Download - Compression Member 2013
Structural Steel DesignCompression members
Team TeachingStructural Design
Civil Engineering Department2014
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
Fy 22
rL
EF cr
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 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
kcr
Af
P
Ef
rL
LEIP
2
2
2
1
Column Slenderness
Euler equation
Where Lk = Effective length = K LK = buckling coefficient fy = yield stress of material
• The AISC specifications for column design are based on several years of research.
c Pn = 0.9 Ag Fcr
Column Design Strength cPn
5.1EF
rKL y
c ycr FF c
2
658.0
5.1EF
rKL y
c
y
c
FFcr 2877.0
Elastic buckling
Inelastic buckling
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
yFE71.4
rKL
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
yFE71.4
rKL
1.5Ef
rL yk
c 1
ycr FF c2
658.0
y
c
FFcr 2877.0
Inelastic buckling elastic
buckling
Design Strength Chart
c Pn = 0.9 Ag Fcr
Column Design Strength AISC 2005cPn
Elastic buckling
Inelastic buckling
yFE
rKL 71.4
yFF
cr FF e
y
658.0
yFE
rKL 71.4
ecr FF 877.0
ye FF 44.0
ye FF 44.0 or
or
2
2
rKL
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
yFE71.4
rKL
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
yFE71.4
rKL
Max KL/r = 200
Inelastic buckling elastic
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)
XY
Z
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
• 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.
REFERENSI
• William T Segui, “ Steel Design”• SNI Baja • Hands Out and Video from Purdue University