Structural Steel Design Compression members

Team TeachingStructural Design

Civil Engineering Department2010

Where, f is assumed to be uniform over the entire cross-section.

Introduction

Compression Members: Structural elements that are subjected to axial compressive forces only are called columns. Columns are subjected to axial loads thru the centroid.

Stress:A

Pf s

If an axial load P is applied and increased slowly, it will ultimately reach a value Pcr that will cause buckling of the column.

Column Buckling

Pcr

Pcr

P

P

(a) (b)Pcr

Pcr

P

P

P

P

(a) (b)

22

LK

IEPcr

Pcr = critical buckling load of the column.

Euler Formula

y

xy

P P x

Py

y

M=Py

L

Pcr

(a)

(b) (c)

d

P

Buckled deflected shape

Figure 5-4 Buckling of a column

P

kxBkxAyEI

Pk

kydx

ydy

EI

P

dx

yd

EI

M

dx

ydPyM

cossin

0022

2

Differential equation solution

BC’s: x=0 y=0 x=L y=0

From 2 boundary Conditions : B=0 and AsinkL=0

solution: A=0 trivial solution sinkL=0 non trivial solution kL=nл

2

22222

22222

L

EInPnL

EI

P

nLkEI

Pk

Fundamental mode n=1,

22

2

2

rL

EF

L

EIP crcr

Euler Equation

where, r2 = I/Ag

Euler Formula

Fy 22

rL

EFcr

Rasio kelangsingan, L/r

Fcr

Formula Euler is based on assumption : Steel material is in linear elastic stage There is no residual stress The column is perfectly straight The load is axial thru the centered, with no

eccentricity. The column in pinned at both ends

Ideal Column

This ideal state is never reached. The stress-state will be non-uniform due to:

• Accidental eccentricity of loading with respect to the centroid

• Member out-of –straightness (crookedness), or

• Residual stresses in the member cross-section due to fabrication processes (cooling process)

Rasio kelangsingan, L/r

Pcr

PyRange of test results

Actual Column• Nonlinear Stress - Strain• There is residual stress• Initial Slenderness• Non- homogenous material• Initial eccentricity • Inaccurate Dimension • Restrain Condition

Actual Column ≠ Ideal Column

(Fcr) < (Fcr)ideal

ge ArL

EP

2

2

)/(

Elastic Buckling :

gc

ycr

ykc

k

cr

Af

P

E

f

r

L

L

EIP

2

2

2

1

Column Slenderness

Euler equation

Where Lk = Effective length = K L K = buckling coefficient fy = yield stress of material

The AISC specifications for column design are based on several years of research.

fc Pn = 0.9 Ag Fcr

Column Design Strength fcPn

5.1E

F

r

KL yc

ycr FF c2

658.0

5.1E

F

r

KL yc

y

c

FFcr 2

877.0

Elastic

buckling

Inelastic buckling

fc Pn = 0.9 Ag Fcr

Column Design Strength AISC 2005fcPn

Elastic buckling

Inelastic buckling

yF

E

r

KL71.4

yF

F

cr FF e

y

658.0

yF

E

r

KL71.4

ecr FF 877.0

ye FF 44.0

ye FF 44.0 or

or

2

2

r

KL

EFe

Design Strength Chart

Fcr/Fy

1.0

0.39

Fcr/Fy

1.0

0.39 Fcr = Fcr =

Fcr =Fcr =

e

y

F

F

658.0 Fy

eF877.0

yF

E71.4

r

KL

Fcr/Fy

1.0

0.39

Fcr/Fy

1.0

0.39 Fcr = Fcr =

Fcr =Fcr =

e

y

F

F

658.0 Fy

e

y

F

F

658.0 Fy

eF877.0

yF

E71.4

r

KL

Max KL/r = 200

0.50 0.70 1.0 1.0 2.0 2.0

0.65 0.80 1.0 2.10 2.01.2

Theoritical K value

Recommended designvalue when ideal conditions are approximate

End condition code

Buckled shape ofcolumn is shown bydashed line

Rotation fixed and translation fixed

Rotation free and translation fixed

Rotation fixed and translation free

Rotation free and translation free

Effective length factors for idealized column end conditions. Courtesy theAmerican Institute of Steel Construction, Inc.

