Warren truss by the method of joints

Before any member force, pin down the supports. Three 20 kN loads hang from the bottom chord, so the two ends together must push up with 60 kN. The truss is symmetric and so is its loading, so there is nothing to break the tie — each end carries exactly half:

With the reactions known, every joint has at most two unknown member forces — the truss is statically determinate (15 members + 3 reactions = 18 = 2 × 9 joints), so the method of joints will unwind it completely.

First, the geometry that serves the whole walk. Every diagonal in this Warren truss runs 2 m across and 3 m up, so each is long, giving and . Compute those two numbers once — they appear in every equation below.

The method of joints wants a pin with at most two unknowns, and joint 1 is exactly that: the 30 kN reaction pushes up, and only the end diagonal and the first bottom-chord member meet there. The diagonal is the only member with a vertical component, so it alone must cancel the reaction:

Let the signs do the talking: we write every force as tension-positive, so the negative answer means the end diagonal is in compression — it leans into the support like a strut — while it shoves the bottom chord outward into tension.

Move up to joint 6, which carries no load. Its two diagonals must balance vertically, and the top chord closes the horizontal:

Joint 2 hangs the first 20 kN load. Its vertical equation gives the inner diagonal, its horizontal equation the bottom-chord center:

Notice the rhythm the walk has settled into: at an unloaded top joint the diagonals simply hand their vertical components across; at a loaded bottom joint the pair must also swallow the 20 kN hanging there. Zig, zag, repeat — that alternation is the Warren truss’s whole story, and Step 5 reads it in full.

Continuing to joint 7 and applying the same two equations gives the top-chord center F₇₋₈ = −53.33 kN (C) and confirms the inner diagonals at ±12.02 kN. And that is where the hand work ends: the truss and its loading are both symmetric about midspan, so every member on the right must carry exactly what its mirror twin on the left carries. Walking past the centerline would only re-derive numbers we already own — all 15 member forces are on the table.

Step back and the whole force field reads like a sentence. The bottom chord is in tension, growing from +20 kN at the ends to +46.67 kN at the center; the top chord is in compression, from −40 kN to −53.33 kN — the flanges of the beam-in-disguise. The diagonals alternate tension and compression as they zig and zag, shedding the shear that decreases toward midspan (hence the small ±12.02 kN inner diagonals versus the ±36.06 kN at the ends).

The results figure below draws every one of those numbers on the frame — members colored by sign (red tension, blue compression, exactly as the Truss Calculator colors them), the support reactions, and the exaggerated deflected shape dashed over the truss.