Calculation of maximum drape
Assume that the maximum drape occurs at mid-span. Using the equation of a parabola:
y = /cr (s-x)
from the tendon profile calculation (Appendix C), we know that:
P 125.32 P
Figure A3: Transverse tendon profile.
also s = 3600mm
when x = 800mm y = 87.16mm At this stage losses are assumed as follows:
At transfer 10% of the jacking load At service 20% of the jacking load.
A thorough check will be carried out after the stress calcu- lations to check that these initial assumptions of 10% and 20% are within reason. If they are not, another estimate should be made and the procedure repeated.
Initial prestress
The initial prestress force, i.e. the jacking force, has been taken to be 70% of the characteristic strength (see National Annex to Eurocode 2 Part 1.1).
For the transverse direction, the tendons will be stressed along gridline A only.
Calculation of P,,
Jacking force = 0.7 x 186 = 130.20kNhendon Prestress force at transfer (1 0% losses) = 1 1 7.1 8kNhendon Prestress force at service (20% losses) = 104.16kNhendon Next the value of prestress force required in each span is calculated. This is done using the chosen balanced load of 8.6kN/m2 (the dead load), the distance between points of inflection, s, and the drape, a, as shown in Figure A4.
Q Q
Figure A4: Drape for load balancing.
k = 2.69 x 105
61
Post-tensioned concrete floors: Design handbook
-
(n = 11) n x
The prestress force is obtained from the following equation, which assumes a parabolic profile.
C Span B B Span A
1289.0 1289.0 1289.0 1289.0 1289.0 1289.0
ws2 I 8a
- 'rqd -
Short length tendons (n = 15) n x 'a, (m)
For span CB,
P r q d = 8.6 x 7 x 36002 / (8 x 87.16 x 1000) = 1119kN
- - 1757.7 1757.7 1757.7 1757.7
Therefore number of tendons = 1119/104.16 = 10.7.
Try 11 tendons per panel.
-
For span BA,
'rqd = 8.6 x 7 x 56002 I (8 x 87.16 x 1000) = 27079kN
- - 25.3 25.3 -87.2 18.3
- - 900 1400 5600 1400
0 0 439.2 181.5 -39.1 131.3
233.0 -69.4 761.3 3 14.6 -67.8 227.6
Therefore number of tendons = 27071104.16 = 25.99.
Try 26 tendons per panel.
Full length tendons n x p,, (W
(n = 11)
As the longer span requires more tendons than the shorter span, 15 of the tendons will be stopped off at the point of inflection in span CB, next to support B. When accurate losses are calculated, the different force profile of these shorter tendons must be taken into account.
C Span B B Span A
1145.8 1145.8 1145.8 1145.8 1145.8 1145.8
The effect of the tendons on the slab is modelled by means of equivalent loads, as shown below. Equivalent loads are discussed in more detail in Section 5.4 and Appendix D. It
'Short length tendons (n = 15) n x Pav (W)
should be noted that the portions of the cable from the edges of the slab to gridlines A and C are horizontal and so do not contribute to the equivalent loads.
- - 1562.4 1562.4 1562.4 1562.4
The equivalent load, w, between any two points of inflection for the chosen number of tendons is given by:
w = 8anP,,/s2 where
n = number oftendons
a = drape at the point considered
s = as shown in Figure A.4
P,, = average force provided by each tendon.
W
/ I
Figure A5: Calculation of equivalent loads due to tendon forces.
Table A I : Calculations of equivalent loads due to transverse tendons, at transfer and after all losses.
Equivalent loads at transfer Full length tendons I
18.3 -87.2 25.3 25.3 -87.2 18.3
900 3600 900 1400 5600 1400
233.0 -69.4 322.1 133.1 -28.7 96.3
a (mm>
s (mm) w (kN1m) Total w (kN/m)
18.3 -87.2 25.3 25.3 -87.2 18.3
900 3600 900 1400 5600 1400
207.1 -61.7 286.3 118.3 -25.5 85.6
- 25.3 25.3 -87.2 18.3
- 900 1400 5600 1400
' w (kN1m) 0 390.4 161.3 -34.8 116.7
'Total w (kNlm) -61.7 676.7 279.7 -60.2 202.3
62
Appendix A: Examples of calculations
When tendons are anchored within the span, as in this example, additional equivalent loads may be generated by the end condition. These must be included in the frame analysis when obtaining the bending moments and shear force diagrams. The forces consist of a vertical and horizontal component of the tendon force applied at the anchor.
Figure A6 shows the effect of an anchorage in terms of additional equivalent loads on the slab.
