CEE 330 — Critical Load Positions

Why This Chapter Exists

Every beam problem so far handed you the loads and their positions. Real bridges are not so polite. A train, a truck, a crane trolley — the load walks the whole length of the span, and every step of that walk is a different structure to analyse.

⚡

The design question changes shape. It is no longer "what is the moment at C?" — there is no single answer. It is "what is the largest moment C will ever see, over every position the train can occupy?" That is a search problem hiding inside a statics problem, and it has two halves: find the critical position, then compute the value there. Half one is the hard half, and it is what this entire chapter attacks.

📸

Photographing a sprinter. You do not want the average runner, and you do not want a random frame. You want the one frame at the finish line. Analysing a fixed load is taking a photo of a statue — any moment will do. Analysing a moving load is photographing motion: the whole skill is in the timing of the shutter. Every formula below is a way of knowing when to press it.

Influence lines told you half the story. An IL answers "if a unit load sits at x, what is the reaction?" Perfect for one wandering load. But a locomotive is fourteen wheels locked into a rigid pattern — you cannot place them independently. The IL still governs the physics; this chapter is the bookkeeping that finds where to drop the whole train onto it.

Sometimes you really can just look

For a small, simple group, inspection settles it. A four-wheel electric crane (two axles, 12 ft apart) runs on a 40 ft beam AB with a section X–X 10 ft from A:

Maximum reaction at A: equal wheel loads, so get them as close to A as possible — wheel 1 directly over A, wheel 2 on the span 12 ft in.

Maximum positive shear at X–X: put the wheels on the right of the section, with the first wheel as close to the section as it can get. Anything left of the section subtracts.

Maximum moment: with two equal wheels rigidly framed together, one of them sitting at the centre gives the greatest moment at that point.

Reach for inspection first — it is free. It runs out the moment you meet a locomotive with fourteen unequal wheels and thirteen different spacings, which is exactly where the rest of this chapter starts.

The train we will use throughout (a steam locomotive on a 60 ft span, wheel 1 leading, moving right → left):

loads (kips) : 10 10 36 36 36 36 10 10 20 20 20 20 20 20 (Σ = 304^k)
spacings (ft) : 5 8 6 6 6 8 4 9 4 4 12 4 4

The Sawtooth

Bring the locomotive on from the right and watch the reaction at A. Nothing, then a trickle as wheel 1 crosses onto the span, then a steady climb as the whole group creeps toward A. Then wheel 1 reaches A, rolls past it, and drops off the end — and R_A falls off a cliff, instantly, by the full weight of that wheel.

Then it climbs again. Then wheel 2 drops off. Climb, cliff, climb, cliff.

🏊

A relay team running toward you. The closer the pack gets, the louder they sound — that is the steady climb. But every time the lead runner crosses the line and vanishes behind you, the noise drops abruptly by one runner's worth. The volume never peaks between departures; it peaks in the instant just before each one leaves. Same for R_A: the maximum is always the moment a wheel is sitting exactly over the support, an instant before it falls off.

The one idea this chapter runs on

Reaction and shear are saw-toothed functions of train position: continuous climbs separated by sudden drops. A maximum can only ever sit at the top of a tooth — and the top of a tooth is precisely the instant a wheel is at the critical point:

Reaction at A

A wheel over the support

The cliff happens when a wheel leaves the span at A.

Shear at a section

A wheel at the section

The cliff happens when a wheel crosses the section.

Moment at a section

A wheel at the section

No cliff here — but the slope flips there. Section 6.

⚡

This collapses an infinite search into fourteen candidates — and in practice into two or three. Instead of sliding the train continuously and computing forever, you only ever test "wheel 1 at the critical point", "wheel 2 at the critical point", and so on. Better still: you do not even compute the function at each. You compute only the change from one candidate to the next, and stop the moment that change turns negative. Two or three lines of arithmetic replaces the whole search.

Reactions: Gain vs Loss

Take the train from "wheel 1 over A" to "wheel 2 over A". That is a move to the left of exactly d₁, the spacing between wheels 1 and 2. Two things happen to R_A, and they pull in opposite directions.

The gain

Everyone shuffles closer

Every load still on the span moves d₁ nearer to A. A load at distance x from A hands A a share (L−x)/L of itself, so each one gains P·d₁/L. Add them up: +ΣP·d₁/L.

