Trusses, beams, deflection & buckling

The method of sections exploits the fact that any free body of a structure in equilibrium must itself be in equilibrium. We pass an imaginary cut through the truss, slicing at most three members whose forces are unknown, and consider the equilibrium of everything on one side of the cut. Because a rigid body in a plane has exactly three independent equations, one cut through three members lets us solve all three — or any one of them in a single equation by choosing a clever moment centre.

To isolate one diagonal, sum vertical forces — the horizontal chords contribute nothing — so the diagonal carries the net shear across the cut. To isolate a chord, take moments about the joint where the diagonal and the other chord meet; their lines of action pass through that point and drop out, leaving one equation in the chord force. The cut force is drawn as tension (arrow pulling away from the free body); a negative result means compression.

The method is the engineer's shortcut: to grade one member of a large bridge truss you need not march through every joint — one well-placed section gives the answer immediately. Here the cut is shown as a dashed line, the chosen free body is shaded, and the chosen member force is solved and cross-checked against the full joint solution.

Cut the beam at a distance x from the left and consider the left free body. The shear force V(x) is the algebraic sum of all transverse forces to the left; the bending moment M(x) is the sum of their moments about the cut. With the sagging-positive convention, V and M obey the fundamental differential relations:

So under a uniform load w the shear varies linearly and the moment parabolically; under no load the shear is constant and the moment linear. A point load P makes a jump of −P in V; an applied couple M0 makes a jump in M; the moment is locally maximum where V passes through zero (dM/dx = 0). Integrating from the left:

For a simply supported span the two reactions follow from sum of moments = 0 and sum of vertical forces = 0; for a cantilever the fixed end supplies a vertical reaction and a fixing moment. The solver samples x finely along the span, accumulates V and M exactly using these rules, and draws both diagrams with their extreme values marked.

For small deflections the curvature of the neutral axis is proportional to the bending moment. The Euler-Bernoulli beam equation relates the transverse deflection y(x) to M(x):

Integrating once gives the slope EI·dy/dx and again gives EI·y, each step introducing a constant of integration. The Macaulay method writes the moment with bracket terms (x − a) that are taken as zero when x is less than a, so one single expression covers the whole beam and integrates termwise — a point load contributes a bracket of power one, a UDL a bracket of power two:

The constants C1 and C2 follow from the boundary conditions: zero deflection at both supports of a simply supported beam, or zero deflection and zero slope at the fixed end of a cantilever. The maximum deflection occurs where the slope y' is zero; the solver scans for that point. Results scale inversely with the flexural rigidity EI, so doubling EI halves the deflection.

An influence line plots the value of one structural response (say the moment at a section C) as a function of the position of a single moving unit load. It answers a different question from a bending-moment diagram: the BMD shows M at every section for a fixed load, while the influence line shows one quantity as the load travels. Influence lines are the basis of design for moving loads — vehicles on bridges, gantry cranes.

For a simply supported span of length L with a unit load at position a from the left, equilibrium gives the reactions, and these define the lines directly:

For shear and moment at a section a distance c from the left support, evaluating V and M as the load passes left and right of C gives two straight segments meeting at C:

By the Muller-Breslau principle the influence line for any reaction or internal action has the shape of the deflected structure produced by releasing that action and giving it a unit displacement — which is exactly the piecewise-linear shapes above for a determinate beam. The solver sweeps the unit load, evaluates the chosen response at each step, traces the influence line, and reports its peak value and the critical load position.

A straight elastic column under axial compression remains straight until the load reaches a critical value, at which it suddenly bows sideways — buckling. Euler solved the governing differential equation EI y'' = −P y for a pin-ended strut and obtained the smallest non-trivial load:

The effective length factor K captures the end restraints: K = 1 for pinned-pinned, K = 0.5 for fixed-fixed, K = 2 for fixed-free (a flagpole), and K is about 0.7 for fixed-pinned. Dividing by area and writing the slenderness ratio lambda = K L / r gives the critical (buckling) stress:

Euler's hyperbola sigma(cr) = pi² E / lambda² is valid only while sigma(cr) stays below the yield (or proportional) limit; for stocky columns (small lambda) the column yields or crushes before it buckles, so the curve is capped at the yield stress. The boundary slenderness is lambda(c) = pi · sqrt(E / sigma(y)). The solver draws this curve, marks the operating point for the chosen geometry, and animates the corresponding buckled mode shape.