10- Step Design of Post-Tensioned Floors - PT-Structurespt-structures.com/.../08/10_Steps_PT_Floor_Design_SI_International... · One-way beam and slab design ... 10- Steps Design

10- Step Design ofPost-Tensioned Floors

Dr Bijan O AalamiProfessor Emeritus,

San Francisco State UniversityPrincipal, ADAPT Corporation; [email protected]

www.adaptsoft.comwww.adaptsoft.comPTPT--Structures copyright 2015Structures copyright 2015

301 Mission StreetSan Francisco, California

High Seismic Force Region

Four Seasons Hotel; Florida

High Wind Force Region Residential/Office Post-Tensioned Building in Dubai

Column supported multistory building

Two-way flat slab construction

Multi-level parking structures

One-way beam and slab design

Example of a Floor System using the Unbonded Post-tensioning System

An example of a grouted

system hardware with flat duct

Post-Tensioning SystemsGrouted System

Example of a Floor System Reinforcedwith Grouted Post-Tensioning System

Preliminary ConsiderationsDesign of Post-Tensioned Floors

Dimensions (sizing)Optimum spans; optimum thickness

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

Boundary conditions; connections Service performance; strength condition

Design strips

Design sections; design values


Dimensions (sizing)Optimum spans; optimum thickness

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

PT amount in service condition is governed mostly by:

Hypothetical tensile stressesTendon spacing

Spans: between 7 m – 9 m ft

Span/thickness ratios 40 for interior35 for exterior with no overhang

One-way and two-way systemsSkeletal Systems share loadSlab systems may not share load, depending on reinforcement layout


F

F

(a) ONE-WAY BEAM SYSTEM

(b) TWO-WAY BEAM SYSTEM

C

A D

B

B

F

Skeletal System

One-way and two-way systemsSkeletal Systems share loadSlab systems may not share loaddepending on reinforcement layout


PL/4

(1-a)P

BEAM

(b) BEAMS ON FOUR COLUMNS

L

(iii)

(ii)

A,C

aP

PB,D

B C

(i)A

L

C

A LOAD

P

B

D

(PL/4)

2

D

COLUMN

Capacity required for 100% of load in each direction, if reinforcement is placed along AD,

and AC

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


h < 2t b > 3h

b

h

SLAB BAND

Slab band is treated as part ofa two-way system. One-way sheardesign provisions meant for beams

do not apply to slab bands

Selection of load path for two-waysystems – Design Strips


Subdivide the structure into design strips in two orthogonal directions

(Nahid slab)

XY

G

F

E

D

C

B

A

31 2 4 5

Subdivide the floor along supportlines in design strips


XY

1 2 3 4 5

F

E

D

C

B

A

Subdivide slab along support lines in design 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. No portion of the slab should be left unassigned.


Design sectionsDesign sections extend over the entire

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


Design valuesActions, such as moments at each design

section are reduced to a “single”representative value to be used for design

559 k-ft is the area (total) of bendingmoment at face of support

10- StepsDesign of Post Tensioned Floors

1. Geometry and Structural System2. Material Properties3. Loads4. Design Parameters5. Actions due to Dead and Live Loads6. Post-Tensioning7. Code Check for Serviceability8. Code Check for Strength9. Check for Transfer of Prestressing10. Detailing

Select design strip and IdealizeExtract; straighten the support line;square the boundary

Step 1 Geometry and Structural System

Select design strip and IdealizeExtract; straighten the support line;square the boundaryModel 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

Step 1 Geometry and Structural System

View of idealized slab-frame

Step 2Material Properties

ConcreteWeight 24 kN/m328 day cylinder 40 MPaElastic modulus 29,725 MPaLong-term deflection factor 2

Non-Prestressed reinforcementfy 460 MPaElastic modulus 200,000 MPa

PrestressingStrand diameter 13 mmStrand area 99 mm2Ultimate strength 1,860 MPaEffective stress 1,200 MPaElastic modulus 200,000 MPa

Step 3Dead and Live Loads

SelfweightBased on member volume

Superimposed dead loadMin (partitions) 2.00 kN/m2

Live loadResidential 3.00 kN/m2

Step 4Design Parameters

Applicable codeACI 318-14IBC 2012 ?Local codes, such as California Building

Code (CBC 2011) ?

Cover for protection against corrosion

Cover to rebarNot exposed to weather 20 mmExposed to weather 50 mm

Cover to tendonNot exposed to weather 20 mmExposed to weather 25 mm


Cover for fire resistivity

Identify “restrained” and “unrestrainedpanels.”

Restrained orUnrestrained

Aggregate Type Cover Thickness, mmfor Fire Endurance

1 hr 1.5 hr 2 hr 3 hr 4 hr

Unrestrained CarbonateSiliceous

Lightweight

---

---

383838

515151

-

Restrained CarbonateSiliceous

Lightweight

---

---

191919

252525

323232

--

For 2-hour fire resistivityRestrained 20 mmUnrestrained 38 mm


Cover for fire resistivityIdentify “restrained” and “unrestrainedpanels.”


