Structure selection and design Copyright Prof Schierle 2012 1

Structure selection and design Copyright Prof Schierle 2012 1

Selection criteria: 

1 Morphology 

2 Capacity Limits 

3 Code Requirements 

4 Cost 

5 Load Conditions 

6 Resources and Technology 

7 Sustainability 

8 Synergy 


B 

Morphology 

Vertical / 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 


Morphology 

Correlation of functional and structural morphologies 

should be considered in selecting structural systems 

• Shear walls complement apartments and hotels 

to serve also as sound barriers between units 

• Moment frames complement office buildings 

to 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 systems 

1 Flush framing (joist top flush with beam) 

Expensive joist/beam connections 

Ducts below beam increase total depth 

2 Layered framing (joists on top of beams) 

Low-cost joist/beam connections 

Reduced total depth assuming: 

Ducts between beams 

Feeders between joists 

Less depth cuts curtain wall and AC costs 

3 S-shape joist and wide flange beams 

4 Moment joint 

5 Twin channels allow pipe passage 

6 Tubular beam and column 

7 Castillated beam (cut from T-shapes) 

increases strength and stiffness 

A Concrete slab on metal deck 

B Joist (S-shape, usual spacing ~ 10’) 

C Beam (wide flange, usual spacing ~ 30’) 

D Spandrel beam 

E Wide flange column (usually W14) 


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Capacity Limits 

• Beam depth and weight increase with span 

• very long beam fails under its own weight 

• 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 spans 

For example: 

• Cost of joints makes short trusses expensive 

• Cost of fittings makes short cables expensive 


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 = 10 for trusses and suspension cables 

• L/d = 20 for beams 

• L/d = 30 for slabs and decks 

Chinese carpentry proportions are documented in 

Building Standards by Li Chieh, construction 

superintendent of Emperor Hui-tsung (1101-1125); 

considering beauty and structure as unified theme. 


Truss 

L/d = 10 


Girder / beam / joist 

L/d = 20 


Concrete slab 

L/d = 30 


Horizontal elements 

Span ranges and span / depth ratios of structure elements 

Span L 

Structure selection and design Copyright Prof Schierle 2012 11 

Depth D

Horizontal systems 

Span ranges, span/depth ratios, span/thickness ratios of systems 

Span L 


Depth D

ype of Construction examples: 

ype I 

ype V 

Code Requirements 

Good design practice starts with code check: 

Building codes define Type of Construction 

based on fire resistance 

Floor area and height limits are based on 

Type of Construction and Occupancy 

• 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 


Rupture Length 

defines the 

stress/weight ratio 

Cost 

Cost is a major and often a deciding factor in 

selecting 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) 


Earthquake 

Snow 

Wind 

Load Conditions 

Location and occupancy define load conditions 

• 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 psf 

Library stack room LL = 125 psf 

• Light-weight structures are effective in areas of 

earthquakes, 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 


Residential and schools 

Manufacturing 

ASCE 7, page 10 

ASCE 7 Table 4.1 excerpts of common live loads 

Assembly 

fixed seating = 60 psf 

movable seating = 100 psf 

light = 125 psf 

heavy = 250 psf 

reading room = 60 psf 

stack room = 150 psf 


Office 

Library 

40 psf 

50 psf 

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

Platform framing Heavy timber 

Heavy steel Light gauge steel 

Site cast concrete Precast concrete 

Masonry Membrane 

Resources & Technology 

Available resources and technology are 

important for structure systems and material 

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 


Sustainability 

Sustainability is an important 

to reduce future cost and 

negative impact on the environment 

• Wood is the only renewable material 

with the lowest energy consumption 

for 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. 

Pragmatic example 

NO Synergy 

Synergy 

Comparing wood beams of 1”x12” boards. 

Stiffness is defined by the moment of inertia I: 

1 board, I = 12x13 /12 I = 1 

10 boards, I = 10 (12x13 /12) I = 10 

10 boards glued, I = 12x103 /12 I = 1000 

Strength is defined by Section modulus S = I/c: 

1 board, S = 1/o.5 S = 2 

10 boards, S = 10/0.5 S = 20 

10 boards, glued, S =1000/5 S = 200 

Note: 

The same amount of material is 

100 times stiffer 

10 times stronger 

when glued to engages fibers in tension and compression 


Columns define plan and circulation 

Vaults define spatial character 

Design Synergy – Gothic cathedral example 


Buttresses resist vault thrust and define facade

Olympic Dome, Rome 

Architect: Piacentini 

Engineer: Nervi 

Synergy 

Prefab ribs: 