The loss

The front wheel walks out

Wheel 1 was sitting on the support, giving A all of itself. One step later it is off the span entirely, giving nothing. That is a clean loss of −P₁.

The change formula

\[ \Delta R \;=\; \underbrace{\frac{\Sigma P\, d_1}{L}}_{\text{everyone creeps closer}} \;-\; \underbrace{P_1}_{\text{front wheel lost}} \]

ΣP = the loads that are on the span at the start and stay on through the move — do not include the one that walks off. d₁ = distance moved (the spacing to the next wheel). P₁ = the load that was over the reaction and is moved off. L = span.

🛒

Queuing at a shop counter. The queue shuffles forward one place. Everybody still in line gets marginally closer to being served — small gains, but there are a lot of them. Meanwhile the person at the front is served and leaves. If the crowd behind is large and the departing customer small, the queue as a whole got closer. If the person leaving was the heavy one, the queue got worse. ΔR just asks which of those two effects won.

Reading the sign — the entire procedure

ΔR positive → keep going. The move improved things; step to the next wheel and test again.
ΔR negative → stop, and go back one. You have just walked down from the peak, so the previous position was the maximum. Do not make the move.

You never have to compute R_A itself until you have found the position. That is the whole economy of the method.

The correction: wheels rolling on at the far end

While the front of the train walks off at A, the tail may be rolling onto the span at B. A load that arrives partway through the move contributes for only part of it:

\[ \Delta R \;=\; \frac{\Sigma P\, d_1}{L} \;+\; \frac{P' e}{L} \;-\; P_1 \]

where P′ is a load that comes on during the movement and e is the distance it has come onto the span when the move finishes. These terms are usually tiny — in the worked example they never change the decision — but include them, because "usually" is not "always".

Simulator: Drive the Locomotive

The top panel is the girder with the train on it. The bottom panel plots your chosen function against train position for the entire traverse — that is the sawtooth, drawn exactly (these functions are piecewise linear, so the plot is not an approximation). The dot is where you are; the marker is the true maximum.

What are we maximising? Section at c = 20 ft from A (used for shear & moment) Train position — drag, or use the buttons

Try this

See the cliffs

Stay on "Reaction at A" and drag the position slider slowly. Watch the plot climb and drop. Count the teeth — one per wheel.

Try this

Walk the method

Press Next wheel repeatedly from the start. The panel shows the gain, the loss and the net Δ. Stop when Δ goes red — that is the answer, and it matches Section 7.

Try this

Move the section

Switch to Shear and drag c from 20 ft to 10 ft. The critical wheel stays wheel 3, but the peak jumps from 75.6 to 112.4 kips — and new terms appear as loads run off the left end.

Shear: Same Trick, New Cliff

Shear at a section is the reaction minus whatever load has already gone past:

\[ V \;=\; R_A \;-\; \Sigma P_{\text{left of section}} \]

So as the train rolls left, V climbs for exactly the reason R_A climbs — and it drops off a cliff not when a wheel leaves the span, but when a wheel crosses the section, moving from the helpful side to the harmful side. Different cliff, identical algebra:

Change in shear

\[ \Delta V \;=\; \frac{\Sigma P\, d_1}{L} \;-\; P_1 \qquad\text{and with a load coming on:}\qquad \Delta V \;=\; \frac{\Sigma P\, d_1}{L} \;+\; \frac{P' e}{L} \;-\; P_1 \]

Here P₁ is the wheel that passes the section, and d₁ is the spacing to the wheel behind it. Same sign rule: keep moving while ΔV is positive; the maximum is the last position before it turns negative.

⚡

Why the two formulas are identical is worth a moment. V and R_A differ only by the loads sitting left of the section. During a move in which nothing else changes hands, that difference is constant — so the change in shear is the change in reaction, minus the one wheel that switched sides. One derivation, two functions.

The subtle case: loads running off the left end

Put the section close to A and something new happens. A load sitting between the section and A is being subtracted from the shear — it produces negative shear at that section. When that load rolls off the left end of the span, the penalty vanishes, so its departure increases the shear.

