s. k. ghosh associates inc. - seaooseaoo.org/downloads/presentations_conf/2010_presentation_1... ·...

1

S. K. Ghosh Associates Inc.

www.skghoshassociates.com

FREQUENTLY MISUNDERSTOOD

IBC/ASCE 7 STRUCTURAL PROVISIONS

S. K. Ghosh and Susan Dowty


Palatine, IL and Aliso Viejo, CA


All sections

referenced are from

ASCE 7-05,

unless otherwise

noted.

PROVISION#1

Enclosure ClassificationFor Wind Design

Enclosure Classification

Section 6.2 Definitions

Open

Partially Enclosed (can experience

“ballooning” or suction effects caused by the

build-up of internal pressure)

Enclosed

Internal Pressure

2





OPEN

A building having each wall at least 80% open.

Ao ≥ 0.8 Ag

for EACH side of the building



AO = A1 + A2 + A3 Ag = H ×××× W





Do stacks of hay

obstruct flow of wind?



PARTIALLY ENCLOSED

1.Ao ≥ 1.10Aoi

2.Ao > 4 sq ft AND > 0.01Ag

3. Aoi/Agi ≤ 0.20



Ao = total area of openings in a wall that receives positive external pressure

Aoi = sum of the areas of openings in the building envelope (walls and roof) not including Ao

Ag = gross area of that wall in which Ao is identified

Agi = the sum of the gross surface areas of the building envelope not including Ag

3





Note: Ao, Ag refer to wall that

receives positive external pressure

Aoi , Agi refer to building envelope

(walls and roof)



• Openings : apertures or holes in the

building envelope which allow air to

flow through the building envelope

and which are designed as “open”

during design winds

Q: A: for Enclosure Classification

Q: Is a fixed glazed opening

considered an opening?

A: NO.



ENCLOSED

A building that does not qualify as

OPEN or PARTIALLY ENCLOSED.




Figure 6-5 Internal Pressure Coefficients,

GCpi

4



Basic Wind Equation

• For buildings with External and

Internal Pressure:

p = qGCp – qiGCpi

qi = Velocity pressure calculated for

internal pressure.

ASCE 7-98

Positive Internal Pressure

ASCE 7-98

Negative Internal Pressure Q: A: for Enclosure Classification

Q: Why does a building need to be

enclosed to use the Simplified Procedure?

A: See C6.4. GCpi = ±0.18 is assumed in the

tables. In a simple diaphragm building,

internal pressures cancel out for the walls

but not for the roof.

Q: A: for Enclosure Classification

Q: Should we treat roll-down doors and

operable louvers as openings?

A: Yes and No.

PROVISION#2

Seismic and Wind Design of Parapets

5



Common Earthquake Damage to

Parapets

13.3.1 Nonstructural Component

Seismic Design Force

Fp (min) = 0.3 SDS Ip Wp for SDS = 1.00,

Fp = 0.30 IpWp

Fp (max) = 1.6 SDS Ip Wp for SDS = 1.00,

Fp = 1.60 IpWp

Fp =0.4 ap SDS

(Rp / Ip)1 +

2z

hWp

13.1.3Nonstructural Component

Importance Factor, IpIp is based on

1. Whether component must function after

the design earthquake or

2. Occupancy Category or

3. Whether component contains hazardous

materials.

Parapets: Ip is based on Occupancy

Category

Nonstructural Component ap and

Rp

The values of ap range from 1.0 to 2.5 and

can be taken as less than 2.5 based on

dynamic analysis.

Rp values range from 1.0 to 12.0 (Tables

13.5-1 and 13.6-1).

Table 13.5-1 ap and Rp for

Architectural Components

Architectural

Component

ASCE 7-05

ap Rp

Cantilever Parapets 2.5 2.5

0.4SDS

1.2SDS

AA= 0.4SDS (1+2zA/h)

AB = 0.4SDS (1+2zB/h)

A

B

zA

zB

h

Floor

Acceleration

Distribution

Explanation of Fp Equation

Z = h for parapet design

6



The 7-in. concrete parapet shown forms

part of a building assigned to SDC D with

a component importance factor of 1.0.

SDS= 1.0g at the site. Determine the

strength-level seismic design moment in

the parapet.

Example: Seismic Design of

Parapets


Parapets

Weight of the parapet per linear foot is

Wp = 150 x 3 x 7/12 = 262.50 lb/ft

The seismic lateral force acting at the

centroid of the parapet is given by

ASCE Equation (13.3-1) as

Fp = (0.4apSDSIp / Rp)(1 + 2z/h )Wp

Where Ip = component importance

factor = 1.0


Parapets

SDS = 1.0g

Wp = weight of parapet = 262.5 lb/ft

ap = component amplification factor from ASCE

Table 13.5-1 = 2.5

h = height of roof above the base = 20 ft

z = height of parapet at point of attachment =

20 ft


Parapets

Rp = component response modification factor

from ASCE Table 13.5-1 = 2.5

Fp = (0.4 x 2.5 x 1.0 1.0 / 2.5) (1 + 2 x 20/20)Wp

= 1.2Wp = 315 lb/ft

Neither ASCE Equation (13.3-2) nor (13.3-3)

governs, and the bending moment at the

base of the parapet is

Mp = 1.5 Fp = 472.5 lb-ft/ft


ParapetsWind Forces on Parapets

ASCE 7-05 Figure C6-12

7



Wind Forces on Parapets

ASCE 7-05 C6.5.11.5 For simplicity, the

front and back pressures on the parapet

have been combined into one coefficient

for MWFRS design.

Wind Forces on Parapets

ASCE 7-05 Figure C6-12

Design Example

The main wind force-resisting system of a

5-story reinforced concrete office building

is designed following the requirements of

the 2009 IBC/ASCE 7-05 wind provisions.

Example Building

Example Building3 ft parapet

Design Criteria

Location of building: Los Angeles, California

V = 85 mph (ASCE 7-05 Fig. 6-1)

Building is enclosed per definition under ASCE

7-05 Sec. 6.2

Assume Exposure B (ASCE 7-05 Sec. 6.5.6.3)

Occupancy Category: II, I = 1.0 (ASCE 7-05

Table 6-1)

8



Effects of Parapets on MWFRS loads

ASCE 7-05 Section 6.5.12.2.4:

pp = qpGCpn (ASCE 7-05 Eq. 6-20)

qp = velocity pressure evaluated at the top of

the parapet

= 0.00256 Kz Kzt Kd V2 I

GCpn = combined net pressure coefficient

= +1.5 for windward parapet

= -1.0 for leeward parapet

At top of parapet, h = 67.5 + 3 = 70.5 ft

Kz = 2.01(z/zg)2/α = 2.01(70.5/1200)2/7 = 0.894

(from ASCE 7-05 Table 6-3)

(α = 7, zg = 1200 ft for Exposure B from

ASCE 7-05 Table 6-2)

Velocity Pressure Exposure

Coefficient, Kz

Topographic Effect Factor, Kzt

Wind Directionality Factor, Kd

Kzt = 1.0

(Assuming the example building to be situated

on level ground, i.e., with H, as shown in ASCE

7-05 Fig. 6-4, equal to zero).

