ISEGORIA / MATH ENCYCLOPEDIA
Structural mechanics: how bridges carry load
A bridge, a crane or a building frame is a machine for moving load to the ground through the members that can carry it. A truss does it with bars that only push or pull, so every force follows from the balance of each joint, and the whole problem is a linear system. A beam does it by bending: shear force and bending moment diagrams say where it is working hardest, and the depth of its section sets both the stress and, cubically, the sag. A slender column does it only up to a critical load, past which it bows sideways in a shape fixed by how its ends are held and braced. Three ideas, equilibrium, curvature and stability, carry most of structural design.
Before you begin: Vectors and matrices, equilibrium of forces, calculus
Predict, manipulate, then test yourself: every experiment ends with practice questions that are generated fresh at three levels. Graphs illustrate the mathematics; they do not replace a proof.
1. A truss bridge, joint by joint
A truss is a stack of triangles made of bars pinned at their ends, so every bar carries only a push or a pull along its own length. That turns the whole problem into linear algebra: the forces on each joint must add to zero, two equations per joint, and a determinate truss with \(m\) bars, \(j\) joints and three support reactions has exactly \(m+3=2j\) unknowns and equations. This is the method of joints, solved here for Pratt, Howe and Warren trusses. Drag the load along the deck and watch bars trade tension for compression; raise the truss and watch the chord forces fall. Pick a joint to see its forces close into a polygon.
Worked example. Take a Pratt truss of six 4 m panels, 4 m high, with 100 kN on joint L2 (x = 8 m). The reactions are \(R_A=100\times16/24=66.7\) kN and \(R_B=33.3\) kN. Next to the load the beam moment is \(M=66.7\times8=533\) kN·m, so the top chord U1U2 is squeezed and the bottom chord L2L3 stretched by \(M/h=133\) kN. The two diagonals left of the load carry the shear \(V=66.7\) kN at 45°: \(66.7/\sin45^\circ=94.3\) kN of tension. In the panel just right of the load the shear has reversed to \(-33.3\) kN, so its diagonal U2L3 is in compression, 47.1 kN: a Pratt diagonal is a tie only while the shear keeps the sign it was designed for.
Watch out. Real joints are bolted or welded, not pinned, and bars also bend under their own weight; the pin-jointed model is accurate for slender bars loaded at the joints. The deck load is shared between the two joints either side by the lever rule, and self-weight is ignored. A real bridge is designed for the worst position of a moving load, which is what the Drive button sweeps through. Bar thickness in the picture follows the square root of the force, as if every bar were sized to the same stress.
Why does a taller truss need less steel in its chords?
Because the chords resist the bending moment as a couple: an equal pull in the bottom chord and push in the top chord, a distance \(h\) apart, give a moment \(F h\). The load fixes \(M\) at each section, so \(F=M/h\), and doubling \(h\) halves every chord force. The price is longer diagonals and verticals and more wind area, so there is a best depth, usually between a sixth and a tenth of the span.
2. Shear, moment and deflection
A beam resists load by bending. Cut it anywhere and the two halves are held in balance by a shear force \(V\) and a bending moment \(M\) across the cut: \(V\) is the net vertical force of everything to its left, \(M\) its net turning effect, and \(V\) is the slope of \(M\). Euler and Bernoulli added one assumption, that cross-sections stay flat and square to the bent axis, which makes the curvature proportional to the moment. The stress then varies linearly through the depth, zero on the neutral axis and largest on the faces. Choose how the beam is held, put loads on it, and read \(V\), \(M\), the deflection and the stress at any section; the numbers come from a finite-element solution that matches the textbook closed forms.
Worked example. A 6 m steel beam, 150 mm wide and 300 mm deep, has \(I=bh^3/12=3.375\times10^{-4}\) m⁴ and \(EI=67\,500\) kN·m². It is simply supported and carries 40 kN at 2 m. The reactions are \(40\times4/6=26.7\) kN and 13.3 kN, the largest moment is under the load, \(M=26.7\times2=53.3\) kN·m, and the largest stress is \(6M/bh^2=6\times53.3\times10^3/(0.15\times0.3^2)=23.7\) MPa. The deflection under the load is \(Pa^2b^2/(3EIL)=40\times4\times16/(3\times67\,500\times6)=2.11\) mm, one 2850th of the span. Making the beam 400 mm deep instead raises \(EI\) by \((4/3)^3=2.37\) and cuts the stress by \((3/4)^2\).
Watch out. Euler–Bernoulli theory ignores the deformation caused by shear, which matters for short deep beams (span under about ten times the depth), and it assumes small slopes. The drawing exaggerates the deflection by the factor in the readout so that it can be seen; every number is the true one. The material is elastic steel (E = 200 GPa) and the stress is not checked against yield. The shear stress shown is the parabola for a rectangular section.
Why does bending stress vanish on the neutral axis while the shear stress is largest there?
Bending stress is proportional to the distance from the neutral axis, so it is zero there. Shear stress is what stops the layers above a given level from sliding over the layers below it, and it equals the rate at which the bending force in the part of the section above that level grows along the beam. That part is largest when the level is the neutral axis, so \(\tau\) peaks there, at \(1.5V/A\) for a rectangle.
3. Buckling modes and bracing
Squeeze a slender rod along its axis and it stays straight until a critical load, then bows sideways. Euler found that load by asking when a bent shape could be in balance: the deflection obeys \(w^{\prime\prime}+k^2w=0\) with \(k^2=P/EI\), a sine and a cosine plus a straight line, and the end conditions decide which combinations are allowed. Only certain values of \(\kappa=kL\), the roots of a determinant, allow a bowed shape at all. They are the buckling loads, and the shapes are the modes. A pinned column gives \(\kappa=\pi,2\pi,3\pi\); clamping or freeing an end moves the roots, which is what the effective length \(KL\) records. A rigid lateral brace is one more constraint: put it at the node of the second mode and the column carries four times the load.
Worked example. The steel tube is 60 mm across with a 4 mm wall, so \(I=2.77\times10^{-7}\) m⁴ and \(EI=55.5\) kN·m², and it is 3 m long. Pinned at both ends it buckles at \(\pi^2EI/L^2=60.8\) kN, a stress of only 86 MPa, a third of yield. Fixed at the base and free at the top, \(\kappa=\pi/2\) and the load falls to 15.2 kN (\(K=2\)). Fixed at both ends, \(\kappa=2\pi\) and it rises to 243 kN (\(K=0.5\)); fixed–pinned gives 124 kN (\(K=0.699\)). A brace at mid-height on the pinned column also lifts the load to 243 kN, because the first bowed shape now has a node there. A brace at 0.3L is worth a factor 3.2.
Watch out. These are the loads for a perfectly straight, elastic column. A real column is crooked and slightly off axis and fails below Euler’s load; a stocky one (\(KL/r\) below about 89 for this steel) yields before it buckles, and the readout says so. Linear theory fixes the shape but not the size of the deflection, so the amplitude drawn is a fixed small amount. The brace is perfectly rigid; a real brace needs enough stiffness, and a flexible one gives a load between the braced and unbraced values.
Why does a brace right next to one end of a pinned column help so much?
Because the bowed shape must now be zero at the end and again at the brace, a short distance away, so it has almost no slope there: the brace stops the end from turning, like a clamp. A pinned column with a brace next to one end buckles like a fixed–pinned one, at about 2.05 times the load, with \(K\approx0.7\). The most effective place is mid-height, where the brace sits at the node of the second mode and the gain is 4.