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10- Step Design of Post-Tensioned Floors
Dr Bijan O Aalami Professor Emeritus, San Francisco State University Principal, ADAPT Corporation; [email protected]
301 Mission Street San Francisco, California High Seismic Force Region
www.adaptsoft.com PTPT-Structures copyright 2015
Four Seasons Hotel; Florida High Wind Force Region
Residential/Office PostTensioned Building in Dubai
Multi-level parking structures Column supported multistory building
One-way beam and slab design
Two-way flat slab construction
Hybrid Construction Post-tensioned podium slab supporting light framed structure above Santana Row; San Jose, California
Post-Tensioned ground supported slab (SOG) is the largest application of posttensioning in USA
Fortaleza, Brazil
KSA
POST-TENSIONED MAT FOUNDATION USING UNBONDED TENDONS
POST-TENSIONED MAT FOUNDATION USING GROUTED TENDONS
Post-Tensioning Systems
12.7 mm (0.5"*)
Unbonded System
WIRE
CORROSION INHIBITING COATING
PLASTIC SHEATHING
NOTE: * NOMINAL DIAMETER
(a) STRAND
(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
Example of a Floor System Reinforced with Grouted Post-Tensioning System An example of a grouted system hardware with flat duct
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
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 ¾ Tendon spacing
Spans: between 7 m – 9 m ft Span/thickness ratios
¾ 40 for interior ¾ 35 for exterior with no overhang
¾ Skeletal Systems share load ¾ Slab systems may not share load, depending on reinforcement layout
One-way and two-way systems ¾ Skeletal Systems share load ¾ Slab systems may not share load depending on reinforcement layout
F
A
(ii) A C
D
COLUMN
BEAM
C
B L
aP
(a) ONE-WAY BEAM SYSTEM F
P
L
(i)
B
LOAD
D 2
F
A
(P L/ 4)
One-way and two-way systems
Preliminary Considerations Design of Post-Tensioned Floors
(1 -a )P
Preliminary Considerations Design of Post-Tensioned Floors
C
B
D A,C
B (iii)
P
B,D PL/4
(b) BEAMS ON FOUR COLUMNS
(b) TWO-WAY BEAM SYSTEM
Skeletal System
Capacity required for 100% of load in each direction, if reinforcement is placed along AD, and AC
Preliminary Considerations Design of Post-Tensioned Floors
Structural system
¾ One-way/two-way; slab band
h b
h < 2t
b > 3h
SLAB BAND Slab band is treated as part of a two-way system. One-way shear design provisions meant for beams do not apply to slab bands
Preliminary Considerations Design of Post-Tensioned Floors
Boundary conditions; connections ¾ Service performance ¾ Strength performance
Detailing for service performance, such as the one shown below is to mitigate cracking from shortening of slab
Preliminary Considerations Design of Post-Tensioned Floors Selection of load path for two-way systems – Design Strips
Preliminary Considerations Design of Post-Tensioned Floors Assumption of releases at connections, or reduced stiffness for selected members is made prior to analysis to achieve a more economical design. In the following, the assignment of reduced stiffness for the uppermost columns, or hinge assumption at connection is not uncommon
Preliminary Considerations Design of Post-Tensioned Floors Subdivide the floor along support lines in design strips 2
1
3
A B
C
D
Subdivide the structure into design strips in two orthogonal directions (Nahid slab)
E F Y X
G
4
5
Preliminary Considerations Design of Post-Tensioned Floors Subdivide slab along support lines in design strips in the orthogonal direction 2
1
3
4
5
A
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
B C
D
E F Y X
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.
10- Steps 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 k-ft is the area (total) of bending moment at face of support
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
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
View of idealized slab-frame
Step 2 Material Properties
Step 3 Dead and Live Loads
Concrete
¾ Weight 24 kN/m3 ¾ 28 day cylinder 40 MPa ¾ Elastic modulus 29,725 MPa ¾ Long-term deflection factor 2
Non-Prestressed reinforcement ¾ fy 460 MPa ¾ Elastic modulus 200,000 MPa
¾ Based on member volume
Superimposed dead load ¾ Min (partitions)
13 mm 99 mm2 1,860 MPa 1,200 MPa 200,000 MPa
2.00 kN/m2
Live load
¾ Residential
Prestressing
¾ Strand diameter ¾ Strand area ¾ Ultimate strength ¾ Effective stress ¾ Elastic modulus
Selfweight
3.00 kN/m2
Step 4 Design Parameters
Applicable code
¾ ACI 318-14 ¾ IBC 2012 ? ¾ Local codes, such as California Building Code (CBC 2011) ?
