Overview of Presentation Basic Concepts in Ductile ... · Basic Concepts in Ductile Detailing of Steel Structures Michael D. Engelhardt University of Texas at Austin Overview of Presentation

Basic Concepts in Ductile Basic Concepts in Ductile Detailing of Steel StructuresDetailing of Steel Structures

Michael D. EngelhardtMichael D. Engelhardt

University of Texas at AustinUniversity of Texas at Austin

Overview of Presentation

• What is Ductility ?

• Why is Ductility Important ?

• How Do We Achieve Ductility in Steel Structures ?

What is Ductility ?

Ductility: The ability to sustain large inelastic deformations without significant loss in strength.

Ductility = inelastic deformation capacity

- material response

- structural component response (members and connections)

- global frame response

Ductility:

F

∆

F

∆

F

∆

F

∆

Fyield

Ductility

displacementrotation

curvaturestrain

etc.

F

∆

Fyield

Ductility

M

θ

Ductility: Qualitative Description

More Ductile

Less DuctileNo Ductility

Mθ

M

θ

Ductility: Quantitative Descriptions

Mp

θyield θmax


M

θ

Mp

θyield θmax

Ductility Factor: µ =θmax

θyield


M

θ

Mp

θyield θmax

θp

Plastic Rotation Angle: θp = θmax - θyield


M

θ

Mp

θyield θmax

θp

Rotation Capacity: R =θp

θyield= µ - 1


M

θ

Mp

θyield θmax

Ductility:Ductility: ductility factor µplastic rotation angle θp

rotation capacity Retc.

Based on:

θyield

θmax

Ductility: Difficulties with Quantitative Descriptions

M

θ

Consider a more realistic load - deformation response......

M

θ

What is θyield

θyield

?

M

θ

What is θyield

θyield

?

M

θ

What is θmax

θmax

?

MmaxM

θ

What is θmax

θmax

?

0.8 Mmax

M

θ

What is θmax

θmax

?

Mp


M

θ

Many definitions of ductility

Many definitions of θyield and θmax


Ductility under cyclic loading.....

∆θ

-40000

-30000

-20000

-10000

0

10000

20000

30000

40000

-0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08

Rotation Angle (rad)

Ben

ding

Mom

ent (

kip-

inch

es)


Ductility under cyclic loading.....

How should ductility be measured ??

What is Ductility ?

Ductility = inelastic deformation capacity

Many ways to quantify ductility

When quantifying ductility.......

Clearly define measure of ductility

Clearly define θyield and θmax

Use consistent definitions when describing ductility demand and ductility supply

F

∆

F

∆

F

∆

F

∆

Ductility = Yielding

How is ductility developed in steel structures ?

Loss of load carrying capability:

Instability

Fracture

Why is Ductility Important?

Permits redistribution of internal stresses and forces

Increases strength of members, connections and structures

Permits design based on simple equilibrium models

Results in more robust structures

Provides warning of failure

Permits structure to survive severe earthquake loading

Why Ductility ?







Example: Plate with hole subjected to tension

PP

P 1/2" x 6"L1" dia. hole

σ

ε

50 ksi X

Material "A"

σ

ε

50 ksi

Material "B"

6"

Example:

PP 50 ksi

σmax = 2.57 σavg

50 ksi = 2.57 x P2.5 in2

Pmax = 49 k

σ

ε

50 ksi X

Material "A"

σmax

Example:

PP

σ

ε

50 ksi

Material "B"

50 ksi

Example:

PP

σ

ε

50 ksi

Material "B"

50 ksi

50 ksi = P2.5 in2

Pmax = 125 k

PP

σ

ε

50 ksi X

Material "A"

σ

ε

50 ksi

Material "B"

Pmax = 49k Pmax = 125k

Example: Flexural Capacity

σ

ε

50 ksi X

Material "A"

σ

ε

50 ksi

Material "B"

MM

4"

12"

σ

ε

50 ksi X

Material "A"

4"

12"

