aisc night school 9 session 2

8/17/2019 AISC Night School 9 Session 2

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Application of the AISC Seismic Design M

Session 2: General Design Requirement

Copyright © 2015

American Institute of Steel Construction

AISC Night School – Seismic Design Manual

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Copyright Materials

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display and use of the presentation without written permission of AISC is prohibited.

© The American Institute of Steel Construction 2015

The information presented herein is based on recognized engineering principles and is for general

information only. While it is believed to be accurate, this information should not be applied to any

specific application without competent professional examination and verification by a licensed

professional engineer. Anyone making use of this information assumes all liability arising from

such use.


Session 2: General Design Requirements Part 2

September 28, 2015

Load combinations for seismic design will be discussed. The session will

present an overview of some of the 2010 Seismic Provisions including

application of the overstrength factor, member requirements, stability bracing

of beams and drift requirements. Examples from the Seismic Design Manual

will be presented to demonstrate concepts discussed in the session.

Course Description


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• Become familiar with load combinations considered for seismic design.

• Gain an understanding of the stability bracing requirements of beams per

the AISC Seismic Provisions.

• Gain an understanding of the application of the overstrength factor.

• Become familiar with the member design requirements of the AISC Seismic

Provisions through demonstrated design examples.

Learning Objectives

AISC Night School – Seismic Design Manual 8

Presented by

Thomas A. Sabol, Ph.D., S.E.

Principal at Englekirk Institutional

Los Angeles, CA

Application of the AISC Seismic Design Manual

Session 2: General Design Requirements Part 2


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Application of the

AISC Seismic Design Manual

Session 2


Last Session• Seismic Performance Goals

• Seismic Design Categories

• Seismic Performance Factors (e.g., R , O )

• Organization of AISC 341 Seismic Provisions

• Steel Material Properties (e.g., yield strength,R y )

• Welding Filler Metal Properties (e.g., Charpy V-Notch)


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B1 General Seismic Design Requirements

Seismic Provisions defer to applicable building code

for:

Required seismic strength with some exceptions (e.g.,

where expected strength is used to determine demand

on one member caused by another member )

Determination of Seismic Design Categories

Limitations on height and irregularities

Design story drift limits

12


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B2 Loads and Load Combinations

Applicable Building Code determines:

Loads and load combinations for required strength of

steel seismic systems

Examples in SDM use “First Printing” of ASCE 7-10 and

may be different from your copy of ASCE 7-10

13


“QE” has both apositive andnegative sign

B2 Loads and Load CombinationsApplicable Building Code determines:

Loads and load combinations for required strength of

steel seismic systems

Example basic LRFD seismic load combinations from

ASCE 7 (ASD similar)

• (1.2 + 0.2S DS )D + ρQ E +0.5L + 0.2S

•

(0.9 - 0.2S DS )D +ρ

Q E + 1.6H

Taking QE with a negative signis assumed to create the

critical case wheninvestigating net tension

14


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Note: L may be taken as0.5L for most areas whereL o ≤ 100 psf

B2 Loads and Load Combinations

When “amplified seismic load” is required :

Use system overstrength factor, o, from ASCE 7 Table

12.2-1 unless otherwise defined by Seismic Provisions

Example load combinations with o

• (1.2 + 0.2S DS )D + oQ E + L + 0.2S

• (0.9 - 0.2S DS )D + oQ E + 1.6H

15


B3 Design Basis

Required strength shall be greater of:

Required strength from application of structural

analysis using loads from the building code

Required strength from Seismic Provisions [e.g.,

expected strength of a member or amplified seismic

load (i.e., seismic load effect with overstrength from

building code)]

16


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B3 Design Basis

Available strength (e.g., design strength, R n, or

allowable strength, R n / ) shall be:

Obtained from LRFD or ASD Specification

As modified by the Seismic Provisions (there aren’t too

many)

17


Example 3.4.2

Moment Frame Column Design (using R = 3 approach)

18


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Example 3.4.2

Given:

Refer to Column CL-1 in Figure 3-2. Verify that a

W12×87 ASTM A992 W-shape is sufficient to

resist the following required strengths between

the base and second levels. The applicable

building code specifies the use of ASCE/SEI 7

for calculation of loads.

