S. K. Ghosh Associates Inc.
FREQUENTLY MISUNDERSTOOD
IBC/ASCE 7 STRUCTURAL PROVISIONS
All sections
referenced
are from
ASCE 7-05,
unless
otherwise
noted.
S. K. Ghosh and Susan Dowty
S. K. Ghosh Associates Inc.
Palatine, IL and Aliso Viejo, CA
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PROVISION
#1
Enclosure Classification
For Wind Design
Internal Pressure
Enclosure Classification
Section 6.2 Definitions
Open
Partially
Enclosed (can experience
“ballooning” or suction effects caused by the
build-up of internal pressure)
Enclosed
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Enclosure Classification
Section 6.2 Definitions
Enclosure Classification
Section 6.2 Definitions
OPEN
A building having each wall at least 80% open.
Ao ≥ 0.8 Ag
for EACH side of the building
AO = A1 + A2 + A3
Enclosure Classification
Section 6.2 Definitions
Ag = H × W
Enclosure Classification
Section 6.2 Definitions
Do stacks of hay
obstruct flow of wind?
Enclosure Classification
Section 6.2 Definitions
Enclosure Classification
Section 6.2 Definitions
PARTIALLY ENCLOSED
Ao = total area of openings in a wall that receives
positive external pressure
1.Ao ≥ 1.10Aoi
Aoi = sum of the areas of openings in the building
envelope (walls and roof) not including Ao
2.Ao > 4 sq ft AND > 0.01Ag
Ag = gross area of that wall in which Ao is
identified
3. Aoi/Agi ≤ 0.20
Agi = the sum of the gross surface areas of the
building envelope not including Ag
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Enclosure Classification
Section 6.2 Definitions
Enclosure Classification
Section 6.2 Definitions
Note: Ao, Ag refer to wall that
receives positive external pressure
• Openings : apertures or holes in the
building envelope which allow air to
Aoi , Agi refer to building envelope
(walls and roof)
flow through the building envelope
and which are designed as “open”
during design winds
Enclosure Classification
Section 6.2 Definitions
Q: A: for Enclosure Classification
Q:
Is a fixed glazed opening
ENCLOSED
considered an opening?
A:
A building that does not qualify as
OPEN or PARTIALLY ENCLOSED.
NO.
Enclosure Classification
Enclosure Classification
Section 6.2 Definitions
Figure 6-5 Internal Pressure Coefficients,
GCpi
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ASCE 7-98
Positive Internal Pressure
Basic Wind Equation
• For buildings with External and
Internal Pressure:
p = qGCp – qiGCpi
qi = Velocity pressure calculated for
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
PROVISION
#2
Q: Should we treat roll-down doors and
operable louvers as openings?
Seismic and Wind
Design of Parapets
A: Yes and No.
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13.3.1 Nonstructural Component
Seismic Design Force
Common Earthquake Damage to
Parapets
Fp =
0.4 ap SDS
(Rp / Ip)
1+
2z
h
Wp
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
13.1.3Nonstructural Component
Importance Factor, Ip
Nonstructural Component ap and
Rp
Ip is based on
The values of ap range from 1.0 to 2.5 and
1. Whether component must function after
the design earthquake or
can be taken as less than 2.5 based on
2. Occupancy Category or
dynamic analysis.
3. Whether component contains hazardous
Rp values range from 1.0 to 12.0 (Tables
materials.
13.5-1 and 13.6-1).
Parapets: Ip is based on Occupancy
Category
Table 13.5-1 ap and Rp for
Architectural Components
Explanation of Fp Equation
1.2SDS
Component
Cantilever Parapets
AB = 0.4SDS (1+2zB/h)
ASCE 7-05
Architectural
ap
Rp
2.5
2.5
B
Floor
Acceleration
Distribution
AA= 0.4SDS (1+2zA/h)
0.4SDS
Z = h for parapet design
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h
zB
A
zA
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Example: Seismic Design of
Parapets
Example: Seismic Design of
Parapets
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
Example: Seismic Design of
Parapets
Weight of the parapet per linear foot is
SDS = 1.0g
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
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
Example: Seismic Design of
Parapets
Wind Forces on 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
ASCE 7-05 Figure C6-12
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Wind Forces on Parapets
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.
ASCE 7-05 Figure C6-12
Design Example
Example Building
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.
Design Criteria
Example Building
3 ft parapet
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,
Table 6-1)
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I = 1.0 (ASCE 7-05
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Velocity Pressure Exposure
Coefficient, Kz
Effects of Parapets on MWFRS loads
ASCE 7-05 Section 6.5.12.2.4:
At top of parapet, h = 67.5 + 3 = 70.5 ft
pp = qpGCpn (ASCE 7-05 Eq. 6-20)
qp = velocity pressure evaluated at the top of
Kz = 2.01(z/zg)2/α = 2.01(70.5/1200)2/7 = 0.894
the parapet
(from ASCE 7-05 Table 6-3)
= 0.00256 Kz Kzt Kd V2 I
(α = 7, zg = 1200 ft for Exposure B from
GCpn = combined net pressure coefficient
= +1.5 for windward parapet
ASCE 7-05 Table 6-2)
= -1.0 for leeward parapet
Topographic Effect Factor, Kzt
Wind Directionality Factor, Kd
Effects of Parapets on MWFRS loads
qp = 0.00256 Kz Kzt Kd V2 I =
Kzt = 1.0
0.00256 × 0.894× 1 × 0.85 × 852 × 1 = 14.06 psf
(Assuming the example building to be situated
• For windward parapet:
on level ground, i.e., with H, as shown in ASCE
pp = qpGCpn = 14.06 × 1.5 = 21.1 psf
7-05 Fig. 6-4, equal to zero).
