CSD365 Lecture 10 DCT Combined Actions Member Restraints

7/23/2019 CSD365 Lecture 10 DCT Combined Actions Member Restraints

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Lecture 10: Design Capacity Tables

- Combined Actions

- Angles as Beams

Member Restraints

Kerri Bland

Civil and Structural Design 365

Steel Design


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Civil and Structural Design 365 (Steel)

Lecture 10 - DCT Combined Actions, Angles. Member Restraints 2

COMMONWEALTH OF AUSTRALIA

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WARNING

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behalf of Curtin University of Technologypursuant to Part VB

of the Copyright Act 1968 (the Act)

The material in this communication may be subject to copyright

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Do not remove this notice

References:

AISC, Design Capacity Tables for structural steel, Volume 1: Open Sections, Third Edition

Standards Australia, AS4100 Steel Structures

Geoff Boughton, Steel Design to AS4100 without Computers, Curtin University, Third Edition


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Members Carrying Combined Actions

Members subject to a single type of action have beenconsidered:

Tension, compression, bending

Many structural elements carry a combination of axial and

bending effects

M M

Truss with load between chord

points (both bending and

compression in the top chord)

Building frame under gravity actions

and wind actions (bending and

compression in beam and columns)


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Members Carrying Combined Actions Often it is obvious that one effect is the dominant action

eg: a beam with a large bending moment and a little bit of axial load

Design for the dominant action, then check for the effect of the

combination of loads.

Sometimes it is not as obvious that one action is dominant

design for one action

choose a bigger member to try to account for both actions

Check for the effect of the combination of loads

Two effects need to be considered:

1. Interaction between load effects

(second order effects)

2. Addition of stresses


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1. Interaction between load effectsBending and Compression

The induced moment is a second order effect not a direct effect from the

applied actions.

First order structural analysis (eg: moment distribution) does not account for

these 2nd order effects

The induced moment (2nd order effect) must be added to the first order moment

Causes

deflection

Members Carrying Combined Actions

Induced

moment

M=P.

P

Bending moment

in member

Axial load on bent member

induces additional moment


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Members Carrying Combined Actions1. Interaction between load effects:Bending and Compression

Moment AmplificationIn order to account for the interaction between load effects and the consequent

second order effects, Clause 4.4.2of the steel code requires amplification of

the elastic first order analysis moments if bending moments and compression

forces occur together in a member:

Where:

M* = the moment to be used in all subsequent design calculations

M1* = first order analysis moment (before amplification)

m= amplification factor 1.0

(varies depending on whether the member is braced or not: bor s)

is a function of the Euler Buckling load calculated for an effective length

Le L for braced members

Le> L for unbraced members

M* = mM1*


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Moment Amplification

Braced members: Force and reaction co-linear

use m= b

where

Nomb= elastic buckling load =

with Leas found previously for compression members

cmis a moment distribution factor = 0.6 - 0.4m 1.0depends on shape of BMD

refer to and of the code

=

omb

mb

N

*N1

c

2

e

2

L

EI


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Moment Amplification

Unbraced (sway) members: Force and reaction not co-linear

use m= the maximum ofsand b

bas for braced frame (but Le>L)

srequires a buckling analysis of the whole frame

DCTs (section 4) provide a set of flow charts outlining the

procedure for determining band s, and give a couple of

examples showing how to determine the design actions of

specified members in structures.


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Moment Amplification: DCTs (section 4)

Figures from DCT.Refer to slide 2

for copyright warning.


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Moment Amplification: DCTs (section 4)

Determination of is based on relative stiffness ofadjacent members, or of a whole frame.

Cant be determined without knowing all member sizes.

Initial first order analysis

Preliminary member design

Analyse again taking second order effects into

account (ie: by calculating and applying momentamplification factors)

check member sizes for the amplified moments.


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Members Carrying Combined Actions2. Addition of Stresses

Compression/Bending (either axis)

Tension/Bending (either axis)

Biaxial Bending with or without axial loading

Cant have M* = Msand N* = Ns

occuring at the same time.

N*

(compression)

-

axial

bending

total

M*

+

+

+

-


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Members Carrying Combined Actions2. Addit ion of Stresses: Section Capacity

Member can achieve section capacity if there are sufficient lateral braces

preventing buckling from compression and bending, in all directions.

Always need to check section capacity.

