In Westergaard analysis the edge load stress is about 1.5 times the interior load stress, and the corner stress about 1.3 times. Yet rigid pavement slabs most often crack at the corners. The reason is that

at night the slab warps upward and the corners lose their support, so a wheel there acts on an unsupported cantilever

the corner is where the concrete is weakest because of poor compaction during construction

the Westergaard corner formula underestimates the stress by a factor of two

corners carry more traffic than edges because vehicles cut across joints

Which of the following correctly distinguish flexible from rigid pavements?

A flexible pavement transmits load to the subgrade by spreading it through layers; a rigid pavement carries it largely in slab bending

A flexible pavement is designed on subgrade CBR and traffic in MSA; a rigid pavement on flexural stress against the modulus of rupture

A weak patch of subgrade shows as rutting in a flexible pavement and as cracking in a rigid one

A flexible pavement transmits load to the subgrade by spreading it through layers; a rigid pavement carries it largely in slab bending

A flexible pavement is designed on subgrade CBR and traffic in MSA; a rigid pavement on flexural stress against the modulus of rupture

A weak patch of subgrade shows as rutting in a flexible pavement and as cracking in a rigid one

A rigid pavement requires a stronger subgrade than a flexible one of the same traffic

The equivalent single wheel load of a dual-wheel assembly, each wheel carrying P, is taken as

P near the surface, rising log-linearly with depth to 2P below a depth of twice the wheel spacing

2P at all depths, since both wheels are always present on the pavement

P at all depths, since a pavement is designed for one wheel at a time

2P near the surface, falling to P at depth as the stresses dissipate

Fatigue cracking in a flexible pavement is controlled in design by limiting

the horizontal tensile strain at the bottom of the bituminous layer

the vertical compressive strain at the top of the subgrade

the shear stress at the interface of the base and sub-base

the vertical deflection at the pavement surface under the wheel

Highway Materials and Pavement Design: Flexible, Rigid, and the Fourth-Power Law

The two pavement types in this chapter are not two ways of doing the same thing; they fail differently, and are therefore designed on different quantities. A flexible pavement has no bending strength worth counting, so it spreads a wheel load through successive layers by shear and friction until the stress reaching the subgrade is small enough for that soil to bear. Its design variable is therefore a thickness that gets the stress down, and its critical quantity is the strength of the subgrade — which is why the IRC method takes the subgrade CBR and the traffic in standard axles and reads a total thickness off a chart. A rigid pavement does the opposite: the concrete slab is stiff enough to act as a beam, so it carries the load in bending and distributes it over a wide area almost regardless of what is underneath. Its design variable is a flexural stress compared against the concrete’s modulus of rupture, and the soil below appears only as a spring constant K. Two consequences follow that are worth fixing before any arithmetic. In a flexible pavement a weak patch of subgrade shows up as a rut; in a rigid pavement it shows up as a crack, because a slab bridging a soft spot bends more than it was designed to. And the single most important number in either design is not in the pavement at all — it is the traffic, because damage goes as the fourth power of axle load, so a 12-tonne axle does five times the damage of an 8-tonne one and a small amount of overloading destroys a road built for the legal limit.

Materials, and what each test is actually asking

Aggregate tests: each one models a different way the stone will be attacked
Test	The property it measures	The field condition it stands for
Crushing value	Resistance to a slowly applied compressive load	A standing wheel — static crushing under a stationary load
Impact value	Resistance to sudden shock	A wheel dropping into a pothole, or construction rolling
Abrasion value (Los Angeles)	Resistance to rubbing and grinding	Years of tyres turning on the surface course
Soundness	Resistance to weathering (sulphate crystallisation)	Freeze–thaw and wet–dry cycles over the pavement’s life
Shape (flakiness, elongation)	How far from cubical the particles are	Flaky stone breaks under rolling and interlocks poorly

Bitumen is graded by penetration or viscosity, and its awkward property is that it is viscoelastic — stiff and brittle when cold, soft and flowing when hot — so a single grade cannot be right for every climate. Hence the four tests that matter and what each is protecting against: penetration (consistency at 25 °C), softening point by ring and ball (protects against rutting in summer), ductility (protects against cracking as the binder ages and stiffens), and flash point (protects the people mixing it). The desirable properties of a bituminous mix are then a set that pull against one another, which is exactly why mix design is a design and not a recipe: it must be stable enough not to rut, durable enough to resist ageing and water, flexible enough not to crack, and workable enough to lay — and more bitumen improves durability and flexibility while destroying stability.

