How does a bridge carry a moving load?

Take the load for a ride

Imagine carrying a bag across a bridge. The total weight can stay the same while different parts of the bridge respond. Move the carriage and watch green members pull and coral members push. Gray means zero axial force in this particular case.

Lift the deck and find the handoff

A load between two joints first enters a small ideal deck panel. That panel has an upward reaction at each end. It pushes down on the corresponding truss joints. The two shares add up to the original load and preserve its turning effect.

Look at one connection

At a stationary ideal joint, all horizontal forces balance and all vertical forces balance. Draw its arrows head to tail: the shape closes. That force polygon explains the colored members; it is not a picture of material flowing along the bars.

Follow a diagonal through a switch

At the quarter-span joint B, diagonal FC pushes. Move the same load to the center C and FC pulls. Its sign depends on the loading pattern, not just its diagonal shape.

Try removing a quiet member

Under the deck-only load, CG carries zero axial force. Removing it leaves G without first-order vertical restraint. A separate downward load at G reveals the missing force path. Adding a horizontal restraint at a distant support does not repair it.

Try a different kind of bridge

A beam develops shear and bending moment. A three-hinged arch also develops horizontal thrust at its supports. A point-loaded arch can still bend between its hinges. Different load paths do not by themselves tell us which structure is stronger.

Declared geometry and units

The planar teaching truss has A(0,0), B(4,0), C(8,0), D(12,0), E(16,0), F(4,h), G(8,h), H(12,h), in metres. Its bars are AB, BC, CD, DE, AF, FG, GH, HE, BF, CG, DH, FC and CH. The chosen depth h spans 2–6 m. A supplies Ax and Ay; E initially supplies Ey. Neither support supplies a moment.

What P represents

P is a force assigned to the front truss, 0–30 kN. The contextual cart is not a measured vehicle mass, and the lesson does not infer how a real deck distributes a whole vehicle load between two side trusses. A separately identified 0–20 kN laboratory load can act downward at G.

Panel load transfer preserves force and moment

For a load at x between joints xᵢ and xᵢ+4, η=(x−xᵢ)/4. The downward truss loads are pᵢ=P(1−η) and pᵢ₊₁=Pη. Thus Σp=P and Σxᵢpᵢ=Px. A load directly at A or E can go directly to that support; it need not load the bars.

Tension-positive joint equilibrium

Each straight, massless, pin-connected bar is a two-force member. Its positive unknown N acts from a joint toward the bar’s other end. Reversing the isolated object reverses the force arrows: tension pulls apart an isolated member’s ends. The 16 joint equations use unit direction vectors and support columns.

An independently checked analytical solution

With downward bottom loads pA…pE and G load W, Ay=Σpᵢ(16−xᵢ)/16+W/2 and Ey=Σpᵢxᵢ/16+W/2. Let Aₙ=Ay−pA, Eₙ=Ey−pE and s=h/√(16+h²). Then FC=(Aₙ−pB)/s, CH=(Eₙ−pD)/s, CG=−W, BF=pB and DH=pD. A separate pivoted matrix solution is checked against the full analytical member solution.

Influence lines answer a specific question

A unit downward deck load gives a member-force influence trace. Between panel endpoints it is linear. For FC, N·s/P at x=0,4,8,12,16 m is 0,−¼,½,¼,0. With G unloaded and P positive, its interior crossing is 16/3 m; CH crosses at 32/3 m. Depth changes force magnitudes but not these crossings. Adding W changes the mixed-load crossings.

Counting is not a geometry test

For equilibrium matrix A, its rank determines independent constraints. First-order mechanisms = 16−rank; self-equilibrated force modes = number of unknowns−rank. Removing CG and adding Ex leaves 16 unknowns and 16 equations but rank 15: one mechanism and one force mode. A complete two-pin truss instead has rank 16 and one force mode, with no first-order mechanism.

A compatible load is not adequate restraint

With CG missing, a bottom-only load can still lie in the matrix column space. A downward G load cannot: it increases the augmented rank. The renderer withholds unsupported unique force colors rather than drawing an approximate solution as balanced. The small G arrow denotes a first-order permitted direction, not finite displacement or collapse.

Beam shear and bending

For the separate simply supported beam, V(z)=Ay−P·I(z>x) and M(z)=Ay·z−P·max(0,z−x). At the load show left and right shear limits. M is continuous, with peak Px(16−x)/16. On the isolated left beam segment, positive shear acts downward at its right cut; positive sagging moment acts counterclockwise there.

