architectural structural course notes
DESCRIPTION
architectural structural course notesTRANSCRIPT
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Elastic Buckling
Illustrations: Daniel L. Schodek:Structures,
fifth edition; Pearson Prentice-Hall, 2004
Schodek fig. 7.1
Illustrations: Daniel L. Schodek:Structures,
fifth edition; Pearson Prentice-Hall, 2004
Schodek fig. 7.1
Metastable
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Column Buckling Formula
Pcr = 2EI
L2Pcr= Critical force that initiates buckling failure
E = Modulus of elasticity of material (material stiffness)
I = Moment of inertia (geometric stiffness)
L = Unbraced column length
Notice it does NOT matter what the STENGTH
of the material is!
Its all about STIFFNESS
(Euler Buckling Equation)
Schodek fig. 7.5
EffectiveLengthFactors
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Illustrations: Daniel L. Schodek:Structures,
fifth edition; Pearson Prentice-Hall, 2004
Schodek fig. 7.7
Real
Columns:Design Considerations
Have significant architectural impactdue to how they affect space
Establish an organizing grid (structuralbay)
Define lengths of primary girders andbeams framing into them
Columns:Design Consid
Most critical str
Typically no red
Remove a colusupports will fa
Controlled demof columns
Loads to columby tributary are
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A B C D E F G H I J K L
23
1
4
5
6
7
89
10
30 9 30 9 30 8 11 8 16 16 8
21
25
7
7
21
Col Tributary Area = (9+30)/2+(21+7)/2=546 ft2(2 floors) = 1092 ft2
Columns:Design Consid
Layout of columspace planning
Can be difficult varying needs iespecially if par Sometimes use
floors above
Expensive optiodeeper membe
Columns:Intermediate-Length
Intermediate-length columns arenormally what is actually used in actualconstruction
Failure mode is a combination ofcrushing action and buckling actionsimultaneously
Computing the allowable axial stress forthese involves complex equations
Columns:Intermediat
Fortunately, thesimplified to a tthe column slensteel, see table
For pinned-endcan be looked urelate member relative to the h
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Illustrations: Daniel L. Schodek:Structures,
fifth edition; Pearson Prentice-Hall, 2004
Schodek fig. 7.3 Columns:Influences on B
Ratio of length slenderness
End support coIntermediate br 2% rule of thu
most cases braonly 2% of thecolumn!
Elastic bucklingstrength. It is r
Columns:Influences on Buckling Capacity
Eccentric loading dramatically reduces capacity
Creates a moment at the top of column that causesstress needing to be resisted in addition to axialload.
P- (P-Delta) Effect:
As building moves laterally, this induces aneccentric loading, even on columns concentricallyloaded.
For some materials (e.g. concrete), design fora minimal eccentricity is mandatory by code.
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Columns:Transfer Girder
TransferGrider
Column V
BCE Place, Toronto, Ontario, Canada
Santiago Calatrava
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Bath Hou
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Hurva Synagogue, Jerusalem, Israel (unbuilt project) Louis Kahn Dulles
Illustration: Understanding Structures, Fuller Moore, WCB/ McGraw-Hill, 1999
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Stuttgart Airport, Germany Von
Gerkan Maarg
Rose Center for Earth & Space, NYC
Polsheck & Partners
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Stanstead Airport, Essex, England
Norman Foster
Stanstead Airport, Essex, England Sir Norman Foster
Student Model by James Fickes, Philadelphia University
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Hong Kong & Shanghai Bank, Hong Kong, China Norman Foster
Illustration: Understanding Structures, Fuller Moore, WCB/ McGraw-Hill, 1999
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Alamillo Bridge, Near Barcelona, Spain Santiago Calatrava
Library, Phillips Exeter Academy, Exeter, NU
Louis I Kahn
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Sendai Mediatheque, Sendai, Japan
Toyo Ito
Tokyo International Forum, Tokyo, Japan
Rafael Violy
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Concrete Construction
Introduction
Background Image: Santiago CalatravaRailway Station, Lyon, France
Basics Moldable Stone
Extreme versatilityvirtually any shape is possib Suitable for solid walls, floor slabs, beams, colum
arches, shells and domes
Initially used by Romans 2000 years ago asPozzalan made from volcanic ash. With fall of Roman empire, secret to maki
concrete was lost until the 19th century. Rediscovered by Englishman Joseph Aspdin
1824, called portland cement after a duralimestone already in use.
Background Image: Heinz I slerShell Structure, Switzerland
Basics
Strength is compressive stress (highstrength mixes can now be as great as12,000 psi in practice, even 20,000 psi inthe lab)
Weakness is low tensile strength (verylittle to speak of)
Background Image: Heinz IslerShell Structure, Switzerland
Physical Makeup
1. Portland Cement (mixture of lime, alumia,silica, various minerals)
Acts as the bonding agent
2. Aggregate:
Fine (Sand)
Coarse (Gravel)
These are added for strength and to reduce vo
of portland cement required3. Water
Background Image: Louis KahnSynagogue (unbuilt project), Jerusalem
Chemical Process Hardens by chemical reaction known as
hydration
Does not dry to become hardwhich is whyconcrete can be placed and harden inunderwater conditions (e.g., In rivers for bridgepiers)
Chemical reactions causes heat (heat ofhydration) Will expand
On reallyhot days sometimes add ice to chill wateror add chemicals to keep cool (admixtures)
Background Image: Student CenterUniversity of California, Berkeley
Chemical Process Curing process takes
28 days to reachdesign strength
Must not dry outwhile curing!
Formwork andpossibly shoring (forslabs, etc.) need tostay in place formuch of that time
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Chemical Process
Amount of water in mix is critical!
Too much and strength can be reduced significantly
Too little and it will not be workable
Amount of water is measured in relationship tocement by weight ratiothe Water-Cement(W/C) ratio
Lower W/C ratio produces stronger concreteactually takes very little water to causehydration to occurmuch less than needed toproduce workable concrete
Background Image: Pier Luigi NerviPapal Audience Hall, Vatican
Chemical Process
Typical W/C ratiois in the range of0.4 to 0.5 or
about 40% to50% of mix iswater.
For a simple span beam, by now the whereshould be obvious to you, right?
Slump test as a measure ofwater/cement ratio
Test cylinder to be crushed totest compressive strength of
mix sample
Various Admixtures
Air-entraining agents (for resistance to freethaw cycles and improved workability)
Accelerators (speeds up curing time)
Retarders (slows down curing to aid placem
Water-reducers (take part of some water toincrease strength)
Various Admixtures
Plasticizers and super-plasticizers (make mixmore fluid to easily level or for pumping)
Pozzalin (ultra fine volcanic ash that improvesworkability and strength)
Many others, and new ones are always beinginvented!
Air-Entrained Concrete Normally used only in exterior environments
but sometimes specified for interior as well
Microscopic air bubbles are drawn (entraineinto batch during mixing
Leave tiny air pockets for water to expand ias it freezes so that it does not build uppressure inside concrete and cause surfacedamage
Unchecked surface damage can lead to furtdeterioration in a cycle that can lead to failu
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Air-EntrainedConcrete
ReinforcedConcrete
Developed in late 19th century toovercome limits of low tensile capacity
Composite material of two constitue
concrete and steel, working together tbest advantage of each
Concrete is good for compression
Steel is good for tension
Reinforced Concrete
Only works because steel is ductile and candeform to absorb stress and
Because coefficient of thermal expansion(amount of volume change due to temperaturechange) is similar for each material
If they were not and R/C member exposed tolarge T, then steel would tear loose fromconcrete.
Coulduse iron for reinforcement, but it is brittleand has much lower tensile strength, so couldeasily fracture and cause whole member to fail.
Concrete Variations
Cast-in-place concrete Site-mixed or trucked in
Poured into forms for beams, slabs, columwalls.
Will take shape of formwork, as well as tex
Advantage is that almost unlimited variety shapes are possible
Great limitation is that formwork must remplace until concrete hardens
Naturally continuous construction
Concrete Variations Precast concrete
Instead of forming on site, precast pieces areformed and cured in a plant
Assembles in a similar manner to steel withbolted and welded connections
Discrete pieces means that lateral bracingmust be considered similar to steeli.e., donot automatically get continuous construction
No shoring is required, so erection is veryrapid
Prestressed Concrete Basic idea is to induce compressive
stresses reverseto those of final loadi
Enables higher capacity members utilihigher strength reinforcing
Precast concrete is very often alsoprestressed
Cast-in-place concrete is often post-tensioned using stranded tendons
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PrecastPrestressedConcrete
Illustrations: Francis D.K. Ching: Building ConstructionIllustrated, third edition; John Wiley & Sons, Inc., 2001pg. 4.08
Post-tensionedConcrete
Illustrations: Francis D.K. Ching: Building ConstructionIllustrated, third edition; John Wiley & Sons, Inc., 2001pg. 4.09
Reinforcing Bars (Rebar)
High strength steel (Fy = 60 ksi, normally)
Key factor is bondof steel to concrete:
Must be able to develop the tensile capacity of the steel
Requires adequate embedment / anchorage
Either by enough rebar length (development length)
Or by using bends and hooks on rebar ends
Also requires adequate cover of conc. over reinf.
