CEE 330 — Influence Lines

The Question SFDs Can't Answer

You can draw a perfect bending moment diagram for a truck parked at midspan. But the truck drives. Somewhere along its journey it produces the worst moment the bridge will ever see — and a diagram drawn for one position cannot tell you where that is.

📷

A panorama versus a CCTV camera. A bending moment diagram is a panorama: one instant, frozen, showing the whole beam at once. An influence line is a fixed security camera pointed at one spot, recording what happens at that spot as the truck drives past. The panorama tells you everything about one moment in time; the camera tells you everything about one place, across all time. To find the worst case you need the camera.

Shear / moment diagram

Load fixed, position varies

Horizontal axis = position along the beam. Ordinate = the value of V or M there, for one specific loading. Answers: "where is this beam worst stressed right now?"

Influence line

Position fixed, load varies

Horizontal axis = position of the unit load. Ordinate = the value of the chosen function at one fixed section. Answers: "where should the load stand to make this section worst?"

⚡

This is the single most common confusion in the course. Both are curves drawn under a beam and they can even look alike, but their horizontal axes mean completely different things. Before you interpret any ordinate, ask yourself: is this axis telling me where the load is, or where I'm looking?

What an Influence Line Actually Is

Definition

An influence line for a quantity (a reaction, or the shear or moment at one chosen section) is a graph whose ordinate at any point x is the value of that quantity when a unit load stands at x.

Two consequences follow immediately, and both are worth stating out loud:

Consequence 1

The ordinate has odd units

Because the load is one unit, an IL ordinate is a value per unit load. For a reaction it is dimensionless; for a moment it has units of length. That is why an IL for M_C can read "2.4 m" — multiply by a load in kN and you get kN·m.

Consequence 2

Superposition is free

Because the response is linear in the load, a load of P at x simply gives P × (ordinate at x). Several loads? Add them. A distributed load? Integrate — which means take the area under the IL. Section 7 turns this into the whole toolkit.

Building One From Scratch

The definition is also the method. Put a unit load at a general position x, solve for the quantity you care about, and plot the answer as a function of x. For a simply supported beam of span L with a section C at distance a from A:

Reactions with the unit load at x. \( R_A = \dfrac{L-x}{L} \), \( R_B = \dfrac{x}{L} \). Those two are the influence lines for the reactions — straight lines from 1 to 0.

Shear at C, load to the right (x > a). Cut at C and look left: only R_A is there, so \( V_C = R_A = \dfrac{L-x}{L} \).

Shear at C, load to the left (x < a). Now the unit load is also on the left segment: \( V_C = R_A - 1 = -\dfrac{x}{L} \).

The jump. Just left of C the ordinate is −a/L; just right it is (L−a)/L. The difference is exactly 1 — the unit load crossing the section. Every shear IL has a unit jump at the section, and that is your check.

Moment at C. Load right: \( M_C = R_A\,a = \dfrac{a(L-x)}{L} \). Load left: \( M_C = R_A a - (a-x) = \dfrac{x(L-a)}{L} \). Two straight lines meeting at C with the peak \( \dfrac{ab}{L} \), where b = L − a.

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Notice what did not happen: nowhere did we need E, I, or the material. Influence lines for a statically determinate structure are pure geometry — always straight-line segments. Curved influence lines only appear for indeterminate structures, and that is a reliable tell.

Watch It Being Traced

Drag the unit load along the beam. The top panel is the real beam with the load where you put it; the three panels below are the influence lines being drawn as you go — the moving dot is the value right now, and the solid part is the history you have already traced.

Unit load position x (m): 4.0 Span L (m) Section C at a (m) Show Draw the complete influence line (ghost)

The Three Shapes to Know

For a simply supported beam these three are worth memorising outright — most exam questions are these with numbers changed.

Reaction R_A

Straight line, 1 → 0

\[ \text{ordinate} = \frac{L-x}{L} \]

Value 1 when the load sits over A, 0 when it sits over B. Obvious once seen: a load directly over a support is carried entirely by it.

Shear V_C

Two parallel lines, unit jump

\[ -\frac{a}{L} \;\to\; +\frac{b}{L} \]

Zero at both supports, discontinuous at C. The two segments are parallel, and the jump across C is always exactly 1.

Moment M_C

Triangle, peak at C

\[ M_{C,\max} = \frac{ab}{L} \]

Zero at both supports, single peak directly under the section. For midspan this gives L/4 — the largest peak any section can have.

Core Idea

Three sanity checks that catch most errors: a reaction IL is 1 over its own support; a shear IL has a jump of exactly 1 at the section; a moment IL peaks at the section and is zero at both supports. If your sketch fails any of these, it is wrong.

Müller-Breslau's Shortcut

The principle

The influence line for any force quantity is, to scale, the deflected shape the structure takes when you remove the restraint corresponding to that quantity and impose a unit displacement in its direction.

For a reaction

Remove the support

Take away that support and push the structure up by 1 there. The shape you get is the IL for that reaction.

For shear at C

Cut and slide

Cut at C, keep the two faces parallel, and slide them past each other by a total of 1. That relative slip is the unit jump you already know about.

For moment at C

Insert a hinge and rotate

Put a hinge at C and impose a unit relative rotation. The kinked shape is the moment IL — and the kink is why the peak sits at C.

✂

It turns algebra into a sketch. Rather than solving for a general load position, you break the structure at the place you care about, move it by one unit, and read off the shape. For determinate structures the released structure is a mechanism, so it moves in straight lines — which is precisely why determinate influence lines are made of straight segments. That single observation lets you sketch most exam ILs in seconds, then fill in ordinates by similar triangles.

