Chapter 2 – Organizations, Codes & Standards
SDR Workbook – 2012 IBC Version
While PBSD has yet to be fully developed, it is expected to be used in future building codes to provide a
methodology to more reliably predict seismic risk in all buildings in terms more useful to building owners
and building users.
Under the ASCE 7-10 seismic design provisions, there exist implied performance levels as demonstrated
by Figure 2.1. For example, Risk Category II (i.e., Ie = 1.0) structures designed to ASCE 7-10 provisions
are expected to meet the following performance levels:
 Collapse prevention for MCER ground motions
 Life-safety for Design ground motions (i.e., 2/3 MCER)
 Immediate occupancy for Frequent ground motions (i.e. , ~ 50% in 50 years / 100 year return
interval)
Risk Category IV (i.e., Ie = 1.5) structures designed to ASCE 7-10 provisions are expected to meet the
following performance levels:
 Life-safety for MCER ground motions
 Immediate occupancy for Design ground motions (i.e., 2/3 MCER)
 Operational for Frequent ground motions
Similarly, Risk Category III (i.e., Ie = 1.25) structures designed to ASCE 7-10 provisions are expected to
meet performance levels that fall between the Risk Category IV and Risk Category II structures as shown
in Figure 2.1 below.
Figure 2.1 – Risk Category vs. Ground Motion
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(Ref. 12 - Figure C11.5-1)
Steven T. Hiner, MS, SE
4-10-14
SDR Workbook – 2012 IBC Version
Chapter 3 – General Provisions & Seismic Design Criteria
Site Class Adjusted MCER Acceleration Parameters
IBC §1613.3.3
SMS & SM1 represent the site class adjusted MCER spectral response acceleration parameters at short
periods and at 1-second period respectively, and are determined by the following equations:
SMS = Fa SS
IBC (16-37)
SM1 = Fv S1
IBC (16-38)
where:
Fa = Site coefficient per IBC Table 1613.3.3(1)
Fv = Site coefficient per IBC Table 1613.3.3(2)
Site coefficient Fa is function of the Site Class and the short period mapped spectral response acceleration
parameter (SS). Site coefficient Fv is a function of the Site Class and the 1-second period mapped spectral
response acceleration parameter (S1).
Fa & Fv are based on the results of empirical analyses of strong-motion data and analytical studies of site
response. In general, softer soils exhibit greater amplification of earthquake ground motions than stiffer
soils, and that amplification can be even more significant for longer period (e.g., taller) structures than for
shorter period structures (e.g., Fv values are greater than Fa values for Site Class C, D & E).
Design Spectral Response Acceleration Parameters
IBC §1613.3.4
SDS & SD1 represent the 5% damped design spectral response acceleration parameters at short periods and
at 1-second period respectively and they are determined by the following equations:
SDS = 2/3 SMS
IBC (16-39)
SD1 = 2/3 SM1
IBC (16-40)
NOTE: Table 3.2 (p. 1-33) and Table 3.3 (p. 1-34) are provided as short cut procedures in determining
SDS & SD1 respectively when SS, S1 & Site Class are known (Site Class D may be assumed unless given).
Figure 3.1 – Design Response Spectrum
(Ref. 9 - ASCE 7 – Figure 11.4-1)
where:
TS 
S D1
S DS
T0  0.2TS
TL = per ASCE 7 –
Figure 22-12
Steven T. Hiner, MS, SE
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Chapter 3 – General Provisions & Seismic Design Criteria
SDR Workbook – 2012 IBC Version
Alternative Seismic Design Category Determination
IBC §1613.3.5.1
Where S1 < 0.75, the Seismic Design Category is permitted to be determined from IBC Table 1613.3.5(1)
alone (i.e, using SDS only) when all of the following apply:
 Ta < 0.8 TS in each of the two orthogonal directions, and
 the fundamental period of the structure used to calculate the story drift - T < TS … in each of
the two orthogonal directions, and
S
 ASCE 7 (12.8-2) is used to determine the seismic response coefficient … CS  DS , and
(R Ie )
 The diaphragms are rigid (per ASCE 7 – §12.3.1) or for diaphragms that are flexible, the
distance between vertical elements of the seismic force-resisting system (SFRS) ≤ 40 feet
Simplified Design Procedure
IBC §1613.3.5.2
Where the alternate Simplified Design Procedure of ASCE 7 – §12.14 is used, the Seismic Design
Category shall be determined in accordance with ASCE 7.
3.7 ASCE 7 Seismic Design Criteria
ASCE 7 – Chapter 11
Scope
ASCE 7 - §11.1.2
Every structure (e.g., buildings and nonbuilding structures), and portion thereof, including nonstructural
components, shall be designed and constructed to resist the effects of earthquake motions as prescribed by
the seismic requirements of ASCE 7.
Applicability
ASCE 7 - §11.1.3
Structures and their nonstructural components shall be designed and constructed in accordance with the
requirements of the following chapters based on the type of structure or component:
 Buildings: ASCE 7 – Chapter 12
 Nonbuilding Structures: ASCE 7 – Chapter 15
 Nonstructural Components: ASCE 7 – Chapter 13
 Seismically Isolated Structures: ASCE 7 – Chapter 17
 Structures with Damping Systems: ASCE 7 – Chapter 18
Seismic Importance Factor, Ie
ASCE 7 - §11.5.1
Each structure shall be assigned an importance factor (Ie) in accordance with ASCE 7 – Table 1.5-2 …
based on the Risk Category of the building (or other structure) from IBC – Table 1604.5.




Risk Category I
Risk Category II
Risk Category III (high occupancy)
Risk Category IV (essential facilities)




I = 1.0
I = 1.0
I = 1.25*
I = 1.5*
The seismic importance factor (Ie) is used in the Seismic Response Coefficient (CS) equations with the
intent to raise the yield level for important structures (e.g., hospitals, fire stations, emergency operation
centers, hazardous facilities, etc.).
Use of an importance factor greater than one is intended to provide for a lower inelastic demand on a
structure which should result in lower levels of structural and nonstructural damage.
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Steven T. Hiner, MS, SE
Chapter 4 – Seismic Design Requirements for Building Structures
4.9
Equivalent Lateral Force Procedure
Seismic Base Shear, V
Strength Design force leveli
SDR Workbook – 2012 IBC Version
ASCE 7 – §12.8
ASCE 7 – §12.8.1
 The seismic base shear in a given direction shall be determined in accordance with the following:
V  C SW
ASCE 7 (12.8-1)
Figure 4.12 – Seismic Base Shear
Seismic Response Coefficient, CS
ASCE 7 – §12.8.1.1
 The seismic response coefficient (CS) shall be determined as follows:
CS 
S DS
(R Ie )
ASCE 7 (12.8-2)
NOTE: ASCE 7 (12.8-2) will always governs when T ≤ TS … which typically occurs with low rise and/or
short period structures (i.e.,  3 stories).
 CS need not exceed the following:
CS 
S D1
for T ≤ TL
T (R Ie )
ASCE 7 (12.8-3)
NOTE: ASCE 7 (12.8-3) typically governs for longer period structures when TS ≤ T ≤ TL … but CS
minimum per ASCE 7 (12.8-5) and (12.8-6) need to be considered.
CS 
S D1TL
for T > TL
T (R Ie )
2
ASCE 7 (12.8-4)
NOTE: ASCE 7 (12.8-4) can apply for very long period (i.e., very tall) structures, when T > TL … but CS
minimum per ASCE 7 (12.8-5) and (12.8-6) will typically govern over ASCE 7 (12.8-4).
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Steven T. Hiner, MS, SE
SDR Workbook – 2012 IBC Version
Chapter 5 – Earthquake Loads and Load Combinations
T = self-straining load (e.g., arising from contraction or expansion resulting from
temperature change, shrinkage, moisture change, creep in component materials,
movement due to differential settlement or combinations thereof)
W = load due to wind pressure
Strength Design or Load & Resistance Factor Design
IBC §1605.2
All applicable Strength Design (SD or LRFD) load combinations must be considered since the most
critical load effect may occur when one or more of the contributing loads (e.g., D, E, L, Lr, S ) are not
acting.

Basic (SD or LRFD) Load Combinations
IBC §1605.2.1
Where Strength Design (or Load and Resistance Factor Design) is used, structures and portions thereof
shall be designed to resist the most critical effects resulting from the equations of IBC §1605.2.1.
There are a total of seven SD/LRFD basic load combinations equations in the IBC. Only the two load
combinations which include the earthquake load (E) are noted below:
 1.2(D + F) + 1.0E + f1L + 1.6H + f2S
or … (1.2 + 0.2SDS)D +  QE + f1L + f2S … when F = 0 & H = 0
IBC (16-5)
 0.9(D + F) + 1.0E + 1.6H
or … (0.9 – 0.2SDS)D –  QE … when F = 0 & H = 0
IBC (16-7)
Exception: Where other factored load combinations are specifically required by the provisions of the
IBC (i.e., Chapter 19 – Concrete) … such combinations shall take precedence.
Allowable Stress Design
IBC §1605.3
All applicable Allowable Stress Design (ASD) load combinations must be considered since the most
critical load effect may occur when one or more of the contributing loads (e.g., D, E, L, Lr, S ) are not
acting.