.Psin a.
tendon
ce
I
Figure A6: Equivalent loads at anchorages.
The vertical component of the tendon force is easily calcu- lated, and should be applied to the slab as a vertical point load at the point where the tendon is anchored. The horizon- tal component forms a positive moment about the centroid, owing to its eccentricity from the centroid of the section, and should be applied in this form to the slab.
It should be noted that the position of the tendon at the anchorage can be arranged so that the tendon is both hori- zontal (no vertical force) and at the centroid of the section (no eccentric moment). In this example the anchorages at the ends of the full-length tendons fulfil this requirement and no additional loads are generated.
Vertical force = P sin a
Eccentric moment at the point of inflection = P cos a x e For a parabolic tendon dyldx = 2mIs2
and so sin a = 2als
Therefore sin a = 2 x 25.32 = 0.1125
cos a = 0.9937
Eccentricity of tendon, e, = -1 12.5 + 25(cover) +
16(diameter of un-tensioned reinforcement) + 8(half the tendon diameter) +
25.32(drape)
= - 3 8 . 1 8 m At transfer:
P = 1757.70kN
P sin a = 197.82kN P e c o s a = -66.68kNm After all losses:
P = 1562.4kN
P s i n a = 175.84kN P e c o s a = -59.276kNm
The equivalent loads from the tendons, the anchors and the superimposed loads are then used to calculate design moments and shears by any convenient method of structural analysis.
This is normally done using an appropriate computer program.
At Serviceability Limit State, an elastic method of analysis should be used for analysing post-tensioned flat slabs, and patterned loading should be used in multi-span situations.
Summary of prestress equivalent loads
Table A2: Summaiy of uniformly distributed equivalent loads from transverse tendons.
227.6
C Span B B Span
Equivalent loads at transfer (kN/m) 233.0 -69.4 761.3 3 14.6 -67.8
I Equivalent loads after all losses (kN1m) 207.1 -61.7 676.7 279.7 -60.2 202.3 I
Table A3: Summary of additional equivalent loads due to internal anchorages.
After all losses 175.8kN 59.3kNm
63
Post-tensioned concretefloors: Design handbook
Check that prestress loads total to zero Upward loads
= (61.7 x 3.6) + (60.2 x 5.6) + 175.8
= 735.8kN Downward loads
= (207.1 + 676.7) x 0.45 + (279.7 + 202.3) x 0.7
= 735.11kN
The small difference between these values is due to earlier approximations. The equivalent loads were altered to total zero at this point to enable consistent calculation of secondary moments.
Calculation of stresses
A , = 7 X 0.225 x 106= 1.575 x Io6mm2
As the section being considered is rectangular and sym- metrical about the centroid, z, and zb are equal.
z, = zb z = bh2 1 6 = 5.91 x 107mm3
As this example is a flat slab, analysed by the equivalent frame method, the allowable stresses are as detailed in Table 4 (Section 5.8.1). To increase ease of construction, un-ten- sioned reinforcement has been deliberately omitted from spans by keeping the tensile stresses below 0.3f,,,,fl (transfer -
where& is replaced by the initial concrete strength) and 0.3f,,,,fl (service).
f ; = PIA, - Pelz, + MAlzt + MJz,
fb = PIA, + Pelzb - MAlzb - MJzb
415.8
Summary. of applied bending momenls \ 294.0
178.5
k-A9+A 112.0
16.8
37.1
107.1
252.0
(a) Self-weight Only (b) Scrvice Load Envelopc
165.2
AL Transfer
(c) Due to equivalent prestress loads
At Service
203.7
89.6 59.5
46.2
93.8 82.6 105.0
At Transfer At Scrvice
(d) Total Applied Load Figure A7: Applied bending moment diagrams.
64
Appendix A: Examples of calculations
Table A4: Stresses at transfer for the transverse direction.
Zone
top bottom top C
CB (hogging)
bottom top bottom top bottom top bottom top bottom top bottom B
B
BA (sagging)
BA (hogging)
A
Stress due to prestress*
(MPa) 1.364 0.272 -0.492 2.128 5.406 -1.538 4.755 -0.887
2.509 1.359 -0.862 4.730 2.504 1.364
* These values include prestress secondary effects
Table AS: Stresses .$er all losses for the transverse direction.
