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7/23/2019 Stiffness and Displacement Method
1/19
.
...
BRIDGE STRUDL M NU L
November 973
: : . : o : . : . : . ~ ; : &Ai , au;
.a
PPENDIX B
STIFFNESS
OR DISPL CEMENT
METHOD
Sect ion
B l
B.2
B.3
ONTENTS
In t roduct ion
Basic
Displacement
Approach
Using
Example
Problem 2.3
Direc t St i f f n e s s
Approach
Using
Example Problem
2.3
Page
B 2
B 8
B 14
B-1
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APPENDIX B
STIFFNESS OR DISPLACEMENT METHOD
B . l In t roduc t ion :
I . Br ie f Discussion
o f
Force o r F l e x i b i l i t y
Method
Inde terminate systems
comprise the
l a rge
major i ty
o f s t ruc tu res to be analyzed and designed
and
hence
the
so lu t ion
process must s a t i s fy
the condi t ions
o f compat ib i l
i t y
and the mater ia l s t r e s s - s t r a i n
behav ior .
Trad i t iona l
methods o f
s t ru c t u ra l
ana lys is
employing
the concept
o f
redundancies
and
cons i s t en t deformations have not proven
to
be as
s imple and d i r e c t
in
appl ica t ion as the s t i f fness
o r
disp lacement
approach
to
be
t r e a t ed
h e re
and
a l so
used
in the
STRUDL
program.
The t r a d i t i o n a l
method
invo lv ing
redundancies has been formal ized in to a matr ix
approach
and
i s
now r e fe r red to as the
fo rce method.
a.
Propped Can t i l ev e r Example:
The fo rce method
o f s t ru c t u ra l
ana lys is
(often re fe r red to as
the
f l e x i b i l i t y
method )
i s
probably
most fami l i a r to us
for the
so lu t ion o f
s t a t i c a l l y indeterminate
s t ruc tu res .
The
propped
can t i l eve r
beam o f Figure
B . l a
prov ides a s imple example o f the use o f
the
fo rce method.
rnA
o l
L
.
t
R
1
Fig B la
A
concen t ra ted
load,
P,
i s
ac t ing
a t
a
dis tance
aL
from
the
l e f t support . This load produces
the reac t ions RA
MA
and
PB
as
shown.
Since we have only two
equat ions
o f
equ i l ib r ium,
LFv=
and l:M
=0
and
t h ree unknown
reac t ions ,
t h i s
beam
i s considered to be inde terminate to the f i r s t degree. To
gain an
add i t iona l
equat ion we cons ider the de f l ec t ions o f
the
s t ruc tu re . The t r a d i t i o n a l
way to approach t h i s problem
i s
to remove one o f
t he
redundant
reac t ions ,
in t h i s case
B 2
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and determine
the
deflec t ion
8
0
a t B
on the s ta t i c a l ly
determinate cant i lever due to the external load P, Figure
B.lb . Since our actual
s t ruc ture does
not have a ver t ica l
deflec t ion
a t B
the
redundant react ion,
~ must
be
of such
Fig.
B.lb
Fig. B.lc
a magnitude tha t t pushes the beam of Figure B. lb upward
with
a displacement equal to
Sc.
I f we apply a uni t value
of the
redundant
to the cant i l eve r shown in Figure B.lc
we
wil l
have a
deflec t ion a t
B upward
equal to
8o
.
Therefore
we can
wri te
This
i s our compat ibi l i ty
equation
saying
tha t
the
def lec t ion
a t
B
i s zero. Here
891
i s
the
ver t i ca l
deflec t ion
a t
B
due
to
a
uni t
load a t
B. We solve Eq. 1
for
~
=
(2)
Having al lows us
to
determine M and
RA
by s ta t i c s .
b . Four
Span Beam
Example: For
a beam
with
a
la rger
number of
redundancies
we
could
proceed
in
a
very
s imi lar manner. For example,
consider
the four span beam of
Figure
B.ld. In t h i s
case
we can
consider
R
1
R
2
R
3
and
R
4
as the redundants, leaving us with the cant i lever beam
of
Figure
B
. l e .
