A rigid pavement works completely differently from a flexible one. A flexible pavement passes the wheel load down through its layers to the soil. A rigid pavement bridges over the soil — the concrete slab is stiff enough to spread the load across a wide area by its own bending strength.
This post covers how that behaviour is modelled: the spring foundation, the radius of relative stiffness, and the three Westergaard stress equations.
How a Rigid Pavement Carries Load
- The rigid pavement is constructed using cement concrete, and its load carrying capacity is primarily due to the rigidity of the slab.
- Rigidity comes from the high modulus of elasticity of concrete.
- The concrete pavement rests on a soil foundation which can be treated as a spring having constant K.
- That K is the modulus of subgrade reaction, found using the plate bearing test.
The key modelling idea: the soil is replaced by a bed of springs. Push down anywhere and the soil pushes back in proportion to how far it is pressed — exactly like a spring, with K as the spring constant.
This is called a Winkler foundation, and Westergaard’s whole analysis rests on it.
Modulus of Subgrade Reaction (K)
K is calculated corresponding to a settlement of 1.25 mm, that is 0.125 cm:
K = p / 0.125 (kg/cm3)
Plate Size Matters
K is not a fixed property of the soil — it depends on the size of plate used. Working from the deflection relations:
Δ = 1.18 p a / Es and Δ = p / K
Equating: p/K = 1.18pa/Es, so K × a = constant
Therefore K75 × 75 = K30 × 30, giving K75 = 0.4 K30
However, IRC recommends K75 = 0.5 K30
This is worth noting carefully, because two different values appear — the theoretical 0.4 from the K × a relationship, and the IRC design value of 0.5. In examinations, use the IRC value unless the question specifically asks for the theoretical derivation.
Normally, rigid pavement design testing is done using a 75 cm plate size.
Relative Stiffness of Slab to Subgrade
The relative stiffness of the slab with respect to the subgrade is represented by the radius of relative stiffness:
l = [ E h3 / (12 (1 − μ2) K) ]1/4 (cm)
| Symbol | Meaning | Value |
|---|---|---|
| E | Modulus of elasticity of cement concrete | 3 × 105 kg/cm2 |
| μ | Poisson’s ratio of concrete | 0.15 |
| h | Slab thickness | cm |
| K | Modulus of subgrade reaction | kg/cm3 |
What l Actually Means
The radius of relative stiffness compares how stiff the slab is against how stiff the soil is.
- A stiff slab on soft soil gives a large l — the load spreads over a wide area.
- A thin slab on firm soil gives a small l — the load stays concentrated.
Note the fourth root and the h3 inside it. Together these mean l varies as h3/4 — so doubling the slab thickness increases l by about 68 percent, not by 8 times. The fourth root heavily damps the effect.
Physically, l is a length, and it tells you the distance over which the slab distributes a concentrated load. It appears in every Westergaard equation for exactly that reason.
Critical Load Positions
The intensity of maximum stress in the slab depends on where the load sits. Three positions are considered critical:
- Interior loading
- Edge loading
- Corner loading
Why Position Matters So Much
The difference comes down to how much slab surrounds the load:
| Position | Support Available | Resulting Stress |
|---|---|---|
| Interior | Slab on all sides | Minimum |
| Edge | Slab missing on one side | Intermediate |
| Corner | Slab missing on two sides — discontinuity in both directions | Maximum |
A load at the interior is helped by concrete in every direction. At a corner, the slab can only push back from one quadrant, so the same wheel produces far higher stress.
Equivalent Radius of Resisting Section
Only a small area of the pavement resists the bending moment of a plate due to loading. Westergaard gave a relation for the radius of that resisting section:
b = √(1.6a2 + h2) − 0.675h when a < 1.724h
b = a otherwise
| Symbol | Meaning |
|---|---|
| a | Radius of wheel load distribution, cm |
| h | Slab thickness, cm |
The condition must always be checked first. Compute 1.724h, compare it with a, and only then decide which expression applies. Skipping this check is a frequent source of lost marks.
Westergaard’s Stress Equations
Westergaard assumed a Winkler foundation (spring foundation) and that the slab is homogeneous and isotropic.
Interior Loading
σinterior = (0.316 P / h2) [ 4 log10(l/b) + 1.069 ]
This is the tensile stress at the slab bottom.
Edge Loading
σedge = (0.572 P / h2) [ 4 log10(l/b) + 0.359 ]
This is the tensile stress at the slab bottom.
Corner Loading
σcorner = (3 P / h2) [ 1 − (a√2 / l)0.6 ]
This is the tensile stress at the TOP of the slab.
| Symbol | Meaning |
|---|---|
| P | Wheel load, kg |
| h | Thickness of slab, cm |
| l | Radius of relative stiffness, cm |
| b | Radius of resisting section, cm |
The Three Equations Compared
| Position | Coefficient | Constant Inside Bracket | Tension Located At |
|---|---|---|---|
| Interior | 0.316 | 1.069 | Bottom |
| Edge | 0.572 | 0.359 | Bottom |
| Corner | 3 | — (different form) | Top |
Two things stand out and both are heavily examined.
First, the coefficients climb: 0.316, 0.572, and then 3. That progression matches the support argument exactly — least stress at the interior, most at the corner.
