Influence lines answered "how bad can it get here?" — but a bridge doesn't crack where you chose to look. This chapter answers the harder question: how bad can it get anywhere, and where is anywhere?
An influence line is built for one section. You pick point C, draw its IL, position the train for the worst case, and out comes the largest moment C will ever feel. Repeat for another section and you need another IL. But the beam has infinitely many sections — and nothing tells you in advance which one is in the most danger.
One IL, one section, worst position found. Answers: "What is the largest moment at C?"
Sweep the section as well as the load. Answers: "What is the largest moment anywhere in the span, and at which section does it occur?" That value is the absolute maximum.
The double sweep has a picture, and it is the most useful diagram in this chapter.
Run the train through every possible position. Each position gives one bending moment diagram. Now, at each section, keep only the largest value ever seen there. Plot those maxima against position and you get the moment envelope — a curve no individual BMD ever reaches, but which every BMD stays inside.
At every section it tells you the moment that section must be able to resist. Compare it against your capacity curve and you can see instantly where the beam is tight.
For a single load the envelope is exactly \(Px(L-x)/L\). For a train it is a parabola with faint kinks where the governing axle changes over.
The envelope's peak is the absolute maximum moment, and it sits slightly away from midspan — on the side the resultant leans toward.
Dispose of shear first, because it needs no theory at all.
The shear IL for any section is largest at the supports, so the worst shear anywhere is the worst reaction. To maximise a reaction, get as much load as close to that support as the train's geometry allows — usually the leading axle sitting right on top of it.
with \(\bar{x}_A\) the distance from support A to the resultant of the axles that are actually on the span. Try pushing the train one axle further too — sometimes letting the first axle run off the end and bringing a heavier one to the support wins.
Moment is the interesting case, and it has an exact answer you can derive in four lines. Two facts make it tractable:
Between axles the BMD is a straight line, so it has no interior maximum. The largest moment must sit at a kink — and every kink is a wheel. So there are only as many candidate sections as there are axles.
For computing reactions, the axles can be replaced by a single force R at the centroid \(\bar{x}\). The individual axles only matter again when you take moments inside the span.
Pick one axle \(P_k\) as the candidate. Let e be the signed distance from \(P_k\) to the resultant R, positive when R lies to the right. Put \(P_k\) at distance x from support A, so R is at \(x+e\).
The formula gives the best moment obtainable for a chosen candidate axle. Different axles give different answers, so the procedure is a short trial, not a single substitution.
It usually wins, because it makes |e| smallest and \(x\) closest to midspan, where the \(Rx^2/L\) term is biggest. In Example 2 the 60 kN axle is 0.83 m from R and does govern.
An axle near the resultant but with a lot of load ahead of it carries a large \(M_{\text{left}}\) penalty. That is why the trailing axle in Example 2 scores only 392 despite a healthy first term. Test them all — it is three lines of arithmetic each.
The top panel is the beam with the train at its current position. The bottom panel shows the bending moment diagram for that position in solid colour, with the full envelope ghosted behind it. Drag the position slider and watch the solid diagram sweep around inside the fixed envelope — then press Snap to absolute maximum and watch it touch the peak.
Set P₃ = 0 for a two-axle train.
Press snap, then move the slider one step either way. The moment falls off in both directions — you are sitting on a genuine maximum.
The peak is barely a bump. Slide the train across a couple of metres and the maximum hardly changes — that is the 0.2% from Section 2.
Now the train is long compared with the beam. Every candidate turns invalid, and the honest answer comes from a single axle at midspan — Section 7.
The derivation quietly assumed the whole train stays on the span while you slide it. On a short span that is impossible, and the rule has to be abandoned rather than trusted.
Everything in this chapter is one sweep in two directions. Move the load to find the worst case at a section; move the section to find the worst case in the beam. The envelope records both sweeps at once, and the bisection rule is simply the shortcut that finds its peak without doing the sweep by hand:
Problem: Axle loads of 60 kN and 40 kN, 4 m apart, cross a 15 m simply supported beam. Find the absolute maximum moment and the absolute maximum shear.
Problem: Axle loads of 40 kN, 60 kN and 20 kN with spacings of 4 m and 3 m cross a 20 m simply supported beam. Find the absolute maximum moment.
Problem: For the Example 2 train, compare the true absolute maximum against the largest moment obtainable at midspan.
1. Why can the absolute maximum moment only occur directly under an axle?
2. The bisection rule positions the train so that:
3. The resultant of a train lies 0.83 m to the left of your candidate axle on a 20 m span. Where does that axle go?
4. Your bisection-rule position puts the rear axle 0.5 m beyond support B. What do you do?
5. Why is the envelope useful even though no single load position ever produces it?