Pratt through-bridge truss

The deck hangs 125 kN on the bottom chord — 25 kN at each interior panel point — and both the truss and its loading are perfectly symmetric about midspan. Neither abutment can be favored, so each simply carries half:

Two numbers, no algebra — but everything that follows traces back to them: every panel shear and every bending moment we take in the later steps starts from that 62.5 kN at the pin.

Stand at the pin, joint 1. Only two members meet it: the end diagonal 1-8 (at 45°) and the bottom chord 1-2 (horizontal). The chord is horizontal, so it can do nothing about the vertical 62.5 kN reaction — the diagonal must swallow the whole thing through its vertical component:

The diagonal is in compression (it props the corner up); its horizontal pull anchors the bottom chord into tension. That is the end panel solved — but grinding joint-by-joint to reach a member at midspan would take a dozen more of these, each one feeding the next. There is a faster road, and it deserves a step of its own.

Slice an imaginary cut straight down through panel 3, between the loaded nodes 3 and 4, and throw the right-hand half of the bridge away. The cut severs exactly three members — the bottom chord 3-4, the diagonal 9-4 and the top chord 9-10 — and their unknown axial forces now act as external forces on the piece we kept. That piece carries the 62.5 kN reaction and the two 25 kN loads at nodes 2 and 3, and like any free body it must balance on its own.

Three unknowns, three equilibrium equations — solvable outright. But the craft of the method is choosing equations that catch one unknown at a time. Take moments about the top node 9: the diagonal and the top chord both pass through that point, so both drop out, and the equation holds only the bottom chord. Take moments about the bottom node 4: now the bottom chord and the diagonal vanish, leaving only the top chord. And vertical equilibrium never sees the horizontal chords at all — it hands over the diagonal by itself. Three lines of statics, three members read directly, however deep inside the span they sit.

Now run the two moment equations from the cut. Moments about the top node 9 (its lever arm from the support is 8 m) isolate the bottom chord:

Moments about the bottom node 4 (12 m along) isolate the top chord above it:

Bottom chord in tension, top chord in compression — the beam-bending signature. Each force fell out of one equation, with no chain of joints in between: the moment in the numerator is exactly the bending moment a plain beam would carry at that cut, so the truss answers like a beam, F = M⁄h.

Vertical equilibrium of the same cut hands you the diagonal. In panel 3 the shear is the reaction minus the two loads already passed, and only the diagonal has a vertical component to balance it:

Every diagonal follows the same rule, — the magnitude shrinks toward midspan as the panel shear does, and the sign tells tension from compression. Panel 2 carries 62.5 − 25 = 37.5 kN of shear, so its diagonal 8-3 carries +53.03 kN (T). The verticals then close out at each joint:

F₂₋₈ hangs the 25 kN deck load, so it is in pure tension. And the center vertical F₄₋₁₀ is a genuine zero-force member: at symmetric load the shear vanishes at midspan, so it carries nothing — it earns its place only when the live load is unbalanced.