"Should I use the stiffness method or finite elements?" comes up in every classroom and every design office, and it is the wrong question. The trouble is that it treats the two as interchangeable alternatives, and they are not.
The stiffness method is the finite element method applied to frame members. What separates a "frame" program from a "finite element" program is the model. A frame element carries a one-dimensional theory of how a member deforms; a shell mesh is a discretised two-dimensional continuum. Choosing between them is an engineering decision, and this post is about when to make it and what it costs to get it wrong.
A straight member needs no mesh
In finite elements, the displacement inside each element is approximated with shape functions. For a member in bending, the usual ones are cubic Hermite polynomials. And the solution of the Euler-Bernoulli beam equation with no load along the span is itself a cubic polynomial, so the approximation coincides with the exact solution.
That is why the stiffness matrix for a prismatic member, the one in any matrix analysis textbook, is exact, and why splitting a member into more elements improves nothing. With a distributed load the same holds at the nodes, as long as the equivalent nodal loads are computed consistently.
| Frame elements in the beam | Midspan deflection [mm] |
|---|---|
| 2 | 1.5625 |
| 4 | 1.5625 |
| 10 | 1.5625 |
| 50 | 1.5625 |
| Closed form 5qL⁴/384EI | 1.5625 |
Two elements or fifty, the engine returns the same number, and it matches the closed form.
Meshing a straight member does not make it more accurate: it makes it slower.
Where they part ways
Euler-Bernoulli theory assumes that plane sections remain plane and perpendicular to the axis, which amounts to ignoring shear deformation. For a slender beam this is a very good approximation. For a deep beam it is not.
We took the same 6 m beam under the same load and increased its depth step by step. Each case was solved twice: with frame elements, as Basic mode does, and as an MITC4 shell meshed in the plane of the beam, which PRO mode can do.
| L/h | Depth h [m] | Frame elements [mm] | Shells [mm] | Frame model low by |
|---|---|---|---|---|
| 20 | 0.30 | 12.500 | 12.567 | 0.5 % |
| 10 | 0.60 | 1.5625 | 1.5961 | 2.1 % |
| 5 | 1.20 | 0.19531 | 0.21206 | 7.9 % |
| 3 | 2.00 | 0.042188 | 0.052145 | 19.1 % |
| 2 | 3.00 | 0.012500 | 0.019017 | 34.3 % |
At L/h = 10, an ordinary beam, the difference is 2%. At L/h = 5 it is almost 8%. At L/h = 3 the frame model gives a deflection 19% smaller, and at L/h = 2, a third smaller.
The frame model always makes the beam stiffer than it really is.
Checking it with a calculator
You do not have to take the shell’s word for it. Timoshenko beam theory includes shear deformation, and for a simply supported beam under uniform load it adds one term to the deflection: q·L² / (8·κ·G·A), with κ = 5/6 for a rectangular section and G = E / (2·(1+ν)).
| Model | Midspan deflection [mm] |
|---|---|
| Frame elements (Euler-Bernoulli) | 0.0422 |
| Frame elements plus the shear term, by hand | 0.0530 |
| MITC4 shells | 0.0521 |
| Plane elasticity, closed form | 0.0523 |
At L/h = 3 that term is 0.0108 mm; added to the 0.0422 mm from Euler-Bernoulli, it gives 0.0530 mm. The shell gives 0.0521 mm, and Timoshenko and Goodier’s closed-form plane-elasticity solution gives 0.0523 mm. The three answers that include shear agree within 2%. The one that ignores it is 19% low.
Do a few hundredths of a millimetre matter?
In that beam on its own, not much. In a statically determinate structure a stiffness error only affects the deflection, and a 2 m deep beam deflects very little anyway. The error starts to matter when stiffness decides how the loads are distributed, which is what happens in any statically indeterminate structure.
The typical case is the shear wall. In a frame model a wall is a very wide column: it gets the wall’s moment of inertia and sits on the wall’s centreline. For a squat wall, long compared with its height, the frame element makes it considerably stiffer than it really is.
| Wall, length × height [m] | Height / length | Frame element [mm] | Shell [mm] | Frame element stiffer by |
|---|---|---|---|---|
| 4 × 3 | 0.75 | 0.0281 | 0.0627 | 2.23 times |
| 3 × 3 | 1.00 | 0.0667 | 0.1129 | 1.69 times |
| 2 × 3 | 1.50 | 0.2250 | 0.2943 | 1.31 times |
| 3 × 6 | 2.00 | 0.533 | 0.625 | 1.17 times |
| 3 × 9 | 3.00 | 1.800 | 1.937 | 1.08 times |
A wall 4 m long and 3 m tall, one storey high, comes out 2.2 times stiffer when modelled with a frame element. One 3 m long and 9 m tall, three storeys, is only 8% stiffer. The difference matters most for squat walls.
And that changes how the shear is shared. Take two walls on the same line, one 4 m and one 2 m long, both 3 m tall and tied together by the slab. With frame elements, the short wall takes 11.1% of the shear. With shells it takes 17.6%, 58% more than the frame model predicts. The extra load the frame model assigns to the long wall comes off the short one, which ends up designed for less shear than it will actually carry.
Questions a frame model cannot answer
So far, these are cases where frame elements give the wrong number. In others they cannot give a number at all, because the question has no place in the model.
- A flat slab. How the moment divides between the column strip and the middle strip, and how much of it concentrates over each column, is two-dimensional information. A beam grillage can approximate it, but that information is not in the model.
- A wall with openings. A frame element has a single cross-section at each point along its axis, so a window in the middle of the wall has nowhere to go.
- The joint. In a frame model the beam-column joint is a point, so the peak moment appears on the centreline. In reality the joint is as big as the column, and the section you design is the one at the column face.
- A concentrated load near a support. Plane sections do not stay plane in that region, for the same reason as in the deep beam.
Where they must agree
The other half of the argument matters just as much. For the slender beam in the first row, at L/h = 20, frame elements and shells differ by 0.5%. That is not zero, because shear is always there, but it is negligible. And it should be: if the two models disagree on a slender member, check the finite element model first, and the mesh in particular.
Both models in one program
Having frame elements and shells in the same program is nothing new. SAP2000, ETABS, Robot and midas combine both in one model, and that is the baseline for professional software. But they are not built to explain what they compute. At the other end, teaching tools such as Ftool and MASTAN2 teach frame analysis very well but stop short of finite elements.
Stabileo has both modes in one program. Basic mode solves with frame elements and walks through the stiffness method in nine steps: degree-of-freedom numbering, local matrices, transformation, assembly, load vector, boundary conditions, solution, reactions and internal forces. PRO mode solves with finite elements, using MITC4 shells and DKT plates alongside frame elements. In the tools we reviewed, we did not find another program that lets you learn how the analysis works and then apply it to a fuller model without switching tools.
In short
- For a prismatic member, the stiffness method is the exact special case of finite elements. What you are choosing is the model.
- A straight member needs no mesh. If splitting it changes the result, something else is wrong.
- Below L/h ≈ 5 the frame model underestimates deflection by more than 8%, and below 3 by more than 19%. It always makes the beam stiffer.
- In an indeterminate structure that error changes how loads are distributed. Check squat walls before beams.
- Flat slabs, walls with openings and loads near supports raise questions a frame model cannot answer.
- Where the member is slender, the two models must agree. If they do not, check the mesh.