Select post-tensioning system For corrosive environment use “encapsulated system.”

For non-corrosive environment, regular hardware may be used


Allowable stresses for two-waysystems

Service conditionTotal (frequent) load case

Tension ---------------------------

Compression --------------------

Sustained (quasi permanent) load case

Tension --------------------------

Compression ------------------

Initial conditionTension ------------------------

Compression ---------------

'0.5 cf

'0.5 cf

'0.25 cf

'0.6 cf

'0.45 cf

'0.6 cf


Allowable deflections (ACI 318 .5)

For visual impact use total deflectionSpan/240Use camber, if necessary

Total deflection subsequent to installation of members that are likely to be damaged

Span/360

Immediate deflection due to live loadSpan/480

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

Step 5Actions due to Dead and Live Loads

Analyze the design strip as a singlelevel frame structure with one row ofsupports above and below, usingIn-house simple frame program(Simple Frame Method; SFM); orin-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.3D FEM software can improve optimization

Step 5Actions due to Dead and Live Loads

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

Moments due to DL (k-ft)

Moments due to LL (k-ft)

Step 6 Post-Tensioning

Selection of design parameters

Selection of PT force and profile

Effective force vs tendon selection optionForce selection option

Calculation of balanced loads;adjustments for percentage of loadbalanced

Calculation of actions due to balanced loads

Step 6 Post-Tensioning

Selection of PT force and profileTwo entry value selections must be made to initiate the computations. Select precompression and % of DL to balance

Step 6Post-Tensioning

Selection of design parametersSelect average precompression 1 MPaTarget 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 6Post-Tensioning

Assume simple parabola forhand calculation

Calculation of balanced loads;adjustment of % of DL balanced

STEP 6Post-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<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 for % of DL balanced


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]

YesNo

No Yes

Is it practical to reduce P/A or tendons?

%DL > 80%?

Rais e tendon to reach%DL ~ 60%

Exit after last span

F121_ACI_2-way_PT_force_082012

Calculation of balanced loadsLateral forced from continuous tendonsLateral force from terminated tendonsMoments from change in centroid of

member

Example of force from continuous tendon


P = 500 ka = 93 mm; b = 186 mm ; L = 9.0 m ; c = {[93/186]0.5/[1 + (93/186)0.5]} * 9.00

= 3.73 m wb = 2 P*a/c2 = (2*193*93/1000)/3.732

= 1.59 kN/m per tendon

Calculation of balanced loadsLateral forced from continuous tendonsLateral force from terminated tendonsMoments from change in centroid of member

Example of force from terminated tendon


L = 10 m ; a = 93 mm; P = 119 kN/tendon c = 0.20*10 = 2.00 m Wb = (2*119*93/1000) / 22 = 5.53 kN/tendon ↓Concentrated force at dead end= 2*119*93/2000

= 11.06 k/tendon ↑

Calculation of balanced loadsLateral forced from continuous tendonsLateral force from terminated tendonsMoments from change in centroid of member

Example of force from shift in membercentroidal axis


Moment at face of drop = MM = P * shift in centroid = P * (Yt-Left – Yt-Right)P = 119 k; Yt-Left = 120 mm; Yt-Right = 146 mmM = 119 (120 – 146)/1000 = - 3.09 kNm

Calculation of actions due to balanced loadsCheck balanced loads for static equilibriumDetermine moments/shears from balanced loads

applied to the frame model that was used for deadand live loads

Note down reactions from balanced loads

Example of Balanced Load Assembly


Calculation of actions due to balanced loadsObtain moments at face-of-supports and mid-spansNote the reactions. The reactions are hyperstatic forces..

Comments:Moments will be used for serviceability check. Reactions will be used for

Strength check.


ACI 318-14 requirements for serviceabilityLoad combinationsStress checkMinimum reinforcementDeflection check.

Load combinationTotal load condition1.00DL + 1.00LL + 1.00PT

Sustained load condition1.00DL + 0.30LL + 1.00PT

Stress check

STEP 7Code Check for Serviceability

Using engineering judgment, select the locations thatare likely to be critical. Typically, these are at the faceof support and for hand calculation at mid-spanAt each section selected for check, use the design actionsapplicable to the entire design section and apply those to the entire cross-section of the design section to arrive at the hypothetical stresses used in code check.