Basic (ASD) Load Combinations
IBC §1605.3.1
Where Allowable Stress Design (or Working Stress Design) … is used, structures and portions thereof
shall be designed to resist the most critical effects resulting from the equations of IBC §1605.3.1.
There are a total of nine Basic (ASD) Load Combination equations in the IBC. Only the three load
combinations which include the earthquake load (E) are noted below:
 D + H + F + (0.6W or 0.7E)
or … (1.0 + 0.14SDS)D + 0.7 QE … when F = 0 & H = 0
IBC (16-12)
 D + H + F + 0.75(0.7E) + 0.75L + 0.75S
or … (1.0 + 0.105SDS)D + 0.75(0.7 QE) + 0.75L + 0.75S
IBC (16-14)
 0.6(D + F) + 0.7E + H
or … (0.6 – 0.14SDS)D – 0.7 QE … when F = 0 & H = 0
IBC (16-16)
Exceptions: see IBC §1605.3.1 for exceptions to crane hook loads, to flat roof snow loads ≤ 30 psf,
roof live loads ≤ 30 psf, flat roof snow loads > 30 psf, etc.
Steven T. Hiner, MS, SE
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Chapter 5 – Earthquake Loads and Load Combinations
SDR Workbook – 2012 IBC Version
5.3 Equivalent Lateral Force Procedure Overview
I, II, III or IV

Ie

Ss & S1

A, B, C, D*, E or F

Fa & Fv

SMS & SM1

SDS & SD1

A*, B*, C, D, E or F

Ta

CS

V = CSW

Fx

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Risk Category
Seismic Importance Factor
Mapped MCER spectral response acceleration parameters
Site Class
Site Coefficients
Soil modified MCER spectral response acceleration parameters
Design spectral response acceleration parameters
Seismic Design Category
Approximate Fundamental Period
Seismic Response Coefficient
Seismic Base Shear at Strength Level (i.e. SD/LRFD)
Vertical Distribution of Seismic Force
Steven T. Hiner, MS, SE
SDR Workbook – 2012 IBC Version
Chapter 6 – Seismic Design Requirements for Nonstructural Components
Chapter 6
Seismic Design Requirements for Nonstructural Components
6.1 ASCE 7 – Chapter 13 Overview
Scope
ASCE 7 – §13.1.1
ASCE 7 – Chapter 13 establishes minimum design criteria for nonstructural components that are
permanently attached to structures, and for their supports and attachments.
A nonstructural component is a part or element of an architectural, electrical or mechanical system.
Seismic Design Category
ASCE 7 – §13.1.2
Nonstructural components shall be assigned to the same Seismic Design Category (SDC) as the structure
that they occupy, or to which they are attached.
Component Importance Factor, Ip
ASCE 7 – §13.1.3
All components shall be assigned a component importance factor (Ip), which will be equal to 1.5 or 1.0.
Use an Ip = 1.5 if any of the following conditions apply:
1. The component is required to function for life-safety purposes after an earthquake, including fire
protection sprinkler systems and egress stairways
2. The component conveys, supports, or otherwise contains toxic, highly toxic, or explosive
substances …
3. The component is in (or attached to) a Risk Category IV structure (i.e., essential facility), and it is
needed for continued operation of the facility or its failure could impair the continued operation of
the facility
4. The component conveys, supports, or otherwise contains hazardous substances …
All other components shall be assigned an Ip = 1.0
Exemptions
ASCE 7 – §13.1.4
The following nonstructural components are exempt from the requirements of ASCE 7 – Chapter 13:
1. Furniture (except storage cabinets > 5 feet tall)
2. Temporary or movable equipment
3. Architectural components in SDC = B (other than parapets supported by bearing walls or shear
walls) provided Ip = 1.0
4. Mechanical and electrical components in SDC = B
5. Mechanical and electrical components in SDC = C provided that Ip = 1.0
6. Mechanical and electrical components in SDC = D, E, or F where Ip = 1.0, the component is
positively attached to the structure, flexible connections are provided between component and
associated ductwork, piping, and conduit and either:
 Component weighs ≤ 400 lbs and has a C.M. located ≤ 4 feet above the adjacent floor
level; or
Steven T. Hiner, MS, SE
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SDR Workbook – 2012 IBC Version
Chapter 6 – Seismic Design Requirements for Nonstructural Components
Specific Requirements & References
Table 6.2 – Specific Mechanical & Electrical Component References
Item
Section Reference
Mechanical Components
Electrical Components
Component Supports
Utility and Service Lines
HVAC Ductwork
Piping Systems
Boilers and Pressure Vessels
Elevator & Escalator Design Requirements
Other Mechanical & Electrical Components
ASCE 7 – §13.6.3
ASCE 7 – §13.6.4
ASCE 7 – §13.6.5
ASCE 7 – §13.6.6
ASCE 7 – §13.6.7
ASCE 7 – §13.6.8
ASCE 7 – §13.6.9
ASCE 7 – §13.6.10
ASCE 7 – §13.6.11
6.6 Structural Walls and their Anchorage
Design for Out-of-Plane Forces
ASCE 7 – §12.11
ASCE 7 – §12.11.1
Structural walls and their anchorage shall be designed for a force normal to the surface (e.g., out-of-plane)
equal to:
Fp  0.4S DS I eWw
 0.10 Ww minimum
Figure 6.2 – Out-of-Plane Forces on Structural Walls
Wall with Cantilever Parapet
Steven T. Hiner, MS, SE
Wall without Parapet
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Chapter 6 – Seismic Design Requirements for Nonstructural Components
SDR Workbook – 2012 IBC Version
Anchorage of Structural Walls and Transfer
of Design Forces into Diaphragms
ASCE 7 – §12.11.2
Wall Anchorage Forces, Fp
ASCE 7 – §12.11.2.1
The anchorage of structural walls to supporting construction (e.g., roof or floor diaphragms) shall provide
a direct connection capable of resisting the following:
Fp  0.4 S DS K a I eW p
ASCE 7 (12.11-1)
 0.2 K a I eW p minimum
where:
K a  1 .0 
Lf
≤ 2.0 maximum
100
Fp = the design force in the individual anchors
ASCE 7 (12.11-2)
SDS = short period design spectral response acceleration parameter
Ie = seismic importance factor
Lf = flexible diaphragm span (feet), use 0 for rigid diaphragms
Wp = the weight of the masonry or concrete wall tributary to the anchor
= Wwall (hw / 2 + hp) (anchor spacing) … for (one-story) walls with a parapet
= Wwall (hw / 2) (anchor spacing) … for (one-story) walls without a parapet
Where the anchorage is not located at the roof and all diaphragms are not flexible, the value from ASCE 7
(12.11-1) is permitted to be multiplied by (1 + 2z/h) / 3, where z is the height of the anchor above the base
of the structure and h is the height of the roof above the base (e.g., h = hn).
NOTE: Structural walls shall be designed to resist bending between anchors where anchor spacing
exceeds 4 feet (i.e., purlin anchors > 4’-0” o.c.).
Refer to Figure 6.3 below for examples of the Lf and Ka values that would be appropriate to determine the
wall anchorage force for the structural walls on each wall line depicted for Building A & Building B.
Figure 6.3 – Wall Anchorage Examples
a. Building A
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b. Building B
Steven T. Hiner, MS, SE
SDR Workbook – 2012 IBC Version
Chapter 6 – Seismic Design Requirements for Nonstructural Components
The following values of Lf and Ka would apply for structural wall anchorage for Building A shown in
Figure 6.3a (Building A) As Flexible Diaphragm:
 Wall anchorage for line A & B: Lf = 150 feet → Ka = 2.0 maximum
 Wall anchorage for line 1 & 2: Lf = 80 feet → Ka = 1.8
As Rigid Diaphragm:
 Wall anchorage for ALL lines: Lf = 0 feet → Ka = 1.0
The following values of Lf and Ka would apply for structural wall anchorage for Building B shown in
Figure 6.3b (Building B) As Flexible Diaphragm:
 Wall anchorage for line A & B (line 1 to 2): Lf = 100 feet → Ka = 2.0
 Wall anchorage for line A & B (line 2 to 3): Lf = 50 feet → Ka = 1.5
 Wall anchorage for line 1, 2 & 3: Lf = 80 feet → Ka = 1.8
As Rigid Diaphragm:
 Wall anchorage for ALL lines: Lf = 0 feet → Ka = 1.0
Additional Requirements for SDC = C, D, E, or F
ASCE 7 – §12.11.2.2
Diaphragms for structures assigned to SDC = C, D, E, or F shall meet the following additional
requirements:
1. Transfer of Anchorage Forces -
 Diaphragms shall be provided with continuous ties (or struts) between diaphragm chords to
distribute these anchorage forces into the diaphragm(s)
 Diaphragm connections shall be positive, mechanical, or welded
 Added chords are permitted to be used to form subdiaphragms (see p. 1-150) to transmit
the anchorage forces to the main continuous cross-ties
 The maximum length-to-width ratio of the structural subdiaphragm shall be 2.5 to 1
 Connections and anchorages capable of resisting the prescribed forces shall be provided
between the diaphragm and the attached components
 Connections shall extend into the diaphragm a sufficient distance to develop the force
transferred into the diaphragm
2. Steel Elements - the strength design forces for steel elements (i.e., purlin anchors), with the
exception of anchor bolts and reinforcing steel, shall be increased by 1.4 times the forces
otherwise required by ASCE 7 – §12.11 (i.e., use 1.4 Fp anchorage force for strap purlin anchors)
3. Wood Diaphragms -
 The continuous ties or struts shall be in addition to the diaphragm sheathing
 Anchorage shall not be accomplished by use of toe nails or nails subject to withdrawal
 Wood ledgers or framing shall not be used in cross-grain bending or cross-grain tension
4. Metal Deck Diaphragms - metal deck shall not be used as the continuous ties (or struts) …
required by ASCE 7 – §12.11.2.2 in the direction perpendicular to the deck span.
Steven T. Hiner, MS, SE
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SDR Workbook – 2012 IBC Version
Chapter 7 – Seismic Design Requirements for Nonbuilding Structures
Determination of Seismic Factors - R, 0 & Cd
R, 0 & Cd are to be determined from ASCE 7 – Table 15.4-2 for nonbuilding structures NOT similar to
buildings.
In general, R & 0 values assigned to these nonbuilding structures are less than those assigned to building
structures (and nonbuilding structures similar to buildings). This is because buildings tend to have
structural redundancy due to multiple bays and frame lines and contain nonstructural and non-considered
resisting elements that effectively give buildings greater damping, strength and ductility during an
earthquake.
Seismic Base Shear, V
ASCE 7 – §12.8.1
Strength Design force level
 When the fundamental period is greater than or equal to 0.06 second (i.e., T ≥ 0.06 second), the
seismic base shear in a given direction shall be determined in accordance with the following:
V  CSW
ASCE 7 (12.8-1)
Seismic Response Coefficient, CS
ASCE 7 – §12.8.1.1
 The Seismic Response Coefficient (CS) shall be determined as follows:
CS 
S DS
(R Ie )
ASCE 7 (12.8-2)
NOTE: ASCE 7 (12.8-2) typically governs for nonbuilding structures, when T < TS
 CS need not exceed the following:
CS 
S D1
for T ≤ TL
T (R I )
ASCE 7 (12.8-3)
CS 
S D1TL
for T > TL
T (R Ie )
ASCE 7 (12.8-4)
and …
2
 CS shall not be less than:
CS  0.044 S DS I e  0.03 minimum
ASCE 7 (15.4-1)
(see ASCE 7-05 Supplement No. 2)
 In addition, for nonbuilding structures where S1 ≥ 0.6 … CS shall not be less than:
CS 
0.8S1
* minimum
(R Ie )
ASCE 7 (15.4-2)
*NOTE: See ASCE 7 – 15.4.1, item 2 for an exception for Tanks and Vessels.
Steven T. Hiner, MS, SE
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Chapter 8 – Diaphragm Design & Wall Rigidity
SDR Workbook – 2012 IBC Version
8.2 Diaphragm Types
Diaphragm Flexibility
ASCE 7 – §12.3.1
The structural analysis shall consider the relative stiffnesses of diaphragms (floor and/or roof), and the
vertical elements of the seismic force-resisting system (e.g., shear walls, braced frames, moment frames).
Unless a diaphragm can be idealized as either flexible or rigid … the structural analysis shall explicitly
include consideration of the stiffness of the diaphragm (i.e., semi-rigid modeling assumption).
Flexible Diaphragm Condition
ASCE 7 – §12.3.1.1
Diaphragms constructed of untopped steel decking (i.e., metal deck without concrete fill) or wood
structural panels (WSP) are permitted to be idealized as flexible if any of the following conditions exist:
 Structures where the vertical elements are steel braced frames; steel and concrete composite
braced frames; or concrete, masonry, steel, or steel and concrete composite shear walls
 One- and two-family dwellings
 Structures of light-frame construction (i.e., wood studs or metal studs) where all of the following
conditions are met:
1. Topping of concrete or similar materials is not placed over wood structural panel (WSP)
diaphragms except for nonstructural toppings  1½ inches thick
2. Each line of vertical elements of the seismic force-resisting system (SFRS) complies with
the allowable story drift of ASCE 7 Table 12.12-1
Rigid Diaphragm Condition
ASCE 7 – §12.3.1.2
Diaphragms of concrete slabs (or concrete filled metal deck) with span-to-depth ratios of 3 to 1 or less
(i.e., L/d ≤ 3) in structures having no horizontal irregularities are permitted to be idealized as rigid.
Calculated Flexible Diaphragm Condition
ASCE 7 – §12.3.1.3
Diaphragms not satisfying the conditions to be considered flexible or rigid … 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 (see Figure 8.4).
Figure 8.4 – Calculated Flexible Diaphragm Condition
    min 
 max  2 max

