Worked structural analysis example

Pratt Truss Analysis Worked Example

Solve a 12 m Pratt truss step by step, including support reactions, axial member forces, zero-force members and deflection.

12 m span 3 m height 3 × 16 kN loads 13 members
First-order linear elastic analysis Pin-jointed truss Axial-only members
Open in 2D Calculator Download Worked Example PDF
Model OB-EX-TRUSS-001Geometry and applied loading
Metric units
Four-panel Pratt truss geometry and loading A 12 metre, four-panel Pratt truss with a pin at the left, a vertical roller at the right and three 16 kilonewton downward joint loads. M1 M2 M3 M4 M5 M6 M7 M8 M9 M10 M11 M12 M13 16 kN 16 kN 16 kN 24 kN 24 kN N1 N2 N3 N4 N5 N6 N7 N8
Applied load Support reaction Pin at N1 · vertical roller at N5
Left vertical reaction
24.000 kN
Right vertical reaction
24.000 kN
Maximum compression
33.941 kN
Maximum tension
24.000 kN
Centre vertical displacement
0.656 mm downward

Pratt truss fundamentals

What is Pratt truss analysis?

A Pratt truss is a triangulated structure with top and bottom chords connected by vertical and diagonal web members. Its characteristic interior diagonals slope down toward the centre of the span. Under typical downward gravity loading, those longer diagonals usually act in tension, while the top chord and some vertical or end members act in compression. The exact force pattern still depends on the geometry, supports and load positions.

Pratt truss analysis converts the truss geometry, support conditions and joint loads into support reactions, axial member forces and, when stiffness properties are included, nodal deflections. A statically determinate pin-jointed Pratt truss can be solved by equilibrium using the method of joints and method of sections; a direct-stiffness solver provides the complete force and displacement response.

Geometry and load path

How to recognize a Pratt truss

  • A top chord and bottom chord connected into triangular panels.
  • Interior diagonals that incline down toward the centre of the span.
  • Loads applied at panel points in the ideal pin-jointed analysis model.
  • Members classified by axial tension, compression or zero force.
Solved-analysis workflow

How to analyze a Pratt truss

  1. Draw the free-body diagram and calculate the pin-and-roller support reactions from global equilibrium.
  2. Check static determinacy and stability using m + r = 2j for this planar truss.
  3. Identify zero-force members before solving the remaining member forces.
  4. Use the method of joints for consecutive joint equilibrium and the method of sections for selected member forces.
  5. Calculate deflection by virtual work or solve the complete model with the direct stiffness method.

Which Pratt truss members are in tension or compression?

For this symmetric gravity-load example, the bottom chord and centre-sloping interior diagonals are in tension, while the top chord, end posts and centre vertical are in compression. The two unloaded outer verticals are zero-force members. Reversing or relocating loads can change that pattern.

How is Pratt truss deflection calculated?

The centre vertical deflection is checked by the unit-load method, also called virtual work, using the real and virtual axial forces in each member. The calculator independently obtains every nodal displacement with the matrix direct stiffness method.

Is Pratt truss analysis the same as truss design?

No. Analysis determines reactions, axial forces and displacements for an idealized model. Structural design must additionally check member yielding and buckling, connections, bracing, load combinations, serviceability and the applicable design code.

01 · Problem definition

A symmetric four-panel Pratt truss under joint loading

A simply supported, four-panel Pratt truss spans 12 m and is 3 m high. Three 16 kN downward loads act at the upper panel points. All members use the same steel modulus and cross-sectional area.

Span12 mfour 3 m panels
Height3 mSpan-to-depth ratio 4.0:1
SupportsPin + roller3 restrained translations
MaterialSteelE = 200 GPa
Member area5,000 mm²Uniform for all members
Loading3 × 16 kNDownward at upper joints
Input table

Nodes and supports

Coordinates in metres
NodeXYSupportRestrained DOFs
N1 0.0 0.0 Pin UX, UY
N2 3.0 0.0 Free —
N3 6.0 0.0 Free —
N4 9.0 0.0 Free —
N5 12.0 0.0 Vertical roller UY
N6 3.0 3.0 Free —
N7 6.0 3.0 Free —
N8 9.0 3.0 Free —
Input table

Joint loads

Global axes
LoadNodeFXFYMoment
L1 N6 0.000 kN -16.000 kN 0.000 kN·m
L2 N7 0.000 kN -16.000 kN 0.000 kN·m
L3 N8 0.000 kN -16.000 kN 0.000 kN·m

Sign convention: Positive axial force is tension. Negative axial force is compression. Positive global Y is upward.