30
Fig. B.ld
Fig. B.le
B-3
4
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The appl ied loads
produce
de f l ec t ions 8
10
,8
20'8
30
and8
4
o.
As before , these de f l ec t ions do not represen t the
t r u e
s t a t e
o f our s t ruc tu re so
w
must cons ider t h a t the
redundants
push
upward j u s t enough
to
e l imina te these displacements . In t h i s
ins tance
we
s h a l l a r r i v e
a t four
compat ib i l i ty cond i t ions .
For example,
applying
a u n i t
load
a t
suppor t
1 yie lds d e f l ec
t i ons
811
t
821
' 831
and841
( see
Figure
B . l f ) .
Simi la r ly ,
a
un i t load
a t poin t
2 y ie lds 8 2, 8
22 '8
32,
8
42.
1
Fig. B.lf
Fig.
8.1g
We could cont inue
applying
the u n i t load a t each p o i n t
and
determining the def lec t ions .
Our
co mp a t i b i l i t y equat ions
become
1o
+
R1
&11
+
R2
~ 2
+ R3 ~ 1 3 + R4
s14
=
0
d20
+ R1
S21
+
R2
~
+
R3
523
+
R4 S 24
=
0
(
3)
~ 3
+
R1
b31
+ R2
~ 3 2
+ R3
33
+ R4 634
=
0
b4
+ Rl
0
41
+ R2 S42
+
R3
s43
+
R
4
S44
=
0
Note
t h a t 8ij - i s
the de f l ec t ion a t support
due
to
a
u n i t
load a t
suppor t
j . The so lu t ion o f these four equat ions gives
values fo r the redundants R
1
, R
2
, and R
4
One
t h ing
t h a t
3
we can observe s t h a t
in
order to determine t h e
redundants
we
must ca lcu la te
t he
de f l ec t ions a t a l l o f
t he
redundant
poin t s
fo r a l l pos i t ions o f
t he
u n i t l o a d - . -
I I . B r i e f Discussion o f t h e St i f fnes s o r Displacement
Methods
The prev ious discuss ion has
d ea l t
with the f l e x i b i l
i t y
o r
fo rce
method o f ana lys i s .
vl can
handle
the same
problems by cons ider ing the s t i f fn e s s o r
di sp lacement
method
of ana lys is . In t h i s
case
we
t ake
the
unknown di sp lacements
of the
s t r u c tu r e
as t h e redundants .
a .
Propped
Can t i l ev e r Example: Again cons ider
the
propped can t i l ever beam o f
Figure
B. lh .
B-4
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Fig. B lh
F.ig B lc
In
t h i s
case
the only unknown
displacement i s
the
r o t a t i o n ,
a a t end B. We then say t h a t t h i s
s t r u c tu r e
only has one
degree o f
freedom. In
order to e l imina te
t h i s
unknown d i s -
placement we clamp
the
end. The app l ied loads then produce
f ixed end
moments MA
and
as shovm in Figure
B
li
How-
ever , we know t h a t
t h i s
i s not
the ac tua l cond i t ion o f our
s t ruc tu re .
The
redundant
ro ta t ion ,
Be
,
produces
a
moment
o f
magnitude
equal to M b u t o f o p p o s ~ t e
d i r ec t i o n .
We can
cons ider the
e f f e c t
of t h i s
r o t a t i o n by
cons ider ing the e f f e c t
of a
un i t ro ta t ion
a t
end
B
Figure B . l j .
1
.
~
Fig.
B lj
Then our
compat ib i l i ty
condi t ion becomes
e + mss e =o
3)
e ~
4)
ms
where ~ B i s
the
moment a t
B
due to
a
u n i t r o t a t i o n
a t
B.