⚡

Where it really earns its keep: indeterminate structures. There, solving for a general unit-load position is genuinely painful, but Müller-Breslau still gives you the shape immediately — a smooth curve rather than straight lines. Even without ordinates, knowing the shape tells you where to put the load for the worst effect, which is often all you need.

Using an Influence Line

Having drawn it, three rules extract every answer you need.

Rule 1 — point load

\[ \text{effect} = P \times y \]

y = IL ordinate under the load. For the maximum, stand the load at the peak.

Rule 2 — distributed load

\[ \text{effect} = w \times A \]

A = area under the IL over the loaded length. For the maximum, load only where the IL has the sign you want.

Rule 3 — several loads

\[ \text{effect} = \sum P_i y_i \]

Add each load times its own ordinate. For a wheel train, try each axle in turn at the peak and take the largest total.

⚡

Rule 2 has teeth. A UDL does not have to cover the whole span — a train can occupy part of a bridge. For maximum positive shear at C you load only the positive region, and Worked Example 2 shows that loading the whole span instead gives 10 kN rather than 18 kN. Partial loading is not a trick question; it is how the worst case actually arises.

Influence Lines for Trusses

A truss carries load only at its joints, so a vehicle on the deck cannot apply load between panel points. The load reaches the truss through the floor system: stringers span between floor beams, and floor beams sit at the panel points.

The consequence

Straight between panel points

Compute the member force with the unit load at each panel point in turn, plot those values, and join them with straight lines. The segments between panel points are straight because the stringer distributes the load linearly to the two adjacent floor beams.

The method

Sections, one position at a time

For a chord or diagonal member, take a section through it and use ΣM or ΣF_y for each unit-load position. Tedious but mechanical — and you only need the panel points, not a general x.

Core Idea

The truss IL is the beam IL evaluated only at panel points, then connected with straight lines. Peaks therefore always occur at a panel point, never between — which makes finding the critical load position much easier than for a beam.

Worked Examples

All three use the same beam: simply supported, L = 10 m, section C at a = 4 m from A (so b = 6 m). Set the simulator above to these values to follow along.

Example 1

Maximum Moment at C

Problem: Find the maximum moment at C from (a) a single 50 kN point load anywhere on the span, and (b) a 10 kN/m UDL covering the whole span.

Peak ordinate: \( \dfrac{ab}{L} = \dfrac{4 \times 6}{10} = 2.4\text{m} \), occurring with the load directly at C.

(a) Point load — Rule 1, standing it at the peak: \( M_C = 50 \times 2.4 = 120\text{kN·m} \)

(b) UDL — Rule 2. The IL is a triangle of base 10 m and height 2.4 m, so area \( = \tfrac12(10)(2.4) = 12\text{m}^2 \).

\( M_C = 10 \times 12 = 120\text{kN·m} \)

Answer: Both give 120 kN·m — a coincidence of these numbers, not a rule. Note the UDL needs no "where should I put it?" thinking: the IL is positive everywhere, so the full span is the worst case.

Example 2

Maximum Shear at C — Where Partial Loading Matters

Problem: For the same beam, find the maximum positive shear at C under a 10 kN/m UDL that may cover any part of the span.

The IL: −a/L = −0.4 just left of C, +b/L = +0.6 just right, zero at both supports.

Positive area (C to B, 6 m long, height 0.6): \( \tfrac12(6)(0.6) = 1.8\text{m} \)

Negative area (A to C, 4 m long, height −0.4): \( \tfrac12(4)(-0.4) = -0.8\text{m} \)

Load only the positive region: \( V_C = 10 \times 1.8 = 18\text{kN} \)

Load the whole span instead: \( V_C = 10(1.8 - 0.8) = 10\text{kN} \)

Answer: 18 kN, with the UDL only from C to B. Covering the whole span gives just 10 kN, because the left portion actively cancels shear. More load is not always worse — that is the whole point of influence lines.

Example 3

A Two-Axle Wheel Train

Problem: Two 60 kN axles, 3 m apart, cross the same beam. Find the maximum moment at C.

The IL for M_C: rises as \( x b/L = 0.6x \) up to the peak 2.4 at x = 4 m, then falls as \( a(L-x)/L = 0.4(10-x) \).

Trial 1 — leading axle at C. Axles at x = 4 m and 7 m. Ordinates: 2.4 and \( 0.4(3) = 1.2 \). \( M_C = 60(2.4) + 60(1.2) = 144 + 72 = 216\text{kN·m} \)

Trial 2 — trailing axle at C. Axles at x = 1 m and 4 m. Ordinates: \( 0.6(1) = 0.6 \) and 2.4. \( M_C = 60(0.6) + 60(2.4) = 36 + 144 = 180\text{kN·m} \)

Compare: 216 > 180, so Trial 1 governs.

Answer: 216 kN·m, with the leading axle standing at C. The rule for wheel trains is exactly this: put each axle at the peak in turn and take the largest total. There is no shortcut — but there are only as many trials as there are axles.

Quick Reference & Quick Check

IL definitionordinate at x = value when unit load is at x

IL axis meansposition of the load, not of the section

IL for R_A(L − x)/L, straight, 1 at A

IL for V_C−a/L to +b/L, jump = 1 at C

IL for M_Ctriangle, peak ab/L at C

Max M peak (midspan)L/4

Point loadeffect = P × y

Distributed loadeffect = w × area under IL

Wheel trainΣP_iy_i; try each axle at the peak

Determinate ILstraight-line segments

Indeterminate ILsmooth curves

Müller-Breslaurelease, impose unit movement, read the shape

1. What does the horizontal axis of an influence line represent?

2. For maximum positive shear at C under a UDL that can cover any part of the span, you should:

3. A shear influence line must have which feature at the section?

4. You sketch an IL for a determinate beam and get a smooth curve. What does that tell you?

5. Müller-Breslau's principle says the IL for the moment at C is obtained by: