APPENDIX C Slab Calculation 2

Slab Calculation 2

Now try calculating the same slab, with the minimum recommended balanced load of 50% of self-weight. (As opposed to 50% of DL LL used in the first calculation). The same thickness of slab will be used, with the same layout

Given:-- 4 bays each way 8000 x 7000 Ult. Load factors 1.0 kPa finish

1.8 kPa partitions 2.5 kPa live load

Slab will be protected from weather Internal columns 500 x 500 External columns 300 x 500 Wind loads taken bv shear walls

DL 1.2 and 1.5 as SABS 0160-1989

LL 1.6

Floor to floor 3.0m

Assume 15.2mm Dia cables, low relaxation Breaking Strength 260.5 KN Area 140 sq. mm. E= 198 GPa Relaxation 1.5% at 1000 hrs

CALCULATION

Assume a 210 thick slab.

Balance 50% of Permanent Load 8.05kPa /2for internal spans

For external 8M spans, balance same proportion of load as previously, i.e 80/66x 0.5 =.606 Permanent load From the formula, the slab should be somewhat thicker than 230mm, but take 210 as minimum practicable.

COMMENT

See diagrams in first part of calculation Formula would give 230 slab then self DL Total perm = Service (1.1) Live load TOTAL 5.25 kPa 8.05 8.85 2.50 11.35 Factored 9.66 4.0 13.66 revised DL 12 factor to allow for extra shear in first int. Check for shear as before, 210 is a bit thin, but acceptable with capitals

As contractor has column shutters with capitals, use capitals:

For analysis assume E of slab = code value: i.e. (20 + 0.2 x 30) = 26.0 GPa

From the cross-sections, take the effective top cover as 47mm to C/L cable for 7m span, and (63 + 31)/2 = 47mm for 8 m span. As bottom cover is 41 for 8 m span, effective drape - 210 - 88mm = 122 mm.

For 7m span, drape is 210 - 98 =112 mm. (See Diag. Calculation 1)

From the diagrams of the cable shapes in the end spans (the cables in centre spans will have the same positions as at the ends of outer spans), the properties of the parabolas may be deduced, and then the losses due to curvature. Referring to the formula in Appendix A, the following table may be calculated, as before.

Span 8.0 end 8.0 cent 7.0 end 7.0 cent a

l

0.4 0.4 0.35 0.35 a2 0.4 0.4 0.35 0.35 0.1., 0.163 0.13 0.163 b2 .041 .041 .051 .051 b3 0.163 0.163 0.163 0.163 L 8.0 8.0 7.0 7.0 1 0.0330 0.0 .0330 0.0 m 1.34 1.854 1.039 149 n -5.411 -7.418 -3.677 -5.214 X 3.7018 4.0 3.2114 3.5 cl .0096 .0122 .0086 .0112 c2 .0114 .0122 .0103 .0112 Drape .0944 .1098 .0853 .1008

CALCULATION

Loads to be Balanced:—

For 8m span, balance 0.606 x permanent load in external span: For normal external bay 7 x 8.05 x .606 = 34.15 kN/m For 8 m internal span, balance 0.5 of permanent load i.e. 7 x 8.05 x 5 = 28.17 kN/m

For 7m span, balance 0.50x permanent load in external & internal span. For internal and external bay, 8 x 8.05 x .5 = 32.20kN

COMMENT

See page 1 For first internal bay, increase cables to allow for increased load at first internal support

CALCULATION

~rom the loads to be balanced, the required prestress force = H > I2

8xDrape

This is the final force after

COMMENT

internal spans would have smaller prestress force for same losses, and assuming a loss of say 18% for 8m spans, and 16% for 7m spans, the initial prestress can be calculated, load balanced. The angle which the cable 'rotates' through is also required for the friction calculation. They are calculated from the because of formula for a parabola. larger drape. e.g. for the 8m end span O = Arctan (2 x 78.4/3268.4) x 2 + Arctan( 2 x 113.5/3931.6) x 2 = .2113 Radians

The following table shows the calculation results.

