Interlaminar Shear Stress⚠ unverified
Mechanical / Composites · Compute the interlaminar (transverse) shear stress in a beam section
Parameters
| Input | Symbol | Unit | Default | Description |
|---|---|---|---|---|
| V | V | N | 1.0 | Transverse shear force |
| Q | Q | m^3 | 1.0 | First moment of area about the neutral axis |
| I | I | m^4 | 1.0 | Second moment of area of the section |
| t | t | m | 1.0 | Width of the section at the point of interest |
| Output | Symbol | Unit | Description |
|---|---|---|---|
| result | τ | Pa | Shear stress, in pascals (Pa); returns 0.0 when ``I * t`` is non-positive |
The science & history
Understanding the Parameters
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Shear force $V$ — the transverse shear at the cross-section (from a bending load). Interlaminar shear is largest where $V$ is largest — near supports and load points of a beam, which is why delamination often starts there.
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First moment $Q$ — of the area between the plane of interest and the outer surface. It is maximum at the neutral axis and zero at the top/bottom, so the interlaminar shear peaks at mid-thickness — the opposite of bending stress, which peaks at the surfaces.
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Section stiffness $I$ and width $t$ — normalise the shear flow into a stress; a thicker or wider section lowers it.
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Why it governs composites — in a metal beam this shear is usually benign, but a laminate is only as strong in through-thickness shear as its matrix, so the interlaminar shear strength (ILSS, measured by the short-beam-shear test) is often the limiting design allowable, especially for short, thick beams.
Derivation (Approaching a Proof)
Consider a beam element of length $dx$ under a varying bending moment (hence a shear force $V = dM/dx$). Isolate the portion of the cross-section above a horizontal plane at height $y_1$. The bending stresses on its two faces differ (because $M$ changes over $dx$), leaving an unbalanced horizontal force that a shear stress on the plane must equilibrate.
Summing the bending-stress difference over the isolated area and setting it equal to the shear force on the plane ($\tau\,t\,dx$):
$$\tau\,t\,dx = \frac{dM}{dx}\,dx\,\frac{\int y\,dA}{I} = V\,dx\,\frac{Q}{I} \;\Longrightarrow\; \tau = \frac{V Q}{I t},$$
where $Q = \int y\,dA$ is the first moment of the isolated area about the neutral axis. This is the standard transverse-shear result; in a laminate the plane of interest is a ply interface, and the computed $\tau$ is compared against the interlaminar shear strength.
Dimensional check. $\tau = \dfrac{V Q}{I t} = \dfrac{\text{N}\cdot\text{m}^3}{\text{m}^4\cdot\text{m}} = \dfrac{\text{N}}{\text{m}^2} = \text{Pa}$ — a stress, as required.
History and Development
The $\tau = VQ/It$ transverse-shear formula (Jourawski/Collignon, 19th century) is a foundation of beam theory. Its role becomes governing in laminated composites, where the through-thickness (matrix-dominated) strength is far below the in-plane strength, making delamination a primary failure mode. The short-beam-shear (ASTM D2344) test deliberately maximises interlaminar shear to measure the ILSS allowable that this stress is checked against.
Related Concepts: Shear Stress, Transverse Strength, Tsai Hill Criterion, Section Modulus, Composite Laminate Theory, Rule of Mixtures transverse
Notes: Peaks at mid-thickness (max $Q$), opposite to bending stress (surfaces). Largest near supports/ load points (max $V$). Governs delamination — compare to interlaminar shear strength (ILSS, short-beam-shear test). $Q$ is the first moment of the area outboard of the plane.