Deflection of Beams

Mechanics of Materials · FE Reference Handbook section

Mechanics of Materials

6 formulas

10 exam-style examples

~57 min

All Mechanics of Materials lectures

Learning objectives

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

This chapter section covers Deflection of Beams within Mechanics of Materials. 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 deflection of beams describes physically and when it applies.
State every one of the 6 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: psi vs ksi and kip vs lb decide the answer choice.

Lecture

Why this section exists. Deflection of Beams is the part of Mechanics of Materials that lets you connect an axially loaded, bent or twisted member 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 a stress or deformation at one point of one member. 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. psi vs ksi and kip vs lb decide the answer choice. 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: deflection of beams.
Wikimedia Commons, CC BY-SA 4.0

Mechanics of Materials — Deflection of Beams: 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 an axially loaded, bent or twisted member. 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 6 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

Using 1/ρ	Quantity produced by "Using 1/ρ = M/(EI)," — read its definition and unit from the handbook line directly above the equation.
EI	Quantity produced by "EI = M, differential equation of deflection curve" — read its definition and unit from the handbook line directly above the equation.
EI 3	Quantity produced by "EI 3 = dM ^ x h /dx = V" — read its definition and unit from the handbook line directly above the equation.
EI 4	Quantity produced by "EI 4 = dV ^ x h /dx =− w" — read its definition and unit from the handbook line directly above the equation.
EI (dy/dx)	Quantity produced by "EI (dy/dx) = ∫M(x) dx" — read its definition and unit from the handbook line directly above the equation.
EIy	Quantity produced by "EIy = ∫[ ∫M(x) dx] dx" — 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.

d 2y
dx 2
d 3y
d 4y
Determine the deflection curve equation by double integration (apply boundary conditions applicable to the deflection and/or
slope).
The constants of integration can be determined from the physical geometry of the beam.

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

Maximum deflection of a uniformly loaded simple beam — Deflection of Beams

A simply supported steel beam spans 21 ft and carries a uniform service load of 1.2 kip/ft. With E = 29,000 ksi and I = 819 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 1.2 kip/ft
L = 21 ft
E = 29,000 ksi
I = 819 in⁴

Find

M_max, Δ_max and the L/360 check

Start with the thinking

Deflection of beams varies with the fourth power of the span — doubling the span multiplies deflection sixteenfold.
Units must be consistent: convert the span to inches and the load to kip/in.

Step-by-step solution

Formula
$M_{m} a x = w L^{2} /8$
Substituting — M = 1.2(21)²/8 = 66.15 kip·ft
Unit conversion
$w = 1.2/12 = 0.1000 k i p / in, L = 252 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.1000) (252)^{4} / [384 (29, 000) (819)]$
Evaluate
$Δ = 0.2211 in$
Limit
$L /360 = 252/360 = 0.700 in \to a cce pt ab l e$

Answer: M = 66.1 kip·ft, Δ = 0.221 in versus a 0.700 in limit

Why the other options are there

0.00013 in (units not converted)
0.0442 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Deflection of Beams

Example 2

Maximum deflection of a uniformly loaded simple beam — Deflection of Beams (2)

A simply supported steel beam spans 27 ft and carries a uniform service load of 2.4 kip/ft. With E = 29,000 ksi and I = 219 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 2.4 kip/ft
L = 27 ft
E = 29,000 ksi
I = 219 in⁴

Find

M_max, Δ_max and the L/360 check

Start with the thinking

Deflection of beams varies with the fourth power of the span — doubling the span multiplies deflection sixteenfold.
Units must be consistent: convert the span to inches and the load to kip/in.

Step-by-step solution

Formula
$M_{m} a x = w L^{2} /8$
Substituting — M = 2.4(27)²/8 = 218.7 kip·ft
Unit conversion
$w = 2.4/12 = 0.2000 k i p / in, L = 324 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.2000) (324)^{4} / [384 (29, 000) (219)]$
Evaluate
$Δ = 4.5186 in$
Limit
$L /360 = 324/360 = 0.900 in \to t oo f l e x ib l e$

Answer: M = 218.7 kip·ft, Δ = 4.519 in versus a 0.900 in limit

Why the other options are there

0.00261 in (units not converted)
0.9037 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Deflection of Beams

Example 3

Maximum deflection of a uniformly loaded simple beam — Deflection of Beams (3)

A simply supported steel beam spans 33 ft and carries a uniform service load of 1.0 kip/ft. With E = 29,000 ksi and I = 932 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 1.0 kip/ft
L = 33 ft
E = 29,000 ksi
I = 932 in⁴

Find

M_max, Δ_max and the L/360 check

Start with the thinking

Deflection of beams varies with the fourth power of the span — doubling the span multiplies deflection sixteenfold.
Units must be consistent: convert the span to inches and the load to kip/in.