Span Load Drape Eff. Length Angle to C/L Angle C/L to Total angle (rad) end (Cumul) 8 m e n d span 34.15 .0944 7.2 .09615 .1135 .20965 8m 2nd span 28.17 .1098 7.2 .122 .122 .45365 8m 3rd span -do- -do- -do- -do- -do- .69765 8m last span as 1st .9073 rad 7m 1st span 32.2 .0853 6.3 .09838 .1182 .21658 7m 2nd span 32.2 .1008 6.3 .128 .128 .4726 7m 3rd span -do- -do- .7286 7m 4th span as 1st span .9452 rad

factor = e-

0025

*

32

-° - -

06

* -

9073

= .8742

i.e. 12.58 % losses 7m spans

The total loss due to friction and wobble is i.e. 11.9 % losses.

factor = e-'

0025 x 28

-° - -

06 x a9452

= .8810

CALCULATION

8 m spans: end span

The prestress required to balance 59 kN/m with a drape of .0951, and an effective length of parabola of 8 - 0.8

34.15

x 12

2

7.2 m is given by = 2344 kN

8 x .0944

allowing an estimated 18% losses, P—, = 2859 kN

If cables of 15.2mm dia (ult. strength 260 kN) are stressed to 80% of ult. the no. of cables required

2859

0.8 x 260

= 13.7 say 14 cables

COMMENT

Although one should calculate the losses from the exponential equation, it is quite acceptable to calculate the loss at the end, and interpolate. An even number is better if stressed from both ends. 8m spans: 2nd span

The prestress required to balance 28.17 kN with drape of .1122 =

28.17 x 7.2

2

= 1627 kN

8 x .1122

Allowing the same losses, Pinit( = 1984 kN. and no of cables = ,

1984

208

= 9.5 cables

say 10.

Assume that the cables in the end span are taken to 1.5 m past the 1st support, and that half the cables are stressed from each end, the diagrams on the next page may be drawn. Although strictly one should calculate the losses from the exponential equation it is quite acceptable to use a linear method.

An even number is better if stressed from both ends. Losses:

8 m spans

7 m Spans Effects of friction

CALCULATION

For 7m external span, force to balance 32.20 kN/m, with drape of .0853, and effective length of 6.3m.=1873 kN With 16% losses , no of cables = 10.8 say 12

For internal span, force to balance34.15 kN/m with drape of .1008, Pfma] = 1585 kN -> 1887 kN before losses

No. of cables = 9.0 say 10

Take losses due to friction and pull-in as before (see first calculation) The results are diagrammed on pages 5 and 6

COMMENT

Diagrams such as shown make it easy to know what the force at any section is.

Other Losses: Elastic Shortening 8m spans

Average stress in end 9.5 m =(2590+2699)7(7 x .21 x 2)= 1.80 MPa

In Centre part ((1927+1949)/2 x3.52)+ (1949)x2.98)/6.5( 7 x .21) = 1.32 MPa Losses in end span =1.80 x 1987(26 x 2) = 6.85 MPa

Average loss over whole span = (1.80 x 9.5 x 2 + 1.32 x 13)/(2 x 32m)x 198726 = 6.11 MPa Loss per cable = 140 x 10"6 x 6.85 x 1000 kN = .96kN in end spans (x 4 = 3.8kN)

= " x6.11 x " = 0.86 kN in centre spans (x 10=8.6kN) Total in end spans 12.4 kN

Because the cables are stressed one at a time, the average loss is 0.5 times the Ratios of E moduli x stress 7m spans

Average stress whole length =((2223 + 2313.6)x 8.5+(1925+1956)/2 x 2.87 +1956 x 1.63) /(14x8x.21)= 1.19 MPa In end average force =((2223 + 2313.6)/2=2268.2—>1.35MPa

Losses 10 cables per cable = 1.19x 1987(26 x 2) x 140 x 10-6 x 1000 = .63 kN

Loss in 2 cables = 1.35x 140x 10'6x 198726/2=0.72 kN 10 x .63 + 2 x .72 = 7.8kN total

