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M
Ductility: Quantitative Descriptions
M p
yield max
Ductility: Quantitative Descriptions
M
M p
yield max
Ductility Factor: µ = max
yield
Ductility: Quantitative Descriptions
M
M p
yield max
p
Plastic Rotation Angle: p = max - yield
Ductility: Quantitative Descriptions
M
M p
yield max
p
Rotation Capacity: R = p
yield
= µ - 1
Ductility: Quantitative Descriptions
M
M p
yield max
Ductility: Ductility: ductility factor µ
plastic rotation angle p
rotation capacity R
etc.
Based on:
yield
max
Ductility: Difficulties with Quantitative Descriptions
M
Consider a more realistic load - deformation response......
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M
What isyield
yield
?
M
What isyield
yield
?
M
What is max
max
?
M ma x M
What is max
max
?
0.8 M max
M
What ismax
max
?
M p
Ductility: Difficulties with Quantitative Descriptions
M
Many definitions of ducti lity
Many definitions of yield and max
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Ductility: Difficulties with Quantitative Descriptions
Ductility under cyclic loading.....
∆
-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)
B e n d i n g M o m e n t ( k i p - i n c h e s )
Ductility: Difficulties with Quantitative Descriptions
Ductility under cyclic loading.....
How should ductili ty be measured ??
What is Ductility ?
Duct i l i ty = inelast ic deformat ion capaci ty
Many ways to quantify ducti lity
When quantifying ducti lity.......
Clearly define measure of ductil ity
Clearly define yield and max
Use consistent definitions when describi ng
ductili ty demand and ductil ity 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 mo dels
Results in more robust structures
Provides warning of failure
Permits structur e 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 mod els
Results in more robust structures
Provides warning of failure
Permits structure to su rvive severe earthquake
loading
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Example: Plate wi th ho le subjected to tens ion
PP
P 1/2" x 6"L1" dia. hole
ε
50 ksi X
Material " A"
ε
50 ksi
Material " B"
6"
Example:
PP50 ksi
max = 2.57 avg
50 ksi = 2.57 xP
2.5 in2
Pmax = 49 k
ε
50 ksi X
Material " A"
max
Example:
PP
ε
50 ksi
Material " B"
50 ksi
Example:
PP
ε
50 ksi
Material " B"
50 ksi
50 ksi =P
2.5 in2
Pmax = 125 k
PP
ε
50 ksi X
Material " A"
ε
50 ksi
Material " B"
Pmax = 49k Pmax = 125k
Example: Flexural Capacity
ε
50 ksi X
Material " A"
ε
50 ksi
Material " B"
MM
4"
12"
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ε
50 ksi X
Material " A"
4"
12"
max = 50 ksi
ksi 50 S
M ma x =
S = 96 in 3
M max = 96 in 3 x 50 ksi = 4800 k-in
4"
12"
max = 50 ksi
ε
50 ksi
Material " B"
4"
12"
50 ksi
ε
50 ksi
Material " B"
50 ksi
ksi 50 Z
M ma x =
Z = 144 in 3
M max = 144 in 3 x 50 ksi = 7200 k-in
ε
50 ksi X
Material " A"
ε
50 ksi
Material " B"
MM
4"
12"
Mmax = 4800 k-in Mmax = 7200 k-in
Example: Beam Capacity
L = 30 ft.
w
M500 k-ft.
Beam "A"
500 k-ft.
M
Beam "B"
Example: Beam Capacity
w
M500 k-ft.
Beam "A"
M
500 k-ft
250 k-ft
8 wL
2
ft k
2
750 8
wL −=
w max = 6.67 k / ft.
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Example: Beam Capacity
w
M
500 k-ft
250 k-ft
500 k-ft
8
wL2
ft k
2
1000 8
wL −=
w max = 8.89 k / ft.
500 k-ft.
M
Beam "B"
L = 30 ft.
w
M500 k-ft.
Beam "A"
500 k-ft.
M
Beam "B"
w max = 6.67 k / ft . w max = 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 mo dels
Results in more robust structures
Provides warning of failure
Permits structur e to survive severe earthquake
loading
Lower Bou nd Theorem of Plast ic Analysis
A li mit load based on an in ternal s tres s or for ce
distribution that satisfies:
1. Equi librium
2. Material Strength Limits for Ductile Response
( ≤F y , M≤ M p , P ≤ P y , 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 duc tile materials and
components:
Designs satisfying equilibr ium and material
strength lim its are safe.
As a desig ner , as long as we sati sfy equili br ium
(i.e. provide a load path), a ductile s tructur e will
redistribut e internal stresses and forces so as to
find the available load path.
