312 COMPUTER AIDED ENGINEERING DESIGN
Adding Eqs. (11.5a) and (11.5b) yieldskk
kk k k
kku
u
uf
ff
fF
F
Fpp
pp q q
qqi
j
ki
j l
ki
j
k–0
- –
0–
- –
= + =⎡⎣⎢
⎢
⎢⎤⎦⎥
⎥
⎥⎡⎣⎢
⎢
⎢⎤⎦⎥
⎥
⎥⎡⎣⎢
⎢
⎢⎤⎦⎥
⎥
⎥⎡⎣⎢
⎢
⎢⎤⎦⎥
⎥
⎥(11.5c)or in compact form
KU = FwhereF =
F
F
Fi
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
kuk
kk
kuk
kuF
F
Fp
pip
pq
qjq
qki
j
k- 0
++0- =
⎡⎣⎢
⎢
⎢⎤⎦⎥
⎥
⎥⎡⎣⎢
⎢
⎢⎤⎦⎥
⎥
⎥⎡⎣⎢
⎢
⎢⎤⎦⎥
⎥
⎥⎡⎣⎢
⎢
⎢⎤⎦⎥
⎥
⎥(11.5d)Assuming that node i is fixed so that ui = 0, Eq. (11.5d) becomes
+0- =
k
kk
kuk
kuF
F
Fp
pq
qjq
qki
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
kuk
kuF
Fpq
qjq
qkj
k++- =
⎡
⎣⎢⎤
⎦⎥⎡
⎣⎢⎤
⎦⎥⎡
⎣⎢⎤
⎦⎥ orkk k
kku
uF
Fpq q
qqj
kj
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
ukk k
kkF
F kkkkkk
kkkF
Fj
kpq q
qqj
k pqqqqq
qpqj
k⎡
⎣
⎢⎤
⎦
⎥⎡
⎣⎢⎤
⎦⎥⎡
⎣
⎢⎤
⎦
⎥⎡
⎣⎢⎤
⎦⎥⎡
⎣
⎢⎤
⎦= ⎥+ –=^1
{( + ) – } +–1
2or u
k
j FF
p
= j k(^1) ( + )