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
Ductility: Quantitative Descriptions
M
θ
Mp
θyield θmax
Ductility Factor: µ =θmax
θyield
Ductility: Quantitative Descriptions
M
θ
Mp
θyield θmax
θp
Plastic Rotation Angle: θp = θmax - θyield
Ductility: Quantitative Descriptions
M
θ
Mp
θyield θmax
θp
Rotation Capacity: R =θp
θyield= µ - 1
Ductility: Quantitative Descriptions
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
Ductility: Difficulties with Quantitative Descriptions
M
θ
Many definitions of ductility
Many definitions of θyield and θmax
Ductility: Difficulties with Quantitative Descriptions
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: Difficulties with Quantitative Descriptions
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 ?
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
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 ?
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
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 ?
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 ?
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
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
double angle tension member
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
double angle tension member
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
double angle tension member
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
double angle tension member
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
double angle tension member
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.
double angle tension member
PP
Pyield ≤ Pbolt-fracture
Pyield = 3.87 in2 x 54 ksi = 209k
Pbolt-fracture = 158k
Pyield ≤ Pbolt-fracture
Bolt fracture will occur before yield of angles non-ductile behavior
double angle tension member
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
double angle tension member
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
Achieving Ductile Response....
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
Achieving Ductile Response....
Connection response is generally non-ductile.....
Connections should be stronger than connected members
Achieving Ductile Response....
Be cautious of high-strength steels
Ref: Salmon and Johnson - Steel Structures: Design and Behavior
General Trends:
As Fy
Elongation (material ductility)
Fy / Fu
Achieving Ductile Response....
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
Achieving Ductile Response....
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
λp λrWidth-Thickness Ratio (b/t)
Mp
Plastic Buckling
Inelastic Buckling
Elastic Buckling
λps
Duct
ility
Slender Element Sections
Mr
Mom
ent C
apac
ity
λp λrWidth-Thickness Ratio (b/t)
Mp
Plastic Buckling
Inelastic Buckling
Elastic Buckling
λps
Duct
ility
Noncompact Sections
Mr
Mom
ent C
apac
ity
λp λrWidth-Thickness Ratio (b/t)
Mp
Plastic Buckling
Inelastic Buckling
Elastic Buckling
λps
Duct
ility
Compact Sections
Mr
Mom
ent C
apac
ity
λp λrWidth-Thickness Ratio (b/t)
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 ⎟⎟⎠
⎞⎜⎜⎝
⎛≤
Achieving Ductile Response....
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 ?