displacement method of analysis

7/25/2019 Displacement Method of Analysis

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Slope Deflection Method for the

Analysis of Indeterminate Structures

ByProf. Dr. Wail Nourildean Al-Rifaie


2/52

All structures must satisfy:

oad-displacement relationship

!"uili#rium of forces

$ompati#ility of displacements


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%sin& the principle of superposition #yconsiderin& separately the moments

de'eloped at each support of a typical

prismatic #eam (AB) sho*n in +i&. ,(a) of acontinuous #eamdue to each of the

displacements and the

applied loads. Assume cloc*ise moments

are /i'e.


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,. Assume ends A and B are fi0ed i. e. the

rotations . 1his means that *e

ha'e to apply countercloc*ise moment at

end A and cloc*ise moment at end B

due to the applied loads to cause 2ero

rotation at each of ends A and B. 1a#le (,)

&i'es for different loadin& conditions.


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1a#le (,)


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3. Release end A a&ainst rotation at end A (rotates to

its final position ) #y applyin& cloc*ise

moment *hile far end node B is held fi0ed as

sho*n in +i&. ,.

4. No* the cloc*ise moment - rotation

relationship is:


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5. 1he carry o'er moment at end B is:

6. In a similar manner if end B of the #eam rotates to

its final position *hile end A is held fi0ed. 1he

cloc*ise moment 7 rotation relationship

is:


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8. 1he carry o'er moment at end A is:

9. If node B is displaced relati'e to as sho*n in +i&.

(,) so that the cord of the mem#er rotates

cloc*ise i. e. positi'e displacement and yet #oth

ends do not rotate then e"ual #ut anticloc*ise

moments are de'eloped in the mem#er as sho*n

in the fi&ure.


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Slope-Deflection !"uation

oad-displacement relationship

If the end moments due to each displacement and

the loadin& are added to&ether the resultant

moments at the ends may then #e *ritten as:


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+or prismatic #eam element e"uation (,) may #e

*ritten as:


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1he slope deflection e"uations (, or 3) relatethe unno*n moments applied to the nodes

to the displacements of the nodes for

any span of the structure. 1o summari2e application of the slope-

deflection e"uations consider thecontinuous #eam sho*n in +i&. (3) *hich has

four de&rees of freedom. No* e"uation (3) can #e applied to each of

the three spans.


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+i&. (3)


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+rom +i&.(3):

!"uili#rium conditions $ompati#ility conditions


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1hese e"uations *ould in'ol'e the four unno*n

rotations . Sol'in& for these four unno*n rotations. It may #e

noted that there is no relati'e deflection #et*een the

supports so that 1he 'alues of the o#tainedrotations may then #e su#stituted in to the slopedeflection e"uations to determine the internal

moments at the ends of each mem#er. If any of the results are ne&ati'e they indicate

countercloc*ise rotation.

DCBA ,,,


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!0ample (,)

Solution

Dra* the shear and moment dia&rams for the #eam

sho*n in +i&.(4). !I is constant.,. %sin& the formulas for the ta#ulated in 1a#le

(,) for the &i'en loadin&s:


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+i&. (4)


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3. 1here are t*o slopes at B and $ i. e.

are unno*ns. Since end A is fi0ed Also

since the supports do not settle nor are they

displaced up or do*n


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No* #y applyin& the e"uili#rium conditions:


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Su#stitutin& the computed 'alues in to momente"uations (a) (#) (c) and (d):


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+i&. (5) Shearin& +orce = Bendin& Moment Di&rams


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!0ample (3)

Determine the internal moments at the supports of thebeam shown in Fig. (5). The support at B is displaced

(settles) 12 mm.


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Solution

1. Two spans must be considered. FEs are determined

using Table (1).

