Symmetric Two-Bar Truss

staticsstressstabilitymass-cost

Hang a weight from two bars pinned to a wall, or prop it up on two struts — either way the load splits between the members, and simple statics tells you exactly how. This is the smallest real truss, and the cleanest place to see the single most important idea in structural analysis: determinacy. Two members, two unknown forces, and the joint gives you two equilibrium equations. The count matches, so equilibrium alone fixes the forces — you never have to ask how stiff the bars are or how far the joint moves.

$F_m = \frac{P}{2\cos\alpha}$

Here $\alpha$ is measured from the vertical (the axis the load points along) — worth stating out loud, because reading it from the horizontal silently swaps every $\cos$ for a $\sin$ and inverts the whole story. Read this formula and watch what a shallow truss costs: as $\alpha \to 90^\circ$ the bars flatten toward horizontal, lose their vertical bite, and the member force runs away to infinity. A nearly-flat cable holding a tightrope walker carries a tension many times their weight; the truss is the same trap. Drive the $\alpha$ knob toward $90^\circ$ in the widget and the force readouts blow up — until the geometry degenerates and the page refuses to answer, because at $\alpha = 90^\circ$ there is no upward component left to carry anything.

Why this is the determinate case — and where Phase 3 begins

The member force above needed nothing but equilibrium. Add a third bar straight down from the same joint and the tidy count breaks: three unknown forces, still only two joint equations. Equilibrium can no longer decide how the load shares out — the stiff bars steal more of it, and you must bring in compatibility (all three members meet at one joint, so their elongations are geometrically linked) and solve equilibrium and compatibility together. That statically indeterminate truss is exactly the kind of coupled linear system the solveLinear engine (Phase 3) exists to handle; this page is the determinate rung right below it, built so the jump is visible. Everything here falls out of back-substitution with no solver at all.

The joint deflection — and the material’s first appearance

Nothing above involved the modulus $E$: the forces and stresses are material-blind (swap steel for titanium and $F_m$, $\sigma = F_m/A$ don’t move by a newton). The material makes its entrance in the deflection. Each bar stretches by its Hooke’s-law amount $e = F_m L/(AE)$, and the joint drops by the projection of that stretch back onto the vertical:

$\delta = \frac{P L}{2 A E \cos^2\alpha}$

The exponent is two, not three — one $\cos\alpha$ rides in through the member force (which already grows as the truss flattens), and a second comes from projecting the joint’s drop onto the inclined member ($e = \delta\cos\alpha$). This is the unit-load theorem made visible, re-derived two independent ways in the tests. Now the material does matter: a titanium truss ($E \approx 114$ GPa) deflects almost twice as far as a steel one ($E \approx 200$ GPa) under the same load, even though both carry the identical member force — stiffness and strength are independent axes, and the deflection is where stiffness shows up. As $\alpha \to 90^\circ$ the deflection diverges too, and the small-displacement assumption (member angles taken fixed under load) breaks down first — the widget warns when $\delta > L/10$.

Compression adds a second way to fail

Flip the load so it presses the joint instead of pulling it and the members go into compression — where a slender bar has a second failure mode the tension member never faces: it can buckle. Each bar is pinned at both ends, so it is an ideal Euler strut ($K = 1$), and its critical load is taken verbatim from the Euler Column page:

$P_{cr} = \frac{\pi^2 E I}{L^2}, \qquad \mathrm{SF}_{buck} = \frac{P_{cr}}{F_m}$

At the default geometry the buckling margin is smaller than the yield margin, so buckling — not yield — governs the compression member. That is why real trusses put their stubby members in compression and their thin ones in tension. Toggle to the tension configuration and the buckling readouts are withheld (a tension member cannot buckle); drive the bar short enough and they are withheld again, this time because Euler overpredicts a short column — the inelastic Johnson regime governs there, and the Euler Column page carries that hand-off.

Mass, and a merit index in waiting

The mass $m = 2\rho A L$ adds the third independent material axis — density, which touches neither the stiffness nor the strength story. Put the three together and a design question appears: for a tension member that must not yield at the least weight, the bar that wins maximizes the specific strength $\sigma_y/\rho$. That merit index — and the map of material classes it draws — is the Ashby work of Phase 5; this page is one of the many that feed it.

