The Design and Construction of

Precast Concrete Structures

Precast & Prestressed Floors

and Composite Slabs
 Precast & Prestressed Floors and Composite
Slabs

Hollow core floor units & slab fields

Double tee units
Half-slab (precast + insitu topping)
Composite floors
Load v span data
Spreadsheets for design of units
and slab fields
Dutch trade association claim fixing
rates of 2000 sq.m per week
40 x 11 feet wide hollow core onto precast walls
at MGM hotel, Las Vegas, 1992
European production = 25 million m2 per year
75% is 150 – 250 mm deep
Double tee units – twice the price but up to 4
capacity than hollow core
Prestressed half-slab popular for housing
and awkward shapes
Propping required over 5 m
 PRESTRESSED HOLLOW CORE FLOOR UNITS

400 – 3600 mm wide; typically 1200 mm

90 – 730 mm deep; typically 150, 200, 250, 300 mm
self weight 1.5 to 5.0 kN/m2
void ratio 40 – 60 % of solid section
spans 6 – 20 m (economical range)
Longitudinal 30 mm
pretensioning No shear or flanges and
strand or wire torsion links webs
Shear key
profile

1195 mm
Are the deeper units ‘beams’ or ‘slabs’?
Should they abide by normal RC rules?
Extrusion or slipformed
100-150 m long bed; no-slump mix grade C50 – 60
Extrusion - rotating screws turn opposite hands
Extrusion – circular mandrels make the holes
Extrusion – circular mandrels make the holes
Slipforming – shear compactor (hammers down
flanges and webs)
Curing temperature contours (c Branco, Lisbon)
Splitting cracks due to sawing
restraints as prestress in transferred
At 16 – 18 hours, circular saw cuts to length
+10 to – 15 mm
400 – 450 deep units have a new market
for 16 m long clear span car parks
Too much plasticiser !! in 450 deep units

Actually air-entrainment agent is used by

several producers as a plasticiser
Single storey supermarket podium and car park
Span/depth ratio = 40
Italian variation (c. ASSAP)
600 mm wide prestressed units in Budapest
7-wire helical strand (1750 MPa) gives good
bond in the important transmission zone

Locks in
Strand pull-in:
An important indicator
of success
Should be about 1 mm,
irrespective of length

Theoretical pull-in limit

for zero prestress
PL
AE

Use a linear scale

between the extremes
700 mm deep units in Italy, often used at 25 m
span for tunnel cut-and-cover
For units > 500 mm deep, mesh is rolled out to
reinforce the outer webs
Bearing onto neoprene or mortar for spans
more than about 15 m onto insitu or masonry

20 m long ASSAP unit

 Section Analysis for Prestressed
- Pretension and losses (about 18 – 25%)
- Service moment (bottom tension critical)
- Ultimate moment (usually  Msr x 1.5)
- Ultimate shear uncracked & flexurally cracked
- End bearing and transmission length
- Deflection and camber (long-term, creep)
- Live load deflection after installation

Msr = (fbt + fct)Zb

Mur = 0.87 fpb Aps (d – dn)

Vco = 0.67 bvh  ft2 + 0.8 fcpx ft

Imposed load

Allowable span
 Bearing limit
Imposed load

Handling limit
span/depth = 50

Span
 Bearing limit
Imposed load

Handling limit
span/depth = 50

Span
 Bearing

Shear
Imposed load

Service
moment

Deflection
Handling

Span
 Service moment
Possibly shear ? control

Deflection
control
Routine bending tests to 25% overload
with 95% recovery
Flexural cracks extend rapidly through the
section due to narrow webs
 Approx
200 mm spacing
 Effective stiffness changes
with increasing load

Load
Deflection
2 to 6
EIeff = EIc + (EIu - EI c ) (M/ Mcrack )
 Example 1
Calculate Msr for the 203 mm deep Class 2 prestressed HCU shown
below. The initial prestressing force may be taken as 70% of
characteristics strength of the standard 7-wire helical strand.
Manufacturer’s data gives relaxation as 2.5%. Geometric and data
given by the manufacturer are as follows:
Ac = 135  103 mm2, I = 678  106 mm4, yt = 99 mm, fcu = 50 N/mm2, Ec =
30 kN/mm2, fci = 35 N/mm2, Eci = 27 kN/mm2, fpu = 1750 N/mm2, Eps = 195
kN/mm2, Aps = 94.2 mm2 per strand and cover = 40 mm.
Section Properties
Zb = I/yb = 678  106/104 = 6.519  106 mm3
Zt = I/yt = 678  106/99 = 6.848  106 mm3

