Moving Concentrated Load Sets

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 Moving Concentrated Load Sets 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 moving concentrated load sets 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. Moving Concentrated Load Sets 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: moving concentrated load sets.
Wikimedia Commons, CC BY-SA 4.0

Structural Analysis — Moving Concentrated Load Sets: 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.

P1 R P2
The absolute maximum moment produced in a beam by a set of "n" moving loads occurs when the resultant "R" of the load
set and an adjacent load are equal distance from the centerline of the beam. In general, two possible load set positions must be
considered, one for each adjacent load.

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

Absolute maximum moment from a moving concentrated load set — Moving Concentrated Load Sets

A simple span of 55 ft carries a moving load set of 28 kips and 23 kips spaced 8 ft apart. Locate the resultant, then compute the absolute maximum moment under the heavier wheel.

Given

L = 55 ft
P₁ = 28 kips, P₂ = 23 kips
Spacing = 8 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
$R = 28 + 23 = 51 k i p s$
Formula — x̄ = ΣP d / ΣP (measured from P₁)
Substituting
$x ˉ = 23 (8) /51 = 3.608 f t$
Critical offset
$e = x ˉ /2 = 1.804 f t f r o mmi d s p an$
Formula
$M_{m} a x = (R / L) (L /2 - e)^{2}$
Substituting — M = (51/55)(27.5 − 1.804)² = 612.3 kip·ft

Answer: M_abs,max ≈ 612.3 kip·ft

Why the other options are there

701.3 kip·ft (single load at midspan)
350.6 kip·ft (uniform-load formula)

Reference: FE Reference Handbook — Structural Analysis → Moving Concentrated Load Sets

Example 2

Absolute maximum moment from a moving concentrated load set — Moving Concentrated Load Sets (2)

A simple span of 80 ft carries a moving load set of 11 kips and 16 kips spaced 10 ft apart. Locate the resultant, then compute the absolute maximum moment under the heavier wheel.

Given

L = 80 ft
P₁ = 11 kips, P₂ = 16 kips
Spacing = 10 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
$R = 11 + 16 = 27 k i p s$
Formula — x̄ = ΣP d / ΣP (measured from P₁)
Substituting
$x ˉ = 16 (10) /27 = 5.926 f t$
Critical offset
$e = x ˉ /2 = 2.963 f t f r o mmi d s p an$
Formula
$M_{m} a x = (R / L) (L /2 - e)^{2}$
Substituting — M = (27/80)(40 − 2.963)² = 463.0 kip·ft

Answer: M_abs,max ≈ 463.0 kip·ft

Why the other options are there

540.0 kip·ft (single load at midspan)
270.0 kip·ft (uniform-load formula)

Reference: FE Reference Handbook — Structural Analysis → Moving Concentrated Load Sets

Example 3

Absolute maximum moment from a moving concentrated load set — Moving Concentrated Load Sets (3)

A simple span of 73 ft carries a moving load set of 14 kips and 21 kips spaced 13 ft apart. Locate the resultant, then compute the absolute maximum moment under the heavier wheel.

Given

L = 73 ft
P₁ = 14 kips, P₂ = 21 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
$R = 14 + 21 = 35 k i p s$
Formula — x̄ = ΣP d / ΣP (measured from P₁)
Substituting
$x ˉ = 21 (13) /35 = 7.800 f t$
Critical offset
$e = x ˉ /2 = 3.900 f t f r o mmi d s p an$
Formula
$M_{m} a x = (R / L) (L /2 - e)^{2}$
Substituting — M = (35/73)(36.5 − 3.900)² = 509.5 kip·ft

Answer: M_abs,max ≈ 509.5 kip·ft

Why the other options are there

638.8 kip·ft (single load at midspan)
319.4 kip·ft (uniform-load formula)

Reference: FE Reference Handbook — Structural Analysis → Moving Concentrated Load Sets

Example 4

Absolute maximum moment from a moving concentrated load set — Moving Concentrated Load Sets (4)

A simple span of 81 ft carries a moving load set of 11 kips and 16 kips spaced 12 ft apart. Locate the resultant, then compute the absolute maximum moment under the heavier wheel.

