APPENDIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND NATIONAL HIGHWAY DEVELOPMENT
1
i
APPENDIXA [
DESIGN OF A RIGID PAVEMENT i
RURAL ROAD i
AND i j
NATIONAL HIGHWAY !
i
DEVELOPMENT
236
APPEND! XA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND NATIONAL HIGHWAY DEVELOPMENT
DESIGN OF A CEMENT CONCRETE PAVEMENT FOR RURAL ROAD (IRC: SP: 202002 / IRC: SP: 622004) A cement concrete pavement is to be designed for a Rural Road in Gujarat State having a traffic volume of upto 500 vehicles per day consisting vehicles, like, agricultural tractors/trailers, light goods vehicles, heavy trucks, buses, animal drawn vehicles, motorized twowheels and cycles. The soil has a soaked CBR value of 2%, 4% and 6% for 30kN wheel load. Table AA1 Design of CC Pavement for Rural Roads Design Parameters: Sample D1 (6% CBR30kN Wheel Load) Traffic Volume (A)
 UP TO 500 cvpd (Assume)
Concrete Grade (fc)
= 25 N/mm2
Characteristic Compressive Cube Strength
= 40.33 N/mm2 at 28 Days Actual Compressive Strength
Flexural Strength ( ff)
= 5.00 N/mm2 [50.00 kg/cm2]
90 days Flexural strength
= 6.10 N/mm2 [61.00 kg/cm2] .
Soaked CBR Value (%)
= 0.06(6%)
Modulus of Subgrade Reaction (k)
= 45 (N/mm2/mm)*10'3
Effective K Value (20% more)
= 54 (N/mm2/mm)* 10"3
Elastic modulus of Concrete (Ec) (As per Actual
= 28,417 N/mm2
Calculation) Poisson’s ratio (p)
= 0.15
Coefficient of thermal coefficient of concrete (a)
= 0.00001/°C
Design Wheel Load (P)
= 30 kN
Tyre pressure (q)
= 0.5 N/mm2 [5 kg/cm2]
Spacing of Contraction Joints (L)
= 3.75m [3750 mm]
Width of Slab (W)
= 3.75m [3750 mm]
Radius of load contact (assumed circular), (a)
=13.82 cm
Trial Thickness for Slab, h = 150mm.
237
APPENDIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND NATIONAL HIGHWAY DEVELOPMENT [
Check for Temperature Stresses:
'
■
j
Assuming a contraction joint spacing of 3.75 m and 3.75m width 1. Temperature Stress (ote):
!
The temperature differential (At) for Gujarat for a slab thickness of 150mm is 12.5°C.
The Radius of Relative Stiffness, 1 = *1 ’
h.
AJ 12 (1 n
Hence, 1 = 623.79mm.
!
L/l = 3750 / 623.79= 6.0
j
W/l = 3750 / 623.79= 6.0
I
i
j
Both values are same, if not then adopt greater one. Bradbury’s Coefficient, C = 0.942 (from figure 1, pg. 9) [Value of C can be ascertained directly from Bradbury’s chart against values of L/l and W/l]
j
Temperature Stress in edge region, ote = Ea^AtHence, ote = 1.63 N/mmI2.
c

 2. Edge Load Stress (ole): From Page: 12, Edge Load Stress,
■
Radius of equivalent distribution of pressure (bj, b = a(if(a/h>= 1.724); (b) = Vl.6 a2 + h2
 0.675 h if (a/h < 1.724),
For slab thickness of 150mm; Edge Load Stress, ole, is 3.39 N/mm2 (3.39 MPa). j
Total Stress = Edge Load Stress + Temperature j Stress = 3.39 + 1.63 = 5.02 N/mm2, which I
n
is less than the allowable flexural strength of 6.10 N/mm . Hence, assumed thickness of slab = 150mm, is OK. [As per Temperature Stress Criteria]
I
1
238
APPENDIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND NATIONAL HIGHWAY DEVELOPMENT Check for Corner Stresses (alc): From
Fig.
