Stranalysis SD METHOD

8/12/2019 Stranalysis SD METHOD

1/78

Chapter-2:Slope Deflection Method

Example: Analyze the propped cantilever shown by using slope defectionmethod. Then draw Bending moment and shear force diagram.

Solution:End A is fied hence A !"

End B is #inged hence B $"

Assume both ends are fied and therefore fied end moments are

%&

w'()

%&

w'(

&

BA

&

AB +==

The Slope deflection e*uations for final moment at each end are

( )

( )

+&,'

E-

%&

w'

&'

E-&(M

+%,'

E-&

%&

w'

&'

E-&(M

B

&

ABBABA

B

&

BAABAB

+=

++=

+

=

++=

-n the above e*uations there is only one un/nown B .

To solve we have boundary condition at B0

Since B is simply supported) the BM at B is zero

ie. MBA!".

iseanticloc/wisrotationtheindicatessignve12

w'E-

"'

E-

%&

w'M,&+e*uation(rom

3

B

B

&

BA

=

=+=

1


2/78

Substituting the value of BE- in e*uation ,%+ and ,&+ we have end moments

"

2

w'

'

%&

w'M

iseanticloc/wismomentindicatessignve12

w'

2

w'

'

&

%&

w'M

3&

BA

&3&

AB

=

++=

=

+=

MBAhas to be zero) because it is hinged.

4ow consider the free body diagram of the beam and find reactions using

e*uations of e*uilibrium.

w'2

3

w'2

5w'6w'6

w'66

"7

w'2

56

w'25

&'w'

2w'

&

'w'M'6

"M

AB

BA

A

&

ABA

B

=

==

=+=

=

+=+=

+=

=

8roblem can be treated as

2


3/78

The bending moment diagram for the given problem is as below

The ma BM occurs where S(!". 9onsider S( e*uation at a distance of

from right support

3


4/78

&

&

:ma

:

w'%&2

;

'2

3

&

w'

2

3w'

2

3MM

Bsupportfrom'2

3atoccursBMmathe#ence

'2

3:

"w:w'2

3S

=

==

=

=+=

And point of contra fleure occurs where BM!") 9onsider BM e*uation at

a distance of from right support.

'3:

"&

:ww':

2

3M

&

:

=

==

(or shear force diagram) consider S( e*uation from B

w'2

5S'S

w'2

3S"S

w:w'2

3S

A:

B:

:

+===

+===

+=

Example:Analyze two span continuous beam AB9 by slope deflection method.

Then draw Bending moment < Shear force diagram. Ta/e E- constant

Solution: (ied end moments are=

4


5/78

>4M?@.%%&

5&"

%&

w'(

>4M?@.%

%&

5&"

%&

w'(

>4M2;.22?

&%""

'

ba(

>4M.?

&%""

'

ab(

&&

9B

&&

B9

&

&

&

&

BA

&

&

&

&

AB

=

=+

=

=

=

=

+=

+=+=

=

=

=

Since A is fied "A= ) )")" 9B

Slope deflection e*uations are=

( )

( )

+&,E-3

&2;.22

?

&E-&2;.22

&'

E-&(M

+%,E-3

%.

?

E-&

.

&'

E-&(M

B

B

ABBABA

B

B

BAABAB

+=

++=

++=

+=

+=

++=

( )

( )

( )

( )

+,E-5

&

5

E-?@.%

&5

E-&?@.%

&'

E-&(M

+3,E-5

&E-

5

?@.%

&5

E-&?@.%

&

'

E-&(M

B9

B9

B99B9B

9B

9B

9BB9B9

++=

+++=

++=

++=

++=

++=

-n all the above four e*uations there are only two un/nown B and 9 .

And accordingly the boundary conditions are

5


6/78

i 1MBA1MB9!"

MBAMB9!"

ii M9B!" since 9 is end simply support.

+?,"E-5

E-

5

&?@.%M

+5,"E-5

&E-

%5

&&&&.@

E-5

&E-

5

?@.%E-

3

&2;.22MM4ow

9B9B

9B

9BBB9BA

>=++=

>=++=

+++=+

Solving simultaneous e*uations 5 < ? we get

E- B ! C &".23 6otation anticloc/wise.

E- 9 ! C %.?@ 6otation anticloc/wise.

Substituting in the slope definition e*uations

MAB ! C . ( ) >4M32.5%23.&"3

%=

MBA ! 22.2; ( ) >4M"".@523.&"3

&+=

MB9 ! C %.?@ ( ) ( ) >4M"".@5?@.%5

&23.&"

5

=+

M9B ! %.?@ ( ) ( ) "?@.%5

23.&"

5

&=+

Reactions:9onsider the free body diagram of the beam.

6


7/78

(ind reactions using e*uations of e*uilibrium.

Span AB: MA ! " 6B? ! %""@515%.32

6B! @".?" >47 ! " 6A6B ! %"">4

6A! %""1@".?"!&;." >4

Span BC: M9 ! " 6B5 ! &"52

5@5

6B! ?5 >4

7!" 6B69 ! &"5 ! %"">4

69! %""1?5 ! 35 >4

Fsing these data BM and S( diagram can be drawn.

7


8/78

Max BM:

Span AB: Ma BM in span AB occurs under point load and can be found

geometrically

Mma!%%3.3315%.32 1( )

>4M&".??

32.5%@5=

Span BC:Ma BM in span B9 occurs where shear force is zero or

changes its sign. #ence consider S( e*uation w.r.t 9

S ! 351&"! " &"

35= !%.@5m

Ma BM occurs at %.@5m from 9

Mma! 35 %.@5 C &"&

@5.% & ! 3".?&5 >4M

8


9/78

Example:Analyze continuous beam AB9D by slope deflection method and thendraw bending moment diagram. Ta/e E- constant.

Solution:

")")" 9BA ===

(EMS M>4.1

?

&%""

'

ab(

&

&

&

&

AB =

==

>4M22.22?

&%""

'

ba(

&

&

&

&

BA +=

+=+=

>4M%.?@1%&

5&"

%&

w'(

&&

B9 =

==

>4M%.?@%&

5&"

%&

w'(

&&

9B +=

+=+=

M>43"15.%&"(9D ==

Slope deflection e*uations=

( ) ( )%111111111E-3

%.&

'

E-&(M BBAABAB >+=++=

( ) ( )&111111111E-3

&2;.22&

'

E-&(M BABBABA >++=++=

( ) ( )311111111E-5

&E-

5

?@.%&

'

E-&(M 9B9BB9B9 >++=++=

( ) ( )11111111E-5&E-5?@.%&'E-&(M B9B99B9B >+++=++=>4M3"M

9D =

-n the above e*uations we have two un/nown rotations 9B and ) accordingly

the boundary conditions are=

9


10/78

"MM

"MM

9D9B

B9BA

=+

=+

( )511111111"E-5

&E-

%5

&&&&.@

E-5

&E-

5

?@.%E-

3

&2;.22MM)4ow

9B

9BBB9BA

>=++=

+++=+

( )?E-5

E-

5

&?@.%%

3"E-5

&E-

5

?@.%MM)And

9B

B99D9B

>++=

+++=+

Solving ,5+ and ,?+ we get

cloc/wiseBG6otation@5.%E-

iseanticloc/[email protected]&E-

9

B

+==

Substituting value of BE- and 9E- in slope deflection e*uations we have

( )