An arch needs its own support model

For the separate three-hinged parabolic arch, y(z)=4hz(16−z)/16², H=Mbeam(8)/h=P·min(x,16−x)/(2h), and March=Mbeam−H·y. Both bases provide horizontal restraint. The shortcut PL/(4h) applies only to a central point load. Base and crown moments vanish; bending between hinges generally does not.

Force, stress and strength are different

A bar force has units of force. Stress also needs area, and bending/deflection need section and stiffness information. Buckling, connections, fatigue, dynamics, self-weight, eccentricity and lateral stability need further models. Metallic section shapes in the original assembly do not supply those omitted engineering properties.

Bridge inspection begins with a load path

Decks, floor beams, side members, bearings and foundations have different jobs. The photo inspection prompts help identify layers while keeping measured condition and calculated capacity separate.

A moving load can set the critical case

Influence lines identify where a specified response becomes positive, negative or large. Real design adds traffic load models, combinations, dynamics and capacity checks; a single attractive force picture does not finish that job.

Trusses beyond roads

Roofs, towers and other frameworks use connected members to transfer forces. The same equilibrium ideas apply when their geometry, joints, supports and loading assumptions are appropriate.

Why engineers compare more than shape

A beam, truss and arch may solve different site and support problems. Available foundation restraint, materials, span, construction, maintenance and architecture all matter. This lesson compares force paths, not universal winners.

A 10 kN load is halfway along the ideal panel from B to C. What reaches the two truss joints?

• 5 kN downward at each
• 10 kN downward at each
• 5 kN upward at each

With h=4 m, P=10 kN and G unloaded, FC is compressed at B. At the center C it is…

• Still compressed
• In tension
• Zero because the load is centered

Place the entire ideal P directly at A, with G unloaded. What happens?

• Every bar carries P
• A reacts with P and the bars carry zero
• Half must travel to E

A bar is in tension. Its force on an isolated joint points…

• Toward the bar’s other end
• Away from the bar’s other end
• Always downward

CG is absent. Adding a horizontal restraint at E fixes the missing vertical restraint at G.

• True: the count now matches
• False: G is still unrestrained vertically
• True whenever the deck load is zero

The complete truss with two pins has no first-order mechanism but one extra force unknown. What is missing?

• Any possible equilibrium
• Information beyond equilibrium to determine the extra force
• Proof that it must collapse

A point-loaded three-hinged parabolic arch has zero moment at its hinges. Must moment be zero everywhere?

• Yes, every arch is compression only
• No, bending can occur between the hinges
• Only if the arch is painted gray

The model reports a compressive bar force. What else is needed before claiming that a real bridge can carry it?

• Only a brighter red color
• Material, sections, connections, stability, loads and appropriate engineering checks
• Nothing: equilibrium proves strength

Ideal axial members, joint equilibrium and the 16 m benchmark

MIT 16.001, Radovitzky, Fall 2021, lectures 6–7, PDF pp.10–21. The three 10 kN bottom-load reference is reproduced numerically. Joint letters are renamed. Lecture photographs and figures are not copied.

Method of joints and conditional zero-force members

Udoeyo, Structural Analysis, chapter5 §§5.4 and5.6. Geometry/rank checks supplement the simple count. Original lesson diagrams use equations; restricted book figures are not adapted.

Triangular arrangements and structural equilibrium

ETH Zürich Structural Design II, Trusses. The rank/nullspace numerical cases are our independently computed examples.

Beam shear, moment and sign conventions

Udoeyo chapter4 §§4.3–4.4: internal force cuts, sagging-positive moment and point-load jumps.

Three-hinged arch equilibrium

Udoeyo chapter6 §§6.1.2–6.1.2.1: crown moment condition, horizontal thrust and corresponding-beam comparison.

Real deck-to-truss transfer and connection dependence

FHWA Covered Bridge Manual, 2005, chapter12: floor beams, stringers and combined truss/arch analysis. Historical explanatory source, not current design or rating instructions.

Real connection behavior needs separate evidence

FHWA-HRT-14-063 gusset-plate research uses experiments and analysis. Ideal bar-force equilibrium alone cannot establish actual connection capacity.

Actual truss bridge photograph and reuse

USGS UAS Aerial Imagery Testing Day; Nick Giro and Chris Lewis. Public Domain on owner page, published March30,2026. The full-width service derivative is unchanged locally; exact exposure date is not separately verified.

Actual arch bridge context and reuse

National Park Service via USGS, Delaware River arch bridge in Narrowsburg, approximately2010. Public Domain. Unchanged full-width service derivative; exact hinges and geometry unverified.