5
(Page
12),
Comer
Load
Stress
for
wheel
load
of
30kN,
for k = 54.0 (N/mm2/mm)*10'3 = 0.054 N/mm2/mm = 0.054 N/mm2/mm (Approx.) and slab thickness of 150mm is 3.20 N/mm2 (3.20 MPa). [Temperature Stress in the corner region is negligible, as the comers are relatively free to warp, hence it can be ignored.] Hence,
239
APPENDIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND NATIONAL HIGHWAY DEVELOPMENT DESIGN OF A CEMENT CONCRETE PAVEMENT FOR RURAL ROAD (IRC: SP: 202002 / IRC: SP: 622004) A cement concrete pavement is to be designed for a Rural Road in Gujarat State having a traffic volume of upto 500 vehicles per day consisting vehicles, like, agricultural tractors/trailers, light goods vehicles, heavy trucks, buses, animal drawn vehicles, motorized twowheels and cycles. The soil has a soaked CBR value of 2%, 4% and 6% for 5 lkN wheel load. Table AB1 Design of CC Pavement for Rural Roads Design Parameters: Sample D1 (6% CBR51kN Wheel Load) Traffic Volume (A)
= UP TO 500 cvpd (Assume)
Concrete Grade (fc)
= 25 N/mm2
Characteristic Compressive Cube Strength
j
= 40.33 N/mm2 at 28 Days Actual Compressive Strength
Flexural Strength (ff)
= 5.00 N/mm2 [50.00 kg/cm2]
90 days Flexural strength
= 6.10 N/mm2 [61.00 kg/cm2]
Soaked CBR Value (%)
= 0.06 (6%)
Modulus of Subgrade Reaction (k)
= 45 (N/mm2/mm)*10'J
Effective K Value (20% more)
= 54 (N/mm2/mm)* 10‘a
Elastic modulus of Concrete (Ec) (As per Actual
= 28,417 N/mm2
Calculation) Poisson’s ratio (p)
= 0.15
Coefficient of thermal coefficient of concrete (a)
= 0.00001/°C
Design Wheel Load (P)
= 51 kN
Tyre pressure (q)
= 0.7 N/mm2 [7 kg/cm2]
Spacing of Contraction Joints (L)
= 3.75m [3750 mm]
Width of Slab (W)
= 3.75m [3750 mm]
Radius of load contact (assumed circular), (a) ;
=15.23 cm
Trial Thickness for Slab, h = 150mm.
240
APPENDIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND NATIONAL HIGHWAY DEVELOPMENT Check for Temperature Stresses: Assuming a contraction joint spacing of 3.75 m and 3.75m width 3. Temperature Stress (ote):
i 

'
The temperature differential (At) for Gujarat jfor a slab thickness of 150mm is 12.5°C.
The Radius of Relative Stiffness, 1= 4 i 12 (l)iz) k Hence, 1 = 623.79mm.
;
L/l = 3750 / 623.79= 6.0

W/l = 3750 / 623.79= 6.0
!
Both values are same, if not then adopt greater one. 
Bradbury’s Coefficient, C = 0.920 (from figure 1, pg. 9) [Value of C can be ascertained directly from Bradbury’s chart against values of L/l and W/l]

Temperature Stress in edge region, ate = Hence, ote= 1.63 N/ram2.
C
j
4. Edge Load Stress (ole): From Page: 12, Edge Load Stress,

Radius of equivalent distribution of pressure (b), b = a(if(a/h>= 1.724); (b) = Vl.6 a2 + h2
j
 0.675 h if (a/h < 1.724), J 
For slab thickness of 150mm; Edge Load Stress, ole, is 3.39 N/mm2 (3.39 MPa). I
j
Total Stress = Edge Load Stress + Temperature)Stress = 3.39 + 1.63 = 5.02 N/mm2, which is less than the allowable flexural strength of 6.10 N/mm . Hence, assumed thickness of slab = 150mm. [As per Temperature Stress Criteria]
241
APPENDIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND
Check for Corner Stresses (olc): From
Fig.
