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. Note that these are in the link local coordinate system, not the joint local coordinate system.
- 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).
- IMPORTANT: The given vectors { Fi } and { Ui } and matrix [Kj j ] are typically defined in a right-handed coordinate system, such that the F2-M3 and F3-M2 coupling terms have the opposite sign. However, the link stiffness properties are specified in terms of the internal forces and deformations, which follow the designers' sign convention, as does the frame object. This means that the V2-M3 and V3-M2 coupling terms have the same sign. More specifically, V2 = -F2. For this reason, the values specified for 5th row and 5th column should each be multiplied by -1. The 5th diagonal term retains its positive sign since it is multiplied twice, once for the row and once for the column.
- 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 ].
Relationship to a one-joint (grounded) link
A one-joint link provides a 6x6 stiffness between a single joint j and the ground. This is the same is the same as using a two-joint link and fully restraining joint i. The stiffness for the one-joint link is thus given by [Kj j ]. This means that the same link property can be used for a one-joint link or a two-joint link that has its first joint fully restrained. The same consideration for the sign of the 5th row and 5th column applies due to V2 = -F2.
See Also
Analysis Reference Manual (The Link/Support Element - Basic) – further information on the general use of links
- Derivation of link equations article