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01 February 2013 1
Additional rules for precast concrete, EN 1992-1-1 chapter 10 (with reference to 8)
01 February 2013 2
Precast structural systems
Foundation blockFor column
Precast frames
01 February 2013 3
Precast concrete systems
Large floor bays, composed of precast prestressed hollow core slabs
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01 February 2013 4
Special aspects of precast concrete structures
• Special production methods• Bearings• Tolerances• Joints• Foundations• Prestressing with pretensionedsteel
01 February 2013 5
Special production methods
( ) ( ( ))ctm cc ctmf t t fαβ= ⋅
Strength-properties at other ages than t = 28 days
Example: tensile strength fctm
Concrete hardening at normal temperatures:
1/ 228( ) exp{ [1 ( ) ]}
cct s
tβ = −
where
where t = age of concrete in days
S = coefficient depending on type of concrete
α= 1 for t < 28 days
α = 2/3 for t ≥ 28 days
01 February 2013 6
Special production methods
( ) ( ( ))ctm cc ctmf t t fαβ= ⋅
1/ 228( ) exp{ [1 ( ) ]}
cct s
tβ = −
Strength-properties at other ages than t = 28 days
Example: tensile strength fctm
Concrete hardening at increased temperatures:
where
where t = age of concrete in days
S = coefficient depending on concrete type
α= 1 for t < 28 days
α = 2/3 for t ≥ 28 days
(4000 /[273 ( )] 13,65)
0,
1
i
nT t
T T i
i
t t e t− + ∆ −
=
= ⋅ ⋅∆∑
T(∆ti) is temperature in 0C during the interval ∆ti
∆ti number of days with temperature T
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01 February 2013 7
a + ∆a2 2a
1
a
a + ∆a3 3
b1
a1
Bearings: definitions (10.9.5)
a = a1 + a2 + a3 + 2 2
2 3a a∆ ∆+
a1 nett bearing length = FEd / (b1 fRd), but ≥ minimum value FEd design value of support reactionb1 nett value of bearing lengthfRd design value of maximum bearing pressure
a2 ineffective length from edge of bearing a3 same for supporting member ∆a2 allowable tolerance for distance between bearings (Table 10.5)
∆a3 allowable tolerance in length Ln of precast member: ∆a3 = Ln/2500
01 February 2013 8
a + ∆a2 2a
1
a
a + ∆a3 3
b1
a1
Bearings (10.9.5)
a = a1 + a2 + a3 + 2 2
2 3a a∆ ∆+
Limits to support pressure
fRd = 0,4 fcd for dry connections fRd = fbed ≤ 0,85fcd for all other cases
wherefcd lowest value of design strength of supporting member
fbed design strength of bearing material
01 February 2013 9
a + ∆a2 2a
1
a
a + ∆a3 3
b1
a1
Bearings (10.9.5)
a = a1 + a2 + a3 + 2 2
2 3a a∆ ∆+
Minimum value a1 in mm
Relative bearing pressure, σEd/fcd ≤ 0,15 0.15-0,4 > 0,4
Line support (floors, roofs) 25 30 40
Ribbed floors and purlins 55 70 80
Concentrated supports (beams) 90 110 140
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01 February 2013 10
a + ∆a2 2a
1
a
a + ∆a3 3
b1
a1
Bearings (10.9.5)
a = a1 + a2 + a3 + 2 2
2 3a a∆ ∆+
Distance a2 in mm
Bearing material and – type σEd/fcd ≤ 0,15 0,15 – 0.4 > 0.4
Steel line support
concentrated
0
5
0
10
10
15
Reinforced concrete ≥ C30 line support
concentrated
5
10
10
15
15
25
Plain and reinforced line support
concrete < C30 concentrated
10
20
15
25
25
35
Masonry line support
concentrated
10
20
15
25
(-)
(-)
01 February 2013 11
a + ∆a2 2a
1
a
a + ∆a3 3
b1
a1
Bearings (10.9.5)
a = a1 + a2 + a3 + 2 2
2 3a a∆ ∆+
Distance a3 in mm
Reinforcement detailing Line support Concentr. support
Continuous bars over support 0 0
Straight bars, horizontal loops at end of
member
5 15, but ≥ end cover
Tendons or straight bars exposed at end of
element
5 15
Vertical loop reinforement 15 Cover + inner radius of
bending
01 February 2013 12
a + ∆a2 2a
1
a
a + ∆a3 3
b1
a1
Bearings (10.9.5)
a = a1 + a2 + a3 + 2 2
2 3a a∆ ∆+
Allowance ∆a2 for tolerances for the clear distance between the faces of the supports:
Bearing material ∆a2
Steel or precast concrete 10 ≤ L/1200 ≤ 30 mm
Masonry or in situ concrete 15 ≤ L/1200 + 5 ≤ 40 mm
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01 February 2013 13
Example for design of bearing
Concentrated support beam (statically determinate), L = 20m, b = 400 mm, fck = 40
N/mm2 , Fed = 1000 kN, neoprene bearing, concrete strength class C40
▪ Allowable bearing pressure fRd = 0,85⋅fcd = 0,85⋅(40/1,5) = 22,7 N/mm2