19


Example 3.4.2

The load combinations that include seismic

effects are:

20

LRFD ASD

LRFD Load Combination 5

from ASCE/SEI 7 Section

12.4.2.3

(including the 0.5 load factor

on L permitted in ASCE/SEI 7

Section 12.4.2.3)

ASD Load Combination 5

from ASCE/SEI 7 Section

12.4.2.3

1.2 0.2 ρ 0.5 0.2E DS S D Q L S + + + 1.0 0.14

0.7ρ

DS

E

S D H

F Q

+ + +

+


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Example 3.4.2

21

From ASCE/SEI 7, this structure is assigned toSeismic Design Category C (ρ = 1.0) and S DS =

0.352.

Given in the problemstatement


Example 3.4.2

The required strengths of Column CL-1

determined by a second-order analysis

including the effects of P -δ and P -Δ with

reduced stiffness as required by the direct

analysis method are:

22

LRFD ASD

P u = 233 kips

V u = 35.0 kipsM u top = 201 kip-ft

M u bot = −320 kip-ft

P a = 165 kips

V a = 23.4 kipsM a top = 131 kip-ft

M a bot = −210 kip-ft


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Example 3.4.2

There are no transverse loadings between thefloors in the plane of bending, and the beams

framing into the column weak axis are pin-

connected and produce negligible moments.

23


Example 3.4.2

Solution:From AISC Manual Table 2-4, the material

properties are as follows:

ASTM A992

F y = 50 ksi

F u = 65 ksi

24

From AISC Manual Table 1-1, the geometric


W12×87

r x = 5.38 in. r y = 3.07 in.


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Example 3.4.2

Available Compressive Strength of Column CL-1

Because the member is being designed using the

direct analysis method, K is taken as 1.0.

25

1.0 14.0 ft 12.0 in./ft

5.38 in.

31.2

x

x

KL

r =

=

1.0 14.0 ft 12.0 in./ft

3.07 in.54.7

y

y

KL

r

=

=

governs


Example 3.4.2

From AISC Manual Table 4-1, the available

compressive strength is:

26

LRFD ASD

925 kipsc nP =φ 616 kipsΩ

n

c

P =


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Example 3.4.2

Available Flexural Strength of Column CL-1

Check the unbraced length for flexure

From AISC Manual Table 3-2:

L p = 10.8 ft

Lr = 43.1 ft

L p < Lb = 14.0 ft < Lr

27


Example 3.4.2

Therefore, the member is subject to lateral-

torsional buckling.

Calculate C b using AISC Specification Equation

F1-1.

28


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LRFD ASD

201 kip-ft320 kip-ft

utop

ubot

M M

=

= −

( )

201kip-ft 320kip-ft201kip-ft

14.0ft

201kip-ft 37.2kips

top bot top

M M M x M x

L

x

x

−

= −

= −

= −

131 kip-ft

210 kip-ft

atop

abot

M

M

=

= −

( )

131kip-ft 210kip-ft131kip-ft

14.0ft

131kip-ft 24.4kips

top bot top

M M M x M x

L

x

x

−

= −

= −

= −


LRFD ASD

Quarter point moments are: Quarter point moments are:

3.50ft

201 kip-ft

37.2kips 3.50ft

70.8 kip-ft

AM x M =

=

−

=

3.50ft

131 kip-ft

24.4kips 3.50ft

45.6 kip-ft

AM x M =

=

−

=

7.00ft

201 kip-ft37.2kips 7.00ft

59.4 kip-ft

BM x M =

=

−

=

7.00ft

131 kip-ft24.4kips 7.00ft

39.8 kip-ft

BM x M =

=

−

=


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LRFD ASD

10.5ft

201 kip-ft

37.2kips 10.5ft

190 kip-ft

320kip-ft

C

max

M x M

M

= =

=

−

=

=

12.5

2.5 3 4 3

12.5 320

2.5 320 3 70.8 4 59.4 3 190

2.20

max b

max A B C

M C

M M M M =

+ + +

=

+ + +

=

10.5ft131 kip-ft

24.4kips 10.5ft

125 kip-ft

210kip-ft

C

max

M x M

M

= =

=

−

= −

=

12.5

2.5 3 4 3

12.5 210

2.5 210 3 45.6 4 39.8 3 125

2.19

max b

max A B C

M C

M M M M =

+ + +

=

+ + +

=


Example 3.4.2

From AISC Manual Table 3-10, with the availableflexural strength of a W12×87 is:

