Within SAP2000, CSiBridge, and ETABS, a link object may be used to manually input a known 12x12 stiffness matrix which represents the connection between two joints.
Modeling procedure
A two-joint link may be modeled and assigned a 12x12 stiffness matrix as follows:
- Draw a two-joint link object which connects the two points. The first joint is denoted i and the second joint is j.
- Carefully note the local coordinate system of the link object. If the link is of finite length L, then the local-1 axis is directed from joint i to joint j. You can change the orientation of the local-2 and -3 axes as desired. If the link is of zero length, then the local-1, -2, and -3 axes are parallel to global-X, -Y, and -Z, respectively, though this orientation may be changed as well.
- Transform the given stiffness matrix to the link local coordinate system as necessary.
- Partition the stiffness matrix as follows:
where:
- { Fi } and { Fj } are the 6 forces and moments at joints i and j.
- { Ui } and { Uj } are the 6 displacements and rotations at joints i and j.
- [Ki i ], [Ki j ], [Kj i ], and [Kj j ] are 6x6 sub-matrices of the full 12x12 stiffness matrix.
- The degrees of freedom (DOF) are ordered as (U1, U2, U3, R1, R2, R3) at each joint.
- { Fi } and { Fj } are the 6 forces and moments at joints i and j.
- Define a link property, then set its type to Linear.
- Activate all 6 DOF and set the two shear distances to zero.
- For the link stiffness properties, use the values from the [Kj j ] sub-matrix. Due to symmetry, only the upper triangle of the sub-matrix needs to be entered (21 values).
- Assign this link property to the link object.
Justification
This procedure works because there is a lot of redundancy in the 12x12 matrix due to both symmetry and the requirement that no forces be generated under rigid-body motion of the link object. For a connected (non-grounded) two-joint object, the force vectors { Fi } and { Fj } should be in equilibrium under any displaced configuration. In the simplest of terms, { Fi } and { Fj } should be equal and opposite except for the moments of the shears, which are affected by the length L and the shear lengths dj2 and dj3.
Kinematics
Internally, the link element automatically does the following:
- For arbitrary { Uj }, the equilibrium relationship between { Fi } and { Fj } enables the determination of [Ki j ] from the given [Kj j ].
- By symmetry, [Kj i ] = [Ki j ]T.
- For arbitrary { Ui }, the equilibrium relationship between { Fi } and { Fj } enables the determination of [Ki i ] from the given [Kj i ].
See Also
- Unknown macro: {new-tab-link}Analysis Reference Manual (The Link/Support Element - Basic, page 229) – further information on the general use of links
CSI