Engineering question: How can we design, simulate and fairly compare trusses before they are fabricated?
This lesson moves from the prescribed-force calculations in 2.9 to engineering decisions based on competing performance measures.
A truss solver models axial-only members connected at idealized joints. It combines node coordinates, member properties, restraints and external loads to calculate the forces and motion needed for equilibrium.
Axial force is not failure capacity. Even with small simulated displacement, a real member may buckle, crack or fail at a connection. A 2D side-frame model omits the deck, lateral bracing, joints, manufacturing tolerance and three-dimensional effects.
Do not treat the simulated result as a safety certification or a complete check of the later Unit 2 bridge challenge.
E is elastic modulus (N/mm²); A is cross-sectional area (mm²); L is member length (mm). Truss force and deformation depend on geometry, loading, and stiffness under the chosen assumptions.
Use the five-joint, seven-member truss below for your baseline. All coordinates are measured in millimeters. This is an educational side-truss model, not an approved finished bridge.
| Joint | x (mm) | y (mm) | Function |
|---|---|---|---|
| A | 0 | 0 | Pin |
| B | 200 | 0 | 100 N down |
| C | 400 | 0 | Roller |
| D | 100 | 90 | Upper joint |
| E | 300 | 90 | Upper joint |
Members: AB, BC, AD, DB, BE, EC, DE. Design B changes only upper joint height to 140 mm.
| Input | Classroom value |
|---|---|
| Span | 400 mm |
| Load | 100 N downward at B |
| Supports | A pin (x/y); C roller (y only) |
| Elastic modulus E | 3,000 N/mm² |
| Each member area A | 25 mm² |
| Change between designs | Rise: 90 mm vs. 140 mm |
These material properties are hypothetical teaching values. They are not certified properties of your manufactured bridge or a guarantee of printable member strength.
Before running: For a centered 100 N load, static equilibrium predicts 50 N upward at each support. Use that result as a software sanity check.
Watch the SkyCiv lesson first to see a truss defined and solved. Use the next lesson to review how to interpret axial-force results. The third video is an optional professional-software example.
While watching: Write the order in which the instructor specifies nodes, members, supports, applied forces and solution outputs.
Watch on YouTube ↗While watching: Identify two ideal-truss assumptions and explain how tension differs from compression.
Watch on YouTube ↗While watching: Observe where material settings, constraints and result plots appear. ANSYS software is not required for this lesson.
Watch on YouTube ↗The engineering workflow is transferable. Detailed finite-element meshing and professional ANSYS interfaces are beyond today's requirements.
Use the built-in model below or an approved online solver. Free tool features can differ, especially whether nodal displacements are supported.
Build 2D pin-jointed trusses and inspect reactions and axial forces. Check whether your chosen edition reports displacement.
Open SkyCiv Truss Calculator ↗ SkyCiv's truss modeling walkthrough ↗A browser-based 2D truss solver that reports member forces, support reactions and nodal displacement under specified properties.
Open StructureCalcs ↗Review support reactions and truss-member equilibrium to check the simulator against hand calculations.
Read Method of Joints ↗The method of sections from 2.9 can independently validate the force in a selected member of a more complex model.
Open Lesson 2.9 ↗Adjust the rise of each truss. Keep the shared loading and stiffness properties the same between A and B. This is a working 2D linear-elastic truss stiffness solver, not a hand-animated guess.
Gold dashed lines exaggerate calculated displacement for visibility. The physical movement is reported numerically below, not shown to scale.
| Output | Design A | Design B | What it tells you |
|---|---|---|---|
| Support reactions | — | — | Upward force at A / C (N) |
| Downward displacement at B | — | — | Center joint vertical motion (mm) |
| Maximum tension | — | — | Largest member tension (N) |
| Maximum compression | — | — | Largest compression magnitude (N) |
| Total ideal member length | — | — | Sum of seven member lengths (mm) |
| Member-volume proxy | — | — | Length × member area (cm³), no fasteners |
| Member | Design A | Design B |
|---|---|---|
| AB | — | — |
| BC | — | — |
| AD | — | — |
| DB | — | — |
| BE | — | — |
| EC | — | — |
| DE | — | — |
Positive member force = tension; negative = compression. Forces are in N.
With P = 100 N, E = 3,000 N/mm² and area = 25 mm², compare a rise of 90 mm to 140 mm. Which design bends less? Which uses more ideal material volume?
The 140 mm-rise truss predicts smaller center deflection and lower maximum compression under this model, but uses more total ideal member length and volume for the same cross-section. This is a trade-off, not proof of greater real-world capacity.
Keep the geometry and load fixed and double the shared E value. Predict what happens to displacement and member axial forces.
In this statically determinate, linear-elastic truss, doubling E halves displacement while axial member-force results remain unchanged when geometry and load are fixed.
A good engineering report compares software results to at least one independent hand calculation.
For P = 100 N, expect 50 N up at each end and zero horizontal reaction. If the software disagrees, check the support and load definitions.
Prepare two versions of the five-joint truss with 400 mm span, 100 N at the center and identical supports and E/A. Change one geometry feature, such as upper-joint rise, and defend your choice using calculated evidence.
Write a sentence or short calculation for each question in your notebook.
A 400 mm truss carries a centered 120 N downward point load. What are its vertical reactions for ideal pin/roller supports?
Each reaction is 60 N upward, obtained from moment equilibrium about either support and vertical-force balance.
Why might a solver report a singular model if the right support does not prevent vertical motion?
The structure may have a rigid-body degree of freedom or insufficient restraint, so the stiffness equations are not solvable.
A truss remains geometrically unchanged, with the same load and cross-sections. What happens to its predicted displacement if E doubles?
With linear-elastic behavior it becomes half as large. In this determinate model member axial forces are unchanged.
The coordinates are entered in meters but E is entered in N/mm² without adjusting solver unit conventions. What is wrong?
The model mixes unit systems and may predict severely incorrect stiffness/displacement. Convert all quantities to consistent units before solving.
The solver reports 80 N compression in a truss member. Does that tell you it cannot buckle?
No. Buckling capacity depends on geometry, EI, end restraint and imperfections, so force demand alone cannot prove safety.
A student changes both the load and material between Designs A and B. Can they attribute displacement changes only to truss geometry?
No. Keep common loads, properties, member area, span and restraints; change a single design factor at a time.
A taller truss bends less but has longer total members. What trade-off is involved?
Lower predicted displacement can require greater member length and nominal volume; selection should balance multiple outcomes rather than one best number.
A side-truss simulation has less than 10 mm center displacement. Does that prove the class bridge meets every design requirement?
No. It excludes full 3D construction, width, mass, joints, buckling, deck loads and real-world testing. It is a preliminary analysis, not complete compliance evidence.
Complete the POE 2.10 Truss Design & Simulation comparison using the worksheet, notebook or approved PDF. The on-page simulation is acceptable as an educational model; the teacher may request external-tool confirmation.
Due date: See the Daily Agenda and follow your teacher's submission instructions.
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