Straight-pipe end forces to stress — SWBPIPE verification

Part VI · VerificationVI–C3

Part VI

Verification

C. Stress recovery · Straight-pipe end forces to stress

A straight pipe 5.0 m long follows global x from node 0 to node 1. Node 1 moves 0.01 m along the pipe and rotates 0.02 rad about its axis. What axial force and torque act at that element end? What axial and torsional shear stresses do these resultants produce?

The case tests transfer from element end forces to stress recovery. Only axial and torsional terms are active.

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 mechanics stress components. Component recovery does not establish a combined stress maximum or code stress.

This benchmark recovers forces from a supplied displacement vector. It does not exercise current admission of imposed boundary motion.

Independent numerical comparison. The recorded test exercises a benchmark crate. It does not exercise the desktop product route.

Fig. VI–C3.—Straight-pipe end forces 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
Elastic modulus	E	2000.0 Pa
Shear modulus	G	800.0 Pa
Area	A	4.0 m²
Torsion constant	J	2.0 m⁴
Torsion radius	r	0.5 m
Length	L	5.0 m
Node 1 axial displacement	u_x	0.01 m
Node 1 torsional rotation	θ_x	0.02 rad

2.Method.

Element axial stiffness is EA/L, and torsional stiffness is GJ/L. Multiplication by the imposed stretch or twist gives force or moment at end j. Recovery uses these end values directly, without a sign change.

Axial stress equals end force divided by area. Torsional shear stress equals end torque times torsion radius divided by torsion constant.

k_x = EA / L = 2000.0 × 4.0 / 5.0 = 1600.0(1)

F_x,j = k_x u_x = 1600.0 × 0.01 = 16.0 N(2)

k_t = GJ / L = 800.0 × 2.0 / 5.0 = 320.0(3)

M_x,j = k_t θ_x = 320.0 × 0.02 = 6.4 N·m(4)

σ_axial = F_x,j / A = 16.0 / 4.0 = 4.0 Pa(5)

τ = M_x,j r / J = 6.4 × 0.5 / 2.0 = 1.6 Pa(6)

3.Results.

Expected end forces and stresses at end j
Quantity	Expected
End-j axial force, F_x,j	16.0 N
End-j torsional moment, M_x,j	6.4 N·m
Axial normal stress, σ_axial	4.0 Pa
Torsional shear stress, τ	1.6 Pa

The case record describes these checks:

The end forces are reported for end j of the element.
The end axial force matches 16.0 N and the end torsional moment matches 6.4 N·m.
The axial normal stress matches 4.0 Pa and the torsional shear stress matches 1.6 Pa.
The stress recovery completes without being blocked.
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 checked force, moment and stress must match the hand calculation to within an absolute difference of 1.0 × 10⁻⁹ in its own units. Yes-or-no conditions, such as the end label and the review flag, 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 archived source at 462f66975d36cd10d5b8455193ada9517d28d353. Use a checkout of that exact revision. This edition did not run the command. The recorded result does not establish current product support.

git checkout --detach 462f66975d36cd10d5b8455193ada9517d28d353
cd projects/chirality-piping
cargo test --manifest-path validation/benchmarks/stress/Cargo.toml integrated_straight_pipe_resultants_feed_stress_recovery_fixture

Hand calculation: validation/hand_calcs/stress/integrated_straight_pipe_resultants.md. Test record, with the recorded run of 2026-07-10: stress-integrated-straight-pipe-original.md.