Finite Element Methods · Chapter 5 of 10 · Intermediate
Shape Functions and Interpolation
Shape functions are how an element fills in the field between its nodes. They are the quiet engine behind every stiffness matrix, and a few simple properties pin them down completely.
Readiness check
This chapter is about interpolation. Tick only what you can do closed-notes.
- Interpolate linearly between two values.
- Differentiate a simple function.
- Recall that strain is a derivative of displacement.
- Map an interval to another with a linear change of variable.
- Evaluate a function at a fraction of its range.
The core idea
An element approximates the field inside it by interpolating the nodal values: u(x) = Σ Ni(x) ui. The shape functions Ni are fixed by a few properties, and their derivatives give the strains.
u(x) = N₁u₁ + N₂u₂ΣNi = 1, Ni = 1 at node iε = du/dx = B uBetween its nodes an element does not know the true field, so it assumes one: the value at any interior point is a weighted blend of the nodal values, with the shape functions as the weights. For a linear bar the weights are N₁ = 1 − x/L and N₂ = x/L, straight lines that hand off from one node to the other. Two properties pin every shape function down: each equals one at its own node and zero at the others (the Kronecker-delta property), and they sum to one everywhere (partition of unity), so a uniform field is reproduced exactly. Differentiating the shape functions gives the strain-displacement matrix B, which converts nodal displacements into strains and, ultimately, the element stiffness. Working in natural coordinates, a parent element on [−1, 1], makes the same functions reusable for every element.
The skills, taught in order
Five skills cover interpolation, the linear functions, their properties, natural coordinates, and the strain-displacement matrix.
5.1 The interpolation idea
An element represents the unknown field as u(x) = Σ Ni(x) ui, a blend of the nodal values weighted by shape functions. This is the single assumption that converts a continuous problem into a finite set of nodal unknowns.
5.2 Linear shape functions
For a two-node bar, the shape functions are the straight lines N₁ = 1 − x/L and N₂ = x/L. They give a linear displacement field and a constant strain, which is why the simplest bar and triangle elements are constant-strain elements.
5.3 The required properties
Two properties define valid shape functions: the Kronecker-delta property (Ni = 1 at node i and 0 at every other node) ensures the interpolation passes through the nodal values, and partition of unity (Σ Ni = 1) ensures a constant field is reproduced exactly, a condition for convergence.
| Property | Statement | Why it matters |
|---|---|---|
| Kronecker delta | Ni = 1 at node i, 0 at others | interpolation hits nodal values |
| Partition of unity | Σ Ni = 1 | reproduces a constant field |
| Smoothness | continuous across elements | compatible deformation |
5.4 Natural coordinates
Mapping each element to a parent element on the interval [−1, 1] lets one set of shape functions, N₁ = (1 − ξ)/2 and N₂ = (1 + ξ)/2, serve every element. The mapping's Jacobian relates derivatives and lengths in the two coordinates, and this idea is the basis of isoparametric elements.
5.5 The strain-displacement matrix
Differentiating the shape functions gives the strain: ε = du/dx = Σ (dNi/dx) ui = B u, where B is the strain-displacement matrix. B is the bridge from nodal displacements to strains and stresses, and it appears in every element stiffness integral.
Engineering connection: the B matrix and natural coordinates introduced here are exactly what the plane-stress and isoparametric chapters build on.
Worked example 1: interpolating with linear shape functions
A two-node bar element has nodal displacements u₁ = 2 mm and u₂ = 6 mm. Using the linear shape functions, find the displacement at one quarter of the length (x = L/4), and confirm partition of unity there.
- ProblemFind u at x = L/4 for the element in Figure 1 and check partition of unity.
- Given / findu₁ = 2 mm, u₂ = 6 mm, evaluate at x = L/4. Find u and check N₁ + N₂.
- AssumptionsLinear two-node bar element.
- ModelN₁ = 1 − x/L, N₂ = x/L; interpolate u = N₁u₁ + N₂u₂.
- EquationsN₁(L/4) = 0.75, N₂(L/4) = 0.25u = N₁u₁ + N₂u₂
- Solveu = 0.75(2) + 0.25(6) = 1.5 + 1.5 = 3 mm. Partition of unity: N₁ + N₂ = 0.75 + 0.25 = 1.
- CheckThe result lies between the nodal values and is weighted toward node 1, as a point near node 1 should be. The strain du/dx = (u₂ − u₁)/L = 4/L is constant, the hallmark of a linear element.
