Mohr's Circle – Stress, 2D — FE Civil Lecture

Mohr's Circle – Stress, 2D

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 Mohr's Circle – Stress, 2D 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 mohr's circle – stress, 2d 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. Mohr's Circle – Stress, 2D 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: mohr's circle – stress, 2d.
Wikimedia Commons, public domain

Mechanics of Materials — Mohr's Circle – Stress, 2D: 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

σa	Quantity produced by "σa = C + R" — read its definition and unit from the handbook line directly above the equation.
σb	Quantity produced by "σb = C – R" — read its definition and unit from the handbook line directly above the equation.
σ1 − σ3	Quantity produced by "σ1 − σ3" — read its definition and unit from the handbook line directly above the equation.
τmax	Quantity produced by "τmax = 2 ." — 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.

To construct a Mohr's circle, the following sign conventions are used.
1. Tensile normal stress components are plotted on the horizontal axis and are considered positive. Compressive normal stress
components are negative.
2. For constructing Mohr's circle only, shearing stresses are plotted above the normal stress axis when the pair of shearing
stresses, acting on opposite and parallel faces of an element, forms a clockwise couple. Shearing stresses are plotted below
the normal axis when the shear stresses form a counterclockwise couple.
The circle drawn with the center on the normal stress (horizontal) axis with center, C, and radius, R, where
d n + x 2xy
vx + v y vx - v y
2 2
The two nonzero principal stresses are then:
in R
y, xy
b a
x, xy
Crandall, S.H., and N.C. Dahl, An Introduction to Mechanics of Solids, McGraw-Hill, New York, 1959.

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

Principal stresses from a plane stress state

At a point, σx = 80 MPa, σy = −20 MPa and τxy = 30 MPa. Find the principal stresses and the maximum in-plane shear.

Given

σx = 80 MPa
σy = −20 MPa
τxy = 30 MPa

Find

σ₁, σ₂, τ_max

Start with the thinking

Average stress is the circle centre; the radius is the maximum shear.
Sign of τxy affects the angle, not the magnitudes.

Figure for Principal stresses from a plane stress state

Step-by-step solution

Centre — σ_avg = (σx + σy)/2 = (80 − 20)/2 = 30.0 MPa
Half-difference — (σx − σy)/2 = 50.0 MPa
Radius — R = √(50² + 30²) = √3,400 = 58.3 MPa
Principal stresses
$σ_{1} = 30.0 + 58.3 = 88.3 M P aan d σ_{2} = 30.0 - 58.3 = - 28.3 M P a$
Maximum shear
$τ_{m} a x = R = 58.3 M P a$

Answer: σ₁ = 88.3 MPa, σ₂ = −28.3 MPa, τ_max = 58.3 MPa

Why the other options are there

σ₁ = 110 MPa (σy sign dropped)
τ_max = 30 MPa (applied shear reported)

Reference: FE Reference Handbook — Mechanics of Materials — Mohr's circle

Example 2

Principal stresses from a plane-stress state — Mohr's Circle – Stress, 2D

An element carries σx = 6 ksi, σy = 4 ksi and τxy = 8 ksi. Find the principal stresses and the maximum in-plane shear.

Given

σx = 6 ksi
σy = 4 ksi
τxy = 8 ksi

Find

σ₁, σ₂ and τ_max

Start with the thinking

The centre of Mohr's circle is the average normal stress.
The radius is the maximum in-plane shear stress.

