The 10 Steps in Design of
The 10 Steps in Design of
Post
Post
-
-
Tensioned Floors
Tensioned Floors
Dr Bijan O Aalami
Professor Emeritus,
San Francisco State University
Principal, ADAPT Corporation; [email protected]
www.adaptsoft.com
www.adaptsoft.com
301 Mission Street San Francisco, California High Seismic Force Region
Four Seasons Hotel; Florida High Wind Force Region
Column supported multistory
building
Multi-level parking structures
One-way beam and slab design
(a) STRAND (b) TENDON
NOTE: * NOMINAL DIAMETER
(0 .5 " *) CORROSION INHIBITING 1 2 .7 m m SHEATHING (c) ANCHORAGE ASSEMBLY STRAND PLASTIC WIRE COATING TUBE
GREASE FILLED PLASTIC CAP
SHEATH
Post-Tensioning Systems
Unbonded System
Example of a Floor System using the Unbonded Post-tensioning System
An example of a grouted
system hardware with flat duct
Post-Tensioning Systems
Grouted System
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
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 interior¾
35 for exterior with no overhang
Selection of load path for two-way
systems –
Design Strips
Preliminary Considerations
Design of Post-Tensioned Floors
Subdivide the structure into design strips in two orthogonal directions
X Y G F E D C B A 3 1 2 4 5
Subdivide the floor along support
lines in design strips
Preliminary Considerations
Design of Post-Tensioned Floors
X Y 1 2 3 4 5 F E D C B A
Subdivide slab along support lines in design
stripsin the orthogonal direction
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 andmid-span
Preliminary Considerations
Design of Post-Tensioned Floors
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
1.
Geometry and Structural System
2.
Materia
l 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
Select design strip and
Idealize
¾
Extract; straightenthe support line;squarethe boundary
Step 1
Geometry and Structural System
Select design strip and Idealize
¾
Extract; straighten the support line; square the boundary¾
Model the slab frame with a row of supports above and below. Thisrepresents an upper level of multi-story concrete frame.
Assume rotational fixity at the far ends;
Assume rollersupport at the far endsStep 1
Geometry and Structural System
View of idealized slab-frame
Step 2
Material Properties
Concrete
¾
Weight 24 kN/m3¾
28 day cylinder 40 MPa¾
Elastic modulus 35,220 MPa¾
Long-term deflection factor 2
Non-Prestressed reinforcement
¾
fy 460 MPa¾
Elastic modulus 200,000 MPa
Prestressing
¾
Strand diameter 13 mm¾
Strand area 99 mm2¾
Ultimate strength 1,860 MPa¾
Effective stress 1,200 MPaStep 3
Loads
Selfweight
¾
Based on member volume
Superimposed dead load
¾
Min (partitions) 1 kN/m2
Live load
¾
Residential 2.0 kN/m2¾
Office 2.5 kN/m2¾
Shopping mall 3.5 kN/m2¾
Parking structure 2.0 kN/m2
Lateral loads¾
Wind¾
Earthquake
Example assumes ¾ Superimposed DL SDL= 2 kN/m2 ¾ Live load LL = 3 kN/m2Step 4
Design Parameters
Applicable code
¾
ACI 318-11; EC2 EN 1992-1-1:2004¾
IBC 2012¾
Local codes, such as California Building Code (CBC 2011); or otherwise
Cover for protection against corrosion
Cover to rebar
¾
Not exposed to weather 20 mm¾
Exposed to weather 50 mm
Cover to tendon
¾
Not exposed to weather 20 mm¾
Exposed to weather 25 mmStep 4
Design Parameters
Cover for fire resistivity
Identify “
restrained
” and “
unrestrained
panels.”