4

4

101.0

12

012.00

10667.618

0012.0

=

=

=

=

x

x

BC

AB


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3. %sin& e"uation 3:

)(0)101.032()12

2(

)()101.032()12

2(

)()10667.632()18

2(

)()10667.63()18

2(

4

4

4

4

lxxEI

M

kxxEI

M

jxxEIM

ixxEI

M

BCBCCB

CBBCBC

BABBA

BABAB

=++=

++=

=

=


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!. E"uilibrium condition#

00 == CB MandM

)(0)101.032()122(

)(0)101.032()12

2()10667.632()

18

2(

)()10667.63()18

2(

4

44

4

pxxEI

nxxEI

xxEI

mxxEI

M

BCBC

CBBCBAB

BABAB

=++

=+++

=


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$n order to obtained the rotations e"uations

(n) % (p) ma& then be sol'ed simultaneousl& it ma& be

noted that since is fi*ed support. Thus

CB and

0=A

.1047647.2.1065294.4 44

radxandradxCB

=+=

+ubstituting these 'alues into e"uations (i to l) &ields


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!0ample (4)

$f end in e*ample (1) is simpl& supported and b&appl&ing the compatibilit& condition their will be three

un,nown rotations

-ow

),,(CBA

)(62.1)2()6.3

2(

)(62.1)2()6.3

2(

)(5.4)2()4.2

2(

)(5.4)2()4.2

2(

dEI

M

cEI

M

bEI

M

aEIM

BCBCCB

CBBCBC

BABBA

BAABAB

++=

+=

+=

+=


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Applyin& the e"uili#rium conditions:

)3(062.1)2()6.3

2(

)2(062.1)2()6.3

2(5.4)2()

4.2

2(

)1(05.4)2()4.2

2(

=++=

=+++

=+=

BCBCCB

CBBCBAB

BAABAB

EIM

EIEI

EIM


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By sol'in& e"uations (, 3 = 4) for and

su#stitute the 'alues into e"uations (a # c d):

CBA ,,

0

.158.4

.158.40

=

=

+=

=

CB

BC

BA

AB

M

iseanticlockwmkNM

clockwisemkNMM


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+hearing force and bending moment diagrams are shown

in the following figure.


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E*ample ()

Determine the moments at each /oint of the frame shown in

Fig.(0). E$ is constant.

Fig. (0)


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MkNFEM

MkNFEM

CB

BC

.8096

)8)(24(5)(

.8096

)8)(24(5)(

2

2

+=+=

==


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Because ends and D are fi*ed supports.

and since no sideswa& will occur.

0=== CDBCAB

CCCDDC

CCCDCD

BCBCBCCB

CBCBBCBC

BBABBA

BBABAB

EIl

EIM

EIEI

M

EIEI

EI

M

EIEIEI

M

EIEI

M

EI

EI

M

1667.0)()2

(

3334.0)2()12

2(

8025.05.080)2()8

2

(

8025.05.080)2()8

2(

3334.0)2()12

2(

1667.0)()12

2

(

==

==

++=++=

+=+=

=+=

=+=


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E"uilibrium conditions#

)2(08025.08334.0

0803334.025.05.0

0

)1(08025.08334.0

08025.05.03334.0

0

=++

=+++

=+

=+

=++

=+

BC

CBC

CDCB

CB

CBB

BCBA

EI

OrEIEIEI

MM

EIEI

Or

EIEIEI

MM


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+ol'ing simultaneousl& &ields

EIand

EI CB

1.1371.137==


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Therefore

seanticlokwimkNM

iseanticlockwmkNM

clockwisemkNM

iseanticlockwmkNM

clockwisemkNM

clockwisemkNM

DC

CD

CB

BC

BA

AB

.9.22

.7.45

.7.45

.7.45

.7.45

.9.22

=

=

=

=

=

=


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Fig.()

The bending moment diagram is shown in Fig.().


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E*ample (5)

Determine the internal moments at each of the frame

shown in Fig.().

