1 · Free-body diagram
Show the total 100 N qualification force, supports and external reactions for the entire bridge.
Principles of Engineering • Unit 2 • Lesson 2.16
Bridge Engineering Analysis
Engineering question: Do calculations support our selected bridge concept—and where must it be reinforced, braced or revised?
Learning mission
Analyze, diagnose, revise
Design selection was a hypothesis. This lesson tests that hypothesis using mechanics and identifies the evidence still needed for the real, three-dimensional bridge.
2 · Member-force analysis
Analyze a suitable determinate side-truss model using joint equilibrium or an approved solver, marking tension and compression.
3 · Engineering risk
Evaluate compression stability, load-sharing, critical joints and preliminary mass, then propose changes.
Review & draw8–10 minutes
Worked mechanics12–15 minutes
Apply to your bridge20–25 minutes
Risk & revisions15 minutes / homework
Suggested pacing: two 45–50 minute periods for full individual-team analysis and documented revisions.
01 / Baseline bridge rules
Numbers every design must address
400 mm
Clear support-to-support span
≤ 65 mm
Overall width, including hardware
100 N
Centered qualification force
≤ 10 mm
Measured midspan deflection at 100 N
≤ 150 g
Completed structural mass
Wood-based
Approved primary members only
Construction limits still apply: approved wood and plywood members, permitted local gussets and printed fittings, PVA/CA glue or legal joins, and M3 nut-and-bolt fasteners only. No acrylic, solid plywood plates, screws, hot glue, epoxy, or 3D-printed chords. Review the complete manufacturing rules ↗
02 / External force equilibrium
Start with the entire bridge, not individual bars
A correct free-body diagram shows the bridge separated from its surroundings, with only the applied force and reactions. Assume an ideal left pin and right roller, a 400 mm span and a centered 100 N downward load.
Step 1 · Moments about left support A
ΣMA = Cᵧ(400) − 100(200) = 0
Cᵧ = 50 N upward.
Step 2 · Vertical balance
ΣFᵧ = Aᵧ + Cᵧ − 100 = 0
Aᵧ = 50 N upward. For no other horizontal loads, Aₓ = 0.
Common errors to avoid
- Dividing 100 N by two before drawing the bridge-wide FBD.
- Placing the entire 100 N independently on each of two side trusses.
- Assuming every side gets exactly 50 N without checking the crossmember and load eccentricity.
- Using centimeters in one equation and millimeters in another.
Even if the real bridge has two side frames, the whole bridge carries 100 N total. Only a symmetric and adequately connected ideal model allows an assumed 50/50 side split.
How the load can reach both trusses
Preliminary side-frame approximation: Under symmetric load sharing, each side carries 50 N downward, with 25 N upward at each end of that side. This approximation must be identified and reconsidered if the center crossmember, side stiffness, joints, or alignment are not symmetric.
03 / Learn from demonstrations
Video library: solving, modeling and interpreting failures
Understanding and Analysing Trusses — The Efficient Engineer
While watching: Which joints could be solved with two or fewer unknown member forces? How do negative solutions indicate compression?
Watch on YouTube ↗Free Online Truss Calculator Tutorial — SkyCiv Engineering
While watching: Identify nodes, members, pin/roller restraints, load locations and the force-result diagram.
Watch on YouTube ↗Understanding Buckling — The Efficient Engineer
While watching: Why is compression force alone insufficient to declare a long, slender member safe?
Watch on YouTube ↗The last video extends the buckling concepts from 2.12. Watch the first two before attempting the independent analysis. Direct YouTube links are included if embedded video is blocked.
04 / From bridge to joint
How to solve member forces without guessing
1. LabelGive joints letters and members endpoints
2. IsolateDraw the complete truss FBD
3. SolveUse moments and joint ΣFₓ = ΣFᵧ = 0
4. VerifyBalance a final joint, compare to solver
Method of joints
Choose a joint with no more than two unknown member forces. Draw unknown forces away from the joint (assume tension). Resolve horizontal and vertical components, solve equilibrium, then move to the adjacent joint.
Positive: tension. Negative: compression for the chosen sign convention.