▪ Support length a1=Fed/bfRd = 1000/(400⋅22,7) = 110 mm▪ Strength class supporting concrete C40 → a2 = 25mm
▪ Vertically bent reinforcement in beam (∅ 32mm, end cover 25mm)
→ a3 = 25 + 2,5⋅32 = 105 mm▪ ∆a2 = L/1200 + 5 = 20000/1200 + 5 = 22 mm▪ ∆a3 = L/2500 = 20000/2500 = 8 mm
a = a1 + a2 + a3 + √(∆a22 + ∆a32) = 110 + 25 + 105 + √(222 + 82)= 263 mm
a + ∆a2 2a
1
a
a + ∆a3 3
b1
a1
01 February 2013 14
Anchorage of reinforcement at support(10.9.5)
d = c + ∆a2 + r in case of vertically bent bars
In the beam and the supporting element thereinforcement should be detailed in such a way that
the node of the truss is in equilibrium, taking due
account of the deviations ∆a2 en ∆a3
For the determination of the anchorage length the
correct geometry should be used. E.g. for thesupporting element the following relation applies:
d = c + ∆a2 in case of horizontal loops or bars with end anchorage
r
d c> a + ∆a1 3
c
r
> a + ∆a1 2 d
a1
01 February 2013 15
Pocket foundations with keyed surface areas (10.9.6.2)
▪ To be treated as a monolythic system
▪ Lapped length at least anchoragelength bar + s, where s = distance between vertical bars
▪ Sufficient horizontal transversereinforcement should be appliedfor lapped splice
▪ Punching cone to be ssumedaccording to dotted inclined lines
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01 February 2013 16
Pocket foundations with smooth surfaces (10.9.6.3)
▪ Equilibrium of forces throughF1, F2 and F3 with correspondingfriction components
▪ Requirement: l ≥ 1,2h ▪ Coefficient of friction µ ≤ 0,3▪ Punching cone to be assumedaccording to dotted inclinedlines
01 February 2013 17
Wall-floor connections (10.9.2)
• No calculated reinforcement is necessary if the vertical load per unit length is smaller than or equal to 0,5h⋅fcd
• This load may be increased to 0,6h⋅fcd if reinforcement is applied according to the figure, where the bar diameter is at least ∅ ≥ 6mm and s is not larger than h or 200mm.
• For higher loads the walls should be provided with reinforcement, assuming eccentric or concentrated forces at the end of the wall.
Walls:
01 February 2013 18
Wall-floor connections (10.9.2)
▪ The effect of clamped moments in the floor can be compensated by top reinforcement in the floor elements or, in the case of hollow core slabs, by coupling bars in cut-outs.
▪ In the latter case it should be realized that this reinforcement should as wel
l contribute to the transmission of shear forces in the plane of the floor
(diaphragm action.)
Floors:
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01 February 2013 19
Connections transmitting compressive forces
(10.9.4.3)
Concentrated transmission
Expansion of soft joint material
For a soft joint material (B) the reinforcement, in case no more accurate calculation is made, may be taken equal to:
As = 0,25 (t/h) Fed/fyd
where :t = thickness of joint materialh = dimension joint material in
direction of reinforcement
FEd = design compressive forceon joint
A B
Reinforcement according to Cl. 6.5
01 February 2013 20
Shear resisting connections
Grouted joints Welded or boltet joints Structural topping
For floors subjected to uniformly distributed loading, if no more accurate calculation is
made to determine the shear force to be transmitted through the longitudinal joint, the
design shear stress per unit length may assumed to be equal to vEd=qEd⋅be/3where qEd = design value of uniformly distributed load
be = width of element
01 February 2013 21
Transmission of forces in end-regions of
beams in case of prestressing with
pretensioned steel (8.10.2)
1. Radial tensile stresses2. Splitting tensile stresses 3. End face tensile stresses
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01 February 2013 22
Introduction of prestressing force for prestressing with pretensioned steel (wires or strands)
The following defined lenghts are distinghuished:
- lpt = transmission length, necessary to develop the prestressing force
- ldisp = Dispersion length, necessary to develop a lineair distribution of the stresses
over the height of the cross-section
- lbpd = anchorage length, necessary to anchor the force Ap⋅σpd .