Cover for protection against corrosion
Cover to rebar
¾ Not exposed to weather ¾ Exposed to weather
Step 4 Design Parameters
Cover for fire resistivity
Identify “restrained” and “unrestrained panels.”
Restrained or Unrestrained
Unrestrained
20 mm 50 mm
Restrained
Cover Thickness, mm for Fire Endurance
Aggregate Type 1 hr
1.5 hr
2 hr
3 hr
Carbonate Siliceous Lightweight
-
-
38 38 38
51 51 51
Carbonate Siliceous Lightweight
-
-
19 19 19
25 25 25
Cover to tendon
¾ Not exposed to weather 20 mm ¾ Exposed to weather 25 mm
Step 4 Design Parameters
Cover for fire resistivity
Identify “restrained” and “unrestrained panels.”
For 2-hour fire resistivity ¾ Restrained ¾ Unrestrained
20 mm 38 mm
Step 4 Design Parameters
Select post-tensioning system
¾ For corrosive environment use “encapsulated system.”
¾ For non-corrosive environment, regular hardware may be used
4 hr -
32 32 32
Step 4 Design Parameters
Step 4 Design Parameters
Allowable deflections (ACI 318 .5)
Allowable stresses for two-way systems
For
Service condition ¾ Total (frequent) load case ' Tension ---------------------------0.5 f c '
Compression -------------------- 0.6 f c
¾ Sustained (quasi permanent) load case Tension -------------------------- 0.5 f c' '
Compression ------------------ 0.45 fc
Initial condition
visual impact use total deflection ¾ Span/240 ¾ Use camber, if necessary
Total deflection subsequent to installation of members that are likely to be damaged ¾ Span/360
Immediate deflection due to live load ¾Span/480 Long-term deflection magnifier 2. This brings the total long-term deflection to 3,
Tension ------------------------ 0.25 f c' Compression --------------- 0.6 f c'
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
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.
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. 3D FEM software can improve optimization
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
Step 6 Post-Tensioning Selection of PT force and profile
¾ Two entry value selections must be made to initiate the computations. Select precompression and % of DL to balance
Effective force vs tendon selection option ¾ Force selection option
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 for % of DL balanced
adjustment of % of DL balanced F121_ACI_PT_2_way_082012
F121_ACI_2-way_PT_force_082012
Assume P/A =150psi[1MPa] 1
Select max drape using tendons from critical span
Select critical span 2
Select max drape
Calculate %of DL balanced (%DL)
Calculate %of DL balanced (%DL)
Yes
No
%DL < 50%?
No
Yes %DL > 80%?
No
P /A<300ps i [2MP a]?
%DL > 80%?
No
Yes
Yes
Increase P/A
Is it practical to reduce P/A or tendons?
P/A>125psi [0.8MPa]? Yes
No Reduce drape
No
Yes
Reduce P/A
Rais e tendon to reach %DL ~ 60%
Reduce P/A or tendons to %DL balanced ~ 60% ; P/A >= 125 psi [0.8 MPa]
Go to next span
Move to next span
Exit after last span
Member with widely different spans
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 balanced loads ¾ Lateral forced from continuous tendons ¾ Lateral force from terminated tendons ¾ Moments from change in centroid of member
Example of force from terminated tendon
Example of force from continuous tendon
P = 500 k a = 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
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
↑
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 to the frame model that was used for dead and live loads Note down reactions from balanced loads
Example of force from shift in member centroidal axis
Example of Balanced Load Assembly
Moment at face of drop = M M = P * shift in centroid = P * (Yt-Left – Yt-Right) P = 119 k; Yt-Left = 120 mm; Yt-Right = 146 mm M = 119 (120 – 146)/1000 = - 3.09 kNm
STEP 6 Post-Tensioning Calculation of actions due to balanced loads ¾ Obtain moments at face-of-supports and mid-spans ¾
STEP 7 Code Check for Serviceability ACI 318-14 requirements for serviceability ¾ ¾ ¾ ¾
Note the reactions. The reactions are hyperstatic forces..
Comments: Moments will be used for serviceability check. Reactions will be used for Strength check.