Kd = 0.85

(from ASCE 7-05 Table 6-4 for main wind force-

resisting system)


qp = 0.00256 Kz Kzt Kd V2 I =

0.00256 × 0.894× 1 × 0.85 × 852 × 1 = 14.06 psf

• For windward parapet:

pp = qpGCpn = 14.06 × 1.5 = 21.1 psf

Force = 21.1 × 3 × 66 / 1000 = 4.18 kips

• For leeward parapet:

pp = qpGCpn = 14.06 × (-1.0) = -14.06 psf

Force = -14.06 × 3 × 66 / 1000 = -2.78 kips


At the roof level, 4.18 + 2.78 = 6.96 kips is to be

added to the design wind force for MWFRS

computed from the windward and leeward walls

Design of Parapets as Component

ASCE 7-05 Section 6.5.12.4.4:

p = qp(GCp – GCpi) (ASCE 7-05 Eq. 6-24)

qp = velocity pressure evaluated at the top of

the parapet

= 0.00256 Kz Kzt Kd V2 I = 14.06 psf

GCp = External pressure coefficient from Figs.

6-11 through 6-17

GCpi = Internal pressure coefficient from Fig.

6-5

9



Effective wind area of the parapet:

Span = 3 ft

Width = 66 ft (> span/3)

A = 3x66 = 198 ft2

External Pressure Coefficient, GCp

Load Case A (ASCE 7-05 Section 6.5.12.4.4)

Positive wall GCp = 0.68 (Figure 6-17 Zones 4

and 5)

Applied to front surface of the parapet



Negative roof edge GCp = -1.76 (Figure 6-17

Zone 2*)

Applied to back surface of the parapet

*Corner Zone 3 is treated as Zone 2 because the parapet is 3 ft high

(Figure 6-17 Note 7)



GCp = 0.68 – (-1.76) = 2.44

GCpi = -0.18 for enclosed building (uniform

porosity)

However, internal pressures on both surfaces of

the parapet cancel each other out.

p = 14.06 x 2.44 = 34.31 psf


Load Case B (ASCE 7-05 Section 6.5.12.4.4)

Effective wind area = 198 ft2

Positive wall GCp = 0.68 (Figure 6-17 Zones 4 and

5)

Applied to back surface of the parapet



Negative wall GCp = -0.76 (Figure 6-17 Zone 4)

= -1.23 (Figure 6-17 Zone 5)

Applied to front surface of the parapet


10




GCp = 0.68 – (-0.76) = 1.44 (For Zone 4)

= 0.68 – (-1.23) = 1.91 (For Zone 5)

GCpi = -0.18 for enclosed building (uniform

porosity)

However, internal pressures on both surfaces of

the parapet cancel each other out.

p = 14.06 x 1.44 = 20.24 psf (Zone 4)

= 14.06 x 1.91 = 26.85 psf (Zone 5)


Clearly, Load Case A governs

Thus, design uniform wind pressure on the

whole width of the parapet

p = 34.31 psf

Design of Parapets as Component

Q: A: for Wind Design of Parapet

Q: In Section 6.5.12.4.4 (parapets for

C&C), the definition for the factor GCpi is

based on the porosity of the parapet

envelope. How is the porosity of the

parapet determined?


A: In the case of parapets, it is expected

most cases to have uniform porosity, so the

"enclosed" classification (+0.18, - 0.18)

would be appropriate. However, if the two

surfaces of the parapet are very different

(one has openings, the other is fully

sealed), the partially-enclosed case might

be relevant.


Q: What is the wind load on the parapet using

Method 1, Simplified Procedure?

A: There is no clear answer. Some

jurisdictions do not allow Method 1 to be

used for buildings with parapets.

PROVISION#3

Torsion, Torsional Irregularity and Direction

of Seismic Loading

11



QEδδδδxe

V

ASCE 7-05 12.8.4 Horizontal

Distribution of Forces

• Rigid diaphragms

– Seismic story shear is to be distributed to elements

of seismic-force-resisting system based on stiffness

of vertical-resisting elements

• Flexible diaphragms

– Seismic story shear is to be distributed to elements

of seismic-force-resisting system based on tributary

areas

• Torsion

– Torsional moment due to difference in location of center of

mass and center of resistance

must be considered for rigid diaphragms

• Accidental torsion

– For rigid diaphragms, must be included in addition to the

torsional moment

• Displacement of center of mass = 5% building

dimension perpendicular to direction of applied forces

ASCE 7-05 12.8.4 Horizontal

Distribution of Forces Failure – Torsion

1976 Philippines

12



Amplification of Torsion

For structures* assigned to SDC C, D, E, or F

without flexible diaphragm and with horizontal

irregularity Type 1a or 1b (Torsional Irregularity

or Extreme Torsional Irregularity), the

accidental torsion Mta at each floor level needs

to be amplified by a factor:

Ax = δmax

1.2δavg

2

≤ 3.0

*Not applicable to light-frame construction

Amplification of Torsion

δA and δB computed assuming Ax = 1.0

Adapted from ASCE 7-05 Figure 12.8-1

Torsional Irregularity Torsional Irregularity

Referenced in:

Section 12.3.3.4 – 25% increase in seismic forces in

connections in diaphragms and collectors

Table 12.6-1 – Permitted analytical procedure

Section 12.7.3 – 3-D structural model required

Section 12.8.4.3 – Amplification of accidental torsion

Section 12.12.1 – Design story drift based on largest difference in deflection

Section 16.2.2 - 3-D structural model required in nonlinear response history procedure

Extreme Torsional Irregularity Extreme Torsional Irregularity

Referenced in:

Section 12.3.3.1 – Prohibited in SDC E and F

Section 12.3.3.4 – 25% increase in seismic forces in connections in diaphragms and collectors

Section 12.3.4.2 (Table 12.3-3) – ρ = 1.3

Table 12.6-1 – Permitted analytical procedure

Section 12.7.3 – 3-D structural model required

Section 12.8.4.3 – Amplification of accidental torsion

Section 12.12.1 – Design story drift based on largest difference in deflection

Section 16.2.2 - 3-D structural model required in nonlinear response history procedure

13



Q: A: for Torsional Irregularity

Q: Do the torsional irregularity provisions

apply to light-frame constructions?

A: Most likely, no. The torsional irregularity

definition applies to diaphragms that are

rigid or semirigid, which is typically not the

case for light-frame construction.

Section 12.5.2 Direction of

Loading

12.5.2 SDC B.

The design seismic forces are permitted to be

applied independently in each of two

orthogonal directions and orthogonal

interaction effects are permitted to be

neglected.


Loading

12.5.3 SDC C.

Structures that have horizontal structural

irregularity Type 5 of Table 12.3-1, shall use

one of the following procedures.