Having BB
,
we can
determine
a l l
other moments
and
r eac t ions .
For
example,
the
moment
a t
A, MA i s given by
M = F M , m
5)
A
B
B
where
mAB i s the moment
a t
A due to a u n i t r o t a t i o n
a t
B.
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b . Four Span Beam Example: This may
b e
a s t r an g e
way
to look a t t h i s problem because it s normally more
d i f f i c u l t t o c a l cu l a t e
the reac t ion
caused by
a
u n i t dis- .
placement than to ca l cu l a t e the displacement
caused
by a
u n i t
reac t ion .
Hm.,ever, we sha l l
soon
see t h a t t he re s
an
advantage in looking
a t
the
problem from t h i s p o in t
o f
view.
Consider
again
the cont inuous beam
o f
Figure B. lk . e see
t h a t
t he re a re fOUr
Unknown rota t iOnS,
81 2 I 3
1
4
Fig.
B.lk
e beg in
by
f ix ing
a l l
suppor t s
aga ins t
ro ta t ion and
d e t e r
mine t he
FEM's
(Figure B. lL) .
A ___,
8' c
f -,, , / ~ - , -'- ' , S f
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Ne can cont:.J.nue to apply the
un i t
ro ta t ion and ge t
t h r ee
addi t iona l
co mp a t i b i l i t y equat ions fo r example, a t j o i n t 2
o =
e
M
8
+
m
e +
m
e +
m
(7)
20
20 2 1 22 2 23 93
Solving
t h i s system
o f equat ions
g ives va lues fo r 8 82 8
and
8
The th ing to note i s t h a t these equat ions only
involve the e f f ec t s produced by members
ad jacen t to
the j o i n t
in
quest ion . In othe r -v1ords
we
do
not have
to determine
ef fec t s on the s t r u c tu r e due to ro t a t ions a t d i s t a n t p o in t s .
This a l lows an e f f i c i e n t
manner o f s to r ing the
problem in
the
computer
and al lows the
computer
t ime to
b e
reduced .
Furthermore , the
method
i s app l icab le to determinate
and
inde terminate systems wi th
equal
ease
and
the
degree o f
indeterminancy
need
not even be determined .
The
preced ing
discuss ion
o f
s t i f fn e s s
method
was
presen ted
to
give
an
overview of the method. We
s h a l l
nex t
cons ider
a more
d e t a i l ed
app l i ca t ion of the s t i f f ne s s o r d i s -
placement approach.
B-7
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.
B.2 Basic Displacement Approach Using Example Problem 2.3
Inde terminate Truss
~ ~ ~ ~ P 4 _ a
Fs Us
t ~ t ~ t . tR3,A3 '
I
8'
. JOINT
LOADS
BAR FffiCES
GEOMETRY
AND
DISPLACEMENTS
AND ELQNGATIONS
fig. B 2a
-
.
Note t ha t the
exte rna l
appl ied loads P, have a
one-to-one
correspondence
with
the
ex te rna l
j o i n t
d isp lace
ments, X, and the
i n t e r n a l
b a r forces
F, have
a one- to -one
correspondence with
the
member
e longa t ions u.
Also note
the one-to-one correspondence between unknown reac t ion com
ponents ,
R and known support
displacements
(not neces sa r i ly
zero) .
a . Equi l ibr ium Matr ix: Rewri t ing the equi l ibr ium
matr ix
shown
on
Page
A-10,
to
inc lude
the add i t i ona l b a r gives
F11
F12l
3 ..
0._
0
-1
0
0
0
: 8
F21
F22
-4. -10.
1.
0
0
0
0
.6
F31 F32
tP}
=
[A)
( F}
o
-4.
=
0
1
0
.8
0
.o
0.
o
0
0
0 0
1
.8
F41
F42
-10
0:
0
0 1
.6.
