Part VI
Verification
C. Stress recovery · Partial-span load to stress
A straight pipe 4.0 m long is fixed at node 0 and free at node 1. A uniform load of −2.0 N/m acts only from one-quarter to three-quarters of its length. What bending moment acts at midspan? What bending and axial stresses follow?
This page preserves a recorded benchmark and its expected values. The dated run below is not a new run of the current product.
These calculations recover separate stress components, not code stress. The current ordinary preview reports supported straight circular normal-stress extrema. Torsional shear remains separate, and arc maxima are withheld. Component recovery in this benchmark does not establish the current maximum or its coverage.
1.Inputs.
Illustrative values, taken from no standard and chosen so the arithmetic can be followed by hand. Most do not describe a real pipe; read them in any consistent set of units.
| Quantity | Symbol | Value |
|---|---|---|
| Length | L | 4.0 m |
| Elastic modulus | E | 1000.0 Pa |
| Shear modulus | G | 400.0 Pa |
| Area | A | 3.0 m2 |
| Second moment about local y | Iy | 1.5 m4 |
| Second moment about local z | Iz | 2.0 m4 |
| Torsion constant | J | 1.0 m4 |
| Uniform local y load | q | −2.0 N/m |
| Load span | a/L to b/L | 0.25 to 0.75 |
| Station | x/L | 0.5 |
| Section modulus about y | Zy | 2.5 m3 |
| Section modulus about z | Zz | 2.0 m3 |
2.Method.
The free-end equivalent loads are Fy,j = −2.0 and Mz,j = 1.8333333333333333. Free-end bending stiffness follows from EIz = 2000.0 and L = 4.0. Solution of the two unknowns gives deflection −0.014 m and rotation −0.004333333333333333 rad.
Recovery from these movements gives fixed-end shear Vy,i = 4.0 and moment Mz,i = 8.0. At x = 2.0 m, the active load extends from 1.0 m to 2.0 m. The lever integral gives its moment about the station.
3.Results.
| Quantity | Expected |
|---|---|
| Midspan bending moment, Mz | 1.0 N·m |
| Bending stress about z | 0.5 Pa |
| Axial normal stress | 0.0 Pa |
The case record describes these checks:
- The midspan stress recovery completes without being blocked and is labeled as this case’s midspan station.
- The midspan bending moment matches 1.0 N·m.
- The bending stress about z matches 0.5 Pa.
- The axial normal stress is exactly zero.
- The result carries the review flags the program attaches to every result.
What it shows. Compared with an independent hand calculation.
Path exercised. The benchmark calls the solver’s components directly: elements, loads, frame solver and stress recovery. It does not go through the program’s own model-to-solve path.
Agreement. The moment and stresses must match the hand calculation to within an absolute difference of 1.0 × 10−9 in their own units. The zero axial stress, station label and review flag are checked exactly. Long values are shown here to seven significant figures; the tests compare the full values in the record.
Only load between the fixed end and station enters the I-side free body. Here that load extends one meter, from 1.0 m to 2.0 m. The lever integral gives the moment of this loaded length about the cut.
4.Recorded command.
This command belongs to the linked case record and its recorded execution path. This website update did not rerun it. A historical command or result does not establish current product support.
cd projects/chirality-piping cargo test --manifest-path validation/benchmarks/stress/Cargo.toml partial_span_load_to_stress_fixture_recovers_midspan_bending_from_spanned_load
Hand calculation:
validation/hand_calcs/stress/tp_phys_006_partial_span_load_to_stress.md.
Test record, with the recorded run of 2026-07-10:
stress-tp-phys-006-partial-span-load-to-stress.md.