Effective Length for different Restraint Condition

..\hasil download purdue univ\column buckling.mpg

Major axis means axis about which it has greater moment of inertia (Ix > Iy)

W12 x 50: E = 29000 ksi Ix = 391 in4. Iy = 56.3 in4

Major X axis : pin-pin Kx = 1.0 (theory ) Kx = 1.0 (recommended)

Unsupported length Lx = 20 ft.

Effective length : Kx Lx = 1.0 x 20 = 20 ft. = 240 in.

Minor Y axis : pin-fix Ky = 0.7 (theory) Ky = 0.8 (recommended) Unsupported length Ly = 20 ft. Effective length f: Ky Ly = 0.8 x 20 = 16 ft. = 192 in.

Example : (1) Determine the buckling strength (Pcr) of a W 12 x 50 column. Its length is 20 ft. For major axis buckling, it is pinned at both ends. For minor buckling, is it pinned at one end and fixed at the other end.

x

y

Critical load buckling about x – axis Pcr-x = Pcr-x = 1942.9 kips

buckling about y-axis Pcr - y= Pcr-y = 437.12 kips

Buckling strength of the column : Pcr = 437.12 kips

Minor (y) axis buckling governs.

22

yy

y

LK

IEPcr

2

2

192

3.5629000

2

2

240

39129000

a) Major axis buckling; (b) minor axis buckling

..\hasil download purdue univ\slenderness ratio.mpg

rx = 6.04 ry = 2.48 Ag = 21.8 in2

Kx = Ky = 1.0 (pin end) Lx = Ly = 20 x 12 = 240 in.

Slenderness ratio KxLx/rx = 240/6.04 = 39.735

KyLy/ry = 240/2.48 = 96.77 (govern)

Cek the limit :

Example : (2) Calculate the design strength of W14 x 74 with length of 20 ft and pinned ends. A36 steel is used.

ksi

rKL

EFe 56.30

77.96

29000*2

2

2

2

68.13336

2900071.471.4

yF

E

yF

E

r

KL71.4

yF

F

cr FF e

y

658.0

Fcr = 21.99 ksi

fcPn = 0.85 (Ag Fcr) = 0.85 (21.8 x 21.99) = 431.4 kips

Design strength of column = 431 kips (inelastic buckling)

Column Design Strength based on Empirical Formula

225,12,1

67,06,1

43,12,125,0

125,0

9.09.0

cc

cc

c

ycr

ygcrgn

ff

fAfAP

Ag = gross area, mm2

fcr = critical stress, MPafy = yield stress, MPaω = buckling coefficient depend on slenderness ratio

Comparison Graph LRFD TCPSBuBG vs AISC

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2 2.2 2.4 2.60

50

100

150

200

250

300

FcrT c( )

FcrA c( )

c

BJ41, Fy=250 MPa

Perbandingan persamaan kekuatan tekuk Fcrantara TCPSBuBG dan AISC untuk BJ41

ω Graph

0 1 20

5

8

0

c( )

2.600 c

0 20 40 60 80 100 120 140 160 180 2000

50

100

150

200

250

300

350

400

450

500

BJ34BJ37BJ41BJ50BJ55

Slenderness ratio kL/r

Cri

tica

l st

ress

Fcr

(M

Pa)

Critical column stress Fcr vc Slenderness ratio according to Load and Resistance Factor Design, for various yield stresses.

AISC assume that column buckling is the governing limit state for column strength.

Column section made from thin (slender) plate elements can fail due to local buckling of the flanges or the webs.

If all the elements of the cross-section have slenderness (b/t) ratio less than lr then the local buckling limit state will not control.

Hence, to prevent local buckling : < l lr Then, use compact or non compact section (no local buckling) The minimum slenderness ratio of compression member :

Local Buckling Limit State

200r

Lk

Width-Thickness Parameters for hot-rolled I and H shapes 

Element < p 

Flange <

Web <

f

f

t

b

2 Fy

E56.0

wt

h

Fy

E49.1

Slenderness Ratio

For other shapes, please see Figure 4.9 W.T Segui 4th edition 

Local STABILITY

The strength corresponding to any buckling mode cannot be developed, however, if the elements of cross section are so thin that local buckling occur. They are :

- flange local buckling (FLB), - web local buckling (WLB).

This buckling strength will depend on the width-thickness ratio of the compression elements of the cross section.

