3 Matrix Displacement Formulation 3.1 INTROD INTRODUCT UCTION ION Though mathematicians, physicists and stress analysts wored independently in the !ield o! "#$, it is the matri% displacement !ormulation o! the stress analysts which lead to !ast de&elopment o! "#$. In!ac t till the word "#$ 'ecame popular, stress stress analyst wored in this !ield in in the name o! matri% displacement displacement method. In matri% displacement method sti!!ness matri% o! an element is assem'led ' y direct approach while in "#$ though d irect sti!!ness matri% may 'e treated as an approach !or assem'ling element properties (sti!!ness matri% as !ar as stress analysis is concerned), it is the energy approached which has re&olutioni*ed entire "#$. +ence in this chapter, a 'rie! e%planation o! matri% displacement method is presented and solution techniues !or simultaneous euations are discussed 'rie!ly.
3.- $TRI/ $TRI/ DI02C#$#NT DI02C#$#NT #UTION0 #UTION0 The standard !orm o! matri% displacement euation is, 456 {8} = {F} where 46 is sti!!ness matri% 789 is displacement &ector and {F\ is {F\ is !orce &ector in the coordinate directions The element k.. o! k.. o! sti!!ness matri% may'e de!ined as the !orce at coordinate i due to unit displacement in coordinate direction j. direction j. The direct method o! assem'ling sti!!ness matri% !or !ew standard cases is 'rie!ly gi&en in this article.
1. :ar #lement Common pro'lems in this category are the 'ars and columns with &arying cross section su';ected to a%ial !orces as shown in "ig. 3.1. "or such 'ar with cross section , shortening 8 is gi&en 'y
8 =
PL EA
T L,
4
(a) ()
A, E 1
55
L L _
!" 1~*
P N- - - - - P
I- - - - -► I1
I
()
(a)
"ig. 3.1 (c)
"ig. 3. EA ? — o £
EA If 5 = 1, P =
L
:y gi&ing unit displacement in coordinate direction 1, the !orces de&elopment in the coordinate direction 1 and - can 'e !ound ("ig. 3.- (')). +ence !rom the de!inition o! sti!!ness matri%, EA , EA , EA EA — and and 5-i =~ — 0imilarly gi&ing unit displacement in coordinate direction - (re!er "ig. 3.- (c)), we get
8 =
EA L
M
Thus,
EA 1 EA 1
B1
~L B1 ~L B1
1
...(3.8)
-. Truss #lement $em'ers o! the trusses are su';ected to a%ial !orces only, 'ut their orientation in the plane may 'e at any angle to the coordinate directions selected. "igure 3.3 shows a typical case in a plane truss. "igure 3.@ (a) shows a typical mem'er o! the truss with
(i) Unit Unit displa displacem cement ent o! end end 1 in %Bdir %Bdirect ectio ion. n. Due to this, displacement along the a%is is 1 % cos as shown in "ig. 3.@ ('). +ence !orces de&elopment at the ends are as shown in !igure. EA EA P E E Fcos 9 L "rom the de!inition o! elements o! sti!!ness matri%, we get EA k u-P cos cos 9 - F - F cos- 9 11 L EA k,, - P sinO sinO E Fcos sin L
8 =
-1
) N I
"ig. 3.@ (ii) Unit displa displacemen cementt in coordinat coordinatee direction direction -G -G This case is shown in "ig. 3.@ (c). In this case a%ial de!ormation is 1 % sin 8 and the !orces de&eloped at each end #ire as shown in the !igure.
EA . n P E E Fsin 6 L
. EA . E P E P cos cos EBBBBBBBsin cos 51i
. B EA B EA . 2 k 12 E P sm sm E Fsin 8 12 E P H
HH
EA k ,, ,, E —P E —P cos cos EBBBBBBBBBBsin cos 3 L KA k d2 sinO sinO EBBBBBBBBBBsin- 8 d2 - —P @ L (iii) (iii) Unit displaceme displacement nt in coordinate coordinate direction direction 3, #%tension along the a%is is 1 % sin and hence the !orces de&eloped are as shown in the "ig. 3.@ (d) EA n P E E Fcos L EA h-i = —P cos cos EBBBBBBBBBcos- 8
8 =
EA ,, E —P E —P sin sin EBBBBBBBBBcos sin -3 L FA FA 2 E P costJ costJ EBBBBBBBcos 8 33 L 1
-n .
n
.. EA . P sin0 sin0 EBBBBBBBBBBBBBBBBB cos smd
i
K A Bi E
L (&i) Due to unit displacement in coordinate direction @, #%tension o! the 'ar is eual to l% sin, and hence the !orces de&eloped are as shown in "ig. 3.@ (e). EA . n P E E Fsin 6 L
I@
EA . n n -P co costJ E suitJ costJ L
EA sin EBBBBBBBBsin- 8 h,A = —P sin -@ L . EA . %3@ E P E P cos8 cos8 E Fsind cose 7
7
.