5
(Page
12),
Comer
Load
Stress
for
wheel
load
of
30kN,
for k = 54.0 (N/mm2/mm)*10'3 = 0.054 N/mm2/mm = 0.054 N/mm2/mm (Approx.) and slab thickness of 190mm is 3.20 N/mm2 (3.20 MPa). [Temperature Stress in the comer region is negligible, as the comers are relatively free to warp, hence it can be ignored.]
j
Hence, oIc = 3.20 N/mm2, which is less than the allowable flexural strength of 6.10 N/mm2. i
So, the slab thickness of 150mm is Safe. The calculations presented above are sample calculations. Similar calculations are done using various values of flexural strengths of concrete.
242
APPENDIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND NATIONAL HIGHWAY DEVELOPMENT DESIGN OF A NATIONAL HIGHWAY ROAD PAVEMENT (IRC: 582002) A cement concrete pavement is to be designed for a two lane twoway National Highway in Gujarat State. The total twoway traffic is 3000 commercial vehicles per day (evpd) at the end of the construction period. Design parameters are provided in Table AC1 and the traffic axle load spectrum is given Table AC2.
TABLE AC1 Design of CC Pavement for Two Lane TwoWay National Highway Design parameters: Sample G5 (6% CBR100DLC3000CVPD)
Present Traffic
=3000 cvpd
Design life
=20 yrs.
Compressive Strength (fck)
= 56.63 N/mm2 = 566.3 kg/cm2
Flexural strength of cement concrete
= 8.15 N/mm2 = 81.50 kg/cm2
(Modulus of rupture) CBR
= 6%
Dry Lean Concrete (DLC)
=100 mm
Effective modulus of subgrade reaction of the DLC
= 18.7 kg/cm3
subbase (k) Elastic modulus of concrete (E)
= 40417 N/mm2
Poisson’s ratio (p)
= 0.15
Coefficient of thermal coefficient of concrete (a)
= 10 x 10‘6/°C
Tyre pressure (q)
= 8 kg/cm2
Rate of traffic increase (r)
= 0.075
Spacing of contraction joints (L)
= 4.5m
Width of slab (b)
= 3.5m
Load safety factor (LSF)
= 1.2
Wheel load (P)
= 8000 kg
C/C distance between two tyres (S)
= 31 cm
Joint width (z)
= 2.0 cm
243
APPENDIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND NATIONAL HIGHWAY DEVELOPMENT
The axle load spectrum obtained from axle load survey is given in the following: Table AC2 Axle Load Spectrum Obtained From Axle Load Survey Single Axle Loads
Tandem Axle Loads
Axle load class,
Percentage of axle
Axle load class,
Percentage of axle
tons
loads
tons
loads
1921
0.6
3438
0.3
1719
1.5
3034
0.3
1517
4.8
2630
0.6
1315
10.8
2226
1.8
1113
22.0
1822
1.5
911
23.3
1418
0.5
Less than 9
30.0
Less than 14
2.0
Total
93.0
Total
7.0
365 X Cumulative repetition in 20 yrs. =
{ (1 + r)2 — 1} r
= 47,418,626 commercial vehicles Design traffic = 25 per cent of the total repetitions of commercial vehicles = 11,854,657 Front axles of the commercial vehicles carry much lower loads and cause small flexural stress in the concrete pavements and they need not be considered in the pavement design. Only the rear axles, both single and tandem, should be considered for the design. In the example, the total number of real axles is, therefore, 11,854,657. Assuming that the midpoint of the axle load class represents the group, the total repetitions of the single axle and tandem axle loads are as follows:
244
APPENDIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND NATIONAL HIGHWAY DEVELOPMENT
Table AC3 Total Repetitions of the Single Axle and Tandem Axle Loads Single Axle Load in tonnes Expected repetitions 20 71127 18 177820 16 569023 14 1280303 12 2608024 10 27622135 Less than 10 3556397 Trial Thickness
