Influence lines and moving loads

In short

An influence line diagram (ILD) shows how one chosen response — a reaction, a shear force, a bending moment or a member force at one fixed section — varies as a unit load travels across the structure. It is the standard tool for structures carrying moving loads: bridge girders and trusses, gantry girders in workshops, crane runways and conveyor supports, where the worst load position is not obvious. Once the ILD is drawn, any load train is handled by simple multiplication (point load × ordinate) or integration (UDL × area). It is a guaranteed question in GATE Civil, state PSC/AE exams and university structural analysis papers, and it is day-to-day work for bridge design engineers in NHAI/MoRTH consultancies, RITES, RDSO-linked railway design cells, L&T, AFCONS and design firms applying IRC:6 and the IRS Bridge Rules.

Explained

What an influence line is — and how it differs from a BMD

• ILD: response at a fixed section, plotted against the position of a unit load. Abscissa = load position; ordinate = value of that response.
• BMD/SFD: response at all sections for a fixed set of loads. Abscissa = section position.
• Ordinate units: reaction and shear ILDs are dimensionless (force per unit force); bending moment ILDs have units of length (m); member-force ILDs are dimensionless.
• Use of the ILD (valid only while the structure is linear elastic and the principle of superposition holds):
  • Point load W at a position where the ordinate is y → response = W·y.
  • Several point loads → Σ W·y (one ordinate per load, read at that load's position).
  • UDL of intensity w over a stretch → w × (area of ILD under that stretch).
  • Couple M applied at a point → M × (slope of ILD at that point).
• For determinate structures ILDs are made of straight lines; for indeterminate structures they are curved.

ILDs for determinate beams (the examinable set)

For a simply supported beam AB of span L, section C at distance a from A, b = L − a:

• Reaction R_A: straight line, ordinate 1 at A, 0 at B; R_B is its mirror image.
• Shear force at C: two parallel lines. Ordinate jumps from −a/L (load just left of C) to +b/L (load just right of C) — a unit discontinuity at C. Positive area = b²/2L, negative area = a²/2L.
• Bending moment at C: triangle with apex at C of ordinate ab/L; zero at both supports. Total area = ab/2.
• Cantilever (fixed at A, free end, length L): for a section at distance a from the fixed end, BM ILD is zero up to the section and rises linearly to −(L − a) at the free end; shear ILD is 0 then constant 1 (sign per convention). Fixed-end moment ILD is a triangle of peak L at the tip.
• Overhanging beam: the same straight lines, extended beyond the supports; ordinates go negative over the overhang. The R_A line, for example, continues above 1 over the left overhang and below 0 over the right overhang.
• Internal hinge (determinate compound beams, e.g. a propped span with a hinge): the ILD has a kink at the hinge; the response ILD is zero at a hinge only if the response is the BM at that hinge.
• Always locate the section first, then write the response as a function of load position z in two parts (z < a and z > a) and plot.

Müller-Breslau principle

• Statement: the ILD for any force response (reaction, shear, moment, member force) is given by the deflected shape of the structure when that restraint is released and a unit displacement is imposed in the direction of the released force, with the ordinates scaled by that unit displacement.
• How the release is made:
  • Reaction: remove the support, give unit vertical settlement/lift at that point.
  • Bending moment at C: insert a hinge at C, apply a unit relative rotation between the two faces.
  • Shear at C: insert a shear release (sliding/roller cut) at C, apply a unit relative vertical slip, keeping the two faces parallel.
  • Axial force in a truss member: cut the member, impose unit relative axial displacement.
• For determinate structures the released structure becomes a mechanism → the deflected shape is piecewise straight (hence straight-line ILDs, obtainable from geometry alone).
• For indeterminate structures the released structure is still stable and bends → smooth curved ILDs; ordinates come from the deflected shape (computed by consistent deformation, moment distribution, stiffness/matrix methods or standard charts such as Pucher/Anger influence tables).
• Why it works: it is Betti–Maxwell's reciprocal theorem written for a unit load and a unit displacement — the deflection at the load point due to the unit release equals the response at the release due to the unit load.
• Practical consequence for continuous bridges: a span-moment ILD changes sign from span to span, so alternate spans are loaded; a support-moment ILD keeps one sign over both spans adjacent to that support and reverses only beyond them, so those two spans plus alternate spans further out are loaded. Either way the worst case is pattern loading, never all spans loaded.

