A Rigid pavement is a pavement consisting of a relatively thin slab of Portland cement concrete overlying a sub-grade or base-course or a regulating layer.
Unlike flexible pavements, rigid pavements are not considered as a multi-layer structure because the high stiffness of the slab helps the slab to distribute load over a relatively wide area of the pavement by beam action.
This results in the stresses that are imposed on the pavement surface being reduced to a level acceptable to the sub-grade. Thus in rigid pavements, a major portion of the structural capacity of the pavement is contributed by the slab itself.
Rigid pavements may be designed as one of the following types:
- Jointed plain concrete (JPCP), i.e, no steel reinforcement.
- Jointed reinforced concrete (JRCP) and
- Continuous reinforced concrete (CRCP)
Jointed Plain Concrete Pavements (JPCP) do not contain any steel reinforcement but contain enough joints to control the location of all the expected natural cracks. However, smooth steel bars are placed at transverse joints and deformed steel bars at longitudinal joints.
The concrete cracks at the joints and not elsewhere in the slabs. The spacing between transverse joints is typically about 4.6m.
Jointed Reinforced Concrete Pavements (JRCP) contain steel mesh reinforcement (sometimes called distributed steel), purposely to increase joint spacing and to hold together intermediate cracks in the slab. The spacing between transverse joints is 9m or more.
Continuously Reinforced Concrete Pavements (CRCP) have continuous longitudinal steel reinforcement and no intermediate transverse expansion or contraction joints other than the construction joints.
Such pavements are allowed to crack in a random transverse cracking pattern with the cracks held tightly together by the continuous steel reinforcement. Transverse cracks are expected in the slab, usually at intervals of 1-1.6m. CRCPs are designed with enough steel, about 0.6-0.7% by cross-sectional area so that cracks are held together tightly.
Slab thickness for rigid pavements generally varies from 200mm to 300mm and is not influenced by the presence or absence of steel reinforcement.
Rigid pavements have the following advantages over asphalt pavements:
- Rigid pavements are generally stronger
- They are more durable
However, they have the following disadvantages:
- They are rougher to drive on
- They have a higher initial cost
- They crack continuously
- They are more difficult to repair in a satisfactory manner.
Causes of stresses in rigid pavements
Rigid pavements may come under stresses induced by a variety of causes including the following:
- Cyclic changes in temperature
- Changes in moisture
- Wheel loads
- Volumetric changes in sub-grade or base course
The above changes tend to deform the rigid slab and in the process cause a variety of stresses to be developed. Also, the continuity of sub-grade support is very important in affecting the intensity of stresses in rigid pavements.
Where sub-grade support has been lost by pumping and/or plastic (permanent) deformation of the sub-grade, high-intensity stresses may develop and cause local rapture of the rigid slab. In principle, rigid pavements resist loads by bending but when badly cracked, they lose this ability and then tend to behave somewhat similar to flexible pavements.
Types of stresses in Rigid pavements
- Curling (warping) stresses
When a temperature gradient exists through the depth of a rigid slab (i.e there is a difference in temperature between the top and the bottom of the slab) the slab will tend to curl or warp and take the shape depicted in the image below provided there is no restraint to the deformation. If however, curling is restrained by the weight of the slab which is always the case, curling stresses would develop.
Tt = temperature at the top of slab
Tb= temperature at the bottom of slab
Consider a finite slab of dimensions shown in Fig. 2.3 in which curling stresses have developed.
Taking X as the longitudinal direction and Y the transverse direction for the slab, the curling stresses in the two directions are given by the following equations:
σx = EαΔT/2(1-μ2)[Cx+μCy]
σy = EαΔT/2(1-μ2)[Cy+μCx]
E = modulus of elasticity of concrete
ΔT = temperature difference between surface and bottom of slab
α = coefficient of thermal expansion
μ = Poisson’s ratio
Cx Cy are correction factors (see image below) dependent on Lx/l or Ly/l ratio where l is the radius of relative stiffness of the slab given by the expression
k= modulus of sub-grade reaction
h= slab thickness
Curling stresses at mid-span of the slab edge are obtained by putting μ = 0 in the corresponding stress equation, i.eσ
σedge(y) = (EαΔT/2)Cy
σedge(x) = (EαΔT/2)Cx
Curling stresses at the corner of the slab is obtained as
σcorner = (EαΔT/3√1-μ2)(√(a/l))
The modulus of sub-grade reaction is the parameter that measures the supporting strength of soils for rigid pavements. Its value is affected by density and moisture condition of the soil.
The parameter is evaluated in the filed through a plate loading test in which pressure is applied to the plate and the corresponding deflections measured.
The modulus is evaluated as the slope of the pressure (P)-deflection (Δ) curve which is approximated to a straight line, i.e
The modulus has units of kN/m3
2. Stresses from loading
Wheel load position determines the magnitude of stress developed from vehicle loading on rigid slabs. Wheel load position may be any one of the following ( See image below):
Westergaard provided empirical expressions for evaluating stresses relating to various load positions as follows:
a. Corner loading
σc = 3P/h2[1-(a√2/l)0.6]
σc = corner stress
a = radius of load contact area
P = load
b. Interior loading
σi = 0.316P/h2[4log10(l/b)+1.069]
σi = interior stress
b = a, when a ≥ 1.724h
b= √(1.6a2+h2) – 0.675h … when a < 1.724h
h= thickness of slab
c. Edge loading
σe = 0.572P/h2[4log10(l/b)+0.359]
σe = edge stress
3. Friction stresses
Stresses can develop in rigid pavements as a result of moisture or temperature changes that cause the slab to contract or expand. If movement caused by contraction or expansion is restrained by the friction between the underside of the slab and the underlying material, friction stresses will develop which in long slabs may be sufficient to cause cracks to develop.