- ConclusionShape functions interpolate the field and, differentiated, give a constant strain. Partition of unity guarantees a uniform field is captured exactly.
Worked example 2: natural coordinates and the Jacobian
An element spans physical coordinates x = 2 to x = 6. Using the parent-element shape functions N₁ = (1 − ξ)/2 and N₂ = (1 + ξ)/2, find the physical position at ξ = 0.5 and the Jacobian of the mapping.
- ProblemFind the physical position at ξ = 0.5 and the Jacobian for the mapping in Figure 2.
- Given / findx₁ = 2, x₂ = 6, evaluate at ξ = 0.5. Find x(ξ) and J = dx/dξ.
- AssumptionsTwo-node isoparametric mapping; same functions for geometry and field.
- Modelx = N₁x₁ + N₂x₂; the Jacobian is dx/dξ = (x₂ − x₁)/2.
- EquationsN₁(0.5) = 0.25, N₂(0.5) = 0.75x = N₁x₁ + N₂x₂, J = (x₂ − x₁)/2
- Solvex = 0.25(2) + 0.75(6) = 0.5 + 4.5 = 5. J = (6 − 2)/2 = 2.
- Checkξ = 0.5 is three quarters of the way from node 1 to node 2, and x = 5 is indeed three quarters along [2, 6]. The Jacobian 2 equals L/2, the stretch from the unit-length parent to the physical element.
- ConclusionNatural coordinates let one parent element serve all elements; the Jacobian carries the size and shape, the foundation of the isoparametric method in Chapter 7.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Functions violating partition of unity | A constant field is not reproduced | "Do the shape functions sum to one?" | Use functions with Σ Ni = 1 everywhere. |
| Forgetting the Jacobian | Wrong derivatives or integrals | "Did I include dx/dξ?" | Convert derivatives and integrals through the Jacobian. |
| Expecting varying strain from a linear element | Linear element used where strain varies | "Is the strain constant in this element?" | Linear elements give constant strain; refine or use higher order. |
| Mixing coordinates | Shape functions evaluated in the wrong frame | "Am I in x or in ξ?" | Keep natural and physical coordinates straight. |
Practice ladder
For a linear bar, evaluate N₁ and N₂ at the midpoint x = L/2.
Show answer
N₁ = 1 − 0.5 = 0.5, N₂ = 0.5. They sum to one, as required.
With u₁ = 4 and u₂ = 10 mm, find the displacement at x = 3L/4.
Show answer
N₁ = 0.25, N₂ = 0.75, so u = 0.25(4) + 0.75(10) = 1 + 7.5 = 8.5 mm.
An element spans x = 1 to x = 9. Find the physical position at ξ = −0.5 and the Jacobian.
Show answer
N₁(−0.5) = 0.75, N₂ = 0.25, so x = 0.75(1) + 0.25(9) = 0.75 + 2.25 = 3. J = (9 − 1)/2 = 4.
Explain why partition of unity is necessary for an FEM solution to converge, using the idea of reproducing a constant field.
What good work looks like
The argument that if shape functions sum to one, a uniform displacement (rigid-body or constant strain) is represented exactly, a requirement for the approximation to improve with refinement.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Take an element, write its shape functions, verify the two required properties, interpolate the field at an interior point, and compute the strain-displacement matrix B.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Write the interpolation.
u(x) = Σ Ni(x) ui, a shape-function blend of nodal values.
2. Give the linear bar shape functions.
N₁ = 1 − x/L and N₂ = x/L.
3. State the two required properties.
Kronecker delta (1 at own node, 0 elsewhere) and partition of unity (Σ Ni = 1).
4. What are the parent-element functions?
N₁ = (1 − ξ)/2 and N₂ = (1 + ξ)/2 on [−1, 1].
5. What is the B matrix?
The strain-displacement matrix from differentiating the shape functions: ε = B u.
Textbook mapping
| Item | Mapping |
|---|---|
| Primary source | Logan, A First Course in the Finite Element Method, Chapters 3 and 10 (Shape Functions, Natural Coordinates) |
| Cross-reference | Reddy, Ch. 3 · Math: Numerical Methods |
| Core topics | 5.1 Interpolation · 5.2 Linear functions · 5.3 Required properties · 5.4 Natural coordinates · 5.5 Strain-displacement matrix |
| Engineering connection | The B matrix and natural coordinates build the plane-stress and isoparametric elements. |
| Read next | Chapter 6: Plane Stress and the Constant-Strain Triangle. |