Truss Analysis Calculator — Axial Forces, Stress & Member Sizing

Q: How do I know if my truss is statically determinate?

For a 2D truss: $ m + r = 2j $ where $ m $ = members, $ r $ = reactions, $ j $ = joints. If $ m + r > 2j $, the truss is indeterminate (the solver handles this). If $ m + r < 2j $, it is a mechanism and will fail.

Q: What is a zero-force member?

A member carrying zero axial force under the given loading. The solver identifies these automatically. They often occur at unloaded joints where geometry creates equilibrium without force.

Q: Can I use this for 3D space trusses?

The current module supports 2D planar trusses only. For 3D space trusses, decompose into planar sub-trusses or use dedicated 3D software.

Q: How does the solver handle thermal loads?

Thermal loads are not currently supported. To approximate thermal effects, compute the equivalent member force $ N = EA\alpha\Delta T $ and apply it as a pre-load externally.

Q: What sign convention is used for member forces?

Positive force indicates tension (member being stretched); negative indicates compression. This follows standard structural engineering convention.

Check	IND	US	EU	ISO
Member axial utilization	implemented	—	—	—

How engineers analyze trusses

Trusses are structures composed of straight members connected at pin joints, designed so that all loads are applied only at joints. Under these idealized conditions, every member is a two-force element carrying only axial force — either tension or compression. This makes trusses extremely efficient: material is used purely in direct stress with no bending waste.

Engineers analyze trusses to determine which members are in tension, which in compression, and whether peak axial stress exceeds the allowable for the chosen section. The classical methods of joints (equilibrium at each pin) and method of sections (free-body cuts) work for simple determinate trusses. For indeterminate trusses or complex topologies, the direct stiffness method with bar elements is the standard computational approach, assembling axial stiffness $E A / L$ per member.

The PhyCalcPro trusses module implements bar-element FEM: users define nodes, members, and supports; the solver returns member forces, stress utilizations, and identifies zero-force members automatically.

Truss configurations

Truss Type	Geometry	Application
Pratt	Diagonals slope toward center	Roof trusses, short-span bridges
Warren	Alternating diagonal directions	Highway bridges, floor trusses
Howe	Diagonals slope away from center	Heavy timber trusses, older bridges
K-truss	Members form K-shape in panels	Tall bridge trusses reducing member length
Vierendeel	Rigid joints, no diagonals	Architectural facades (behaves as frame)
Space truss	3D tetrahedral/octahedral	Roof structures, tower masts

Engineering workflow

Define joint positions (node coordinates) in the 2D plane.
Connect members between nodes; assign cross-sectional area $A$ and modulus $E$ to each.
Apply support conditions: at least three restraints to prevent rigid-body motion.
Apply loads at joints only (convert distributed member loads to equivalent nodal forces).
Run the bar-element FEM solver.
Identify tension (+) and compression (-) members from signed axial forces.
Check axial stress utilization $U = ∣ σ ∣/ σ_{allow}$ for each member.
Flag zero-force members (may be needed for stability but carry no load under this case).
For compression members, verify buckling capacity using the Column Buckling module.

Key quantities and formulas

Bar element stiffness (local coordinates):

\frac{E A}{L} [1 - 1 - 1 1] {u} = {F}

Axial stress and utilization:

σ = \frac{N}{A}, U = \frac{∣ σ ∣}{σ _{allow}}

Member force recovery from nodal displacements:

N = \frac{E A}{L} (u_{j} - u_{i}) \cdot cos θ + (v_{j} - v_{i}) \cdot sin θ)

Member elongation:

δ = \frac{N L}{E A}

Worked example

Problem: A simple Warren truss with 3 panels spans 9 m (3 m per panel) and has a height of 2 m. Two equal 40 kN loads act at the bottom chord interior joints. All members have $A = 12$ cm $^{2}$ , $E = 210$ GPa. Find the maximum member force and stress.

Step 1 — Reactions:

Total load = 80 kN, symmetric. $R_{A} = R_{B} = 40$ kN.

Step 2 — Method of sections (center panel diagonal):

Cut through the center panel. Taking moments about the top chord joint above the cut:

N_{bottom} = \frac{M}{h} = \frac{40 \times 4.5 - 40 \times 1.5}{2.0} = \frac{120}{2.0} = 60 kN (tension)

Maximum diagonal force (from joint equilibrium):

N_{diag} = \frac{40}{sin θ} = \frac{40}{sin ( 33.7° )} = 72.1 kN

where $θ = arctan (2/3) = 33.7°$ .