ILDs for trusses, floor-beam systems and three-hinged arches

• Through/deck truss with floor beams: the moving load reaches the truss only at panel points, so the ILD is a straight line between panel points — a transition line replaces the sharp discontinuity inside the panel in which the section lies.
• Chord member force: use the method of sections about the opposite joint — the ILD looks like the BM ILD at that joint divided by the lever arm (depth h): peak = (a·b/L)/h.
• Diagonal (web) member force: obtained from the panel shear; the ILD resembles the shear ILD with a transition line across the panel. Diagonals of a Pratt truss may reverse sign for some load positions — hence counter-bracing.
• Three-hinged arch, span L, central rise h (crown hinge at mid-span):
  • Horizontal thrust ILD = (simple-beam mid-span BM ILD)/h → triangle with peak L/(4h) at mid-span.
  • BM at any section X = (beam moment at X) − H·y_X; its ILD is the difference of two triangles and changes sign — so the worst moment in an arch often comes from partial loading (quarter-span loading is a classic case).
• Gantry/crane girders (steel buildings): the crab wheels are a moving load system; design uses the same ILD logic with vertical, lateral surge and longitudinal drag forces per IS 875 (Part 2) and IS 800:2007, plus an impact allowance.

Moving load analysis — rules for the worst position

• Single point load W on a simply supported span:
  • Max reaction/max shear = W (load over the support).
  • Max BM at a section C = W·ab/L; absolute maximum = WL/4 at mid-span.
• UDL longer than the span (w): load the whole span for maximum BM; for maximum positive shear at C load only the span to the right of C.
• UDL shorter than the span (length d): position it so that the section divides the load in the same ratio as it divides the span (left part = d·a/L). For maximum positive shear at C, place one end of the load at C with the whole load lying on the positive (right) side.
• Two or more point loads: the maximum occurs with one of the loads at the section; test each load in turn at the section and take the largest Σ W·y. Usually, but not always, it is the heavier load.
• Criterion for the critical load (series of loads), max BM at section C: advance the train until the average load on the left of C just exceeds the average load on the whole span, i.e. ΣW_left/a changes from < W_total/L to > W_total/L. The load crossing C at that instant is the critical one.
• Maximum shear at C: place a heavy load just to the right of C with as much of the train as possible to the right of C (the positive side); maximum shear in the span occurs at the support.
• Absolute maximum bending moment (point-load train): occurs under one of the heavy loads near mid-span, when that load and the resultant R of all loads on the span are equidistant from mid-span; check each candidate heavy load.
• Absolute maximum BM for a UDL shorter than the span: load placed symmetrically about mid-span; value = w·d(2L − d)/8.
• Equivalent UDL (EUDL): the uniform load that produces the same maximum effect as the actual train — w_e = 8M_max/L² for moment, 2V_max/L for end shear. Railway practice (IRS Bridge Rules) tabulates EUDL against span, which is why railway designers rarely roll wheel trains by hand.
• Why moving loads need extra care: dynamic effects (impact), stress reversal and hence fatigue (critical for steel bridges and gantry girders), and the fact that several responses peak at different load positions — each response needs its own load case.

