Post-tension Flat Slab Design Example u5b20

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Design of Post-Tension Slab

Hand Calculation for Typical Flat Slab Design

Based on: BS 8110 1997 TR 43 – 1st Edition

Prepared by:

Nov. 2008

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Two Way Post-tension Flat Slabs: A post-tensioned prestressed banded flat slab floor system for an office complex is shown in figure.1 together with atypical sub-frame section. The structure is checked both at serviceability and ultimate limit states. These checks are carried out at serviceability and ultimate limit states. These checks are carried out at transfer and under working loads conditions.

1.

Input design data:

1.1

Material: Figure 1: structural plan

-

Concrete: Unit weight = 24 KN/m3 fck = 40 MPa fcu = 50 MPa fcui = 25 MPa Ec = 30 GPa Eci =25 GPa

-

(cylinder strength at 28 day) (cube strength at 28 day) (strength at transfer) (elastic modulus at 28 day) (elastic modulus at transfer)

Bonded reinforcement: fy = 460 Mpa

-

Prestressing steel: 12.9 mm diameter super strand placed in metal ducts. Pk = 186 KN (characteristic strength of strand) Aps = 100 mm2 (area of strand) fpu = Pk/Aps = 1860 MPa (characteristic strength of prestress steel) Eps= 195 GPa (elastic modulus)

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1.2

Loading: -

Imposed loading: (according to architectural drawing) Partition = 2.0 KN/m2 Finishing = 1.5 KN/m2 Services and false ceiling = 0.5 KN/m2 Total S.I.D.L = 4.0 KN/m2

-

Live load: Typical offices building = 3.0 KN/m2

-

1.3

Total imposed loading = 4.0 + 3.0 = 7.0 KN/m2

Serviceability classifications: -

BS 8110 code, TR 43 - 1st Eedition. Class 3.

2.

Design step:

2.1

Preliminary designs: For normal slab load, and initial span, we can take span/depth = 42 10,000 ≈ 240mm 42 Punching shear affected by size of column and slab thickness. However to reduce shear reinforcement requirement a depth of 250 mm is chosen.

-

Self weight = 6.0 KN/m2

-

Total dead load = 4.0 + 6.0 =10.0 KN/m2

-

Total live load = 3.0 KN/m2

-

Balanced load:

In these example a balanced load consisting 60% of all dead load is chosen 0.6 X 10 = 6 KN/m2.

2.2

Tendon profile: Nominal cover requirement in accordance with BS 8110-1, sec.3.4: For 2 hour fire resistance take cover = 25 mm.

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Take nominal cover to be 25 mm refer to figure.2.

(a) Transverse direction

(b) Longitudinal direction

Figure 2:Tendon and reinforcing steel postion

Note:

The positioning of reinforcement must be considered at this stage, so as to obtain practical arrangement of the steel at . Based on tendon eccentricities show in figure.3 and the position of inflection point (0.1 times the span from the center of the s) the tendon profile can be calculated.

Figure 3:Transfer tendon profile

At these stage losses are assumed as follows: -

At transfer 10% of the jacking load. At service 20% of the jacking load.

A through check will be carried out after the stress calculates to check that these initial assumptions are reasonable.

2.3

Initial Prestressing force: The initial prestress force i.e jacking force, has been taken to be 75% of the characteristic strand strength (BS8110-1:1197, sec 4.7.1). Calculation of Pavg: -

Jacking force (Pj) = 0.75 X 186 = 139.5 KN/strand. Prestress force at transfer (Po) = 125.55 KN/strand. Prestress force at service (Pe) =111.6 KN/strand.

(10% losses) (20% losses)

Next the value of prestress force required in each span is calculate, this is done using the chosen balanced load of 6.0 KN/m2 (60% of dead load) the distance between points of inflection, s, and the drape, a, as shown in figure.4, (refer to appendix A for profile calculation)

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Figure 4 : Draps for load balanced

The prestress force is obtained from the following equation which assumes a parabolic profile. Prqd =

ws 2 8a

For span A-B and C-D: -

Prqd = (6.0 X 7.0 X 64002) / (8.0 X 67.5 X 1000) = 3185 KN Therefore number of the tendons = Prqd / effective force = 3185 / 111.6 = 28.54 Try 30 tendons per “10 X 3S” or “6 X 5S” (more economical).

For span B-C: -

Prqd = (6.0 X 7.0 X 80002) / (8.0 X 120 X 1000) = 2800 KN Therefore number of the tendons = 2800 / 111.6 = 25 Try 30 tendons per as before “6 X 5S”.

The effect of the tendon in the slab is modeled by mean of equivalent as shown below. It should be noted that the portions of cable from the edges of the slab to grid lines A and D are horizontal and so do not contribute to the equivalent load. The equivalent load w, between any two points of inflection for the chosen number of tendons is given by: W=

8 a n Pav s2

Where: n: number of strand. a: drab at the point considered (up +, down -). s: distance between point of inflection.

Figure 5 : Calculation of equivalent load due to tendon forces

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At transfer: Pav = 125.55 KN n = 30 Table 1 : Calculations of equivalent loads due to transverse tendons

n. Pav (KN) a (mm) s (mm) w (KN/m)

A 3766.5 18.6 1600 218.9

span 3766.5 -67.5 6400 -49.7

B 3766.5 26.3 1600 309.6

B 3766.5 30 2000 226

span 3766.5 -120 8000 -56.5

C 3766.5 30 2000 226

C 3766.5 26.3 1600 309.6

span 3766.5 -67.5 6400 -49.7

D 3766.5 18.6 1600 218.9

C 3348 30 2000 200.9

C 3348 26.3 1600 275.2

span 3348 -67.5 6400 -44.1

D 3348 18.6 1600 194.6

At service: Pav = 111.6 KN n = 30 Table 2 : Calculations of equivalent loads due to transverse tendons

A 3348 18.6 1600 194.6

n. Pav (KN) a (mm) s (mm) w (KN/m)

2.4

span 3348 -67.5 6400 -44.1

B 3348 26.3 1600 275.2

B 3348 30 2000 200.9

span 3348 -120 8000 -50.2

Stresses Calculations: ft =

P P.e M A M S − + + Ac Zt Zt Zt

fb =

P P.e M A M S + A c Zb Zb Zb

Where: MA = moment due live and dead load. MS = moment due secondary effect. Zt = top section modulus. Zb = bottom section modulus. Ac = 7.0 X 0.25 X 106 =1.75 X 106 mm2 As the section being considered is rectangular and symmetrical abut about the centroid, Zt and Zb are equal. Zt =Zb = Z = bh2/6 = 7.29 X 107mm3

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2.4.1

Allowable stresses:

Maximum allowable concrete compressive and tensile stresses for floor with banded tendon are given in BS 8110, part 1, section 4.3.4.2 and 4.3.4.3. -

Allowable Stresses at transfer: Compression = 0.5 f ci = 0.5 × 25 = 12.5 MPa Tension = 0.36 f ci = 0.36 × 25 = 1.8 MPa

-

Allowable Stresses at service: Compression = 0.4 f cu = 0.4 × 50 = 20 MPa Tension: tensile stresses calculated to limit the crack width see table 3.

Table 3 : Tensile stresses for class 3

Limiting crack width (mm) Grouted post-tension tendons

2.4.2

0.1 0.2

Design stresses for concrete grade 40 N/mm2 50 and over 30 N/mm2 N/mm2 3.2 4.1 4.8 3.8 5.0 5.8

Applied loads and moment:

Own weight only: wo,wt = 6 X 7 = 42.0 KN/m Figure 6 : moment digram for own weight

Service load: wservice = (6 + 4) X 7+ 3 X 7 = 70 KN/m + 80 KN/m

Figure 7 : moment diagram for service load

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2.4.3

Stresses at transfer:

Po = 30 X 125.55 = 3766.5 KN Table 4 : Stresses at transver for the transverse direction

ZONE

FIBER

e(mm)

STRESS DUE PRESTRESS (MPa)

A

top bottom top bottom top bottom top bottom top bottom top bottom top bottom top bottom top bottom

0 0 80 80 -70 -70 -70 -70 80 80 -70 -70 -70 -70 80 80 0 0

-2.152 -2.152 1.98 -6.29 -5.77 1.47 -5.77 1.47 1.98 -6.29 -5.77 1.47 -5.77 1.47 1.98 -6.29 -2.152 -2.152

