Stress and strain tensors

Start with force on a cut

Imagine an inclined cut through a continuous material. A small patch of that cut has a current area and an oriented unit normal n. Its traction t is force per current area, so it has units Pa = N/m². Traction can point partly across the cut and partly along it; one scalar pressure cannot describe both parts.

Use right-handed Cartesian axes, with x to the right and y upward. For this supplied example, the in-plane Cauchy stress is σ = [[6, 2], [2, 3]] MPa. The cut has n = (4/5, 3/5). These are stipulated numbers, not measurements of a material.

Here σ_ij means force component i on a face whose normal points in coordinate direction j. With this force-first convention, t = σn: multiply each matrix row by the normal column. Thus t_x = 6(4/5) + 2(3/5) = 6 MPa and t_y = 2(4/5) + 3(3/5) = 17/5 MPa.

For traction uniform over a current patch of area A = 10⁻⁴ m², the resulting force is At = (600, 340) N, because 1 MPa = 10⁶ N/m². If traction varies over a patch, its force is the area integral of traction; multiplying one local value by the whole area need not work.

Separate normal and shear

The normal stress on this cut is the scalar σ_nn = n·t, where a dot product multiplies corresponding components and adds them. Positive σ_nn means tension in our convention. The normal traction vector is t_n = σ_nn n.

Choose the positive unit tangent m = (−3/5, 4/5), perpendicular to n. The signed shear coefficient is τ = m·t. Its sign tells us which way the tangential traction points relative to this chosen m. The shear vector is t_s = τm, and t = t_n + t_s.

Here σ_nn = 171/25 = 6.84 MPa and τ = −22/25 = −0.88 MPa. The negative shear coefficient gives t_s = (66/125, −88/125) MPa, opposite the chosen positive tangent. Its magnitude is 0.88 MPa; a magnitude has no negative sign.

The two component vectors are perpendicular. Their sum reproduces the original traction, even though neither component need point along a coordinate axis.

Change basis, not the material

A stress tensor is the linear map from a cut normal to its traction at a point. Its matrix is the set of components in a chosen basis. Changing that basis changes the matrix entries while preserving the physical map.

Let Q have the new unit basis vectors n and m as its columns: Q = [[4/5, −3/5], [3/5, 4/5]]. The same stress in that basis is σ′ = QᵀσQ = [[171/25, −22/25], [−22/25, 54/25]] MPa. The first column contains the normal and signed tangent components on the n-normal cut. This is a passive coordinate change; the material has not been rotated or deformed.

The principal directions are unit directions e for which σe = λe, so traction on that normal is parallel to the normal and has no shear component. In this example the in-plane principal stresses are 7 MPa and 2 MPa, along (2, 1)/√5 and (−1, 2)/√5. The trace 9 MPa and determinant 14 MPa² are unchanged by the passive basis change.

We separately stipulate plane stress: σ_xz = σ_yz = σ_zz = 0. The full three-dimensional stress therefore has a third principal stress of 0, along z. This is a condition on stress, not a declaration that the out-of-plane strain vanishes.

Symmetric Cauchy stress belongs to the classical continuum scope used here: angular-momentum balance without independent body couples or couple stresses. A continuum model with such additional rotational effects needs its own balance laws; symmetry is not an assumption to transfer to every model.

Describe a separate displacement

Now prescribe a different example: a reference square has metre-valued coordinates X = (X, Y), with vertices (0, 0), (1, 0), (1, 1), (0, 1) m. Its displacement is u(X) = HX, where H = [[0.002, 0.004], [−0.001, −0.001]] is dimensionless. No material law connects this displacement to the earlier stress.

For example, at (1, 1) m the displacement is (0.006, −0.002) m. The actual position is X + u = (1.006, 0.998) m. H records how displacement changes with reference position: its entry H_ij is ∂u_i/∂X_j.

The deformation gradient is F = I + H = [[1.002, 0.004], [−0.001, 0.999]], with I the identity matrix [[1, 0], [0, 1]]. F maps reference line vectors to their actual deformed line vectors. This affine example has the same H everywhere on the square.

Adding the same translation to every point would change position and displacement but not H, F or strain. A displacement value alone does not specify a local shape change.

Keep the half factor

For small displacement gradients, the infinitesimal strain is ε = (H + Hᵀ)/2. The remaining skew part Ω = (H − Hᵀ)/2 describes infinitesimal rotation. Their sum is H. They are dimensionless, although the displacement itself is measured in metres.

For the supplied H, ε = [[0.002, 0.0015], [0.0015, −0.001]] and Ω = [[0, 0.0025], [−0.0025, 0]]. The x direction extends to first order, the y direction contracts to first order, and the skew part corresponds to a small clockwise rotation. Ω is not the exact matrix of a finite rotation.

The engineering shear is γ_xy = H_xy + H_yx = 2ε_xy = 0.003. For initially perpendicular coordinate line elements, positive γ_xy is the first-order decrease of their right angle, expressed in radians. Tensor shear ε_xy = 0.0015 is half that engineering value.

For any reference unit direction a, aᵀεa gives the first-order fractional change in length. In the direction n = (4/5, 3/5), it is 59/25000 = 0.00236. A negative value would describe contraction to first order.

Read the cut and the magnified square

The upper panel draws the traction decomposition in the original x-right, y-up basis. The short line is the cut. The violet normal arrow and the orange shear arrow add head-to-tail to the blue traction arrow. All three use the same 36 pixels per MPa, so the shorter shear arrow reflects its smaller magnitude.