Indian codes and practice

• Highway bridges — IRC:6 "Loads and Load Combinations" (Section II of the Bridge Code; 2017 edition, with periodic amendments — verify the current amendment slip before design):
  • Class A train — total 554 kN: axles 27, 27, 114, 114, 68, 68, 68, 68 kN at specified spacings; the normal loading for all permanent bridges on roads with legal traffic.
  • Class B — lighter version, for temporary/specified structures.
  • Class AA — tracked 700 kN (two tracks of 350 kN, 3.6 m contact length) and wheeled 400 kN; mandatory for bridges in municipal/industrial areas and specified routes.
  • Class 70R — tracked 700 kN and wheeled train of 1000 kN (7 axles, 80–170 kN); generally governs over Class AA.
  • Special Vehicle (SV) loading for expressways and designated corridors (a very heavy multi-axle trailer) was introduced in the 2017 edition.
  • Impact (Clause 208): for Class A/B, I = 4.5/(6 + L) for RC bridges and I = 9/(13.5 + L) for steel bridges (L in m), with upper and lower cut-offs for very short and very long spans; tracked and wheeled vehicles use tabulated percentages instead of the formula.
• Railway bridges — IRS Bridge Rules (RDSO): standard BG/MG loading with tabulated EUDL for bending moment and shear; dynamic effect is the Coefficient of Dynamic Augment (CDA), for broad gauge single track CDA = 0.15 + 8/(6 + L) subject to code limits; plus longitudinal (tractive/braking) and centrifugal forces.
• Buildings and industrial structures: IS 875 (Part 2) imposed loads including crane and gantry effects; IS 800:2007 covers fatigue checks for members under moving loads; IRC:112 and IRC:24 are used for concrete and steel bridge member design once the load effects are known.
• Transverse distribution of a wheel load across multiple girders (Courbon, grillage, finite element) is a separate design step carried out after the longitudinal ILD analysis.

Key points

• Influence line – variation of ONE response at ONE section as a unit load moves; not a BMD.
• Determinate ILD – always straight-line segments; indeterminate ILD – curved.
• Reaction ILD – unity at its own support, zero at the other, linear in between.
• Shear ILD at C – ordinates −a/L and +b/L with a unit jump at C.
• Moment ILD at C – triangle of peak ab/L, area ab/2; units are metres.
• Müller-Breslau – ILD = deflected shape after releasing that restraint and imposing a unit displacement.
• Point load – response = W × ordinate; UDL – response = w × area; couple – M × slope.
• Single rolling load – max shear = W at support; absolute max BM = WL/4 at mid-span.
• UDL shorter than span – position so the section divides the load as it divides the span.
• Absolute max BM (train) – under a heavy load placed so that the load and the resultant straddle mid-span equally.
• Critical load criterion – advance until ΣW_left/a just exceeds W_total/L.
• Equivalent UDL – w_e = 8M_max/L², the basis of IRS EUDL tables.
• Truss with floor beams – transition line across the panel containing the cut.
• Three-hinged arch – thrust ILD peak = L/4h; partial loading often governs arch moments.
• Continuous spans – pattern loading, not all spans: alternate spans for a span moment; the two adjacent spans (then alternate ones) for a support moment.
• Impact – IRC:6 formulas 4.5/(6+L) for RC and 9/(13.5+L) for steel (Class A/B); railways use CDA.
• Fatigue and stress reversal – the real penalty of moving loads in steel bridges and gantry girders.
• Standard load trains – IRC Class A (554 kN), Class AA, Class 70R, SV; railway BG loading via EUDL.

Formulas and examples

Example 1 — ILDs and a UDL longer than the span. Span L = 12 m, section C at a = 4 m (b = 8 m), UDL w = 20 kN/m covering the whole span.

• BM ILD peak = ab/L = 4 × 8/12 = 2.667 m; area = ab/2 = 4 × 8/2 = 16 m².
• M_C = 20 × 16 = 320 kN·m. Check directly: M = wa b/2 = 20 × 4 × 8/2 = 320 ✓
• Shear ILD: −a/L = −0.333, +b/L = +0.667. Positive area = b²/2L = 64/24 = 2.667 m; negative area = a²/2L = 16/24 = 0.667 m.
• Max positive shear (load only right of C) = 20 × 2.667 = 53.33 kN; max negative = 20 × 0.667 = 13.33 kN; whole span loaded → 53.33 − 13.33 = 40 kN (check: 120 − 20×4 = 40 ✓).

Example 2 — two wheel loads, maximum BM at a given section. Span 20 m; loads 100 kN (leading) and 60 kN, 4 m apart; section C at a = 8 m (b = 12 m). ILD peak at C = 8 × 12/20 = 4.8 m.

