Mechanics of Materials · Chapter 9

Stress Transformation

How the stress components at a point change with the element's orientation — principal stresses, maximum shear, and Mohr's circle.

Chapter Objectives

Explain why one physical stress state has different components on rotated planes.
Select a stress transformation, principal-stress, or Mohr-circle method from the requested result.
Calculate transformed stresses and orientations using a declared coordinate and shear convention.
Distinguish maximum in-plane shear from absolute maximum shear in a plane-stress part.
Verify results with invariants, complementary shear, and a material-appropriate failure check.

9.1Plane-Stress Transformation

Plane Stress

The general state of stress at a point has six components, but most engineering loadings are coplanar, leaving only two normal components and one shear component: \(\sigma_x\), \(\sigma_y\), \(\tau_{xy}\), acting on four faces of the element. This is plane stress.

Why Transformation?

The same physical state of stress at a point looks different on elements of different orientation: an element rotated by \(\theta\) carries different values \(\sigma_{x'}\), \(\sigma_{y'}\), \(\tau_{x'y'}\) that are equivalent to the originals. Stress is force per area on a particular plane, not an internal force resultant. Both the traction direction and the plane normal rotate, producing the \(2\theta\) terms.

9.2General Equations of Plane-Stress Transformation

Sign Convention

A stress component is positive when it acts in the positive coordinate direction on the face whose outward normal is positive (so \(\sigma\) positive = tension; \(\tau_{xy}\) positive = acting in +y on the +x face). The angle θ runs from the +x axis to the +x′ axis, positive counterclockwise (right-hand rule about z).

First isolate a differential wedge. Tractions on its faces are stress times the projected areas; force equilibrium in \(x'\) and \(y'\), followed by division by the inclined area, gives the equations below. They apply to a small, continuous region away from a notch, contact edge, or load introduction where plane stress is a defensible surface idealization.

Plane-Stress Transformation Equations

\[ \sigma_{x'} = \frac{\sigma_x + \sigma_y}{2} + \frac{\sigma_x - \sigma_y}{2}\cos 2\theta + \tau_{xy}\sin 2\theta \] \[ \tau_{x'y'} = -\frac{\sigma_x - \sigma_y}{2}\sin 2\theta + \tau_{xy}\cos 2\theta \] \[ \sigma_{y'} = \frac{\sigma_x + \sigma_y}{2} - \frac{\sigma_x - \sigma_y}{2}\cos 2\theta - \tau_{xy}\sin 2\theta \]

\(\sigma_{y'}\) follows from the first equation with θ + 90° substituted.
A useful check: \(\sigma_{x'} + \sigma_{y'} = \sigma_x + \sigma_y\) — the sum of normal stresses is invariant under rotation.

Problem-solving strategy

Use equilibrium of the whole member and a cut to obtain the local \(\sigma_x,\sigma_y,\tau_{xy}\); transformation does not replace that step.
Draw the positive faces, state tension-positive normal stress and positive \(\theta\) counterclockwise, then identify the requested physical plane.
Use direct equations for one specified angle; use principal relations or Mohr's circle when an extreme value or orientation is requested.
Sort principal values with \(\sigma_1\geq\sigma_2\), substitute an angle back to pair it with its stress, and compare the third stress when absolute shear matters.
Audit units, trace invariance, zero shear on principal planes, and the relevant yield or fracture limit.

9.3Principal Stresses and Maximum In-Plane Shear Stress

Since \(\sigma_{x'}\) and \(\tau_{x'y'}\) depend only on θ, the orientations that extremize them follow from setting the derivatives to zero.

In-Plane Principal Stresses and Principal Planes

\[ \tan 2\theta_p = \frac{\tau_{xy}}{(\sigma_x - \sigma_y)/2} \qquad\qquad \sigma_{1,2} = \frac{\sigma_x + \sigma_y}{2} \pm \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} \]

\(\sigma_1 \ge \sigma_2\) are the principal stresses — the maximum and minimum normal stresses at the point.
The two roots \(\theta_{p1}, \theta_{p2}\) are 90° apart, and no shear stress acts on the principal planes.

Maximum In-Plane Shear Stress and Average Normal Stress

\[ \tan 2\theta_s = \frac{-(\sigma_x - \sigma_y)/2}{\tau_{xy}} \qquad \tau_{\substack{max\\ in\text{-}plane}} = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} \qquad \sigma_{avg} = \frac{\sigma_x + \sigma_y}{2} \]

\(\tan 2\theta_s\) is the negative reciprocal of \(\tan 2\theta_p\): the maximum-shear element sits 45° from the principal element.
Unlike the principal planes, the planes of maximum shear also carry the normal stress \(\sigma_{avg}\).

