Lecture 11: Simple States of Stress

Overview

Steven M. Wise reviews traction, Cauchy's stress theorem, and equilibrium, then develops simple stress states and how they determine tractions on surfaces. He defines spherical, uniaxial, pure shear, and plane stress; uses the eigenvalues and eigenvectors of a symmetric stress tensor to identify principal stresses and maximum normal and shear stresses; and decomposes stress into spherical and trace-free deviatoric parts.

Key takeaways

Cauchy's postulate makes traction a local function of position and surface normal, and Cauchy's theorem expresses that dependence as t(n) = S n.
A symmetric stress tensor has three real principal stresses and orthogonal principal directions; a surface normal aligned with one of those directions produces zero shear traction.
For uniaxial stress S = σd⊗d, planes perpendicular to d have zero traction, while planes with normal d carry traction σd.
Pure shear has equal off-diagonal components in its aligned frame, S₁₂ = S₂₁ = τ, and zero diagonal entries, making its trace zero.
Given ordered principal stresses σ₁ > σ₂ > σ₃, maximum shear stress is |σ₁ − σ₃|/2 and occurs on planes with normals halfway between e₁ and e₃.
The decomposition S = −pI + Sᵈ separates pressure-driven volume change from trace-free deviatoric stress associated primarily with shape change.

Chapters

0:00 Traction, Cauchy's Postulate, and Action-Reaction

A cut divides a body into B− and B+; traction is the force per unit area exerted by B+ on B−.
Cauchy's postulate says traction at a point depends on the point and the surface normal, not on the cut's curvature.
Action-reaction requires opposite tractions on the two sides of an internal surface, whose outward normals are n and −n.

8:00 Cauchy's Stress Theorem and Local Equilibrium

Cauchy's stress theorem represents traction as a linear function of the unit normal: t(n) = S n, where S is the Cauchy stress tensor.
Force equilibrium on every open subset gives div S + ρb = 0, where ρ is mass density and b is specific body force.
Stress has units of pascals, or N/m²; its divergence and ρb both have units of N/m³.
Moment equilibrium implies that S is symmetric in the equilibrium setting discussed here.

12:00 Spherical Stress as Uniform Compression or Tension

A spherical stress state has S = −pI, with p a scalar and I the identity tensor.
For any unit normal n, the traction is −pn, so it is normal to the surface and has the same magnitude in every direction.
When p is positive, the traction points inward, as with pressure deep underwater.

16:00 Uniaxial Stress Acts Along One Direction

A uniaxial stress state has S = σ d⊗d, where d is a unit vector and σ is a scalar.
The traction on a plane with normal n is t = σ(d·n)d, so it always lies along d.
If n is perpendicular to d, traction is zero; if n = d, traction is σd.
Positive σ gives tension in the lecture's sign convention, while negative σ gives compression.

21:00 Pure Shear and Its Surface Tractions

Pure shear is defined by S = τ(d⊗m + m⊗d), with orthogonal unit vectors d and m.
For a surface normal n, traction is t = τ[(m·n)d + (d·n)m].
With n = d or n = m, the traction is tangential to the surface; reversing the normal reverses the traction.
The illustrated shear loading distorts a small cube toward a prism by changing its right angles.

27:00 Pure Shear Components and the Trace-Free Pattern

In the right-handed frame (d, m, d×m), pure shear has matrix components S₁₂ = S₂₁ = τ and all other entries zero.
The diagonal entries vanish, so the trace of this pure shear tensor is zero.
Wise connects trace-free, or deviatoric, tensors to later fluid mechanics topics, including Navier-Stokes stress.

31:00 Plane Stress Includes Uniaxial Stress and Pure Shear

Plane stress has a frame in which only the upper-left 2×2 block of the stress matrix can be nonzero.
The third row and third column vanish in that frame, leaving no stress components involving the third direction.
Pure shear is a special plane-stress state with zero diagonal entries; uniaxial stress also fits the plane-stress form.

33:00 Principal Stresses from Symmetric Stress Tensors

For a symmetric stress tensor, the spectral theorem provides three real eigenvalues, counting multiplicities, and an orthonormal eigenvector basis.
The eigenvalues σ₁, σ₂, and σ₃ are principal stresses; their associated eigenvectors e₁, e₂, and e₃ are principal stress directions.
Each pair satisfies S eᵢ = σᵢeᵢ, with no summation over i in this eigenvalue equation.
Eigenvalue ordering and a right-handed basis are separate choices, so the numbered directions may need reordering to form the desired frame.