Input table

Member connectivity

Pin-ended axial members
M1N1 → N23.000 m
M2N2 → N33.000 m
M3N3 → N43.000 m
M4N4 → N53.000 m
M5N6 → N73.000 m
M6N7 → N83.000 m
M7N1 → N64.243 m
M8N6 → N23.000 m
M9N6 → N34.243 m
M10N7 → N33.000 m
M11N8 → N34.243 m
M12N8 → N43.000 m
M13N8 → N54.243 m

02 · Step-by-step hand calculation

Reactions, zero-force members and key axial forces

The determinate truss can be solved from equilibrium alone. The selected joint and section checks below demonstrate the sign and magnitude of key member forces without reproducing every joint calculation.

1
Determinacy check

m + r = 2j

13 + 3 = 2(8) = 16

The count m + r = 2j satisfies the necessary condition for static determinacy. Inspection of the triangulated geometry and pin-and-roller support arrangement confirms that the truss is stable and has no internal or external mechanism.

2
Support equilibrium

Calculate the reactions

  1. Let A = N1 and B = N5. The geometry and loading are symmetric, so the two vertical reactions are equal.
  2. ΣFy = 0: RAy + RBy - 3(16) = 0.
  3. RAy = RBy = 48 / 2 = 24.000 kN.
  4. ΣFx = 0: RAx = 0.000 kN.
3
Inspection rule

Identify zero-force members

M8 and M12 = 0.000 kN

At N2 and N4, two collinear bottom-chord members meet one non-collinear vertical member with no external joint load. The non-collinear member is therefore a zero-force member.

4
Method of joints

Joint N1

  1. The end post M7 is inclined 45°.
  2. ΣFy = 0: 24 + M7 sin 45° = 0, so M7 = -33.941 kN (compression).
  3. ΣFx = 0: M1 + M7 cos 45° = 0, so M1 = +24.000 kN (tension).
5
Method of sections

Section through M2, M5 and M9

  1. Take the left portion of the truss and assume the cut-member forces act in tension.
  2. ΣM about N6 = 0: -24(3) + M2(3) = 0, so M2 = +24.000 kN.
  3. ΣFy = 0: 24 - 16 - M9 sin 45° = 0, so M9 = +11.314 kN.
  4. ΣFx = 0: M2 + M5 + M9 cos 45° = 0, so M5 = -32.000 kN.
M2 +24.000 kN Tension
M5 -32.000 kN Compression
M9 +11.314 kN Tension
6
Virtual work

Vertical displacement at N7 by the unit-load method

δN7 = Σ(Ni ni Li / EA) = 0.655529 mm ↓

Apply a 1 kN downward virtual load at N7. Using the real member forces Ni, virtual member forces ni, member lengths Li and EA = 1,000,000 kN, the virtual-work sum gives the displacement in the direction of the unit load.

The result is positive in the direction of the downward unit load; in global coordinates, uy = -0.655529 mm because global +Y is upward.

View unit-load calculation table

Symmetric members with identical Ni, ni and Li are grouped. The contribution includes every member listed in the row.

Unit-load contributions to the vertical displacement at N7
Members Ni (kN) ni (kN) Li (m) Group contribution (mm)
M1-M4 +24.000 +0.500000 3.000 0.144000
M5-M6 -32.000 -1.000000 3.000 0.192000
M7, M13 -33.941 -0.707107 4.243 0.203647
M8, M12 0.000 0.000000 3.000 0.000000
M9, M11 +11.314 +0.707107 4.243 0.067882
M10 -16.000 -1.000000 3.000 0.048000
Total displacement in the direction of the unit load 0.655529

03 · Optimal Beam Solver results

Axial-force and displacement results

The Optimal Beam Solver uses the direct stiffness method to reproduce the determinate force solution and calculate the displacement response from the assigned EA values.

StabilityStableNo solver warnings
Equilibrium residual2.91e-11Relative numerical residual
Governing memberM7 / M1333.941 kN compression
Maximum resultant movement0.671 mmAt centre upper node N7
Axial-force diagram

Tension, compression and zero-force members

Tension (+)Compression (−)Zero
Pratt truss axial-force diagram Tension members are blue, compression members are orange and zero-force members are grey. M1 · 24.0 M2 · 24.0 M3 · 24.0 M4 · 24.0 M5 · -32.0 M6 · -32.0 M7 · -33.9 M8 · 0.0 M9 · 11.3 M10 · -16.0 M11 · 11.3 M12 · 0.0 M13 · -33.9 N1 N2 N3 N4 N5 N6 N7 N8
Optimal Beam Solver output

Support reactions

Global axes
NodeRXRY
N10.000 kN24.000 kN
N50.000 kN24.000 kN
Optimal Beam Solver output

