202
multi-degree of freedom linear structures
The constant kl} is defined as the force that is required at node i to counteract a unit elastic displacement ^SJ = 1 imposed at node j, under the condition that ail displacements Ç3l = 0 for i / j. In this single displacement case,
_ fctJ(l) = kij, i ± j. If such a displacement condition is applied sequentially
to each node, then the net force at each node j can be obtained by superposition. For instance, the elastic force qsl at node 1 is a linear combination of the
forces at node 1 resulting from the unit displacements at ail nodes, or
qsl = fcllGl + l'iîC* +-----+■
(8.13)
Likewise, superposition gives the elastic force qS2 at node 2 as
<7s2 = ^21€sl + ^22^S2 d-----------^27v€sN
By induction, the force qsl resulting from deflections at ail nodes is
N
qsi = fcilGl + ki2^s2 d-------b kijZsj d-------h ^iN^sN =
(^.15)
J=1
Equation (8.15) can be expressed in matrix notation in the following two ways:
9si
qS2
fcll
fc2l
^12
•
^22
■
■
klN
•
k2y
' î.i ’
€s2
QsN _
^N2
• •
kNN _ . ZsN .
(8.16)
(8.17)
Example Problem 8.4. In practice, a reliable structural computer package
(SAP 2000, for instance) is used to compute the K matrix. This general procedure is illustrated for the simple jacket template structure of Example Problem
8.1, shown in Figure 8.1. A plane frame, static force-displacement analysis is
made in the two stages depicted in Figures 8.2. In the first stage, Figure 8.2a, a
unit displacement
= 1 is imposed at level 1, and £s2 = 0 is imposed at level
2. The corresponding forces feu and
necessary to achieve those displacements are computed. In the second stage, Figure 8.2b, £sl = 0 and ^s2 = 1, for
which the respective forces at levels 1 and 2 are fc12 and k22. Each such force is
assumed to be positive in the direction of its coordinate. If a force is opposite
to the direction of its coordinate, then that force is a négative number. In this
problem, it is intuitive that A42 (= /c2i) is a négative number. The stiffness
matrix for this example is
K -
^11
k\2
/C21
k22
(8.18)
In the computer analysis of such a frame, a typical structural package requires such input data as the coordinate locations of the junctions between the
multi-degree of freedom linear structures
The constant kl} is defined as the force that is required at node i to counteract a unit elastic displacement ^SJ = 1 imposed at node j, under the condition that ail displacements Ç3l = 0 for i / j. In this single displacement case,
_ fctJ(l) = kij, i ± j. If such a displacement condition is applied sequentially
to each node, then the net force at each node j can be obtained by superposition. For instance, the elastic force qsl at node 1 is a linear combination of the
forces at node 1 resulting from the unit displacements at ail nodes, or
qsl = fcllGl + l'iîC* +-----+■
(8.13)
Likewise, superposition gives the elastic force qS2 at node 2 as
<7s2 = ^21€sl + ^22^S2 d-----------^27v€sN
By induction, the force qsl resulting from deflections at ail nodes is
N
qsi = fcilGl + ki2^s2 d-------b kijZsj d-------h ^iN^sN =
(^.15)
J=1
Equation (8.15) can be expressed in matrix notation in the following two ways:
9si
qS2
fcll
fc2l
^12
•
^22
■
■
klN
•
k2y
' î.i ’
€s2
QsN _
^N2
• •
kNN _ . ZsN .
(8.16)
(8.17)
Example Problem 8.4. In practice, a reliable structural computer package
(SAP 2000, for instance) is used to compute the K matrix. This general procedure is illustrated for the simple jacket template structure of Example Problem
8.1, shown in Figure 8.1. A plane frame, static force-displacement analysis is
made in the two stages depicted in Figures 8.2. In the first stage, Figure 8.2a, a
unit displacement
= 1 is imposed at level 1, and £s2 = 0 is imposed at level
2. The corresponding forces feu and
necessary to achieve those displacements are computed. In the second stage, Figure 8.2b, £sl = 0 and ^s2 = 1, for
which the respective forces at levels 1 and 2 are fc12 and k22. Each such force is
assumed to be positive in the direction of its coordinate. If a force is opposite
to the direction of its coordinate, then that force is a négative number. In this
problem, it is intuitive that A42 (= /c2i) is a négative number. The stiffness
matrix for this example is
K -
^11
k\2
/C21
k22
(8.18)
In the computer analysis of such a frame, a typical structural package requires such input data as the coordinate locations of the junctions between the