Tandem Axle Expected Load in tonnes repetitions 35564 36 35564 32 71128 : 28 213384 24 177820 1 20 59273 16 Less than 16 237093 i
= 19 cm Table AC4 Cumulative Fatigue Life
Axle load (AL), tonnes
AL x 1.2
Stress, kg/cm2 from charts
Stress ratio
Expected Repetitions, n
Fatigue life, N
Fatigue life consumed
(1)
(2)
(3)
(4)
(5)
(6)
Ratio (5)/(6)
20 18 16 14
24.0 21.6 19.2 16.8
44.80 41.00 37.40 35.63
36
43.2
31.68
Single axle 12.38 x 104 71128 0.55 64.30 x 104 177820 0.50 15.48 x 106 569024 0.46 47.22 x 10* 1280303 0.44 Tandem axle Infinity 35564 0.39 Cumulative fatigue life consumed
0.57 0.28 0.04 0.00 0.00 0.89
The cumulative fatigue life consumed being less than 1; the design is safe from fatigue considerations. Check for Temperature Stresses: Eat
Edge warping stress (Ste) =
2
Radius of relative stiffness (/) = 59.62 cm
245
APPENDIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND NATIONAL HIGHWAY DEVELOPMENT (see below under comer stress) Therefore, L// = 450 / 59.62 = 7.5 Bradbury’s Coefficient, which can be ascertained directly from Bradbury’s chart against values of U l and B/1, (C) = 1.054 from fig.2. (IRC: 582002) The temperature differential was taken as 12.98°C for the Gujarat region. Eat „ ...... ^
■
Edge warping stress
=
2
i ry
=27.65 kg/cm
Total of temperature warping stress and the highest axle load stress = 44.835 + 27.65 = 72.48 kg/cm2 which is less than 81.50 kg/cm2, the flexural strength. So the pavement thickness of 19 cm is safe under the combined action of wheel load and temperature. Check for Corner Stresses: Comer stress is not critical in a dowelled pavement. The comer stress can be calculated value from the following formula: Comer stress Sc = j l — (^)
' j
The 98 percentile axle load is 16 tonnes. The wheel load, therefore, is 8 tonnes. Radius of relative stiffness (/) = 4
Eh3 12(1 ii2)k
59.62 cm
a = radius of area of contact of wheel. i
Considering a single axle dual wheel, a=()1/2
V pn / i
a = 0.8521 x ——h (—l—)0'5 qXn
Comer stress S0=
a = 26.52 cm
n\ 0.5227 xq )
^ l — (~j~) j =28.38 kg/cm2
The comer stress is less than the flexural strength of the concrete, i.e., 81.50 kg/cm2 and the pavement thickness of 19 cm assumed is safe. 246
APPENDIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND NATIONAL HIGHWAY DEVELOPMENT
Design of Dowel Bars Table AC5 Design Parameters for Dowel Bars Diameter of the dowel bar (b)
= 3.2 cm (assumed)
Modulus of Dowel/Concrete interaction (Dowel support) (K)
= 41500 kg/cm2/cm
Modulus of the elasticity of the Dowel, kg/cm2
= 2.0 X 106 kg/cm2
Dowel/Concrete interaction (E) = 5.147 cm4
Moment of Inertia of Dowel (I)
Design wheel load (P) = 98 percentile axle load is 16 tonne. The wheel load, therefore, is 8000 kg (dual wheel load) Percentage of load transfer
= 40 %
Permissible bearing stress in concrete is calculated as under: „ _ (l0.166)/cfc
Fb—:
fCk = characteristic compressive strength of concrete cube (15 cm) after 28 days curing concrete = 566.30kg/cm2
Fb;
(10.16—2.5) 566.30 9.525
: 455.42 kg/cm~
Assumed spacing between the dowel bars
= 19 cm
First dowel bar is placed at a distance
= 15 cm from the pavement edge
Assumed length of the dowel bar
= 50 cm
Dowel bars upto a distance of 1.0 x radius of relative stiffness, from the point of load application are effective in load transfer. Number of dowel bars participating in load transfer when wheel load is just over the dowel bar close to the edge of the slab =1+1/ spacing = 1+ 59.62 / 19 = 4 dowels.
