PP3-Design of CC Pavement

7/31/2019 PP3-Design of CC Pavement

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Design of Cement ConcretePavements for Rural Roads

BySri P. Ganeswara Rao,

M. Tech.,(Structures)

Chief Engineer, PR, Vig & QC


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Factors governing design

Wheel load Legal axle load in India 102 KN. Hence, the design wheel load

is 51 KN

For link roads, wheel load 30 KN (Where only agriculture tractorsand trailers and light commercial vehicles traffic only).

Tyre Pressure 0.7 Mpa for wheel load 51 KN 0.5 Mpa for wheel load 30 KN

Design Period: 20 years

Characteristics of the Sub Grade The strength of the sub grade strength is expressed in terms of

modulus of sub grade reaction k which is determined by platebearing test (k750 = 0.5k300(1)) or from CBR values using thefollowing table.


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Approximate k valuecorresponding to CBR values

Soaked CBR

%

2 3 4 5 7 10 15 20 50

k value in

N/mm2/

mmx10-3

21 28 35 42 48 5 62 69 140

Note: k value may be taken 20% more than the k value of sub grade when proper sub base is provided


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Sub-base:

The provision of sub base below theconcrete has the following advantages. It provides a uniform and reasonable

firm support.

It prevents mud pumping on sub gradeof clays and silts.

It acts as levelling course on distortednon uniform and undulating sub grade.

It acts as a capillary cut off.


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Sub base materials:The following materials may be used as sub base

materials. W B M. Granular Sub base or gravel with CBR more than

20%.

Soil cement or soil lime with CBR more than 20%.

Minimum sub base thickness 150mm for wheel load of 51 KN 75mm for wheel load of 30 KNNote: i) Sub base surface shall be finished smooth.

ii) A polythene sheet of 125 microns thickness shallbe provided over the sub base to act as aseparation layer between the sub base andconcrete slab.


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Concrete Strength:

The design is based on flexural strength of the concrete

ft

= 0.7 fc

(2)where ft = Flexural Strength, N/mm2

fc = Characteristic compressive cube strength, N/mm2

Note: i) If the flexural strength observed from laboratory tests is higherthan that given by the above formula, the same may be used.

ii) The 90 day strength be used for design instead of the 28 day

strength (the 90 day flexural strength may be taken as 1.2times the 28 day flexural strength or as determined fromlaboratory tests).

iii) Heavy traffic should not be allowed for 90 days.

iv) For pavement construction, the characteristic 28 daycompressive strength should be atleast 30 Mpa and the

characteristic 28 day flexural strength shall be atleast 3.8 Mpa.


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Design of Slab Thickness:

Critical Stress ConditionThe slab is subjected to mainly two types of stresses.

(1) Due to wheel load

(2) Due to temperature maximum differential during day betweentop and bottom of the slab

The total stress = stress due to wheel load + stress due to temperaturedifferential

The bottom edge of the slab is subject to maximum tensile stressdue to wheel load and due to temperature differential. Hence, it is acritical section.

The top corners of the slab are subject to maximum tensile stressdue to wheel load. But due to temperature differential the stress iszero at corner. Hence the corners of slab are to be checked forstress due to wheel load.


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Calculation of Stresses:The stresses at edges or at corners can be computed using formulae or charts as follows:

Edge Stresses:

(a) Due to Load: The load stress in the critical edge region may be obtained as per Westergaard

analysis as modified by Teller and Sutherland from the following correlation (metric units).le = 0.529 P (1+0.54) (4log10 + log10 b 0.4048) (3)h2 b

where le = Load stress in the edge region, MPaP = Design Wheel load, N

h = Pavement slab thickness, mm

= Poissons ratio for concreteE = Modulus of elasticity for concrete, MPa

k = Modulus of sub grade reaction of the pavement foundation,N/mm3x10-3

= radius of relative stiffness, mm= 4 Eh3 (4)

12(1-2)kb = radius of equivalent distribution of pressure

= a for a 1.724

h= (1.6a2 + h2) 0.675h for a < 1.724 (5)h

and a = radius of load contact assumed circular, mm.

= ( P ) where p is tyre pressure

p


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(b)Due to Temperature: The temperature stress at the critical edgeregion may be obtained as per Westergaards analysis, usingBradburys coefficient from the following correlation:

te = Et C (6)

2where, te = temperature stress in the edge region, MPa

t = maximum temperature differential during daybetween top and bottom of the slab, oC

= coefficient of thermal expansion of concrete, /oC

C = Bradburys coefficient, which can be ascertaineddirectly from Bradburys chart against values of L/and W/

L = slab length or spacing between consecutivecontraction joints, m

W = slab width , m

= radius of relative stiffness, m


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Corner Stresses:

As per Westergaards analysis as modified

by Kelley

c

= 3P ( 1 ( a2 )1.2 ) (7)

h2

c = load stress in the corner region


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Design ChartsCHART FOR DETERMINATION OF COEFFICIENT C

L/l or C L/l or C

W/l W/l

1 0 7 1.03

2 0.04 8 1.077

3 0.175 9 1.08

4 0.44 10 1.075

5 0.72 11 1.05

6 0.92 12 &

above

1


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Recommended Temperature Differentials for ConcreteSlabs

Zone States Temperature DifferentialoC in Slabs of thickness

150mm 200mm 300mm

I Punjab, U.P., Uttaranchal, Rajasthan, Haryana,

Gujarat and North M.P., excluding hilly regions

12.5 13.1 14.3

II Bihar, Jharkand, West Bengal, Assam and EasternOrissa excluding hilly regions and coastal areas.

