Calculator · Structural Mechanics
2D Truss Solver Calculator
Define a planar pin-jointed truss joint by joint and solve it by the method of joints: every joint satisfies ΣFx = 0 and ΣFy = 0. Member forces are reported signed — positive in tension, negative in compression — together with the support reactions and the equilibrium residuals that prove the solution.
Truss
Loads
Joints
Coordinates are authoritative: +x to the right, +y upwards. Up to 24 joints in this version. Removing a joint also removes the members, support and loads attached to it.
Members
Area is optional: leave it at 0 and the member force is still solved, only its stress is withheld. Up to 48 members in this version.
Supports
A truss support carries translational reactions only — an ideal pin joint transmits no moment. One support per joint; a planar truss needs at least three restrained directions.
Nodal loads
Loads act at joints only. Angles are measured counter-clockwise from +x; magnitude cannot be negative. Several loads at one joint add, and a load at a support joint is valid.
Results
Classification
Stable, statically determinate; rank 6 of 6 unknowns.
Largest member force |N|
Magnitude only; signed tension/compression values are shown below.
Largest joint residual
4.8e-16 of the force scale — numerical round-off.
Support reactions
| Joint | Support | Component | Reaction |
|---|---|---|---|
| A | Pin | Rx | ≈ 0 N |
| A | Pin | Ry | +5000 N |
| B | Roller ↑ | Ry | +5000 N |
Member axial forces
| Member | i–j | L [m] | N | State | σ = N/A |
|---|---|---|---|---|---|
| AB | A–B | 4 | +3333.33 N | T — tension | 6.6667 MPa |
| AC | A–C | 3.606 | -6009.25 N | C — compression | -15.023 MPa |
| BC | B–C | 3.606 | -6009.25 N | C — compression | -15.023 MPa |
Equilibrium check
| Classification | Stable and statically determinate: m + r = 3 + 3 = 2j = 6, and the equilibrium matrix has full rank 6. |
|---|---|
| Global ΣFx of loads and reactions | +5.22959e-12 N |
| Global ΣFy of loads and reactions | +1.81899e-12 N |
| Global ΣMz about the origin | -3.638e-12 N·m |
| Sensitivity κ₂ of the equilibrium matrix | 4.378 |
Assumptions
- · Planar, pin-jointed truss; members carry axial force only.
- · Loads are applied at the nodes; member self-weight is not included.
- · Small displacements: equilibrium is written on the undeformed geometry.
- · Member force is positive in tension: a positive force pulls each joint towards the other end of the member.
- · Supports carry translational reactions only; an ideal pin joint transmits no moment.
- · Statically determinate trusses only: no stiffness, displacement or deflected-shape analysis is performed.
Open in the Lab
The whole truss travels with you — joints, members, areas, supports and loads — where you can inspect a single joint's free body and every member in turn.
Technical reference
at every joint: ΣFx = 0, ΣFy = 0
Fx = F cos θ, Fy = F sin θ; θ positive counter-clockwise from +x
u_ij = ((xj − xi)/L, (yj − yi)/L), L = √((xj − xi)² + (yj − yi)²)
member e at joint i contributes +N_e u_ij ; at joint j −N_e u_ij
necessary count: m + r = 2j
σ = N / A
- N
- member axial force, positive in tension[N]
- m
- number of members[—]
- r
- number of scalar support reactions[—]
- j
- number of joints[—]
- A
- member cross-sectional area, optional[m²]
- σ
- member normal stress, same sign as N[Pa]
The truss is planar and pin-jointed: joints are ideal frictionless pins, members are straight two-force members carrying axial force only, and loads are applied at joints. No member bending, no joint moment and no member self-weight are included.
A positive member force means tension, and a tensile force pulls each joint towards the other end of its member. A negative force is compression and pushes each joint away.
m + r = 2j is necessary for static determinacy but NOT sufficient. The solver also takes the rank of the actual joint-equilibrium matrix: a count-balanced truss whose geometry is a mechanism is rank deficient and is reported as unstable rather than solved.
Fewer unknowns than equations is a mechanism; more unknowns than equations is statically indeterminate. This version solves statically determinate trusses only and refuses the indeterminate case rather than picking one arbitrary solution.
Member stress σ = N/A is reported only for the members whose area was supplied. A missing area withholds that one stress and never blocks the force solution.
No displacement, elongation or deflected shape is computed: that needs member stiffness and is outside this version. Nothing here is a code check, a member sizing or a connection design.
Related tools
Statics & Equilibrium treats the whole truss as one rigid body and answers what the external reactions are. This solver goes inside it: the same equilibrium applied joint by joint, which is what reveals the internal member forces.