Total in centre span = 6.3 kN

CALCULATION

Long term losses

Relaxation 1.5% x 2 = 3%

For 8m spans: end span 3% x (2578+2687)/2 kN = 79 kN

centre 3% x( (1918+1940)/2x3.52 +(1940 x 2.98/6.5))=1934kN x 3%=58kN For 7m spans: end span 3% x 2252 = 67.6kN

centre 3% x 1950 = 58.5 kN Creep. (Same creep factor as 1st calc)

If stressed at 3 days (common for prestressed flat slabs) and with a humidity ol 45%, the creep for a 210 slab, interpolating between 150 and 300 thick slabs given in the code (BS8110) gives a creep factor of 3.7

Then loss = stress in concrete x ratio of moduli x 3.7

8m spans: End Section conc, stress = (2.578 MN + 2.687)/(2*7*.21) = 1.79MPa Loss of steel stress = 1.79 * 3.7 * 198/26 = 50.5MPa

Per cable 50.5 x .140(Area) = 7.06 kN

Centre Section( (1.918+1.94)x 3.52+1.94x 2.98)/6.5/(7 x .21) x 198/26 x 3.7 = 37.1 MPa Per cable 37. 1x .14 = 5.19 kN Therefore average loss in long cables

=(5.19x6.5+7.06x9.5)/16m = 6.30 kN Total creep loss: End 7.06 x 4 + 6.3 x 10 = 91.2 kN

Centre 6.3x10 = 63kN

7m spans End Section Conc stress after elastic losses(2215.1 + 2305.8)/(2x 8 x.21)= 1.345MPa Centre ((1918.7 + 1947.7)/2 x 2.87 + 1949.7x 1.63) /(4.5 x 8 x 2.1)= 1.154 MPa Average(1.154 x 5.5 + 1.345 x 8.5)/14 = 1.27MPa

Loss of steel stress : end 1.345 x 3.7 x 198/26= 37.9MPa Per cable 37.9 x .14 = 5.306 kN

Centre loss of steel stress 1.27 x 3.7 x 198/26 = 35.78MPa loss/cable = 35.7 x .14 = 5.0kN

Losses in centre 10 x 5.0 = 50kN

Losses in end section 12 x 5.306 = 63.7 kN

7m spans Relaxation end span 2 cables 3% x (373.4 + 388.6V2 = 11.43kN Total 10 cables (1956x 11.37 + 1950 x 2.63)/14--> 1912.8 kN 1912.8 x 3% = 57.4 kN

Total for 12 cables = 68.8 kN Shrinkage as before: for 1 cable 10.26 kN

8m spans 14 cables —>143.6 kN 10 cables —> 102.6 kN 7m spans 12 cables —> 123.1 kN

10 cables —>102.6 kN

COMMENT

The code gives a factor of 2.0

on the 1000 hour value for class 1 BS5896 Varies with initial prestress, temp etc. Should strictly be taken as stress at the centroid of the cables. Total loss Cause No of cables Initial force (208kN/cable) Elastic Relaxation Shrinkage Creep TOTAL

Percentage of init. force loss after friction loss %Loss due to friction Total % loss of initial

8m End span 14 2912 kN 12.4 kN 79 kN 143.6 91.2 321.2 11.1 11.4 22.5 8m Centre 10 2080 8.6 58 102.6 63 232 11.1 7.45 18.5 7m End span 12 2496 7.8 67.6 123.1 63.7 262 10.5 11.08 21.6 7m Center 10 2080 6.3 58.5 102.6 50 217.4 10.4 8.31 18.7

The losses are somewhat lower than with the higher prestress but still appreciably higher than the 16% (or less) assumed by some commercial designers.

Position 8m end span 8m 2nd span

7m end span 7m 2nd span

Initial prestress (after elastic losses) 2578 kN 2669 2687 1918 1940 1940 2215 2289.5 2306 1919 1948 Loss 311 311 223 223 255.6 -do-210 210 Final prestress 2267 kN 2358 2376 1695 1717 1717 1959 2034 2050 1709 1738 ANALYSIS

The slabs are analysed by the equivalent frame method, using Long's method to calculate the equivalent stiffness of the columns. Three loading cases are needed: Dead load on all spans, Line load on even spans, and live load on odd spans.

These may be combined together with the load factors to give the desired bending moments diagrams.

