Structural Design 266 Steel Design
Topic 2Compression Members (1)
Nominal Section Capacity
Kerri Bland
Compression Members (1) 2
Structural Design 266 (Steel)
References:References:
•• OneSteelOneSteel, Hot Rolled and Structural Steel Products, Third Edition, Hot Rolled and Structural Steel Products, Third Edition
•• Standards Australia, AS4100Standards Australia, AS4100--1998 : Steel Structures1998 : Steel Structures
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Compression Members (1) 3
Structural Design 266 (Steel)
Steel Structures Code (AS 4100):Philosophy
Calculate Limit State Actions
Anticipate all failure mechanisms
Calculate failure load of each possible failure mechanism
Design capacity = φ * lowest failure load
Design capacity ≥ ultimate actions
Compression Members (1) 4
Structural Design 266 (Steel)
Steel Structures Code (AS 4100):Philosophy
Failure LoadA function of yield stress or ultimate tensile stressYield and ultimate tensile stress for different types of sections and steels given in <Table 2.1> of AS4100Yield and ultimate tensile stress for standard sections are given in Onesteel Hot Rolled and Structural Steel Products Manual
Note that yield and ultimate stresses for different plate thicknesses comply with AS4100
Capacity Factor φ Different capacity factors for different types of actions, members and components given in <Table 3.4> of AS4100
Compression Members (1) 5
Structural Design 266 (Steel)
AS4100-1998: Steel Structures Code
Design of Compression membersSection 6: Members subject to Axial
Compression
Compression Members (1) 6
Structural Design 266 (Steel)
Design of: Compression MembersServiceability limit state
Actions of this magnitude can occur more than once during the life of the structureSteel remains elastic Elastic deformation
axial loads: (shortening)
Appropriate deflection limits limited by structural formAEPL
=δ
Compression Members (1) 7
Structural Design 266 (Steel)
Design of: Compression MembersStrength (ultimate) limit state
Expect structure to just support this loadThree possible failure mechanisms
Squashing Yielding of steel in the memberPlastic deformationAll steel is at yield stress (fy)
CrunchingLocal plate bucklingSquat sections (large plane area compared to length) but with thin plates that buckle
BucklingGlobal member bucklingSlender member – whole member buckles
Compression Members (1) 8
Structural Design 266 (Steel)
Design of: Compression MembersStrength (ultimate) limit state
Three Failure Mechanisms (compression):SquashingCrunching Buckling
Compression failure due to squashing or tension failure due to gross yielding of section (depends on fy)
Compression failure due to buckling of plate elements (depends on fy and section geometry) or member buckling (depends on member slenderness)
Tension failure due to fracture through holes (depends on fu)
Compression Members (1) 9
Structural Design 266 (Steel)
Design of: Compression MembersSquash load
Can only be achieved if plates are thick enough so that they don’t buckle
ynfA=
An=?
Compression Members (1) 10
Structural Design 266 (Steel)
Design of: Compression MembersAn defined in <6.2.1>
An = net area= gross area – cross section of unfilled holes(holes with bolts are considered to be filled holes)
Where unfilled holes reduce the section by less than 100{1-[fy/(0.85fu)]}% (ie: reduced by less than ≈ 20-30%)use An=Ag (most of the time)
Compression Members (1) 11
Structural Design 266 (Steel)
Design of: Compression MembersSquash load
Can only be achieved if plates are thick enough so that they don’t buckle
Nominal Section Capacity (Ns)Crunching load – limited by local buckling of plate elementskf : form factor; based on effective areakf takes plate element slenderness into account
If plates are thick (ie: won’t buckle before full yielding occurs) then kf =1If plates are just too thin to allow full yielding to occur then kf is just less than 1If plates are very thin then kf is much less than 1
kf : <6.2.2>
ynfA=
ynf fAk=<6.2.1>
Compression Members (1) 12
Structural Design 266 (Steel)
Calculating kf : Consider each individual plate in the section separately
Plates with both ends stiffened (ie: web) buckle less readily than those with only one end stiffened (ie: flange outstand)
Buckling is easier for plates with high b/t ratio
Thus slenderness for any given plate
Design of: Compression Members
Plate with stiffened edges
Plate with only one edge stiffened
b
ttb
e ∝λ
Compression Members (1) 13
Structural Design 266 (Steel)
Calculating kf : For each plate:
Calculate plate element slenderness :for each plate element<6.2.3> (for a flat plate element)
Calculate effective width for each plate element:
<6.2.4>
where λey = yield slenderness limit obtained from <T6.2.4>
Yield slenderness limit is the slenderness at which full yielding of the plate element without buckling just occurs. If the slenderness is any higher (ie: more slender) the plate will buckle before yielding occurs.