Figure for Principal stresses from a plane-stress state — Mohr's Circle – Stress, 2D

Step-by-step solution

Centre — σ_avg = (σx + σy)/2
Substituting
$σ_{a} v g = (6 + 4) /2 = 5.00 k s i$
Radius — R = √[((σx − σy)/2)² + τxy²]
Substituting
$R = (1.00)^{2} + (8)^{2} = 8.06 k s i$
Principal
$σ_{1} = 13.06 k s i, σ_{2} = - 3.06 k s i$
Max shear
$τ_{m} a x = R = 8.06 k s i$

Answer: σ₁ = 13.06 ksi, σ₂ = -3.06 ksi, τ_max = 8.06 ksi

Why the other options are there

σ₁ = 14 ksi (shear added directly)
τ_max = 4.03 ksi (radius halved)

Reference: FE Reference Handbook — Mechanics of Materials → Mohr's Circle – Stress, 2D

Example 3

Principal stresses from a plane-stress state — Mohr's Circle – Stress, 2D (2)

An element carries σx = 16 ksi, σy = -8 ksi and τxy = 10 ksi. Find the principal stresses and the maximum in-plane shear.

Given

σx = 16 ksi
σy = -8 ksi
τxy = 10 ksi

Find

σ₁, σ₂ and τ_max

Start with the thinking

The centre of Mohr's circle is the average normal stress.
The radius is the maximum in-plane shear stress.

Figure for Principal stresses from a plane-stress state — Mohr's Circle – Stress, 2D (2)

Step-by-step solution

Centre — σ_avg = (σx + σy)/2
Substituting
$σ_{a} v g = (16 + - 8) /2 = 4.00 k s i$
Radius — R = √[((σx − σy)/2)² + τxy²]
Substituting
$R = (12.00)^{2} + (10)^{2} = 15.62 k s i$
Principal
$σ_{1} = 19.62 k s i, σ_{2} = - 11.62 k s i$
Max shear
$τ_{m} a x = R = 15.62 k s i$

Answer: σ₁ = 19.62 ksi, σ₂ = -11.62 ksi, τ_max = 15.62 ksi

Why the other options are there

σ₁ = 26 ksi (shear added directly)
τ_max = 7.81 ksi (radius halved)

Reference: FE Reference Handbook — Mechanics of Materials → Mohr's Circle – Stress, 2D

Example 4

Principal stresses from a plane-stress state — Mohr's Circle – Stress, 2D (3)

An element carries σx = 14 ksi, σy = 8 ksi and τxy = 7 ksi. Find the principal stresses and the maximum in-plane shear.

Given

σx = 14 ksi
σy = 8 ksi
τxy = 7 ksi

Find

σ₁, σ₂ and τ_max

Start with the thinking

The centre of Mohr's circle is the average normal stress.
The radius is the maximum in-plane shear stress.

Figure for Principal stresses from a plane-stress state — Mohr's Circle – Stress, 2D (3)

Step-by-step solution

Centre — σ_avg = (σx + σy)/2
Substituting
$σ_{a} v g = (14 + 8) /2 = 11.00 k s i$
Radius — R = √[((σx − σy)/2)² + τxy²]
Substituting
$R = (3.00)^{2} + (7)^{2} = 7.62 k s i$
Principal
$σ_{1} = 18.62 k s i, σ_{2} = 3.38 k s i$
Max shear
$τ_{m} a x = R = 7.62 k s i$

Answer: σ₁ = 18.62 ksi, σ₂ = 3.38 ksi, τ_max = 7.62 ksi

Why the other options are there

σ₁ = 21 ksi (shear added directly)
τ_max = 3.81 ksi (radius halved)

Reference: FE Reference Handbook — Mechanics of Materials → Mohr's Circle – Stress, 2D

Example 5

Principal stresses from a plane-stress state — Mohr's Circle – Stress, 2D (4)

An element carries σx = 16 ksi, σy = -8 ksi and τxy = 4 ksi. Find the principal stresses and the maximum in-plane shear.

Given

σx = 16 ksi
σy = -8 ksi
τxy = 4 ksi

Find

σ₁, σ₂ and τ_max

Start with the thinking

The centre of Mohr's circle is the average normal stress.
The radius is the maximum in-plane shear stress.