Eccentricity, e = 104 – 40 – 12.5/2 = 57.7 mm

Initial prestress in tendons, fpi = 0.70  1750 = 1225 N/mm2
Initial prestressing force, Pi = fpi  Aps = 1225  7  94.2  10-3 = 807.8 kN
Check Stress at Transfer
fbc = (807.8  103/135  103) + (807.8  103  57.75/6.519 106)
= +13.14 N/mm2 (compression)  0.5fci (17.5 N/mm2)
OK
ftt = (807.8  103/135  103)  (807.8  103  57.75/6.848 106)
=  0.83 N/mm2 (tension)  0.45fci (2.66 N/mm2)

Prestress at level of centroid of the strands:

fcc’ = Pi/A + Pie2/I = +9.96 N/mm2
Losses
Elastic loss = 9.96  195/27 = 71.9 N/mm2 (5.87% loss)
Creep loss = 1.8  5.87 = 10.56% loss
Shrinkage loss = 300  10-6  195  103 = 57.5 N/mm2 (4.69% loss)
Relaxation loss = 1.2 2.5 = 3.0% loss

Total losses = 5.87 + 10.56 + 4.69 + 3.0 = 24.12%

or
100% – 24.12% = 75.88% of the initial prestress values
Final Prestress in Bottom and Top
fbc = +13.14  0.7588 = +9.97 N/mm2 (compression)
ftt = 0.83  0.7588 = 0.63 N/mm2 (tension)

Service Moment
At the bottom fibre, Msr is limited by the tensile stress limit of
0.45fcu = 3.2 N/mm2
Msr = (fbc + 0.45fcu)Zb = (9.97 + 3.2)  6.519  106 10-6
= 85.8 kNm  MORE CRITICAL

At the top fibre, Msr is limited by the compressive stress limit of

0.33fcu = 16.5 N/mm2
Msr = (fbc + 0.33fcu)Zt = (0.63 + 16.5)  6.848  106 10-6
= 117.3 kNm > 85.8 kNm
 Example 2
Calculate Mur for the section given in Example 1. Is the unit critical at
service or at ultimate limit state? Manufacturer’s data gives the breadth
of the top of the HCU as b = 1168 mm.

d = 203 – 40 – 12.5/2 = 156.75 mm

fpuAps/fcubd = (1750  659.4) / (50 1168  156.75) = 0.126

From Example 1; fpe/fpu = 0.7  0.7588 = 0.531

From Table 4.4; fpb/0.87fpu = 0.966

 Ultimate force in the strands, Fst = 0.87  0.966  1750  (7  94.2)  10-3
= 969.8 kN
Using equilibrium;
Fcc = Fst
0.45fcub(0.9X) = 969.8  103 N  X = 41.0 mm  28 mm (upper flange)
Distance to the centroid of the compression block, dn = 41.0/2 = 20.5 mm

Mur = 0.87 fpb Aps (d – dn)

= 969.8  103  (156.75 – 20.5)  10-6 = 132.1 kNm
Shear failure in 400 deep units
(c. Engstrom, Sweden)

Shear searches
out the
weakest web
Shear capacity of single webs more reliable

Shear bond crack stops

 Example 3

Calculate the uncracked shear capacity for the section in Example 1

using (a) the rectangular area term 0.67bvh and (b) using the term Ibv/Ay’.
Use the bearing length = 100 mm.

From Example 1, prestress after losses at the centroidal axis, fcp = 4.54 N/mm2
Critical shear point is at 100 (bearing length) + 104 (height to centroidal axis) =
204 mm
Development length, lp = greater of 203 mm (slab depth) or 240  12.5/35 =
507 mm
x/lp = 204/507 = 0.402, then fcpx = 2.92 N/mm2
Total breadth at centroid, bv = 1168 – (6  150) = 268 mm
ft = 0.24fcu = 0.2450 = 1.7 N/mm2
(a) Using 0.67bvh
Vco = 0.67  268  203  1.72 + 0.8  2.92  1.7  10-3 = 95.5 kN

(a) Using Ibv/Ay’

Area above the centroidal axis, A = 67500 mm2
y’ = 64.5 mm (calculate from geometry)
Vco = 678  106  268  1.72 + 0.8  2.92  1.7  10-3 = 109.3 kN
67500  64.5
Watch out for flexural-shear (Vcr) failure !
 Decompression
Vco point

Vcr

Vcr = 0.55 (1 – loss) vc bv d + Mo Vu

Mu
 P

UDL

SF diagram

Vcr failure
here
Failure of solid prestressed unit during
pouring of insitu topping
 Example 4
Calculate the minimum value of Vcr for the slab in Example 1.