Very important to check
Design of: Compression Members
⎟⎟⎠
⎞⎜⎜⎝
⎛=
250f
tb y
eλ
bbbe
eye ≤⎟⎟
⎠
⎞⎜⎜⎝
⎛=
λλ
b2
t2
t1b1
Compression Members (1) 14
Structural Design 266 (Steel)
Design of: Compression Members
Note how flat plates with both longitudinal edges supported have a higher yield slenderness limit,ie: they can reach a higher stress (higher load for the same area) before buckling will occur as they have more support, thus less tendency to buckle
Table taken from AS4100 1998Refer to slide 2 for copyright warning
Compression Members (1) 15
Structural Design 266 (Steel)
Calculating kf :
<6.2.2>
Where:Ag = gross area of the sectionAe = effective area
Design of: Compression Members
g
ef A
Ak =
be.flange outstand
tflange
tweb
be.web ( )( ) webweb.eflangewebdtanflangeouts.e
e
t.bt.tb22tb
++=
= ∑
Compression Members (1) 16
Structural Design 266 (Steel)
Design of: Compression MembersNominal Section Capacity: Ns = kf An fy
⇒ Ns is determined by Ae.fy
To rephrase: if a section has slender plate elements, such that they buckle before they yield, in order to calculate the section capacity (the amount of compression force it can take before failure) it would be logical to use the full cross sectional area and multiply it by it’s “plate buckling stress” (a bit tricky to find). However, in order to simplify calculations, the code uses the yield stress (easy to find) and a reduced area (effective area), such that: section capacity = Ag.plate buckling stress = Ae.yield stress = kf Anfy (as before) (Note: plate buckling stress = kf.yield stress)
The relationship between the buckling stress and the yield stress is a function of the slenderness of the plates (cross sectional geometry), and that is used in the determination of the effective plate widths and thus kf
enfgng
ef AA.k then ,AA usually and
AAk As ===
Compression Members (1) 17
Structural Design 266 (Steel)
Design of: Compression MembersExample
Determine the nominal section capacity of a 610 UB 101section.
1. Determine form factor kf:
Table and image taken from Onesteel Hot Rolled and Structural Steel Products ManualRefer to slide 2 for copyright warning.
Compression Members (1) 18
Structural Design 266 (Steel)
Design of: Compression MembersExample
Determine the nominal section capacity of a 610 UB 101 section.1. Determine form factor kf → Find effective width of plate elements
WEB
61250320
6.10572
250f
tb y
e
=
⎟⎠⎞
⎜⎝⎛=
⎟⎠
⎞⎜⎝
⎛=λ =14.8
=572
=228
=602=10.6
45ey =λ
Web slenderness:
Web yield slenderness limit:
mm4226145572
bbbe
eye
=⎟⎠⎞
⎜⎝⎛=
≤⎟⎟⎠
⎞⎜⎜⎝
⎛=
λλ
Effective web width:
(<b)
Table 16 and image taken from Onesteel Hot Rolled and Structural Steel Products ManualTable 6.2.4 from AS4100-1998Refer to slide 2 for copyright warning.