. EA . - k AA = P sind E Fsm 8 L
The sti!!ness matri% is cos- 8 #$% 8 sin 8 sin Bcos- 8 cos8 cos8 sin8 sin8 sin EA L
8 Bcos- 8 Bcos 8 sin K
_ EA ~L
Bcos8 Bcos8 sin8 sin8 -#$% 8 s 8 siin
-
cos
sin- 8
P> &
&!
#$% 8 sin K ' 2
Bsin- 8 cos 8 sin- 8
& -
-'
!2 —
'
&
—
-!2
&
!
2
Bcos 8 sin
-
.(3.K)
Lhere > and ! are the direction cosines o! the mem'er i.e. > E cos 8 and ! = cos = cos (M B 8) E sin 8 " (&)
:eam #lement In the analysis o! continuous 'eams normally a%ial de!ormation is negligi'le (small de!lection theory) and hence only two unnowns may 'e taen at each end o! a element ("ig. 3.8). Typical element and the coordinates o! displacements selected are shown in "ig. 3.8 ('). The end !orces
de&eloped due to unit displacement in all the !our coordinate directions are shown in "ig. 3.K (a, ', c, d). 1-#H
1
y (a) 1- E' 1- E' L
)
AE&
3
8
A K E'
H
L
M 2
2
(a)
6 E&
)
K E'
L
()
L2 1- E' 1- E'
6E&
2E&
1}
K E' L2
L
. ......................
! A $l
6E& 2
.
.
1
+ A.
3
11-H "ig. 3.8
yB K E'
AE& &
"ig. 3.K "rom the de!inition o! sti!!ness matri% and looing at positi&e senses indicated, we can write (a) Due to unit displacement in coordinate direction 1, ,
, ? 12E& k u-—-
, _6E&
, ? 12E& 531? 6E&
k A/ -
2E& A3@ VV L 4 (')%! Due to unit unit displac displacement ement in in coordinate coordinate directi direction on -, & ' & P 1"
A@@
A-@
&F
K>H 0E& &
E
, r _6E& , F& , F&
2E& k
02 ~
? AE& 6E& & '& 0L ? AE& 21 S1- ~ S3- B ' '& 1 '& (c) Due to unit1"displacemen displ acement t in coordinate ate directi direction on " coordin L L & '& 21 01 3, G2 S-- B —
-
-
L
LB ,?
?
1-H
S13
, ? 6E&
, ? 1-H
S-3
S33
?
L
?
L
...(3.H) _ 6 E' L2
~y~ L
(d) Due to unit displacement in coordinate direction @ , I! a%ial de!ormations in the 'eam elements are to 'e considered as in case o! columns o! !rames, etc. ("ig. 3.H), it may 'e o'ser&ed that a%ial !orce do not a!!ect &alues o! 'ending moment and shear !orce and &ice &ersa is also true. +ence sti!!ness matri% !or the element shown in "ig. 3.Q is o'tained 'y com'ining the sti!!ness matrices o! 'ar element and 'eam element element and arranging in proper locations. "or "or this case
EA L
# #
2E& 1} 6E L
-
6 2
L 0
L
2
6
L2 6
L2 2
L2
L
E A E A
1-
K
1} 6E
L
2
L
L
2
2E
6 L 0
L2 6E L2
L
...(3.Q)
EA L
# #
11 (')
(a)
"ig. 3.H
1 3
B
The !ollowing special !eatures o! matri% displacement euations are worth notingW (i) The matri% matri% is ha&ing diagonal diagonal dominanc dominancee and is positi&e positi&e de!inite. de!inite. +ence +ence in the solution solution process there there is no need to rearrange the euations to get diagonal dominance. (ii) The matri% is symmetric. It is o'&ious !rom $a%well=s reciprocal theorem. +ence only upper or lower lower triangular elements may 'e !ormed and others o'tained using symmetry. (iii) The matri% is ha&ing 'anded nature nature i.e. the non*ero elements o! sti!!ness matri% are concentrated near the diagonal o! the matri%. The elements away !ro m the diagonal are *ero. Considera'le s a&ing is e!!ected in storage reuirement o! s ti!!ness matri% in the memory o! computers 'y a&oiding storage o! *ero &alues o! sti!!ness matrices. The 'anded nature o! matri% is shown in "ig. 3.M. "ig. 3.M
In this case instead o! storing storing / si*e si*e matri% only only % 3 % 3 si*e si*e matri% can 'e stored.