Code gives different equations for checking the section capacity based on a

range of different configurations:

Code allows for a higher Mrxif the section is compact

Formula in code also modified depending on form factor (kf) (equal to or less thanone), and whether member axial actions are tensile or compressive (Section 8)

DCT [tables 8.1-] help determine section capacity under combined actions using the

appropriate code equations for standard sections (see marked up copy of Table 8.1-5

following)

When considering section capacity, there is no differentiation between tensile and

compressive axial loads fail either way at the same stress

=

=

s

*

sx

rx

N

N1M

forceaxialbyreducedcapacitymomentsectionM


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Members Carrying Combined Actions2. Addition of Stresses

Compactness

kf

CC

C

C

C

C

C

C

C

C

C

C

N

C

C

N

CN

C

C

N

N

C

C

C

C

C

C

0.9500.926

0.888

0.928

0.902

0.979

0.948

0.922

0.938

0.913

0.996

0.963

0.930

0.991

0.952

0.915

1.001.00

0.949

1.00

1.00

1.00

0.990

1.00

1.00

1.00

1.00

1.00

Ns= kfAnfy(section capacity of

compression

member)

Nt= Agfy

or = 0.85ktAnfy(section capacity of

tension member)

**See note (4)

For a Compact section with k f


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Members Carrying Combined Actions2. Addition of Stresses: Member Capacity

Takes member buckling into consideration.

Member capacity usually limits performance of a structural member under

combined actions (section capacity can limit always check)

Member capacity > section capacity

When considering member capacity, there are different equations for tensionand compression forces.

Compression exacerbates buckling instabilities

Tension tends to reduce buckling instabilities

Need to already have a member size to determine the capacity (too difficult to

use tables to determine member size required based on loads too manyvariables). Usually determine member size based on the most significant

member effect, and check for combined actions.

Need to use code to determine the correct equations to use, then use the DCT

to determine some of the variables in the equations.


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Combined axial compression and bending

Minor axis bending (in plane) (cant buckle)

Major axis bending (in plane) (FLR-cant buckle)

=

cy

*

syiy*

N

N1MMM

Section bending capacity about y-axis

Tables 5.2-Compression capacity based on effective

length of column in the y-axis

Tables 6-_ (B)

=

cx

*

sxix*

N

N1MMM

Section bending capacity about x-axis

Tables 5.2-Compression capacity based on

effective length of column in the x-axis

Tables 6-_ (A)


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Combined axial compression and bending

Major axis bending (out of plane) (not FLR - can buckle) (always check in planealso, even if not FLR)

=cy

*

bxox*N

N1MMM

Member bending capacity about x-axis

based on Leand m

Tables 5.3-

Compression capacity based on

effective length of column in the y-axis

Tables 6-_ (B)


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Combined axial tension and bending

Major axis bending (out of plane) (not FLR - can buckle)

Only use this capacity if it is certain that the tensile force will always be there to

reduce the buckling instability effects.

Very important to do section capacity check (Mrx) as this could very easilygovern in this case.

rx

t

*

bxox* M

N

N1MMM

+=

Member bending capacity about x-axis

based on Leand m

Tables 5.3-

Tensile capacity

Tables 7-

Section moment capacity

reduced by axial force

Tables 8.1-


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Members Carrying Combined ActionsAngles as beams

Angles loaded parallel to legs have biaxial bending (ie: combined actions)

With FLR (eg: cast into slab)

Has moments about each principal axis

Constrained to deflect in one direction only

Lateral restraint induces lateral loads (which is what constrains the deflections)

Need to combine these effects to determine the angle capacity difficult

DCT tabulate strength limit state and serviceability limit state

T8.2-

Gives the factored load on the whole span (W*or Ws) same as for FLR

bending (assuming UDL)

Tables for EA and UA, with the vertical leg up or down


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Members Carrying Combined ActionsAngles as beams

Angles loaded parallel to legs have biaxial bending (ie: combined actions)

Without FLR (eg: as lintels for masonry masonry cant provide lateral restraint)

Has moments about each principal axis

Second order moments due to deflections

Torsion applied as load not aligned with shear centre

Second order effects between all of these actions

Lateral torsional buckling also an issue

DCT tabulate strength limit state and serviceability limit state

T8.3-

Gives the factored load on the whole span (W*or Ws) (assuming UDL)

Tables for EA and UA, with the vertical leg up or down


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Member Restraints

Lateral Restraints

Shorten effective length of member

Increases member efficiency

Possible member size reduction

Improves the performance of:

Axial compression members

Bending members (major axis bending)

Lateral restraints require design


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Member Restraints

Restraint of Axial Compression Members

Restraint provided to any point in the cross section

May be provided by members already performing another function.

Maybe specifically added to provide lateral restraint.

Restraints must be designed to carry:

Any forces applied externally to the restraint

0.025*axial compression in the restrained member (2.5%)

Providing lateral restraint

about the y-axis


about the y-axis


about the y-axis


about the x-axis


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Member Restraints


Example:

A column is subject to an axial compression load of 1000kN.

It has been designed assuming a lateral restraint is able to be provided at

mid-height.

Determine the restraining force that the lateral restraint needs to bedesigned for:

1000 kN

Restraining force that restraints need to be able to

provide = ?25 kN


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Member Restraints

Restraint of Axial Compression MembersExample:

Determine the member sizes required (EA) to provide the design

restraining force in each of the following scenarios (assume the restraintsare pinned at both ends, and that M16 bolts are used for the connections):

A column is subject to an axial compression load of 1000kN.

It has been designed assuming a lateral restraint is able to be provided at mid-height.

