Two forces act in global Y at the same node, one of +8.0 N and one of −3.0 N. A member also carries a uniform weight of 1.25 N/m in global Z, and another node is given an imposed displacement of −0.02 m in Z. Before any solution, does the solver add the two forces and keep the weight and the displacement as separate contributions?

This is a check of load bookkeeping rather than of structural response.

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.

2.Method.

Forces at the same node and in the same direction add. The element weight and the imposed displacement are different kinds of input and pass through unchanged.

3.Results.

The tests check that:

• Load preparation completes without blocking.
• It records two nodal loads, one element uniform load and one imposed displacement.
• The global load vector at node 1 in Y matches the summed force.
• The weight intensity and the imposed displacement pass through with their values unchanged.

What it shows. Checks the loads or boundary conditions prepared for the solver against a hand calculation. No system is solved.

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 computed value must match the reference within an absolute difference of 1.0 × 10⁻⁹ in the case’s own units; counts and structural outcomes must match 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/mechanics/Cargo.toml primitive_load_fixture_prepares_all_contribution_types

Hand calculation: validation/hand_calcs/mechanics/primitive_load_preparation.md. Test record, with the recorded run of 2026-07-10: mech-primitive-load-prep.md.