ASCE 7-05 12.5.2 Direction of

Loading

12.5.3 SDC C.

a. Orthogonal Combination Procedure.

ELF, modal response spectrum, or linear

response history analysis, with loading applied

independently in any two orthogonal

directions…

100% + 30%


Loading

12.5.3 SDC C.

b. Simultaneous Application of Orthogonal

Ground Motion.

Linear or nonlinear response history

analysis, with orthogonal pairs of ground

motion acceleration histories applied

simultaneously.

14




Loading

12.5.4 SDC D, E or F.

The orthogonal combination procedure … shall

additionally be required for any column or

wall that forms part of two or more

intersecting seismic-force-resisting systems

and is subjected to axial load due to seismic

forces acting along either principal plan axis

equaling or exceeding 20% of the axial load

design strength of the column or wall.

PROVISION#4

REDUNDANCY

REDUNDANCY FACTOR, ρρρρ

ASCE 7-05 Section 12.3.4

Note that ρρρρ = 1.0 when the SIMPLIFIED

PROCEDURE of Section 12.14 is used.

SDC ρρρρ

A NA

B & C 1.0

D, E & F 1.0 or 1.3

REDUNDANCY FACTOR, ρρρρ


12.3.4 Redundancy

12.3.4.1 Conditions Where

Value of ρρρρ is 1.0

12.3.4.2 Redundancy

Factor, ρρρρ, for SDC

D through F

Section 12.3.4.1

ρ for SDC D - F

ρ = 1.0 for the following:

1. Structures assigned to SDC B and C.

2. Drift calculation and P-delta effects.

3. Design of nonstructural components.

4. Design of nonbuilding structures, not similar to buildings.

15



Section 12.3.4.1

ρ for SDC D - F

ρ = 1.0 for the following:

5. Design of collector elements, splices and their connections for which load combinations with overstrength are used.

6. Design of members or connections where load combinations with overstrength are required for design.

7. Diaphragm loads determined using Eq. 12.10-1.

8. Structures with damping systems designed in accordance with ASCE 7-05 Chapter 18.

Section 12.3.4.2

ρ for SDC D - F

ρ = 1.0 or 1.3

ρ = 1.3 unless ONE of two conditions is met.

If Condition # 1 is met, then ρρρρ = 1.0

If Condition #2 is met, then ρρρρ = 1.0

Both conditions do NOT need to be met

for ρρρρ = 1.0

Section 12.3.4.2

ρ for SDC D - F

CONDITION #1 and CONDITION #2 only need to

be checked at each story resisting more than

35% of the base shear.

Section 12.3.4.2

ρ for SDC D - F

CONDITION #1:

Can an individual element be removed from

the lateral-force-resisting-system without:

• Causing the remaining structure to suffer a reduction of story strength >

33%, or

• Creating an extreme torsional

irregularity?

TABLE 12.3-3 REQUIREMENTS FOR EACH STORY RESISTING MORE

THAN 35% OF THE BASE SHEAR

CONDITION #2

If a structure is regular in plan and there are at

least 2 bays of seismic force-resisting perimeter

framing on each side of the structure in each

orthogonal direction at each story resisting >

35% of the base shear.

Section 12.3.4.2

ρ for SDC D - F

16



Section 12.3.4.2

ρ for SDC D - F

Seismic Force-Resisting

Perimeter Framing

Two Bays

Q: A: for Redundancy

Q: How many bays are in a shear wall?

A: Length of shear wall/ story height…or for

light-framed construction (defined in

Section 11.2), 2 x length of shear wall/

story height


Q: Does the redundancy factor apply

to the design of foundations?

A: Yes.

Q: I am using “Condition #1” to determine

ρ for a wood-frame building. All of the

shear walls are relatively long; in other

words, the height of each shear wall (8’)

is less than its length (9’, 10’, 12’). Can I

assign ρ = 1.0 because there are no

shear walls with a h/l ratio > 1.0?


A: Yes. Q: A: for Redundancy

Q: We have a building that is 325 feet tall (31

stories) with shear walls. We are using

Condition #1 to determine ρ. When Table

12.3-3 uses the phrase “height-to-length”

ratio, is that the height-to-length ratio “within

any story”? Or is it referring to the overall

height-to-length ratio which, for our building,

would mean a height of 325 feet.

17




A: The h/l ratio is intended to be

story height-to-length ratio.

llll < 10 ft

h = 10 ft


Q: Can the value of ρ be different at

each level of the same building?


A: No, ρ cannot be different at each

level of the same building. However,

depending on the structural system, ρ

can be different in the two orthogonal

directions of the same building if

Condition #1 is used.


Q: Why doesn’t Table 12.3-3

address dual systems? If you have

a dual system, can you assume ρ =

1.0?

A: No….. Q: A: for Redundancy

“Braced frame, moment frame and shear wall

systems have to conform to redundancy

requirements. Dual systems are also included,

but in most cases are inherently redundant.

Shear walls with a height-to-length aspect ratio

greater than 1.0 have been included, even

though the issue has been essentially solved by

requiring collector elements and their connection

to be designed for Ω0 times the design force.”

18




Q: Do you need to determine the

redundancy factor for “nonbuilding

structures similar to buildings” or can you

assume the redundancy factor equals 1?


A: You need to determine the redundancy

factor. If the code did not intend that the

redundancy factor be determined for

“Nonbuilding structures similar to

buildings”, there would be an exception to

Section 15.5.1 as is done in Section 15.6.


Q: Does the redundancy factor need

to be determined if dynamic

analysis is used?

A: Yes.

REDUNDANCY EXAMPLE

a

a

Wall A

Stiffness Ka

Wall B

Stiffness Kb

Wall C

Stiffness Kc

Wall D

Stiffness Kd

Wall E

Stiffness Ke

Wall F

Stiffness Kf

Wall G

Stiffness Kg

Wall H

Stiffness Kh

REDUNDANCY EXAMPLE

• SDC D

• one story concrete shear wall building

• Ka = Kb = Kc = Kd = Ke = Kf = Kg = Kh = K

• All walls have the same nominal shear

strength, Vn.

• The story height is 18 feet.

• The length of each shear wall is 15 feet.

Let a denote the horizontal dimension of

this building.

19



REDUNDANCY EXAMPLE

CONDITION #1 CONDITION #2

Step 1:

Remove shear wall and see

if story strength is reduced

by more than 33%.

Step 2:

See if there is an extreme

torsional irregularity created.

Step 1:

Check if structure is regular

in plan.

Step 2:

Are there at least 2 bays

of….on each side in each

orthogonal direction?

REDUNDANCY EXAMPLE

Check Condition #2 first (it’s easier)

Q: How many bays are in a shear wall?

A: Length of shear wall/ story height…or

for light-framed construction, 2 x length

of shear wall/ story height.

For example: (15/18)(2) = 1.67 < 2

REDUNDANCY EXAMPLE

CONDITION #1:

Removal of a shear wall or wall pier with a

height-to-length ratio greater than 1.0 within

any story, or collector connections thereto,

would not result in more than a 33% reduction

in story strength, nor does the resulting system

have an extreme torsional irregularity

(horizontal structural irregularity Type 1b).

