2 Slope Deflection Method for Statically Indeterminate Portal Frame

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TO

TOPIC

PIC 2

2 Slope Deflection Method

Slope Deflection Method

for Statically

Indetermin

ate

Portal Frames

by: by:

SALIA!ATI "I#TI $AMA%&AMA# SALIA!ATI "I#TI $AMA%&AMA# CI'IL (#)I#((%I#) D(PA%TM(#T CI'IL (#)I#((%I#) D(PA%TM(#T

1 1

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Learnin* O+tcome

&pon completion of this topic, st+dents sho+ld be

able to:

-2 2 . .

(/plain the portal frame

0

Disc+ss the factors that ca+se frame to s1ay or not to s1ay

&nno1ns +sin* the slope deflection method

Dra1 shear force and bendin* moment dia*ram

2

&nderstand the basic concept of slope deflection method for sol3in*

the indeterminate beams:

-- Calc+late the internal moment at portal frame s+pport andCalc+late the internal moment at portal frame s+pport and

4

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Learnin* O+tcome

&pon completion of this topic, st+dents sho+ld be

able to:

-2 2 . .

(/plain the portal frame

0

Disc+ss the factors that ca+se frame to s1ay or not to s1ay

&nno1ns +sin* the slope deflection method

Dra1 shear force and bendin* moment dia*ram

2

&nderstand the basic concept of slope deflection method for sol3in*

the indeterminate beams:

-- Calc+late the internal moment at portal frame s+pport andCalc+late the internal moment at portal frame s+pport and

4

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250 Frames

-3 3 •

• +sed in b+ildin*s and are composed of beams and col+mns that+sed in b+ildin*s and are composed of beams and col+mns that

are either pin or fi/ed connected

are either pin or fi/ed connected •

• lie tr+sses, frames e/tend in t1o or three dimensions5lie tr+sses, frames e/tend in t1o or three dimensions5

•

• the loadin* on a frame ca+ses is *enerally 6indeterminate7 from athe loadin* on a frame ca+ses is *enerally 6indeterminate7 from a

standpoint of

standpoint of analysianalysis5s5 •

• the stren*th of s+ch a frame is deri3ed from the momentthe stren*th of s+ch a frame is deri3ed from the moment

interactions bet1een the beams and the col+mns at the ri*id

4

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250 Frames

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• Types of frames:

i5 Symmetrical portal frames 8non9s1ay ii5 &nsymmetrical portal frames 8s1ay

Symmetrical portal frames

• Both the columns are of the

same length , having similar end conditions(i.e either

hinged or xed)

• Subected to s!mmetrical

loading

• "he oints of such a #ortal

frame $ill not be subected

Unsymmetrical portal frames

• either the columns are not

s!mmetrical , or the frame is not s!mmetricall! loaded

• Subected to some hori%ontal

movement, &no$n as s$a!, to one side or the other

• "he rigid oints bet$een the

columns and beams $ill have motion of translation

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252 Symmetrical Portal Frames

(Non-sway)

-'

A symmetrical portal frame is that in 1hich; a "oth col+mns are of the same len*th5

b <a3in* similar end conditions 8hin*ed or fi/ed, moment of inertia, and mod+l+s of elasticity5

c S+b4ected to symmetrically load5

*_{symmetrical portal frame, the joints will not be subjected}

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252 Symmetrical Portal Frames

(Non-sway)

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25- &ns

ymmetrical Portal Frames

(sway)

-

• S+b4ected to some hori=ontal mo3ement no1n,

as s1ay, to one side or the other5

• !hich either ;

i the col+mn are not symmetrical, or

ii the frame is not symmetrically loaded5

• %i*id 4oints bet1een the col+mns and beam

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25- &ns

ymmetrical Portal Frames

(sway)

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25- &ns

ymmetrical Portal Frames

(sway)

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25. Side S1ay

-1

Any appreciable lateral or side91ard mo3ement of top of a 3ertical member relati3e to its bottom is called side s1ay, s1ay or lateral drift5

Portal frame

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25. Side S1ay

-11

•

Causes of side sway:

i) Length of dierent columns in a

frame structure is dierent!

ii) "ue to un-symmetrical loading!

iii) Lateral loads are acting!

i#) "ierent end conditions of the

columns

#) Non-uniform section of the mem$ers

(Sections of columns ha#ing dierent cross sectional properties! For e%ample if moment of inertia of one #ertical

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25. Side S1ay

-12

Side s1ay may be pre3ented in a frame by:

• Pro3idin* shear or partition 1alls5

• Fi/in* the top of frame 1ith ad4oinin* ri*id str+ct+res5

• Pro3ision of properly desi*ned shear 1alls act as

bacbone of the str+ct+re and red+cin* the lateral deflections5

• Pro3ision of lateral bracin* 1hich may be dia*onal

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25. Analysis of Portal Frames

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• All the symmetrical portal frames are analysed by

openin* them o+t, and threatin* them e/actly lie a three span contin+o+s beam

• The 4+nction of the beams and col+mn beha3es

lie a ri*id 4oint5

* Calculate the the internal moment for_{portal frame}

>

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25? FI@(D (#D MOM(#T 8F(M

-14

(/ample 250:

30 kN A A B

Calc+late the fi/ed end moments for each member: a

C

2m 2m

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25? FI@(D (#D MOM(#T 8F(M

-1' Answer: a 30 kN A FEM AB FEM_BA A 2m 2m B F(M_A" > 9 0? #m; F(M_"A >  0? #m S#an -B A FEM BC FEM_CB B C F(M_"C > B #m; F(M_C" > B #m S#an B 3m

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25? FI@(D (#D MOM(#T 8F(M

-1

(/ercise 250:

5 kN A A B

Calc+late the fi/ed end moments for each member: a C 4m 2m 1m 9 kN/m

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25? FI@(D (#D MOM(#T 8F(M

-(/ercise 250:

13 kN A A B

Calc+late the fi/ed end moments for each member: b C 3m 2m 2m 25 kN/m Ans: FEM _AB = - 18.7kNm; FEM _BA = + 18.7 kNm; FEM _BC = - ! kNm; FEM _B" = -#. kNm; D 2m 2EI EI 2EI

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25? FI@(D (#D MOM(#T 8F(M

-(/ercise 250:

12 kN A A B

Calc+late the fi/ed end moments for each member: c C 3m 3m 4m 9 kN/m Ans: FEM _AB = - 1#. kNm; FEM _BA = + 1$. kNm; FEM _BC = - #.7 kNm; FEM _CB = +#.7 kNm; FEM _B" = -% kNm; FEM _"B = +% kNm; FEM _CE = - 2.22 kNm; D 2m 3EI 2EI 2EI 1m 8 kN 5 kN 3m E 2EI

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25 SLOP( D(FL(CTIO# M(T<OD

-1+

Step &: 'rite Slope "eection *uation

Step +: Calculate Fi%ed nd ,oment (F,) for each mem$er

Step : Sol#e *uili$rium *uation and sol#e for un.nown

Step /: Su$stitute into Slope deection

*uation and calculate the internal moment Step 0: Calculate reaction and draw SF" and 1,"

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25 SLOP( D(FL(CTIO# M(T<OD

-2

(/ample 252:

5 kN A A B

Calc+late internal moment for portal frames s+b4ected to a load as fi*+re belo1 by +sin* slope deflection method, then dra1 SFD and "MD a C 4m 2m 1m 9 kN/m

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25 SLOP( D(FL(CTIO# M(T<OD

-21

ST(P 0: Calc+late Fi/ed (nd Moment F(M_A" > 9 02 #m;

F(M_"A >  02 #m; F(M_"C > 9 2522 #m; F(M_C" >  0500 #m

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25 SLOP( D(FL(CTIO# M(T<OD

-22

ST(P 2 : Slope Deflection (+ation

Since the beam is fi/ed at A and C, therefore slope at A and C, and

9999999999999999999999999 0

9999999999999999999999999 2

9999999999999999999999999

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25 SLOP( D(FL(CTIO# M(T<OD

-23

ST(P - : Moment (+ilibri+m At (ach Eoint 5

[

]

EI EI EI EI B B B B 7 34 . 29 3 7 0 22 . 2 3 4 12 − = = =     − + + θ θ θ θ

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25 SLOP( D(FL(CTIO# M(T<OD

-24

ST(P . : 'al+e Of Internal Moment

Su$stituting the #alue of in e*uations (i)2 (ii and (i#) 3

Final ,oments at support 42 12 and C3

kNm EI EI M kNm EI EI M kNm EI EI M kNm EI EI M CB BC BA AB 68 . 1 11 . 1 7 34 . 29 3 2 81 . 7 22 . 2 7 34 . 29 3 4 81 . 7 12 7 34 . 29 1 . 14 12 7 34 . 29 2 − = + − × = − = − − × = + = + − × = − = − − × = EI B 7 34 . 29 − = θ

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25 SLOP( D(FL(CTIO# M(T<OD

-2'

ST(P ? : Shear Force 8#  "endin* Moment 8#m

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25 SLOP( D(FL(CTIO# M(T<OD

-2 14.1 .4 .*1 .* 1.+ 1* 1.31 1.* 4.' 0 0    1+.' .' 1.' 1.42 2.1 0  0 SFD "MD 2.4 m x x x x x 17 . 2 57 . 19 28 . 78 43 . 16 43 . 16 4 57 . 19 = − = − = 96 . 10 2 81 . 7 1 . 14 = + m x x x x 47 . 2 43 . 23 49 . 9 68 . 1 3 81 . 7 = = − = m a a 65 . 4 47 . 2 81 . 7 47 . 1 = =

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25 SLOP( D(FL(CTIO# M(T<OD

-2

(/ercise 252:

30 kN A A B

Calc+late internal moment for portal frames s+b4ected to a load as fi*+re belo1 by +sin* slope deflection method, then dra1 SFD and "MD a C 2m 2m 3m Ans'er: M _AB = - 18.21 kNm; M _BA = + 8.7 kNm; M _BC = - 8.7 kNm; M _CB = - &.2% kNm ( _A = 17.&1 kN; ( _B1= 12.% kN; ( _B2 = &.2% kN; ( _C = -&.2% kN

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25 SLOP( D(FL(CTIO# M(T<OD

-2* 1*.21 1.1 *.' *.' 13.3+ 3 4.2+ 0    1.41 4.2+ 12.'+ 0  0 SFD _"MD 0

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25 SLOP( D(FL(CTIO# M(T<OD

-(/ercise 252:

13 kN A A B

Calc+late internal moment for portal frames s+b4ected to a load as fi*+re belo1 by +sin* slope deflection method, then dra1 SFD and "MD c C 3m 2m 2m 25 kN/m Ans: M _AB = - .#$ kNm; M _BA = + &&.%% kNm; M _B" = +.!1 kNm; M _"B = ! kNm; ( _A = 2&.$8 kN; ( _B1= !.#2 kN; ( _B2 = .2 kN; ( _"= 7.7 kN; ( _B$= ! kN D 2m 2EI EI 2EI

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25 SLOP( D(FL(CTIO# M(T<OD

-3 13 1'.'1 12.' '.3 0 2*.13   24.3* '.2  0 SFD _"MD '  '.2' .' 4'.1 ' '.1 .1+