Column Slender

7/29/2019 Column Slender

1/11

Interaction Diagram 2 side reinforcement

Faimun (ITS Surabaya) Rekayasa Struktur Semester Genap 2011/2012 1 / 11
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2/11

Interaction Diagram 4 side reinforcement

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3/11

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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4/11

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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5/11

Effective Length Factor, k

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6/11

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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7/11

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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8/11

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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9/11

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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10/11

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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11/11

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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Column Slender