Load combination ¾ Total load condition 1.00DL + 1.00LL + 1.00PT ¾
Load combinations Stress check Minimum reinforcement Deflection check.
Sustained load condition 1.00DL + 0.30LL + 1.00PT
Stress check
Using engineering judgment, select the locations that are likely to be critical. Typically, these are at the face of support and for hand calculation at mid-span At each section selected for check, use the design actions applicable 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/A S = I/Yc ; I = second moment of area; Yc = distance to farthest tension fiber
STEP 6 Post-Tensioning
STEP 6 Post-Tensioning
ACI 318-14 Minimum Reinforcement
ACI 318-14 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? No
5
ft ? tension stress
8
11
Yes 12
No added rebar required
6 7
10
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 6 Post-Tensioning ACI 318-14 Minimum Reinforcement 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
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
As = Area of steel required Acf = Larger of cross-sectional area of the strip in direction of analysis and orthogonal to it.
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/240 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/480 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/360
h = member thickness; b = design section width
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
STEP 8 Strength Check Determination of Hyperstatic actions ¾ Direct Method – based on reactions from balanced loads
Load combinations U1 = 1.2DL + 1.6LL + 1.0HYP U2 = 1.4DL + 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 ¾ The figure below shows the forces on a PT member. In calculating the force from PT tendons, use either the ACI 318 or the following simplified procedure, based on parametric study of common building structures can be used for slabs
¾ f’c ≥ 28 MPa; P/A ≤ 1.7 MPa ¾ c/dt ≤ 0.375 ; dt is distance from compression fiber to farthest tension rebar ¾ Tendon Length ≤ 38 m for single end stressing; if length ≤ 76 m double end stressing ¾ fps is conservatively 1480 MPa if span is less than 10 m ¾ fps is conservatively 1340 MPa if span is greater than 10 m
STEP 8 Strength Check
STEP 8 Strength Check
Verify adequacy (detail) of the design for
Check for adequate ductility ¾ Ductility is deemed adequate, if c/dt <= 0.345 dt This condition guarantees that steel will yield, before concrete in compression crushes.
transfer of unbalanced moment at supports ¾ 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. ¾ The reinforcement associated with the transfer of unbalanced moment must be placed over a narrow band at the support (next slide) ¾ In most cases, this provision leads to a “detailing” requirement, as opposed to added rebar, since the reinforcement for slab design is in excess of that needed for transfer of unbalanced moment. MOMENTS Mc
COLUMN LINE
SLAB/BEAM CENTER- LINE
STEP 8 Strength Check
Punching Shear Design
Transfer of unbalanced moment
PUNCHED OUT COLUMN REGION
SHEAR STRESS DUE TO kM u
Mu
D108/SLIDES/060591
SHEAR STRESS DUE TO Vu
Vu
CRITICAL SURFACE
TWO-WAY SLAB
ILLUSTRATION OF CRITICAL SURFACE
Place the reinforcement within the narrow band identified as rebar strip
FOR THE EVALUATION OF PUNCHING SHEAR STRESSES
Definition based on ACI 318
STEP 9 Check for Transfer of Prestressing
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
At stressing check extreme fiber stresses ¾ Add rebar when “representative”
(hypothetical) tension stresses exceed a threshold
Hence the member is likely to experience stresses more severe than when in service
¾ Do not exceed “representative”
hypothetical compressive stresses. Wait until concrete gains adequate strength
Local failure at stressing
STEP 10 Detailing
STEP 9 Check for Transfer of Prestressing Load combination ¾
U = 1.00*Selfweight + 1.15*PT
SUPPORT
Allowable stresses Tension 0.25√f’ci
¾
Compression 0.60f’ci
¾ If tension exceeds, provide rebar in tensile zone to resist N
EQ. EQ.
MID-SPAN SEE PLAN EQ. EQ.
SUPPORT EQ. EQ.
