Struktur Beton 2-P-Delta Effect [Compatibility Mode]

7/24/2019 Struktur Beton 2-P-Delta Effect [Compatibility Mode]

1/19

1

Struktur Beton 2Slender Columns

dan P-delta effectSjahril A. Rahim

Departemen Teknik Sipil FTUI

2015

A

Po,max

Po Material Failure

Tn

(Pb,Mb)

M=Pe

Strength of Short Columns

Slender Columns

A

B

e

e

Mc=P(e+)

PMe=Pe

LM=Pe P

Reduction in axial capacity

P

P

P

Short columnInteraction

diagram

Load-maximumMoment curve

Mode of Failure of Columns

A

B

e

Mc=P(e+)

Me=Pe

P

Me=Pe

(Pb,Mb)

Short column

Material failure

P

M

L

e

P

Stability failureC

Load-maximumMoment curve


2/19

2

Mode of Failure

Short column:

Load-moment curve: O-A, Me=PeMaterial failure

Intermediate column:

Load-moment curve: O-B, Mc=P(e+)Reduction of failure load

Material failure

Long column:

Load-moment curve: O-C, Mc=P(e+), unstableM/P Stability failure

Slenderness

Short Column: klu/r 34-12M1b/M2b

klu/r 22 Medium Column: Metode pembesaran

moment

Long Column: klu/r > 100 Second order

analysis

Mome n

t,M

Curvature

Moment-Curvature diagrams for a column cross section

P=Pb

,

P < Pb

P > Pb

M EIM

Load-moment behavior for hinged columns

subjected to sustained loads

AB

CD

Failure

Moment, MMoment, M

Axialload,P

Axialload,P

FailureAD

(a) Rapid-sustained-rapid loading history (b) Creep buckling

PA

PC

P

tPA

PC

P

t

PA > 0.70 PD

Loading history Loading history


3/19

3

Analysis and design compressive members

Second order analysis

Moment magnifier method



Nonlinear material properties

Cracking

Pengaruh kelengkungan komponen tekan

dan gaya lateral, durasi beban

Rangkak dan susut

Interaksi dengan pondasi

Gaya dalam, momen berfaktor

Metode pembesaran momen

Analisis dengan orde pertama

Momen rencana: Momen hasil analisis x

faktor momen (berfaktor) dan Pu


4/19

4

(1) Slender Columns

For long column the following factor should be

considered: buckling, elastic shorthening, andsecondary moment due to lateral deflection

The extra length will cause a reduction in the

column strength that varies with the column

effective height, width the section, the

slenderness ratio, and the column end

conditions;

The slenderness ratio: l/r, where r=(I/A)

Radius of gyration

hhrxA

Ir

bhA

bhI

xx

x

3.0288.0

12

1 3

DrrA

Ir

DA

DII

yxx

x

yx

25.0

4

642

4

b

h

D

A

Ir

A

Ir

y

y

xx

Slender columns

Long with relatively high slenderness ratio,where lateral bracing or shear walls are

required; Long with a medium slenderness ratio that

cause a reduction in the column strength.Lateral bracing may not be required, butstrength reduction must be considered;

Short where the slenderness ratio is relativelysmall, causing a slight reduction in strength andmay be neglected

(2) Effective column length (Klu)

The unsupported length, lu, represents the

unsupported height of the column between

two floors;

The effective length factor, K, represent

the ratio of the distance between points of

0 moment in the column and the

unsupported height of the column in one

direction


5/19

5

The unsupported length

lu lu lulu

The effective length factor (K)

(a) Braced (b) Unbraced

Klu

Inflection points

Klu

Effective lengths of columns and length factor K:

0.50 0.70 1.0 1.0 2.0 2.0

0.65 0.80 1.0 2.10 2.01.2

Theoritical Kvalue

Recommendeddesignvaluewhenideal conditionsareapproximate

Endconditioncode

Buckledshapeofcolumnisshownbydashedline

Rotationfixedandtranslationfixed

Rotationfreeandtranslationfixed

Rotationfixedandtranslationfree

Rotationfreeandtranslationfree

Effectivelengthfactors for idealizedcolumnendconditions.CourtesytheAmericanInstituteof Steel Construction,Inc.