( )

( ) ( )

( ) ( )

>4M3"M

>4M"".3"[email protected]&5

&@5.%

5

?@.%M

>4M%%.?@@5.%5

&[email protected]&

5

?@.%M

>4M%%.?@[email protected]&3

&2;.22M

>4M"".?%[email protected]&&

%.M

9D

9B

B9

BA

AB

=

+=+++=

=+=+=

+=++=

=+=

Reactions:9onsider free body diagram of beam AB) B9 and 9D as shown

10


11/78

ABSpan

>43%.3&6%""6

>4?;.?@6

?%%%.?@%""?6

BA

B

B

==

=+=

B9Span

>4&.5@65&"6

>452.&6

%%.?@3"5&

5&"56

BB

9

9

===

+=

Maimum Bending Moments=

Span AB= Hccurs under point load

== >4M&?.?2

?

?%%%.?@?%33.%33Ma

Span B9= where S(!") consider S( e*uation with 9 as reference

11


12/78

m%3.&&"

52.&

"&"52.&S:

==

==

M>4&?.%53"&

%3.&&"%3.&52.&M

&

ma ==

Example: Analyse the continuous beam AB9D shown in figure by slope

deflection method. The support B sin/s by %5mm.

Ta/e ?&5 m%"%&"-andmI>4%"&""E ==

Solution=

-n this problem A !") B ") 9 ") !%5mm

(EMs=

>4M.'

ab(

&

&

AB ==

>4M2;.22'

ba( &

&

BA +=+=

>4M?@.%2

w'(

&

B9 ==

>4M?@.%2

w'(

&

9B +=+=

(EM due to yield of support B

(or span AB=

12


13/78

>4M?%"""

%5%"%&"%"

?

&""?

'

E-?mm

?5

&

&baab

=

=

==

(or span B9=

>4M?.2%"""

%5%"%&"%"

5

&""?

'

E-?mm

?5

&

&cbbc

+=

+=

+==

Slope deflection e*uation

( )

( )

( )

( )

( )

( )

( )

( )5111111111>4M3"M

111111111E-5

&E-

5

3%.5"

?.2&E-5

&

%.?@

'

E-?+&,

'

E-&(M

3111111111E-5

&E-

5

"3.33

?.2&E-5

&%.?@1

'

E-?+&,

'

E-&(M

&111111111E-3

&2;.2&

?E-3

&22.2;

'

E-?+&,

'

E-&(M

%111111111E-3

%.5"

?E-3

%.1

'E-?&

'E-(

+'

3&,

'

E-&(M

9D

B9

B9

&B99B9B

9B

9B

&9BB9B9

B

B

&ABBABA

B

B

&BAAB

BAABAB

>=

>+++=

++++=

+++=

>++=

+++=

+++=

>++=

++=

++=

>+=

+=

++=

++=

13


14/78

There are only two un/nown rotations B and 9 . Accordingly the boundary

conditions are

4ow)

"E-5

E-

5

&3%.&"MM

"E-

5

&E-

%5

&&2?.;MM

"MM

"MM

9B9D9B

9BB9BA

9D9B

B9BA

=++=+

=++=+

=+

=+

Solving these e*uations we get

oc/wise Anticl@%.;E-

oc/wise Anticl35.3%E-

9

B

=

=

Substituting these values in slope deflections we get the final moments=

( )

( )( ) ( )

( ) ( )

>4M3"M

>4M"".3"35.3%5

&@%.;

5

3%.5"M

>4M;;.?%@%.;5

&35.3%

5

"3.33M

>4M;;.?%35.3%3

&2;.2&M

>4M2;.?"35.3%3

%.5"M

9D

9B

B9

BA

AB

=

+=+++=

=++=

+=++=

=+=

9onsider the free body diagram of continuous beam for finding reactions

Reactions:

Span AB=6B ? ! %"" ?%.;; C ?".2;

6B! ??.25

6A! %"" C 6B

!33.%5 >4

14


15/78

Span B9=

6B 5 ! &" 5 &

5 ?%.;; C 3"

6B! 5?." >4

69! &" 5 1 6B

!3.?" >4

Example:Three span continuous beam AB9D is fied at A and continuous over

B) 9 and D. The beam subJected to loads as shown. Analyse the beam by slope

deflection method and draw bending moment and shear force diagram.

Solution:

15


16/78

Since end A is fied ")")")" DcBA ==

(EMs=

>4M3"12

?"

2

l(AB =

==

>4M3"2

?"

2

l

(BA +=

+=+=

>4M%&.5

M(B9 +=+=

>4M%&.5

M(9B +=+=

>4M3%3.31%&

%"

%&

wl(

&&

9D =

==

>4M%3.33%&

%"

%&

wl(

&&

D9 +=

+=+=


( )BAABAB &'

E-&(M ++=

( )"

E-&3"1 B++=

( )%11111111E-".53"1 B >+=

( )ABBABA &'

E-&(M ++=

( )"&

E-&3" B++=

( )&111111111E-3" B >++=

( )9BB9B9 &'

E-&(M ++=

( )&

E-&%&.5 9B ++=

( )3111111111E-5."E-%&.5 9B >++=

( )B99B9B &'

E-&(M ++=

( )&

E-&5.%& B9 ++=

( )111111111E-5."E-%&.5 B9 >++=

( )D99D9D &'

E-&(M ++=

( )&

E-&%3.331 D9 ++=

( )51111111111E-5."E-33.%3 D9 >++=

16


17/78

( )9DD9D9 &'

E-&(M ++=

( )&

E-&%3.33 9D ++=

( )?1111111111E-E-5."%3.33 D9 >++=

-n the above E*uations there are three un/nowns) E- D9B E-=++

By solving ,@+) ,2+ < ,;+) we get

;".%2E-

%5.%%E-

".&E-

D

9

B

=

+==

By substituting the values of DcB and) in respective e*uations we get

( )

( )

( ) ( )

( )

( )( ) ( ) >4M";".%2%5.%%5."33.%3M

>4M?3.%%;".%25."%5.%%33.%3M

>4M?3.%%".&5."%5.%%5.%&M

>4M5.;?1%%.%5.5"&."1%&.5M

>4M;?.5".&3"M

>4M"&.&".&5."3"M

D9

9D

9B

B9

BA

AB

=+++==++=

+=+++==++++=

+=++==+=

Reactions:9onsider the free body diagram of beam.

17


18/78

Beam AB:

>4"%5.3"6?"6

>4;25.&"

"&.&;?.5&?"6

BA

B

==

=+

=

Beam BC:

downwardis6>4;&.%366

>4;&.%3

;?.55"?3.%%6

B9B

9

==

=+

=

Beam CD:

>4;%.&&6%"6

>4";.%@

?3.%%&%"6

D9

D

==

=

=

18


19/78

Example:Analyse the continuous beam shown using slope deflection method.

