architectural structural course notes

5/26/2018 architectural structural course notes

1/242

Elastic Buckling

Illustrations: Daniel L. Schodek:Structures,

fifth edition; Pearson Prentice-Hall, 2004

Schodek fig. 7.1



Schodek fig. 7.1

Metastable


2/242

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


3/242



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


4/242

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


5/242



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.


6/242

Columns:Transfer Girder

TransferGrider

Column V

BCE Place, Toronto, Ontario, Canada

Santiago Calatrava


7/242

Bath Hou


8/242

Hurva Synagogue, Jerusalem, Israel (unbuilt project) Louis Kahn Dulles

Illustration: Understanding Structures, Fuller Moore, WCB/ McGraw-Hill, 1999


9/242

Stuttgart Airport, Germany Von

Gerkan Maarg

Rose Center for Earth & Space, NYC

Polsheck & Partners


10/242

Stanstead Airport, Essex, England

Norman Foster

Stanstead Airport, Essex, England Sir Norman Foster

Student Model by James Fickes, Philadelphia University


11/242

Hong Kong & Shanghai Bank, Hong Kong, China Norman Foster

Illustration: Understanding Structures, Fuller Moore, WCB/ McGraw-Hill, 1999


12/242

Alamillo Bridge, Near Barcelona, Spain Santiago Calatrava

Library, Phillips Exeter Academy, Exeter, NU

Louis I Kahn


13/242

Sendai Mediatheque, Sendai, Japan

Toyo Ito

Tokyo International Forum, Tokyo, Japan

Rafael Violy


14/242


15/242

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


16/242

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


17/242

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


18/242

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)


19/242

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


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.


20/242

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


21/242


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


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


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


22/242

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


23/242

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


24/242

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


25/242

Continuous Slab on Beams (sim. to cont. beam on cols.)

Continuous BeamMoment & Shear

Coefficients


26/242

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


27/242

Formulas for Flexure, Shear and Deflection Formulas for Flexure, Shear and Deflection

Formulas for Flexure, Shear and Deflection


28/242

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


29/242

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


30/242

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


31/242

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


32/242

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


33/242

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


34/242

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


35/242

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


36/242

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.


37/242

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


38/242

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)


39/242

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


40/242

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


41/242

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)


42/242

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:


43/242

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


44/242

Lateral Force Transfer

Francis D.K. Ching, Building Construction Illustrated

Vertical SupportStructural Patterns

Daniel Schodek, Structures


45/242

Placement of Lateral Force Resisting Elements In Plan




46/242

Concrete Shear Wall

Ligh t Wood-Framed Shear Wall


47/242


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.


48/242

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.


49/242

Steel Rigid Frame


50/242


51/242


52/242

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


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


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


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!


53/242

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


54/242

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


55/242

2-D Floor Isolation System

3-D Floor Isolation System

Cosine Cur


56/242

Los Angele

(Photos courtesy Brad Aagaard, USGS Postdoctoral Scholar at

Los Angeles City Hall Retrofit Los Ange


57/242

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


58/242

Seismic Loads

Origins

Plate Structure of Earths Crust

Plate Boundary


59/242

Pangaea

ca. 500,000,000 B.C.

Gap

Overlap

Edge ofContinentalShelf


60/242

San AndreasFault From Air


61/242

P (Primary) Waves (Mostly Direct, Push-Pull in nature)

S (Secondary) Waves (Mostly Reflected, Side-to-side in nature)

S

S


62/242

3 Types ofSeismic Wave

Actions

Types of

http://www.analog.com/library/ana

P-waves

Rayleigh-waves

Terraforming:

Effects of Earthquakeson the Landscape


63/242

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


64/242

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


65/242

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


66/242


67/242


68/242

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


69/242

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


70/242

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


71/242

Shear and Moment Diagrams

file:///C|/Users/1230875789C/Desktop/ARCH 412Technology III/Shear_and_Moment_Diagrams.htm[7/7/2010 11:34:34 AM]

This presentation contains content that your browser may not be able to show properly. This presentation was opt

for more recent versions of Microsoft Internet Explorer.

If you would like to proceed anyway, click here.
http://c%7C/Users/1230875789C/Desktop/ARCH%20412%E2%80%94Technology%20III/Shear_and_Moment_Diagrams_files/frame.htmhttp://c%7C/Users/1230875789C/Desktop/ARCH%20412%E2%80%94Technology%20III/Shear_and_Moment_Diagrams_files/frame.htm


72/242

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


73/242

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


74/242


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!


