Partial-span load to stress — SWBPIPE verification

Part VI · VerificationVI–C9

Part VI

Verification

C. Stress recovery · Partial-span load to stress

A straight pipe 4.0 m long is fixed at node 0 and free at node 1. A uniform load of −2.0 N/m acts over its middle half only, from a quarter to three quarters of the length. What bending moment acts at midspan, and what bending and axial stresses follow from it?

Fig. VI–C9.—Partial-span load 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
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 local y	I_y	1.5 m⁴
Second moment about local z	I_z	2.0 m⁴
Torsion constant	J	1.0 m⁴
Uniform local y load	q	−2.0 N/m
Load span	a/L to b/L	0.25 to 0.75
Station	x/L	0.5
Section modulus about y	Z_y	2.5 m³
Section modulus about z	Z_z	2.0 m³

2.Method.

The partial-span load is replaced by equivalent loads at the free end, F_y,j = −2.0 and M_z,j = 1.8333333333333333. With EI_z = 2000.0 and L = 4.0 the free-end bending stiffness terms follow. Solving the two-unknown system gives a free-end deflection of −0.014 m and a rotation of −0.004333333333333333 rad.

From these the solver recovers the fixed-end shear V_y,i = 4.0 and moment M_z,i = 8.0. At the station x = 2.0 m, the load is active only from 1.0 m to 2.0 m. Its moment about the station enters through the lever integral.

k_vv = 12 EI_z / L³ = 375.0, k_vθ = −6 EI_z / L² = −750.0, k_θθ = 4 EI_z / L = 2000.0(1)

lever = x · (active length) − (active end² − span start²) / 2 = 2.0 × 1.0 − (2.0² − 1.0²) / 2 = 0.5(2)

M_z(x) = M_z,i − V_y,i x − q · lever = 8.0 − 4.0 × 2.0 − (−2.0 × 0.5) = 1.0 N·m(3)

σ_b,z = M_z / Z_z = 1.0 / 2.0 = 0.5 Pa(4)

σ_axial = N / A = 0.0 / 3.0 = 0.0 Pa(5)

3.Results.

Expected midspan values
Quantity	Expected
Midspan bending moment, M_z	1.0 N·m
Bending stress about z	0.5 Pa
Axial normal stress	0.0 Pa

The tests check that:

The midspan stress recovery completes without being blocked and is labelled as this case’s midspan station.
The midspan bending moment matches 1.0 N·m.
The bending stress about z matches 0.5 Pa.
The axial normal stress is exactly zero.
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. The moment and stresses must match the hand calculation to within an absolute difference of 1.0 × 10⁻⁹ in their own units. The zero axial stress, station label and review flag are checked exactly. Long values are shown here to seven significant figures; the tests compare the full values in the record.

For the student

Only the part of the load between the fixed end and the station belongs to the free body cut at the station. Here that is the one metre from 1.0 m to 2.0 m. The lever integral is simply the moment of that loaded length about the cut.

4.Run it yourself.

cd projects/chirality-piping
cargo test --manifest-path validation/benchmarks/stress/Cargo.toml partial_span_load_to_stress_fixture_recovers_midspan_bending_from_spanned_load

Hand calculation: validation/hand_calcs/stress/tp_phys_006_partial_span_load_to_stress.md. Test record, with the recorded run of 2026-07-10: stress-tp-phys-006-partial-span-load-to-stress.md.