PCI 6 th Edition Preliminary Component Selection.
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Transcript of PCI 6 th Edition Preliminary Component Selection.
PCI 6th EditionPCI 6th Edition
Preliminary Component Selection
Presentation OutlinePresentation Outline
• Building optimization• Preliminary sizing• Load tables• Additional gravity loading considerations• Fire Resistance considerations• Vibration considerations• Thermal considerations
Building OptimizationBuilding Optimization
• Maximize repetitive and modular dimensions
• Use simple spans• Standardize openings• Use local component sizes• Minimize component types and sizes
Building OptimizationBuilding Optimization
• Consider tolerances in connection design• Avoid over specifying design requirements
– Allowable stresses– Allowable cambers– Allowable Deflections– Coatings on reinforcing steel– Embedded hardware– Loose Connection hardware
Building OptimizationBuilding Optimization
• Use of exterior wall panels as load bearing components and structural walls
• Maximize form use and minimize form differences / variation
• Contact a local producer as early as possible during the design development stages of a project for assistance and answering questions
Why estimate component size?Why estimate component size?
• To verify the product fits the application
• Establish floor to floor height
• Establish floor area
• To estimate project cost
Preliminary Analysis Should Include:Preliminary Analysis Should Include:
• Framing dimensions• Span-to-depth ratios• Connection concepts• Gravity and lateral load resisting system• Mechanisms for the control of volume
changes
Preliminary Span to Depth RatiosPreliminary Span to Depth Ratios
• Hollow-core– Floor slabs 30 to 40– Roof slabs 40 to 50
• Stemmed Components and Solid Slabs– Floor 25 to 35– Roof 35 to 40
• Beams 10 to 20
Load Table AssumptionsLoad Table Assumptions
• Flexural strength
• Shear strength
• Release stresses
• Stress limits under service loads
Flexural Strength ControlFlexural Strength Control
• Equivalent uniform load
• Evaluated at critical moment sections
• Load factors: 1.2D + 1.6L
• Strength reduction factor: = 0.9
Release Stress LimitsRelease Stress Limits
• Compression limit:
• Tension limit:
7fc̀
6 fc̀
Service Loading Stress LimitsService Loading Stress Limits
• Extreme fiber in compression – Prestress plus sustained loads: 0.45f’c
– Prestress plus total load: 0.6f’c
• Extreme fiber in tension – Double tees and beams: – Flat Deck Members:
12 fc̀
7.5 fc̀
Shear Strength ControlShear Strength Control
• Load Factor: 1.2D + 1.6L
• Strength Reduction Factor: = 0.75
Beam Load TablesBeam Load Tables
• Loading is uniform
• Strength design - same as double tees
• Stress limits - same as double tees
Load Table In-depthLoad Table In-depth
PCI Design Handbook, 6th Edition Page 2-16
Load Table In-depthLoad Table In-depth
• Section dimensions• Section properties• Material properties• Strand geometry• Depression points
Load TableLoad Table
• Section dimensions• Section properties• Material properties• Strand geometry• Depression points
Load TableLoad Table
• Section dimensions• Section properties• Material properties• Strand geometry• Depression points
Load TableLoad Table
• Section dimensions• Section properties• Material properties• Strand geometry• Depression points
Load TableLoad Table
• Section dimensions• Section properties• Material properties• Strand geometry• Depression points
Load Table ExampleLoad Table Example
Given:Section geometry and material properties• 10DT24 roof tee• Lightweight concrete• f’c = 5000 psi• f’ci min = 3500 psi• Design length = 58’-6”
Load Table ExampleLoad Table Example
Given:Design Loads• Superimposed Dead Load = 10 psf (roofing)• Superimposed Live Load = 30 psf (snow)Total Service Load = 40 psf (for design tables)
Problem:• Determine a suitable strand pattern from load
tables
PCI Design Handbook, 6th Edition Page 2-16
Load Table ExampleLoad Table Example
Resulting Strand PatternResulting Strand Pattern
• (8) ½” 270ksi strands
• Straight strand pattern
• 2.6” Camber at erection
• 2.2” Camber long term
• Assumptions– Initial pull = 0.75fpu
– Initial losses = 10%
– Total losses = 20%
• Always develop a final design based on specific conditions
Load TablesLoad Tables
• Limitations– Special materials
• Concrete• Strand
– Unique geometry• Pie shaped pieces• Blockouts
– Special or unique loading conditions
– Fire truck loading
Additional Loading ConsiderationsAdditional Loading Considerations
• Snow
• Drifting loads
Additional Loading ConsiderationsAdditional Loading Considerations
• Snow
• Drifting loads
• Corridor loads
• Walkways
Additional Loading ConsiderationsAdditional Loading Considerations
• Snow
• Drifting loads
• Corridor loads
• Walkways
• Impact
• Combination of load
Additional Loading ConsiderationsAdditional Loading Considerations