σmax = 50 ksi

ksi50SM

max ==σ

S = 96 in3

Mmax = 96 in3 x 50 ksi = 4800 k-in

4"

12"

σmax = 50 ksi

σ

ε

50 ksi

Material "B"

4"

12"

50 ksi

σ

ε

50 ksi

Material "B"

50 ksi

ksi50ZM

max ==σ

Z = 144 in3

Mmax = 144 in3 x 50 ksi = 7200 k-in

σ

ε

50 ksi X

Material "A"

σ

ε

50 ksi

Material "B"

MM

4"

12"

Mmax = 4800 k-in Mmax = 7200 k-in

Example: Beam Capacity L = 30 ft.

w

θ

M500 k-ft.

Beam "A"

500 k-ft.

M

Beam "B"

θ

Example: Beam Capacity w

θ

M500 k-ft.

Beam "A"

M

500 k-ft

250 k-ft

8wL2

ftk2

7508

wL −=

wmax = 6.67 k / ft.

Example: Beam Capacity w

M

500 k-ft

250 k-ft

500 k-ft

8wL2

ftk2

10008

wL −=

wmax = 8.89 k / ft.

500 k-ft.

M

Beam "B"

θ

L = 30 ft.

w

θ

M500 k-ft.

Beam "A"

500 k-ft.

M

Beam "B"

θ

wmax = 6.67 k / ft. wmax = 8.89 k / ft.

Why Ductility ?







Lower Bound Theorem of Plastic Analysis

A limit load based on an internal stress or force distribution that satisfies:

1. Equilibrium

2. Material Strength Limits for Ductile Response(σ ≤ Fy , M ≤ Mp, P ≤ Py , etc)

is less than or equal to the true limit load.

Lower bound theorem only applicable for ductile structures

Implications of the lower bound theorem ............Implications of the lower bound theorem ............For a structure made of ductile materials and components:

Designs satisfying equilibrium and material strength limits are safe.

As a designer, as long as we satisfy equilibrium (i.e. provide a load path), a ductile structure will redistribute internal stresses and forces so as to find the available load path.

Example of lower bound theorem: Beam Capacity L = 30 ft.

w

Mp = 500 k-ft.

M

Ductile flexural behavior

θ

What is the load capacity for this beam ?? wmax = 8.89 k / ft.

L = 30 ft.

w

What is the load capacity for this beam ??By lower bound theorem:

Choose any moment diagram in equilibrium with the applied load.

The moment cannot exceed Mp at any point along the beam.

The resulting load capacity "w" will be less than or equal to the true load capacity.

L = 30 ft.

w

8wL2

Moment diagram in equilibrium with applied load "w"

Possible lower bound solutions......

L = 30 ft.

w

8wL2

M

500 k-ft

ftk2

5008

wL −= w = 4.44 k / ft. (≤ 8.89 k / ft. )

L = 30 ft.

w

8wL2

M

500 k-ft

ftk2

5008

wL −= w = 4.44 k / ft. (≤ 8.89 k / ft. )

L = 30 ft.

w

8wL2M

500 k-ft

ftk2

7508

wL −= w = 6.67 k / ft. (≤ 8.89 k / ft. )

250 k-ft

L = 30 ft.

w

8wL2

M

500 k-ft

ftk2

10008

wL −= w = 8.89 k / ft. (= true wmax )

500 k-ft

Examples of lower bound theorem

Flexural capacity of steel section:Fy

Fy

σ ≤ Fy

C

T

Equilibrium: C = T

d

Mn = C * d = Z Fy

Examples of lower bound theoremFlexural capacity of a composite section:

Fy

σsteel ≤ Fy

Equilibrium: C = T

Mn = C * d

0.85 fc'

C

Td

σconc ≤ 0.85 fc'

Why Ductility ?







Why Ductility ?







Ground Acceleration

Building:

Mass = m

Building Acceleration

F = ma

Earthquake Forces on Buildings:

Inertia Force Due to Accelerating Mass

Conventional Building Code Philosophy for Conventional Building Code Philosophy for EarthquakeEarthquake--Resistant DesignResistant Design

Objective: Prevent collapse in the extremeearthquake likely to occur at a building site.

