Design of Seismic Resistant Steel Building Structures-Concentrically_Braced_Frames

1Design of SeismicDesign of Seismic--Resistant SteelResistant Steel

Building StructuresBuilding Structures

Prepared by:Djoni Simanta

Concentrically Braced Frames

Concentrically Braced FramesConcentrically Braced Frames

Description and Types of Concentrically BracedFrames

Basic Behavior of Concentrically Braced Frames SNI Seismic Provisions for Special Concentrically

Braced Frames




Braced Frames

Concentrically Braced Frames (CBFs)Concentrically Braced Frames (CBFs)Beams, columns and braces arranged to form avertical truss. Resist lateral earthquake forces bytruss action.

Develop ductility through inelastic action in braces.- braces yield in tensionbraces buckle in compression

Advantages- high elastic stiffness

Disadvantages- less ductile than other systems (SMFs, EBFs, BRBFs)

- reduced architectural versatility

2Types of CBFsTypes of CBFs

Single Diagonal Inverted V- Bracing V- Bracing

X- Bracing Two Story X- Bracing

4Concentrically Braced FramesConcentrically Braced Frames



Braced Frames

Inelastic Response of CBFs under Earthquake LoadingInelastic Response of CBFs under Earthquake Loading


Tension Brace: Yields(ductile)

Compression Brace: Buckles(nonductile)

Columns and beams: remain essentially elastic


Compression Brace(previously in tension):Buckles(nonductile)

Tension Brace (previously incompression): Yields(ductile)

Columns and beams: remain essentially elastic

6Developing Ductile Behavior in CBFsDeveloping Ductile Behavior in CBFs

Design brace connections for maximumforces and deformations imposed by braceduring cyclic yielding/buckling

General ApproachGeneral Approach

Developing Ductile Behavior in CBFsDeveloping Ductile Behavior in CBFs

Design beams and columns (and columnsplices and column bases) for maximumforces imposed by braces

General ApproachGeneral ApproachDeveloping Ductile Behavior in CBFsDeveloping Ductile Behavior in CBFs

General ApproachGeneral Approach

Design braces based oncode specifiedearthquake forces.

Design all other frameelements for maximumforces that can bedeveloped by braces.

7Maximum Forces Developed by BracesMaximum Forces Developed by Braces

Braces in TensionBraces in Tension -- Axial Force:Axial Force:

P

P

Pmax = PyFor design:

Take Pmax = Ry Fy Ag

Maximum Forces Developed by BracesMaximum Forces Developed by BracesBraces in CompressionBraces in Compression -- Axial ForceAxial Force

P

P

Pmax

For design:

Take Pmax = 1.1 Ry Pn( Pn = Ag Fcr )

Take Presidual = 0.3 Pn

Presidual 0.3 Pcr

Maximum Forces Developed by BracesMaximum Forces Developed by Braces

Braces in CompressionBraces in Compression -- Bending Moment:Bending Moment:

PP

MM

For "fixed" end braces: flexural plastic hinges will form atmid-length and at brace ends. Brace will impose bendingmoment on connections and adjoining members.

Plastic Hinges

For design:

Take Mmax = 1.1 Ry Fy Zbrace (for critical buckling direction)

Maximum Forces Developed by BracesMaximum Forces Developed by Braces

Braces in CompressionBraces in Compression -- Bending Moment:Bending Moment:

For "pinned" end braces: flexural plastic hinge will form atmid-length only. Brace will impose no bending moment onconnections and adjoining members.

Must design brace connection to behave like a "pin"

PP

PP

Plastic Hinge

8Maximum Forces in Columns and BeamsMaximum Forces in Columns and Beams

To estimate maximum axial forcesimposed by braces on columns andbeams:

Braces in tension:Braces in tension:Take P = Ry Fy Ag

Braces in compression:Braces in compression:

Take P = 1.1 Ry Pn or P = 0.3 Pnwhichever produces critical design case

ExampleExample

Find maximum axialcompression in column.

Tension Braces:Take P = Ry Fy Ag

Compression Braces:Take P = 0.3 Pn

ExampleExample

Ry Fy Ag

Ry Fy Ag

Ry Fy Ag

0.3 Pn

0.3 Pn

Column Axial Compression =

[ (Ry Fy Ag ) cos + (0.3 Pn) cos ] + Pgravity

(sum brace forces for all levelsabove column)

0.3 Pn

ExampleExample

Find maximum axialtension in column.