2


Diaphragm Deflection
ASCE 7 – §12.12.2
The deflection of the diaphragm … shall not exceed the permissible deflection of the attached elements
(e.g., bearing walls) … which shall be that deflection that will permit the attached element to maintain its
structural integrity under the individual loading and continue to support the prescribed loads.
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Steven T. Hiner, MS, SE
SDR Workbook – 2012 IBC Version
Chapter 8 – Diaphragm Design & Wall Rigidity
Figure 8.8 below demonstrates the comparison that is often made between the analysis of a uniformly
loaded flexible diaphragm (on the left) and a uniformly loaded simply supported beam (on the right). For
the design of flexible diaphragms, the shear diagram can be used to determine the maximum unit shear at
the end supports (e.g., shear walls). The moment diagram can be used to determine the maximum chord
force, or the chord force at a specific point on the chord boundary member (see Figure 8.10, p. 1-113).
Figure 8.8 – Flexible Diaphragm Loading
Steven T. Hiner, MS, SE
(Ref. 19)
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SDR Workbook – 2012 IBC Version
Chapter 8 – Diaphragm Design & Wall Rigidity
The Drag Force should always be equal to zero at each end of the collector. Numerical rounding of the
calculated unit diaphragm shear and unit wall shear may result in Drag Forces that do not quite “zero” at
the ends of the collector.
Typically, the wall top plates (at the diaphragm boundary) are used as the chords and collectors for wood
framed stud wall construction.
Collector Elements
ASCE 7 – §12.10.2
Collector elements shall be provided that are capable of transferring the seismic forces originating in other
portions of the structure to the element providing resistance to those forces (e.g., shear walls, braced
frames, moment resisting frames, etc.).
Seismic Design Category C, D, E or F
ASCE 7 – §12.10.2.1
In structures assigned to SDC = C, D, E or F - collector elements and their connections including
connections to vertical elements (e.g., shear walls, braced frames, moment resisting frames, etc.) shall be
designed to resist the maximum of the forces determined from ASCE 7 – §12.10.2.1 items 1, 2 and 3 (see
p. 1-77), which considers the seismic load effects including overstrength factor (0) of ASCE 7 – §12.4.3.
ASCE 7 – §12.10.2.1 allows two exceptions: (drag) forces calculated from Fpx maximum per ASCE 7
equation (12.10-3) using ASCE 7 – §12.4.2.3 load combinations; and for structures braced entirely by
light-frame shear walls (i.e., wood studs or cold formed steel studs) … with drag forces determined from
the governing Fpx force per ASCE 7 (12.10-1), (12.10-2) and (12.10-3).
8.6 Masonry & Concrete Wall Rigidity
Cantilever Shear Wall – Deflection
Figure 8.12 – Cantilever Shear Wall / Pier
 Total   Flexure   Shear
C 
F H3
3EI

1 .2 F H
AG
where:
F = force at top of wall
H = height of wall to force, F
E = modulus of elasticity
G = shear modulus
A = area = t D
I = moment of inertia = t D3/12
Wall Rigidity, R
Rigidity is proportional to the reciprocal of deflection, and is essentially the relative stiffness of the lateral
resisting element (i.e., shear wall).
Masonry & concrete shear walls resist lateral loads in proportion to their Rigidities (R), therefore only
"relative rigidities" are needed.
The relative cantilever wall rigidities are determined for each shear wall using their respective H/D ratios
and Table D1 - Relative Rigidity of Cantilever Shear Walls / Piers (Appendix D, p. 5-20).
Steven T. Hiner, MS, SE
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Chapter 8 – Diaphragm Design & Wall Rigidity
SDR Workbook – 2012 IBC Version
Figure 8.13 – Cantilever Shear Wall Relative Rigidity
 R1 

V1  F 
 R1  R2 
 R2 

V2  F 
 R1  R2 
Fixed Shear Wall – Deflection
Figure 8.14 – Fixed Shear Wall / Pier
 Total   Flexure   Shear
F 
F H3
12 EI

1.2 F H
AG
where:
F = force at top of wall
H = height of wall to force, F
E = modulus of elasticity
G = shear modulus
A = area = t D
I = moment of inertia = t D3/12
Shear Wall with Openings
This method can be used to approximate the shear force in a particular pier when the lateral force (F ) to
the total wall is known, such as for a flexible diaphragm building. The assumption is that the lighter
shaded portion of wall above and below the openings will provide fixity of the darker shaded individual
piers. Determine the "Fixed" Rigidity (RF) of each of the individual piers using their respective H/D ratios
and Table D2 - Relative Rigidity of Fixed Shear Walls / Piers (Appendix D, p. 5-21).
Figure 8.15 – Shear Wall with Fixed Piers

RF1
Force to Pier 1, F1  F 
 RF1  RF 2  RF 3  RF 4
1-116



Steven T. Hiner, MS, SE
SDR Workbook – 2012 IBC Version
Chapter 9 – IBC Chapter 23 – Wood
where:
E  E h  Ev
The weight of the wall (WW) can be considered a dead load effect (D). While the weight of the wall might
be considered in the determination of the total seismic force on the shear wall (i.e., V1 + CSWW), it will not
be used to determine Ev because the dead load effect (D) does not contribute to the determination of the
calculated unit wall shear.
NOTE: The Redundancy factor () shall be considered in the design of the shear walls.
Therefore, E = E h   QE   V1 …
So the ASD calculated unit wall shear:
ASD wall  
 (0.7)(total wall shear)

shear wall width

 (0.7 V1 )
b
(units of plf)
NOTE: The equation above is used to determine the drag force, and may be used for shear wall design
when the wall weight (Ww) is not significant, not given in a problem statement, or when the diaphragm
design force (e.g., ws = fpx = Fpx / L) includes all perimeter walls of the building … essentially when the
base shear (V ) is used to design the diaphragm.
Otherwise, the weight of the shear wall (Ww) can be included in the ASD calculated unit wall shear:
ASD calculated wall  
 (0.7)(V1  C SWW )
b
(units of plf)
Using the ASD calculated wall  and SDPWS Table 4.3A (see Table 9.5), choose the appropriate:




WSP panel grade
WSP nominal panel thickness
Fastener (nail) size
Fastener (nail) spacing
Such that the ASD allowable wall  ≥ ASD calculated wall 
Unit Wall Shear Reduction
SDPWS Table 4.3.4, footnote 1
The nominal unit shear capacities of SDPWS Table 4.3A will require a reduction for walls resisting
seismic loads for any individual shear wall aspect ratio that exceeds 2:1 (but is less than 3.5:1) –


h/b ≤ 2:1 → use nominal unit shear capacities from SDPWS Table 4.3A 3 with no reduction
2:1 < h/b ≤ 3.5:1 → use nominal unit shear capacities from SDPWS Table 4.3A 3 multiplied by
2b/h
NOTE: For h/b = 3.5:1 … this will result in 2b/h = 0.57, or a 43% reduction in the nominal unit shear
capacities noted in SDPWS Table 4.3A when resisting seismic loads (not wind loads).
For 2:1 < h/b ≤ 3½:1 … the reduction factor 2b/h will be somewhere between 1.00 and 0.57, resulting in a
0% to 43% reduction in the nominal unit wall shear values noted in SDPWS Table 4.3A when resisting
seismic loads (not wind loads).
Steven T. Hiner, MS, SE
1-139
SDR Workbook – 2012 IBC Version
Chapter 10 – Other Material Chapters
NOTE: Following the January 17, 1994 Northridge Earthquake, over 100 steel buildings with welded
moment-resisting frames were found to have beam-to-column connection fractures. Usually the fractures
initiated at the complete joint penetration weld between the beam bottom flange and the column flange.
Investigators have suggested that the fractures may be due to a number of factors such as notch effects
created by a left in place weld backing bar; sub-standard welding (excessive porosity, slag inclusions,
incomplete fusion); and/or pre-earthquake fractures due to initial shrinkage of the weld during cool-down.
In September 1994, the International Conference of Building Officials (ICBO) adopted an emergency
code change to the 1994 Uniform Building Code (UBC). This code change omitted the pre-qualified
connection (see Figure 10.7) and required that connections be designed to sustain inelastic rotation and
develop the strength criteria as demonstrated by cyclic testing or calculation.
In November 2000, the SAC Joint Venture finalized the welded steel moment frame issues which were
published by the Federal Emergency Management Agency (FEMA) in the following documents:
 FEMA 350 – Recommended Seismic Design Criteria for New Steel Moment-Frame Buildings
 FEMA 351 – Recommended Seismic Evaluation and Upgrade Criteria for Existing Welded Steel
Moment-Frame Buildings
 FEMA 352 – Recommended Post-Earthquake Evaluation and Repair Criteria for Welded Steel
Moment-Frame Buildings
 FEMA 353 – Recommended Specifications and Quality Assurance Guidelines for Steel MomentFrame Construction for Seismic Applications
Doubler Plates
Doubler plates are sometimes added to one (or both sides) of the column web and are intended to increase
the columns web shear strength and web crippling capacity within the panel zone.
Continuity Plates
Continuity plates are sometimes added to provide “continuity” of the intersecting beam (or girder) flanges
across the column web. The requirement for continuity plates is a function of the column yield strength,
flange width, and flange thickness, and the intersecting beam yield strength, flange width, and flange
thickness.
Intermediate Moment Frames (IMF)