32

LRFD ASD

Check yielding (plastic moment)

limit state, using AISC Manual

Table 3-2,

Check yielding (plastic moment)

limit state, using AISC Manual

Table 3-2,

2.20 477 kip-ft

1,050 kip-ft

b nM =

=

φ

2.19 318 kip-ftΩ

696 kip-ft

n

b

M =

=

495 kip-ft 1,050 kip-ftb pM =


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Example 3.4.2Interaction of Flexure and Compression in Column CL-1

Using AISC Specification Section H1, check the

interaction of compression and flexure in Column CL-1, as follows:

33

LRFD ASD

Because P r /P c > 0.2, use AISCSpecification Equation H1-1a.

Because P r /P c > 0.2, use AISCSpecification Equation H1-1a.

, as determined previously

925 kips

233 kips

925 kips

0.252

c c n

r

c

P P

P

P

=

=

=

=

φ , as determined previouslyΩ

616 kips

165 kips

616 kips

0.268

nc

c

r

c

P P

P

P

=

=

=

=


Example 3.4.2

34

LRFD ASD

81.0 Eq. H1-1a

9

8 320 kip-ft0.252 0 0.827

9 495 kip-ft

0.827 1.0 o.k.

ry r rx

c cx cy

M P M Spec.

P M M

+ + ≤

+ + =

<

81.0 Eq. H1-1a

9

8 210 kip-ft0.268 0 0.835

9 329 kip-ft

0.835 1.0 o.k.

ry r rx

c cx cy

M P M Spec.

P M M

+ + ≤

+ + =


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Example 3.4.2

Available Shear Strength of Column CL-1

From AISC Manual Table 3-2, the available shear

strength of a W12×87 is:

35

LRFD ASD

193 kips 35.0 kips o.k.v nV = >φ / Ω 129 kips 23.4 kips o.k.n v V = >


Example 3.4.2

The W12x87 is adequate to resist the required

strengths given for Column CL-1.

Note: Load combinations that do not include

seismic effects must also be investigated.

36


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Example 3.4.2

Moment Frame Column Design (using R = 3 approach)

37

End of Example


Example 4.3.1

SMF Story Drift and Stability Check

38


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Given:Refer to the floor plan shown in Figure 4-7 and the

SMF elevation shown in Figure 4-8. Determine

if the frame satisfies the ASCE/SEI 7 drift and

stability requirements based on the given

loading.

The applicable building code specifies the use of

ASCE/SEI 7 for calculation of loads.

39


SMF floor plan

40

SMF elevation


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Example 4.3.1

The seismic design story shear acting between

the second and third levels, V x , is 140 kips asdefined in ASCE/SEI 7 Section 12.8.4.

From an elastic analysis of the structure that

includes second-order effects and accounts for

panel-zone deformations, the maximum

interstory drift occurs between the third and

fourth levels:

δ xe = δ4e− δ3e = 0.482 in.

41


δ xe = δ4e− δ3e = 0.482 in.

42

This is the difference in

displacement (drift)between two adjacent

floors. The “e” signifiesthat these displacementswere obtained from an

elastic analysis.

Story Drift Determination between Levels 3 and 4

δ3e

δ4e

Partial Frame Elevation

Level 4

Level 3

Undeformedframe

Deformedframe


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In this example, the stability check will be made at

the second level. The story drift between the

second and third levels is 0.365 in.

(δ3e− δ2e) = 0.365 in.

43

Solution:

From AISC Manual Table 1-1, the geometric


W24x76bf = 8.99 in.


Reduced-beam-section (RBS) connections are

used at the frame beam-to-column connections

and the flange cut will reduce the stiffness of

the beam.

Example 4.3.3 illustrates the design of the RBS

geometry and the flange cut on one side of the

web is c = 2 in.

44

RBS (plan view)


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Some analysis programs allow for

direct input of RBS dimensions fromwhich the reduced stiffness can becalculated. This isn’t always practicalfor preliminary designs because you

must know the dimensions of the RBScut.

Section 5.8, Step 1, of ANSI/AISC 358 states that the

calculated elastic drift, based on gross beam

section properties, may be multiplied by 1.1 for

flange reductions up to 50% of the beam flange

width in lieu of specific calculations of effective

stiffness.

