Structural Steel DesignCompression members

Team TeachingStructural Design

Civil Engineering Department2015

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:APfs

• 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

LKIEPcr

Pcr = critical bucklingload of the column.

Euler Formula

y

xy

P P x

Py

yM=Py

L

Pcr

(a)

(b) (c)d

P

Buckled def lected shape

Figure 5- 4 Buckl ing of a column

P

kxBkxAyEIPk

kydx

ydyEIP

dxyd

EIM

dxydPyM

cossin

0022

2

Differential equation solution

BC’s: x=0 y=0x=L y=0

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

solution: A=0 trivial solutionsinkL=0 non trivial solution

kL=nл

2

22222

22222

LEInPnL

EIP

nLkEIPk

Fundamental mode n=1,

22

2

2

rL

EFLEIP crcr

Euler Equation

where, r2 = I/Ag

Euler Formula

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 tothe 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 ArLEP 2

2

)/(

Elastic Buckling :

gc

ycr

ykc

kecr

Af

P

Ef

rL

LEIPP

2

2

2

1

Column Slenderness

Euler equation

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

Euler stress in the elastic range :

Critical Stress in the elastic range is reduced to account for the effect of initial crookedness

Elastic Buckling :

For inelastic columns, the tangent modulus equation, is replaced by the exponential equation

Critical Stress in the inelastic range

Inelastic Buckling :

The boundary between inelastic and elastic columns, give the same value of Fcr. This occurs when KLr is pproximately

yFE

rKL 71.4

c Pn = 0.9 Ag Fcr

Column Design Strength cPn (AISC)

Elastic buckling

Inelastic buckling

yFE

rKL 71.4

yFF

cr FF e

y

658.0

yFE

rKL 71.4 ecr FF 877.0

or

or

2

2

rKL

EFe

25.2e

y

FF

25.2e

y

FF

Design Strength Chart

Max KL/r = 200

elastic buckling

Inelastic buckling

0.50 0.70 1.0 1.0 2.0 2.0

0.65 0.80 1.0 2.10 2.01.2

Theorit ical K value

Recommended designvalue when ideal condit ions are approximate

End condit ion code

Buckled shape ofcolumn is shown bydashed l ine

Rotat ion f ixed and t ranslat ion f ixedRotat ion free and t ranslat ion f ixedRotat ion f ixed and t ranslat ion freeRotat ion free and t ranslat ion free

Effect ive length factors for idealized column end condit ions. Courtesy theAmerican Inst itute of Steel Const ruct ion, Inc.

Effective Length for different Restraint Condition

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

Different Effective length due to Lateral Support on Minor Axis

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

• W12 x 50: E = 29000 ksiIx = 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-fixKy = 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

LKIE

Pcr

2

2

1923.5629000

2

2

24039129000

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 = 39.735KyLy/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 56.3077.9629000*

2

2

2

2

68.13336

2900071.471.4 yF

E

yFE

rKL 71.4

yFF

cr FF e

y

658.0

Fcr = 21.99 ksi

cPn = 0.9 (Ag Fcr) = 0.9 (21.8 x 21.99) = 431.4 kips

Design strength of column = 431 kips (inelastic buckling)

yFE

rKL 71.4

A992  Fy = 50ksiFu = 65ksi

Ag = 17 inc2 

68.13336

2900071.471.4 yF

E

yFF

cr FF e

y

658.0

ksi

rKL

EFe 08.9655.54

29000*2

2

2

2

Fcr = 40.21 ksi

cPn = 0.9 (Ag Fcr) = 0.9 (17x40.21) = 615.3 Kips ((inelastic buckling on major axis)

225,12,1

67,06,143,12,125,0

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

5.1EF

rKL y

c

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

Crit

ical

stre

ss F

cr (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 r then the local buckling limit state will not control.

• Hence, to prevent local buckling : r

• Then, use compact or non compact section (no local buckling)

• The minimum slenderness ratio of compression member :

Local Buckling Limit State

200rLk

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

Element < p

Flange <

Web <

f

f

tb2 Fy

E56.0

wth

FyE49.1

Slenderness Ratio

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

edition

Local STABILITY

The strength corresponding to any buckling modecannot be developed, however, if the elements ofcross 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 thecross section.

The strength must be reduced if the shape hasany slender elemen

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

https://www.google.com/search?q=lateral+torsional+buckling+example&sa=X&biw=1280&bih=606&tbm=isch&tbo=u&source=univ&ei=X6ntVLDQI8fhuQTmo4LgAw&ved=0CCUQsAQ#imgdii=_&imgrc=3pmUf0TVJswp_M%253A%3BUe6PAdEpYDJu9M%3Bhttp%253A%252F%252Fi.ytimg.com%252Fvi%252FGUyX9nToAAQ%252Fhqdefault.jpg%3Bhttp%253A%252F%252Fwww.youtube.com%252Fwatch%253Fv%253DGUyX9nToAAQ%3B480%3B360

• 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

LIE

LIE

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 f rame pada kondisi instabil itas(b) Panjang tekuk kolomdipengaruhi kekakuanbatang ynag bertemudit it ik A dan B

b

b

c

c

LIE

LIE

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

LIE

LIE

G

Alignment Charts for effective column length in a continuous frame

Design Concept : LRFD TCPSBuBG 2002

ncu PP

Where :Nu = ultimate axially loaded fatorNn = compressive nominal strength = Agfcrfcr = critical buckling stressc = 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

021.1360.6493.6

1220723

1218723

1212425

1210425

LILI

G

b

b

c

c

A

835.0360.6

3125.5

1220723

1218723

1215425

1212425

LILI

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.

Tekuk lentur-torsi

Unsur tekan yang terdiri dari siku ganda atau profil berbentuk T, denganelemen‐elemen penampang mempunyai rasio lebar‐tebal, λr lebih kecildaripada yang ditentukan dalam Tabel 7.5‐1, harus memenuhi

nltnu NN

dimana:

2

411

2

85,0

crzcry

crzcrycrzcryclt

cltgnlt

n

ffHff

Hff

f

fAN

Ef

rL

ff

ryxH

yxA

IIr

rAGJf

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 T

Untuk tekuk lentur terhadap sumbu lemah y-yLky = panjang tekuk dalam arah sumbulemah y-y

5.7 Komponen struktur tersusun dihubungkan dengan pelatmelintang

(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

rL

dimana:

minrL

rL

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

LI

aI

10

Dimana: Ip = momen inersia pelat kopel, untuk tebal t dantinggi 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 pelatmelintang:

nilai m; komponen struktur tersusun

a a a a

a a a

x x x 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=2a

1 1

Potongan 1-1

Gambar Pelat kopel

(4) Koefisien Tekuk dan kuat nominal tekan:

Koefisien tekuk ωx dan ωy ditentukan oleh harga‐harga λx danλiy, 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

ah

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 antara samadengan 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 perludihitung 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 padaarah sumbu bahan dan arah sumbu bebas bahan;

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

REFERENSI

• William T Segui, “ Steel Design”• SNI Baja • Hands Out and Video from Purdue University