1000 kN

6m

4.5m

(a)

1000 kN

6m

4.5m

(b) (c)

1000 kN

6m

4.5m

(d)

1000 kN

6m

4.5m

Refer to Blackboard for capacities of angles with effective lengths longer than those in the

DCTs: Lecture 12 -Minor axis buckling capacity for long anglesin compression

C S 36 (S )


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Member Restraints

Restraint of Axial Compression Members Must be designed to carry:



Can have restraints reducing the effective length of a number of parallel

compression members.

The likelihood of every column needing to resist the ultimate load at the

same time is very remote

So the restraint force required for the group of columns can be reduced to:

(0.025*load) in one member+ (0.0125*load) in up to s ix more members

P P P P P P P P P P

2.5%P

2.5%P

+

1(1.25%P)=

3.75%P

2.5%P

+

2(1.25%P)=

5.0%P

2.5%P

+

3(1.25%P)=

6.25%P

2.5%P

+

4(1.25%P)=

7.5%P

2.5%P

+

5(1.25%P)=

8.75%P

2.5%P

+

6(1.25%P)=

10%P

2.5%P

+

6(1.25%P)=

10%P

2.5%P

+

6(1.25%P)=

10%P

2.5%P

+

6(1.25%P)=

10%P

Ci il d St t l D i 365 (St l)


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Member Restraints


Must be designed to carry:



Where restraints are closer than necessary (more restraints provided

than necessary), the design restraining force for each restraint can be

reduced :

restraint force in restraining member =

Say a column section has been selected such that the column will not buckle

about its strong axis at ultimate strength loads

Lateral restraints will be required at third points along the column to prevent

buckling about its weak axis at ultimate strength loads (ie: require two restraints)

If we happened to have lateral restraints at quarter points (ie: providing three

restraints) along the column (to suit other structural requirements perhaps), it

means each lateral restraint only needs to designed for a load of

0.025*(2/3)*compression load in member

member)restrainedtheinforce(*n

n025.0

actual

min



26/30



Member Restraints

Restraint of Bending Members

Restraint provided to the critical flangeof the member

May be provided by members already performing another function.

Maybe specifically added to provide lateral restraint.

Restraints must be designed to carry:


0.025*maximum axial force in the critical flanges of adjacentsegments of the restrained member (2.5%)



27/30



Member RestraintsRestraint of Bending Members

Determining the axial force in the flange:

Determine the elastic moment of the beam: Mel = fyZx

If M* < Mel, then stress distribution is linear (elastic behaviour)

If M* > Mel, then stress distribution is not linear (plastic behaviour)

Average stress =

Critical flange area

A = tf*bf

Critical flange force = A

y

=I

My

Average stress =

Critical flange area

A = tf*bf

Critical flange force = yA

y



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Member Restraints


Example:

A 7m long simply supported beam is subject to an ultimate UDL of 20 kN/m

(load applied to the top flange).

Select an appropriate UB beam (from DCT) assuming a lateral restraint is to

be provided to the critical flange at mid-span.

Determine the restraining force that the lateral restraint needs to be

designed for:

7m

20 kN/m

Lateral restraint at mid-span

7.2 kN

360UB44.7

Ci il and Str ct ral Design 365 (Steel)


29/30



Member Restraints

Restraint of Bending Members Must be designed to carry:


0.025*maximum axial force in the critical flange (2.5%)

Can have restraints reducing the effective length of a number of parallel bending

members.

The likelihood of every beam needing to resist the ultimate load at the same

time is very remote

So the restraint force required for the group of beams can be reduced to:

(0.025*cf force) in one member+ (0.0125*cf force) in up to six more members

2.5%F

2.5%F

+

1(1.25%F)

=

3.75%F

2.5%F

+

2(1.25%F)

=

5.0%F

2.5%F

+

3(1.25%F)

=

6.25%F

2.5%F

+

4(1.25%F)

=

7.5%F

2.5%F

+

5(1.25%F)

=

8.75%F

2.5%F

+

6(1.25%F)

=

10%F

2.5%F

+

6(1.25%F)

=

10%F

2.5%F

+

6(1.25%F)

=

10%F

2.5%F

+

6(1.25%F)

=

10%F

Where F = Critical Flange Force Plan view of a series

of parallel beams



30/30


Member Restraints


Must be designed to carry:



Where restraints are closer than necessary (more restraints provided

than necessary), the design restraining force for each restraint can be

reduced :

restraint force in restraining member =

We have a beam section that is able to support the ultimate strength loads

with a mid-span lateral restraint (ie: one restraint sufficient). If we happened to have lateral restraints at quarter points (ie: providing three

restraints) along the beam (to suit other structural requirements perhaps), it

means each lateral restraint only needs to designed for a load of

0.025*(1/3)*axial compression force in the critical flange of the member

force)flangecritical(n

n025.0

actual

min

CSD365 Lecture 10 DCT Combined Actions Member Restraints

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