How layered-elastic response enters pavement design

Depth spreads the story

Double the load and the response doubles here. Double the depth and it does not simply halve.

TRH4 provides a South African structural-design framework. Use the selected design method consistently with its assumptions and applicability. (CSRA, 1996)

Start with a stress solution you can check

Governing relationship: σz(0,z) = q[1 − (1 + a²/z²)^(−3/2)].

P = 40 kN; a = 0.15 m; q = 565.884 kPa.

z is depth below the circular load centre (m). Stress is positive in compression.

The film first plots pressure (kPa), then follows the calculation and plots vertical stress (kPa). At z = 0.30 m, σz = 160.97 kPa. Load spreading reduces centreline stress with depth in this reference problem.

Elastic reference experiment

Stress spreading beneath a circular load

Drawn layer depths are reference locations; this is not a multilayer solution.

Download example data (CSV)

Model, calculations and transcriptEquations, worked steps, source notes and a timed reading guide

Understand the model

A finite circular load spreads stress with depth in a homogeneous elastic half-space. This reference is a solver benchmark, not a layered pavement design.

Work through the inputs and equations below, then compare the calculated output with the film. All numerical inputs in this study are synthetic.

Inputs and symbol key

P; a

Applied force (N); circular contact radius (m).

q

Uniform contact pressure (Pa).

z; σz

Depth on the centreline (m); vertical compressive stress (Pa).

Values used in this film

• P = 40 kN; a = 0.15 m; q = 565.884 kPa.
• z is depth below the circular load centre (m). Stress is positive in compression.

The mathematical relationships

1. Equation 1

   Source and scope
   $$q=P/(\pi a^2)$$

   Compute the uniform contact pressure using force divided by circular area.

2. Equation 2

   Source and scope
   $$\sigma_z=q[1-(1+a^2/z^2)^{-3/2}]$$

   Multiply pressure by a dimensionless influence factor. Depth and radius must use the same length unit. Stress approaches q near the loaded surface and zero at great depth.

Work through the example

Use the relationships above with consistent units. These substitutions are our worked example, not values quoted from the source.

1. Radius-depth ratio

   a/z = 0.15/0.30 = 0.5

   dimensionless

2. Influence factor

   1 − (1 + 0.5²)^(−1.5) = 0.284458

   dimensionless

3. Contact pressure

   40,000 / (π × 0.15²) = 565.884

   kPa

4. Vertical stress

   565.884 × 0.284458 = 160.970

   kPa, rounded

Read the output

At z = 0.30 m, σz = 160.97 kPa. Load spreading reduces centreline stress with depth in this reference problem.

Check your understanding

Confirm that halving the load halves stress at every plotted point. Check the analytical centreline against the area-integrated kernel and verify the surface and far-field limits without evaluating a singular formula at z = 0.

Assumptions and limits

Drawn layer depths are reference locations; this is not a multilayer solution.

Return to the full references and method notes to check the cited edition and its applicability.

Timestamped film transcript 29 short entries · select a time to revisit the film

1. 0:00

   Stress spreading beneath a circular load. 40 kN over radius 0.15 m; homogeneous elastic half-space.

2. 0:01

   MATHEMATICS · define the relationship and the meaning of each symbol before calculating.

3. 0:07

   P = 40 kN; a = 0.15 m; q = 565.884 kPa. z is depth below the circular load centre (m). Stress is positive in compression.

4. 0:14

   INPUT · uniform radial contact pressure · read the axis units and the numerical values.

5. 0:23

   40 kN over radius 0.15 m; homogeneous elastic half-space.

6. 0:31

   At z = 0.15 m: σz = 365.8 kPa. Depth increases away from the loaded surface.

7. 0:34

   At z = 0.30 m: σz = 161.0 kPa. Depth increases away from the loaded surface.

8. 0:37

   At z = 0.60 m: σz = 49.2 kPa. Depth increases away from the loaded surface.

9. 0:40

   At z = 1.20 m: σz = 13.0 kPa. Depth increases away from the loaded surface.

10. 0:43

    OUTPUT · attenuation with depth · read the axis units and the numerical values.

11. 0:52

    At z = 0.30 m, σz = 160.97 kPa. Load spreading reduces centreline stress with depth in this reference problem.

12. 0:59

    CALCULATION · substitute the defined inputs; keep units through each intermediate step.

13. 1:05

    Radius-depth ratio: a/z = 0.15/0.30 = 0.5 dimensionless

14. 1:08

    Influence factor: 1 − (1 + 0.5²)^(−1.5) = 0.284458 dimensionless

15. 1:11

    Contact pressure: 40,000 / (π × 0.15²) = 565.884 kPa

16. 1:14

    Vertical stress: 565.884 × 0.284458 = 160.970 kPa, rounded

17. 1:17

    SENSITIVITY · change one declared input · read the axis units and the numerical values.

18. 1:26

    Depth 0.15 m: 365.814 kPa vertical stress. 40 kN uniform circular load; radius stays 0.15 m.