STAGGER
STAGGER
¾
Position of rebar
TOP REBAR AT SUPPORT TYP. WALL
DROP CAP COLUMN
BOTTOM
PLAN
As = N / (0.5 fy)
9 If compression exceeds, wait until concrete
Lc/6
gains adequate strength
Lc/6 POST-TENSIONED SLAB
* DROP COLUMN
Lc/3 Lc SUPPORT LINE
ELEVATION
Ld
Thank you for listening.
www.adaptsoft.com [email protected]
Dr Bijan O Aalami Professor Emeritus, San Francisco State University Principal, ADAPT Corporation; [email protected]
301 Mission Street San Francisco, California High Seismic Force Region
www.adaptsoft.com PTPT-Structures copyright 2015
Four Seasons Hotel; Florida High Wind Force Region
Residential/Office PostTensioned Building in Dubai
Multi-level parking structures Column supported multistory building
One-way beam and slab design
Two-way flat slab construction
Hybrid Construction Post-tensioned podium slab supporting light framed structure above Santana Row; San Jose, California
Post-Tensioned ground supported slab (SOG) is the largest application of posttensioning in USA
Fortaleza, Brazil
KSA
POST-TENSIONED MAT FOUNDATION USING UNBONDED TENDONS
POST-TENSIONED MAT FOUNDATION USING GROUTED TENDONS
Post-Tensioning Systems
12.7 mm (0.5"*)
Unbonded System
WIRE
CORROSION INHIBITING COATING
PLASTIC SHEATHING
NOTE: * NOMINAL DIAMETER
(a) STRAND
(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
Example of a Floor System Reinforced with Grouted Post-Tensioning System An example of a grouted system hardware with flat duct
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
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 ¾ Tendon spacing
Spans: between 7 m – 9 m ft Span/thickness ratios
¾ 40 for interior ¾ 35 for exterior with no overhang
¾ Skeletal Systems share load ¾ Slab systems may not share load, depending on reinforcement layout
One-way and two-way systems ¾ Skeletal Systems share load ¾ Slab systems may not share load depending on reinforcement layout
F
A
(ii) A C
D
COLUMN
BEAM
C
B L
aP
(a) ONE-WAY BEAM SYSTEM F
P
L
(i)
B
LOAD
D 2
F
A
(P L/ 4)
One-way and two-way systems
Preliminary Considerations Design of Post-Tensioned Floors
(1 -a )P
Preliminary Considerations Design of Post-Tensioned Floors
C
B
D A,C
B (iii)
P
B,D PL/4
(b) BEAMS ON FOUR COLUMNS
(b) TWO-WAY BEAM SYSTEM
Skeletal System
Capacity required for 100% of load in each direction, if reinforcement is placed along AD, and AC
Preliminary Considerations Design of Post-Tensioned Floors
Structural system
¾ One-way/two-way; slab band
h b
h < 2t
b > 3h
SLAB BAND Slab band is treated as part of a two-way system. One-way shear design provisions meant for beams do not apply to slab bands
Preliminary Considerations Design of Post-Tensioned Floors
Boundary conditions; connections ¾ Service performance ¾ Strength performance
Detailing for service performance, such as the one shown below is to mitigate cracking from shortening of slab
Preliminary Considerations Design of Post-Tensioned Floors Selection of load path for two-way systems – Design Strips
Preliminary Considerations Design of Post-Tensioned Floors Assumption of releases at connections, or reduced stiffness for selected members is made prior to analysis to achieve a more economical design. In the following, the assignment of reduced stiffness for the uppermost columns, or hinge assumption at connection is not uncommon
Preliminary Considerations Design of Post-Tensioned Floors Subdivide the floor along support lines in design strips 2
1
3
A B
C
D
Subdivide the structure into design strips in two orthogonal directions (Nahid slab)
E F Y X
G
4
5
Preliminary Considerations Design of Post-Tensioned Floors Subdivide slab along support lines in design strips in the orthogonal direction 2
1
3
4
5
A
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
B C
D
E F Y X
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.
10- Steps 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 k-ft is the area (total) of bending moment at face of support
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
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
View of idealized slab-frame
Step 2 Material Properties
Step 3 Dead and Live Loads
Concrete
¾ Weight 24 kN/m3 ¾ 28 day cylinder 40 MPa ¾ Elastic modulus 29,725 MPa ¾ Long-term deflection factor 2
Non-Prestressed reinforcement ¾ fy 460 MPa ¾ Elastic modulus 200,000 MPa
¾ Based on member volume
Superimposed dead load ¾ Min (partitions)
13 mm 99 mm2 1,860 MPa 1,200 MPa 200,000 MPa
2.00 kN/m2
Live load
¾ Residential
Prestressing
¾ Strand diameter ¾ Strand area ¾ Ultimate strength ¾ Effective stress ¾ Elastic modulus
Selfweight
3.00 kN/m2
Step 4 Design Parameters
Applicable code
¾ ACI 318-14 ¾ IBC 2012 ? ¾ Local codes, such as California Building Code (CBC 2011) ?