Any method of analysis may be used: moment distribution if a computer is not available, or a frame analysis if one is available. The hogging moments for the ultimate limit state may then be reduced by 15%, and the sagging moments increased accordingly to maintain equilibrium.

The design moment is at the face of the column or capital, but the the total statically required moment is: W(L-2D/3) /8

If a computer program is used, there is an advantage in arranging a node at 1/3 of the column or capital dimension from the centreline of the column, as the moment at that point will not be less than the moment given for statically required moment.

Loads (As for previous calculation) LOAD Finishes Partitions Self. Wt Total DL Live load Service (1.1DL + LL) 1.l0kPa 1.98 5.77 8.85 2.50 Ultimate (1.2DL + 1.6LL) 1.20kPa 2.16 6.30 9.66 4.00

Ultimate Dead Load (1.5 DL)

l.SkPa

2.7 kPa

7.87kPa 12.07kPa

Total loads on spans 8m Spans DL 8m spans LL 7m spans DL 7m spans LL Service 61.95kn/m 17.5 70.8 20.0 Ultimate 67.62kN/m 28.0 77.28 32.0 Self only 36.75 36.75 42.0 42.0

NOTE: As the Ultimate Deadload case of 1.5 DL at 12.07 kPa is less than the Ultimate DL + LL of 13.66 kPa, it may be effectively ignored, although there may be some places where the moments could be fractionally higher. A construction loading at initial prestress may need to be calculated if the propping is not adequately arranged.

Loads due to prestress

For a parabola, the equivalent uniform load caused by a tension in the cable is given by w L2 /8 = P h where h is the drape

for the cables. The drapes can be read from the table on page 1

The loads will not be uniform, but will be trapezoidal if the variation of prestress along the beam is taken into account

In addition the moments due to eccentricity at the ends must be taken into account in the analysis if the cables are not exactly central. In our case we have assumed a 20mm eccentricity upwards at the end'

The moment at the end of the 8m spans will be 2267 x .02 =56.7 kNm The moment at the end of the 7m spans will be 1959 x .02 = 49.0kNm

The loads given are for final prestress. Initial prestress (after friction and elastic losses) combined with deadload only may be a critical state, but it is probably sufficient to take the stresses due to final prestress and multiply them by an average factor.

Note: The forces at intermediate points are interpolated Position 8m 1st. span L end 8m 1st start of sag 8m 1st . end of sag 8m 1st span RH end 8m 2nd span L end -do- end of cable -do- beyond end 8m 2nd RH end 7m 1st span LH 7m 1st: start of sag 7m 1st end of sag 7m 1st RH 7m 2nd LH end 7m 2nd end of cable 7m 2nd beyond end cable 7m 2nd Rh end Prestress 2267 2272 2353 2358 2354 2376 1717 1717 1959 1966 2026 2034 2034 2050 1709 1738 Drape .0096 .0944 .0944 .0114 .0122 .1098 .1098 .0122 .0086 .0853 .0853 .0103 .0122 .1008 .1008 .0122 Length 0.8 7.2 7.2 0.8 0.8 7.2 7.2 0.8 0.7 6.3 6.3 0.7 0.7 6.3 6.3 0.7 Lateral load 272.0 kN/m down 33.1kN/m up 34.28 -do-336.0 kN/m down 359.6 down 40.26 up 29.1 up 261.8 kN/m down 275.1 down 33.80 up 34.83 up 342.0 down 405.1 down 41.65 up 34.72 up 346.2 kN/m down

272 kN/m

336 359.6

261.6

0.4

Loads on 8m spans due to final prestress

275.1

kN/m

342 405.1

346.2

Loads on 7m spans due to f i n a l prestress

Columns: Stiffness by Long's method (See Sample Calculation 1) 7m spans Exterior: E (30 MPa)=26GPa

fe =.5 x .33 /12 = 1.125 x 10-3 m units

Kc = 4EI/LC

Kc = Kc /(1+.1272KC L/EhJc)

