Truss Solver (Method of Joints) mechanism

Model scope

The equilibrium formulation and stated application are correct for ideal, stable, statically determinate 2D pin-jointed trusses. Verdict is NEEDS_REVIEW because the specification omits required unit conventions, member-force sign convention, ideal-truss assumptions, and a stability/singularity test; the m+r=2j honest label must not be treated as a sufficient validity check.

• The stated equilibrium equations are the standard method, but m+r=2j is only a necessary determinacy count, not a sufficient stability test; the solver must also detect mechanisms, geometric instability, duplicate or dependent equilibrium equations, and singular matrices.
• The specification does not explicitly state the member-force sign convention, assumed here as positive=tension and negative=compression.
• The method assumes an ideal pin-jointed planar truss: loads and reactions act at joints, members are straight two-force members, joints are frictionless pins, and self-weight is either omitted or converted to equivalent joint loads.
• The reaction equations and support constraints must be assembled consistently with the support types; the formula description does not state how pin, roller, or other restraints map into unknown reactions.
• A singular or ill-conditioned linear system must be rejected rather than solved as a valid truss result.

Units and conventions

• All joint coordinates must use one consistent length unit, and all applied loads must use one consistent force unit; member forces and reactions are returned in the force unit of the loads.
• Direction cosines are dimensionless, so coordinates in millimetres produce the same force ratios as equivalent coordinates in metres, provided all coordinates use the same length unit.
• Moments used for reaction checks have force-length units, such as N·mm or kN·m, with consistent conversion.
• Test vectors use positive x to the right, positive y upward, counterclockwise positive moments, and positive member force for tension.

Inspectable vector

One of 4 source-bound review vectors:

{
  "inputs": {
    "joints": {
      "A": [
        0,
        0
      ],
      "B": [
        4,
        0
      ],
      "C": [
        2,
        3
      ]
    },
    "members": {
      "AB": [
        "A",
        "B"
      ],
      "AC": [
        "A",
        "C"
      ],
      "BC": [
        "B",
        "C"
      ]
    },
    "supports": {
      "A": {
        "type": "pin"
      },
      "B": {
        "type": "roller",
        "restrained_direction": "y"
      }
    },
    "loads": {
      "C": {
        "Fx": 0,
        "Fy": -10
      }
    },
    "coordinate_unit": "m",
    "force_unit": "kN",
    "sign_convention": "positive=tension",
    "determinacy_check": "m+r=2j"
  },
  "expected_outputs": {
    "member_forces": {
      "AB": "10/3 kN (tension)",
      "AC": "-5*sqrt(13)/3 kN (compression)",
      "BC": "-5*sqrt(13)/3 kN (compression)"
    },
    "reactions": {
      "A": {
        "Rx": "0 kN",
        "Ry": "5 kN"
      },
      "B": {
        "Ry": "5 kN"
      }
    },
    "determinacy": {
      "m": 3,
      "r": 3,
      "j": 3,
      "m_plus_r": "6",
      "two_j": "6"
    },
    "joint_residuals": {
      "A": [
        "0 kN",
        "0 kN"
      ],
      "B": [
        "0 kN",
        "0 kN"
      ],
      "C": [
        "0 kN",
        "0 kN"
      ]
    }
  },
  "reference_data": {
    "member_lengths": {
      "AB": "4 m",
      "AC": "sqrt(13) m",
      "BC": "sqrt(13) m"
    },
    "direction_cosines": {
      "AC": {
        "from_A_to_C": [
          "2/sqrt(13)",
          "3/sqrt(13)"
        ]
      },
      "BC": {
        "from_B_to_C": [
          "-2/sqrt(13)",
          "3/sqrt(13)"
        ]
      }
    }
  },
  "working": "Global reactions: sum M_A=0 gives 4*B_y-10*2=0, so B_y=5 kN. Sum F_y=0 gives A_y+5-10=0, so A_y=5 kN. Sum F_x=0 gives A_x=0. At joint C, let F_AC=F_BC by horizontal equilibrium: (-2/sqrt(13))F_AC+(2/sqrt(13))F_BC=0. Vertical equilibrium gives (-3/sqrt(13))F_AC+(-3/sqrt(13))F_BC-10=0. Therefore F_AC=F_BC=-5*sqrt(13)/3 kN. At joint A, F_AB+(2/sqrt(13))F_AC=0, so F_AB-10/3=0 and F_AB=10/3 kN. Residuals: A: (0+10/3-10/3, 5-5)=(0,0); B: (-(10/3)+10/3, 5-5)=(0,0); C: (10/3-10/3, 5+5-10)=(0,0)."
}

Failure boundary

BREAK_CLAIMED · SELF_FLAGGED

A coincident pair of joints creates a zero-length member whose direction cosines contain 0/0, so the solver must reject the input rather than pass it to the matrix solver. Sign and directional conventions also require explicit handling to avoid reversed tension/compression and reaction signs.

Other records

Sources and evidence

These pointers bind the page to evidence. They do not grant authority or collapse verification into deployment.

• VERIFIED_LEDGER.json
• evidence-tranche8/tranche8-results.json
• BOUNDARY_HAZARD_CATALOG.json