Restrained or
Unrestrained Aggregate Type Cover Thickness, mm. forFire Endurance of 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 20 mm.¾
Unrestrained 40 mmStep 4
Design Parameters
Cover for fire resistivity
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 TensionCompression 0.60f’c
¾
Sustained load case Tension 0.5√f’c Compression 0.60 f’c
Based on EC2
¾
Frequent load caseTension Ft = 0.30 fck(2/3) Compression 0.60fck
¾
Quasi permanent load case Tension Ft = 0.30 fck(2/3) Compression 0.45fck¾
Crack width normal exposure (?) UnbondedBonded
0.5
√f’
cStep 4
Design Parameters
Allowable deflections
(EC2/ approx. ACI)
For visual impact use total deflection¾ Span/250
¾
Use camber, if necessary
Total deflection subsequent to installation of members that are likely to be damaged¾
Span/350
Immediate deflection due to live load¾
Span/500
Long-termdeflection 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, 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 EFMis 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 variesStep 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
Effective force/tendon selection option - force selection
Calculation of balanced loads; adjustments for percentage of loadbalanced
Calculation of actions due to balanced loadsStep 6
Post-Tensioning
Selection of PT force and profile¾Two entry value assumptions must be made to
initiate the computations. Select
precompressionand % of DL to balance
Step 6
Post-Tensioning
Selection of
design parameters
¾
Select average precompression 1 MPa¾
Target to balance 60% of DL
Selection of PT
force
and
profile
¾
Assume simple parabola mapped within the bounds of top and bottom coversForce diagram of simple parabola
Step 6
Post-Tensioning
¾ Assume simple parabola for
Calculation of balanced loads; adjustment of % of DL balancedSTEP 6
Post-Tensioning
Select critical span
Select max drape Calculate %of DL balanced
(%DL) %DL < 50%? Reduce P/A Reduce drape Yes Increase P/A No Yes No No Yes Yes No Assume P/A =150psi[1MPa]
P /A<300ps i [2MP a]? %DL > 80%? P/A>125psi [0.8MPa]? Go to next span 1 2 F121_ACI_PT_2_way_082012
Member with widely different spans
Calculation of balanced loads; adjustment of % of DL balanced
STEP 6
Post-Tensioning
Select max drape using tendons from critical span
Calculate %of DL balanced (%DL)
Move to next span
Reduce P/A or tendons to %DL balanced ~ 60% ; P/A >= 125 psi [0.8 MPa] Yes No No Yes Is it practical to reduce P/A or tendons? %DL > 80%? Ra is e te ndon to re a c h % DL ~ 60%
Exit after last span
F121_ACI_2-way_PT_force_082012
Calculation of balanced loads¾ Lateral forced from continuous tendons
¾ Lateral force from terminated tendons
¾ Moments from change in centroid of member
Example of force from continuous tendon
STEP 6
Post-Tensioning
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
Calculation of balanced loads¾ Lateral forced from continuous tendons
¾ Lateral force from terminated tendons
¾ Moments from change in centroid of member
Force from terminated tendon
STEP 6
Post-Tensioning
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 ↓
Calculation of balanced loads¾ Lateral forced from continuous tendons
¾ Lateral force from terminated tendons
¾ Moments from change in centroid of member
Example of force from change in member centroid
STEP 6
Post-Tensioning
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
Calculation of actions due to balanced loads¾ Check balanced loads for static equilibrium
¾ Determine moments/shearsfrom balanced loads applied the frame used for dead and
live loads
¾ Note down reactionsfrom balanced loads
STEP 6
Post-Tensioning
Calculation of actions due to balanced loads¾ 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 6
Post-Tensioning
Code requirements for serviceability¾ Load combinations
¾ Stress/crack width check
¾ Minimum reinforcement
¾ 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
STEP 7
Code Check for Serviceability
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
ACI
Minimum ReinforcementSTEP 7
Code Check for Serviceability
Yes
6
8
ACI Minimum Rebar for two-way systems
F114_041112
Bonded Unbonded
At supports As = 0.0075Acf
Add rebar to resist
force in tensile zone No added rebarrequired
EXIT
Add rebar to increase Moment capacity to 1.2 Mcr 7 5 4 3 2 1 11 10 9 PT system? ` ` No ft ? tension stress In span calculate hypothetical tension stress ft ft > 2 root 'c
[ft > 0.17 root f'c ] [ ft =< 0.17 root f'c ]ft =< 2 root 'c
Does Mcr exceeed 1.2xmoment capacity?
