Archive · 12 October 2026 · Station resultants to stress

Part VI · VerificationVI–C7

Part VI

Verification

C. Stress recovery · Station resultants to stress

A straight pipe 4.0 m long is fixed at node 0 and free at node 1. It carries −2.0 N/m along its length and a point force of −4.0 N at midspan. The companion mechanics case gives free-end deflection and rotation. What internal forces act at midspan? What axial, bending and torsional stresses do these forces produce?

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–C7.—Station resultants to stress

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
Second moment about local z	I_z	2.0 m⁴
Uniform local y load	q	−2.0 N/m
Point local y force	P	−4.0 N
Point force station	a/L	0.5
Free-end deflection	v(L)	−0.04533333 m
Free-end rotation	θ(L)	−0.01466667 rad
Stress area	A	3.0 m²
Section modulus about y	Z_y	2.5 m³
Section modulus about z	Z_z	2.0 m³
Torsion constant	J	1.0 m⁴
Torsion radius	r	0.5 m

2.Method.

From known free-end movement, recovery gives fixed-end shear V_y,i = 12.0 and moment M_z,i = 24.0. Recovery then proceeds to x = 2.0 m and includes each load encountered. The point force is exactly at the station and is included.

Midspan axial force, torque and moment about y are zero. Stress formulas use force divided by area, moment divided by section modulus, and torque times radius divided by torsion constant. This case does not include transverse shear stress.

V_y(x) = V_y,i + q x + P = 12.0 + (−2.0 × 2.0) + (−4.0) = 4.0 N(1)

M_z(x) = M_z,i − V_y,i x − q x²/2 − P(x − a) = 24.0 − 12.0 × 2.0 − (−2.0 × 2.0²/2) − (−4.0 × 0.0) = 4.0 N·m(2)

σ_b,z = M_z / Z_z = 4.0 / 2.0 = 2.0 Pa(3)

σ_axial = N / A = 0.0 / 3.0 = 0.0 Pa(4)

3.Results.

Expected midspan values (the test checks the moment, the bending stress about z and the axial stress)
Quantity	Expected
Axial force, N	0.0 N
Shear force, V_y	4.0 N
Bending moment, M_z	4.0 N·m
Axial normal stress	0.0 Pa
Bending stress about z	2.0 Pa
Bending stress about y	0.0 Pa
Torsional shear stress	0.0 Pa

The case record describes these checks:

The midspan stress recovery completes without being blocked and is labeled as the midspan station.
The midspan bending moment matches 4.0 N·m.
The bending stress about z matches 2.0 Pa.
The axial normal stress is exactly zero.

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 bending stress must match the hand calculation to within an absolute difference of 1.0 × 10⁻⁹ in their own units. The zero axial stress and the station label are checked exactly. Long values are shown here to seven significant figures; the tests compare the full values in the record.

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 load_to_resultant_station_stress_fixture_recovers_bending_component

Hand calculation: validation/hand_calcs/stress/tp_phys_004_load_to_resultant_stress.md (with validation/hand_calcs/mechanics/tp_phys_004_load_to_resultant.md). Test record, with the recorded run of 2026-07-10: stress-tp-phys-004-load-to-resultant.md.