Influence Lines for Beams and Trusses

Structural Analysis · FE Reference Handbook section

Structural Analysis

0 formulas

10 exam-style examples

~45 min

All Structural Analysis lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Influence Lines for Beams and Trusses within Structural Analysis. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what influence lines for beams and trusses describes physically and when it applies.
State every one of the 0 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: keep kip-ft consistently; EI in kip-in² needs a 1,728 factor.

Lecture

Why this section exists. Influence Lines for Beams and Trusses is the part of Structural Analysis that lets you connect a determinate or one-degree indeterminate structure to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as reactions, an influence-line ordinate, or one deflection. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. keep kip-ft consistently; EI in kip-in² needs a 1,728 factor. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Interior of a steel and glass pedestrian bridge showing the structural framing. — Photo 1. Where this shows up in practice: influence lines for beams and trusses.
Wikimedia Commons, CC BY-SA 4.0

Structural Analysis — Influence Lines for Beams and Trusses: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a determinate or one-degree indeterminate structure. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 0 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

An influence line shows the variation of an effect (reaction, shear and moment in beams, bar force in a truss) caused by moving
a unit load across the structure. An influence line is used to determine the position of a moveable set of loads that causes the
maximum value of the effect.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

This section is conceptual; there are no equations to memorise.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Truss member force by the method of joints

A symmetric pin-jointed truss of span 12 m and height 4 m carries 60 kN at the apex. Find the force in each of the two inclined top members.

Given

Span = 12 m, height = 4 m
Apex load = 60 kN
Symmetric geometry

Find

Force in the inclined members

Start with the thinking

Symmetry means each inclined member carries half the vertical load.
Resolve at the apex joint using the member slope.

Figure for Truss member force by the method of joints

Step-by-step solution

Member length
$L = (6^{2} + 4^{2}) = 7.21 m$
Vertical component per member
$F v = 60/2 = 30.0 k N$
Slope ratio
$L / h e i g h t = 7.21/4 = 1.803$
Member force
$F = F v (L / h e i g h t) = 30.0 (1.803)$
Result
$F = 54.1 k N co m p r ess i o nin e a c hin c l in e d m e mb er$

Answer: 54.1 kN compression

Why the other options are there

30.0 kN (vertical component reported as member force)
45.0 kN (horizontal projection used)

Reference: FE Reference Handbook — Statics — Trusses

Example 2

Influence line ordinate for a reaction

For a 20 m simple span, what is the influence-line ordinate for R_A when a unit load sits 6 m from B, and what is R_A for a 150 kN axle there?

Given

L = 20 m
Load position: 6 m from B (14 m from A)

Find

Ordinate and R_A

Start with the thinking

The influence line for R_A is a straight line from 1.0 at A to 0 at B.
Ordinate equals the distance from B divided by the span.

Step-by-step solution

Ordinate
$η = (L - x) / L w i t h x m e a s u r e df r o m A$
Substitute
$η = (20 - 14) /20 = 0.300$
Reaction
$R_{A} = P η = 150 (0.300)$
Result
$R_{A} = 45.0 k N$
Check — R_B = 150 − 45.0 = 105 kN, consistent with the load sitting near B ✓

Answer: η = 0.300, R_A = 45.0 kN

Why the other options are there

η = 0.700 (measured from the wrong end)
R_A = 75 kN (symmetric assumption)

Reference: FE Reference Handbook — Structural Analysis — Influence lines

Example 3

Moment from an influence-line ordinate — Influence Lines for Beams and Trusses

For a 54 ft simple span, a single 27 kip axle stands 19 ft from the left support. Use the influence line for moment at that point to find the moment.

Given

L = 54 ft
a = 19 ft
P = 27 kip

Find

Moment at the section

Start with the thinking

The influence-line ordinate at the section itself is a(L − a)/L.
Moment equals load times ordinate.

Step-by-step solution

Ordinate
$y = a (L - a) / L$
Substituting
$y = 19 (54 - 19) /54 = 12.315 f t$
Moment — M = P·y
Substituting — M = 27(12.315) = 332.5 kip·ft

Answer: M ≈ 332.5 kip·ft

Why the other options are there

1,458 kip·ft (span used as the ordinate)
364.5 kip·ft (midspan assumed)

Reference: FE Reference Handbook — Structural Analysis → Influence Lines for Beams and Trusses

Example 4

Absolute maximum moment from a moving concentrated load set — Influence Lines for Beams and Trusses

A simple span of 72 ft carries a moving load set of 29 kips and 30 kips spaced 6 ft apart. Locate the resultant, then compute the absolute maximum moment under the heavier wheel.