AISC 341-10 – §E.2
R = 4½ for steel IMF*
IMF’s are expected to withstand limited inelastic deformations (in their members and connections) when
subjected to the Design Basis Earthquake ground motions.
Limited special detailing of beam-column joint, therefore limited ductility (i.e., lower R).
Ordinary Moment Frames (OMF)

AISC 341-10 – §E.1
R = 3½ for steel OMF*
OMF’s are expected to withstand minimal inelastic deformations (in their members and connections)
when subjected to the Design Basis Earthquake ground motions.
Very limited special detailing of beam-column joint, therefore very limited ductility (i.e., very low R).
*NOTE: Typically steel IMF’s and OMF’s will not be permitted in structures assigned to Seismic Design
Category D, E or F … with the exception of steel OMF’s meeting the requirements of ASCE 7 –
§12.2.5.6.1 for SDC = D or E or §12.2.5.6.2 for SDC = F, and steel IMF’s meeting requirements of ASCE
7 – §12.2.5.7.1 for SDC = D or §12.2.5.7.2 for SDC = E or §12.2.5.7.3 for SDC = F.
Steven T. Hiner, MS, SE
1-165
SDR Workbook – 2012 IBC Version
Part 2 – Example Problems
SM1 = Fv S1
= 1.3 (0.52) = 0.68
IBC (16-38)
SD1 = 2/3 SM1
= 2/3 (0.68) = 0.45i
IBC (16-40)
NOTE: Alternatively, SDS & SD1 can be quickly determined using Tables 3.2 & 3.3 (p. 1-33 & 34):
Table 3.2 (p. 1-33)  SS = 1.25  Site Class C  SDS = 0.83
Table 3.3 (p. 1-34)  S1 = 0.52  Site Class C  SD1 = 0.45
B.) Seismic Design Category, SDC
S1 = 0.52 < 0.75  therefore, use IBC Table 1613.3.5(1) & Table 1613.3.5(2) to determine SDC
SDS = 0.83 & RC = II  2012 IBC Table 1613.3.5(1)  SDC = D
SD1 = 0.45 & RC = II  2012 IBC Table 1613.3.5(2)  SDC = D
 iSDC = Dii
C.) Approximate Fundamental Period, Ta
Ta  Ct hnx
ASCE 7 (12.8-7)
Steel SCBF  ASCE 7 – Table 12.8-2: all other structural systems – Ct = 0.02 & x = 0.75
Ta
= 0.02 (hn)0.75
= 0.02 (36 feet)0.75 = 0.29 secondi
Or using Table C1 (Appendix C, p. 5-18)  CBF & hn = 36 feet  Ta = 0.29 second
NOTE: T = 0.29 second < TS 
S D1 0.45
= 0.54 second   ASCE (12.8-2) will govern Cs

S DS 0.83
D.) Seismic Response Coefficient, CS
CS 
S DS
(0.83)
=
= 0.138
(R Ie )
(6 1.0)
 governs
ASCE 7 (12.8-2)
CS need not exceed the following,
CS 
S D1
(0.45)
=
= 0.259
0.29(6 1.0)
T (R Ie )
ASCE 7 (12.8-3)
NOTE: ASCE 7 (12.8-4), (12.8-5), and (12.8-6) are not applicable since T < TS (and T << TL)
CS = 0.138i
E.) Seismic Base Shear, V
V = CS W
= 0.138 (280 kips) = 38.6 kipsi
Steven T. Hiner, MS, SE
ASCE 7 (12.8-1)
2-7
SDR Workbook – 2012 IBC Version
Part 2 – Example Problems
SD1 = 2/3 SM1
= 2/3 (0.28) = 0.19i
IBC (16-40)
Table 3.2 (p. 1-33)  SS = 0.73  Site Class B  SDS = 0.49 … by interpolation
Table 3.3 (p. 1-34)  S1 = 0.28  Site Class B  SD1 = 0.19
B.) Seismic Design Category, SDC
S1 = 0.28 < 0.75  therefore, use IBC Table 1613.3.5(1) & Table 1613.3.5(2) to determine SDC
SDS = 0.49 & RC = IV  2012 IBC Table 1613.3.5(1)  SDC = D
SD1 = 0.19 & RC = IV  2012 IBC Table 1613.3.5(2)  SDC = D
 iSDC = Dii
C.) Approximate Fundamental Period, Ta
Ta  Ct hnx
ASCE 7 (12.8-7)
Concrete SMF  ASCE 7 – Table 12.8-2: Concrete moment-resisting frames – Ct = 0.016 & x = 0.9
Ta = 0.016 (hn)0.9
= 0.016 (67 feet)0.9 = 0.70 secondi
Using Table C1 (Appendix C, p. 5-18)  Concrete MRF & hn = 67 feet  Ta = 0.70 second
NOTE: T = 0.70 second > TS 
S D1 0.19
= 0.39 second   ASCE 7 (12.8-2) will not govern Cs

S DS 0.49
D.) Seismic Response Coefficient, CS
CS 
S DS
(0.49)
=
= 0.092
(R Ie )
(8 1.5)
ASCE 7 (12.8-2)
CS need not exceed the following,
CS 
S D1
(0.19)
=
= 0.051
0.70(8 1.5)
T (R Ie )
 governs
ASCE 7 (12.8-3)
CS shall not be less than the following,
CS  0.044S DS I e = 0.044(0.49)(1.5) = 0.032
ASCE 7 (12.8-5)
≥ 0.01 minimum
NOTE: ASCE 7 (12.8-4) is not applicable since T = 0.70 second << TL = 12 seconds & ASCE 7 (12.8-6)
is not applicable since S1 = 0.28 < 0.6
CS = 0.051i
E.) Seismic Base Shear, V
V = Cs W
ASCE 7 (12.8-1)
= 0.051 (5,000 kips) = 255 kipsi
Steven T. Hiner, MS, SE
2-9
SDR Workbook – 2012 IBC Version
Part 2 – Example Problems
SD1 = 2/3 SM1
= 2/3 (1.35) = 0.90i
IBC (16-40)
Table 3.2 (p. 1-33)  SS = 1.75  Site Class D (assumed)  SDS = 1.17
Table 3.3 (p. 1-34)  S1 = 0.90  Site Class D (assumed)  SD1 = 0.90
B.) Seismic Design Category, SDC
S1 = 0.90 > 0.75 & RC = II  2012 IBC §1613.3.5  SDC = E
 iSDC = Ei
C.) Approximate Fundamental Period, Ta
Ta  Ct hnx
ASCE 7 (12.8-7)
WSP shear wall  ASCE 7 – Table 12.8-2: all other structural systems – Ct = 0.02 & x = 0.75
Ta = 0.02 (hn)0.75
= 0.02 (25 feet)0.75 = 0.22 secondi
Using Table C1 (Appendix C, p. 5-18)  Shear Walls & hn = 25 feet  Ta = 0.22 second
NOTE: T = 0.22 second < TS 
S D1 0.90
= 0.77 second   ASCE 7 (12.8-2) will govern Cs.

S DS 1.17
D.) Seismic Response Coefficient, CS
CS 
S DS
(1.17)
=
= 0.180
(R Ie )
(6.5 1.0)
 governs
ASCE 7 (12.8-2)
CS need not exceed the following,
CS 
S D1
(0.90)
=
= 0.629
T ( R I e ) 0.22(6.5 1.0)
ASCE 7 (12.8-3)
NOTE: ASCE 7 (12.8-4), (12.8-5), and (12.8-6) are not applicable since T < TS (and T << TL)
CS = 0.180i
E.) Seismic Base Shear, V
V = CS W
ASCE 7 (12.8-1)
= 0.180 (135 kips) = 24.3 kipsi
NOTE: For low-rise buildings (i.e., ≤ 3 stories), ASCE 7 (12.8-2) will typically govern since these
structures have relatively short periods such that T < TS … therefore, it is typically not necessary to
calculate the structures period from equation ASCE 7 (12.8-7), or seismic response coefficient from
ASCE 7 (12.8-3), (12.8-4), (12.8-5) & (12.8-6) for low-rise buildings.
Steven T. Hiner, MS, SE
2-11
SDR Workbook – 2012 IBC Version
Part 2 – Example Problems
Table 3.2 (p. 1-33)  SS = 1.52  Site Class E  SDS = 0.91 … by interpolation
Table 3.3 (p. 1-34)  S1 = 0.76  Site Class E  SD1 = 1.22 … by interpolation
B.) Seismic Design Category, SDC
S1 = 0.76 > 0.75 & RC = IV  2012 IBC §1613.3.5  SDC = F
 iSDC = Fi
C.) Approximate Fundamental Period, Ta
Ta  Ct hnx
ASCE 7 (12.8-7)
Dual System D.1 - steel EBF & steel SMF  ASCE 7 – Table 12.8-2: Ct = 0.03 & x = 0.75
Ta = 0.03 (hn)0.75
= 0.03 (108 feet)0.75 = 1.01 secondi
Using Table C1 (Appendix C, p. 5-18)  Dual System D.1 & hn = 108 feet  Ta = 1.01 second
NOTE: T = 1.00 second < TS 
S D1 1.22