Step-by-step solution

Formula
$M_{m} a x = w L^{2} /8$
Substituting — M = 1.0(33)²/8 = 136.1 kip·ft
Unit conversion
$w = 1.0/12 = 0.0833 k i p / in, L = 396 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.0833) (396)^{4} / [384 (29, 000) (932)]$
Evaluate
$Δ = 0.9872 in$
Limit
$L /360 = 396/360 = 1.100 in \to a cce pt ab l e$

Answer: M = 136.1 kip·ft, Δ = 0.987 in versus a 1.100 in limit

Why the other options are there

0.00057 in (units not converted)
0.1974 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Deflection of Beams

Example 4

Maximum deflection of a uniformly loaded simple beam — Deflection of Beams (4)

A simply supported steel beam spans 20 ft and carries a uniform service load of 2.8 kip/ft. With E = 29,000 ksi and I = 228 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 2.8 kip/ft
L = 20 ft
E = 29,000 ksi
I = 228 in⁴

Find

M_max, Δ_max and the L/360 check

Start with the thinking

Deflection of beams varies with the fourth power of the span — doubling the span multiplies deflection sixteenfold.
Units must be consistent: convert the span to inches and the load to kip/in.

Step-by-step solution

Formula
$M_{m} a x = w L^{2} /8$
Substituting — M = 2.8(20)²/8 = 140.0 kip·ft
Unit conversion
$w = 2.8/12 = 0.2333 k i p / in, L = 240 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.2333) (240)^{4} / [384 (29, 000) (228)]$
Evaluate
$Δ = 1.5245 in$
Limit
$L /360 = 240/360 = 0.667 in \to t oo f l e x ib l e$

Answer: M = 140.0 kip·ft, Δ = 1.525 in versus a 0.667 in limit

Why the other options are there

0.00088 in (units not converted)
0.3049 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Deflection of Beams

Example 5

Maximum deflection of a uniformly loaded simple beam — Deflection of Beams (5)

A simply supported steel beam spans 39 ft and carries a uniform service load of 3.4 kip/ft. With E = 29,000 ksi and I = 1231 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 3.4 kip/ft
L = 39 ft
E = 29,000 ksi
I = 1231 in⁴

Find

M_max, Δ_max and the L/360 check

Start with the thinking

Deflection of beams varies with the fourth power of the span — doubling the span multiplies deflection sixteenfold.
Units must be consistent: convert the span to inches and the load to kip/in.

Step-by-step solution

Formula
$M_{m} a x = w L^{2} /8$
Substituting — M = 3.4(39)²/8 = 646.4 kip·ft
Unit conversion
$w = 3.4/12 = 0.2833 k i p / in, L = 468 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.2833) (468)^{4} / [384 (29, 000) (1231)]$
Evaluate
$Δ = 4.9575 in$
Limit
$L /360 = 468/360 = 1.300 in \to t oo f l e x ib l e$

Answer: M = 646.4 kip·ft, Δ = 4.958 in versus a 1.300 in limit

Why the other options are there

0.00287 in (units not converted)
0.9915 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Deflection of Beams

Example 6

Maximum deflection of a uniformly loaded simple beam — Deflection of Beams (6)

A simply supported steel beam spans 23 ft and carries a uniform service load of 3.0 kip/ft. With E = 29,000 ksi and I = 1196 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 3.0 kip/ft
L = 23 ft
E = 29,000 ksi
I = 1196 in⁴

Find

M_max, Δ_max and the L/360 check

Start with the thinking

Deflection of beams varies with the fourth power of the span — doubling the span multiplies deflection sixteenfold.
Units must be consistent: convert the span to inches and the load to kip/in.

Step-by-step solution

Formula
$M_{m} a x = w L^{2} /8$
Substituting — M = 3.0(23)²/8 = 198.4 kip·ft
Unit conversion
$w = 3.0/12 = 0.2500 k i p / in, L = 276 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.2500) (276)^{4} / [384 (29, 000) (1196)]$
Evaluate
$Δ = 0.5446 in$
Limit
$L /360 = 276/360 = 0.767 in \to a cce pt ab l e$

Answer: M = 198.4 kip·ft, Δ = 0.545 in versus a 0.767 in limit

Why the other options are there

0.00032 in (units not converted)
0.1089 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Deflection of Beams

Example 7

Maximum deflection of a uniformly loaded simple beam — Deflection of Beams (7)

A simply supported steel beam spans 38 ft and carries a uniform service load of 3.2 kip/ft. With E = 29,000 ksi and I = 1995 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 3.2 kip/ft
L = 38 ft
E = 29,000 ksi
I = 1995 in⁴

Find

M_max, Δ_max and the L/360 check

Start with the thinking

Deflection of beams varies with the fourth power of the span — doubling the span multiplies deflection sixteenfold.
Units must be consistent: convert the span to inches and the load to kip/in.

