A cantilever jib truss on a wall

Every truss so far has stood on two supports with the load between them. A cantilever has no second support out in space — everything anchors to the wall behind it. Two pins, at node 1 (bottom) and node 2 (top), bolt the truss to the wall; the horizontal chord runs from node 2 through node 3 out to the tip at node 4, and two diagonals reach up from node 1 — one to node 3, one all the way to the tip. Name the parts now, because the numbers will confirm the names: the chord is the tie running back to the wall, and the tip diagonal is the strut propping the jib from below.

The count still closes: 4 members + 4 reaction components (two pins) = 8 = 2 × 4 joints — statically determinate. But those four reaction components outnumber the three global equilibrium equations, so for the first time we cannot find the reactions first. No matter: the method of joints works from any joint with at most two unknowns, and the tip is exactly that joint. We solve the members outward-in, and the reactions will fall out at the end.

One triangle serves every equation: the tip diagonal runs 6 m across and 2.5 m down, so it is long, giving and .

Joint 4 hangs the 18 kN and meets only two members — the chord back to node 3 and the long diagonal down to node 1. Be careful with direction before you write anything: the diagonal runs down toward node 1, so tension in it would pull the tip down-and-left. But the 18 kN load needs an upward component at the tip, and tension in a member sloping down cannot give one — so before any algebra we know the diagonal is in compression: it props the tip from below. Writing every force tension-positive, the vertical equation says exactly that:

The horizontal equation then hands over the chord: the strut’s horizontal component shoves the tip toward the wall… no — read the signs — it shoves the tip outward, and the chord must haul it back:

The physical picture is now complete at the tip: the strut carries the load down its slope into the bottom of the wall, and in doing so pushes the tip away from the wall with 43.2 kN — which the top chord resists in tension, pulled toward the wall. Tie above, strut below: the cantilever in one sentence.

Move back to joint 3 and something classic happens. The two chord members 2–3 and 3–4 are collinear — both horizontal — the only other member is the diagonal 1–3, and no load hangs at the joint. Resolve perpendicular to the chords: the chords, being collinear, have no perpendicular component at all, so the diagonal is the only member in that equation — and it must therefore vanish:

That is the general zero-force rule: two collinear members plus one extra member at an unloaded joint ⇒ the extra member carries nothing, because the equilibrium equation perpendicular to the collinear pair has exactly one member with a perpendicular component. The chord force simply passes straight through, 43.2 kN all the way to the wall.

So why is member 1–3 there at all? Because the rule holds only for this load case. This is the geometry of a jib crane, and a crane’s trolley rolls along the chord: park the load over node 3 instead of the tip and joint 3 is loaded — member 1–3 wakes up and carries it. Zero-force members are case-by-case, not useless: they wait for the load positions that need them (and meanwhile brace the chord against buckling).

Every member is known, so the wall pins read straight off their joints. At the bottom, node 1: the strut arrives carrying −46.8 kN and pushes into the pin with its two components — the wall bearing takes H₁ = +43.2 kN (the node is pushed into the wall) and V₁ = +18 kN, the entire vertical load. At the top, node 2: only the chord arrives, pulling the pin toward the jib with 43.2 kN — so the wall must pull back: H₂ = −43.2 kN of pull-out, and V₂ = 0. Check the whole thing as a couple — the horizontal pair, 2.5 m apart, must balance the tip load’s overturning moment:

Pull-out force × wall depth exactly equals load × reach. The 18 kN never overturns the truss because the wall answers it with an equal and opposite couple — tension at the top anchor, compression at the bottom.