additional rules for precast concrete, en 1992-1-1 … and reinforced ... d = c + ∆a2 in case of...

1

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

2

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

3

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

4

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

5

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

6

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:

7

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

8

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


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))

9

01 February 2013 25

Introduction prestressing force for

pretensioned steel

Steel stressA


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


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

10

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


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


11

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


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


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


12

01 February 2013 34

σ - ε relation prestressing steel


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


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


13

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


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.


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


14

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


additional rules for precast concrete, en 1992-1-1 … and reinforced ... d = c + ∆a2 in case of...

Documents