Trap 1

Do not put it in ΣP

ΣP is "loads on the span that stay on". A wheel that leaves during the move is not one of them — leave it out or you double-count it.

Trap 2

Add its departure as a gain

Add + P·x/L, where x is how far that load still had to travel to reach A when the move began. Positive, because losing a load that was hurting you helps.

In the section-at-10-ft example this shows up as the innocuous-looking + 10 × 5/60 and + 10 × 2/60 terms — wheel 1 and wheel 2 leaving the span while wheels 3 and 4 come up to the section.

⚖

Shear is a tug of war across the cut. Loads to the right of the section pull your way through the reaction; loads to the left pull against you. A wheel crossing the section is a player switching teams — which is why it costs you the full P₁ and not half of it. A wheel dropping off at A is an opposing player leaving the field, which is why it is a gain.

Moment: The Average-Load Criterion

Moment has no cliffs. Nothing switches sides discontinuously — a load crossing the section contributes zero moment about it at that instant, so M is perfectly continuous. Yet the answer is still discrete: at maximum moment a wheel is at the section. Understanding why gives you the most useful criterion in the chapter.

Read it off the moment influence line: a triangle with its peak i under the section C, rising over the left arm a and falling over the right arm b.

Moving everything left: the right-hand loads climb, the left-hand loads descend

W₁ is the resultant of everything left of C, W₂ of everything right of it. Move the train left and the two resultants slide along the two slopes in opposite directions. The moment peaks where their effects cancel.

The change in moment

\[ \Delta M \;=\; I - D \;=\; W_2\frac{i}{b} \;-\; W_1\frac{i}{a} \]

I = increase from the loads on the right climbing toward the peak. D = decrease from the loads on the left sliding down toward A. Positive ΔM means keep moving left.

Set ΔM = 0 and the criterion falls out

\[ W_1\frac{i}{a} = W_2\frac{i}{b} \;\;\Longrightarrow\;\; \frac{W_1}{a} = \frac{W_2}{b} = \frac{W_1+W_2}{a+b} = \boxed{\;\frac{W}{L}\;} \]

In words: the moment at C is a maximum when the average load per foot to the left of C equals the average load per foot to the right — which, by the same step, equals the average load per foot over the whole span. That last form is the one everybody quotes, because W and L are trivially known.

⚖

Balancing load density, not load. Not "equal weight on both sides" — that would ignore the arms. It is equal weight per foot. If the right-hand stretch is more densely loaded than the span average, dragging more of it toward C still pays. Once the left is the denser side, you have overshot. Maximum moment is the crossover, and the span average W/L is the neutral reading you compare against.

Why a wheel must be sitting at the section

W₁ and W₂ only change when a load crosses C. Between crossings both are constant, so ΔM holds one fixed sign and the moment marches steadily in one direction. It can therefore only turn from rising to falling at the instant a load passes from the right of C to the left. Same conclusion as the sawtooth, reached through a slope change instead of a cliff.

How you actually test it

Put wheel k at the section. The wheel itself is ambiguous — it is on the boundary — so test it both ways:

Count wheel k on the right: the left-hand average must be less than the span average → W₁/a < W/L

Count wheel k on the left: the left-hand average must now be greater → W₁/a > W/L

Both hold → wheel k satisfies the criterion. Only one holds → it does not; move on.

⚡

Two warnings the textbook is emphatic about. (1) More than one wheel can satisfy the criterion — it happens whenever loads pass off at the left or come on at the right during the move. When it does, compute M for each candidate and take the larger; the answers are usually within a percent or two, but "usually" does not sign a drawing. (2) Since maximum moment wants as much load on the span as possible, do not waste time testing the first wheel or two — start where the heavy axles are, and let the criterion do the rest.

Worked Examples

All four use the 14-wheel locomotive from Section 1 on a 60 ft span. Every position and value below can be reproduced in the simulator.

Example 1

Maximum Reaction at A

Problem: Find the wheel position giving the greatest reaction at A, and its value.

