Composite Sections

Mechanics of Materials · FE Reference Handbook section

Mechanics of Materials

7 formulas

10 exam-style examples

~59 min

All Mechanics of Materials lectures

Learning objectives

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

This chapter section covers Composite Sections 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 composite sections describes physically and when it applies.
State every one of the 7 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. Composite Sections 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.

Concrete cylinder under axial load in a compression testing machine. — Photo 1. Where this shows up in practice: composite sections.
Wikimedia Commons, public domain

Mechanics of Materials — Composite Sections: 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 7 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

σ1	Quantity produced by "σ1 = -nMy/IT" — read its definition and unit from the handbook line directly above the equation.
σ2	Quantity produced by "σ2 = -My/IT" — read its definition and unit from the handbook line directly above the equation.
IT	Quantity produced by "IT = moment of inertia of the transformed section" — read its definition and unit from the handbook line directly above the equation.
n	Quantity produced by "n = modular ratio E1/E2" — read its definition and unit from the handbook line directly above the equation.
E1	Quantity produced by "E1 = elastic modulus of Material 1" — read its definition and unit from the handbook line directly above the equation.
E2	Quantity produced by "E2 = elastic modulus of Material 2" — read its definition and unit from the handbook line directly above the equation.
y	Quantity produced by "y = distance from the neutral axis to the fiber location above or below the neutral axis" — 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 bending stresses in a beam composed of dissimilar materials (Material 1 and Material 2) where E1 > E2 are:
where
The composite section is transformed into a section composed of a single material. The centroid and then the moment of inertia
are found on the transformed section for use in the bending stress equations.
COMPOSITE TRANSFORMED
SECTION SECTION
MATERIAL 1 E1, A1 E2, nA1
NEUTRAL
AXIS
MATERIAL 2 E2, A2 E2, A2
b b

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 — Composite Sections

A simply supported steel beam spans 31 ft and carries a uniform service load of 2.6 kip/ft. With E = 29,000 ksi and I = 524 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 2.6 kip/ft
L = 31 ft
E = 29,000 ksi
I = 524 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 = 2.6(31)²/8 = 312.3 kip·ft
Unit conversion
$w = 2.6/12 = 0.2167 k i p / in, L = 372 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.2167) (372)^{4} / [384 (29, 000) (524)]$
Evaluate
$Δ = 3.5553 in$
Limit
$L /360 = 372/360 = 1.033 in \to t oo f l e x ib l e$

Answer: M = 312.3 kip·ft, Δ = 3.555 in versus a 1.033 in limit

Why the other options are there

0.00206 in (units not converted)
0.7111 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Composite Sections

Example 2

Maximum deflection of a uniformly loaded simple beam — Composite Sections (2)

A simply supported steel beam spans 28 ft and carries a uniform service load of 2.7 kip/ft. With E = 29,000 ksi and I = 804 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 2.7 kip/ft
L = 28 ft
E = 29,000 ksi
I = 804 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 = 2.7(28)²/8 = 264.6 kip·ft
Unit conversion
$w = 2.7/12 = 0.2250 k i p / in, L = 336 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.2250) (336)^{4} / [384 (29, 000) (804)]$
Evaluate
$Δ = 1.6015 in$
Limit
$L /360 = 336/360 = 0.933 in \to t oo f l e x ib l e$

Answer: M = 264.6 kip·ft, Δ = 1.601 in versus a 0.933 in limit

Why the other options are there

0.00093 in (units not converted)
0.3203 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Composite Sections

Example 3

Maximum deflection of a uniformly loaded simple beam — Composite Sections (3)

A simply supported steel beam spans 23 ft and carries a uniform service load of 3.6 kip/ft. With E = 29,000 ksi and I = 511 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 3.6 kip/ft
L = 23 ft
E = 29,000 ksi
I = 511 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 = 3.6(23)²/8 = 238.0 kip·ft
Unit conversion
$w = 3.6/12 = 0.3000 k i p / in, L = 276 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.3000) (276)^{4} / [384 (29, 000) (511)]$
Evaluate
$Δ = 1.5296 in$
Limit
$L /360 = 276/360 = 0.767 in \to t oo f l e x ib l e$

Answer: M = 238.0 kip·ft, Δ = 1.530 in versus a 0.767 in limit

Why the other options are there

0.00089 in (units not converted)
0.3059 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Composite Sections

Example 4

Maximum deflection of a uniformly loaded simple beam — Composite Sections (4)

A simply supported steel beam spans 12 ft and carries a uniform service load of 1.9 kip/ft. With E = 29,000 ksi and I = 433 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 1.9 kip/ft
L = 12 ft
E = 29,000 ksi
I = 433 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 = 1.9(12)²/8 = 34.20 kip·ft
Unit conversion
$w = 1.9/12 = 0.1583 k i p / in, L = 144 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.1583) (144)^{4} / [384 (29, 000) (433)]$
Evaluate
$Δ = 0.0706 in$
Limit
$L /360 = 144/360 = 0.400 in \to a cce pt ab l e$

