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Mechanics of Materials · Chapter 7 of 10 · Advanced

Stress Transformation and Mohr's Circle

The same stress state looks different on a rotated plane. Transformation finds the orientation of the largest stress, the principal stress that actually drives failure, and Mohr's circle shows it at a glance.

Readiness check

This chapter rotates a stress state, so it leans on trigonometry. Tick only what you can do closed-notes.

Recall normal and shear stress on an element.
Use sine, cosine, and the inverse tangent.
Find the hypotenuse of a right triangle.
Read coordinates on a circle.
Recall bending and torsion stresses.

0 or 1 weak itemsContinue with this chapter.

2 weak itemsRefresh trigonometry in trigonometry.

3 or more weak itemsRevisit the stress element in Chapter 1 first.

The core idea

Stress at a point depends on the plane you look at; rotating the element finds the principal planes, where shear vanishes and the normal stress is largest or smallest.

σ_avg = (σ_x+σ_y)/2R = √[((σ_x−σ_y)/2)² + τ_xy²]σ_1,2 = σ_avg ± R, τ_max = R

A point in a loaded part carries σ_x, σ_y, and τ_xy on a chosen element. Rotate the element and these values change, tracing a circle. The maximum and minimum normal stresses, the principal stresses σ₁ and σ₂, occur on planes with zero shear; the maximum shear stress equals the circle's radius R. Because materials fail on the plane of greatest stress, these extremes, not the values on your arbitrary axes, govern design.

The skill works when: you compute σ_avg and R and read off σ_1,2 = σ_avg ± R and τ_max = R.

The skill breaks down when: failure is judged from σ_x alone, ignoring a larger principal stress on a rotated plane.

The concept. Mohr's circle plots the stress state. Its center is the average stress, its radius is the maximum shear, and its crossings of the σ-axis are the principal stresses σ₁ and σ₂.

The skills, taught in order

Transformation is trigonometry on a stress element, and Mohr's circle is its picture. Five skills cover the state, the equations, the principal values, the circle, and its use.

7.1 Plane stress

At a point, the in-plane stress state is three numbers: normal stresses σ_x and σ_y and shear τ_xy, drawn on a small square element. Many real cases (free surfaces, thin members) are plane stress, with no stress on the third face.

7.2 The transformation equations

Rotate the element by θ and the stresses become σ_x' = σ_avg + ((σ_x−σ_y)/2)cos2θ + τ_xysin2θ, with a companion equation for τ_x'y'. They show every component varies sinusoidally with 2θ, which is why Mohr's circle works.

7.3 Principal stresses and maximum shear

The extreme normal stresses are σ_1,2 = σ_avg ± R, occurring where the shear is zero, on planes at θ_p given by tan2θ_p = 2τ_xy/(σ_x−σ_y). The maximum in-plane shear is τ_max = R, on planes 45° from the principal ones.

Quantity	Formula
Average stress	σ_avg = (σ_x+σ_y)/2
Radius	R = √[((σ_x−σ_y)/2)² + τ_xy²]
Principal stresses	σ_1,2 = σ_avg ± R
Maximum shear	τ_max = R

7.4 Mohr's circle

Plot σ horizontally and τ vertically. The circle has center (σ_avg, 0) and radius R. Points on the circle give the stresses on every plane; the σ-axis crossings are the principal stresses, and the top and bottom are ±τ_max. Angles on the circle are twice the physical rotation.

7.5 Application to combined states

A shaft under bending and torsion has, at its surface, a normal stress from bending and a shear from torsion. Transformation combines them into principal stresses and a maximum shear, the values a failure theory compares against the material's strength.

Engineering connection: brittle materials fail on the maximum normal (principal) stress, ductile ones on maximum shear; both come from transformation, feeding the combined-loading analysis of Chapter 8.

Worked example 1: principal stresses

A point is in plane stress with σ_x = 80 MPa, σ_y = −40 MPa, and τ_xy = 30 MPa. Find the principal stresses, the principal angle, and the maximum in-plane shear.

Figure 1. The element on the x-y axes (left) transforms to a principal element rotated 13.3°, where the shear is zero and the normal stresses reach 87.1 and −47.1 MPa.

ProblemFind σ₁, σ₂, θ_p, and τ_max for the state in Figure 1.
Given / findσ_x = 80, σ_y = −40, τ_xy = 30 MPa. Find the principal stresses, angle, and maximum shear.
AssumptionsPlane stress, standard sign convention (tension positive).
ModelCompute the average stress and the radius, then the principal stresses and angle.
Equationsσ_avg = (σ_x+σ_y)/2 R = √[((σ_x−σ_y)/2)² + τ_xy²] σ_1,2 = σ_avg ± R
Solveσ_avg = (80 − 40)/2 = 20 MPa. R = √[(60)² + (30)²] = √4500 = 67.1 MPa. So σ₁ = 87.1 MPa, σ₂ = −47.1 MPa, τ_max = R = 67.1 MPa. θ_p = ½ tan⁻¹(60/120) = 13.3°.
CheckThe principals straddle the average (20 ± 67.1), as they must, and σ₁ + σ₂ = 40 = σ_x + σ_y, an invariant of transformation, confirming the work.
ConclusionThe largest tension (87.1 MPa) exceeds σ_x (80), so designing on σ_x alone would be unconservative. Transformation finds the true worst stress and the plane it acts on.

Result. σ₁ = 87.1 MPa, σ₂ = −47.1 MPa, τ_max = 67.1 MPa at θ_p = 13.3°.

Worked example 2: a shaft under bending and torsion

At the surface of a shaft, bending gives a normal stress σ_x = 60 MPa and torsion gives a shear τ_xy = 40 MPa (σ_y = 0). Use Mohr's circle to find the principal stresses and the maximum shear.

