S t r u c t u ra lE n g i n e e r i n g E xa m R e v i e w C o u rs e
L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2) Structural Engineering Review Course
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L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Lesson Overview Chapter 7: Lateral Forces •
•
Flexible, Semi-Rigid, and Rigid Diaphragms
•
Collectors
•
Subdiaphragms
•
Approximate Methods of Analysis
•
•
Lateral Design Force on Walls and Parapets
Seismic Forces on Nonstructural Components Anchorage of Structural Walls to Flexible and Rigid Diaphragms
Note: For clarity and completeness, some of the information from the previous presentations on Wind Loads and Seismic Loads is repeated.
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Distribution of Lateral Forces (Part 2)
Learning Objectives You will learn… •
•
•
diaphragm theory for flexible and rigid diaphragms how to distribute seismic and wind loads to structural sub-systems and components (ex: diaphragms, chords, etc.) how to navigate the ASCE/SEI77-10 and IBC 2012 design codes for lateral loads, choose variables, and use tables and figures
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L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Learning Objectives You should already be familiar with… •
layout of the referenced design codes (IBC 2012 & ASCE/SEI77-10)
•
load application by tributary areas
•
linear interpolation
•
calculating weighted averages
•
•
•
•
common terms from seismic and wind loading (ex: center of mass, rigidity, windward, leeward, etc.)
•
typical LFRS (braced frames, moment frames, shear walls, etc.) typical building components (braces, beams, trusses, etc.) theory of earthquake forces (inertial forced induced by ground shaking) seismic and wind loads lessons
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Distribution of Lateral Forces (Part 2)
Referenced Codes and Standards •
•
International Building Code (IBC, 2012) Minimum Design Loads for Buildings and Other Structures (ASCE/SEI7, 2010)
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Distribution of Lateral Forces (Part 2)
Flexible Diaphragms flexible diaphragm
Figure 7.19 Flexible Diaphragm
diaphragm whose lateral deformation under a lateral load is more than twice the average story drift of the adjoining vertical elements of the LFRS [ASCE/SEI7 Sec. 12.3.1.3] To qualify as a flexible diaphragm, ΔD>2Δavg.
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Distribution of Lateral Forces (Part 2)
Flexible Diaphragms •
When subjected to a transverse force, a flexible diaphragm undergoes a lateral displacement without rotation.
•
Lateral forces are distributed to the LFRS by tributary areas.
•
Flexible diaphragms cannot transmit torsion.
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Distribution of Lateral Forces (Part 2)
Flexible Diaphragms •
In accordance with ASCE/SEI7 Sec. 12.3.1.1, the following types of diaphragms may be considered flexible. •
•
•
untopped steel decking or wood structural panels supported by vertical elements of steel or composite braced frames, or by concrete, masonry, steel or composite shear walls untopped steel decking or wood structural panels in one- and two-family residential buildings of light frame construction
Diaphragms satisfying these criteria can be classified as flexible diaphragms with no calculation or analysis required.
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Distribution of Lateral Forces (Part 2)
Flexible Diaphragms •
Light-frame construction diaphragms of untopped steel decking or wood structural panels are considered flexible, provided all of the following are true. •
•
•
•
topping of concrete or similar materials not placed over wood structural panel diaphragms (except for nonstructural toppings ≤ 1.5 in thick). each line of LFRS complies with allowable story drift of ASCE/SEI7 Table 12.12-1
Diaphragms satisfying these criteria can be classified as flexible diaphragms with no calculation or analysis required. There is no torsional shear stress from eccentric mass placement in either the walls or diaphragm because flexible diaphragms are not considered capable of distributing torsion.
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Distribution of Lateral Forces (Part 2)
Example – Flexible Diaphragm For a given lateral loading, displacements at ends of the diaphragm are Δ1 = 0.25 in and Δ2 = 0.75 in. The maximum displacement in diaphragm is 1.65 in at the midspan.