19. 1:29

    Depth 0.30 m: 160.97 kPa vertical stress. 40 kN uniform circular load; radius stays 0.15 m.

20. 1:33

    Depth 0.60 m: 49.1893 kPa vertical stress. 40 kN uniform circular load; radius stays 0.15 m.

21. 1:36

    Depth 1.20 m: 13.0085 kPa vertical stress. 40 kN uniform circular load; radius stays 0.15 m.

22. 1:39

    Check: Confirm that halving the load halves stress at every plotted point. Check the analytical centreline against the area-integrated kernel and verify the surface and far-field limits without evaluating a singular formula at z = 0.

23. 1:45

    ONE ROW · Double the load and the response doubles here. Double the depth and it does not simply halve.

24. 2:03

    PARAMETER MAP · 6161 evaluated cells. The colour limits stay fixed across every row.

25. 2:13

    The load axis makes linearity visible while the depth axis exposes the nonlinear geometric attenuation. This centreline result integrates a uniformly loaded circular area. It is a

26. 2:23

    useful reference check for a stress calculation before adding layer contrasts. The 3D height is compressive stress, not vertical displacement.

27. 2:30

    VALUE LANDSCAPE · Height and colour encode Vertical stress (kPa). This is a graph over the two input axes, not automatically a physical pavement shape.

28. 2:42

    The equation describes a homogeneous half-space. It cannot show stress redistribution caused by different layer stiffnesses or calculate fatigue life.

29. 2:51

    Drawn layer depths are reference locations; this is not a multilayer solution.

Download the timed captions (VTT)

The transcript records the on-screen wording. Formula provenance and interpretation are explained in the source note above.

Separate inputs, response and interpretation

For an educational model, list layer geometry, material behaviour, boundary conditions and loading first. A response calculation then produces stresses, strains or displacements. A separate calibrated relationship may relate selected responses to performance.

Inspect a layer stack

Change the illustrative layer thickness below. The drawing changes geometry only; it does not calculate stress or predict life. This deliberate limit prevents a convincing visual from being mistaken for an engineering solver.

Where the half-space stops helping

Real materials and interfaces may depart from idealised assumptions. Never infer a validated design from the appearance of a deformed 3D mesh.

Three questions to keep separate

What has been idealised? Name the layers, their thicknesses, the assumed material behaviour, interfaces and loading. An idealisation is useful because it simplifies a complicated pavement, but its simplifications must remain visible.

What did the model calculate? Label a stress, strain or displacement with its location, direction and units. The response is not automatically a prediction of field performance.

How was performance inferred? A separate relationship may connect a selected response to an expected outcome within a defined calibration context. State that relationship and its applicability. A polished three-dimensional animation cannot replace these records.

The diagram here answers only a geometry question: what happens to a drawn layer when its illustrative thickness changes? It is an explanatory step towards understanding a model, not a hidden model implementation.

Do not let a layer stack imply a validated response

Take an illustrative stack with 40 mm surfacing, a 150 mm base and a 200 mm subbase. The drawn stack totals 40 + 150 + 200 = 390 mm above the supporting material. Increasing the base to 200 mm changes that geometric total to 440 mm, an increase of 50 mm or 12.8% of the original total.

Case review

First pass the reference check

Does the stress calculation reproduce a simple solution before layers are added?

20 kN · 80.485 Centreline stress (kPa)

The discriminating check

Verify units, pressure, geometry and the reference solution before interpreting results from a more complex model.

Evidence supports the comparison

After the reference check, examine layer assumptions, interfaces and the specific response quantity required.

Evidence is missing or incompatible

If the reference fails, correct the implementation or inputs before attaching a design interpretation.

What you can conclude. Passing a half-space check is necessary evidence of implementation consistency; it does not validate a layered design model.

Original worked case. Read the source editions and method notes.

Those calculations establish a change in geometry. They do not establish a percentage improvement in life, a proportional decrease in strain, or a particular structural capacity. A mechanical response depends on the model's material properties, load representation, interfaces and boundary conditions as well as thickness.

TRH4 supplies a South African structural-design context. The article's geometry example is not a substitute for a selected design method or a calibrated performance relationship. (CSRA, 1996)

A technical explanation should label each calculated response with its location and direction. A tensile strain at one point is not interchangeable with a compressive strain at another. Similarly, a deflection is not a modulus. The location, quantity and units are part of the result.

The sensitivity experiment below changes depth beneath a fixed load and computes the resulting vertical stress. The layer-thickness inventory above remains a separate geometry example: those thicknesses are not inputs to the homogeneous reference solution.

Start with a solvable stress benchmark

The film calculates centreline vertical stress below a uniformly loaded circular area in a homogeneous elastic half-space. With q = P/(πa²), σz(0,z) = q[1 − (1 + a²/z²)^(−3/2)]. The inputs are P = 40 kN and a = 0.15 m. Stress is reported in kPa and depth in metres. The curve is evaluated at increasing depths rather than drawn as an arbitrary bulb. (Rocscience, n.d.)

