Static Condensation

7 min read • 1,280 words

This example uses the getTangent method to perform static condensation.

This example replicates the classic cantilever‐beam problem where:

We have two beam elements forming a cantilever of total length $L$ .
Nodes are arranged along the beam so that Node 1 is fixed, and Nodes 2 and 3 are free.
Each free node has two DOFs: transverse displacement $(u)$ and rotation $(\theta)$ .

Accordingly, the system has 4 active DOFs. We will:

Construct the model in OpenSees using ndm=2, ndf=3.
Extract the global stiffness and mass matrices by calling model.getTangent(k=1) and model.getTangent(m=1).
Condense out the rotational DOFs to see how to statically reduce a 4×4 matrix down to 2×2.

1) Static Condensation

Below is a small helper function condense() that takes a square matrix $\boldsymbol{K}$ (stiffness or mass) and an array of indices ic identifying which DOFs to keep. All other DOFs are statically condensed out using

\boldsymbol{K}_{\mathrm{condensed}} = \boldsymbol{K}_{tt} - \boldsymbol{K}_{t0} \; \boldsymbol{K}_{00}^{-1} \; \boldsymbol{K}_{0t}

where $(t)$ are the retained DOFs and $(0)$ are the ones to be condensed out.

def condense(K, ic):
    """
    Perform static condensation on matrix K, returning a smaller matrix
    that retains only the DOFs in the list 'ic' and statically eliminates
    all other DOFs.
    
    Returns
    -------
    Kc : 2D numpy array
         The condensed matrix of size len(ic) x len(ic).
    """
    import numpy as np
    
    # Convert list to np.array for indexing
    ic = np.array(ic, dtype=int)
    
    # All DOFs = 0..N-1
    N = K.shape[0]
    ix = np.setdiff1d(range(N), ic)  # DOFs to be condensed out
    
    # Partition:
    K_tt = K[np.ix_( ic, ic )]
    K_t0 = K[np.ix_( ic, ix )]
    K_0t = K[np.ix_( ix, ic )]
    K_00 = K[np.ix_( ix, ix )]

    # If there are no DOFs to condense out, just return the original
    if len(ix) == 0:
        return K

    # Perform the usual static condensation formula
    # K_condensed = K_tt - K_t0 * inv(K_00) * K_0t
    Kc = K_tt - K_t0 @ np.linalg.solve(K_00, K_0t)

    return Kc

2) A Cantilever Model

We now create a small function build_cantilever_model() that sets up:

A 2D model (ndm=2) with 3 DOFs per node: $(u_x, u_y, \theta_z)$ .
Three nodes along the $x$ -axis, with Node 1 fixed in all DOFs, Node 2 and Node 3 each having $u_x$ fixed (to ensure pure bending), but allowing $u_y$ and $\theta_z$ .
Two elastic beam elements, each of length $L/2$ .
Lumped masses at Nodes 2 and 3 in the transverse direction only, matching $\frac{mL}{4}$ and $\frac{mL}{2}$ respectively.

import opensees.openseespy as ops

def build_cantilever_model(L=1.0, E=1.0, I=1.0, m=1.0):
    """
    Builds a 2D cantilever of total length L, split into two beam elements.
    Node 1: fully fixed in (ux, uy, rz).
    Node 2 and 3: fix only ux, keep (uy, rz) free.
    
    The beam has E, I for bending. 
    We assign lumped masses at node 2,3 in the y-direction only, with:
      node2 mass = (0, mL/4, 0)
      node3 mass = (0, mL/2, 0)
    """
    # Start the Model: 2D problem, ndf=3
    model = ops.Model(ndm=2, ndf=3)

    # Create 3 nodes at 0, L/2, L along x-axis
    model.node(1, 0.0,  0.0)
    model.node(2, L/2,  0.0)
    model.node(3, L,    0.0)

    # Fix Node 1 in all DOFs (ux, uy, rz)
    model.fix(1, 1, 1, 1)

    # Node 2, 3: fix only ux => (1, 0, 0)
    model.fix(2, 1, 0, 0)
    model.fix(3, 1, 0, 0)

    # Geometric transformation for a linear beam in 2D
    transf_tag = 1
    model.geomTransf("Linear", transf_tag)

    # Define a 2D elastic beam (elasticBeamColumn)
    A  = 1.0   # cross-sectional area (not crucial for pure bending)
    Iz = I     # moment of inertia
    # E is user-specified
    # We'll ignore axial effects or keep them large to approximate pure bending

    # Element 1: Node 1 -> Node 2
    model.element("elasticBeamColumn", 1, 1, 2, A, E, Iz, transf_tag)
    # Element 2: Node 2 -> Node 3
    model.element("elasticBeamColumn", 2, 2, 3, A, E, Iz, transf_tag)

    # Lumped mass at Node 2 and Node 3 in the y-direction only
    model.mass(2, 0.0, m*L/4, 0.0)
    model.mass(3, 0.0, m*L/2, 0.0)

    return model

3) Stiffness & Mass Matrices

Once the model is built, we obtain the global tangent matrices. The method:

model.getTangent(k=1)  # for stiffness
model.getTangent(m=1)  # for mass

returns the entire system matrix, including constrained DOFs. We must then extract the sub-block for the free DOFs, which are:

Node 2: DOFs (u_x=4, u_y=5, $\theta$ =6) but we fixed the x-displacement, so the “active” DOFs are 5 and 6.
Node 3: DOFs (u_x=7, u_y=8, $\theta$ =9) but we fixed the x-displacement, so the “active” DOFs are 8 and 9.

