BITUMINOUS MATERIALS
WORKED EXAMPLE
10.2 Calculate the combined load and temperature stress for the longitudinal end condition and corner region on the basis of the following data :
1. Design wheel load = 5100 kg.
2. Thickness of pavement = 20 cm
3. Modulus of sub-grade reaction = 15 kg/cm3
4. Modulus of Electricity of concrete = 3.0 ×105kg/cm2 5. Poissons ratio of concrete = 0.15
6. Equivalent radius of contact area = 15 cm 7. Slab-dimensions = 4.5m ×3.5 m
8. Thermal coefficient of concrete ( ) = 10×10-6/oc.
9. Temperature difference = 180C.
10. Safe flexural strength of concrete mix = 38.5 kg/cm2
SOLUTION :
Since a/h = 15/20 = 0.75 is less than 1.724
Radius of resisting section = b = 1 6. a2 +h2 0 675. h
= 1 6. ×152+202 0 675 20. ×
= 14.0 cm Stress developed due to edge loading
Se= 0 572.h2 p
[
4 log10( )
l b/ +0 359.]
Radius of relative stiffness =1=
(
Eh)
k Warping stress along the longitudinal edge is given bySce = CxE t/2
small, since the slab is a short slab and hence neglected. Combined stress = load stress + warping stress
= 24.0 + 27.54 - 0.81 = 50.73 kg/cm2 Corner region Sc= 3P/h2 1
(
a 2 /l)
0 6.= 3×5100/202 1
(
15 2 60 8/ .)
0 6. =28 0. Kg cm / 2Max. warping stress = E t/3(1-µ) a /l
=
[ (
3 10× 5×10 10× 6×18 3 1 015)
/(
.) ]
15/60 8 9 15. = . kg cm / 2Frictional stress = Zero at corner region
Combined stress = Load stress + Warping stress
= 28.0+9.15 = 37.15 kg/cm2 10.5. I.R.C. RECOMMENDATIONS : 10.5.1. DESIGN PARAMETERS :
The following design parameters may be assumed.
1. Wheel load = 5100 kg; equivalent circular are of 15 cm 2. Tyre pressure = 6.3 to 7.3 kg/cm2
3. Traffic value may be projected to for 20 years. Based on traffic intensity corrections have to be carried out to the designed thickness. These corrections are given in IRC : 58 - 1974, Guidelines for Design of Rigid pavements.
4. Temperature differentials for calculating the warping stresses have been recommended based on thickness of slab end region and are available in IRC: 58-1974.
5. Modulus of sub-grade reaction K is to be determined using standard plate of 75cm diameter at 0.125 cm deflection. The minimum K - value of 5.5. kg/cm3 is specified for laying cement concrete pavement. In clayey sub-grades a suitable sub-base course may be provided to increase the K -value.
6. The flexural strength of concrete used in the pavement should not be less than 40 kg/cm2. µ= 0.15 = 10 ×10-5 per 0C.
E = 3 ×10+5 kg/cm2
10.5.2. CALCULATION OF STRESSES :
1. The wheel load stresses at the edge and corner regions are calculated using the equations (10.8) and (10.9), respectively.
2. Temperature stresses are calculated using the equations (10.11) and (10.13).
3. The critical combination of stresses in summer is obtained and the flexural stress so obtained should be less than 40 kg/cm2for the designed thickness of slab.
4. The design thickness is adjusted for traffic intensity as explained in 10.4.1. above.
10.5.3. SPACING OF JOINTS :
The following maximum spacing are recommended.
Nature of
Foundation is rough for slabs of all thickness for 25mm wide expansion joint
For smooth foundation surfaces of slabs constructed in summer for slab thickness upto 20 cm.
For slabs up to thickness of 25 cm.
When the construction is carried out in winter For unreinforced slabs of all thickness
10.6. JOINTS IN CONCRETE PAVEMENTS :
Joints may be broadly divided into transverse and longitudinal joints. Transverse joints may be conveniently classified into four groups - expansion joints, contraction joints, warping joints, and construction joints. Longitudinal joints are required in concrete roads more than 4.5 m wide to allow for transverse warping and to allow for uneven settlement of the sub-grade.