Stresses due to friction are evaluated as
σc = γcLfa/2
σc = friction stress in concerete
γc = unit weight of concrete
L = Length of slab
fa = friction coefficient between the underside of slab and underlying material (fa=1.5)
The equation above indicates that the friction stress developed in the concrete is independent of the thickness of the slab but dependent on the length.
Therefore, longer slabs have a higher tendency to suffer cracks that shorter slabs from high friction stresses. This explains why the lengths of plain concrete rigid pavements are made as short as possible.
Use of steel in rigid pavements slabs
Steel is used in rigid pavements in three modes (see image below)
- transverse and longitudinal reinforcements
- dowel bars
- tie bars
These reinforcements are assumed not to increase the structural capacity of the slab but are used for the purpose of
- tying concrete together should it crack
- preventing faulting
- maintaining load transfer across cracks through aggregate interlock
- increasing joint spacing
Wire fabric or mat reinforcement may be used in rigid pavement slabs to control temperature cracks. It is assumed that all the tensile stresses from friction are taken by steel alone.
However, because of the amount of steel depends on the length of the slab, for very short slabs (6m or less) steel reinforcement may be omitted.
Typically, rigid pavement reinforcing steel constitutes about 0.6 – 0.7 percent of the pavement cross-sectional area and is placed at mid-depth or close to the slab top as possible while maintaining minimum concrete coverage over the reinforcement.
At least 63mm (2.5inches) of PCC cover should be maintained over the reinforcing steel to minimize the potential of steel corrosion by chlorides.
Longitudinal steel is used to control concrete volume changes due to temperature and moisture variations and to keep transverse cracks tightly closed.
Assuming that a crack will develop and resistance to movement will be overcome by tension in steel, the steel required per unit width of slab is given as
As = γchLfa/2fs
As= area of steel per unit width of slab
fs= allowable steel stress
L= the length of slab for longitudinal reinforcement
h= the thickness of slab
To get the total area of steel for the cross-section, multiply As by the width of the slab, i.e.,
Once the area of steel has been determined, the minimum number of bars required for a selected bar size and the spacing can be computed as,
n = At/Ab
s = W/n
At= total area of steel for longitudinal reinforcement
Ab= cross sectional area of steel bar selected
W= width of slab
n=number of bars
s= spacing of bars
For jointed reinforced concrete pavements, the steel is discontinued at the joint.
Transverse reinforcement reduces the risk of random longitudinal cracks opening up and thus reduces the potential of punch-outs.
They keep longitudinal cracks and joints closed. Many designers take the amount of steel required for transverse reinforcement as half that for longitudinal reinforcement. Transverse bar spacing should be no closer than 0.91m(36in) nor greater than 1.5m (60 in).
Tie bars are either deformed steel bars or connectors used to restrain the faces of abutting (adjacent) slabs in contact at longitudinal joints. The objective is to tie the slabs together so that the joint will be tightly closed. Although tie bars may provide some minimal amount of load transfer, they are not designed to act as such. The amount of steel required for tie bars is evaluated as
As = γchLwfa/fs
As= area of steel required per unit length of slab
Lw= lane width or distance from the longitudinal joint to the free edge where no tie bars exist.
For a given area of steel, it is advisable to choose the smallest size bars. For a given bar diameter (d), the length of tie bar is
tl = 1/2(fsd/μ)
tl= length of tie bar
d=diameter of bar
u= allowable bond strength
For simplicity of construction, it is standard practice for most agencies to use 12.5mm (1 1/2 inch) diameter bars that are between 0.6 and 1.0 m long and spaced about 0.9m on centers as tie bars.
Dowle bars are placed across transverse joints to allow load transfer from one slab to the adjoining slab without restricting horizontal joint movement.
Faulting or differential vertical displacement at joints and cracks between abutting slabs may result when there is load transfer across the crack or joint. (see image).
Dowels increase load transfer efficiency by allowing the leave slab to assume some of the load before the load is actually over it (see image). This reduces joint deflection and stress in the approach and leave slabs.
image.. Load transfer at transverse joint provided by dowel
When dowels are placed, one end of the bar is fixed and the other end into the adjoining slab is lubricated to allow for free longitudinal movements of the slabs relative to one another caused by expansion or contraction.
Lubrication of dowel bars may be achieved by any of the following means:
- Coating one end of the bar with a de-bonding material to prevent bonding to the PCC.
- Wrapping one of the bars in a sheath of polythene material.
- Encapsulating one end of the bar in PVC pipe.
Dowel bars are typically 32 to 38 mm (1.25 to 1.5 inches) in diameter, 460mm (18 inches) long, are to be placed at mid-depth of the slab (see image), and spaced 30cm (12in) on centers.
In plain concrete pavements, the length of the slab or joint spacing is dictated by the amount of joint opening or the shrinkage characteristics of the concrete rather than the stress induced by the shrinkage/expansion. The amount of joint opening may be estimated by the following formula
ΔL = CL(αΔT + ε)
ΔL = Joint opening
L = joint spacing or length of slab
ΔT =change in temperature between placement temperature and the lowest minimum daily temperature
α = coefficient of thermal expansion of concrete (8-12×10-6/C)
ε = drying shrinkage coefficient of concrete (1×10-4)
C = adjustment factor due to friction between slab and underlying material (0.6 for stabilized base, 0.8 for granular base)
Longer spacing causes the joint to open wider and decrease the efficiency of load transfer