Force = 21.1 × 3 × 66 / 1000 = 4.18 kips
Kd = 0.85
• For leeward parapet:
(from ASCE 7-05 Table 6-4 for main wind force-
pp = qpGCpn = 14.06 × (-1.0) = -14.06 psf
resisting system)
Force = -14.06 × 3 × 66 / 1000 = -2.78 kips
Effects of Parapets on MWFRS loads
Design of Parapets as Component
At the roof level, 4.18 + 2.78 = 6.96 kips is to be
ASCE 7-05 Section 6.5.12.4.4:
added to the design wind force for MWFRS
p = qp(GCp – GCpi)
computed from the windward and leeward walls
(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
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External Pressure Coefficient, GCp
External Pressure Coefficient, GCp
Load Case A (ASCE 7-05 Section 6.5.12.4.4)
Effective wind area of the parapet:
Positive wall GCp = 0.68 (Figure 6-17 Zones 4
Span = 3 ft
and 5)
Width = 66 ft (> span/3)
Applied to front surface of the parapet
A = 3x66 = 198 ft2
External Pressure Coefficient, GCp
External Pressure Coefficient, GCp
Load Case A (ASCE 7-05 Section 6.5.12.4.4)
Load Case A (ASCE 7-05 Section 6.5.12.4.4)
Negative roof edge GCp = -1.76 (Figure 6-17
Zone 2*)
GCp = 0.68 – (-1.76) = 2.44
Applied to back surface of the parapet
GCpi = -0.18 for enclosed building (uniform
porosity)
*Corner Zone 3 is treated as Zone 2 because the parapet is 3 ft high
(Figure 6-17 Note 7)
However, internal pressures on both surfaces of
the parapet cancel each other out.
p = 14.06 x 2.44 = 34.31 psf
External Pressure Coefficient, GCp
External Pressure Coefficient, GCp
Load Case B (ASCE 7-05 Section 6.5.12.4.4)
Load Case B (ASCE 7-05 Section 6.5.12.4.4)
Effective wind area = 198
ft2
Negative wall GCp = -0.76 (Figure 6-17 Zone 4)
Positive wall GCp = 0.68 (Figure 6-17 Zones 4 and
= -1.23 (Figure 6-17 Zone 5)
5)
Applied to back surface of the parapet
Applied to front surface of the parapet
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External Pressure Coefficient, GCp
Design of Parapets as Component
Load Case B (ASCE 7-05 Section 6.5.12.4.4)
GCp = 0.68 – (-0.76) = 1.44 (For Zone 4)
Clearly, Load Case A governs
= 0.68 – (-1.23) = 1.91 (For Zone 5)
GCpi = -0.18 for enclosed building (uniform
Thus, design uniform wind pressure on the
porosity)
whole width of the parapet
However, internal pressures on both surfaces of
the parapet cancel each other out.
p = 34.31 psf
p = 14.06 x 1.44 = 20.24 psf (Zone 4)
= 14.06 x 1.91 = 26.85 psf (Zone 5)
Q: A: for Wind Design of Parapet
Q: A: for Wind Design of Parapet
A: In the case of parapets, it is expected
Q: In Section 6.5.12.4.4 (parapets for
most cases to have uniform porosity, so the
C&C), the definition for the factor GCpi is
"enclosed" classification (+0.18, - 0.18)
based on the porosity of the parapet
would be appropriate. However, if the two
surfaces of the parapet are very different
envelope. How is the porosity of the
(one has openings, the other is fully
parapet determined?
sealed), the partially-enclosed case might
be relevant.
Q: A: for Wind Design of Parapet
Q:
PROVISION
#3
What is the wind load on the parapet using
Method 1, Simplified Procedure?
Torsion, Torsional
Irregularity and Direction
of Seismic Loading
A: There is no clear answer. Some
jurisdictions do not allow Method 1 to be
used for buildings with parapets.
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δxe
QE
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
ASCE 7-05 12.8.4 Horizontal
Distribution of Forces
Failure – Torsion
• 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
1976 Philippines
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Amplification of Torsion
Amplification of Torsion
For structures* assigned to SDC C, D, E, or F
δA and δB computed assuming Ax = 1.0
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
Adapted from ASCE 7-05 Figure 12.8-1
*Not applicable to light-frame construction
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
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Section 12.5.2 Direction of
Loading
Q: A: for Torsional Irregularity
Q:
Do the torsional irregularity provisions
12.5.2 SDC B.
apply to light-frame constructions?
