Timoshenko Beam Theory
Timoshenko Beam Theory: Theoretical Foundations
Differences from Euler-Bernoulli Beam
Professor, what's the difference between Timoshenko beam and Euler-Bernoulli beam?
The biggest difference is that it considers shear deformation. The Euler-Bernoulli beam assumes "cross-sections remain perpendicular to the neutral axis," but the Timoshenko beam relaxes this assumption. Cross-sections are allowed to tilt relative to the neutral axis.
Mathematically:
- EB Beam: Rotation angle = derivative of deflection โ $\theta = dw/dx$
- Timoshenko Beam: Rotation angle โ derivative of deflection โ $\theta \neq dw/dx$, the difference is the shear deformation
Shear strain $\gamma$:
$dw/dx$ is the slope of the beam axis, $\theta$ is the rotation of the cross-section. This difference is the shear deformation, right?
Exactly. The EB beam enforced $\gamma = 0$ (no shear deformation). The Timoshenko beam relaxes this constraint to describe more general beam behavior.
Governing Equations
Please tell me the differential equations for the Timoshenko beam.
They become two coupled differential equations:
Here $A_s = \kappa A$ is the effective shear area, $\kappa$ is the shear correction factor.
What is the shear correction factor $\kappa$?
Beam theory assumes uniform shear stress across the section, but actual shear stress distribution is parabolic. $\kappa$ is a coefficient that corrects for this difference.
| Cross-section Shape | $\kappa$ |
|---|---|
| Rectangular section | 5/6 โ 0.833 |
| Circular section | 6/7 โ 0.857 |
| Thin-walled circular tube | 1/2 = 0.5 |
| I-section (web shear) | $A_w / A$ (web area / total area) |
$\kappa$ is less than 1... so the effective shear area is smaller than the total cross-sectional area.
Yes. For I-beams, shear is mostly carried by the web, so the effective shear area is close to the web area. Flanges contribute to bending but hardly to shear.
Decomposition of Deflection
Can the deflection of a Timoshenko beam be decomposed into bending and shear components?
Yes. Total deflection is:
For a simply supported beam with a central point load:
The first term is the EB beam deflection, and the second term is the additional shear contribution.
Ratio of shear deflection:
Here $r = \sqrt{I/A}$ is the radius of gyration of the section. The larger $r/L$ is (thicker, shorter beams), the greater the contribution of shear deformation.
For steel ($E/G \approx 2.6$) with a rectangular section ($\kappa = 5/6$) and $L/h = 10$, what percentage of the total deflection is shear deflection?
Calculating gives about 3%. For $L/h = 5$, about 12%. For $L/h = 3$, about 32%. The rule of thumb that shear deformation cannot be ignored for $L/h < 5$ comes from this calculation.
Timoshenko Beam Element in FEM
What's special about the Timoshenko beam element in FEM?
The Timoshenko beam element treats $w$ and $\theta$ as independent variables. The EB beam element had the constraint $\theta = dw/dx$, but the Timoshenko beam element has no such constraint.
However, caution is needed. With standard 2-node elements (linear interpolation), shear locking occurs.
Shear locking?
It's the opposite phenomenon of shear locking in EB beams. When trying to represent pure bending deformation with a Timoshenko beam element, parasitic shear strain occurs and the element locks. As a result, deflection is underestimated.
Countermeasures:
- Reduced integration (1-point Gauss integration) โ avoids shear locking
- Assumed Strain method โ independent approximation of shear strain
- Higher-order elements (3 nodes or more) โ locking is naturally eliminated
EB beam elements have shear locking, Timoshenko beam elements have shear locking... opposite problems occur, huh?
Yes. It's an eternal theme in FEM element design. EB beams ignore shear to avoid locking, Timoshenko beams include shear and fight locking. The practical solution is reduced integration.
Summary
Let me organize the Timoshenko beam theory.
Key points:
- Considers shear deformation โ $\gamma = dw/dx - \theta \neq 0$
- Shear correction factor $\kappa$ โ depends on cross-section shape. 5/6 for rectangle
- Significant for $L/h < 5$ โ thick/short beams, sandwich panels, composite girders
- Beware of shear locking โ avoid with reduced integration
- Superset of EB beam โ Timoshenko beam converges to EB beam as $L/h \to \infty$
So, if in doubt, just use the Timoshenko beam?
Exactly. The Timoshenko beam encompasses the EB beam. For slender beams with large $L/h$, it gives the same result as the EB beam, so there's no problem always using the Timoshenko beam. That's why the default beam elements in Abaqus and Ansys are Timoshenko beams.
The Birth of Timoshenko Beam Theory
Stephen Timoshenko introduced shear deformation into beam theory in his 1921 paper "On the correction factor for shear of the differential equation for transverse vibrations of prismatic bars." He is also known for revolutionizing engineering education at Stanford University after emigrating to the US following the Russian Revolution.
Numerical Solution and Implementation
Formulation of Timoshenko Beam Element
Please tell me the stiffness matrix for the Timoshenko beam element.
For a 2-node Timoshenko beam element with bending degrees of freedom only ($w_1, \theta_1, w_2, \theta_2$):
Here $\Phi = 12EI/(GA_s L^2)$ is the shear deformation parameter.
If $\Phi = 0$ (no shear deformation), it matches the EB beam stiffness matrix!
Perfect observation. $\Phi$ can be thought of as the correction amount from the EB beam. The larger $\Phi$ is (thicker, shorter beams), the greater the influence of shear deformation and the larger the difference from the EB beam.
Details of Shear Locking
Could you explain the mechanism of shear locking in a bit more detail?
Consider representing pure bending ($M$ = constant) with a 2-node linear interpolation element. If the bending moment is constant, then $\theta$ should be constant ($d\theta/dx = M/(EI)$), but with linear interpolation of $w$, $dw/dx$ is also constant. Then $\gamma = dw/dx - \theta = \text{constant} \neq 0$.
Shear strain appears even in pure bending... this is parasitic shear, right?
Yes. The energy from this shear strain is stored unnecessarily, causing deformation to be underestimated (response is too stiff).
Why reduced integration (1-point Gauss) solves it: $\gamma$
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