247
APPENBIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND NATIONAL HIGHWAY DEVELOPMENT »a—b———iLi.i'.——TTwm~~—nmmr—MrifffMiM—■——w—
Assuming that the load transferred by the first dowel is Pt and assuming that the load on dowel bar at a distance of 1 from the first dowel to be zero, the total transferred by dowel bar system = (1+^ + ^+L^)Pt= 2.09Pt
Load carried by the outer dowel bar, P, = 'MS*‘ 1“J m =1532.55 kg
Check for Bearing Stress: 7Tb*
Moment of Inertia of Dowel,’ I = — 64 T_ 7TX2.54 _ , n„___ 4
1 64 1.92 cm Relative stiffness of dowel bar embedded in concrete (P) =
4 /~Kb~ y 4 El
Where;
P
41500 X 2.5 y4X2X 10s X 1.92
0.29
Bearing stress in dowel bar = (Pt X k) X (2+Pz) / (4p3EI) = 452.50 kg/cm2 which is less than 455.42 kg/cm2 Hence, the dowel bar spacing and diameter assumed are safe. Design of Tie Bars: Table AC6 Design Parameters for Tie Bars Slab thickness (h)
= 19 cm
Lane width (b)
= 3.5 m
Coefficient of friction (f)
= 1.5
Density of concrete (W)
= 2400 kg/m'3
Allowable tensile stress in plain bars (S), (As per IRC: 212000)
= 1250 kg/m2
248
APPENDIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND NATIONAL HIGHWAY DEVELOPMENT Allowable tensile stress in deformed bars (S), (As per IRC; 212000)
= 2000 kg/m2
Allowable bond stress in plain tie bars (B)
= 17.5 kg/m2
Allowable bond stress in deformed tie bars (B)
= 24.6 kg/m2
Diameter of tie bar (d)
= 12 mm
1. Spacing and length of the plain bar Area of steel bar per metre width ofjoint to resist the frictional force at slab bottom
As = ~ = 1.92 cm2 Assuming a diameter of tie bar of 12 mm, the cross sectional area 7t d2
Cross sectional area of tie bar (A) = — 4
A=1.13 sq.cm. Perimeter of tie bar (P) = n d = jc x 1.2 = 3.77 cm Spacing of tie bars = A / As =59.02 cm Provide at a spacing of 59 cm c/c Length of tie bar (L) = Cross sectional area of tie bar (A) = 1.13 sq.cm. Perimeter of Tie Bar (P) = 3.77 cm Length of tie bar (L) =
BXP
= 42.86 cm
Increase length by 10 cm for loss of bond due to painting and another 5 cm for tolerance in placement. Therefore, the length is 42.86 + 10 + 5 = 57.86 cm, Say 58 cm
2. Spacing and length of the deformed bar Area of steel bar per metre width ofjoint to resist the frictional force at slab bottom .
bfW
.
