Wind Load Asce 7-05 Presentation

1S. K. Ghosh Associates Inc.

www.skghoshassociates.com

FREQUENTLY MISUNDERSTOOD IBC/ASCE 7 STRUCTURAL PROVISIONS

S. K. Ghosh and Susan DowtyS. K. Ghosh Associates Inc.

Palatine, IL and Aliso Viejo, CA


All sections referenced are from ASCE 7-05, unless otherwise noted.

PROVISION#1

Enclosure ClassificationFor Wind Design

Enclosure ClassificationSection 6.2 Definitions

Open Partially Enclosed (can experience

ballooning or suction effects caused by the build-up of internal pressure)

Enclosed

Internal Pressure




OPENA building having each wall at least 80% open.

Ao 0.8 Agfor EACH side of the building


AO = A1 + A2 + A3 Ag = H W



Do stacks of hay obstruct flow of wind?


PARTIALLY ENCLOSED1.Ao 1.10Aoi2.Ao > 4 sq ft AND > 0.01Ag3. 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 AoAg = 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




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.


ENCLOSEDA building that does not qualify as OPEN or PARTIALLY ENCLOSED.


Enclosure ClassificationFigure 6-5 Internal Pressure Coefficients,

GCpi



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



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 + 2zh Wp

13.1.3Nonstructural Component Importance Factor, Ip

Ip is based on 1. Whether component must function after

the design earthquake or2. Occupancy Category or3. 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

hFloor

Acceleration Distribution

Explanation of Fp Equation

Z = h for parapet design



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


Weight of the parapet per linear foot isWp = 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 )WpWhere Ip = component importance factor = 1.0


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


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


Wind Forces on Parapets

ASCE 7-05 Figure C6-12




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.


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 CriteriaLocation of building: Los Angeles, CaliforniaV = 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)



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/ = 2.01(70.5/1200)2/7 = 0.894 (from ASCE 7-05 Table 6-3)

( = 7, zg = 1200 ft for Exposure B from ASCE 7-05 Table 6-2)

Velocity Pressure Exposure Coefficient, Kz

Topographic Effect Factor, KztWind 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)

Effects of Parapets on MWFRS loadsqp = 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

Effects of Parapets on MWFRS loadsAt 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 ComponentASCE 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



Effective wind area of the parapet:Span = 3 ftWidth = 66 ft (> span/3)A = 3x66 = 198 ft2

External Pressure Coefficient, GCpLoad 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

External Pressure Coefficient, GCp

Load Case A (ASCE 7-05 Section 6.5.12.4.4)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)

External Pressure Coefficient, GCpLoad Case A (ASCE 7-05 Section 6.5.12.4.4)

GCp = 0.68 (-1.76) = 2.44GCpi = -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

External Pressure Coefficient, GCpLoad Case B (ASCE 7-05 Section 6.5.12.4.4)

Negative wall GCp = -0.76 (Figure 6-17 Zone 4)= -1.23 (Figure 6-17 Zone 5)

Applied to front surface of the parapet


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S. K. Ghosh Associates Inc.


Load Case B (ASCE 7-05 Section 6.5.12.4.4)

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?

Q: A: for Wind Design of ParapetA: 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: A: for Wind Design of Parapet

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

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QE xe

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 resistancemust 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

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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 = max

1.2avg

2

3.0

*Not applicable to light-frame construction

Amplification of Torsion

A and B 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 collectorsTable 12.6-1 Permitted analytical procedureSection 12.7.3 3-D structural model requiredSection 12.8.4.3 Amplification of accidental torsionSection 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 FSection 12.3.3.4 25% increase in seismic forces in connections in diaphragms and collectorsSection 12.3.4.2 (Table 12.3-3) = 1.3Table 12.6-1 Permitted analytical procedureSection 12.7.3 3-D structural model requiredSection 12.8.4.3 Amplification of accidental torsionSection 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

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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.