ΣFₓ = 0 · ΣFᵧ = 0
Know the model assumptions
- Members are treated as straight two-force axial elements.
- Joints are ideal pins (no moment transfer) and forces act at joints.
- Support restraints are correctly modeled; the truss is stable.
- An ideal 2D truss ignores real wood joints, out-of-plane sway, eccentric load transfer, connector bending and imperfections.
Important: m + r = 2j is a useful necessary count for a simple determinate planar truss, but not proof of stability or unique solvability. Check connectivity and geometry too.
05 / Fully worked example
Three-member triangle: a small solvable model
Consider one idealized side truss, distinct from your Pratt, Warren or Howe concept. Its joints are A = (0,0), B = (200,100) and C = (400,0) mm. It carries its assumed 50 N share of the bridge's total 100 N center load at joint B.
Geometry
θ = arctan(100/200) ≈ 26.565°
LAB = LBC = √(200²+100²) ≈ 223.61 mm
Support forces on this one side: Aᵧ = Cᵧ = 25 N, Aₓ = 0.
At apex joint B
FAB = FBC = −50/(2 sin θ) ≈ −55.90 N
Both diagonals are in compression because the negative sign reverses the assumed tension arrows.
At left joint A
FAC = (50/2) / tan θ = +50.00 N
The lower chord AC is in tension. Use the opposite side's joint C to verify force balance. Record signs and units.
Interpret the real question
A −55.90 N axial member demand does not tell you whether a wooden bar or connection can carry it. The member may be slender, imperfect, weak in compression, or poorly connected.
This is a demonstration of method, not a force diagram for every Pratt, Howe or Warren bridge. Transfer the method—not the numerical results—to your selected concept.
06 / Interactive mechanics lab
Explore geometry, compression and possible instability
This calculator is an exact method-of-joints solution for the idealized three-member side-triangle, combined with a separate theoretical Euler buckling check for each sloping bar. It does not analyze your full two-sided bridge, joints or actual midspan deflection.
The 50/50 load split is a simplifying assumption. The slider demonstrates how unequal side sharing changes demand; it does not calculate the actual distribution from your bridge geometry.
Tension and compression color key
■ Teal = tension ■ Orange = compression
The geometry updates with the height slider; straight line widths are schematic and do not depict deformation.
Calculated teaching-model outputs
50.00 NApplied load on chosen side
25.00 NReaction at each end of chosen side
−55.90 NAB and BC axial force (compression)
+50.00 NAC axial force (tension)
148.04 NIdeal Euler critical load of each diagonal (K = 1)
0.38Axial compression demand ÷ ideal Euler load
1.40 N/mm²Absolute axial stress, diagonals (F/A)
78.89 gVery rough two-side member mass + allowance
Theoretical model interpretation appears here.
Model caution: Euler's Pcr = π²EI/(KL)² is only a theoretical elastic instability estimate, using K = 1 and a hypothetical E. It is not real failure capacity, safety factor or bridge pass/fail evidence. The model omits member defects, imperfect ends, adhesive connections, buckling outside the assumed plane and actual supported-load distribution. The rough mass uses illustrative density 0.50 g/cm³ for ideal wood bars plus a user-adjustable connection/bracing allowance. It does not prove the ≤150 g final mass or ≤10 mm deflection requirements.
What if the truss becomes taller?
Hold the total bridge load, side share and material section constant. Increase triangle rise from 100 to 140 mm. Which axial force changes?
Reveal explanation
A taller triangle has a larger diagonal angle θ. For the simplified central-apex load, |FAB| = Pside/(2sinθ) decreases, and the bottom tie tension Pside/(2tanθ) decreases. However, diagonal length becomes longer, which can reduce Euler buckling capacity: lower axial demand does not automatically mean a safer real member.
What if one side carries more load?
Change the side-share slider from 50% to 65%. Are the total bridge reactions still 50 N at each end when the 100 N applied load remains centered?
Reveal explanation
For the ideal complete bridge with a centered 100 N load, yes, the bridge-wide end reactions remain 50 N each. But if one side carries 65 N and the other carries 35 N, the assumed reactions on each side differ (32.5 N at each end versus 17.5 N); the actual load split must be evaluated from the 3D connection and stiffness.