Tension Braces:Take P = Ry Fy Ag

Compression Braces:Take P = 0.3 Pn

9ExampleExample

Ry Fy Ag

0.3 Pn

Ry Fy Ag

Ry Fy Ag

0.3 Pn

0.3 Pn

Column Axial Tension =

[ (Ry Fy Ag ) cos + (0.3 Pn) cos ] - Pgravity

(sum brace forces for all levelsabove column)

ExampleExample

Find maximum axialcompression in column.

Tension Brace:Take P = Ry Fy Ag

Compression Brace:Take P = 0.3 Pn

ExampleExample

Ry Fy Ag0.3 Pn Column Axial Compression =

(Ry Fy Ag ) cos + (0.3 Pn) cos + Pgravity

Note

Based on elastic frame analysis:

Column Axial Force = Pgravity

ExampleExample

Find maximum bendingmoment in beam.


Compression Brace:Take P = 0.3 Pn

10

ExampleExample

Ry Fy Ag0.3 Pn

ExampleExample

( Ry Fy Ag - 0.3 Pn ) sin

Compute moment in beamresulting from application ofconcentrated load at midspanof ( Ry Fy Ag + 0.3 Pn ) sin

and add moment due togravity load

Note

Based on elastic frame analysis:

Moment in beam 0

ExampleExample

Find maximum axialtension and compressionthat will be applied togusset plate.


Compression Brace:Take P = 1.1 Ry Pn

ExampleExample

1.1 Ry PnRy Fy Ag

Check gusset buckling,beam web crippling, etc.

Check gusset yield, gussetnet section fracture, gussetblock shear fracture, localbeam web yielding, etc.

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Braced Frames

SNISNI Seismic ProvisionsSeismic Provisions

Section F1 Ordinary Concentrically Braced Frames (OCBF)

Section F2 Special Concentrically Braced Frames (SCBF)

Section F1Ordinary Concentrically Braced Frames

(OCBF)Scope

Basis of Design

Analysis

System Requirements

Members

Connections

SNI Steel Seismic Provisions - OCBFF1.1 Scope

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Section F2Special Concentrically Braced Frames (SCBF)

Scope

Basis of Design

Analysis

System Requirements

Members

Connections

SNI Steel Seismic Provisions - SCBFF2.1 Scope

Special concentrically braced frames (SCBF) areexpected to withstand significant inelastic deformationswhen subjected to the forces resulting from the motionsof the design earthquake.

SNI Steel Seismic Provisions - SCBFF2.5 MembersSlenderness

Bracing members shall have:yF

E4r

KL

Fy = 36 ksi: KL/r 114

Fy = 42 ksi: KL/r 105

Fy = 46 ksi: KL/r 100

Fy = 50 ksi: KL/r 96

13

Slenderness

Bracing members shall have:yF

E4r

KL

Exception:

Braces with: 200r

KLFE4

y

are permitted in frames in which the availablestrength of the columns is at least equal to themaximum load transferred to the columnconsidering Ry times the nominal strengths ofthe brace elements.

ExampleExample

Find required axialcompression strength ofcolumn.

ExampleExample

Ry Fy Ag

Ry Fy Ag

Ry Fy Ag

0.3 Pn

0.3 Pn

Required column axial compressionstrength =

[ (Ry Fy Ag ) cos + (0.3 Pn) cos ]+ [(1.2 + 0.2SDS) D + 0.5L]

0.3 Pn

OR

0 QE+ [(1.2 + 0.2SDS) D + 0.5L]

yFE4

rKL

All bracing members:

Note: 0 = 2 for SCBF and OCBF

ExampleExample

Ry Fy Ag

Ry Fy Ag

Ry Fy Ag

0.3 Pn

0.3 Pn

Required column axial compressionstrength =

[ (Ry Fy Ag ) cos - (0.3 Pn) cos ]+ [(1.2 + 0.2SDS) D + 0.5L]

0.3 Pn

0 QE+ [(1.2 + 0.2SDS) D + 0.5L]

Bracing members with: 200rKL

FE4

y

NOT PERMITTED

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F2.5 MembersBasic Required Strength

Where the effective net area of bracing members isless than the gross area, the required tensilestrength of the brace, based on a limit state offracture of the net section shall be at least Ry Fy Ag ofthe bracing member.