(this is also needed to protect steel from corrosion) Virtually always use deformed rods to increase
bond by adding mechanical resistance to frictionalresistance
Reinforcing Bars (Rebar)
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Flexural Reinforcement
Okay, so we use steel to take the tension inconcrete memberbut how muchand wher
For a simple span beam, by now the whereshould be obvious to you, right?
N
N A
Think back to flexural
behavior in a solid,prismatic section:
yc
Neutral Axisc = Distance from N.A. touter surface extremefiber distance
Flexural Stress Formula:
Fb = My/I or, in case of extreme fiber stress:
Fb = Mc/I = M/S, where S = I/c = Section Modulus
See examples, pp. 245-245
Flexural Reinforcement
Alright, so thats the where? (for starters), butwhat about the how much? question??
N
AN A
Allowable Stress Design Metho
Steel and wood design methods wevestudied are known as Allowable Stressdesign methods.
Work by taking capacity of material anadding a factor of safety by reducing tstress for design.
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For Structural Steel:Allowable stress in bending = 2/3 Fy
Fb = 2/3(36 ksi) = 24 ksi (= 167MPa)(Allowable bending stress)
FB = 24 ksi(167 MPa)
Fy = 36 ksi
(250 MPa)
Allowable Stress and Factor of Safety
Figure 5.22, p. 282: Barry Onouyeand KevinKane: Statics and Strength of Materials for
Architecture and Building Construction, secondedition; Prentice-Hall, 2001 Elastic Curves for Various Structural Materials
Two things make concrete very different from steel or w1) It has no defined yield point, so how do we know
Elastic Modulus (E) by definition = Stress/Strain
For concrete, E is taken at a strain of 0.003, or 0.3% fc is concrete strength at 0.3% strain
2) Concrete is nothomogeneous in cross section: Tensioforces are concentratedin steel reinforcement
Ultrimate Strength Design Method
Steel and wood design methods weve studiedare known as Allowable Stress designmethods.
Work by taking capacity of material and addinga factor of safety by reducing the stress fordesign.
USD works in the opposite manner, by notreducing the stresses on the material, but byincreasing the loads.
This is because we know the capacity of amaterial much better than we know thevariability of loads.
More realistic stress distribution in concrete recognizesnonlinear stress distribution and cracked concrete secti
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Flexural Reinforcement
The amount of steel in a given crosssection is known as the steel ratio
Similar to the water/cement ratio, it is a
percentage calculation, but this time doneby cross-sectional areas:
Steel Ratio, = As/Ac(is lower case Greek letter Rhobook uses p)
As = Ac
Flexural Reinforcement
So in the balanced condition, the amount oftensile steel exactly balances the compressicapacity of the concrete
This varies depending on the member crosssection and the strength of concrete and ste
Balanced steel ratio is b But this is unsafe!Code limits to specifica
prevent possibility of compressive failure ofconcrete before yielding of steel
max = 0.75 b
Flexural Reinforcement
Because even beams with no load (asidefrom self-weight) undergo volumechanges with drying shrinkage andtemperature fluctuations, the code alsoprescribes a minimum amount of steel:
min = 200 / Fy
min = 200 / 60,000 = 0.0033 for 60 ksi(temperature and shrinkage steel)
Ultimate Strength Design Meth
For R/C design, we use two primaryfactors, one for DL and one for LL
The resulting load is known as theUltimate Load and is defined as:
U = 1.4DL + 1.7LL
The ultimate load is then used as thebasis of design and members areproportioned to carry this load.
Ultimate Strength Design Method
Capacity Reduction Factor:
In addition to the ultimate design load, forsafety since there is some variability inworkmanship and material quality, afurther safety factor is implemented onthe material side of the equation.
For flexure, =0.9
For shear, =0.85
Ultimate Strength Design Meth
Book derived equation for strength ofconcrete:
Based on internal resisting moment cou
M = C Moment Arm = T M.A
C = T = AsFyMu = AsFyjud = AsFy(d-a/2)
a =AsFy/ 0.85fcb
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0.85 fc
a
b = Beam Width
T = AsFy
PCA Simplified Design Method
Developed in 1980s as response to ever-increasingly complex design theories
Mostbuildings are of moderate size andheight, and dont substantially benefitfrom a more rigorous design approach
Assumes fc = 4000 psi, Fy = 60,000 psi
= 0.75 max =0.5(0.75 b)=0.375 b) AS = Mu/4d
Within about 20% of more precise calc.
General Design Approach
Iterative by nature
Self-weight of beam is typically large ashouldnt be ignored, but
Dont know the size of a beam to startwith, so need to make best guess initiathen check
General Design Approach
Start with minimums for depth
Must make allowances for concrete coverover rebar
Sometimes based on code limits for longterm creep deflections
Many solutions are possible for a givenspan!
Any variation of depth and reinforcingsteel that gives appropriate capacity isacceptable
Shear Reinforcement
in Concrete Beams Remember how shear behaves:
In steel beams, the consideration isbuckling of the web due to compressiprinciple stress, in wood its horizontalsplitting along weak cell boundaries
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Shear Reinforcement Wood is affected by the longitudinal
component of shear, where it tends to causesplitting along the grain
Large Concentrated Load
Wood Beam
ShearSplitting
Shear Reinforcement Steel is affected by the compression
component of shear, (tension is no problem
Solution: WStiffener
Problem: WeCrippling
C
C
Shear Reinforcement
Solution: Shear Reinf.Problem: Diagonal Cracking
T
T
So concrete is affected by the (surprise!)tensile component of principal shear stress(compression is no problem!):
Shear Reinforcement
It should be obvious that the pattern opotential diagonal cracks in concrete isanalogous, but exactly the reversedirection, to the pattern of web cripplinin steel
Any time you see diagonal cracking in
concrete or masonry, you should bethinking shear stress
Shear Reinforcement
Sometimes flexural steel is bent up toform shear reinforcement at end of beam:
But long bent bars like this arecumbersome and awkward to work with,so vertical stirrups are used typically
Shear
Stirrups
Shear Reinforcement
Stirrup spacing often varies along the lengththe member, depending on the magnitude oshear forces present in member:
V+-
Closer spacingat high shear
Wider spacingat lower shear
None required in areaof very low shear
Uniformly-Loaded Member
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Typical Beam Cross Section
Shear Stirrups
Hooks are often usedto fully develop shearstirrup capacity
Smaller longitudinal bars at topof member aid construction, orsometimes larger bars used to
add more strength to helpagainst compressive stress
or
may be M- reinforcement
Primary tensilereinforcementmay have
hooked ends to aid inbond if not sufficientdevelopment length
Space equally, may be
several layers of bars.Smallest spacing mustaccommodate largestcoarse aggregate size.
Concrete cover providesbond and protects steel
Typical Simple-Span Beam
Shear ForceResisted by Each
Shear Stirrup
Shear In Concrete Beam
Capacity is shared between conc. &
Code allows shear to be computeddistance d from face of support
Concrete shear capacity is a functio
fv = 2fc (for fc = 4 ksi fv =
Vc = fvbd
v = Shear stress in excess of fv (v
Shear stirrup capacity: V = AvFy
Shear stirrup spacing:
s = AvFy/ vb
Max spacing of stirrups:
s = d/2 for v 4fc (= 253 psi fo
s = d/4 for v > 4fc
First stirrup at s/2 from support
No stirrups required where V < Vc
Extend stirrups a distance d beyotheoretical cutoff point
~
~
~
~
v~{ ~
extend stirrupsdist nce d beyondtheoretic l cutoff point
Continuous Beams
Unlike steel and wood, concrete iscontinuous by nature
Hard to notmake a moment connection!
But, still need to reinforce properly
Primary effect is on placement of steel formomentbecause of contraflexure, M-
develops tension in top of beam or slab
M+
-
M+ in SpanM- at Support
(Top Tension) (Bottom Tens
Point of Contraflexure (zeromoment, zero slopechanges from
concave up to concave down)
Continuous Beam on Columns
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Continuous Slab on Beams (sim. to cont. beam on cols.)
