structure selection

Structure selection and design Prof Schierle 1


1 Morphology

2 Capacity Limits

3 Code Requirements

4 Cost

5 Load Conditions

6 Resources and Technology

7 Sustainability

8 Synergy

Selection criteria:


MorphologyVertical / lateral systems:

1 Shear walls are least flexible but good at apartments and hotels with party walls

2 Cantilevers provide the least intrusion at ground floor

3 Moment frames are most flexible, good for office buildings

4 Braced frames are more flexible than walls but less than moment frames Bracing is usual around central cores

A Concrete moment joint: Rebars extend through beam & column

B Steel moment joint: Beam flanges welded to column flanges Stiffener plates between column flanges resist bending stress of beam flanges

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MorphologyCorrelation of functional and structural morphologiesshould be considered in selecting structural systems

• Shear walls complement apartments and hotelsto serve also as sound barriers between units

• Moment frames complement office buildingsto allow flexible plans for changing rental needs

• Long span structures complement exhibit halls,auditoria, etc. for unobstructed view

Morphology can also be applied to elements:

• Trusses restrict duct passage

• Vierendeel girders allow duct passage


Horizontal steel systems1 Flush framing (joist top flush with beam)

Expensive joist/beam connectionsDucts below beam increase total depth

2 Layered framing (joists on top of beams)Low-cost joist/beam connections Reduced total depth assuming:Ducts between beamsFeeders between joistsLess depth cuts curtain wall and AC costs

3 S-shape joist and wide flange beams4 Moment joint5 Twin channels allow pipe passage6 Tubular beam and column7 Castillated beam (cut from T-shapes)

increases strength and stiffnessA Concrete slab on metal deckB Joist (S-shape, usual spacing ~ 10’)C Beam (wide flange, usual spacing ~ 30’)D Spandrel beamE Wide flange column (usually W14)


Rupture Lengthdefines the stress/weight ratio

CostCost is a major and often a deciding factor inselecting system and material. Cost of several options are usually compared

General rules:• Wood structures cost less than other

materials but are limited to low-rise

• Wood platform framing is more common and costs less than heavy timber framing

• Short spans cost less than long spans

• Low-rise costs less than high-rise

• Simple structures cost less than complex ones

• Wall structures cost less than moment frames and braced frames

Rupture length = length a material can hang without breaking under own weight (compression for concrete and masonry)


Capacity Limits

• Economic limit is reached before most capacity is required to support self-weight

• Short spans cost less than long spans

• Like beams, other structures have capacity limits

• Capacity limits also include minimum spansFor example:

• Cost of joints makes short trusses expensive

• Cost of fittings makes short cables expensive

• Beam depth and weight increase with span• very long beam fails under its own weight


Span / depth ratios Structure elements and systems have optimal L/d (span / depth) ratios that may be defined as

10–20–30 rule:

• L/d = 20 for beams

• L/d = 30 for slabs and decks

Chinese carpentry proportions are documented inBuilding Standards by Li Chieh, constructionsuperintendent of Emperor Hui-tsung (1101-1125);considering beauty and structure as unified theme.

• L/d = 10 for trusses and suspension cables


Horizontal elementsSpan ranges and span/depth ratios of structure elements

Structure and Design, page 600

Span Range:Recommended min.

and max. spans

Span/Depth ratio:For simple supports,

use average ratioAdjust Depth:

• Increase @ heavy loadand wide spacing

• Double @ cantilevers• Decrease @ light load

and close spacing• Decease @ elements

with overhangs

Span L

Depth

D


Horizontal systemsSpan ranges, span/depth ratios, span/thickness ratios of systems

Structure and Design, page 600

Span Range:Recommended min.

and max. spans

Span/Depth ratio:For simple supports,

use average ratioAdjust Depth:

•Increase @ heavy loadand wide spacing

• Decrease @ light loadand close spacing

Span/Thickness ratio:Use average ratioAdjust thickness:

• Decrease @ light loadand close spacing

• Increase @ heavy loadand wide spacing

Span L

Depth

D


TrussL/d = 10


Girder / beam / joist

L/d = 20


Concrete slab

L/d = 30


Code RequirementsGood design practice starts with code check:Building codes define Type of Constructionbased on fire resistanceFloor area and height limits are based on Type of Construction and Occupancy

Type of Construction examples:Type I

Type V

• Type I may be steel, concrete, or masonry with no height and floor area limitation

• Type II may be steel, concrete, or masonry with some height and floor area limitation

• Type III, IV, and V may be of any material but subject to limited height and floor area

• Codes define allowable stress for material: wood, steel, concrete, and masonry

• Codes define floor live loads based on occupancy and roof load based on location

• Codes define design methods and required loads for gravity, wind, and earthquakes


Load ConditionsLocation and occupancy define load conditions

Snow

Wind

Earthquake

• Roof live load = 20 psf in areas without snow, but may be up to 400 psf in mountains

• Floor load depends on occupancy, for example: Office LL = 50 psfLibrary stack room LL = 125 psf

• Light-weight structures are effective in areas ofearthquakes, since seismic forces are mass times acceleration (f = m a)

• Ductile steel and wood structures are effective to absorb dynamic seismic load

• Stiff concrete shear walls are effective to resist wind load but they increase seismic load

• Heavy structures are effective in areas of strong wind load, like Florida

• Thermal loads are critical in areas of great temperature variation, like Chicago


reading room = 60 psfstack room = 150 psfLibrary

light = 125 psf heavy = 250 psf Manufacturing

fixed seating = 60 psf movable seating = 100 psf Assembly

50 psfOffice

40 psf Residential and schools

ASCE 7 Table 4.1 excerpts of common live loads

ASCE 7, page 10

Live load reduction Since large members are unlikely fully loaded, ASCE 7 allows live load reductions (except for public spaces and LL ≥ 100 psf):

For members supporting ≥ 600 sq. ft.Reduction shall not exceed 50% for members supporting 1 floor, 60 % for members supporting 2 or more floors


Resources & TechnologyAvailable resources and technology areimportant for structure systems and material

Platform framing Heavy timber

Heavy steel Light gauge steel

Site cast concrete Precast concrete

Masonry Membrane

Resources:

• Wood structures require forests

• Steel structures require iron ore

• Concrete and masonry are widely available

Technology:

• Platform framing is very common in the US but unfamiliar in some other countries

• Steel structures are common in the US Concrete is common in Asia and Europe.

• Precast concrete requires nearby plant to reduce transportation cost

• Masonry: old technology, labor intensive

• Membrane structures: new technology


SustainabilitySustainability is an important to reduce future cost and negative impact on the environment

• Wood is the only renewable materialwith the lowest energy consumptionfor production

• But wood is combustible and not allowed for type I & II construction)

• Steel can be recycled but not renewed and requires more energy for production than wood

• Concrete can be recycled but requires more energy for production than wood

• Masonry can be recycled but requires more energy for production than wood


Synergy - a system, greater than the sum of its parts.

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rgy

10 boards, I = 10 (12x13/12) I = 1010 boards glued, I = 12x103/12 I = 1000Strength is defined by Section modulus S = I/c:1 board, S = 1/o.5 S = 210 boards, S = 10/0.5 S = 2010 boards, glued, S =1000/5 S = 200Note:The same amount of material is 100 times stiffer10 times stronger

when glued to engages fibers in tension and compression

Comparing wood beams of 1”x12” boards.Stiffness is defined by the moment of inertia I:1 board, I = 12x13/12 I = 1

Pragmatic example


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Olympic Dome, RomeArchitect: PiacentiniEngineer: Nervi

a stroke of genius

SynergyPrefab ribs:

• Resist buckling• Provide lighting• Enhance acoustics• Define gestalt


Conceptual design / analysis and testing methods


Schematic designGlobal moment and shear may be used to analyze elements like beam, truss, cable, archThey all resist the global moment by a horizontalcouple. The product of couple force F and its leverarm d resist the global moment:

M = F d hence F = M / d

Designation of force F varies: T (tension), C (compression), H (horizontalreaction). For simple support and uniform load:

M = w L2 / 8V = w L / 2M = max. global momentV = maximum global shearw = uniform load per unit lengthL = span

For other load or support conditions M and V are computed as for equivalent beams.