Cover for protection against corrosion
Cover to rebar
¾ Not exposed to weather ¾ Exposed to weather
Step 4 Design Parameters
Cover for fire resistivity
Identify “restrained” and “unrestrained panels.”
Restrained or Unrestrained
Unrestrained
20 mm 50 mm
Restrained
Cover Thickness, mm for Fire Endurance
Aggregate Type 1 hr
1.5 hr
2 hr
3 hr
Carbonate Siliceous Lightweight
-
-
38 38 38
51 51 51
Carbonate Siliceous Lightweight
-
-
19 19 19
25 25 25
Cover to tendon
¾ Not exposed to weather 20 mm ¾ Exposed to weather 25 mm
Step 4 Design Parameters
Cover for fire resistivity
Identify “restrained” and “unrestrained panels.”
For 2-hour fire resistivity ¾ Restrained ¾ Unrestrained
20 mm 38 mm
Step 4 Design Parameters
Select post-tensioning system
¾ For corrosive environment use “encapsulated system.”
¾ For non-corrosive environment, regular hardware may be used
4 hr -
32 32 32
Step 4 Design Parameters
Step 4 Design Parameters
Allowable deflections (ACI 318 .5)
Allowable stresses for two-way systems
For
Service condition ¾ Total (frequent) load case ' Tension ---------------------------0.5 f c '
Compression -------------------- 0.6 f c
¾ Sustained (quasi permanent) load case Tension -------------------------- 0.5 f c' '
Compression ------------------ 0.45 fc
Initial condition
visual impact use total deflection ¾ Span/240 ¾ Use camber, if necessary
Total deflection subsequent to installation of members that are likely to be damaged ¾ Span/360
Immediate deflection due to live load ¾Span/480 Long-term deflection magnifier 2. This brings the total long-term deflection to 3,
Tension ------------------------ 0.25 f c' Compression --------------- 0.6 f c'
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
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.
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. 3D FEM software can improve optimization
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
Step 6 Post-Tensioning Selection of PT force and profile
¾ Two entry value selections must be made to initiate the computations. Select precompression and % of DL to balance
Effective force vs tendon selection option ¾ Force selection option
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 for % of DL balanced
adjustment of % of DL balanced F121_ACI_PT_2_way_082012
F121_ACI_2-way_PT_force_082012
Assume P/A =150psi[1MPa] 1
Select max drape using tendons from critical span
Select critical span 2
Select max drape
Calculate %of DL balanced (%DL)
Calculate %of DL balanced (%DL)
Yes
No
%DL < 50%?
No
Yes %DL > 80%?
No
P /A<300ps i [2MP a]?
%DL > 80%?
No
Yes
Yes
Increase P/A
Is it practical to reduce P/A or tendons?
P/A>125psi [0.8MPa]? Yes
No Reduce drape
No
Yes
Reduce P/A
Rais e tendon to reach %DL ~ 60%
Reduce P/A or tendons to %DL balanced ~ 60% ; P/A >= 125 psi [0.8 MPa]
Go to next span
Move to next span
Exit after last span
Member with widely different spans
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 balanced loads ¾ Lateral forced from continuous tendons ¾ Lateral force from terminated tendons ¾ Moments from change in centroid of member
Example of force from terminated tendon
Example of force from continuous tendon
P = 500 k a = 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
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
↑
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 to the frame model that was used for dead and live loads Note down reactions from balanced loads
Example of force from shift in member centroidal axis
Example of Balanced Load Assembly
Moment at face of drop = M M = P * shift in centroid = P * (Yt-Left – Yt-Right) P = 119 k; Yt-Left = 120 mm; Yt-Right = 146 mm M = 119 (120 – 146)/1000 = - 3.09 kNm
STEP 6 Post-Tensioning Calculation of actions due to balanced loads ¾ Obtain moments at face-of-supports and mid-spans ¾
STEP 7 Code Check for Serviceability ACI 318-14 requirements for serviceability ¾ ¾ ¾ ¾
Note the reactions. The reactions are hyperstatic forces..
Comments: Moments will be used for serviceability check. Reactions will be used for Strength check.