where L=8m, LC =3, h=.21. and c= .3

Then equivalent stiffness = stiffness x .645 orequiv. I = 725 x 10"6

8m spans interior : Ic = .54/12 = 5.208 x 10-3 , k = .0564

c= 0.5

Then equiv. stiff = stiff x .596 Equiv I = 3.10 x 10-3

7m exterior Ic = .3 x -53/12 = :- .125 x 10-3

c=0.5 k=.1272

Then equiv stiff = stiff x .555 Equiv I = 1.73 x 10-3

7m interior I = 5.20?, x 10-3 . c=.5. k= .0564

Then equiv. stiff = stiff x .628 Equiv I = 3.27 x 1-3

Slab Stiffness The moment of Inertia of the 8m spans is 7 x .213 /12 = .0054 m

The moment of inertia of the 7m spans is 8 x .213 /12 = .00617 m

The E is taken as the same as the columns, and the creep factor (for calculation deflections due to long term loads) as 3.5 The results of the analysis are given below: Because the structure is symmetrical, only the first two spans are shown.

It should be realised that as the loads are calculated for an interior span, the first interior column band should be reinforced for approximately a 7% greater load.

i.e. for this design, 1 extra prestressing cable, and some additional reinforcement over the columns. MOMENTS Position 8m 1st span Left 8m 1st span centre 8m 1st span right 8m 2nd span left 8m 2nd span centre 8m 2nd span right 7m 1st span left 7m 1 st span centre 7m 1st span right 7m 2nd span left 7m 2nd span centre 7m 2nd span right ULT (1.2DL +1.6L1) -22.9 kNm 367.0 -557.3 -476.8 215.5 -343.1 -40.8 293.7 -445.4 -400.9 185.9 -287.3 ULT (adjusted 15%) -14.1 403.1 -441.7 -378.4 264.1 -264.7 -25.6 326.5 -352.0 -314.1 225.9 -217.5 Service (1.1DL +LL) -17.0 301.4 -457.9 -396.2 170.9 -285.1 -31.1 240.8 -369.9 -333.1 146.6 -238.7 DL self only - 6.0 134.4 -211.6 -183.3 67.5 -131.9 -10.5 106.8 -170.8 -154.1 56.8 -110.4 Final Prestress +63.4 -99.6 167.9 148.1 -54.1 105.7 48.8 -69.7 125.1 118.6 -49.7 98.0 DEFLECTIONS (allowing 3.5 creep factor on permanent loads)

Assuming an uncracked section (see later for correction) Position 8m 1st span 8m 2nd span 7m 1st span 7m 2nd span Service Load (1.1 DL +LL) 28.7 mm 9.0 mm 14.9 mm 5.0 mm Final Prestress -10.7 mm - 3.7 mm - 4.6 mm - 2.0 mm Net Deflection 18.0 mm 6.3 mm 10.3 mm 3.0 mm

Serviceability Limit State

For controlling cracking, it has been traditional to limit tensile stresses. For the sake of completeness, this will be done, but the incremental stress method is considered better.. (Use the program described in Appendix E) The permissible tensile stress is given as 0.45Ö fcor 2.46 MPa

4.O4 Mfti

\^

\

T = 4.O4x75.2/2 \ =151.9 kN ^ 4.P4 \ / 1 x 210 mm/ 4.04 -i- 7.24 / / = 75.2 mm / / \ t 752. mm » h = 21O mm \• V^ 4.0 7.24 MRi

Serviceability Tensile stress Position 8m 1st L 8m 1st C 8m 1st R 8m 2nd L 8m 2nd C 8m 2nd R 7m 1st L 7m 1st C 7m 1st R 7m 2nd L 7m 2nd C 7m 2nd R Net moment (service-prestress) 46.4 201.8 -290.0 -248.1 116.8 179.4 17.7 172.1 -244.8 -214.5 96.9 -140.7 Comp. kN 2267 2358 2376 2376 1717 1717 1959 2034 2050 2050 1709 1738 Stress (moment) ±0.90 3.93 5.64 4.83 2.27 3.49 0.30 2.93 4.16 3.65 1.65 2.39 Stress (comp) 1.54 1.57 1.60 1.60 1.17 1.17 1.17 1.19 1.21 1.21 1.02 1.03 Net tensile stress (MPa) 2.36 4.04* 3.23* 1.10 2.32 -1.74 2.95* 2. 44 0.63 1.36 Z of section (M units) .0514 .0588 Slab Area (M units) 1.47 1.68