The design seismic forces are permitted to be
A:
applied independently in each of two
Most likely, no. The torsional irregularity
orthogonal directions and orthogonal
definition applies to diaphragms that are
interaction effects are permitted to be
rigid or semirigid, which is typically not the
neglected.
case for light-frame construction.
Section 12.5.2 Direction of
Loading
ASCE 7-05 12.5.2 Direction of
Loading
12.5.3 SDC C.
12.5.3 SDC C.
a. Orthogonal Combination Procedure.
Structures that have horizontal structural
irregularity Type 5 of Table 12.3-1, shall use
ELF, modal response spectrum, or linear
one of the following procedures.
response history analysis, with loading applied
independently in any two orthogonal
directions…
100% + 30%
Section 12.5.2 Direction of
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.
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Section 12.5.2 Direction of
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.
REDUNDANCY FACTOR, ρ
ASCE 7-05 Section 12.3.4
PROVISION
#4
REDUNDANCY
SDC
ρ
A
NA
B&C
1.0
D, E & F
1.0 or 1.3
Note that ρ = 1.0 when the SIMPLIFIED
PROCEDURE of Section 12.14 is used.
Section 12.3.4.1
ρ for SDC D - F
REDUNDANCY FACTOR, ρ
ASCE 7-05 Section 12.3.4
12.3.4
Redundancy
12.3.4.1
Conditions Where
1. Structures assigned to SDC B and C.
Value of ρ is 1.0
2. Drift calculation and P-delta effects.
Redundancy
3. Design of nonstructural components.
Factor, ρ, for SDC
4. Design of nonbuilding structures, not
similar to buildings.
12.3.4.2
ρ = 1.0 for the following:
D through F
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Section 12.3.4.2
ρ for SDC D - F
Section 12.3.4.1
ρ for SDC D - F
ρ = 1.0 or 1.3
ρ = 1.0 for the following:
5. Design of collector elements, splices and their
connections for which load combinations with
overstrength are used.
ρ = 1.3 unless ONE of two conditions is met.
If Condition # 1 is met, then ρ = 1.0
6. Design of members or connections where load
combinations with overstrength are required for design.
If Condition #2 is met, then ρ = 1.0
7. Diaphragm loads determined using Eq. 12.10-1.
Both conditions do NOT need to be met
8. Structures with damping systems designed in
accordance with ASCE 7-05 Chapter 18.
for ρ = 1.0
Section 12.3.4.2
ρ for SDC D - F
Section 12.3.4.2
ρ for SDC D - F
CONDITION #1:
CONDITION #1 and CONDITION #2 only need to
be checked at each story resisting more than
Can an individual element be removed from
35% of the base shear.
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
Section 12.3.4.2
ρ for SDC D - F
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.
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Section 12.3.4.2
ρ for SDC D - F
Q: A: for Redundancy
Q: How many bays are in a shear wall?
Seismic Force-Resisting
Perimeter Framing
Two Bays
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: A: for Redundancy
Q: A: for Redundancy
Q: I am using “Condition #1” to determine
Q: Does the redundancy factor apply
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?
to the design of foundations?
A: Yes.
A:
Q: A: for Redundancy
Yes.
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.
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Q: A: for Redundancy
Q: A: for Redundancy
A: The h/l ratio is intended to be
Q:
Can the value of ρ be different at
each level of the same building?
story height-to-length ratio.
l < 10 ft
h = 10 ft
Q: A: for Redundancy
A:
Q: A: for Redundancy
No, ρ cannot be different at each
Q:
Why doesn’t Table 12.3-3
address dual systems? If you have
a dual system, can you assume ρ =
1.0?
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: A: for Redundancy
A: No…..
“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.”
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Q: A: for Redundancy
Q:
Q: A: for Redundancy
A:
Do you need to determine the
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.
redundancy factor for “nonbuilding
structures similar to buildings” or can you
assume the redundancy factor equals 1?
Q: A: for Redundancy
Q:
Does the redundancy factor need
to be determined if dynamic
analysis is used?
A:
Yes.
REDUNDANCY EXAMPLE
Wall E
Stiffness Ke
REDUNDANCY EXAMPLE
Wall F
Stiffness Kf
Wall A
Stiffness Ka
Wall C
Stiffness Kc
Wall B
Stiffness Kb
Wall D
Stiffness Kd
•
•
•
•
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.
a
Wall G
Stiffness Kg
Wall H
Stiffness Kh
a
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REDUNDANCY EXAMPLE
CONDITION #1
CONDITION #2
Step 1:
Step 1:
Remove shear wall and see
Check if structure is regular
if story strength is reduced
in plan.
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
by more than 33%.
for light-framed construction, 2 x length
Step 2:
Step 2:
See if there is an extreme
Are there at least 2 bays
torsional irregularity created.
of….on each side in each
of shear wall/ story height.
For example: (15/18)(2) = 1.67 < 2
orthogonal direction?