AB(sgging) B B BC(sgging) C C CD(sgging) D

STRESS DUE TO SELF WEIGHT (MPa) 3.00 -3.00 -1.54 1.54 3.13 -3.13 4.79 -4.79 -2.41 2.41 2.41 -2.41 3.13 -3.13 -1.54 1.54 3.00 -3.00

TOTAL STRESS (MPa)

ALLAWABLE STRESS (MPa)

STATE

0.85 -5.152 0.44 -4.75 -2.64 -1.66 -0.98 -3.32 -0.43 -3.88 -3.66 -0.94 -2.64 -1.66 0.44 -4.75 0.85 -5.15

1.8 -12.5 -12.5 1.8 1.8 -12.5 1.8 -12.5 -12.5 1.8 1.8 -12.5 1.8 -12.5 -12.5 1.8 1.8 -12.5

OK OK OK OK OK OK OK OK OK OK OK OK OK OK OK OK OK OK

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2.4.4

Stresses at service:

Po = 30 X 111.6 = 3348 KN Table 5 : Stresses after all stresses for the transverse direction

ZONE

FIBER

e(mm)

STRESS DUE PRESTRESS (MPa)

A

top bottom top bottom top bottom top bottom top bottom top bottom top bottom top bottom top bottom

0 0 80 80 -70 -70 -70 -70 80 80 -70 -70 -70 -70 80 80 0 0

-1.913 -1.913 1.76 -5.59 -5.12 1.30 -5.12 1.30 1.76 -5.59 -5.12 1.30 -5.12 1.3 1.76 -5.59 -1.913 -1.913

AB(sgging) B B BC(sgging) C C CD(sgging) D

2.5

Ultimate limit state:

2.5.1

Determination of hyperstatic action:

STRESS DUE TO SELF WEIGHT (MPa) 6.55 -6.55 -3.35 3.35 6.76 -6.76 10.38 -10.38 -5.23 5.23 10.38 -10.38 6.76 -6.76 -3.35 3.35 6.55 -6.55

TOTAL STRESS (MPa)

ALLAWABLE STRESS (MPa)

STATE

4.64 -8.46 -1.59 -2.24 1.64 -5.46

5.8 -20.0 -20.0 4.8 5.8 -20.0 5.8 -20.0 -20.0 4.8 5.8 -20.0 5.8 -20.0 -20.0 4.8 5.8 -20.0

OK OK OK OK OK OK OK OK OK OK OK OK OK OK OK OK OK OK

-9.08 -3.47 -0.36 5.26 -9.08 1.64 -5.46 -1.59 -2.24 4.64 -8.46

Hyperstatic moments can be calculated either directly or indirectly (Alami, 1998b) for skeletal , such as beams and floor systems that are modeled as strips of isolated slab frames , hyperstatic actions can be successfully calculated using both methods. In this example we will use the indirect method (covenantal method) in calculation of hyperstatic action. Indirect method is based on the following relationship: Mhyp = Mbal – P.e Where: e = eccentricity of post-tensioning with respect to the neutral axis of the section (positive if CGS is above the neutral axes otherwise negative). Mhyp =hyperstatic moment. Mbal = balanced moment due to balanced load. P = post-tensioning force “positive” 9 www.jea.org.jo   

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Balanced loads: As described before: Wbal =

8 a n Pav s2 Figure 8 : moment diagram for balance load

-

At A: Mbal= 232 KN.m , Pe = 3348 KN , e = 0.0 mm Mhyp= (232) + (3348 X 0 / 1000) = 232 KN.m

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At center of span AB: Mbal= -117.7 KN.m , Pe = 3348 KN , e = 45-125 = -80 mm Mhyp= (-117.7) + (3348 X 80 / 1000) = 150.14 KN.m

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At left of B: Mbal= 238.1 KN.m , Pe = 3348 KN , e = -70 mm Mhyp= (238.1) - (3348 X 70 / 1000) = 3.74 KN.m

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At right of B: Mhyp= (417.8) - (3348 X 70 / 1000) = 183.44 KN.m

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At center of span B-C: Mhyp= (-209.7) + (3348 X 80 / 1000) = 58.14 KN.m

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At left of span C: Mhyp= (417.8) - (3348 X 70 / 1000) = 183.44 KN.m

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At right of C: Mhyp= 3.74 KN.m

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At center of C-D: Mhyp= 150.14 KN.m

