March 6, 2017 | Author: Khaled Abdelbaki | Category: N/A
The 10 Steps in Design of Post-Tensioned Floors
Dr Bijan O Aalami Professor Emeritus, San Francisco State University Principal, ADAPT Corporation;
[email protected]
www.adaptsoft.com
301 Mission Street San Francisco, California High Seismic Force Region
Column supported multistory building Four Seasons Hotel; Florida High Wind Force Region
Two-way flat slab construction
Post-Tensioning Systems
12.7 mm (0.5"*)
Unbonded System
WIRE
CORROSION INHIBITING COATING
PLASTIC SHEATHING
NOTE: * NOMINAL DIAMETER
(a) STRAND
Multi-level parking structures One-way beam and slab design
(b) TENDON
GREASE FILLED PLASTIC CAP SHEATH
TUBE
STRAND
(c) ANCHORAGE ASSEMBLY
Example of a Floor System using the Unbonded Post-tensioning System
Post-Tensioning Systems Grouted System
An example of a grouted system hardware with flat duct
Example of a Floor System Reinforced with Grouted Post-Tensioning System
Preliminary Considerations Design of Post-Tensioned Floors
Dimensions (sizing)
¾ Optimum spans; optimum thickness
Structural system
¾ One-way/two-way; slab band
Boundary conditions; connections
¾ Service performance; strength condition
Load Path; Design strips Design sections; design values
Preliminary Considerations Design of Post-Tensioned Floors
Dimensions (sizing)
¾ Optimum spans; optimum thickness
Preliminary Considerations Design of Post-Tensioned Floors Selection of load path for two-way systems – Design Strips
An optimum design is one in which the reinforcement determined for “service condition” is used in its entirety for “strength condition.”
PT amount in service condition is
governed mostly by: ¾ Hypothetical tensile stresses (USA), or crack width (EC2) ¾ Tendon spacing (USA)
Common spans: 25 – 30 ft (8 – 9 m) Span/thickness ratios
¾ 40 - 45 for interior ¾ 35 for exterior with no overhang
Subdivide the structure into design strips in two orthogonal directions (Nahid slab)
Preliminary Considerations Design of Post-Tensioned Floors Subdivide the floor along support lines in design strips
Preliminary Considerations Design of Post-Tensioned Floors Subdivide slab along support lines in design strips in the orthogonal direction 2
1 2
1
3
4
3
4
5
5
A A
B B
C C
D D
E F
E
Y X
F Y X
G
An important aspect of load path selection in a two-way system is that every point of the slab should be assigned to a specific design strip. No portion of the slab should be left unassigned.
Preliminary Considerations Design of Post-Tensioned Floors
Design sections
¾ Design sections extend over the entire
design strip and are considered at critical locations, such as face of support and mid-span
Design values
¾ Actions, such as moments at each design section are reduced to a “single” representative value to be used for design
559 is the area (total) value of bending moment at face of support
10 Steps Design of Post Tensioned Floors
Step 1 Geometry and Structural System
Select design strip and Idealize 1. Geometry and Structural System 2. Material Properties 3. Loads 4. Design Parameters 5. Actions due to Dead and Live Loads 6. Post-Tensioning 7. Code Check for Serviceability 8. Code Check for Strength 9. Check for Transfer of Prestressing 10. Detailing
Step 1 Geometry and Structural System
Select design strip and Idealize
¾ Extract; straighten the support line;