2
As = —= 1.20 cm
Assuming a diameter of tie bar of 12 mm, the cross sectional area Cross sectional area of tie bar (A) = A=1.13 sq.cm. Perimeter of Tie Bar (P) = rc d = tc x 1.2 = 3.77 cm Spacing of tie bars = A / As =94.44 cm
249
APPENDIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND NATIONAL HIGHWAY DEVELOPMENT Provide at a spacing of 94 cm c/c Length of tie bar (L) =
j
Cross sectional area of tie bar (A) =1.13 sq.cml Perimeter of Tie Bar (P) = 3.77 cm Length of tie bar (L) =
BXP
= 48.78 cm
Increase length by 10 cm for loss of bond due to painting and another 5 cm for tolerance in placement. Therefore, the length is 48.78 + 10 + 5 = 63.78 cm, Say 64 cm
250
APPENDIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND NATIONAL HIGHWAY DEVELOPMENT DESIGN OF A NATIONAL HIGHWAY ROAD PAVEMENT (IRC: 582002) A cement concrete pavement is designed for a two lane twoway National Highway in Gujarat State. The total twoway traffic is 3000 commercial vehicles per day (cvpd) at the end of the construction period. Design parameters are provided in Table AD1 and traffic axle load spectrum is given Table AD2. Table AD1 Design of CC Pavement for Two Lane TwoWay National Highway Design parameters: Sample G5 (6% CBR150DLC3000CVPD)
Present Traffic
=3000 cvpd
Design life
=20 yrs.
Compressive Strength (fck)
= 56.63N/mm2 = 566.3 kg/cm2
Flexural strength of cement concrete
= 8.15 N/mm2 = 81.5 kg/cm2
(Modulus of rupture) CBR
= 6%
Dry Lean Concrete (DLC)
=150 mm
Effective modulus of subgrade reaction of the DLC
= 24.25 kg/cm3
subbase (k) Elastic modulus of concrete (E)
= 40417 kg/cm2
Poisson’s ratio (p)
= 0.15
Coefficient of thermal coefficient of concrete (a)
= 10 x 10'b/'C
Tyre pressure (q)
= 8 kg/cm2
Rate of traffic increase (r)
= 0.075
Spacing of contraction joints (L)
= 4.5m
Width of slab (b)
= 3.5m
Load safety factor (LSF)
= 1.2
Wheel load (P)
= 8000 kg
C/C distance between two tyres (S)
= 31 cm
Joint width (z)
!
= 2.0 cm
251
APPENDIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND NATIONAL HIGHWAY DEVELOPMENT The axle load spectrum obtained from axle load survey is given in the following:
Table AD2 Axle Load Spectrum Obtained From Axle Load Survey Single Axle Loads
Tandem Axle Loads
Axle load class,
Percentage of axle
Axle load class,
Percentage of axle
tons
loads
tons
loads
1921
0.6
3438
0.3
1719
1.5
3034
0.3
1517
4.8
2630
.0.6
1315
10.8
2226
1.8
1113
22.0
1822
1.5
911
23.3
1418
0.5
Less than 9
30.0
Less than 14
2.0
Total
93.0
Total
7.0
Cumulative repetition in 20 yrs. =
365 Xi4 { (1 + r)2 — 1} r
= 47,418,626 commercial vehicles Design traffic = 25 per cent of the total repetitions of commercial vehicles = 11,854,657 Front axles of the commercial vehicles carry much lower loads and cause small flexural stress in the concrete pavements and they need not he considered in the pavement design. Only the rear axles, both single and tandem, should be considered for the design. In the example, the total number of real axles is, therefore, 11,854,657. Assuming that midpoint of the axle load class represents the group, the total repetitions of the single axle and tandem axle loads are as follows:
252
APPENDIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND '_______________________ NATIONAL HIGHWAY DEVELOPMENT
Table AD3 Total Repetitions of the Single Axle and Tandem Axle Loads Single Axle Load in tonnes
Tandem Axle
Expected
Load in tonnes
Expected repetitions
repetitions 20
71127
36
35564
18
177820
32
35564
16
569023
28
71128
14
1280303
24
213384
12
2608024
20
177820
10
27622135
16
59273
Less than 10
3556397
Less than 16
237093
Trial Thickness
.