🧠 The Marshall volumetrics, in one worked set

Take a mix with bulk specific gravity G_mb = 2.40, theoretical maximum G_mm = 2.55 and 5% bitumen by weight of mix at G_b = 1.02. The four quantities follow in order and each has a physical meaning. Air voids V_a = (1 − G_mb/G_mm) × 100 = 5.88% — the space filled with nothing, and the target is 3 to 5% because too few voids leave the mix nowhere to expand into on a hot day (it bleeds and ruts) and too many let in water and air (it strips and ages). Bitumen volume V_b = G_mb × P_b/G_b = 2.40 × 5/1.02 = 11.76%. Voids in the mineral aggregate VMA = V_a + V_b = 17.65% — the total space between the stones, which must be large enough to hold enough binder. Voids filled with bitumen VFB = V_b/VMA × 100 = 66.67% exactly. Note the relation the question usually hides: V_a, V_b and VMA are not independent, so being given any two of the four fixes the rest, and a candidate who tries to compute VMA from the aggregate gradation has taken a much longer road to the same number.

Wheel loads: ESWL, and the fourth-power law

A truck does not put a single wheel on the road; it puts a dual wheel on each side of an axle, and how the pavement feels that pair depends entirely on depth. Close to the surface the two stress bulbs do not overlap, so each wheel acts alone and the equivalent single wheel load is simply P. Deep down the bulbs have merged completely and the pair acts as a single load of 2P. Between those two depths the ESWL rises log-linearly, and the two boundaries are conventionally taken as z = d/2 (half the clear gap) and z = 2s (twice the centre-to-centre spacing). For two 2000 kg wheels at d = 20 cm and s = 30 cm: at 10 cm the ESWL is 2000 kg, at 20 cm it is 2615 kg, at 30 cm 3059 kg, and at 60 cm and below the full 4000 kg. Which is why the same axle is a mild load on a surface course and a severe one on a subgrade — the deeper layer is carrying both wheels at once.

🎯 Damage goes as the fourth power, and that changes everything

Traffic for pavement design is not counted in vehicles, or even in tonnes — it is counted in standard axles, that is in the equivalent single axle load each real axle represents, and the conversion is the vehicle damage factor VDF = (axle load / 8.16 t)⁴. The exponent is empirical, from the AASHO Road Test, and it is the single most consequential number in this chapter. Work it: an 8 t axle has VDF = 0.92, a 10.2 t axle 2.44, and a 12 t axle 4.68. So one 12-tonne axle does five times the damage of an 8-tonne one, and — the number that matters for enforcement — overloading a 10.2 t axle to 12 t, a rise of only 17.6 per cent in load, multiplies its damage by 1.92. Two conclusions follow that are not obvious from a linear intuition. First, cars are irrelevant: a 1-tonne axle has a VDF of about 0.0002, so a million cars do less damage to a pavement than a few thousand loaded trucks, and a traffic count that does not separate commercial vehicles is useless for design. Second, overloading is not a marginal problem but the dominant one — a fleet running 20% over the legal axle limit roughly doubles the design traffic, halving the life of a pavement built exactly to specification. Which is why axle-load surveys, and not just vehicle counts, are part of the IRC method, and why a road that fails early is more often a traffic-estimation failure than a materials failure.

Flexible pavement design by the IRC method

Measure the subgrade CBR — the California Bearing Ratio, a penetration test whose value is the load to push a plunger a given depth as a percentage of the load for a standard crushed stone. It is a strength index, not a modulus, and its virtue is repeatability rather than theory.
Count the commercial vehicles per day, apply a growth rate over the design life, and convert to million standard axles with the vehicle damage factor, the lane distribution factor and the directional distribution.
Read the total thickness from the IRC design catalogue against CBR and MSA, then compose it from a bituminous surfacing, a granular base and a granular sub-base — the composition matters as much as the total, because the layers must get progressively weaker downward and never the reverse.

