Post Tension BIJAN AALAMI

8/13/2019 Post Tension BIJAN AALAMI

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The 10 Steps in Design ofThe 10 Steps in Design of

PostPost--Tensioned FloorsTensioned Floors

Dr Bijan O Aalami

Professor Emeritus,

San Francisco State University

Principal, ADAPT Corporation; [email protected]

www.adaptsoft.comwww.adaptsoft.com

301 Mission Street

San Francisco, California

High Seismic Force Region

Four Seasons Hotel; Florida

High Wind Force Region

Column supported multistory

building

Two-way flat slab construction


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Multi-level parking structures

One-way beam and slab design

(a) STRAND (b) TENDON

NOTE: * NOMINAL DIAMET

(0.5

"*)

CORROSIONINHIBITING

12.7mm

SHEATH

(c) ANCHORAGE ASSEMBLY

STRAND

PLASTIC

WIRECOATING

TUBE

GREASE FILLED PLASTIC CAP

SHEATH

Post-Tensioning SystemsUnbonded 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


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Example of a Floor System Reinforced

with Grouted Post-Tensioning System

Preliminary Considerations

Design of Post-Tensioned Floor

Dimensions (sizing) Optimum spans; optimum thickness

Structural system One-way/two-way; slab band

Boundary conditions; connections Service performance; strength condit

Load Path; Design strips

Design sections; design values


Design of Post-Tensioned Floors

Dimensions (sizing) Optimum spans; optimum thickness

An optimum design is one in which thereinforcement determined for servicecondition is used in its entirety forstrength condition.

PT amount in service condition is

governed mostly by: Hypothetical tensile stresses (USA), orcrack width (EC2)

Tendon spacing (USA)

Common spans: 25 30 ft (8 9 m)

Span/thickness ratios 40 - 45 for interior

35 for exterior with no overhang

Selection of load path for two-wasystems Design Strips



Subdivide the structure into design

strips in two orthogonal directions(Nahid slab)


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X

Y

G

F

E

D

C

B

A

31 2 4 5

Subdivide the floor along support

lines in design strips



X

Y

1 2 3 4 5

F

E

D

C

B

A

Subdivide slab along support lines in desig

strips in 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.

portion of the slab should be left unassigned.



Design sections Design sections extend over the entire

design strip and are considered at criticallocations, such as face of support andmid-span



Design valuesActions, such as moments at each de

section are reduced to a singlerepresentative value to be used for des

559 is the area (total) value of bending

moment at face of support


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10 Steps

Design of Post Tensioned Floors

. Geometry and Structural System

2. Material Properties. Loads

. Design Parameters

. Actions due to Dead and Live Loads

. Post-Tensioning

. Code Check for Serviceability

. Code Check for Strength

. Check for Transfer of Prestressing

0. Detailing

Select design strip and Idealize Extract; straighten the support line;

square the boundary

Step 1

Geometry and Structural System

Select design strip and Idealize Extract; straighten the support line;

square the boundary

Model the slab frame with a row ofsupports above and below. Thisrepresents an upper level of multi-storyconcrete frame.

Assume rotational fixity at the farends;

Assume rollersupport at the farends

Step 1

Geometry and Structural System

View of idealized slab-frame

Step 2

Material Properties

ConcreteWeight 24 kN/m3

28 day cylinder 40 MPa

Elastic modulus 35,220 MPa

Long-term deflection factor 2

Non-Prestressed reinforcement fy 460 MPa

Elastic modulus 200,000 M

Prestressing Strand diameter 13 mm

Strand area 99 mm2

Ultimate strength 1,860 MPa

Effective stress 1,200 MPa

Elastic modulus 193,000 M


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Step 3

Loads

Selfweight Based on member volume

Superimposed dead load Min (partitions) 1 kN/m2

Live load Residential 2.0 kN/m2

Office 2.5 kN/m2

Shopping mall 3.5 kN/m2

Parking structure 2.0 kN/m2

Lateral loads

Wind

Earthquake

Example assumes Superimposed DL SDL= 2 kN/m2

Live load LL = 3 kN/m2

Step 4

Design Parameters

Applicable codeACI 318-11; EC2 EN 1992-1-1:2004

IBC 2012

Local codes, such as California BuildiCode (CBC 2011); or otherwise

Cover for protection against corros

Cover to rebar Not exposed to weather 20 mm Exposed to weather 50 mm

Cover to tendon

Not exposed to weather 20 mm Exposed to weather 25 mm

Step 4

Design Parameters

Cover for fire resistivity

Identify restrained and unrestrainedpanels.