• Case 1 — 100 kN at C (z = 8), 60 kN trailing at z = 4 → ordinate = 4 × 12/20 = 2.4 m.

M_C = 100(4.8) + 60(2.4) = 480 + 144 = 624 kN·m.

• Case 2 — 60 kN at C, 100 kN ahead at z = 12 → ordinate = 8(20 − 12)/20 = 3.2 m.

M_C = 60(4.8) + 100(3.2) = 288 + 320 = 608 kN·m.

• Governing M_C = 624 kN·m (Case 1). Verification: R_A = [100(12) + 60(16)]/20 = 108 kN; M_C = 108(8) − 60(4) = 864 − 240 = 624 ✓

Example 3 — absolute maximum BM for the same two loads. Resultant R = 160 kN, located 60 × 4/160 = 1.5 m behind the 100 kN load.

• Put the 100 kN load at x from A with R at x − 1.5; equidistant from mid-span: x − 10 = 10 − (x − 1.5) → x = 10.75 m.
• R_A = [100(20 − 10.75) + 60(20 − 6.75)]/20 = [925 + 795]/20 = 86 kN.
• M = 86(10.75) − 60(4) = 924.5 − 240 = 684.5 kN·m (under the 100 kN load) — larger than the 624 kN·m at the fixed section, as expected.
• Equivalent UDL for moment: w_e = 8(684.5)/20² = 13.69 kN/m.
• IRC:6 impact, 20 m RC span, Class A: I = 4.5/(6 + 20) = 0.173 → design moment = 684.5 × 1.173 = 803 kN·m. For a steel span: I = 9/(13.5 + 20) = 0.269 → 684.5 × 1.269 = 869 kN·m.

Example 4 — UDL shorter than the span. Span 20 m, UDL w = 30 kN/m over d = 8 m; section C at a = 8 m (b = 12 m); ILD peak 4.8 m.

• Position for maximum M_C: left part = d·a/L = 8 × 8/20 = 3.2 m, right part = 4.8 m → load from z = 4.8 m to z = 12.8 m.
• Ordinates at load ends: at z = 4.8 → 4.8 × 12/20 = 2.88 m; at z = 12.8 → 8(20 − 12.8)/20 = 2.88 m.
• ILD area = ½(2.88 + 4.8)(3.2) + ½(4.8 + 2.88)(4.8) = 12.288 + 18.432 = 30.72 m².
• M_C = 30 × 30.72 = 921.6 kN·m. Check: W = 240 kN at centroid 8.8 m → R_A = 134.4 kN; M_C = 134.4(8) − 30(3.2²)/2 = 1075.2 − 153.6 = 921.6 ✓ (formula: w·a·b·d(2L − d)/(2L²) = 30×8×12×8×32/800 = 921.6 ✓)
• Absolute maximum (load symmetric about mid-span, z = 6 to 14): M = w·d(2L − d)/8 = 30 × 8 × (40 − 8)/8 = 960 kN·m.

Example 5 — maximum positive shear at a section (two loads). Span 20 m, section at a = 8 m; shear ILD ordinates −0.4 (left of C) and +0.6 (right of C); ordinate at z = 4 is −0.2, at z = 12 is +0.4.

Maximum positive shear at C = 76 kN — with the *lighter* load at the section, because the heavier one then sits on the positive side.

Where it is used

• Highway bridge design (NHAI/MoRTH, state PWDs and consultants such as RITES, STUP, TCE): longitudinal girder moments and shears for IRC Class A, Class AA, 70R and SV load trains, with ILDs generated per girder and combined with Courbon/grillage distribution.
• Railway bridge design and rating (RDSO, Indian Railways bridge cells, RVNL): EUDL and CDA from the IRS Bridge Rules for plate girders, open-web girders and ballasted-deck slabs; load rating of existing spans before allowing heavier axle loads (e.g. 25 t axle routes).
• Industrial steel buildings: gantry/crane girders and crane runway brackets in factories, steel plants and ports — vertical wheel loads plus surge, with fatigue checks per IS 800:2007.
• Metro and flyover viaducts: precast segmental and U-girder spans designed for launching-gantry travel and for metro train axle loads; construction-stage moving loads (launching girders, trailers) are a routine check.
• Assessment and strengthening: permissible-load certificates for over-dimensional cargo (ODC) movements over bridges; ILDs decide where the trailer axles may stand.
• Job roles: bridge/structural design engineer, proof-checking consultant, bridge inspector and load-rating engineer, software user of STAAD.Pro, MIDAS Civil, LUSAS and CSiBridge (all of which run moving-load trains over user-defined influence surfaces).