The lower panel uses reference coordinates at 360 pixels per metre. Its dashed square is the reference shape; the solid outline and square markers show X + 20u(X). Only displacement is multiplied by 20 to make the small change visible. The drawing is not the physical deformed square, and its area must not be read as the physical area change.

The table separates the physical displacement and position from the magnified drawing. Both panels use supplied examples; the force arrows do not predict the lower square’s motion.

Check the finite metric

Small strain drops terms quadratic in H. A finite metric check retains them: the Green–Lagrange strain is E_G = (FᵀF − I)/2 = ε + HᵀH/2. It measures change in squared lengths of reference line elements; it is not itself their engineering fractional length change.

Here E_G = [[0.0020025, 0.0015045], [0.0015045, −0.0009915]]. The correction HᵀH/2 is positive semidefinite. Its trace is 11/10⁶, so in this example 0 ≤ aᵀ(E_G − ε)a ≤ 11/10⁶ for every unit direction a. This is an absolute bound, not a guarantee of small relative error when the linear prediction is near zero.

A second exact check is the square’s physical area ratio: det F = 1.001002, giving area change 0.001002 times its reference area. The linear area prediction is trace ε = 0.001. These are two-dimensional area statements; they do not determine a three-dimensional volume change.

A rigid 90° rotation preserves every length, so FᵀF = I and E_G = 0. Yet inserting its large H = F − I into the infinitesimal formula gives ε = −I. The linear approximation is outside its small-gradient domain, even though the actual strain is zero. Small change of length alone does not make a large rotation suitable for linear strain.

What the two tensors tell us

Stress describes local force per current area on differently oriented cuts. Strain describes local change of shape or length relative to a chosen configuration. To infer deformation from stress, a material model, constraints and suitable boundary information are needed. The two supplied matrices alone do not identify an elastic modulus, a failure threshold or a safe load.

Passive basis changes act on both tensors by the same second-order transformation rule, QᵀTQ for the column-basis convention used here. Active deformation changes F and the material’s geometry. Re-expressing a tensor in new coordinates does not perform that deformation.

This lesson introduces matrix multiplication and projections locally. For a return path, the published lessons Vectors and vector spaces and Linear systems and numerical eigenvalue problems give more practice with the operations and principal directions. Understanding the cut decomposition does not require solving a full continuum boundary-value problem.

Worked examples

Try it yourself

1. At the same point, reverse the cut normal from n to −n without changing σ. Find the traction on the oppositely oriented face and its normal stress.

   Show the answer and reasoning

   The traction is (−6, −17/5) MPa. The normal stress remains 171/25 MPa.

   Linearity gives t(−n) = −t(n). Projecting on the reversed normal gives (−n)·(−t) = n·t, so tensile normal stress keeps its sign even though the force direction reverses.

2. If the same traction is uniform on a current area 2 × 10⁻⁴ m², what is the resultant force? Does doubling area change the supplied stress tensor?

   Show the answer and reasoning

   (1200, 680) N. The stress tensor stays as supplied.

   Current area doubles while force per area is held fixed, so At doubles. This is a stipulated uniform-traction comparison; it does not infer how a real specimen’s stress would respond to a geometric change.

3. For the supplied H, calculate ε_xy, engineering γ_xy and Ω_xy. Are they the same quantity?

   Show the answer and reasoning

   ε_xy = 0.0015, γ_xy = 0.003 and Ω_xy = 0.0025; they are distinct dimensionless quantities.

   The symmetric half-sum is (0.004 − 0.001)/2. Engineering shear is the full sum, and the skew half-difference is (0.004 + 0.001)/2. The first two describe shear deformation with different conventions; the third describes infinitesimal rotation.

4. The stress components become [[171/25, −22/25], [−22/25, 54/25]] MPa in the n, m basis. Has the material undergone the magnified deformation in the lower panel?

   Show the answer and reasoning

   No. This is a passive change of stress components; it does not specify a deformation.

   QᵀσQ re-expresses the same stress map. The separate displacement example requires H and F. Its shape cannot be inferred from this coordinate change or from the stress matrix alone.

5. Take the finite rotation F = [[0, −1], [1, 0]]. Calculate E_G and the infinitesimal ε formed from H = F − I. Which identifies its unchanged lengths?

   Show the answer and reasoning

   E_G = 0, while ε = −I. The finite metric correctly identifies unchanged lengths.

   FᵀF = I gives zero Green–Lagrange strain. The symmetric part of H is −I because this H is large. Applying the linear formula outside its small-gradient domain produces a spurious strain for a large rigid rotation.

6. In the stipulated plane-stress example, what is the third principal stress? Can you also conclude that ε_zz = 0?

   Show the answer and reasoning

   The third principal stress is 0. The out-of-plane strain is not determined.

   The full stress has a zero z row and column, so z is a principal direction with stress 0. Plane strain is a different condition. A relation between stress and strain would require additional material and constraint information.

What to carry forward

• Declare which stress index labels force and which labels face normal before using a matrix.
• Normal stress and a fixed-tangent shear coefficient are signed projections; their vectors add to traction.
• A passive basis change preserves the physical tensor, while active deformation changes geometry.
• Infinitesimal strain is the symmetric half-gradient; engineering shear is twice the tensor shear component.
• Finite metric checks and explicit display magnification keep small-strain and visual interpretations within their domains.