15.6 16.4 16.6

III Maharashtra, Karnataka, South M.P., Chattisgarh,

Andhra Pradesh, Western Orissa and North Tamil

Nadu excluding hilly regions and coastal areas.

17.3 19 20.3

IV Kerala and South Tamil Nadu excluding hilly regions

and coastal areas.

15 16.4 17.6

V Coastal areas bounded by hills. 14.6 15.8 16.2

VI Coastal areas unbounded by hills. 15.5 17 19


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CONCRETE PAVEMENT THICKNESS FOR RURALROADS

28-Day Concrete

Strength

(Compressive)

(MPa)

Pavement Thickness (mm)

Low Traffic Heavy Traffic

(Wheel load-30 kN) (Wheel load-51 kN)

Zone-

I

Zone-II,

IV, V, VI

Zone-

III

Zone

-I

Zone-II,

IV, V, VI

Zone-

III

Temperature differential,0C

15.0

0

15.1o

to

17.0o

17.1o

to

20.0o


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DesignProcedure

1. Select design wheel load, concrete flexural strength, modulus of sub gradereaction, modulus of elasticity of concrete, Poissons ratio, coefficient ofthermal expansion of concrete.

2. Decide Joint Spacing and lane width.

3. Select tentative design thickness of slab, based on defined design load, kvalue/CBR and flexural strength of concrete.

4. Ascertain maximum temperature stress for the critical edge region fromEquation (6).

5. Calculate the residual available strength of concrete for supporting trafficloads.

6. Ascertain edge load stress from Equation (3) or Fig. 2 or Fig.4 as relevantand calculate the factor of safety.

7. In case the available factor of safety is less than or far in excess of 1, adjustthe tentative slab thickness and repeat steps 3 to 6 till the factor of safety is1 or slightly more.

8. Check the adequacy of thickness in the corner region by ascertaining cornerload stress from Equation (7) or fig. 3 or Fig. 5 as relevant and readjust thethickness if inadequate.


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Example:

A cement concrete pavement is to be designed for arural road in A.P. having a traffic volume of 150 vehiclesper day consisting vehicles like agriculturaltractors/trailers, light good vehicles, heavy trucks, buses,animal drawn vehicles, motorized two wheels andcycles. Design the pavement. The soil has a soakedCBR value of 4.

Design:Wheel load = 51 KN

k value corresponding to CBR value of 4 is 35x10-3N/mm2/mm (from table).

Sub base

Provide 150mm granular sub base

Effective k value = 1.2x35x10-3 = 42x10-3 N/mm2/mm(since required sub base is provided)


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Concrete Strength:

Adopt M30 i.e., 28 day compressive strength

of 30 MPa.Flexural strength = 0.7 fc = 0.7 30Therefore, 28 day Flexural Strength=3.834

MPaThickness

Try a thickness of 150mm

Edge load stressFrom Fig. 4, for k= 42x10-3 N/mm3 andthickness 150mmm

Edge load stress = 4.5 MPa


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Temperature Stress:From the table, the temperature differential for AP for a slab thickness of 150mm is 17.3 0C.

Assuming a contraction joint spacing of 3.75m and 3.75m width, the radius of relative stiffness is as under.

L = 3750mm

B = 3750mm = radius of relative stiffness

= 4 Eh312(1-2)k

E = 3x104 N/mm2

h = 150mm

= 0.15

k = 42x10-3 N/mm3

= 4 3x104x1503x103

12 (1- 0.152)x42

= 673.3 mm

L/ = 3750/673.3 = 5.57W/ = 3750/673.3 = 5.57Both values are same.

For L/ = 5.57 from table, Bradburys coefficient C = 0.834

Using chart at Fig. 1, for C = 0.834 & temp. differential 17.30

C, the temp. stress te = 2.1 MPaTotal Stress = Edge load stress +Temperature stress = 4.5 + 2.1 = 6.6 MPaThis is greater than the allowable flexural strength of 4.6MPa.

So thickness of 150mm assumed is inadequate.

Try a thickness of 190mm.


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Edge load stress

From Fig.4, k= 42x10-3 N/mm3 and thickness = 200mm, the edge load stress=2.9MPa.

Temperature stress:

From Table, the temperature differential for AP for a slab thickness of 190mm is

18.70C.Radius of relative stiffness = 4 Eh3

12(1-2)k

= 4 3x104x1903x103

12 (1- 0.152)x42

= 804 mm

L/ = 3750/804 = 4.66W/ = 3750/8045 = 4.66Both values are same.

For L/ = 4.66 from tables Bradburys coefficient C = 0.625Using chart at Fig.1, for C=0.625 and temperature difference 18.70C, temperaturestress te = 1.6MPa

The total stress is less than 4.6MPa, hence, the assumed thickness of 190mm is OK.Corner Stress:

From Fig. 5, Corner load stress for wheel load of 51kN, K=42x10-3 N/mm2 and slabthickness of 190mm, Corner Stress c = 3.4 MPaThe corner stress is less than 4.6 MPa, hence, the thickness of 190mm assumed issafe.


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PP3-Design of CC Pavement