10.6.1. SPACING OF EXPANSION JOINTS :
The width of the gap in the expansion joint depends upon the length of slab.
Expansion joint spacing is designed based on the maximum temperature variations expected and width of the joint. Dowel bars are provided at expansion joints for load transfer form one slab to the other. It is recommended not to have a gap more than 2.5 cm for an expansion joint. If $’ is the maximum expansion in a slab of length Lewith a temperature rise from t1to t2in degrees centigrade, then the spacing of the expansion joints is given by
Le(metres) =
( )
$1
100 C t2 t1 10.17
where C is the coefficient of expansion of concrete $’ expansion of the slab in cm. It is assumed that the joint filler may be compressed upto 50% of its thickness and therefore the expansion joint gap should be twice the allowable expansion in concrete (i.e.,). 2 $’.
Fig 10.3. TYPICAL EXPANSION JOINT 10.6.2. SPACING OF CONTRACTION JOINTS :
The slab contracts due to the fall of slab temperature below the construction temperature. This movement is resisted by the sub-grade drag or friction between the bottom fiber of the slab and the sub-grade. Length of slab to resist the frictional drag, that is spacing of contraction joints,
LC= 2
104 Sc
Wf × 10.18
Here
LC= spacing between the contraction joints, m.
f = Coefficient of friction (1.5)
W = Unit weight of slab, kg/m3(2400 kg/m3)
Sc = Allowable stress in tension in cement concrete.
10.6.3. WARPING JOINTS :
If expansion and contraction joints are properly designed and constructed there is no need of providing warping joints in addition.
Construction aspects of these joints are discussed n Unit No.12.
10.7. SELF - ASSESSMENT QUESTIONS :
1) Find out the spacing of expansion and contraction joints given the following data.
Expansion gap. 2.5 cm.
Laying temperature of concrete = 100C Slab temperature in summer = 540C.
Coefficient of thermal expansion of concrete = 10 ×10-6 /0C.
Coefficient of friction = 1.5
Ultimate tensile stress in concrete = 1.6 kg/cm2 Factor of safety = 2.
NOTE : $1in the formula = 2.5 / 2 = 1.25 cm.
(Ans : Spacing of Expansion joint = 28.5m, spacing of contraction joint = 4.44m).
2) Explain the terms (a) Modulus of sub-grade reaction (b) Radius of relative stiffness and (c) radius of resisting section.
3) Distinguish between warping stresses and frictional stresses.
10.8. SUMMARY :
Stresses are set up in concrete road slabs by wheel loading and by changes in temperature. The wheel load stresses are calculated for three critical load positions, namely, interior, edge and corner positions of the road slab, based on Westergaurds theory. The equations given by Estergaurd require considerable amount of trails in their solution, if the slab thickness has to be determined. Bradbury suggested a simplified procedure in terms of stress coefficients. The IRC considers the edge and corner loading positions only and recommends the use of Westergaurds edge load formula modified by Teller and Sutherland and the Westergaurd corner load formula modified by Kelley for calculating these stresses.
Temperature tends to produce two types of stresses in a concrete road slab. They are warping stresses and frictional stresses. Whenever the top and bottom surfaces of a concrete pavement simultaneously possess different temperatures, the slabs tends to warp downwards or upwards inducing warping stresses. These warping stresses are calculated by the equations developed by Bradbury. The increase or decrease of pavement average temperature causes expansion or contraction of pavement slab. When these movements are restrained frictional stresses are developed in the slabs. For designing the pavement slab critical combination of these stresses have to be obtained. For Indian conditions, critical combination of tresses take place on a summer mid-day at the edges.
Joints are provided in concrete roads to allow for the movements of the slab due to changes in temperature and moisture content, but the number of joints should be a minimum.
This lesson discusses spacing of expansion and contraction joints.
10.9. REFERENCES :
1. Bindra, S.P. (1977) - A Course in High Way Engineering - Dhanpathy Rai and Sons, Delhi.
2. Khanna, Dr. S.K. and Justo, Dr. C.E.G. (1991) - Highway Engineering - Nem Chand and Bros, Roorkee.
3. - (1955) Concrete Roads, H.M.S.O. Publication.
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TRANSPORTATION ENGINEERING