Example of lower bound theorem: Beam Capacity
L = 30 ft.
w
M p
= 500 k-ft.
M
Ductile flexural behavior
What is the load capacity for this beam ?? w max = 8.89 k / ft.
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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 M p 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
8
wL2
Moment diagram in equilibrium with
applied load "w"
Possible lower bound solut ions......
L = 30 ft.
w
8
wL 2
M
500 k-ft
ft k
2
500
8
wL −= w = 4.44 k / ft. (≤ 8.89 k / ft. )
L = 30 ft.
w
8
wL2
M
500 k-ft
ft k 2
500 8
wL −= w = 4.44 k / ft. (≤ 8.89 k / ft. )
L = 30 ft.
w
8
wL2 M
500 k-ft
ft k
2
750 8
wL −= w = 6.67 k / ft. (≤ 8.89 k / ft. )
250 k-ft
L = 30 ft.
w
8
wL2
M
500 k-ft
ft k
2
1000 8
wL −= w = 8.89 k / ft. (= true w max )
500 k-ft
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Examples of lower bound theorem
Flexural capacity of steel section:
F y
F y
≤F y
C
T
Equilibrium: C = T
d
M n = C * d = Z F y
Examples of lower bound theorem
Flexural capacity of a composite section:
F y
steel ≤F y
Equilibrium: C = T
M n = C * d
0.85 f c '
C
Td
conc ≤0.85 f c '
Why Ductility ?
Permits redistribution of internal stresses and
forces
Increases strength of members, connections and
structures
Permits design based on simple equilibrium mo dels
Results in more robust structures
Provides warning of failure
Permits structur e 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 mo dels
Results in more robust structures
Provides warning of failure
Permits structur e to survive severe earthquake
loading
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Ground
Acceleration
Building:
Mass = m
Building
Acceleration
F = ma
Earthquake Forces
on Buildings:
Inertia Force Due to
Accelerating Mass
Conventional Building Code Philosophy forConventional Building Code Philosophy for
EarthquakeEarthquake--Resistant DesignResistant Design
Objective: Prevent collapse in the extreme
earthquake 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 Behavior Design for Ductile Behavior
H
HDuctility = Inelastic Deformation
HH
Required Strength
MAX
Helastic
3/4 *Helastic
1/2 *Helastic
1/4 *Helastic
Available Ductility
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How Do We Achieve Ductility inSteel Structures ?
Achieving Ducti le Response....
Ductile Limit States Must Precede Brittle Limit States
ExampleExample
double angle tension member
gusset plate
PP
Du ct il e L im it State: Gro ss -s ec ti on y ie ld in g of t en si on membe r
B ri tt le Li mi t States : Ne t- sec ti on f ra cture o f te ns io n 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 guss et
double angle tension member
PP
Example: Gross-section y ie ld ing of tens ion
member must precede net section
fracture of tension member
Gross-section yield: P yield = Ag F y
Net-section fracture: P fracture = Ae F u
double angle tension member
PP
P yield ≤P fracture
Ag F y ≤Ae F u
u
y
g
e
F
F
A
A≥
The required strength for brittle limit
states is defined by the capacity of the
ductile element
u
y
F
F = yield ratio
Steels with a low yield ratio are
preferable for ductile behavior
double angle tension member
PP
Example: Gross-section yield ing of tension member
must precede bolt shear fracture
Gross-section yield: P yield = Ag F y
Bolt shear fracture: P bolt-fracture = n b n s Ab F v F v=
0.4 F u-bolt -N
0.5 F u-bolt -X
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double angle tension member
PP
P yield ≤P bolt-fracture The 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 in 2 F v = 0.5 x 120 ksi = 60 ksi
P bolt-fracture = 3 x 0.44 in 2 x 60 ksi x 2 = 158k
Ang les: 2L 4 x 4 x 1/4 A36
Ag = 3.87 in 2
P yield = 3.87 in2 x 36 ksi = 139k
double angle tension member
PP
P yield ≤P bolt-fracture
P yield = 139k P bolt-fracture = 158 k 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 A36angle is 54 ksi.
double angle tension member
PP
P yield ≤P bolt-fracture
P yield = 3.87 in2 x 54 ksi = 209k
P bolt-fracture = 158 k
P yield ≤P bolt-fracture
Bolt fracture will occur before yield of angles non-ductile behavior
double angle tension member
PP
P yield ≤P brittle
Stronger is not b etter 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
P yield ≤P brittle
The required strength for brittle limit
states is defined by the expected
capacity of the ductile element (not
minimum specified capacity)
P yield = Ag R y F y R y F y = expected yield stress of angles
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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 yi eld ratio, F y / F u are preferable
for ductil e elements
Achieving Ducti le Response....