Figure 7.19 Flexible Diaphragm
Should the diaphragm be classified as flexible?
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Distribution of Lateral Forces (Part 2)
Example – Flexible Diaphragm For a given lateral loading, displacements at ends of the diaphragm are Δ1 = 0.25 in and Δ2 = 0.75 in. The maximum displacement in diaphragm is 1.65 in at the midspan. Should the diaphragm be classified as flexible?
Calculate average displacement of the diaphragm.
avg
1 2 2 0.25 in 0.75 in 2
0.50 in Calculate relative maximum displacement of the diaphragm at midspan.
D
max
avg
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Distribution of Lateral Forces (Part 2)
Example – Flexible Diaphragm For a given lateral loading, displacements at ends of the diaphragm are Δ1 = 0.25 in and Δ2 = 0.75 in. The maximum displacement in diaphragm is 1.65 in at the midspan. Should the diaphragm be classified as flexible?
Check if the diaphragm is rigid using the previously presented criteria.
D 2 avg Convert to simpler equation. D 1.15 in avg 0.5 in 2.3 2.0 [diapragm is flexible]
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Distribution of Lateral Forces (Part 2)
Framing Terminology Blocking usually frames into joists or sub-purlins .
SEIS Figure 7.5 Framing Members
Joists frame into purlins. Purlins frame into beams. Beams frame into girders. Girders frame into walls.
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Distribution of Lateral Forces (Part 2)
Semirigid Diaphragms Diaphragms are considered semirigid if they do not satisfy the requirements of ASCE/SEI7 Sec. 12.3.1.1 – Sec. 12.3.1.3 and IBC Sec. 1613.6.1. Analysis of structures with semirigid diaphragms must include consideration of the actual stiffness of the diaphragm system in determining deformations as well as load distribution to the LFRS.
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Distribution of Lateral Forces (Part 2)
Rigid Diaphragms rigid diaphragm •
diaphragm that, for the purposes of distributing story shear and torsional moment, has a lateral deformation ≤ twice the average story drift [IBC 202]
D 2 avg •
Diaphragms of concrete slabs or concrete filled metal deck with span-to-depth ratios ≤ 3 for structures that have no horizontal irregularities are considered rigid. [ASCE/SEI7 Sec. 12.3.1.2]
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Distribution of Lateral Forces (Part 2)
Rigid Diaphragms Rigid diaphragms distribute lateral forces to the vertical seismic-load-resisting elements based on •
•
the relative rigidity of these elements the torsional displacements produced by the rigid-body rotation of the diaphragm
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Distribution of Lateral Forces (Part 2)
Rigid Diaphragms To calculate the torsional displacements, the following must be known. •
center of rigidity
•
center of mass
•
Figure 7.20 Rigid Diaphragm
rigidity, orientation, and location of each vertical lateral force resisting element
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Distribution of Lateral Forces (Part 2)
Rigid Diaphragms center of rigidity Figure 7.20 Rigid Diaphragm
point about which a structure rotates when subjected to a torsional moment center of mass
point through which the lateral force, V, acts
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Distribution of Lateral Forces (Part 2)
Rigid Diaphragms For the structure in Fig. 7.20, torsional moment acting on the diaphragm is defined as T = Ve.
Figure 7.20 Rigid Diaphragm
V = lateral force e = eccentricity of the center of mass with respect to the center of rigidity
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Distribution of Lateral Forces (Part 2)
Rigid Diaphragms Where the center of mass and center of rigidity are not coincident, •
•
displacement of building consists of a translation in the direction of the lateral force coupled with a rotation about the center of rigidity forces due to translational movement are portioned to the LFRS in proportion to their relative stiffness In plane shear force in wall i, FSi
•
VR Ri
Translational forces only affect LFRS parallel to the direction of the lateral force being resisted and the relative stiffness is only with respect to parallel LFRS.