Then draw bending moment and shear force diagram.

Solution:-n this problem fiedisAend)"A =

(EMs=

M>453.331%&

2%"

%&

wl(

&&

AB =

==

>4M53.33%&wl(

&

BA +=+=

>4M&&.5"12

?3"

2

l(B9 =

==

>4M&&.5"2

'(9D +=+=


( )BAABAB &'

E-&(M ++=

( )"2-3E&53.331 B++=

( )%11111111E-

353.331 B >+=

( )ABBABA &'

E-&(M ++=

( )"&2

-3E&53.33 B+

++=

19


20/78

( )&11111111E-&

353.33 B >+=

( )9BB9B9 &'E-&(M ++=

( )&?

-&E&&&.51 9B ++=

( )311111111E-3

&E-

3

&&.51 9B >++=

( )B99B9B &'

E-&(M ++=

( )&?

-&E&&&.5 B9 +++=

( )11111111E-3

&

E-3

&&.5 B9 >+++=-n the above e*uation there are two un/nown 9B and ) accordingly the

boundary conditions are=

"Mii

"&MMi

9B

B9BA

=

=

( )5"E-3

&E-

?

%@23.5

&E-3

&E-

3

5.&&E-

&

333.53&MM)4ow

9B

9BBB9BA

>=+++=

++++=+

"E-3&E-

35.&&Mand B99B =++=

,?+11111111111E-3

%&5.%%E-

3

&B9 >=

Substituting in e*n. ,5+

cloc/wiseantirotation3&.%@%5

?52.E-

"E-?

%5.52

"E-3

%&5.%%E-

?

%@23.5

B

B

BB

==

=++

=+

from e*uation ,?+

20


21/78

( )

iseanticloc/wrotation%5;.2

3&.%@3

%&5.%%

&

3E- 9

=

=

Substituting %5;.2E-and3&.%@E- 9B == in the slope deflection e*uationwe get (inal Moments=

( )

( ) >4M%2.&@3&.%@&

333.53M

>4M1??."%@.3&1

333.53M

BA

AB

+=++=

=+=

( ) ( )

( ) ""."+3&.%@,3

&%5;.2

3

5.&&M

>4M%2.5%%5;.23

&3&.%@

3

5.&&M

9B

B9

=+++=

=++=

Reactions:9onsider free body diagram of beams as shown

Span AB:

>[email protected]%"6

>4%3.352

2%"".??%2.&@6

BA

B

==

=+

=

Span BC:

>4@.?63"6

>453.&3?

33"%2.5%6

B9

B

==

=+=

21


22/78

Max BM

Span AB:Ma BM occurs where S(!") consider S( e*uation with A as origin

>[email protected]??&

2@.%"[email protected]@.M

m2@.

"%"[email protected]

&

ma

==

=

==

Span BC:Ma BM occurs under point load

M>4%.%;&

%2.5%5MB9 ma ==

Example:Analyse the beam shown in figure. End support 9 is subJected to an

anticloc/wise moment of %& >4M.

22


23/78

Solution:-n this problem fiedisend)"A =

(EMs=

>4M?@.&?%&

&"

%&

wl(

&&

B9 =

==

>4M&?.?@%&

wl(

&

9B +=+=


( )BAABAB &'

E-&(M ++=

( )"

-&E&" B++=

( )%111111111E-B

>=

( )ABBABA &'

E-&(M ++=

( )"& -&E&" B++=( )&111111111E-& B >=

( )9BB9B9 &'

E-&(M ++=

( )&

-5.%E&&?.?@1 9B +

+=

( )3111111111E-

3E-

&

3&?.?@1 9B >++=

( )B99B9B &'

E-&(M ++=

( )&

-5.%E&&?.?@ B9 +++=

( )111111111E-

3E-

&

3&?.?@ B9 >+++=

-n the above e*uation there are two un/nowns 9B and ) accordingly the

boundary conditions are

"%&M

"MM

9B

B9BA

=+

=+

,5+111111111"?@.&?E-

3E-

&

@

E-

3E-

&

3?@.&?E-&MM)4ow

9B

9BBB9BA

>=+=

++=+

,?+111111111"E-&

3E-

[email protected]

%&E-

3E-

&

3?@.&?%&M)and

9B

B99B

>=++=

+++=+

(rom ,5+ and ,?+

23


24/78

@&.%&5

2?E-

"?E-2

&5

"33.%;E-

3E-

2

3

"?@.&?E-

3E-

&

@

B

B

9B

9B

+=+=

=

=++

=+

(rom ,?+

( )

iseanticloc/wrotationindicatessignve1%.33

@&.%

[email protected]

3

&E- 9

=

+=

e*uationsdeflectionslopeisE-andE-ngSubstituti 9B

( )

>4M%&+@&.%,

3+%.33,

&

3?@.&?M

>4M.&;%.33

3+@&.%,

&

3?@.&?M

>4M&.&;+@&.%,&E-&M

>4M@&.%E-M

9B

B9

BBA

BAB

=+++=

=++====

+==

Reaction:9onsider free body diagrams of beam

Span AB:

24


25/78

>4".%%66

>4".%%

.&;@&.%6

BA

B

==

=+

=

Span BC:

>4?.&;6&"6

>43?.5"

&&"%&.&;

6

B9

B

===

++

=

Example:Analyse the simple frame shown in figure. End A is fied and ends B 4M%"2

'(

>4M%"2

&"

2

'(

>4M?@.&?%&

&"

%&

wl

(

>4M?@.&?%&

&"

%&

wl(

>4M33.53?

&%&"

'

ba(

>4M?@.%"??

&%&"

'

ab(

DB

9D

&&

9B

&&

B9

&

&

&

&

BA

&

&

&

&

AB

==

+=

+=+=

+=

+=+=

=

==

+=

+=+=

=

==

Slope deflections are

( )

( ) +%,E-

3

&?@.%"?

?

-&E&?@.%"?

&'

E-&(M

BB

BAABAB

>+=+=

++=

26


27/78

( )

( )

( )

( )

( )

( ) +,E-

3E-

&

3?@.&?&

&

-3

E&?@.&?

&'

E-&(M

+3,E-

3E-

&

3?@.&?&

&

-3

E&?@.&?

&'

E-&(M

+&,E-3

33.53&

?

-&E&33.53

&'

E-&(M

B9B9

B99B9B

9B9B

9B9BB9

BB

BBBABA

>+++=+++=

++=

>++=++=

++=

>++=++=

++=

( )

( )

( )

( ) +?,E-&

%E-%"&

E-&%"

&'E-&(M

+5,E-&

%E-%"&

E-&%"

&'

E-&(M

BDBD

BDDBDB

DBDB

DBBDBD

>++=++=

++=

>+++=+++=

++=

-n the above e*uations we have three un/nown rotations B ) 9 ) Daccordingly we have three boundary conditions.

"MMM BDB9BA =++

"M9B = Since 9 and D are hinged

"MDB =4ow

,;+11111"E-E-&

%%"M

,2+11111"E-&

3E-

3?.?@&M

,@+11111"E-&

%E-

3E-

?