Given

L = 72 ft
P₁ = 29 kips, P₂ = 30 kips
Spacing = 6 ft

Find

Resultant position and absolute maximum moment

Start with the thinking

For a moving concentrated load set, the absolute maximum moment occurs when the span centreline bisects the distance between the resultant and the wheel considered.
Influence-line reasoning locates the critical position before any statics is done.

Step-by-step solution

Resultant
$R = 29 + 30 = 59 k i p s$
Formula — x̄ = ΣP d / ΣP (measured from P₁)
Substituting
$x ˉ = 30 (6) /59 = 3.051 f t$
Critical offset
$e = x ˉ /2 = 1.525 f t f r o mmi d s p an$
Formula
$M_{m} a x = (R / L) (L /2 - e)^{2}$
Substituting — M = (59/72)(36 − 1.525)² = 973.9 kip·ft

Answer: M_abs,max ≈ 973.9 kip·ft

Why the other options are there

1,062 kip·ft (single load at midspan)
531.0 kip·ft (uniform-load formula)

Reference: FE Reference Handbook — Structural Analysis → Influence Lines for Beams and Trusses

Example 5

Moment from an influence-line ordinate — Influence Lines for Beams and Trusses (2)

For a 60 ft simple span, a single 11 kip axle stands 27 ft from the left support. Use the influence line for moment at that point to find the moment.

Given

L = 60 ft
a = 27 ft
P = 11 kip

Find

Moment at the section

Start with the thinking

The influence-line ordinate at the section itself is a(L − a)/L.
Moment equals load times ordinate.

Step-by-step solution

Ordinate
$y = a (L - a) / L$
Substituting
$y = 27 (60 - 27) /60 = 14.850 f t$
Moment — M = P·y
Substituting — M = 11(14.850) = 163.4 kip·ft

Answer: M ≈ 163.4 kip·ft

Why the other options are there

660.0 kip·ft (span used as the ordinate)
165.0 kip·ft (midspan assumed)

Reference: FE Reference Handbook — Structural Analysis → Influence Lines for Beams and Trusses

Example 6

Absolute maximum moment from a moving concentrated load set — Influence Lines for Beams and Trusses (2)

A simple span of 83 ft carries a moving load set of 10 kips and 14 kips spaced 13 ft apart. Locate the resultant, then compute the absolute maximum moment under the heavier wheel.

Given

L = 83 ft
P₁ = 10 kips, P₂ = 14 kips
Spacing = 13 ft

Find

Resultant position and absolute maximum moment

Start with the thinking

For a moving concentrated load set, the absolute maximum moment occurs when the span centreline bisects the distance between the resultant and the wheel considered.
Influence-line reasoning locates the critical position before any statics is done.

Step-by-step solution

Resultant
$R = 10 + 14 = 24 k i p s$
Formula — x̄ = ΣP d / ΣP (measured from P₁)
Substituting
$x ˉ = 14 (13) /24 = 7.583 f t$
Critical offset
$e = x ˉ /2 = 3.792 f t f r o mmi d s p an$
Formula
$M_{m} a x = (R / L) (L /2 - e)^{2}$
Substituting — M = (24/83)(41.5 − 3.792)² = 411.2 kip·ft

Answer: M_abs,max ≈ 411.2 kip·ft

Why the other options are there

498.0 kip·ft (single load at midspan)
249.0 kip·ft (uniform-load formula)

Reference: FE Reference Handbook — Structural Analysis → Influence Lines for Beams and Trusses

Example 7

Moment from an influence-line ordinate — Influence Lines for Beams and Trusses (3)

For a 36 ft simple span, a single 34 kip axle stands 11 ft from the left support. Use the influence line for moment at that point to find the moment.

Given

L = 36 ft
a = 11 ft
P = 34 kip

Find

Moment at the section

Start with the thinking

The influence-line ordinate at the section itself is a(L − a)/L.
Moment equals load times ordinate.