Answer: M = 34.2 kip·ft, Δ = 0.071 in versus a 0.400 in limit

Why the other options are there

0.00004 in (units not converted)
0.0141 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Composite Sections

Example 5

Maximum deflection of a uniformly loaded simple beam — Composite Sections (5)

A simply supported steel beam spans 24 ft and carries a uniform service load of 2.8 kip/ft. With E = 29,000 ksi and I = 985 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 2.8 kip/ft
L = 24 ft
E = 29,000 ksi
I = 985 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(24)²/8 = 201.6 kip·ft
Unit conversion
$w = 2.8/12 = 0.2333 k i p / in, L = 288 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.2333) (288)^{4} / [384 (29, 000) (985)]$
Evaluate
$Δ = 0.7317 in$
Limit
$L /360 = 288/360 = 0.800 in \to a cce pt ab l e$

Answer: M = 201.6 kip·ft, Δ = 0.732 in versus a 0.800 in limit

Why the other options are there

0.00042 in (units not converted)
0.1463 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Composite Sections

Example 6

Maximum deflection of a uniformly loaded simple beam — Composite Sections (6)

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 = 1542 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 = 1542 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
$Δ = 5 (0.2833) (468)^{4} / [384 (29, 000) (1542)]$
Evaluate
$Δ = 3.9577 in$
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 = 3.958 in versus a 1.300 in limit

Why the other options are there

0.00229 in (units not converted)
0.7915 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Composite Sections

Example 7

Maximum deflection of a uniformly loaded simple beam — Composite Sections (7)

A simply supported steel beam spans 21 ft and carries a uniform service load of 1.1 kip/ft. With E = 29,000 ksi and I = 1606 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 1.1 kip/ft
L = 21 ft
E = 29,000 ksi
I = 1606 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 = 1.1(21)²/8 = 60.64 kip·ft
Unit conversion
$w = 1.1/12 = 0.0917 k i p / in, L = 252 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.0917) (252)^{4} / [384 (29, 000) (1606)]$
Evaluate
$Δ = 0.1033 in$
Limit
$L /360 = 252/360 = 0.700 in \to a cce pt ab l e$

Answer: M = 60.6 kip·ft, Δ = 0.103 in versus a 0.700 in limit

Why the other options are there

0.00006 in (units not converted)
0.0207 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Composite Sections

Example 8

Maximum deflection of a uniformly loaded simple beam — Composite Sections (8)

A simply supported steel beam spans 39 ft and carries a uniform service load of 1.6 kip/ft. With E = 29,000 ksi and I = 1284 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 1.6 kip/ft
L = 39 ft
E = 29,000 ksi
I = 1284 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 = 1.6(39)²/8 = 304.2 kip·ft
Unit conversion
$w = 1.6/12 = 0.1333 k i p / in, L = 468 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.1333) (468)^{4} / [384 (29, 000) (1284)]$
Evaluate
$Δ = 2.2366 in$
Limit
$L /360 = 468/360 = 1.300 in \to t oo f l e x ib l e$

Answer: M = 304.2 kip·ft, Δ = 2.237 in versus a 1.300 in limit

Why the other options are there

0.00129 in (units not converted)
0.4473 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Composite Sections

Example 9

Maximum deflection of a uniformly loaded simple beam — Composite Sections (9)

A simply supported steel beam spans 34 ft and carries a uniform service load of 2.8 kip/ft. With E = 29,000 ksi and I = 507 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 2.8 kip/ft
L = 34 ft
E = 29,000 ksi
I = 507 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(34)²/8 = 404.6 kip·ft
Unit conversion
$w = 2.8/12 = 0.2333 k i p / in, L = 408 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.2333) (408)^{4} / [384 (29, 000) (507)]$
Evaluate
$Δ = 5.7260 in$
Limit
$L /360 = 408/360 = 1.133 in \to t oo f l e x ib l e$

Answer: M = 404.6 kip·ft, Δ = 5.726 in versus a 1.133 in limit

Why the other options are there

0.00331 in (units not converted)
1.1452 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Composite Sections

Example 10

Maximum deflection of a uniformly loaded simple beam — Composite Sections (10)

A simply supported steel beam spans 22 ft and carries a uniform service load of 3.1 kip/ft. With E = 29,000 ksi and I = 1851 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 3.1 kip/ft
L = 22 ft
E = 29,000 ksi
I = 1851 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 = 3.1(22)²/8 = 187.6 kip·ft
Unit conversion
$w = 3.1/12 = 0.2583 k i p / in, L = 264 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.2583) (264)^{4} / [384 (29, 000) (1851)]$
Evaluate
$Δ = 0.3044 in$
Limit
$L /360 = 264/360 = 0.733 in \to a cce pt ab l e$

Answer: M = 187.6 kip·ft, Δ = 0.304 in versus a 0.733 in limit

Why the other options are there

0.00018 in (units not converted)
0.0609 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Composite Sections

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

Composite Sections contains 7 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.