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7/29/2019 Column Slender
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Interaction Diagram 2 side reinforcement
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Interaction Diagram 4 side reinforcement
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Slender or Long Column
If column slenderness ratio exeedsthe limit for short column, thecompression member will buckleprior to reach its limit state ofmeterial failure.
The effective length klu is used asthe modified length of the columnto account for end restraints otherthan pinned.
The value ofk:Both ends fixed k= 0.5Both ends fixed, lateral motion exist k= 1.0Both ends pined, no leteral motion k= 1.0One ends fixed, other end free k= 2.0
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Column in structural frame
1 Braced compressive members.
k= 0.7 + 0.05(A + B) 1.0k= 0.85 + 0.05min 1.0
where:
=EI/lucolumns
EI/lnbeams
lu is the unsupported length of column; ln is clear beam span.2 Unbraced compression members restrained at both end.
For m < 2 : k=20 m
20 1 + m
For m 2 : k= 0.9
1 + m
where: m is the average of the value at the two ends.3 Unbraced compression members hinged at one end.
k= 2.0+ 0.3
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Effective Length Factor, k
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Moment Magnification: First-Order Analysis
Properties of member in a structure:1 Modulus of Elasticity: Ec = 4700
fc
2 Moment InertiaBeams 0.35Ig
Columns 0.70IgWall : uncrack 0.7IgWall : crack 0.35IgFlat plates and flat slabs 0.25Ig
3 Area: 1.0As4 Radius of giration: r = 0.30h for rectangular member, r = 0.25D
for circular members.
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Moment Magnification in Nonsway Frames
Structure is nonsway if:
Q =Pu0Vulc
0.05
Where: Pu and Vu are the total vertical load and the story shear, and 0 is the relative deflectionof that story due to VuThe slenderness effect can be disregarded if
klur 34 12
M1M2
andM1M2
0.5
M1/M2 is positive if single curvature, and negative if double curvature.The magnified moment becomes
Mc = nsM2
where
ns =Cm
1 (Pu/0.75Pc) 1.0
Pc =2EI
(klu)2
and
EI=0.2EcIg + EsIse
1 + dor EI=
0.4EcIg
1 + d
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Moment Magnification in Nonsway Frames (cont.)
Cm a factor relating the atual moment diagram to an equivalentuniform moment diagram. For member without transverse load, thatis, subjected to end loads only,
Cm = 0.6 +M1
M2
0.4 where M2 M1
M1/M2 is positif if column is bent in single curvature. For memberswith transverse load between support, Cm = 1.0Minimum factored moment M2 is
M2,min = Pu(15 + 0.03h), where h in mm
IfM2,min exceeded the applied moment M2, the value ofCm = 1
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Moment Magnification in Sway Frames
The slenderness effect can be disregarded if
klur< 22
The end moment M1 and M2 should be magnify as follows
M1 = M1ns + sM1s
M2 = M2ns + sM2s
On the assumtion that M2 > M1, the design moment should be
Mc = M2ns + sM2s
Magnified sway moment can be calculated as
sMs =Ms
1 (Pu/0.75Pc) Ms, where s 2.5
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Example:1. Sway frame
Compute slenderness column A-B, with single curvature moment 75 KN-m , as picture
300x450mm 300x450
300x600 300x600
300x500
A
B
6000mm 7000
3500
3000
Inertia :Top Col : 112 300 500
3 = 3125000000 mm4
Top Beam: 112 300 4503 = 2278125000 mm4Bot Beam: 112 300 600
3 = 5400000000 mm4
A =0.73125000000
30000.352278125000
6000 +0.352278125000
7000
= 2.95
B =0.73125000000
3000 +0.73125000000
35000.3554000000006000 + 0.3554000000007000
= 2.31
from alignment chart k= 1.74.
lu = 3000
450
2+
600
2
= 2475
klur
= 1.74 24750.3 500
= 28.71
SNI code for shot column
klur 34 12
75
75
= 22
Hence column A-B is slender columnFaimun (ITS Surabaya) Rekayasa Struktur Semester Genap 2011/2012 10 / 11
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Example:2. NoSway
2.5 m
2.5 m
90 kN
650 kN
650 kN
Primary moment due to lateral load, Ec = 25000 MPa
Mn =90 5
4= 112.5 kN-m
Pc =2EI
(klu)2(euler bucling load)
=
2 25000 112 300 3003
(1.0 5000)2
= 6662 kN
Magnified moment
Mc
= Mn
1
1 PPc
= 112.51
1 6506662= 112.5 1.11 = 124.875 kN-m
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