01 February 2013 23
Introduction of prestressing force for
pretensioned steel
Stress in strandA
B Distance to end of element
lpt = basic value transmission length :
bptpmpt fl /021 φσαα=
α1 = 1,0 for slow release of stressat prestressing1,25 for fast release of stress
α2 = 0,25 for wires0,19 for 3 and 7 wire strands
∅ = nominal diameter wire/strand σpm0 = stress in prestressing steel
just after transmission ofprestress
fbpt = bond strength
(Eq. 8.16)
01 February 2013 24
Introduction of prestressing for
pretensioned steel
Steel stressA
B Distance to end of element
Bond stress fbpt formula 8.15 fbpt = ηp1⋅η1⋅fctd(t)
ηp1 = 2,7 for indented wires3,2 for 3 and 7-wire strands
η1 = 1,0 for good bond conditions0,7 other cases
fctd(t) = design tensile strength concreteat prestressing
= αct⋅0,7⋅fctm(t)/γcαct = sustained loading factor
(3.1.6(2)), for fctm(t), see (3.1.2(8))
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01 February 2013 25
Introduction prestressing force for
pretensioned steel
Steel stressA
B Distance to end of element
lpt1 = 0,8lpt (most unfavourable situation for control of splitting and end-face cracking).
lpt2 = 1,2lpt (most unfavourable situation for control of shear tension)
Eq. 8.17 en 8.18
01 February 2013 26
Anchorage of wires and strands
Steel stressA
B Distance to end of element
lbpd = anchorage length
lbpd =lpt2 + α2∅(σpd - σ pm∞)/fbpd
σpd = stress in prestressing steelin ULS in area cracked in bending
σpm∞ = working prestress
fbpd = ηp2η1fctdηp2 = 1,4 for deformed wires
1,2 for 7-wire strandsη1 (see sheet 24)
01 February 2013 27
Introduction prestressing force in case of pretensioned steel
Dispersion length ldisp
It may be assumed that the stresses in the concrete have a linear
distribution over the height of the cross section after a distance:
ldisp = √(lpt2 + d2)
lptldisp
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01 February 2013 28
Worked example prestressed hollow core slab
Data
▪ Concrete strength class C45/55
▪ h = 0,265 m, b = 1,196 m, ep = 0,092mm, Ac = 0,172 m2, Ic = 1,5⋅10-3 m4, S = 7,55⋅10-3 m3 ,
Σbw = 0,231 m, W = Ic/(h/2) = 11,32⋅10-3 m3
▪ 8 strands 3/8” FeP 1770
(8∅ 9,5mm, Ap = 8⋅51 = 408 mm2, fp0,1k =
1520 N/mm2, fpd = fp0,1k/γs = 1520/1,1 = 1382
fpk/γs = 1610 N/mm2 , initial stress σpm0 = 1290
N/mm2, working prestress σpm∞ = 1050 N/mm2
▪ Cylinder strength at prestressing 30 N/mm2,
fctm = 2,9 N/mm2, fctk = 2,0 N/mm2,
fctd = fctk/γc = 2,0/1,5 = 1,33
01 February 2013 29
DETERMINATION OF
TRANSMISSION LENGTH
Bond strength:
fbpt = ηp1⋅η1⋅fctd(t)
= 3,2⋅1,0⋅1,33 = 4,26 N/mm2
Basic transmission length:
lpt = α1⋅α2⋅∅⋅σmp0/fbpt
= 1,25⋅0,19⋅9,5⋅1290/4,26 = 683 mm
σpm0 = 1290 N/mm2
lpt
fbpt
Worked example prestressed hollow core slab
01 February 2013 30
Bond strength for anchorage in ULS:
fbpd = ηp2⋅η1fctd = 1,2⋅1,0⋅1,8 =
2,16 N/mm2
Total anchorage length of strand
with maximum stress fpd = 1321 N/mm2
lbpd = lpt2 + α2⋅∅(σpd - σpm∞)/fbpd
= 1,2⋅683 + 0,19⋅9,3⋅(1321 – 1050)/2,16
= 1041 mm
σpm∞ = 1050 N/mm2
Envelope anchorage force
fpd = 1321 N/mm2
lbpd=1041 mmlpt=683 mm
σpm0 = 1290 N/mm2
Worked example prestressed hollow core slab
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01 February 2013 31
What is the maximum allowable length of the slab, if it is required that it should
not crack in the SLS?