= 1.34 second   ASCE 7 (12.8-2) will govern Cs
S DS 0.91
D.) Seismic Response Coefficient, CS
CS 
(0.91)
S DS
= 0.170
=
(R Ie )
(8 1.5)
 governs
ASCE 7 (12.8-2)
CS need not exceed the following,
CS 
S D1
(1.22)
= 0.229
=
T ( R I e ) 1.00(8 1.5)
ASCE 7 (12.8-3)
NOTE: ASCE 7 (12.8-4), (12.8-5), and (12.8-6) are not applicable since T < TS (and T << TL)
CS = 0.170i
E.) Seismic Base Shear, V
V = CS W
ASCE 7 (12.8-1)
= 0.170 (24,000 kips) = 4,080 kipsi
NOTE: This scenario is somewhat unusual in that ASCE 7 (12.8-2) ended up as the governing equation
for the seismic response coefficient (CS) even though the building was an 8-story structure … simply
because T = Ta < TS. The reason that ASCE 7 (12.8-3) did not end up governing is because the
structure is supported by soft soil (Site Class E). The presence of soft soil will “push” TS toward the
longer period range of the design response spectrum to where taller/longer period structures will still
be governed by ASCE 7 (12.8-2).
Steven T. Hiner, MS, SE
2-13
SDR Workbook – 2012 IBC Version
Part 2 – Example Problems
B.) N-S DIRECTION: Diaphragm span = L1 = L2 = L / 2 = 62.5′, d = 50′
r
1. Maximum Unit (Roof) Diaphragm Shear,
The addition of the interior shear wall on line 2 will create two diaphragms that span equal amounts
(e.g., L1 = L2 = L / 2). One diaphragm spans from line 1 to line 2, and the other diaphragm spans
from line 2 to line 3.
ws = V / L = (35,000 lbs) / 125′ = 280 plf
V1 = V3 = ws (L1 / 2) = ws (L2 / 2) = (280 plf )(62.5′) / 2 = 8,750 lbs
roof 1 = 3 = V1 / d = (8,750 lbs) / 50′ = 175 plfi
From Part A.1 - roof 1 = 3 = 350 plf →
2. Maximum Unit Wall Shear,
(SD force level)
 50% reduction in max. unit (roof) diaphragm sheari
w
By inspection, the maximum unit wall shear will occur on line 2 & 3  Wall Line 1: V1 = V / 4 & total shear wall length, Σb1 = 30′
wall 1 = V1 / Σw = (8,750 lbs) / 30′ = 291 plfi
(SD force level)
 Wall Line 2: V2 = V / 2 & total shear wall length, Σb2 = 50′
wall 1 = V1 / Σw = (17,500 lbs) / 50′ = 350 plfi ← governs
(SD force level)
 Wall Line 3: V3 = V / 4 & total shear wall length, Σb3 = 25′
wall 3 = V3 / Σw = (8,750 lbs) / 25′ = 350 plfi ← governs
From Part A.2 - wall 3 = 700 plf →
(SD force level)
 50% reduction in max. unit wall sheari
3. Maximum Chord Force on lines A & B, CF
max. CF = ws (L1)2 / 8d
= (280 plf )(62.5′)2 / 8(50′) = 2,730 lbsi
From Part A.3 - max. chord force CF = 10,940 lbs →
(SD force level)
 75% reduction in maximum chord forcei
4. Maximum Drag Force, Fd
Drag Force - Line 1
Steven T. Hiner, MS, SE
Drag Force - Line 2
Drag Force - Line 3
2-33
SDR Workbook – 2012 IBC Version
Part 2 – Example Problems
4. Drag Force Diagram on lines A & B, Fd
roof A = B = 494 plf
Wall Line A: Fd = 0 lbs
Wall Line B: Fd = (494 plf)(20′) = 9,880 lbs
(SD/LRFD force level)


Drag Force – Line A
Drag Force – Line B
B.) E-W DIRECTION: L = 40′, d = 70′
1. Design Seismic Force to Diaphragm, ws = fp1 = Fp1/L
roof DL + 20% snow
exterior walls
ws = fp1 = 0.190 [(16 psf + 20%·100 psf)(70′) + (85 psf)(14′/2 + 2′)(2 walls)] = 770 plfi
2. Unit Roof Shear on lines 1 & 2, r
V1 = V2 = ws L / 2 = (770 plf)(40′/2) = 15,400 lbs
Roof 1 = 2 = V1 / d = (15,400 lbs) / 70′ = 220 plfi
(SD/LRFD force level)
3. Maximum Chord Force on lines A & B, CF
max. M = ws L2 / 8 = (770 plf)(40′)2 / 8 = 154,000 lb-ft
max. CF = (154,000 lb-ft) / 70’ = 2,200 lbsi
(SD/LRFD force level)
4. Shear Force to walls 1A & 1B
Relative Rigidities: assume cantilever walls, Table D1 - Relative Rigidity of Cantilever Shear
Walls / Piers (Appendix D, p. 5-20)

Table D1 (p. 5-20)
 R1A = 0.833
Wall 1B: H/D = 14′/22′ = 0.64 
Table D1 (p. 5-20)
 R1B = 3.369
Wall 1A: H/D = 14′/11′= 1.27
R =
R1A + R1B = 0.833 + 3.369 = 4.202
Steven T. Hiner, MS, SE
2-37
Part 2 – Example Problems
SDR Workbook – 2012 IBC Version
B.) Center of Rigidity, CR
Shear Wall Rigidities: (assume cantilever walls, Table D1 - Relative Rigidity of Cantilever Shear
Walls / Piers, Appendix D, p. 5-20)
Wall A : H/D = 15′/30′ = 0.50

Table D1 (p. 5-20)

RA = 5.0
Wall B : H/D = 15′/20′ = 0.75

Table D1 (p. 5-20)

RB = 2.54
Table D1 (p. 5-20)

RC = 7.49
R
Y
 R A  R B = 5.0 + 2.54 = 7.54
Wall C : H/D = 15′/40′ = 0.375 
Wall D : RD = RB = 2.54
Wall E : RE = RD = 2.54
R
X
 RC  R D  R E = 7.49 + 2.54 + 2.54 = 12.57
R x
R
R y