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%


12.5.3 SDC C.b. Simultaneous Application of Orthogonal

Ground Motion.

Linear or nonlinear response historyanalysis, with orthogonal pairs of groundmotion acceleration histories appliedsimultaneously.

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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, ASCE 7-05 Section 12.3.4

Note that = 1.0 when the SIMPLIFIED PROCEDURE of Section 12.14 is used.

SDC A NA

B & C 1.0

D, E & F 1.0 or 1.3

REDUNDANCY FACTOR, ASCE 7-05 Section 12.3.4

12.3.4 Redundancy12.3.4.1 Conditions Where

Value of is 1.012.3.4.2 Redundancy

Factor, , for SDC D through F

Section 12.3.4.1 for SDC D - F

= 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.

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= 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.


= 1.0 or 1.3

= 1.3 unless ONE of two conditions is met.

If Condition # 1 is met, then = 1.0

If Condition #2 is met, then = 1.0

Both conditions do NOT need to be met

for = 1.0


CONDITION #1 and CONDITION #2 only need to be checked at each story resisting more than 35% of the base shear.


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 #2If a structure is regular in plan and there are at

least 2 bays of seismic force-resisting perimeterframing on each side of the structure in each orthogonal direction at each story resisting > 35% of the base shear.


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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 heightor 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 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 = 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 . 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.

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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 be different at each level of the same building?


A: No, cannot be different at each level of the same building. However, depending on the structural system, can be different in the two orthogonal directions of the same building if Condition #1 is used.


Q: Why doesnt Table 12.3-3 address dual systems? If you have a dual system, can you assume = 1.0?

A: No.. Q: A: for RedundancyBraced 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 0 times the design force.

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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 AStiffness Ka

Wall BStiffness Kb

Wall CStiffness Kc

Wall DStiffness Kd

Wall EStiffness Ke

Wall FStiffness Kf

Wall GStiffness Kg

Wall HStiffness 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.

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REDUNDANCY EXAMPLECONDITION #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 (its easier)Q: How many bays are in a shear wall?A: Length of shear wall/ story heightor

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 EXAMPLECONDITION #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 AStiffness Ka

Wall BStiffness Kb

Wall DStiffness Kd

Wall EStiffness Ke

Wall FStiffness Kf

Wall GStiffness Kg

Wall HStiffness Kh

REDUNDANCY EXAMPLE

Definition of Extreme TorsionalIrregularity 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.

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REDUNDANCY EXAMPLE

The torsional stiffness about the center of rigidity (CR) is determined as:

=

REDUNDANCY EXAMPLEThe determination of extreme torsional

irregularity requires the evaluation of the story drifts a and b, as shown below.

a/3 2a/3

a/6CR

CM

V

a b

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 , and that is the rotation caused by the torsion T, then

This ratio is less than 2.33 only if /(a) is larger than 1.08.

REDUNDANCY EXAMPLE

Therefore, no extreme torsional irregularity is created and = 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 factor of 1.0.

PROVISION#5

SEISMIC ANALYSISPROCEDURE SELECTION

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Seismic Analysis Procedure Selection

STATIC ANALYSISPROCEDURES

ASCE 7-05SECTION

Simplified DesignProcedure

12.14

Equivalent Lateral Force Procedure

12.8

Seismic Analysis Procedure SelectionDYNAMIC ANALYSISPROCEDURES

ASCE 7-05SECTION

Modal Response Spectrum Analysis

12.9

Linear Response History Analysis

16.1

Nonlinear Response History Analysis

16.2

Simplified Design ProcedureException 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 ProcedureSection 12.14

1. Occupancy Category I or II2. Site Class A, B, C, or D3. Three stories or less in height4. Bearing wall system or building

frame system5. through 12

Q: A: for Simplified Design ProcedureQ: What are the benefits of using the

Simplified Design Procedure?