Objectives are not to:

- limit damage- maintain function- provide for easy repair

To Survive Strong Earthquake without Collapse:

Design for Ductile BehaviorDesign for Ductile Behavior

H

HDuctility = Inelastic Deformation

HH

Required Strength

MAX

Helastic

3/4 *Helastic

1/2 *Helastic

1/4 *Helastic

Available Ductility

How Do We Achieve Ductility in Steel Structures ?

Achieving Ductile Response....

Ductile Limit States Must Precede Brittle Limit States

ExampleExample

double angle tension membergusset plate

PP

Ductile Limit State: Gross-section yielding of tension member

Brittle Limit States: Net-section fracture of tension member

Block-shear fracture of tension member

Net-section fracture of gusset plate

Block-shear fracture of gusset plate

Bolt shear fracture

Plate bearing failure in double angles or gusset

double angle tension member

PP

Example: Gross-section yielding of tension member must precede net section fracture of tension member

Gross-section yield: Pyield = Ag Fy

Net-section fracture: Pfracture = Ae Fu


PP

Pyield ≤ Pfracture

Ag Fy ≤ Ae Fu

u

y

g

e

FF

AA

≥

The required strength for brittle limit states is defined by the capacity of the ductile element

u

y

FF

= yield ratio Steels with a low yield ratio are preferable for ductile behavior


PP

Example: Gross-section yielding of tension member must precede bolt shear fracture

Gross-section yield: Pyield = Ag Fy

Bolt shear fracture: Pbolt-fracture = nb ns Ab Fv Fv =0.4 Fu-bolt -N

0.5 Fu-bolt -X


PP

Pyield ≤ Pbolt-fractureThe required strength for brittle limit states is defined by the capacity of the ductile element

The ductile element must be the weakest element in the load path


PP

Example: Bolts: 3 - 3/4" A325-X double shear Ab = 0.44 in2 Fv = 0.5 x 120 ksi = 60 ksiPbolt-fracture = 3 x 0.44 in2 x 60 ksi x 2 = 158k

Angles: 2L 4 x 4 x 1/4 A36

Ag = 3.87 in2

Pyield = 3.87 in2 x 36 ksi = 139k


PP

Pyield ≤ Pbolt-fracture

Pyield = 139k Pbolt-fracture = 158k OK

What if the actual yield stress for the A36 angles is greater than 36 ksi?

Say, for example, the actual yield stress for the A36 angle is 54 ksi.


PP


Pyield = 3.87 in2 x 54 ksi = 209k

Pbolt-fracture = 158k


Bolt fracture will occur before yield of angles non-ductile behavior


PP

Pyield ≤ Pbrittle

Stronger is not better in the ductile element

(Ductile element must be weakest element in the load path)

For ductile response: must consider material overstrength in ductile element


PP

Pyield ≤ Pbrittle

The required strength for brittle limit states is defined by the expected capacity of the ductile element (not minimum specified capacity)

Pyield = Ag RyFy Ry Fy = expected yield stress of angles


Ductile Limit States Must Precede Brittle Limit States

Define the required strength for brittle limit states based on the expected yield capacity for ductile element

The ductile element must be the weakest in the load path

Unanticipated over strength in the ductile element can lead to non-ductile behavior.

Steels with a low value of yield ratio, Fy / Fu are preferable for ductile elements


Connection response is generally non-ductile.....

Connections should be stronger than connected members


Be cautious of high-strength steels

Ref: Salmon and Johnson - Steel Structures: Design and Behavior

General Trends:

As Fy

Elongation (material ductility)

Fy / Fu


Be cautious of high-strength steels

High strength steels are generally less ductile (lower elongations) and generally have a higher yield ratio.