Continuous BeamMoment & Shear
Coefficients
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Deflection Considerations
An important consideration for serviceability
Excessive deflections (especially under LiveLoad) can lead to:
Floors and roofs that flex too muchrigidlyattached materials (walls, plaster ceilings,windows) can be cracked.
Roofs that sag too much under rain can haveprogressive deflection, possibly leading tocollapse.
Perceptionby occupants that a structure isunsafe or is failing
Deflection Considerations
Typically is a checkafter a member isselected.
Can be a controlling factor for longer
members, so in some cases members aredesigned for deflection and checked forflexural and shear stresses.
Limits are set based on a ratio to the spanlength. Limits depend on usage and loadingconditions.
l = Span Length
Common Deflection Limits
Examples: a) For span of 30 and all=l/240: all = 30ft(12in/ft)/240 = 1.5
b) If act= 0.75 on a span of 24, ratio = 24ft(12in/ft)/0.75 = 384 = l/384
Deflection is controlled by four factors: Load
Span
Elastic Modulus (Material Stiffness)
Moment of Inertia (Geometric Stiffness)
Actual exact deflection computationsare complex and cumbersome,involving computing moments of the
M/EI diagramwell use basic chartformulas.
Deflection Computations
For uniformly loaded simple spanbeam: = 5wl4/384EI
Note that stress level has no bearing indeflection calcs!
Watch your units in deflectioncalculations!!! Veryeasy to makecomputational errors in these calcs.
Deflection Computations
Formulas for Flexure, Shear and Deflection
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Formulas for Flexure, Shear and Deflection Formulas for Flexure, Shear and Deflection
Formulas for Flexure, Shear and Deflection
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Introduction to Lateral ForcesIntroduction to Lateral Forces
Lateral Forces
Typically considered to be those whichact parallel to the ground plane
May occur at many angles other thanperfectly horizontal
Generally considered to acttransversely to the primary structuralsystem
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Whats the Big Deal?
Essential for a structure to have lateralresistance
Buildings cant stand against wind,seismic or other lateral forces otherwise
More than any other structuralcomponent, the lateral force-resistingstructure has significant impact on space
planning
The Right Way
Theres a right way and a wrong way togo about it
The right way is to recognize that it iscritical to consider lateral forces from thevery start, and
Integrate lateral force-resisting structurewithin initial schematic design
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The Wrong Way
The wrong way is to leave it until the endfor the structural engineer to work out
You might get lucky and all will be fine,or
You may perhaps end up with a conflictof necessary cross bracing that needs tobe in exactly the wrong place.
The Wrong Way
Plan and elevation configuration may evencause difficulty for an engineer to make asuitable structural system work properly,efficiently and economically
In the worst case scenario, there are numerous
structural disasters that have resulted not somuch by poor engineering as simply poorly-conceived buildings that were essentially forcedto work structurally
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Take Note!
The larger the lateral forces are(whether from wind or seismic forces),the bigger the structural impact andthe more crucial it becomes for thearchitect to consider lateral forcesfrom the earliest planning time!!
Types of Lateral Forces
Wind and seismic forces are the mostfundamental lateral forces that an architectmust be familiar with
Most architects at some point need to dealwith one or more other types of lateral forces,
so it is important to at least be familiar withthem.
Lateral forces can be internal to a structure orexternally acting outside of it
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Internal Lateral Forces
Those which occurfrom the nature ofthe structure itself,such as the thrust ofan arch, vault orshell, or the tensionpull from a cable or
membrane
Other Internal Lateral Forces
Restrained thermal movementassociated with temperature change If prevented from expanding or contracting, a
material will undergo internal forces and stress indirect proportion to its coefficient of thermalexpansion and the degree of temperature change
Finger plate expansion jointin bridge deck
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Other Internal Lateral Forces
Volumetric changes e.g., control joints are required in concrete
slabs because as concrete cures it loosesmoisture and contracts
Without enforcing (hence, controlling)where the crack occurs, it will crack in anunappealing random pattern that is alsomore deleterious to the surface than acontrol joint.
External Lateral Forces
Most familiar are wind and seismicforces, but there are others:Fluid pressure from water and otherliquidsSoil against a basement or retaining
wall, or perhaps retained materials suchas sand, grain, or even coal or woodchips in a power plant storage bin
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Building ConstructionIllustrated, p. 3.10
External Lateral Forces
Flood waters can produce devastatinglylarge lateral pressures and scour awayat foundations, potentially underminingthe stability of a building or a bridgesupport pier
A "rolling force"is generated on bridgegirders from other large objects likemovable cranes on rails Occurs when a massive object (truck, train
or crane, etc.) is decelerated
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Gantry Crane
Wind and Seismic Loads
Most fundamental lateral forces that anarchitect must be familiar withMay be so small as to be unnoticed, orlarge enough to level citiesOccur simultaneously with gravity loads
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Wind Loads
Wind is really a very complexphenomena with a complex interactionon a building structure
It is influenced greatly by local terrainWhen contacting a building, it canproduce pressures and suction forceson any surface of a building, plus
internal pressures that tend to balloonthe building outward
Wind Loads
Can be thought of against a buildinglike the way an airplane wing behavesAs air moves over the curved surface of
the wing, the molecules separate and thenrejoin.
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Wind Loads
Air over the top of the wing moves faster.The Bernoulli effect says this creates lowerpressure, which becomes lift that keeps theplant aloft
Wind Loads
Similarly for a building: Windward face will experience pressure
forces
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Wind Loads
Leeward face will experience suction
Wind LoadsRoof: Flat roof will experience suction Pitched roof will experience suction if wind
parallel to ridge (similar to a flat roof)
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Wind Loads
Pitched Roof: Lee side will experience suctionif wind perpendicular to ridge
Windward side may experience suction orpressure, depending on steepness of slope(pressure only at pitch of about 9:12)
Wind Pressure and Suction
Wind: Actual Behavior
Wind: Effects
Sliding Overturning
Wind:BuildingCodes
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What do you already knowabout seismic loads?
Lets test your intuition
This building was damaged by anearthquake.
How did it happenTake 2 minutes to talk with each otherand make a list
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Seismic Loads
Motion originates outside of a buildingEffect is internal (c.f., external wind)Forces generated by inertia of buildingmass as ground moves below thestructure
Ground Motion (Action)
Building Motion (Reaction)
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Seismic Loads
Generates forces in direct proportionto the building's mass and stiffness
A massless building would in fact haveno seismic forces with at all!By altering the building's stiffness, asubstantial change to seismic force ispossible (basis for some design
approaches)
How are Lateral Forces Resisted?
Most of the building components thatcomprise the gravity-resistingstructure are also those whichcomprise the lateral force-resistingstructure, except that the forces aremoving differently
Easiest to visualize in terms of windloads, though seismic is similar:
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James Ambrose, Building Structures
Lateral Load Propagation in a Basic Box Structure
WindLoad
Principal Vertical-Plane Lateral Framing Structures
In-planeDiaphragm Action
Triangulation(Vertical Truss)
Moment ResistantJoints
Edward Allen,Architects Studio Companion
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Lateral Force Transfer
Francis D.K. Ching, Building Construction Illustrated
Vertical SupportStructural Patterns
Daniel Schodek, Structures
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Placement of Lateral Force Resisting Elements In Plan
Daniel Schodek, Structures
Daniel Schodek, Structures
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Concrete Shear Wall
Ligh t Wood-Framed Shear Wall
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Daniel Schodek, Structures
Diagonal
Cross
Bracing:
These
slender rod
bracing
members can
take only
tension, while
the heavier
members on
the opposite
corner canwork in both
tension and
compression.
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Inverted K-Bracing:
The members in this
arrangement always resist
compression since they provide
a mid-span support for attached
beams. Lateral loads will either
add or subtract from that
compressive force depending
on the direction.
Diagonal Bracing:
This arrangement with heavy
diagonal members is capable of
resisting both tension and
compression.