+olution

Fig.()


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1. Fi*ed end moments#

12)(,

12)(:

12

)5.1()(,

12

)5.1()(:

22

22

wLFEM

wLFEMBCSpan

LwFEM

LwFEMABSpan

CBBC

BAAB

+==

+==

8)(,8)(:

Pl

FEM

Pl

FEMBDSpan BDDB +==


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2. 3oint moments#

)(8

)2()2

(

)(8

)2()2

(

)(12

)2()2

(

)(12)2()2

(

)(12

)5.1()2()

5.1

2(

)(12

)5.1()2()5.1

2(

2

2

2

2

nPL

L

EIM

mPL

L

EIM

lwL

L

EIM

kwL

L

EIM

jLw

L

EIM

iLwL

EIM

BDBDDB

DBBDBD

BCBCCB

CBBCBC

ABABBA

BAABAB

+=

++=

++=

+=

++=

+=


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!. E"uilibrium conditions#

+ol'ing e"uations (12!) simultaneousl& &ields

)4(8

)()2

(

)3(012

)2()2

(

)2(08

)2()2

(

12)2()

2(

12

)5.1()2()

5.1

2(

0

)1(012

)5.1(

)2()5.1

2

(

2

22

2

PL

L

EIM

wL

L

EIM

PL

L

EI

wL

L

EILw

L

EI

M

Lw

L

EI

M

BBDDC

BCBCCB

DBBD

CBBCABAB

B

BAABAB

=

=++=

=++

+++++

=

=+=

)4(8

)()2

(

)3(012

)2()2

(

)2(08

)2()2

(

12)2()

2(

12

)5.1()2()

5.1

2(

0

)1(012

)5.1(

)2()5.1

2

(

2

22

2

PL

L

EIM

wL

L

EIM

PL

L

EI

wL

L

EILw

L

EI

M

Lw

L

EI

M

BBDDC

BCBCCB

DBBD

CBBCABAB

B

BAABAB

=

=++=

=++

+++++

=

=+=

CBA ,,+ol'ing e"uations (12!) simultaneousl& &ields

+ubstituting the rotation 'alues into e"uations (i to n) to

determine the /oint moments.


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E*ample (4)

Determine the /oint internal moments of the frame shown in

Fig.(1) both ends and D are fi*ed.ssume 5.1)(1)()( === CDBCAB

L

EIand

L

EI

L

EI

Fig.(1)


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+olution

1. Fi*ed end moments#

mkNFEM

mkNFEMBCSpan

mkNFEM

mkNFEMABSpan

CB

BC

BA

AB

.96.1212

)2.7(3)(

.96.1212

)2.7(3)(:

.0.8)4.5(

)6.3)(8.1(10)(

.0.4)4.5(

)8.1)(6.3(10)(:

2

2

2

2

2

2

+=+=

==

=+=

==


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$t is assumed that the a*ial deformation is neglected so that== OO CCBB as shown in the following figure.


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$t ma& be noted that

2. 3oint moments#

05.1 ====DABCABCD

and

)(75.65.1

)5.4(5.1

)(75.63

)5.42(

)(96.12)2(5.1)(96.12)2(

)(8)2()5.1

2(

)(4)3(

n

M

m

M

lMkM

jL

EIM

iM

ABC

ABCDC

ABC

ABCCD

BCCB

CBBC

ABBABBA

ABBAB

=

=

=

=

++=

+=

+=

=


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!. E"uilibrium conditions#

)2(96.1275.65

0:int

)1(96.434

0:int

=+

=+

=+

=+

ABCB

CDCB

ABCB

BCBA

Or

MMCo

Or

MMBo


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+ince a hori6ontal displacemententire frame in the *7direction. This &ields

occurs the summing forces on the

)3(667.1075.425.2

0

6.34.53

1010

6.3

4.53

10

:

010:0

=++

=+

++

++

+=

++=

=+=+

ABCB

DCCDBAAB

DCCDD

BAABA

DA!

Or

MMMM

MM"

and

MM"

w#ic#In

""F


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+ol'ing e"uations (1 2 !) &ields9194.1,565.0,8208.2 +==+= ABCB

B& substituting these 'alues into moment e"uations (i to n)#

mkNM

mkNM

mNkM

mkNM

mkNM

mkNM

DC

CD

CB

BC

BA

AB

.8035.13

.6509.14

..6509.14

.8833.7

.8833.7

.9374.6

=

=

+=

=

+=

=


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Bending moment diagram is plotted in the following figure.

displacement method of analysis

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