Interview questions

• Q: How does an influence line differ from a bending moment diagram? A: An influence line gives one response at one fixed section as a unit load moves; a BMD gives the moment at every section for one fixed set of loads.
• Q: State Müller-Breslau's principle. A: The influence line for a force quantity is the deflected shape produced when that restraint is released and a unit displacement is imposed in the direction of the released force, with ordinates scaled to that unit displacement.
• Q: Why are influence lines for determinate structures straight lines? A: Releasing one restraint turns a determinate structure into a mechanism, so the displaced shape consists of rigid straight segments.
• Q: What are the units of a bending-moment influence line ordinate? A: Length (metres) — it is moment per unit load, so multiplying by a force in kN gives kN·m.
• Q: Where does the absolute maximum bending moment occur in a simply supported span under a moving load train? A: Under one of the heavy loads near mid-span, positioned so that the load and the resultant of all loads on the span are equidistant from mid-span.
• Q: For a single rolling point load W on span L, what is the absolute maximum BM and shear? A: M = WL/4 at mid-span and shear = W with the load over a support.
• Q: How do you place a UDL shorter than the span for maximum moment at a section? A: So that the section divides the loaded length in the same ratio in which it divides the span.
• Q: Why is the heaviest load not always the critical one for maximum shear at a section? A: The shear ILD has opposite signs on the two sides of the section, so the arrangement that keeps the rest of the train on the favourable side can govern even with a lighter load at the section.
• Q: What is an equivalent UDL and why do railway codes use it? A: A uniform load giving the same maximum moment or shear as the actual wheel train (w_e = 8M_max/L²); the IRS Bridge Rules tabulate it so designers need not roll wheel loads for every span.
• Q: Which IRC load classes must a highway bridge be checked for, and how is impact handled? A: Class A/B, Class AA and Class 70R trains (plus SV loading on designated corridors) per IRC:6; impact for Class A/B is 4.5/(6+L) for concrete and 9/(13.5+L) for steel, with tabulated percentages for tracked and wheeled vehicles.
• Q: Why is pattern loading used for continuous bridge girders? A: The influence line changes sign along the beam, so loading every span cancels part of the effect: alternate spans give the maximum span moment, while a support moment needs the two spans adjacent to that support loaded, then alternate spans beyond.
• Q: What is a transition line in a truss influence line? A: Where loads enter the truss only at panel points, the ILD runs as a straight line across the panel containing the cut, replacing the sharp discontinuity of a beam ILD.

Common mistakes

• Treating the ILD as a BMD — reading the ordinate at the section instead of at the load's position.
• Forgetting that a bending-moment ILD has units of metres, and reporting kN·m for the ordinate itself.
• Multiplying a UDL by the ILD ordinate instead of by the ILD area (and vice versa for point loads).
• Taking the whole span loaded as always worst for shear — the negative part of the shear ILD must be excluded for maximum positive shear; similarly, loading all spans of a continuous girder instead of the critical pattern.
• Assuming the absolute maximum BM occurs exactly at mid-span, or always under the heaviest axle, without testing each candidate load.
• Drawing straight-line ILDs for indeterminate structures, or missing the unit jump in a shear ILD and the kink at an internal hinge.
• Omitting impact/CDA, or applying the Class A formula to tracked vehicles, and ignoring stress reversal and fatigue in steel members under moving loads.

Related topics

Structural analysis: determinate and indeterminate structures · Bridge engineering · Steel structures design (IS 800) · Matrix methods of structural analysis · Structural dynamics basics · Tall buildings, shear walls and lateral loads