Truss member forces in a plane

A member pinned at both ends, with no load along it, can carry a force along its own axis and nothing else. Each member therefore stands for one unknown, its axial force N, positive in tension. Every joint, with its loads and the pull or push of each member that meets there, must be in equilibrium: the forces on it add up to zero in x and in y. With j joints that is 2j equations, and the unknowns are the m member forces and the r reactions of the supports.

When m + r = 2j and the truss is stable, there are as many equations as unknowns and equilibrium alone gives every force. That is what the method of joints does by hand, one joint at a time; here all the joints are solved together. The section and the modulus of a member appear nowhere in that solution: a statically determinate truss carries the same forces in steel as in aluminium, and in any size of section. They are asked for only to work out stress, elongation and displacement afterwards.

When m + r is greater than 2j the truss is statically indeterminate: equilibrium no longer says how the load is shared. The direct stiffness method adds the compatibility that is missing. The unknowns are the displacements of the joints; a member whose ends move apart along its axis by δ carries N = (EA/L)δ, for a length L, a modulus E and an area A; and the stiffness matrix of the truss is the sum of the members’ own. Solving that system gives the displacements, then every N. The forces now depend on how the members’ EA/L compare with one another: raising every EA by the same factor changes no force. If only some members are given a section, the calculator takes EA as the same in all of them and says so.

A pin holds its joint against moving in either direction and so gives two components of reaction. A roller holds it against moving along one line only, and gives one force along that line. A roller of any angle is treated exactly, by taking its joint’s displacement along the line it leaves open; no stiff spring stands in for it.

Before any solving, the calculator asks whether the truss could shift while every member keeps its length, which depends on the geometry and the supports and never on E and A. If it can, because a panel has no diagonal, or the supports hold it in too few directions, or part of it can swing on its own, no force is given: the joints that can move are named. Counting m + r against 2j does not settle it, as a truss can meet that count and still be a mechanism.

The page also lists the members that carry nothing by inspection. At an unloaded joint with no support, two members that are not on one line both carry zero; and when an unloaded joint with no support has two members along one line and a third across them, that third carries zero. Those members are then set aside and the rule is applied again at the next joint, so the list can run on.

Equilibrium of the whole is checked on the answer: the loads and the reactions together are summed in x, in y and in moment about the centre of the truss, and the largest force left unbalanced at any joint is found. Each is rounding, and an answer for which they are not is withheld.

The shapes offered to start from (triangle, king post, queen post, Pratt, Howe, Warren with and without verticals, Fink, and a cantilever from a wall) fill the form with joints, members, supports and loads for a span, a depth and a number of panels. The bottom chord lies on y = 0. The load you enter is the one on each joint along the top chord, with half of it on the joint at either end, where the chord has one; the supports are either a pin on the left with a roller on the right, or its mirror, or a pin on each side. In a Pratt truss the diagonals slope down towards the middle and in a Howe truss up towards it. A value a shape cannot be built with, such as an odd number of panels for a Pratt truss, is replaced by the nearest it can, and the form says what it changed. Everything is editable afterwards: a shape is a starting point and not a recommendation.

Two cases with a closed form, to hold the answers against.

A triangle of span L and height h, with a load P at the apex and a pin and a roller at the ends: each reaction is P/2; each rafter is in compression by Ps/(2h), where s = √((L/2)² + h²) is its length; the tie along the bottom is in tension by PL/(4h).

A vertical bar and two bars inclined at an angle θ either side of it, all of the same EA, joining a loaded joint to three pinned supports on a line: with a load P down on that joint, the vertical bar carries P/(1 + 2cos³θ) and each inclined one Pcos²θ/(1 + 2cos³θ), all in tension. Statics alone cannot give these three numbers; the shorter bar takes the most.