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APPENDIX-A DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD ANDNATIONAL HIGHWAY DEVELOPMENT

1i

APPENDIX-A[

DESIGN OF A RIGID PAVEMENTi

RURAL ROADi

ANDij

NATIONAL HIGHWAY!i

DEVELOPMENT

236

APPEND! X-A DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD ANDNATIONAL HIGHWAY DEVELOPMENT

DESIGN OF A CEMENT CONCRETE PAVEMENT FOR RURAL ROAD

(IRC: SP: 20-2002 / IRC: SP: 62-2004)

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 two-wheels and cycles. The soil has a soaked CBR value of 2%, 4% and 6% for

30kN wheel load.

Table AA-1

Design of CC Pavement for Rural Roads

Design Parameters: Sample D1 (6% CBR-30kN 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

Calculation)= 28,417 N/mm2

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


[ ' ■

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 ji

W/l = 3750 / 623.79= 6.0 Ij

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^At- c

Hence, ote = 1.63 N/mmI 2. |

|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, whichI 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]

I1

238


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, <rlc = 3.20 N/mm2, which is less than the allowable flexural strength of 6.10

N/mm2.

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.

239


DESIGN OF A CEMENT CONCRETE PAVEMENT FOR RURAL ROAD

(IRC: SP: 20-2002 / IRC: SP: 62-2004)

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 two-wheels and cycles. The soil has a soaked CBR value of 2%, 4% and 6% for

5 lkN wheel load.

Table AB-1

Design of CC Pavement for Rural Roads

Design Parameters: Sample D1 (6% CBR-51kN 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

Calculation)= 28,417 N/mm2


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


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= 4i 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 = C

jHence, ote= 1.63 N/ram2.

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); j(b) = Vl.6 a2 + h2 - 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

jTotal 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

APPENDIX-A 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.

iSo, 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


DESIGN OF A NATIONAL HIGHWAY ROAD PAVEMENT (IRC: 58-2002)

A cement concrete pavement is to be designed for a two- lane two-way National Highway

in Gujarat State. The total two-way traffic is 3000 commercial vehicles per day (evpd) at

the end of the construction period. Design parameters are provided in Table AC-1 and the

traffic axle load spectrum is given Table AC-2.

TABLE AC-1

Design of CC Pavement for Two- Lane Two-Way National Highway

Design parameters: Sample G5 (6% CBR-100DLC-3000CVPD)

Present Traffic =3000 cvpd

Design life =20 yrs.

Compressive Strength (fck) = 56.63 N/mm2 = 566.3 kg/cm2

Flexural strength of cement concrete

(Modulus of rupture)

= 8.15 N/mm2 = 81.50 kg/cm2

CBR = 6%

Dry Lean Concrete (DLC) =100 mm

Effective modulus of subgrade reaction of the DLC

sub-base (k)

= 18.7 kg/cm3

Elastic modulus of concrete (E) = 40417 N/mm2


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


The axle load spectrum obtained from axle load survey is given in the following:

Table AC-2

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

19-21 0.6 34-38 0.3

17-19 1.5 30-34 0.3

15-17 4.8 26-30 0.6

13-15 10.8 22-26 1.8

11-13 22.0 18-22 1.5

9-11 23.3 14-18 0.5

Less than 9 30.0 Less than 14 2.0

Total 93.0 Total 7.0

365 X { (1 + r)2 — 1}Cumulative repetition in 20 yrs. = 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


Table AC-3

Total Repetitions of the Single Axle and Tandem Axle Loads

Single Axle Tandem AxleLoad in tonnes Expected

repetitionsLoad in tonnes Expected

repetitions20 71127 36 3556418 177820 i 32 3556416 569023 : 28 7112814 1280303 24 21338412 2608024 1 20 17782010 27622135 16 59273

Less than 10 3556397 Less than 16 237093

Trial Thickness = 19 cmTable AC-4

Cumulative Fatigue Life

Axle load (AL), tonnes

AL x 1.2

Stress, kg/cm2

from charts

Stressratio

Expected Repetitions, n

Fatigue life, N

Fatigue life consumed

(1) (2) (3) (4) (5) (6) Ratio (5)/(6)

Single axle20 24.0 44.80 0.55 71128 12.38 x 104 0.5718 21.6 41.00 0.50 177820 64.30 x 104 0.2816 19.2 37.40 0.46 569024 15.48 x 106 0.0414 16.8 35.63 0.44 1280303 47.22 x 10* 0.00

Tandem axle36 43.2 31.68 0.39 35564 Infinity 0.00

Cumulative fatigue life consumed 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


(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: 58-2002)

The temperature differential was taken as 12.98°C for the Gujarat region.

Eat „■ ...... ^ i ry

Edge warping stress = 2 =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 Eh312(1 -ii2)k 59.62 cm

a = radius of area of contact of wheel.

iConsidering a single axle dual wheel,

a=(-)1/2

V pn /i

a = 0.8521 x —-—h -(—l—)0'5 a = 26.52 cmq X n n\ 0.5227 xq )

Comer stress S0= ^ |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


Design of Dowel Bars

Table AC-5

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

Dowel/Concrete interaction (E)

= 2.0 X 106 kg/cm2

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 %

Permissible bearing stress in concrete is calculated as under:

„ _ (l0.16-6)/cfcFb—:

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

APPENBIX-A DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD ANDNATIONAL HIGHWAY DEVELOPMENT

»a—b———iL-i.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„___ 41------ ---------1.92 cm64

