Load over part of a span — SWBPIPE verification

Part VI · VerificationVI–B12

Part VI

Verification

B. Elements and loads · Load over part of a span

A cantilever pipe four meters long carries −2.0 N/m only over its middle half, from one-quarter to three-quarters of its length. What equivalent end loads does this patch produce? How does the free end move? What shear and moment occur at midspan, where part of the patch lies between the fixed end and the section?

This page preserves a recorded benchmark and its expected values. The dated run below is not a new run of the current product.

Fig. VI–B12.—Load over part of a span

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
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 Y	q	−2.0 N/m
Loaded span start	a/L	0.25
Loaded span end	b/L	0.75
Resultant station	x/L	0.5
Supports	—	Node 0 fixed, node 1 free

2.Method.

The equivalent end loads come from integrating the cubic beam shape functions over the loaded interval only.

With node 0 fixed, two unknowns remain at node 1: the displacement and the rotation. The hand calculation writes the 2 × 2 bending stiffness for them and solves it directly against the free-end loads.

The midspan resultants follow from equilibrium, starting from the fixed-end shear of 4.0 N and moment of 8.0 N·m. Only the part of the patch between 1.0 m and 2.0 m counts.

F_y,i = qL ∫_a^b N_i dr = −2.0 N, M_z,i = qL² ∫_a^b (r − 2r² + r³) dr = −11/6 N·m(1)

K_free = [375, −750; −750, 2000], F_free = [−2.0, 11/6](2)

u_y,1 = −7/500 = −0.014 m, θ_z,1 = −13/3000 = −0.004333333333333333 rad(3)

V_y(x) = 4.0 + (−2.0 × 1.0) = 2.0 N(4)

M_z(x) = 8.0 − 4.0 × 2.0 − (−2.0 × 0.5) = 1.0 N·m(5)

3.Results.

Equivalent end loads, free-end movement and midspan resultants
Quantity	Expected
Equivalent load, node 0, U_y	−2.0 N
Equivalent load, node 0, R_z	−1.833333 N·m
Equivalent load, node 1, U_y	−2.0 N
Equivalent load, node 1, R_z	1.833333 N·m
Free-end displacement in Y	−0.014 m
Free-end rotation about Z	−0.004333333 rad
Midspan shear V_y	2.0 N
Midspan bending moment M_z	1.0 N·m

The case record describes these checks:

The equivalent force and moment at both nodes match the hand values.
The free-end displacement and rotation match the hand values.
The midspan 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.

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/mechanics/Cargo.toml tp_phys_006_partial_span_load_to_resultant_fixture_routes_spanned_loads

Hand calculation: validation/hand_calcs/mechanics/tp_phys_006_partial_span_load_to_resultant.md. Test record, with the recorded run of 2026-07-10: mech-tp-phys-006-partial-span-load-to-resultant.md.