Step 3 — Stress in critical member:

σ = \frac{N}{A} = \frac{72.1 \times 1 0 ^{3}}{12 \times 1 0 ^{- 4}} = 60.1 MPa

Utilization against allowable 165 MPa (S275 with SF=1.67): $60.1/165 = 0.36$ — adequate.

Common mistakes and checks

Applying loads between joints: Truss theory assumes loads act only at nodes. Distributed loads on a top chord must be resolved to the adjacent nodes before analysis.
Ignoring compression buckling: Tension members fail by yielding, but compression members fail by buckling at loads well below yield. Always check compression members with the Column Buckling module.
Insufficient supports: A 2D truss needs at least 3 independent restraints (e.g., pin + roller). Too few produces a singular stiffness matrix; too many produces an indeterminate truss (which the solver handles).
Misidentifying zero-force members: Zero-force members under one load case may carry force under others. Do not remove them without checking all load combinations.
Neglecting connection eccentricity: Real truss connections have gusset plates with eccentricities that introduce secondary bending; the idealized pin-joint model ignores this.
Using member self-weight incorrectly: Distributed self-weight must be split equally to end nodes; applying it as a member load is incorrect for truss analysis.

FAQ

How do I know if my truss is statically determinate?

For a 2D truss: $m + r = 2 j$ where $m$ = members, $r$ = reactions, $j$ = joints. If $m + r > 2 j$ , the truss is indeterminate (the solver handles this). If $m + r < 2 j$ , it is a mechanism and will fail.

What is a zero-force member?

A member carrying zero axial force under the given loading. The solver identifies these automatically. They often occur at unloaded joints where geometry creates equilibrium without force.

Can I use this for 3D space trusses?

The current module supports 2D planar trusses only. For 3D space trusses, decompose into planar sub-trusses or use dedicated 3D software.

How does the solver handle thermal loads?

Thermal loads are not currently supported. To approximate thermal effects, compute the equivalent member force $N = E A α Δ T$ and apply it as a pre-load externally.

What sign convention is used for member forces?

Positive force indicates tension (member being stretched); negative indicates compression. This follows standard structural engineering convention.

Use the PhyCalcPro calculator

Open the Truss Analysis calculator

Purpose

Determine axial forces in two-dimensional pin-jointed truss members under nodal loading. Identifies tension and compression members, flags zero-force links, and reports axial stress utilization against allowable values for preliminary truss sizing.

Physics & theory

Truss members are two-force elements carrying only axial force along the member axis. At each pin joint, equilibrium ( $\sum F_{x} = 0$ , $\sum F_{y} = 0$ ) holds. The structure stiffness matrix involves only translational DOFs. Axial stress is $σ = N / A$ . Compression members require separate buckling checks.

Governing equations

\frac{E A}{L} [1 - 1 - 1 1] {u} = {F}, σ = \frac{N}{A}, U = \frac{∣ σ ∣}{σ _{allow}}

Numerical method

Bar-element FEM: each member contributes axial stiffness in global coordinates after direction-cosine transformation. The assembled system is solved for nodal displacements; member forces are recovered from relative end displacements.

Inputs

Parameter	Description
Nodes	$x, y$ coordinates, support conditions
Members	End nodes, cross-sectional area $A$ , elastic modulus $E$
Loads	Nodal force components
Allowable stress	For utilization screening

Outputs

Member axial force (signed tension/compression)
Axial stress and utilization ratio per member
Reaction forces at supports
Zero-force member identification
Deformed shape (optional)

Design codes & checks

Indicative: Member axial utilization
US: AISC 360 tension/compression member context (screening)
EU: EN 1993-1-1 member rules (screening)

Assumptions & limitations

Pin joints, members connected at centroidal axes.
No joint eccentricity, secondary bending, or in-module buckling check.
2D planar truss; no 3D spatial truss.
Linear elastic; no cable slack or compression-only release logic.
Loads must be applied at nodes only.

Verification

CI: verification benchmarks in src/data/verification/ where available
Engineer sign-off: validation-master-checklist.md

References

Hibbeler, R. C. Structural Analysis, 10th ed. Pearson.
AISC. Steel Construction Manual, 16th ed.
EN 1993-1-1:2005. Eurocode 3 — Tension and compression members.
Kassimali, A. Structural Analysis, 6th ed. Cengage.
ISO 10721:1997. Steel structures — Static analysis and design.

Truss Analysis: Pin-Jointed Structure Engineering Guide

Assumptions

Limitations

Engineering checks