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Distribution of Lateral Forces (Part 2)
Rigid Diaphragms Where the center of mass and center of rigidity are not coincident •
•
Torsional forces resulting from eccentricity between center of mass and center of rigidity also produce torsional forces in the LFRS of the structure Forces are distributed based on their torsional stiffness as Tr R Torsional shear force in wall i,FTi i i J J = Σ r i 2 Ri r = perpendicular distance between shear wall i and center of rigidity
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Distribution of Lateral Forces (Part 2)
Rigid Diaphragms accidental eccentricity •
•
built-in factor of safety to account for limited information and variability of as-built conditions consideration of an accidental eccentricity is required to account for uncertainty in determining the location of the center of rigidity and center of mass
accidental torsion •
•
torsional effects resulting from accidental eccentricity to account for accidental torsion, center of mass assumed to be displaced each way from its actual location by a distance equal to 5% of the building dimension perpendicular to the direction of the applied force [ASCE/SEI7 Sec. 12.8.4.2]
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Distribution of Lateral Forces (Part 2)
Rigid Diaphragms A building in seismic design categories C through F has a torsional irregularity as defined in ASCE/SEI7 Table 12.3-1 (horizontal structural irregularity Type 1a or 1b). •
•
Amplify the accidental torsion as specified in ASCE/SEI7 Sec. 12.8.4.3. 2 max The amplification factor, Ax ASCE/SEI7 Eq. 12.8-14 1.2 avg
δmax = maximum displacement at level x δavg = average displacement of ends of diaphragm Structures in seismic design categories E and F with horizontal structural irregularity Type 1b are not permitted [ASCE/SEI7 Sec. 12.3.3.1].
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Distribution of Lateral Forces (Part 2)
Example – Rigid Diaphragms Example 7.5
Figure 7.20 Rigid Diaphragm
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Distribution of Lateral Forces (Part 2)
Example – Rigid Diaphragms Example 7.5
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Distribution of Lateral Forces (Part 2)
Example – Rigid Diaphragms
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Distribution of Lateral Forces (Part 2)
Example – Rigid Diaphragms
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Distribution of Lateral Forces (Part 2)
Example – Rigid Diaphragms SEIS Example 7.5
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Distribution of Lateral Forces (Part 2)
Example – Rigid Diaphragms
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Distribution of Lateral Forces (Part 2)
Example – Torsion Forces A two-story structure with a wood structural panel diaphragm supported by concrete shear walls is shown. The center of mass is at the center of the building. The relative rigidity per unit length of each shear wall is equal, and the center of rigidity is at a 10 ft eccentricity. Ignoring accidental eccentricity, what are the relative amplification in loads in the LFRS due to torsional forces in the shear walls?
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Distribution of Lateral Forces (Part 2)
Example – Torsion Forces In accordance with ASCE/SEI7 Sec. 12.3.1.1, wood structural supported by walls vertical elementspanels of composite shear may be considered flexible diaphragms. Since the diaphragm is flexible, it cannot transmit torsion. Therefore, amplification in loads in the LFRS due to torsional effects is zero.
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Distribution of Lateral Forces (Part 2)
Collectors collector (also known as a drag strut) •
•
•
SEIS Figure 7.6 Use of a Collector
horizontal diaphragm element, parallel and in line with the applied force collects and transfers diaphragm shear forces to the vertical elements of the LFRS required where shear walls or braced frames are internal to a diaphragm, or terminate along the boundary of a diaphragm.
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Distribution of Lateral Forces (Part 2)
Collector Force collector force •
•
•
product of the diaphragm load (per unit area) and the collector tributary areas tributary areas usually taken as some fraction of the diaphragm span areas located on either side of the collector not the same along the length of the collector, but increases to a maximum at the point where the collector frames into the shear wall
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Distribution of Lateral Forces (Part 2)
Columns Supporting Flexible Diaphragms •
•
•
Interior or exterior columns may be used to support the vertical (gravity) loads from the floor or roofs. These columns typically do not provide lateral support when the diaphragm is flexible. A girder that frames into such a column at one girder end and a shear wall at the other end almost always acts as a collector for seismic forces parallel to the girder.