&33?.??

E-&

%E-%"E-

3E-

&

3?@.&?E-

3

33.53MMM

DBDB

9B9B

D9B

DB9BBBDB9BA

>=++=

>=++=

>=+++=

++++++=++

Solving e*uations @) 2) < ; we get

%.%E-

3?.%3E-

23.2E-

D

9

B

+=

=

=

Substituting these values in slope e*uations

27


28/78

"+23.2,&

%+%.%,%"M

>4M32.2+%.%,&

%+23.2,%"M

"+23.2,

3+3?.%3,

&

3?@.&?M

>4M;.;+3?.%3,

3+3.2,

&

3?@.&?M

>4M5?.%+23.2,3

33.53M

>4M5?.%%&+23.2,3

&?@.%"?M

DB

BD

9B

B9

BA

AB

=++=

=+++=

=+++=

=++=

=+=

=+=

Reactions:9onsider free body diagram of each members

28


29/78

Span AB:

>423.;%6%&"6

>4%@.&2?

&%&"5?.%%&5?.%6

BA

B

==

=+

=

Span BC:

>45%5.&@6&"6

>425.5&

&&";.;6

B9

B

==

=+

=

Column BD:

[ ]&"##>4@2.%

>4;&.@

33.2&&"#

DAB

D

=+=

=

=

29


30/78

Example:Analyse the portal frame shown in figure and also drawn bending

moment and shear force diagram

Solution:Symmetrical problem1 Sym frame Sym loading

")")")" D9BA ==

(EMS

>4M%"?.?@1?

&2"

?

&2"

'

cd

'

ab(

&

&

&

&

&

&

&

&

&

%B9

=

=

=

>4M?@.%"?'

dc

'

ba(

&

&&

&

&

9B +=++=

30


31/78


( ) ( ) ( )%11111111E-&

%"

E-&"&

'

E-&(M BBBAABAB >=++=++=

( ) ( ) ( )&1111111E-"&

E-&"&

'

E-&(M

BBABBABA >=++=++=( )

( )3111111E-3

&E-

3

?@.%"?+&,

?

-&E&?@.%"?

&'

E-&(M

9B9B

9BB9B9

>++=++=

++=

( )

( )111111E-3

&E-

3

?@.%"?+&,

?

-&E&?@.%"?

&'

E-&(M

B9B9

B99B9B

>+++=+++=

++=

( )

( )51111111E-+"&,

E-&"

&'

E-&(M

99

D99D9D

>=++=

++=

( )

( )?1111111E-&

%+",

E-&"

&'

E-&(M

99

9DD9D9

>=++=

++=

-n the above e*uation there are two un/nown rotations. Accordingly the boundary

conditions are

"MM

"MM

9D9B

B9BA

=+ =+

4ow ,@+1111111"E-3

&E-

3

@?@.%"?MM 9BB9BA >=++=+

,2+1111111"E-3

@E-

3

&?@.%"?MM 9B9D9B >=+++=+

Multiply by ,@+ and ,2+ by &

9loc/wise?5

3"3.;?"E-

"E-3

5;?"."31

subtracts

"E-3

%E-

3

3.&%3

"E-

3

%E-

3

;?;.@?

B

B

9B

9B

+=+=

=+

=+++

=++

31


32/78

Fsing e*uation ,@+

c/wise Anticlo??3

@?@.%"?

&

31

E-3

@?@.%"?

&

3E- B9

=

+=

+=

#ere we find 9B = . -t is obvious because the problem is symmetrical.

aremoments(inal

( )

( )

>4M13&?&

%M

>4M?M

>4M??3

&+?,

3

?@.%"?M

>4M??

3

&?

3

?@.%"?M

>4M?M

>4M3&&

?M

D9

9D

9B

B9

BA

AB

==

=

+=+++=

=++=

=

+=+=

9onsider free body diagramKs of beam and columns as shown

32


33/78

By symmetrical we can write

>4M2"66

>4M?"66

9D

BA

==

==

4ow consider free body diagram of column AB

Apply

>4

3&?#

"M

A

A

B

=

+=

=

Similarly from free body diagram of column 9D

Apply

>4

3&?#

"M

D

A

9

=+=

=

Check:

"##

"#

DA =+

=

#ence o/ay

33


34/78

4ote= Since symmetrical) only half frame may be analysed. Fsing first threee*uations

and ta/ing 9B =

Example:Analyse the portal frame and then draw the bending moment diagram

34


35/78

Solution:

This is a symmetrical frame and unsymmetrically loaded) thus it is an

unsymmetrical problem and there is a sway

Assume sway to right.

#ere ")")")" DBDA ===

(EMS=

>4M@5.;32

352"

'

ba

(

>4M&5.5?2

352"

'

ab(

&

&

&

&

9B

&

&

&

&

B9

+=

+=+=

=

==

Slope deflection e*uations

( )

( )&11111111E-2

3E-

3"&

&E-"

'

3&

'

E-&(M

%11111111E-2

3E-

&

%

3"

&E-"

'

3&

'

E-&(M

BB

ABBABA

BB

BAABAB

>=

++=

++=

>=

++=

++=

35


36/78

( )

( ) ( )

( )( ) ( )

( )

( )?111111111E-23

E-&

%

3

"

&E-

"

'

3&

'

E-&(M

5111111111E-2

3E-

3"&

&E-"

'

3&

'

E-&(M

111111111E-

%E-

&

%@5.;3&

2

&E-@5.;3

&'

E-&

(M

3111111111E-

%E-

&

%&5.5?&

2

&E-&5.5?

&'

E-&(M

99

9DD9D9

99

D99D9D

B9B9

B99B9B

9B9B

9BB9B9

>=

++=

++=

>=

++=

++=

>++=+++=

++=

>++=++=

++=

-n the above e*uation there are three un/nowns and) 9B ) accordingly theboundary conditions are)

"MMMM

"

MM

MM

)e.i

conditionShear111"8##

"MM

conditionsintLo"MM

D99DBAAB

D99DBAAB

#DA

9D9B

B9BA

=+++=

++

+>=++

=+>=+

( )@"E-2

3E-

%E-

&

3&5.5?

"E-

%E-

&

%&5.5?E-

2

3E-MM)4ow

9B

9BBB9BA

>=++=

=++=+

( )2"E-2

3E-

&

3E-

%@5.;3

"E-2

3E-E-

%E-

&

%@5.;3MM)And

9B

9B99D9B

>=++=

=+++=+

( );"E-&

3E-

&

3E-

&

3

E-2

3E-

&

%

E-23E-E-

23E-E-

23E-

&%MMMM)And

9B

9

9BBD99DBAAB

>=+=

+

++=+++

36


37/78

,2+


38/78

Reactions:consider the free body diagram of beam and columns

9olumn AB=

>45.&&

3%.55?;.3#A =

+=

Span B9=

%@.5%62"6

>423.&22

32"?;.?3%.556

B9

B

==

=+

=

9olumn 9D=

5.&&

3%.&5?;.?#D =+=

Check:

# ! "#A #D! "&&.5 C &&.5 ! "#ence o/ay

38


39/78

Example:(rame AB9D is subJected to a horizontal force of &" >4 at Joint 9 asshown in figure. Analyse and draw bending moment diagram.

