Pt , which must be equal to
one-half of the applied load based on 100% joint efficiency, or
Pt = 4500/3 .27 = 1376 lb (6.1 kN) .
Id = π(0.75) 4/64 = 0.0155 in . 4 (6450 mm4 ) .
β = [1.5 X106X0.75/(4 x 29x106x0.0155)]025 = 0.889 in. (22.6 mm).
σb = 1.5x106X 1376(2+0.889x0.2)/(4x0.703x29x106X0.0155)=3556 psi
(24.5MPa)
For a 3000-psi (20 .7-MPa) concrete, the allowable bearing stress is
fb = (4 – 0 .75) X 3000/3 = 3250 psi (22 .4 MPa) .
Because the actual bearing stress is about 10% greater than the
allowable, the design is not considered satisfactory .

Example 6 :
A 241-mm slab resting on a foundation with k = 13.6 MN/m3. Twelve
dowels at 305 mm on centers are placed at the joint on the 3 .66-m
lane. Two 40-kN wheel loads are applied at points A and B. Determine
the maximum load on one dowel
40-kN 40-kN

152 mm 1.83 m

241 mm

Solution:
l = [27.6x (0.241)3/(12 x 0 .9775 x 13.6)] 0.25 = 1.25 m
so 1.8 l = 2.24 m

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

40-kN 40-kN

152 mm 1.83 m

241 mm

2.24 m

First, consider the 40-kN load at A

If the dowel at A has a load factor of 1, the load factors at other
dowels can be determined from similar triangles.
The sum of these factors results in 4 .18 effective dowels, so the load
carried by the dowel at A is 20/4 .18 or 4 .8 kN .
The loads carried by other dowels can be determined by proportion

40-kN 40-kN

152 mm 1.83 m
241 mm

2.24 m 2.24 m

consider the 40-kN load at B

If the dowel at B has a load factor of 1, the load
factors at other dowels can be determined from the
triangular distribution
The sum of these factors results in 7 .08 effective
dowels.
Note that the dowels on the other side of the
longitudinal joint are not considered effective in
carrying the load .
The load carried by the dowel at B is 20/7 .08 or 2 .8 kN

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CE414_SP2017 Stresses and Deflections in Rigid Pavements

40-kN 40-kN

152 mm 1.83 m

241 mm

2.24 m

Superposition (Summation) of both forces due to A

and B

Figure d shows the forces on each dowel due to the

combined effect of both loads .
It can be seen that the dowel nearest to the
pavement edge is the most critical and should be
used for design purposes .
The load carried by this dowel can be determined
directly by Pt = 20/4 .18 + 0.18 x 20/7 .08 = 5 .3 kN
Pt = 1077+ 114= 1191 lb.

38