R
X CR 
Y
= [5.0 (0′) + 2.54 (80′)] / (7.54) = 26.9 feeti
Y
Y CR
X
= [7.49 (40′) + 2.54 (10′) + 2.54 (0′)] / (12.57) = 25.9 feeti
X
2-54
Steven T. Hiner, MS, SE
SDR Workbook – 2012 IBC Version
4.19
What is the approximate fundamental period (Ta) of a 110 foot tall steel eccentrically braced
frame (steel EBF) building?
a.
b.
c.
d.
4.20
1.04 second
1.56 second
1.90 second
2.12 second
What is the approximate fundamental period (Ta) of a 35 foot tall Dual System building (w/
steel SMF’s & reinforced concrete shear walls)?
a.
b.
c.
d.
4.23
0.50 second
0.43 second
0.64 second
0.89 second
What is the approximate fundamental period (Ta) of a 195 foot tall special steel moment frame
(steel SMF) building?
a.
b.
c.
d.
4.22
0.68 second
1.02 second
1.20 second
1.54 second
What is the approximate fundamental period (Ta) of a 5 story building using entirely
intermediate steel moment frames (steel IMF), with all story heights greater than 10 feet?
a.
b.
c.
d.
4.21
Part 3 – Multiple Choice Problems
0.48 second
0.39 second
0.35 second
0.29 second
Given TS = 0.6 second & Seismic Design Category D (SDC = D), which of the following
structures would not be permitted the use of the Equivalent Lateral Force (ELF) procedure?
a.
b.
c.
d.
e.
A 200 foot tall regular structure with T = 2.4 seconds
A 3-story light-frame irregular structure of Risk Category II
A 150 feet tall structure with T = 2.0 seconds and reentrant corner irregularity
All of the above
Both a & c
An owner proposes to construct an office building of Seismic Design Category D using steel
special concentrically braced frames (SCBF) as the vertical seismic force-resisting elements
(in both principal directions). Answer questions 4.24 to 4.27.
4.24
What response modification coefficient (R) should be used for seismic design?
a.
b.
c.
d.
8
7
6
3¼
Steven T. Hiner, MS, SE
3-19
Part 3 – Multiple Choice Problems
5.3
A lateral analysis of a 2-story office building determines that a steel braced frame column has
the following axial loads: D = 20 kips, L = 15 kips, Lr = 0 kips and E = 25 kips. Assume  =
1.0 to determine the maximum axial compression force in this column using the Strength
Design (SD or LRFD) load combinations of IBC §1605.2.
a.
b.
c.
d.
5.4
SDR Workbook – 2012 IBC Version
45.0 kips
52.9 kips
56.5 kips
61.2 kips
In ASCE 7 – §12.4.2.1, the symbol QE represents which of the following?
a. The effects of the horizontal seismic forces from the seismic base shear (V )
b. The effects of the horizontal seismic forces from the nonstructural component seismic
design force (Fp)
c. The seismic design base shear (V )
d. Both a & b
5.5
Which of the following conditions would require the use of the load combinations with
overstrength factor (Ω0) of ASCE 7 – §12.4.3.2?
I. Vertical structural irregularity type 4 (ASCE 7 – Table 12.3-2)
II. Vertical structural irregularity type 5a (ASCE 7 – Table 12.3-2)
III. Horizontal structural irregularity type 4 (ASCE 7 – Table 12.3-1)
a.
b.
c.
d.
5.6
I
I & III
II & III
I, II & III
Use of a redundancy factor ( ρ) greater than 1.0 is intended to:
a.
b.
c.
d.
reduce the inelastic response and ductility demand of a structure
increase the seismic base shear (V) on a structure
decrease the calculated story drift within a structure
Both a & b
A column of a steel special concentrically braced frame (SCBF), of a single-story Medical
Office building (SDC = D), is determined to support the following axial load effects: dead
load - D = 35 kips, floor live load - L = 0 kips, roof live load - Lr = 15 kips, and horizontal
seismic load effect - QE = 15 kips. Given SDS = 1.25, overstrength factor Ω0 = 2, and
redundancy factor  = 1.3, answer questions 5.7 through 5.13.
5.7
What is the vertical seismic load effect (Ev) axial force in this column?
a.
b.
c.
d.
5.8
What is the horizontal seismic load effect (Eh) axial force in this column?
a.
b.
c.
d.
3-34
0 kips
± 7.0 kips
± 8.8 kips
± 11.7 kips
± 15.0 kips
± 19.5 kips
± 24.4 kips
± 11.7 kips
Steven T. Hiner, MS, SE
SDR Workbook – 2012 IBC Version
5.9
What is the maximum axial compression force in this column using the Strength Design (SD or
LRFD) load combinations of IBC §1605.2?
a.
b.
c.
d.
5.10
50.8 kips
63.2 kips
72.0 kips
80.8 kips
What is the minimum axial compression force in this column using the strength design load
combinations with overstrength factor of ASCE 7 – §12.4.3.2?
a.
b.
c.
d.
5.14
± 15.0 kips
± 20.0 kips
± 30.0 kips
± 37.5 kips
What is the maximum axial compression force in this column using the strength design load
combinations with overstrength factor of ASCE 7 – §12.4.3.2?
a.
b.
c.
d.
5.13
3.2 kips
7.3 kips
9.1 kips
11.7 kips
What is the horizontal seismic load effect with overstrength factor (Emh) axial force in this
column?
a.
b.
c.
d.
5.12
53.0 kips
61.5 kips
66.9 kips
70.3 kips
What is the minimum axial compression force in this column using the Strength Design (SD or
LRFD) load combinations of IBC §1605.2?
a.
b.
c.
d.
5.11
Part 3 – Multiple Choice Problems
10.2 kips
5.0 kips
1.5 kips
– 7.2 kips
What is the maximum redundancy factor ( ρ) that needs to be considered for a structure
assigned to Seismic Design Category E (SDC = E)?
a.
b.
c.
d.
1.0
1.25
1.3
1.5
Steven T. Hiner, MS, SE
3-35
Part 3 – Multiple Choice Problems
6.3
SDR Workbook – 2012 IBC Version
A 2,000 pound sign is to be attached to a three-story medical office building. The sign is to be
attached to the exterior 8″ thick concrete wall, 5 feet above the top of the 3rd floor (i.e., Level 2)
as shown in the figure below. Given an SDS = 0.93, what would be the horizontal seismic force
for anchorage?
a.
b.
c.
d.
500 lbs
1,000 lbs
1,700 lbs
2,400 lbs
Section
6.4
Using the figure in Problem 6.3 (above), what would be the horizontal seismic force for
anchorage if the sign was to be anchored at the top of the parapet (i.e., 4 feet above the roof
level)?
a.
b.
c.
d.
6.5
What component response modification factor (Rp) should be used for the seismic bracing of a
suspended ceiling system in a building?
a.
b.
c.
d.
6.6
Element of a structure
Architectural component
Mechanical component
Electrical component
A chimney of a single-family wood framed house is considered a/an:
a.
b.
c.
d.
3-38
1
1½
2½
3
A heat exchanger attached to a building structure is considered a/an:
a.
b.
c.
d.
6.7
500 lbs
1,900 lbs
2,100 lbs
2,400 lbs
Element of a structure
Architectural component
Mechanical component
Electrical component
Steven T. Hiner, MS, SE
SDR Workbook – 2012 IBC Version
6.8
Given a 2000 lbs storage tank mounted on the roof of a Risk Category II building where
SDS = 0.73, ap = 1 & Rp = 2½ … what is the seismic design force, Fp?
a.
b.
c.
d.
6.9
Part 3 – Multiple Choice Problems
700 lbs
1400 lbs
2000 lbs
2700 lbs
Below is a cross section through a large pipe supported from the concrete roof slab of a
Hospital where SDS = 0.95, Ip = 1.5, ap = 2½ & Rp = 9. The pipe and its contents weigh 100 plf
and have lateral bracing at a spacing of 8 feet (i.e., 8'-0" o.c.). Calculate the axial force in the
brace due to the seismic design force, Fp?
a.
b.
c.
d.
50 lbs
80 lbs
380 lbs
540 lbs
Section
6.10
What component amplification factor (ap) should be used to design the required steel
reinforcement size and spacing in a 4 foot tall concrete parapet, braced 3 feet above the roof
level?
a.
b.
c.
d.
6.11
Given a mechanical component with an operating weight (Wp) of 1500 lbs and component
stiffness (Kp) of 3000 lbs/inch, what is the component period (Tp)?
a.
b.
c.
d.
6.12
1
1¼
1½
2½
0.01 second
0.10 second
0.23 second
0.78 second
For masonry or concrete wall buildings, anchor straps (i.e., purlin anchors) at the boundaries of
floor and roof diaphragms are utilized to:
a.
b.
c.
d.
transfer gravity loads to the wall
transfer diaphragm shear to the shear walls
reduce shrinkage of wood joists
secure the walls to the diaphragm
Steven T. Hiner, MS, SE
3-39
Part 4 – Multiple Choice Solutions
Problem
Answer
SDR Workbook – 2012 IBC Version
Reference / Solution
0.9( D  F )  1.0 E  1.6 H
IBC (16-7)
minimum axial, P = 0.9(35 kips + 0) + 1.0(– 28.3 kips) + 1.6(0)
= 3.2 kips 
5.11
c
p. 1-74, Horizontal Seismic Load Effect w/ Overstrength Factor & ASCE 7-10
p. 69, §12.4.3.1
ASCE 7 (12.4-7)
Emh = ± Ω0 QE
Emh = ± 2 (15 kips) = ± 30.0 kips 
5.12
d
p. 1-76, Basic (SD or LRFD) Load Combinations w/ Overstrength Factor &
ASCE 7-10 p. 69, §12.4.3.2
0.2SDS = 0.2(1.25) = 0.25
Ω0 QE = + 30.0 kips … use “+” to determine maximum compression.
The following strength design load combination including overstrength factor
(Ω0) will govern for maximum axial compression –
5. (1.2  0.2 S DS ) D   0QE  L  0.2 S
(1.2 + 0.25)(35 kips) + 30.0 kips + 0 + 0.2(0) = 80.8 kips 
5.13
d
p. 1-76, Basic (SD or LRFD) Load Combinations w/ Overstrength Factor &
ASCE 7-10 p. 69, §12.4.3.2
0.2SDS = 0.2(1.25) = 0.25
Ω0 QE = – 30.0 kips … use “–” to determine minimum compression.
The following strength design load combination including overstrength factor
(Ω0) will govern for mimimum axial compression –
7. (0.9  0.2 S DS ) D   0QE
(0.9 – 0.25)(35 kips) + (– 30.0 kips) = – 7.2 kips 
5.14
c
p. 1-72 to 73, Redundancy Factor & ASCE 7-10 p. 67, §12.3.4.2
For structures assigned to SDC = D … E or F –  = 1.3 shall be used unless
one of the following two conditions is met, where  = 1.0 is permitted …
 maximum ρ = 1.3 
5.15
d
p. 1-59, Overturning Moment
OTMbase = 6 kips (12) + 12 kips (24) = 360 kip-ft
TA = CB = (360 kip-ft) / (16′) = 22.5 kips
 TA = 22 kips 
5.16
b
By observation, the lateral forces shown in the given direction will result in
the frame deflecting to the right which will “stretch” the braces X1 & X2, and
“compress” the braces Y1 & Y2. Therefore braces X1 & X2 will be in
tension and braces Y1 & Y2 will be in compression. Because the rod braces
cannot resist compression, only braces X1 & X2 will be effective to resist the
forces in the given direction.
 X1 & X2 
4-28
Steven T. Hiner, MS, SE
SDR Workbook – 2012 IBC Version
Part 4 – Multiple Choice Solutions
Problem
Answer
Reference / Solution
5.17
c
X2 is the brace is the 2nd story and it will carry the entire 12 kip force as the
horizontal component in that brace. The resultant axial (tension) force can be
determined from Trigonometry. With a 12 story height and 16 bay spacing,
the brace length will be 20 (i.e., 3-4-5 triangle … or 12-16-20).
 TX2 = (12 kips)(20) / (16) = 15 kips 
5.18
c
p. 1-72, Horizontal Seismic Load Effect & ASCE 7-10 p. 69, §12.4.2.1
QE = effects of horizontal seismic forces from V or Fp …
The axial force in brace X2 is due to F2 (but not equal to F2), and the Fx
forces are due to the base shear V, therefore the axial force in brace X2 is
from V (but certainly not equal to V ).
 QE 
5.19
d
X1 is the brace is the 1st story and it will carry the entire (12 kip + 6 kip) force
as the horizontal component in that brace. The resultant axial (tension) force
can be determined from Trigonometry. With a 12 story height and 16 bay
spacing, the brace length will be 20 (i.e., 3-4-5 triangle … or 12-16-20).
 TX1 = (18 kips)(20) / (16) = 22.5 kips ≈ 22 kips 
5.20
c
Per 5.19, brace X1 will carry the entire 18 kips of horizontal seismic force in
the given direction, and that entire force will be transferred to support A.
 Shear at A, VA = 18 kips 
5.21
a
Per 5.19, brace Y1will carry none of the horizontal seismic force in the given
direction because brace Y1 cannot resist compression forces.
 Shear at B, VB = 0 kips 
5.22
a
p. 1-72, Vertical Seismic Load Effect & ASCE 7-10 p. 69, §12.4.2.2
Ev = ± 0.2SDS D
ASCE 7 (12.4-4)
 Ev = ± 0.2(0.72)(110 kips) = ± 15.8 kips ≈ ± 16 kips 
5.23
d
p. 1-72, Horizontal Seismic Load Effect & ASCE 7-10 p. 69, §12.4.2.1
Per 5.18 and 5.19, the axial force in brace X1 (TX1) is a QE force and is equal
to 22.5 kips.
QE = TX1 = 22.5 kips
ρ = 1.3 (given)
Eh = ρ QE
ASCE 7 (12.4-3)
 Eh = 1.3(22.5 kips) = 29.2 kips ≈ 29 kips 
6.1
d
p. 1-84 & ASCE 7-10 p. 91, Table 13.5-1 - Architectural Component
Cantilever Elements (Unbraced …) - parapets – ap = 2½ 
6.2
b
ASCE 7-10 p. 91 & 93, Tables 13.5-1 and 13.6-1
Footnote c – overtsrength as required for anchorage to concrete 
6.3
c
p. 1-82, Seismic Design Force & ASCE 7-10 p. 88 & 89, §13.3.1
(continued)
Steven T. Hiner, MS, SE
4-29
Part 4 – Multiple Choice Solutions
Problem
Answer
SDR Workbook – 2012 IBC Version
Reference / Solution
SDS = 0.93 (given)
ap = 2½ & Rp = 3 – Table 13.5-1 - Signs & Billboards
z = 12′ + 10′ + 5′ = 27′
h = hn = 12′ + 2(10′) = 32′
(z / h) = 27′ / 32′ = 0.84
Ip = 1.0 – ASCE 7-10 – §13.1.3
Rp /Ip = (3 / 1.0) = 3

0.4a p S DSW p 
1  2 z 
Fp 
ASCE 7 (13.3-1)
RP I P  
h 
= 0.4(2.5)(0.93) Wp [1 + 2 (0.84)] / (3) = 0.83 Wp  (governs)
maximum Fp  1.6 S DS I pW p
ASCE 7 (13.3-2)
= 1.6(0.93)(1.0) Wp = 1.49 Wp
minimum Fp  0.3S DS I pW p
ASCE 7 (13.3-3)
= 0.3(0.93)(1.0) Wp = 0.28 Wp
 Fp = 0.83(2,000 lbs) = 1,660 lbs ≈ 1,700 lbs 
6.4
b
p. 1-82, Seismic Design Force & ASCE 7-10 p. 88 & 89, §13.3.1
SDS = 0.93 (given)
ap = 2½ & Rp = 3 – Table 13.5-1 - Signs & Billboards
z = h + 4′ parapet? … no, because ASCE 7-10 – §13.3.1 states that “the value
of z / h need not exceed 1.0”
 use (z / h) = 1.0
Ip = 1.0 – ASCE 7-10 – §13.1.3
Rp/Ip = (3 / 1.0) = 3