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Distribution of Lateral Forces (Part 2)
Example – Collectors SEIS Example 7.3
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Distribution of Lateral Forces (Part 2)
Example – Collectors
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Distribution of Lateral Forces (Part 2)
Example - Collectors Example 7.6
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Distribution of Lateral Forces (Part 2)
Example - Collectors
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Distribution of Lateral Forces (Part 2)
Example - Collectors
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Distribution of Lateral Forces (Part 2)
Example - Collectors
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Distribution of Lateral Forces (Part 2)
Example - Collectors
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Distribution of Lateral Forces (Part 2)
Example - Collectors
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Distribution of Lateral Forces (Part 2)
Approximate Methods of Analysis •
•
several methods available to estimate forces in a rigid frame subjected to lateral loads portal method is simplest •
•
quick for simple frames, but can be applied to larger and more complex systems as well requires relatively “regular” member distributions (all same size/type of columns, all same size/type of beams, consistent bay spacing and story heights, etc.)
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L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Portal Method portal method assumes 1. point of inflection occurs at mid-height of each column 2. point of inflection occurs at midpoint of each beam or girder 3. shear and moment in each interior column twice that of exterior column 4. axial force in an interior column is zero
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L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Example: Portal Method Example 7.1
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Distribution of Lateral Forces (Part 2)
Example: Portal Method Example 7.1
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Distribution of Lateral Forces (Part 2)
Example: Portal Method
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Distribution of Lateral Forces (Part 2)
Example: Portal Method
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Distribution of Lateral Forces (Part 2)
Example: Portal Method
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Distribution of Lateral Forces (Part 2)
Example: Portal Method
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Distribution of Lateral Forces (Part 2)
Moment Distribution Procedure moment distribution procedure •
•
most accurate method for approximate determination of member forces in rigid frame neglects axial forces
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Distribution of Lateral Forces (Part 2)
Moment Distribution Procedure Skew symmetry may be used to simplify the process. •
modified stiffness for columns: EI/L
•
modified stiffness for beams: 6EI/L
•
carryover factor between columns: −1
•
carryover factor in beams: 0
•
Figure 7.6 Sway Distribution for Single Bay Frame
initial fixed-end moments obtained by imposing unit virtual sway displacements on each story in turn
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Distribution of Lateral Forces (Part 2)
Principle of Multiples principle of multiples •
•
•
superposition method for determining forces in frame members simplifies and expedites moment distribution method used with multi-bay frames (repeating or symmetric systems)
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Distribution of Lateral Forces (Part 2)
Principle of Multiples •
•
•
•
srcinal frame replaced by two substitute frames
Figure 7.7 Principle of Multiples
ratio of applied load to member stiffness the same for each frame joint rotations and sway displacements are the same in substitute frames and srcinal frame member forces in substitute frames summed to determine corresponding member forces in the srcinal frame
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Distribution of Lateral Forces (Part 2)
Principle of Multiples •
•
frame shown in Fig. 7.7 has relative stiffness circled for each member
Figure 7.7 Principle of Multiples
principle of multiples divides multibay frame on left into two simpler frames for analysis
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Distribution of Lateral Forces (Part 2)
Example: Principle of Multiples Example 7.2
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Distribution of Lateral Forces (Part 2)
Example: Principle of Multiples Example 7.2
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Distribution of Lateral Forces (Part 2)
Example: Principle of Multiples
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Distribution of Lateral Forces (Part 2)
Example: Principle of Multiples
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Distribution of Lateral Forces (Part 2)
Example: Principle of Multiples
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Distribution of Lateral Forces (Part 2)
Cantilever Method cantilever method •
SEIS Figure 8.2 Cantilever Method Frame Deflection
assumes entire floor works together as a unit (as opposed to portal method, which treats each bay independently from others)
•
preferred method for tall (> 25 stories) buildings
