Equilibrium

A bridge is a machine designed to do nothing, and the whole of structural analysis begins by taking that seriously. Classical mechanics predicts motion from forces; a structure is the case where the motion has to come out zero, and the same second law read backwards then tells you what the forces must have been.

Statics is a special case, not a separate subject

Newton's second law says $\sum F = m a$ for a particle, and for an extended body it comes with a rotational partner, $\sum M = I α$ about the centre of mass. A structure that stays put has $a = 0$ and $α = 0$ , so both right-hand sides vanish:

\sum F = 0 \sum M = 0

Nothing has been added. Every result in this course is that pair of conditions, applied to smaller and smaller pieces of the object until the pieces are small enough to compare against a material test.

In a plane the two vector conditions become three scalar equations: forces sum to zero horizontally, forces sum to zero vertically, and moments sum to zero about any point. Three equations means at most three unknowns can be found, which is the constraint that shapes everything that follows. In three dimensions there are six, three force components and three moment components, and the same arithmetic applies with larger numbers.

The second condition is worth pausing on, because it is easy to think it is only about spinning. A book resting on a table has no tendency to spin, yet moment balance is what decides where the table's reaction acts. Moment equilibrium is the statement that the forces are distributed correctly, not merely that they cancel in total. Two equal and opposite forces offset by a distance sum to zero and still tear a structure apart.

The free body diagram

The only technique in statics is to draw a boundary, delete everything outside it, and replace what was deleted by the forces it exerted. What is left is a free body, and it must satisfy the three equations on its own.

The boundary can be drawn anywhere. Around the whole bridge, and the forces crossing it are the loads and the reactions from the abutments. Around one pin joint, and the forces crossing it are the members meeting there. Around a slice through a beam, and what crosses the cut is what the beam is carrying internally, which is the subject of a later lesson. Choosing the boundary well is the entire skill, and the rule of thumb is to cut through the thing you want to know about, since a force only appears on a diagram if it crosses the boundary.

Two errors account for most bad free bodies. The first is drawing internal forces: if two members inside the boundary push on each other, those forces cancel in pairs by Newton's third law and must not be shown. The second is forgetting self-weight, which for a steel beam might be five per cent of the load and for a concrete arch might be ninety.

Every force on the diagram needs a magnitude, a direction and a point of application. The point matters because it enters the moment equation, and a load applied at the wrong place gives the right total and the wrong distribution. A distributed load of $w$ kilonewtons per metre acting over a length $L$ can be replaced, for the purpose of finding external reactions, by a single force $w L$ acting at the centroid of the loaded area. That substitution is legal for the whole-body equations and illegal the moment you cut the beam open, a distinction the third lesson will need.

Moments, and why the point does not matter

The moment of a force about a point is $M = F d$ , where $d$ is the perpendicular distance from the point to the line of action. Sign convention is a free choice as long as it is kept: this course takes anticlockwise as positive.

If a body is in equilibrium, moments sum to zero about every point, not just about a convenient one. The proof is short. Suppose moments balance about a point $O$ and forces balance too. Take any other point $P$ , displaced from $O$ by a vector $r$ . The moment of a force $F_{i}$ acting at $x_{i}$ about $P$ is its moment about $O$ plus a correction $- r \times F_{i}$ . Summing over all forces, the corrections add up to $- r \times \sum F_{i}$ , which is zero because the forces balance. So the moments about $P$ sum to whatever they summed to about $O$ , namely zero.

This is a practical gift rather than a curiosity. Choosing the moment point on the line of action of an unknown force removes that unknown from the equation, so a well-chosen point turns three simultaneous equations into three separate ones. It is also the cheapest possible check on arithmetic: take moments about a second point, and if the answer is not zero, something above it is wrong.

What a support can resist

A support is idealised by what it prevents. If it prevents a movement, it supplies a force in that direction; if it prevents a rotation, it supplies a moment.

A roller prevents movement perpendicular to its surface only, so it supplies one force, normal to the surface, and no moment. A pin prevents movement in both directions in the plane but allows rotation, so it supplies two force components and no moment. A fixed support, or built-in end, prevents both movements and the rotation, so it supplies two forces and a moment: three unknowns from one support. A cable is a support that can only pull, and a smooth wall is one that can only push.

These are idealisations and they are honest ones only within limits. A real bolted steel connection is neither a perfect pin nor a perfect fixed end; it has some rotational stiffness and it is normally modelled as a pin because assuming less restraint is the safe error for bending. A bridge bearing is built as a roller for a reason: without it, a 100 metre steel deck warming by 30 degrees Celsius would try to grow about 36 millimetres, and preventing that generates forces far larger than the traffic.

Example. A beam spans $6$ m between a pin at $A$ and a roller at $B$ . A point load of $12$ kN acts $2$ m from $A$ . Find the reactions.

