Steel compression lecture

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

COMMONWEALTH OF AUSTRALIA Copyright Regulation 1969WARNING This material has been copied and communicated to you by or on behalf of Curtin University of Technology pursuant to Part VB of the Copyright Act 1968 (the Act) The material in this communication may be subject to copyright under the Act. Any further copying or communication of this material by you may be the subject of copyright protection underthe Act. Do not remove this notice



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



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



AS4100-1998: Steel Structures Code

Design of Compression membersSection 6: Members subject to Axial

Compression



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



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




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)



Design of: Compression MembersSquash load

Can only be achieved if plates are thick enough so that they don’t buckle

ynfA=

An=?



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)



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>



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



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


⎟⎟⎠

⎞⎜⎜⎝

⎛=

250f

tb y

eλ

bbbe

eye ≤⎟⎟

⎠

⎞⎜⎜⎝

⎛=

λλ

b2

t2

t1b1




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



Calculating kf :

<6.2.2>

Where:Ag = gross area of the sectionAe = effective area


g

ef A

Ak =

be.flange outstand

tflange

tweb

be.web ( )( ) webweb.eflangewebdtanflangeouts.e

e

t.bt.tb22tb

++=

= ∑



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 ===



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.




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.




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)




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

=

==





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.





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.




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

Documents

Steel compression lecture