• Resist buckling 

• Provide lighting 

• Enhance acoustics 

• Define gestalt 

a stroke of genius 


Conceptual design / analysis 

and testing methods 


Schematic design 

Global moment and shear may be used to analyze 

elements like beam, truss, cable, arch 

They all resist the global moment by a horizontal 

couple. The product of couple force F and its lever 

arm d resist the global moment: 

M = F d hence 

F = M / d 

Designation of force F varies: 

T (tension), C (compression), H (horizontal 

reaction). For simple support and uniform load: 

M = w L2 / 8 

V = w L / 2 

M = max. global moment 

V = maximum global shear 

w = uniform load per unit length 

L = span 

For other load or support conditions M and V are 

computed as for equivalent beams. 


Assuming simply supported condition and uniform 

load, the max global shear occurs at supports and 

max global moment at mid-span 

Beams resist bending by a couple, with 2/3 beam 

depth d as lever arm; compression C on top and 

tension T on bottom. 

Trusses resist the global moment by top bar 

compression and bottom bar tension, with truss 

depth as lever arm (max chord forces @ max M) 

C = T = M / d 

Web bars resist shear (max force @ max shear) 

Suspension cables resist the global moment 

by horizontal reaction H times sag f as lever 

arm (max cable force at supports, where H, R 

and cable tension T form equilibrium vectors: 

T = (H 2 + R 2 ) 1/2 

Arches resist global moments like cables, but in 

compression instead of tension: 

C = (H 2 +R 2 ) 1/2 


Radial pressure 

Referring to A: 

Radial pressure per unit length acting 

on a circular ring yields ring tension 

T = R p 

T = ring tension 

R = radius of ring curvature 

p = 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 center 

would reverse tension to compression. 

Proof 

Referring to ring segment B: 

• T acts normal to radius R 

• p acts normal to ring segment of length 1 

Referring to ring segment B and polygon C: 

• p and T represent equilibrium at o 

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

T = R p 


Prestress 

The effect of prestress on cable structures is 

shown 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 only 

Wire force F = P 

2 Wire with prestress PS 

resists load P in upper and lower link. 

Upper link increases: F = PS + P/2 

Lower link decreases: F = PS – P/2 

Prestress reduces deflection � to half 

3 Stress/strain diagram 

A Stress/strain without prestress 

B Stress/strain with prestress 

C Prestress reduced to zero (PS = 0) 

D Prestressed wire after PS = 0 

Note: 

Prestress should be half the stress under 

load + a reserve for thermal variation 


Static test model 


Static Model 

Static models may resist axial stress (truss), 

Bending stress (beam) or 

Combined stress (moment frame) 

Static models have three scales: 

Geometric Scale: S g= L m/L o (length scale) 

Force Scale: S f = P m/P o (force scale) 

Strain Scale: S s= � m/� o (strain scale) 

The strain scale may amplify small deflections, 

but should be 1:1 to avoid errors @ large strain 

The following derivations assume: 

A = Cross-section area 

E = Modulus of elasticity 

I = Moment of inertia 

m = Subscript for model 

o = Subscript for original structure 

Axial stress model 

Unit Strain � = �L/L 

�L = P L / (AE) hence 

Force P=A E �L/L= A E � hence 

S f = A mE m/(A oE o) S S 

S f = A mE m/(A oE o) If S S = 1 

S f = A m/A o = S g 2 If = E m = E o 


Bending stress model 

Unit Strain � = � / L 

Strain � = kPL 3 / (EI) 

Force P = EI� / (kL 3 ) = EI / (kL 2 ) � 

Force Scale S f = P m / P o 

S f = [E mI m / (E oI o)] (k o / k m ) (L o 2 / Lm 2 ) (�m / � o) 

Since model and original have the same load 

and supports, the constants of integration 

k m = k o, hence k o/k m may be ignored; and 

L o 2 / Lm 2 = 1 / SG 2 

� m / � o = S S 

Therefore the force scale is: 

S f = E mI m/(E oI o) (1/S g 2 ) Ss 

S f = E mI m/(E oI o) (1/S g 2 ) If Ss = 1 

S f = I m/I o (1/S g 2 ) If Em = E o 

S f = S g 2 If details are @ geometric scale 

Combined axial and bending resistance 

Models of both axial and bending resistance, 

such as moment frames, require all details 

are in geometric scale since 

• Cross section area increases linear with depth 

• Moment of Inertia increases at 3 rd power 



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Structure selection and design Copyright Prof Schierle 2012 1

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