Shearing Force and Bending Moment Sign Conventions

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 Shearing Force and Bending Moment Sign Conventions 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 shearing force and bending moment sign conventions 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. Shearing Force and Bending Moment Sign Conventions 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: shearing force and bending moment sign conventions.
Wikimedia Commons, CC BY-SA 4.0

Mechanics of Materials — Shearing Force and Bending Moment Sign Conventions: 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

dV ^ x h	Quantity produced by "dV ^ x h" — read its definition and unit from the handbook line directly above the equation.
w ^ x h	Quantity produced by "w ^ x h =− dx" — read its definition and unit from the handbook line directly above the equation.
dM ^ x h	Quantity produced by "dM ^ x h" — read its definition and unit from the handbook line directly above the equation.
V	Quantity produced by "V = 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.

1. The bending moment is positive if it produces bending of the beam concave upward (compression in top fibers and tension
in bottom fibers).
2. The shearing force is positive if the right portion of the beam tends to shear downward with respect to the left.
POSITIVE BENDING NEGATIVE BENDING
POSITIVE SHEAR NEGATIVE SHEAR
Timoshenko, S., and Gleason H. MacCullough, Elements of Strengths of Materials, K. Van Nostrand Co./Wadsworth Publishing Co., 1949.
The relationship between the load (w), shear (V), and moment (M) equations are:
2 x

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 section — Shearing Force and Bending Moment Sign Conventions

A 9 in. × 19 in. rectangular beam carries a moment of 306 kip·ft. Find the maximum bending stress.

Given

M = 306 kip·ft
b = 9 in.
h = 19 in.

Find

σ_max

Start with the thinking

Flexure formula needs I and the extreme-fibre distance.
Convert kip·ft to kip·in before substituting.

Figure for Bending stress in a rectangular section — Shearing Force and Bending Moment Sign Conventions

Step-by-step solution

Moment of inertia
$I = b h^{3} /12 = 9 (19)^{3} /12 = 5, 144 i n^{4}$
Extreme fibre
$c = h /2 = 9.5 in$
Moment in kip·in — M = 306 × 12 = 3672 kip·in
Flexure
$σ = M c / I$
Substituting
$σ = 3672 (9.5) /5, 144 = 6.78 k s i$

Answer: σ_max ≈ 6.78 ksi

Why the other options are there

0.57 ksi (kip·ft not converted)
13.56 ksi (full depth used for c)

Reference: FE Reference Handbook — Mechanics of Materials → Shearing Force and Bending Moment Sign Conventions

Example 2

Bending stress in a rectangular section — Shearing Force and Bending Moment Sign Conventions (2)

A 10 in. × 25 in. rectangular beam carries a moment of 68 kip·ft. Find the maximum bending stress.

Given

M = 68 kip·ft
b = 10 in.
h = 25 in.

Find

σ_max

Start with the thinking

Flexure formula needs I and the extreme-fibre distance.
Convert kip·ft to kip·in before substituting.

Figure for Bending stress in a rectangular section — Shearing Force and Bending Moment Sign Conventions (2)

Step-by-step solution

Moment of inertia
$I = b h^{3} /12 = 10 (25)^{3} /12 = 13, 021 i n^{4}$
Extreme fibre
$c = h /2 = 12.5 in$
Moment in kip·in — M = 68 × 12 = 816 kip·in
Flexure
$σ = M c / I$
Substituting
$σ = 816 (12.5) /13, 021 = 0.78 k s i$

Answer: σ_max ≈ 0.78 ksi

Why the other options are there

0.07 ksi (kip·ft not converted)
1.57 ksi (full depth used for c)

Reference: FE Reference Handbook — Mechanics of Materials → Shearing Force and Bending Moment Sign Conventions

Example 3

Bending stress in a rectangular section — Shearing Force and Bending Moment Sign Conventions (3)

A 10 in. × 20 in. rectangular beam carries a moment of 135 kip·ft. Find the maximum bending stress.

Given

M = 135 kip·ft
b = 10 in.
h = 20 in.

Find

σ_max

Start with the thinking

Flexure formula needs I and the extreme-fibre distance.
Convert kip·ft to kip·in before substituting.

