Assemble element stiffness matrices into the global system and solve:
$$[K]\{u\}=\{F\}$$Member axial force from nodal displacements:
$$f = \frac{EA}{L}\bigl[(u_j-u_i)\cos\theta + (v_j-v_i)\sin\theta\bigr]$$Element stiffness: $k = EA/L$, transformed to global DOFs via rotation matrix.
While paused, moving a slider updates the results immediately.
What is the Direct Stiffness Method for Trusses?
Physical Model & Key Equations
The fundamental system equation assembled from all truss members. Each member acts as a spring oriented in 2D space, contributing to the overall stiffness matrix $[K]$.
$$[K]\{u\}= \{F\}$$$[K]$: Global stiffness matrix (assembled from all elements).
$\{u\}$ : Vector of all unknown nodal displacements ($u_i, v_i$).
$\{F\}$: Vector of applied nodal forces.
Solving this gives you how much every node in the truss moves.
Once displacements are known, the axial force in any member (between nodes i and j) is calculated. This is the formula that determines the tension (positive) or compression (negative) visualized in the simulator.
$$f = \frac{EA}{L}\bigl[(u_j-u_i)\cos\theta + (v_j-v_i)\sin\theta\bigr]$$$E$: Young's Modulus (material stiffness, set by the slider).
$A$: Cross-sectional Area (member thickness, set by the slider).
$L, \theta$: Member length and angle from horizontal.
$u, v$: Nodal displacements from the system solution.
The term in brackets $[...]$ is essentially the member's elongation projected along its axis.
Real-World Applications
Bridge Design & Optimization: Engineers use this exact analysis to size members in steel truss bridges. By running analyses with different load cases (like a truck on the bridge), they can identify which members are in high compression and need to be reinforced, directly using the force values $f$ calculated by this formula.
Roof and Tower Structures: The analysis of transmission towers, radio masts, and warehouse roofs relies on planar truss models. The color-coding is vital for quickly spotting overstressed members—compression members (blue) often fail by buckling, which requires a different check than tension members (red).
CAE Software Verification: The formulation here is identical to elements like ANSYS LINK180 or Abaqus T2D2. This tool is perfect for performing a "sanity check" on a small model before building a complex, time-consuming FEM analysis in commercial software, ensuring your boundary conditions and loads make sense.
Educational Hand Calculations: In structural engineering courses, students learn to solve small trusses by hand. This simulator validates those manual matrix solutions instantly. Changing E and A shows the principle of load path and how stiffness distribution affects force flow in statically indeterminate structures.
Common Misunderstandings and Points to Note
When you start using this simulator, there are a few points you should be aware of. First, we often hear comments like, "The displacement is too small to see! Is the calculation wrong?" For an actual steel bridge (E=205 GPa), if you apply a 10 kN load to a member with a cross-sectional area of 1000 mm², the elongation is on the order of mere microns. Remember that the simulator exaggerates displacements for visibility. While you can trust these calculation results in practice, pay close attention to the modeling of "support conditions". Here we use simple pin supports, but in actual structures, rotation and movement are often somewhat restrained, which can significantly change the results.
Next, regarding the setting of parameters "E" and "A". For example, if you make "A" extremely small, the axial force remains the same but the displacement becomes enormous and unrealistic. Conversely, if you set "E" to a wood value (approx. 10 GPa), you get about 20 times the displacement under the same load compared to steel (approx. 200 GPa). While evaluating the effect of changing materials is good, note that if a member becomes extremely slender, buckling—a different phenomenon—becomes dominant. You cannot make safety judgments based solely on this tool's results in such cases.
Finally, be mindful of overlooking "zero-force members". At specific joints with no applied load, some members will have zero axial force due to force equilibrium. The simulator might display zero axial force in black, but in design, these are often members necessary to maintain the structural shape, not "unnecessary members". Try checking this in a K-truss, for instance.
How to Use
- Define member properties: enter Young's modulus (eVal in GPa), member length (the slider in m), and cross-sectional area (aVal in mm²)
- Specify node coordinates and apply loads: input horizontal (fx1, fx2, …) and vertical (fy1, fy2, …) forces in kN at each joint
- Run the Direct Stiffness Method solver—the simulator assembles the global stiffness matrix, solves for nodal displacements, and computes member axial forces
- Review results: max tension/compression in kN, max displacement in mm, and member force distribution with red (tension) and blue (compression) color coding
Worked Example
A simple 2-member truss: steel (E=200 GPa), each member 3 m long, 500 mm² cross-section. Node 1 fixed at origin, node 2 at (3 m, 0), node 3 at (1.5 m, 2.598 m). Apply vertical load of 50 kN at node 3. The solver yields: member 1–3 tension = 57.7 kN (red), member 2–3 tension = 57.7 kN (red), node 3 vertical displacement = 1.44 mm, strain = 7.2×10⁻⁴.
Practical Notes
- Verify reaction forces sum to applied loads; use this check to detect incorrect boundary conditions or load input
- For slender members (L/r > 100), verify buckling limits independently—this solver assumes no lateral instability
- High displacement relative to member length indicates geometric nonlinearity; results become inaccurate beyond ~5% strain
- Aluminum trusses (E=69 GPa) show ~3× larger displacements than steel at identical geometry and load