Solution:

(rame is Symmetrical and unsymmetrical loaded hence there is a sway.

Assume sway towards right

(EMS"(((((( D99D9BB9BAAB ======

Slope deflection e*uations are

39


40/78

( )

( )&E-3

&E-

3

3

3&

3

E-&

'

3&

'

E-&(M

%111111111E-

3

&E-

3

&

3

3

3

E-&

'

3&

'

E-&(M

B

B

ABBABA

B

B

BAABAB

>=

=

++=

>=

=

++=

( )

( )

( )

( )

( )

( )

( )5E-3

&E-

3

3

3&

3

&E-

'

3&

'

E-&(M

11111111E-5."E-

&

E-&

&'

E-&(M

311111111E-5."E-

&

E-&

&'

E-&(M

9

9

D99D9D

B9

B9

B99B9B

9B

9B

9BB9B9

>=

=

++=

>+=

+=

++=

>+=

+=

++=

( )?111111111E-3

&E-

3

&

3

3

3

E-&

'

3&

'

E-&(M

9

c

9DD9D9

>=

=

++=

The un/nown are


41/78

"?"MMMM

"&"3

MM

3

MM)e.i

"&"##.---

"MM.--

"MM.-

D99DBAAB

D99DBAAB

DA

9D9B

B9BA

=+++

=+

++

=+

=+

=+

( )@"E-3

&E-5."E-

3

@

E-5."E-E-3

&E-

3

MM4ow

9B

9BBB9BA

>=+=

++=+

( )2"E-3

&E-

3

@E-5."

E-3

&E-

3

E-5."E-MMand

9B

9B99D9B

>=+=

++=+

( );"?"E-3

2E-&E-&

?"E-3

&E-

3

&E-

3

&

E-3

E-

3

&E-

3

E-

3

&E-

3

&?"MMMMand

9B

9

9BBD99DBAAB

=+=

+

++=+++

Solving ,@+.,2+ < ,;+ we get

@@.3E-

)%2.2E-

)%2.2E-

9

B

==

=

Substituting the value of and) 9B in slope deflection e*uations

( ) ( )

( ) ( )

( )( )

( ) ( )

( ) ( ) >4M@3.%@@@.33

&%2.2

3

&M

>4M&@.%&@@.33

&%2.2

3

M

>4M&@.%&%2.2%2.25."M>4M&@.%&%2.25."%2.2"M

>4M&@.%&@@.33

&%2.2

3

M

>4M@3.%@@@.33

&%2.2

3

&M

D9

9D

9B

B9

BA

AB

+==

+==

== =+=

+==

+==

41


42/78

Reactions:9onsider the free body diagram of the members

Member AB:

>4%"3

&@.%&@3.%@#A =

+=

Member BC:

downwards6ofdirectionindicatessignve1>4%35.?66

>4%35.?

&@.%&&@.%&6

B9B

9

==

=+

=

Member CD:

righttoleftis#ofdirectiontheindicatessignve1>4%"3

&@.%&@3.%@# DD =

=

42


43/78

Check: # ! "

#A #D 8 ! "

%" %" C &" ! "

#ence o/ay

Example:Analyse the portal frame subJected to loads as shown. Also draw

bending moment diagram.

The frame is symmetrical but loading is unsymmetrical. #ence there is a sway.

Assume sway towards right. -n this problem ")")")" D9BA ==

(EMs=

>4M%3.331%&

%"

%&

wl(

&&

AB =

==

>4M%3.33%&

%"

%&

wl(

&&

BA +=+=+=

>4M%%&.512

%";"

2

wl(B9 =

==

>4M%%&.52

%";"

2

wl(9B +=

==


43


44/78

+++=

'

3&

'

E-&(M BAABAB

3"

E-&%3.331 B

++=

( )%111111111E-3@5."E-5."%3.331 B >+=

+++=

'3&

'E-&(M ABBABA

3"&

E-&%3.33 B

++=

( )&111111111E-3@5."E-%3.33 B >+=

( )9BB9B9 &'

E-&(M ++=

( )&%"

-3E&%%&.51 9B ++=

( )3111111111E-?."%.&E-%%&.51 9B >++=

( )B99B9B &'E-&(M ++=

( )&%"

-3E&%%&.5 B9 +++=

( )111111111E-?."%.&E-%%&.5 B9 >++=

++='

3&

'

E-&(M D99D9D

3"&

E-&" 9

++=

( )5111111111E-3@5."E- 9 >=

++=

'3&

'E-&(M 9DD9D9

3&"

E-&" 9

++=

( )?111111111E-3@5."".5E- 9 >=

E-andE-E-un/nowns3areThere 9B) ) accordingly the boundary conditionsare

""##

"MM

"MM

DA

9D9B

B9BA

=++

=+

=+

MM#

MM#and

2"MM#

&

%"MM##ere

D99DD

B99DD

BAABA

BAABA

+=

+=

+=

+=

44


45/78

"2"MMMM

""

MM

2"MM

D99DBAAB

D99DBAAB

=++++

=++

++

4ow MBA MB9! "

( ) ( )

,2+11111111"E-3@5."E-?."E-&.&5.%%&

"E-3@5."E-E-?."E-&.%5.%%&

5"MMand

,@+11111111"%@.;;E-3@5."E-?."E-&.&

"E-?."E-&.%5.%%&E-3@5."E-33.%3

B9

9B9

D99B

9B

9BB

>=++=+++

+=+>=+

=++

,;111111111"2"E-1%.5E-%.5E-5.%

"2"E-3@5."E-5."

E-3@5."E-E-3@5."E-33.%3E-3@5."E-5."33.%3

"2"MMMMalso

9B

9

9BB

D99BBAAB

>=++

=++

++++

=++++

By solving ,@+) ,2+ and ,;+ we get

3.??E-

?.5;E-

?5.@&E-

9

B

+=

=

=

(inal moments=( ) ( )

( )( ) ( )

( ) ( )

>4M@".5+3.??,3@5."+?.5;,5."M

>4M5&.2+3.??,3@5."?.5;M

>4M5&.2?5.@&?."?.5;&.%5.%%&M

>4M%".?%?.5;?."?5.@&&.%5.%%&M

>4M%".?%3.??3@5."?5.@&M

>4M1%.22??.33@5."?5.@&5."33.%3M

D9

9D

9B

B9

BA

AB

====

=++==+=+=

=+==+=

45


46/78

Reactions:9onsider the free body diagrams of various members

Member AB:

lefttorightfromis#ofdirectionindicatessignve1>4%;5.5

&%"22.%%".?%#

A

A

=

=

46


47/78

Member BC:

>43.326;"6

>43.@%"

5;"%".?%5&.26

9B

9

==

=+

=

Member CD

>42%.3

@.55.2#D =

+=

Check

# ! "

#A #D%" ! "15.&" 1 3.2% " ! "

#ence o/ay

Example:Analyse the portal frame and then draw the bending moment diagram

47


48/78

Solution:

Since the columns have different moment of inertia) it is an unsymmetrical

frame. Assume sway towards right

(EMS=

>4M?"2

'(

>4M?"2

?2"

2

'(

9B

B9

+=+=

=

==

#ere ")" DA ==


( )

( )&11111111E-2

3E-

3"&

&E-"

'

3&

'

E-&(M

%11111111E-2

3E-

&

%

3"

&E-"

'

3&

'

E-&(M

BB

ABBABA

BB

BAABAB

>=

++=

++=

>=

++=

++=

48


49/78

( )

( ) ( )

( )

( ) ( )

( )

( )?111111111E-

3E-

3"

&E&-"

'

3&

'

E-&(M

5111111111E-

3E-&

3"&

&E&-"

'

3&

'

E-&(M

111111111E-3

E-

3

&?"&

?

