Free engineering calculators

First version

2D truss

Paste nodes and bars, define supports and nodal loads, and solve axial forces and displacement.

MPa
m²

One row per item. Up to 40 nodes and 100 members; coordinates ±100 m. pinned restrains x and z; rollerX restrains z; rollerZ restrains x. Loads are signed in global x/z. Mechanisms are rejected by the solver.

Deflection0.000209737 m

Support reactions

IDRx (kN)Rz (kN)My (kN·m)
1-0.00000000000000710543150
50150

Member axial forces

Tension (+) · Compression (−)

N1: 15 kNN2: 15 kNN3: 15 kNN4: 15 kNN5: -21.2132 kNN6: -20 kNN7: -20 kNN8: -21.2132 kNN9: 0.00000000000001807 kNN10: -10 kNN11: 0.00000000000001807 kNN12: 7.07107 kNN13: 7.07107 kN

Engine vs formula: Relative difference 1.4 × 10−13 %

N1: Engine 15 / Classical method 15 kN · 1.2 × 10−13 %

N2: Engine 15 / Classical method 15 kN · 5.9 × 10−14 %

N3: Engine 15 / Classical method 15 kN · 1.2 × 10−14 %

N4: Engine 15 / Classical method 15 kN · 1.1 × 10−13 %

N5: Engine -21.2132 / Classical method -21.2132 kN · 5 × 10−14 %

N6: Engine -20 / Classical method -20 kN · 0 %

N7: Engine -20 / Classical method -20 kN · 0 %

N8: Engine -21.2132 / Classical method -21.2132 kN · 1.2 × 10−13 %

N10: Engine -10 / Classical method -10 kN · 7.1 × 10−14 %

N12: Engine 7.07107 / Classical method 7.07107 kN · 1.4 × 10−13 %

N13: Engine 7.07107 / Classical method 7.07107 kN · 1.4 × 10−13 %

Open in Stabileo ↗

Node 1 has only truss connections and no rotational restraint — instability risk

Node 2 has only truss connections and no rotational restraint — instability risk

Node 3 has only truss connections and no rotational restraint — instability risk

Node 4 has only truss connections and no rotational restraint — instability risk

Node 5 has only truss connections and no rotational restraint — instability risk

Node 6 has only truss connections and no rotational restraint — instability risk

Node 7 has only truss connections and no rotational restraint — instability risk

Node 8 has only truss connections and no rotational restraint — instability risk

Dense solver (13 free DOFs)

Residual 4.50e-15 (13 free DOFs)

Closed form / Classical method

Method of joints: ΣFx = ΣFz = 0 at every node (determinate trusses).

Assumptions

Linear elastic, small displacements. Prismatic Euler–Bernoulli frame members; truss bars carry axial force only with ideal pin joints. Self-weight is excluded. No shear deformation, geometric nonlinearity or design checks.

Technical details

Production Stabileo solver · planar Euler–Bernoulli frame elements.

Total nodal DOFs: 24