square the boundary ¾ Model the slab frame with a row of supports above and below. This represents an upper level of multi-story concrete frame. Assume rotational fixity at the far ends; Assume roller support at the far ends
¾ Extract; straighten the support line; square the boundary
Step 2 Material Properties
Concrete
¾ Weight 24 kN/m3 ¾ 28 day cylinder 40 MPa ¾ Elastic modulus 35,220 MPa ¾ Long-term deflection factor 2
Non-Prestressed reinforcement ¾ fy ¾ Elastic modulus
460 MPa 200,000 MPa
Prestressing
¾ Strand diameter ¾ Strand area ¾ Ultimate strength ¾ Effective stress ¾ Elastic modulus
View of idealized slab-frame
13 mm 99 mm2 1,860 MPa 1,200 MPa 193,000 MPa
Step 3 Loads
Step 4 Design Parameters
Selfweight
Applicable code
¾ Based on member volume
¾ ACI 318-11; EC2 EN 1992-1-1:2004 ¾ IBC 2012 ¾ Local codes, such as California Building
Superimposed dead load ¾ Min (partitions)
1 kN/m2
Code (CBC 2011); or otherwise
Live load
¾ Residential ¾ Office ¾ Shopping mall ¾ Parking structure
Cover for protection against corrosion
2.0 kN/m2 2.5 kN/m2 3.5 kN/m2 2.0 kN/m2
Cover to rebar
¾ Not exposed to weather ¾ Exposed to weather
Lateral loads ¾ Wind ¾ Earthquake
Cover to tendon
¾ Not exposed to weather 20 mm ¾ Exposed to weather 25 mm
Example assumes ¾ Superimposed DL ¾ Live load
SDL= 2 kN/m2 LL = 3 kN/m2
Step 4 Design Parameters
Step 4 Design Parameters
Cover for fire resistivity
Cover for fire resistivity
Identify “restrained” and “unrestrained panels.”
Restrained or Unrestrained
Cover Thickness, mm. for Fire Endurance of
Aggregate Type 1 hr
1.5 hr
2 hr
3 hr
4 hr
Unrestrained
Carbonate Siliceous Lightweight
-
-
40 40 40
50 50 50
-
Restrained
Carbonate Siliceous Lightweight
-
-
20 20 20
25 25 25
30 30 30
For 2-hour fire resistivity ¾ Restrained ¾ Unrestrained
20 mm 50 mm
20 mm. 40 mm
Identify “restrained” and “unrestrained panels.”
Step 4 Design Parameters
Allowable Stresses or Crack Width Selection for Two-Way System Based on ACI
¾ Total load case
Tension 0.5√f’c 0.60f’c Compression ¾ Sustained load case Tension 0.5√f’c Compression 0.60 f’c
Based on EC2
¾ Frequent load case Tension Ft = 0.30 fck (2/3) Compression 0.60fck ¾ Quasi permanent load case Tension Ft = 0.30 fck (2/3) Compression 0.45fck ¾ Crack width normal exposure (?) Unbonded Bonded
Step 5 Actions due to Dead and Live Loads
Analyze the design strip as a single
level frame structure with one row of supports above and below, using
In-house simple frame program
(Simple Frame Method; SFM); or in-house Equivalent Frame Program (EFM); Specialty commercial software
All the three options yield safe designs.
But, each will give a different amount of reinforcement. The EFM is suggested by ACI-318. To some extent, it accounts for biaxial action of the prototype structure in the frame model. Accuracy and ability of commercial software for optimization varies
Step 4 Design Parameters
Allowable deflections (EC2/ approx. ACI)
For
visual impact use total deflection ¾ Span/250 ¾ Use camber, if necessary
Total deflection subsequent to installation of members that are likely to be damaged ¾ Span/350
Immediate deflection due to live load ¾Span/500 Long-term deflection magnifier 2. This brings the total long-term deflection to 3,
Step 5 Actions due to Dead and Live Loads
Analyze the design strip as a single level frame structure with one row of supports above and below.