&E&-?"

&'

E-&(M

3111111111E-3

&E-

3

?"&

?

&E&-?"

&'

E-&(M

99

9DD9D9

99

D99D9D

9BB9

B99B9B

9B9B

9BB9B9

>=

++=

++=

>=

++=

++=

>+++=++=

++=

>++=++=

++=


"MMMM

"

MM

MM)e.i

conditionShear111"##

conditionsintLo"MM

"MM

D99DBAAB

D99DBAAB

DA

9D9B

B9BA

=+++

=+

++

>=+>=+

=+

( )@"E-2

3E-

3

&E-

3

@?"

"E-3

&E-

3

?"E-

2

3E-MM)4ow

9B

9BBB9BA

>=++=

=++=+

( )2"?"E-3E-

3%"E-

3&

"E-

3E-&E-

3

E-

3

&?"MM)And

9B

99B9D9B

>=++=

=++++=+

( );"E-

;E-3E-

&

3

E-

3E-

E-

3E-&E-

2

3E-E-

2

3E-

&

%MMMM)And

9B

9

9BBD99DB9AB

>=+=

+

++=+++

49


50/78

,@+inE-ofvaluengSubstituti

E-3E-&

3

;

E-,;+(rom 9B

+=

( )%"1111111"?"E-?

%E-

%&

&5

"?"E-&

%E-

%E-

3

&E-

3

@

"?"E-3E-&

3

;

2

3E-

3

&E-

3

@

9B

9B9B

9B9B

>=+

=+

=

++

Substituting value of E-in ,2+

( )%%1111111"?"E-3

@E-

?

%

"?"E-E-&

%E-

3

%"E-

3

&

"?"E-3E-&

3

;

2

3

E-3

%"

E-3

&

9B

9B9B

9B9B

>=++

=++

=+

++

Solving ,%"+ < ,%%+ we get E- B !3%."3

By E*uation ,%%+

3.&@

?"E-

?

%

@

3E- B9

=

+=

4ow

55.%?E-3E-&

3

;

E- 9B =

+=

4ow

E- B !3%."3) 3.&@E- 9 = ) E- 55.%?=Substituting these values in slope deflection e*uations)The final moments are=

50


51/78

( ) ( )

( )

( ) ( )

( ) ( )

( ) >4M5&.%555.%?

3;3.&@M

>4M5.3+55.%?,

3+;3.&@,&M

>4M3.3;3.&@3

"3.3%

3

&?"M

>4M&5.3@;3.&@3

&"3.3%

3

?"M

>4M&.3@55.%?2

3"3.3%M

>4M@&.&%55.%?2

3"3.3%

&

%M

D9

9D

9B

B9

BA

AB

==

==

+=+++=

=++=

==

==


9olumn AB=

>4@.%

@&.&%&5.3@#A =+=

Beam B9=

"3.%62"6

>4;@.32?

32"5.3&5.3@6

B9

B

==

=+

=

51


52/78

9olumn 9D=

>4@.%

5&.%55.3#D =

+=

Check:

# ! "#A #D! "%.@1%.@!"

#ence o/ay

Ex:8ortal frame shown is fied at ends A and D) the Joint B is rigid and Joint 9 ishinged. Analyse the frame and draw BMD.

Solution:

(EMKs=

52


53/78

")")")")"#ere

>4M?"2

?2"

2

'(

>4M?"2

?2"

2

'(

9D9BBDA

9B

B9

==

+=

=+=

=

==

Since 9 is hinged member 9B and 9D will rotate independently. Also the

frame is unsymmetrical) will also have sway. 'et the sway be towards right.

The slope deflections are=

+%,E-23E-

&%

3"

E-&"

'

3&

'

E-&(M

B

B

BAABAB

>=

++=

++=

53


54/78

( )

( )

( )

+?,E-2

3E-

&

%

3"

E-&"

'

3&

'

E-&(M

+5,E-2

3E-

3"&

E-&"

'

3&

'

E-&(M

+,E-3

&E-

3

?"

&?

-&.E&?"

&'

E-&(M

+3,E-3

&E-

3

?"

&?

-&.E&?"

'

3&

'

E-&(M

+&,E-

2

3E-

3"&

E-&"

'

3&

'

E-&(M

9D

9D

9DDD9D9

9D

9D

D9D9D9D

B9B

B9B

B9B9B9B

9BB

9BB

9BB9B9

B

B

ABBABA

>=

++=

++=

>=

++=

++=

>+++=

+++=

++=

>++=

++=

++=

>=

++=

++=

-n the above e*uations areand)) 9D9BB un/nowns. According theboundary conditions are

-. MBAMB9 ! ")

--. M9B! ")---. M9D! ")

-7. #A#D! "

"MMMM

"

MM

MM)e..i

D99DBAAB

D99DBAAB

=+++

=+

++

54


55/78

4ow using the boundary conditions=

+%,"?"E-3

&E-

%5

&?

?"E-3

&E-

5

3

3

@

"?"E-5

2

2

3

E-3

&

E-3

@

MM

+@,E*uationinngSubstituti

+%3,E-5

2E-

&

3

%5

%?E-gives+%&,E*uation

+%&,"E-%?

%5E-

&

3

"E-&

3E-

2

3

&

3E-

&

3MMMM

+%",inSub

+%%,E-2

3E-+;,(rom

+%","E-&

3E-

&

3E-

&

3

2

3E-

&

%E-

2

3E-E-

2

3E-E-

2

3E-

&

%MMMM

+;,"E-2

3E-M

+2,"?"E-3

E-

3

&M

+@,"?"E-2

3E-

3

&E-

3

@

E-3

&E-

3

?"E-

2

3E-MM

9BB

9BB

B9BBB9BA

BB

B

BD99DBAAB

cD

9DB

9D9DBBD99DBAAB

9D9D

9BB9B

9BB

9BBBB9BA

>=+=

+

=

=

+=+

>==

>==

=

+=+++

>=

>=+=

+++=+++

>==

>=++=

>=+=

++=+

havewe&by,%+e*uationgmultiplyinand+2,E*uationinngSubstituti

&;.?&

%5%2"E-

NNNNNNNNNNNNNNNNNNNNN

"%2"E-%5

&

NNNNNNNNNNNNNNNNNNNNN

"%&"E-

3

E-

%5

5&

"?"E-3

E-

3

&

B

B

9BB

9BB

==

=+

=+

=++

55


56/78

2?.%"&&;.?5

2E-

5

2E-(rom,%3+ B ===

[email protected]

3E-,%%+(rom 9D ==

%?5.@@

?"2?.%"&2

3&;.?