Given

L = 81 ft
P₁ = 11 kips, P₂ = 16 kips
Spacing = 12 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
$R = 11 + 16 = 27 k i p s$
Formula — x̄ = ΣP d / ΣP (measured from P₁)
Substituting
$x ˉ = 16 (12) /27 = 7.111 f t$
Critical offset
$e = x ˉ /2 = 3.556 f t f r o mmi d s p an$
Formula
$M_{m} a x = (R / L) (L /2 - e)^{2}$
Substituting — M = (27/81)(40.5 − 3.556)² = 455.0 kip·ft

Answer: M_abs,max ≈ 455.0 kip·ft

Why the other options are there

546.8 kip·ft (single load at midspan)
273.4 kip·ft (uniform-load formula)

Reference: FE Reference Handbook — Structural Analysis → Moving Concentrated Load Sets

Example 5

Absolute maximum moment from a moving concentrated load set — Moving Concentrated Load Sets (5)

A simple span of 62 ft carries a moving load set of 21 kips and 16 kips spaced 16 ft apart. Locate the resultant, then compute the absolute maximum moment under the heavier wheel.

Given

L = 62 ft
P₁ = 21 kips, P₂ = 16 kips
Spacing = 16 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
$R = 21 + 16 = 37 k i p s$
Formula — x̄ = ΣP d / ΣP (measured from P₁)
Substituting
$x ˉ = 16 (16) /37 = 6.919 f t$
Critical offset
$e = x ˉ /2 = 3.459 f t f r o mmi d s p an$
Formula
$M_{m} a x = (R / L) (L /2 - e)^{2}$
Substituting — M = (37/62)(31 − 3.459)² = 452.6 kip·ft

Answer: M_abs,max ≈ 452.6 kip·ft

Why the other options are there

573.5 kip·ft (single load at midspan)
286.8 kip·ft (uniform-load formula)

Reference: FE Reference Handbook — Structural Analysis → Moving Concentrated Load Sets

Example 6

Absolute maximum moment from a moving concentrated load set — Moving Concentrated Load Sets (6)

A simple span of 40 ft carries a moving load set of 26 kips and 12 kips spaced 8 ft apart. Locate the resultant, then compute the absolute maximum moment under the heavier wheel.

Given

L = 40 ft
P₁ = 26 kips, P₂ = 12 kips
Spacing = 8 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
$R = 26 + 12 = 38 k i p s$
Formula — x̄ = ΣP d / ΣP (measured from P₁)
Substituting
$x ˉ = 12 (8) /38 = 2.526 f t$
Critical offset
$e = x ˉ /2 = 1.263 f t f r o mmi d s p an$
Formula
$M_{m} a x = (R / L) (L /2 - e)^{2}$
Substituting — M = (38/40)(20 − 1.263)² = 333.5 kip·ft

Answer: M_abs,max ≈ 333.5 kip·ft

Why the other options are there

380.0 kip·ft (single load at midspan)
190.0 kip·ft (uniform-load formula)

Reference: FE Reference Handbook — Structural Analysis → Moving Concentrated Load Sets

Example 7

Absolute maximum moment from a moving concentrated load set — Moving Concentrated Load Sets (7)

A simple span of 54 ft carries a moving load set of 22 kips and 26 kips spaced 7 ft apart. Locate the resultant, then compute the absolute maximum moment under the heavier wheel.