Objective: yield of gross section ofbrace prior to fracture of net section

ExampleExample

double angle bracing membergusset plate

Check double angle bracing member forlimit state of net section fracture

15

Pu= Ry Fy Ag

Required axial tension strengthof brace for limit state offracture of the net section

Pu= Ry Fy Ag

Critical Net Section

Ae = U AnAe < Ag due to:

bolt hole (An < Ag ), and

shear lag (U < 1)

Pu= Ry Fy Ag

Limit state: fracture of net section

Pn = (0.75) Ae (Rt Fu)

Per Section 6.2: use expected tensile strength Rt FU when checking net sectionfracture of bracing member, since Ry Fy of the same member is used tocomputed the required strength

Pu= Ry Fy Ag

Limit state: fracture of net section(0.75) Ae (Rt Fu) Ry Fy Ag

utyy

g

e

FR75.0FR

AA

OR:

16

Pu= Ry Fy Ag

Limit state: fracture of net section

utyy

g

e

FR75.0FR

AA

For A36 Angles:

03.1ksi582.175.0ksi365.1

AA

g

e

For A572 Gr. 50 Angles:

03.1ksi651.175.0ksi501.1

AA

g

e

Need to Reinforce Net Section (Ae need not exceed Ag )

Pu= Ry Fy Ag

Also check block shear rupture of bracing member....

Pn = (0.75) Ubs Ant Rt Fu + lesser of0.6 Anv Rt Fu0.6 Agv Ry Fy

Reinforcing net section of bracing member.... ExampleExample gusset platerectangular HSS bracingmember

Check HSS bracing member for limit stateof net section fracture

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Pu= Ry Fy Ag

Required axial tension strengthof brace for limit state offracture of the net section

Pu= Ry Fy Ag

Critical Net Section

Ae = U AnAe < Ag due to:

slot (An < Ag ), and

shear lag (U < 1)

Pu= Ry Fy Ag

Limit state: fracture of net section(0.75) Ae (Rt Fu) Ry Fy Ag

utyy

g

e

FR75.0FR

AA

OR:For A500 Gr B rectangular HSS:

14.1ksi583.175.0ksi464.1

AA

g

e

Need to Reinforce Net Section (Ae need not exceed Ag )

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Pu= Ry Fy Ag


Pn = (0.75) Ubs Ant Rt Fu + lesser of0.6 Anv Rt Fu0.6 Agv Ry Fy

Ant 0

L t = design wall thickness of HSS

Ant = Agv = 4 L t

For A500 Gr B rectangular HSS: Rt Fu = 1.3 x 58 ksi = 75.4 ksi

Ry Fy = 1.4 x 46 ksi = 64.2 ksi

Pu= Ry Fy Ag


L t = design wall thickness of HSS

Pn = (0.75) ( 4 L t x 0.6 x 64.2 ksi) 1.4 x 46 ksi x Ag

tA557.0

L g

= minimum length of welded overlapneeded based on block shearrupture in HSS bracing member

Reinforcing net section of bracing member.... F2.4.4a Lateral Force Distribution

Along any line of bracing, braces shall be deployedin opposite directions such that, for either directionof force parallel to the bracing, at least 30 percent butnot more than 70% of the total horizontal force alongthat line is resisted by braces in tension..

19

Deploy braces so that about half are in tension (and theother half in compression)

All braces in tension (or compression) NG

OK

F2.4.4a Lateral Force Distribution Width-Thickness Limitations

Columns and braces shall meet requirements ofSection D1.1.b.

i.e. columns and braces must be seismicallycompact : ps

Width-Thickness Limitations

Columns: ps


Braces: form plastic hinge during buckling

P

plastic hinge

With high b/t's - local buckling and possiblyfracture may occur at plastichinge region

20


Bracing Members: ps

For rectangular HSS (A500 Gr B steel):

1.16ksi46

ksi2900064.0FE64.0

tb

y

21

SNI Steel Seismic Provisions - SCBFF2.6.c. Required Strength of Bracing ConnectionsRequired Tensile Strength

The required tensile strength of bracing connections (includingbeam-to-column connections if part of the bracing system) shallbe the lesser of the following:

1. Ry Fy Ag of the bracing member.

2. The maximum load effect, indicated by analysis thatcan be transferred to the brace by the system.

Few practical applications of Item 2.

