180 CHAPTER 6 Matrix Methods
directioncosinesλandμtakedifferentvaluesforeachofthethreemembers,sorememberingthatthe
angleθismeasuredanticlockwisefromthepositivedirectionofthexaxis,wehavethefollowing:
Member θ λ μ
1–2 0 1 0
1–3 90 0 1
2–3 135 −1/
√
21/
√
2Thememberstiffnessmatricesaretherefore
[K 12 ]=
AE
L
⎡
⎢
⎢
⎣
10 − 10
00 00
−10 10
00 00
⎤
⎥
⎥
⎦ [K^13 ]=
AE
L
⎡
⎢
⎢
⎣
0000
010 − 1
0000
0 −10 1
⎤
⎥
⎥
⎦
[K 23 ]=
AE
√
2 L
⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
1
2 −1
2 −1
21
2
−^121212 −^12−^121212 −^12
1
2 −1
2 −1
21
2⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦
(i)Thenextstageistoaddthememberstiffnessmatricestoobtainthestiffnessmatrixforthecomplete
framework.Sincetherearesixpossiblenodalforcesproducingsixpossiblenodaldisplacements,the
complete stiffness matrix is of the order 6×6. Although the addition is not difficult in this simple
problem,caremustbetaken,whensolvingmorecomplexstructurestoensurethatthematrixelements
areplacedinthecorrectpositioninthecompletestiffnessmatrix.Thismaybeachievedbyexpanding
eachmemberstiffnessmatrixtotheorderofthecompletestiffnessmatrixbyinsertingappropriaterows
andcolumnsofzeros.Suchamethodis,however,timeandspaceconsuming.Analternativeprocedure
issuggestedhere.ThecompletestiffnessmatrixisoftheformshowninEq.(ii)
⎧
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎨
⎪⎪⎪
⎪⎪
⎪⎪
⎪⎩
Fx,1
Fy,1
Fx,2
Fy,2
Fx,3
Fy,3⎫
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎬
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎭
=
⎡
⎢
⎣
k 11 k 12 k 13
k 21 k 22 k 23
k 31 k 32 k 33⎤
⎥
⎦
⎧
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎨
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎩
u 1
v 1
u 2
v 2
u 3
v 3⎫
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎬
⎪⎪
⎪⎪
⎪⎪
⎪⎪
⎭
(ii)Thecompletestiffnessmatrixhasbeendividedintoanumberofsubmatricesinwhich[k 11 ]isa2× 2
matrixrelatingthenodalforcesFx,1,Fy,1tothenodaldisplacementsu 1 ,v 1 ,andsoon.Itisasimple
mattertodivideeachmemberstiffnessmatrixintosubmatricesoftheform[k 11 ],asshowninEqs.(iii).
AllthatremainsistoinserteachsubmatrixintoitscorrectpositioninEq.(ii),addingthematrixelements