Wheel 1 over A → wheel 2 over A (move 5 ft). On the span and staying on: 234^k. Nothing new comes on.

\[ \Delta R = 234\times\tfrac{5}{60} - 10 = +9.5 \quad\text{increase} \]

Wheel 2 → wheel 3 (move 8 ft). ΣP = 224^k; wheel 12 (20^k) rolls on by e = 1 ft.

\[ \Delta R = 224\times\tfrac{8}{60} + 20\times\tfrac{1}{60} - 10 = +20.2 \quad\text{increase} \]

Wheel 3 → wheel 4 (move 6 ft). ΣP = 208^k; wheel 13 rolls on by e = 3 ft. Now we are losing a 36^k axle instead of a 10^k one:

\[ \Delta R = 208\times\tfrac{6}{60} + 20\times\tfrac{3}{60} - 36 = -14.2 \quad\text{decrease} \]

Negative → do not make that move. The previous position governs. Summing the reaction directly with wheel 3 at A: R_A = 120.7 + 9.5 + 20.2 = 150.4^k.

Answer: maximum R_A = 150.4 kips, with wheel 3 directly over A. Note what did the work: the loss term jumped from 10 to 36 the moment a driver axle reached the support, and that alone flipped the sign. Note also that dropping the two small "rolling on" terms would not have changed any decision — but you cannot know that until you have written them down.

Example 2

Maximum Positive Shear at a Section 20 ft from A

Problem: Same train, same span. Find the position for greatest positive shear at the section, and its value.

Wheel 1 at section → wheel 2 at section (5 ft). ΣP = 174^k; wheel 8 (10^k) comes on by 2 ft.

\[ \Delta V = 174\times\tfrac{5}{60} + 10\times\tfrac{2}{60} - 10 = +4.83 \quad\text{increase} \]

Wheel 2 → wheel 3 (8 ft). ΣP = 184^k; wheel 9 (20^k) comes on by 1 ft.

\[ \Delta V = 184\times\tfrac{8}{60} + 20\times\tfrac{1}{60} - 10 = +14.87 \quad\text{increase} \]

Wheel 3 → wheel 4 (6 ft). ΣP = 204^k; wheel 10 comes on by 3 ft; a 36^k axle now crosses the section.

\[ \Delta V = 204\times\tfrac{6}{60} + 20\times\tfrac{3}{60} - 36 = -14.6 \quad\text{decrease} \]

With wheel 3 at the section, R_A = 95.57^k and the loads left of the section total 20^k (wheels 1 and 2), so V = 95.57 − 20.

Answer: maximum V = 75.6 kips, with wheel 3 at the section. Same critical wheel as Example 1, and for the same reason — the first heavy axle arriving at the critical point is what turns the sign.

Example 3

Shear at a Section 10 ft from A — Loads Running Off

Problem: Move the section to 10 ft from A. Now wheels reach the end of the span while later wheels are still coming up to the section, so the Article 64 correction bites.

Wheel 1 → wheel 2 at the section (5 ft). ΣP = 184^k, wheel 9 comes on by 3 ft. Nothing leaves the span yet.

\[ \Delta V = 184\times\tfrac{5}{60} + 20\times\tfrac{3}{60} - 10 = +6.33 \quad\text{increase} \]

Wheel 2 → wheel 3 (8 ft). Wheel 1 now runs off at A, so it is excluded from ΣP (204 − 10 = 194^k) and its departure is added as a gain, +10×5/60, because it had 5 ft left to travel. Wheels 10 and 11 (40^k together) come on by 7 and 3 ft — average 5.

\[ \Delta V = 194\times\tfrac{8}{60} + 40\times\tfrac{5}{60} + 10\times\tfrac{5}{60} - 10 = +20.03 \quad\text{increase} \]

Wheel 3 → wheel 4 (6 ft). Wheel 2 runs off with 2 ft to travel; ΣP = 224^k; a 36^k axle crosses the section.

\[ \Delta V = 224\times\tfrac{6}{60} + 10\times\tfrac{2}{60} - 36 = -13.27 \quad\text{decrease} \]

With wheel 3 at the section, R_A = 122.4^k and only wheel 2 (10^k) is left of the section.

Answer: maximum V = 112.4 kips, again with wheel 3 at the section. Compare with Example 2's 75.6 kips: the closer the section is to the support, the larger the shear — exactly what the shear influence line predicts. The two +P·x/L terms are small here, but the wheels they represent had to be kept out of ΣP, and that is a 10-kip error each if you forget.

