Loads on a turned pipe — SWBPIPE verification

Part VI · VerificationVI–B11

Part VI

Verification

B. Elements and loads · Loads on a turned pipe

A cantilever pipe four metres long runs along global Y, fixed at node 0 and free at node 1. It carries a uniform load of −2.0 N/m and a midspan point force of −4.0 N, both entered in global X. The loads must pass into the pipe’s own axes and the results back out again. Does the solver get the magnitudes and the signs right?

Fig. VI–B11.—Loads on a turned pipe

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
Node 0	—	[0, 0, 0]
Node 1	—	[0, 4, 0]
Length	L	4.0 m
Elastic modulus	E	1000.0 Pa
Shear modulus	G	400.0 Pa
Area	A	3.0 m²
Second moment about y	I_y	1.5 m⁴
Second moment about z	I_z	2.0 m⁴
Torsion constant	J	1.0 m⁴
Uniform load, global X	q	−2.0 N/m
Point force, global X	P	−4.0 N
Point-force position	a/L	0.5

2.Method.

The local axes in global terms are x = [0, 1, 0], y = [1, 0, 0] and z = [0, 0, −1]. A force in global X is therefore a local y force of the same sign. A local moment about z becomes a global moment about Z with the opposite sign.

In local axes the problem is the ordinary cantilever: equivalent end loads from the uniform load and the shape functions, free-end movement from the cantilever formulas, and midspan resultants from equilibrium. The answers are then turned back into global axes.

u_x(L) = −0.032 + −0.013333333333333334 = −0.04533333333333334 m(1)

θ_z,global(L) = −(−0.010666666666666666 + −0.004) = 0.014666666666666668 rad(2)

V_y(x) = 12.0 + (−2.0 × 2.0) + −4.0 = 4.0 N(3)

M_z(x) = 24.0 − 12.0 × 2.0 − (−2.0 × 2.0²/2) − (−4.0 × 0.0) = 4.0 N·m(4)

3.Results.

Global assembled loads, free-end movement and local midspan resultants
Quantity	Expected
Assembled load, node 0, U_x	−6.0 N
Assembled load, node 0, R_z	4.666667 N·m
Assembled load, node 1, U_x	−6.0 N
Assembled load, node 1, R_z	−4.666667 N·m
Free-end displacement in global X	−0.04533333 m
Free-end rotation about global Z	0.01466667 rad
Midspan local shear V_y	4.0 N
Midspan local bending moment M_z	4.0 N·m

The tests check that:

The assembled global force and moment at the fixed node match the hand values, including sign.
The free-end displacement in global X and rotation about global Z match the hand values.
The midspan local shear and bending moment match the hand values.
All of these quantities are finite.

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 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.

For the student

Compare the signs with the unturned pipe. The displacement keeps its sign, but the rotation and end moments change sign. Local z points along negative global Z, so a rotation that is negative in the pipe’s own axes is positive in global axes.

4.Run it yourself.

cd projects/chirality-piping
cargo test --manifest-path validation/benchmarks/mechanics/Cargo.toml tp_phys_005_oriented_load_to_resultant_fixture_transforms_global_loads

Hand calculation: validation/hand_calcs/mechanics/tp_phys_005_oriented_load_to_resultant.md. Test record, with the recorded run of 2026-07-10: mech-tp-phys-005-oriented-load-to-resultant.md.