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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
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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
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Analysis and design compressive members
Second order analysis
Moment magnifier method
Second order analysis
Second order analysis
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
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(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
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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
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The effective length factor (K)
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
/
/
The effective length factor (K)
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
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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
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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.
Step-by-step prosedur untuk kolom yang
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
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Step-by-step prosedur untuk kolom yang
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
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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
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Pembesaran Momen Rangka Portal
Bergoyang (unbraced)
Step-by-step prosedur untuk kolom yang
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
Step-by-step prosedur untuk kolom yang
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
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Second order analysis
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
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Effect of sustained loads Methods of second-order analysis
Iterative P- analysis
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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
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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 ,
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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
Step-by-step prosedur untuk kolom yang
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
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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.