9.4Mohr's Circle — Plane Stress

Squaring and adding the two transformation equations eliminates θ and leaves the equation of a circle in the σ–τ plane:

Mohr's Circle

\[ \left(\sigma_{x'} - \sigma_{avg}\right)^2 + \tau_{x'y'}^2 = R^2 \qquad\text{with}\quad \sigma_{avg} = \frac{\sigma_x + \sigma_y}{2},\;\; R = \sqrt{\left(\frac{\sigma_x - \sigma_y}{2}\right)^2 + \tau_{xy}^2} \]

Axes: σ positive to the right, τ positive downward (so counterclockwise rotation of the element corresponds to counterclockwise rotation on the circle).
Center at \(C(\sigma_{avg}, 0)\); the reference point \(P(\sigma_x, \tau_{xy})\) represents θ = 0.

Mohr's circle: center at σavg, radius R. The circle's intersections with the σ axis are the principal stresses; its bottom (or top) point gives the maximum in-plane shear.

FIG. 9-1 (original diagram)

Reading the Circle

A rotation θ of the physical element corresponds to a rotation 2θ in the same sense around the circle from the reference point P. The circle's intersections with the σ axis are the principal stresses \(\sigma_1 = \sigma_{avg} + R\) and \(\sigma_2 = \sigma_{avg} - R\) (where τ = 0); the extreme vertical points give \(\tau_{max} = R\) acting together with \(\sigma_{avg}\) — and they sit 90° around the circle (45° on the element) from the principal points. The circle makes every transformation result visible at a glance.

9.5Absolute Maximum Shear Stress

Looking Out of the Plane

Ductile materials fail by shear, so the absolute maximum shear matters. A plane-stress element with principal stresses \(\sigma_1, \sigma_2\) (and \(\sigma_3 = 0\) out of plane) can be viewed in three planes — x–y, x–z, y–z — each with its own Mohr's circle. The absolute maximum shear is the radius of the largest of the three circles.

Absolute Maximum Shear Stress (Plane Stress)

\[ \sigma_1, \sigma_2\ \text{same sign:}\quad \tau_{\substack{abs\\max}} = \frac{\sigma_1}{2} \qquad\qquad \sigma_1, \sigma_2\ \text{opposite signs:}\quad \tau_{\substack{abs\\max}} = \frac{\sigma_1 - \sigma_2}{2} \]

Same sign (both tension or both compression): the governing circle spans from 0 to σ1 — the critical shear acts out of plane, on an element rotated 45° about an in-plane axis, and exceeds the in-plane maximum.
Opposite signs: the in-plane circle is the largest, so the absolute maximum equals the in-plane maximum.

EXAMPLE 1Stress on an Inspection Plane

At a smooth plate surface, a local plane-stress element has \(\sigma_x=72\ \text{MPa}\), \(\sigma_y=18\ \text{MPa}\), and \(\tau_{xy}=24\ \text{MPa}\). Find the normal and shear stresses on an inspection plane whose outward normal is rotated \(30^\circ\) counterclockwise from \(+x\).

Recognition strategy

A specified physical plane and all three in-plane components signal direct plane-stress transformation, not a principal-stress search.

Given information

Stresses are in MPa; tension is positive, and positive shear acts in \(+y\) on the \(+x\) face.

Model and interpretation

Use a differential surface element, with no out-of-plane traction. The requested normal defines \(x'\), so \(\theta=+30^\circ\).

Step-by-step solution

1. Establish terms. The invariant center and half-difference are \(\sigma_{\mathrm{avg}}=45\ \text{MPa}\) and \(D=27\ \text{MPa}\); use \(2\theta=60^\circ\).

2. Transform normal stress. Substitute the inclined-plane equilibrium relation: \(\sigma_{x'}=45+27\cos60^\circ+24\sin60^\circ=79.3\ \text{MPa}\).

3. Transform shear stress. With the stated positive shear convention, \(\tau_{x'y'}=-27\sin60^\circ+24\cos60^\circ=-11.4\ \text{MPa}\).

4. Recover the companion normal stress. Trace invariance gives \(\sigma_{y'}=90-79.3=10.7\ \text{MPa}\), completing the rotated element.

Verification

Units: Every term in each transformation is MPa because trigonometric factors are dimensionless; no force or area unit has been mixed into the local stress calculation.

Signs and direction: The negative \(\tau_{x'y'}\) means its actual direction is opposite the assumed positive direction on the \(+x'\) face; complementary shear has the matching opposite-face sense.

Plausibility: \(79.3+10.7=90\ \text{MPa}\), matching the original trace, and the transformed normal lies between the principal extremes rather than implying a new load.

Final result: \(\boxed{\sigma_{x'}=79.3\ \text{MPa},\ \tau_{x'y'}=-11.4\ \text{MPa},\ \sigma_{y'}=10.7\ \text{MPa}}\)

Engineering interpretation

The plane is predominantly tensile but carries reversing-sense shear. This local result can guide a seam or crack-plane assessment, provided notch effects and the actual three-dimensional surface condition are separately evaluated.

EXAMPLE 2Principal Stress at a Bracket Fillet

A finite-element stress extraction at a smooth location beside a bracket fillet reports \(\sigma_x=-40\ \text{MPa}\), \(\sigma_y=80\ \text{MPa}\), and \(\tau_{xy}=30\ \text{MPa}\). Determine the in-plane principal stresses, the orientation for \(\sigma_1\), and maximum in-plane shear.