It may be seen that 3 of the stresses are more than the Report 25 stresses, and reinforcement is required. The tension to be taken by reinforcement is calculated by simply taking the force from the tensile stress diagram as in the sketch at a stress of 0.58 fv It should be

noted that it is assumed that the moments are evenly distributed across the section. This is clearly incorrect. 151.9kN /(.58 x 450) ->582sq.mm/m ->Y12 @ 180

To calculate the steel required to control crackwidth by a more logical method, the formula in BS8007 is used. Moments are taken about the level of the tension steel reinforcement, i.e.

The required area of steel is given by calculating the tension required for a moment of M + P(d-h/2), and then reducing the tension by P. i.e. transferring the compressive force to the reinforcement level.. 75% of the total hogging moment and 55% of the total sagging moment should be taken in the column band for the purpose of calculating crackwidths

The following table was calculated using a computer program for crackwidths to BS8007 It may be seen that areas are similar to the areas required for ultimate load, which supports the Report 25 stresses. SABS 0100 seems to give slightly higher areas if the tensile stress in the concrete is taken into account. Hogging moments are taken as 75% of total in column bands, and 55% for sagging moments

The result is slightly more reinforcement in the position which requires more than nominal reinforcement

Required area of reinforcement for cracking (0.15% = 315 sq.mm/m) Position 8m 1st L 8m 1st C 8m 1st R 8m 2nd L 8m 2nd C 8m 2nd R 7m 1st L 7m 1st C 7m 1st R 7m 2nd L 7m 2nd C 7m 2nd R Net. Moment (service-prestress) 46.4 kNm 201.8 -290.0 -248.1 116.8 -179.4 -17.7 172.1 -244.8 -214.5 96.9 -140.7 Moment /m in band 9.9 31.7 62.! -18.3 38.4 -23.6 45.9 -13.3 26.4 Prestress MPa 1.54 1.57 1.60 1.17 1.17 1.17 1.19 1.21 1.21 1.02 1.03 Steel stress MPa -238 246 360 271 Bar Dia. 10@300 16 @ 200 12@ 120 nom 10@250 12@180 nom d 180 166 Area (sq.mm/m) nil nil 992 946 nil Nominal(297) nil Nominal 625 Nominal Nominal (Calculated using SABS0100 with 0.2mm crackwidth. See Appendix E) The areas are slightly greater than for the Report 25 Calculation (e.g Y16 at 200 or Y12 @ 1?0 as compared with Y12 at 180)

Ultimate Load Limit State.

15% reduction of hogging moments ( with corresponding increase of sagging moments) is allowed and has been taken.

The decompression moment, equal to the prestress x Z, is the moment required to reduce the compression on the extreme fibre to zero. If the applied moment (Ultimate dead + Live + prestress) is less than the decompression moment, no reinf. is required. If it is greater, an equivalent moment M' is calculated, by displacing the prestress to the level of the reinforcement, and adding a moment of P(d-h/2). The tension calculated from this moment then has the prestress force deducted, to obtain the net tension on the reinforcement M7Jd-P. The required steel area, is then (M;/Jd-P)/fs

exceeds this. The lever arm as a proportion of effective depth is given approximately by

d-h/2 = 65mm for the 8m spans, and 49mm for the 7m spans.

The approximate moment /metre taken by Nominal reinforcement is 391 MPa x 0.75 d x .15% x .21m= Required Area of reinforcement for Ultimate Limit State

16.8 kNmford=.182 mm 15.3kNm/m for d=. 166 Position 8m 1st L 8m 1st C 8m 1st R 8m 2nd L 8m 2nd C 8m 2nd R 7m 1st L 7m 1st C 7m 1st R 7m 2nd L 7m 2nd C 7m 2nd R Net. Moment (Ult-prestress) 49.3 303.5 -273.8 -230.3 210.0 -159.0 -23.2 256.8 -226.9 176.2 -119.5 Net Moment /m in band (A) 10.6 47.7 -58.7 33.0 -34.1 4.3 35.3 -42.5 24.2 -22.4 Mom. /m P(d-h/2) (B) 21.0 21.4 21.8 16.0 16.0 '2.0 12.2 12.4 10.5 10.6 DecomMom/m Prestress x Z 11.3 8.6 Design Moment/m (A+B) — 69.1 80.5 49.0 50.1 — 47.5 54.9 34.7 33.0 Net Tension kN/m — 116.6 155.4 47.1 105.4 Bar Dia & spacing Y10@250 Y10@250 Y10@200 Y10@250 Area (sq.mm/m) Nom 298 398 Nom 121 Nom Nom 270 Nom Nom