Step-by-step solution

Ordinate
$y = a (L - a) / L$
Substituting
$y = 11 (36 - 11) /36 = 7.639 f t$
Moment — M = P·y
Substituting — M = 34(7.639) = 259.7 kip·ft

Answer: M ≈ 259.7 kip·ft

Why the other options are there

1,224 kip·ft (span used as the ordinate)
306.0 kip·ft (midspan assumed)

Reference: FE Reference Handbook — Structural Analysis → Influence Lines for Beams and Trusses

Example 8

Absolute maximum moment from a moving concentrated load set — Influence Lines for Beams and Trusses (3)

A simple span of 77 ft carries a moving load set of 21 kips and 20 kips spaced 6 ft apart. Locate the resultant, then compute the absolute maximum moment under the heavier wheel.

Given

L = 77 ft
P₁ = 21 kips, P₂ = 20 kips
Spacing = 6 ft

Find

Resultant position and absolute maximum moment

Start with the thinking

For a moving concentrated load set, the absolute maximum moment occurs when the span centreline bisects the distance between the resultant and the wheel considered.
Influence-line reasoning locates the critical position before any statics is done.

Step-by-step solution

Resultant
$R = 21 + 20 = 41 k i p s$
Formula — x̄ = ΣP d / ΣP (measured from P₁)
Substituting
$x ˉ = 20 (6) /41 = 2.927 f t$
Critical offset
$e = x ˉ /2 = 1.463 f t f r o mmi d s p an$
Formula
$M_{m} a x = (R / L) (L /2 - e)^{2}$
Substituting — M = (41/77)(38.5 − 1.463)² = 730.4 kip·ft

Answer: M_abs,max ≈ 730.4 kip·ft

Why the other options are there

789.3 kip·ft (single load at midspan)
394.6 kip·ft (uniform-load formula)

Reference: FE Reference Handbook — Structural Analysis → Influence Lines for Beams and Trusses

Example 9

Moment from an influence-line ordinate — Influence Lines for Beams and Trusses (4)

For a 50 ft simple span, a single 11 kip axle stands 20 ft from the left support. Use the influence line for moment at that point to find the moment.

Given

L = 50 ft
a = 20 ft
P = 11 kip

Find

Moment at the section

Start with the thinking

The influence-line ordinate at the section itself is a(L − a)/L.
Moment equals load times ordinate.

Step-by-step solution

Ordinate
$y = a (L - a) / L$
Substituting
$y = 20 (50 - 20) /50 = 12.000 f t$
Moment — M = P·y
Substituting — M = 11(12.000) = 132.0 kip·ft

Answer: M ≈ 132.0 kip·ft

Why the other options are there

550.0 kip·ft (span used as the ordinate)
137.5 kip·ft (midspan assumed)

Reference: FE Reference Handbook — Structural Analysis → Influence Lines for Beams and Trusses

Example 10

Absolute maximum moment from a moving concentrated load set — Influence Lines for Beams and Trusses (4)

A simple span of 56 ft carries a moving load set of 22 kips and 20 kips spaced 7 ft apart. Locate the resultant, then compute the absolute maximum moment under the heavier wheel.

Given

L = 56 ft
P₁ = 22 kips, P₂ = 20 kips
Spacing = 7 ft

Find

Resultant position and absolute maximum moment

Start with the thinking

For a moving concentrated load set, the absolute maximum moment occurs when the span centreline bisects the distance between the resultant and the wheel considered.
Influence-line reasoning locates the critical position before any statics is done.

Step-by-step solution

Resultant
$R = 22 + 20 = 42 k i p s$
Formula — x̄ = ΣP d / ΣP (measured from P₁)
Substituting
$x ˉ = 20 (7) /42 = 3.333 f t$
Critical offset
$e = x ˉ /2 = 1.667 f t f r o mmi d s p an$
Formula
$M_{m} a x = (R / L) (L /2 - e)^{2}$
Substituting — M = (42/56)(28 − 1.667)² = 520.1 kip·ft

Answer: M_abs,max ≈ 520.1 kip·ft

Why the other options are there

588.0 kip·ft (single load at midspan)
294.0 kip·ft (uniform-load formula)

Reference: FE Reference Handbook — Structural Analysis → Influence Lines for Beams and Trusses

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a determinate or one-degree indeterminate structure, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Influence Lines for Beams and Trusses contains 0 relations; you must be able to find this page in under 15 seconds.
Exam style: reactions, an influence-line ordinate, or one deflection.
Unit rule: keep kip-ft consistently; EI in kip-in² needs a 1,728 factor.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

keep kip-ft consistently; EI in kip-in² needs a 1,728 factor
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.