Station sweep in requested order — SWBPIPE verification

Part VI · VerificationVI–C10

Part VI

Verification

C. Stress recovery · Station sweep in requested order

A straight pipe 4.0 m long has known I-end shear of 4.0 N and bending moment of 8.0 N·m. A uniform load of −2.0 N/m acts over its middle half. Stations are requested at 0.75, 0.25, 0.5 and 1.0 of the length, deliberately out of order. Does recovery return the correct shear, moment and bending stress at each station in that order?

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.

Fig. VI–C10.—Station sweep in requested order

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
I-end shear	V_y,i	4.0 N
I-end bending moment	M_z,i	8.0 N·m
Uniform local y load	q	−2.0 N/m
Load span	a/L to b/L	0.25 to 0.75
Requested station fractions	x/L	0.75, 0.25, 0.5, 1.0
Area	A	3.0 m²
Section modulus about z	Z_z	2.0 m³

2.Method.

The load runs from a = 1.0 m to b = 3.0 m. For a station at distance x from the I end, only the part of the load at or before x is active. The shear and moment follow from the I-end values and the active load.

The axial force is zero at every station, so the axial stress is zero. The bending stress is the moment divided by the section modulus. Pressure is not supplied.

active length = max(0, min(x, b) − a)(1)

lever = x · (active length) − (active end² − a²) / 2(2)

V_y(x) = V_y,i + q · (active length)(3)

M_z(x) = M_z,i − V_y,i x − q · lever(4)

σ_b,z = M_z / Z_z, σ_axial = 0.0 / 3.0 = 0.0(5)

3.Results.

Expected values in the requested order
Index	Station x/L	V_y	M_z	Bending stress about z
0	0.75	0.0 N	0.0 N·m	0.0 Pa
1	0.25	4.0 N	4.0 N·m	2.0 Pa
2	0.5	2.0 N	1.0 N·m	0.5 Pa
3	1.0	0.0 N	0.0 N·m	0.0 Pa

The case record describes these checks:

Both the internal-force sweep and the stress sweep return four stations.
Each result carries its requested fraction, in the order 0.75, 0.25, 0.5, 1.0, and is labeled by its position in the request.
The shear, moment and bending stress at each station match the table.
The axial stress is exactly zero at every station.
No station result is blocked, and each carries the standard review flags.

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. Each fraction, force, moment and stress must match the hand calculation to within an absolute difference of 1.0 × 10⁻⁹ in its own units. Counts, labels, the zero axial stress and the review flags are checked exactly. Long values are shown here to seven significant figures; the tests compare the full values in the record.

For the student

At 0.75 of the length, the load ends. Shear and moment become zero and stay zero to the free end. No load acts beyond that station, so the free body on that side carries no load. This station sweep tests that boundary condition.

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 station_sweep_stress_fixture_preserves_order_and_recovers_stresses

Hand calculation: validation/hand_calcs/stress/tp_phys_007_station_sweep_stress.md. Test record, with the recorded run of 2026-07-10: stress-tp-phys-007-station-sweep-stress.md.