The reinforcement areas required for ultimate load by this method are a bit smaller than those required for crack control Compare the above calculation with the Report 25 method of calculating the reinforcement required for ultimate

moments:-This method does not take account of the moments due to prestress, but effectively assumes the cables are bonded, and that the moments are taken over the full width of the section.

At the first interior support, of the 8m spans

f, = 2.358lMN/( 14 x 140. x 10-6) =1203.1 MPa (Effective pe prestress) f pu = 260.5 MN/140. x 10-6=1860.7 MPa -. Aps = 14x 140 x10-6 =l.96x 10-6

Effective depth of cables =.163m (see page 3 of calcs) From equation 52 of BS8110

for l/d =16/0.163 =98, f pb = 1203.1 + 61.2= 1264 MPa

0.7 f = 1302 MPa therefore use 1264MPa

For 7m end spans , 1=28/2=14, and for 7m internal spans, 1=28/3=9.33 For prestress only, Tult = 2.478 MN

Then N.A. depth = 2.478/(7 x .45 x 30) = 26.2 mm and lever arm = .163 -.026/2 = .150 m

M.R. Prestress = .150x 2.478= 371.4 kNm

Actual ultimate moment 441.7kNm. Therefore additional steel is required. dr e i n f=.182and dcable = .163

Say approximately Tensionreinff= (441.7- 371.4)7371.4 x 2478kN = 469

At 391 MPa (450/1.15), the area required = 1392 sq.mm over 3.5 m width = 343 sq. mm/m (Y10 @ 220 ) Try Y10 @ 200 : Area = 1374 sq.mm Tensile capacity =537 kN

See diagram, Tension = 2478+537 kN = 3015kN. N.A depth = 3015/C7 - .45 x 30)=31.9mm MR =537x .16 6+ 2478 x .147 = 453kNm O.K.

The other positions are checked in the table below 0.15% =315sqmm/m-->Y10 @ 250

For end 8m spans (SABS 0100) 1=32/2=16 For internal spans, 1= 32/3 = 10.67 (for 8m int 1295MPa, for 7m end 1274. for 7m internal f b = 1302MPa)

Position 8m 1st L 8m 1st C 8m 1st R 8m 2nd C Ult. M -14.1 403.1 -441.5 264.1 fpb x!40. x 10-6 x No. cables 2.478 MN -do-1.813 MN Eff.d .130 .163 -do-N.A. depth 27mm -do-19.3mm Lever arm 116 149 153 M.R. 287 371 371 279

DM

0 32 70.5 0 As reqd. 0 (.15% min) Nom Y10 @200 Nominal

Position 8m 2nd R 7m 1st L 7m 1st C 7m 1st R 7m 2nd C 7m 2nd R Ult. M -264.7 - 25.6 326.5 -352.0 225.9 -217.5 fpb x!40. x 10-6 x No. cables 1.813 MN 2.138 MN -do-1.813MN -do-Eff.d -do-.130 .163 -do-N.A. depth 19.8mm -do-16.8mm Lever arm 120 153 154.6 M.R. 268 257 327.3 -do-282

-do-DM

0 25 0 As reqd. Y10 @250 Y10@250 Y10@250

The reinforcement required for ultimate moments by this method are slightly less than required for cracking, and slightly less than the reinforcement required by the suggested method..