07 / Apply this to your own bridge
Analyze the 2.15 concept you selected
Copy the chosen concept into the analysis packet. Use a dimensioned 2D side-truss model if pin-jointed analysis is appropriate. When a concept has continuous beams, a load between nodes or moment-resisting joints, document the limitation and use a teacher-approved more suitable method instead of pretending the pin-truss method applies.
Set up a valid analysis
- Label every modeled joint and enter approximate coordinates in mm.
- List all connected members and show actual proposed bracing (or separately identify omitted 3D elements).
- Represent one end as a pin and the other as a roller for the idealized support model, if appropriate.
- Show the complete 100 N bridge-wide load first. State and justify any side-frame load-split assumption.
- Check ideal determinacy and connectivity; solve support reactions.
- Find at least three relevant axial member forces by hand or with an approved model; mark positive tension and negative compression.
- Check at least one result by an independent equilibrium equation.
Then evaluate the risks
- Which compressed member has the longest unsupported length and smallest weak-axis stiffness?
- Where could buckling occur before a member crushes?
- Does the 100 N test force really reach the model's joint locations?
- Could gussets, printed fittings, M3 bolts or adhesive joints become the weak link?
- How will out-of-plane sway or twisting be prevented?
- What model or actual test is needed to evaluate the 10 mm deflection criterion?
- Does the full two-sided bridge mass budget remain below 150 g including glue and hardware?
Data from earlier work: Your 2.13 material testing produced actual bending stiffness and observations—not directly measured fracture or compressive strengths. Use the appropriate data where justified, report measurement limitations and do not treat k (N/mm) as elastic modulus E (N/mm²). Your 2.14 budget provides the initial mass estimate.
08 / Critical joint and failure assessment
Engineering risk register
For each possible failure location, record a qualitative severity and likelihood from 1 (low) to 5 (high). The product provides a discussion priority score, not a statistically calibrated failure probability or a safety certification. Explain evidence and proposed mitigation in your submitted analysis.
| Potential weakness | Severity | Likelihood | Priority score |
|---|---|---|---|
| Long compression diagonalPossible global buckling of a slender member | — | ||
| Central load-transfer jointLoad distribution, eccentricity and local crushing | — | ||
| Gusset or adhesive connectionConnection shear, splitting or peel | — | ||
| Lateral bracingOut-of-plane sway or twist | — | ||
| End support interfaceLocal bearing, sliding or rotation | — |
Set ratings after reviewing the evidence, then document mitigation.
Critical joint 1 · Center load entry
Analyze how the load transfers from the loading contact to the two side frames. Sketch joint interfaces, forces, crossmember geometry and potential slip, bending or peel.
Critical joint 2 · End support or bracing
Analyze how support reactions enter the structure or how a brace holds the two sides together. Identify permitted fittings/fasteners and a method to validate the joint.
09 / Revise before CAD
Document your design decision and what remains unverified
Selected concept record
Numerical concept check
Enter preliminary mass, width and span to screen basic requirements. A blank deflection prediction remains unverified.
Update recommendation: If a compression member, joint or bracing layout is concerning, change member geometry or load path and then re-run relevant calculations. Do not simply add unlimited material.
Preliminary screening is not fabrication approval. Passing geometric or estimated mass checks does not demonstrate the actual 100 N capacity or 10 mm deflection limit. Do not claim your bridge meets the load requirement until it has been analyzed appropriately and physically qualified under teacher supervision.
10 / Independent checks
Answer, then reveal the solution
1 · Whole structure reactions
A 100 N force acts at the center of a 400 mm simply supported bridge. Compute Aᵧ and Cᵧ.
Reveal explanation
Aᵧ = Cᵧ = 50 N upward. Moments about A: Cᵧ(400) = 100(200). Balance vertical forces to obtain Aᵧ.
2 · Side-frame load
If symmetry justifies a 50/50 load share, what load is applied to each side truss and what are its two ideal end reactions?