High strength steels are generally undesirable for ductile elements


Use Sections with Low Width-Thickness Ratios and Adequate Lateral Bracing

M

θ

Mp

Increasing b / t

Effect of Local Buckling on Flexural Strength and Ductility

Mθ

Mr

Mom

ent C

apac

ity

λp λrWidth-Thickness Ratio (b/t)

Mp

Plastic Buckling

Inelastic Buckling

Elastic Buckling

λps

Duct

ility

Mr

Mom

ent C

apac

ity


Mp

Plastic Buckling

Inelastic Buckling

Elastic Buckling

λps

Duct

ility

Slender Element Sections

Mr

Mom

ent C

apac

ity


Mp

Plastic Buckling

Inelastic Buckling

Elastic Buckling

λps

Duct

ility

Noncompact Sections

Mr

Mom

ent C

apac

ity


Mp

Plastic Buckling

Inelastic Buckling

Elastic Buckling

λps

Duct

ility

Compact Sections

Mr

Mom

ent C

apac

ity


Mp

Plastic Buckling

Inelastic Buckling

Elastic Buckling

λps

Duct

ility

SeismicallyCompact Sections

Local buckling of noncompact and slender element sections

Local buckling of a seismically compact moment frame beam.....

-5000

-4000

-3000

-2000

-1000

0

1000

2000

3000

4000

5000

-0.05 -0.04 -0.03 -0.02 -0.01 0 0.01 0.02 0.03 0.04 0.05Drift Angle (radian)

Ben

ding

Mom

ent (

kN-m

)

RBS Connection

Mp

Mp

Local buckling of a seismically compact EBF link.....

-200

-150

-100

-50

0

50

100

150

200

-0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08

Link Rotation, γ (rad)

Link

She

ar F

orce

(kip

s)

Effect of Local Buckling on DuctilityFor ductile flexural response:

Use compact or seismically compact sections

Example: W-Shape

bf

tf

h

tw

y

s

f

f

FE38.0

t2b

≤

Beam Flanges

Beam Web

y

s

w FE45.2

th

≤

Compact:

y

s

f

f

FE30.0

t2b

≤Seismically Compact:

Compact:

Seismically Compact:

y

s

w FE76.3

th

≤

Lateral Torsional BucklingLateral Torsional Buckling

Lateral torsional buckling controlled by:

y

b

rL

Lb = distance between beam lateral braces

ry = weak axis radius of gyration

Lb Lb

Beam lateral braces

M

θ

Mp

Increasing Lb / ry

Effect of Lateral Torsional Buckling on Flexural Strength and DuEffect of Lateral Torsional Buckling on Flexural Strength and Ductility:ctility:

Mθ

Effect of Lateral Buckling on Ductility

For ductile flexural response:

Use lateral bracing based on plastic designrequirements or seismic design requirements

Plastic Design: yy2

1b r

FE

MM076.012.0L ⎟

⎟⎠

⎞⎜⎜⎝

⎛

⎥⎥⎦

⎤

⎢⎢⎣

⎡⎟⎟⎠

⎞⎜⎜⎝

⎛+≤

Seismic Design:y

yb r

FE086.0L ⎟⎟⎠

⎞⎜⎜⎝

⎛≤


Recognize that buckling of a compression member is non-ductile

Pcr

P

Pcr

δ

δ

PExperimental Behavior of Brace Under Cyclic Axial LoadingExperimental Behavior of Brace Under Cyclic Axial Loading

δP

W6x20 Kl/r = 80

How Do We Achieve Ductile Response in Steel Structures ?

• Ductile limit states must precede brittle limit statesDuctile elements must be the weakest in the load path

Stronger is not better in ductile elements

Define Required Strength for brittle limit states based on expected yield capacity of ductile element

• Avoid high strength steels in ductile elements

• Use cross-sections with low b/t ratios

• Provide adequate lateral bracing

• Recognize that compression member buckling is non-ductile

• Provide connections that are stronger than members

How Do We Achieve Ductile Response in Steel Structures ?

Overview of Presentation Basic Concepts in Ductile ... · Basic Concepts in Ductile Detailing of Steel Structures Michael D. Engelhardt University of Texas at Austin Overview of Presentation

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