This benchmark isolates stress spreading. It does not assign separate stiffnesses to the surfacing, base and subgrade in the article's geometry example. A layered solution must additionally satisfy the chosen interface and boundary conditions. Use the benchmark to check units, symmetry and load scaling before interpreting more complicated numerical output.

Calculation laboratory

How does load spread with depth in the reference solution?

Calculation step	Substitution or result	Units and interpretation
Radius-depth ratio	a/z = 0.15/0.30 = 0.5	dimensionless
Influence factor	1 − (1 + 0.5²)^(−1.5) = 0.284458	dimensionless
Contact pressure	40,000 / (π × 0.15²) = 565.884	kPa
Vertical stress	565.884 × 0.284458 = 160.970	kPa, rounded

The influence factor is dimensionless and lies between zero and one beneath the centre of this uniformly loaded circular area. Its limiting values provide useful checks: approaching the loaded surface gives contact pressure, while increasing depth indefinitely gives vanishing vertical stress. A heatmap must use a fixed pressure scale across the sensitivity cases so that changing colour actually represents changing stress.

At large depth, expanding the expression gives σz approximately 3P/(2πz²), the centreline point-load limit. At shallow depth, that approximation is unsuitable because the finite contact area matters. The reference solution therefore provides both a numerical benchmark and a way to identify where a simpler approximation begins to depart.

A layered pavement calculation introduces material properties and interface conditions that this reference does not contain. Vertical stress cannot be converted to an arbitrary strain simply by dividing by one modulus while ignoring the remaining stress components and constitutive relation. Similarly, a stress contour is not a fatigue-life contour. A design analysis must declare its layer system, bonding assumptions, load representation and transfer functions separately. The homogeneous circular-load solution is used here as a check before trusting a more complex solver. (Rocscience, n.d.) (University of Cambridge, 2005)

Controlled sensitivity experiment

Controlled case	Calculated output	What changes
Depth 0.15 m	365.813950 kPa vertical stress	40 kN uniform circular load; radius stays 0.15 m.
Depth 0.30 m	160.970440 kPa vertical stress	40 kN uniform circular load; radius stays 0.15 m.
Depth 0.60 m	49.189321 kPa vertical stress	40 kN uniform circular load; radius stays 0.15 m.
Depth 1.20 m	13.008511 kPa vertical stress	40 kN uniform circular load; radius stays 0.15 m.

CALCULATION LABORATORY

How does load spread with depth in the reference solution?

Compare the cases on a fixed scale. Select a case to inspect its input change and calculated output.

Sensitivity case1. Depth 0.15 m2. Depth 0.30 m3. Depth 0.60 m4. Depth 1.20 m

Depth 0.15 m: 365.81395 kPa vertical stress. 40 kN uniform circular load; radius stays 0.15 m.

Reproduce and challenge the result

Confirm that halving the load halves stress at every plotted point. Check the analytical centreline against the area-integrated kernel and verify the surface and far-field limits without evaluating a singular formula at z = 0.

These are original synthetic calculations. The sensitivity cases vary the stated input while retaining the other assumptions; they are not field observations or treatment thresholds. The film’s calculation and sensitivity chapters use the same numerical cases as this table.

Separate load scaling from depth attenuation

The load axis makes linearity visible while the depth axis exposes the nonlinear geometric attenuation. This centreline result integrates a uniformly loaded circular area. It is a useful reference check for a stress calculation before adding layer contrasts. The 3D height is compressive stress, not vertical displacement.

Reproduce the experiment. σz = P/(πa²)·[1−(1+a²/z²)^(-3/2)], a=0.15 m. The downloadable table contains 6,161 evaluated cells, with the two input coordinates and output value retained on every row. These are parameter combinations, not that many independent field observations. The equation extends the worked example above; it is original teaching arithmetic, not a new standard or calibrated prediction.

Computed parameter atlas · 6,161 cells

Depth spreads the story

Double the load and the response doubles here. Double the depth and it does not simply halve.

Download every calculated cell (CSV) · Open full-size explorer · Equation source and tests (ZIP)

The equation describes a homogeneous half-space. It cannot show stress redistribution caused by different layer stiffnesses or calculate fatigue life.

References and method notes

1. CSRA (1996). TRH4: Structural Design of Flexible Pavements for Interurban and Rural Roads. 1996 edition. Original source (Accessed: 5 September 2026).Design framework; confirm project applicability
2. Rocscience (n.d.). Settle3: Settlement and Consolidation Theory Manual. Theory manual. Original source (Accessed: 5 September 2026).Section 2.3.1: Boussinesq stress kernel and integration over a loaded area
3. University of Cambridge (2005). Engineering Tripos Part IIA: 3D5 Geotechnical Engineering. Examination paper and formula information, 2005. Original source (Accessed: 5 September 2026).Data sheet Sections 3.1 and 3.3, printed pp. 6 and 8: circular-load stress and surface settlement

Use the source edition and project requirements that apply to your work. Educational examples are not site-specific design advice.