Zone
C
top bottom top bottom top bottom top bottom top bottom top bottom top bottom CB (sagging)
CB (hogging)
B
B
BA (sagging)
A
Stress due to prestress*
(MPa) 1.27 0.18 -0.23 1.68 0.83 0.62 2.67 0.77 5.3 1 -1.88 -0.77 4.2 1 3.13 0.30
Stress due to Total stress self-weight
-0.280 1.084
0.280 0.552
0.534 0.043
-0.534 1.594
-1.891 3.515
1.891 0.355
-1.429 3.326
1.429 0.542
-0.154 2.355
0.154 1.513
1.790 0.928
0.006 1.370
Stress due to Total stress self-weight
-0.77
1.26
1.49 I
-1.49 0.19
-1.3 1 -0.48
1.31 1.93
-2.67 0.00
2.67 3.44
-7.04 -1.73
7.04 5.16
4.27 3.50
-4.27 -0.06
-2.93 0.20
2.93 I 2.23
Allowable stress (MPa)
7.5 -2.65 -1.06 10 7.5 -2.65 7.5 -2.65 10 -1.06 -1.06
10 7.5 -2.65
Allowable stress (MPa) -3.31
10.5 14 -1.32 -1.32 14 -3.31
10.5 -3.31 10.5 14 -1.32 -3.32 10.5 Hogging and sagging values are given where they both occur
in one zone. Each span is split into three zones, from the end to 2L/10, from 2L/10 to 8L/10 and from 8L/10 to L.
In this example, the construction load is smaller than the load at service and larger than that at transfer. This means that the construction case is not likely to be a governing situation and so the stresses are not calculated.
6.5
~
Post-tensioned concrete floors: Design handbook
A.1.3 Loss calculations 6) Losses due to wedge set
At this stage the losses should be calculated accurately to check that the initial assumptions of 10% at transfer and 20%
at service were reasonable. The method for calculating the various steps of loss is given in Appendix B.
Force loss at anchorage, 6Pw = 2p’I’
where
p ’ = slope of force profile
I’ = length of tendon effected by draw-in
= x Eps x ApJP’1 Full-length tendons
Short-term losses Take wedge draw-in,
A = 6mm Eps = 195kN/mm2 A,,= 100mm2 p ’ =
=
( P A -PC)’(LI + L2)
(130.2 - 121.07) / (7 + 4.5) a) Losses due to friction
Table B1 gives recommended values for the coefficients p and a:
Hence
I’ = d[(6 x 10T3 x 195 x 100)/0.79]
= 12.14m p=O.O6 and a = O.OSrads/m
Deviated angle per metre, a’ = (16 x total drape)/L2
As I’ is greater than the length of the tendon, at stressing anchorage:
Total drape = (18.27 + 25.32)/2 + 87.16 = 108.96mm (the same €or both spans)
6Pw= (A x Eps x Aps)/I + (P’x I )
= ( 6 ~ 1 9 5 ~ 1 0 0 ) / ( 1 1 . 5 ~ 1O3)+O.79x 11.5= 19.30kN Span CB
a’ = (16 x 108.96 x 10-3)/4.52 = 0.086rad/m
and at dead end:
Span BA
a’ = (16 x 108.96 x 10-3)/72 = 0.036radm 6Pw= (A X Eps X Aps)/l - (P’ x I )
= (6 x 195 x 100)/(11.5 x 103) - 0.79 x 11.5 = 1.04kN Jacking force = 130.2kN
Forces after friction losses and wedge set (see Figure A8):
Forces after friction losses (see Figure A8) are:
PA =
PB =
130.2 - 19.30 = 110.9kN
125.58 - [(19.30 - 1.04) x 4.5/11.5] - 1.04
= 117.4kN
Pc = 121.07 - 1 . 0 4 ~ 120.0kN
P A = 130.2kN
c) Losses due to early thermal shrinkage
Q
Stresslng force
where
100 x10“ (see Section 3.3) -
‘etsh -
I 130.2
‘125.6
8 121.1
,114.5 117.2]
. 108.0 “etsh = 100 xlOd x 195 x 100 = 1.95kN
- 94.5
Force after all losses \ I Stresslng
0 end
, Force at transfer / :
d) Elastic losses Figure A8: Force profiles for full-length tendons.
where
‘es - - 0.5 xfco / Eci
66
Appendix A: Examples of calculations
f,, is the stress in the concrete adjacent to the tendon. Since this is unlikely to be critical, the stress is calculated at a representative point and will be taken as uniform over the whole tendon length.
