Basic Concepts in Ductile Detailing of Steel Structures

Overview of Presentation Basic Concepts in Ductile Detailing of Steel Structures

• What is Ductility ? • Why is Ductility Important ? • How Do We Achieve Ductility in Steel Structures ?

Michael D. Engelhardt University of Texas at Austin

F What is Ductility ? ∆ F F

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

Ductility Ductility = inelastic deformation capacity F

Fyield Ductility: - material response - structural component response (members and connections) - global frame response ∆

Ductility: Qualitative Description Ductility F θ

Fyield M

M More Ductile ∆ No Ductility Less Ductile displacement rotation curvature strain etc. θ Ductility: Quantitative Descriptions Ductility: Quantitative Descriptions

M M

Mp Mp

θ θ θ θ θ θ yield max yield max θ max Ductility Factor: µ = θ yield

Ductility: Quantitative Descriptions Ductility: Quantitative Descriptions

M θ M θ p p

Mp Mp

θ θ θ θ θ θ yield max yield max

θ p Plastic Rotation Angle: θ = θ - θ Rotation Capacity: = µ -1 p max yield R = θ yield

Ductility: Quantitative Descriptions Ductility: Difficulties with Quantitative Descriptions

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

Mp M

θ θ θ yield max Ductility: ductility factor µ Based on: θ θ θ plastic rotation angle p yield θ rotation capacity R max etc. M M

θ θ θ θ yield yield

θ θ What is yield ? What is yield ?

M M Mmax 0.8 Mmax

θ θ θ θ max max

θ θ What is max ? What is max ?

Ductility: Difficulties with Quantitative Descriptions M M

θ θ θ max

Many definitions of ductility

θ θ θ Many definitions of yield and max What is max ? Ductility: Difficulties with Quantitative Descriptions Ductility: Difficulties with Quantitative Descriptions

Ductility under cyclic loading..... Ductility under cyclic loading.....

θ ∆

40000

30000

20000

10000

-10000

-20000

-30000

Bending Moment (kip-inches) Bending Moment How should ductility be measured ?? -40000 -0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08 Rotation Angle (rad)

How is ductility developed in steel structures ? What is Ductility ? F ∆ F F Ductility = inelastic deformation capacity ∆

Many ways to quantify ductility ∆

When quantifying ductility...... Ductility = Yielding F Loss of load carrying Clearly define measure of ductility capability: θ θ Clearly define yield and max Instability Fracture Use consistent definitions when describing ductility demand and ductility supply ∆

Why is Ductility Important? Why Ductility ?

Permits redistribution of internal stresses and Permits redistribution of internal stresses and forces forces Increases strength of members, connections and Increases strength of members, connections and structures structures Permits design based on simple equilibrium models Permits design based on simple equilibrium models Results in more robust structures Results in more robust structures Provides warning of failure Provides warning of failure Permits structure to survive severe earthquake Permits structure to survive severe earthquake loading loading Example: Plate with hole subjected to tension Example:

σ 50 maxksi P 6" P P P

PL 1/2" x 6" 1" dia. hole σ σ σ max = 2.57 avg σ σ 50 ksi X 50 ksi = 2.57 x P 50 ksi X 50 ksi 2 Material "A" 2.5 in Material "A" Material "B" k ε Pmax = 49 ε ε

Example: Example:

50 ksi 50 ksi P P P P

σ σ P 50 ksi 50 ksi 50 ksi = 2.5 in2 Material "B" Material "B" P = 125 k ε ε max

Example: Flexural Capacity

4" P P

M M 12"

σ σ 50 ksi X 50 ksi

Material "A" Material "B" σ σ ε ε 50 ksi X 50 ksi

k k Material "A" Material "B" Pmax = 49 Pmax = 125 ε ε 4" 4"

12" 12"

σ σ max = 50 ksi max = 50 ksi M σ = = 50 ksi max S σ σ S = 96 in3 50 ksi 50 ksi X 3 k-in Mmax = 96 in x 50 ksi = 4800 Material "B" Material "A" ε ε

4" 4" 50 ksi

M M 12" 12"

σ σ 50 ksi 50 ksi X 50 ksi M σ = = 50 ksi Material "A" Material "B" max Z σ ε ε Z = 144 in3 50 ksi

3 k-in k-in k-in Material "B" Mmax = 144 in x 50 ksi = 7200 Mmax = 4800 Mmax = 7200

Example: Beam Capacity Example: Beam Capacity w L = 30 ft.

250 k-ft

M wL2 8

500 k-ft

M M M 2 wL − 500 k-ft. 500 k-ft. 500 k-ft. = 750 k ft 8 Beam "A" Beam "B" Beam "A" wmax = 6.67 k / ft. θ θ θ Example: Beam Capacity L = 30 ft. w w

500 k-ft 250 k-ft wL2 M 8 M M 500 k-ft. 500 k-ft.