A: Here are the benefits: = 1, o = 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

Seismic Analysis Procedure SelectionTable 12.6-1

*

**

* 12.3-1** 12.3-2

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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.)



Dynamic Analysis is required if a building meets all of the following conditions:

SDC D, E, or FNot of light-frame constructionContains one of the following

irregularities:Torsional or Extreme TorsionalStiffness-Soft Story, Stiffness Extreme Soft Story, Weight (Mass) or Vertical Geometric


Seismic Analysis Procedure Selection Table 12.6-1

All structures of light frame construction, irrespective of height

Dynamic analysis never required

Seismic Analysis Procedure Selection Table 12.6-1

Occupancy Category I or II buildings of other construction not exceeding two stories in height

Dynamic analysis not required

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Seismic Analysis Procedure SelectionQ: 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

QE xe

V

Drift DeterminationSection 12.8.6

Step 1: Determine xe at each floor level where xe is the lateral deflection at floor level x determined by elastic analysis under code-prescribed seismic forces.

Step 2: Multiply xe by Cd given in Table 12.12-1, the product representing the estimated design earthquake displacement.

Drift DeterminationSection 12.8.6

Step 3: Divide xeCd by I, Importance Factor: x = xeCd /I

Step 4: Determine design story drift:x = x (x 1)

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Drift DeterminationSection 12.8.6 Story Drift

x = x x-1 < ax = Cd xe / 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 forceresisting 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 / for any story. 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 xeare 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.

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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 SeparationSection 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.


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.


Code Requirement 2006 IBC/ ASCE 7-052009 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 (x) 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 Separation2009 IBC Section 1613.6.7Code Requirement 2006 IBC/ ASCE 7-05

2009 IBC/ ASCE 7-05

BUILDING SEPARATIONS (paraphrased):All structures shall be separated from adjoining structures. Separations shall allow for the maximum inelastic displacement M (including torsion). Adjacent buildings on the same property shall be separated by at least MT 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, M, 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

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Building Separation2009 IBC Section 1613.6.7

Note difference between x and MThe deflections of Level x at the center of

mass (12.8.6),x = Cd xe / I

M = Cd max / I (Equation 16-44)max = the maximum displacement at Level x

computed assuming Ax = 1 (12.8.4.3)



Separation of Two Adjacent Buildings


10 separation

M1 at edge = 6

M2 at adjacent edge = 8

Q:A: for Building SeparationQ: 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 M1 and M2. To avoid damaging contact, ASCE 7 allows the separation distance to be the statistical sum of M1 and M2, which is less than the arithmetic sum.

Q:A: for Building Separation

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



Q:A: for Building Separation

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 0 Values forHorizontal and Vertical

Combinations

R, Cd and o 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 o coefficients shall apply to each system, including the limitations on system use contained in Table 12.2-1.

Section 12.2.2 Combinations of Framing Systems in Different

Directions

R =5Cd = 5o = 2

R = 8, Cd = 5 , o = 3

Section 12.2.3 Combinations of 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 downCd and 0: Cannot decrease as you go down

R=8, Cd=5.5, 0=3

R=5, Cd=5, 0=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, 0=2.5

R=8, Cd=5.5, 0=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, 0=2.5

R=8, Cd=5.5, 0=3

5, 5.5, 3

5, 5.5, 3

5, 5.5, 3

5, 5.5, 3

R=8, Cd=5.5, 0=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 Horizontal Combinations12.2.3.2 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, 0 , 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 0-values of all the individual structural systems shall be used.


The other possible interpretation is that the Cd-and 0-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.

12.2.3.2 Horizontal CombinationsSecond, large values of xe 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 0-values.

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

Different Structural Systems Used in Same DirectionR 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.

o value The o 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 o shall be used.