3

@

&

3

?"E-2

3E-

3

@

&

3E-+@,(rom B9B

=

=

=

2?.%"&E-)[email protected])%?5.@@E-)&;.?E- 9D9BB ====

(inal Moments are( ) ( )

( )

( ) ( )

( ) ( )

( )

( ) ( ) >4M&;.%;2?.%"&2

[email protected]

&

%M

"2?.%"&2

[email protected]

"&;.?3

&%?5.@@

3

?"M

>4M@&.&5%?5.@@3&&;.?

3?"M

>4M@&.&52?.%"&2

3&;.?M

>4M&.?2?.%"&2

3&;.?

&

%M

D9

9D

9B

B9

BA

AB

==

==

=+++=

=++=

==

==

Reactions:9onsider the free body diagram of various members

56


57/78

Column AB:

>42&5.

&.?@&.&5#A =

=

Beam BC:( )

>4@%.35&;.2"6

>4&;.?

32"@&.&56

9

B

==

=+

=

Column CD:

>42&.

&2.%;#D ==

Check:# ! "

#A#D ! "

#ence o/ay.

57


58/78

Example:Analyse the portal frame shown in figure the deflection method andthen draw the bending moment diagram

58


59/78

(ig

Solution:

The frame is unsymmetrical) hence there is a sway. 'et the sway be

towards right.

")")")" D9BA ==

(EMS=

>4M3"&%5(

>4M?@.%%&

5&"(

>4M?@.%%&

5&"(

9E

&

9B

&

B9

==

+=

+=

=

=


( )

( )&11111111E-3@5."E-

3"&

&E-"

'

3&

'

E-&(M

%11111111E-3@5."E-5."

3"

&E-"

'

3&

'

E-&(M

BB

ABBABA

BB

BAABAB

>=

++=

++=

>=

++=

++=

59


60/78

( )

( ) ( )

( )

( ) ( )

( )

( )?111111111E-3@5."E-5."

3"

&E-"

'

3&

'

E-&(M

5111111111E-3@5."E-

3"&

&E-"

'

3&

'

E-&(M

111111111E-?."E-&.%?@.%&5

%.5-&E?@.%

&'

E-&(M

3111111111E-?."E-&.%?@.%&5

%.5-&E?@.%

&'

E-&(M

99

9DD9D9

99

D99D9D

B9B9

B99B9B

9B9B

9BB9B9

>=

++=

++=

>=

++=

++=

>++=+++=

++=

>++=+

+=

++=


"MMMM)e.i

"##

"MMM

"MM

D99DBAAB

DA

9E9D9B

B9BA

=+++

=+

=++

=+

4ow)

( )@"?@.%E-3@5."E-?."E-&.&

"E-?."E-&.%?@.%%E-3@5."E-

"MM

9B

9BB

B9BA

>=+=

=++=+

( )2"?@.%%E-3@5."E-&.&E-?."

"3"E-3@5."E-E-?."E-&.%?@.%MM)And

9B

9B99D9B

>=++=

=+++=+

( );"E-5.%E-5.%E-5.%

"E-3@5."E-5."E-3@5."E-&E-3@5."E-E-3@5."E-5."

"MMMM

9B

99BB

D99DB9AB

>=+

=+++

=+++

Solving the above e*uations

we get) E- ;2.&3B= ) E- ?&.%E-)3?.;9 ==

Substituting these values in slope deflection e*uations) we have

60


61/78

( ) ( )( )

( ) ( )( ) ( )

>4M3"M

>4M%?.%"?&.%3@5."3?.;5."M

>4M2.%+?&.%,3@5."3?.;M

>4M23.;2.&3?."3?.;&.%?@.%M

>4M5%.%23?.;?.";2.&3&.%?@.%M

>4M5".%2?&.%3@5.";2.&3M

>4M5".??&.%3@5.";2.&35."M

9E

D9

9D

9B

B9

BA

AB

===

=+=+=+++==++=

+====


61


62/78

9olumn AB=

>4&5.?

5.?5.%2#A =

+=

Span B9=

@3.65&"6

>4&@.555

5.&5&"5.%223.6

9B

9

==

=+=

9olumn 9D=

&5.?

2.%%?.%"#D =

+=

Check:

# ! "#A #D! " ! "

#ence o/ay

62


63/78

Example:Analyse the portal frame shown and then draw bending momentdiagram.

Solution:

-t is an unsymmetrical problem hence there is a sway be towards right")")")" D9BA ==

(EMs=

>4M%.?@1%&

5&"

%&

wl(

&&

B9 =

==

>4M%.?@%&

5&"

%&

wl(

&&

9B +=

==


++=

'

3&

'

E-&(M BAABAB

3

3"

3

E-&" B

++=

( )%111111111E-3

&E-

3

& B >=

++=

'

3&

'

E-&(M ABBABA

3

3"&

3

E-&" B

++=

( )&111111111E-3

&E-

3

B >=

++='

3&

'

E-&(M 9BB9B9

( )&5

-5.%E&%.?@1 9B +

+=

63


64/78

( )3111111111E-5

3E-

5

?%.?@1 9B >++=

++='

3&

'

E-&(M B99B9B

( )"&5

-5.%E&?@.% B9 +

+=

( )111111111E-?."%.&E-%.?@ B9 >++=

++='

3&

'

E-&(M D99D9D

3"&

E-&" 9

++=

( )51111111111E-3@5."E- 9 >=

++='

3&

'

E-&(M 9DD9D9

3"

E-&" 9

++=

( )?1111111111E-3@5."".5E- 9 >=

-n the above e*uations there are three un/nown and) 9B and accordinglythe Boundary conditions are=

"+MM,3+M,M

"

MM

3

MMi.e

"##

"MM

"MM

D99DBAAB

D99DBAAB

DA

9D9B

B9BA

=+++

=+

++

=+

=+

=+

64


65/78

4ow

+2,"E-3@5."E-?."E-&.&?@.%

"E-3@5."E-E-?."E-&.%?@.%

"MM

+@,"?@.%E-3

&

E-5

3

E-53.&

?@.%E-5

3E-

5

?E-

3

&E-

3

"MM

B9

9B9

9D9B

9B

9BB

B9BA

>=++

=+++

=+>=+

++

=+

"

MM

3

MM D99DBAAB =+

++

[ ]

+;,"[email protected]

"E-&5.&E-5.E-32E-

3%?E-

32E-

32

"E-3@5."E-5."E-3@5."E-3

E-3

&E-

3

E-

3

&E-

3

&

9B

9BB

99

BB

>=+

=++

=+

+

+

By solving ,@+) ,2+ and ,;+ we get

2.%&E-

%@.&3E-

?.&5E-

9

B

+=

=

+=

(inal moments=

( )

( ) ( )

>4M?5.%?2".%&3@5."@".&35."M

>4M5".&2+2".%&,3@5."@".&3M

>4M5".&2?.&"?"."%@.&3&.%?@.%M

>4M".&5?@.%%@.&35

3?.&5

5

?M

M>4".&52.%&3

&?.&5

3

M

>4M2.2.%&3

&?.&5

3

&M

D9

9D

9B

B9

BA

AB

==

==

=++=

=+=

==

==

65


66/78

Reactions:9onsider the free body diagram

Member AB:

>4&2.%%3

.2".&5#A =

+=

Member BC:

>43?.2?.5%5&"6

>4?.5%&

&

55&"3".&"5.&2

6

B

9

==

=+

=

Member CD:

66


67/78

#D!