Recognition strategy

Because the requested normal extremes and zero-shear planes are unknown orientations, use the principal-stress radius and then verify the angle.

Given information

The reported components use MPa and the local axes are right handed; the point is treated as plane stress.

Model and interpretation

The continuum result is local, not a section resultant. Take positive angle counterclockwise from \(x\) toward \(x'\) and use \(\operatorname{atan2}\) for quadrant control.

Step-by-step solution

1. Compute circle data. \(\sigma_{\mathrm{avg}}=20\ \text{MPa}\), \(D=(-40-80)/2=-60\ \text{MPa}\), and \(R=\sqrt{(-60)^2+30^2}=67.1\ \text{MPa}\).

2. Obtain principal values. The zero-shear normal stresses are \(\sigma_1=20+67.1=87.1\ \text{MPa}\) and \(\sigma_2=20-67.1=-47.1\ \text{MPa}\).

3. Resolve the principal orientation. \(\,2\theta_p=\operatorname{atan2}(30,-60)=153.4^\circ\), hence \(\theta_{p1}=76.7^\circ\); its perpendicular plane carries \(\sigma_2\).

4. Identify the shear condition. The maximum in-plane shear is \(R=67.1\ \text{MPa}\), on planes \(45^\circ\) from either principal plane with normal stress \(20\ \text{MPa}\).

Verification

Units: Squaring MPa components gives \(\text{MPa}^2\), so the radius returns to MPa; the angle is dimensionless and reported in degrees only for orientation.

Signs and direction: Substitution at \(76.7^\circ\) gives essentially zero shear and the larger positive normal stress, confirming that this root belongs to \(\sigma_1\), not \(\sigma_2\).

Plausibility: The mixed tensile-compressive state must have principal values of opposite sign, and \((87.1-(-47.1))/2=67.1\ \text{MPa}\) independently recovers the radius.

Final result: \(\boxed{\sigma_1=87.1\ \text{MPa}\text{ at }76.7^\circ,\ \sigma_2=-47.1\ \text{MPa},\ \tau_{\max,\mathrm{in}}=67.1\ \text{MPa}}\)

Engineering interpretation

The tensile principal direction is a useful crack-initiation orientation, while the mixed signs elevate shear. A fillet mesh result should be checked for convergence and compared with fatigue allowables rather than treated as a certified margin.

EXAMPLE 3Absolute Shear in a Pressurized Shell

At a thin pressure-vessel shell location away from a nozzle, membrane analysis gives principal in-plane stresses \(\sigma_1=150\ \text{MPa}\) and \(\sigma_2=60\ \text{MPa}\). Treating radial stress as zero, determine the absolute maximum shear stress and compare it with the in-plane maximum shear.

Recognition strategy

Known principal stresses eliminate rotation equations. The phrase absolute maximum requires comparison of all three principal stresses, including \(\sigma_3=0\).

Given information

Both membrane stresses are tensile in MPa; thin-wall plane stress is assumed only at a smooth shell location.

Model and interpretation

Order the three principal stresses as \(150\), \(60\), and \(0\ \text{MPa}\). Maximum shear is half the difference between the extreme principal stresses.

Step-by-step solution

1. State the three-dimensional set. Plane stress supplies \(\sigma_3=0\ \text{MPa}\); this is a stress boundary condition and does not mean the out-of-plane strain is zero.

2. Calculate in-plane shear. The circle joining the two membrane stresses has radius \((150-60)/2=45.0\ \text{MPa}\).

3. Compare the remaining circles. The radii joining \(150\) to \(0\) and \(60\) to \(0\) are \(75.0\ \text{MPa}\) and \(30.0\ \text{MPa}\), respectively.

4. Select the governing condition. The largest radius is the \(150\)-to-zero circle, so the critical shear occurs on an out-of-plane orientation rather than on the shell surface.

Verification

Units: Differences and halves of stresses remain MPa; no shell force resultant, thickness, or pressure belongs in this already-resolved point-stress comparison.

Signs and direction: All three ordered principal stresses are nonnegative, so the governing pair is the largest tensile stress and the zero radial stress, not the two in-plane values.

Plausibility: \(75.0\ \text{MPa}\) exceeds the in-plane value \(45.0\ \text{MPa}\), as expected whenever both in-plane principal stresses are tensile and the third principal stress is zero.

Final result: \(\boxed{\tau_{\max,\mathrm{abs}}=75.0\ \text{MPa};\quad \tau_{\max,\mathrm{in}}=45.0\ \text{MPa}}\)

Engineering interpretation

A ductile-yield screening based only on in-plane shear would be nonconservative here. Vessel acceptance still requires the governing code stress classification, corrosion allowance, weld efficiency, cyclic pressure history, and a nozzle discontinuity assessment.

Chapter recap

Transform stresses only after obtaining a defensible local state from equilibrium or analysis.
Use the trace and radius to select principal stresses, maximum in-plane shear, and orientations.
For plane stress, include the zero third principal stress before claiming an absolute maximum shear.
Before moving on: draw a signed stress element, choose the proper method, calculate an orientation, and audit it by substitution and physical limits.