Effective length and K fo r Braced and Unbraced Frame

kL>2L

L


6/19

6


The effective length of columns can be

estimated by using the alignment chartshown the following figure;

To find the effective length factor K, it is

necessary first to calculate the end restraint

factors A and B at the top and bottom ofthe column, respectively, where

beamsoflEI

columnsoflEIc

/

/


For a hinged end, is infinite and may be

assumed to be 10,0; For a fixed end, is zero and may be

assumed to be 1,0;

There are two monograms, one for braced

frames where side sway is prevented, and

second for unbraced frame where

sidesway is not prevented.

Alternative for evaluating the effective length

factor, K

For braced compression members, an upper

bound of K may be taken as smaller of the

following two expressions:

0.105.085.0

0.1)(05.07.0

min

K

K BA

Where A and B are the values of at the ends of the column and min isthe smaller of the two values


7/197

Alternative for evaluating the effective length

factor, K

For unbraced compression members

restrained at both end, the effective lengthfactor, K, may be assumed as follows:

219.0

220

20

mm

mm

forK

forK

Where m is the average of the values at the two ends of thecompression members.

Members Stiffness (EI)

Loading Model Response

Stiffness

Cracking

Members Stiffness (EI)

'

'5.1

4700

043.0)(

cc

ccc

fE

fwE

(1) Elasticity modulus of concrete:

(2) Effective moment of inertia:

Beam 0,35 Ig

Column 0,70 Ig

Walls (uncracked) 0,70 Ig

(cracked) 0,35 Ig

Flat plates and flat slabs 0,25 Ig

(3) Area 1,0 Ag

(4) The moment of inertia shall be divided by (1+d) when sustainedlateral loads act on the structure of for stability check, where

)(

)(max

loadaxialfactoredtotal

loadaxialsustainedfactoredimumd

Stiffness kolom untuk evaluasi Eulers

buckling load

Eulers bucling load dari sebuah kolomdengan ujung sendi-sendi:

2

2

)(klu

EIPc


8/198

Limitation of the slenderness ratio (Klu/r),

Nonsway frames

Limitation of the slenderness ratio (Klu/r),

Nonsway frames

The effect of slenderness may be neglected and thecolumn may be design as short column when klu/r 34-

12M1/M2 The ratio M1/M2 is considered positive if the member is

bent in single curvature and negative for doublecurvature

The ratio (34-12M1/M2) shall not be greater than 40

If the factored column moments are zero or e=Mu/Pu 1.0

Step-by-step prosedur untuk kolom yang

tidak goyang sbb:

d

gc

d

sesgc

u

c

IEEI

IEIEEI

Kl

EIP

1

4.0

1

2.0

)( 2

2

dimana beban kriitis Euler:

dan EI ditentukan oleh rumus berikut ini:

d= 1.2PDL/(1.2PDL+1.6PLL)

Note: (1) Jika Pu >0.75Pc, ns akan bernilai negative, kolom beradadalam kondisi unstabel

(2) Jika ns > 2.0 perbesar ukuran kolom, karena perhitunganmenjadi sangat sensitif terhadap asumsi yang dibuat.


tidak goyang sbb:

Nilai Cm adalah suatu faktor yang menghubungkan

diagram momen aktual dgn suatu diagram momen

merata equivalen

Cm=0.6+0.4M1/M2 > 0.40

Cm=1 bilamana ada beban transversal di kolom

0M1/M21 -1M1/M20M1 < M2

Single

curvatureDouble curvature


11/1911


tidak goyang sbb:

Dalam hal M2min > M2, Cm=1.0 atau

Cm=0.6+0.4M1/M2

Braced frames

i i

Definition on nonsway and sway frames: Stability Index

Completely braced frames and completely unbraced frames

Completely braced in given direction, if lateral stability provided by

walls, bracing or buttress design to resist all the lateral forces

Completely unbraced if in a given plane if all resistance if allresistance to lateral loads comes from bending of the columns

