Frame Deflection by Unit Load Method

Structural Analysis · FE Reference Handbook section

Structural Analysis

15 formulas

10 exam-style examples

~60 min

All Structural Analysis lectures

Learning objectives

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

This chapter section covers Frame Deflection by Unit Load Method 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 frame deflection by unit load method describes physically and when it applies.
State every one of the 15 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. Frame Deflection by Unit Load Method 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: frame deflection by unit load method.
Wikimedia Commons, CC BY-SA 4.0

Structural Analysis — Frame Deflection by Unit Load Method: 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 15 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.

Notation used in this section

i	Quantity produced by "i= 1 i" — read its definition and unit from the handbook line directly above the equation.
∆	Quantity produced by "∆ = displacement at point of application of unit load (+ in direction of unit load)" — read its definition and unit from the handbook line directly above the equation.
mi	Quantity produced by "mi = moment equation in member i caused by the unit load" — read its definition and unit from the handbook line directly above the equation.
Mi	Quantity produced by "Mi = moment equation in member i caused by loads applied to frame" — read its definition and unit from the handbook line directly above the equation.
Li	Quantity produced by "Li = length of member i" — read its definition and unit from the handbook line directly above the equation.
Ii	Quantity produced by "Ii = moment of inertia of member i" — read its definition and unit from the handbook line directly above the equation.
0	Quantity produced by "0 = ∆ BB + δ BB" — read its definition and unit from the handbook line directly above the equation.
∆ BB	Quantity produced by "∆ BB = 8 EI" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

The displacement of any point on a frame caused by external loads is found by applying a unit load at that point that cor-
responds to the desired displacement:
members
where
If either the real loads or the unit load cause no moment in a member, that member can be omitted from the summation.
Elementary Statically Indeterminate Structures by Force Method of Analysis
The force method is typically used to solve for elements or structures with a single degree of indeterminacy. The method states
that the deflection resulting from the removal of a redundant support is equal and opposite to the deflection that the redundant
reaction causes, resulting in a net zero deflection.
A B C
L/2 L/2
− 5wL4
∆ BB 384 EI
L/2 L/2
δ BB ByL3
48 EI
L/2 L/2
A B
–wL 4
ByL3
3 EI

Core formulas for this FE topic

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

Worked exam-style examples

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

Example 1

Midspan deflection of a uniformly loaded beam

A simply supported steel beam spans 8.0 m and carries 15 kN/m. With E = 200 GPa and I = 250 × 10⁶ mm⁴, find the midspan deflection.

Given

L = 8.0 m
w = 15 kN/m
E = 200 GPa
I = 250 × 10⁶ mm⁴

Find

Δ_max

Start with the thinking

Use consistent N and mm throughout.
L⁴ dominates — an error in L is a factor-of-16 error.

Figure for Midspan deflection of a uniformly loaded beam

Step-by-step solution

Formula
$Δ = 5 w L^{4} / (384 E I)$
Units
$w = 15 N / mm, L = 8, 000 mm, E = 200, 000 M P a, I = 250 \times 1 0^{6} m m^{4}$
Numerator
$5 (15) (8, 000)^{4} = 5 (15) (4.096 \times 1 0^{15}) = 3.072 \times 1 0^{17}$
Denominator
$384 (200, 000) (250 \times 1 0^{6}) = 1.92 \times 1 0^{16}$
Result
$Δ = 16.0 mm$

Answer: Δ = 16.0 mm (L/500, acceptable for L/360)

Why the other options are there

3.2 mm (L in metres)
80 mm (384 replaced with 77)

Reference: FE Reference Handbook — Mechanics of Materials — Beam deflection

Example 2

Midspan deflection of a simple beam — Frame Deflection by Unit Load Method

A 20 ft simple span carries a 32 kip load at midspan. With E = 29,000 ksi and I = 1,400 in⁴, find the midspan deflection.

Given

P = 32 kip
L = 20 ft
I = 1,400 in⁴
E = 29,000 ksi

Find

Δ at midspan

Start with the thinking

Deflection scales with L³ — convert to inches before cubing.
Compare against L/360 for serviceability.

Step-by-step solution

Formula
$Δ = P L^{3} / (48 E I)$
Span in inches
$L = 20 \times 12 = 240 in$
Substituting
$Δ = 32 (240)^{3} / (48 \times 29, 000 \times 1, 400)$
Evaluate
$Δ = 0.2270 in$
Serviceability
$L /360 = 0.667 in . \to a cce pt ab l e$

Answer: Δ ≈ 0.227 in.

