Rigid Pavement Design Concepts

Rigid pavements are constructed using Portland cement concrete (PCC) slabs resting on a subbase or directly on the subgrade. Unlike flexible pavements, which distribute loads through a layered system, rigid pavements distribute traffic wheel loads over a wide area through the bending stiffness (flexural rigidity) of the concrete slab itself. On the PE Civil Transportation exam, rigid pavement design is based on the AASHTO 1993 rigid pavement design methodology.

1. Concrete Slab Load Distribution

The concrete slab has a very high modulus of elasticity ($E_c \approx 3 \cdot 10^6$ to $5 \cdot 10^6\text{ psi}$), which allows it to act as a structural beam on an elastic foundation. This causes wheel load stresses to be distributed widely, resulting in very low pressures reaching the underlying base and subgrade layers.

Wheel Load (P)Concrete Slab (PCC)Base / Subbase CourseSubgradeSlab Thickness (D)← Wide, low-pressure distribution

2. AASHTO 1993 Rigid Pavement Design Equation

The objective of rigid pavement design is to determine the required slab thickness ($D$, in inches). The AASHTO design equation is:

$$\log_{10}(W_{18}) = Z_R S_0 + 7.35 \log_{10}(D + 1) - 0.06 + \frac{\log_{10}\left(\frac{\Delta PSI}{4.5 - 1.5}\right)}{1.0 + \frac{1.624 \cdot 10^7}{(D + 1)^{8.46}}} + (4.22 - 0.32 p_t) \log_{10}\left[ \frac{S'_c C_d (D^{0.75} - 1.13)}{215.63 J \left( D^{0.75} - \frac{18.42}{(E_c / k)^{0.25}} \right)} \right]$$

Where:

  • $W_{18} = \text{design traffic demand}$ (total 18-kip Equivalent Single Axle Loads (ESALs) over the design life).
  • $Z_R = \text{standard normal deviate}$ corresponding to the design Reliability ($R$, %).
  • $S_0 = \text{overall standard deviation}$ (typically $0.30$ to $0.40$ for rigid pavements; $0.35$ is standard).
  • $D = \text{thickness of the concrete slab (in inches)}$.
  • $\Delta PSI = \text{design Serviceability Loss} = p_i - p_t$.
    • $p_i = \text{initial serviceability index}$ (typically $4.5$ for rigid pavements).
    • $p_t = \text{terminal serviceability index}$ (typically $2.5$ for major highways).
  • $S'_c = \text{concrete modulus of rupture}$ (flexural strength of concrete in psi).
  • $C_d = \text{drainage coefficient}$ (adjusts slab performance based on drainage quality).
  • $J = \text{load transfer coefficient}$ (accounts for the ability of joints or shoulders to transfer load).
  • $E_c = \text{modulus of elasticity of concrete (in psi)}$.
  • $k = \text{modulus of subgrade reaction (in psi/in or pci)}$.

3. Modulus of Subgrade Reaction ($k$)

The modulus of subgrade reaction ($k$) represents the stiffness of the soil foundation supporting the concrete slab. It is measured in the field using a Plate Bearing Test (using a $30\text{-inch}$ diameter steel plate):

$$k = \frac{p}{\delta}$$

Where:

  • $p = \text{contact pressure applied by the plate (psi)}$
  • $\delta = \text{deflection of the plate (inches)}$
  • Units: $\text{psi/in}$ or $\text{lb/in}^3$ (often written as pci - pounds per cubic inch).

The design $k$-value is adjusted to account for the presence of a subbase layer, the depth to bedrock, and seasonal moisture variations. It is also corrected for the Loss of Support (LS), which is a factor ($0.0$ to $3.0$) that accounts for the potential erosion of subbase material beneath the slab.

4. Concrete Material Properties

Modulus of Rupture ($S'_c$)

Concrete is weak in tension, and rigid pavements fail primarily through flexural fatigue cracking at the bottom of the slab. Thus, the design strength is characterized by the flexural strength or modulus of rupture ($S'_c$), measured using a third-point beam loading test (ASTM C78).

$$S'_c \approx 8 \text{ to } 10 \sqrt{f'_c} \quad \text{(in psi)}$$

Where $f'_c$ is the compressive strength of the concrete in psi.

Modulus of Elasticity ($E_c$)

The concrete modulus of elasticity is calculated from its compressive strength:

$$E_c = 57,000 \sqrt{f'_c} \quad \text{(in psi)}$$

5. Joint Design and Load Transfer

To control cracking caused by thermal expansion, contraction, and moisture gradients, rigid pavements are divided into individual slabs by joints.