Connection respons e is generally non-duc tile.....
Connections should be stronger than connected
members
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Achieving Ductile Response....
Be cautious of high-strength steels
Ref: Salmon and Johnson -Steel
Structu res: Design and Behavior
General Trends:
As F y
Elongation (material
ductility)
F y / F u
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Achieving Ductile Response....
Be cautious of high-strength steels
High strength steels are generally less ductil e
(lower elongations) and generally have a higher
yield ratio.
High strength steels are generally undesirable
for ductile elements
Achieving Ducti le Response....
Use Sections wi th Low Width-Thickness Ratios and
Adeq uate Lateral Br acing
M
M p
Increasing b / t
Effect of Local Buckling on Flexural Strength and Ductility
M
Mr
M o m e n t C a p a c i t y
λp λr Width-Thickness Ratio (b/t)
Mp
Plastic Buckling
Inelastic Buckling
Elastic Buckling
λps
D u c t i l i t y
Mr
M
o m e n t C a p a c i t y
λp λr Width-Thickness Ratio (b/t)
Mp
Plastic Buckling
Inelastic Buckling
Elastic Buckling
λps
D u c t i l i t y
Slender Element Sections
Mr
M
o m e n t C a p a c i t y
λp λr Width-Thickness Ratio (b/t)
Mp
Plastic Buckling
Inelastic Buckling
Elastic Buckling
λps
D u c t i l i t y
Noncompact Sections
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Mr
M o m e n t C a p a c i t y
λp λr Width-Thickness Ratio (b/t)
Mp
Plastic Buckling
Inelastic Buckling
Elastic Buckling
λps
D u c t i l i t y
Compact Sections
Mr
M o m e n t C a p a c i t y
λp λr Width-Thickness Ratio (b/t)
Mp
Plastic Buckling
Inelastic Buckling
Elastic Buckling
λps
D u c t i l i t y
Seismically
Compact 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 0.05
Drift Angle (radian)
B e n d i n g M
o m e n t ( k N - m )
RBS Connection
Mp
Mp
Local buckling of a seismically compact EBF link.....
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-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, (rad)
L i n k
S h e a r F o r c e
( k i p s )
Effect of Local Buckling on Ductility
For ductile flexural response:
Use compact or seismically compact sections
Example: W-Shape
b f
t f
h
t w
y
s
f
f
F
E 38 .0
t 2
b ≤
Beam Flanges
Beam Web
y
s
w F
E 45 .2
t
h ≤
Compact:
y
s
f
f
F
E 30 .0
t 2
b ≤
Seismically Compact:
Compact:
Seismically Compact:
y
s
w F
E 76 .3
t
h ≤
Lateral Torsional Bu cklingLateral Torsional Bu ckling
Lateral torsional buckling
controlled by:
y
b
r
L
Lb = distance between beam lateral braces
r y = weak axis radius of gyration
L b L b
Beam lateral braces
M
M p
Increasing Lb / r y
Effect of Lateral Torsional Buckling on Flexural Strength and DuEffect of Lateral Torsional Buckling on Flexural Strength and Ductility:ctility:
M
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Effect of Lateral Buckling on Ductility
For ductile flexural response:
Use lateral bracing based on plastic design
requirements or seismic design requirements
Plastic Design: y
y 2
1
b r
F
E
M
M 076 .0 12 .0 L
⎟⎠
⎞
⎝
⎥⎦
⎤
⎢⎣
⎡⎟⎠
⎞ ⎝
+
Seismic Design:y
y
b r F
E 086 .0 L
⎟⎠
⎞
⎝
≤
Achieving Ducti le Response....
Recognize that buckling o f a compression member is
non-ductile
Pcr
P
Pcr
δ
δ
P
Experimental Behavior of Brace Under Cyclic Axial LoadingExperimental Behavior of Brace Under Cyclic Axial Loading
δ
P
W6x20 Kl/r = 80
How Do We Achieve Ductile Response in Steel Structures ?
• Ductile limit states must precede brittle limit states
Ductile 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
• Avo id h igh str engt h st eels in du cti le elements
• Use cross-sections with low b/t ratios
• Provide adequate lateral bracing
• Recognize that compression member buckling is no n-ductile
• Provide connections that are stronger than members
How Do We Achieve Ductile Response in Steel Structures ?
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