Try it

Governing relations

Derivation

Steps marked modeling step are where physics enters by citation — every other line is machine-proven to follow from them, and the cited models are independently re-derived in the test pipeline where possible. See what is and isn't machine-verified.

How it fails

The clean split of load between two members is an idealization — pin-jointed, weightless, straight, two-force bars loaded exactly through the joint. Real trusses depart from every one of those assumptions, and the departures are where they actually fail:

• Compression members buckle, and buckling is sudden. A tension member yields gracefully, gives warning, and holds a predictable load; a compression member can find a cheaper bent shape and go all at once, with no nearby equilibrium to catch it. That asymmetry is why design codes put heavier safety factors on compression members and why this page refuses the Euler readouts below the transition slenderness — where the elastic formula overpredicts a short column on the dangerous side. The full short-vs-slender hand-off (Johnson’s parabola, tangency at $\sigma_y/2$) lives on the Euler Column page.
• The joint is the usual culprit. Members are strong; the connection that gathers them — the pin, the gusset plate, the weld — concentrates load into a small volume and is where cracks start and bolts shear. This page models the members and treats the joint as an ideal frictionless pin (its detailed design is out of scope), but in practice a truss is only as good as its joints, and fatigue at a connection under repeated loading kills more members than static overload ever does.
• A load that misses the joint centroid bends the members. The two-force-member picture holds only if the resultant passes cleanly through the pin. Offset it — an eccentric bracket, a cable pulling off-axis — and each member carries bending on top of its axial force, deflections grow from the first newton with no clean bifurcation, and the eccentric column secant story takes over. A member’s own weight does the same thing in miniature, sagging it into a shallow beam.
• Shallow trusses are a trap. The $1/\cos\alpha$ in the member force is not a curiosity: a truss built close to horizontal to save headroom carries member forces many times the applied load, and its joint deflection diverges even faster ($1/\cos^2\alpha$). Push $\alpha$ far enough and the geometry itself changes appreciably under load — the small-displacement assumption fails, the linear result understates the deflection, and a very shallow two-bar frame can snap through to an inverted shape. That large-displacement instability is beyond the linear model here; the page warns as $\delta$ passes $L/10$ and refuses outright at $\alpha = 90^\circ$.
• Rigid frames are not trusses. If the joints resist rotation (welded, bolted moment connections) the members carry bending as well as axial load and the determinate force split no longer holds — the structure is a frame, closer in spirit to a cantilever beam at each joint than to a pin-jointed truss. Assuming pins when the joints are actually stiff underpredicts the bending stresses that then govern.
• Buckling picks the weakest axis. The round bars here dodge the issue by symmetry — every bending axis is alike — but a rectangular or angle section buckles about its thin direction at a load the in-plane picture never shows. Bracing that kills the weak-axis shape can multiply a compression member’s capacity.

Sources

• Gere, J. M., & Goodno, B. J., Mechanics of Materials, 9th ed., Cengage, 2018 — axially loaded members and trusses (§2.1–2.3: two-force members, statically determinate joints), displacements by the unit-load/energy method (§9.8–9.9), and the small-displacement assumption underlying linear truss analysis. Euler buckling of pinned columns is §11.3, reused from the Euler Column page.
• Budynas, R. G., & Nisbett, J. K., Shigley's Mechanical Engineering Design, 10th ed., McGraw-Hill, 2015 — §4-11–§4-12 (Euler long columns and the Johnson parabola for intermediate columns, tangent to the Euler curve at the limiting slenderness √(2π²CE/S_y)). Cited here for the transition slenderness λ_T and the compression-member hand-off to the Johnson regime, which this page cross-links to Euler Column rather than re-implementing.
• Timoshenko, S. P., Strength of Materials, Part I: Elementary Theory and Problems, 3rd ed., Van Nostrand, 1955 — statically determinate trusses and joint displacements by the elongation geometry (the compatibility triangle), the classical treatment cross-checked here.