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Distribution of Lateral Forces (Part 2)
Cantilever Method •
•
•
•
analysis begins at roof, moves down
SEIS Figure 8.2 Cantilever Method Frame Deflection
floors remain plane (though not horizontal) forces in columns proportional to distance from frame’s center of gravity inflection points assumed at midheight of columns, mid-length of girders
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Distribution of Lateral Forces (Part 2)
Example: Cantilever Method SEIS Example 8.2
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Distribution of Lateral Forces (Part 2)
Example: Cantilever Method SEIS Example 8.2
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Distribution of Lateral Forces (Part 2)
Example: Cantilever Method
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Distribution of Lateral Forces (Part 2)
Example: Cantilever Method
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Distribution of Lateral Forces (Part 2)
Poll Question: Approximate Methods of Analysis For tall and slender structures, which approximate method of analysis is the simplest way to find internal forces and reactions in a frame? (A) portal method (B) moment distribution procedure (C) principle of multiples (D) cantilever method
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Distribution of Lateral Forces (Part 2)
Poll Question: Approximate Methods of Analysis For tall and slender structures, which approximate method of analysis is the simplest way to find internal forces and reactions in a frame?
The cantilever method best approximates the behavior of tall and slender structures.
(A) portal method
Moment distribution is more likely to be tedious and time-consuming.
(B) moment distribution procedure
The answer is (D).
(C) principle of multiples (D) cantilever method
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Distribution of Lateral Forces (Part 2)
Poll Question: Approximate Methods of Analysis Which approximate method of analysis is the most accurate way to find internal forces and reactions in a frame? (A) portal method (B) moment distribution procedure (C) principle of multiples (D) cantilever method
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Distribution of Lateral Forces (Part 2)
Poll Question: Approximate Methods of Analysis Which approximate method of analysis is the most accurate way to find internal forces and reactions in a frame? (A) portal method (B) moment distribution proce dure
Although it tends to be tedious and difficult, the moment distribution procedure typically provides the most accurate solution.
The answer is (B).
(C) principle of multiples (D) cantilever method
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Distribution of Lateral Forces (Part 2)
Poll Question: Approximate Methods of Analysis For short and wide structures, which approximate method of analysis is the simplest way to find internal forces and reactions in a frame? (A) portal method (B) moment distribution procedure (C) principle of multiples (D) cantilever method
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Distribution of Lateral Forces (Part 2)
Poll Question: Approximate Methods of Analysis For short and wide structures, which approximate method of analysis is the simplest way to find internal forces and reactions in a frame?
The portal method best approximates the behavior of short structures.
The answer is (A).
(A) portal method
(B) moment distribution procedure (C) principle of multiples (D) cantilever method
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Distribution of Lateral Forces (Part 2)
Lateral Design Force on Walls and Parapets •
ASCE/SEI7 Sec. 12.11.1 specifies the out-of-plane seismic force on a wall as provides an amplification factor based on the height of the component
Fp 0.40I e SDSW p
0.10W p •
•
Parapets in seismic design category B and above must be designed as architectural components. ASCE/SEI7 Equations 13.3-1 through 13.3-3 are used to determine the lateral earthquake force for design of nonstructural components.
0.4a p S DS I p z 1 2 W p R h p
Fp
1.6 SDS I pW p
ASCE/SEI7 Eq. 13.3-2
0.3S DS I pW p
ASCE/SEI7 Eq. 13.3-3
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Distribution of Lateral Forces (Part 2)
Lateral Forces on N onstructural Components The value of SDS is the same as that used for the building. The value of Ip = 1.5 for the following cases. (Otherwise Ip = 1.0) •
•
anchoring machinery and equipment required for life-safety systems, tanks containing toxic and explosive substances , or equipment containing hazardous substances
provides an amplification factor based on the height of the component
0.4a p S DS I p z 1 2 W p R h p
Fp
ASCE/SEI7 Eq. 13.3-1
1.6 SDS I pW p
ASCE/SEI7 Eq. 13.3-2
0.3S DS I pW p
ASCE/SEI7 Eq. 13.3-3
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Distribution of Lateral Forces (Part 2)
Component Amplification Factor •
natural period of the elements, components and/or equipment not a direct factor when calculating the seismic forces on nonstructural components (indirectly accounted for through the component amplification factor, ap)
•
component amplification factor,ap given in ASCE/SEI7 Tables 13.5-1 and 13.6-1. •
•
Dynamic properties or empirical data may be used to determine this factor, but 2.5 ≥ ap ≥ 1.0. Rigid components experience little amplification and are assigned ap = 1.0.
