312 COMPUTER AIDED ENGINEERING DESIGN
Adding Eqs. (11.5a) and (11.5b) yields
kk
kk k k
kk
u
u
u
f
ff
f
F
F
F
pp
pp q q
qq
i
j
k
i
j l
k
i
j
k
–0
- –
0–
- –
= + =
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
(11.5c)
or in compact form
KU = F
whereF =
F
F
F
i
j
k
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
is the vector of net external forces acting on the nodes with respective subscripts,
K the global stiffness matrix and U the global displacement vector. Note that the element stiffness
properties are inherited by the global stiffness matrix K, in that the latter is also singular, symmetric
and positive semi-definite. Singularity of the stiffness matrix implies that the linear system in Eq.
(11.5c) has at least one rigid-body degree of freedom and the system cannot be solved unless some
displacements are known or constrained a priori. We can further simplify Eq. (11.5c) as
k
ku
k
kk
k
uk
k
u
F
F
F
p
pi
p
pq
q
jq
q
k
i
j
k
- 0
+
+
0
- =
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
(11.5d)
Assuming that node i is fixed so that ui = 0, Eq. (11.5d) becomes
+
0
- =
k
kk
k
uk
k
u
F
F
F
p
pq
q
jq
q
k
i
j
k
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
(11.5e)
Fi represents the reaction force at node ithat depends on displacements uj and uk (only uj in this case).
To determine only the displacements, we need to solve
kk
k
u
k
k
u
F
F
pq
q
j
q
q
k
j
k
+
+
- =
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
⎥ or
kk k
kk
u
u
F
F
pq q
qq
j
k
j
k
+ –
=
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
⎥
Alternatively, the above set of equations can also be obtained by eliminating the first row (entirely)
and first column of the coefficient/stiffness matrix (corresponding to the fixed degree of freedom ui)
in Eq. (11.5c). Further solving gives
u
u
kk k
kk
F
F kkkk
kk
kkk
F
F
j
k
pq q
qq
j
k pqqq
qq
qpq
j
k
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
⎥
⎡
⎣
⎢
⎤
⎦
= ⎥
+ –
=^1
{( + ) – } +
–1
2
or u
k
j FF
p
= j k
(^1) ( + )