0
0
Fs1
F;;2
(8)
F51
Fs2
F11.
F12
-1 0 0
- .6
0
0
11
R12
F21
F22
R21 . R22
=
0
0
"()
-.
8
-1
0
F31
F32
R31
R32
0
0 .1
0
0 - .6
{Rj
[AR] [F
F41
F42
Fs1
Fs2
Fs1
F52
B-8
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b .
Compat ib i l i ty Matr ix : The equ i l ib r ium
matr ix
i s no longer
square
A
5
x
6
} and
hence cannot be
i nve r t ed .
The degree
o f
indeterminacy
i s
NF - NP
=
6 - 5
=
1
Therefore , we must apply the s t r e s s - s t r a in r e la t ionsh ips and
the
condi t ions
of
compat ib i l i ty .
The
s t r e s s - s t r a i n
assumption
fo r
ax i a l l y
loaded
members i s
s imply
r
X
= Eex or Fx = E
Ux
or
(9)
Ax
L
One
such
equat ion can be wri t t en
for
each
ba r , hence
:: .
tF) 6x2
[5]
6x6
{u
1
6x2
l
u11
u12
EA1 L1
u21
u22
EA2 L2
EA3 L3
ZEROS
u31
u32
EA41L4
(10)
u41
u42
tFl
=
EA
5
!L
5
EA
6
EROS
u51
u52
us1 u62
Applying
compat ib i l i ty condi t ions to a
t r u s s
means
s imply t h a t the member
elongat ions
u ~ must be cons i s ten t
with the
j o i n t
displacements [x and ~ }
.
u1
=
b11x1 b12x2
b15x5
b 1 6 ~ 1
b1a ~ 3
.
u2
=
b21x1
b22x2
b25x5
b 2 6 ~ 1
u3
=
b31xl
b36 ~ 1
1 1
u4
=
b41x1 b46
~ 1
us
=
b51x1
b 5 6 ~ 1
us
=
bslxl
b62x2
bssxs
b66
~ 1
bsa D-3
(12)
B-9
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What
we
a re
saying i s t ha t
b a r e longat ions
are some l i ne a r
combination
o f
the
exte rna l
j o in t displacements .
To obta in
the
co e f f i c i en t s
o f [ B] def in ing
the
compat ib i l i ty
matr ix we may apply a un i t displacement in
t he
d i rec t ion o f each of the
exte rna l j o in t
disp lacements .
For
example,
xl =1 x
' 2
ul
=
bll
u2
b21
u3 b31
etc
t4
ul
bll
0
(small
deflections)
u2
=
bzl
-1.0
1
us
=
bs1
=
-0.8
u3
=
u
4
=
u
5
=
0
Fig. 8.2b
as another example s e t
1;
x
1
. =x
2
- A Ax
3
x A
4
=ul
=u2
= ~
=
u2
=
b25
=
0
(small
deflections)
u3
b35
=
+1.
u4
=
b45
=
+O.S
ul
= u5 = us
=
o.
Fig.
8.2c
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Each column
may b e determined success ive ly
t o y i e ld
0
1.
0 0
1.
0 0
-1.
0
1.
0 0
0 0
0
0
0 0 0 0
-1.
[8]=
0
0
.8
0
.6
[BRJ=
-. 6
-.
8 0
(13)
0 0 0 1.
0 0 - 1.
0
.8
.6
0
.8
0
0
0 -.6
(we've
j u s t
done columns
1
and
5)
c .