0.4a p S DSW p 
1  2 z 
Fp 
ASCE 7 (13.3-1)
RP I P  
h 
= 0.4(2.5)(0.93) Wp [1 + 2(1.0)] / (3) = 0.93 Wp  (governs)
maximum Fp  1.6 S DS I pW p
ASCE 7 (13.3-2)
= 1.6(0.93)(1.0) Wp = 1.49 Wp
minimum Fp  0.3S DS I pW p
ASCE 7 (13.3-3)
= 0.3(0.93)(1.0) Wp = 0.28 Wp
 Fp = 0.93(2,000 lbs) = 1,860 lbs ≈ 1,900 lbs 
6.5
c
p. 1-85 & ASCE 7-10 p. 91, Table 13.5-1 – Architectrural Components
Ceilings (All) – Rp = 2½ 
6.6
c
p. 1-86 & ASCE 7-10 p. 93, Table 13.6-1 – Mechanical & Electrical
Components
Heat exchanger = mechanical component 
4-30
Steven T. Hiner, MS, SE
SDR Workbook – 2012 IBC Version
Part 4 – Multiple Choice Solutions
Problem
Answer
Reference / Solution
6.7
b
p. 1-85 & ASCE 7-10 p. 91, Table 13.5-1 – Architectrural Components
Chimney (as part of a “house”) = architectural component 
6.8
a
p. 1-82, Seismic Design Force & & ASCE 7-10 p. 88 & 89, §13.3.1
SDS = 0.73, ap = 1 & Rp = 2½ (given)
z/h = 1.0 – since tank is anchored on the roof (i.e., z = h)
Ip = 1.0 – ASCE 7-10 – §13.1.3
Rp/Ip = (2½ / 1.0) = 2½

0.4a p S DSW p 
1  2 z 
ASCE 7 (13.3-1)
RP I P  
h 
= 0.4(1)(0.73) Wp [1 + 2(1.0)] / (2½) = 0.35 Wp  (governs)
Fp 
maximum Fp  1.6 S DS I pW p
ASCE 7 (13.3-2)
= 1.6(0.73)(1.0) Wp = 1.17 Wp
minimum Fp  0.3S DS I pW p
ASCE 7 (13.3-3)
= 0.3(0.73)(1.0) Wp = 0.22 Wp
 Fp = 0.35(2,000 lbs) = 700 lbs 
6.9
d
p. 1-82, Seismic Design Force & ASCE 7-10 p. 88 & 89, §13.3.1
SDS = 0.95, Ip = 1.5, ap = 2½ & Rp = 9 (given)
z/h = 1.0 – since pipe is suspended from, and braced to, the roof (i.e., z = h)
Rp/Ip = (9 / 1.5) = 6.0

0.4  a p S DSW p 
1  2 z 
ASCE 7 (13.3-1)
RP I P  
h 
= 0.4(2.5)(0.95) Wp [1 + 2 (1.0)] / (6.0) = 0.48 Wp  (governs)
Fp 
maximum Fp  1.6 S DS I pW p
ASCE 7 (13.3-2)
= 1.6(0.95)(1.5) Wp = 2.28 Wp
minimum Fp  0.3S DS I pW p
ASCE 7 (13.3-3)
= 0.3(0.95)(1.5) Wp = 0.43 Wp
 Fp = 0.48(100 plf) = 48 plf
Horizontal component in brace, H = Fp (8′ o.c.) = 48 plf (8′) = 384 lbs/brace
Resultant axial force in brace  R  2 H  2 (384 lbs/brace) = 540 lbs/brace 
6.10
a
p. 1-85 & ASCE 7-10 p. 91, Table 13.5-1 – Architectural Component
Cantilever elements (braced above its C.M.) - parapets – ap = 1 
6.11
c
p. 1-86 & ASCE 7-10 p. 94, §13.6.2 – Component Period
Wp = 1500 lbs & Kp = 3000 lbs/inch (given)
(continued)
Steven T. Hiner, MS, SE
4-31
Part 4 – Multiple Choice Solutions
Problem
SDR Workbook – 2012 IBC Version
Answer
Reference / Solution
g = 386.4 in/sec2
Wp
Tp  2
Kpg
 2
ASCE 7 (13.6-1)
1500
= 0.23 second 
(3000)(386.4)
6.12
d
p. 1-92, Figure 6.6b
Purlin anchors secure the walls to the diaphragm to resist out-of-plane loads
6.13
a
p. 1-86 & ASCE 7-10 p. 93, Table 13.6-1 – Mechanical & Electrical
Components
Boiler – ap = 1 & Rp = 2½ 
6.14
d
p. 1-88 & ASCE 7-10 p. 76, §12.11.2.1 – Wall Anchorage Forces
Fp shall not be less than 0.2 Ka Ie Wp 
6.15
b
p. 1-88 & ASCE 7-10 p. 76, §12.11.2.1 – Wall Anchorage Forces
Structural walls shall be designed to resist bending between anchors where
the anchor spacing exceeds 4 feet 
6.16
c
p. 1-87 & ASCE 7-10 p. 76, §12.11.1 – Design for Out-of-Plane Forces
Ie = 1.5 – ASCE 7-10 p. 4, Table 1.5-2 for Police Station (RC = IV)
Structural walls shall be designed for a force normal to the surface (i.e., outof-plane load) of:
Fp = 0.4 SDS Ie Wwall = 0.4(0.62)(1.5)(100 psf) = 37 psf  governs
 0.10 Wwall minimum = 0.10(100 psf) = 10 psf
 Fp = 37 psf 
6.17
c
p. 1-88 & ASCE 7-10 p. 76, §12.11.2.1 – Wall Anchorage Forces
SDS = 1.13 (given)
Lf = 0 for rigid diaphragms
L
K a  1.0  f = 1.0 + (0 / 100) = 1.0
100
Ie = 1.25 – ASCE 7-10 p. 4, Table 1.5-2 for Jails (RC = III)
Wp = Wwall hw 2  h p  … for (one-story) walls with a parapet
= (85 psf)(14′/2 + 3′) = 850 plf
Fp  0.4 S DS K a I eW p
ASCE 7 (12.11-1)
= 0.4(1.13)(1.0)(1.25)(850 plf) = 480 plf  governs
Fp  0.2 K a I eW p minimum
= 0.2(1.0)(1.25)(850 plf) = 212 plf minimum
 Fp = 480 plf 
4-32
Steven T. Hiner, MS, SE
Part 4 – Multiple Choice Solutions
Problem
Answer
SDR Workbook – 2012 IBC Version
Reference / Solution
ap = 1 & Rp = 2½ – ASCE 7-10 – Table 13.6-1 - Compressors
z = h1 = 10′ … since 2nd floor level = Level 1 (i.e., first level above the base)
h = hn = 10 stories (10′ / story) = 100′
z/h = 10′ / 100′ = 0.10
Ip = 1.0 – ASCE 7-10 – §13.1.3
Rp/Ip = (2½ / 1.0) = 2½

0.4a p S DSW p 
1  2 z 
Fp 
ASCE 7 (13.3-1)
RP I P  
h 
= 0.4(1)(0.73) Wp [1 + 2(0.10)] / (2½) = 0.14 Wp
maximum Fp  1.6 S DS I pW p
ASCE 7 (13.3-2)
= 1.6(0.73)(1.0) Wp = 1.17 Wp
minimum Fp  0.3S DS I pW p
ASCE 7 (13.3-3)
= 0.3(0.73)(1.0) Wp = 0.22 Wp  governs
 Fp = 0.22(10,000 lbs) = 2,200 lbs 
6.24
c
p. 1-82, Seismic Design Force & ASCE 7-10 p. 88 & 89, §13.3.1
SDS = 0.83 (given)
ap = 2½ & Rp = 3 – ASCE 7-10 – Table 13.5-1 – Signs & Billboards
z/h = 1.0 – since Billboard is anchored to the roof (i.e., z = h)
Ip = 1.0 – ASCE 7-10 – §13.1.3
Rp/Ip = (3 / 1.0) = 3

0.4a p S DSW p 
1  2 z 
Fp 
ASCE 7 (13.3-1)
RP I P  
h 
= 0.4(2½)(0.83) Wp [1 + 2(1.0)] / (3) = 0.83 Wp  governs
maximum Fp  1.6 S DS I pW p
ASCE 7 (13.3-2)
= 1.6(0.83)(1.0) Wp = 1.33 Wp
minimum Fp  0.3S DS I pW p
ASCE 7 (13.3-3)
= 0.3(0.83)(1.0) Wp = 0.25 Wp
 Fp = 0.83(10 kips) = 8.3 kips 
6.25
c
p. 1-84 & ASCE 7-10 p. 50, §11.2 – Component, Rigid
Nonstructural component having a fundamental period less than or equal to
0.06 s.
  0.06 second 
6.26
d
ASCE 7-10 p. 91, §13.5.6.1 – Seismic Forces
The weight of the ceiling, Wp , ... shall be taken as not less than 4 psf.
 4 psf 
6.27
c
p. 1-81, Component Importance Factor & ASCE 7-10 p. 87, §13.1.3 - item 1
(continued)
4-34
Steven T. Hiner, MS, SE
Part 4 – Multiple Choice Solutions
Problem
Answer
SDR Workbook – 2012 IBC Version
Reference / Solution
 Wall A - Design ex = 3′ – 7.5′ =  4.5 feet 
8.26
d
p. 1-120, Design Torsional Moment & p. 122, Torsion (Y-Direction)
accidental ex =  7.5′
Wall B - Design ex = 3′ + 7.5′ = +10.5′
MT = Mt + Mta = (500 kips)(10.5′) = + 5,250 kip-ft 
8.27
a
p. 1-118 to 119 & ASCE 7-10 p. 73, §12.8.4.1 – Inherent Torsion
inherent (i.e., calculated) ey = 0′
M t  Vx e y = (500 kips)(0′) = 0 kip-ft 
8.28
c
p. 1-119 & ASCE 7-10 p. 73, §12.8.4.2 – Accidental Torsion
accidental ey =  5% Ly =  0.05(80′) =  4.0′
Mta = Vx ( 0.05 Ly) = (500 kips)(  4.0′) =  2,000 kip-ft 
8.29
c
p. 1-120, Design Torsional Moment & p. 121, Torsion (X-Direction)
Rd2 = 3 (52)2 + 2 (78)2 + 4 (40)2 (2) = 33,080 ft2
MT = Mt + Mta = 0 + 2,000 = 2,000 kip-ft
V1 = direct shear + torsional shear
= (500 kips)(4) / (4+4) + 2,000(4)(40) / 33,080 = 250 +10 = 260 kips 
8.30
a
p. 1-112, Figure 8.9
Diaphragm sheathing is analogous to the web of a steel beam and resists the
shear stresses in the diaphragm.
 the web of a steel beam 
8.31
b
p. 1-119, Accidental Torsional Moment & p. 122, Torsion (Y-Direction)
Mta = Vy ( 0.05 Lx) = 100 kips ( 5.0′) =  500 kip-ft
Rd2 = R (50′)2 + R (50′)2 = 5,000 R
torsional shear, VT = MT R d / Rd2 = ( 500)(R)(50′)/5,000 R =  5.0 kips 
8.32
b
p. 1-48, Table 4.1 & ASCE 7-10 p. 66, Table 12.3-1
1.2 avg = 1.2 (0.25" + 0.75") / 2 = 0.60"
1.4 avg = 1.4 (0.25" + 0.75") / 2 = 0.70"
max = 0.75"  1.4 avg = 0.70" → Extreme Torsional Irregularity exists
 II 
8.33
c
p. 1-119 & ASCE 7-10 p. 73, §12.8.4.3 – Amplification of Accidental
Torsional Moment
accidental ey =  5% (200′) =  10′
Mta = Vy ( 0.05 Lx) = 200 kips ( 10′) =  2,000 kip-ft
avg = (0.25" + 0.75") / 2 = 0.50"
Ax = (max / 1.2 avg )2 = [(0.75") / 1.2(0.50")]2 = 1.56
Amplified Mta = Ax Mta = 1.56 ( 2,000 kip-ft) =  3,120 kip-ft 
4-40
Steven T. Hiner, MS, SE
SDR Workbook – 2012 IBC Version
Part 4 – Multiple Choice Solutions
Problem
Answer
Reference / Solution
9.9
b
p. 1-144, Shear Wall Overturning / Hold-Downs
 = 1.0 (given)
V1 = V2 = V / 2 = 33.6 kips / 2 = 16.8 kips
 0.7  (V1 h)  0.7(1.0)(16.8)(10' )
= – 3.92 kips
T