500 k-ft Beam "A" Beam "B"

M 2 wL − 500 k-ft. = 1000 k ft θ θ 8 Beam "B" wmax = 6.67 k / ft. wmax = 8.89 k / ft. wmax = 8.89 k / ft. θ

Why Ductility ? Lower Bound Theorem of Plastic Analysis

Permits redistribution of internal stresses and A limit load based on an internal stress or force forces distribution that satisfies: Increases strength of members, connections and structures 1. Equilibrium Permits design based on simple equilibrium models 2. Material Strength Limits for Ductile Response σ≤ ≤ ≤ ( Fy , M Mp, P Py , etc) Results in more robust structures is less than or equal to the true limit load. Provides warning of failure Permits structure to survive severe earthquake loading Lower bound theorem only applicable for ductile structures

Example of lower bound theorem: Beam Capacity Implications of the lower bound theorem ...... L = 30 ft. For a structure made of ductile materials and components: w

Designs satisfying equilibrium and material strength limits are safe. M k-ft. Mp =500 As a designer, as long as we satisfy equilibrium (i.e. provide a load path), a ductile structure will Ductile flexural behavior redistribute internal stresses and forces so as to θ find the available load path.

What is the load capacity for this beam ?? wmax = 8.89 k / ft. L = 30 ft. L = 30 ft.

w w

What is the load capacity for this beam ?? By lower bound theorem: wL2 8 Choose any moment diagram in equilibrium with the applied load. Moment diagram in equilibrium with The moment cannot exceed Mp at any point along the beam. applied load "w" The resulting load capacity "w" will be less than Possible lower bound solutions...... or equal to the true load capacity.

L = 30 ft. L = 30 ft.

w w

500 k-ft

wL2 8 M M

wL2 8 k-ft 2 500 wL − = 500 k ft w = 4.44 k / ft. (≤ 8.89 k / ft. ) 8 2 wL − = 500 k ft w = 4.44 k / ft. (≤ 8.89 k / ft. ) 8

L = 30 ft. L = 30 ft.

w w

250 k-ft 500 k-ft wL2 2 M wL M 8 8 500 k-ft 500 k-ft

2 2 wL − wL − = 750 k ft w = 6.67 k / ft. (≤ 8.89 k / ft. ) = 1000 k ft w = 8.89 k / ft. (= true w ) 8 8 max Examples of lower bound theorem Examples of lower bound theorem Flexural capacity of a composite section: Flexural capacity of steel section: ' 0.85 fc Fy C C d d T T

Fy Fy σ≤F y σ ≤ σ ≤ ' steel Fy conc 0.85 fc Equilibrium: C = T Equilibrium: C = T

Mn = C * d = Z Fy Mn = C * d

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 ?

Building: F = ma Mass = m Earthquake Forces on Buildings: Inertia Force Due to Accelerating Mass

Ground Acceleration

Conventional Building Code Philosophy for To Survive Strong Earthquake Earthquake-Resistant Design without Collapse:

Objective: Prevent collapse in the extreme Design for Ductile Behavior earthquake likely to occur at a building site.

Objectives are not to: - limit damage - maintain function - provide for easy repair

H H H Helastic

3/4 *Helastic Available Ductility Ductility = Inelastic Deformation H 1/2 *H elastic Required Strength

1/4 *Helastic

MAX Achieving Ductile Response....

Ductile Limit States Must Precede Brittle Limit States How Do We Achieve Ductility in Steel Structures ?

Example double angle tension member gusset plate double angle tension member

P P P P

Example: Gross-section yielding of tension Ductile Limit State: Gross-section yielding of tension member member must precede net section fracture of tension member Brittle Limit States: Net-section fracture of tension member

Block-shear fracture of tension member Gross-section yield: Pyield = Ag Fy Net-section fracture of gusset plate Net-section fracture: Pfracture = Ae Fu Block-shear fracture of gusset plate Bolt shear fracture Plate bearing failure in double angles or gusset

double angle tension member double angle tension member

P P P P

Example: Gross-section yielding of tension member The required strength for brittle limit ≤ must precede bolt shear fracture Pyield Pfracture states is defined by the capacity of the ductile element ≤ Ag Fy Ae Fu Gross-section yield: Pyield = Ag Fy 0.4 Fu-bolt -N Bolt shear fracture: Pbolt-fracture = nb ns Ab Fv Fv = A Fy Fy Steels with a low yield ratio are e ≥ = yield ratio 0.5 Fu-bolt -X preferable for ductile behavior Ag Fu Fu double angle tension member double angle tension member

P P P P

Example: Bolts: 3 - 3/4" A325-X double shear The required strength for brittle limit ≤ 2 Pyield Pbolt-fracture states is defined by the capacity of the Ab = 0.44 in Fv = 0.5 x 120 ksi = 60 ksi ductile element 2 k Pbolt-fracture = 3 x 0.44 in x 60 ksi x 2 = 158

