Archive · 12 October 2026 · Combined axial and bending stress

Part VI · VerificationVI–C12

Part VI

Verification

C. Stress recovery · Combined axial and bending stress

A straight pipe 4.0 m long carries an axial effect of 120.0 N. Known I-end shear is 4.0 N, and I-end bending moment is 8.0 N·m. A uniform load of −2.0 N/m acts along the full length. What axial force and bending moment occur at midspan? What axial and bending stresses follow? No pressure is supplied.

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–C12.—Combined axial and bending 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
Pipe area	A	6.0 m²
Axial-effect force	F_a	120.0 N
Displacements for the axial effect	—	all zero
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
Station	x/L	0.5
Stress area	A	6.0 m²
Section modulus about y	Z_y	3.0 m³
Section modulus about z	Z_z	2.0 m³

2.Method.

The axial effect is recovered as in the thermal axial case: with no displacement, the I-end axial force is N_i = 120.0. That force is combined with the given I-end shear and moment.

At midspan, x = 2.0 m, the load acting on the I-side free body runs from 0.0 m to 2.0 m. The axial force is unchanged along the pipe. The two stresses follow from force over area and moment over section modulus.

lever = x · (active length) − (active end)² / 2 = 2.0 × 2.0 − 2.0² / 2 = 2.0(1)

M_z(x) = M_z,i − V_y,i x − q · lever = 8.0 − 4.0 × 2.0 − (−2.0 × 2.0) = 4.0 N·m(2)

σ_axial = N_station / A = 120.0 / 6.0 = 20.0 Pa(3)

σ_b,z = M_z / Z_z = 4.0 / 2.0 = 2.0 Pa(4)

3.Results.

Expected midspan values
Quantity	Expected
Midspan axial force, N	120.0 N
Midspan bending moment, M_z	4.0 N·m
Axial normal stress	20.0 Pa
Bending stress about z	2.0 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 axial force matches 120.0 N and the bending moment matches 4.0 N·m.
The axial stress matches 20.0 Pa and the bending stress about z matches 2.0 Pa.
No hoop or longitudinal pressure stress is reported.
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. Each force, moment and stress must match the hand calculation to within an absolute difference of 1.0 × 10⁻⁹ in its own units. Labels, flags and the absence of pressure stresses 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 combined_axial_effect_and_bending_resultants_feed_station_stress_recovery

Hand calculation: validation/hand_calcs/stress/tp_phys_009_combined_axial_bending_to_stress.md. Test record, with the recorded run of 2026-07-10: stress-tp-phys-009-combined-axial-bending-to-stress.md.