Why the other options are there

392.2 in. (span left in feet)
0.142 in. (uniform-load formula used)

Reference: FE Reference Handbook — Structural Analysis → Frame Deflection by Unit Load Method

Example 3

Fixed-end moments for a loaded member — Frame Deflection by Unit Load Method

A member of a rigid frame spans 28 ft and carries 1.75 kip/ft. What are the fixed-end moments, and what is the midspan moment once they are applied?

Given

w = 1.75 kip/ft
L = 28 ft
Both ends fixed

Find

FEM and midspan moment

Start with the thinking

Fixed-end moment for a uniform load is wL²/12.
The midspan value is wL²/24 with fixed ends.

Step-by-step solution

Fixed-end moment
$F E M = w L^{2} /12$
Substituting — FEM = 1.75(28)²/12 = 114.3 kip·ft at each end
Midspan
$M_{m} i d = w L^{2} /24$
Substituting — M_mid = 1.75(28)²/24 = 57.2 kip·ft
Check — simple-span value wL²/8 = 171.5 kip·ft is the sum of the end and mid effects ✓

Answer: FEM = 114.3 kip·ft; midspan = 57.2 kip·ft

Why the other options are there

171.5 kip·ft (simple-span value)
686.0 kip·ft (cantilever value)

Reference: FE Reference Handbook — Structural Analysis → Frame Deflection by Unit Load Method

Example 4

Midspan deflection of a simple beam — Frame Deflection by Unit Load Method (2)

A 15 ft simple span carries a 12 kip load at midspan. With E = 29,000 ksi and I = 1,400 in⁴, find the midspan deflection.

Given

P = 12 kip
L = 15 ft
I = 1,400 in⁴
E = 29,000 ksi

Find

Δ at midspan

Start with the thinking

Deflection scales with L³ — convert to inches before cubing.
Compare against L/360 for serviceability.

Step-by-step solution

Formula
$Δ = P L^{3} / (48 E I)$
Span in inches
$L = 15 \times 12 = 180 in$
Substituting
$Δ = 12 (180)^{3} / (48 \times 29, 000 \times 1, 400)$
Evaluate
$Δ = 0.0359 in$
Serviceability
$L /360 = 0.500 in . \to a cce pt ab l e$

Answer: Δ ≈ 0.036 in.

Why the other options are there

62.05 in. (span left in feet)
0.022 in. (uniform-load formula used)

Reference: FE Reference Handbook — Structural Analysis → Frame Deflection by Unit Load Method

Example 5

Fixed-end moments for a loaded member — Frame Deflection by Unit Load Method (2)

A member of a rigid frame spans 24 ft and carries 3.50 kip/ft. What are the fixed-end moments, and what is the midspan moment once they are applied?

Given

w = 3.50 kip/ft
L = 24 ft
Both ends fixed

Find

FEM and midspan moment

Start with the thinking

Fixed-end moment for a uniform load is wL²/12.
The midspan value is wL²/24 with fixed ends.

Step-by-step solution

Fixed-end moment
$F E M = w L^{2} /12$
Substituting — FEM = 3.50(24)²/12 = 168.0 kip·ft at each end
Midspan
$M_{m} i d = w L^{2} /24$
Substituting — M_mid = 3.50(24)²/24 = 84.0 kip·ft
Check — simple-span value wL²/8 = 252.0 kip·ft is the sum of the end and mid effects ✓

Answer: FEM = 168.0 kip·ft; midspan = 84.0 kip·ft

Why the other options are there

252.0 kip·ft (simple-span value)
1,008 kip·ft (cantilever value)

Reference: FE Reference Handbook — Structural Analysis → Frame Deflection by Unit Load Method

Example 6

Midspan deflection of a simple beam — Frame Deflection by Unit Load Method (3)

A 27 ft simple span carries a 26 kip load at midspan. With E = 29,000 ksi and I = 1,550 in⁴, find the midspan deflection.

Given

P = 26 kip
L = 27 ft
I = 1,550 in⁴
E = 29,000 ksi

Find

Δ at midspan

Start with the thinking

Deflection scales with L³ — convert to inches before cubing.
Compare against L/360 for serviceability.

Step-by-step solution

Formula
$Δ = P L^{3} / (48 E I)$
Span in inches
$L = 27 \times 12 = 324 in$
Substituting
$Δ = 26 (324)^{3} / (48 \times 29, 000 \times 1, 550)$
Evaluate
$Δ = 0.4099 in$
Serviceability
$L /360 = 0.900 in . \to a cce pt ab l e$

Answer: Δ ≈ 0.410 in.