The strength must be reduced if the shape has any slender elemen

..\hasil download purdue univ\local buckling.mpg

When individual column is part of a frame, their ends are connected to other members (beams etc.).

Effective length factor K depend on the restraint offered by the other members connected at the ends.

Effective length factor K depend on the relative rigidity (stiffness) of the members connected at the ends.

Effective length factor for columns in frames : 1. Check whether the column is part of a braced or unbraced frame.

Braced frame : 0.5 < K ≤ 1 Unbraced frame : 1 < K ≤ ∞

2. Determine the relative rigidity factor G for both ends of the column

Effective Length of Column in Frame

b

b

c

c

L

IEL

IE

G

KL for Braced Frame, Unbraced Frame

kL>2L

L<kL<2L

0,7L<kL<L

0,5L<kL<0,7L

P P P P

P P P P(a) Braced Frame, hinged base

(c) Braced Frame, fixed base

(b) Unbraced Frame, hinge based

(d) Unbraced Frame, fixed base

L

L

Alignment Chart to calculate K

A

B

A

B

G

G

A

B

L

(a) Deformasi frame pada kondisi instabilitas(b) Panjang tekuk kolomdipengaruhi kekakuanbatang ynag bertemudititik A dan B

b

b

c

c

L

IEL

IE

G G : the ratio of the summation of the rigidity (EI/L) of all columns coming together at an end to the summation of the rigidity (EI/L) of all beams coming together at the same end.

Boundary Condition

• Pin ended : Σ(IBB/LBB) = 0 GB - ~ This ideal state is never reached, Recomended value : GB= 10

• Perfectly fixed end : Σ(IBB/LBB) = ~ GB 0, This ideal state is never reached, Recommended value : GB= 1

b

b

c

c

L

IEL

IE

G

Alignment Charts for effective column length in a continuous frame

Design Concept : LRFD TCPSBuBG 2002

ncu PP

Where : Nu = ultimate axially loaded fator Nn = compressive nominal strength = Agfcr

fcr = critical buckling stress fc = 0,9

Unbraced frame. W 12 x 79 : Ix = 425 in4

W14x68 Ix = 723 in4

Lx = Ly = 12 ft. Ky = 1.0 Kx depends on boundary conditions,

which involve restraints due to beams and columns connected to the ends of column AB.

Calculate the effective length factor for the W12 x 79 column AB of the frame shown below. Assume that the column is oriented in such a way that major axis bending occurs in the plane of the frame. Assume that the columns are braced at each story level for out-of-plane buckling. The same column section is used for the stories above and below.

10 ft.

10 ft.

12 ft.

15 ft.

20 ft.18 ft.18 ft.

W14 x 68

W14 x 68

W14 x 68

B

A

W12

x 7

9

W12

x 7

9

W12

x 7

9

10 ft.

10 ft.

12 ft.

15 ft.

20 ft.18 ft.18 ft.

W14 x 68

W14 x 68

W14 x 68

B

A

W12

x 7

9

W12

x 7

9

W12

x 7

9

021.1360.6

493.6

1220

723

1218

7231212

425

1210

425

L

IL

I

G

b

b

c

c

A

835.0360.6

3125.5

1220

723

1218

7231215

425

1212

425

L

IL

I

G

b

b

c

c

B

from Alignment Chart Kx=1.3

KyLy = 1.0 x 12

= 12 ft. Kx Lx = 1.3 x 12

= 15.6 ft.

5.6 Tekuk lentur-torsi

Unsur tekan yang terdiri dari siku ganda atau profil berbentuk T,dengan elemen-elemen penampang mempunyai rasio lebar-tebal,λr lebih kecil daripada yang ditentukan dalam Tabel 7.5-1, harus memenuhi

nltnu NN

dimana:

2

411

2

85,0

crzcry

crzcrycrzcryclt

cltgnlt

n

ff

Hff

H

fff

fAN

E

f

r

L

ff

r

yxH

yxA

IIr

rA

GJf

y

y

kyc

ycry

yx

crz

20

20

20

20

20

20

20

1

Keterangan:ř0 = jari-jari girasi polar terhadap pusat geserx0,y0 = koordinat pusat geser terhadap titik berat, x0=0 untuk profil siku ganda dan profil TUntuk tekuk lentur terhadap sumbu lemah y-yLky = panjang tekuk dalam arah sumbu lemah y-y

