Truss Deflection by Unit Load Method

Structural Analysis · FE Reference Handbook section

Structural Analysis

12 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 Truss 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 truss deflection by unit load method describes physically and when it applies.
State every one of the 12 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. Truss 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: truss deflection by unit load method.
Wikimedia Commons, CC BY-SA 4.0

Structural Analysis — Truss 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 12 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

! fi ^ DLhi	Quantity produced by "! fi ^ DLhi" — read its definition and unit from the handbook line directly above the equation.
i	Quantity produced by "i= 1" — read its definition and unit from the handbook line directly above the equation.
∆joint	Quantity produced by "∆joint = joint 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.
fi	Quantity produced by "fi = force in member i caused by unit load (+ tension)" — read its definition and unit from the handbook line directly above the equation.
(∆L)i	Quantity produced by "(∆L)i = change in length caused by external effect (+ for increase in member length):" — read its definition and unit from the handbook line directly above the equation.
(α	Quantity produced by "(α = coefficient of thermal expansion)" — read its definition and unit from the handbook line directly above the equation.
L, A	Quantity produced by "L, A = member length and cross-sectional area" — read its definition and unit from the handbook line directly above the equation.
E	Quantity produced by "E = member elastic modulus" — 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 a truss joint caused by external effects (truss loads, member temperature change, member misfit) is found
by applying a unit load at the point that corresponds to the desired displacement.
members
where
AE i

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

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

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 3

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 4

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

A 27 ft simple span carries a 14 kip load at midspan. With E = 29,000 ksi and I = 950.0 in⁴, find the midspan deflection.

Given

P = 14 kip
L = 27 ft
I = 950.0 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
$L = 27 \times 12 = 324 in$
Substituting
$Δ = 14 (324)^{3} / (48 \times 29, 000 \times 950.0)$
Evaluate
$Δ = 0.3601 in$
Serviceability
$L /360 = 0.900 in . \to a cce pt ab l e$

Answer: Δ ≈ 0.360 in.

Why the other options are there

622.2 in. (span left in feet)
0.225 in. (uniform-load formula used)

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

Example 5

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

A 17 ft simple span carries a 14 kip load at midspan. With E = 29,000 ksi and I = 550.0 in⁴, find the midspan deflection.

Given

P = 14 kip
L = 17 ft
I = 550.0 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
$L = 17 \times 12 = 204 in$
Substituting
$Δ = 14 (204)^{3} / (48 \times 29, 000 \times 550.0)$
Evaluate
$Δ = 0.1552 in$
Serviceability
$L /360 = 0.567 in . \to a cce pt ab l e$

Answer: Δ ≈ 0.155 in.

Why the other options are there

268.3 in. (span left in feet)
0.097 in. (uniform-load formula used)

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

Example 6

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

A 29 ft simple span carries a 33 kip load at midspan. With E = 29,000 ksi and I = 300.0 in⁴, find the midspan deflection.

Given

P = 33 kip
L = 29 ft
I = 300.0 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
$L = 29 \times 12 = 348 in$
Substituting
$Δ = 33 (348)^{3} / (48 \times 29, 000 \times 300.0)$
Evaluate
$Δ = 3.3304 in$
Serviceability
$L /360 = 0.967 in . \to e x cee d s t h e l imi t$

Answer: Δ ≈ 3.330 in.

Why the other options are there

5,755 in. (span left in feet)
2.081 in. (uniform-load formula used)

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

Example 7

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

A 17 ft simple span carries a 24 kip load at midspan. With E = 29,000 ksi and I = 1,350 in⁴, find the midspan deflection.

Given

P = 24 kip
L = 17 ft
I = 1,350 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
$L = 17 \times 12 = 204 in$
Substituting
$Δ = 24 (204)^{3} / (48 \times 29, 000 \times 1, 350)$
Evaluate
$Δ = 0.1084 in$
Serviceability
$L /360 = 0.567 in . \to a cce pt ab l e$

Answer: Δ ≈ 0.108 in.

Why the other options are there

187.4 in. (span left in feet)
0.068 in. (uniform-load formula used)

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

Example 8

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

A 34 ft simple span carries a 22 kip load at midspan. With E = 29,000 ksi and I = 1,950 in⁴, find the midspan deflection.

Given

P = 22 kip
L = 34 ft
I = 1,950 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
$L = 34 \times 12 = 408 in$
Substituting
$Δ = 22 (408)^{3} / (48 \times 29, 000 \times 1, 950)$
Evaluate
$Δ = 0.5505 in$
Serviceability
$L /360 = 1.133 in . \to a cce pt ab l e$

Answer: Δ ≈ 0.550 in.

Why the other options are there

951.2 in. (span left in feet)
0.344 in. (uniform-load formula used)

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

Example 9

Midspan deflection of a simple beam — Truss Deflection by Unit Load Method (6)

A 26 ft simple span carries a 30 kip load at midspan. With E = 29,000 ksi and I = 1,250 in⁴, find the midspan deflection.

Given

P = 30 kip
L = 26 ft
I = 1,250 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
$L = 26 \times 12 = 312 in$
Substituting
$Δ = 30 (312)^{3} / (48 \times 29, 000 \times 1, 250)$
Evaluate
$Δ = 0.5236 in$
Serviceability
$L /360 = 0.867 in . \to a cce pt ab l e$

Answer: Δ ≈ 0.524 in.

Why the other options are there

904.9 in. (span left in feet)
0.327 in. (uniform-load formula used)

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

Example 10

Midspan deflection of a simple beam — Truss Deflection by Unit Load Method (7)

A 19 ft simple span carries a 25 kip load at midspan. With E = 29,000 ksi and I = 1,650 in⁴, find the midspan deflection.

Given

P = 25 kip
L = 19 ft
I = 1,650 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
$Δ = 25 (228)^{3} / (48 \times 29, 000 \times 1, 650)$
Evaluate
$Δ = 0.1290 in$
Serviceability
$L /360 = 0.633 in . \to a cce pt ab l e$

Answer: Δ ≈ 0.129 in.

Why the other options are there

222.9 in. (span left in feet)
0.081 in. (uniform-load formula used)

Reference: FE Reference Handbook — Structural Analysis → Truss 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

Truss Deflection by Unit Load Method contains 12 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.