• Snow• Drifting loads• Corridor loads• Impact• Combination
of load• Beware of piling snow
– Add Snow gate/chute
Fire Resistance ConsiderationsFire Resistance Considerations
• Time Ratings• Based on
– Square footage– Building type– Cover
Requirements
Fire Resistance ConsiderationsFire Resistance Considerations
• Three Methods to Determine Fire Resistance Rating– Testing (§703) – ASTM E
119– Prescriptive (§720)– Calculated (§721)
Office Building ExampleOffice Building Example
Given:– Exterior bearing wall system– Floor system
Assumptions– Unlimited area potential– Using prescriptive methods
Problem:– Determine required wall and floor resistance
requirements and reinforcing cover
Solution StepsSolution Steps
Step 1 - Determine group classification
Step 2 - Determine construction type based on building area and available footprint
Step 3 - Determine component resistance requirements
Step 4 - Determine reinforcing cover requirements
Step 1 – Group Occupancy ClassificationStep 1 – Group Occupancy Classification
• Section 303– IBC 2003
• Group Classification– Group A – Assembly– Group B – Business– Group E – Educational– Group F – Factory– Group H – High Hazard– Group I – Institutional– Group M – Mercantile– Group R – Residential– Group S – Storage– Group S-2 – Parking Garage– Group U – Other / Utility
Step 2 – Determine Construction TypeStep 2 – Determine Construction Type
• Table 503 IBC 2003– Allowable Height and
Building Areas
• Based on– Building Group– Building Size
• Group – B• Unlimited Footprint• Construction Type
Type I - B
Step 3 – Wall Resistance RequirementsStep 3 – Wall Resistance Requirements
• Table 602– IBC 2003
• Function of Building Element and Construction Type
• Example – – Exterior Bearing Wall– Type I B 2 hour
• Table 602– IBC 2003
• Function of Building Element and Construction Type
• Example – – Floor
Construction– Type I B
2 hour
2 hour
Step 3 – Wall Resistance RequirementsStep 3 – Wall Resistance Requirements
Step 4 – Thickness of Insulating Material - WallStep 4 – Thickness of Insulating Material - Wall
• Table 720.1– IBC 2003
Step 4 – Thickness of Insulating Material - FloorStep 4 – Thickness of Insulating Material - Floor
• Table 720.1– IBC 2003
Example ConclusionExample Conclusion
• Office– Unlimited Area– Maximum 11 Stories / 160 ft
• Type IB Construction• Exterior Bearing Wall
– 2 hours– 1 ½” Cover
• Floor System– 2 hours– 2 ½” Cover
Code Endurance Table ExampleCode Endurance Table Example
Given:The following Double Tee (page 9-49)Assumptions– Strands are ½” diameter– Siliceous aggregate– Normal weight concrete– Topped System – Restrained
Problem:For a 2 hr rating determine– The strand cover required– Floor Thickness required
Solution StepsSolution Steps
Step 1 – Determine effective cross sectional area and associated cover
requirements
Step 2 – Compare to cover provided
Step 3 – Determine heat transfer requirements and compare to provided conditions
Step 1 – Effective Area and Required CoverStep 1 – Effective Area and Required Cover
• Table 9.3.7.1(5) (Page 49)• Average Stem Width
(3.75 + 5.75)/2 = 4.75 in• Effective Flange width = 3 x Avg Stem
3(4.75) = 14.25 in.• Aeff = 22(4.75)+14.25(5)=175.75 in2
Step 2 – Supplied CoverStep 2 – Supplied Cover
• Side cover provided [3.75 + (3.5/22)(5.75–3.75) – 0.5]/2 = 1.78 in
• Bottom cover provided2 – 0.5/2 = 1.75 in
• Both exceed 1 ½ in OK
Step 3 – Heat TransferStep 3 – Heat Transfer
• Per Figure 9.3.6.15 in Minimum
• Provided 2 in tee flangeand3 in topping= 5 in total
Vibration ConsiderationsVibration Considerations
• Causes– Machinery– Exercise– Cars– Walking– Impact
Minimize Vibration or AffectsMinimize Vibration or Affects
The goalDecrease the amplitude of the vibration
OR
Decrease the Systems Natural Frequency
• Decrease Span• Increase Mass
Vibration SolutionVibration Solution
• Based on minimum natural frequencies
• Calculations are approximate as estimation of damping and human response is varied
Types of Analysis MethodsTypes of Analysis Methods
• Based on excitation– Walking– Rhythmic Activities– Mechanical Equipment
Natural FrequencyNatural Frequency
• Vibration limits are a function of Natural Frequency, fn
Where
g – Acceleration due to gravity
– Displacement of system
fn∝
g
⎛
⎝⎜
⎞
⎠⎟
Minimum fnMinimum fn
• Floors with natural frequencies lower than 3 Hertz are not recommended
• People may more readily synchronize their actions at lower frequencies
DampingDamping
• Handbook references are based on modal damping
• Highly dependent on the non-structural items– Partitions– Ceilings– Furniture
HarmonicsHarmonics
• Higher modes of vibration
• Deal with – Rhythmic Activities – Oscillating Equipment
Office DT ExampleOffice DT Example
Given:– 10DT32+2
• Page 2-18
– Open office area– 60-ft span
Problem:– Check for vibration
caused by walking.