-

At D: Mhyp = 232 KN.m

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Figure 9 : Moment diagram for hiperstatic

Figure 10 : Moment diagram for dead load

Figure 11 : Moment diagram for live load

2.5.2

Total ultimate moment: The load combination for ultimate strength design: MU = 1.4 MD.L + 1.6 ML.L + 1.0 Mhyp

Figure 12 : Moment diagram for ultimate load pear

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2.5.3

Design section: Section analyses may be carried out in accordance with clause 3.7 of BS 8110, part 1, the moment capacity of the section can calculated using following equation. Mu = fPb Aps (d- dn) Where: Mu = design moment of resistance of the section. fPb = design tensile strength in the tendon. Aps = area of prestressing tendon in the tension zone. d = effective depth to the centroid of the steel area. dn = depth of centroid of the compression zone = 0.45x. x = depth of neutral axis. b = effective width of the section. - When d = 125 mm. f pu Aps 1860 × 30 × 100 = = 0.127 f cu bd 50 × 7000 × 125

1116 = 0.6 f pk 1860 From table 4.4 BS 8110. Sec 4.3.7.3 finds value of fPb and x. fPb = 1731.7 MPa x = 36.25 mm dn = 0.45 x 36.25 = 16.3 mm Mu = -1731.7 x 30 x 100 x (125-16.3) = -564 KN.m f pe

=

When d = 205 mm. f pu Aps 1860 × 30 ×100 = = 0.078 f cu bd 50 × 7000 × 205 f pe 1116 = = 0.6 f pk 1860 fPb = 1767 MPa x = 35.9 mm dn = 0.45 x 35.9 = 16.14 mm Mu = 1767 x 30 x 100 x (205-16.14) = 1001 KN.m - When d = 195 mm. f pu Aps 1860 × 30 ×100 = 0.08 = f cu bd 50 × 7000 ×195 f pe 1116 = = 0.6 f pk 1860 fPb = -1767 MPa x = 35.1 mm dn = 0.45 x 35.1 = 15.8 mm Mu = -1767 x 30 x 100 x (195-15.8) = -950 KN.m

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Engineers Training Center   Table 6 : Comparison between applied moment and moment capacity at ultimate limit state

ZONE A AB B B BC C C CD D

M (KN.m)/ (applied) -411 503.42 -708.62 -911.0 609.8 -911.0 -708.62 503.42 -411

Mu (KN.m)/ (capacity) -564 1001 -950 -950 1001 -950 -950 1001 -564

Because the moment capacity at ultimate limit state > applied moment, no un-tensioned reinforcement is required.

2.5.4

Minimum un-tensioned reinforcement requirement:

Reinforcement required = 0.075% Ac Ac = hb = 250 X 7000 = 1.75 X 106 mm2 As = 0.00075 X 1.75 X 106 = 1312.5 mm2 Use 7 T 16 = 1407 mm2 The reinforcement should extend into the span by 0.2 × span measured from the centerline of the column and the width of strip is the column breadth plus 3 times the slab depth as shown in figure 13.

Figure 13 : Un-tensioned reinforcement details

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2.6

Post-tension losses:

2.6.1 Short-term losses: 2.6.1.1 Friction losses:

Px = Po × e - μx(α +ϖ ) Where: Px = Force at distance x from stressed end. Po = 139.5 KN (Stressing force (at anchor)). μ = 0.2 (friction coefficient). α = angle change in tendon from anchor to point considered (radians). ω = 0.0085 (“wobble” factor (radians/m)) Total drape from figure.3 (refer to appendix A for profile calculation) -

Total drab for span A-B and span C-D: (18.6+26.3)/2 + 67.5 = 89.95 mm

-

Total drab for span B-C: (30+30)/2 + 120 = 150 mm

16 × total drape L2 (16 × 89.95 × 10 −3 ) α1 = α 3 = = 0.023 rad / m 82 (16 × 150 × 10 −3 ) α2 = = 0.024 rad / m 10 2

α=

PA = 139.5 KN PB = (139.5) e-(0.2)(8)(0.023+0.0085) = 132.64 KN PC = (132.64) e-(0.2)(10)(0.024+0.0085) = 124.3 KN PD = (124.3) e-(0.2)(8)(0.023+0.0085) = 118.2 KN