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Steel Rigid Frame
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Analysis and Designof Spanning Members
Returning full circle to beginning of classwhen we looked at framing patterns
Taking a focused look at solid cross-sectionspanning members (beams, girders, joists,purlins, girts, etc.) since they are one ofthe most ubiquitous of constructionelements
Looking to understand actual internalbehavior so that rational analysis anddesign procedures can be employed
Analysis and Designof Spanning Members
Three primary criteria:1. Flexural Stress
2. Shear Stress3. Deflection
Additional criteria (not alwaysapplicable, depending on conditions):
1. Lateral-torsional buckling
2. Bearing stress
3. Local web crippling
4. Torsion
Analysis and Designof Spanning Members
Each criteria needs to be addressed foreach member
Different spans, locations and loadingconditions will influence which factors willcontrol
Typically flexural stress will govern thedesign but
Short spans frequently controlled by shearstress
Long spans frequently controlled bydeflection limitations
Analysis and Designof Spanning Members
The actual analysis or design (i.e.selection) of a member is a relativelystraightforward process using asimple equation, however
However, MOST of the effort isinvolved with:
Determining loads
Tracing the load paths and creating beamFBDs
Generation of shear and moment diagrams Computation of maximum shear and moment
Differences Between AxialStress and Flexural Stress
Magnitude varies along the length ofthe span
Magnitude varies through the depthof the member, and reverses sense(tension to compression or vice-versa)
W
R=W
Basic axial stress is uniformand easy to compute:
W
f = W/A
Not so with flexural stress!
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Onouye, Ch. 8
1. Principle of Flexural Stress
yc
Neutral Axisc = Distance from N.A.to outer surfaceextreme fiberdistance
Flexural Stress Formula:
fb = My/I or, in case of extreme fiber stress:
fb = Mc/I = M/S, where S = I/c = Section Modulus
Differences Between AxialStress and Flexural Stress
Magnitude varies along the length ofthe span
Magnitude varies through the depthof the member, and reverses sense(tension to compression or vice-versa)
Shape mattersA lot!
Deflections vary along length of span
Shear stresses develop
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Seismic Base Isolation
Seismic Acceleration Graph
Building Acceleration with Seismic BaseIsolator
(University of Brighton, U.K.: http://www.brighton.ac.uk/environment/mscce/kpresponse.htm)(Hart-Weidlinger Associates:
http://www.wai.com/Hart/Images/base
%20isolator%20San%20Fran.jpg)
Base Isolator
The technology that is recently drawing conside
safeguard building by isolating them from heavy
earthquake isolation device is mounted betweetremor of a base at t he time of an earthquake. T
laminated rubber that supports the load of a strufurther reduces vibration by keeping the relative
LFPS (Friction Pendulum System)
2-D Floor Isolation System 3
FPS (Friction Pendulum System) LRB (Lead R
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2-D Floor Isolation System
3-D Floor Isolation System
Cosine Cur
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Los Angele
(Photos courtesy Brad Aagaard, USGS Postdoctoral Scholar at
Los Angeles City Hall Retrofit Los Ange
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Los Angeles City Hall Retrofit
Wind Motion Attenua
(Aurotech web site: http://www.robot.com.tw/tmd.htm)
400 Ton (!) Concrete Tuned Mass Damper atop Citicorp Building
(Online Ethics Center For Engineering and Science
http://onlineethics.org/images/moral/LeMessurier/25.gif)
730 Ton (!!!) Spherical
Steel Tuned Mass
Damper atop Taipei101 Building
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Seismic Loads
Origins
Plate Structure of Earths Crust
Plate Boundary
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Pangaea
ca. 500,000,000 B.C.
Gap
Overlap
Edge ofContinentalShelf
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San AndreasFault From Air
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P (Primary) Waves (Mostly Direct, Push-Pull in nature)
S (Secondary) Waves (Mostly Reflected, Side-to-side in nature)
S
S
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3 Types ofSeismic Wave
Actions
Types of
http://www.analog.com/library/ana
P-waves
Rayleigh-waves
Terraforming:
Effects of Earthquakeson the Landscape
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1964 Earthquake nearAnchorage, Alaska
Sand Boil (note
Niigata, Japan, 1964 Kawagishi-Cho Apartment Buildings Collapse
Liquefaction Same type buildup of water pressure in soil that causes sand
boils creates a weakening of the soil and loss of bearing capacityby dispersing soil particles and turning moist soil into mud
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What Produces the DamagingForces In Structures?
Ground motion below building
Results in inertial reaction of buildingtrying to stay still by Newtons 1st lawof motion
Which leads to lateral displacementbehaving as a lateral structural force
What ProduForces
Dynamic Forces
Motion: F= MA F=WA/g or F=W
W= Building wei A = Ground acc g = Gravitationa
c = Seismic basforce is a %
Other Factors AffectingSeismic Loads on Structures
Magnitude of Ground Acceleration
Building Inertia (directly proportional to mass)
Natural Vibrational Period of Building
Natural Vibrational Period of Soil
Nature of Structural Framing System
Other FaSeismic Lo
If building infinacceleration, t
But real buildin
extent, the accbut steady, ancomplex variat
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Code-Prescribed SeismicLoad Equation
Allows us to look at the actual dynamicforce on the structure as a static load.
Based on previous equation, F=Wc,with additional modifiers
Z = Seismic Zone C
I = Importance Facto
C = Coefficient for gconsideration the intbuilding vibration pe
W = Building Weight
Rw = Building Frame
Old Method based
Code-PreLoa
V
1994 UBC Seismic Zone MapCurrent (200
Map of P
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Occurs simultaneous with flexuralbehavior, always.
Typically is a check after a member isselected based on flexural stressconsiderations.
Sometimes is a controlling factor forshorter span members or those withlarge concentrated loads near asupport.
2. Principle of Shear Stress Principle of Shear Stress
Transverse Shear Force:F = 0 (V = RA in this case)
Transverse Shear Stress :fv = V/A
Shear Force vs. Shear Stress
Equivalence of Horizontal Shear, Vertical Shear andDiagonal Tension / Compression Principal Stresses
Horizontal & Vertical Shear Stress Principal Stress
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Horizontal shear and vertical shear happen atsame time! Cant have one without the other.
Different materials will respond differently tovarious components of shear stress.
Shear in Wood
Wood is affected by the longitudinalcomponent of shear, where it tends to causesplitting along the grain
Large Concentrated Load
Wood Beam
ShearSplitting
Shear in Steel
Steel is affected by the compression component ofshear, (tension is no problem!):
Solution: WebStiffeners
Problem: WebCrippling
Shear in Concrete
Solution: Shear Reinf.Problem: Diagonal Cracking
Concrete is affected by the tensile component ofprincipal shear stress (compression is no problem!):
Shear in Concrete
It should be obvious that the pattern ofpotential diagonal cracks in concrete isanalogous, but exactly the reversedirection, to the pattern of webcrippling in steel
Any time you see diagonal cracking inconcrete or masonry, you should bethinking shear stress
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Principle of Shear Stress
Shear Stress Formula:
General Equation: fv = VQ/Ib
Q = First moment of area above N.A. (seeSchodek pg. 257)
I = Moment of Inertia
b = Beam width (breadth)
For rectangularsection: fv = 3/2(V/bd)
Approximation for SteelOnly: fv = V/dtwShear Stress Distribution in a Rectangular Section
DepthBelow
N.A.
DepthAboveN.A.
Magnitude of Stress
fv(max) = 3/2(V/b
Shear Stress Distribution in a Wide Flange Section(Actual vs. Approximation)
fv(avg) = V/dtw
d
tw
Shear Stress Distribution in a Tee Section
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Shear and Moment Diagrams
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Connections:Connections:
The Poetics of StructureThe Poetics of Structure
Connection Concepts
Represent the greatest opportunity forarchitectural expression of structure
Literally are a point of energy transfer,as well as figuratively, metaphorically
God is in the details Mies van der Rohe
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Connection Concepts
Most vital aspect of a structure
Lose a connection, lose everything itsresponsible for carrying
Transfer of force depends on howstructure was modeled
Roller
Pin
Fixed
Illustrations: Daniel L. Schodek,Strcutures, 5th ed., 2004 Pearson/Prentice-HallSchodek Fig. 2.15
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Illustrations: Daniel L. Schodek,Strcutures, 5th ed., 2004 Pearson/Prentice-HallSchodek Fig. 3.9
Connection Concepts
Design details depend on type ofmaterial
Ideal connection types are onlyapproximated by actual construction
e.g. bolted connections in steel normallyconsidered to deform under load, act aspin connection.
Very hard to achieve true fixity!
