From loads to resultants — SWBPIPE verification

Part VI · VerificationVI–B10

Part VI

Verification

B. Elements and loads · From loads to resultants

A straight pipe four metres long is fixed at one end and free at the other. It carries a uniform load of −2.0 N/m over its whole length and a point force of −4.0 N at midspan, both across the pipe. The solver must turn both loads into equivalent end forces and moments, solve for the movement of the free end, and recover the shear and bending moment at midspan.

Fig. VI–B10.—From loads to resultants

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
Point force, global Y	P	−4.0 N
Point-force position	a/L	0.5
Supports	—	Node 0 fixed, node 1 free

2.Method.

Each load is turned into equivalent end forces and moments. The uniform load gives qL/2 and qL²/12 at each end. The point load is shared out by the cubic shape functions of a beam element, evaluated at its position.

The free-end displacement and rotation follow from the standard cantilever formulas for a uniform load and for a point load, added together. The midspan shear and moment follow from equilibrium, starting from the shear of 12.0 N and moment of 24.0 N·m at the fixed end.

F_y,i = qL/2 = −4.0 N, M_z,i = qL²/12 = −2.6666666666666665 N·m (uniform load)(1)

h_i = 1 − 3r² + 2r³ = 0.5, θ_i = L(r − 2r² + r³) = 0.5 (point load, r = 0.5)(2)

v(L) = qL⁴/(8EI_z) + Pa²(3L − a)/(6EI_z) = −0.032 + −0.013333333333333334 = −0.04533333333333334 m(3)

θ(L) = qL³/(6EI_z) + Pa²/(2EI_z) = −0.010666666666666666 + −0.004 = −0.014666666666666668 rad(4)

V_y(x) = V_y,i + qx + P = 12.0 + (−2.0 × 2.0) + −4.0 = 4.0 N(5)

M_z(x) = M_z,i − V_y,ix − qx²/2 − P(x − a) = 4.0 N·m(6)

3.Results.

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

The tests check that:

The assembled force and moment at the fixed node 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.

For the student

The point force sits exactly at the station asked for, and the solver counts it as already passed. That is why the midspan shear is 4.0 N and not 8.0 N. At the free end the same bookkeeping gives zero shear and zero moment, as a free end must.

4.Run it yourself.

cd projects/chirality-piping
cargo test --manifest-path validation/benchmarks/mechanics/Cargo.toml tp_phys_004_load_to_resultant_fixture_assembles_solves_and_recovers

Hand calculation: validation/hand_calcs/mechanics/tp_phys_004_load_to_resultant.md. Test record, with the recorded run of 2026-07-10: mech-tp-phys-004-load-to-resultant.md.