Relative stiffness of dowel bar embedded in concrete (P) =4 /~Kb~ y 4 El

Where;

P41500 X 2.5 0.29-y4X2X 10s X 1.92

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 AC-6

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: 21-2000) = 1250 kg/m2

248


Allowable tensile stress in deformed bars (S), (As per IRC; 21-2000) = 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 of joint to resist the frictional force at slab bottom

As = ~ = 1.92 cm2

Assuming a diameter of tie bar of 12 mm, the cross sectional area7t 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) = = 42.86 cmBXP

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 of joint to resist the frictional force at slab bottom. bfW . 2As = —= 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


249



Length of tie bar (L) = j

Cross sectional area of tie bar (A) =1.13 sq.cml


Length of tie bar (L) = = 48.78 cmB X P


placement. Therefore, the length is 48.78 + 10 + 5 = 63.78 cm, Say 64 cm

250


DESIGN OF A NATIONAL HIGHWAY ROAD PAVEMENT (IRC: 58-2002)

A cement concrete pavement is designed for a two- lane two-way National Highway in

Gujarat State. The total two-way traffic is 3000 commercial vehicles per day (cvpd) at the

end of the construction period. Design parameters are provided in Table AD-1 and traffic

axle load spectrum is given Table AD-2.

Table AD-1

Design of CC Pavement for Two- Lane Two-Way National Highway

Design parameters: Sample G5 (6% CBR-150DLC-3000CVPD)

Present Traffic =3000 cvpd

Design life =20 yrs.

Compressive Strength (fck) = 56.63N/mm2 = 566.3 kg/cm2

Flexural strength of cement concrete

(Modulus of rupture)

= 8.15 N/mm2 = 81.5 kg/cm2

CBR = 6%

Dry Lean Concrete (DLC) =150 mm

Effective modulus of subgrade reaction of the DLC

sub-base (k)

= 24.25 kg/cm3

Elastic modulus of concrete (E) = 40417 kg/cm2


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


The axle load spectrum obtained from axle load survey is given in the following:

Table AD-2

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

19-21 0.6 34-38 0.3

17-19 1.5 30-34 0.3

15-17 4.8 26-30 .0.6

13-15 10.8 22-26 1.8

11-13 22.0 18-22 1.5

9-11 23.3 14-18 0.5

Less than 9 30.0 Less than 14 2.0

Total 93.0 Total 7.0

365 Xi4 { (1 + r)2 — 1}Cumulative repetition in 20 yrs. = 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 mid-point

of the axle load class represents the group, the total repetitions of the single axle and

tandem axle loads are as follows:

252

APPENDIX-A DESIGN OF A RIGID PAVEMENT FOR RURAL ROAD AND '_______________________ NATIONAL HIGHWAY DEVELOPMENT

Table AD-3

Total Repetitions of the Single Axle and Tandem Axle Loads

Single Axle Tandem Axle

Load in tonnes Expected

repetitions

Load in tonnes Expected

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 AD-4

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)

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

Tandem axle

36 43.2 30.24 0.37 ; 35564 Infinity 0.00

Cumulative fatigue liiFe consumed 0.28

253


ifcl iiihiliaMaWiMWHW—I—————

The cumulative fatigue life consumed being less than 1; the design is safe from fatigue

considerations.

Check for Temperature Stresses:

EatEdge warping stress (Ste) = 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: 58-2002)

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


a = 0.8521 x ——Iqxn

S / P n V 0.5227 xq

a = 26.52 cm '

Comer stress Sc= ^ j = 25.29 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.

Design of Dowel Bars

Table AD-5

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

Dowel/Concrete interaction (E)i

= 2.0 x 106 kg/cm2

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

(10.16-b)fckb 9.525 i

f0k = characteristic compressive strength of concrete cube (15 cm) after 28 days curing

concrete jI

= 566.3 kg/cm2

Fb = (10'16~g225) 56-6~ = 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


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 transferLoad earned by the outer dowel bar, Pt =------- ;---------------—----- ;-----------------

= (8000x0.4)/1.48 =2167.39 kg jCheck for Bearing Stress:

Ttb^Moment of Inertia of Dowel, I = —

T JT X 3.2* c-i.n 4I =-------- = 5.147 cm

64

Relative stiffness of dowel bar embedded in concrete (P):

Where:

P =41500 X 3.2 = 0.244 X 2 X 10s X 5.147

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


Design of Tie Bars:

Table AD-6

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: 21-2000) = 1250 kg/m2

Allowable tensile stress in deformed bars (S), (As per IRC: 21-2000) = 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 of joint 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 arean dz i


A=1.13 sq.cm.

Perimeter of tie bar (P) = n d = rc x 1.2 = 3.77 cm



Length of tie bar (L) = ~~~~ jJti a r

Cross sectional area of tie bar (A) = 1.13 sq.cm.

Perimeter of Tie Bar (P) = 3.77 cm j

Length of tie bar (L) = = 42.86 cmJ3 b i



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2. Spacing and length of the deformed bar

Area of steel bar per metre width of joint to resist the frictional force at slab bottom. bfW . ... 2As = -^- = 1.20 cm i

Assuming a diameter of tie bar of 12 mm, the cross sectional area


A=1.13 sq.cm.

Perimeter of Tie Bar (P) = it d = n x 1.2 = 3.77 cm



Length of tie bar (L) = 2*^p-

Cross sectional area of tie bar (A) =1.13 sq.cm.


Length of tie bar (L) = ^^^ = 48.78 cmB X P



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