σ = (MD + ML + MPT)/S + P/AS = I/Yc ; I = second moment of area;Yc = distance to farthest tension fiber

ACI 318-14 Minimum Reinforcement


Yes

6

8

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

7

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 cracking moment Mcr at

supports and spans

12

ACI 318-14 Minimum ReinforcementRebar over support is function of geometry of the

design strip and the strip in the orthogonal directionRebar in span is a function of the magnitude of

the hypothetical tensile stress


As = 0.00075*Acf

As = Area of steel requiredAcf = Larger of cross-sectional area of the strip in direction of analysis and orthogonal to it.

ACI 318-14 Minimum ReinforcementRebar over support is function of geometry of the

design strip and the strip in the orthogonal directionRebar in span is a function of the magnitude of

the hypothetical tensile stress

In span, provide rebar if the hypothetical tensile stress exceeds 0.16√f’c

The amount of reinforcement As is given by:As = N / (0.5fy)

where N is the tensile force in tension zone


h = member thickness; b = design section width

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

Total Deflection(1 + 2)(ΔDL + ΔPT + 0.3 ΔLL ) + 0.7 ΔLL < span/240This 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

Immediate deflection from live loadΔimmediate = 1.00ΔL < span/480This check is applicable, where non-structural members are likely to be damaged. Otherwise, span/240 applies

Presence of members likely to be damaged from sustained deflection (1+ 2)(0.3 ΔLL ) + 0.7 ΔLL < span/360

STEP 7Deflection Check

Steps in strength checkLoad combinations

Determination of hyperstatic actionsCalculation of design moments (Mu)Calculate capacity/rebar for design moment MuCheck for punching shearCheck/detail for unbalanced moment at support

Load combinationsU1 = 1.2DL + 1.6LL + 1.0HYPU2 = 1.4DL + 1.0HYPwhere, HYP is moment due to hyperstatic actions from prestressing

Determination of Hyperstatic actionsDirect Method – based on reactions from balanced loadsIndirect Method – Using primary and post-tensioningmoments

STEP 8Strength Check

Determination of Hyperstatic actionsDirect Method – based on reactions from balanced loads


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

and negative moment capacity along the member length

Rebar needs to be added, where capacity falls short of demand

First, find the capacity and compare it with demand


The figure below shows the forces on a PT member.In calculating the force from PT tendons, use eitherthe ACI 318 or the following simplified procedure, based on parametric study of common building structures can be used for slabs


f’c ≥ 28 MPa; P/A ≤ 1.7 MPac/dt ≤ 0.375 ; dt is distance from compressionfiber to farthest tension rebar Tendon Length ≤ 38 m for single end stressing;if length ≤ 76 m double end stressingfps is conservatively 1480 MPa if span is less than 10 m fps is conservatively 1340 MPa if span is greaterthan 10 m

Check for adequate ductilityDuctility is deemed adequate, if c/dt <= 0.345 dtThis condition guarantees that steel will yield, before concrete in compression crushes.


Verify adequacy (detail) of the design fortransfer of unbalanced moment at supports

Unbalanced moment (Mc)is defined as the difference between the design moments on the opposite sides of a column support. This is the moment that is resisted by the support.The reinforcement associated with the transfer of unbalanced moment must be placed over anarrow band at the support (next slide) In most cases, this provision leads to a “detailing”requirement, as opposed to added rebar, sincethe reinforcement for slab design is in excess of that needed for transfer of unbalanced moment.


LINECENTER-

SLAB/BEAM LINECOLUMN

M

MOMENTS

c


Transfer of unbalanced moment

Place the reinforcement within the narrow band identified as rebar strip

FOR THE EVALUATION OF PUNCHING SHEAR STRESSESILLUSTRATION OF CRITICAL SURFACE

D108/SLID

ES/060591 Vu

COLUMN REGIONPUNCHED OUT

TWO-WAY SLAB

CRITICAL SURFACE

SHEAR STRESSDUE TO Vu

M uDUE TOSHEAR STRESS

kMu

Punching Shear Design

Definition based on ACI 318

STEP 9Check 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


At stressing check extreme fiber stresses

Add rebar when “representative”(hypothetical) tension stresses exceeda thresholdDo not exceed “representative”hypothetical compressive stresses. Waituntil concrete gains adequate strength

Local failure at stressingLocal failure at stressing


Load combination

U = 1.00*Selfweight + 1.15*PT

Allowable stresses

Tension 0.25√f’ci

Compression 0.60f’ci

If tension exceeds, provide rebar in tensile zone to resist N

As = N / (0.5 fy)

If compression exceeds, wait until concrete gains adequate strength

Position of rebar

STEP 10Detailing

STAGGER

SEE PLANEQ.

MID-SPAN

STA

GG

ER

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/3Lc

SLAB

Lc/6

*

d

BOTTOM

PLAN

ELEVATION

Thank you for listening.

[email protected]

Documents

10- Step Design of Post-Tensioned Floors - PT-Structurespt-structures.com/.../08/10_Steps_PT_Floor_Design_SI_International... · One-way beam and slab design ... 10- Steps Design