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Image University of California, Berkeley, Godden Slide Library: http://nisee.berkeley.edu/jpg/6257_3021_0122/IMG0033.jpg
Roller & Rocker-Type Connections
Image University of California, Berkeley, Godden Slide Library: http://nisee.berkeley.edu/jpg/6257_3021_0122/IMG0018.jpg
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Image University of California, Berkeley, Godden Slide Library: http://nisee.berkeley.edu/jpg/6257_3021_0122/IMG0029.jpg
Image University of California, Berkeley, Godden Slide Library: http://nisee.berkeley.edu/jpg/6257_3021_0122/IMG0022.jpg
Rome International Airport
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Image University of California, Berkeley, Godden Slide Library: http://nisee.berkeley.edu/jpg/6257_3021_0122/IMG0023.jpg
Rome International AirportOverhanging Beam Roller Joint
Image University of California, Berkeley, Godden Slide Library: http://nisee.berkeley.edu/jpg/6257_3021_0122/IMG0060.jpg
Pinned-Type Connections
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Image University of California, Berkeley, Godden Slide Library: http://nisee.berkeley.edu/jpg/6257_3021_0122/IMG0062.jpg
Renzo PianoPA Technology Centre, Princeton, NJ
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Image Deborah J. Oakley
Santiago CalatravaMilwaukee Art Museum Addition
Image Deborah J. Oakley
Santiago CalatravaMilwaukee Art Museum Addition
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Various Knife-Plate Connections Cutler Anderson Architects
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Gates Residence, Bellvue, Washington Bohlin Cywinski Jackson, Architects
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Grace Episcopal Church, Bainbridge Island, WA Cutler Anderson Architects
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Capitol Hill Library, Seattle, WA Cutler Anderson Architects
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Truss Pedestrian Bridge
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Joint Point of Entry,Orovile, WA & Osoyoos, BC
Cutler Anderson Architects
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Continental Train Platform, Waterloo Station, English Channel TunnelNicholas Grimshaw & Parners
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Traversina Footbrige, Viamala Gorge, Switzerland
Conzett Bronzini Gartmann, Engineer
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Original Mt. PalomarObservatory Telescope MirrorTension Hanger Connection
Corning Museum of Glass,Corning, NY
Smith-Miller & Hawkinson,Architects
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Framed Beam Connection:Steel to Concrete Using
Embedded Plate
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Thorncrown Chapel, Arkansas, Fay Jones Architect
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Philip Merrill Environmental Center, Chesapeake Bay Foundation,Annapolis, Maryland Smith Group, Architects
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Broadgate Office Building, London, Skidmore, Owings & Merril
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Illustration Source: http://www.ce.berkeley.edu/~boza/courses/cee122/lectures/lecture2/connect-brace.jpg
Concentric Bracing Lines of Action in a Truss
Illustration Source: http://www.ce.berkeley.edu/~boza/courses/cee122/lectures/lecture2/connect-brace.jpg
Concentric Bracing Lines of Action in a Truss No Moment in Joint
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Hartford Civic Center Roof Failure
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e
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Design of Bolted Connections
Shear Connections in Tension
Hanger rods
Cable anchors
Diagonal braces
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Riveted Connections Traditional Steel Construction
Connection Method
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Rivet Installation
Red Hot Rivet
Clamping ForceDeveloped asRivet Cools andShrinks
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Tension Hanger Connection
Corning Museum of Glass,Corning, NY
Smith-Miller & Hawkinson,Architects
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Bolted Connection Failure Modes
Illustrations: Daniel L. Schodek,Strcutures, 5th ed., 2004 Pearson/Prentice-HallSchodek Fig. 16.4
1) Shear Failure of Bolt
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2) Bearing Failureof Connected Material
3) Tensile Failure of Connected Material
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Block Shear Failure
4) Tearing (Block Shear) Failureof Connected Material
Full Moment Connections
Fixed connections that do not allowrotation
Reinforced concrete is naturally fixed
Steel can readily be made fixed
Wood is almost never used in fullmoment connectionachieve fixity inwood by knee bracing.
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FIX
Illustrations: Barry Onouye and Kevin Kane:Statics and Strength of Materials forArchitecture and Building Construction, second edition; Prentice-Hall, 2001
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What Makes a Connection Fixed and When is it Pinned?
Pinned connectionallows free rotation
End rotation of a simply- supportedbeam with a uniform load:
w L3
24EI =
( in radians)
Rigid connectionStays 90
Conclusion: It takes a VERY small amount of rotation to make a connection pinned!
Example:
For a 20 long W 24x104with w=4.2 k/ft load:
= 0.01 rad = 0.43
d = 24
0.43
~ 0.09 2.3 mm
Bm.Centerline
12
Illustrations: Daniel L. Schodek,Strcutures, 5th ed., 2004 Pearson/Prentice-HallSchodek Fig. 3.9
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Illustrations: Barry Onouye and Kevin Kane:
Statics and Strength of Materials for Architectureand Building Construction, second edition;Prentice-Hall, 2001
Illustrations: Barry Onouye and Kevin Kane:Statics and Strength of Materials forArchitecture and Building Construction, second edition; Prentice-Hall, 2001
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Illustrations: Barry Onouye and Kevin Kane:
Statics and Strength of Materials for Architectureand Building Construction, second edition;Prentice-Hall, 2001
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Flanges must be connected
for fixed joint to occur!
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Illustrations: Francis D.K. Ching: Building Construction Illustrated, third edition;John Wiley & Sons, Inc., 2001
Stockley Park, Middlesex, UK
Norman Foster and Partners
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Stockley Park, Middlesex, UK
Norman Foster and Partners
Corning Museum of Glass,Corning, NY
Smith-Miller & Hawkinson,Architects
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Original Mt. Palomar Observatory Telescope MirrorBase Support Frame for Tension Hanger Connection
Corning Museum of Glass, Corning, NY
Smith-Miller & Hawkinson, Architects
Office Building, Palo Alto, CAImage University of California, Berkely, Godden Slide Library: http://nisee.berkeley.edu
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Discovery Museum, Bridgeport, Connecticut Chan Krieger Sieniewicz Architects
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Michaels / Sisson Residence, Mercer Island, Washington, 1998; Miller | Hull Partnership
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UNIVERSITY of MARYLANDSchool of Architecture, Planning and Preservation
ARCH 412 | Technology III
Prof. Deborah Oakley
Step-by-Step Outline of Beam Design Procedure: The Least You Need to Know
This guideline summarizes the basic analysis and design procedure covered this semester,as well as highlighting the essential concepts. Please also review the last two pages of the
handout from chapter 8 of the Onouye text, which also summarizes key concepts.
1. Assess the framing layout overall, as well as problem requirements:a. What is the framing hierarchy? Are you dealing with a beam or a girder?
What is the loading pattern (uniform, concentrated at mid span,
concentrated at the 1/3 points are common)
b. What materials required or identified to be used? This directly determines theallowable stresses for flexure and shear (see the Properties of Various
Structural Materials table). If given a beam to check (a na lysis) look up the
dimensional and structural shape properties from the appropriate tables, orcompute section properties if specified in problem.
c. What limits (if any) are stipulated for deflection? Limits are most commonlyspecified in the form a ratio to the span length.
2. Draw free-body diagrams for the loading of each member under consideration. (ifyou need a refresher on this, see homework #7 from Tech II, as well as the Chapter
3 (section 3.5) readings in the Schodek Struc ture stextbook (which you should have
in your notes from Tech II). This is also covered in the Onouye text in Chapter 4. Both
of these books are on reserve in the library.
3.
Based on this FBD, compute the loads by tributary width to each member foruniform loads. Apply any special concentrated loads that might be indicated on
the framing plan (e.g., large piece of machinery, etc.)
4. Compute the end reactions. If these are beams framing into girders, then these endreactions become concentrated loads on the girders. (Again, see material from
Tech II if you need a refresher). These end reactions typically constitute the
maximum shear values for checking shear stresses.