Variable load: qQk = 3 kN/m2 or for a slab (b = 1,2m): qQk = 3,6 kN/m’
Dead load: qGk = Acgc = 0,172⋅25 = 4,3 kN/m’Maximum load BGT: qQk + qGk = 7,9 kN/m’
Concrete compressive stress at bottom due to prestressing force Fpm∞:
σcb = -Fpm∞/Ac - Fpm∞⋅ep/Wc =
-0,428/0,172 – 0,428⋅0,092/(11,32⋅10-3) = -5,96 MN/m2
The flexural tensile strength of the concrete is
defined as (3.1.8):
fctm,fl = (1,6 – h/1000)fctm ≥ fctm with h[m], so
fctm,fl = (1,6 – 0,265)⋅3,8 = 1,335⋅3,8 = 5,07 MPafctk,fl = 0,7⋅5,07 = 3,55 MPa
MQ+G
Qk+Gk
Fpm∞ep
Worked example prestressed hollow core slab
01 February 2013 32
The bending moment for which at mid-span at the bottom of the cross-section the
characteristic tensile strength of the concrete is reached is:
Mcrack = (σcp + fctk,fl)⋅Wc = (5,97 + 3,55)⋅11,32 = 107,8 kNm
For a maximum load qQk + qGk = 7,9 kN/m’
it is found that:
(1/8)(qQk + qGk)⋅lmax2 = 107,8 which leads to
lmax = 10,45 m
Worked example prestressed hollow core slab
01 February 2013 33
A slab length is assumed equal to 10 m
Control if the structural safety criteria with regard to
the following failure modes are met:
- Bending resistance
- Shear resistance
- Anchorage capacity
Worked example prestressed hollow core slab
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01 February 2013 34
σ - ε relation prestressing steel
Worked example prestressed hollow core slab
01 February 2013 35
Flexural resistance
The design ultimate moment is:
Md = 1/8⋅(γGqGk + γQqQk)⋅l2
= (1/8)⋅(1,2⋅4,3 + 1,5⋅3,6)⋅102
= 132 kNm
λx
z
ηfcd
40mm
Ncd
40mm
Fpd
Design tensile force of prestressing steel: Fpd = Ap⋅{(fpd + fpk/γp)/2}= 408⋅{(1382+1610)/2}⋅10-3
= 610 kN
Horizontal equilibrium requires: Ncd = FpdThis results in:
(ηfcd)⋅(λx)⋅btot = 610000Or: (1,0⋅30)⋅(0,8x)⋅1196 = 610000 → x = 21,3 mmThe ultimate bending moment is then: Mu = z⋅Fpd = (265 – 40 – 21,3.0,8/2)⋅610⋅10-3 = 132 kNm,
OK
Worked example prestressed hollow core slab
01 February 2013 36
Md = 126 kNm
A
Areas cracked (B) and uncracked (A) in flexure in ULS.
Area A, uncracked in flexure: control for shear tension and anchorage failure.
qd = 10,6 kN/m
Md = 132 kNm
Cracking moment 107.8 kNm
2859 mm 2859 mm2141 mm 2141 mm
AA B A
Worked example prestressed hollow core slab
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01 February 2013 37
Control for shear flexure capacity in region B (cracked in flexure)
2859 mm 2141 mm
Vd = 53,0 kN
22,7 kN
In the region cracked in flexure Vd,max = 22,7 kN. According to Eq. (6.2.a) :
VRd,c = [0,12k(100 ρlfck)1/3 + 0,15⋅σcp]⋅bwd = 0,12⋅1,94⋅(100⋅0,0078⋅30)1/3 + 0,15⋅2,5] 231⋅225
= 54,1 kN > 22,7 kN, so OK
Worked example prestressed hollow core slab
01 February 2013 38
Controle afschuifdraagvermogen in gebied B
lxlpt2
ctdcplctdw
cRd ffS
bIV σα+
⋅= 2
, )(
lx distance between intersection pointdescribed at left hand side and end of slab
lpt2 upper vlaue of transmission length,assumed to be 1,2⋅lpt
αl = lx/lpt2 ≤ 1,0 for strands and wires
Cross sections which are nearer to the support
than the intersection point between the axis of
gravity and the line from the inner edge of the
support inclined under 450 don’t have to
controlled for shear failure.
Worked example prestressed hollow core slab
01 February 2013 39
Controle afschuifdraagvermogen in gebied B
lxlpt2
ctdcplctdw
cRd ffS
bIV σα+
⋅= 2
, )(
lx= s + d/2 = 40 + 112 = 152 mmlpts = 1,2⋅lpt = 1,2⋅683 = 820 mmαl = lx/lpt2 = 152/820 = 0,185
VRd,c = (1,5⋅10-3⋅0,231/7,55⋅10-3)⋅√(1,82 + 0,185⋅2,5⋅1,80)
= 0,093 MN = 93 kN > 53 kN OK
Worked example prestressed hollow core slab
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01 February 2013 40
Control of anchorage length
In the USL the first bending crack islocated at a distance 2,86 m from thesupport. This is far outside the areawhere the reduction of steel stressapplies (until 1041 mm from the endof the slab).
OK
σpm∞ = 1050 N/mm2
Envelope for anchorage force
fpd = 1321 N/mm2
lbpd=1041 mmlpt=683 mm
σpm0 = 1290 N/mm2
Worked example prestressed hollow core slab