Stresses in Beams

Mechanics of Materials · FE Reference Handbook section

Mechanics of Materials

15 formulas

10 exam-style examples

~60 min

All Mechanics of Materials lectures

Learning objectives

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

This chapter section covers Stresses in 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 stresses in beams 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: psi vs ksi and kip vs lb decide the answer choice.

Lecture

Why this section exists. Stresses in 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: stresses in beams.
Wikimedia Commons, CC BY-SA 4.0

Mechanics of Materials — Stresses in 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 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

σx	Quantity produced by "σx = –My/I" — read its definition and unit from the handbook line directly above the equation.
M	Quantity produced by "M = moment at the section" — read its definition and unit from the handbook line directly above the equation.
I	Quantity produced by "I = moment of inertia of the cross section" — 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.
c	Quantity produced by "c = distance from the neutral axis to the outermost fiber of a symmetrical beam section" — read its definition and unit from the handbook line directly above the equation.
s	Quantity produced by "s = I/c: the elastic section modulus of the beam" — read its definition and unit from the handbook line directly above the equation.
τxy	Quantity produced by "τxy = VQ/(Ib)" — read its definition and unit from the handbook line directly above the equation.
V	Quantity produced by "V = shear force" — read its definition and unit from the handbook line directly above the equation.
Q	Quantity produced by "Q = Al y l = first moment of area above or below the point where shear stress is to be determined" — read its definition and unit from the handbook line directly above the equation.
A′	Quantity produced by "A′ = area above the layer (or plane) upon which the desired transverse shear stress acts" — read its definition and unit from the handbook line directly above the equation.
y l	Quantity produced by "y l = distance from neutral axis to area centroid" — read its definition and unit from the handbook line directly above the equation.
b	Quantity produced by "b = width or thickness or the cross-section" — 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 normal stress in a beam due to bending:
where
The maximum normal stresses in a beam due to bending:
where
where
Transverse shear stress:
where
Hibbeler, Russel C., Mechanics of Materials, 10th ed., Pearson, 2015, pp. 386 –387.
where
Transverse shear flow:

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

Bending stress in a rectangular beam

A 200 mm wide × 400 mm deep timber beam carries a maximum moment of 60 kN·m. What is the extreme fibre bending stress?

Given

b = 200 mm, h = 400 mm
M = 60 kN·m

Find

σ_max

Start with the thinking

Section modulus S = bh²/6 for a rectangle.
Convert kN·m to N·mm before dividing.

Step-by-step solution

Section modulus
$S = b h^{2} /6 = 200 (400)^{2} /6 = 5.33 \times 1 0^{6} m m^{3}$
Moment in N·mm — M = 60×10⁶ N·mm
Flexure formula
$σ = M / S$
Substitute
$σ = 60 \times 1 0^{6} /5.33 \times 1 0^{6}$
Result
$σ = 11.3 M P a$

Answer: σ_max = 11.3 MPa

Why the other options are there

22.5 MPa (bh²/3 used)
5.6 MPa (depth and width swapped)

Reference: FE Reference Handbook — Mechanics of Materials — Stresses in beams

Example 2

Maximum transverse shear stress

The same 200 × 400 mm beam carries a shear force of 90 kN. What is the maximum transverse shear stress?

Given

V = 90 kN
Rectangular section 200 × 400 mm

Find

τ_max

Start with the thinking

For a rectangle the parabolic distribution peaks at 1.5 V/A.
Average shear stress is not the answer the exam wants.

Step-by-step solution

Area
$A = 200 (400) = 80, 000 m m^{2}$
Average shear
$V / A = 90, 000/80, 000 = 1.13 M P a$
Rectangular peak
$τ_{m} a x = 1.5 V / A$
Substitute
$τ_{m} a x = 1.5 (1.13)$
Result
$τ_{m} a x = 1.69 M P aa tt h e n e u t r a l a x i s$

Answer: τ_max = 1.69 MPa

Why the other options are there

1.13 MPa (average reported)
2.25 MPa (factor 2 used instead of 1.5)

Reference: FE Reference Handbook — Mechanics of Materials — Transverse shear

Example 3

Maximum deflection of a uniformly loaded simple beam — Stresses in Beams

A simply supported steel beam spans 22 ft and carries a uniform service load of 1.1 kip/ft. With E = 29,000 ksi and I = 571 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 1.1 kip/ft
L = 22 ft
E = 29,000 ksi
I = 571 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(22)²/8 = 66.55 kip·ft
Unit conversion
$w = 1.1/12 = 0.0917 k i p / in, L = 264 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.0917) (264)^{4} / [384 (29, 000) (571)]$
Evaluate
$Δ = 0.3501 in$
Limit
$L /360 = 264/360 = 0.733 in \to a cce pt ab l e$

Answer: M = 66.6 kip·ft, Δ = 0.350 in versus a 0.733 in limit

Why the other options are there

0.00020 in (units not converted)
0.0700 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Stresses in Beams

Example 4

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

A simply supported steel beam spans 31 ft and carries a uniform service load of 2.5 kip/ft. With E = 29,000 ksi and I = 1320 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 2.5 kip/ft
L = 31 ft
E = 29,000 ksi
I = 1320 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 = 2.5(31)²/8 = 300.3 kip·ft
Unit conversion
$w = 2.5/12 = 0.2083 k i p / in, L = 372 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.2083) (372)^{4} / [384 (29, 000) (1320)]$
Evaluate
$Δ = 1.3571 in$
Limit
$L /360 = 372/360 = 1.033 in \to t oo f l e x ib l e$