TRANSVERSE JOINT(Dowel Bars)Slab 1Slab 2Dowel Bar (Smooth)LONGITUDINAL JOINT(Tie Bars)Lane 1Lane 2Tie Bar (Deformed)

Dowel Bars (Transverse Joints)

  • Purpose: Smooth round steel bars placed across transverse joints to transfer shear loads from one slab to the next.
  • Mechanism: They allow the slabs to expand and contract horizontally along the length of the bar (since one side is lubricated/capped), but prevent vertical differential movement (faulting).
  • Impact on Design: The use of dowel bars increases load transfer efficiency, which lowers the concrete stress and reduces the required slab thickness ($D$).

Tie Bars (Longitudinal Joints)

  • Purpose: Deformed steel rebars placed across longitudinal joints to hold adjacent traffic lanes or shoulders together.
  • Mechanism: Unlike dowel bars, tie bars are bonded to the concrete and do not allow joint opening or horizontal movement. They prevent lane separation.

Load Transfer Coefficient ($J$)

The $J$ parameter accounts for the presence of load transfer devices (dowel bars) and tied concrete shoulders:

  • Lower $J$ values (2.5 - 3.2): Indicate excellent load transfer (doweled transverse joints and/or tied concrete shoulders).
  • Higher $J$ values (3.8 - 4.5): Indicate poor load transfer (undoweled joints and aggregate interlock only, with asphalt shoulders).

6. Worked Examples

Worked Example 1: Concrete Properties

A rigid pavement is specified to be constructed using concrete with a $28\text{-day}$ compressive strength ($f'_c$) of $4500\text{ psi}$.

Goal: Estimate the concrete’s modulus of rupture ($S'_c$) using the NCEES standard relation ($S'_c = 9 \sqrt{f'_c}$) and its modulus of elasticity ($E_c$).

Solution:

  1. Calculate Modulus of Rupture ($S'_c$): $$S'_c = 9 \sqrt{f'_c} = 9 \sqrt{4500} = 9 \cdot 67.08 = 603.7\text{ psi}$$
  2. Calculate Modulus of Elasticity ($E_c$): $$E_c = 57,000 \sqrt{f'_c} = 57,000 \sqrt{4500} = 57,000 \cdot 67.08 = 3.82 \cdot 10^6\text{ psi}$$

Result: The modulus of rupture is $604\text{ psi}$ and the modulus of elasticity is $3.82 \cdot 10^6\text{ psi}$.

Worked Example 2: Modulus of Subgrade Reaction ($k$)

A plate bearing test is conducted on a prepared subgrade. A load of $14,130\text{ lb}$ is applied to a $30\text{-inch}$ diameter plate, resulting in a deflection of $0.05\text{ inches}$.

Goal: Calculate the modulus of subgrade reaction ($k$) of the subgrade in pci.

Solution:

  1. Calculate Plate Area ($A$): $$D = 30\text{ inches} \implies R = 15\text{ inches}$$ $$A = \pi R^2 = \pi (15)^2 = 225\pi \approx 706.86\text{ in}^2$$
  2. Calculate Applied Pressure ($p$): $$p = \frac{\text{Load}}{A} = \frac{14,130\text{ lb}}{706.86\text{ in}^2} = 19.99\text{ psi} \approx 20.0\text{ psi}$$
  3. Calculate Modulus of Subgrade Reaction ($k$): $$k = \frac{p}{\delta} = \frac{20.0\text{ psi}}{0.05\text{ inches}} = 400.0\text{ pci}$$

Result: The modulus of subgrade reaction is $400\text{ pci}$.

7. Exam Pitfalls and Tips

  • Dowel Bars vs. Tie Bars: Do not confuse these two joint reinforcements:
    • Dowel bars are smooth, placed at transverse joints, and transfer shear loads while allowing horizontal thermal expansion/contraction.
    • Tie bars are deformed, placed at longitudinal joints, and hold the lanes together (preventing horizontal movement).
  • Initial Serviceability ($p_i$): In the rigid pavement design formula, the default initial serviceability is $p_i = 4.5$. Do not use the flexible pavement default value ($4.2$) unless explicitly instructed.
  • Standard Deviation ($S_0$): Use $S_0 = 0.35$ (or within $0.30 - 0.40$) for rigid pavement problems. Ensure you do not use the flexible pavement value ($0.45$).
  • Units of $k$: Remember that $k$ has units of pressure over length ($\text{psi/in}$ or $\text{pci}$). If a problem states the foundation stiffness in $\text{pci}$, it is providing the $k$-value directly.