Why the other options are there

708.2 in. (span left in feet)
0.256 in. (uniform-load formula used)

Reference: FE Reference Handbook — Structural Analysis → Frame Deflection by Unit Load Method

Example 7

Fixed-end moments for a loaded member — Frame Deflection by Unit Load Method (3)

A member of a rigid frame spans 17 ft and carries 1.75 kip/ft. What are the fixed-end moments, and what is the midspan moment once they are applied?

Given

w = 1.75 kip/ft
L = 17 ft
Both ends fixed

Find

FEM and midspan moment

Start with the thinking

Fixed-end moment for a uniform load is wL²/12.
The midspan value is wL²/24 with fixed ends.

Step-by-step solution

Fixed-end moment
$F E M = w L^{2} /12$
Substituting — FEM = 1.75(17)²/12 = 42.1 kip·ft at each end
Midspan
$M_{m} i d = w L^{2} /24$
Substituting — M_mid = 1.75(17)²/24 = 21.1 kip·ft
Check — simple-span value wL²/8 = 63.2 kip·ft is the sum of the end and mid effects ✓

Answer: FEM = 42.1 kip·ft; midspan = 21.1 kip·ft

Why the other options are there

63.2 kip·ft (simple-span value)
252.9 kip·ft (cantilever value)

Reference: FE Reference Handbook — Structural Analysis → Frame Deflection by Unit Load Method

Example 8

Midspan deflection of a simple beam — Frame Deflection by Unit Load Method (4)

A 33 ft simple span carries a 17 kip load at midspan. With E = 29,000 ksi and I = 800.0 in⁴, find the midspan deflection.

Given

P = 17 kip
L = 33 ft
I = 800.0 in⁴
E = 29,000 ksi

Find

Δ at midspan

Start with the thinking

Deflection scales with L³ — convert to inches before cubing.
Compare against L/360 for serviceability.

Step-by-step solution

Formula
$Δ = P L^{3} / (48 E I)$
Span in inches
$L = 33 \times 12 = 396 in$
Substituting
$Δ = 17 (396)^{3} / (48 \times 29, 000 \times 800.0)$
Evaluate
$Δ = 0.9480 in$
Serviceability
$L /360 = 1.100 in . \to a cce pt ab l e$

Answer: Δ ≈ 0.948 in.

Why the other options are there

1,638 in. (span left in feet)
0.592 in. (uniform-load formula used)

Reference: FE Reference Handbook — Structural Analysis → Frame Deflection by Unit Load Method

Example 9

Fixed-end moments for a loaded member — Frame Deflection by Unit Load Method (4)

A member of a rigid frame spans 29 ft and carries 2.50 kip/ft. What are the fixed-end moments, and what is the midspan moment once they are applied?

Given

w = 2.50 kip/ft
L = 29 ft
Both ends fixed

Find

FEM and midspan moment

Start with the thinking

Fixed-end moment for a uniform load is wL²/12.
The midspan value is wL²/24 with fixed ends.

Step-by-step solution

Fixed-end moment
$F E M = w L^{2} /12$
Substituting — FEM = 2.50(29)²/12 = 175.2 kip·ft at each end
Midspan
$M_{m} i d = w L^{2} /24$
Substituting — M_mid = 2.50(29)²/24 = 87.6 kip·ft
Check — simple-span value wL²/8 = 262.8 kip·ft is the sum of the end and mid effects ✓

Answer: FEM = 175.2 kip·ft; midspan = 87.6 kip·ft

Why the other options are there

262.8 kip·ft (simple-span value)
1,051 kip·ft (cantilever value)

Reference: FE Reference Handbook — Structural Analysis → Frame Deflection by Unit Load Method

Example 10

Midspan deflection of a simple beam — Frame Deflection by Unit Load Method (5)

A 19 ft simple span carries a 9 kip load at midspan. With E = 29,000 ksi and I = 1,300 in⁴, find the midspan deflection.

Given

P = 9 kip
L = 19 ft
I = 1,300 in⁴
E = 29,000 ksi

Find

Δ at midspan

Start with the thinking

Deflection scales with L³ — convert to inches before cubing.
Compare against L/360 for serviceability.

Step-by-step solution

Formula
$Δ = P L^{3} / (48 E I)$
Span in inches
$L = 19 \times 12 = 228 in$
Substituting
$Δ = 9 (228)^{3} / (48 \times 29, 000 \times 1, 300)$
Evaluate
$Δ = 0.0589 in$
Serviceability
$L /360 = 0.633 in . \to a cce pt ab l e$

Answer: Δ ≈ 0.059 in.

Why the other options are there

101.9 in. (span left in feet)
0.037 in. (uniform-load formula used)

Reference: FE Reference Handbook — Structural Analysis → Frame Deflection by Unit Load Method

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

Frame Deflection by Unit Load Method contains 15 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.