Second, the corner equation is different in kind. It has no logarithm, and crucially its tension appears at the top of the slab rather than the bottom.
Why Corner Tension Is at the Top
Picture a load at the corner. The corner region tries to deflect downwards, but it is held by the rest of the slab behind it — so the slab bends like a cantilever sticking out. In a cantilever, the top fibre goes into tension.
At the interior and edge, by contrast, the slab is supported all around and sags under the load like a beam on springs, putting the bottom fibre in tension.
This distinction matters practically — it determines where reinforcement or thickening would be needed, and it explains why corner cracking appears from the surface downwards.
Formula Summary
| Quantity | Expression or Value |
|---|---|
| Modulus of subgrade reaction | K = p / 0.125 kg/cm3 |
| Settlement used | 1.25 mm = 0.125 cm |
| Plate relation, theoretical | K × a = constant, giving K75 = 0.4 K30 |
| Plate relation, IRC | K75 = 0.5 K30 |
| Radius of relative stiffness | l = [Eh3/(12(1 − μ2)K)]1/4 |
| E for concrete | 3 × 105 kg/cm2 |
| Poisson’s ratio | 0.15 |
| Radius of resisting section | b = √(1.6a2 + h2) − 0.675h if a < 1.724h; else b = a |
| Interior stress | (0.316P/h2)[4log10(l/b) + 1.069] |
| Edge stress | (0.572P/h2)[4log10(l/b) + 0.359] |
| Corner stress | (3P/h2)[1 − (a√2/l)0.6] |
Quick Revision Notes
- Rigid pavement carries load primarily by the rigidity of the slab, arising from the high modulus of elasticity of concrete.
- The soil foundation is treated as a spring with constant K — a Winkler foundation.
- K corresponds to a settlement of 1.25 mm (0.125 cm).
- K × a = constant, so theoretically K75 = 0.4 K30, but IRC recommends K75 = 0.5 K30.
- Rigid pavement testing normally uses a 75 cm plate.
- l = [Eh3/(12(1 − μ2)K)]1/4, with E = 3 × 105 kg/cm2 and μ = 0.15.
- Three critical load positions: interior, edge, corner.
- b = √(1.6a2 + h2) − 0.675h when a < 1.724h, otherwise b = a.
- Coefficients: interior 0.316, edge 0.572, corner 3.
- Interior and edge stresses are tensile at the slab bottom; corner stress is tensile at the top.
- Westergaard assumed a Winkler (spring) foundation and a homogeneous, isotropic slab.
Mistakes Students Commonly Make
- Forgetting to check the condition a < 1.724h before choosing the expression for b.
- Saying corner tension occurs at the bottom. It is at the top of the slab.
- Swapping the constants inside the brackets. Interior takes 1.069; edge takes 0.359.
- Using 0.4 instead of the IRC value 0.5 for the K75 to K30 conversion.
- Taking the fourth root of only part of the expression for l. The entire bracket is raised to the power 1/4.
- Omitting the √2 in the corner stress equation.
- Using a settlement of 0.25 cm for K. It is 0.125 cm.
- Assuming corner stress is smallest because the corner carries less slab. It is the largest, precisely because support is missing in two directions.
Conclusion
Rigid pavement design begins by replacing the soil with a bed of springs and then asking how a stiff concrete slab bends on that bed. The radius of relative stiffness l measures how widely the slab spreads a load, and the radius of resisting section b measures how small an area actually resists the bending. Westergaard’s three equations then give the stress at each critical position — smallest at the interior with support all round, largest at the corner where support is missing in two directions, and uniquely tensile at the top there because the corner behaves like a cantilever.
Frequently Asked Questions
How does a rigid pavement carry load?
Primarily through the rigidity of the concrete slab, which comes from the high modulus of elasticity of concrete. The slab spreads the load rather than passing it directly down.
How is the soil foundation modelled?
As a spring with constant K, the modulus of subgrade reaction. This is called a Winkler foundation, and Westergaard’s analysis assumes it.
At what settlement is K determined?
1.25 mm, that is 0.125 cm.
What is the relation between K measured with 75 cm and 30 cm plates?
Theoretically K × a is constant, giving K75 = 0.4 K30. However IRC recommends K75 = 0.5 K30.
What is the radius of relative stiffness?
l = [Eh³/(12(1 − μ²)K)]1/4, which represents the stiffness of the slab relative to the subgrade and indicates the distance over which the slab distributes a concentrated load.
What values of E and Poisson’s ratio are used for concrete?
E = 3 × 105 kg/cm² and Poisson’s ratio μ = 0.15.
What are the three critical load positions?
Interior loading, edge loading and corner loading.
What is the equivalent radius of resisting section?
b = √(1.6a² + h²) − 0.675h when a is less than 1.724h, and b = a otherwise.
Where does tension occur for each loading position?
At the slab bottom for interior and edge loading, and at the top of the slab for corner loading.
Why is corner tension at the top?
Because the corner region behaves like a cantilever, held by the rest of the slab while trying to deflect downwards, which puts the top fibre in tension.
Which loading position gives the maximum stress?
Corner loading, because there is discontinuity in both directions and the slab has support from only one quadrant.