In actually, there is no such think as a completely braced frame, and

no clear-cut boundary exist between braced and unbraced frames

Compare the moment in the end of column using the first order

analysis and the second order analysis, if different more than 5%,

the frames considered as unbraced frame

Alternatively, ACI Section 10.11.4.2 allows designers to assume that

a story in a frame is non sway if stability index is less than or equal

to 0.05

Example-nonsway column

Ditanya: apakah kolom tersebut cukup kuat

ml u 4 .5 0Panjang kolom tidak ditopang:

Kolom merupakan bagian dari frame yang ditopang (braced) dandilenturdalamsingle

curvaturepada sumbu utamanya

0.65Kolom dengan sengkang biasa:

MPaf y 40 0

MPafc 28Mutu material:mmh 550

Ukuran kolom: mmb 350kN mM L 13 0kNP L 4 98Beban hidup:kN mM D 1 42

kNP D 55 6Beban mati:

Modul 4: Kolom langsing dengan Faktor pembesaran momenData:

Sjahril A. RahimStruktur beton 2


12/1912

Interaksi kolom 350 x 550 mm, tulangan dua sisi

2 x 2580 mm2

------------> kolom langsing

Batas 22B at as 3 4 1 2M1b

M2b

M2b Mu

M1b Mu

A

B

Ma

Mb

lc

Ma

Mb

kelangsingan 28.343kelangsingan K lu

r1000

r

1

12b h

3

b hJari-jari inersia

(conservative)K 1Braced frame----------->

(2) Cek apakah termasuk kolom panjang:

kN mMu 378.4Mu 1.2 MD 1.6 ML

kNP u 1 .4 6 4 1 03P u 1 .2 P D 1.6 PL

(1) Hitung beban ultimate y ang diperlukan:

Step-step penyelesaian:

b 1.105b

Cm

1Pu 1000

Pc

0.65Faktor pembesaran momen

P c 2 .3 6 1 1 07Pc

2 EI

K lu 1000( ) 2

Hitung besarnya Euler bucling load:

E I 4.844 1013EI

0.2 Ec Ig Es Ise

1 d

MPaE s 200000

d 0.456d

1.2 PD

1.2 PD 1.6 PL

Rasio beban mati:

Ise 2.319 108Ise 2 As

h

2d

2

mmd 63

As Asmm2A s 2 58 0

I g 4 .8 53 1 09Ig

1

12b h3

Momen inersia:

E c 2.487 104E c 4 7 00 f c

Cm 1C m 0 . 6 0 .4M1b

M2b

(3) Hitung faktor pembesaran momen

(4) Hitung momen rencana dan gaya aksial rencana

PnPu

P n 2 . 25 2 1 03

Mnb Mu

M n 6 4 3 .5 4 5

(5)Cek terhadapinteractiondiagram


13/1913

Pembesaran Momen Rangka Portal

Bergoyang (unbraced)


bergoyang sbb:

Panjang kolom: tentukan panjang kolom

tidak didukung lu Panjang efektif kolom: klu, Tentukan

panjang efektif kolom k dengan bantuanalligment chart, K>1,0

Hitung radius of gyration

Batas kelangsingan: pengaruhkelangsingan dapat diabaikan bilamana:klu/r 22


bergoyang sbb

Hitung momen yang tidak diperbesar, Mns,

akibat beban yang tidak menimbulkan

pergoyangan (sway). Analisis elastik ordepertama dengan stiffness member sesuai SNI

10.1.1. Untuk kombinasi beban U=1,2D+1,0L+E,

momen Mns hasil dari kombinasi beban

1,2D+1,0L

Hitung sway momen yang diperbesar, sMs:Beban yang menimbulkan momen Ms adalah

dari E

P- Moments and P- Moments


14/1914


Load level for the analysis:

1.2D+L+E

Stiffness of the member:

Ultimate limit state

Serviceability limit state

Foundation rotation:

Stiffness reduction factors for flexure

based on ACI318-08*Members Service State Ultimate State

Beams 0,50EcIg 0,35EcIg

Columns 1,00EcIg 0,70EcIg

Walls

Uncracking 1,00EcIg 0,70 Ec Ig

Cracking 0,50EcIg 0,35EcIg

Flat Plate 0,35EcIg 0,25EcIg

Area 1,0Ag 1,0Ag

* ACI318-08

1,4 times the stiffness used for analysis under factored lateral load, ACI318-08 Chapter 8.8 ACI318-08 Chapter 10.11