75/242

Image University of California, Berkeley, Godden Slide Library: http://nisee.berkeley.edu/jpg/6257_3021_0122/IMG0033.jpg

Roller & Rocker-Type Connections



76/242



Rome International Airport


77/242


Rome International AirportOverhanging Beam Roller Joint


Pinned-Type Connections


78/242


Renzo PianoPA Technology Centre, Princeton, NJ


79/242

Image Deborah J. Oakley

Santiago CalatravaMilwaukee Art Museum Addition

Image Deborah J. Oakley

Santiago CalatravaMilwaukee Art Museum Addition


80/242

Various Knife-Plate Connections Cutler Anderson Architects


81/242

Gates Residence, Bellvue, Washington Bohlin Cywinski Jackson, Architects


82/242


83/242

Grace Episcopal Church, Bainbridge Island, WA Cutler Anderson Architects


84/242

Capitol Hill Library, Seattle, WA Cutler Anderson Architects


85/242

Truss Pedestrian Bridge


86/242

Joint Point of Entry,Orovile, WA & Osoyoos, BC

Cutler Anderson Architects


87/242


88/242

Continental Train Platform, Waterloo Station, English Channel TunnelNicholas Grimshaw & Parners


89/242

Traversina Footbrige, Viamala Gorge, Switzerland

Conzett Bronzini Gartmann, Engineer


90/242

Original Mt. PalomarObservatory Telescope MirrorTension Hanger Connection

Corning Museum of Glass,Corning, NY

Smith-Miller & Hawkinson,Architects


91/242

Framed Beam Connection:Steel to Concrete Using

Embedded Plate


92/242

Thorncrown Chapel, Arkansas, Fay Jones Architect


93/242

Philip Merrill Environmental Center, Chesapeake Bay Foundation,Annapolis, Maryland Smith Group, Architects


94/242

Broadgate Office Building, London, Skidmore, Owings & Merril


95/242

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


96/242

Hartford Civic Center Roof Failure


97/242

e


98/242

Design of Bolted Connections

Shear Connections in Tension

Hanger rods

Cable anchors

Diagonal braces


99/242

Riveted Connections Traditional Steel Construction

Connection Method


100/242

Rivet Installation

Red Hot Rivet

Clamping ForceDeveloped asRivet Cools andShrinks


101/242

Tension Hanger Connection




102/242

Bolted Connection Failure Modes


1) Shear Failure of Bolt


103/242

2) Bearing Failureof Connected Material

3) Tensile Failure of Connected Material


104/242

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.


105/242

FIX

Illustrations: Barry Onouye and Kevin Kane:Statics and Strength of Materials forArchitecture and Building Construction, second edition; Prentice-Hall, 2001


106/242

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



107/242

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


108/242

Illustrations: Barry Onouye and Kevin Kane:

Statics and Strength of Materials for Architectureand Building Construction, second edition;Prentice-Hall, 2001


109/242

Flanges must be connected

for fixed joint to occur!


110/242

Illustrations: Francis D.K. Ching: Building Construction Illustrated, third edition;John Wiley & Sons, Inc., 2001

Stockley Park, Middlesex, UK

Norman Foster and Partners


111/242

Stockley Park, Middlesex, UK

Norman Foster and Partners




112/242


113/242

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


114/242

Discovery Museum, Bridgeport, Connecticut Chan Krieger Sieniewicz Architects


115/242

Michaels / Sisson Residence, Mercer Island, Washington, 1998; Miller | Hull Partnership


116/242


117/242


118/242


119/242


120/242

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


121/242

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


122/242


123/242


124/242


125/242

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?


If some reinforcing steel in a beam is good,

then more must be better, right?

Whats WRONG about this?


How long does concrete taketo reach full strength?


When concrete is being mixed and it begins toharden, why is adding water NOT a good idea?


What is a slump test and what critical propertydoes it tell about a concrete mix?


Identify the three primarycomponents of concrete.


What is air entrained concrete and why is itused? What types of construction would it mos

likely employ it?


126/242

Tech III oncept ard


Concrete does not dry to become hard. What

is the curing process called and what is oneimportant characteristic of it?


What would potentially happen

if the coefficient of thermal expansion

for concrete and reinforcing steel

were not approximately the same?


Where should primary tension steel be placedon a continuous span flexural member?


Where is shear reinforcement normally locatedin a flexural member?


What is the purpose of the bumps(deformations) on typical

reinforcing steel rods?