The two failure modes the method is guarding against are worth naming, because they are what the two design charts represent. Rutting is permanent deformation accumulating in the subgrade, controlled by limiting the vertical compressive strain on top of it. Fatigue cracking starts at the bottom of the bituminous layer, where bending puts it in tension, and is controlled by limiting the horizontal tensile strain there. Notice that they pull in opposite directions on one variable: a thicker bituminous layer reduces the subgrade strain but is itself stiffer and can attract more tensile strain if it is thin and stiff rather than thick and stiff. That is why the IRC catalogue gives compositions rather than a single number, and why substituting a thinner bituminous layer over a thicker granular base is not the free trade it appears to be.

Rigid pavement: the slab, the spring, and the edge that governs

Westergaard modelled the concrete slab as a plate on a bed of independent springs — a Winkler foundation, whose stiffness is the modulus of subgrade reaction K in kg/cm³. The scale of the problem is set by the radius of relative stiffness, l = [Eh³/(12(1 − μ²)K)]^0.25, which is not a physical dimension of anything but the distance over which the slab shares a load with the ground. A stiffer or thicker slab has a larger l and spreads the load further; a stiffer subgrade has a smaller l and lets the slab do less work. Load stresses are then computed at three positions, and the practical result is which of the three is largest.

Westergaard stresses: E = 3 × 10⁵ kg/cm², h = 25 cm, μ = 0.15, K = 8 kg/cm³, a = 15 cm, P = 5100 kg
Position	Load stress	Relative to interior
Interior	10.63 kg/cm²	1.00
Edge	15.92 kg/cm²	1.50 — the largest
Corner	13.77 kg/cm²	1.30

⚠️ Load stress and temperature stress do not both peak in the same place

The edge carries the highest load stress, 50 per cent above the interior, because the slab there is supported on one side only and has no neighbouring concrete to share with — which is why edges are thickened, tied to a shoulder, or given dowels. But a rigid pavement is stressed by more than wheels. The slab also warps: sun on the top face makes it hotter than the bottom, the slab tries to curl downwards at its edges, the self-weight and the subgrade resist, and a temperature stress results. The critical combination is therefore not simply "the largest of the three". In daytime, when the top is hotter, the warping stress is tensile at the bottom in the interior and the critical case is edge load plus edge warping. At night the gradient reverses, the corners lift clear of the subgrade, and a wheel arriving at an unsupported corner produces the critical corner case even though its computed load stress is lower than the edge’s. This is the reason a pavement is checked for more than one combination, why joints are placed where they are, and why the corner — despite ranking second in the table above — is where slabs actually break. Note also that l behaves counter-intuitively with K: a stiffer subgrade reduces l, concentrating the load over a smaller area, so improving the ground does less for a rigid pavement than intuition suggests, and thickening the slab does much more.