Ultimate MR of section = 537x .166 + 2473 x .147= 453 kNm

It is also desirable to check the ultimate load case with half of the cables in the end span removed, with DL + .25 LL. in the end spans only. The MR due to half the number of cables may be taken as half the M.R. in the table above if Report 25 method is used, or the moments calculated for prestress may be halved and deducted from the moments for DL + .25LL

The second method will be used Position 8m 1st L 8m 1st C 8m 1st R 7m 1st L 7m 1st C 7m 1st R Ult. M (DL + .25LL) - 8.2 kNm 172.8 -267.9 - 15.4 137.6 -190.0 PSMom/2 36.7 -49.8 83.9 27.0 -24.4 62.6 Moment/m PS (d-h/2)/2 10.7 10.7 4.3 4.3 4.3 Ult Mom/m 27.1 57.4 2.9 18.9 35.6 Net. M/m for tension 37.8 68.1 7.2 23.2 39.9 Required Area (per m col band)

-169. sq mm 622. sq mm -355. sq mm Norn Y12 at 180 Nom

It may be seen that the reinforcement for the 1.2DL + 1.6LL ultimate load case is adequate everywhere except the first interior support of the 8m spans

Deflections:

The deflections in the table above were calculated for a non-cracked slab

If the sections are cracked, the Moment of Inertia would be 2.56 x 10-3' instead of the uncracked value of 5.4 x 10-3

Using the ACI formula , and taking moment at first cracking as the moment at which the tensile stress reaches 2MPa, the

cracking moment in the 8m 1st span is (1.54+ 2) x 7 x .212/6 = 182 kNm. Taking the design moments as the serviceability moments, and taking the moments due to prestress into account, the moment at midspan of the 8 m 1st span is 201.8 kNm. i.e. cracked, and at the first internal support

290 kNm

Then Ie midspan = 4.13 xl0- 3

I Support = 3.26 x 10-3 using the service load moments calculated above

The actual moment over the support is more concentrated, and there may be more cracking. The net equivalent M. of I. is then.85 x 4.13 + .15 x 3.26 =4.0

The calculated deflection will then be 5.4/4.0 x the calculated one.

Now the calculated deflection assumes that the slab acts as a band spanning in one direction, and neglects the span in the other direction. If one adds half the calculated deflection in the short direction to the calculated deflection in the long direction, this will probably be a reasonable estimate.

The total deflection is then (18.0 + 5.15) x 5.4/4.0 = 31mm

This is 1/223 of the short span, and may be acceptable if there are no rigid partitions. It should be noted that if there are reasons the slab might be more cracked, e.g. temperature stresses or shrinkage, the deflections could be considerably greater.

Shear: The calculated ultimate shear at the first interior support ( at the face of the supports) is (from the computer calculations) 841.95 kN (as compared to the simply supported load of 765.9 kN) . The shear assumed in the preliminary calculations was 860 kN

The design shear is 1.15 times this, or 968. kN Alternatively, from the code, Veff = Vt (1 + 1.5M/Vt x)

Mt, the moment transmitted to the column, is 226.0 kNm, and x is 1.5 + .18 x 1.5 (The column capital +1.5 slab depths)

then Veff = 1.227 x 841.95 =1033.4 kN

From this may be deducted the vertical component of the prestressing cables

At 1.06 m from the c/1 of column, the slope of the cables is .0529 from the properties of the parabola in the 8m span vertical component = 4/14x 2267 x .0529 =34 kN x 2 = 68 kN

In the 7m span the slope is .0575

Component = 5/12 x .0525 x 2056 kN =44.9 kN x 2 =89.8 kN

Total vert. component on 4 sides =157.8 kN and net design shear = 1033.4 - 157.8=875 kN

Area of reinforcement (Y10@200) in one direction, and Y12 @ 180 in the other. Average is 510 sq mm/m Average % = 100(1860 x 4,5 x 140.x 10"6 + 510x1.71 x 10-6 x 450)/(1.71 x .21x 450) = 0.97%

Permissible shear stress = .0.70 MPa

Actual shear stress = 851/(4 x 2.04 x .167) = .63 MPa. Therefore no shear reinf. reqd

At other columns, shear is less, and moment transferred to column is smaller. No shear reinforcement required.

One should also calculate that the width of slab at the external columns is adequate to transfer the moment, but with Long's method, the moment transferred is quite small, and with the column capitals, it is not necessary.