5.7 Komponen struktur tersusun dihubungkan dengan pelat melintang

(1) Komponen tersusun yang disatukan pada seluruh panjang,boleh dihitung sebagai batang tunggal

(2) Dihubungkan dengan pelat melintang pada tempat-tempat tertentu: Kekuatan: Terhadap sumbu bahan Terhadap sunmbu bebas bahan(3) Kelangsingan:

(a) tegak lurus sumbu x-x:

(b) tegak lurus sumbu bebas bahan y-y(kelangsiangan ideal):

22

2 lyiy

x

kxx

m

r

L

dimana:

minr

L

r

L

ll

y

kyy

Keterangan:m = konstanta seperti pada gambarLky = panjang tekuk komponen struktur tersusun pada tegak lurussumbu y-yry = jari-jari girasi terhadap sumbu y-yLl = spasi antara pelat kopelrmin = jari-jari girasi elelemn komponen struktur terhadap sumbuyang memberikan nilai terkecil

Persyaratan dipakainya rumus kelangsingan ideal:

(a) Pelat-pelat kopel membagi komponen struktur tersusun menjadi sama panjang

(b) Pembagian minimum 3(c) Hubungan pelat kopel dengan elemen kaku(d) Pelat kopel cukup kaku:

l

lp

L

I

a

I10

Dimana: Ip = momen inersia pelat kopel, untuk tebal t dan tinggi h, maka Ip = 2x1/12 t h3, mm4. Il = momen inersia elemen komponen terhadap sumbu l-l, mm4. a = jarak dua pusat titik berat komponen,mm.

Potongan struktur tersusun, dihubungkan dengan pelat melintang:

Gambar 5.13 Komponen struktur tersusun, nilai m

a a a a

a a a

x xx x x x x x

x x x x

y

y

y y y

y

l

l

l l l

l lm=2 m=2 m=2 m=2

m=3 m=4

l

L1/2

L1/2

DL1/2

D/2

D/2

T

T

Ll

a

h

a

a

x x

y l

m=2

a

1 1

Potongan 1-1

Gambar 5.14 Pelat kopel

(4) Koefisien Tekuk dan kuat nominal tekan:

Koefisien tekuk ωx dan ωy ditentukan oleh harga-harga λx dan λiy, dan kuat tekan nominal diambil sebagai nilai terkecil daridiantara:

iy

ygn

x

ygn

fAN

fAN

dan

(5) Selanjutnya perencanaan komponen struktur dihitung dari:

ncu NN

(6) Persyaratan kestabilan elemen-elemen penampang:

50

2,1

2,1

l

liy

lx

dan

(7) Pelat-pelat kopel direncanakan terhadap gaya lintang yang bekerja pada seluruh panjang, sebesar:

uu ND 02,0

Dengan Nu = kuat tekan perlu komponen struktur tersusun akibatbeban-beban terfaktor.Gaya yang bekerja pada kopel ditunjukkan pada Gambar 5.x.(b).

Gaya akibat gaya lintang Du:

T

T

D/2 D/2

D/2 D/2

T= D.L1/a

L1

L1

L1

D

D

D

Nu

Nu

a

h

Gambar 5.15 Gaya pada pelat kopel

Gaya pada pelat kopel

a

x x

y l

m=2

L1/2

L1/2

DL1/2

D/2D/2

T

T

Gambar 5.16Gaya pelat kopel

5.8 Komponen struktur tersusun dengan jarak antarasama dengan tebal pelat kopel

O

O

O

O

X

X

Y

Y

X

X

Y

Y

X X

Y

Y

X X

Y

Y

l

l

l

l

l

l

l

l

(a) (b) (c) (d)

Gambar 5.17

(1) Baja siku dobel seperti Gambar 5.17 (a) dan (b), hanya perlu dihitung terhadap tekuk pada arah sumbu bahan x-x;

(2) Untuk baja siku tidak sama kaki pada Gambar 5.17 (b), persamaan pendekatan rx = 0,87 r0;

(3) Gambar 5.17 (c) dan (d), perlu dihitung terhadap tekuk pada arah sumbu bahan dan arah sumbu bebas bahan;

(4) Untuk Gambar 5.17 (c) dan (d), λ iy dapat diambil λy;

4.5.5 4.5.8 4.5.10

pr