12x12 stiffness of two-joint links

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:

1. Draw a two-joint link object which connects the two points. The first joint is denoted i and the second joint is j.
   
   
2. 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.
   
   
3. Transform the given stiffness matrix to the link local coordinate system as necessary.
   
   
4. Partition the stiffness matrix as follows:
   
   
   where:
   
   
   • { F_i } and { F_j } are the 6 forces and moments at joints i and j.
     
     
   • { U_i } and { U_j } are the 6 displacements and rotations at joints i and j.
     
     
   • [K_{i i} ], [K_{i j} ], [K_{j i} ], and [K_{j 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.
     
     
5. Define a link property, then set its type to Linear.
   
   
6. Activate all 6 DOF and set the two shear distances to zero.
   
   
7. For the link stiffness properties, use the values from the [K_{j j} ] sub-matrix. Due to symmetry, only the upper triangle of the sub-matrix needs to be entered (21 values).
   
   
8. 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 { F_i } and { F_j } should be in equilibrium under any displaced configuration. In the simplest of terms, { F_i } and { F_j } 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 { U_j }, the equilibrium relationship between { F_i } and { F_j } enables the determination of [K_{i j} ] from the given [K_{j j} ].

• By symmetry, [K_{j i} ] = [K_{i j} ]^T.

• For arbitrary { U_i }, the equilibrium relationship between { F_i } and { F_j } enables the determination of [K_{i i} ] from the given [K_{j i} ].

See Also

• Unknown macro: {new-tab-link}

  CSI

  Analysis Reference Manual (The Link/Support Element - Basic, page 229) – further information on the general use of links