2d Truss Stiffness Matrix

[Alysa Guillama]

We will look at the development of the matrix structural analysis method for the simple case of a structure made only out of truss elements that can only deform in one direction. This will allow us to get a taste of how matrix structural analysis works without having to learn about all of the details and complexities that are present in beam and frame systems. As we will see, since we have only one-dimensional truss elements, each node in the system only has one possible degree of freedom (translation along the axis of the structural members).

These two equations define the force/deflection behaviour of the truss at both nodes simultaneously. They are only a function of displacements of the nodes (the nodal displacements) and the forces applied to the nodes (the nodal forces).

where the matrix on the left of the equal sign is called the force vector, the large central matrix is called the stiffness matrix and the smaller matrix on the right with the displacements is called the displacement vector. The term vector just means a matrix with only one column. If we multiply the large central matrix by the vector of displacements on the right, we get:

which are the same as equations \eqrefeq:Truss1D-Mat-Line1 and \eqrefeq:Truss1D-Mat-Line2. This matrix equation constitutes a complete model for the behaviour of a one-dimensional truss element. The large matrix in the middle is called the stiffness matrix of the element because it contains all of the stiffness terms. It contains the most important information for the model, and it is useful to think about it as a separate element:

This is the stiffness matrix of a one-dimensional truss element. Other types of elements have different types of stiffness matrices. For a truss element in 2D space, we would need to take into account two extra degrees of freedom per node as well as the rotation of the element in space. Beam elements that include axial force and bending deformations are more complex still. For real physical systems, stiffness matrices are always square and symmetric about the diagonal axis of the matrix.

If we have a structural analysis problem with multiple one-dimensional truss elements, we must first define the stiffness matrices for each individual element as described in the previous section. After we define the stiffness matrix for each element, we must combine all of the elements together to form on global stiffness matrix for the entire problem. This matrix defines all of the interconnections between the elements and includes all of the information related to the stiffness of each element for each degree-of-freedom.

Once the stiffness matrix is formed, the full system of equations in the form shown in equation \eqrefeq:truss1D-Full-System may be solved. At each nodal DOF (each row), we must either know the external force or the nodal deflection. Then, we can solve only those rows where we don't know the deflection. Once we have all of the nodal deflections, we can solve for the nodal forces. Then, using the individual element stiffness matrices, we can solve for the internal force in each element.

The full process for a matrix structural analysis for a one dimensional truss will be demonstrated using the simple example shown in Figure 11.2. This is a one dimensional structure, meaning that all of the nodes are only permitted to move in one direction. This structure consists of four different truss elements which are numbered one through four as shown in the figure. These elements are connected at four different nodes, also numbered one through four as shown. The left end of the structure, at node 1 is restrained and cannot move. Nodes 2 and 4 have external loads, and Node 3 has an imposed displacement of $13\mathrm\,mm$ to the right (positive).

This is a system of four equations and four unknowns. This is a system that is easily solved using a computer. For this example, since there are only two free displacement degrees of freedom, we can expand the second and fourth rows (the second and fourth equations) to get:

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The extra 1's don't seem to be physically correct according to my intuition. They say the element becomes stiffer when it's under tension. That suggests the force/displacement relationship is not linear in a seemingly arbitrary way.

EDIT: Using the full diagonal matrix with pre-stressed frequency analysis leads to what seems like an incorrect result - a 1D spring-mass system's natural frequency increases with tension on the spring. In other words, this common physics concept of gravity not influencing the motion described here is violated. The same problem occurs with solid elements in Calculix. Are they really wrong?

EDIT 2: Example showing that the full-diagonal matrix does not work for pre-stressed frequency analysis. It's a 1D spring with length 1, spring constant k, mass m concentrated at each node and no constraints.

The natural frequency should be independent of P. We can see this is true when using the 2nd geometric stiffness matrix above by solving the eigenvalue equation$$(\textbf K_e + \textbf K_g - \omega^2 \textbf M) \textbf u = 0$$I form the 1D matrix by taking on the 1st and 3rd rows and columns from the 2D matrix above.

which has the non-zero eigenvalue $ \omega^2=\frac2km $. This is consistent with a simple spring-mass oscillator. The factor of 2 appears because it's really two 1-dof oscillators with half the spring each.

This has a non-zero eigenvalue of $ \omega^2=\frac2(k+P)m $. It's apparently wrong because it's a function of P but it should be independent of P. It shows that a spring is stiffened by increased tension. Conversely, if P is negative, it also shows that compression softens it and it even becomes unstable when P=-k. This indicates axial buckling (not Euler column buckling) under compression, which seems to be unphysical. Even if $\Delta L$ is not zero, it can't make $\textbf K_g$ zero, so there will always be some discrepancy.

So I'm wondering if the matrix with the full diagonal above is really correct. At least it appears to be wrong when used for frequency or linear buckling analysis. Solid elements in Calculix follow this "incorrect" behavior for frequency analysis, which adds to the confusion.

If your objective is to perform a geometrically nonlinearanalysis of truss structures where the elements are allowedto undergo arbitrarily large rotations, then your first formof the geometric stiffness matrix with the "extra" ones is the correct one.

To see this, it is useful to step back to the internal force vector for thistruss element. There are several ways to derive this vector but a simple wayis to just write the equilibrium equations for the member in its deformedposition. That gives

where $\theta$ is the angle of rotation between the undeformed and deformedtruss element.This equation is on page 547 of Bathe. In a nonlinear analysis, the need fora "stiffness matrix" is the result of solving the nonlinear equations bythe Newton-Raphson method; the derivative of the residual vector with respect to to theunknowns is needed. Simple differentiation of the internal force vectorabove with respect to the nodal displacements yields a tangent-K that includes the first form of the $k_g$ matrix.That is what Bathe does on page 547. The nonlinear mechanics is all containedin the internal force vector and the tangent-K is just a straightforward differentiation. The first form of $k_g$ is correct in that sense.

In the notes you refer to by Gavin, he describes the approximations hemakes in deriving the simpler matrix. These approximations yield a $k_g$matrix that is fine for most engineering linearbuckling and pre-stressed vibration analysis but not large-rotation nonlinear analysis.

"Most of the early work, as well as the confusion alluded to by Martin, pertains to what is now called thegeometric stiffness matrix. The early name for it was initial stress matrix. A bar geometric stiffness wasfirst presented in [767]. If this is compared to (8.26) one may observe that half of the nonzero entries aremissing. Consequently that inaugural KG is not invariant with respect to the choice of coordinate frames,and has the wrong rank."

Here are a few comments on your example problems. First, the classicallinear spring is a very simplistic model of an elastic solid so it ishardly surprising that a more sophisticated continuum mechanics modelgives different results. A similar comment can be made about the"classical" $k_g$ matrix derivation. It is based on an analysis ofthin, pre-stressed members where only transverse deflections are considered;axial deflections are explicitly ignored. So it is hardly surprising thatthe two $k_g$ matrices give different results in problems where onlyaxial effects are considered.

The goal of this example is to analytically relate the displacement d to the uniform cross-section area parameter A for a particular bar in the cantilever truss structure. The governing equation is represented by Md2xdt2+Kx=F. Here, d results from a load at the upper-right joint of the cantilever. The truss is attached to the wall at the leftmost joints.

The truss bars are made of a linear elastic material with uniform density. The cross-section area A of the bar highlighted in blue (see figure) can vary. All other parameters, including the uniform cross-section areas of the other bars, are fixed. Obtain the displacement of the tip by using small and linear displacement assumptions.

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