Rela t ionsh ip
o f Compat ib i l i ty
and
Equil ibr ium
Matrices : t i s
ext remely
i n t e r e s t i n g
and
s i g n i f i c a n t to
note a t t h i s t ime the t ranspose re l a t ionsh ip
between
the
equ i l ib r ium and
t he compa t ib i l i t y
matr ices o r
[8]
=
[A]T
[8R] = [ARJT (14)
This re l a t ionsh ip always holds fo r l i n e a r l y
e l a s t i c
s t r u c -
t u re s
and can
be
proved
by
the
p r in c ip l e o f
v i r t u a l
work.
d . System
S t i f f n e s s Matr ix:
In
summary
{P)
=
[A]
{F)
Equilibrium
R}
=
~ ] { }
(15)
{F)
=
[S] { }
Stress-Strain
(16)
{u}
=
[A]T
{xJ
[ARJT
Compatibili_ty
(17)
S u b s t i t u t e (17)
i n to
(16) to obta in
(18)
B-11
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Then s u b s t i t u t e
18 i n t o
15
P}
=
[ASAT
J {x} + [ A S A ~ ] fa} . (19)
{Pj - [ A S A ~ ] {a) = [K] (xj
{x} = [K]-1 [
PJ
[ A S A ~ ] ] (20)
1
J
=
[ARSAT] fxJ + A R S A ~
J fa)
(21)
f
a l l suppor t d isplacements a re zero the
bas ic so lu t ion
process fo r the example i s
(22)
(x}5x2 =
[ASATJ5!s (Plsx2
= [KJ5 s
{PJsx2
]
6x2
=
[SAT] 6x5
fxlsx2
2.3)
0 EA
1
0
0 0
Ll
-EA
2
0
EA
2
o.
0
0
0 0
0
EA
3
L3
[SAT]=
24)
0 0
.8EA
4
0 .6EA
4
L4
L4
0
0
0
EA
5
0
Ls
-..
8EA
6
.6EA
6
0
.8EA6
0
s
L
Ls
6
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nd
As
A
0
~ ~ - 6 ~
48
.
L Ls
Ls
L
-.
48 __
A
{I-+
.36
~ : - . 4 8 ~ - -
Ls
Ls
IK E) .
0
.48
L
(2
6
.
0
_
0
4 8 L '
. 4
.
.____...,___ ,
,___
These two matr ices
plus the load
matr ix
{PJ
a re what i s
required to so lve fo r t he displacements and fo rces in the
s t ru c t u ra l
system
under
inves t iga t ion .
The method i s
genera l ly
re fe r red to
as
the
displacement
method
because
disp lacements are the pr imary unknown quan t i t i e s .
Also
note
the symmetr ical condi t ion of the s t i f f ne s s
matr ix .
This
i s
proved by the r ec ip roc i ty
theorem.
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B.3 "Direc t
St i f fne ss" Approach
Using Example Problem 2 .3 ,
Inde terminate Truss
D i rec t
s t i f f n e s s simply
impl ies t h a t
one i s
going
to
ob ta in the
s t i f f n e s s
matr ix [ ~
withou t
genera t ing
the [A]
[s] ,
and
[B]
matr ices
and
t hen pe r fo rming
the matr ix
mul t i -
p l i ca t i o n
opera t ions .
This
method i s
much
more e f f i c i e n t
computa t ional ly
and
requ i res cons iderab ly l e s s
e f fo r t
in the
prepara t ion o f
da ta
i n p u ~
a .
Discuss ion
o f
the development o f the system
s t i f fn e s s matr ix d i r e c t l y from
phys ica l
co n s i d e ra t i o n s
To
motivate the development cons ider
the i n d e t e r -
minate t r u s s j u s t
i nves t iga t ed and
wri t e t h e bas i c s t i f f ne s s
equat ions as
Pl
=
K11X1
+ K12X2 + K13X3 +
K14X4
+
K1sXs
p2
=
K21X1 +
K22X2
.K23X3 +
K24X4
+
K2sXs
p3
=
K31X1
+
K32X2 +
K33X3
+
K34X4
+
K35X5
26)
p4
K41X1
+
K42X2
+ K43X3
+
K44X4
+
K 4 5 X ~
p
K51X1
+ K52X2 +
K53X3
+
Ks4X4
+
KssXs
5
Again these
co e f f i c i en t s
may
be
determined by
def in ing a se t
o f
values fo r the independent va r i ab le s x] ,
in
order
to
i s o l a t e
one
column
of
the
matr ix .