Amplification of drift values for cuts less than the

maximum may be linearly interpolated.

45


Example 4.3.1

For bf = 8.99 in., the maximum cut is:

0.5(8.99 in.) = 4.50 in.

Thus, the total 4-in. cut is:

(4.00 in./4.50 in.)100 = 88.9% of the maximum cut

The calculated elastic drift needs to be amplified

by 8.89% (say 9%).

46

Sum of maximumcuts on both sides

of flange

c = 2” Total cutis 2x2” = 4”

This amplificationaccounts for the fact

that the analytical modelused gross sections


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Drift Check

From an elastic analysis of the structure that

includes second order effects, the maximum

interstory drift occurs between the third and

fourth levels. The effective elastic drift is:

47

4 3δ δ δ

0.482 in.

xe e e−

=

δ 1.09δ

1.09 0.482 in.

0.525 in.

xe RBS xe

=

=

Amplification ofdrfit by 9% due

to RBS cut


Example 4.3.1

Per the AISC Seismic Provisions Section B1, the

design story drift and the story drift limits are

those stipulated by the applicable building code.

ASCE/SEI 7 Section 12.8.6 defines the design

story drift, Δ, computed from δ x , as the

difference in the deflections at the center of

mass at the top and bottom of the story under

consideration, which in this case is the thirdlevel.

48


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C d = 5.5 for SMFper ASCE 7, Table

12.2-1

Example 4.3.1

49

δΔ ASCE / SEI 7 Eq. 12.8-15

5.5 0.525 in.

1.0

2.89 in.

d xe

e

C

I =

=

=

C d amplifies the elasticdrift (calculated usingreduced forces) into anestimate of the (actual)

inelastic drift


Example 4.3.1

From ASCE/SEI 7 Table 12.12-1, the allowable

story drift at level x , Δa, is 0.020hsx , where hsx is the story height below level x .

(Although not assumed in this example, ∆a can be

increased to 0.025hsx if interior walls,

partitions, ceilings and exterior wall systems

are designed to accommodate these increasedstory drifts.)

50


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For ρ = 1.3, this provision

has the effect of reducingthe allowable drift (i.e., thestructure would have to bestiffer than if ρ = 1.0).

Example 4.3.1

ASCE/SEI 7 Section 12.12.1.1 requires for seismicforce resisting systems comprised solely of

moment frames in structures assigned to

Seismic Design Category D, E or F, that the

design story drift shall not exceed (∆a /ρ) for

any story.

Determine the allowable story drift as follows:

51


Example 4.3.1

52

Δ 0.020

ρ ρ

0.020(12.5 ft)(12 in./ft)

1.0

3.00 in.

a sx h=

=

=

Δ 2.89 in. a< ∆

o.k

The frame satisfies the drift requirements.

Story heightbelow Level 3

In this example,because ρ = 1.0,this provision hasnot impact on the

design


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Example 4.3.1

Frame Stability Check

ASCE/SEI Section 12.8.7 provides a method for

the evaluation of the P -∆ effects on moment

frames based on a stability coefficient θ, which

should be checked for each floor. For the

purposes of illustration, this example checks

the stability coefficient only for the third level.

53


Afloor = Aroof ≈ 75 ft(120 ft) = 9,000 ft2

Dfloor = 9,000 ft2(85 psf)/1,000 lb/kip

= 765 kips

Droof = 9,000 ft2 (68 psf)/1,000 lb/kip)= 612 kips

Dwall = 175 lb/ft[2(75 ft + 120 ft)]/(1,000 lb/kip)

= 68.3 kips per level

The stability coefficient, θ, is determined as follows:

54

“D ” and “L” arethe dead andlive loads,

respectively.

P x is totalvertical loadacting on agiven story

Δθ (ASCE/SEI 7 Eq. 12.8-16)

x e

x sx d

P I

V h C =


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29,000 ft 20 psf / 1,000 lb/kip

180 kips

roof L =

=

Lfloor = 9,000 ft2(50 psf)/(1,000 lb/kip)

= 450 kips


ASCE/SEI 7 does not explicitly specify load

factors to be used on the gravity loads for

determining P x , except that Section 12.8.7

does specify that no individual load factor

need exceed 1.0.

This means that if the combinations of ASCE/SEI 7

Section 2.3 are used, a factor of 1.0 can be

used for dead load rather than the usual 1.2factor used in the LRFD load combination, for

example.