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Distribution of Lateral Forces (Part 2)
Component Response Modification Factor Values of the component response modification factor,Rp, are given in ASCE/SEI7 Table 13.5-1, Coefficients for Architectural Components. •
The maximum value of Rp is 3.5 and is reserved for penthouses and elements that allow high deformability.
•
Most components and equipment are given a Rp of 2.5.
•
A Rp of 1.5 is used for equipment that only allows low levels of deformability.
•
For fasteners of exterior nonstructural components, Rp is 1.0.
Values of Rp also given in ASCE/SEI7 Table 12.6-1.
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Distribution of Lateral Forces (Part 2)
Force Distribution on Nonstructural Components Lateral forces on nonstructural components are driven by inertial effects, so forces should be distributed based on the mass of the component (results in uniformly distributed loading for components with uniformly distributed mass).
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Distribution of Lateral Forces (Part 2)
Considerations for Exterior Cladding Panels exterior cladding panels •
•
•
must resist seismic force, Fp (calculated from the equations for nonstructural components) must accommodate relative seismic displacements of the structure (calculated from ASCE/SEI7 Sec. 13.3.2 or 0.5 in., whichever is greater) panel connectors (bolts, inserts, welds, dowels, etc.) designed for seismic force, Fp (calculated from the equations for nonstructural components)
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Distribution of Lateral Forces (Part 2)
Lateral Design Force on Walls and Parapets Parapets are designed for a higher load than walls because of poor seismic performance and lack of redundancy. •
•
minimizes public safety hazard resulting component amplification factor,ap = 2.5
When analyzing parapets for lateral forces using the equations on the previous slide, the lateral forces should be considered uniformly distributed over the height.
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Distribution of Lateral Forces (Part 2)
Anchorage of Structural Walls to Flexible Diaphragms •
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A major cause of failure in past earthquakes has been the separation of flexible diaphragms from concrete and/or masonry load-bearing walls, due partially to diaphragm flexibility relative to shearwall causing large in-elastic demands on diaphragm to wall anchorage. Supporting walls must be securely anchored to subdiaphragm ties to prevent separation. Where wall anchor spacing > 4 ft, the wall must be designed to span between anchors. [ASCE/SEI7 Sec. 12.11.2.1] Steel elements in the anchorage system are required to resist 1.4 times the calculated anchorage force. [ASCE/SEI7 Sec. 12.11.2.2.2]
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L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Example – Analysis of Out of Plane W all Forces SEIS Example 6.14
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L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Example – Analysis of Out of Plane W all Forces
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L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Example – Analysis of Out of Plane W all Forces
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L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Example – Analysis of Out of Plane W all Forces
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S t r u c t u ra lE n g i n e e r i n g E xa m R e v i e w C o u rs e
L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Example – Analysis of Out of Plane W all Forces
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L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Example – Analysis of Out of Plane W all Forces
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L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Example – Analysis of Out of Plane W all Forces
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L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Example – Analysis of Out of Plane W all Forces
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L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Example – Analysis of Out of Plane W all Forces
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L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Anchorage of Structural Walls to Flexible Diaphragms For buildings assigned to seismic design categories B through F, anchors must be designed for the following force. [ASCE/SEI7 Sec. 12.11.2.1] Fp 0.4SDS ka I eW p
ASCE/SEI7 Eq. 12.11-1
Lf
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amplification factor for diaphragm flexibility, ka 1.0
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Lf = span (in ft) of a flexible diaphragm measured between vertical elements that
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ASCE/SEI7 Eq. 12.11-2