REDUNDANCY EXAMPLE
REDUNDANCY EXAMPLE
CONDITION #1:
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).
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
Wall E
Stiffness Ke
REDUNDANCY EXAMPLE
Definition of Extreme Torsional
Irregularity in ASCE 7-05 Table 12.3-1:
Wall F
Stiffness Kf
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.
Wall A
Stiffness Ka
a
Wall B
Stiffness Kb
Wall D
Stiffness Kd
Wall G
Stiffness Kg
Wall H
Stiffness Kh
a
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REDUNDANCY EXAMPLE
REDUNDANCY EXAMPLE
The determination of extreme torsional
The torsional stiffness about the center
of rigidity (CR) is determined as:
irregularity requires the evaluation of the
story drifts δa and δb, as shown below.
δb
δa
a/3
CR
2a/3
CM
a/6
=
V
θ
REDUNDANCY EXAMPLE
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
According to ASCE 7-05 Table 12.3-1,
extreme torsional irregularity does not exist
when
This can be transformed to
This ratio is less than 2.33 only if δ/(aθ) is
larger than 1.08.
REDUNDANCY EXAMPLE
PROVISION
#5
Therefore, no extreme torsional irregularity is created
and ρ = 1.0.
(Note that the term 0.05a is for accidental torsion)
SEISMIC ANALYSIS
PROCEDURE SELECTION
Thus, the horizontal structural irregularity Type 1b does
not exist and the configuration qualifies for a ρ factor
of 1.0.
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Seismic Analysis Procedure Selection
STATIC ANALYSIS
PROCEDURES
ASCE 7-05
SECTION
Simplified Design
Procedure
Equivalent Lateral Force
Procedure
12.14
Seismic Analysis Procedure Selection
12.8
Simplified Design Procedure
Exception to Section 12.1
DYNAMIC ANALYSIS
PROCEDURES
ASCE 7-05
SECTION
Modal Response
Spectrum Analysis
Linear Response History
Analysis
12.9
Nonlinear Response
History Analysis
16.2
16.1
Simplified Design Procedure
Section 12.14
1. Occupancy Category I or II
2. Site Class A, B, C, or D
3. Three stories or less in height
4. Bearing wall system or building
frame system
5. through 12……
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
Seismic Analysis Procedure Selection
Table 12.6-1
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
* 12.3-1
*
**
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** 12.3-2
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Seismic Analysis Procedure Selection
Seismic Analysis Procedure Selection
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.)
Seismic Analysis Procedure Selection
Table 12.6-1
Seismic Analysis Procedure Selection
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”
Seismic Analysis Procedure Selection
Table 12.6-1
Seismic Analysis Procedure Selection
Table 12.6-1
All structures of light frame
Occupancy Category I or II buildings of
construction, irrespective of height
other construction not exceeding two
stories in height
– Dynamic analysis never required
– Dynamic analysis not required
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Seismic Analysis Procedure Selection
Seismic Analysis Procedure Selection
Q:
A:
In Table 12.6-1, it states “Regular
Structures with T < 3.5Ts and all structures of
It does not mean "and." It means
"or." Regular Structures with T <
light frame construction” are permitted to use
3.5Ts are permitted to use Equivalent
an Equivalent Lateral Force Analysis. Does
this mean that the building must meet both
Lateral Force Analysis. All structures of
conditions (regular with T < 3.5Ts and light
light frame construction, irrespective of
frame construction), or does only one of these
height, are also permitted to
two characteristics need to be satisfied?
use Equivalent Lateral Force Analysis.
δxe
QE
PROVISION
#6
DRIFT AND BUILDING
SEPARATION
V
Drift Determination
Section 12.8.6
Drift Determination
Section 12.8.6
Step 1: Determine δxe at each floor level
where δxe is the lateral deflection at floor
level x determined by elastic analysis
Step 3: Divide δxeCd by I, Importance
Factor: δ x = δxeCd /I
under code-prescribed seismic forces.
Step 4: Determine design story drift:
Step 2: Multiply δxe by Cd given in Table
∆x = δx – (δx – 1)
12.12-1, the product representing the
estimated design earthquake
displacement.
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Drift Determination
Section 12.8.6
Story Drift
• ∆x = δx − δx-1 < ∆a
δx = Cd δxe / I
Cd = deflection amplification factor
Allowable Drift – Additional
Requirement
Allowable Drift
ASCE 7-05 Table 12.12-1
ASCE 7-05 Section 12.12.1.1
Occupancy Category
Structure
I or II
III
• For seismic force–resisting systems comprised
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
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.
Q: A: for Drift Determination
Q:
Q: A: for Drift Determination
Q:
Why is drift divided by the Importance
allowable stress design (ASD) than for
strength design (SD)?
Factor?
A:
Is drift determined differently for
Because the forces under which δxe
A:
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.
No. The same procedure is used
regardless of whether ASD or SD is
used.
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Q: A: for Drift Determination
Q: A: for Drift Determination
Q: Does minimum base shear need to be
Q: Does the upper-bound limitation on
considered for drift determination?
A:
period T need to be considered for drift
determination?
