Stiffness and Displacement Method

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


2/19

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


3/19

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


4/19

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


5/19

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.

B-5


6/19

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


7/19

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


8/19

.

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


9/19

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


10/19

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

B-10


11/19

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


12/19

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

B-12


13/19

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.

B-13


14/19

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


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


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


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


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


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


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

Documents

Stiffness and Displacement Method