Tension
Compression
Near-zero
Applied Load
Warren Truss — method of joints force analysis
Member force diagram (tension +, compression −)
A truss bridge distributes loads through a network of triangular frames, converting bending forces into purely axial tension and compression in each member. The method of joints — solving the equilibrium equations ΣF_x = 0 and ΣF_y = 0 at every node — is the classic technique for finding these member forces. Understanding whether each member is in tension (being pulled apart) or compression (being squeezed) is critical for selecting appropriate materials: steel handles both well, but concrete is strong in compression and weak in tension, while timber is the reverse. Real bridge failures, such as the Quebec Bridge collapse of 1907, were traced to underestimated compressive loads causing Euler buckling.
Choose between Warren, Pratt, Howe, and K-Truss configurations, then set the point load magnitude and position and the cross-sectional area of members. The canvas colour-codes each member — red for tension, blue for compression — with width proportional to force magnitude. A bar chart shows all member forces simultaneously, and the safety factor (yield stress divided by maximum stress) warns when members are at risk.
What is the method of joints?
The method of joints is a technique for finding the axial force in every member of a statically determinate truss. At each joint (node), you apply the two equilibrium equations ΣF_x = 0 and ΣF_y = 0. Starting from a joint with only two unknown member forces (typically a support node), you solve two equations in two unknowns, then move to adjacent joints. The process is repeated until all member forces are found.
What is the difference between tension and compression in a truss?
A member in tension is being pulled at both ends — it resists elongation. A member in compression is being pushed inwards — it resists shortening. The sign convention in structural analysis typically assigns positive values to tension and negative to compression. Compressive members are susceptible to buckling (sudden lateral deflection), quantified by Euler's formula F_cr = π²EI/(KL)², where E is Young's modulus, I is second moment of area, and L is member length.
How does a Warren truss differ from a Pratt truss?
In a Warren truss, diagonals alternate direction (forming equilateral triangles with no verticals), distributing loads more evenly and using fewer members. In a Pratt truss, vertical members are in compression and diagonals are in tension under typical downward loading — this suits steel (which handles tension well) and was the dominant 19th-century railway bridge design. The Howe truss reverses this: vertical members in tension, diagonals in compression — better suited to timber verticals.
A plane truss is statically determinate when the number of members m and joints j satisfy m = 2j − 3 (with 3 reaction forces). This means every member force can be found from equilibrium alone, without needing the material stiffness. If m > 2j − 3, the truss is statically indeterminate (redundant members), requiring compatibility equations and knowledge of material properties to solve. If m < 2j − 3, the truss is a mechanism and will collapse.
The safety factor (SF) is the ratio of the material's yield strength to the maximum actual stress: SF = σ_yield / σ_max. Steel bridges typically require SF ≥ 1.5 to 2.0 under normal service loads, with higher factors (2.5–3.0) for structures where failure consequences are severe. Eurocode 3 (structural steel design) specifies partial safety factors for loads and materials separately, typically giving an effective overall SF of around 1.5. An SF below 1.0 indicates certain failure.
Euler buckling is the sudden lateral deflection of a slender compressive member when the axial load exceeds the critical Euler load F_cr = π²EI/(KL)², where K is the effective length factor (1.0 for pin-ended, 0.5 for both ends fixed). Slender members with low I (second moment of area) or long effective length buckle at loads far below the material's compressive yield strength. This is why compressed top chords of bridges are often box sections or wide flanges — to increase I without adding much weight.
A triangle is the only polygon that is inherently rigid — it cannot deform into another shape without changing the length of at least one member. A quadrilateral frame without diagonals is a mechanism that folds easily. By triangulating a structure, every external load is resolved into tension or compression along straight members, avoiding the bending moments that occur in beams and that require far more material to resist.
Modern bridge trusses are almost exclusively structural steel (typically S355 grade in Europe, with yield strength 355 MPa), often hot-dip galvanised or painted for corrosion protection. High-strength steel cables (σ_y ≈ 1600 MPa) are used in suspension and cable-stayed bridges. Older bridges used wrought iron, cast iron, or timber; some modern pedestrian bridges use aluminium alloys or carbon-fibre composites to minimise self-weight.
Stress σ = F / A, where F is the axial force in the member (in Newtons, positive for tension) and A is the cross-sectional area in m². The simulator applies this formula for each member using the forces computed by the method of joints and the cross-sectional area you set. For a steel member with A = 0.01 m² and F = 250 kN, stress = 250,000 / 0.01 = 25 MPa — well below the yield strength of 355 MPa for S355 steel.
The K-truss (or Belgian truss) uses diagonal members that meet at the mid-point of vertical members, forming K-shaped panels. It is efficient for long-span bridges because the shorter K-shaped diagonals reduce buckling length. The Forth Railway Bridge in Scotland (completed 1890) uses a cantilever truss arrangement, and many early 20th-century American rail bridges employed Pratt and Warren K-trusses. Today, K-truss bridges are still built for railway and highway viaducts where spans exceed 60–80 m.