Figure for Bending stress in a rectangular section — Shearing Force and Bending Moment Sign Conventions (3)

Step-by-step solution

Moment of inertia
$I = b h^{3} /12 = 10 (20)^{3} /12 = 6, 667 i n^{4}$
Extreme fibre
$c = h /2 = 10.0 in$
Moment in kip·in — M = 135 × 12 = 1620 kip·in
Flexure
$σ = M c / I$
Substituting
$σ = 1620 (10.0) /6, 667 = 2.43 k s i$

Answer: σ_max ≈ 2.43 ksi

Why the other options are there

0.20 ksi (kip·ft not converted)
4.86 ksi (full depth used for c)

Reference: FE Reference Handbook — Mechanics of Materials → Shearing Force and Bending Moment Sign Conventions

Example 4

Bending stress in a rectangular section — Shearing Force and Bending Moment Sign Conventions (4)

A 9 in. × 28 in. rectangular beam carries a moment of 368 kip·ft. Find the maximum bending stress.

Given

M = 368 kip·ft
b = 9 in.
h = 28 in.

Find

σ_max

Start with the thinking

Flexure formula needs I and the extreme-fibre distance.
Convert kip·ft to kip·in before substituting.

Figure for Bending stress in a rectangular section — Shearing Force and Bending Moment Sign Conventions (4)

Step-by-step solution

Moment of inertia
$I = b h^{3} /12 = 9 (28)^{3} /12 = 16, 464 i n^{4}$
Extreme fibre
$c = h /2 = 14.0 in$
Moment in kip·in — M = 368 × 12 = 4416 kip·in
Flexure
$σ = M c / I$
Substituting
$σ = 4416 (14.0) /16, 464 = 3.76 k s i$

Answer: σ_max ≈ 3.76 ksi

Why the other options are there

0.31 ksi (kip·ft not converted)
7.51 ksi (full depth used for c)

Reference: FE Reference Handbook — Mechanics of Materials → Shearing Force and Bending Moment Sign Conventions

Example 5

Bending stress in a rectangular section — Shearing Force and Bending Moment Sign Conventions (5)

A 8 in. × 13 in. rectangular beam carries a moment of 239 kip·ft. Find the maximum bending stress.

Given

M = 239 kip·ft
b = 8 in.
h = 13 in.

Find

σ_max

Start with the thinking

Flexure formula needs I and the extreme-fibre distance.
Convert kip·ft to kip·in before substituting.

Figure for Bending stress in a rectangular section — Shearing Force and Bending Moment Sign Conventions (5)

Step-by-step solution

Moment of inertia
$I = b h^{3} /12 = 8 (13)^{3} /12 = 1, 465 i n^{4}$
Extreme fibre
$c = h /2 = 6.5 in$
Moment in kip·in — M = 239 × 12 = 2868 kip·in
Flexure
$σ = M c / I$
Substituting
$σ = 2868 (6.5) /1, 465 = 12.73 k s i$

Answer: σ_max ≈ 12.73 ksi

Why the other options are there

1.06 ksi (kip·ft not converted)
25.46 ksi (full depth used for c)

Reference: FE Reference Handbook — Mechanics of Materials → Shearing Force and Bending Moment Sign Conventions

Example 6

Bending stress in a rectangular section — Shearing Force and Bending Moment Sign Conventions (6)

A 6 in. × 28 in. rectangular beam carries a moment of 138 kip·ft. Find the maximum bending stress.

Given

M = 138 kip·ft
b = 6 in.
h = 28 in.

Find

σ_max

Start with the thinking

Flexure formula needs I and the extreme-fibre distance.
Convert kip·ft to kip·in before substituting.

Figure for Bending stress in a rectangular section — Shearing Force and Bending Moment Sign Conventions (6)

Step-by-step solution

Moment of inertia
$I = b h^{3} /12 = 6 (28)^{3} /12 = 10, 976 i n^{4}$
Extreme fibre
$c = h /2 = 14.0 in$
Moment in kip·in — M = 138 × 12 = 1656 kip·in
Flexure
$σ = M c / I$
Substituting
$σ = 1656 (14.0) /10, 976 = 2.11 k s i$

Answer: σ_max ≈ 2.11 ksi

Why the other options are there

0.18 ksi (kip·ft not converted)
4.22 ksi (full depth used for c)

Reference: FE Reference Handbook — Mechanics of Materials → Shearing Force and Bending Moment Sign Conventions

Example 7

Bending stress in a rectangular section — Shearing Force and Bending Moment Sign Conventions (7)

A 13 in. × 26 in. rectangular beam carries a moment of 261 kip·ft. Find the maximum bending stress.

Given

M = 261 kip·ft
b = 13 in.
h = 26 in.

Find

σ_max

Start with the thinking

Flexure formula needs I and the extreme-fibre distance.
Convert kip·ft to kip·in before substituting.