SolutionSolution
• Find the systems minimum Natural Frequency, fn and compare it to the Fundamental Frequency, f
f ≥ fn
Solution StepsSolution Steps
Step 1 – Establish natural frequency equation
Step 2 – Determine natural frequency parameters
Step 3 – Calculate the effective weight
Step 4 – Calculate natural frequency
Step 5 – Determine expected system frequency
Step 1 – Establish Natural FrequencyStep 1 – Establish Natural Frequency
• Systems Natural Frequency - Empirical formula based on walking
• Represents smallest natural frequency to prevent disturbance
• Equation 9.7.6.1
Where:
W – Effective Weight
K, – factor representing structure type and damping
fn≥2.86 ln
K
W
⎛
⎝⎜⎞
⎠⎟⎡
⎣⎢⎢
⎤
⎦⎥⎥
Units of 2.86 are 1/sec
Step 2 – Natural Frequency ParametersStep 2 – Natural Frequency Parameters
• K and – Table 9.7.6.1 (pg 9-69)
Footnotes: a. For floors with few non-structural components and furnishings, open work area, and churches b. For floors with non-structural components and furnishings, cubicles c. For floors with full-height partitions
Step 3 – Calculate Effective WeightStep 3 – Calculate Effective Weight
• Effective Weight (W) W = w · (Effective Area)
w = supported Load
• Effective Area = [60%(L)] · L = 0.6L2
W=w·0.6l2
Step 3 – Calculate Effective WeightStep 3 – Calculate Effective Weight
• Supported Load– Superimposed Load - Assume 10 psf– Dead Load - Double Tee + Topping
• Effective Weight
W=w⋅0.6l2 =99psf ⋅0.6 ⋅ 60ft( )
2=214kips
w =64psf +2in
12 inft
⋅150pcf +10psf =99psf
Step 4 – Calculate Natural FrequencyStep 4 – Calculate Natural Frequency
• Recall
fn≥2.86 1
secln
13kips
0.2 ⋅214kipskips
⎛
⎝⎜⎞
⎠⎟⎡
⎣⎢⎢
⎤
⎦⎥⎥
=3.18hz
fn≥2.86 ln
K
W
⎛
⎝⎜⎞
⎠⎟⎡
⎣⎢⎢
⎤
⎦⎥⎥
Step 5 – Expected Fundamental FrequencyStep 5 – Expected Fundamental Frequency
• Fundamental Frequency of selected floor unitsFigure 9.7.4.1 (Page 9-68)
3.8
Example ConclusionExample Conclusion
• Calculated Minimum Natural Frequencyfn = 3.18hz
• Expected Fundamental Frequencyf = 3.8hz
f ≥ fn - Therefore OK
Thermal ConsiderationsThermal Considerations
• Thermal Calculations– Thermal Resistance: R -Values– Heat Transmittance: U - Values
• Thermal Lag or Storage –– Significant benefit of Precast Concrete
Construction
• Moisture Control• Thermal Bridging
Thermal CalculationsThermal Calculations
• Thermal Resistance: R -Values– Example Table 9.1.4.1– Page 9-7
Thermal CalculationsThermal Calculations
• Heat Transmittance:– U - Values
U =
1R∑
Thermal Lag or StorageThermal Lag or Storage
• Thermal Lag or Storage –– Significant benefit of
Precast Concrete Construction
Moisture Control DiscussionMoisture Control Discussion
• Air Barriers
• Vapor Retarder
Air BarriersAir Barriers
• Continuity throughout the building envelope
• Ability to support a differential air pressure
• Must be virtually air impermeable• Must be durable
Vapor Retarder Vapor Retarder
• Low permeability materials
• Stop or retard the passage of moisture
Typical Wythe Connection MethodsTypical Wythe Connection Methods
• Wythe Ties
• Solid Zones
PurposePurpose
• To create larger section properties for structural considerations
• Maintain structural integrity between the multiple wythes of concrete
Thermal BridgingThermal Bridging
• Direct Link / Bridge through a primary insulating material connecting two or more non-insulating materials
Thermal BridgingThermal Bridging
• Effect of these Bridges– Cold Spots– Condensation Buildup
• Calculating R and U values– Zonal Method – Metal Ties– Characteristic Section Method – Solid Zones
Thermal Bridging CalculationsThermal Bridging Calculations
• Calculating R and U values– Zonal Method – Metal Ties– Characteristic Section Method – Solid
Zones
Preliminary Design BenefitsPreliminary Design Benefits
• Accurate early planning for smooth running schedule and cost control
• Accurate Material Requirements
• Accurate Section Geometry
• Precast design team can provide benefit to the entire project design team
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Questions?Questions?