Figure for Principal stresses from a plane-stress state — Mohr's Circle – Stress, 2D (4)

Step-by-step solution

Centre — σ_avg = (σx + σy)/2
Substituting
$σ_{a} v g = (16 + - 8) /2 = 4.00 k s i$
Radius — R = √[((σx − σy)/2)² + τxy²]
Substituting
$R = (12.00)^{2} + (4)^{2} = 12.65 k s i$
Principal
$σ_{1} = 16.65 k s i, σ_{2} = - 8.65 k s i$
Max shear
$τ_{m} a x = R = 12.65 k s i$

Answer: σ₁ = 16.65 ksi, σ₂ = -8.65 ksi, τ_max = 12.65 ksi

Why the other options are there

σ₁ = 20 ksi (shear added directly)
τ_max = 6.32 ksi (radius halved)

Reference: FE Reference Handbook — Mechanics of Materials → Mohr's Circle – Stress, 2D

Example 6

Principal stresses from a plane-stress state — Mohr's Circle – Stress, 2D (5)

An element carries σx = 19 ksi, σy = 7 ksi and τxy = 5 ksi. Find the principal stresses and the maximum in-plane shear.

Given

σx = 19 ksi
σy = 7 ksi
τxy = 5 ksi

Find

σ₁, σ₂ and τ_max

Start with the thinking

The centre of Mohr's circle is the average normal stress.
The radius is the maximum in-plane shear stress.

Figure for Principal stresses from a plane-stress state — Mohr's Circle – Stress, 2D (5)

Step-by-step solution

Centre — σ_avg = (σx + σy)/2
Substituting
$σ_{a} v g = (19 + 7) /2 = 13.00 k s i$
Radius — R = √[((σx − σy)/2)² + τxy²]
Substituting
$R = (6.00)^{2} + (5)^{2} = 7.81 k s i$
Principal
$σ_{1} = 20.81 k s i, σ_{2} = 5.19 k s i$
Max shear
$τ_{m} a x = R = 7.81 k s i$

Answer: σ₁ = 20.81 ksi, σ₂ = 5.19 ksi, τ_max = 7.81 ksi

Why the other options are there

σ₁ = 24 ksi (shear added directly)
τ_max = 3.91 ksi (radius halved)

Reference: FE Reference Handbook — Mechanics of Materials → Mohr's Circle – Stress, 2D

Example 7

Principal stresses from a plane-stress state — Mohr's Circle – Stress, 2D (6)

An element carries σx = 7 ksi, σy = 0 ksi and τxy = 13 ksi. Find the principal stresses and the maximum in-plane shear.

Given

σx = 7 ksi
σy = 0 ksi
τxy = 13 ksi

Find

σ₁, σ₂ and τ_max

Start with the thinking

The centre of Mohr's circle is the average normal stress.
The radius is the maximum in-plane shear stress.

Figure for Principal stresses from a plane-stress state — Mohr's Circle – Stress, 2D (6)

Step-by-step solution

Centre — σ_avg = (σx + σy)/2
Substituting
$σ_{a} v g = (7 + 0) /2 = 3.50 k s i$
Radius — R = √[((σx − σy)/2)² + τxy²]
Substituting
$R = (3.50)^{2} + (13)^{2} = 13.46 k s i$
Principal
$σ_{1} = 16.96 k s i, σ_{2} = - 9.96 k s i$
Max shear
$τ_{m} a x = R = 13.46 k s i$

Answer: σ₁ = 16.96 ksi, σ₂ = -9.96 ksi, τ_max = 13.46 ksi

Why the other options are there

σ₁ = 20 ksi (shear added directly)
τ_max = 6.73 ksi (radius halved)

Reference: FE Reference Handbook — Mechanics of Materials → Mohr's Circle – Stress, 2D

Example 8

Principal stresses from a plane-stress state — Mohr's Circle – Stress, 2D (7)

An element carries σx = 15 ksi, σy = 5 ksi and τxy = 11 ksi. Find the principal stresses and the maximum in-plane shear.

Given

σx = 15 ksi
σy = 5 ksi
τxy = 11 ksi

Find

σ₁, σ₂ and τ_max

Start with the thinking

The centre of Mohr's circle is the average normal stress.
The radius is the maximum in-plane shear stress.