Given

L = 54 ft
P₁ = 22 kips, P₂ = 26 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
$R = 22 + 26 = 48 k i p s$
Formula — x̄ = ΣP d / ΣP (measured from P₁)
Substituting
$x ˉ = 26 (7) /48 = 3.792 f t$
Critical offset
$e = x ˉ /2 = 1.896 f t f r o mmi d s p an$
Formula
$M_{m} a x = (R / L) (L /2 - e)^{2}$
Substituting — M = (48/54)(27 − 1.896)² = 560.2 kip·ft

Answer: M_abs,max ≈ 560.2 kip·ft

Why the other options are there

648.0 kip·ft (single load at midspan)
324.0 kip·ft (uniform-load formula)

Reference: FE Reference Handbook — Structural Analysis → Moving Concentrated Load Sets

Example 8

Absolute maximum moment from a moving concentrated load set — Moving Concentrated Load Sets (8)

A simple span of 88 ft carries a moving load set of 15 kips and 29 kips spaced 9 ft apart. Locate the resultant, then compute the absolute maximum moment under the heavier wheel.

Given

L = 88 ft
P₁ = 15 kips, P₂ = 29 kips
Spacing = 9 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
$R = 15 + 29 = 44 k i p s$
Formula — x̄ = ΣP d / ΣP (measured from P₁)
Substituting
$x ˉ = 29 (9) /44 = 5.932 f t$
Critical offset
$e = x ˉ /2 = 2.966 f t f r o mmi d s p an$
Formula
$M_{m} a x = (R / L) (L /2 - e)^{2}$
Substituting — M = (44/88)(44 − 2.966)² = 841.9 kip·ft

Answer: M_abs,max ≈ 841.9 kip·ft

Why the other options are there

968.0 kip·ft (single load at midspan)
484.0 kip·ft (uniform-load formula)

Reference: FE Reference Handbook — Structural Analysis → Moving Concentrated Load Sets

Example 9

Absolute maximum moment from a moving concentrated load set — Moving Concentrated Load Sets (9)

A simple span of 41 ft carries a moving load set of 21 kips and 26 kips spaced 9 ft apart. Locate the resultant, then compute the absolute maximum moment under the heavier wheel.

Given

L = 41 ft
P₁ = 21 kips, P₂ = 26 kips
Spacing = 9 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
$R = 21 + 26 = 47 k i p s$
Formula — x̄ = ΣP d / ΣP (measured from P₁)
Substituting
$x ˉ = 26 (9) /47 = 4.979 f t$
Critical offset
$e = x ˉ /2 = 2.489 f t f r o mmi d s p an$
Formula
$M_{m} a x = (R / L) (L /2 - e)^{2}$
Substituting — M = (47/41)(20.5 − 2.489)² = 371.9 kip·ft

Answer: M_abs,max ≈ 371.9 kip·ft

Why the other options are there

481.8 kip·ft (single load at midspan)
240.9 kip·ft (uniform-load formula)

Reference: FE Reference Handbook — Structural Analysis → Moving Concentrated Load Sets

Example 10

Absolute maximum moment from a moving concentrated load set — Moving Concentrated Load Sets (10)

A simple span of 49 ft carries a moving load set of 22 kips and 24 kips spaced 7 ft apart. Locate the resultant, then compute the absolute maximum moment under the heavier wheel.

Given

L = 49 ft
P₁ = 22 kips, P₂ = 24 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
$R = 22 + 24 = 46 k i p s$
Formula — x̄ = ΣP d / ΣP (measured from P₁)
Substituting
$x ˉ = 24 (7) /46 = 3.652 f t$
Critical offset
$e = x ˉ /2 = 1.826 f t f r o mmi d s p an$
Formula
$M_{m} a x = (R / L) (L /2 - e)^{2}$
Substituting — M = (46/49)(24.5 − 1.826)² = 482.6 kip·ft

Answer: M_abs,max ≈ 482.6 kip·ft

Why the other options are there

563.5 kip·ft (single load at midspan)
281.8 kip·ft (uniform-load formula)

Reference: FE Reference Handbook — Structural Analysis → Moving Concentrated Load Sets

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

Moving Concentrated Load Sets 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.