REDUNDANCY EXAMPLE

CONDITION #1

Step 1:

Remove shear wall and see if

story strength is reduced by

more than 33%.

Step 2:

See if there is an extreme

torsional irregularity created by

doing so.

REDUNDANCY EXAMPLE

a

a

Wall A

Stiffness Ka

Wall B

Stiffness Kb

Wall D

Stiffness Kd

Wall E

Stiffness Ke

Wall F

Stiffness Kf

Wall G

Stiffness Kg

Wall H

Stiffness Kh

REDUNDANCY EXAMPLE

Definition of Extreme Torsional

Irregularity in ASCE 7-05 Table 12.3-1:

Extreme Torsional Irregularity exists where

the maximum story drift, computed

including accidental torsion, at one end of

the structure transverse to an axis is more

than 1.4 times the average of the story

drifts at the two ends of the structure.

20



REDUNDANCY EXAMPLE

The torsional stiffness about the center

of rigidity (CR) is determined as:

=

REDUNDANCY EXAMPLE

The determination of extreme torsional

irregularity requires the evaluation of the

story drifts δa

and δb, as shown below.

a/3 2a/3

a/6

CRCM

V

δaδb

θ

REDUNDANCY EXAMPLE

According to ASCE 7-05 Table 12.3-1,

extreme torsional irregularity does not exist

when

This can be transformed to

REDUNDANCY EXAMPLE

Assume that the story drift caused only by

the lateral force V is equal to δ, and that θ

is the rotation caused by the torsion T, then

This ratio is less than 2.33 only if δ/(aθ) is

larger than 1.08.

REDUNDANCY EXAMPLE

Therefore, no extreme torsional irregularity is created

and ρ = 1.0.

(Note that the term 0.05a is for accidental torsion)

Thus, the horizontal structural irregularity Type 1b does

not exist and the configuration qualifies for a ρ factor

of 1.0.

PROVISION#5

SEISMIC ANALYSISPROCEDURE SELECTION

21



Seismic Analysis Procedure Selection

STATIC ANALYSIS

PROCEDURES

ASCE 7-05

SECTION

Simplified Design

Procedure

12.14

Equivalent Lateral Force

Procedure

12.8


DYNAMIC ANALYSIS

PROCEDURES

ASCE 7-05

SECTION

Modal Response

Spectrum Analysis

12.9

Linear Response History

Analysis

16.1

Nonlinear Response

History Analysis

16.2

Simplified Design Procedure

Exception to Section 12.1

EXCEPTION: As an alternative, the simplified design procedure of Section 12.14 is permitted to be used in lieu of the requirements of Sections 12.1 through 12.12, subject to all of the limitations contained in Section 12.14.

Note: Section 12.13 is Foundation Design

Simplified Design Procedure

Section 12.14

1. Occupancy Category I or II

2. Site Class A, B, C, or D3. Three stories or less in height

4. Bearing wall system or building

frame system5. through 12……

Q: A: for Simplified Design Procedure

Q: What are the benefits of using the

Simplified Design Procedure?

A: Here are the benefits:

• ρ = 1, Ωo = 2.5.

• No period (T) determination.

• No triangular distribution of seismic forces.

• Determination of Fa simplified; Ss need not

exceed 1.5g.

• Drift need not be calculated


Table 12.6-1

*

**

* 12.3-1

** 12.3-2

22




Table 12.6-1

If a building is assigned SDC D, E, or F

and has a T ≥ 3.5 Ts, then dynamic

analysis procedure must be used.

(Ts is the period at which the flat-top portion of

the response spectrum transitions to the

descending (period-dependent) branch.)



Table 12.6-1

Dynamic Analysis is required if a building

meets all of the following conditions:

SDC D, E, or F

Not of light-frame construction

Contains one of the following

irregularities:

“Torsional” or “Extreme Torsional”

“Stiffness-Soft Story”, “Stiffness – Extreme Soft

Story”, “Weight (Mass)” or “Vertical Geometric”



Table 12.6-1

All structures of light frame

construction, irrespective of height

– Dynamic analysis never required


Table 12.6-1

Occupancy Category I or II buildings of

other construction not exceeding two

stories in height

– Dynamic analysis not required

23




Q: In Table 12.6-1, it states “Regular

Structures with T < 3.5Ts and all structures of

light frame construction” are permitted to use

an Equivalent Lateral Force Analysis. Does

this mean that the building must meet both

conditions (regular with T < 3.5Ts and light

frame construction), or does only one of these

two characteristics need to be satisfied?


A: It does not mean "and." It means

"or." Regular Structures with T <

3.5Ts are permitted to use Equivalent

Lateral Force Analysis. All structures of

light frame construction, irrespective of

height, are also permitted to

use Equivalent Lateral Force Analysis.

PROVISION#6

DRIFT AND BUILDINGSEPARATION

QEδδδδxe

V

Drift Determination

Section 12.8.6Step 1: Determine δxe at each floor level

where δxe is the lateral deflection at floor

level x determined by elastic analysis

under code-prescribed seismic forces.

Step 2: Multiply δxe by Cd given in Table

12.12-1, the product representing the

estimated design earthquake

displacement.

Drift Determination

Section 12.8.6

Step 3: Divide δxeCd by I, Importance

Factor: δ x = δxeCd /I

Step 4: Determine design story drift:

∆x = δx – (δx – 1)

24



Drift Determination

Section 12.8.6 Story Drift

• ∆x = δx − δx-1 < ∆a

δx = Cd δxe / I

Cd = deflection amplification factor

Allowable Drift

StructureOccupancy Category

I or II III IV

Structures, other than masonry shear wall

structures, 4 stories or less with interior walls, partitions, ceilings, and exterior wall systems

that have been designed to accommodate the

story drift.

0.025hsx 0.020hsx 0.015hsx

Masonry cantilever shear wall structures 0.010hsx 0.010hsx 0.010hsx

Other masonry shear wall structures 0.007hsx 0.007hsx 0.007hsx

All other structures 0.020hsx 0.015hsx 0.010hsx

ASCE 7-05 Table 12.12-1

• For seismic force–resisting systems comprised

solely of moment frames in structures assigned

to Seismic Design Categories D, E, or F, the

design story drift shall not exceed ∆a /ρ for any

story. ρ shall be determined in accordance with

Section 12.3.4.2.

Allowable Drift – Additional

Requirement

ASCE 7-05 Section 12.12.1.1

Q: A: for Drift Determination

Q: Why is drift divided by the Importance

Factor?

A: Because the forces under which δxe

are computed are already amplified by I,

and the drift limits set forth in Table

12.12-1 are more restrictive for higher

occupancy category buildings.


Q: Is drift determined differently for

allowable stress design (ASD) than for

strength design (SD)?

A: No. The same procedure is used

regardless of whether ASD or SD is

used.

25




Q: Does minimum base shear need to be

considered for drift determination?

