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. Note that these are in the link local coordinate system, not the joint local coordinate system.
     
     
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
• Derivation of link equations