Vertical combinations

R: Cannot increase as you go downCd and 0: Always correspond to R

ASCE 7-10 Section 12.2.3.1 Vertical Combinations

R = 8, Cd = 5.5, 0 = 3

R = 5, Cd = 5, 0 = 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, 0 = 2.5

R = 8, Cd = 5.5, 0 = 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, 0 = 2.5

R = 8, Cd = 5.5, 0 = 3

5, 5, 2.5

5, 5, 2.5

5, 5, 2.5

5, 5, 2.5

R = 8, Cd = 5.5, 0 = 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 CombinationsQ: 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

CodeMaster2009 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

ProvisionsBuilding Code Minimum 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 Shear

Section 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. 2Supplement 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]CS = 0.03 0.044SDSI 0.03 (Eq. 15.4-1) [applicable to nonbuilding structures not similar to buildings]CS = 0.01 0.044SDSI 0.01 (Eq. 15.4-3) [applicable to an exception for nonbuilding structures not similar to buildings]

32



Minimum Seismic Base Shearvs. Ground Motion

0.044SDSI 0.01 or SDSI 0.227or SS value given in table

Site Class I = 1 I = 1.25 I = 1.5

A 0.426 0.341 0.284B 0.341 0.273 0.227C 0.284 0.227 0.189D 0.213 0.170 0.142E 0.136 0.109 0.091

Minimum Seismic Base Shear

Minimum Seismic Base Shear

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 = 1.0Height = 120 ft, SDS = 1.00, SD1 = 0.40Ta = 0.016(120)0.9 = 1.19 sec

GovernsDoes 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 1602Definition for Diaphragms

DIAPHRAGMDiaphragm, blockedDiaphragm boundaryDiaphragm chordDiaphragm, flexibleDiaphragm, rigidDiaphragm, 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.

ASCE 7-05 Section 12.3.1.1Definition 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.

ASCE 7-05 Section 12.3.1.3Definition for Flexible Diaphragms

by Calculation

ASCE 7-05 Figure 12.3-1 Definition for Flexible Diaphragm

by Calculation

MAXIMUM DIAPHRAGMDEFLECTION (MDD)AVERAGE DRIFT OFVERTICAL ELEMENT

(ADVE)

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

S

De

SEISMIC LOADING

2006 and 2009 IBC Section 1613.6.1Definition 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


Condition #1:Toppings of concrete or similar materials

are not placed over wood structural panel diaphragms except for nonstructural toppings no greater than 1 inches thick.


Condition #2:Each line of vertical elements of the

lateral force-resisting system complies with the allowable story drift of Table 12.12-1.


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




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].

ASCE 7-05 Section 12.3.1.2Definition 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 1602.1Definition 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


(ADVE)

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)

ASCE 7-05 Section 12.3.1Definition 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 framesComposite steel and concrete braced framesConcrete, 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

NN

Y

Y

Y

N

N

STAR

T

36




(ADVE)

Is MDD > 2 (ADVE)?

S

De

SEISMIC LOADING

Assume Flexible

YAssume 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 COMBINATIONSis replaced with

LOAD COMBINATIONS WITH OVERSTRENGTH FACTORSof 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 factorsone 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 = 0 QE + 0.2 SDSD, whileE = QE + 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 + 0QE + L + 0.2S(0.9 0.2SDS)D + 0QE + 1.6H


Basic Combinations for Allowable Stress Design with Overstrength Factor(1.0 + 0.14SDS)D + H + F + 0.70QE(1.0 + 0.105SDS)D + H + F + 0.5250QE + 0.75L

+ 0.75(Lr or S or R)(0.6 0.14SDS)D + 0.70QE + H


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, and3. 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?


Cantilever Column Systems 12.2.5.2SDC 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.


A: Elements Supporting Discontinuous Walls or Frames

12.3.3.3SDC B - F



39




A: Collector Elements 12.10.2.1 (SDC C - F)


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!!