Angles: 2L 4 x 4 x 1/4 A36 The ductile element must be the 2 weakest element in the load path Ag = 3.87 in 2 k Pyield = 3.87 in x 36 ksi = 139

double angle tension member double angle tension member

P P P P

≤ ≤ Pyield Pbolt-fracture Pyield Pbolt-fracture

k k 2 k Pyield = 139 Pbolt-fracture = 158 OK Pyield = 3.87 in x 54 ksi = 209

k Pbolt-fracture = 158 What if the actual yield stress for the A36 angles is greater than 36 ksi? ≤ Pyield Pbolt-fracture Say, for example, the actual yield stress for the A36 angle is 54 ksi. Bolt fracture will occur before yield of angles non-ductile behavior

double angle tension member double angle tension member

P P P P

≤ ≤ Pyield Pbrittle Pyield Pbrittle

Stronger is not better in the ductile element The required strength for brittle limit (Ductile element must be weakest element in the load path) states is defined by the expected For ductile response: must consider material overstrength in capacity of the ductile element (not minimum specified capacity) ductile element

Pyield = Ag RyFy Ry Fy = expected yield stress of angles Achieving Ductile Response.... Achieving Ductile Response....

Ductile Limit States Must Precede Brittle Limit States Connection response is generally non-ductile.....

Define the required strength for brittle limit states based on the expected yield capacity for ductile element Connections should be stronger than connected members

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.... General Trends:

As F Be cautious of high-strength steels y

Elongation (material ductility)

Fy / Fu

Ref: Salmon and Johnson - Steel Structures: Design and Behavior Achieving Ductile Response.... Achieving Ductile Response....

Be cautious of high-strength steels Use Sections with Low Width-Thickness Ratios and Adequate Lateral Bracing 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

Effect of Local Buckling on Flexural Strength and Ductility

θ Plastic Buckling M Mp Inelastic Buckling M Mr Elastic Buckling

Mp Moment Capacity

λ λ λ Width-Thickness Ratio (b/t) ps p r Increasing b / t Ductility

Plastic Buckling Plastic Buckling

Mp Inelastic Buckling Mp Inelastic Buckling

Mr Elastic Buckling Mr Elastic Buckling

Moment Capacity Slender Element Sections Moment Capacity Noncompact Sections

λ λ λ Width-Thickness Ratio (b/t) λ λ λ Width-Thickness Ratio (b/t) ps p r ps p r Ductility Ductility Plastic Buckling Plastic Buckling

Mp Inelastic Buckling Mp Inelastic Buckling

Mr Elastic Buckling Mr Elastic Buckling

Compact Sections Seismically Moment Capacity Moment Capacity Compact Sections

λ λ λ Width-Thickness Ratio (b/t) λ λ λ Width-Thickness Ratio (b/t) ps p r ps p r Ductility Ductility

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

Local buckling of noncompact and slender element sections

5000 Local buckling of a seismically compact EBF link.....

4000 Mp 3000

2000

1000

-1000

-2000 Bending Moment (kN-m) Moment Bending

-3000 Mp -4000 RBS Connection -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) Effect of Local Buckling on Ductility

200 For ductile flexural response:

150 Use compact or seismically compact sections

100

50 bf

-50 Example: W-Shape tf -100

-150 h Link Shear Force (kips) -200 -0.08 -0.06 -0.04 -0.02 0 0.02 0.04 0.06 0.08 tw Link Rotation, γ (rad)

Lateral Torsional Buckling Beam Flanges Lateral torsional buckling L b E controlled by: b Compact: f ≤ 0.38 s r 2tf Fy y

Lb = distance between beam lateral braces Seismically Compact: b E f ≤ s 0.30 ry = weak axis radius of gyration 2tf Fy

Beam lateral braces Beam Web h E ≤ s Compact: 3.76 tw Fy

h Es Seismically Compact: ≤ 2.45 Lb Lb tw Fy

Effect of Lateral Torsional Buckling on Flexural Strength and Ductility:

θ M M

Increasing Lb / ry

θ Effect of Lateral Buckling on Ductility Achieving Ductile Response....

For ductile flexural response: Recognize that buckling of a compression member is Use lateral bracing based on plastic design non-ductile requirements or seismic design requirements

⎡ ⎛ M ⎞⎤⎛ E ⎞ Plastic Design: L ≤ 0.12 + 0.076⎜ 1 ⎟ ⎜ ⎟r b ⎢ ⎜ ⎟⎥⎜ ⎟ y ⎣⎢ ⎝ M2 ⎠⎦⎥⎝ Fy ⎠

⎛ E ⎞ Seismic Design: L ≤ 0.086⎜ ⎟r b ⎜ ⎟ y ⎝ Fy ⎠

Experimental Behavior of Brace Under Cyclic Axial Loading P cr W6x20 Kl/r = 80

δ P

Pcr

δ P

How Do We Achieve Ductile Response in Steel Structures ? 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

• Provide connections that are stronger than members

• 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