A: Yes, Section 12.8.6.1 requires that all

of the requirements of Section 12.8 be

satisfied for the purpose of computing

drift.


Q: Does the upper-bound limitation on

period T need to be considered for drift

determination?

A: No. Section 12.8.6.2 does not require

the period to be subject to the upper

limit of CuTa for the purpose of drift

determination.

Building Separation

Section 12.12.3

12.12.3 Building Separation. All portions

of the structure shall be designed and

constructed to act as an integral unit in

resisting seismic forces unless

separated structurally by a distance

sufficient to avoid damaging contact

under total deflection as determined in

Section 12.8.6.

Building Separation

Section 12.12.3

This section applies to SDCs B

through F.

Does not address adjacent

buildings on the same property.

Does not address minimum

setback distance from property line.

Building Separation

Section 12.12.3

Code Requirement2006 IBC/

ASCE 7-05

2009 IBC/

ASCE 7-05

PORTIONS OF THE SAME STRUCTURE:

All portions of the structure shall be designed

and constructed to act as an integral unit in

resisting seismic forces unless separated

structurally by a distance sufficient to avoid

damaging contact under total deflection (δx) as

determined in Section 12.8.6.

ASCE 7-05

Section 12.12.3

(applies to all

SDCs)

ASCE 7-05

Section 12.12.3

(applies to all

SDCs)

Continued on next page

Building Separation

2009 IBC Section 1613.6.7

Code Requirement2006 IBC/

ASCE 7-052009 IBC/

ASCE 7-05

BUILDING SEPARATIONS (paraphrased):

All structures shall be separated from adjoining structures.

Separations shall allow for the maximum inelastic

displacement δM (including torsion).

Adjacent buildings on the same property shall be separated

by at least δMT where

When a structure adjoins a property line not common to a

public way, that structure shall also be set back from the

property line by at least the displacement, δM, of that

structure.

Exception: Smaller separations or property line setbacks

shall be permitted when justified by rational analyses based

on maximum expected ground motions.

No

requirement

2009 IBC

Section

1613.6.7

26



Building Separation


Note difference between δx and δM

The deflections of Level x at the center of

mass (12.8.6),

δx = Cd δxe / I

δM = Cd δmax / I (Equation 16-44)

δmax = the maximum displacement at Level x

computed assuming Ax = 1 (12.8.4.3)

Building Separation


Building Separation


Separation of Two Adjacent Buildings

Building Separation


10″ separation

δM1 at edge = 6″

δM2 at adjacent edge = 8″

Q:A: for Building Separation

Q: ASCE 7-05 Section 12.12.3 contains the

language “sufficient to avoid damaging

contact.” What is damaging contact?

A: To avoid any contact at all, the separation

distance would have to be the arithmetic sum of

δM1 and δM2. To avoid damaging contact, ASCE

7 allows the separation distance to be the

statistical sum of δM1 and δM2, which is less than

the arithmetic sum.


Q: I do not understand the logic of

requiring less separation between two

buildings on the same property than

between two identical buildings on different

sides of the property line.

27




A: The first provision is concerned with

damaging contact from pounding of

buildings belonging to presumably the

same owner. The property line setback

requirement is based on consideration

that one owner should not encroach

onto another property.

PROVISION#7

R, Cd and Ω0 Values forHorizontal and Vertical

Combinations

R, Cd and Ωo Values for Horizontal

and Vertical Combinations

Horizontal Combinations can be

either….

• In different directions

• In same direction

Section 12.2.2 Combinations of

Framing Systems in Different

Directions

• Different seismic force-resisting systems

may be used to resist seismic forces along

each of two orthogonal plan axes.

• The respective R, Cd, and Ωo coefficients

shall apply to each system, including the

limitations on system use contained in

Table 12.2-1.


Framing Systems in Different

Directions

R =5Cd = 5Ωo = 2½

R = 8, Cd = 5 ½, Ωo = 3


Framing Systems in the Same

DirectionWhere different seismic force-resisting systems

are used in combination to resist seismic

forces in the same direction of structural

response, other than those combinations

considered as dual systems, the more

stringent system limitation contained in Table

12.2-1 shall apply and the design shall comply

with the requirements of this section.

28



12.2.3.1 Vertical Combinations

R: Cannot increase as you go down

Cd and ΩΩΩΩ0: Cannot decrease as you go down

R=8, Cd=5.5, Ω0=3

R=5, Cd=5, Ω0=2.5

8, 5.5, 3

8, 5.5, 3

8, 5.5, 3

5, 5.5, 3

5, 5.5, 3

5, 5.5, 3

R=5, Cd=5, Ω0=2.5

R=8, Cd=5.5, Ω0=3

5, 5, 2.5

5, 5, 2.5

5, 5, 2.5

5, 5.5, 3

5, 5.5, 3

5, 5.5, 3

R=5, Cd=5, Ω0=2.5

R=8, Cd=5.5, Ω0=3

5, 5.5, 3

5, 5.5, 3

5, 5.5, 3

5, 5.5, 3

R=8, Cd=5.5, Ω0=3 8, 5.5, 3

8, 5.5, 3

12.2.3.2 Horizontal Combinations

12.2.3.2 Where a combination of different structural systems is utilized to resist lateral

forces in the same direction, value of R used for design in that direction shall not be greater than the least value of R for any of the systems

utilized in that direction.


12.2.3.2 Resisting elements are permitted to be designed using the least value of R for the

different structural systems found in each independent line of resistance if the following three conditions are met: 1) Occupancy Category

I or II building, 2) two stories or less in height, and 3) use of light frame construction or flexible diaphragms.

The value of R used for design of diaphragms in such structures shall not be greater than the least

value for any of the systems utilized in that same direction.


The deflection amplification factor, Cd, and the

system overstrength factor, ΩΩΩΩ0 , in the direction

under consideration at any story shall not be

less than the largest value of this factor for the

R factor used in the same direction being

considered.


The second paragraph of ASCE 7-05 Section

12.2.3.2 is far from clear.

One possible interpretation is that when different

structural systems are combined in the same

direction of a building or other structure, the

largest Cd- and ΩΩΩΩ0-values of all the individual

structural systems shall be used.


The other possible interpretation is that the Cd-

and ΩΩΩΩ0-values shall correspond to the least R-

value of all the individual structural systems.

The second interpretation appears to be the

more logical in view of the following example

(discussion is continued in terms of Cd alone).

29




Consider a rather extreme example where a

prestressed masonry shear wall (R = 1.5, Cd =

1.75) is combined with a special steel moment-

resisting frame (R = 8, Cd = 5.5). There is no

question that the R-value is 1.5. The question is

whether the Cd-value is 1.75 or 5.5. 5.5 does not

seem logical – for two reasons.


First, the combined system is much more rigid

than the special steel moment frame itself.

Until the prestressed masonry shear wall

hinges at its base, which is extremely unlikely

in view of the large design forces that would

result from an R = 1.5, large inelastic

displacements do not seem to be possible.