Short tendons
Friction, early thermal shrinkage and elastic losses are the same as for the full-length tendons, as are the long-term losses. The effect of wedge set is different as the tendon length is different and must be recalculated.
f,, = 1.984MPa
E,, = 0.5 x 1.984/(21.7 x 103) = 4.57 x 10-5
Spes = 4.57 x 10-5 x 195 x 100 = 0.89kN Force at dead end = 125.58e4.45 o.06(o.086 + 0.05) = 125.14kN Losses due to wedge set
6P, at dead end = (A x E,, x Ap,)/l - @’x r )
= 6 x 195 x 100/7450 - (130.2 - 125.14) = 10.641cN Prestress at transfer
Prestress force at A = 110.9 - 1.95 - 0.89 = 108.0kN Prestress force at B = 117.4 - 1.95 - 0.89 = 114.5kN
Prestress force at C = 120.0 - 1.95 - 0.89 = 117.2kN SP, at stressing anchorage = (A x Eps x APJI + (p’x r)
= 6 x 195 x 100/7450 + (130.2 - 125.14) = 20.77kN Long-term losses
Forces after friction losses and wedge set:
P, = 130.2 - 20.77 = 109.4kN
PB = 25.58 - 45(20.77 - 1064)/745 - 10.64 = 114.3kN a) Relaxation of steel
6Pr = 1000-hour relaxation value x relaxation factor
x prestress force at transfer Prestress at transfer
Prestress force at A = 109.4 - 0.89 - 1.95 Prestress force at B = 114.3 - 0.89 -1.95
= 106.6kN
= 111.5kN From Table B2 in Appendix B values are taken for an initial
jacking force equal to 70% of the characteristic strength.
Prestress after all losses
Prestress force at A = 106.6 - 4.07 - 5.85 -3.57 = 92.5kN Prestress force at B = 11 1.5 - 4.25 - 5.85 -3.57 = 97.8kN loss due to relaxation = 2.5%
relaxation factor = 1.5
i Therefore, SP, = 2.5% x 1.5 x P, = 3.75% x P,
9
I I
I Stressing force
= 0.0375 x 110.01 = 4.13kN
@rB = 0.0375 x 116.50 = 4.37kN SPrC = 0.0375 x 119.14 = 4.47kN
I I
I : Force at transfer I I
I b) Shrinkage of concrete
111.8 w- 111.5 - - - - - - 1- - - -
106.6 Force after all losses
I %h = ‘sh Eps
‘sh = 300 x 104 (see Sec,tion 3.3)
~ “sh = 3 0 0 ~ 1 0 - 6 ~ 1 9 5 ~ 100 = 5.85kN
c) Creep of concrete 7450
1 c
Figure A9: Force profiles for short tendons.
Check of the assumed losses against the actual looses where
9 = creep coefficient (see Appendix B) = 2.0 At transfer
Average short-term loss for span CB (full length tendons only)
f,, = 1.984MPa
‘cr = (1.984 x 2) / (21.7 x 103) = 1.83 x 104 6Pcr = 1.83 x 104x 195 x 100 = 3.57kN
= [(130.2 - 117.2)/130.2 + (130.2 - 114.5)/130.2]/2 x 100
= 1.0%
Average short-term loss for span BA (both tendon lengths) Prestress after all losses
Prestress force at A = 108.0 - 4.13 - 5.85 - 3.57 = 94.5kN Prestress force at B = 114.5 - 4.37 - 5.85 - 3.57 = 100.7kN Prestress force at C = 117.2 - 4.47 - 5.85 - 3.57 = 103.3kN
’
= 100[(11/26)(130.2 - 114.5 + 130.2 0150 108.0) + (15/26)(130.2 - 111.5 + 130.2 - 106.6)]/(130.2 X 2) 15.5%
67
Post-tensioned concrete floors: Design handbook
all columns 400 x 40
at 8.5m x 8.5m centres I
Average overall short term loss = (11 + 15.5)/2 = 13.2%
Afer all losses
Average long-term loss for span CB (full length tendons only)
= [(130.2 - 103.3)/130.2 + (130.2 - 100.7)/130.2]/2 X 100
= 21.6%
The following design loadings will be used:
Self weight - Calculated on the basis of normal weight concrete.
Imposed dead - Allow 1 .8kN/m2 (to include services, finishes, partitions etc).
Imposed live - Allow 5.0kN/m2 (based on required use).
Edge loading - Allow 7.5kN/m run (Dead) for cladding etc.
Average long-term loss for span BA (both tendon lengths)
It is assumed that the structure is for use as an office.
= 100[(11/26)(130.2 - 100.7 + 130.2 - 94.5) +
(15/26)(130.2 - 97.8 + 130.2 - 92.5)]/(130.2 x 2)
= 26.6%
Average overall long term loss = (21.6 + 26.6)/2= 24.1%
Although the assumed losses of 10% and 20%, respectively, have been exceeded recalculation is not considered neces- sary, as it will not cause an increase in the number of tendons.
Also the calculation of stresses for the correct losses are unlikely to exceed the allowable values.