5. Compute the maximum moment and shear based on the loading type. Typicalconditions include:
a. Uniform load: M = wL2/8 V = wL/2 (same as end reactions for #4 above)b. Single concentrated load at mid-span: M = PL/4 V = P/2c. Two concentrated loads at 1/3 points: M = PL/3 V = P
6. Based on the maximum moment either compute the maximum flexural stress(A na lysis) or select a beam to satisfy the maximum moment (Design), or possibly
check for its allowable capacity (Ca p a c ity Ra t ing ):
a. Analysis: fb= Mmax/Sxb. Design: Sreqd= Mmax/ Fbc. Capacity rating: Mallowable= SxFb
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ARCH 412 | Technology III Beam Design Outline
7. Check shear stresses:a. Steel: Compute average shear as V/(d)(tw)b. Wood: Compute maximum shear as 1.5V/A
(where A is member cross sectional area)
8. Check deflections against allowable limits. Deflection formulae vary depending onthe loading, just as do moment formulae. Following are several typical loading
cases. See references for other loading conditions:
a. Uniform load: = 5wL4/384EIb. Single concentrated load at mid-span: = PL3/48EIc. Two concentrated loads at 1/3 points: = 23PL3/648EI
Check computed deflections against stipulated criteria (typically expressed as a
ratio of the span length. (See table 8.1 in Onouye Chapter 8 readings)
Common Symbols
Geometric Properties:
A Cross sectional area
NA Neutral axis (also centroidal axis)
d Member depth (also frequently listed as h with same meaning)
b Member width (breadth)
bf Width of flange (for flanged members)
tw Web thickness (for flanged members)Ix Moment of Inertia (for a rectangular section, Ix=bh3/12
Sx Section modulus about major axis (Sx = Ix/c where c = distance from neutral axis to
extreme outer surface (fiber) of the member)
Mechanical:
M Moment
V Shear
Deflection
E Modulus of elasticity
Fb Allowable bending stress
fb Actual calculated bending stress
Fv Allowable bending stress
fv Actual calculated bending stress
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Tech III oncept ard
Subject: Concrete Construction
What is and what does it represent?
Tech III oncept ardSubject: Concrete Construction
What is meant by a balanced steel ratio?
Tech III oncept ardSubject: Concrete Construction
If some reinforcing steel in a beam is good,
then more must be better, right?
Whats WRONG about this?
Tech III oncept ardSubject: Concrete Construction
How long does concrete taketo reach full strength?
Tech III oncept ardSubject: Concrete Construction
When concrete is being mixed and it begins toharden, why is adding water NOT a good idea?
Tech III oncept ardSubject: Concrete Construction
What is a slump test and what critical propertydoes it tell about a concrete mix?
Tech III oncept ardSubject: Concrete Construction
Identify the three primarycomponents of concrete.
Tech III oncept ardSubject: Concrete Construction
What is air entrained concrete and why is itused? What types of construction would it mos
likely employ it?
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Tech III oncept ard
Subject: Concrete Construction
Concrete does not dry to become hard. What
is the curing process called and what is oneimportant characteristic of it?
Tech III oncept ardSubject: Concrete Construction
What would potentially happen
if the coefficient of thermal expansion
for concrete and reinforcing steel
were not approximately the same?
Tech III oncept ardSubject: Concrete Construction
Where should primary tension steel be placedon a continuous span flexural member?
Tech III oncept ardSubject: Concrete Construction
Where is shear reinforcement normally locatedin a flexural member?
Tech III oncept ardSubject: Concrete Construction
What is the purpose of the bumps(deformations) on typical
reinforcing steel rods?
Tech III oncept ardSubject: Concrete Construction
What is meant by development lengthof reinforcing bars?
Tech III oncept ardSubject: Concrete Construction
Why are reinforcing bars sometimesbent into hooks?
Tech III oncept ardSubject: Concrete Construction
What is the purpose of bar supportsfor reinforcing steel?
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Tech III oncept ard
Subject: Concrete Construction
What is the purpose of post-tensioning or
prestressing a concrete member? How does isdiffer from using conventional reinforcing steel?
Tech III oncept ardSubject: Soils and Foundations
What is the purpose of sieve analysis and what
property of soils does it provide?
Tech III oncept ardSubject: Soils and Foundations
What characteristics distinguish coarse-grainedsoils from fine-grained soils?
Tech III oncept ardSubject: Soils and Foundations
At a minimum, where should soil borings betaken when a new building is being planned?
Tech III oncept ardSubject: Soils and Foundations
What type of soil must you NEVER build on?
Tech III oncept ardSubject: Soils and Foundations
What is the purpose of dewatering?
Tech III oncept ardSubject: Soils and Foundations
What is the function of a caisson bell?
Tech III oncept ardSubject: Soils and Foundations
What are two principle differences betweenpiles and caissons?
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Tech III oncept ard
Subject: Soils and Foundations
What is a grade beam and what is its function?
Tech III oncept ardSubject: Soils and Foundations
What types of information are
provided on a boring log?
Tech III oncept ardSubject: Soils and Foundations
What are two potential problems when buildingon clay soils?
Tech III oncept ardSubject: Soils and Foundations
Why must exterior footings be placedbelow the frost line?
Tech III oncept ardSubject: Soils and Foundations
What is a mat foundation?
Tech III oncept ardSubject: Soils and Foundations
A given volume of soil can be thought of ingeneral as consisting of what three constituents
Tech III oncept ardSubject: Soils and Foundations
Pile foundations often cannot be used in denseurban areas for what reason?
Tech III oncept ardSubject: Soils and Foundations
What information does the Standard PenetratioTest (SPT) provide?
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Tech III oncept ard
Subject: Soils and Foundations
In what primary way do shallow foundations
differ from those which are called deep?
Tech III oncept ardSubject: Soils and Foundations
Describe the function of a pile cap.
Tech III oncept ardSubject: Lateral Forces
What are the three primary vertical lateral forceresisting structure types?
Tech III oncept ardSubject: Lateral Forces
What is the primary difference betweenthe way in which wind and seismic forces
are generated in a building?
Tech III oncept ardSubject: Lateral Forces
Identify two different sources of lateral forces,
aside from wind and seismic loads.
Tech III oncept ardSubject: Lateral Forces
What is the name of the science that studies the
movement of earths surface structure (themotion of which is the source for earthquakes
and volcanoes)?
Tech III oncept ardSubject: Lateral Forces
Under a wind load, which faces on a typicalbuilding structure have pressure forces
and which are suction forces?
Tech III oncept ardSubject: Lateral Forces
What is the function of a horizontalfloor or roof diaphragm?
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Tech III oncept ardSubject: Lateral Forces
Explain the concept of the beam analogy with
regard to lateral forces.
Tech III oncept ardSubject: Lateral Forces
Conceptually, what is the difference
between the normal force method andprojected area method of computing
wind pressures on a building?
Tech III oncept ardSubject: Lateral Forces
Why are asymmetric building plans
or elevations undesirable in locationssubject to high seismic forces?
Tech III oncept ardSubject: Lateral Forces
What is the purpose of base isolation?
Tech III oncept ardSubject: Lateral Forces
What is one way of dealing with an
irregular building plan that will addressconcerns for high seismic loads?
Tech III oncept ardSubject: Lateral Forces
How does a flexible floor or roof diaphragm
differ from one that is rigid?
Tech III oncept ardSubject: Lateral Forces
At a minimum, how many lateral force-resistingwalls or frames are required to resist
torsional forces on a diaphragm?
How must they be arranged?
Tech III oncept ardSubject: Lateral Forces
What is a soft story and why is it a problem?
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Tech III oncept ardSubject: Lateral Forces
With respect to lateral forces, what is meant
by the concept of continuity in a building
structural system?
Tech III oncept ardSubject: Lateral Forces
What materials can be used to
construct shear walls?
Tech III oncept ardSubject: Lateral Forces
How may a flexible diaphragm be made into arigid one? (Hint: This is frequently done in
conventional construction practice)
Tech III oncept ardSubject: Lateral Forces
What are the special wind region areas shown
shaded on the wind . What does this mean and as designer how should you go about determining wha
the site-specific wind velocity is?
Tech III oncept ardSubject: Lateral Forces
What is a tuned mass damper and what is its
function relative to lateral forces.
Tech III oncept ardSubject: General Structures
Modulus of elasticity and moment of inertia
both are measures of what structural property?How do they differ?
Tech III oncept ardSubject: General Structures
What is the structural definition of moment?
Tech III oncept ardSubject: General Structures
What is the definition of slenderness ratio?What is the implication in column design?
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Tech III oncept ardSubject: General Structures
What are the three fundamental support types?
What forces do they each resist?
Tech III oncept ardSubject: General Structures
What are the two fundamental
equations of equilibrium?
Tech III oncept ardSubject: General Structures
What is the fundamental definition of stress?
Tech III oncept ardSubject: General Structures
Describe three ways in which momentcouples can be found in buildingstructures at varying scales.
Tech III oncept ardSubject: General Structures
What are the limits to the applicability of a two-
way structure versus a one-way structure?
Tech III oncept ardSubject: General Structures
What is an indeterminate structural member
and how does it differ from one that isstatically determinate?
Tech III oncept ardSubject: Concrete Construction
Why are there different load factors in concrete
design for dead loads versus live loads? Why is
the live load factor higher?
Tech III oncept ardSubject: Concrete Construction
The building code prescribes a minimum
amount of reinforcing (0.33%, for grade 60
regbar). What purpose does this minimum steeserve?