= 19 cm Table AD4 Cumulative Fatigue Life
Axle load
AL x
(AL),
1.2
Stress
Expected
Fatigue
Fatigue life
ratio
Repetitions, n
life, N
consumed
(4)
(5)
(6)
Ratio
from charts
tonnes (1)
Stress, kg/cm2
(2)
(3)
(5)/(6) Single axle 20
24.0
42.30
0.52
71128
35.00 x 10b
0.21
18
21.6
39.00
0.48
177820
26.43 x 10s
0.07
16
19.2
35.50
0.44
569024
22.68 x 10y
0.00
14
16.8
33.80
0.41
1280303
Infinity
0.00
Infinity
0.00
Tandem axle 36
43.2
30.24
0.37 ;
35564
Cumulative fatigue liiFe consumed
0.28
253
APPENDIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND NATIONAL HIGHWAY DEVELOPMENT ifcl iiihiliaMaWiMWHW—I—————
The cumulative fatigue life consumed being less than 1; the design is safe from fatigue considerations. Check for Temperature Stresses: Edge warping stress (Ste) =
Eat 2
L
Radius of relative stiffness (/) = 55.87 cm (see below under comer stress) Therefore, U l = 450 / 55.87 = 8.1 Bradbury’s Coefficient, which can be ascertained directly from Bradbury’s chart against values of L/1 and B/1, (C) = 1.077 from fig.2. (IRC: 582002) The temperature differential was taken as 12.98°C for the Gujarat region. Eat _
Edge warping stress
=
2
= 28.25 kg/cm
Total of temperature warping stress and the highest axle load stress = 42.33 + 28.25 = 70.59 kg/cm2 which is less than 81.50 kg/cm2, the flexural strength. So the pavement thickness of 19 cm is safe under the combined action of wheel load and temperature. Check for Corner Stresses: Comer stress is not critical in a dowelled pavement. The comer stress can be calculated value from the following formula: Comer stress Sc =
l 1 — (^r)
]
The 98 percentile axle load is 16 tonnes. The wheel load, therefore, is 8 tonnes. I
pf.3
Radius of relative stiffness (l) = 4 I———= 55.87 cm
a = radius of area of contact of wheel. Considering a single axle dual wheel,
254
APPENDIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND NATIONAL HIGHWAY DEVELOPMENT
a = 0.8521 x ——I qxn
S/
P
n V 0.5227 xq
a = 26.52 cm
'
j = 25.29 kg/cm2
Comer stress Sc= ^
The comer stress is less than the flexural strength of the concrete, i.e., 81.50 kg/cm2 and the pavement thickness of 19 cm assumed is safe. Design of Dowel Bars Table AD5 Design Parameters for Dowel Bars Diameter of the dowel bar (b)
= 3.2 cm (assumed)
Modulus of Dowel/Concrete interaction (Dowel support) (K)
= 41500 kg/cm2/cm
Modulus of the elasticity of the Dowel, kg/cm2
= 2.0 x 106 kg/cm2
Dowel/Concrete interaction (E) i
Moment of Inertia of Dowel (I)
= 5.147 cm4
Design wheel load (P) = 98 percentile axle load is 16 tonne. The wheel load, therefore, is 8000 kg (dual wheel load) Percentage of load transfer
= 40 % i
Permissible bearing stress in concrete is calculated as under: i
(10.16b)fck b
i
9.525
f0k = characteristic compressive strength of concrete cube (15 cm) after 28 days curing concrete
j I
= 566.3 kg/cm2 Fb = (10'16~g225) 566~ = 413.80kg/cm2

Assumed spacing between the dowel bars
= 23.5 cm
First dowel bar is placed at a distance
= 15 cm from the pavement edge
255
APPENDIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND NATIONAL HIGHWAY DEVELOPMENT Assumed length of the dowel bar
= 50 cm
Dowel bars upto a distance of 1.0 x radius of relative stiffness, from the point of load application are effective in load transfer. Number of dowel bars participating in load transfer when wheel load is just over the dowel bar close to the edge of the slab =1+1/spacing = 1+ 55.87 / 23.5 = 3 dowels. I
Assuming that the load transferred by the first dowel is Pt and assuming that the load on dowel bar at a distance of 1 from the first dowel to be zero, the total transferred by dowel bar system
[
= (1+^ + i=r+L7^)Pt = 1.48Pt
;
T
,
. .,
x
,
..