Nodal displacement

Translations in millimetres
NodeUXUY
N10.0000.000
N20.072-0.444
N30.144-0.608
N40.216-0.444
N50.2880.000
N60.240-0.444
N70.144-0.656
N80.048-0.444
Complete result table

Member axial forces

Positive = tension · negative = compression
M1 N1–N2 24.000 kN Tension
M2 N2–N3 24.000 kN Tension
M3 N3–N4 24.000 kN Tension
M4 N4–N5 24.000 kN Tension
M5 N6–N7 -32.000 kN Compression
M6 N7–N8 -32.000 kN Compression
M7 N1–N6 -33.941 kN Compression
M8 N6–N2 0.000 kN Zero
M9 N6–N3 11.314 kN Tension
M10 N7–N3 -16.000 kN Compression
M11 N8–N3 11.314 kN Tension
M12 N8–N4 0.000 kN Zero
M13 N8–N5 -33.941 kN Compression

04 · Hand-versus-solver verification

Independent equilibrium checks match the production result

The reaction, selected member-force and unit-load displacement calculations are independently stated above, then compared with the stored Optimal Beam Solver snapshot.

All declared checks passedVerified July 26, 2026
CheckHand calculationOptimal BeamDifferenceStatus
Left vertical reaction R_Ay 24.000000 kN 24.000000 kN 0.000% Pass
Right vertical reaction R_By 24.000000 kN 24.000000 kN 0.000% Pass
Bottom chord M2 24.000000 kN 24.000000 kN 0.000% Pass
Top chord M5 -32.000000 kN -32.000000 kN 0.000% Pass
Diagonal M9 11.313708 kN 11.313708 kN 0.000% Pass
Centre vertical displacement N7 -0.655529 mm -0.655529 mm 0.000% Pass
Vertical equilibrium 48.000 kN load 48.000 kN reaction 0.000% Pass
Analysis engineOptimal Beam 2D Structural Analysis
Solver calculation packageDownload full 29-sheet PDF
Declared tolerance1×10−8 absolute + relative
Numerical residual2.910e-11

05 · Structural idealization

What this model represents—and what it does not

The analytical model isolates the in-plane axial response of an ideal pin-jointed truss. It is an analysis benchmark, not a completed member or connection design.

Included in the model

  • Members are straight, prismatic and connected by ideal pins at the nodes.
  • Loads are applied only at panel points, so the members carry idealized axial force.
  • The material is linear elastic and all members use E = 200 GPa.
  • All members use a uniform cross-sectional area of 5000 mm².
  • The left support restrains global X and Y; the right support restrains global Y only.
  • Member self-weight is omitted.

Outside the analysis scope

  • Connection stiffness, gusset-plate behavior and joint eccentricity.
  • Member buckling, yielding, fracture or code-based resistance checks.
  • Geometric nonlinearity, P-Delta effects and large-displacement behavior.
  • Out-of-plane response, lateral bracing and three-dimensional load paths.
  • Construction tolerances, residual stress and fabrication effects.
Engineering boundary

Use the calculated forces to inform design checks, but separately verify buckling, yielding, connections, bracing and code compliance for the real structure.

06 · Common modeling mistakes

Configuration errors that change the structural problem

These are not cosmetic differences. Each one changes the idealization, restraint count, load path or interpretation of the result.

01

Applying a distributed load directly to an ideal truss member instead of converting it to panel-point loads.

02

Using moment-resisting member behavior when an axial-only pin-jointed idealization is intended.

03

Drawing crossing members without creating a shared node at the intersection.

04

Restraining both supports horizontally and unintentionally introducing an extra reaction.

05

Treating a negative axial-force result as an error instead of compression.

06

Using the solver result as a strength or buckling design check.

Continue exploring

Open the benchmark, then test a controlled variation

Each variation starts from the same verified model. Change one modeling decision at a time so the resulting force and displacement changes remain explainable.

Editable variation

Double all joint loads

Open the base model, then change each 16 kN load to 32 kN.

Open base model Editable variation

Reverse the diagonal orientation

Compare this Pratt arrangement with a Howe-style web under the same loading.

Open base model Editable variation

Change the member area

Keep the determinate member forces unchanged and observe how axial displacement varies with EA.

Open base model

Truss guide library

Move between verified examples built to the same span, depth, supports and loading basis.

3 worked examples

Use the exact benchmark

Review the calculation—or continue in the calculator

The page, editable calculator model, structural diagram, tables, downloadable JSON and regression test all use this same reviewed example record.

Open in 2D Calculator Download Worked Example PDF Download solver calculation package Download model JSON

Engineering authorship

Prepared and reviewed by Tamer Hijjawi, P.Eng.

Analysis scope: First-order linear elastic 2D analysis. Last model verification: July 26, 2026.

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