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UNIVERSITY of MARYLAND
School of Architecture, Planning and PreservationARCH 412 | Tech IIIProfessor Deborah Oakley
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Multiframe Analysis of Rigid FramDesign Trials Record
Allowable Beam max(inches)= L/
Allowable Frame Drift max(inches)= H/
Trial
No.
Overall
Frame
Drift
Total Steel
Weight
Beam
Size
Beam
Max
fb + fa(ksi)
Column
Size
ry(in)
L/ry Fa(table
C-36)
1
2
3
4
5
Final Beam Size: Final Column Size:
L =
H1
=
H2
=
H
=
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Our design for this library was heavily inu-
enced by the desire for structural simplicity, clear
site lines, and solar efciency. The structure is
extremely straighforward with a twenty-ve foot
grid creating six blocks in two oors. The excep-
tion to the structural rigidity is the overhanging
path coming off the elevator and service corridor
facing the double story space of the atrium. The
services, storage and circulation cores are all
stacked on the west side of the building. The place-
ment of the elevator and the egress stairs in the
southwest corner allowed for the insertion of the
shear walls and rigid frame around those services.
The structural function of the corner is expressed
in the exterior with a blank volume. A correspond-
ing volume in the northeast corner also presents
the blank corner. Together the two corners con-
trast the dominant glazing of the other modules
of the facade. Of course, with the large amounts
of glass in two-thirds of the walls, the glass, the
openings are protected by brise soleil throughout.
In the interior there was the intention to main-
tain as many spaces open and seamless as possible.
With the stacked service cores, the only enclosed
zone is the reference area, which is pushed against
the blank walls, which also provide protection forthe books in the room. The main reading room is
on the second oor, as is the kids area, allowing
for an expansive open space that creates a com-
fortable educatonal and leisure environment. All
in all the library is a simple and straighforward
building that makes the most of the basic goals we
set out from the beginning.
Brise Soleil
Detail
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First FLoor Plan
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Second FLoor Plan
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South Elevation
North Elevation
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West Elevation
East Elevation
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Floor Plan - Structural
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Roof Plan - Structure
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Section E-W
Section N-S
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DetailRo
of/WallSection
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Framing Plan Axonometric
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Gravity Loads:
ROOF
Membrane = 2.0 psf
Sheathing = 3.0 psf
Insulation = 1.5 psf
Decking = 2.0 psf
Ceiling = 1.0 psf
Mechanical = 5.0 psf
Dead Load = 14.5 psf
Live Load = 30.0 psf
Total Roof Load = 44.5 psf
FLOOR
Floor finish = 2.0 psf
Concrete slab = 50.0 psf
Steel decking = 8.0 psf
Ceiling = 2.0 psf
Mechanical = 5.0 psf
Dead Load = 67.0 psf
Live Load = 150.0 psf
Total Floor Load = 217 psf
Allowable frame drift = h/200
frame = (14 ft + 12 ft) ( 12 in / ft ) / 200 = 1.56 in
Allowable beam deflection = L/240beam = 25 ft ( 12 in / ft ) / 240 = 1.25 in
Use A572 steel:Fy= 50 ksiFb= 30 ksiFv= 20 ksiFt= 30 ksiE = 30,000 ksi
Case Western Reserve University Library
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Typical Beam Calculations
Beam: Roof beam with total load of 44.5 psf
W = 5 ft ( 44.5 lb/ft2
) = 222.5 lb/ft = 0.2 k/ft
Mmax= wl28 = 0.2 k/f ( 25 ft )
28 = 15.625 kft
Vmax= wl2 = 0.2 k/f ( 25 ft )2 = 2.5 k
Sreq= Mmax/ Fb= 15.625 kft (12 in/ft )30 k/in2 = 6.25 in3
Ireq= (5wl4)(384Eallowable)
= [ 5 (0.2 k/ft1ft/12in) (25ft12 in/ft )4][ 384 (30,000 k/in2) 1.25 in ]
= 46.875 in4
(dtw)req= VmaxFv= 2.5 k20 k/in2 = 0.125 in2
Requires: Sx> 6.25 in3 , I x> 46.875 in
4 and dtw> 0.125 in2
Possible beam size for roof from Wide Flange Shapes chart
Member size Sxin3
Ixin4
din
twin
dtwin2
W8 x 15 11.8 48.0 8.11 0.245 1.987
W10 x 12 10.9 53.8 9.87 0.190 1.875
W12 x 14 14.9 88.6 11.91 0.200 2.382
W10 x 12 is the lightest of all possible beam sizes, so use W10 x 12 for roof beam.End reaction on typical roof beam transfers as point loads on girder below.
Pon roof girder= wl / 2 = 222.5 lb/ft (25ft) / 2 = 2781.25 lb = 2.8 k
Beam: Floor beam with total load of 217 psf
W = 5 ft ( 217 lb/ft2
) = 1085 lb/ft = 1.1 k/ft
Mmax= wl28 = 1.1 k/f ( 25 ft )
28 = 85.9 kft
Vmax= wl2 = 1.1 k/f ( 25 ft )2 = 13.75 k
Sreq= Mmax/ Fb= 85.9 kft (12 in/ft )30 k/in2
= 34.36 in3
Ireq= (5wl4)(384Eallowable)
= [ 5 (1.1 k/ft1ft/12in) (25ft12 in/ft )4][ 384 (30,000 k/in2) 1.25 in ]
= 258 in4
(dtw)req= VmaxFv= 13.75 k20 k/in2 = 0.688 in2
Requires: Sx> 34.36 in3 , I x> 258 in
4 and dtw> 0.688 in2
Possible beam size for floor from table A3
Member size Sxin3
Ixin4
din
twin
dtwin2
W8 x 67 60.4 272 9.00 0.570 5.130
W10 x 49 54.6 272 9.98 0.340 3.393
W12 x 35 45.6 285 12.50 0.300 3.750W14 x 30 42.0 291 13.84 0.270 3.737
W16 x 26 38.4 301 15.69 0.250 3.923
W18 x 35 57.6 510 17.70 0.300 5.310W21 x 44 81.6 843 20.66 0.350 7.231
W24 x 55 114.0 1350 23.57 0.395 9.310
W16 x 29 is the lightest of all possible beam sizes, so use W16 x 29 for floor beam.End reaction on typical floor beam transfers as point loads on girder below.
Pon floor girder= wl / 2 = 1085 lb/ft (25ft) / 2 = 13562.5 lb = 13.6 k
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Typical Girder Calculations
Girder: Roof girder with total load of 44.5 psfActual loading condition:
Since four concentrated loads are equally distributed on 25 feet long girder a uniformlydistributed load condition is used in calculation.
W = 25 ft ( 44.5 lb/ft 2 ) = 1112.5 lb/ft = 1.1 k/ft
Mmax= wl28 = 1.1 k/f ( 25 ft )28 = 85.9 kft
Vmax= wl2 = 1.1 k/f ( 25 ft )2 = 13.75 k
Sreq= Mmax/ Fb= 85.9 kft (12 in/ft )30 k/in2 = 34.36 in3
Ireq= (5wl4)(384Eallowable)
= [ 5 (1.1 k/ft1ft/12in) (25ft12 in/ft )4][ 384 (30,000 k/in2) 1.25 in ]
= 258 in4
(dtw)req= VmaxFv= 13.75 k20 k/in2
= 0.688 in2
Requires: Sx> 34.36 in3
, I x> 258 in4
and dtw> 0.688 in2
Possible girder size for roof from table A3
Member size Sxin3
Ixin4
din
twin
dtwin2
W8 x 67 60.4 272 9.00 0.570 5.130
W10 x 49 54.6 272 9.98 0.340 3.393
W12 x 35 45.6 285 12.50 0.300 3.750
W14 x 30 42.0 291 13.84 0.270 3.737
W16 x 26 38.4 301 15.69 0.250 3.923W18 x 35 57.6 510 17.7 0.300 5.310
W21 x 44 81.6 843 20.66 0.350 7.231W24 x 55 114.0 1350 23.57 0.395 9.310
W16 x 29 is the lightest of all possible girder sizes, so use W16 x 29 for roof girder.