Reveal explanation
50 N per side truss, with 25 N upward at each end for each symmetric side model. Adding reactions across both side frames gives 100 N in total.
3 · Triangle axial force
A = (0,0), B=(200,100), C=(400,0) mm, 50 N downward at B. Determine the axial force in AC.
Reveal explanation
θ = arctan(100/200). At the left joint A, AC = 25/tanθ = +50 N tension.
4 · Interpretation
The computed axial force in a diagonal is −55.9 N. Is it in tension or compression, and does the number prove safety?
Reveal explanation
It is compression under the tension-positive convention. Force demand alone does not establish allowable stress, buckling capacity, local strength or connection adequacy.
5 · Buckling and geometry
For the same E, I and end factor K, what happens to theoretical Euler buckling load if unsupported length doubles?
Reveal explanation
The theoretical Euler critical load becomes one-quarter of its previous value because Pcr ∝ 1/L².
6 · False deflection claim
A truss solver returns all axial forces and support reactions but no displacement results. Can a student claim compliance with ≤10 mm deflection?
Reveal explanation
No. You need an appropriate displacement model with defensible E and member area/connection assumptions, and eventually a physical test; force equilibrium alone cannot establish deformation.
7 · Actual joints
Why might glued wooden gussets invalidate the exact behavior of an ideal pin-jointed truss?
Reveal explanation
Real joints can transfer some moments, bend, slip, split or fail in adhesive shear/peel. Loads may also enter a member between joints. These effects are absent from an ideal two-force bar model.
8 · Manufacturing screen
Your bridge design is 62 mm wide and 140 g, but uses acrylic gussets and M4 bolts. Can it be approved?
Reveal explanation
No. Acrylic and M4 bolts violate the manufacturing rules even though the width and mass values are below the numeric caps.
Reference desk
Engineering resources for your independent analysis
Engineering Statics · Method of Joints
Free online textbook section with joint FBD examples, tension sign convention and independent checking.
Read the Method of Joints ↗SkyCiv · Truss Hand Calculations
Explains supported models, axial force outputs and checks for unstable or indeterminate trusses.
Read annotated calculation steps ↗SkyCiv · Method of Sections Tutorial
How to analyze selected members without calculating every bar, extending the tools from 2.9.
Open Method of Sections guide ↗SkyCiv · Free Truss Calculator
Alternative software for joint/member model checking when your selected concept is appropriately simplified.
Try the free truss calculator ↗Student submission
Exactly what to submit for POE 2.16
Submit a Bridge Engineering Analysis and Risk Review as one organized PDF, worksheet packet, or teacher-approved notebook record. Include the updated selected concept from Lesson 2.15.
Required evidence
- Selected concept ID/name and a dimensioned side/top drawing showing all modeled joints, the center load and support locations.
- One bridge-wide FBD with 100 N at midspan, correct support reaction calculations and units.
- A stated and justified load-share assumption for the two side trusses, if using 2D side models.
- A labeled member-force diagram for the chosen concept: at least three axial forces (tension or compression), with equations or approved solver evidence.
- At least one independent equilibrium check and a list of model assumptions/limitations.
- One compression/buckling-risk evaluation with dimensions, assumptions, an equation or clear qualitative rationale.
- A critical-joint analysis for two different joints, including load paths, permitted connection details and potential failure modes.
- A preliminary mass/width/span screen, plus an explanation of how the 10 mm deflection limit will be checked.
- A risk register with severity/likelihood priorities and a written plan for at least two targeted revisions or verification steps.
Submission quality check
- 100 N is the total bridge load, not 100 N applied independently to each side.
- All measurements and calculations use compatible units and clear force signs.
- Simulation screenshots identify the model geometry, members, restraints and loading.
- Buckling/strength predictions are not mislabeled proven load capacity.
- Real 2.13 test readings are not confused with assumed modulus values.
- Full finished-mass estimate includes wood, bracing, glue, gussets, M3 hardware and printed fittings.
- Proposed construction follows Unit 2 fabrication rules.
- The final recommendation identifies what needs correction before Fusion modeling in Lesson 2.17.
Due date: See the Daily Agenda and teacher submission directions.
Review Lesson ↑