Restrained or

UnrestrainedAggregate Type Cover Thickness, mm. for

Fire Endurance of

1 hr 1.5 hr 2 hr 3 hr 4 hr

Unrestrained Carbonate

SiliceousLightweight

-

--

-

--

40

4040

50

5050

-

Restrained CarbonateSiliceousLightweight

---

---

202020

252525

303030

--

For 2-hour fire resistivity Restrained 20 mm.

Unrestrained 40 mm

Step 4

Design Parameters

Cover for fire resistivity

Identify restrained and unrestraipanels.


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Step 4

Design Parameters

Allowable Stresses or Crack WidthSelection for Two-Way System

Based on ACI Total load case

Tension

Compression 0.60fc Sustained load case

Tension 0.5fcCompression 0.60 fc

Based on EC2 Frequent load case

Tension Ft = 0.30 fck(2/3)

Compression 0.60fck Quasi permanent load case

Tension Ft = 0.30 fck(2/3)

Compression 0.45fck Crack width normal exposure (?)

Unbonded

Bonded

0.5fc

Step 4

Design Parameters

Allowable deflections (EC2/ approx. A

For visual impact use total deflection Span/250

Use camber, if necessary

Total deflection subsequent to installationmembers that are likely to be damaged

Span/350

Immediate deflection due to live load

Span/500

Long-term deflection magnifier 2. This brinthe total long-term deflection to 3,

Step 5

Actions due to Dead and Live Loads

Analyze the design strip as a singlelevel frame structure with one row ofsupports 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 ofreinforcement.

The EFM is suggested byACI-318. To someextent, it accounts for biaxial action of theprototype structure in the frame model.

Accuracy and ability of commercial softwarefor optimization varies

Step 5

Actions due to Dead and Live Loa

Analyze the design strip as a single levelframe structure with one row of supportsabove and below.


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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 balancedloads

Step 6

Post-Tensioning

Selection of PT force and profileTwo entry value assumptions must be mad

initiate the computations. Selectprecompression and % of DL to balance

Step 6

Post-Tensioning

Selection of design parameters Select average precompression 1 MPa

Target to balance 60% of DL

Selection of PT force and profileAssume simple parabola mapped within

the bounds of top and bottom covers

Force diagram of simple parabola

Step 6

Post-Tensioning

Assume simple parabola forhand calculation


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Calculation of balanced loads;

adjustment of % of DL balanced

STEP 6

Post-Tensioning

Select critical span

Select max drape

Calculate %of DL balanced (%DL)

%DL < 50%?

Reduce P/AReduce drape

Yes

Increase P/A

NoYes

No

No Yes

Yes No

Assume P/A =150psi[1MPa]

P /A 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 drap e 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]

YesNo

No Yes

Is it practical to reduceP/A or tendons?

%DL > 80%?

Raise tendon to reach%DL ~ 60%

Exit after last span

F121_ACI_2-way_PT_force_0

Calculation of balanced loads Lateral forced from continuous tendons

Lateral force from terminated tendons

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 loads Lateral forced from continuous tendons


Moments from change in centroid of mem

Force from terminated tendon

STEP 6

Post-Tensioning

L = 10 m ; a = 93 mm ; P = 119 kN; c =0.20*10 = 2.00

Wb = (3 * 119.0 * 2 * 93 / 1000) / 2.02 16.60 kN/m

Concentrated force at dead end = 2*16.60 = 33.20 k


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Calculation of balanced loads

Lateral forced from continuous tendons


Moments from change in centroid of member

Example of force from change in membercentroid

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 load

Check balanced loads for static equilibrium

Determine moments/shears from balanced loads

applied the frame used for dead and

live loads

Note down reactions from balanced loads

STEP 6

Post-Tensioning

Calculation of actions due to balanced loads Obtain moments at face-of-supports and mid-spans

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 serviceability Load combinations

Stress/crack width check

Minimum reinforcement

Deflection check.

Load combination

Frequent (Total) load condition

1.00DL + 0.50LL + 1.00PT

Quasi Permanent (Sustained) load condition

1.00DL + 0.30LL + 1.00PT

Stress check

STEP 7

Code Check for Serviceabil ity

Using engineering judgment, select the locatio

likely to be cr itical. Typically, these are at the f

and for hand calculation at mid-span

At each section selected for check, use the de

applicable to the entire design section and app

the entire cross-section of the design section

the hypothetical stresses used in code check.