Pavg =

(139.5 + 132.64 + 124.3 + 118.2) = 128.66 KN 4

Po - Pavg = 139.5 – 128.66 = 10.84 KN

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2.6.1.2

Wedge set losses:

Because Length of the tendon 26 m < 30 m so use one live end. δPw = 2 L’ P’ Where: δPw = Force losses due draw-in. ΔE PS Α PS L' = (Length of the tendon affected by draw-in). P' (PA - PD ) (Slop of the force profile). P' = (L1 + L 2 + L 3 ) Δ = 6 mm (Wedge draw-in). Eps = Modulus of elasticity of the tendon. Aps = Area of the tendon. P' =

L' =

(139.5 - 118.2) = 0.82 KN (8 + 10 + 8) (6 X 10 -3 X 195 X 100) = 11.95 m < Length of the tendon 0.82

So δPw = 2 P’ L’ = 2 X 0.82 X 11.95 = 19.6 KN

2.6.1.3

Elastic shortening losses:

δPes = εes Eps Aps Where: δPes = Force losses due Elastic shortening. εes = 0.5 fco/ Eci fco = n Po /bh (stress in concrete adjacent to the tendon after transfer). Eci = Modulus of elasticity of concrete at time of transfer. fco = (5 X 139.5 X 103)/ (1500 X 250) = 1.86 MPa εes = 0.5 X1.86/ 2500 = 3.72 X 10-5 δPes = 3.72 X 10-5 X 195000 X 100 / 1000 = 0.725 KN

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2.6.2

Long-term losses:

2.6.2.1

Creep of concrete losses:

δPcr = εcr Eps Aps Where: δPcr = Force losses due creep. εcr = φ fco/ Eci φ = 2.9 (creep coefficient (BS 8110, part 2, figure.1)) εcr = 2.9 X 1.86 / 25000 = 2.16 X 10-4 δPcr = (2.16 X 10-4 X 195000 X 100) / (1000) = 4.21 KN 2.6.2.2

Shrinkage of concrete losses:

δPsh = εsh Eps Aps εsh = 300 X 10-6 δPcr = (300 X 10-6X 195000 X 100) / (1000) = 5.85 KN 2.6.2.3

Relaxation of the tendon losses:

δPr = 1000 hour relaxation value x relaxation factor x the pre-stressed force at transfer. Pi =139.5 -10.84 – 19.6 – 0.725 = 108.3 KN δPr = 0.035 X 1.5 X 108.3 = 5.69 KN 2.6.3

Summary for P.T losses:

-

Short-term losses: 1- Friction losses = 10.84 KN ………………….…… 7.7% 2- Wedge set losses = 4.48 KN ……………………... 3.2% 3- Elastic shortening losses = 0.725 KN …………..... 0.5%

Total short-term losses = 16.1 KN ………...………….... 11.5% -

long-term losses: 1- Creep of concrete losses = 4.21 KN ………….……. 3.0% 2- Shrinkage of concrete losses = 5.85 KN ……….….. 4.2% 3- Relaxation of the tendon losses = 5.69 KN …….….. 4.0%

Total long-term losses = 15.75 KN ………………….….. 11.2% Total losses = 31.85 KN ………………..……………….... 22.8% 16 www.jea.org.jo   

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Figure 14 : force profile for full-length longitudinal tendons

► APPENDIX A: Calculation of Tendon Geometry:

Figure 15 : Tendon geometry

jX + mX+n = 0 m = (p2-2L)(q1 – q2 ) + p1 (q3- q2) = (800 - 2 X 8000) (125-45) + 800 (195 - 45) = -1096000 n = (q1-q2)(L-P2)L = (125-45)(8000-800)8000 =4608X106 L’ =(-m ± √ [ m2 – 4 j n ] )/ 2 j = (1096000 ± 1577973.384 ) /-140 = -19094 , 3442.67 L’ = 3442.67 mm a1 = [(q1 – q2)P1]/ L’ = [80 X 800] / 3442.67 = 18.6 mm a2 = [(q3 – q2)P2]/ [L-L’]= [150 X 800] / 4557.33 = 26.33 mm - for span B-C: q1 = q3 = 195 mm, P1=1000 , L’= 5000 a4 = [195 - 45] X 1000 / 5000 = 30 mm

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