provide lateral support to the diaphragm in the direction considered •
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Fp = 0 [for rigid diaphragms] L = 0 for rigid diaphragm
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L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Anchorage of Structural Walls to Rigid Diaphragms For buildings assigned to seismic design categories B through F with rigid diaphragms, anchors, other than those at roof level, must be designed for the following force. [ASCE/SEI7 Sec. 12.11.2.1] z Wp Fp 0.40 I e S DS 1 2 h 3
0.20I eW p Anchorage force for rigid diaphragms at roof level determined with ka = 1.0 from Fp 0.40 I e SDSW p
ASCE/SEI7 Eq. 12.11-1
0.20 I eW p
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L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Subdiaphragms •
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In seismic design categories C through F, continuous ties must be provided across the complete depth of the diaphragm. [ASCE/SEI7 Sec. 12.11.2.2.1 ] These ties transfer the wall anchorage forces across the depth of the diaphragm and prevent the walls and diaphragm from separating. To reduce the number of full depth ties required, subdiaphragms and added chords are used to span between the full depth ties. The maximum permitted length-to-width ratio of the subdiaphragm is 2.5 to 1.0.
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L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Subdiaphragms •
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Figure 7.21 shows a typical arrangement of subdiaphragms and crossties
Figure 7.21 Subdiaphragms and Crossties
All structural walls must be anchored to the diaphragm to prevent separation of the walls from the diaphragm. [IBC Sec. 1604.8.2] Wall anchors are provided at the end of the subdiaphragm ties and must have a capacity to resist the minimum horizontal force. plan view
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L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Subdiaphragms •
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For walls of structures assigned to seismic design category A, the wall anchor design force, Fp = 0.20Wp ≥ 5 lbf/ft2. [ASCE/SEI7 Sec. 1.4.5] Subdiaphragm chords resist the local bending in the subdiaphragm that spans between the continuous ties, while carrying loads applied by the wall anchorage ties. Continuous ties must be in addition to plywood sheathing or metal deck that is considered ineffective in providing the ties. [ASCE/SEI7 Sec. 12.11.2.2.3]
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L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Provisions for Deformation Compatibility For elements required to meet the deformation compatibility provisions of ASCE/SEI7 Sec. 12.12.5, consider •
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P-delta effects on elements expected deformations (should be the greater of the story drift or maximum inelastic displacement) additional deformations that may result from foundation flexibility and diaphragm deflections stiffening effect of these elements (when determining forces in the structure)
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L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Provisions for Deformation Compatibility For structures in seismic design categories D, E or F, structural framing elements and connections subjected to the deformations resulting from seismic forces must be designed and detailed to maintain support of design gravity (dead plus live) loads under expected seismic deformations. The provisions of ASCE/SEI7 Sec. 12.12.5 are more rigorous than previous building code specifications.
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L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Learning Objectives You have learned… •
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diaphragm theory for flexible and rigid diaphragms how to distribute seismic and wind loads to structural sub-systems and components (ex: diaphragms, chords, etc.) how to navigate the ASCE/SEI77-10 and IBC 2012 design codes for lateral loads, choose variables, and use tables and figures
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L a t e r a l F o r c e s : D i s t r i b u t io no f L a t e ra l F o r c e s (P a r t 2 )
Distribution of Lateral Forces (Part 2)
Lesson Overview Chapter 7: Lateral Forces •
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Flexible, Semi-Rigid, and Rigid Diaphragms
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Collectors
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Subdiaphragms
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Approximate Methods of Analysis
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Seismic Forces on Nonstructural Components Anchorage of Structural Walls to Flexible and Rigid Diaphragms
Lateral Design Force on Walls and Parapets
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