56


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This also means that the vertical component

0.2S DS

D need not be considered here.

Therefore, for this example, the load combination

used to compute the total vertical load on a

given story, P x , acting simultaneously with the

seismic design story shear, V x , is 1.0D + 0.5L

based on ASCE/SEI 7 Section 2.3 including the

0.5 factor on L permitted by Section 2.3, where L

is the reduced live load.

57


Note that consistent with this, the same

combination was used in the second order

analysis for this example for the purpose of

computing the fundamental period, base

shear, and design story drift.

58


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Example 4.3.1

The total dead load supported by the columns onthe second level, assuming that the columns

support the equivalent of two floors worth of

curtain wall in addition to other dead loads, is:

59

1.0 1.0[612 kips 2(765 kips) 2(68.3 kips)]

2,280 kips

DP = + +

=

DFloorDRoof DWall


LFloor

Example 4.3.1

The total live load supported by the columns on

the second level is:

60

0.5 0.5 2 450 kips 180 kips

540 kips

LP = +

=

LRoof


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Example 4.3.1

Therefore, the total vertical design load carried bythe columns on the second level is:

61

2,280 kips 540 kips

= 2,820 kips

x P = +


Example 4.3.1

The seismic design story drift at the top of the

second level, including the 9% amplification

on the drift, is:

62

δΔ from ASCE / SEI 7 Eq. 12.8-15

5.5(1.09)(0.365 in.)

1.02.19 in.

d xe

e

C

I =

=

=


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Example 4.3.1

From an elastic analysis of the structure, the

seismic design story shear acting at the third

level under the story drift loading using the

equivalent lateral force procedure is V x = 140

kips and the floor-to-floor height is hsx = 12.5 ft.

Therefore, the stability coefficient is:

63

2,820 kips 2.19 in. 1.0θ

140 kips 12.5 ft 12 in./ft 5.5

0.0535

=

=


Example 4.3.1

Because a second-order analysis was used to

compute the story drift, θ is adjusted as

follows to verify compliance with θmax , per

ASCE/SEI 7 Section 12.8.7.

64

θ 0.0535

1 θ 1 0.0535

0.0508

=

+ +

=


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Second-order effects include P-δ andP-Δ. Δ is the first order interstorydrift due to lateral loads. δ is thelocal deformation of the column due

to these loads, initial columnimperfections, etc.

Example 4.3.1

According to ASCE/SEI 7, if θ is less than or equal

to 0.10, second-order effects need not be

considered for computing story drift.

Note that whether or not second-order effects on

member forces must be considered per

ASCE/SEI 7 has to be verified, as it was in this

example; however, Chapter C of the AISC

Specification requires second order effects be

considered in all cases in the analysis used

for member design.

65


Example 4.3.1

Check the maximum permittedθ

The stability coefficient may not exceed θmax . In

determining θmax , β is the ratio of shear demand

to shear capacity for the level being analyzed,

and may be conservatively taken as 1.0.

66

0.5θ 0.25 ASCE / SEI 7 Eq. 12.8-17

β0.5

1.0(5.5)

0.0909 0.25

d C

= ≤

=

= ≤

max


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Example 4.3.1

The adjusted stability coefficient satisfies themaximum:

0.0508 < 0.0909 o.k.

The moment frame meets the allowable story drift

and stability requirements for seismic loading.

67


Example 4.3.1

Comments:

There are a total of six bays of frames in the SMF

direction in this example. Considering the

relative expense of SMF connections, it is

probably more cost-effective to reduce the

number of bays to four, and increase member

sizes to satisfy the strength and stiffness

requirements.

68


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Intent of this chapter is to provide analytical requirements

for use in designing structural steel seismic systems.

Currently, there is little prescriptive material in the

provisions section, but there are analytical and modeling

recommendations in the Commentary.

SDM contains discussions and examples illustrating some

of the aspects of seismic system analysis. See Example

5.3.2 as an illustration.