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2005EditionofASCE7MinimumDesignLoadsforBuildingandOtherStructures

SupplementNo.2

Supplement No. 2 of ASCE 705 revises the minimum base shear equations for bothbuildings and nonbuilding structures. The need for this change was indicated by theresults from the75%DraftofATC63,QuantificationofBuildingSystemPerformanceandResponseParameters,whichindicatethattallbuildingsmayfailatanunacceptablylowseismiclevelandthereforetheminimumbaseshearequationforbuildingsisbeingrestoredtothatwhichappearedinthe2002editionofASCE7.

Becausenonbuildingstructuresnotsimilar tobuildingshave lowRvaluescomparedtothespecialreinforcedconcretemomentframesstudiedinATC63,theASCE7standardscommitteechosenottorestorethehighminimumbaseshearsfornonbuildingstructuresnotsimilar tobuildings found inASCE702. Inmanycases, thesepreviousminimumbaseshearsgavemanynonbuildingstructuresnotsimilartobuildingseffectiveRvaluesless than 1.0. Therefore, the Seismic Subcommittee believes that the minimum baseshear equation of 0.044SDSI used for buildings should also be applied to nonbuildingstructuresnotsimilartobuildings.

SupplementNo.2modifiesthreeequationsofthestandard(Eq.12.85,15.41and15.43)asshownbelow:

SupplementNo.2toASCE705:

Revise Equation 12.85 of Section 12.8.1.1 of ASCE 705 as shownbelow:

12.8.1.1CalculationofSeismicResponseCoefficient. Theseismicresponsecoefficient,Cs,shall bedeterminedinaccordancewithEq.12.82.

=

IR

SC DSs (Eq.12.82)

where:

SDS= thedesignspectralresponseaccelerationparameterintheshortperiodrangeasdeterminedfromSection11.4.4

R = theresponsemodificationfactorinTable12.21,and

I = theoccupancyimportancefactordeterminedinaccordancewithSection11.5.1

Thevalueof CscomputedinaccordancewithEq.12.82neednotexceedthefollowing:

L1D

s TTfor

IR

T

SC

= (Eq.12.83)

L2

L1Ds TTfor

IR

T

TSC >

= (Eq.12.84)

Csshallnotbelessthan

Cs=0.01 0.044SDSI0.01 (Eq.12.85)

Inaddition,forstructureslocatedwhereS1 isequaltoorgreaterthan0.6g,Csshallnotbelessthan

=

I

RS.

Cs150 (Eq.12.86)

whereIandRareasdefinedin Section12.8.1.1and

SD1= thedesignspectralresponseaccelerationparameterataperiodof1.0sec,asdeterminedfromSection11.4.4

T = thefundamentalperiodofthestructure(sec)determinedinSection12.8.2

TL = longperiodtransitionperiod(sec)determinedinSection11.4.5S1 = themappedmaximumconsideredearthquakespectral

responseaccelerationparameterdeterminedinaccordancewithSection11.4.1

ReviseEquations 15.41 and15.42 ofSection15.4.1, item2,as shownbelow:

2. Fornonbuildingsystemsthathavean RvalueprovidedinTable15.42,theseismicresponsecoefficient(Cs)shallnotbetakenlessthan

Cs=0.030.044SDSI 0.03 (15.41)

andfornonbuildingstructureslocatedwhereS10.6g,Csshallnotbetakenlessthan

=

IRS

Cs18.0 (15.42)

EXCEPTION:TanksandvesselsthataredesignedtoAWWAD100,AWWAD103,API650AppendixE,andAPI620AppendixLasmodifiedbythisstandard,shallbesubjecttothelargeroftheminimumbaseshearvaluesdefinedbythereferencedocumentorthefollowingequations:

Cs=0.010.044SDSI 0.01 (15.43)

andfornonbuildingstructureslocatedwhereS10.6g,Csshallnotbetakenlessthan

=

IRS

Cs15.0 (15.44)

Minimumbaseshearrequirementsneednotapplytotheconvective(sloshing)componentofliquidintanks.

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