No added rebar required
Calculate the cracking moment Mcr at supports and spans
12
ACI 318-11 Minimum Reinforcement¾ 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
STEP 7
Code Check for Serviceability
As = 0.00075 * Acf
As = Area of steel required
Acf = Larger of cross-sectional area of the strip in
direction of analysis and orthogonal to
ACI 318-11 Minimum Reinforcement¾ 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
STEP 7
Code Check for Serviceability
h = member thickness; b = design section width
EC2 Minimum Reinforcement¾ Minimum based on cross-sectional area
¾ Minimum based on hypothetical tensile stress
•
Based on cross-sectional “area” of sectionAsmin ≥ (0.26* fctm *bt*d / fyk) ≥ 0.0013* bt *d
STEP 7
Code Check for Serviceability
•
Based on value of hypothetical tensile stresses for crack control
Check probable crack width
Read deflections from the frame analysis of thedesign 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 camberto 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 7
Deflection Check
Steps in strength check¾ Load combinations
¾ Determination of hyperstatic actions
¾ Calculation of design moments (Mu)
¾ Calculate capacity/rebarfor design moment Mu
¾ Check for punching shear
¾ 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
Determination of Hyperstatic actions
¾ Direct Method – based on reactions from balanced loads
STEP 8
Strength Check
A comment on capacityversus 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
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.
STEP 8
Strength Check
¾ USING EC2
¾ Assume tendon stress under service condition 1,200 MP
¾ Assume tendon stress at ultimate limit state 1300 MPa
¾ USING ACI
¾ Tendon Length ≤ 38 m for single end stressing; ; length 35 m ≤ length ≤ 75 m double end stressing
¾ fps is conservatively 1,480 MPa if span is less than 11 m
¾ fps is conservatively 1,340 MPa if span is greater than
Check for adequate ductility ACI
¾ Ductility is deemed adequate, if c/dt <= 0.345
STEP 8
Strength Check
EC2
¾Ductility is deemed adequate, if c/h<= 0.43 h = member thickness
This condition guarantees that steel will yield, before concrete in compression crushes.
FOR THE EVALUATION OF PUNCHING SHEAR STRESSES ILLUSTRATION OF CRITICAL SURFACE
D 1 08/S L IDE S /060591 Vu COLUMN REGION PUNCHED OUT TWO-WAY SLAB CRITICAL SURFACE SHEAR STRESS DUE TO Vu M u DUE TO SHEAR STRESS kM u
Punching Shear Design
Refer to ACI 318-11 section 11.11
STEP 9
Check for Transfer of Prestressing
At stressing:
Tendon has its maximum force;
Concrete is at its weakest strength; and
Live load to counteract prestressing is absentHence the member is likely to experience stresses more severe than when in service
¾Add rebar when “representative” (hypothetical) tension stresses exceed
a threshold
¾Do not exceed “representative” hypothetical compressive stresses
STEP 9
Check for Transfer of Prestressing
Load combination
¾
U = 1.00*Selfweight + 1.15*PT
Check for allowable stresses
¾
Tension stress
¾
Compression stress
9 If tension exceeds, provide rebar in tensile zone to resist Nc
9
If compression exceeds, wait until concrete gains adequate strength
Position of rebar
STEP 10
Detailing
STAGGER SEE PLAN EQ. MID-SPAN S T A G G E R DROP CAP COLUMN SUPPORT EQ. EQ. EQ. SUPPORT TYP. TOP REBAR AT SUPPORT EQ. WALL EQ. SUPPORT LINE POST-TENSIONED COLUMN DROP Lc/6 L Lc/3 Lc SLAB Lc/6 * d BOTTOMPLAN
ELEVATION
Thank you for listening.
www.adaptsoft.com; [email protected]