Answer: M = 300.3 kip·ft, Δ = 1.357 in versus a 1.033 in limit

Why the other options are there

0.00079 in (units not converted)
0.2714 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Stresses in Beams

Example 5

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

A simply supported steel beam spans 29 ft and carries a uniform service load of 1.1 kip/ft. With E = 29,000 ksi and I = 1010 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 1.1 kip/ft
L = 29 ft
E = 29,000 ksi
I = 1010 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(29)²/8 = 115.6 kip·ft
Unit conversion
$w = 1.1/12 = 0.0917 k i p / in, L = 348 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.0917) (348)^{4} / [384 (29, 000) (1010)]$
Evaluate
$Δ = 0.5977 in$
Limit
$L /360 = 348/360 = 0.967 in \to a cce pt ab l e$

Answer: M = 115.6 kip·ft, Δ = 0.598 in versus a 0.967 in limit

Why the other options are there

0.00035 in (units not converted)
0.1195 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Stresses in Beams

Example 6

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

A simply supported steel beam spans 16 ft and carries a uniform service load of 1.3 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 = 1.3 kip/ft
L = 16 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 = 1.3(16)²/8 = 41.60 kip·ft
Unit conversion
$w = 1.3/12 = 0.1083 k i p / in, L = 192 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.1083) (192)^{4} / [384 (29, 000) (524)]$
Evaluate
$Δ = 0.1261 in$
Limit
$L /360 = 192/360 = 0.533 in \to a cce pt ab l e$

Answer: M = 41.6 kip·ft, Δ = 0.126 in versus a 0.533 in limit

Why the other options are there

0.00007 in (units not converted)
0.0252 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Stresses in Beams

Example 7

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

A simply supported steel beam spans 34 ft and carries a uniform service load of 1.1 kip/ft. With E = 29,000 ksi and I = 669 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 1.1 kip/ft
L = 34 ft
E = 29,000 ksi
I = 669 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(34)²/8 = 159.0 kip·ft
Unit conversion
$w = 1.1/12 = 0.0917 k i p / in, L = 408 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.0917) (408)^{4} / [384 (29, 000) (669)]$
Evaluate
$Δ = 1.7048 in$
Limit
$L /360 = 408/360 = 1.133 in \to t oo f l e x ib l e$

Answer: M = 159.0 kip·ft, Δ = 1.705 in versus a 1.133 in limit

Why the other options are there

0.00099 in (units not converted)
0.3410 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Stresses in Beams

Example 8

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

A simply supported steel beam spans 34 ft and carries a uniform service load of 3.6 kip/ft. With E = 29,000 ksi and I = 1130 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 3.6 kip/ft
L = 34 ft
E = 29,000 ksi
I = 1130 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(34)²/8 = 520.2 kip·ft
Unit conversion
$w = 3.6/12 = 0.3000 k i p / in, L = 408 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.3000) (408)^{4} / [384 (29, 000) (1130)]$
Evaluate
$Δ = 3.3031 in$
Limit
$L /360 = 408/360 = 1.133 in \to t oo f l e x ib l e$

Answer: M = 520.2 kip·ft, Δ = 3.303 in versus a 1.133 in limit

Why the other options are there

0.00191 in (units not converted)
0.6606 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Stresses in Beams

Example 9

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

A simply supported steel beam spans 40 ft and carries a uniform service load of 3.5 kip/ft. With E = 29,000 ksi and I = 1438 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 3.5 kip/ft
L = 40 ft
E = 29,000 ksi
I = 1438 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 = 3.5(40)²/8 = 700.0 kip·ft
Unit conversion
$w = 3.5/12 = 0.2917 k i p / in, L = 480 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.2917) (480)^{4} / [384 (29, 000) (1438)]$
Evaluate
$Δ = 4.8343 in$
Limit
$L /360 = 480/360 = 1.333 in \to t oo f l e x ib l e$

Answer: M = 700.0 kip·ft, Δ = 4.834 in versus a 1.333 in limit

Why the other options are there

0.00280 in (units not converted)
0.9669 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Stresses in Beams

Example 10

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

A simply supported steel beam spans 19 ft and carries a uniform service load of 3.1 kip/ft. With E = 29,000 ksi and I = 815 in⁴, compute the maximum moment, the midspan deflection, and compare it with an L/360 limit.

Given

w = 3.1 kip/ft
L = 19 ft
E = 29,000 ksi
I = 815 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(19)²/8 = 139.9 kip·ft
Unit conversion
$w = 3.1/12 = 0.2583 k i p / in, L = 228 in$
Formula
$Δ = 5 w L^{4} / (384 E I)$
Substituting
$Δ = 5 (0.2583) (228)^{4} / [384 (29, 000) (815)]$
Evaluate
$Δ = 0.3846 in$
Limit
$L /360 = 228/360 = 0.633 in \to a cce pt ab l e$

Answer: M = 139.9 kip·ft, Δ = 0.385 in versus a 0.633 in limit

Why the other options are there

0.00022 in (units not converted)
0.0769 in (coefficient dropped)

Reference: FE Reference Handbook — Mechanics of Materials → Stresses in 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

Stresses in Beams contains 15 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.