30'
b
 for ASD, uplift T = 4.0 kips 
9.10
b
p. 1-130, Wood Structural Panel Diaphragms
V = CS W = 0.196 W (given)
For a single-story building – ws = fp1 = Fp1 / L = CS wp1
East-West: ws = 0.196 [(25 psf)(75′) + (15 psf)(12′/2) 4 walls] = 438 plf
Vmax = ws L / 2 = (438 plf)(40′) / 2 = 8,760 lbs
for ASD, roof  = (0.7 Vmax) / d = 0.7 (8,760 lbs) / 75′ = 82 plf  80 plf 
9.11
c
p. 1-135, Table 9.5 & SDPWS p. 31, Table 4.3A
3/8″ rated sheathing w/ 8d common @ 2″ o.c. → s = 1060 plf
15/32″ Structural I w/ 10d common @ 6″ o.c. → s = 680 plf NG!
15/32″ Structural I w/ 10d common @ 4″ o.c. → s = 1020 plf NG!
15/32″ Structural I w/ 10d common @ 3″ o.c. → s = 1330 plf OK
15/32″ Structural I w/ 10d common @ 2″ o.c. → s = 1740 plf OK
 use 15/32″ Structural I w/ 10d common @ 3″ o.c. = 1330 plf  1060 plf 
9.12
d
p. 1-131 to 132, Table 9.2 - 9.3 & SDPWS p. 18-20, Table 4.2A & 4.2C
Load parallel to continuous panel joints = CASE 3 (weak direction)
15/32″ sheathing w/ 8d @ 6″ o.c. unblocked → Table 9.3 (4.2C) →
s = 360 plf → ASD  = s / 2.0 = (360 plf) / 2.0 = 180 plf < 275 plf NG!
15/32″ sheathing w/ 10d @ 6″ o.c. unblocked → Table 9.3 (4.2C) →
s = 380 plf → ASD  = s / 2.0 = (380 plf) / 2.0 = 190 plf < 275 plf NG!
15/32″ sheathing w/ 8d @ 6″ o.c. blocked → Table 9.2 (4.2A) →
s =540 plf → ASD  = s / 2.0 = (540 plf) / 2.0 = 270 plf < 275 plf NG!
15/32″ sheathing w/ 10d @ 6″ o.c. blocked → Table 9.2 (4.2A) →
s = 580 plf → ASD  = s / 2.0 = (580 plf) / 2.0 = 290 plf  275 plf OK
 use 15/32″ sheathing w/ 10d common @ 6″ o.c. 
9.13
d
p. 1-110, Flexible Diaphragm Analysis
V = CS W
For a single-story building – ws = fp1 = Fp1 / L = CS wp1
N-S: ws = 0.20[20 psf (120′) + (15 psf) (16′/2 + 2.5′) 2 walls] = 543 plf
 ws = 540 plf 
9.14
c
p. 1-127, Diaphragm Aspect Ratios
East-West: L / d = 120′ / 40′ = 3:1 
Steven T. Hiner, MS, SE
4-47
Part 4 – Multiple Choice Solutions
SDR Workbook – 2012 IBC Version
Problem
Answer
Reference / Solution
9.15
a
p. 1-113, Chord Force
Maximum chord force (CF) occurs at the point of largest moment and least
diaphragm depth. This maximum moment occurs on lines A & B (East-West
loading) at the center of the diaphragm span (i.e., maximum CF = ws L2/8d
where L is maximum & d is minimum).
a 
9.16
c
p. 1-130, Wood Structural Panel Diaphragms
V1 = V5 = ws L / 2 = (350 plf)(120′/2) = 21,000 lbs
For ASD, roof 1 = 5 = (0.7 V1) / d = 0.7 (21,000 lbs) / 40′ = 370 plf 
9.17
c
p. 1-132, Table 9.3 & SDPWS p. 18-20, Table 4.2A (blocked diaphragms)
Load perpendicular to continuous panel joints = CASE 1 (strong direction)
15/32″ Structural I w/ 10d @ 6″ o.c. blocked → Table 9.2 (4.2A) →
s = 640 plf → ASD  = s / 2.0 = (640 plf) / 2.0 = 320 plf < 400 plf NG!
15/32″ Structural I w/ 10d @ 4″ o.c. blocked → Table 9.2 (4.2A) →
s = 850 plf → ASD  = s / 2.0 = (850 plf) / 2.0 = 425 plf > 400 plf OK
Stop … no need to check 10d @ 2½″ o.c. or 10d @ 2″ o.c.
 Diaphragm boundary nailing = 10d @ 4″ o.c. 
9.18
c
p. 1-114, Drag Force
Maximum drag force on line 5 occurs at the left end of the 20′ shear wall (i.e.,
at “y”) –
F5 = (400 plf)(20′) = 8,000 lbs = 8 kips 
9.19
d
p. 1-139, Wood Structural Panel Shear Walls
V5 = ws L / 2 = (350 plf)(120′/2) = 21,000 lbs
 (0.7 Vmax ) 1.0(0.7)(21,000)

= 740 plf 
For ASD, wall  5 
20'
b
9.20
a
p. 1-135, Table 9.5 & SDPWS p. 31, Table 4.3A
15/32″ Structural I w/ 10d common @ 6″ o.c. → s = 680 plf →
ASD  = s / 2.0 = (680 plf) / 2.0 = 340 plf << 850 plf NG!
15/32″ Structural I w/ 10d common @ 4″ o.c. → s = 1020 plf →
ASD  = s / 2.0 = (1020 plf) / 2.0 = 510 plf << 850 plf NG!
15/32″ Structural I w/ 10d common @ 3″ o.c. → s = 1330 plf →
ASD  = s / 2.0 = (1330 plf) / 2.0 = 665 plf << 850 plf NG!
15/32″ Structural I w/ 10d common @ 2″ o.c. → s = 1740 plf →
ASD  = s / 2.0 = (1740 plf) / 2.0 = 870 plf > 850 plf OK
 shear wall edge nailing = 10d common @ 2″ o.c. 
4-48
Steven T. Hiner, MS, SE
Part 4 – Multiple Choice Solutions
Problem
Answer
SDR Workbook – 2012 IBC Version
Reference / Solution
19/32″ sheathing w/ 10d @ 6″ o.c. unblocked → Table 9.3 (4.2C) →
s = 570 plf → ASD  = s / 2.0 = (570 plf) / 2.0 = 285 plf
 allowable unit (roof) shear = 285 plf 
9.31
d
p. 1-135, Table 9.5 & SDPWS p. 31, Table 4.3A
h/b = 12′/8.5′ = 1.4:1 < 2:1 … unit shear values do not require a reduction
factor for seismic
15/32″ Structural I w/ 10d common at 4″ o.c. → Table 9.5 (4.3A) →
s = 1020 plf → ASD  = s / 2.0 = (1020 plf) / 2.0 = 510 plf
 allowable unit wall shear = 510 plf 
9.32
b
p. 1-135, Table 9.5 & SDPWS p. 31, Table 4.3A
h/b = 12′/4.25′ = 2.82:1 > 2:1 … unit shear values will require a reduction
factor for seismic …
reduction factor = 2b/h = 2(4.25′/12′) = 0.71
15/32″ Structural I w/ 10d common @ 3″ o.c. → Table 9.5 (4.3A) →
s = 1330 plf → ASD  = s / 2.0 = (1330 plf) / 2.0 = 665 plf
 allowable unit wall shear = (0.71)(665 plf) = 470 plf 
9.33
b
p. 1-135, Table 9.5 & SDPWS p. 31, Table 4.3A
h/b = 12′/3.42′ = 3.5:1 > 2:1 … unit shear values will require a reduction
factor for seismic
reduction factor = 2b/h = 2(3.42′/12′) = 0.57
15/32″ Structural I w/ 10d common @ 3″ o.c. → Table 9.5 (4.3A) →
s = 1330 plf → ASD  = s / 2.0 = (1330 plf) / 2.0 = 665 plf
 allowable unit wall shear = (0.57)(665 plf) = 380 plf 
9.34
d
p. 1-125 & 2012 IBC p. 453, §2301.2 - items 1, 2 & 3
The design of structural elements or systems constructed partially or wholly
of wood or wood-based products, shall be in accordance with one of the
following methods: Allowable Stress Design (ASD), Load and Resistance
Factor Design (LRFD), Conventional Light-Frame Construction … etc.
 I, II & III 
9.35
c
p. 1-152 & ASCE 7-10 p. 76, §12.11.2.2.1
SDC = C, D, E or F - subdiaphragms are to be designed for the …
wall anchorage force per ASCE 7-10 – §12.11.2 
9.36
a
p. 1-134, Table 9.4
I. Wood structural panels - permitted in all SDC’s
II. Gypsum wallboard, etc. - Not permitted for seismic in SDC = E & F
III. Particleboard (blocked) - Not permitted for seismic in SDC = D, E & F
I 
4-50
Steven T. Hiner, MS, SE