Influence Lines & Rolling Loads

Influence line — the idea

A bending moment diagram shows the moment at every section for loads in fixed positions. An influence line diagram (ILD) shows how a response (reaction, shear, moment or member force) at one fixed section changes as a unit load moves across the structure.

The ordinate of an ILD at a point = value of the response at the chosen section when a unit load stands at that point. ILDs are essential for bridges, gantry girders and any structure carrying moving loads, because they tell us where to place the loads for the worst effect.

ILDs for a simply supported beam (span $L$)

Let a unit load be at distance $x$ from the left support A.

Reactions

Both ILDs are triangles with a maximum ordinate of 1 at their own support.

Shear force at a section C (distance $a$ from A, $b$ from B)

• Load to the right of C: $V_C=R_A=\frac{L-x}{L}$ (positive).
• Load to the left of C: $V_C=-R_B=-\frac{x}{L}$ (negative).

The ILD has a jump of 1 at C: ordinate $-a/L$ just left and $+b/L$ just right; the two lines are parallel.

Bending moment at section C

• Load right of C: $M_C=R_{A}a$; load left of C: $M_C=R_{B}b$.

The ILD is a triangle with maximum ordinate $ab/L$ at C.

Cantilever (fixed at B, section C at distance $c$ from the free end region)

For a section C of a cantilever, a unit load between C and the free end produces $V_C=1$ (constant) and $M_C=-(\text{distance from C})$; a load between C and the fixed end produces nothing at C.

Overhanging beams

ILD lines for reactions and for sections within the span are simply extended straight into the overhangs, changing sign where they cross the support line.

Using an ILD

Rolling (moving) loads — maximum values

Single point load $W$

• Max positive shear at C: load just right of C → $Wb/L$.
• Max negative shear at C: load just left of C → $-Wa/L$.
• Max BM at C: load at C → $Wab/L$. Absolute maximum BM (anywhere on the span): load at mid-span → $WL/4$.

UDL longer than the span

Cover the whole span for maximum BM; cover the part of the span giving area of one sign for maximum shear:

• $M_{C,max}=w\cdot \frac{ab}{2}$ (area of triangle $=\frac{1}{2}L\cdot ab/L$).
• $V_{C,max}^{+}=w\cdot \frac{b^2}{2L}$ (load on BC only); $V_{C,max}^{-}=-w\frac{a^2}{2L}$ (load on AC only).

UDL shorter than the span (length $d$)

For maximum BM at C, place the load so that C divides the load in the same ratio as it divides the span:

Equivalently, the ILD ordinates at the two ends of the load are equal.

Series of concentrated loads (wheel loads)

Maximum BM at a given section C: the maximum occurs with one of the loads at C, placed such that the average load per unit length on the left of C equals that on the whole span (the load at C is counted on the side that makes the inequality change sign). Try the heavy wheels over C.

Maximum shear at C: usually with the leading heavy wheel just to one side of C; check by computing $\sum W_i{y}_i$ for trial positions.

Absolute maximum bending moment (Barré's rule)

For a train of loads on a simply supported span, the absolute maximum BM occurs under one of the loads (usually the one nearest the resultant) when that load and the resultant of all loads on the span are equidistant from the mid-span.