Key takeaways

Flexible pavements spread load and are designed on thickness against CBR; rigid pavements bend and are designed on flexural stress.
VDF = (axle load/8.16)⁴: a 12 t axle does 5× the damage of an 8 t axle, and a 17.6% overload nearly doubles it.
ESWL is P near the surface and 2P deep down, rising log-linearly between z = d/2 and z = 2s.
Marshall: V_a = (1 − G_mb/G_mm), V_b = G_mb·P_b/G_b, VMA = V_a + V_b, VFB = V_b/VMA — only two are independent.
Westergaard edge load stress is 1.5× interior; but at night the corner lifts, so the corner case can still govern.
A stiffer subgrade reduces l, concentrating the load — improving the ground helps a rigid pavement less than expected.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Using the fourth-power law with a standard axle of 8.16 tonnes, the vehicle damage factor of a 12-tonne single axle is ______ (round to three decimal places)
Numerical answer — type the value.
Show answer
Answer: 4.677
4.677. VDF = (12/8.16)⁴ = (1.47059)⁴ = 4.6770. One overloaded axle counts as nearly five standard axles in the design traffic, and the number matters far more than the formula. Set it beside two others: an 8 t axle gives (8/8.16)⁴ = 0.924, and a 10.2 t axle gives 2.441. So going from 10.2 t to 12 t — an increase in load of only 17.6 per cent — multiplies the damage by 4.677/2.441 = 1.916, very nearly doubling it. That is the whole argument for axle-load enforcement in one line, and it is why the marginal tonne on an already-loaded truck is the expensive one. Two consequences follow that a linear intuition never reaches. First, passenger cars are effectively invisible: a 1-tonne axle has VDF = (1/8.16)⁴ ≈ 0.00023, so four thousand cars do about as much damage as one legally loaded truck axle, and a traffic survey that does not separate commercial vehicles is worthless for pavement design. Second, the design is far more sensitive to the axle-load spectrum than to the vehicle count, so an axle-load survey is part of the IRC method rather than an optional refinement. A road built exactly to specification for a legally loaded fleet, and then used by a fleet running 20% over, reaches its design life in roughly half the time — and that failure will look like bad construction while being an arithmetic consequence of the traffic.
A concrete slab has E = 3 × 10⁵ kg/cm², thickness 25 cm and Poisson ratio 0.15, resting on a subgrade of modulus of subgrade reaction K = 8 kg/cm³. The radius of relative stiffness in cm is ______ (round to two decimal places)
Numerical answer — type the value.
Show answer
Answer: 84.07
84.07 cm. l = [Eh³/(12(1 − μ²)K)]^0.25. The numerator is 3 × 10⁵ × 25³ = 4.6875 × 10⁹; the denominator is 12 × (1 − 0.0225) × 8 = 93.84; the quotient is 4.9952 × 10⁷, and its fourth root is 84.07 cm. The quantity is not a physical dimension of the slab or the load — it is the length scale over which the slab shares a load with the ground, and reading it that way makes its behaviour predictable. A stiffer or thicker slab has a larger l and distributes the wheel load over a wider area; the h³ inside a fourth root means l ∝ h^0.75, so thickening the slab by 20% widens the distribution by about 15%. The counter-intuitive half is K. A stiffer subgrade gives a smaller l — l ∝ K^(−0.25) — because a stiff ground pushes back locally and lets the slab spread the load less. Doubling K from 8 to 16 reduces l only to 70.7 cm, a 16% change for a doubled subgrade modulus. That is the honest reason why improving the subgrade does far less for a rigid pavement than for a flexible one: a flexible pavement passes the load to the subgrade and depends on its strength directly, while a rigid slab carries the load in bending and treats the ground as a bed rather than a support. The design lever for a rigid pavement is slab thickness, and everything else is secondary.
In Westergaard analysis the edge load stress is about 1.5 times the interior load stress, and the corner stress about 1.3 times. Yet rigid pavement slabs most often crack at the corners. The reason is that
1. at night the slab warps upward and the corners lose their support, so a wheel there acts on an unsupported cantilever
2. the corner is where the concrete is weakest because of poor compaction during construction
3. the Westergaard corner formula underestimates the stress by a factor of two
4. corners carry more traffic than edges because vehicles cut across joints
Show answer
Answer: A — at night the slab warps upward and the corners lose their support, so a wheel there acts on an unsupported cantilever
At night the corners lift and lose their support. The load stresses in the table are computed for a slab resting fully on its foundation, and at the corner that assumption is the one that fails. A concrete slab warps with the temperature gradient through its depth. By day the top is hotter, the slab tries to expand more at the top than the bottom, and it presses down at its edges and corners — support is maintained, and the day-time critical case is edge load combined with edge warping. At night the gradient reverses: the top cools faster than the bottom, the slab curls upward at its extremities, and the corners physically lift clear of the subgrade. A wheel arriving at such a corner is loading an unsupported cantilever projecting from two free edges, and the actual stress is far above what the Westergaard corner formula — which assumes contact — predicts. That is why corners are where slabs break, why dowel bars are provided at transverse joints to transfer load across, and why the corner region is often thickened. The general lesson is worth more than the fact: the critical case in a rigid pavement is a combination of load and warping stress, not the largest load stress on its own, and which combination governs changes between day and night. Option (b) blames workmanship for a phenomenon that occurs in perfectly built pavements. Option (c) invents an error in a formula that is correct within its own assumption. Option (d) has the traffic argument backwards — edges carry more wheel passes than corners, not fewer.
A bituminous mix has bulk specific gravity 2.40 and theoretical maximum specific gravity 2.55. Its air void content, as a percentage, is ______ (round to two decimal places)
Numerical answer — type the value.
Show answer
Answer: 5.88
5.88%. V_a = (1 − G_mb/G_mm) × 100 = (1 − 2.40/2.55) × 100 = (1 − 0.94118) × 100 = 5.88%. The two gravities differ only because of the air: G_mm is what the mix would weigh per unit volume with no voids at all, and G_mb is what it actually weighs compacted, so the shortfall is the void fraction directly. The number matters because the design target is narrow — typically 3 to 5% — and both directions of failure are real. Too few voids and the mix has nowhere to expand on a hot day: the binder is squeezed to the surface, the mix bleeds, loses skid resistance and ruts under traffic. Too many and water and air get in, the binder oxidises and hardens, and the mix strips as water breaks the bond between bitumen and aggregate. At 5.88% this mix is just above the usual range, which points to slightly too little binder or insufficient compaction — and note that compaction is the variable a site can still change after the mix is made. The other volumetrics follow from the same sample and are worth computing together: at 5% bitumen by weight with G_b = 1.02, V_b = G_mb·P_b/G_b = 11.76%, so VMA = V_a + V_b = 17.65% and VFB = V_b/VMA = 66.67% exactly. Only two of those four are independent, which is why a question can give any two and expect the rest.
Which of the following correctly distinguish flexible from rigid pavements?
1. A flexible pavement transmits load to the subgrade by spreading it through layers; a rigid pavement carries it largely in slab bending
2. A flexible pavement is designed on subgrade CBR and traffic in MSA; a rigid pavement on flexural stress against the modulus of rupture
3. A weak patch of subgrade shows as rutting in a flexible pavement and as cracking in a rigid one
4. A rigid pavement requires a stronger subgrade than a flexible one of the same traffic
Show answer
Answer: A — A flexible pavement transmits load to the subgrade by spreading it through layers; a rigid pavement carries it largely in slab bending; B — A flexible pavement is designed on subgrade CBR and traffic in MSA; a rigid pavement on flexural stress against the modulus of rupture; C — A weak patch of subgrade shows as rutting in a flexible pavement and as cracking in a rigid one
(a), (b) and (c). (a) is the structural difference everything else follows from. A flexible pavement has no useful bending stiffness, so each layer spreads the stress a little further until what reaches the subgrade is small enough for that soil; a concrete slab is stiff enough to act as a plate on an elastic foundation and carries the load in bending over a distance set by the radius of relative stiffness. (b) is the design consequence: the flexible method is governed by the strength of the ground (CBR) and the traffic in standard axles, while the rigid method compares a computed flexural stress with the concrete’s modulus of rupture, and the ground enters only as a spring constant K. (c) is the failure signature, and it is the most useful of the three in the field: a soft spot under a flexible pavement deflects locally and accumulates permanent deformation — a rut; the same soft spot under a slab is bridged, and the slab bends more than it was designed to and cracks. Diagnosis follows structure. (d) is wrong, and it inverts the actual relationship. Because a rigid slab distributes load so widely, it is less dependent on subgrade strength, not more — which is precisely why concrete pavements are used over weak or variable soils where a flexible pavement would need an uneconomic thickness. The arithmetic supports it: l ∝ K^(−0.25), so doubling the subgrade modulus changes the load distribution by only 16%. What a rigid pavement does demand of the ground is not strength but uniformity: a slab tolerates a uniformly weak subgrade far better than a patchy one, because differential support is what bends it.
The equivalent single wheel load of a dual-wheel assembly, each wheel carrying P, is taken as
1. P near the surface, rising log-linearly with depth to 2P below a depth of twice the wheel spacing
2. 2P at all depths, since both wheels are always present on the pavement
3. P at all depths, since a pavement is designed for one wheel at a time
4. 2P near the surface, falling to P at depth as the stresses dissipate