Compression Members (1) 19
Structural Design 266 (Steel)
Design of: Compression MembersExample
Determine the nominal section capacity of a 610 UB 101 section.1. Determine form factor kf → Find effective width of plate elements
FLANGE OUTSTAND
8250300
8.147.108
250f
tb y
e
=⎟⎠⎞
⎜⎝⎛=
⎟⎟⎠
⎞⎜⎜⎝
⎛=λ
=14.8
=572
=228
=602=10.6
16ey =λ
Flange outstand slenderness:
Flange outstand yield slenderness limit:
7.108bb
bmm4.2168
167.108
bbb
e
e
eye
==∴
>=⎟⎠⎞
⎜⎝⎛=
≤⎟⎟⎠
⎞⎜⎜⎝
⎛=
λλEffective flange
outstand width:
Tabl
e 16
and
imag
e ta
ken
from
One
stee
lHot
Rol
led
and
Stru
ctur
al S
teel
Pro
duct
s M
anua
lTa
ble
6.2.
4 fro
m A
S41
00-1
998
Ref
er to
slid
e 2
for c
opyr
ight
war
ning
.
=108.7
Slenderness of element less than yield slenderness limit, therefore flange won’t buckle before yielding, so effective width equals actual width (don’t need to reduce area to take buckling into account)
Compression Members (1) 20
Structural Design 266 (Steel)
Design of: Compression MembersExample
Determine the nominal section capacity of a 610 UB 101 section.1. Determine form factor kf
Find effective width of plate elementsWeb effective width = 422 mmFlange effective width = 228 mm
Find effective area of section
Find form factor kf
( )2
e
mm112206.10*4228.14*228*2A
=
+=
Imag
e an
d Ta
ble
15 ta
ken
from
One
stee
lHot
Rol
led
and
Stru
ctur
al S
teel
Pro
duct
s M
anua
lR
efer
to s
lide
2 fo
r cop
yrig
ht w
arni
ng.
=14.8
=572
=228
=602=10.6
=108.7
863.0k1300011220
AAk
f
g
ef
=
==
Compression Members (1) 21
Structural Design 266 (Steel)
Design of: Compression MembersExample
Determine the nominal section capacity of a 610 UB 101 section.1. Determine form factor kf
863.0kf =
Note:As kf is based upon section geometry and geometry for standard sections is known, kf has been calculated for all standard sections and is given in the manuals…Therefore, it is not necessary to calculate kf for standard sections. This example was a demonstration only.kf in manual (0.888) is not exactly the same as what was calculated (0.863). Manual used a different method (more accurate but more comlex) to calculate kf. The kf calculated here is fairly close and slightly conservative –so is acceptable for use. However, would not normally calculate kf for standard sections.
Tabl
e 16
take
n fro
m
One
stee
lHot
Rol
led
and
Stru
ctur
al S
teel
Pro
duct
s M
anua
lR
efer
to s
lide
2 fo
r co
pyrig
ht w
arni
ng.
Compression Members (1) 22
Structural Design 266 (Steel)
Design of: Compression MembersExample
Determine the nominal section capacity of a 610 UB 101 section.1. Determine form factor kf
2. Calculate Nominal Section Capacity
863.0kf =
kN 3366300*13000*863.0
fAkN ynfs
==
=
Tables 15 & 16 taken from Onesteel Hot Rolled and Structural Steel Products ManualRefer to slide 2 for copyright warning.
Compression Members (1) 23
Structural Design 266 (Steel)
Design of: Compression MembersStrength (ultimate) limit state
Squashing = AnfyYielding of steel in the memberPlastic deformationAll steel is at yield stress (fy)
Crunching = kf AnfyLocal plate bucklingSquat sections (large plane area compared to length) but with thin plates that buckle
Ns (nominal section capacity) = kf Anfy <6.2.1>
Encompasses squashing and crunching failure mechanisms.Section will fail by crunching unless the local plates are stocky enough to not buckle, in which case kf = 1 and Ns = squash load