Second, large values of δδδδxe would automatically

result from the low value of R used in design.

These, multiplied by the Cd of 5.5 would yield

unrealistically large total displacements. Cd of

1.75 appears to be much more logical.

This second interpretation was implicit in the

1997 Uniform Building Code, where 0.7R was

used in place of Cd.


The proposed rewrite provides clarification of the

second paragraph of ASCE 7-05 Section

12.2.3.2. The rewrite also offers clarification

concerning another complication that may

arise, which is that different structural systems

having the same R-value sometimes have

different Cd- and ΩΩΩΩ0-values.

12.2.3.2 Horizontal CombinationsTable 12.2.3.2 R, Cd, and Ωo Values for Combination of

Different Structural Systems Used in Same Direction

R value The least value of R for any of the systems used.

Exception: Resisting elements are permitted to be designed using the

least value of R for the different structural systems found in each

independent line of resistance if the following three conditions are met: 1)

Occupancy Category I or II building, 2) two stories or less in height, and 3)

use of light frame construction or flexible diaphragms.

Cd value The Cd value corresponding to the system with the least value of R for any

of the systems used. In the case where two or more systems have the

same least value of R, the largest of the corresponding values of Cd shall be

used.

Ωo value The Ωo value corresponding to the system with the least value of R for any

of the systems used. In the case where two or more systems have the

same least value of R, the largest of the corresponding values of Ωo shall be

used.

Vertical combinations

R: Cannot increase as you go downCd and ΩΩΩΩ0: Always correspond to R

ASCE 7-10 Section 12.2.3.1 Vertical Combinations

R = 8, Cd = 5.5, Ω0 = 3

R = 5, Cd = 5, Ω0 = 2.5

8, 5.5, 3

8, 5.5, 3

8, 5.5, 3

5, 5, 2.5

5, 5, 2.5

5, 5, 2.5

R = 5, Cd = 5, Ω0 = 2.5

R = 8, Cd = 5.5, Ω0 = 3

5, 5, 2.5

5, 5, 2.5

5, 5, 2.5

5, 5, 2.5

5, 5, 2.5

5, 5, 2.5

R = 5, Cd = 5, Ω0 = 2.5

R = 8, Cd = 5.5, Ω0 = 3

5, 5, 2.5

5, 5, 2.5

5, 5, 2.5

5, 5, 2.5

R = 8, Cd = 5.5, Ω0 = 3 8, 5.5, 3

8, 5.5, 3

30



12.2.4 Combination Framing

Detailing Requirements

Structural components common to different

framing systems used to resist seismic

motions in any direction shall be designed

using the detailing requirements of Chapter 12

required by the highest response modification

coefficient, R, of the connected framing

systems.

Q: A: for Combinations

Q: I am designing a building that has a combination of special

reinforced masonry shear walls and special steel braced frames

in the same direction. R values are 5.5 and 6, respectively. I

understand that I should design the building with the smaller R

= 5.5 for both the masonry shear walls and steel braced frames

in this direction for seismic design in accordance with ASCE 7-

05 Section 12.2.3.2. But someone told me that the building

should be designed by analyzing the entire building twice: use

R = 5.5 for the entire building to analyze and design the

masonry shear walls and use R = 6 for the entire building to

design the steel frames. I don't think this is right. What is your

thought on this?

Q: A: for Combinations

A: What you understand is correct. The latter

interpretation is unfamiliar and incorrect.

PROVISION#8

MINIMUM SEISMICBASE SHEAR

CodeMaster

2009 IBC Seismic Design

ASCE 7-05 12.8.1

Design Base Shear

31



ASCE 7-05 12.8 Equivalent Lateral

Force Procedure Revisions to ASCE 7-05 Seismic

Provisions

Building CodeMinimum Base Shear

Applicable in All SDCs

1997 UBC Vmin = 0.11 Ca I W

2000 & 2003 IBC Vmin = 0.044 SDS I W

2006 IBC & ASCE 7-05 w/ Supplement No. 1 Vmin = 0.01 W

2009 IBC & ASCE 7-05 w/ Supplement Nos. 1

and 2

Vmin = 0.044 SDS I W ≥

0.01W

ASCE 7-05 w/ Supplement No. 2

Design Base ShearSection 1613

Earthquake Loads

• Section 1613.1 references ASCE 7

• Chapter 35 entry for ASCE 7 reads as

follows:

ASCE 7-05 Minimum Design Loads for Buildings

and Other Structures including Supplements No. 1 and 2, excluding Chapter 14 and Appendix 11A

ASCE 7-05 Including

Supplement No. 1

“Including Supplement No. 1”

ASCE 7-05 Supplement No. 2

Supplement No. 2 modifies Eqs. 12.8-5,

15.4-1 and 15.4-3 as shown below:

CS = 0.01 0.044SDSI ≥ 0.01 (Eq. 12.8-5) [applicable

to buildings]


to nonbuilding structures not similar to buildings]


to an exception for nonbuilding structures not

similar to buildings]

32



Minimum Seismic Base Shear

vs. Ground Motion

0.044SDSI ≥ 0.01 or SDSI ≥ 0.227

or SS ≥ value given in table

Site Class I = 1 I = 1.25 I = 1.5

A 0.426 0.341 0.284

B 0.341 0.273 0.227

C 0.284 0.227 0.189

D 0.213 0.170 0.142

E 0.136 0.109 0.091



For special reinforced concrete moment

frames,

0.016h0.9 ≥ 9/8

h ≥ (70.31)1/0.9 = 113 ft

Minimum Seismic Base Shear -

Example

Concrete SMRF building - R = 8, I = 1.0

Height = 120 ft, SDS = 1.00, SD1 = 0.40

Ta = 0.016(120)0.9 = 1.19 sec

Governs

Does not govern

ASCE 7-05 Supplement No. 2

Q: Where does ASCE officially announce ASCE 7-

05 Supplement No. 2?

A: http://content.seinstitute.org/files/pdf/

SupplementNo2ofthe2005EditionofASCE7.pdf

Q: A: for ASCE 7-05 Supplement

No. 2

Q: Does ASCE 7-05 Supplement No. 2

minimum base shear need to be considered

for drift determination?

A: Yes, Section 12.8.6.1 requires that all of the

requirements of Section 12.8 be satisfied for

the purpose of computing drift.

33



Q: A: for ASCE 7-05 Supplement

No. 2

Q: Is ASCE 7-05 going to be published with

errata and Supplement No. 2 incorporated?

A: “Additional printings will, to the extent

possible, include as extra pages the

supplements and errata. However there's a

conscious decision not to integrate them

directly into the text so as to minimize

confusion between one book and another.”

PROVISION#9

FLEXIBLE VS. RIGIDDIAPHRAGMS

2006 and 2009 IBC Section 1602

Definition for Diaphragms

DIAPHRAGM

Diaphragm, blocked

Diaphragm boundary

Diaphragm chord

Diaphragm, flexible

Diaphragm, rigid

Diaphragm, unblocked

FLEXIBLE DIAPHRAGMS

Prescriptive Approach

&

Calculation Approach

2006 and 2009 IBC Section 1602

Definition for Flexible Diaphragm

Diaphragm, flexible.

