A moving load is the awkward part of bridge analysis. Dead load stays where you put it, but a lorry can be anywhere on the span, and the answer to "what is the worst it can do here?" depends entirely on where it stops. An influence line makes that question quick to answer, and it is one of the few pieces of structural analysis you can sketch freehand and still trust. Here is how to build one for a simply supported span and read the governing vehicle position straight off it.
What an Influence Line Shows
A bending moment diagram gives the moment at every section for one fixed arrangement of load. An influence line does the opposite. It fixes one point on the beam and shows how the response at that point changes as a unit load of 1 kN moves across the span. Both diagrams can be triangles on a simply supported beam and still mean completely different things.
Watch the units, because they differ by response. On a reaction or shear influence line the ordinate is response per unit load, which makes it dimensionless: multiply it by the real load in kN and you get kN. On a moment influence line the ordinate carries length, so an ordinate of 4.8 m under a 100 kN axle gives 480 kNm. Mixing those two up is the most common slip.
Influence lines rest on linear elasticity and superposition, which is what lets you scale ordinates by real loads and add the contributions of several axles. The Müller-Breslau principle gives a quick check: the influence line for a response is the shape the beam adopts if you release the restraint corresponding to that response and impose a unit displacement. For a simply supported beam that produces straight lines and triangles, exactly as the algebra does.
Setting Up the Beam and Its Three Key Lines
Call the span L, with supports A at the left-hand end and B at the right. Put the unit load at distance x from A. For a section of interest C, let a be its distance from A and b = L − a its distance from B. Three relationships do all the work.
Reaction at the supports
Taking moments about B gives RA = (L − x)/L: a straight line falling from 1.0 at A to zero at B. The reaction at B is its mirror image, x/L. The two ordinates always sum to one.
Shear at C
With the load left of C, the shear at the section is VC = −x/L, running from zero at A to −a/L at C. With the load right of C, VC = (L − x)/L, running from +b/L at C to zero at B. You get two parallel straight lines with a step of exactly 1.0 at C — the unit load crossing, contributing its full value.
Bending moment at C
With the load at or left of C, MC = x·b/L. With the load at or right of C, MC = a(L − x)/L. Both branches are linear and meet at C, so the diagram is a triangle with a peak of ab/L. At midspan of a 20 m span that peak is 5.0 m; move the section a metre off centre and it falls to 4.95 m. The peak can never exceed L/4.
A Worked Example: 20 m Span, Section 8 m from A
Take a = 8 m and b = 12 m. Sketching the three diagrams gives:
- Reaction at A: 1.0 at A, 0.6 under C, zero at B.
- Shear at C: −8/20 = −0.40 just left of C, +12/20 = +0.60 just right.
- Moment at C: zero at both supports, peaking at 8 × 12 / 20 = 4.80 m at C.
Now roll a two-axle vehicle across — 100 kN per axle, axles 4 m apart. The moment at C is simply the sum of each load times the ordinate beneath it.
Trial one, leading axle at C: that axle contributes 100 × 4.80 = 480 kNm, and the trailing axle at x = 4 m sits under an ordinate of 4 × 12 / 20 = 2.40 m, adding 240 kNm. Total 720 kNm.
Trial two, trailing axle at C: still 480 kNm from the axle at C, but the leading axle now sits at x = 12 m under an ordinate of 8 × 8 / 20 = 3.20 m, adding 320 kNm. Total 800 kNm, which governs.
A third position with axles at 6 m and 10 m gives 3.60 + 4.00 = 7.60 m, or 760 kNm. Lower. That is not luck. Because the influence line is piecewise linear, the running total of load times ordinate is also piecewise linear between the points where an axle crosses the peak or reaches a support, so the maximum always sits at one of those breakpoints.
Finding the Worst-Case Vehicle Position
- Decide which response you are checking and draw the matching influence line. Label the ordinates at every position an axle could occupy.
- Cut a paper strip the length of the span and mark the ordinate values on it. Slide the strip under a drawing of the vehicle and read off the sum of load times ordinate. It looks like a student trick; for two or three axles it beats a spreadsheet.
- Test each axle over the peak in turn, then test the vehicle pushed hard against each support.
- Repeat with the minimum ordinates if hogging moment or reversed shear matters.
- Check that every axle is actually on the span at the governing position. A vehicle that has run off the end is not a valid case.
Distributed Loads and Real Assessment Loading
For a UDL the method changes shape and becomes more useful. The response is the load intensity times the area under the influence line, and because you choose where the load sits, you load only the regions giving the sign you want. For maximum sagging moment at C you load the whole span, since that triangle is positive throughout. For maximum positive shear just right of C you load only the part of the span to the right of C and leave the rest clear — a pattern a bending moment diagram alone will never show you.
UK highway load models follow the same logic. HA loading combines a uniformly distributed load with a knife edge load, so it is placed by loading the positive or negative regions of the influence line. HB loading is a notional heavy vehicle defined by axle loads and spacings, so it is placed by testing breakpoints as above. The influence line tells you where to put the load; the assessment standard tells you how large it is and which factors apply.
One caveat. This is a single-line beam analysis. Load on a real deck spreads transversely between beams, and no influence line captures that. Apply a distribution factor, or model the deck as a grillage, before treating the result as the effect on any one girder.
Four Checks and a Habit Worth Keeping
Seconds of checking catch most errors:
- Shear ordinates either side of C must differ by exactly 1.0.
- Reaction ordinates must sum to 1.0 at every position, and neither may exceed 1.0.
- Moment ordinates must be zero at both supports and peak at ab/L — never above L/4.
- Moment ordinates are in metres; shear and reaction ordinates are dimensionless.
Then make the sketch before you reach for software. Estimate the worst case by eye from the diagram, run the numbers, and compare. When the two disagree you have found either an arithmetic slip or a load position you had not considered, and both are worth the minute it takes.
Photo: k_r_craft / Pixabay