V and M is not required to determine the minimum value of Vcr. Also, the
distance to the commencement of the flexurally cracked region need not to
be known.
100As/bd = (100  7  94.2) / (268  156.75) = 1.57
From Table 3.8: BS 8110: Part 1;
vc = 1.08 N/mm2
Mo = 0.8 Zb fbc = 0.8  6.519  106  9.97  10-6 = 52.0 kNm
Mu from Example 2 = 140.7 kNm
fpe/fpu = 0.7  0.7588 = 0.531
Vcr, min = [(1 – 0.55  0.531) 1.08  268  156.75  10-3) +
[52.0  Vcr, min/140.7]  Vcr, min = 51.0 kN
 Bearing Capacity

Refers to contact bearing pressure at the bearing ledge

Non-isolated – components are connected to other components with a
secondary means of support
Isolated – components rely entirely on their own bearing for total support

Ultimate bearing capacity, Fb = fb lb lw

Ultimate bearing stress, fb
Effective bearing length, lb (Perpendicular to the
floor span)
 Net bearing width, lw  40 mm (non-solated) or 60 mm (isolated)

lw = Nominal bearing width – spalling tolerances – allowances for

constructional inaccuracies
 Example 5
Calculate the bearing capacity of the HCU in Example 1. The unit has an
actual 75 mm dry bearing width onto a reinforced concrete beam where
the cover to the bars in the bearing ledge is 30 mm and the depth of the
bearing ledge is 250 mm. Assumed that the unit has secondary
support.

Ultimate bearing stress, fb = 0.4  50 = 20 N/mm2

Net bearing width for non-isolated component, lw = 75 (actual) – 30 (cover) –
0 (ineffective HCU bearing length distance) = 45 mm  40 mm OK
Effective bearing length, lb = least of (i) 1200 mm; (ii) 700 mm; (iii) 600 mm
Use lb = 600 mm
Fb = 20  45  600  10-3 = 540 kN > Vco (109.3 kN) OK
Cover to the tendons

100 mm
50-55 mm
Aswad tests on edge loads (PCI JOURNAL)
Very brittle in this mode !
Reinforced Hollow Core: 600 mm wide
Housing and office spans up to about 5.5 m
Self weight approx 270 kg/m
Must use partial cracked stiffness and quasi-
permanent live load (f = 0.3) for deflections
Made upside down. 6 no. T8-T20 bars
Two pass of concrete, grade C40
Cores withdrawn immediately. Cycle takes
less than 3 minutes. Equipment maintenance
approx. 50% of extrusion m/c costs
24 hour drying and humid curing
Composite floors required for double tee,
but optional for hollow core
Surface laitence due to cutting slurry
More than 0.5 mm thick
 Smooth Rough Smooth Rough Smooth Rough
1.8

1.6
Interface shear strength (N/mm2)

1.4

1.2

1.0
Average line
0.8 BS 8110 washed
BS 8110 roughened
0.6 BS 8110 as-cast

0.4

0.2

0.0
Dry Ponded Optimum wet

Relationship of interface shear strength with different surface conditions

(I S Ibrahim & K S Elliott, 2007)
50 mm minimum at the highest point, increasing
(with slab and beam cambers) to about 80 mm
 Composite Construction

 Minimum thickness of topping  40 mm

 Average depth of topping allowances for camber should be made –
allowing span/300 will suffice
 Concrete grade – C25 to C30
 Minimum mesh reinforcement area = 0.13%  Concrete area
 Advantages to composite construction:
(a) Increase bending resistance and flexural siffness
(b) Improve vibration, thermal & acoustic performance
(c) Provide horizontal diaphragm action
(d) Provide horizontal stability ties across floors
(e) Provide a continuous and monolithic floor finish
with 75 mm
topping
Composite design – 2 stage approach
Stage • Selfweight of slab + in-situ
concrete topping + construction
traffic allowance (1.5 kN/m2)
1 • Section properties = Precast unit