Example 4

Maximum Moment at C, 20 ft from A

Problem: Find the position of the train for maximum bending moment at C (a = 20 ft, b = 40 ft, L = 60 ft), and the moment there.

Maximum moment wants as much load on the span as possible, so start at wheel 3 — the first driver axle. Test each candidate against W/L.

Wheel 3 at C. W = 204^k on the span, so W/L = 204/60 = 3.40.
Counted right: W₁/a = 20/20 = 1.00 → 3.40 > 1.00 ✓
Counted left: W₁/a = 56/20 = 2.80 → 3.40 > 2.80 ✗ (needs to be less)
Wheel 3 fails. Too little load is behind C — keep rolling.

Wheel 4 at C. W = 224^k, W/L = 3.733.
Counted right: 56/20 = 2.80 → 3.733 > 2.80 ✓ Counted left: 92/20 = 4.60 → 3.733 < 4.60 ✓
Wheel 4 satisfies the criterion.

Wheel 5 at C. Wheel 1 has run off, wheel 2 now sits exactly over A, and wheel 11 has come on. A load standing on the support contributes nothing to the moment at C and is about to leave, so it takes no part in the criterion: W = 224^k, W/L = 3.733.
Counted right: 72/20 = 3.60 → 3.733 > 3.60 ✓ Counted left: 108/20 = 5.40 → 3.733 < 5.40 ✓
Wheel 5 also satisfies it — precisely the multiple-solution case the criterion warns about, caused by wheels leaving at A. Wheel 6 fails, so compute both and compare.

Wheel 4 at C. R_A = 2(10)(56.5/60) + 4(36)(37/60) + 2(10)(18/60) + 2(20)(5/60) = 18.83 + 88.80 + 6.00 + 3.33 = 116.97^k
M_C = 116.97(20) − [2(10)(16.5) + 36(6.0)] = 2339.3 − 546 = 1793.3 ft-kips

Wheel 5 at C. R_A = 4(36)(43/60) + 2(10)(24/60) + 3(20)(9/60) = 103.2 + 8.0 + 9.0 = 120.2^k
M_C = 120.2(20) − 2(36)(9) = 2404 − 648 = 1756.0 ft-kips

Answer: maximum M_C = 1793.3 ft-kips with wheel 4 at the section. Wheel 5 gives 1756 — only 2% less, which is exactly why the textbook says the difference between competing candidates is usually small, and exactly why you still check both. (Each grouped term above is a set of equal wheels multiplied by their average distance from the far support — a shortcut, not a new principle.)

A note on the printed figure. The original text gives 1792 ft-kips here. It rounds each reaction term to one decimal (18.8 instead of 18.833, 3.3 instead of 3.333) before summing, which loses 0.07^k off R_A and 1.3 ft-kips off the moment. The exact value is 1793.3, which is what the simulator reports. Neither the critical wheel nor the design decision changes — but it is worth seeing how quickly rounded intermediates drift.

Quick Reference & Quick Check

The core insightmaxima occur with a wheel at the critical point

Critical point, reactionover the support

Critical point, shear & momentat the section

Change in reactionΔR = ΣPd₁/L − P₁

Change in shearΔV = ΣPd₁/L − P₁

Load coming onadd + P′e/L

Load running off at Aadd + P·x/L (shear only)

ΣP meansloads on the span that stay on

P₁ meansthe load leaving the span / passing the section

Stop rulemake moves while Δ > 0; stop when Δ < 0

Change in momentΔM = I − D = W₂(i/b) − W₁(i/a)

Moment criterionW₁/a = W₂/b = W/L

Criterion in wordsaverage load left = average load right

Tiesseveral wheels may qualify — compute each

Worked answersR_A 150.4^k · V 75.6^k · M 1793 ft-k

Simple load groupsinspection beats algebra — use it

1. Why does the maximum reaction always occur with a wheel exactly over the support?

2. You compute ΔR for a move and get a negative number. What do you do?

3. A wheel is going to roll off the left end of the span during your move. In ΔV for a section 10 ft from A, that wheel:

4. The criterion W₁/a = W₂/b = W/L says the moment at C is a maximum when:

5. Moment is a continuous function of train position — no cliffs. So why is the critical position still "a wheel at the section"?