Figure for Principal stresses from a plane-stress state — Mohr's Circle – Stress, 2D (7)

Step-by-step solution

Centre — σ_avg = (σx + σy)/2
Substituting
$σ_{a} v g = (15 + 5) /2 = 10.00 k s i$
Radius — R = √[((σx − σy)/2)² + τxy²]
Substituting
$R = (5.00)^{2} + (11)^{2} = 12.08 k s i$
Principal
$σ_{1} = 22.08 k s i, σ_{2} = - 2.08 k s i$
Max shear
$τ_{m} a x = R = 12.08 k s i$

Answer: σ₁ = 22.08 ksi, σ₂ = -2.08 ksi, τ_max = 12.08 ksi

Why the other options are there

σ₁ = 26 ksi (shear added directly)
τ_max = 6.04 ksi (radius halved)

Reference: FE Reference Handbook — Mechanics of Materials → Mohr's Circle – Stress, 2D

Example 9

Principal stresses from a plane-stress state — Mohr's Circle – Stress, 2D (8)

An element carries σx = 20 ksi, σy = 4 ksi and τxy = 12 ksi. Find the principal stresses and the maximum in-plane shear.

Given

σx = 20 ksi
σy = 4 ksi
τxy = 12 ksi

Find

σ₁, σ₂ and τ_max

Start with the thinking

The centre of Mohr's circle is the average normal stress.
The radius is the maximum in-plane shear stress.

Figure for Principal stresses from a plane-stress state — Mohr's Circle – Stress, 2D (8)

Step-by-step solution

Centre — σ_avg = (σx + σy)/2
Substituting
$σ_{a} v g = (20 + 4) /2 = 12.00 k s i$
Radius — R = √[((σx − σy)/2)² + τxy²]
Substituting
$R = (8.00)^{2} + (12)^{2} = 14.42 k s i$
Principal
$σ_{1} = 26.42 k s i, σ_{2} = - 2.42 k s i$
Max shear
$τ_{m} a x = R = 14.42 k s i$

Answer: σ₁ = 26.42 ksi, σ₂ = -2.42 ksi, τ_max = 14.42 ksi

Why the other options are there

σ₁ = 32 ksi (shear added directly)
τ_max = 7.21 ksi (radius halved)

Reference: FE Reference Handbook — Mechanics of Materials → Mohr's Circle – Stress, 2D

Example 10

Principal stresses from a plane-stress state — Mohr's Circle – Stress, 2D (9)

An element carries σx = 9 ksi, σy = 6 ksi and τxy = 6 ksi. Find the principal stresses and the maximum in-plane shear.

Given

σx = 9 ksi
σy = 6 ksi
τxy = 6 ksi

Find

σ₁, σ₂ and τ_max

Start with the thinking

The centre of Mohr's circle is the average normal stress.
The radius is the maximum in-plane shear stress.

Figure for Principal stresses from a plane-stress state — Mohr's Circle – Stress, 2D (9)

Step-by-step solution

Centre — σ_avg = (σx + σy)/2
Substituting
$σ_{a} v g = (9 + 6) /2 = 7.50 k s i$
Radius — R = √[((σx − σy)/2)² + τxy²]
Substituting
$R = (1.50)^{2} + (6)^{2} = 6.18 k s i$
Principal
$σ_{1} = 13.68 k s i, σ_{2} = 1.32 k s i$
Max shear
$τ_{m} a x = R = 6.18 k s i$

Answer: σ₁ = 13.68 ksi, σ₂ = 1.32 ksi, τ_max = 6.18 ksi

Why the other options are there

σ₁ = 15 ksi (shear added directly)
τ_max = 3.09 ksi (radius halved)

Reference: FE Reference Handbook — Mechanics of Materials → Mohr's Circle – Stress, 2D

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

Mohr's Circle – Stress, 2D 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.