What is meant by development lengthof reinforcing bars?


Why are reinforcing bars sometimesbent into hooks?


What is the purpose of bar supportsfor reinforcing steel?


127/242

Tech III oncept ard


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?


What characteristics distinguish coarse-grainedsoils from fine-grained soils?


At a minimum, where should soil borings betaken when a new building is being planned?


What type of soil must you NEVER build on?


What is the purpose of dewatering?


What is the function of a caisson bell?


What are two principle differences betweenpiles and caissons?


128/242

Tech III oncept ard

Subject: Soils and Foundations

What is a grade beam and what is its function?


What types of information are

provided on a boring log?


What are two potential problems when buildingon clay soils?


Why must exterior footings be placedbelow the frost line?


What is a mat foundation?


A given volume of soil can be thought of ingeneral as consisting of what three constituents


Pile foundations often cannot be used in denseurban areas for what reason?


What information does the Standard PenetratioTest (SPT) provide?


129/242

Tech III oncept ard

Subject: Soils and Foundations

In what primary way do shallow foundations

differ from those which are called deep?


Describe the function of a pile cap.

Tech III oncept ardSubject: Lateral Forces

What are the three primary vertical lateral forceresisting structure types?


What is the primary difference betweenthe way in which wind and seismic forces

are generated in a building?


Identify two different sources of lateral forces,

aside from wind and seismic loads.


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)?


Under a wind load, which faces on a typicalbuilding structure have pressure forces

and which are suction forces?


What is the function of a horizontalfloor or roof diaphragm?


130/242


Explain the concept of the beam analogy with

regard to lateral forces.


Conceptually, what is the difference

between the normal force method andprojected area method of computing

wind pressures on a building?


Why are asymmetric building plans

or elevations undesirable in locationssubject to high seismic forces?


What is the purpose of base isolation?


What is one way of dealing with an

irregular building plan that will addressconcerns for high seismic loads?


How does a flexible floor or roof diaphragm

differ from one that is rigid?


At a minimum, how many lateral force-resistingwalls or frames are required to resist

torsional forces on a diaphragm?

How must they be arranged?


What is a soft story and why is it a problem?


131/242


With respect to lateral forces, what is meant

by the concept of continuity in a building

structural system?


What materials can be used to

construct shear walls?


How may a flexible diaphragm be made into arigid one? (Hint: This is frequently done in

conventional construction practice)


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?


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?


What is the structural definition of moment?


What is the definition of slenderness ratio?What is the implication in column design?


132/242


What are the three fundamental support types?

What forces do they each resist?


What are the two fundamental

equations of equilibrium?


What is the fundamental definition of stress?


Describe three ways in which momentcouples can be found in buildingstructures at varying scales.


What are the limits to the applicability of a two-

way structure versus a one-way structure?


What is an indeterminate structural member

and how does it differ from one that isstatically determinate?


Why are there different load factors in concrete

design for dead loads versus live loads? Why is

the live load factor higher?


The building code prescribes a minimum

amount of reinforcing (0.33%, for grade 60

regbar). What purpose does this minimum steeserve?


133/242


134/242


135/242


136/242


137/242


138/242


139/242

UNIVERSITY of MARYLAND

School of Architecture, Planning and PreservationARCH 412 | Tech IIIProfessor Deborah Oakley


140/242


141/242


142/242

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

=


143/242


144/242

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


145/242

First FLoor Plan


146/242

Second FLoor Plan


147/242

South Elevation

North Elevation


148/242

West Elevation

East Elevation


149/242

Floor Plan - Structural


150/242

Roof Plan - Structure


151/242

Section E-W

Section N-S


152/242

DetailRo

of/WallSection


153/242

Framing Plan Axonometric


154/242

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


155/242

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



= 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


156/242

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



= 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



= 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.


157/242

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


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


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


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


158/242

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


159/242

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


160/242

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


161/242


162/242

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


163/242


164/242

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


165/242

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


166/242


167/242

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


168/242


169/242

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


170/242


171/242


172/242


173/242


174/242


175/242


176/242


177/242


178/242


179/242


180/242


181/242


182/242


183/242


184/242


185/242


186/242


187/242


188/242


189/242


190/242


191/242


192/242


193/242


194/242


195/242


196/242


197/242


198/242


199/242


200/242


201/242


202/242


203/242


204/242


205/242


206/242


207/242


208/242


209/242


210/242

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

Documents

architectural structural course notes