Show answer
Answer: A — P near the surface, rising log-linearly with depth to 2P below a depth of twice the wheel spacing
P near the surface, rising to 2P at depth. The idea is entirely about overlap of stress bulbs. Just below the surface each wheel’s bulb of significant stress is narrow and the two do not intersect, so a point under one wheel feels only that wheel: the equivalent single wheel load is P. As depth increases the bulbs widen and begin to overlap; a point on the centreline now receives a contribution from both, and the ESWL rises. Far enough down the two bulbs have merged into one, and the pair is indistinguishable from a single load of 2P. The conventional boundaries are z = d/2, half the clear gap between the tyres, and z = 2s, twice the centre-to-centre spacing, with a log-linear interpolation between — a straight line on a log–log plot of ESWL against depth. Worked for two 2000 kg wheels with d = 20 cm and s = 30 cm: at 10 cm the ESWL is 2000 kg, at 20 cm 2615 kg, at 30 cm 3059 kg, and at 60 cm and below the full 4000 kg. The design consequence is what makes this examinable rather than a curiosity: the upper layers of a pavement are designed for a single wheel and the lower layers for the pair, so the same axle is a mild load on the surface course and a severe one on the subgrade. Option (b) over-designs everything near the surface, (c) ignores the pair entirely and under-designs the subgrade, and (d) has the physics exactly reversed — stresses spread with depth, they do not concentrate.
For the mix of question 4 (G_mb = 2.40, G_mm = 2.55) with 5% bitumen by weight of mix and bitumen specific gravity 1.02, the voids filled with bitumen (VFB), as a percentage, is ______ (round to two decimal places)
Numerical answer — type the value.
Show answer
Answer: 66.67
66.67%. Build it in three steps, each of which is a volume expressed as a percentage of the total compacted volume. The bitumen volume is V_b = G_mb × P_b/G_b = 2.40 × 5/1.02 = 11.76%. The air voids, from question 4, are V_a = 5.88%. The voids in the mineral aggregate — the total space between the stones, whether it holds binder or air — is therefore VMA = V_a + V_b = 5.88 + 11.76 = 17.65%. And the fraction of that space actually occupied by binder is VFB = V_b/VMA × 100 = 11.76/17.65 = 66.67% exactly. Reading those four as a set rather than four formulae is what makes mix design intelligible. VMA is a property of the aggregate skeleton — how much room the stones leave between them — and it must be large enough to accommodate enough binder for durability without leaving so much that the mix is unstable, which is why specifications set a minimum VMA that depends on the nominal maximum aggregate size. VFB then says how that room has been divided between bitumen and air, and its usual target of 65 to 75% is a restatement of the air-void target from the other side. Note the redundancy the exam exploits: V_a, V_b, VMA and VFB contain only two independent quantities, so a question can supply any two and demand the others, and recognising that saves you from hunting for data that was never needed.
Fatigue cracking in a flexible pavement is controlled in design by limiting
1. the horizontal tensile strain at the bottom of the bituminous layer
2. the vertical compressive strain at the top of the subgrade
3. the shear stress at the interface of the base and sub-base
4. the vertical deflection at the pavement surface under the wheel
Show answer
Answer: A — the horizontal tensile strain at the bottom of the bituminous layer
The horizontal tensile strain at the bottom of the bituminous layer. A flexible pavement is checked against two distress mechanisms, and the marks are in keeping them apart, because each has its own critical strain at its own location. Fatigue cracking begins where the bituminous layer is in tension. Under a wheel the bound layer bends like a beam over the softer material below, so its underside stretches — and repeated stretching under millions of load cycles propagates a crack upward from the bottom of the layer until it appears at the surface, typically as the interconnected pattern known as alligator cracking. That is why the criterion is the horizontal tensile strain at the bottom of the bituminous layer and not at the top. Rutting, by contrast, is permanent deformation accumulating in the subgrade, and it is controlled by option (b) — the vertical compressive strain on top of the subgrade. Both criteria appear in the IRC method, and the design thickness is whichever satisfies both. The pair is worth understanding together because they pull in opposite directions: a thicker bituminous layer reduces the subgrade strain and so resists rutting, but a layer that is stiff and thin can attract more tensile strain than one that is stiff and thick, so simply adding stiffness is not automatically an improvement. That interaction is exactly why the IRC catalogue prescribes layer compositions rather than a single total thickness, and why a design cannot be adjusted by moving thickness freely between layers. Option (c) is not a design criterion in the method. Option (d) is a measured field quantity (Benkelman beam deflection) used for evaluating and strengthening existing pavements, not the mechanistic criterion for fatigue.