Design load (P)x Percentage of load transfer
Load earned by the outer dowel bar, Pt = ;— ;= (8000x0.4)/1.48
=2167.39 kg
j
Check for Bearing Stress: Ttb^
Moment of Inertia of Dowel, I = — T
JT X 3.2*
ci.n
I = = 5.147 cm
4
64
Relative stiffness of dowel bar embedded in concrete (P): Where:
P=
41500 X 3.2 4 X 2 X 10s X 5.147
= 0.24 I
Bearing stress in dowel bar = (Pt x k) x (2+pz)'/ (4p3EI) = 399.80 kg/cm2 which is less than 413.80 kg/cm2 Hence, the dowel bar spacing and diameter assumed are safe.
256
APPENDIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND NATIONAL HIGHWAY DEVELOPMENT
Design of Tie Bars: Table AD6 Design Parameters for Tie Bars Slab thickness (h)
i
= 19 cm
Lane width (b)
= 3.5 m
Coefficient of friction (f)
= 1.5
Density of concrete (W)
;
= 2400 kg/mJ
Allowable tensile stress in plain bars (S), (As per IRC: 212000)
= 1250 kg/m2
Allowable tensile stress in deformed bars (S), (As per IRC: 212000)
= 2000 kg/m2
Allowable bond stress in plain tie bars (B)
= 17.5 kg/m2
Allowable bond stress in deformed tie bars (B)
= 24.6 kg/m2
Diameter of tie bar (d)
= 12 mm
1. Spacing and length of the plain bar Area of steel bar per metre width ofjoint to resist the frictional force at slab bottom
As = y = 1.92 cm Assuming a diameter of tie bar of 12 mm, the cross sectional area Cross sectional area of tie bar (A) =
n dz
i
A=1.13 sq.cm. Perimeter of tie bar (P) = n d = rc x 1.2 = 3.77 cm Spacing of tie bars = A / As =59.02 cm Provide at a spacing of 59 cm c/c Length of tie bar (L) = ~~~~ Jti r
j
a
Cross sectional area of tie bar (A) = 1.13 sq.cm. Perimeter of Tie Bar (P) = 3.77 cm Length of tie bar (L) =
J3
= 42.86 cm b
j i
Increase length by 10 cm for loss of bond due to painting and another 5 cm for tolerance in placement. Therefore, the length is 42.86 + 10 + 5 = 57.86 cm, Say 58 cm
257
APPENDIXA DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND NATIONAL HIGHWAY DEVELOPMENT
2. Spacing and length of the deformed bar Area of steel bar per metre width ofjoint to resist the frictional force at slab bottom .
bfW
. ...
2
As = ^ = 1.20 cm
i
Assuming a diameter of tie bar of 12 mm, the cross sectional area Cross sectional area of tie bar (A) = A=1.13 sq.cm. Perimeter of Tie Bar (P) = it d = n x 1.2 = 3.77 cm Spacing of tie bars = A / As =94.44 cm Provide at a spacing of 94 cm c/c Length of tie bar (L) = 2*^pCross sectional area of tie bar (A) =1.13 sq.cm. Perimeter of Tie Bar (P) = 3.77 cm Length of tie bar (L) = ^^^ = 48.78 cm BXP
Increase length by 10 cm for loss of bond due to painting and another 5 cm for tolerance in placement. Therefore, the length is 48.78 + 10 + 5 = 63.78 cm, Say 64 cm
258