Note that oQE is NOT an acceptable method to establish"maximum load effect"

Ry Fy Ag

22

Pu = Ry Fy Ag

Pu cos

Pu sin

Consider load paththrough connection region Pu = Ry Fy Ag

Pu cos

Pu sin

Consider load paththrough connectionregion:

Uniform Force Method -Vertical Component of Putransferred to column.

VucVub

Vuc + Vub = Pu sin Vuc is transferred directly to column

Vub is transferred indirectly to columnthrough beam and beam tocolumn connectionVub

Pu = Ry Fy Ag

Pu cos

Pu sin

Huc

Hub

Huc + Hub = Pu cos Hub is transferred directly to beam

Huc is transferred indirectly to beamthrough column and beam tocolumn connection

Huc

Consider load paththrough connectionregion:

Uniform Force Method -Horizontal Component ofPu transferred to beam.

Pu = Ry Fy Ag

Pu cos

Pu sin

Consider load path throughconnection region:

Use caution in use of boltsand welds.

Section 7.2:"Bolts and welds shall not bedesigned to share force in ajoint or the same forcecomponent in a connection."

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Pu = Ry Fy Ag

Pu cos

Pu sin

If designed by uniform forcemethod - this connectionviolates Section 7.2

Bolts and welds must transfer sameforce components.

SNI Steel Seismic Provisions - SCBFF2.6.c. Required Strength of Bracing ConnectionsRequired Flexural Strength

The required flexural strength of bracing connections is1.1 Ry Mp of bracing member.

P

MM

For "fixed" end braces: flexural plastic hinges will form atmid-length and at brace ends. Brace will impose bendingmoment on connections and adjoining members.

Plastic Hinges

Mu = 1.1 Ry Mp = 1.1 Ry Fy Zbrace(for critical buckling direction)

1.1 Ry Mp-brace

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SNI Steel Seismic Provisions - SCBFF2.6.c.Required Strength of Bracing ConnectionsRequired Flexural Strength

The required flexural strength of bracing connections is1.1 Ry Mp of bracing member.

Exception:

Brace connections that can accommodate theinelastic rotations associated with brace post-buckling deformations need not meet thisrequirement.

For "pinned" end braces: flexural plastic hinge will form atmid-length only. Brace will impose no bending moment onconnections and adjoining members.

Must design brace connection to behave like a "pin"

PP

PP

Plastic Hinge

Buckling perpendicularto gusset plate

Line of rotation ("foldline") when the bracebuckles out-of-plane(thin direction of plate)

To accommodate brace end rotation: provide "fold line"

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2t

2tConcrete floor slab

2t

26

Concrete floor slab

Styrofoam

2t

27

> 2t

>2t

> 2t

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SNI Steel Seismic Provisions - SCBFF2.6.c Required Strength of Bracing ConnectionsRequired Compressive Strength

The required compressive strength of bracing connections shallbe at least 1.1 Ry Pn

Pn = Ag Fcr of bracing member(per Chapter E of AISC Main Specification)

1.1 Ry Pn

Check:- buckling of gusset plate

- web crippling for beam and column

SNI Steel Seismic Provisions - SCBF

F2.4a V-Type and Inverted V-Type Bracing

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(1) Design beams for unbalanced load that will occurwhen compression brace buckles and tension braceyields.

Take force in tension brace: Ry Fy AgTake force in compression brace: 0.3 Pn

Assume beam has no vertical supportbetween columns.

Ry Fy Ag0.3 Pn

wgravity = (1.2 + 0.2 SDS) D + 0.5L

ExampleL

Beam-to-column connections:simple framing

wgravity = (1.2 + 0.2 SDS) D + 0.5L

Example

L

( Ry Fy Ag - 0.3 Pn ) sin ( Ry Fy Ag + 0.3 Pn ) cos

Forces acting on beam:



(2) Both flanges of beams must be provided with lateralbraces with a maximum spacing of Lpd

and

Both flanges of the beam must be braced at the pointof intersection of the braces.

Per Main AISC Specification (Appendix 1):

yy2

1pd rF

EMM076.012.0L

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SNI Seismic Provisions - SCBF

F2.4b K-Type Bracing

K-Type Braces are not Permitted for SCBF

Design of Seismic Resistant Steel Building Structures-Concentrically_Braced_Frames

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