The unknowns are the two components at $A$ and the vertical force at $B$ . No horizontal load acts, so the horizontal component at $A$ is zero. Take moments about $A$ , which eliminates both components there: $R_{B} \times 6 - 12 \times 2 = 0$ , giving $R_{B} = 4$ kN. Vertical forces then give $R_{A} = 12 - 4 = 8$ kN. Check by taking moments about $B$ : $8 \times 6 - 12 \times 4 = 48 - 48 = 0$ .

Now you. The same arrangement spans $8$ m with a point load of $20$ kN acting $3$ m from $A$ . Find both reactions and check them.

Answer

Moments about $A$ give $R_{B} \times 8 = 20 \times 3$ , so $R_{B} = 7.5$ kN and $R_{A} = 12.5$ kN. Moments about $B$ : $12.5 \times 8 - 20 \times 5 = 100 - 100 = 0$ .

Two loads at once, and a distributed one

Real beams carry several things, and because the equilibrium equations are linear the contributions simply add.

Example. A beam spans $10$ m between a pin at $A$ and a roller at $B$ . It carries a uniformly distributed load of $4$ kN m⁻¹ over its whole length, plus a point load of $30$ kN at $7$ m from $A$ . Find the reactions.

The distributed load totals $4 \times 10 = 40$ kN and acts, for this purpose, at midspan. Moments about $A$ : $R_{B} \times 10 = 40 \times 5 + 30 \times 7 = 200 + 210 = 410$ , so $R_{B} = 41$ kN. Vertical balance gives $R_{A} = 40 + 30 - 41 = 29$ kN. Check about $B$ : $29 \times 10 - 40 \times 5 - 30 \times 3 = 290 - 200 - 90 = 0$ .

Now you. A cantilever of length $3$ m is built into a wall at its left end and carries a uniformly distributed load of $5$ kN m⁻¹ over its whole length. Find the vertical reaction and the fixing moment at the wall.

Answer

The total load is $5 \times 3 = 15$ kN, acting at $1.5$ m from the wall. Vertical balance gives a reaction of $15$ kN upwards. Moments about the built-in end give a fixing moment of $15 \times 1.5 = 22.5$ kN m, acting to hold the beam up.

The determinacy count

Count the unknown reaction components and compare with the number of equilibrium equations available. Call them $r$ and $e$ , with $e = 3$ for a single rigid body in a plane.

If $r < e$ the body is a mechanism: there are not enough restraints, and it moves. Three rollers all pointing the same way leave the body free to slide sideways, and no arrangement of loads will hold it. If $r = e$ the structure is statically determinate, and the three equations give the three unknowns. If $r > e$ it is statically indeterminate, and equilibrium alone cannot separate the reactions.

A beam on two supports is determinate. Add a third support in the middle and it is indeterminate to the first degree: there are four unknowns and three equations, and the equations are perfectly true but insufficient. A beam built in at both ends has six unknowns against three, indeterminate to the third degree.

The counting is necessary but not sufficient. Three restraints arranged so that all their lines of action pass through one point leave the body free to rotate about that point, no matter what the count says, and the same happens if they are all parallel. Geometry can defeat arithmetic here, which is why the count is a first screen rather than a proof.

Example. A horizontal beam is supported by a pin at one end and a roller at the other, and additionally propped by a vertical rod at midspan. Classify it.

The pin gives two unknowns, the roller one, the rod one: $r = 4$ against $e = 3$ . It is statically indeterminate to the first degree. Statics gives three true equations relating four unknowns, and the fourth relation has to come from somewhere else.

Now you. A signboard is held by a pin at its base and two cables to a wall, both cables horizontal and at different heights. Treating it as one rigid body in a plane, classify it.

Answer

The pin contributes two unknowns and each cable one, so $r = 4$ against $e = 3$ : indeterminate to the first degree. Note also that a cable can only pull, so any solution returning a compressive cable force would have to be rejected and the cable removed from the model.

Where equilibrium stops being enough

Indeterminacy is not a defect to be avoided. Engineers build indeterminate structures deliberately, because the extra restraint gives a second load path: if one support settles or one member cracks, an indeterminate structure redistributes and a determinate one collapses. Continuous beams over several supports are standard for exactly this reason.

The missing equations have a physical source, and it is the one thing equilibrium refuses to look at. A rigid body cannot tell you how the load shares between two props, because a truly rigid body would balance on whichever prop was infinitesimally taller. Real props compress, and the share is decided by how much each one shortens under load. The extra conditions are geometric statements that the deformed parts still fit together, and they cannot be written until deformation has a language.

Building that language takes the next four lessons. Before it, though, comes the more common case, where equilibrium is enough and the object is not one rigid body but a framework of dozens of members. Applying $\sum F = 0$ joint by joint solves those completely, and it is where structural analysis earns its keep.