Yes, Section 12.8.6.1 requires that all
A:
of the requirements of Section 12.8 be
satisfied for the purpose of computing
drift.
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
Building Separation
Section 12.12.3
This section applies to SDCs B
through F.
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.
Does not address adjacent
buildings on the same property.
Does not address minimum
setback distance from property line.
Building Separation
2009 IBC Section 1613.6.7
Building Separation
Section 12.12.3
Code Requirement
2006 IBC/
ASCE 7-05
2009 IBC/
ASCE 7-05
No
requirement
2009 IBC
Section
1613.6.7
BUILDING SEPARATIONS (paraphrased):
Code Requirement
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.
2006 IBC/
ASCE 7-05
2009 IBC/
ASCE 7-05
ASCE 7-05
Section 12.12.3
(applies to all
SDCs)
ASCE 7-05
Section 12.12.3
(applies to all
SDCs)
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.
Continued on next page
Exception: Smaller separations or property line setbacks
shall be permitted when justified by rational analyses based
on maximum expected ground motions.
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Building Separation
2009 IBC Section 1613.6.7
Building Separation
2009 IBC Section 1613.6.7
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
2009 IBC Section 1613.6.7
Building Separation
2009 IBC Section 1613.6.7
10″ separation
δM2 at adjacent edge = 8″
δM1 at edge = 6″
Separation of Two Adjacent Buildings
Q:A: for Building Separation
Q:
Q:A: for Building Separation
ASCE 7-05 Section 12.12.3 contains the
Q:
language “sufficient to avoid damaging
contact.” What is damaging contact?
A:
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.
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.
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Q:A: for Building Separation
A:
The first provision is concerned with
PROVISION
#7
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.
R, Cd and Ω0 Values for
Horizontal and Vertical
Combinations
Section 12.2.2 Combinations of
Framing Systems in Different
Directions
R, Cd and Ωo Values for Horizontal
and Vertical Combinations
• Different seismic force-resisting systems
Horizontal Combinations can be
may be used to resist seismic forces along
either….
each of two orthogonal plan axes.
• The respective R, Cd, and Ωo coefficients
• In different directions
shall apply to each system, including the
• In same direction
limitations on system use contained in
Table 12.2-1.
Section 12.2.2 Combinations of
Framing Systems in Different
Directions
Section 12.2.3 Combinations of
Framing Systems in the Same
Direction
Where different seismic force-resisting systems
are used in combination to resist seismic
forces in the same direction of structural
R =5
Cd = 5
Ωo = 2½
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
R = 8, Cd = 5 ½, Ωo = 3
with the requirements of this section.
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12.2.3.2 Horizontal Combinations
12.2.3.1 Vertical Combinations
R=8, Cd=5.5, Ω0=3
8, 5.5, 3
R=5, Cd=5, Ω0=2.5
5, 5, 2.5
8, 5.5, 3
R=5, Cd=5, Ω0=2.5
8, 5.5, 3
5, 5, 2.5
5, 5.5, 3
5, 5.5, 3
5, 5.5, 3
R=8, Cd=5.5, Ω0=3
5, 5.5, 3
R=8, Cd=5.5, Ω0=3
5, 5.5, 3
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.
8, 5.5, 3
8, 5.5, 3
5, 5, 2.5
R=5, Cd=5, Ω0=2.5
5, 5.5, 3
5, 5.5, 3
R=8, Cd=5.5, Ω0=3
5, 5.5, 3
5, 5.5, 3
5, 5.5, 3
R: Cannot increase as you go down
Cd and Ω0: Cannot decrease as you go down
12.2.3.2 Horizontal Combinations
12.2.3.2 Horizontal Combinations
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 deflection amplification factor, Cd, and the
system overstrength factor, Ω0 , 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
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.
considered.
12.2.3.2 Horizontal Combinations
12.2.3.2 Horizontal Combinations
The other possible interpretation is that the Cd-
The second paragraph of ASCE 7-05 Section
and Ω0-values shall correspond to the least R-
12.2.3.2 is far from clear.
value of all the individual structural systems.
One possible interpretation is that when different
The second interpretation appears to be the
structural systems are combined in the same
more logical in view of the following example
direction of a building or other structure, the
(discussion is continued in terms of Cd alone).
largest Cd- and Ω0-values of all the individual
structural systems shall be used.
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12.2.3.2 Horizontal Combinations
12.2.3.2 Horizontal Combinations
Consider a rather extreme example where a
First, the combined system is much more rigid
prestressed masonry shear wall (R = 1.5, Cd =
than the special steel moment frame itself.
1.75) is combined with a special steel moment-
Until the prestressed masonry shear wall
resisting frame (R = 8, Cd = 5.5). There is no
hinges at its base, which is extremely unlikely
question that the R-value is 1.5. The question is
in view of the large design forces that would
whether the Cd-value is 1.75 or 5.5. 5.5 does not
result from an R = 1.5, large inelastic
seem logical – for two reasons.
displacements do not seem to be possible.