38:00 Principal Directions Produce Zero Shear Traction

On a surface whose normal is a principal direction eᵢ, the traction is t = S eᵢ = σᵢeᵢ.
Because this traction is parallel to the surface normal, its entire contribution is normal traction.
The shear traction on a principal plane is therefore zero.

42:00 Resolving Traction into Normal and Shear Parts

For a surface normal n and traction t, the normal traction vector is tₙ = (t·n)n.
The shear traction is tₛ = t − tₙ, and tₙ + tₛ reconstructs the total traction.
Since n is a unit vector, tₛ·n = 0, so shear traction lies in the surface plane.
The magnitudes |tₙ| and |tₛ| are called normal stress and shear stress on that plane.

49:00 Principal Planes Have No Shear Stress

If n is a principal direction with eigenvalue σ, then t = σn.
The normal component equals the full traction, while the shear component vanishes.
Rotating a cut away from a principal direction generally produces both normal and tangential traction components.

51:00 Maximum Normal and Maximum Shear Stress Results

For distinct ordered principal stresses σ₁ > σ₂ > σ₃, the maximum normal-stress magnitude is max(|σ₁|, |σ₃|).
That maximum occurs when the plane normal is ±e₁ or ±e₃, according to which endpoint eigenvalue has greater magnitude.
The maximum shear stress is |σ₁ − σ₃|/2.
Maximum shear occurs for normals ±(e₁ + e₃)/√2 and ±(e₁ − e₃)/√2; Wise assigns the derivation to the slides and homework.

56:00 Proof That Principal Directions Maximize Normal Stress

Expand a unit normal as n = Σ nᵢeᵢ in the orthonormal principal basis, where Σnᵢ² = 1.
The scalar t·n is q = Σσᵢnᵢ², a weighted average of the principal stresses.
For σ₁ > σ₂ > σ₃, q lies between σ₃ and σ₁, so |q| cannot exceed max(|σ₁|, |σ₃|).
Choosing n = ±e₁ or ±e₃ attains the endpoint value with the larger absolute magnitude.

1:08:00 Spherical and Deviatoric Stress Decomposition

The spherical part of S is Sₛ = (1/3)(tr S)I = −pI, with pressure defined as p = −(1/3)tr S.
Define the deviatoric part as Sᵈ = S − Sₛ; then S = Sₛ + Sᵈ and tr Sᵈ = 0.
For a symmetric stress tensor, tr S equals the sum of its three principal stresses, so p = −(σ₁ + σ₂ + σ₃)/3.
The spherical part tends to change volume without changing shape, while the deviatoric part tends to change shape with little volume change.

1:13:00 Pressure Sign Convention and Transition to Kinematics

Positive p in Sₛ = −pI corresponds to inward, compressive spherical stress.
Wise notes that pressure conventions can differ between elasticity and fluid mechanics contexts.
The lecture closes chapter three and previews examples followed by chapter four on kinematics.

Transcript excerpt

0:04 Welcome back to continue mechanics. Then we are in chapter three, our second and actually last lecture discussing continuing continuum mass and force concepts. I want to recap the harder or more challenging of the things that we discussed last time before we get on to the to the next things. So let's talk about uh traction stress equilibrium and um the um uh action reaction principle. So remember what we're considering here is taking a body B and dividing it up into two uh subbodies B minus and B+.

0:52 Our normal points out of B minus and into B+. The force per unit area exerted by B+ on B minus is the traction that we call TN. So if I make a cut and divide this body into two subbodies B minus B+ and I examine this point X here then the force per unit area that B+ exerts on B minus is attraction. So it's a vector. Maybe it emanates like this TN. Now what we showed or what we um argued

1:43 or I I should say actually what Koshi argued was that this TN could be resolved or expressed as a function of some function. We're we're going to kind of abuse notation here and call it t that only depends on the normal and the point x. So this says that actually the traction that you get this force per unit area that this one body exerts on the other body really doesn't have uh anything to do with the curvature of this cutting plane. It really only has to do with the point you're evaluating it at and the normal at that point. So for example, if I have another

2:30 uh let's say we had three possible ways of cutting that surface, a black, a red surface,

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Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Steven M. Wise.