4節点四面体要素(TET4)
Theory and Physics
What is the TET4 Element?
Professor, is TET4 the most basic element for 3D analysis?
It is the most basic, but it is an element that should not be used in practical work. I'll say this upfront. I'll explain the reasons in detail later, but first, let's understand the theory.
A "do not use" declaration right from the start...?
Learning the theory of TET4 is important. It's the optimal element for understanding the fundamentals of FEM. However, if you want accurate results in practical work, you should use TET10.
Shape Function
TET4 is a tetrahedron with 4 nodes, each with 3 degrees of freedom ($u, v, w$). Total 12 DOF.
TET4 is a tetrahedron with 4 nodes, each with 3 degrees of freedom ($u, v, w$). Total 12 DOF.
The shape functions are expressed using volume coordinates (barycentric coordinates) $L_1, L_2, L_3, L_4$:
Displacement interpolation:
Since the shape functions are linear (first-order), the displacement also changes linearly within the element, right?
Correct. Displacement being linear means strain is constant within the element. This is the fatal weakness of TET4.
The Problem of Constant Strain
What's the problem with strain being constant?
In areas where stress changes rapidly (stress concentration zones, areas with bending stress gradients), it cannot represent stress variation at all.
For example, modeling a beam under bending with TET4:
- Theoretically, the top surface of the cross-section is in tension, the bottom in compression, with a linear distribution
- With TET4, stress is constant within each element, resulting only in a rough, stair-step approximation
- Extremely fine mesh is required to obtain accurate bending stress
With TET10 (quadratic element), strain varies linearly within the element, so it can represent bending.
Exactly. TET10 has 10 nodes with mid-side nodes on each edge and uses quadratic shape functions. Strain varies linearly within the element, allowing it to represent bending stress distribution within a single element.
TET4 Stiffness Matrix
The TET4 element stiffness matrix is 12×12. Since the B matrix is a constant matrix, it can be calculated in closed form without numerical integration:
$V_e$ is the element volume, $[B]$ is the strain-displacement matrix (constant), $[D]$ is the constitutive matrix.
Not needing numerical integration is an advantage, right?
It's fast to compute. However, due to its low accuracy, you ultimately need a finer mesh, which increases the DOF count and offsets the computational time advantage.
Quantitative Accuracy Comparison
Can you show the accuracy difference between TET4 and TET10 numerically?
Comparison of tip deflection of a cantilever beam ($L/h = 10$) with theoretical value:
| Element Type | Mesh | DOF | Deflection Error |
|---|---|---|---|
| TET4 | Coarse (100 elements) | 200 | -40% (underestimated) |
| TET4 | Medium (1,000 elements) | 1,500 | -15% |
| TET4 | Fine (10,000 elements) | 12,000 | -5% |
| TET10 | Coarse (100 elements) | 1,200 | -2% |
| TET10 | Medium (1,000 elements) | 9,000 | -0.3% |
The difference widens further for stress accuracy. Since TET4 stress is constant within an element, the peak stress concentration value converges very slowly to the theoretical value no matter how fine the mesh is.
When Should TET4 Be Used?
Is there any legitimate reason to use TET4?
Almost none. If forced to say:
- Explicit method impact analysis — TET4 has a simple mass matrix, which can be advantageous for the stable time increment in explicit methods (some LS-DYNA cases)
- Coupling with fluid mesh — When the CFD code is TET4-based, the structural side may also use TET4 for consistency
- Preprocessor constraints — Some old automatic mesh generators can only output TET4
However, all of these are "unavoidable circumstances," and if TET10 can be used, you should always use TET10.
Summary
Let me organize the theory of TET4.
Key points:
- 4 nodes, linear shape functions, constant strain within element — The most basic 3D element in FEM
- Low accuracy — Requires 5~10 times more DOF than TET10 to represent bending or stress concentration
- Not used in practice — TET10 is always superior as a replacement
- Optimal for theoretical learning — A textbook element for understanding FEM principles
- "Do not use TET4" is an FEM iron rule — Just knowing this dramatically improves analysis quality
It's interesting to learn it in order to understand the reason "not to use" it.
Knowing the limitations of TET4 is very important for understanding FEM accuracy. An engineer who can explain why TET4 is bad is an engineer who can use FEM correctly.
TET4 Linear Shape Functions and Constant Strain Characteristics
The 4-node tetrahedral element (TET4) has linear shape functions using volume coordinates L1 to L4, resulting in constant strain within the element (Constant Strain Tetrahedron, CST). The concept of triangular elements was presented by R. Courant in 1943, and Turner et al. extended it to 3D tetrahedra in 1956. As the simplest 3D solid element, it is easy to implement, but has the drawback of requiring a huge number of elements to reproduce bending or stress concentration due to its constant strain.