Stage • Superimposed load

• Section properties = Composite
2 section
 Stress diagram at serviceability

0.33fcu limit
Topping + +
_
+

+ + =
Precast

fct limit
_ _
+ _

Prestress Imposed Imposed Total

Stage 1 Stage 2 stresses
stresses stresses
 Stage 1 – Service Moment, M1

Bottom fibre stress:

fb1 = fbc – M1/Zb1  +0.33fcu

Top fibre stress:

ft1 = ftc + M1/Zt1  0.45fcu
 Stage 2 – Service Moment, M2

Bottom fibre stress of the precast unit:

fb2 = – M2/Zb2

Top fibre stress of the precast unit:

ft2 = +M1/Zt2

Top fibre stress of the in-situ topping:

f’t2 = +M1/Z’t2

beff = b  Ec’/Ec where b is the full breadth

Adding Stage 1 + Stage 2:
fb = fbc – M1/Zb1 – M2/Zb2  0.45fcu
ft = ftc + M1/Zt1 + M2/Zt2  +0.33fcu
f’t = + M2/Z’t2  +0.33fcu in-situ

Bottom of the slab:

Ms2 = M2  (0.45fcu + fbc)Zb2 – [M1(Zb2/Zb1)]

Top of the slab:

Ms2 = M2  (0.33fcu – ftc)Zt2 – [M1(Zt2/Zt1)]
 Example 6
Calculate the Stage 2 service bending moment that is available if the
HCU in Example 1 has a 50 mm minimum thickness structural topping.
The floor is simply supported over an effective span of: (a) 4.0 m; (b)
8.0 m. The precamber of the HCU may be assumed as span/300 without
loss of accuracy. Use fcu in-situ = 30 N/mm2 and Ec in-situ = 26 kN/mm2. Self
weight of concrete = 24 kN/m3.
What is the minimum imposed loading for each span?
Effective breadth of topping, beff = 1200  26/30 = 1040 mm
Total depth of composite section = 203 + 50 = 253 mm
Depth to neutral axis of composite section from top, yt2
= (1040  50  25) + (135 000  (50 + 99)) / (135 000 + 1040  50)
= 114.5 mm
Second moment of area of composite section, I2 = 1266 mm4
Then Zb2 = 1266  106/(253 – 114.5) = 9.14  106 mm4
Zt2 = 1266  106/114.5 = 19.63  106 mm4
Zb2/Zb1 = 9.14  106 /6.519  106 = 1.40
Zt2/Zt1 = 19.63  106 /6.848  106 = 2.86
(a) 4.0 m
Precamber = 4000 / 300 = 13 mm
Maximum depth of topping at supports = 50 + 13 = 63 mm
Average depth of topping = (50 + 63)/2 = 57 mm
Self weight of topping = 0.057  24  1.2 = 1.64 kN/m run for 1.2 m wide unit
Self weight of HCU = 3.24 kN/m run
Stage 1 moment, M1 = (3.24 + 1.64)  4.02/8 = 9.76 kNm
Bottom of the slab:
Ms2  (0.45fcu + fbc)Zb2 – [M1(Zb2/Zb1)]
Ms2 = [(3.2 + 9.97)  9.14  106 – 9.76  106  1.40]  10-6 = 106.7 kNm
Top of the slab:
Ms2 = M2  (0.33fcu – ftc)Zt2 – [M1(Zt2/Zt1)]
M2 = [(16.5 – (– 0.63))  19.63  106 – 9.76  106  2.86]  10-6 = 308.3 kNm
The allowable maximum imposed load = 8  106.7/4.02 = 53.4 kN/m

* Additional notes:
Total Ms = Ms1 + Ms2 = 9.76 + 106.7 = 116.5 kNm

Now do for (b) 8.0 m

Ultimate limit state – 2 stage approach
 Ultimate limit state

Total area of reinforcement, Aps = Aps1 + Aps2

Effective breadth, beff = b  fcu in-situ/fcu
Stage 1
fpbAps1 = 0.45fcu (0.9bX1)  but dn1 = 0.45X1
then dn1 = fpbAps1/0.9fcub
then Mu1 = fpbAps1(d – dn1) ---------------- (1)
Stage 2
Aps2 = Aps – Aps1
Mu2 = fpbAps2 (d + hs – dn2) ---------------- (2)
Replacing (1) into (2):
Mu2 = fpb [Aps2 – Mu1/fpb(d – dn1)] (d + hs – dn2)
One Step Approach
Mu = fpbAps (d + hs – dn)
 Example 7
Calculate the imposed ultimate bending moment in Example 1 using a
50 mm minimum thickness structural topping for the following design
approaches: (a) the two-stage approach; and (b) the simplified one-step
approach. The floor is simply supported over an effective span of 8.0 m.
All other details as Example 6.