Hence the 4 free DOFs in the global system are $[5,6,8,9]$ . Let’s see how we do it:

import numpy as np

if __name__ == "__main__":

    # Example parameters
    L = 1.0
    E = 1.0
    I = 1.0
    m = 1.0

    # Build the model
    model = build_cantilever_model(L, E, I, m)

    # Get the full stiffness and mass matrices
    K_full = model.getTangent(k=1)
    M_full = model.getTangent(m=1)

    print("Full system stiffness K_full:\n", K_full, "\n")
    print("Full system mass M_full:\n", M_full, "\n")

    # The 4 active DOFs are [5,6,8,9]
    free_dofs = [5, 6, 8, 9]

    # Extract 4x4 submatrix
    K_4x4 = K_full[np.ix_(free_dofs, free_dofs)]
    M_4x4 = M_full[np.ix_(free_dofs, free_dofs)]

    print("4x4 stiffness matrix (DOFs 5,6,8,9):\n", K_4x4, "\n")
    print("4x4 mass matrix (DOFs 5,6,8,9):\n", M_4x4, "\n")

4) Performing Static Condensation

In the textbook example, the 4 DOFs are labeled:

\mathbf{u} = \{\, u_1,\; u_2,\; u_3,\; u_4 \}

But in our global matrix, we see the order is $(5,6,8,9)$ . Typically, we want to condense out the rotations (call them $u_3, u_4$ in the text) and keep the translations ( $u_1, u_2$ ).

If we interpret dof5 and dof8 as the two translations $u_1, u_2$
and dof6 and dof9 as the two rotations $u_3, u_4$ ,

then we should reorder the 4×4 matrix so that translations come first $[5,8]$ and rotations come second $[6,9]$ . After reordering, we can apply condense() to eliminate the rotation DOFs.

    # Reorder the 4x4 so that translations are in the first block: [5,8], then rotations [6,9].
    reorder = [0, 2, 1, 3]  
    # Explanation: 
    #   0 -> index 0 in K_4x4 is dof5
    #   2 -> index 2 in K_4x4 is dof8
    #   1 -> index 1 in K_4x4 is dof6
    #   3 -> index 3 in K_4x4 is dof9

    K_4x4_reordered = K_4x4[np.ix_(reorder, reorder)]
    print("K_4x4 in 'translation-then-rotation' order:\n", K_4x4_reordered, "\n")

    # DOFs to keep: the first 2 (the translations)
    keep = [0, 1]  # keep indices [0,1] in the reordered matrix
    K_2x2_condensed = condense(K_4x4_reordered, keep)

    print("Condensed 2x2 stiffness (after eliminating rotations):\n", K_2x2_condensed, "\n")

Finally, if desired, we can also find the transformation matrix $\mathbf{T}$ that recovers the condensed DOFs (rotations) in terms of the retained DOFs (translations):

\mathbf{u}_{0} = - \mathbf{K}_{00}^{-1}\,\mathbf{K}_{0t} \;\mathbf{u}_t \quad\Longrightarrow\quad \mathbf{T} = - \mathbf{K}_{00}^{-1}\,\mathbf{K}_{0t}

    # Partition K_4x4_reordered as blocks:
    #  K_tt is the top-left 2x2
    #  K_t0 is the top-right 2x2
    #  K_0t is the bottom-left 2x2
    #  K_00 is the bottom-right 2x2
    K_tt = K_4x4_reordered[:2, :2]
    K_t0 = K_4x4_reordered[:2, 2:]
    K_0t = K_4x4_reordered[2:, :2]
    K_00 = K_4x4_reordered[2:, 2:]

    import numpy as np
    T = -np.linalg.solve(K_00, K_0t) 
    print("Transformation matrix T (rotations vs translations):\n", T, "\n")

    print("--- End of demonstration ---")

Conclusion

We built a two-element cantilever in OpenSees with 3 nodes, 4 active DOFs ( $y$ -translations and rotations at nodes 2 and 3).
We used model.getTangent() to obtain the global stiffness and mass matrices, then sliced out the $4 \times 4$ sub-block for the free DOFs.
We performed static condensation on the $4 \times 4$ matrix to reduce it down to a $2 \times 2$ matrix that retains only the translational DOFs, matching the textbook procedure and final results.

This illustrates how the new OpenSees Python API can be leveraged for matrix-based analysis, including direct access to system tangents and flexible manipulations such as static condensation.

Solid extrusions

Tension test

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