Physical Meaning of Each Term
- Inertia term (mass term): $\rho \ddot{u}$, i.e., "mass × acceleration". Have you ever experienced being thrown forward when slamming on the brakes? That "being carried away" feeling is precisely the inertial force. Heavier objects are harder to set in motion and harder to stop once moving. Buildings shake during earthquakes because the ground moves suddenly and the building's mass "gets left behind". In static analysis, this term is set to zero, which assumes "forces are applied slowly enough that acceleration is negligible". It absolutely cannot be omitted for impact loads or vibration problems.
- Stiffness term (elastic restoring force): $Ku$ or $\nabla \cdot \sigma$. When you stretch a spring, you feel a "force trying to return it", right? That is Hooke's law $F=kx$, the essence of the stiffness term. Now a question — an iron rod and a rubber band, which stretches more under the same force? Obviously the rubber. This "resistance to stretching" is the Young's modulus $E$, which determines stiffness. A common misconception: "high stiffness = strong" is incorrect. Stiffness is "resistance to deformation", strength is "resistance to failure" — different concepts.
- External force term (load term): Body force $f_b$ (gravity, etc.) and surface force $f_s$ (pressure, contact force, etc.). Think of it this way — the weight of a truck on a bridge is a "force acting on the entire contents" (body force), the force of the tires pushing on the road surface is a "force acting only on the surface" (surface force). Wind pressure, water pressure, bolt tightening force... all are external forces. A common mistake here: getting the load direction wrong. Intending "tension" but it becomes "compression" — sounds like a joke, but it actually happens when coordinate systems are rotated in 3D space.
- Damping term: Rayleigh damping $C\dot{u} = (\alpha M + \beta K)\dot{u}$. Try plucking a guitar string. Does the sound continue forever? No, it gradually fades. That's because vibrational energy is converted to heat by air resistance and internal friction in the string. Car shock absorbers work on the same principle — intentionally absorbing vibrational energy to improve ride comfort. What if damping were zero? Buildings would continue shaking forever after an earthquake. Since that doesn't happen in reality, setting appropriate damping is crucial.
Assumptions and Applicability Limits
- Continuum assumption: Treats material as a continuous medium, ignoring microscopic heterogeneity
- Small deformation assumption (for linear analysis): Deformation is sufficiently small compared to initial dimensions, stress-strain relationship is linear
- Isotropic material (unless otherwise specified): Material properties are independent of direction (anisotropic materials require separate tensor definition)
- Quasi-static assumption (for static analysis): Ignores inertial and damping forces, considering only the balance between external and internal forces
- Non-applicable cases: Large deformation/large rotation problems require geometric nonlinearity. Plasticity, creep, and other nonlinear material behaviors require constitutive law extensions
Dimensional Analysis and Unit Systems
| Variable | SI Unit | Notes / Conversion Memo |
|---|---|---|
| Displacement $u$ | m (meter) | When inputting in mm, unify loads and elastic modulus to MPa/N system |
| Stress $\sigma$ | Pa (Pascal) = N/m² | MPa = 10⁶ Pa. Be careful of unit system inconsistency when comparing with yield stress |
| Strain $\varepsilon$ | Dimensionless (m/m) | Note the distinction between engineering strain and logarithmic strain (for large deformation) |
| Elastic modulus $E$ | Pa | Steel: ~210 GPa, Aluminum: ~70 GPa. Note temperature dependence |
| Density $\rho$ | kg/m³ | In mm system: tonne/mm³ (= 10⁻⁹ tonne/mm³ for steel) |
| Force $F$ | N (Newton) | Unify as N in mm system, N in m system |
Numerical Methods and Implementation
TET4 Implementation Details
Is TET4 implementation simpler compared to other elements?
It's the simplest. The B matrix is constant, so no numerical integration is needed. It's optimal as an introduction to FEM programming.
B Matrix Calculation
Calculate the gradient of volume coordinates from the coordinates of the 4 nodes $(x_i, y_i, z_i)$ and construct the B matrix. The TET4 B matrix is a 6×12 constant matrix:
Here $b_i, c_i, d_i$ are constants calculated from nodal coordinates. $V$ is the tetrahedron volume.
How do you calculate $V$?
From the coordinates of the 4 nodes:
If $V > 0$, the node order is correct (right-handed). If $V < 0$, the element is inverted.
Element Names by Solver
| Solver | Element Name | Notes |
|---|---|---|
| Nastran | CTETRA (4-node version) | PSOLID property |
| Abaqus |
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