12.2.3.2 Horizontal Combinations
12.2.3.2 Horizontal Combinations
Second, large values of δxe would automatically
The proposed rewrite provides clarification of the
result from the low value of R used in design.
second paragraph of ASCE 7-05 Section
These, multiplied by the Cd of 5.5 would yield
12.2.3.2. The rewrite also offers clarification
unrealistically large total displacements. Cd of
concerning another complication that may
1.75 appears to be much more logical.
arise, which is that different structural systems
This second interpretation was implicit in the
having the same R-value sometimes have
1997 Uniform Building Code, where 0.7R was
different Cd- and Ω0-values.
used in place of Cd.
12.2.3.2 Horizontal Combinations
Table 12.2.3.2 R, Cd, and Ωo Values for Combination of
Vertical combinations
Different Structural Systems Used in Same Direction
R value
The least value of R for any of the systems used.
R = 8, Cd = 5.5, Ω0 = 3
Exception: Resisting elements are permitted to be designed using the
8, 5.5, 3
least value of R for the different structural systems found in each
independent line of resistance if the following three conditions are met: 1)
R = 5, Cd = 5, Ω0 = 2.5
of the systems used. In the case where two or more systems have the
5, 5, 2.5
5, 5, 2.5
R = 8, Cd = 5.5, Ω0 = 3
5, 5, 2.5
R = 5, Cd = 5, Ω0 = 2.5
used.
5, 5, 2.5
R: Cannot increase as you go down
Cd and Ω0: Always correspond to R
of the systems used. In the case where two or more systems have the
ASCE 7-10 Section 12.2.3.1 Vertical Combinations
same least value of R, the largest of the corresponding values of Ωo shall be
used.
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5, 5, 2.5
5, 5, 2.5
R = 8, Cd = 5.5, Ω0 = 3
same least value of R, the largest of the corresponding values of Cd shall be
Ωo value The Ωo value corresponding to the system with the least value of R for any
8, 5.5, 3
8, 5.5, 3
5, 5, 2.5
5, 5, 2.5
R = 5, Cd = 5, Ω0 = 2.5
The Cd value corresponding to the system with the least value of R for any
R = 8, Cd = 5.5, Ω0 = 3
5, 5, 2.5
5, 5, 2.5
8, 5.5, 3
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
5, 5, 2.5
8, 5.5, 3
5, 5, 2.5
5, 5, 2.5
S. K. Ghosh Associates Inc.
Q: A: for Combinations
12.2.4 Combination Framing
Detailing Requirements
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
Structural components common to different
understand that I should design the building with the smaller R
framing systems used to resist seismic
= 5.5 for both the masonry shear walls and steel braced frames
motions in any direction shall be designed
in this direction for seismic design in accordance with ASCE 7-
using the detailing requirements of Chapter 12
05 Section 12.2.3.2. But someone told me that the building
required by the highest response modification
should be designed by analyzing the entire building twice: use
R = 5.5 for the entire building to analyze and design the
coefficient, R, of the connected framing
masonry shear walls and use R = 6 for the entire building to
systems.
design the steel frames. I don't think this is right. What is your
thought on this?
Q: A: for Combinations
A:
PROVISION
#8
What you understand is correct. The latter
interpretation is unfamiliar and incorrect.
MINIMUM SEISMIC
BASE SHEAR
ASCE 7-05 12.8.1
Design Base Shear
CodeMaster
2009 IBC Seismic Design
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ASCE 7-05 12.8 Equivalent Lateral
Force Procedure
Revisions to ASCE 7-05 Seismic
Provisions
Building Code
1997 UBC
Vmin = 0.11 Ca I W
2000 & 2003 IBC
ASCE 7-05 w/ Supplement No. 2
Design Base Shear
Minimum Base Shear
Applicable in All SDCs
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
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
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
“Including Supplement No. 1”
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]
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Minimum Seismic Base Shear
vs. Ground Motion
Minimum Seismic Base Shear
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
Minimum Seismic Base Shear Example
Minimum Seismic Base Shear
Concrete SMRF building - R = 8, I = 1.0
For special reinforced concrete moment
Height = 120 ft, SDS = 1.00, SD1 = 0.40
frames,
Ta = 0.016(120)0.9 = 1.19 sec
0.016h0.9 ≥ 9/8
h ≥ (70.31)1/0.9 = 113 ft
Governs
Does not govern
ASCE 7-05 Supplement No. 2
Q: A: for ASCE 7-05 Supplement
No. 2
Q: Does ASCE 7-05 Supplement No. 2
Q: Where does ASCE officially announce ASCE 705 Supplement No. 2?
minimum base shear need to be considered
for drift determination?
A:
http://content.seinstitute.org/files/pdf/
A:
SupplementNo2ofthe2005EditionofASCE7.pdf
Yes, Section 12.8.6.1 requires that all of the
requirements of Section 12.8 be satisfied for
the purpose of computing drift.
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Q: A: for ASCE 7-05 Supplement
No. 2
Q: Is ASCE 7-05 going to be published with
PROVISION
#9
errata and Supplement No. 2 incorporated?
