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Structural Analysis

Statics Truss Analyzer (Method of Joints)

Set Warren or Pratt truss panel count, height, and load to automatically calculate all member forces using the method of joints (ΣFx=0, ΣFy=0). Three-tab layout with color-coded truss diagram (tension blue / compression red), member force chart, and stress ratio chart.

Truss Settings

Truss Type
Panel Count N
Panel Width L
m
Height H
m
Joint Load P
kN

While paused, moving a slider updates the results immediately.

Member forces

Results
0.0
Reaction RA (kN)
0.0
Reaction RB (kN)
0.0
Max tension (kN)
0.0
Max compression (kN)
0
Zero-force members
m+r=2j
Members m / determinacy check m+r=2j (method of joints)
Truss

Blue: tension members, red: compression members. Line width is proportional to force magnitude. ▼ marks applied loads, ▲ marks support reactions.

Forces

Axial force in each member. Positive = tension (blue), negative = compression (red).

Stress

Stress for an assumed cross-section A = 10 cm². Dashed lines: allowable tensile stress 200 MPa (green) and allowable compressive stress −150 MPa (orange).

Theory & Key Formulas

$$\sum F_x = 0, \quad \sum F_y = 0 \text{ (each joint)}$$ $$m = 2n - 3 \text{ (determinacy condition)}$$

Conversation: Deepen Your Understanding

🙋
I see trusses all the time in bridges and roofs—why is that triangular arrangement so strong? Wouldn't a square shape work?
🎓
A quadrilateral can distort under diagonal shear and become a parallelogram. A triangle is different: once its three side lengths are fixed, its shape is fixed. That geometric stability is the source of truss strength. By combining triangles, a truss can support large loads with relatively light members.
🙋
When I tried it in the simulator, the top chord (upper horizontal member) turned red (compression) and the bottom chord (lower horizontal member) turned blue (tension). Why is that?
🎓
A simply supported beam is a good analogy: under load, its top face is in compression and its bottom face in tension. A truss works the same way—the top chord acts as a compression strut and the bottom chord as a tension tie, like a bow with its string drawn. That is why high-strength steel is often used for the bottom chord of long-span bridges.
🙋
I see the “static determinacy condition m = 2n - 3”—what happens if it's not satisfied?
🎓
If m < 2n-3, the truss is unstable: it does not have enough members to hold its shape. If m > 2n-3, it is statically indeterminate: there are extra members, so equilibrium equations alone cannot solve it. Real bridges often use indeterminate trusses because the redundancy lets other members take over the load if one fails. Solving them, however, requires taking displacements into account, typically with FEM (the finite element method).
🙋
In a Warren truss, why do the diagonal members alternate between blue (tension) and red (compression)?
🎓
Shear varies symmetrically from the supports toward midspan. Near a support, the upward reaction produces shear and a diagonal carries tension; as you move toward the centre the shear direction changes and the diagonal goes into compression. This alternating pattern is characteristic of the Warren truss, which also has the advantage that all diagonals have the same length, making fabrication easier. Increase the panel count and the alternating pattern becomes clear.
🙋
When trusses are handled in a CAE FEM analysis, is it doing the same calculation as this simulator?
🎓
Essentially yes—it does the same thing in one matrix solve. Each member's stiffness matrix (2×2 to 4×4) is assembled into a global stiffness matrix, boundary conditions are applied, and the system KU=F is solved (K: stiffness matrix, U: displacement vector, F: force vector). The result matches the method of joints, but because it is a matrix method it handles indeterminate and 3D structures in the same way. Trying a truss first in Ansys is a good way to learn the basics of FEM.

Frequently Asked Questions

A truss whose member count satisfies m = 2n - 3 (n = number of joints), so that all member forces are uniquely determined by the force equilibrium equations alone. With fewer members it is unstable (it collapses); with more it is statically indeterminate (it cannot be solved without considering deformation).
The method of joints solves the equilibrium of each joint in turn and is suited to finding all member forces. The method of sections (Ritter method) cuts the truss with an imaginary section and finds specific member forces directly from the equilibrium of that section, which is efficient when only certain members are needed. This simulator uses the method of joints.
Compression members risk buckling—a sudden sideways deformation that leads to collapse. The buckling load is given by Euler's formula Pcr = π²EI/(KL)² and becomes smaller the longer and more slender the member is. A tension member can be used up to its yield strength as long as its cross-section is sufficient, whereas a compression member may fail at a stress far below the yield stress.
The top and bottom chord forces (bending moment ÷ height) become smaller. The deeper the truss, the lower the chord forces and the more material is saved, although self-weight and wind resistance then become issues. In bridge design an economical truss depth of about span/8 to span/12 is a common guideline. You can watch the chord forces change as you vary H in the simulator.
Even members that carry no force under the normal load case are kept for purposes such as: ① reinforcement when the load pattern changes, ② buckling restraint (dividing long compression members into shorter segments to raise their buckling load), and ③ stability during construction. The key judgment is that a member unnecessary for the design load may still be needed under real loads.
Historically, Pratt trusses have been used more often in railway and highway bridges. The reason is efficiency: the diagonals tend to be in tension (so slender high-strength members can be used) while the verticals take compression (they are short, so less prone to buckling). With advances in welding, Warren trusses have also become more common in modern construction.

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How to Use

  1. Choose the truss type (Warren or Pratt) and the panel count N, then set the panel width L (span = N×L) and the height H
  2. Set the joint load P (applied downward at every interior top-chord joint, total (N−1)P)
  3. The method of joints computes the axial force in every member; check them in the truss diagram (blue = tension, red = compression), the member force chart and the stress chart for an assumed section A = 10 cm²
  4. Support reactions and each member's axial force are also listed numerically (Member forces)

Worked Example

Pratt truss with N = 6 panels, panel width L = 4 m (span 24 m), height H = 3 m and joint load P = 20 kN (five interior joints, 100 kN in total): the reactions are RA = RB = 50 kN. The maximum compression is −120 kN in the central top chords U3 and U4 (midspan bending moment 360 kN·m ÷ H = 3 m), the maximum tension is 106.7 kN in the bottom chords L3 and L4, and the end diagonals D1 and D6 carry 83.3 kN in tension. With the assumed section A = 10 cm², the top-chord stress is −120 MPa, 80% of the allowable compressive stress of −150 MPa. Switching to Warren with the same inputs, the maximum tension becomes 120 kN at the central bottom chord and the end diagonals D1 and D6 turn into −83.3 kN compression members that need a buckling check.

Practical Notes