Step 6 Post-Tensioning
Selection of design parameters Selection of PT force and profile
Step 6 Post-Tensioning Selection of PT force and profile
¾ Two entry value assumptions must be made to initiate the computations. Select precompression and % of DL to balance
Effective force/tendon selection option - force selection
Calculation of balanced loads;
adjustments for percentage of load balanced
Calculation of actions due to balanced loads
Step 6 Post-Tensioning
Selection of design parameters
¾ Select average precompression 1 MPa ¾ Target to balance 60% of DL
Selection of PT force and profile
¾ Assume simple parabola mapped within the bounds of top and bottom covers Force diagram of simple parabola
Step 6 Post-Tensioning ¾ Assume simple parabola for hand calculation
STEP 6 Post-Tensioning
STEP 6 Post-Tensioning Calculation of balanced loads;
Calculation of balanced loads; adjustment of % of DL balanced
adjustment of % of DL balanced F121_ACI_PT_2_way_082012
Assume P/A =150psi[1MPa]
F121_ACI_2-way_PT_force_082012
1
Select critical span
Select max drape using tendons from critical span
2
Select max drape
Calculate %of DL balanced (%DL)
Calculate %of DL balanced (%DL) Yes
No
%DL < 50%?
No
P /A 80%?
Yes %DL > 80%?
No
Yes
Yes
Increase P/A P/A>125psi [0.8MPa]? Yes
No Reduce drape
Is it practical to reduce P/A or tendons?
Reduce P/A
No
Rais e tendon to reach %DL ~ 60%
Go to next span
STEP 6 Post-Tensioning ¾ Lateral forced from continuous tendons ¾ Lateral force from terminated tendons ¾ Moments from change in centroid of member
Exit after last span
STEP 6 Post-Tensioning Calculation of balanced loads ¾ Lateral forced from continuous tendons ¾ Lateral force from terminated tendons ¾ Moments from change in centroid of member
Reduce P/A or tendons to %DL balanced ~ 60% ; P/A >= 125 psi [0.8 MPa]
Move to next span
Member with widely different spans
Calculation of balanced loads
Yes
Force from terminated tendon
Example of force from continuous tendon
P = 500 k a = 93 mm ; b = 186 mm ; L = 9 m ; c = {[93/186]0.5/[1 + (93/186)0.5]} * 9.00 = 3.73 m Wb/tendon = 2 P*a/c2 = 119.0 kN * (2*93/1000)/3.732 = 119.0 kN / tendon * 0.013 / m =1.59 kN/m / tendon
L = 10 m ; a = 93 mm ; P = 119 kN; c =0.20*10 = 2.00 m Wb = (3 * 119.0 * 2 * 93 / 1000) / 2.02 16.60 kN/m ↓ Concentrated force at dead end = 2*16.60 = 33.20 k ↑
STEP 6 Post-Tensioning Calculation of balanced loads ¾ Lateral forced from continuous tendons ¾ Lateral force from terminated tendons ¾ Moments from change in centroid of member
STEP 6 Post-Tensioning Calculation of actions due to balanced loads ¾ Check balanced loads for static equilibrium ¾ ¾
Determine moments/shears from balanced loads applied the frame used for dead and live loads Note down reactions from balanced loads
Example of force from change in member centroid
Moment at face of drop = M M = P * shift in centroid =P * (Yt-Left – Yt-Right) P = 23*119 kN; Yt-Left = 120 mm ; Yt-Right = 146 mm M= 23*119(120 – 146)/1000 = -71.16 kNm
STEP 6 Post-Tensioning Calculation of actions due to balanced loads ¾ Obtain moments at face-of-supports and mid-spans ¾
Note the reactions. The reactions are hyperstatic actions.
Comments: Moments and precompression will be used for serviceability check. Reactions will be used for Strength check.
STEP 7 Code Check for Serviceability Code requirements for serviceability ¾ ¾ ¾ ¾
Load combinations Stress/crack width check Minimum reinforcement Deflection check.