What is the maximum ultimate imposed loading? Is the composite slab

critical at service or at ultimate?
Effective depth in HCU, dn1 = 156.7 mm
Effective depth in composite section, dn2 = 156.7 + 50 = 206.7 mm
Effective breadth of topping, beff = 1200  30/50 = 720 mm

Two-stage Approach
Stage 1 moment, Mu1 = 1.4  40.4 = 56.6 kNm
Mu1 = 56.6  106 = (0.87  1750  Aps1 [156.7 – (0.87  1750  Aps1/0.9  50 
1168)]

Hence Aps1 = 227 mm2 and dn1 = 7.2 mm. Then Aps2 = 652 – 227 = 432 mm2
fpuAps/fcubeff(d + hs) = 1750  432 / 50  720  206.7 = 0.10

From Table 4.4; fpb/0.87fpu = 1.0

dn2 = 1.0  0.87  1750  432/0.9  50  720 = 22.2 mm
Then Mu2 = 1.0  0.87  1750  432  (206.7 – 22.2)  10-6 = 132.5 kNm
Ultimate load = 8  132.5/8.02 = 16.5 kN/m run
One-stage Approach
fpuAps/fcubeff(d + hs) = 1750  659 / 50  720  206.7 = 0.155

From Example 1; fpe/fpu = 0.531

From Table 4.3; fpb/0.87fpu = 0.93
X/d = 0.324, hence X = 67 mm and dn = 30 mm
Then Mu = 0.93  0.87  1750  659  (206.7 – 30)  10-6 = 180.0 kNm
Ultimate load = 8  180.0/8.02 = 22.5 kN/m run

Subtract self weight of HCU + topping = 1.4  5.05 = 7.1 kN/m

Leaving imposed load = 22.5 – 7.1 = 15.4 kN/m
PROPPING
 Self weight of slab
++

Self weight of topping, w

wL2/32
_
+ +
Propping reaction

R = 0.625wL

Composite slab +

+0.15625wL2

Superimposed load ++

Ms2 = M2  (0.45fcu + fbc)Zb2 – [M1(Zb2/Zb1)] – 0.15625wL2

 Interface Shear Stress in Composite Slabs

In the neutral axis is below the interface, X  hs

Fv = 0.45fcubeffhs

In the neutral axis is above the interface, X  hs,

Fv = 0.45fcubeff0.9X

Average ultimate shear stress at the interface;

ave = Fv/bLz

b = transverse width of the interface

Lz = distance between the points of minimum and maximum moment

Design shear stress, h = 2ave (uniformly distributed load)

If h  limiting values in Table 5.5: BS 8110: Part 1; provide shear reinforcement

(per 1 m run) projecting from the precast unit into the structural topping. The
amount of steel required:
Af = 1000bh/0.87fy
 Example 8
Calculate the shear reinforcement necessary to satisfy the ultimate
horizontal shear force at the precast – in situ interface in Example 7. the
top surface of the HCU is “as extruded” finish. Use HT reinforcement, fy
= 500 N/mm2 and L = 8.0 m.

From Example 7, X2 = (22.2/0.45) = 49.3 mm  hs = 50 mm

Neutral axis is above the interface;
Fv = 0.45  50  720  0.9  49.3  10-3 = 719 kN
The distance Lz = 8000/2 = 4000 mm
Interface breadth = 1200 mm
ave = 719  103 / 1200  4000 = 0.15 N/mm2

Uniformly distributed load, design shear stress h

= 2  0.15 = 0.30 N/mm2  0.55 N/mm2 from Table 5.5: BS 8110: Part 1
 DOUBLE TEE FLOOR UNITS

2400 – 3000 mm wide

300 – 2000 mm deep; typically 400 - 800 mm
self weight 2.6 to 10 kN/m2
void ratio 70 – 80 % of solid section
spans 8 – 30 m (economical range)
Double tee units – mostly prestressed, but RC if
manufacturer prefers

Bearing pads required 150 x 150 x 10

Double tee long-line casting
Prevent settlement cracks at the top of the web
Double tee - ‘concrete train’ in Germany
Self topped units with 120 mm thick flanges
Double tee – self compacting concrete pour
1 batching + 4 workers = 240 m2 per day
equates to product cost / salary ratio = 20
Half joint box and
confinement U bars
Half joint lowers the centroid of the floor plate
Site weld to adjacent flange

Make a small saw cut

to prevent spalling
MINIMUM STRUCTURAL DEPTH
FOR 6 – 10 m
SAME LOAD AND SPAN
CONDITIONS

FLOOR SPAN
= 1.0 TO 1.5
BEAM SPAN