Part VI
Verification
C. Stress recovery · Torsional shear stress
An invented section with a torsion constant of 80.0 m4 carries a twisting moment of 40.0 N·m. What shear stress does the torque produce at a radius of 2.0 m? The torque, the radius and the torsion constant are all supplied directly, so the case checks the recovery step alone.
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 |
|---|---|---|
| Torsional moment | T | 40.0 N·m |
| Torsion radius | r | 2.0 m |
| Torsion constant | J | 80.0 m4 |
2.Method.
In elastic torsion of a circular or annular section the shear stress grows in proportion to the radius. At radius r it is the torque times r divided by the torsion constant.
3.Results.
| Quantity | Expected |
|---|---|
| Torsional shear stress, τ | 1.0 Pa |
The tests check that:
- The stress recovery completes without being blocked.
- The recovered torsional shear stress matches 1.0 Pa.
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 recovered stress must match the hand calculation to within an absolute difference of 1.0 × 10−9 Pa. Yes-or-no conditions, such as whether the result is blocked, are checked exactly. Long values are shown here to seven significant figures; the tests compare the full values in the record.
4.Run it yourself.
cd projects/chirality-piping cargo test --manifest-path validation/benchmarks/stress/Cargo.toml recovers_torsional_shear_fixture
Hand calculation:
validation/hand_calcs/stress/torsional_shear.md.
Test record, with the recorded run of 2026-07-10:
stress-torsional-shear-original.md.