Figure for Bending stress in a rectangular section — Shearing Force and Bending Moment Sign Conventions (7)

Step-by-step solution

Moment of inertia
$I = b h^{3} /12 = 13 (26)^{3} /12 = 19, 041 i n^{4}$
Extreme fibre
$c = h /2 = 13.0 in$
Moment in kip·in — M = 261 × 12 = 3132 kip·in
Flexure
$σ = M c / I$
Substituting
$σ = 3132 (13.0) /19, 041 = 2.14 k s i$

Answer: σ_max ≈ 2.14 ksi

Why the other options are there

0.18 ksi (kip·ft not converted)
4.28 ksi (full depth used for c)

Reference: FE Reference Handbook — Mechanics of Materials → Shearing Force and Bending Moment Sign Conventions

Example 8

Bending stress in a rectangular section — Shearing Force and Bending Moment Sign Conventions (8)

A 11 in. × 19 in. rectangular beam carries a moment of 365 kip·ft. Find the maximum bending stress.

Given

M = 365 kip·ft
b = 11 in.
h = 19 in.

Find

σ_max

Start with the thinking

Flexure formula needs I and the extreme-fibre distance.
Convert kip·ft to kip·in before substituting.

Figure for Bending stress in a rectangular section — Shearing Force and Bending Moment Sign Conventions (8)

Step-by-step solution

Moment of inertia
$I = b h^{3} /12 = 11 (19)^{3} /12 = 6, 287 i n^{4}$
Extreme fibre
$c = h /2 = 9.5 in$
Moment in kip·in — M = 365 × 12 = 4380 kip·in
Flexure
$σ = M c / I$
Substituting
$σ = 4380 (9.5) /6, 287 = 6.62 k s i$

Answer: σ_max ≈ 6.62 ksi

Why the other options are there

0.55 ksi (kip·ft not converted)
13.24 ksi (full depth used for c)

Reference: FE Reference Handbook — Mechanics of Materials → Shearing Force and Bending Moment Sign Conventions

Example 9

Bending stress in a rectangular section — Shearing Force and Bending Moment Sign Conventions (9)

A 7 in. × 14 in. rectangular beam carries a moment of 91 kip·ft. Find the maximum bending stress.

Given

M = 91 kip·ft
b = 7 in.
h = 14 in.

Find

σ_max

Start with the thinking

Flexure formula needs I and the extreme-fibre distance.
Convert kip·ft to kip·in before substituting.

Figure for Bending stress in a rectangular section — Shearing Force and Bending Moment Sign Conventions (9)

Step-by-step solution

Moment of inertia
$I = b h^{3} /12 = 7 (14)^{3} /12 = 1, 601 i n^{4}$
Extreme fibre
$c = h /2 = 7.0 in$
Moment in kip·in — M = 91 × 12 = 1092 kip·in
Flexure
$σ = M c / I$
Substituting
$σ = 1092 (7.0) /1, 601 = 4.78 k s i$

Answer: σ_max ≈ 4.78 ksi

Why the other options are there

0.40 ksi (kip·ft not converted)
9.55 ksi (full depth used for c)

Reference: FE Reference Handbook — Mechanics of Materials → Shearing Force and Bending Moment Sign Conventions

Example 10

Bending stress in a rectangular section — Shearing Force and Bending Moment Sign Conventions (10)

A 8 in. × 23 in. rectangular beam carries a moment of 158 kip·ft. Find the maximum bending stress.

Given

M = 158 kip·ft
b = 8 in.
h = 23 in.

Find

σ_max

Start with the thinking

Flexure formula needs I and the extreme-fibre distance.
Convert kip·ft to kip·in before substituting.

Figure for Bending stress in a rectangular section — Shearing Force and Bending Moment Sign Conventions (10)

Step-by-step solution

Moment of inertia
$I = b h^{3} /12 = 8 (23)^{3} /12 = 8, 111 i n^{4}$
Extreme fibre
$c = h /2 = 11.5 in$
Moment in kip·in — M = 158 × 12 = 1896 kip·in
Flexure
$σ = M c / I$
Substituting
$σ = 1896 (11.5) /8, 111 = 2.69 k s i$

Answer: σ_max ≈ 2.69 ksi

Why the other options are there

0.22 ksi (kip·ft not converted)
5.38 ksi (full depth used for c)

Reference: FE Reference Handbook — Mechanics of Materials → Shearing Force and Bending Moment Sign Conventions

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

Shearing Force and Bending Moment Sign Conventions 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.