Solve 2D pin-jointed trusses using the Direct Stiffness Method. Select from 5 preset configurations — Pratt, Warren, Cantilever, K-Truss, or Simple Triangle — and instantly visualize member tension/compression forces and deformed shape.
Truss Configuration
Truss Preset
Material & Section
Young's Modulus E
GPa
Cross-Section Area A
cm²
Loading
Load Magnitude P
kN
Load Direction
Display Options
Deformation Scale
×
Results
Tension (+)
Compression (−)
Zero force
Results
—
Max Displacement [mm]
—
Max Tension [kN]
—
Max Compression [kN]
—
Free DOFs
Truss
Scale: 100×
Red = Tension | Blue = Compression | ▲ = Pin support | △ = Roller support
Theory & Key Formulas
Assemble element stiffness matrices and solve the global equation:
What exactly is the "global stiffness matrix" K that this simulator is assembling? It sounds complicated.
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Basically, it's a giant "spring constant" for the entire structure. Each truss member is like a spring. The Direct Stiffness Method is a systematic way to combine all their individual stiffnesses into one big equation for the whole truss. In practice, the simulator does this for you when you pick a "Truss Preset" above.
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Wait, really? So each bar's stiffness depends on its angle? How does that work?
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Exactly! A bar only resists axial force, but in 2D, that force has components in the x and y directions. The stiffness depends on its orientation, captured by direction cosines (c, s). Try changing the "Load Direction" in the simulator. You'll see how forces in slanted members change dramatically because their c and s values change.
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So after solving K*u=f for displacements (u), how do we know if a member is in tension or compression?
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Great question! Once you have the nodal displacements, you can calculate how much each element elongated or shortened. A common case is a bridge truss: the top chords are usually in compression, and the bottom chords in tension. In the simulator, after you click "Solve", look at the color of the members—red for tension, blue for compression. Try increasing the "Load Magnitude P" to see the forces grow.
Physical Model & Key Equations
The fundamental equation solved in any linear static Finite Element Analysis is the equilibrium of internal and external forces, expressed as a system of linear equations.
$$\mathbf{K}\,\mathbf{u}= \mathbf{f}$$
Here, $\mathbf{K}$ is the global stiffness matrix (assembled from all elements), $\mathbf{u}$ is the vector of unknown nodal displacements, and $\mathbf{f}$ is the vector of applied nodal forces.
The stiffness of an individual truss element (bar) in 2D space, accounting for its orientation, is given by the element stiffness matrix.
Where $A$ is cross-sectional area, $E$ is Young's Modulus, $L$ is element length, and $c = \cos\theta$, $s = \sin\theta$ are the direction cosines of the element's angle $\theta$. This matrix relates the forces and displacements at the two ends (four degrees of freedom) of the bar.
Frequently Asked Questions
A positive value indicates that the member is under tensile force, while a negative value indicates compressive force. On the diagram, these are color-coded in blue (tension) and red (compression), allowing you to see at a glance which members are subjected to which direction of force.
The current version only supports five preset shapes, such as Pratt, Warren, and Cantilever. The ability to edit free-form shapes is under consideration for future updates.
Each preset shape comes with default standard fixed supports and pin supports. By clicking the support icon on the screen, you can individually turn on/off constraints in the x and y directions, allowing you to change to roller supports or fixed supports.
The deformation is displayed with an enlarged scale to improve visibility. You can adjust the magnification factor from 0.1x to 10x using the "Deformation Scale" slider at the bottom right of the screen. If you want to compare with the actual deformation amount, please reduce the scale.
Real-World Applications
Bridge Design: Trusses are the backbone of many bridges, from small pedestrian crossings to large railway spans. Engineers use this exact FEM analysis to ensure each member can safely handle the tension or compression from traffic loads, wind, and its own weight, optimizing material use.
Roof and Tower Structures: The triangular frameworks in roof supports and communication towers (like cell phone masts) are classic trusses. Analysis determines critical members that might buckle under compression, guiding where to add bracing or increase thickness.
Construction Cranes: The boom of a tower crane is a truss structure. Calculating member forces is essential for safety, ensuring the crane doesn't collapse when lifting heavy loads at various angles—a scenario you can mimic by changing the load direction in the simulator.
Aircraft and Spaceframe Design: The internal fuselage frames of some aircraft and the chassis of race cars or spaceframes use truss-like principles. Lightweight strength is paramount, and FEM helps find the minimum material needed to withstand operational forces like acceleration and aerodynamic pressure.
Common Misconceptions and Points to Note
When you start playing with this tool, there are a few points you should be careful about. First, don't forget the "pin joint" assumption. This simulator represents a world where member ends are perfectly connected by "pins" that are free to rotate. Therefore, members only develop axial forces (tension/compression). However, actual welded or bolted connections are somewhat "rigid" and transmit bending moments as well. It's dangerous to directly apply results to real-world design without understanding this difference. For example, while this tool can reproduce the risk of slender members buckling under compression, it cannot show local stress concentrations at the joints.
Next, a sense of realism in parameter settings. Sliding the sliders around is fun, but for instance, changing "Young's modulus E" from 10 GPa (a rubber-like value) to 200 GPa (steel) reduces deformation to less than 1/100th. In real structures, changing materials dramatically affects cost and weight, so this intuition is important. Also, making the "Cross-sectional area A" extremely small causes compression members to deform unnaturally large, which indicates the limits of linear analysis. In reality, before such deformation occurs, a nonlinear phenomenon called buckling would happen, causing failure. Don't take the tool's results at face value; maintain a skeptical eye, asking, "Isn't this compression member too slender?"
Finally, interpreting support conditions. The difference between a fixed support (△) and a roller support (▽) is whether horizontal movement is restrained. For example, making one side of a bridge fixed and the other a roller provides room for the bridge to expand and contract with temperature changes. If both sides are fixed, the tool shows increased stiffness and smaller deformation, but in reality, enormous thermal stress would develop. A single support setting fundamentally changes the structure's behavior and load path.
Select a preset truss geometry (5-bar, 10-bar, or Warren configurations) using the dropdown, or input node coordinates and member connectivity manually.
Define material properties: enter Young's modulus (e.g., 200 GPa for steel), cross-sectional area (e.g., 500 mm²), and applied nodal loads in kN.
The solver assembles the global stiffness matrix K using direction cosines for each member, applies boundary conditions (fixed supports), and solves K*u=f via Gaussian elimination.
Review max displacement magnitude in mm, member tensions/compressions in kN, and the count of free degrees of freedom in the results panel.
Worked Example
A 5-member pin-jointed steel truss with E=200 GPa, A=1000 mm², span=3 m, and vertical load of 50 kN at the apex. After assembly with direction cosines and support constraints at base nodes, the solver yields max vertical displacement of 8.3 mm, peak tension in top chord of 71.4 kN, and max compression in diagonals of 38.2 kN across 6 free DOFs (3 nodes × 2 directions minus 2 support reactions).
Practical Notes
Use 45° diagonal members in simple trusses to minimize direction cosine errors; verify cos(θ)² + sin(θ)² = 1 after rotation calculations.
Slender members with L/r > 100 (e.g., 2 m tubular column, 25 mm radius) may require buckling checks beyond this linear elastic solver.
Pin-joint assumption fails for moment-stiff gusset plates; use FEA frame elements (not trusses) when member bending is significant.
Scale factor adjusts visualization only—does not change stiffness matrix; always work in consistent SI units internally.