Post Tension BIJAN AALAMI

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


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.


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



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.”


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


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<300ps i [2MP a]?

No

%DL > 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


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


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 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 <= 0.345 EC2 ¾ Ductility is deemed adequate, if c/h<= 0.43 h = member thickness This condition guarantees that steel will yield, before concrete in compression crushes.

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

Punching Shear Design PUNCHED OUT COLUMN REGION

SHEAR STRESS DUE TO kM u

Mu

D108/SLIDES/060591

SHEAR STRESS DUE TO Vu

Vu

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 absent Hence the member is likely to experience stresses more severe than when in service ¾ Add rebar when “representative”

CRITICAL SURFACE

(hypothetical) tension stresses exceed a threshold ¾ Do not exceed “representative”

hypothetical compressive stresses

TWO-WAY SLAB

ILLUSTRATION OF CRITICAL SURFACE FOR THE EVALUATION OF PUNCHING SHEAR STRESSES

Refer to ACI 318-11 section 11.11

STEP 10 Detailing

STEP 9 Check for Transfer of Prestressing Load combination ¾

SUPPORT

U = 1.00*Selfweight + 1.15*PT

EQ. EQ.

Tension stress

¾

Compression stress

9 If tension exceeds, provide rebar in tensile zone to resist Nc

MID-SPAN SEE PLAN EQ.

EQ.

SUPPORT EQ. EQ.

STAGGER

STAGGER

Check for allowable stresses ¾

Position of rebar

TOP REBAR AT SUPPORT TYP. WALL

DROP CAP COLUMN

BOTTOM

PLAN

9 If compression exceeds, wait until concrete gains adequate strength

Lc/6

POST-TENSIONED SLAB

Lc/6

* DROP COLUMN

Lc/3 Lc SUPPORT LINE

ELEVATION

Thank you for listening.

www.adaptsoft.com; [email protected]

Ld

Post Tension BIJAN AALAMI

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