The simplest truss: one triangle

Look at the frame above. Three straight members, connected to each other only at their ends — and those connections are pins, which behave like door hinges: they let the members swivel freely and cannot twist them. Now think about what one member can actually feel. It is held at its two ends and nowhere else, and the hinges cannot bend it — so the only thing it can carry is force along its own length. Engineers call this a two-force member, but the idea is homelier than the name: the member can pull its two ends together, or push them apart, and that is all it can do.

The other rule of a truss is that loads land only at the joints — here the 20 kN sits right on the apex pin, not partway along a member. Put the two rules together and the whole structure dissolves into something wonderfully simple: no bending anywhere, just pushes and pulls. Solving a truss means deciding, member by member, which one it is — and how hard.

One honest note about the picture, since you will meet it on every truss page here. The joints are drawn the way they are actually built — steel gusset plates with the members bolted onto them — not as literal hinges. A real plate does resist a little twisting. Engineers analyze it as a frictionless pin anyway, because the members are slender and the effect is small, and because the pin assumption is what makes the arithmetic below possible at all. Drawn as built, analyzed as pinned: that single idealisation is the method of joints.

Before we step inside the triangle, stand outside it. The 20 kN acts at the apex, which sits exactly midway between the two supports, and the triangle itself is symmetric — so neither support can carry more than the other. Each takes half:

That is everything the outside world does to this truss: 20 kN down at the top, 10 kN up at each foot. Now we can go joint by joint and find what the members themselves are doing.

The tool is called the method of joints, and it asks very little of you. Zoom in on a single pin. That pin is not moving, so the forces meeting there must balance — horizontally (ΣF_x = 0) and vertically (ΣF_y = 0). Two equations, so pick a joint where no more than two forces are unknown. The apex, joint 3, is perfect: the 20 kN load is known, and only the two rafters meet there — exactly two unknowns.

First, the geometry — and here the numbers are kind. Each rafter runs 3 m across and 4 m up, so its length is : the famous 3-4-5 right triangle. That means of any force F in a rafter, of it acts vertically and horizontally — no calculator, no trig tables.

One bookkeeping habit and we can solve: assume every member is in tension — pulling away from the joint, down along its own line — and let the algebra correct us with a minus sign if we guessed wrong. By symmetry the two rafters must carry the same force F. Balancing the vertical directions at the pin:

Read the equation before the algebra: every term in it points downward — two assumed pulls and the load — and nothing pushes up. That cannot balance, so F had to come out negative: the rafters do not pull on the apex, they push. Each one carries 12.5 kN of compression — it is being squashed between the load pressing down and the support propping it from below, and its 0.8 × 12.5 = 10 kN of vertical push, doubled, is exactly what holds the 20 kN up. (The horizontal parts, 0.6 × 12.5 each way, cancel by symmetry.)

Now slide down to joint 1, the left foot. Three things meet there: the 10 kN reaction pushing up, the rafter 1-3, and the bottom member 1-2. We just learned the rafter is in compression, so it pushes on this joint away from the apex — down and outward, along its own 3-4-5 line. Split that 12.5 kN push into its two parts: 0.8 × 12.5 = 10 kN straight down, and 0.6 × 12.5 = 7.5 kN outward. The vertical part goes straight into the ground:

It matches the reaction exactly — our first free check, and a habit worth keeping. But the horizontal 7.5 kN has nowhere to go. There are no sideways loads on this truss, so the pin has no net sideways reaction to offer — and at the far end the roller could not resist sliding even in principle. Left alone, the two feet would skid apart and the triangle would flatten. The bottom member is what forbids it: it grabs the foot and pulls it back toward the other one,

and this time the sign comes out positive: the bottom member really is being stretched — 7.5 kN of tension. Its whole job is to stop the feet spreading, which is why engineers call it a tie. The figure below paints the answer onto the frame.