15/1915

Effect of sustained loads Methods of second-order analysis

Iterative P- analysis


16/1916

P- effect pada struktur frame:

Struktur gedung yang tingginya diukur dari taraf penjepitan

lateral lebih dari 10 tingkat atau 40 m, harus

diperhitungkan terhadap pengaruh P- effect, yaitu gejala

yang timbul pada struktur yang agak flexible, dimana

simpangan yang besar kesamping akibat beban gempa

lateral menimbulkan beban lateral tambahan akibat

momen guling yang terjadi oleh beban gravitasi yang titik

tangkapnya menyimpang kesamping.

H

Mb

Pg

FE

y

F FFE

FE

Lateral Force,F

Displacement

Mb = FEH +Pg

1

2

3

4

i

i-1

N

i

i-1

ui wui i wui i/hi

wui i/hi

hi

Level

(a)Displacedpositionof storyweight

(b)Additional overturningmoments

or lateral loads

OverturningLoadsduetoTranslationof Storyweight

Direct P- analysis for sway frames Direct P- analysis for sway frames


17/1917

Alternatif dalam menghitung sMs sesuaiSNI 10.3.4

Analisis orde kedua

Direct P- analysis Sway Frame Magnifier (Faktor

pembesaran momen sway frame)

Analisis orde kedua

M1

M2

M1

M2

sMs

Hasil analisis elastis orde kedua berdasarkan nilai kekakuan komponen

struktur sesuai 10.11.(1)

Direct P- analysis

lc

cu

ou

lV

P

o

)(

1

)/()(1 cuou

o

lVP

MsMsM

cu

ou

lV

PQ

ss

ss MQ

MM

1

Momen pada frame proportional langsung dengan defleksi, second order

moment:

Dimana Mo = first order moment

M = second order moment

Jika Q = Stability index

Dengan mengabaikan flexibility factor ,


18/1918

Sway Frame Magnifier

lc

s

c

u

sss

M

P

P

MM

75,0

1

22

)(Klu

EIPc

dengan Pu= Jumlah beban aksial berfaktor untuk semua kolom padatingkat yang ditinjau

= jumlah beban kritis Euler untuk semua kolom pada

tingkat yang ditinjau

d

gc

d

sesgc

IEEI

IEIEEI

1

4,0

1

2,0

K = factor panjang efektif kolom dihitung dari grafik untuk

unbraced frame

tingkatpadagesergayaTotal

tingkatpadatetapgesergayaMaxd

dalam hal untuk angin dan gempa d = 0


bergoyang sbb

Momen pada ujung-ujung kolom

dimana M2 > M1

ssns

ssns

MMM

MMM

222

111

Maximum momen diantara ujung kolom

l1 2 3

12

3

Perilaku restrained kolom sangat langsing yang dilengkung dalam

double curvature

/h

M/PnohDeflection Moments


19/1919

Cek apakah momen maksimum terjadi diantaraujung-ujung kolom.

Umumnya, maksimum momen pada kolomtejadi di ujung-ujung kolom dan kolom dirancangsesuai dengan momen ini.

Akan tetapi dalam hal gaya aksia sangat besardan slenderness melampaui batas yangdiberikan pada (10.13.5) diperlukan untukmengecek pada suatu potongan momen lebihbesar momen-momen di ujung kolom

Dalam hal lu/r > 35/((Pu/fcAg)) momen

pada suatu titik diantara kedua ujung lebihbesar dari momen dikedua ujung.

Dalam hal ini kolom harus dirancang untuk

memikul beban berfaktor Pu dan Mc yang

dihitung sbb:

d

gc

d

sesgc

u

c

c

u

mns

nsc

IEEI

IEIEEI

Kl

EIP

P

P

C

MM

1

4.0

1

2.0

)(

0.1

75.01

2

2

2

dimana

d = ditentukan sesuai kombinasi bebanK=ditentukan menurut butir 10.12.(1), K=1 konservatif atau ditentukan

dengan alligment chart.

Struktur Beton 2-P-Delta Effect [Compatibility Mode]

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