Beams resist bending by a couple, with 2/3 beamdepth d as lever arm; compression C on top andtension T on bottom.

Trusses resist the global moment by top barcompression and bottom bar tension, with trussdepth as lever arm (max chord forces @ max M)

C = T = M / dWeb bars resist shear (max force @ max shear)

Suspension cables resist the global momentby horizontal reaction H times sag f as leverarm (max cable force at supports, where H, Rand cable tension T form equilibrium vectors:

T = (H2 + R2)1/2

Arches resist global moments like cables, but incompression instead of tension:

C = (H2+R2)1/2

Assuming simply supported condition and uniformload, the max global shear occurs at supports andmax global moment at mid-span


Triangular load

Reactions RShear VMoment Mhttp://bendingmomentdiagram.com/solve/

w

R=wL/6 R=wL/3

V=wL/6

V=wL/3X=0.577736L

M=0.064152wL2


Radial pressureReferring to A:Radial pressure per unit length acting on a circular ring yields ring tension

T = R pT = ring tension R = radius of ring curvaturep = uniform radial pressure per unit length

Units must be compatible:• If p is force / ft, R must be in feet• If p is force / m, R must be in meters Reversed pressure toward the ring centerwould reverse tension to compression.ProofReferring to ring segment B:• T acts normal to radius R• p acts normal to ring segment of length 1Referring to ring segment B and polygon C:• p and T represent equilibrium at o

T / p = R / 1 (similar triangles), henceT = R p


PrestressThe effect of prestress on cable structures isshown on a wire with and without prestress, subject to a load P applied at its center. 1 Wire without prestress

resists load P in upper link onlyWire force F = P

2 Wire with prestress PSresists load P in upper and lower link.

Upper link increases: F = PS + P/2Lower link decreases: F = PS – P/2Prestress reduces deflection to half

3 Stress/strain diagramA Stress/strain without prestressB Stress/strain with prestressC Prestress reduced to zero (PS = 0)D Prestressed wire after PS = 0Note:Prestress should be half the stress underload + a reserve for thermal variation


Static test model


Static ModelStatic models may resist axial stress (truss), Bending stress (beam) or Combined stress (moment frame)Static models have three scales:Geometric Scale: Sg= Lm/Lo (length scale) Force Scale: Sf = Pm/Po (force scale)Strain Scale: Ss= m/o (strain scale)The strain scale may amplify small deflections,but should be 1:1 to avoid errors @ large strainThe following derivations assume:A = Cross-section areaE = Modulus of elasticityI = Moment of inertiam = Subscript for modelo = Subscript for original structure

Axial stress modelUnit Strain = L/LL = P L / (AE) henceForce P=A E L/L= A E henceSf = AmEm/(AoEo) SSSf = AmEm/(AoEo) If SS = 1Sf = Am/Ao = Sg

2 If = Em = Eo


Bending stress modelUnit Strain = / LStrain = kPL3 / (EI)Force P = EI / (kL3) = EI / (kL2) Force Scale Sf = Pm / Po Sf = [EmIm / (EoIo)] (ko / km ) (Lo

2 / Lm2) (m / o)

Since model and original have the same loadand supports, the constants of integration km = ko, hence ko/km may be ignored; and Lo

2 / Lm2 = 1 / SG

2

m / o = SSTherefore the force scale is:Sf = EmIm/(EoIo) (1/Sg

2) Ss

Sf = EmIm/(EoIo) (1/Sg2) If Ss = 1

Sf = Im/Io (1/Sg2) If Em = Eo

Sf = Sg2 If details are @ geometric scale

Combined axial and bending resistanceModels of both axial and bending resistance,such as moment frames, require all detailsare in geometric scale since• Cross section area increases linear with depth• Moment of Inertia increases at 3rd power

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