?5.%?5.&2 +! %%.&2 >4

Check:# ! "

#A #D! "Satisfied) hence o/ay

Example:A portal frame having different column heights are subJected for forces

as shown in figure. Analyse the frame and draw bending moment diagram.

Solution:-

-t is an unsymmetrical problem")")")" D9BA == ) hence there is a sway be towards right.

67


68/78

(EMs=

>4M%512

3"

2

l(AB =

==

>4M%52

3"

2

l(BA +=

+=+=

>4M3"12?"

2l(B9 ===

>4M3"2

?"

2

l(9B +=

+=+=

9D( ! D9( ! "


++=

'

3&

'

E-&(M BAABAB

3"

-&E&%51 B ++=

( )%11111111E-@5."E-%51 B >+=

++=

'

3&

'

E-&(M ABBABA

3"&

-&E&%5 B

++=

( )&11111111E-@5."E-&%5 B >++=

( )9BB9B9

&'

E-&(M ++=

( )& -&E&3"1 9B ++=( )3111111111E-E-&3"1 9B >++=

( )B99B9B &'

E-&(M ++=

( )&

-&E&3" B9 +

+=

( )111111111E-&E-3" B9 >++=

++=

'

3&

'

E-&(M

D99D9D

3

3"&3

E-&" 9 ++=

( )5111111111E-3

&E-

3

9 >=

++='

3&

'

E-&(M

9DD9D9

3

3"

3

E-&" 9

++=

68


69/78

( )?111111111E-3

&E-

3

& 9 >=

There are three un/nowns) E-) E-4M"".%;@;5.&"3

&+@%.@,

3

&M

>4M%5.&@;5.&"3

&+@%.@,

3

M

>4M%5.&5@@.;@%.@&3"M

>4M%2.551@.@%1;.5@@&3"1M

>4M55.%2@;5.&"@5."5@@.;&%5M

>4M"%.&%@;5.&"@5."5@@.;%5M

D9

9D

9B

B9

BA

AB

==

===++=

=+==++=

=+=

69


70/78

Reactions:9onsider free body diagrams of the members

70


71/78

Member AB:

>4?%5.%5

&3""%.&%55.%2#A =

=

1ve sign indicates the direction of #Ais from right to left.Member BC:

>4".3%?".&26?"6

>4?".&2

%5.&&?"55.%26

B9

B

===

=+=

Member CD:

>432.%3

%5.&%;#D =

+=

Check:

# ! "

#A #D 3" ! "

1%5.?& C %.32 3" ! "

#ence o/ay

71


72/78

Example: Analyse the frame using slope deflection method and draw the

Bending Moment Diagram.

Solution: Assume sway towards right

-t can be observed from figure in that direction of moments due to sway in

member AB are anticloc/wise and that for member 9D are cloc/wise. ise shall

be ta/en to incorporate the same in the slope deflection e*uation.

72


73/78

(EMS

"#ere

M>43&%&

w-(

M>43&%&

&1

%&

w-(

DA

&

9B

&

&

B9

==

+=+=

=

=

=

Slope deflection e*uations are=

( )

( )&E-3

&E-

3

3

3&

3

E-&

'

3&

'

E-&(M

%1111111E-3

&E-

3

&

3

3

3

E-&

'

3&

'

E-&(M

B

B

ABBABA

B

B

BAABAB

>=

=

++=

>=

=

++=

( )

( )

( )

( )

( )

( )

( )5E-3

&E-

3

3

3&

3

&E-

'

3&

'

E-&(M

11111111E-E-&3&

&

-&E&3&

&'

E-&(M

3111111111E-E-&3&

&

-&E&3&

&'

E-&(M

9

9

D99D9D

B9

B9

B99B9B

9B

9B

9BB9B9

>+=

+=

+++=

>+++=

+++=

++=

>++=

++=

++=

73


74/78

( )?11111111E-

3

&E-

3

&

3

3

3

E-&

'

3&

'

E-&(M

9

9

9DD9D9

>+=

+=

+++=

The un/nown are =+=

++=+

( )2"3&E-3

&E-

3

%"E-

E-3

&E-

3

E-E-&3&MM

9B

9B99D9B

>=+++=

+++++=+

( );"5E-3

E-E-

;"E-3

2E-&E-&

;"E-3

&E-

3

&E-

3

&

E-3

E-

3

&E-

3

E-

3

&E-

3

&;"MMMM

9B

9B

9

9BBD99DBAAB

>=+=

+=

+

+=++

(rom ,@+ < ,;+

( )%"1111111"%";E-3E-3

%@

"5E-3

E-E-

"?E-3

E-&E-

3

&"

9B

9B

9B

>=+

=+

=+

By ,2+ and ,;+

74


75/78

( )%%11111111"%";E-3

%@E-3

"5E-3

E-E-

"?E-3

E-

3

&"E-&

9B

9B

9B

>=++

=+

=+++

By ,%"+ < ,%%+

"@%.%??E-3%@

&"2

"@%.5@E-3E-%@

&@

"%";E-3E-3

%@

B

9B

9B

=

=++

=+

22."&"2

3%@@%.%??E- B +=

=

(rom ,%"+

22."E-3

%@%";

3

%E- B9 =

=

(rom ,;+

[ ]

( )[ ] "@.;5522."22."

3

5E-E-

3E- 9B

+=+=

+=

Thus "@.;5E-)22."E-)22."E- 9B ===

Substituting these values in slope deflection e*uations

( ) ( )

( ) ( )

( ) ( )( ) ( )

( ) ( )

( ) ( ) >4M%&.3?"@.;53

&22."

3

&M

>4M22.2"@.;53

&22."

3

M

>4M22.222."22."&3&M

>4M22.222."22."&3&M

>4M22.2"@.;53

&22."

3

M

>4M%&.3?"@.;53

&22."

3

&M

D9

9D

9B

B9

BA

AB

+=+=

+=+=

=+++==++=

==

==

75


76/78

To find the reaction consider the free body diagram of the frame

Reactions:

9olumn AB

>4%53

%&.3?22.2#A =

+=

Beam AB

76


77/78

>42

&

&22.222.2

6B =++

=

>422&69 ==

9olumn 9D

>4%53

%&.3?22.2#D =

+=

9hec/# ! "#A #D8 ! "1%5 C %5 3" ! "

#ence o/ay

77


78/78

Documents

Stranalysis SD METHOD