Load combination ¾ Frequent (Total) load condition 1.00DL + 0.50LL + 1.00PT ¾
Quasi Permanent (Sustained) load condition 1.00DL + 0.30LL + 1.00PT
Stress check
Using engineering judgment, select the locations th likely to be critical. Typically, these are at the face of and for hand calculation at mid-span At each section selected for check, use the design a applicable to the entire design section and apply tho the entire cross-section of the design section to arri the hypothetical stresses used in code check. σ = (MD + 0.5ML + MPT) / S + P/A S = I/Yc ; I = second moment of area of ; Yc = distance to farthest tension fiber
STEP 7 Code Check for Serviceability
STEP 7 Code Check for Serviceability
ACI Minimum Reinforcement
ACI 318-11 Minimum Reinforcement F114_041112
ACI Minimum Rebar for two-way systems `
1
¾ Rebar over support is function of geometry of the design strip and the strip in the orthogonal direction ¾ Rebar in span is a function of the magnitude of the hypothetical tensile stress
PT system? 2
9 Unbonded
At supports As = 0.0075Acf
3
Bonded
Calculate the cracking moment Mcr at supports and spans
`
In span calculate hypothetical tension stress ft
4
Does Mcr exceeed 1.2xmoment capacity?
10
11
No 5
ft ? tension stress
12 No added rebar required
6 7
8
Yes
ft > 2 root 'c [ft > 0.17 root f'c ]
ft =< 2 root 'c [ ft =< 0.17 root f'c ]
Add rebar to resist force in tensile zone
No added rebar required
Add rebar to increase Moment capacity to 1.2 Mcr
As = 0.00075 * Acf EXIT
STEP 7 Code Check for Serviceability
As = Area of steel required Acf = Larger of cross-sectional area of the strip in direction of analysis and orthogonal to
STEP 7 Code Check for Serviceability
ACI 318-11 Minimum Reinforcement ¾ Rebar in span is a function of the magnitude of the hypothetical tensile stress
In span, provide rebar if the hypothetical tensile stress exceeds 0.166√f’c The amount of reinforcement As is given by: As = N / (0.5*fy) where N is the tensile force in tension one
h = member thickness; b = design section width
EC2 Minimum Reinforcement ¾ Minimum based on cross-sectional area ¾ Minimum based on hypothetical tensile stress
•
Based on cross-sectional “area” of section
Asmin ≥ (0.26* fctm *bt*d / fyk) ≥ 0.0013* bt *d
•Based on value of hypothetical tensile stresses for crack control Check probable crack width Add rebar based using the code
STEP 7 Deflection Check Read deflections from the frame analysis of the design strip for dead, live and PT; (ΔDL , ΔLL , and ΔPT ). . Make the following load combinations and check against the allowable values for each case ¾ Total Deflection (1 + 2)(ΔDL + ΔPT + 0.3 ΔLL ) + 0.7 ΔLL < span/250 This is on the premise of sustained load being 0.3 time the design live load. It is for visual effects; Provide camber to reduce value, where needed and practical ¾ Immediate deflection from live load Δimmediate = 1.00ΔL < span/500 This check is applicable, where non-structural members are likely to be damaged. Otherwise, span/240 applies ¾ Presence of members likely to be damaged from sustained deflection (1+ 2)(0.3 ΔLL ) + 0.7 ΔLL < span/350
STEP 8 Strength Check Determination of Hyperstatic actions ¾ Direct Method – based on reactions from balanced loads
STEP 8 Strength Check Steps in strength check ¾ ¾ ¾ ¾ ¾ ¾
Load combinations Determination of hyperstatic actions Calculation of design moments (Mu) Calculate capacity/rebar for design moment Mu Check for punching shear Check/detail for unbalanced moment at support
Load combinations (EC2) U1 = 1.35DL + 1.50LL + 1.0HYP where, HYP is moment due to hyperstatic actions from prestressing Determination of Hyperstatic actions ¾ Direct Method – based on reactions from balanced loads ¾ Indirect Method – Using primary and post-tensioning moments
STEP 8 Strength Check A comment on capacity versus demand ¾ Post-tensioned members possess both a positive and negative moment capacity along the member length ¾ Rebar needs to be added, where capacity falls short of demand ¾ First, find the capacity and compare it with demand
STEP 8 Strength Check
STEP 8 Strength Check
¾ The figure below shows the forces on a PT member. In calculating the force from PT tendons, use either the code formulas or the following simplified procedure, based on parametric study of common building structures can be sued.
Check for adequate ductility ACI ¾ Ductility is deemed adequate, if c/dt