Step-by-step solution

Formula
$M_{m} a x = w L^{2} /8$
Substituting — M = 3.2(38)²/8 = 577.6 kip·ft
Unit conversion
$w = 3.2/12 = 0.2667 k i p / in, L = 456 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.2667) (456)^{4} / [384 (29, 000) (1995)]$
Evaluate
$Δ = 2.5949 in$
Limit
$L /360 = 456/360 = 1.267 in \to t oo f l e x ib l e$

Answer: M = 577.6 kip·ft, Δ = 2.595 in versus a 1.267 in limit

Why the other options are there

0.00150 in (units not converted)
0.5190 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Deflection of Beams

Example 8

Maximum deflection of a uniformly loaded simple beam — Deflection of Beams (8)

A simply supported steel beam spans 29 ft and carries a uniform service load of 2.8 kip/ft. With E = 29,000 ksi and I = 1219 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 2.8 kip/ft
L = 29 ft
E = 29,000 ksi
I = 1219 in⁴

Find

M_max, Δ_max and the L/360 check

Start with the thinking

Deflection of beams varies with the fourth power of the span — doubling the span multiplies deflection sixteenfold.
Units must be consistent: convert the span to inches and the load to kip/in.

Step-by-step solution

Formula
$M_{m} a x = w L^{2} /8$
Substituting — M = 2.8(29)²/8 = 294.3 kip·ft
Unit conversion
$w = 2.8/12 = 0.2333 k i p / in, L = 348 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.2333) (348)^{4} / [384 (29, 000) (1219)]$
Evaluate
$Δ = 1.2605 in$
Limit
$L /360 = 348/360 = 0.967 in \to t oo f l e x ib l e$

Answer: M = 294.3 kip·ft, Δ = 1.260 in versus a 0.967 in limit

Why the other options are there

0.00073 in (units not converted)
0.2521 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Deflection of Beams

Example 9

Maximum deflection of a uniformly loaded simple beam — Deflection of Beams (9)

A simply supported steel beam spans 18 ft and carries a uniform service load of 4.0 kip/ft. With E = 29,000 ksi and I = 1077 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 4.0 kip/ft
L = 18 ft
E = 29,000 ksi
I = 1077 in⁴

Find

M_max, Δ_max and the L/360 check

Start with the thinking

Deflection of beams varies with the fourth power of the span — doubling the span multiplies deflection sixteenfold.
Units must be consistent: convert the span to inches and the load to kip/in.

Step-by-step solution

Formula
$M_{m} a x = w L^{2} /8$
Substituting — M = 4.0(18)²/8 = 162.0 kip·ft
Unit conversion
$w = 4.0/12 = 0.3333 k i p / in, L = 216 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.3333) (216)^{4} / [384 (29, 000) (1077)]$
Evaluate
$Δ = 0.3025 in$
Limit
$L /360 = 216/360 = 0.600 in \to a cce pt ab l e$

Answer: M = 162.0 kip·ft, Δ = 0.302 in versus a 0.600 in limit

Why the other options are there

0.00018 in (units not converted)
0.0605 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Deflection of Beams

Example 10

Maximum deflection of a uniformly loaded simple beam — Deflection of Beams (10)

A simply supported steel beam spans 36 ft and carries a uniform service load of 1.7 kip/ft. With E = 29,000 ksi and I = 223 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 1.7 kip/ft
L = 36 ft
E = 29,000 ksi
I = 223 in⁴

Find

M_max, Δ_max and the L/360 check

Start with the thinking

Deflection of beams varies with the fourth power of the span — doubling the span multiplies deflection sixteenfold.
Units must be consistent: convert the span to inches and the load to kip/in.

Step-by-step solution

Formula
$M_{m} a x = w L^{2} /8$
Substituting — M = 1.7(36)²/8 = 275.4 kip·ft
Unit conversion
$w = 1.7/12 = 0.1417 k i p / in, L = 432 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.1417) (432)^{4} / [384 (29, 000) (223)]$
Evaluate
$Δ = 9.9343 in$
Limit
$L /360 = 432/360 = 1.200 in \to t oo f l e x ib l e$

Answer: M = 275.4 kip·ft, Δ = 9.934 in versus a 1.200 in limit

Why the other options are there

0.00575 in (units not converted)
1.9869 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Deflection of Beams

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 an axially loaded, bent or twisted member, 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

Deflection of Beams contains 6 relations; you must be able to find this page in under 15 seconds.
Exam style: a stress or deformation at one point of one member.
Unit rule: psi vs ksi and kip vs lb decide the answer choice.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

psi vs ksi and kip vs lb decide the answer choice
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.