Load combination ¾ Total load condition 1.00DL + 1.00LL + 1.00PT ¾
Load combinations Stress check Minimum reinforcement Deflection check.
Sustained load condition 1.00DL + 0.30LL + 1.00PT
Stress check
Using engineering judgment, select the locations that are likely to be critical. Typically, these are at the face of support and for hand calculation at mid-span At each section selected for check, use the design actions applicable 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/A S = I/Yc ; I = second moment of area; Yc = distance to farthest tension fiber
STEP 6 Post-Tensioning
STEP 6 Post-Tensioning
ACI 318-14 Minimum Reinforcement
ACI 318-14 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? No
5
ft ? tension stress
8
11
Yes 12
No added rebar required
6 7
10
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 6 Post-Tensioning ACI 318-14 Minimum Reinforcement 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
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
As = Area of steel required Acf = Larger of cross-sectional area of the strip in direction of analysis and orthogonal to it.
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/240 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/480 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/360
h = member thickness; b = design section width
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
STEP 8 Strength Check Determination of Hyperstatic actions ¾ Direct Method – based on reactions from balanced loads
Load combinations U1 = 1.2DL + 1.6LL + 1.0HYP U2 = 1.4DL + 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 ¾ The figure below shows the forces on a PT member. In calculating the force from PT tendons, use either the ACI 318 or the following simplified procedure, based on parametric study of common building structures can be used for slabs
¾ f’c ≥ 28 MPa; P/A ≤ 1.7 MPa ¾ c/dt ≤ 0.375 ; dt is distance from compression fiber to farthest tension rebar ¾ Tendon Length ≤ 38 m for single end stressing; if length ≤ 76 m double end stressing ¾ fps is conservatively 1480 MPa if span is less than 10 m ¾ fps is conservatively 1340 MPa if span is greater than 10 m
STEP 8 Strength Check
STEP 8 Strength Check
Verify adequacy (detail) of the design for
Check for adequate ductility ¾ Ductility is deemed adequate, if c/dt <= 0.345 dt This condition guarantees that steel will yield, before concrete in compression crushes.
transfer of unbalanced moment at supports ¾ 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. ¾ The reinforcement associated with the transfer of unbalanced moment must be placed over a narrow band at the support (next slide) ¾ In most cases, this provision leads to a “detailing” requirement, as opposed to added rebar, since the reinforcement for slab design is in excess of that needed for transfer of unbalanced moment. MOMENTS Mc
COLUMN LINE
SLAB/BEAM CENTER- LINE
STEP 8 Strength Check
Punching Shear Design
Transfer of unbalanced moment
PUNCHED OUT COLUMN REGION
SHEAR STRESS DUE TO kM u
Mu
D108/SLIDES/060591
SHEAR STRESS DUE TO Vu
Vu
CRITICAL SURFACE
TWO-WAY SLAB
ILLUSTRATION OF CRITICAL SURFACE
Place the reinforcement within the narrow band identified as rebar strip
FOR THE EVALUATION OF PUNCHING SHEAR STRESSES
Definition based on ACI 318
STEP 9 Check for Transfer of Prestressing
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
At stressing check extreme fiber stresses ¾ Add rebar when “representative”
(hypothetical) tension stresses exceed a threshold
Hence the member is likely to experience stresses more severe than when in service
¾ Do not exceed “representative”
hypothetical compressive stresses. Wait until concrete gains adequate strength
Local failure at stressing
STEP 10 Detailing
STEP 9 Check for Transfer of Prestressing Load combination ¾
U = 1.00*Selfweight + 1.15*PT
SUPPORT
Allowable stresses Tension 0.25√f’ci
¾
Compression 0.60f’ci
¾ If tension exceeds, provide rebar in tensile zone to resist N
EQ. EQ.
MID-SPAN SEE PLAN EQ. EQ.
SUPPORT EQ. EQ.
STAGGER
STAGGER
¾
Position of rebar
TOP REBAR AT SUPPORT TYP. WALL
DROP CAP COLUMN
BOTTOM
PLAN
As = N / (0.5 fy)
9 If compression exceeds, wait until concrete
Lc/6
gains adequate strength
Lc/6 POST-TENSIONED SLAB
* DROP COLUMN
Lc/3 Lc SUPPORT LINE
ELEVATION
Ld
Thank you for listening.
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