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Within [SAP2000|sap2000:home], [CSiBridge|csibridge:home], and [ETABS|etabs:Home], a [link|kb:Link] object may be used to manually input a known 12x12 stiffness matrix which represents the connection between two [joints|kb:Joint].


h2. Modeling procedure

A two-joint link may be modeled and assigned a 12x12 stiffness matrix as follows:
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# Draw a two-joint link object which connects the two points. The first joint is denoted _i_ and the second joint is _j_.
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# 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.
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# Transform the given stiffness matrix to the link local coordinate system as necessary.
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# Partition the stiffness matrix as follows:
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where:
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#* \{ _F{_}{_}{~}i{~}_ \} and \{ _F{_}{_}{~}j{~}_ \} are the 6 forces and moments at joints _i_ and _j_.
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#* \{ _U{_}{_}{~}i{~}_ \} and \{ _U{_}{_}{~}j{~}_ \} are the 6 displacements and rotations at joints _i_ and _j_.
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#* \[_K{_}{_}{~}i i{~}_ \], \[_K{_}{_}{~}i j{~}_ \], \[_K{_}{_}{~}j i{~}_ \], and \[_K{_}{_}{~}j j{~}_ \] are 6x6 sub-matrices of the full 12x12 stiffness matrix.
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#* The degrees of freedom (DOF) are ordered as (U1, U2, U3, R1, R2, R3) at each [joint|kb:Joint]. Note that these are in the link local coordinate system, not the joint local coordinate system.
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# Define a [link|kb:Link] property, then set its type to Linear.
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# Activate all 6 DOF and set the two shear distances to zero.
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# 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).
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# Assign this link property to the link object.


h2. 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 dj{~}2~ and dj{~}3~.


h2. 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 \[_K{_}{_}{~}j i{~}_ \].


h2. See Also

* {new-tab-link:http://www.csiberkeley.com/}CSI{new-tab-link} [_Analysis Reference Manual_|doc:CSI Analysis Reference Manual] (The Link/Support Element - Basic, page 229) -- further information on the general use of links

* [doc:Derivation of link equations] article