3.3 0O2UTION O" $TRI $TRI/ / DI02C#$#NT #UTION0 #UTION0 The matri% displacement euations are linear simultaneous euations. These euations can 'e sol&ed using Xaussian elimination method. 2et the euations to 'e sol&ed 'e
S1 1 S1
S1-
S13
S--
S-3
Z
Sli
S-
k
S
1 S
i
-n
-
5 ;
S 1
k2
?Sn l
" ;kk
S53
;
n2
" ;nk
3
Sn
5 -
%
> k
* k
h
<
n
n
...(3.-)
" S++?
i.e. 4A5 4A 5 {} = 7!t9 The Xauss elimination method consists in reducing matri% to upper triangular matri% and then !inding the &aria'les &aria'les n , % ..., ..., ...., ...., 2 , %, 'y 'ac su'stitution 7( 1: To eliminate eliminate %, in the lower euationsW (i) "irst "irst euat euation ion is is mainta maintaine ined d as it is (ii) (ii) "or eua euatio tions ns 'elo 'elow w 1, 0 ?=;ij-
-@ij
5ii and bjp E !t, B F
ft
t the end o! this, the euations will 'e S1 1
S1-
Si5
S13
01
S- B
S ; i '
5
01
01
03
" 02
k
E
!
/
01
01
" 01
01_
n .
!t(1) i n
The a'o&e process is called pi&otal operation ona n. "or pi&otal operation on ;, no changes are made in 5 row y
ij
atl) a t l )
,(5BD
!or i,j = kB 1, ...,n.
'ut !or the rows 'elow >,(5B1) t
and <- k =
k
<4 A !or* E Y 1, ...,n.
;
kk
!ter n - 1 pi&otal operations, matri% euation is o! the !or m
S 1
;n
S13 S-
S-
S-
Si5 ... S- B a[...
S
;
.. . .. .
+2
? S
5 5 3
;? S- n
\
;
kk
...
3n
;
kn
!lOB1) u
nn
P < ' )
1S
5
<
E B
i!
...(3.3)
* k /
n .
P S P O 1 S
"rom the last euation, and then, n <
i-
* C j ;
---BBBBB ---BBBBBH P Y 1BBBBBBB, i - n F 1, n B -...1 ;C
...(3.@)
Thus the reuired solution is o' tained.
3.@ T#C+NIU#0 O" 0IN 0INX X CO$UT#R $#$OR< $#$OR< R#UIR#$#NT0 R#UIR#$#NT0 In "#$ si*e o! sti!!ness matri% o! si*e 1 % 1 or e&en more is not uncommon. +ence memory reuirement !or storing sti!!ness matri% is &ery high. I! user tries to implement Xaussian elimination straight way as descri'ed a'o&e, ends up with the pro'lem o ! shortage o! memory. The !ollowing techniues are used to reduce memory reuirement !or storing the sti!!ness matricesW (i) Use o! symm symmetry etry and 'anded 'anded nature nature (ii) artitionin artitioning g o! matri% matri% ("rontal ("rontal soluti solution). on). (iii) (iii) 0yline 0yline storage. storage.
(i) Use o! 0ymmetry and :anded Nature 0ince the sti!!ness matri% is always symmetric and 'anded in nature, techniues ha&e 'een de&eloped to store only semi'and width o! nonB*ero elements and get the solution. I! : is the semi'and width o! - - matri% we need to store only only % 3 elements as indicated in "ig. 3.M('). The diagonal o! the gi&en matri% is stored as the !irst column o! the modi!ied matri%. The computer coding is modi!ied to use modi!ied matri% !or the solution o! the gi&en pro'lem. The modi!ication reuired is, @D =
a%
+ j ~ i B 1)
(ii) artitioning o! the $atri% "or larger systems, e&en this method o! storage may 'e inadeuate. In such cases the partitioning o! the matri% is made as shown in "ig. 3.1.
