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instrument implementation / implementation-contract
Shear & Moment Diagram Plotter implementation
A source-bound implementation contract for Shear & Moment Diagram Plotter; deployment and executable qualification remain separate.
Binding contract
This record describes what an executable surface must bind and validate. It does not claim that a public calculator, package, or production route is deployed.
- Record owner
- Guth Instruments
- Lifecycle
- verified-with-binding-errata
- Evidence ID
- t6-034
- Effect
- Computation or source lookup only
Input and validation requirements
A zero-length beam with coincident supports produces a 0/0 reaction calculation and has no determinate beam solution. The instrument must reject that input; accepting it would create undefined reactions and invalid diagrams.
Required corrections
[
{
"vector": 4,
"field": "sampled M@1.75",
"wrong": 2.9375,
"correct": 2.8125,
"cause": "own working computes 6.125-3.0625-0.25=2.8125 then field ships 2.9375"
}
]Inspectable interface fixture
The sample below is evidence for interface and validation tests, not a production implementation.
{
"inputs": {
"beam_length_m": 6,
"supports": [
{
"type": "pin",
"position_m": 0
},
{
"type": "roller",
"position_m": 6
}
],
"loads": [
{
"type": "point_load",
"position_m": 3,
"magnitude_kN": 12,
"direction": "downward"
}
],
"sign_convention": "sagging_M_positive",
"integration": "exact_piecewise",
"sampled_x_m": [
0,
2,
3,
"3+",
"4",
6
]
},
"expected_outputs": {
"reactions": {
"R_A_kN": 6,
"R_B_kN": 6
},
"sampled_points": [
{
"x_m": 0,
"V_kN": 6,
"M_kNm": 0
},
{
"x_m": 2,
"V_kN": 6,
"M_kNm": 12
},
{
"x_m": "3-",
"V_kN": 6,
"M_kNm": 18
},
{
"x_m": "3+",
"V_kN": -6,
"M_kNm": 18
},
{
"x_m": 4,
"V_kN": -6,
"M_kNm": 12
},
{
"x_m": 6,
"V_kN": -6,
"M_kNm": 0
}
],
"Mmax": {
"location_m": 3,
"value_kNm": 18,
"location_note": "continuous M; V changes from +6 to -6 at x=3"
}
},
"working": "Equilibrium: R_A+R_B-12=0. Taking moments about A: 6R_B-12(3)=0, so R_B=36/6=6 kN and R_A=12-6=6 kN. For 0<=x<3, w=0, so V=6 kN and M(x)=6x; M(2)=6(2)=12 kN·m and M(3-)=6(3)=18 kN·m. At x=3, the downward point load causes V(3+)=V(3-)-12=6-12=-6 kN, while M is continuous, so M(3+)=18 kN·m. For 3<x<=6, M(x)=18-6(x-3); M(4)=18-6(1)=12 kN·m and M(6)=18-6(3)=0. Since V is positive before x=3 and negative after it, Mmax=18 kN·m at x=3."
}Other records
Sources and evidence
These pointers bind the page to evidence. They do not grant authority or collapse verification into deployment.
VERIFIED_LEDGER.jsonevidence-tranche6/tranche6-results.jsontranche errata record