For
example,
i
x
1
= = =
o
then P
1
= K
11
, = K
21
,
2
=
x
3
=
x
4
x
5
P
2
P
3
= K
31
, P
4
= K
41
,
and
= K
51
.
Phys ica l ly t h i s means
t h a t
5
for
a
given
s t a t e o f displacement , what
a re
the r e q ~ i r e d
app l ied loads to produce t h i s
s t a t e?
Therefore, the s t i f f ne s s
co e f f i c i en t K
. .
i s def ined to be
t he load
a t
coord ina te
i given
l.J
a u n i t displacement a t coord ina te j , a l l
othe r
displacements
equal to zero .
For t h i s s t a t e o f
displacements
t h e
member e longat ions
are :
ul
=
0;
u2
-1;
u3
=
0;
Fig.
8.3a
u4.
0;
us
=
0;
u
6
=-o.a
B-14
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15/19
27
Hence
the assoc ia ted b a r
forces are
=
-0.8
EAs
Ls
Then
consider the
equ i l ib r ium
o f
the
j o i n t s
Fig. B.3b
EA2
K n + - L
2
EA
) + 0 . 8 - . 8 - - 2 ) = 0
s
EA
K
21
- o .o.s
< 0 . 8 ~ >
= o.
EA
K31 -(-L
2
) -0.8
(0)
=0
28)
2
K41
-0
-0.8 (-0.8 EA2 = 0
L2
K
51
o o.s = o
These
co e f f i c i en t s
a re
the
same as
those
obta ined in
the
f i r s t
column
o f the [K] matr ix
when
the
t r i p l e
matr ix mul t i
p l i ca t ion was
employed. I f
a
s imi la r
operat ion
i s employed
fo r each of the ex te rna l displacement coordinates
the
remaining four columns o f
the
s t i f f ne s s matr ix could b e
obta ined
and
would agree with
those
obtained prev ious ly .
~
Using t h i s concept to
develop the s t i f f ne s s matr ix
i nd ica t e s the
composi t ion
of the indiv idua l terms and a l so
c l e a r l y i de n t i f i e s which members o f the
system
w i l l con t r ibu te
to the indiv idua l coef f i c i en t s .
In p a r t i c u l a r , any given
member w i l l only con t r ibu te to those
co e f f i c i en t s
assoc ia ted
with
the
ex te rna l
coordinates of the
ends
o f the
member. The
co e f f i c i en t s Kii
w i l l
cons i s t
o f con t r ibu t ions from
each
member
framing
i n to
the
j o i n t assoc ia ted with
coordinate i
In more
genera l terms
each member
con t r ibu te s to the s t i f f ne s s o f t he
j o in t s i n to which they frame. This suggests t ha t
the s t i f f
ness matr ix could be genera ted from the s t i f f ne s s proper t i e s
o f
the component
p a r t s
o r as the summation o f the element
s t i f f ne s s matr ices .
B-15
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16/19
F i r s t
t ake
an ind iv idua l
t rus s
ba r sub jec ted to
an ax ia l
fo rce F.
y
GEOMETRY
Fig.
B.3c.
EQUILIBRIUM
STRESS STRAIN
COMPATIBILITY
pl
=
-Fi
cos o
7/23/2019 Stiffness and Displacement Method
17/19
, j
l
-
cos
2
cc:. i
.
,
cos oc::.
l .
s in oc:.
l .
I
cos
2
oc::
l .
cos
ex:.
l .
sinoc.
l .
E l e ~ e
'-.
2
2
-
cos:.
;
s in
oci
s in
ex: cos
oc.
sinoc.
s i n
oc:.
St i f fn
i
l . l . i
2
l.
cos oc:
s in
ex:.
cos
2
o