A: “Additional printings will, to the extent
FLEXIBLE VS. RIGID
DIAPHRAGMS
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.”
2006 and 2009 IBC Section 1602
Definition for Diaphragms
FLEXIBLE
DIAPHRAGMS
DIAPHRAGM
Diaphragm, blocked
Prescriptive Approach
Diaphragm boundary
&
Diaphragm chord
Calculation Approach
Diaphragm, flexible
Diaphragm, rigid
Diaphragm, unblocked
ASCE 7-05 Section 12.3.1.1
Definition for Flexible Diaphragm
(Prescriptive)
2006 and 2009 IBC Section 1602
Definition for Flexible Diaphragm
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.
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.
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ASCE 7-05 Section 12.3.1.3
Definition for Flexible Diaphragms
by Calculation
ASCE 7-05 Figure 12.3-1
Definition for Flexible Diaphragm
by Calculation
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.
De
MAXIMUM DIAPHRAGM
DEFLECTION (MDD)
SEISMIC LOADING
AVERAGE DRIFT OF
VERTICAL ELEMENT
(ADVE)
S
Note: Diaphragm is flexible If MDD > 2 (ADVE).
2006 and 2009 IBC Section 1613.6.1
Definition for Flexible Diaphragm
(Prescriptive)
2006 and 2009 IBC Section 1613.6.1
Definition for Flexible Diaphragm
1613.6.1 Assumption of flexible
diaphragm. Add the following text at the
end of Section 12.3.1.1 of ASCE 7:
Condition #1:
Toppings of concrete or similar materials
are not placed over wood structural
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…
panel diaphragms except for
nonstructural toppings no greater than 1
½ inches thick.
2006 and 2009 IBC Section 1613.6.1
Definition for Flexible Diaphragm
2006 and 2009 IBC Section 1613.6.1
Definition for Flexible Diaphragm
Condition #2:
Condition #3:
Each line of vertical elements of the
Vertical elements of the lateral-force-
lateral force-resisting system complies
resisting system are light-framed walls
with the allowable story drift of Table
sheathed with wood structural panels
12.12-1.
rated for shear resistance or steel
sheets.
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2006 and 2009 IBC Section 1613.6.1
Definition for Flexible Diaphragm
ASCE 7-05 Section 12.3.1.2
Definition for Rigid Diaphragm
(Prescriptive)
12.3.1.2 Rigid Diaphragm
Condition #4:
Condition. Diaphragms of concrete slabs or
Portions of wood structural panel diaphragms that
cantilever beyond the vertical elements of the lateral-
concrete filled metal deck with span-to-depth
force-resisting system are designed in accordance
ratios of 3 or less in structures that have no
with [2006 IBC: Section 2305.2.5] [2009 IBC: Section
horizontal irregularities are permitted to be
4.2.5.2 of AF&PA SDPWS].
idealized as rigid.
2006 and 2009 IBC Section 1602.1
Definition for Rigid Diaphragm
(Calculation)
ASCE 7-05 Figure 12.3-1
Definition for Diaphragm
Diaphragm, rigid
De
A diaphragm is rigid for the purpose of
MAXIMUM DIAPHRAGM
DEFLECTION (MDD)
SEISMIC LOADING
AVERAGE DRIFT OF
VERTICAL ELEMENT
(ADVE)
distribution of story shear and torsional
S
moment when the lateral deformation of the
Note: Diaphragm is flexible If MDD > 2 (ADVE).
diaphragm is less than or equal to two times
Note: Per 2009 IBC Section 1602.1, diaphragm is
rigid if MDD ≤ 2(ADVE)
the average story drift.
Is any of the following true?
ASCE 7-05 Section 12.3.1
Definition for Diaphragms
12.3.1 Diaphragm Flexibility….Unless a
Y
1- & 2-family dwelling of Vertical elements one of the following:
light-frame construction •Steel braced frames
•Composite steel and concrete braced
Four conditions in 2009
frames
IBC Section 1613.6.1
•Concrete, masonry, steel or composite
are met
shear walls
diaphragm can be idealized as either flexible
START
or rigid in accordance with Sections 12.3.1.1,
12.3.1.2, or 12.3.1.3, the structural analysis
Is diaphragm wood
structural panels or
untopped steel
decking?
N
See Next Slide
N
shall explicitly include consideration of the
N
stiffness of the diaphragm (i.e. semirigid
modeling assumption).
Is diaphragm
• Concrete slab?
• Concrete filled metal deck?
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Assume Flexible
Y
Assume Rigid
N
Y
Y
Is span-to-depth ratio
≤ 3 and no horizontal
irregularities?
S. K. Ghosh Associates Inc.
Is MDD > 2 (ADVE)?
De
MAXIMUM DIAPHRAGM
DEFLECTION (MDD)
SEISMIC LOADING
AVERAGE DRIFT OF
VERTICAL ELEMENT
(ADVE)
S
N
Y
Assume Rigid
Assume Flexible
2006 IBC 1605.4 Special Seismic
Load Combinations
PROVISION
#10
Section 1605.4 is deleted in its entirety in
the 2009 IBC.