= (MD + 0.5ML + MPT) / S + P/AS = I/Yc ; I = second moment of area of ;

Yc = distance to farthest tension fiber


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ACI Minimum Reinforcement

STEP 7


Yes

6

ACI Minimum Rebarfor two-way systems

F114_041112

BondedUnbonded

At supportsAs = 0.0075Acf

Add rebar to resistforce in tensile zone

No added rebarrequired

EXIT

Add rebar to increaseMoment capacity to

1.2 Mcr

5

4

3

2

1

11

10

9

PT system?

`

`

No

ft ?tension stress

In span calculatehypothetical tension

stress ft

ft > 2 root 'c[ft > 0.17 root f'c ]

ft =< 2 root 'c[ ft =< 0.17 root f'c ]

Does Mcr exceeed1.2xmoment capacity?

No added rebarrequired

Calculate the crackingmoment Mcr at

supports and spans

12

ACI 318-11 Minimum Reinforcement Rebar over support is function of geometry of t

design strip and the strip in the orthogonal

direction

Rebar in span is a function of the magnitude o

the hypothetical tensile stress

STEP 7


As = 0.00075 * Acf

As = Area of s teel required

Acf = Larger of cross-sectional area of the

strip in

direction of analysis and orthogonal to

ACI 318-11 Minimum Reinforcement Rebar in span is a function of the magnitude of the

hypothetical tensile stress

In span, provide rebar if the hypothetical

tensile

stress exceeds 0.166f cThe amount of reinforcement As is

given by:

As

= N / (0.5*fy

)

where N is the tensile force in tension

e

STEP 7


h = member thickness; b = design section

width

EC2 Minimum Reinforcement Minimum based on cross-sectional area

Minimum based on hypothetical tensile stress

Based on cross-sectional area of section

Asmin (0.26* fctm *bt*d / fyk) 0.0013

*d

STEP 7


Based on value of hypothetical tensile stressesfor crack control Check probable crack width Add rebar based using the code


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Read deflections from the frame analysis of thedesign strip for dead, live and PT; (DL , LL , and PT ).

. Make the following load combinations and

check against the allowable values for each case

Total Deflection

(1 + 2)(

DL +

PT + 0.3

LL ) + 0.7

LL < span/250This 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

Immediate deflection from live load

immediate = 1.00L < span/500

This check is applicable, where non-structural

members are likely to be damaged. Otherwise,

span/240 applies

Presence of members likely to be damaged fromsustained deflection

(1+ 2)(0.3 LL ) + 0.7 LL < span/350

STEP 7

Deflection Check

Steps in strength check Load combinations

Determination of hyperstatic actions

Calculation of design moments (Mu)

Calculate capacity/rebarfor 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

Direct Method based on reactions from balan

loads

Indirect Method Using primary and post-tensi

moments

STEP 8

Strength Check

Determination of Hyperstatic actions

Direct Method based on reactions from balanced

loads

STEP 8

Strength Check

A comment on capacity versus demand Post-tensioned members possess both a positive

and negative moment capacity along the memb

length

Rebar needs to be added, where capacity falls sh

of demand

First, find the capacity and compare it with deman

STEP 8

Strength Check


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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

USING EC2

Assume tendon st ress under serv ice cond it ion 1,200 M

Assume tendon st ress at ul timate limi t s tate 1300 MPa

USING ACI

Tendon Length 38 m for single end stressing; ; length

35 m length 75 m double end stressing

fps is conservatively 1,480 MPa if span is less than 11

fps is conservatively 1,340 MPa if span is greater than

Check for adequate ductility

ACI

Ductility is deemed adequate, if c/dt


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STEP 9

Check for Transfer of Prestressing

Load combination

U = 1.00*Selfweight + 1.15*PT

Check for allowable stresses

Tension stress

Compression stress

If tension exceeds, provide rebar in tensile

zone to resist Nc

If compression exceeds, wait until concrete

gains adequate strength

Position of rebar

STEP 10

Detailing

STAGGER

SEE PLAN

EQ.

MID-SPAN

STAGGER

DROP CAP

COLUMN

SUPPORT

EQ. EQ.EQ.

SUPPORT TTOP REBAR

SUPPORT

EQ.

WALL

EQ.

SUPPORT LINE

POST-TENSIONED

COLUMN

DROP

Lc/6

Lc/3

Lc

SLAB

Lc/6

*

BOTTOM

PLAN

ELEVATION

Thank you for listening.

www.adaptsoft.com;

[email protected]

Post Tension BIJAN AALAMI

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