A diaphragm is flexible for the purpose of distribution

of story shear and torsional moment where so

indicated in Section 12.3.1 of ASCE 7, as modified in

Section 1613.6.1.

12.3.1.1 Flexible Diaphragm

Condition. Diaphragms constructed of untopped

steel decking or wood structural panels are

permitted to be idealized as flexible in structures

in which the vertical elements are steel or

composite steel and concrete braced frames, or

concrete, masonry, steel, or composite shear

walls. Diaphragms of wood structural panels or

untopped steel decks in one- and two-family

residential buildings of light-frame construction

shall also be permitted to be idealized as flexible.


Definition for Flexible Diaphragm (Prescriptive)

34



12.3.1.3 Calculated Flexible Diaphragm Condition. Diaphragms … are permitted to be

idealized as flexible where the computed maximum in-plane deflection of the diaphragm under lateral load is more than two times the

average story drift of adjoining vertical elements of the seismic force –resisting system of the associated story under equivalent

tributary lateral load as shown in Fig. 12.3-1.


Definition for Flexible Diaphragms by Calculation

ASCE 7-05 Figure 12.3-1


by Calculation

MAXIMUM DIAPHRAGM

DEFLECTION (MDD)

AVERAGE DRIFT OF

VERTICAL ELEMENT(ADVE)

Note: Diaphragm is flexible If MDD > 2 (ADVE).

S

De

SEISMIC LOADING

2006 and 2009 IBC Section 1613.6.1

Definition for Flexible Diaphragm(Prescriptive)

1613.6.1 Assumption of flexible diaphragm. Add the following text at the

end of Section 12.3.1.1 of ASCE 7:

Diaphragms constructed of wood structural

panels or untopped steel decking shall

also be permitted to be idealized as flexible, provided four given conditions

are met…

2006 and 2009 IBC Section 1613.6.1


Condition #1:

Toppings of concrete or similar materials

are not placed over wood structural

panel diaphragms except for

nonstructural toppings no greater than 1

½ inches thick.

2006 and 2009 IBC Section 1613.6.1


Condition #2:

Each line of vertical elements of the

lateral force-resisting system complies

with the allowable story drift of Table

12.12-1.

2006 and 2009 IBC Section 1613.6.1


Condition #3:

Vertical elements of the lateral-force-

resisting system are light-framed walls

sheathed with wood structural panels

rated for shear resistance or steel

sheets.

35



2006 and 2009 IBC Section 1613.6.1


Condition #4:

Portions of wood structural panel diaphragms that

cantilever beyond the vertical elements of the lateral-

force-resisting system are designed in accordance

with [2006 IBC: Section 2305.2.5] [2009 IBC: Section

4.2.5.2 of AF&PA SDPWS].


Definition for Rigid Diaphragm

(Prescriptive)

12.3.1.2 Rigid Diaphragm

Condition. Diaphragms of concrete slabs or

concrete filled metal deck with span-to-depth

ratios of 3 or less in structures that have no

horizontal irregularities are permitted to be

idealized as rigid.

2006 and 2009 IBC Section 1602.1

Definition for Rigid Diaphragm

(Calculation)

Diaphragm, rigid

A diaphragm is rigid for the purpose of

distribution of story shear and torsional

moment when the lateral deformation of the

diaphragm is less than or equal to two times

the average story drift.

ASCE 7-05 Figure 12.3-1

Definition for Diaphragm

MAXIMUM DIAPHRAGM

DEFLECTION (MDD)

AVERAGE DRIFT OF


Note: Diaphragm is flexible If MDD > 2 (ADVE).

S

De

SEISMIC LOADING

Note: Per 2009 IBC Section 1602.1, diaphragm is rigid if MDD ≤ 2(ADVE)


Definition for Diaphragms

12.3.1 Diaphragm Flexibility….Unless a

diaphragm can be idealized as either flexible

or rigid in accordance with Sections 12.3.1.1,

12.3.1.2, or 12.3.1.3, the structural analysis

shall explicitly include consideration of the

stiffness of the diaphragm (i.e. semirigid

modeling assumption).

Is diaphragm wood structural panels or

untopped steel decking?

Is any of the following true?

1- & 2-family dwelling of

light-frame construction

Vertical elements one of the following:

•Steel braced frames

•Composite steel and concrete braced

frames

•Concrete, masonry, steel or composite

shear walls

Four conditions in 2009

IBC Section 1613.6.1

are met

Is diaphragm• Concrete slab?

• Concrete filled metal deck?

Is span-to-depth ratio ≤ 3 and no horizontal

irregularities?

Assume Rigid

Assume Flexible

See Next Slide

Y

N

NY

Y

Y

N

N

ST

AR

T

36



MAXIMUM DIAPHRAGM

DEFLECTION (MDD)

AVERAGE DRIFT OF


Is MDD > 2 (ADVE)?

S

De

SEISMIC LOADING

Assume Flexible

Y

Assume Rigid

N

PROVISION#10

Special Seismic Load Combinations

2006 IBC 1605.4 Special Seismic

Load Combinations

Section 1605.4 is deleted in its entirety in

the 2009 IBC.

SPECIAL SEISMIC LOAD COMBINATIONS

is replaced with

LOAD COMBINATIONS WITH OVERSTRENGTH FACTORS

of ASCE 7-05

Why Was 2006 IBC Section 1605.4

Deleted?

To eliminate a disconnect between IBC and

ASCE 7-05:

– 2006 IBC Section 1605.4 had one set of

“special seismic load combinations”

applicable to both ASD and strength design.

– ASCE 7-05 has two sets of “load combinations

with overstrength factors”…one for ASD and

one for strength design.

Why Was 2006 IBC Section 1605.4

Deleted?

To eliminate a disconnect between IBC and ASCE 7-

05 (cont.):

– 2006 IBC had separate, unique load

combinations that were to be applied where

specifically required.

– ASCE 7-05 prescribes an equation for Em that

is to be used in ASCE 7-05 Ch. 2 load

combinations.

37



1605.4 Special Seismic Load

Combinations (2006 IBC)

• 1.2D + f1L + Em (Equation 16-22)

• 0.9D + Em (Equation 16-23)

Em = ΩΩΩΩ0 QE + 0.2 SDSD, while

E = ρρρρQE + 0.2 SDSD

12.4.3.2 Load Combinations with

Overstrength Factor (ASCE 7-05)

Basic Combinations for Strength Design

with Overstrength Factor

(1.2 + 0.2SDS)D + ΩΩΩΩ0QE + L + 0.2S

(0.9 − 0.2SDS)D + ΩΩΩΩ0QE + 1.6H


Overstrength Factor (ASCE 7-05)

Basic Combinations for Allowable Stress Design

with Overstrength Factor

(1.0 + 0.14SDS)D + H + F + 0.7ΩΩΩΩ0QE

(1.0 + 0.105SDS)D + H + F + 0.525ΩΩΩΩ0QE + 0.75L

+ 0.75(Lr or S or R)

(0.6 − 0.14SDS)D + 0.7ΩΩΩΩ0QE + H


Overstrength Factor (ASCE 7-05)Where allowable stress design methodologies are

used, allowable stresses are permitted to be

determined using an allowable stress increase of

1.2.