Girder: Floor girder with total load of 217 psf
W = 25 ft ( 217 lb/ft 2 ) = 5425 lb/ft = 5.4 k/ft
Mmax= wl28 = 5.4 k/f ( 25 ft )28 = 422 kft
Vmax= wl2 = 5.4 k/f ( 25 ft )2 = 67.5 k
Sreq= Mmax/ Fb= 422 kft (12 in/ft )30 k/in2 = 168.8 in3
Ireq= (5wl4)(384Eallowable)
= [ 5 (5.4 k/ft1ft/12in) (25ft12 in/ft )4][ 384 (30,000 k/in2) 1.25 in ]
= 1266 in4
(dtw)req= VmaxFv= 67.5k20 k/in2
= 3.375 in2
Requires: Sx> 168.8 in3
, I x> 1266 in4
and dtw> 3.375 in2
Possible beam size from table A3
Member size Sxin3
Ixin4
din
twin
dtwin2
W12 x 152 209 1430 13.71 0.870 11.928
W14 x 120 190 1380 14.48 0.590 13.032
W18 x 97 188 1750 18.59 0.535 9.946
W21 x 83 171 1830 21.43 0.515 11.037
W24 x 76 176 2100 23.92 0.440 6.125W27 x 84 213 2850 26.71 0.460 12.287
W30 x 99 269 3990 29.65 0.520 15.418
W24 x 76 is the lightest of all possible girder sizes, so use W24 x 76 for floor girder.
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Calculation for Columns
Tributary area for columns on NW corner:A = 1/4 ( 25 ft )2 = 156.25 ft2
Sizing for upper story column on NW cornerRoof load: 44.5 lb/ft2 x 156.25 ft2 = 6953.13 lb = 7k
Requires: 12 feet tall column, allowable axial load > 7k
Possible NW corner column size for upper story from table
Column size Allowable axial loadW4 x 13 28
W5 x 16 54W6 x 9 16
W6 x 12 22W6 x 16 32
W6 x 9 is the lightest of all possible column sizes, so use W6 x 9 for upper story columnat NW corner.
Sizing for lower story column on NW cornerRoof load: 44.5 lb/ft2 x 156.25 ft2 = 6953.13 lb = 7kFloor load: 217 lb/ft
2x 156.25 ft
2= 33906.3 lb = 40k
Total load: 7k + 40k = 47k
Requires: 14 feet tall column, allowable axial load > 40k
Possible NW corner column size for upper story from tableColumn size Allowable axial load
W6 x 15 49W6 x 20 70
W6 x 25 90
W6 x 15 is the lightest of all possible column sizes, so use W6 x 15 for lower storycolumn at NE corner.
Sizing for lower story controlling column B2:Tributary area for upper story = ( 25 ft ) 2 = 625 ft2
Roof load: 44.5 lb/ft2 x 625 ft2 = 27812.5 lb = 28kTributary area for lower story = 3/4 ( 25 ft )
2+ 6ft x 25ft/2 = 543.75 ft
2
Floor load: 217 lb/ft2 x 543.75 ft2 = 117994 lb = 118kTotal load: 28k + 118k = 146k
Requires: 14 feet tall column, allowable axial load > 146k
Possible lower story mid-column on west end
Column size Allowable axial load
W8 x 31 168W8 x 35 190
W10 x 33 171
W8 x 31 is the lightest of all possible controlling column sizes, use W8 x 31 for all lowerstory columns
Sizing for upper story controlling column B2:Tributary area for upper story = ( 25 ft ) 2 = 625 ft2
Roof load: 44.5 lb/ft2
x 625 ft2
= 27812.5 lb = 28k
Requires: 12 feet tall column, allowable axial load > 28k
Possible controlling column size for upper storyColumn size Allowable axial load
W4 x 13 28W5 x 16 54
W6 x 16 32
W4 x 13 is the lightest of all possible controlling column sizes, use W4 x 13 for all upperstory columns
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Calculation for Foundation Footing
Sizing for foundation footing on NW cornerTotal axial load on NW corner = 47kUse allowable soil bearing of 3000 psf = 3k/ft
2
A = P/qs= 47k3k/ft2 = 15.7ft2
Size = (15.7 ft2
)1/2
= 3.96 ft
use 4x 4 square footing assuming 12thick
Sizing for largest foundation footing B2Total axial load on B2 = 146kUse allowable soil bearing of 3000 psf = 3k/ft2
A = P/qs= 146k3k/ft2 = 48.7ft2
Size = (48.7 ft2 )1/2 = 6.98 ft
use 7x 7 square footing assuming 12thick
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Multiframe Analysis
FRAME GRAVITY LOADSRoof girder: W = 44.5 lb/ft 2 (25ft / 2) = 556.25 lb/ft = 0.56 k/ftFloor girder: W = 217 lb/ft2 (25ft / 2) = 2712.5 lb/ft = 2.7 k/ft
Loads used in analysis:NS wind load 2.8k @ roof levelNS wind load 6.1k @ second floor levelEW wind load 1.9k @ roof levelEW wind load 4.0k @ second floor levelWroof= 0.56 k/ftWfloor= 2.7 k/ft
Initial member sizes:roof beam: W10 x 12roof girder: W16 x 29
not available in multiframe use W16 x 31
upper story column: W4 x 13floor beam: W16 x 29floor girder: W24 x 76lower story column: W8 x 31
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first trial conclusion:roof girder: W16 x 29 not available in multiframe use W16 x 31upper story column: W4 x 13floor girder: W24 x 76lower story column: W8 x 31
frame drift 3.479 > 1.56 NG
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second trial conclusion:
roof girder: W16 x 29 not available in multiframe use W16 x 31upper story column: W8 x 13floor girder: W24 x 76lower story column: W12 x 30
Frame drift 1.535 < 1.56 OKGirder deflection 0.345 < 1.25 OKCombined axial + bending stress (sbz bot + sx) = 28.292 + 5.356 = 33.648 > 30
NG
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Third trial:roof girder: W16 x 29 not available in multiframe use W16 x 31upper story column: W8 x 15floor girder: W24 x 76lower story column: W12 x 35Frame drift 1.312 < 1.56 OKGirder deflection 0.322 < 1.25 OK
Combined axial + bending stress (sbz bot + sx) col L2 = 21.958 + 1.731 = 23.689 < 30
Combined axial + bending stress (sbz bot + sx) col L1 = 25.322 + 4.57 = 29.892 < 30GOOD
Final member selection for rigid frame:roof beam: W10 x 12roof girder: W16 x 31upper story column: W8 x 15floor beam: W16 x 29floor girder: W24 x 76lower story column: W12 x 35
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Brace frame analysis
FRAME GRAVITY LOADSRoof girder: W = 44.5 lb/ft 2 (25ft / 2) = 556.25 lb/ft = 0.56 k/ftFloor girder: W = 217 lb/ft
2(25ft / 2) = 2712.5 lb/ft = 2.7 k/ft
Loads used in analysis:NS wind load 2.8k @ roof levelNS wind load 6.1k @ second floor levelEW wind load 1.9k @ roof levelEW wind load 4.0k @ second floor levelWroof= 0.56 k/ftWfloor= 2.7 k/ft
Wind truss forces:
MOT= 4050 lb x 14ft + 1867.5 lb x (12ft + 14ft ) = 105255 lb
ftT = C = M/d = 105255 lbft / 25ft = 4210.2 lb
Initial member sizes used in brace frame analysis:roof girder: W16 x 29 not available in multiframe use W16 x 31upper story column: W4 x 13floor girder: W24 x 76lower story column: W8 x 31upper story brace: 1/4rodlower story brace: 1/4rod
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Final member selection for brace frame:roof girder: W16 x 31upper story column: W8 x 15floor girder: W24 x 76lower story column: W12 x 35upper story brace: 1/2rodlower story brace: 3/8rod
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Shear wall stability:
12CMU shear wall w = 80psf
MOT= 4050 lb x 14ft + 1867.5 lb x (12ft + 14ft ) = 105255 lbft
W = 80 lb/ft2
(25ft x 26ft) = 52000 lb
MR= Wl/2 = 52000 lb (25ft / 2) = 650000 lbft
Factor of SafetyOverTurning= MR / M OT= 650000 / 105255 = 6.18 > 1.5 OK
FR= W= 52000 lb (0.35) = 18200 lb
Factor of SafetySliding = FR/ V = 18200lb / (1867.5lb + 4050lb)= 3.08 > 1.25 OKShear wall stable for overturning and sliding stability
Possible Connection Axon
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FRAME GRAVITY LOADS
Roof girder EW: W = 32psf (25ft / 2) = 400 lb/ft = .40 k/ft
Roof girder NS: W = 32psf (35ft / 2) = 560 lb/ft = .56 k/ft
Floor girder: W = 140psf (25ft / 2) = 1750 lb/ft = 1.75 k/ft
Loads used in analysis: NS wind load 9k @ roof level
NS wind load 12.1k @ second oor level