Special Seismic Load
Combinations
SPECIAL SEISMIC LOAD COMBINATIONS
is replaced with
LOAD COMBINATIONS WITH OVERSTRENGTH FACTORS
of ASCE 7-05
Why Was 2006 IBC Section 1605.4
Deleted?
Why Was 2006 IBC Section 1605.4
Deleted?
To eliminate a disconnect between IBC and ASCE 7-
To eliminate a disconnect between IBC and
05 (cont.):
ASCE 7-05:
– 2006 IBC had separate, unique load
– 2006 IBC Section 1605.4 had one set of
combinations that were to be applied where
“special seismic load combinations”
specifically required.
applicable to both ASD and strength design.
– ASCE 7-05 prescribes an equation for Em that
– ASCE 7-05 has two sets of “load combinations
is to be used in ASCE 7-05 Ch. 2 load
with overstrength factors”…one for ASD and
combinations.
one for strength design.
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1605.4 Special Seismic Load
Combinations (2006 IBC)
12.4.3.2 Load Combinations with
Overstrength Factor (ASCE 7-05)
Basic Combinations for Strength Design
• 1.2D + f1L + Em
(Equation 16-22)
• 0.9D + Em
(Equation 16-23)
with Overstrength Factor
(1.2 + 0.2SDS)D + Ω0QE + L + 0.2S
(0.9 − 0.2SDS)D + Ω0QE + 1.6H
Em = Ω0 QE + 0.2 SDSD, while
E = ρQE + 0.2 SDSD
12.4.3.3 Load Combinations with
Overstrength Factor (ASCE 7-05)
12.4.3.2 Load Combinations with
Overstrength Factor (ASCE 7-05)
Where allowable stress design methodologies are
Basic Combinations for Allowable Stress Design
used, allowable stresses are permitted to be
with Overstrength Factor
determined using an allowable stress increase of
(1.0 + 0.14SDS)D + H + F + 0.7Ω
Ω0QE
1.2.
(1.0 + 0.105SDS)D + H + F + 0.525Ω
Ω0QE + 0.75L
This increase shall not be combined with increases
+ 0.75(Lr or S or R)
in allowable stresses or load combination
reductions …except that combination with the
(0.6 − 0.14SDS)D + 0.7Ω
Ω0QE + H
duration of load increases permitted in AF&PA
NDS is permitted.
1617.1.2 Maximum Seismic Load
Effect, Em (2000, 2003 IBC)
What Takes the Place of Deleted
2006 IBC Section 1605.4?
Where allowable stress design
methodologies are used with the special
New language in 2009 IBC
load combinations of Section 1605.4,
Section 1605.1.
design strengths are permitted to be
determined using an allowable stress
increase of 1.7 ….
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S. K. Ghosh Associates Inc.
What Takes the Place of Deleted
2006 IBC Section 1605.4?
Q: A: for Load Combinations with
Overstrength Factor
1605.1 General. Buildings and other structures and
portions thereof shall be designed to resist:
Q: Section 1605.1 of the 2009 IBC requires
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.
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?
Q: A: for Load Combinations with
Overstrength Factor
Q: A: for Load Combinations with
Overstrength Factor
A:
Cantilever Column Systems
A:
12.2.5.2
SDC B - F
Elements Supporting
Discontinuous Walls or Frames
12.3.3.3
SDC B - F
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.
Q: A: for Load Combinations with
Overstrength Factor
Q: A: for Load Combinations with
Overstrength Factor
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S. K. Ghosh Associates Inc.
Q: A: for Load Combinations with
Overstrength Factor
A:
Collector Elements
Q: A: for Load Combinations with
Overstrength Factor
12.10.2.1 (SDC C - F)
Load Combinations with
Overstrength Factor
Thank You!!
Chapter 18 References:
For more information…
1810.3.6.1 Splices of deep foundation elements,
SDC C through F
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1810.3.9.4 Seismic reinforcement, SDC C and
above, Exception 3
Main Office
334 East Colfax Street, Unit E
Palatine, IL 60067
Phone: (847) 991-2700
Fax: (847) 991-2702
Email: skghoshinc@gmail.com
1810.3.11.2 Deep foundation element resistance to
uplift forces, SDC D through F
1810.3.12 Grade beams, SDC D through F
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43 Vantis Drive
Aliso Viejo, CA 92656
Phone: (949) 215-6560
Email: susandowty@gmail.com
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.
C s = S DS (Eq. 12.8­2) æ R ö
ç ÷
è I ø
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: C s = S D 1 æ R ö
T ç ÷
è I ø
S T C s = D 1 L æ R ö
T 2 ç ÷
è I ø
for T £ T L (Eq. 12.8­3) for T > T L (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
C s = 0 . 5 S 1 æ R ö
ç ÷
è I ø
(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
C s = 0 . 8 S 1 æ R ö
ç ÷
è I ø
(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
C s = 0 . 5 S 1 æ R ö
ç ÷
è I ø
(15.4­4) Minimum base shear requirements need not apply to the convective (sloshing) component of liquid in tanks.