71


Chapter D

72


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Chapter D

General Member and Connection Design

Requirements

Contains provisions that apply to multiple systems

(e.g., member requirements for ductility and bracing at

plastic hinges)

General connection requirements (e.g., bolting and

welding requirements and column splices)

Deformation compatibility

H-piles

73


Chapter D

D1 Member Requirements

Seismic Provisions may require certain members to

be “moderately ductile,”λmd , or “highly ductile,” λ hd

These requirements may be more stringent than

found in Specification Table B4.1

These new designations replace “compact” and

“seismically compact” from earlier editions

These provisions present requirements to limit

(delay) local flange or local web buckling

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Chapter D


“Compactness” describes a section sufficiently

stocky to develop a fully plastic stress distribution

without buckling

Certain members in seismic systems are expected to

delay onset of buckling beyond initial development of

the plastic distribution, so “compactness” isn’t a

very descriptive term for the behavior sought

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Chapter D


Table D1.1 presents λmd and λhd values…there are no

significant technical changes from “compact” and

“seismically compact” values

Table D1.1 contains helpful graphics to make it easier

to understand which value applies for different parts

of structural sections

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Chapter DD1 Member Requirements

Requirements for Width-to-Thickness Ratios

Formerly“seismicallycompact”

Formerly“compact”

λmd = 9.15for

F y = 50

λhd = 7.23for

F y = 50 ksi

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Chapter DD1 Member RequirementsRequirements for Width-to-Thickness Ratios

λmd = 16.10for

F y = 46

λhd = 13.81for

F y = 46 ksi

Eliminates manyrectangular orsquare HSS

sections (e.g.,> HSS12x

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SDM Table 1-A has summary of width-to-thickness by

SFRS compression member type

79

Similartables forangles and

HSS



SDM Tables 1-3 to 1-7 identify members that

may be used in different SFRS

Tables cover:

W-shapes

Angles

Rectangular HSS

Square HSS

Round HSS

Pipe

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W24x162satisfies width-thickness requirements for all

SFRS (shown by “•”)

W24x55 does not satisfy width-thickness requirements for OCBF,SCBF and EBF braces

81

Similar tablesfor other

shapes


Chapter DD1.2 Stability Bracing of Beams

Stability bracing is specified for seismic systems to

control lateral-torsional buckling

Lateral-torsional buckling

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For moderately and highly ductile members:

Both flanges must be braced or the section torsionally

braced

Lateral bracing provided byconcrete structural slab and full-height perpendicular framing

Lateral bracing provided by shallowperpendicular steel framing andstiffener – wood framing was notconsidered adequate

83



For moderately ductile members:

Unbraced length between lateral braces shall not

exceed Lb = 0.17r y E/F y

L b ≤0.17r y E/F y

Lateral bracing fortop and bottomflanges

For F y = 50 ksi,L b ≤ 98.6r y

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For highly ductile members:

Unbraced length between lateral braces shall not

exceed Lb = 0.086r y E/F y

L b ≤0.086r y E/F y

85


Not the same C d in ASCE 7



Beam bracing shall meet requirements of Specification

Appendix 6 for lateral or torsional bracing where the

required strength of the brace is

P rb = 0.02M r C d /ho (Spec . A-6-7)

and

M r = R y F y Z (Provisions D1-1a for LRFD)

86

This is an example of the SeismicProvisions specifying requiredstrength based on expectedstrength – not code demand


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…and the required stiffness of the brace is

(Spec . A-6-8 for LRFD)

where C d = 1.0

ho = distance between flange centroids

87

1 10 r d br

b o

M C

L hβ

φ

=

h o

=

d i s t a n c e

b e t w e e n

f l a n g e

c e n t r o i d s



At plastic hinges (or directly adjacent thereto):

Brace top and bottom flanges or brace against torsional

buckling

Required strength of bracing is P u = 0.06R y F y Z /ho(lateral bracing) or M u = 0.06R y F y Z (torsional bracing)

Bracing stiffness shall satisfy requirements of Appendix

6 of Specification but C d = 1.0 and M r = M u = R y F y Z

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**

**

Lateral bracing for top and bottomflange (not required if there is aconcrete structural slab per AISC358 for SMF and IMF)


At plastic hinges (or directly adjacent thereto):

Bracing adjacentto plastic hinge

Plastic hinge

89


Chapter D

90

Next Session• Example: SMF Beam Stability Bracing

• Protected Zones

• Column Requirements

• Example: SMF Column Strength Check

• Bolted and Welded Joints (General)

• Continuity Plates and Stiffeners

• Column Splice

• Example: Column Splice


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Chapter D

91

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aisc night school 9 session 2

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