Figure 2. The bending normal stress and torsion shear plot as a point; the circle through it gives σ₁ = 80, σ₂ = −20, and τ_max = 50 MPa, a clean 3-4-5 triangle.

ProblemFind the principal stresses and maximum shear for the shaft surface in Figure 2.
Given / findσ_x = 60 MPa (bending), σ_y = 0, τ_xy = 40 MPa (torsion). Find σ₁, σ₂, τ_max.
AssumptionsPlane stress at the surface; bending and torsion stresses computed elsewhere.
ModelCenter the circle at σ_avg, radius from the half-difference and the shear, then read the extremes.
Equationsσ_avg = 30 MPa R = √[(30)² + (40)²] = 50 σ_1,2 = 30 ± 50
Solveσ_avg = 60/2 = 30 MPa. R = √(30² + 40²) = 50 MPa. So σ₁ = 80 MPa, σ₂ = −20 MPa, and τ_max = 50 MPa.
CheckThe 30-40-50 right triangle makes the radius exactly 50. The maximum shear (50) exceeds the torsion shear alone (40), because bending adds to it, the reason combined loading is more severe than either part.
ConclusionCombined bending and torsion is the classic shaft case. Transformation yields the principal stress (for brittle shafts) and the maximum shear (for ductile ones), which Chapter 8 turns into a design check.

Result. σ₁ = 80 MPa, σ₂ = −20 MPa, τ_max = 50 MPa.

Misconceptions and diagnostics

Mistake	Symptom	Diagnostic question	Correction
Designing on σ_x	A larger principal stress missed	"Could a rotated plane have higher stress?"	Use σ₁ = σ_avg + R, the true maximum normal stress.
Angle confusion on Mohr	Principal direction off by a factor of 2	"Is this 2θ on the circle or θ on the part?"	Angles on Mohr's circle are twice the physical rotation.
Dropping τ_xy in R	Radius too small	"Did I include the shear term?"	R combines the half-difference and τ_xy in quadrature.
Forgetting shear adds to bending	Shaft underestimated	"Are bending and torsion combined?"	Transform the combined state; τ_max exceeds either alone.

Practice ladder

Level 1 · Direct skill

For σ_x = 50 MPa, σ_y = 10 MPa, τ_xy = 0, find the principal stresses.

Show answer

With no shear, the axes are already principal: σ₁ = 50, σ₂ = 10 MPa. R = 20, τ_max = 20 MPa at 45°.

Level 2 · Mixed concept

For pure shear τ_xy = 35 MPa (σ_x = σ_y = 0), find the principal stresses.

Show answer

σ_avg = 0, R = 35, so σ₁ = +35, σ₂ = −35 MPa at 45°. Pure shear is equivalent to equal tension and compression on planes rotated 45°, which is why brittle shafts crack on a helix.

Level 3 · Independent problem

A shaft has bending stress 90 MPa and torsion shear 30 MPa at its surface. Find σ₁ and τ_max.

Show answer

σ_avg = 45, R = √(45² + 30²) = √2925 = 54.1. σ₁ = 45 + 54.1 = 99.1 MPa, σ₂ = −9.1 MPa, τ_max = 54.1 MPa. Bending dominates here, so σ₁ is close to the bending stress.

Level 4 · Transfer to real engineering

Find a real part under combined stress (a pressurised pipe with axial load, a shaft with a pulley). Identify σ_x, σ_y, τ_xy, and compute the principal stresses and maximum shear.

What good work looks like

The three components identified from the loads, σ_avg and R computed, principal stresses and τ_max reported, and a note on which governs failure (principal for brittle, shear for ductile).

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that σ₁ + σ₂ equals σ_x + σ_y as an invariant."

"Give me five stress states; I will sketch each Mohr's circle before computing."

"Find the principal stresses." Building σ_avg and R yourself is the skill.

"Is this safe?" Choosing the governing stress (principal or shear) is the point.

Portfolio task

Take one combined-stress point (a shaft surface, a pressure-vessel wall) through transformation: find the principal stresses and τ_max, and state which failure measure governs.

Must include: σ_avg, R, σ_1,2, τ_max, and the invariant check σ₁+σ₂ = σ_x+σ_y.

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write the average stress and the radius.

σ_avg = (σ_x+σ_y)/2; R = √[((σ_x−σ_y)/2)² + τ_xy²].

2. Give the principal stresses and maximum shear.

σ_1,2 = σ_avg ± R; τ_max = R.

3. What is special about a principal plane?

The shear stress is zero and the normal stress is an extreme.

4. How do angles on Mohr's circle relate to the part?

They are twice the physical rotation (2θ on the circle).

5. Which stress governs failure for brittle vs ductile materials?

Maximum principal (normal) stress for brittle; maximum shear for ductile.

TodayFinish this quiz and Levels 1 and 2 of the ladder.

+1 dayRe-derive the principal stresses and a Mohr's circle from a blank page.

+3 daysOne general state and one shaft (bending plus torsion).

+7 daysApply transformation to combined loading, Chapter 8.

+30 daysConnect principal stress to failure theories in design.

Textbook mapping

Item	Mapping
Primary source	Beer, Johnston, DeWolf and Mazurek, Mechanics of Materials (6th ed), Chapter 7 (Transformations of Stress and Strain)
Cross-reference	Hibbeler, Ch. 9 · Gere and Goodno, Ch. 7
Core topics	7.1 Plane stress · 7.2 Transformation equations · 7.3 Principal stresses · 7.4 Mohr's circle · 7.5 Combined states
Engineering connection	Failure planes, brittle vs ductile criteria, and shaft analysis.
Read next	Chapter 8: Combined Loadings.