8節点四辺形要素(QUAD8)
Theory and Physics
Q8 Element — Second-Order Accurate 2D Element
Professor, is Q8 an upgraded version of Q4?
Yes. It's an 8-node quadratic element that adds 4 mid-side nodes at the midpoint of each edge to Q4's 4 vertices. It's the 2D version of HEX20.
Shape Functions
Q8 shape functions are Serendipity-type quadratic polynomials:
Q8 shape functions are Serendipity-type quadratic polynomials:
Vertex Nodes:
Mid-Side Nodes (e.g., $\xi_i = 0$):
So it's Q4's bilinear form plus $\xi^2, \eta^2$ terms. That's why it can accurately represent bending.
Exactly. The "lack of $\xi^2, \eta^2$ terms" which was the cause of shear locking in Q4 is resolved in Q8. Shear locking does not occur in Q8.
Q8 Accuracy
- Displacement: $O(h^3)$ (Q4 is $O(h^2)$)
- Stress: $O(h^2)$ (Q4 is $O(h)$)
Stress convergence is one order faster. Where Q4 needs 100 elements, Q8 can achieve comparable accuracy with 25 elements.
Curved Edge Advantage
The mid-side nodes in Q8 can represent curves, right?
By snapping the mid-side nodes to CAD surfaces, edges can become quadratic curves. The approximation accuracy for circular holes or curved boundaries is significantly higher than Q4 (straight edges).
Is Q8 more advantageous for stress concentration evaluation?
Overwhelmingly so. Comparing stress concentration at a circular hole ($K_t = 3.0$) between Q4 and Q8, Q8 is 5-10% more accurate with the same number of elements. The gap widens further with coarser meshes.
Integration Scheme
| Integration | Gauss Points | Characteristics |
|---|---|---|
| Full Integration (3×3) | 9 | Highest accuracy. No locking |
| Reduced Integration (2×2) | 4 | Shear locking not needed but conventionally used. One hourglass mode |
If Q8 doesn't have shear locking, why use reduced integration?
Reduced integration Q8 (CPS8R, etc.) is strong against volumetric locking. More robust than full integration for problems where $\nu \to 0.5$. Also, computational cost is 4/9 that of full integration.
Practical Recommendation:
- Linear Elasticity ($\nu < 0.45$) → CPS8 (Full Integration) or CPS8R (Reduced Integration) are both OK
- Incompressible Materials ($\nu > 0.45$) → CPS8R (Reduced Integration) is recommended
Summary
Let me organize the theory of Q8.
Key Points:
- 8-node Serendipity-type quadratic element — Upgraded version of Q4
- No shear locking — Because it has $\xi^2, \eta^2$ terms
- Accurately approximates surfaces with curved edges — Advantageous for stress concentration evaluation
- Half the DOF of Q4 for equivalent accuracy — Efficient
- Standard for precise 2D FEM analysis — When in doubt, use Q8
Now I understand the reason stated on the Q4 page that "Q8 is more efficient."
Q4 is the "element for learning the basics," Q8 is the "element for practical use." Understanding both and then using Q8 is best.
Serendipity Shape Functions of the Q8 Element
The 8-node quadrilateral element (Q8) belongs to the Serendipity family and was systematized in 1966 by Ergatoudis, Irons, and Zienkiewicz during research at Lancaster University. The eight shape functions corresponding to the 4 vertex nodes and 4 mid-side nodes contain complete quadratic polynomials and possess high-accuracy characteristics where the error in pure bending problems generally converges in proportion to the cube of the element size h.
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 "feeling of being carried away" 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 while 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 can be ignored". It absolutely cannot be omitted for impact loads or vibration problems.
- Stiffness Term (Elastic Restoring Force): $Ku$ or $\nabla \cdot \sigma$. When you pull a spring, you feel a "force trying to return it", right? That's Hooke's law $F=kx$, and it's the essence of the stiffness term. So here's a question — an iron rod and a rubber band, which stretches more when pulled with 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". Stiffness is "resistance to deformation", strength is "resistance to failure" — they are 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), while 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 typical 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 keep swaying forever after an earthquake. Since that doesn't happen in reality, setting appropriate damping is important.
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 definitions)
- Quasi-Static Assumption (for static analysis): Ignores inertial and damping forces, considers only balance between external and internal forces
- Non-Applicable Cases: Large deformation/large rotation problems require geometric nonlinearity. Nonlinear material behavior like plasticity or creep requires 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 mismatch 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 to N in mm system, N in m system |
Numerical Methods and Implementation
Q8 Implementation Details
Please explain the numerical integration and handling in solvers for Q8.
Q8 is a quadratic element, so the B matrix is a first-order polynomial. $B^T D B$ is quadratic, requiring 3×3 (9-point) Gauss integration for exact integration. Reduced integration approximates it with 2×2 (4 points).
Element Names by Solver
| Variation | Nastran | Abaqus | Ansys |
|---|---|---|---|
| Full Integration | CQUAD8 | CPS8 | PLANE183(full) |
| Reduced Integration | — | CPS8R | PLANE183(red.) |
| Hybrid | — | CPE8H, CPE8RH | — |
Does Nastran's CQUAD8 only have full integration?
Nastran's CQUAD8 internally uses optimized integration. While it doesn't have the same level of refinement as Q4's CQUAD4, its basic accuracy as a quadratic element is high. There is no explicit reduced integration option, but it rarely causes practical problems.
Mid-Side Node Considerations
What should we be careful about when handling mid-side nodes?
Like TET10 and HEX20, Q8 also requires:
- Snap mid-side nodes to CAD curves — Improves approximation accuracy for arcs, etc.
- Do not deviate significantly from the edge midpoint — Keep within 25%–75% of the edge length
- Check that the Jacobian does not become negative — Element degeneration due to incorrect mid-side node placement
Q9 (Lagrange-type) Element
Is there also a Q9 element?
Q9 is a 9-node Lagrange-type quadratic element that adds an element center node to Q8. It contains the $\xi^2\eta^2$ term, making it a complete quadratic polynomial.
The practical difference between Q8 and Q9 is small. Q9 has an extra center node, increasing DOF, but the accuracy improvement is marginal. Nastran's CQUAD9 and Abaqus's CPS9 are available in some solvers, but Q8 is sufficient for most cases.
Mesh Convergence Demo
Can you show the difference in mesh convergence between Q4 and Q8 numerically?
Maximum stress in the Kirsch problem (infinite plate with a circular hole, uniaxial tension, $K_t = 3.0$):
Is there also a Q9 element?
Q9 is a 9-node Lagrange-type quadratic element that adds an element center node to Q8. It contains the $\xi^2\eta^2$ term, making it a complete quadratic polynomial.
The practical difference between Q8 and Q9 is small. Q9 has an extra center node, increasing DOF, but the accuracy improvement is marginal. Nastran's CQUAD9 and Abaqus's CPS9 are available in some solvers, but Q8 is sufficient for most cases.
Can you show the difference in mesh convergence between Q4 and Q8 numerically?
Maximum stress in the Kirsch problem (infinite plate with a circular hole, uniaxial tension, $K_t = 3.0$):
| Mesh | Q4 (CPS4I) | Q8 (CPS8R) | Theoretical Value |
|---|---|---|---|
| Coarse (8 elements around hole) | 2.65 (-12%) | 2.91 (-3%) | 3.00 |
| Medium (16 elements around hole) | 2.88 (-4%) | 2.98 (-0.7%) | 3.00 |
| Fine (32 elements around hole) | 2.96 (-1.3%) | 3.00 (0%) | 3.00 |