This increase shall not be combined with increases

in allowable stresses or load combination

reductions …except that combination with the

duration of load increases permitted in AF&PA

NDS is permitted.

1617.1.2 Maximum Seismic Load

Effect, Em (2000, 2003 IBC)

Where allowable stress design

methodologies are used with the special

load combinations of Section 1605.4,

design strengths are permitted to be

determined using an allowable stress

increase of 1.7 ….

What Takes the Place of Deleted

2006 IBC Section 1605.4?

New language in 2009 IBC

Section 1605.1.

38



What Takes the Place of Deleted

2006 IBC Section 1605.4?

1605.1 General. Buildings and other structures and

portions thereof shall be designed to resist:

1. The load combinations specified in Section 1605.2,

1605.3.1 or 1605.3.2,

2. The load combinations specified in Chapters 18 through

23, and

3. The load combinations with overstrength factor

specified in Section 12.4.3.2 of ASCE 7 where required

by Section 12.2.5.2, 12.3.3.3 or 12.10.2.1 of ASCE 7.

With the simplified procedure of ASCE 7 Section 12.14,

the load combinations with overstrength factor of

Section 12. 14.3.2 of ASCE 7 shall be used.

Q: A: for Load Combinations with

Overstrength Factor

Q: Section 1605.1 of the 2009 IBC requires

buildings and other structures and portions

thereof to be designed to resist the load

combinations with overstrength factor

specified in Section 12.4.3.2 of ASCE 7-05

where required by Section 12.2.5.2, 12.3.3.3, or

12.10.2.1. Can you elaborate?


Overstrength Factor

Cantilever Column Systems 12.2.5.2

SDC B - F

A:

Foundation and other elements used to provide overturning resistance at the base of cantilever column elements shall have the strength to resist the load combinations with over strength factor of Section 12.4.3.2.


Overstrength Factor

A: Elements Supporting

Discontinuous Walls or Frames

12.3.3.3

SDC B - F


Overstrength FactorQ: A: for Load Combinations with

Overstrength Factor

39




Overstrength Factor

A: Collector Elements 12.10.2.1 (SDC C - F)


Overstrength Factor

Load Combinations with

Overstrength Factor

Chapter 18 References:

1810.3.6.1 Splices of deep foundation elements,

SDC C through F

1810.3.9.4 Seismic reinforcement, SDC C and above, Exception 3

1810.3.11.2 Deep foundation element resistance to uplift forces, SDC D through F

1810.3.12 Grade beams, SDC D through F

Thank You!!

For more information…


Main Office

334 East Colfax Street, Unit E

Palatine, IL 60067

Phone: (847) 991-2700

Fax: (847) 991-2702

Email: [email protected]

West Coast Office

43 Vantis Drive

Aliso Viejo, CA 92656

Phone: (949) 215-6560

Email: [email protected]

2005 Edition of ASCE 7 Minimum Design Loads for Building and Other Structures

Supplement No.2

Supplement No. 2 of ASCE 705 revises the minimum base shear equations for both buildings and nonbuilding structures. The need for this change was indicated by the results from the 75% Draft of ATC63, Quantification of Building System Performance and Response Parameters, which indicate that tall buildings may fail at an unacceptably low seismic level and therefore the minimum base shear equation for buildings is being restored to that which appeared in the 2002 edition of ASCE 7.

Because nonbuilding structures not similar to buildings have low Rvalues compared to the special reinforced concrete moment frames studied in ATC63, the ASCE 7 standards committee chose not to restore the high minimum base shears for nonbuilding structures not similar to buildings found in ASCE 702. In many cases, these previous minimum base shears gave many nonbuilding structures not similar to buildings effective Rvalues less than 1.0. Therefore, the Seismic Subcommittee believes that the minimum base shear equation of 0.044SDSI used for buildings should also be applied to nonbuilding structures not similar to buildings.

Supplement No. 2 modifies three equations of the standard (Eq. 12.85, 15.41 and 15.4 3) as shown below:

Supplement No. 2 to ASCE 705:

Revise Equation 12.85 of Section 12.8.1.1 of ASCE 705 as shown below:

12.8.1.1 Calculation of Seismic Response Coefficient. The seismic response coeffi cient, Cs, shall be determined in accordance with Eq. 12.82.

=

I R S

C DS s (Eq. 12.82)

where:

SDS = the design spectral response acceleration parameter in the short period range as determined from Section 11.4.4

R = the response modification factor in Table 12.21, and

I = the occupancy importance factor determined in accordance with Section 11.5.1

The value of Cs computed in accordance with Eq. 12.82 need not exceed the following:

L 1 D

s T T for

I R T

S C ≤

= (Eq. 12.83)

L 2

L 1 D s T T for

I R T

T S C >

= (Eq. 12.84)

Cs shall not be less than

Cs = 0.01 0.044SDSI ≥ 0.01 (Eq. 12.85)

In addition, for structures located where S1 is equal to or greater than 0.6g, Cs shall not be less than

=

I R S . C s 1 5 0 (Eq. 12.86)

where I and R are as defined in Section 12.8.1.1 and

SD1 = the design spectral response acceleration parameter at a period of 1.0 sec, as determined from Section 11.4.4

T = the fundamental period of the structure (sec) determined in Section 12.8.2

TL = longperiod transition period (sec) determined in Section 11.4.5 S1 = the mapped maximum considered earthquake spectral

response acceleration parameter determined in accordance with Section 11.4.1

Revise Equations 15.41 and 15.42 of Section 15.4.1, item 2, as shown below:

2. For nonbuilding systems that have an R value provided in Table 15.42, the seismic response coefficient (Cs) shall not be taken less than

Cs = 0.03 0.044SDSI ≥ 0.03 (15.41)

and for nonbuilding structures located where S1 ≥ 0.6g, Cs shall not be taken less than

=

I R S C s 1 8 . 0 (15.42)

EXCEPTION: Tanks and vessels that are designed to AWWA D100, AWWA D103, API 650 Appendix E, and API 620 Appendix L as modified by this standard, shall be subject to the larger of the minimum base shear values defined by the reference document or the following equations:

Cs = 0.01 0.044SDSI ≥ 0.01 (15.43)

and for nonbuilding structures located where S1 ≥ 0.6g, Cs shall not be taken less than

=

I R S C s 1 5 . 0 (15.44)

Minimum base shear requirements need not apply to the convective (sloshing) component of liquid in tanks.

s. k. ghosh associates inc. - seaooseaoo.org/downloads/presentations_conf/2010_presentation_1... ·...

Documents