The Method of Joints: Solve a Truss One Joint at a Time
A truss is a frame of straight bars pinned together at their ends - think of a bridge or a roof made of triangles. Because the bars are pinned, not welded rigid, and loads act only at the joints, each bar is a two-force member: it can only pull or push along its own length, nothing else. That single fact makes trusses easy to solve. The method of joints uses it directly: cut the truss apart at the pins, and look at one joint at a time. At a joint, every bar force and every external load meet at a single point, so the joint is just a particle in equilibrium - the same ΣFx=0, ΣFy=0 you already know. Solve one joint, carry its bar forces to the next, and march across the whole structure. Start by seeing why each bar carries force only along its line.
The whole method rests on one property of the bars. Press a member to pick it out. A truss bar is pinned at both ends and carries no load in between, so only two forces act on it - one at each pin. For those two forces to balance, ΣF=0 and ΣM=0 on the bar, they must be equal, opposite, and lie along the line joining the two pins. So a truss bar can do exactly one thing: push or pull along its own axis. It cannot carry a sideways force or a bending moment, the way a beam does. This is why a truss is so efficient and so solvable: every one of its bars is reduced to a single unknown - the axial force in it.
A bar's axial force comes in two flavors. Toggle between them. If the two end forces pull the bar outward, the bar is stretched - that is tension, and we call it positive. If they push inward, the bar is squeezed - that is compression, negative. By Newton's third law, the bar pushes or pulls back on the joints with equal and opposite force, so the convention flips at the joint: a tension member pulls its joint toward itself, a compression member pushes its joint away. When you draw a joint, the trick is to assume every unknown bar is in tension - arrow pointing away from the joint, along the bar. If the number comes out negative, the bar is really in compression. Letting the sign decide saves you from guessing directions.
Now isolate a single joint. Pick one. When you cut the bars meeting at a joint, each cut bar leaves behind its axial force, and the external load or support reaction acts there too. Every one of these forces passes through the joint - they are concurrent - so there is no moment equation to write. A joint is just a particle, and a particle in equilibrium obeys ΣFx=0 and ΣFy=0, two equations. That means you can solve a joint that has at most two unknown bar forces. Look at joint C at the top: the load pushes down, and the two diagonal bars push up and outward to hold it - two unknowns, two equations, solvable. The whole truss is just this little problem, repeated at joint after joint.
Put it together on a simple triangular truss: a pin at the left, a roller at the right, and a load hanging at the top joint. Drag the load and watch the bar forces respond. First, the supports: by symmetry each carries half the load, so each vertical reaction is P2. Then go to the top joint, where only two bars are unknown. The two diagonals must push up to carry P, so both are in compression, each P(2 sinθ). Move down to a support joint: the diagonal's horizontal push must be balanced by the bottom bar, which is pulled taut - tension, P(2 tanθ). Notice the pattern: the diagonals squeeze, the bottom chord stretches. Drag P larger and every force scales with it, exactly in proportion.
One last trick saves real work: some bars carry no force at all. Toggle the two cases. First, at a joint where only two bars meet and no load is applied, if the bars are not in a straight line, both must be zero - there is nothing to balance their separate directions. Second, at a joint where two bars are collinear and a third comes in at an angle, with no load, the lone angled bar is zero and the two collinear ones simply pass the force through. Spotting these zero-force members first removes bars from the puzzle before you compute anything. That rounds out the method of joints: find the reactions, start at a joint with two unknowns, write ΣFx=0 and ΣFy=0, carry each result forward, and use symmetry and zero-force members to cut the labor. From here, the same free-body thinking scales up to whole frames and machines.